# The Theorems of Schauder and Peano

In the text on Brouwer’s Fixed Point Theorem, I had confidently stated that Schauder’s Theorem follows from it with less effort and that one may easily conclude things like Peano’s Theorem from there. As a matter of fact, things lie considerably deeper than I had naively thought. If sound proofs are to be given, there is technical work to be done in many instances. The ideas are not tough themselves, but the sheer number of steps to be taken and the methodological machinery cannot be neglected. Let’s see.

We shall follow the lines of Heuser’s books (both this one and this one), as we did before, to collect the ingredients to give a proof of Schauder’s Fixed Point Theorem. It involves a statement about convex sets, on which we will focus first, followed by an excursion on approximation in normed vector spaces. We shall also need the theorem named after Arzelà and Ascoli, being a basic glimpse into the ways of thinking of functional analysis. All of this will allow us to prove Schauder’s Theorem in a rather strong flavour. For the conclusion, we split our path: we give both Heuser’s treatment of Peano’s Theorem in the spirit of functional analysis and Walter’s more elementary approach (which, however, also makes use of the Theorem of Arzelà-Ascoli).

Remember that a set $K$ is called convex, if for any $x,y\in K$ and for any $\alpha\in[0,1]$, we have $\alpha x + (1-\alpha)y\in K$. This formalizes the intuition that the line connecting $x$ and $y$ be contained in $K$ as well.

Lemma (on convex sets): Let $E$ be a normed space and let $x_1,\ldots,x_n\in E$. Let

$\displaystyle\mathrm{conv}(x_1,\ldots,x_n) = \bigcap_{\substack{K\subset E~\mathrm{ convex}\\\{x_1,\ldots,x_n\}\subset K}} K$

the convex hull. Then,

$\displaystyle \mathrm{conv}(x_1,\ldots,x_n) = \left\{v\in E\colon v=\sum_{i=1}^n\lambda_ix_i\text{ with }\sum_{i=1}^n\lambda_i=1; \lambda_i\geq0\right\},\qquad(\diamondsuit)$

and $\mathrm{conv}(x_1,\ldots,x_n)$ is compact.

Proof: Let us first prove the representation of the convex hull. For the “$\subset$“-direction, we will show that the set on the right-hand side of $(\diamondsuit)$ is convex. Let $x = \sum_{i=1}^n\lambda_ix_i$ and $y=\sum_{i=1}^n\mu_ix_i$, with $\sum\lambda_i=\sum\mu_i=1$. Let $\alpha\in[0,1]$, then

$\displaystyle \alpha x + (1-\alpha)y = \sum_{i=1}^n\bigl(\alpha\lambda_i+(1-\alpha)\mu_i\bigr)x_i,$

where $\sum_{i=1}^n\bigl(\alpha\lambda_i+(1-\alpha)\mu_i\bigr) = \alpha+(1-\alpha) = 1$. Hence, $\alpha x+(1-\alpha)y$ is part of the set on the right-hand side of $(\diamondsuit)$ .

We now turn to the “$\supset$“-direction. Let $y_1,\ldots, y_m\in\mathrm{conv}(x_1,\ldots,x_n)$. We show that $\sum_{j=1}^m\mu_jy_j\in\mathrm{conv}(x_1,\ldots,x_m)$ if $\sum_{j=1}^m\mu_j = 1$. That means any point that as a representation as in the right-hand side of $(\diamondsuit)$ must be in $\mathrm{conv}(x_1,\ldots,x_n)$. This is clear for $m=1$. For $m>1$, we take $C:=\sum_{j=1}^{m-1}\mu_j$ to see

\displaystyle \begin{aligned} \sum_{j=1}^m\mu_jy_j &= \sum_{j=1}^{m-1}\mu_jy_j + \mu_my_m\\ &= C\sum_{j=1}^{m-1}\frac{\mu_j}{C}y_j + \mu_my_m\\ &= C\sum_{j=1}^{m-1}\tilde\mu_j y_j + (1-C)y_m. \end{aligned}

Note that $\mu_m = \sum_{j=1}^m\mu_j - \sum_{j=1}^{m-1}\mu_j = 1-C$. By induction,

$\displaystyle \sum_{j=1}^{m-1}\tilde\mu_jy_j\in\mathrm{conv}(x_1,\ldots,x_n),$

since $\sum_{j=1}^{m-1}\tilde\mu_j = \frac1C\sum_{j=1}^{m-1}\mu_j = 1.$ Hence,

$\displaystyle\sum_{j=1}^m\mu_jy_j\in\mathrm{conv}(x_1,\ldots,x_n).$

Finally, we shall prove compactness. Let $(y_k)_k\subset\mathrm{conv}(x_1,\ldots,x_n)$, with a representation $y_k = \sum_{i=1}^n\lambda_i^{(k)}x_i$. The sequences $(\lambda_i^{(k)})_k$ are bounded by $[0,1]$ and hence have convergent subsequences. Choosing subsequences $n$ times (for each $i=1,\ldots,n$), we find a subsequence $(\lambda_i^{(k_\ell)})_\ell$ that converges to some $\lambda_i$ for each $i=1,\ldots,n$. Besides,

$\displaystyle\sum_{i=1}^n\lambda_i = \sum_{i=1}^n\lim_{\ell\to\infty}\lambda_i^{(k_\ell)} = \lim_{\ell\to\infty}1 = 1.$

This yields

$\displaystyle\lim_{\ell\to\infty}y_{k_\ell} = \lim_{\ell\to\infty}\sum_{j=1}^n\lambda_j^{(k_\ell)}x_j = \sum_{j=1}^n\lambda_jx_j\in\mathrm{conv}(x_1,\ldots,x_n).$

q.e.d.

We will now prove a result that extends Brouwer’s Fixed Point Theorem to a more general setting. This is the one that I had skimmed earlier, believing it consisted only of standard arguments; in principle, this is true. But let’s have a closer look at it and how these standard arguments work together.

Theorem (on fixed points in real convex sets): Let $\emptyset\neq K\subset\mathbb{R}^p$ be convex, compact, and let $f:K\to K$ be continuous. Then $f$ has a fixed point.

Proof: 0th step. As $K$ is compact, it is bounded and thus there is some $r>0$ such that $K\subset B_r(0)$.

1st step. Let us construct the best approximation of some $x\in B_r(0)$ within $K$; that means we look for a $z\in K$ with $\left|x-z\right| = \inf_{y\in K}\left|x-y\right|$.

Taking $\gamma:=\inf_{y\in K}\left|x-y\right|$, there is a sequence $(z_n)_n\subset K$ with $\lim_{n\to\infty}\left|x-z_n\right| = \gamma$.

We wish to prove that $(z_n)_n$ is a Cauchy sequence. From the basic properties of any scalar product, we find

\displaystyle \begin{aligned} \left|u+v\right|^2+\left|u-v\right|^2 &= \left\langle u+v,u+v\right\rangle + \left\langle u-v,u-v\right\rangle \\ &= \left|u\right|^2 + \left|v\right|^2 + 2\left\langle u,v\right\rangle + \left|u\right|^2 + \left|v\right|^2 - 2\left\langle u,v\right\rangle \\ &= 2 \left|u\right|^2 + 2 \left|v\right|^2. \end{aligned}

In our case, this shows

\displaystyle \begin{aligned} \left|z_n-z_m\right|^2 &= \left|(z_n-x)-(z_m-x)\right|^2 \\ &= 2\left|z_n-x\right|^2 + 2\left|z_m-x\right|^2- \left|(z_m+z_n)-2x\right|^2 \\ &= 2\left|z_n-x\right|^2 + 2\left|z_m-x\right|^2 - 4\left|\frac{z_m+z_n}2-x\right|^2. \end{aligned}

Since $K$ is convex, $\frac12z_n+\frac12z_m\in K$. Therefore,

$\displaystyle\left|\frac{z_m+z_n}2-x\right|\geq\gamma.$

Thus,

\displaystyle \begin{aligned} \lim_{n,m\to\infty}\left|z_n-z_m\right|^2&\leq 2\lim_{n\to\infty}\left|z_n-x\right|^2 + 2\lim_{m\to\infty}\left|z_m-x\right|^2 - 4\gamma^2 \\ &= 2\gamma^2+2\gamma^2-4\gamma^2 = 0. \end{aligned}

Therefore, $(z_n)_n$ is a Cauchy sequence, having a limit $y$, say. As $K$ is closed, $y\in K$. In total, we have seen (noting that the absolute value is continuous)

$\displaystyle\gamma = \lim_{n\to\infty}\left|z_n-x\right| = \left|y-x\right|.$

$y$ is the best approximation to $x$ within $K$.

2nd step. The best approximation is unique.

If there were two of them, $u$ and $V$, say, then $\left|x-u\right| = \left|x-v\right| = \gamma$. If we consider the sequence $(z_n)_n$ that alternates between $u$ and $V$, we’d find $\left|x-z_n\right| = \gamma$ for all $n$, and hence $(z_n)_n$ is a Cauchy sequence by what we found in step 1. Therefore $z_n$ must be convergent, which implies $u=v$.

3rd step. The mapping $A:B_r(0) \to K$ that takes $x$ to its best approximation, is continuous.

Let $(x_n)_n\subset B_r(0)$ with $\lim_{n\to\infty}x_n =: x$. Let $\varepsilon>0$. For sufficiently large $n$, we have

$\displaystyle \gamma_n :=\inf_{y\in K}\left|x_n-y\right| \leq \inf_{y\in K}\bigl(\left|x_n-x\right|+\left|x-y\right|\bigr) < \varepsilon + \inf_{y\in K}\left|x-y\right| = \varepsilon + \gamma.$

Besides,

$\displaystyle \gamma\leq\left|x - A(x_n)\right|\leq\left|x_n-A(x_n)\right| + \left|x_n-x\right| = \gamma_n + \left|x_n-x\right| < \gamma_n+\varepsilon.$

These inequalities give us

$\displaystyle \gamma\leq\left|x-A(x_n)\right| < 2\varepsilon + \gamma,\qquad\text{ which means }\lim_{n\to\infty}\left|x-A(x_n)\right| = \gamma.$

Hence, $\lim_{n\to\infty}A(x_n)$ is the best approximation to $x$. As this is unique, we have shown that $A$ is sequentially continuous.

4th step. The quest for the fixed point.

The mapping $f\circ A:B_r(0)\to K\subset B_r(0)$ is continuous. By Brouwer’s Fixed Point Theorem, $f\circ A$ has a fixed point: there is some $w\in B_r(0)$ with $f\bigl(A(w)\bigr) = w$. As $f$ only takes images in $K$, we must have $w\in K$. By construction, for points in $K$, the mapping $A$ does not do anything: hence

$\displaystyle w = f\bigl(A(w)\bigr) = f(w).$

q.e.d.

Corollary (on fixed points in convex sets of normed spaces): Let $x_1,\ldots,x_n\in E$ a normed vector space, let $\emptyset\neq K\subset\mathrm{span}(x_1,\ldots,x_n)$ convex, compact, and let $f:K\to K$ continuous. Then $f$ has a fixed point.

Proof: Let us choose a base for $\mathrm{span}(x_1,\ldots,x_n)$ from the $x_1,\ldots,x_n$. We take w.l.o.g. $x_1,\ldots,x_p$ for a certain $p\leq n$. Then any $y\in\mathrm{span}(x_1,\ldots,x_n)$ has a unique representation as $y=\sum_{j=1}^p\beta_jx_j$, and the maping

$\displaystyle A:\mathrm{span}(x_1,\ldots,x_n)\to\mathbb{R}^p,\qquad y\mapsto(\beta_1,\ldots,\beta_p)$

is a bijection. As all norms on $\mathbb{R}^p$ are equivalent, convergence issues are not affected by this bijection. Hence, the theorem and its proof work out in the setting of this corollary, too. q.e.d.

Note that this Corollary may deal with an infinite-dimensional space, however we make use of a finite-dimensional subspace only. This will become relevant in Schauder’s Theorem as well.

Theorem (Arzelà 1895, Ascoli 1884): Let $X\subset\mathbb{R}^d$ compact, let $\mathcal F$ be a family of continuous real-valued functions on $X$, which satisfies two properties:

• it is pointwise bounded: for any $x\in X$, there is some $M(x)\in\mathbb{R}$ with $\left|f(x)\right|\leq M(x)$, for all $f\in\mathcal F$.
• it is equicontinuous: for any $\varepsilon>0$ there is some $\delta>0$ such that for any $x,y\in X$ with $\left|x-y\right|<\delta$ we have $\left|f(x)-f(y)\right|<\varepsilon$, for all $f\in\mathcal{F}$.

Then, $\mathcal F$ is relatively compact, that means every sequence in $\mathcal F$ has a uniformly convergent subsequence.

Note, that we do not demand the limit of the convergent subsequence to be contained in $\mathcal F$; that would mean compact, instead of relatively compact.

Proof: 1st step. We get hold of a countable dense subset of $X$.

For our immediate uses of the theorem, it should suffice to choose $\mathbb{Q}\cap X$, since we will take $X$ to be intervals and there will not be any need for more exotic applications. However, to show up something a little more general, have a look at the sets $\bigl(U_{1/k}(x)\bigr)_{x\in X}$. This is a covering of $X$ and finitely many of them will suffice to cover $X$, for instance $M_k:=\{x_{k1},\ldots,x_{kn}\}$. The set $M:=\bigcup_{k=1}^\infty M_k$ is countable. By construction, for any $x_0\in X$ and any $\varepsilon>0$, we can find some point $y\in M$ that has $\left|x-y\right|<\varepsilon$. Therefore, $M$ is dense in $X$.

2nd step. We construct a certain subsequence to a given sequence $(f_n)_n\subset\mathcal F$.

This step is at the heart of the Arzelà-Ascoli-Theorem, with a diagonal argument to make it work. Let us enumerate the set $M$ from the step 1 as $\{x_1,x_2,\ldots\}$.

As $\mathcal F$ is pointwise bounded, the sequence $\bigl(f_n(x_1)\bigr)_n\subset\mathbb{R}$ is bounded as well. By Bolzano-Weierstrass, it has a convergent subsequence that we will call $\bigl(f_{1,n}(x_1)\bigr)_n$.

If we evaluate this new sequence in $x_2$, we arrive at $\bigl(f_{1,n}(x_2)\bigr)_n\subset\mathbb{R}$, which is bounded as well. Again, we find a convergent subsequence that is now called $\bigl(f_{2,n}(x_2)\bigr)_n$. As this is a subsequence of $\bigl(f_{1,n}(x_1)\bigr)_n$, it converges in $x_1$ as well.

We continue this scheme and we find an array of sequences like this

 $f_{11}$ $f_{12}$ $f_{13}$ $\cdots$ $f_{21}$ $f_{22}$ $f_{23}$ $\cdots$ $f_{31}$ $f_{32}$ $f_{33}$ $\cdots$ $\vdots$ $\vdots$ $\vdots$ $\ddots$

where each row is a subsequence of the row above. Row $K$ is convergent in the point $x_k$ by Bolzano-Weierstrass and convergent in the points $x_1,\ldots,x_{k-1}$ by construction.

Now, consider the sequence $(f_{nn})_n$. It will converge in any point of $M$.

3rd step. Our subsequence of the 2nd step converges uniformly on $X$. We will use equicontinuity to expand the convergence from $M$ to the whole of $X$.

As $\mathcal F$ is equicontinuous, we will find for any $\varepsilon>0$ some $\delta>0$ with $\left|f_{nn}(x)-f_{nn}(y)\right| < \frac\varepsilon3$, for all $n\in\mathbb{N}$, as long as $\left|x-y\right|<2\delta$. Since $X$ is compact, there are some points $y_1,\ldots,y_p\in X$ with $X\subset\bigcup_{j=1}^pU_\delta(y_j)$. And as $M$ is dense in $X$, we can find some $\xi_j\in U_\delta(y_j)\cap M$ for any $j=1,\ldots,p$.

Let $x\in U_\delta(y_j)$, then

$\displaystyle \left|x-\xi_j\right| \leq \left|x-y_j\right|+\left|y_j-\xi_j\right| < 2\delta,$

which shows

$\displaystyle \left| f_{nn}(x)-f_{nn}(\xi_j)\right| < \frac\varepsilon 3\qquad\text{ for any }n\in\mathbb{N}\text{ and }x\in X\cap U_\delta(y_j).$

We have already seen that $(f_{nn})_n$ is convergent on $M$, and hence (convergent sequences are Cauchy-sequences)

$\displaystyle \left|f_{nn}(\xi_j) - f_{mm}(\xi_j)\right| < \frac\varepsilon 3\qquad \text{ for sufficiently large }m,n\text{ and for any }j=1,\ldots,p.$

Now, let $x\in X$, no longer restricted. Then, there is some $j=1,\ldots,p$ such that $x\in U_\delta(x_j)$, and

\displaystyle \begin{aligned} \left|f_{nn}(x)-f_{mm}(x)\right| &\leq \left| f_{nn}(x)-f_{nn}(\xi_j)\right| + \left|f_{nn}(\xi_j) - f_{mm}(\xi_j)\right| + \left|f_{mm}(\xi_j) - f_{mm}(x)\right| \\ &< \frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3 = \varepsilon. \end{aligned}

Thus, $\left\|f_{nn}-f_{mm}\right\|_\infty < \varepsilon$ for sufficiently large $n,m$. This sequence is a Cauchy sequence and hence convergent. q.e.d.

This was our last stepping stone towards Schauder’s Theorem. Let’s see what we can do.

Theorem (Schauder, 1930): Let $E$ be a normed vector space, $\emptyset\neq K\subset E$ convex and closed, let $f:K\to K$ continuous, $f(K)$ relatively compact. Then $f$ has a fixed point.

Proof: 1st step. As $f(K)$ is relatively compact, its closure is compact. We construct a finite approximating subset of $\overline{f(K)}$.

Let $\varepsilon>0$. There are some finitely many points $x_1,\ldots,x_m\in\overline{f(K)}$ with $\overline{f(K)}\subset\bigcup_{j=1}^mU_\varepsilon(x_j)$. In particular, for any $x\in f(K)$, there is some $j=1,\ldots,m$ with $\left|x_j-x\right| < \varepsilon$. Let us consider the function for $x\in f(K)$

$\displaystyle \varphi_j(x):=\mathbf{1}_{\left|x_j-x\right|<\varepsilon}\bigl(\varepsilon-\left|x-x_j\right|\bigr).$

It is obviously continuous, and as $\overline{f(K)}$ is covered by these $U_\varepsilon$,

$\displaystyle \varphi(x)=\sum_{j=1}^m\varphi_j(x) > 0.$

This allows $\psi_j(x):=\frac{\varphi_j(x)}{\varphi(x)}$ to be well-defined, and by construction $\psi(x)=\sum_{j=1}^m\psi_j(x) = 1$. Hence, the function $g:f(K)\to\mathrm{conv}(x_1,\ldots,x_m)$

$\displaystyle g(x):=\sum_{j=1}^m\psi_j(x) x_j$

is continuous (the Lemma on convex sets tells us that this actually maps into the convex hull). Now, let $x\in f(K)$. We find

$\displaystyle g(x)-x = \sum_{j=1}^m\psi_j(x)x_j - x = \sum_{j=1}^m\psi_j(x)\bigl(x_j-x\bigr) = \sum_{\substack{j=1\\\left|x_j-x\right|<\varepsilon}}^m\psi_j(x)\bigl(x_j-x),$

and therefore, for any $x\in f(K)$,

$\displaystyle \left|g(x)-x\right| \leq \sum_{\substack{j=1\\\left|x_j-x\right|<\varepsilon}}^m\psi_j(x)\left|x_j-x\right| < \varepsilon\sum_{j=1}^m\psi_j(x) = \varepsilon.$

This shows that $g$ uniformly approximates the identity on $\mathrm{conv}(x_1,\ldots,x_m)\subset f(K)$. Note that $g$ depends on the choice of $\varepsilon$.

2nd step. Reference to the Theorem on fixed points in convex sets and approximation of the fixed point.

We set $h:=g\circ f$, which is a continuous mapping

$\displaystyle h:K\to\mathrm{conv}(x_1,\ldots,x_m)\subset f(K) \subset K.$

We can restrict it to $\mathrm{conv}(x_1,\ldots,x_m)$ and then re-name it $\tilde h$.  By the Lemma on convex sets, $\mathrm{conv}(x_1,\ldots,x_m)$ is compact, it is finite-dimensional, and by the Corollary on fixed point sets in normed spaces, $\tilde h$ has a fixed point $z$:

$z = \tilde h(z) = g\bigl(f(z)\bigr)\qquad\text{ for some }z\in\mathrm{conv}(x_1,\ldots,x_m).$

Therefore,

$\displaystyle \left|f(z)-z\right| = \left|f(z)-g\bigl(f(z)\bigr)\right| < \varepsilon.$

Note that $z$ depends on $g$ and hence on $\varepsilon$.

3rd step. Construction of the fixed point.

For any $n\in\mathbb{N}$, by step 2, we find some $z_n\in\mathrm{conv}(x_1^{(n)},\ldots,x_{m(n)}^{(n)})\subset f(K)\subset K$ with

$\displaystyle\left|f(z_n)-z_n\right| < \frac1n.$

As $f(K)$ is relatively compact, the sequence $\bigl(f(z_n)\bigr)_n$ has a convergent subsequence: there is some $\tilde z\in\overline{f(K)}$ with $\tilde z=\lim_{k\to\infty}f(z_{n_k})$. As $K$ is closed, we get $\tilde z\in \overline{f(K)}\subset\overline K = K$. Now,

$\displaystyle\left|z_{n_k} - \tilde z\right| \leq \left|z_{n_k} - f(z_{n_k})\right| + \left|f(z_{n_k}) - \tilde z\right| < \frac1{n_k} + \varepsilon_{n_k},$

which means that $z_{n_k}$ and $\tilde z$ get arbitrarily close: $\tilde z = \lim_{k\to\infty}z_{n_k}$. Since $f$ is continuous, we arrive at

$\displaystyle f(\tilde z) = \lim_{k\to\infty}f(z_{n_k}) = \tilde z$. q.e.d.

It is apparent that Schauder’s Theorem already has very general conditions that are tough to weaken further. Obviously the Theorem gets false if $f$ is not continuous. If $K$ were not closed, we’d get the counter-example of $f:(0,1)\to(0,1)$, $x\mapsto x^2$, which doesn’t have any fixed points. If $K$ were not convex, we’d get the counter-example of $f:\partial B_1(0)\to \partial B_1(0)$, $e^{it} \mapsto e^{i(t+\pi)}$. It is hard to give a counter-example if $f(K)$ is not relatively compact – in fact I would be interested to hear of any such counter-example or of the generalization of Schauder’s Theorem to such cases. Which is the most general such fixed point theorem?

Now, we are able to harvest the ideas of all this work and apply it to differential equations. Usually, in courses on ordinary differential equations, the famous Picard-Lindelöf-Theorem is proved, which states that for well-behaved functions $f$ (meaning that they satisfy a Lipschitz-condition), the initial-value problem

$y'(x) = f\bigl(x,y(x)\bigr),\qquad y(x_0) = y_0,$

has a unique solution. This is a powerful theorem which simplifies the entire theory of differential equations. However, a little more holds true: it suffices that $f$ is continuous to guarantee a solution. However, uniqueness is lost in general. While in many applications one can assume continuity of $f$ without remorse (especially in physics), a Lipschitz-condition is much harder to justify. This is not to diminish the usefulness of Picard and Lindelöf, as any model has assumptions to be justified – the Lipschitz-condition is just one of them (if one even bothers to demand for a proper justification of existence and uniqueness – sometimes this would seem obvious from the start).

Let us have a look at what Peano told us:

Theorem (Peano, 1886/1890): Let $f:R\to\mathbb{R}$ be continuous, where

$\displaystyle R:=\bigl\{(x,y)\in\mathbb{R}^2\colon \left|x-x_0\right| \leq a, \left|y-y_0\right|\leq b\bigr\},$

let $M:=\max_{(x,y)\in R}\left|f(x,y)\right|$, $\alpha := \min\bigl(a, \frac bM\bigr)$.

Then, the initial value problem $y'(x)= f\bigl(x,y(x)\bigr)$, $y(x_0)=y_0$, has a solution on the interval $[x_0-\alpha, x_0+\alpha]$.

Concerning the interval on which we claim the solution to exist, have a look at how such a solution $y$ might behave: as we vary $x$, the solution may “leave” $R$ either to the vertical bounds (to left/right) or to the horizontonal bounds (up/down). A solution $y$ can at most have a slope of $\pm M$, and thus, if it leaves on the horizontal bounds, this will happen at $x_0\pm\frac bM$ as the earliest point. If it doesn’t leave there, it will exist until $x_0\pm a$. Of course, it might exist even further, but we have only demanded $f$ to be defined till there. A little more formally, the mean value theorem tells us

$\displaystyle \left|y(x)-y_0\right| = \left|y'(\xi)\right|\left|x-x_0\right| = \left|f\bigl(\xi,y(\xi)\bigr)\right|\left|x-x_0\right| \leq M\alpha\leq b.$

This guarantees that the solution $y$ is well-defined on $R$, because $f$ is defined there.

Proof: 0th step. To simplify notation, let us set

\displaystyle \begin{aligned} J&:=[x_0-\alpha, x_0+\alpha],\\ \mathcal{C}(J)&:=\bigl\{f:J\to\mathbb{R}\text{ continuous}\bigr\},\\ K&:=\bigl\{y\in\mathcal{C}(J)\colon \left|y(x)-y_0\right|\leq b\text{ for any }x\in J\bigr\}. \end{aligned}

1st step. We twist the problem to another equivalent shape, making it more accessible to our tools.

First, let $y(x)$ be a solution to the initial value problem on a sub-interval $I\subset J$. Then, for any $x\in I$,

$\displaystyle y'(x) = f\bigl(x,y(x)\bigr),\quad y(x_0)=y_0,$

and hence

$\displaystyle y(x) = y_0 + \int_{x_0}^x f\bigl(t,y(t)\bigr)dt.\qquad (\heartsuit)$

On the other hand, if we start from this equation and suppose that it holds for any $x\in I$, $y$ must be differentiable with $y'(x) = f\bigl(x,y(x)\bigr)$ and $y(x_0)=y_0$.

We have seen that a function $y$ solves the initial value problem on $J$ if and only if it satisfies the equation $(\heartsuit)$ on $J$.

2nd step. We try to give a representation of the problem as a fixed-point-problem.

Let us consider the mapping

$\displaystyle A:K\to\mathcal{C}(J),~ y\mapsto y_0 + \int_{x_0}^{\text{\large\textbf{.}}} f\bigl(t,y(t)\bigr)dt.$

This is a functional where we plug in a continuous function and where we get a continuous function back. In particular, and to make it even more painfully obvious,

$\displaystyle (Ay)(x) = y_0+\int_{x_0}^xf\bigl(t,y(t)\bigr)dt,\qquad\text{for any }x\in J$.

Therefore, $y$ is a solution to the intial value problem, if it is a fixed point of $A$, meaning $Ay=y$.

3rd step. We show that $A$ maps $K$ to itself. We have defined $A$ only on $K$, so let $y\in K$ and $x\in J$; then:

$\displaystyle \left|(Ay)(x)-y_0\right| = \left|\int_{x_0}^x f\bigl(t,y(t)\bigr) dt\right| \leq M\left|x-x_0\right| \leq \alpha M\leq b.$

The second-to-last inequality follows from $x\in J$, the last one from the definition of $\alpha$.

This shows that $Ay\in K$.

4th step. $K\neq\emptyset$ is obvious, as the constant function $y_0$ is in $K$.

5th step. $K$ is convex. Let $f,g\in K$ and let $\beta\in[0,1]$. Then, for any $x\in J$,

\displaystyle \begin{aligned} \left|(1-\beta)f(x)+\beta g(x) - y_0\right| &= \left|(1-\beta)\bigl(f(x)-y_0\bigr) + \beta\bigl(g(x)-y_0\bigr)\right| \\ &\leq (1-\beta)\left|f(x)-y_0\right| + \beta\left|g(x)-y_0\right| \\ &\leq (1-\beta)b+\beta b = b. \end{aligned}

This proves $(1-\beta)f+\beta g\in K$.

6th step. $K$ is a closed set in $\mathcal{C}(J)$, where we use the topology of uniform convergence.

Consider the sequence $(f_n)_n\subset K$ which converges uniformly to some $f\in\mathcal{C}(J)$. Remember that $\mathcal{C}(J)$ is complete, which is why we can do this. Then, for any $x\in J$,

$\displaystyle\left|f(x)-y_0\right| = \left|\lim_{n\to\infty}f_n(x)-y_0\right| = \lim_{n\to\infty}\left|f_n(x)-y_0\right| \leq \lim_{n\to\infty} b = b.$

This shows that $f\in K$.

7th step. Using the topology of uniform convergence, the mapping $A:K\to K$ is continuous.

Let $\varepsilon>0$. The function $f$ is continuous on the compact set $R$ and hence uniformly continuous. Therefore, there is some $\delta>0$ such that for $\left|u-v\right|<\delta$,

$\displaystyle \left|f(t,u)-f(t,v)\right|<\frac\varepsilon\alpha.$

Now, let $y,z\in K$ with $\left\|y-z\right\|_\infty < \delta$. Then we have just seen that for any $t\in J$

$\displaystyle \left|f\bigl(t,y(t)\bigr)-f\bigl(t,z(t)\bigr)\right| < \frac\varepsilon\alpha.$

That yields

\displaystyle \begin{aligned} \left|(Ay)(x)-(Az)(x)\right| &= \left|\int_{x_0}^xf\bigl(t,y(t)\bigr)dt - \int_{x_0}^xf\bigl(t,z(t)\bigr)dt\right| \\ &< \left|x-x_0\right|\frac\varepsilon\alpha\\ &\leq\varepsilon. \end{aligned}

In particular,

$\displaystyle\left\|Ay-Az\right\|_\infty < \varepsilon.$

8th step. The set $A(K)\subset K$ is relatively compact. Note that $A(K)$ is a set of continuous functions.

Let $y\in K$ and let $x,x_1,x_2\in J$. Then, every function of $A(K)$ is bounded pointwise, since

$\displaystyle \left|(Ay)(x)\right| = \left|y_0+\int_{x_0}^xf\bigl(t,y(t)\bigr)dt\right|\leq \left|y_0\right|+\left|x-x_0\right|M \leq \left|y_0\right|+\alpha M.$

Besides, $A(K)$ is equicontinuous, because of

\displaystyle \begin{aligned} \left|(Ay)(x_1)-(Ay)(x_2)\right| &= \left|\int_{x_0}^{x_1}f\bigl(t,y(t)\bigr)dt - \int_ {x_0}^{x_2}f\bigl(t,y(t)\bigr)dt\right| \\ &= \left|\int_{x_1}^{x_2}f\bigl(t,y(t)\bigr) dt\right| \\ &\leq \left|x_1-x_2\right| M. \end{aligned}

Arzelà and Ascoli now tell us that any sequence in $A(K)$ has a uniformly convergent subsequence.

9th and final step. From Schauder’s Fixed Point Theorem and steps 3 to 8, $A$ has a fixed point in $K$. From step 2, the initial value problem has a solution. q.e.d.

There was a lot of technical work that we have only needed to invoke Schauder’s Theorem. Some of this could have been avoided, if we had a more elementary proof of Schauder’s Theorem. Such a proof exists, however, some of our machinery is still needed – the proof cannot honestly be called elementary. In some way, the proof matches our procedure given above, however, not everything is needed in such a fine manner. Let’s have short look at it; this is taken from Walter’s book.

Proof (Peano’s Theorem in a more elementary fashion): We proceed in two parts. First, we shall prove the weaker statement that if $f$ is continuous and bounded on the (non-compact) set $[x_0,x_0+a]\times\mathbb{R}$, then there is a solution to the intial-value problem on $[x_0,x_0+a]$. Afterwards, we extend this to our compact set $R$. We won’t deal with extending the solution to the left of $x_0$ as it’s neither important nor difficult. In the previous proof we didn’t need to bother about this.

1st step. Let us define a function on $[x_0,x_0+a]$ using some parameter $\alpha\in(0,a]$ by

$\displaystyle z_\alpha(x) = y_0\mathbf{1}_{x\leq x_0} + \left(y_0+\int_{x_0}^xf\bigl(t,z_\alpha(t)\bigr)dt\right)\mathbf{1}_{x>x_0}.$

This is well-defined, since on the sub-interval $[x_0+k\alpha, x_0+(k+1)\alpha)$ we have $t-\alpha \in[x_0+(k-1)\alpha, x_0+k\alpha)$, and thus $z_\alpha(t-\alpha)$ has been recursively defined; hence $z_\alpha$ is defined as well.

Let us denote $\mathcal{F}:=\bigl\{z_\alpha\colon\alpha\in(0,a]\bigr\}\subset\mathcal{C}\bigl([x_0,x_0+a]\bigr).$

2nd step. $\mathcal F$ is equicontinuous. Let $\varepsilon > 0$ and let $x_1,x_2\in[x_0,x_0+a]$. Then we get, as $f$ is bounded by some $M$,

$\displaystyle \left|z_\alpha(x_1)-z_\alpha(x_2)\right| = \left|\int_{x_1}^{x_2} f\bigl(t, z_\alpha(t-\alpha)\bigr)dt\right| \leq \left|x_2-x_1\right|M,$

which doesn’t depend on $\alpha$, $x_1$ or $x_2$ (only on their distance). Hence, if $\left|x_1-x_2\right|<\delta = \frac\varepsilon M$,

$\displaystyle \left|z_\alpha(x_1)-z_\alpha(x_2)\right| \leq \varepsilon.$

3rd step. $\mathcal F$ is pointwise bounded. This is obvious from $\left|z_\alpha(x)\right|\leq\left|y_0\right|+aM$, which doesn’t depend on $\alpha$ or $x$.

4th step. We determine a solution to the intial-value problem.

From steps 2 and 3 and from Arzelà-Ascoli, we know that the sequence $(z_{1/n})_n\subset\mathcal{F}$ has a uniformly convergent subsequence. Let us denote its limit by $y(x)$, which is defined for all $x\in[x_0,x_0+a]$. This allows us to get

\displaystyle \begin{aligned} \left|z_{1/n_k}\left(t-\frac1{n_k}\right) - y(t)\right| &\leq \left| z_{1/n_k}\left(t-\frac1{n_k}\right) - z_{1/n_k}(t)\right| + \left|z_{1/n_k}(t) - y(t)\right| \\ &\leq \frac M{n_k} + \varepsilon\\ &< \overline\delta, \end{aligned}

for any $t\in[x_0,x_0+a]$ and for sufficiently large $K$. It should be clear what we intend to say with $\overline\delta$ (let’s bring in a little sloppiness here, shall we). Since $f$ is continuous in its second component, this proves

$\displaystyle \left|f\left(t, z_{1/n_k}\left(t-\frac1{n_k}\right)\right) - f\bigl(t,y(t)\bigr)\right| < \overline\varepsilon\qquad\text{for any }t\in[x_0,x_0+a].$

Hence, as every participant here converges uniformly,

\displaystyle \begin{aligned} y(x) &= \lim_{k\to\infty}z_{1/n_k}(x) \\ &= \lim_{k\to\infty}\left(y_0 + \int_{x_0}^x f\left(t, z_{1/n_k}\left(t-\frac1{n_k}\right)\right)dt\right)\\ &= y_0 + \int_{x_0}^x \lim_{k\to\infty} f\left(t, z_{1/n_k}\left(t-\frac1{n_k}\right)\right) dt\\ &= y_0 + \int_{x_0}^x f\bigl(t, y(t)\bigr) dt. \end{aligned}

This shows that

$y'(x) = f\bigl(x, y(x)\bigr),\qquad y(x_0) = y_0.$

5th step. Extension to the general case: Let $f$ be defined on the compact rectangle $R$.

We give a continuation of $f$ beyond $[y_0-b,y_0+b]$ for all $x\in[x_0-a, x_0+a]$ via

$\displaystyle \tilde f(x,y) = \begin{cases}f(x, y_0-b) ,&\text{for }y < y_0-b\\ f(x,y), &\text{for }(x,y)\in R\\ f(x,y_0+b),&\text{for }y>y_0+b\end{cases}$

Obviously, $\tilde f$ is continuous and bounded. From Step 1, $y' = \tilde f(x, y)$ has a solution on $[x_0, x_0+a]$. For $\left|x-x_0\right|\leq\frac bM$, we get

\displaystyle \begin{aligned} \left|y(x) - y_0\right| &\leq\left|y'(\xi)\right|\left|x-x_0\right| \\ &= \left|\tilde f\bigl(\xi,y(\xi)\bigr)\right|\left|x-x_0\right| \\ &\leq M\left|x-x_0\right|\leq b \end{aligned}

Therefore, the solution of $y'=f(x,y)$ is well-defined if $\left|x-x_0\right|\leq\frac bM$, and thus the solution is guaranteed to exist for $\left|x-x_0\right|<\alpha = \min\bigl(a,\frac bM\bigr)$. q.e.d.

As a sort of last-minute addendum, I have stumbled upon two articles from the 1970s that shed some more light on the issue of elementary proofs to Peano’s Theorem which completely avoid the technicalties of Schauder’s Theorem and of Arzelà-Ascoli. One is called “There is an Elementary Proof of Peano’s Existence Theorem” by Wolfgang Walter (the author of the book we cited earlier; Amer. Math. Monthly 78 1971, 170-173), the other is “On Elementary Proofs of Peano’s Existence Theorems” by Johann Walter (Amer. Math. Monthly 80, 1973, 282-286). The issue of whether Arzelà-Ascoli can be avoided is solved by both papers positively: they give proofs of Peano’s Theorem which only deal with standard calculus methods. Basically, the employ the Euler polygon method to construct a solution of the intitial value problem. However, again, the proofs are not constructive. Besides, “elementary” is not to be confused with “easy”, Peano’s Theorem is still nothing that lies directly on the surface of things. A brief look at the second of those articles (to the best of my knowledge the identical names of the authors are a coincidence) raises hope that this proof is actually not too hard – it should be understandable with a lot less effort than the proof via Schauder’s Theorem that we gave above in full detail; remember that Schauder itself required many non-standard theorems on its way. The elementary proof will only work for one-dimensional differential equations, but we bothered only with those anyway; it uses monotonicity of its approximating sequence which is only applicable in the real numbers. On the plus side, the proof explicitly constructs a solution via the Euler method.

The papers also shed some light on the history of Peano’s Theorem and the quest for its proof (together with some rather unusual disagreement on whether an earlier proof is valid or not; some interesting lines to read in passing). This should be enough on this matter for now. If the interest holds up (which is, to this extent, rather unlikely), we’ll return to it. But not for now.

# What is a Brachistochrone?

The cycloid was a core object of mathematical studies during the development of calculus and, before that, for geometry. It arises as a special case of curves in astronomy and it has been used as a challenge for competitors during the rise of the analytic method.

Let us have a look at its definition first, and what it describes heuristically. The cycloid is the planar curve parametized via

$\displaystyle\begin{pmatrix}x(t)\\y(t)\end{pmatrix} = \begin{pmatrix}r(t-\sin t)\\r(1-\cos t)\end{pmatrix},\qquad t\in[0,2\pi], \quad r>0.$

It is the curve of a point on the periphery of a circle that is rolled along the $x$-axis. The parametization follows like this:

The circle rolls along the $x$-axis with constant speed, therefore the angle $\sphericalangle PMD = t$. As $\sin t = \frac{PD}{r}$, we get $BA = r\sin t$. Now, the $x$-component of $P$ is

$\displaystyle x = OB = OA-BA = rt-r\sin t = r(t-\sin t).$

For the $y$-component of $P$$BP = AD = AM-DM = r- r\cos t$, and thus

$\displaystyle y = BP = r(1-\cos t).$

The cycloid can be considered as a special case of the epicycloid. Those have been a matter of interest in the pre-Kepler era, when astronomers tried to explain the motion of the planets in the night sky. As they considered perfect circles as orbits only (as opposed to the ellipses that they actually are), and as they postulated the Earth to be in the center of all those orbits, it was tricky to explain away the observations of different arc speeds and loops that the planets sometimes take. The solution was to imagine the planets circling around Earth, but on this circle was the center of another circle, on which the planets moved. Thus, the planets travelled on an epicycle; and sometimes one of those wasn’t enough (“salvation of the phenomena”).

A first simplification of this theory came from Copernicus who dropped the assumption that the Earth would be in the center of all things, but who didn’t get rid of the perfect circles. For the final resolution, the world had to wait for Kepler and Tycho Brahe. But anyway, that’s not why we’re here.

The cycloid is the curve, on which a point of mass will travel the quickest, if it just rolls along it, drawn by gravitation only (if friction is disregarded); in Greek, this is called the “brachistochrone“. It is remarkable that this quickest path is not the shortest path – there are quicker ways than a straight line. There is some sort of trade-off between gaining speed quickly and between keeping the path sufficiently short. We will look at two different approaches to prove that the cycloid can do this trick.

Another remarkable property of the cycloid is being the “tautochrone“: if points of mass are placed anywhere on this curve, they will travel to another point on this curve in exactly the same amount of time. Points that are farther away will gain more speed in order to close the distance. This is a highly interesting property for building a pendulum: no matter how big the amplitude, the frequency will always be the same. This, in turn, is the core feature of an exact clock, which was a sort of holy grail for scientists to find during the 17th century (not just for ship navigation). This property has been found by Huygens, who had not been able to use calculus methods for this (his solution is hidden in quite cumbersome geometry).

More on this curve and some very nice experiments may be seen in this youtube-video from the highly interesting channel vsauce. I especially love the excitement of both guys when they actually see these properties of the cycloid curve in action.

The brachistochrone problem was posed by Johann Bernoulli in a journal as a quest for the most enlightened mathematicians of the world (“acutissimis mathematicis qui toto orbe florent“). We will see his very elegant approach right below. His brother Jacob found a more general approach, but his train of thought is much more cumbersome – we will see a modernized simplification of this later. Both brothers engaged in a non-friendly competition by posing problems like this one to each other, always hoping for each other’s errors to gloat over. In retrospect, both of them advanced the applications of calculus when it was conceived; note however, that very many of the things that are named after Bernoulli (Bernoulli numbers, Bernoulli distribution, the Law of Large Numbers) have come from Jacob, not from Johann. But the other enlightened mathematicians of the time also retrieved the solution, particularly Leibniz and Newton who both are said to have found the solution in a matter of few hours, and both of them appreciating the beauty of the problem.

Now, let’s see how Johann came to his solution. We will look at some physical properties first.

The Speed Lemma: Consider a point of mass $m$ that travels without friction along any sort of curve in $\mathbb{R}^2$, the only force on it being the gravitation. Let $g$ be Newton’s gravitational constant. Then, when it has travelled height $h$, its speed is $v = \sqrt{2gh}$.

Proof: As physics tells us, the sum of kinetic and potential energy is constant. One may prove this mathematically by doing very basic integration and thinking of Newton’s second axiom (the one with force, mass and acceleration); we won’t go into this. Now, the kinetic energy is $\frac12mv^2$ (for physicists that’s the definition, for mathematicians that’s an easy lemma), while the potential energy is $mgh$. Our zero-level for the potential energy is set such that the potential energy vanishes, when the point of mass has travelled height $h$. By our set-up, the point of mass has no speed in the beginning and hence no kinetic energy. We have found

$\displaystyle 0 + mgh = E_{\mathrm{kin}}^{\mathrm{start}} + E_{\mathrm{pot}}^{\mathrm{start}} = E_{\mathrm{kin}}^{\mathrm{end}} + E_{\mathrm{kin}}^{\mathrm{end}} = \frac12mv^2 + 0,$

which means

$\displaystyle v^2 = 2gh,$

which was to be shown. q.e.d.

One might wonder if there is some problem here, that the speed formula does not depend on the kind of curve that the point of mass moves on. Indeed, without friction there is no problem. One can argue in an entirely different way about decomposition of the gravitational force in a force directed along the (derivative of the) curve and a normal force orthogonal to this one. This decomposed force is of course smaller than the gravitation and hence brings less acceleration to our point. In turn, one can compute the time it takes the point to travel to height $h$, and the speed that it has gained by then. As physics is consistent in itself (surprise!), we arrive at the same result that we gained via kinetic and potential energy. Not being a physicist, I can’t tell with certainty if this connection just stems from a little proof that I didn’t see, or if this is some sort of recognition that the world actually behaves responsibly and rationally. I won’t even start to question this here.

The Time Lemma: Consider the same setting as in the Speed Lemma. On top of that, let the curve on which our point travels be given by a differentiable function $y = f(x)$. Let the point travel from $(0,0)$ to $(b,d)$. The time it takes for this is

$\displaystyle T = \int_0^b\sqrt{\frac{1+(f'(x))^2}{2gf(x)}}\mathrm{d}x.$

Proof (by a little hand-waving): Consider any point $(x,y)$ on the curve, with $x\in(0,b)$. The infinitesimal time our point of mass spends in $(x,y)$ is

$\displaystyle \mathrm{d}t = \frac{\mathrm{d}s}{v(x)} = \frac{\sqrt{(\mathrm{d}x)^2 + (\mathrm{d}y)^2}}{\sqrt{2gf(x)}} = \sqrt{\frac{1+(\mathrm{d}y/\mathrm{d}x)^2}{2gf(x)}}\mathrm{d}x.$

Taking a leap of faith and integrating this (which is supposed to amount to the sum of all such infinitesimal times) gives

$\displaystyle T = \int_0^b \sqrt{\frac{1+(f'(x))^2}{2gf(x)}}\mathrm{d}x.$

In a post on physical interprations of mathematics, a little physical computation can’t be too wrong now, can it. q.e.d.

The Reflection Principle: Consider a ray of light travelling in $\mathbb{R}^2$ from point $(0,h_1)$ to point $(a,h_2)$, being reflected somewhere on the $x$-axis. The resulting angles of reflection $\alpha$ and $\beta$ are equal.

Proof: The underlying physical principle is to choose the line of minimal length for the reflection. A mathematician would put this as an axiom, a physicist will consider this granted by the way that nature behaves. Let’s go with it: the length of the chosen path is, as long as the ray of light is reflected in the point $(x,0)$,

$\displaystyle L(x) = \sqrt{x^2+h_1^2} + \sqrt{(x-a)^2+h_2^2},$

hence we look for some $x$ with

$\displaystyle 0 = L'(x) = \frac{x}{\sqrt{x^2+h_1^2}} + \frac{x-a}{\sqrt{(x-a)^2+h_2^2}} = \sin\alpha - \sin\beta.$

As, $\alpha, \beta\in(0,\frac\pi2)$ for obvious reasons, this is the assertion. q.e.d.

The Refraction Lemma: Consider a ray of light changing the medium in which it travels. Let the speeds of light in those media be $c_1$ and $c_2$. The resulting angles of refraction have a constant proportion:

$\displaystyle\frac{\sin\alpha}{\sin\beta} = \frac{c_1}{c_2}.$

Proof: Now, the speed of light gets relevant and the physical principle is to find the path of minimal time. By the basic laws on time and speed we get

$\displaystyle T(x) = \frac{\sqrt{x^2+h_1^2}}{c_1} + \frac{\sqrt{(x-a)^2+h_2^2}}{c_2}$

and we look for some $x$ with

$\displaystyle 0 = T'(x) = \frac1{c_1}\frac{x}{\sqrt{x^2+h_1^2}} + \frac1{c_2}\frac{x-a}{\sqrt{(x-a)^2+h_2^2}} = \frac1{c_1}\sin\alpha - \frac1{c_2}\sin\beta.$

The lemma is proved. q.e.d.

We have all ingredients to follow Johann Bernoulli’s idea to find the brachistochrone now. The basic question is, what is the quickest path for a point of mass to take, if it is to travel from one point in the plane, $(0,0)$ say, to another one $(b,d)$? Johann’s ingenious idea was to compare this to the path that a ray of light will take – as we have postulated, the ray of light will choose the quickest path as well. The acceleration may stem from gravitation or the path may result from the change of the media, but the aim is the same; as Bernoulli wrote: “who would deny us to replace one approach by the other?”

Hence, let us consider a “continous” change of media, for instance by making a limit of finer layers of media for the ray of light to traverse. As the Refraction Lemma showed, we will get a constant quotient of $\frac{\sin\alpha}{v}$. By the Speed Lemma, our point of mass has gained $v=\sqrt{2gy}$, if it has arrived at level $y$ only being accelerated by gravitation.

Now, using the designations of the following picture (note that $\beta=\frac\pi2-\alpha$),

\begin{aligned} \displaystyle \sin\alpha = \cos\beta = \sqrt{\frac{\cos^2\beta}{\cos^2\beta+\sin^2\beta}} = \sqrt{\frac1{1+\tan^2\beta}} = \sqrt{\frac1{1+(\frac{\mathrm{d}y}{\mathrm{d}x})^2}}. \end{aligned}

In particular,

$\displaystyle\sin\alpha = \frac{1}{\sqrt{1+(y')^2}}.$

As $\frac{\sin\alpha}{v}$ is constant, we find the differential equation

$\displaystyle \frac{1}{\sqrt{1+(y')^2}} = k\cdot v = k\sqrt{2gy},$

or equivalently,

$\displaystyle y' = \sqrt{\frac{1}{k^2\sqrt{2gy}^2}-1} = \sqrt{\frac{1-2gk^2y}{2gk^2y}} = \sqrt{\frac{\frac{1}{2gk^2} - y}{y}}.$

By setting $a:=\frac{1}{2gk^2}$ and by separation of variables,

$\displaystyle x+C = \int\sqrt{\frac{y}{a-y}} \mathrm{d}y.$

Then, we substitute $y(s):=a\sin^2s$, yielding $\frac{\mathrm{d}y}{\mathrm{d}s} = 2a\sin s \cos s$,

\displaystyle \begin{aligned} \int\sqrt{\frac{a\sin^2 s}{a-a\sin^2s}}2a\sin s\cos s \mathrm{d}s &= 2a\int\sqrt{\frac{\sin^2s}{1-\sin^2s}}\sin s\cos s \mathrm{d}s \\ &= 2a\int\sin^2s\mathrm{d}s. \end{aligned}

This integral can be readily solved via partial integration:

$\displaystyle \int\sin^2s\mathrm{d}s = -\sin s\cos s + \int\cos^2s\mathrm{d}s = -\sin s\cos s + s - \int\sin^2s\mathrm{d}s,$

meaning

$\displaystyle \int\sin^2s\mathrm{d}s = \frac12(s-\sin s\cos s).\qquad(\spadesuit)$

Altogether we have found (note that we do not re-substitute $y$ for $s$, since we are not interested in a parametrization like $x=x(y)$)

$\displaystyle x+C = 2a\frac12(s-\sin s\cos s) = \frac a2\bigl(2s-\sin(2s)\bigr).$

As we can set our coordinates such that $x(0)=0$ (the point will begin its voyage in $(0,0)$), we get $C=0$. This shows

\displaystyle \begin{aligned} x(s) &= \frac a2\bigl(2s-\sin(2s)\bigr),\\ y(s) &= a\sin^2s = \frac a2\bigl((1-\cos^2s)+\sin^2s\bigr) = \frac a2\bigr(1-\cos(2s)\bigr). \end{aligned}

Setting $r:=\frac a2$ and $t:=2s$, we retrieve the standard parametrization of the cycloid:

$\displaystyle x(t) = r(t-\sin t),\qquad y(t) = r(1-\cos t).$

The brachistochrone must be a cycloid.

But now for a completely different approach. The brachistochrone can also be found via calculus of variations, which is considerably harder, from a technical point of view, than what we did above. On the other hand, these techniques can be applied to a much broader spectrum of problems. We can only sketch many of the issues here.

Historically, the brachistochrone problem has been the start to calculus of variations. Jacob Bernoulli solved the problem with methods like this, much more general but much less elegant than his brother.

At the core is the observation that we wish to minimize a functional

$\displaystyle J(f) = \int_a^bF\bigl(t, f(t), f'(t)\bigr) \mathrm{d}t,$

over the set $M:=\{f:[a,b]\to\mathbb{R}\colon f(a)=c, f(b)=d, f\in\mathcal{C}^2\}$. Is there some $g\in M$ such that $J(g)\leq J(f)$ for all $f\in M$?

We consider the function $F$ to be defined as $F(t,y,p)$. The inputs $y$ and $p$ will play the roles of the solution function and its derivative, respectively.

Notice that we restrict ourselves already to smooth functions $f\in\mathcal{C}^2$. From a physical point of view, there is no reason why the brachistochrone shouldn’t be just continuous. However, tougher mathematics would be necessary to track down this one.

If the space $M$ is well-behaved, usual compactness arguments tell us that there is a minimum. But it is much harder to pinpoint.

Theorem (Euler-Lagrange; tiny special case): A necessary condition for a $\mathcal C^2$-function $g$ to be a solution to the minimization problem is

$\displaystyle \frac{\partial}{\partial y}F\bigl(t,g(t),g'(t)\bigr) - \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p}F\bigl(t,g(t),g'(t)\bigr) = 0.$

In its expanded form, this is (dropping the arguments for reasons of better legibility)

$\displaystyle \frac{\partial F}{\partial y}- \frac{\partial^2F}{\partial t\partial p}-\frac{\partial^2F}{\partial y\partial p}\cdot g'(t)-\frac{\partial^2F}{\partial p^2}\cdot g''(t) = 0.$

Proof: Let $g$ be the minimum and let $\eta\in\mathcal{C}^2$ with $\eta(a)=\eta(b)=0$. We then consider

$\displaystyle \varphi(\varepsilon):=\int_a^bF\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta'(t)\bigr) \mathrm{d}t.$

Since we chose everything to be well-behaved, $\varphi$ will be differentiable. As $g$ minimizes the functional $J$, $\varphi(0) = J(g) \leq \varphi(\varepsilon)$, and hence $\varphi'(0) = 0$. Note that the derivative is a $\frac{\mathrm{d}}{\mathrm{d}\varepsilon}$ here. For the function $g$, the derivative means $\frac{\mathrm{d}}{\mathrm{d}t}$.

Now, let us compute this (calculemus!)

\displaystyle \begin{aligned} \varphi'(\varepsilon) &= \int_a^b\frac{\mathrm{d}}{\mathrm{d}\varepsilon} F\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta'(t)\bigr)\mathrm{d}t \\ &= \int_a^b \left[\frac{\partial}{\partial y}F\bigl(t, g(t)+\varepsilon\eta(t), g'(t) + \varepsilon \eta'(t)\bigr)\eta(t) +\right.\\ &\hphantom{=}\qquad\left.+ \frac{\partial}{\partial p}F\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta'(t)\bigr)\eta'(t) \right]\mathrm{d}t \end{aligned}

Integration by parts yields, together with the fact that $\eta(a) = \eta(b) = 0$,

\displaystyle \begin{aligned} \int_a^b \frac{\partial}{\partial p}F(t,y,p)\eta'(t) \mathrm{d}t &= \left[\frac{\partial}{\partial p}F(t,y,p)\eta(t)\right]_a^b - \int_a^b\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p}F(t,y,p)\eta(t) \mathrm{d}t \\ &= -\int_a^b\eta(t)\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p}F(t,y,p) \mathrm{d}t. \end{aligned}

We conclude

\displaystyle \begin{aligned} \varphi'(\varepsilon) &= \int_a^b \left[\frac{\partial}{\partial y}F\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta'(t)\bigr)\eta(t) \right.\\ &\hphantom{=}\qquad\left.- \eta(t)\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p} F\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta'(t)\bigr) \right]\mathrm{d}t\\ &= \int_a^b\eta(t)\left[\frac{\partial}{\partial y}F\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta'(t)\bigr)\right.\\ &\hphantom{=}\qquad\left. - \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p} F\bigl(t, g(t)+\varepsilon\eta(t), g'(t)+\varepsilon\eta(t)\bigr)\right]\mathrm{d}t. \end{aligned}

This expression must vanish, as we demand $\varphi'(0)=0$, if $g$ is supposed to be a solution to the minimization problem. We have an arbitrary function $\eta$ involved, so the expression in brackets will have to vanish entirely. Formally, one can see this by contradiction: if in some point $t_0$, the bracket-expression did not vanish, we could choose some interval $[t_1,t_2]\subset [a,b]$ where this bracket-expression didn’t vanish at all (it is continuous, after all). On this interval, we set $\eta(t):=(t-t_1)^4(t-t_2)^4$, we find the integrand strictly positive there, and vanishing outside. Contradiction to $\varphi'(0)=0$.

For $\varepsilon=0$, the statement follows. We have thus proved the Euler-Lagrange equation in this particular case. q.e.d.

Notice that we didn’t speak about sufficient conditions. That would overstretch this text by far – let’s ignore this.

The Simplification Lemma: In the special case, when $F$ only depends on $y$ and $p$, and not directly on its first argument $t$, the Euler-Lagrange equation will simplify to the condition

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\left(F\bigl(g(t), g'(t)\bigr)-g'(t)\frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr)\right) = 0.$

Proof: This follows by a straight-forward computation:

\begin{aligned} \displaystyle &\hphantom{=}\frac{\mathrm{d}}{\mathrm{d}t}\left(F\bigl(g(t),g'(t)\bigr)-g'(t)\frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr)\right) \\ &\stackrel{(\circ)}{=} \frac{\partial}{\partial y}F\bigl(g(t),g'(t)\bigr)g'(t) + \frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr)g''(t) +\\ &\hphantom{=}\quad- g''(t)\frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr)-g'(t)\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr) \\ &\stackrel{\hphantom{(\ast)}}{=} \frac{\partial}{\partial y}F\bigl(g(t),g'(t)\bigr)g'(t) - g'(t)\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr)\\ &\stackrel{(\ast)}{=} \left(\frac{\partial^2}{\partial y\partial p}F\bigl(g(t),g'(t)\bigr) g'(t) + \frac{\partial^2}{\partial p^2}F\bigl(g(t),g'(t)\bigr)g''(t)\right) g'(t) +\\ &\hphantom{=}\quad - g'(t)\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial}{\partial p}F\bigl(g(t),g'(t)\bigr)\\ &\stackrel{(\circ)}{=} \frac{\partial^2}{\partial y\partial p}F\bigl(g(t),g'(t)\bigr) \bigl(g'(t)\bigr)^2 + \frac{\partial^2}{\partial p^2}F\bigl(g(t),g'(t)\bigr)g'(t)g''(t) +\\ &\hphantom{=}\quad - g'(t)\left(\frac{\partial^2}{\partial y\partial p}F\bigl(g(t),g'(t)\bigr)g'(t) + \frac{\partial^2}{\partial p^2}F\bigl(g(t),g'(t)\bigr)g''(t)\right)\\ &\stackrel{\hphantom{(\ast)}}{=} \frac{\partial^2}{\partial y\partial p}F\bigl(g(t),g'(t)\bigr) \bigl(g'(t)\bigr)^2 + \frac{\partial^2}{\partial p^2}F\bigl(g(t),g'(t)\bigr)g'(t)g''(t) +\\ &\hphantom{=}\quad - \frac{\partial^2}{\partial y\partial p}F\bigl(g(t),g'(t)\bigr)\bigl(g'(t)\bigr)^2 - \frac{\partial^2}{\partial p^2}F\bigl(g(t),g'(t)\bigr)g'(t)g''(t)\\ &\stackrel{\hphantom{(\ast)}}{=} 0. \end{aligned}

We have used the expanded form of the Euler-Lagrange equation in $(\ast)$ together with the chain-rule and the feature that in the present special case $\frac{\partial}{\partial t}F=0$, and the chain-rule all by itself in $(\circ)$. All over the place, we have used that $g$ is a solution to the Euler-Lagrange equation and thus needs to be plugged into $F$. q.e.d.

Now that we have the ingredients, let’s try and find the brachistochrone by calculus of variations. By the Time Lemma, we want to minimize the expression

$\displaystyle \int_0^b F\bigl(f(x), f'(x)\bigr)\mathrm{d}x,\qquad F(y,p) = \sqrt{\frac{1+p^2}{2gy}}.$

By the Simplification Lemma, any solution $\varphi$ will have

\displaystyle \begin{aligned} c &= \sqrt{\frac{1+(\varphi'(x))^2}{2g\varphi(x)}} - \varphi'(x)\frac{\varphi'(x)}{2g\varphi(x)}\sqrt{\frac{2g\varphi(x)}{1+(\varphi'(x))^2}}\\ &= \sqrt{\frac{1+(\varphi'(x))^2}{2g\varphi(x)}}\left(1-\frac{(\varphi'(x))^2}{1+(\varphi'(x))^2}\right)\\ &= \sqrt{\frac{1+(\varphi'(x))^2}{2g\varphi(x)}}\frac1{1+(\varphi'(x))^2}\\ &= \frac1{\sqrt{2g\varphi(x)}\sqrt{1+(\varphi'(x))^2}}, \end{aligned}

which means

$\displaystyle \varphi(x)\left(1+\bigl(\varphi'(x)\bigr)^2\right) = \frac1{2gc^2} =: C\qquad\qquad(\clubsuit)$

$\varphi$ will be a solution depending on $x$. On the other hand, we look for a parametrization of a curve in $\mathbb{R}^2$, hence we try to find both functions $x(t)$ and $y(t)$, that are connected via $\varphi(x) = \varphi\bigl(x(t)\bigr) = y(t)$. We set, by divine insight,

$\displaystyle y(t) = C\frac{1-\cos t}{2} = C\sin^2\frac t2.$

The chain rule then says $\frac{\mathrm{d}}{\mathrm{d}t}\varphi\bigl(x(t)\bigr) = \frac{\mathrm{d}}{\mathrm{d}x}\varphi(x)\frac{\mathrm{d}}{\mathrm{d}t}x(t)$, and hence

$\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\frac{\mathrm{d}}{\mathrm{d}t}\varphi\bigl(x(t)\bigr)}{\frac{\mathrm{d}}{\mathrm{d}x}\varphi(x)} = \frac{\frac{\mathrm{d}}{\mathrm{d}t}y(t)}{\frac{\mathrm{d}}{\mathrm{d}x}\varphi(x)}.$

By $(\clubsuit)$,

\displaystyle \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\varphi(x) = \frac{\mathrm{d}}{\mathrm{d}x}\varphi\bigl(x(t)\bigr) &= \sqrt{\frac{C}{\varphi(x(t))} - 1} \\ &= \sqrt{\frac{C}{C\sin^2\frac t2}-1} \\ &= \sqrt{\frac{1-\sin^2\frac t2}{\sin^2\frac t2}} \\ &= \cot\frac t2. \end{aligned}

Altogether,

\displaystyle \begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\frac{\mathrm{d}}{\mathrm{d}t}y(t)}{\frac{\mathrm{d}}{\mathrm{d}x}\varphi(x)} &= \frac{C2\frac12\sin\frac t2\cos\frac t2}{\cot\frac t2} \\ &= \frac{C\sin\frac t2\cos\frac t2}{\cos\frac t2}\sin\frac t2 \\ &= C\sin^2\frac t2. \end{aligned}

We already have almost integrated this one before in $(\spadesuit)$, the substitution $s(t)=\frac t2$ yields

\displaystyle \begin{aligned} x(t) = C\int\sin^2\frac t2 \mathrm{d}t = 2C\int\sin^2(s)\mathrm{d}s &= 2C\frac{s-\sin s\cos s}{2} \\ &= C\left[s-\frac12\sin(2s)\right] \\ &= \frac C2(t-\sin t). \end{aligned}

This shows, that any solution to the minimization problem must look like

$\displaystyle\begin{pmatrix}x(t)\\y(t)\end{pmatrix} = \begin{pmatrix}\frac C2(t-\sin t)\\\frac C2(1-\cos t)\end{pmatrix},$

and is hence a cycloid. What we haven’t proved is, that it actually is a solution to the minimization problem – we didn’t speak about the sufficient condition with Euler-Lagrange, not about regularity of our set $M$ and only about $\mathcal C^2$-functions in the first place (I won’t even go into the physical hand-waving). But anyway, the little tricks and the big machinery of technique make both approaches really insightful and interesting. This makes it a good place to end.

# Brouwer’s Fixed Point Theorem

Recently, we have concluded the text on the Transformation formula, remarking that it was a tool in an elementary proof of Brouwer’s Fixed Point Theorem. Let’s have a closer look at that.

Brouwer’s Fixed Point Theorem is at the core of many insights in topology and functional analysis. As many other powerful theorems, it can be stated and understood very easily, however the proof is quite deep. In particular, the conclusions that are drawn from it, are considered even deeper. As we shall see, Brouwer’s theorem can be shown in an elementary fashion, where the Transformation Formula, the Inverse Function Theorem and Weierstrass’ Approximation Theorem are the toughest stepping stones; note that we have given a perfectly elementary proof of Weierstrass’ Theorem before. This makes Brouwer’s theorem accessible to undergraduate calculus students (even though, of course, these stepping stones already mean bringing the big guns to the fight). The downside is that the proof, even though elementary, is quite long-ish. The undergraduate student needs to exercise some patience.

Theorem (Brouwer, 1910): Let $K:=\{x\in\mathbb{R}^p\colon \left|x\right|\leq1\}$ be the compact unit ball, and let $f:K\to K$ be continuous. Then $f$ has a fixed point, i.e. there is some $x\in K$ such that $f(x)=x$.

There are many generalizations of the Theorem, considering more complex sets instead of $K$, and taking place in the infinite-dimensional space. We shall get back to that later. First, we shall look at a perfectly trivial and then a slightly less trivial special case.

For $p=1$, the statement asks to find a fixed point for the continuous mapping $f:[0,1]\to[0,1]$. W.l.o.g. we have shrunk the set $K$ to $[0,1]$ instead of $[-1,1]$ to avoid some useless notational difficulty. This is a standard exercise on the intermediate value theorem with the function $g(x):=f(x)-x$. Either, $f(1)=1$ is the fixed point, or else $f(1)<1$, meaning $g(1)<0$ and $g(0)=f(0)\geq0$. As $g$ is continuous, some point needs to be the zero of $g$, meaning $0=g(\xi) = f(\xi)-\xi$ and hence $f(\xi)=\xi$. q.e.d. ($p=1$)

For $p=2$, things are still easy to see, even though a little less trivial. This is an application of homotopy theory (even though one doesn’t need to know much about it). The proof is by contradiction however. We will show an auxiliary statement first: there is no continuous mapping $h:K\to\partial K$, which is the identity on $\partial K$, i.e. $h(x)=x$ for $x\in\partial K$. If there was, we would set

$\displaystyle H(t,s):=h(se^{it}), \qquad t\in[0,2\pi], s\in[0,1].$

$H$ is a homotopy of the constant curve $h(0) = H(t,0)$ to the circle $e^{it} = h(e^{it}) = H(t,1)$. This means, we can continuously transform the constant curve to the circle. This is a contradiction, as the winding number of the constant is $0$, but the winding number of the circle is $1$. There can be no such $h$.

Now, we turn to the proof of the actual statement of Brouwer’s Theorem: If $f$ had no fixed point, we could define a continuous mapping as follows: let $x\in K$, and consider the line through $x$ and $f(x)$ (which is well-defined by assumption). This line crosses $\partial K$ in the point $h(x)$; actually there are two such points, we shall use the one that is closer to $x$ itself. Apparently, $h(x)=x$ for $x\in\partial K$. By the auxiliary statement, there is no such $h$ and the assumption fails. $f$ must have a fixed point. q.e.d. ($p=2$)

For the general proof, we shall follow the lines of Heuser who has found this elementary fashion in the 1970’s and who made it accessible in his second volume of his book on calculus. It is interesting, that most of the standard literature for undergraduate students shies away from any proof of Brouwer’s theorem. Often, the theorem is stated without proof and then some conclusions and applications are drawn from it. Sometimes, a proof via differential forms is given (such as in Königsberger’s book, where it is somewhat downgraded to an exercise after the proof of Stoke’s Theorem) which I wouldn’t call elementary because of the theory which is needed to be developed first. The same holds for proofs using homology groups and the like (even though this is one of the simplest fashions to prove the auxiliary statement given above – it was done in my topology class, but this is by no means elementary).

A little downside is the non-constructiveness of the proof we are about to give. It is a proof by contradiction and it won’t give any indication on how to find the fixed point. For many applications, even the existence of a fixed point is already a gift (think of Peano’s theorem on the existence of solutions to a differential equation, for instance). On the other hand, there are constructive proofs as well, a fact that is quite in the spirit of Brouwer.

In some way, the basic structure of the following proof is similar to the proof that we gave for the case $p=2$. We will apply the same reasoning that concluded the proof for the special case (after the auxiliary statement), we will just add a little more formality to show that the mapping $g$ is actually continuous and well-defined. The trickier part in higher dimensions is to show the corresponding half from which the contradiction followed. Our auxiliary statement within this previous proof involved the non-existence of a certain continuous mapping, that is called a retraction: for a subset $A$ of a topological space $X$, $f:X\to A$ is called a retraction of $X$ to $A$, if $f(x)=x$ for all $x\in A$. We have found that there is no retraction from $K$ to $\partial K$. As a matter of fact, Brouwer’s Fixed Point Theorem and the non-existence of a retraction are equivalent (we’ll get back to that at the end).

The basic structure of the proof is like this:

• we reduce the problem to polynomials, so we only have to deal with those functions instead of a general continuous $f$;
• we formalize the geometric intuition that came across in the special case $p=2$ (this step is in essence identical to what we did above): basing on the assumption that Brouwer’s Theorem is wrong, we define a mapping quite similar to a retraction of $K$ to $\partial K$;
• we show that this almost-retraction is locally bijective;
• we find, via the Transformation Formula, a contradiction: there can be no retraction and there must be a fixed point.

Steps 3 and 4 are the tricky part. They may be replaced by some other argument that yields a contradiction (homology theory, for instance), but we’ll stick to the elementary parts. Let’s go.

Lemma (The polynomial simplification): It will suffice to show Brouwer’s Fixed Point Theorem for those functions $f:K\to K$, whose components are polynomials on $K$ and which have $f(K)\subset\mathring K$.

Proof: Let $f:K\to K$ continuous, it has the components $f = (f_1,\ldots,f_p)$, each of which has the arguments $x_1,\ldots,x_p$. By Weierstrass’ Approximation Theorem, for any $\varepsilon>0$ there are polynomials $p_k^\varepsilon$ such that $\left|f_k(x)-p_k^\varepsilon(x)\right| < \varepsilon$, $k=1,\ldots,p$, for any $x\in K$. In particular, there are polynomials $\varphi_{k,n}$ such that

$\displaystyle \left|f_k(x)-\varphi_{k,n}(x)\right| < \frac{1}{\sqrt{p}n}\qquad\text{for any }x\in K.$

If we define the function $\varphi_n:=(\varphi_{1,n},\ldots,\varphi_{p,n})$ which maps $K$ to $\mathbb{R}^p$, we get

\displaystyle \begin{aligned} \left|f(x)-\varphi_n(x)\right|^2 &= \sum_{k=1}^p\left|f_k(x)-\varphi_{k,n}(x)\right|^2 \\ &< \frac{p}{pn^2} \\ &= \frac1{n^2},\qquad\text{for any }x\in K \end{aligned}

and in particular $\varphi_n\to f$ uniformly in $K$.

Besides,

$\displaystyle \left|\varphi_n(x)\right|\leq\left|\varphi_n(x)-f(x)\right| + \left|f(x)\right| < \frac1n + \left|f(x)\right| \leq \frac1n + 1 =:\alpha_n.$

This allows us to set

$\displaystyle \psi_n(x) = \frac{\varphi_n(x)}{\alpha_n}$.

This function also converges uniformly to $f$, as for any $x\in K$,

\displaystyle \begin{aligned} \left|\psi_n(x)-f(x)\right| &= \left|\frac{\varphi_n(x)}{\alpha_n} - f(x)\right| \\ &= \frac1{\left|\alpha_n\right|}\left|\varphi_n(x)-\alpha_nf(x)\right|\\ &\leq \frac1{\left|\alpha_n\right|}\left|\varphi_n(x)-f(x)\right| + \frac1{\left|\alpha_n\right|}\left|f(x)-\alpha_nf(x)\right|\\ &< \frac1{\left|\alpha_n\right|}\frac1n + \frac1{\left|\alpha_n\right|}\left|f(x)\right|\left|1-\alpha_n\right|\\ &< (1+\delta)\frac1n + \frac{\delta}{1+\delta}\left|f(x)\right|\\ &< \varepsilon \qquad\text{for }n\gg0. \end{aligned}

Finally, for $x\in K$, by construction, $\left|\varphi_n(x)\right|\leq\alpha_n$, and so $\left|\psi_n(x)\right| = \frac{\left|\varphi_n(x)\right|}{\alpha_n} < 1$, which means that $\psi_n:K\to\mathring K$.

The point of this lemma is to state that if we had shown Brouwer’s Fixed Point Theorem for every such function $\psi_n:K\to\mathring K$, whose components are polynomials, we had proved it for the general continuous function $f$. This can be seen as follows:

As we suppose Brouwer’s Theorem was true for the $\psi_n$, there would be a sequence $(x_n)\subset K$ with $\psi_n(x_n) = x_n$. As $K$ is (sequentially) compact, there is a convergent subsequence $(x_{n_j})\subset(x_n)$, and $\lim_jx_{n_j} = x_0\in K$. For sufficiently large $j$, we see

$\displaystyle \left|\psi_{n_j}(x_{n_j})-f(x_0)\right| \leq\left|\psi_{n_j}(x_{n_j})-f(x_{n_j})\right| + \left|f(x_{n_j})-f(x_0)\right| < \frac\varepsilon2 + \frac\varepsilon2.$

The first bound follows from the fact that $\psi_{n_j}\to f$ uniformly, the second bound is the continuity of $f$ itself, with the fact that $x_{n_j}\to x_0$. In particular,

$\displaystyle x_0 = \lim_{j} x_{n_j} = \lim_{j} \psi_{n_j}(x_{n_j}) = f(x_0).$

So, $f$ has the fixed point $x_0$, which proves Brouwer’s Theorem.

In effect, it suffices to deal with functions like the $\psi_n$ for the rest of this text. q.e.d.

Slogan (The geometric intuition): If Brouwer’s Fixed Point Theorem is wrong, then there is “almost” a retraction of $K$ to $\partial K$.

Or, rephrased as a proper lemma:

Lemma: For $f$ being polynomially simplified as in the previous lemma, assuming $x\neq f(x)$ for any $x\in K$, we can construct a continuously differentiable function $g_t:K\to K$, $t\in[0,1]$, with $g_t(x)=x$ for $x\in\partial K$. This function is given via

$\displaystyle g_t(x) =x + t\lambda(x)\bigl(x-f(x)\bigr),$

$\displaystyle \lambda(x)=\frac{-x\cdot\bigl(x-f(x)\bigr)+\sqrt{\left(x\cdot\bigl(x-f(x)\bigr)\right)^2+\bigl(1-\left|x\right|^2\bigr)\left|x-f(x)\right|^2}}{\left|x-f(x)\right|^2}.$

The mapping $t\mapsto g_t$ is the direct line from $x$ to the boundary of $\partial K$, which also passes through $f(x)$. $\lambda(x)$ is the parameter in the straight line that defines the intersection with $\partial K$.

Proof: As we suppose, Brouwer’s Fixed Point Theorem is wrong the continuous function $\left|x-f(x)\right|$ is positive for any $x\in K$. Because of continuity, for every $y\in \partial K$, there is some $\varepsilon = \varepsilon(y) > 0$, such that still $\left|x-f(x)\right|>0$ in the neighborhood $U_{\varepsilon(y)}(y)$.

Here, we have been in technical need of a continuation of $f$ beyond $K$. As $f$ is only defined on $K$ itself, we might take $f(x):=f\bigl(\frac{x}{\left|x\right|}\bigr)$ for $\left|x\right|>1$. We still have $\left|f(x)\right| < 1$ and $f(x)\neq x$, which means that we don’t get contradictions to our assumptions on $f$. Let’s not dwell on this for longer than necessary.

On the compact set $\partial K$, finitely many of the neighborhoods $U_{\varepsilon(y)}(y)$ will suffice to cover $\partial K$. One of them will have a minimal radius. We shall set $\delta = \min_y\varepsilon(y) +1$, to find: there is an open set $U = U_\delta(0)\supset K$ with $\left|x-f(x)\right| >0$ for all $x\in U$.

Let us define for any $x\in U$

$\displaystyle d(x):=\frac{\left(x\cdot\bigl(x-f(x)\bigr)\right)^2+\bigl(1-\left|x\right|\bigr)^2\left|x-f(x)\right|^2}{\left|x-f(x)\right|^4}$.

It is well-defined by assumption. We distinguish three cases:

$a)$ $\left|x\right|<1$: Then $1-\left|x\right|^2>0$ and the numerator of $d(x)$ is positive.

$b)$ $\left|x\right|=1$: Then the numerator of $d(x)$ is

$\displaystyle \left(x\cdot\bigl(x-f(x)\bigr)\right)^2 = \bigl(x\cdot x - x\cdot f(x)\bigr)^2 = \bigl(\left|x\right|^2-x\cdot f(x)\bigr)^2 = \bigl(1-x\cdot f(x)\bigr)^2,$

where by Cauchy-Schwarz and by assumption on $f$, we get

$\displaystyle x\cdot f(x) \leq \left|x\right|\left|f(x)\right| = \left|f(x)\right| < 1\qquad (\spadesuit).$

In particular, the numerator of $d(x)$ is strictly positive.

$c)$ $\left|x\right|>1$: This case is not relevant for what’s to come.

We have seen that $d(x)>0$ for all $\left|x\right|\leq 1$. Since $d$ is continuous, a compactness argument similar to the one above shows that there is some $V = V_{\delta'}(0)\supset K$ with $d(x)>0$ for all $x\in V$. If we pick $\delta'=\delta$ if $\delta$ is smaller, we find: $d$ is positive and well-defined on $V$.

The reason why we have looked at $d$ is not clear yet. Let us grasp at some geometry first. Let $x\in V$ and $\Gamma_x = \left\{x+\lambda\bigl(x-f(x)\bigr)\colon\lambda\in\mathbb{R}\right\}$ the straight line through $x$ and $f(x)$. If we look for the intersection of $\Gamma_x$ with $\partial K$, we solve the equation

$\displaystyle\left|x+\lambda\bigl(x-f(x)\bigr)\right| = 1.$

The intersection “closer to” $x$ is denoted by some $\lambda>0$.

This equation comes down to

\displaystyle \begin{aligned} && \left(x+\lambda\bigl(x-f(x)\bigr)\right) \cdot \left(x+\lambda\bigl(x-f(x)\bigr)\right) &=1 \\ &\iff& \left|x\right|^2 + 2\lambda x\cdot\bigl(x-f(x)\bigr) + \lambda^2\left|x-f(x)\right|^2 &=1\\ &\iff& \lambda^2\left|x-f(x)\right|^2 + 2\lambda x\cdot\bigl(x-f(x)\bigr) &= 1-\left|x\right|^2\\ &\iff& \left(\lambda+\frac{x\cdot\bigl(x-f(x)\bigr)}{\left|x-f(x)\right|^2}\right)^2 &= \frac{1-\left|x\right|^2}{\left|x-f(x)\right|^2} + \left(\frac{x\cdot\bigl(x-f(x)\bigr)}{\left|x-f(x)\right|^2}\right)^2 \\ &\iff& \left(\lambda+\frac{x\cdot\bigl(x-f(x)\bigr)}{\left|x-f(x)\right|^2}\right)^2 &= \frac{(1-\left|x\right|)^2\left|x-f(x)\right|^2+\left(x\cdot\bigl(x-f(x)\bigr)\right)^2}{\left|x-f(x)\right|^4} \\ &\iff& \left(\lambda+\frac{x\cdot\bigl(x-f(x)\bigr)}{\left|x-f(x)\right|^2}\right)^2 &= d(x). \end{aligned}

As $x\in V$, $d(x)>0$, and hence there are two real solutions to the last displayed equation. Let $\lambda(x)$ be the larger one (to get the intersection with $\partial K$ closer to $x$), then we find

\displaystyle \begin{aligned} \lambda(x) &= \sqrt{d(x)} - \frac{x\cdot\bigl(x-f(x)\bigr)}{\left|x-f(x)\right|^2}\\ &= \frac{-x\cdot\bigl(x-f(x)\bigr)+\sqrt{\left(x\cdot\bigl(x-f(x)\bigr)\right)^2+\bigl(1-\left|x\right|^2\bigr)\left|x-f(x)\right|^2}}{\left|x-f(x)\right|^2}. \end{aligned}

By construction,

$\displaystyle \left|x+\lambda(x)\bigl(x-f(x)\bigr)\right| = 1,\qquad\text{for all }x\in V.\qquad(\clubsuit)$

Let us define

$\displaystyle g_t(x) = x+t\lambda(x)\bigl(x-f(x)\bigr),\qquad t\in[0,1],~x\in V.$

This is (at least) a continuously differentiable function, as we simplified $f$ to be a polynomial and the denominator in $\lambda(x)$ is bounded away from $0$. Trivially and by construction, $g_0(x)=x$ and $\left|g_1(x)\right| = 1$ for all $x\in V$.

For $\left|x\right|<1$ and $t<1$, we have

\displaystyle \begin{aligned} \left|x+t\lambda(x)\bigl(x-f(x)\bigr)\right| &\stackrel{\hphantom{(\clubsuit)}}{=} \left|\bigl(t+(1-t)\bigr)x + t\lambda(x)\bigl(x-f(x)\bigr)\right|\\ &\stackrel{\hphantom{(\clubsuit)}}{=}\left|t\left(x+\lambda(x)\bigl(x-f(x)\bigr)\right)+(1-t)x\right|\\ &\stackrel{\hphantom{(\clubsuit)}}{\leq} t\left|x+\lambda(x)\bigl(x-f(x)\bigr)\right|+(1-t)\left|x\right|\\ &\stackrel{(\clubsuit)}{=} t+(1-t)\left|x\right|\\ &\stackrel{\hphantom{(\clubsuit)}}{<} t+(1-t) = 1\qquad (\heartsuit). \end{aligned}

Hence, $\left|g_t(x)\right|<1$ for $\left|x\right|<1$ and $t\in[0,1)$. This means $g_t(\mathring K)\subset\mathring K$ for $t<1$.

For $\left|x\right|=1$, we find (notice that $x\cdot\bigl(x-f(x)\bigr)>0$ for $\left|x\right|=1$, by $(\spadesuit)$).

\displaystyle \begin{aligned} \lambda(x) &= \frac{-x\cdot\bigl(x-f(x)\bigr)+\sqrt{\left(x\cdot\bigl(x-f(x)\bigr)\right)^2}}{\left|x-f(x)\right|^4} \\ &= \frac{-x\cdot\bigl(x-f(x)\bigr)+x\cdot\bigl(x-f(x)\bigr)}{\left|x-f(x)\right|^4} = 0. \end{aligned}

This is geometrically entirely obvious, since $\lambda(x)$ denotes the distance of $x$ to the intersection with $\partial K$; if $x\in\partial K$, this distance is apparently $0$.

We have seen that $g_t(x)=x$ for $\left|x\right|=1$ for any $t\in[0,1]$. Hence, $g_t(\partial K)=\partial K$ for all $t$. $g_t$ is almost a retraction, $g_1$ actually is a retraction. q.e.d.

Note how tricky the general formality gets, compared to the more compact and descriptive proof that we gave in the special case $p=2$. The arguments of the lemma and in the special case are identical.

Lemma (The bijection): Let $\hat K$ be a closed ball around $0$, $K\subset\hat K\subset V$. The function $g_t$ is a bijection on $\hat K$, for $t\geq0$ sufficiently small.

Proof: We first show that $g_t$ is injective. Let us define $h(x):=\lambda(x)\bigl(x-f(x)\bigr)$, for reasons of legibility. As we saw above, $h$ is (at least) continuously differentiable. We have

$\displaystyle g_t(x) = x+th(x),\qquad g_t'(x)=\mathrm{Id}+th'(x).$

As $\hat K$ is compact, $h'$ is bounded by $\left|h'(x)\right|\leq C$, say. By enlarging $C$ if necessary, we can take $C\geq1$. Now let $x,y\in\hat K$ with $g_t(x)=g_t(y)$. That means $x+th(x)=y+th(y)$ and so, by the mean value theorem,

$\displaystyle \left|x-y\right| = t\left|h(x)-h(y)\right|\leq tC\left|x-y\right|.$

By setting $\varepsilon:=\frac1C$ and taking $t\in[0,\varepsilon)$, we get $\left|x-y\right| = 0$. $g_t$ is injective for $t<\varepsilon$.

Our arguments also proved $\left|th'(x)\right| < 1$. Let us briefly look at the convergent Neumann series $\sum_{k=0}^\infty\bigl(th'(x)\bigr)^k$, having the limit $s$, say. We find

$\displaystyle sth'(x) = \sum_{k=0}^\infty\bigl(th'(x)\bigr)^{k+1} = s-\mathrm{Id},$

which tells us

$\displaystyle \mathrm{Id} = s-s\cdot th'(x) = s\bigl(\mathrm{Id}-th'(x)\bigr).$

In particular, $g_t'(x) = \mathrm{Id}-th'(x)$ is invertible, with the inverse $s$. Therefore, $\det g_t'(x)\neq0$. Since this determinant is a continuous function of $t$, and $\det g_0'(x) = \det\mathrm{Id} = 1$, we have found

$\displaystyle \det g_t'(x) > 0 \text{ for any }t\in[0,\varepsilon),~x\in\hat K$.

Now, let us show that $g_t$ is surjective. As $\det g_t'(x)$ never vanishes on $\hat K$, $g_t$ is an open mapping (by an argument involving the inverse function theorem; $g_t$ can be inverted locally in any point, hence no point can be a boundary point of the image). This means that $g_t(\mathring K)$ is open.

Let $z\in K$ with $z\notin g_t(\mathring K)$; this makes $z$ the test case for non-surjectivity. Let $y\in g_t(\mathring K)$; there is some such $y$ due to $(\heartsuit)$. The straight line between $y$ and $z$ is

$\displaystyle \overline{yz}:=\left\{(1-\lambda)y+\lambda z\colon \lambda\in[0,1]\right\}.$

As $g_t$ is continuous, there must be some point $v\in\partial g_t(\mathring K)\cap\overline{yz}$; we have to leave the set $g_t(\mathring K)$ somewhere. Let us walk the line until we do, and then set

$\displaystyle v=(1-\lambda_0)y+\lambda_0z,\qquad\text{with }\lambda_0=\sup\left\{\lambda\in[0,1]\colon\overline{y;(1-\lambda)y+\lambda z}\subset g_t(\mathring K)\right\}.$

Now, continuous images of compact sets remain compact: $g_t(K)$ is compact and hence closed. Therefore, we can conclude

$\displaystyle g_t(\mathring K)\subset g_t(K)\quad\implies\quad \overline{g_t(\mathring K)}\subset g_t(K)\quad\implies\quad v\in\overline{g_t(\mathring K)}\subset g_t(K).$

This means that there is some $u\in K$ such that $v=g_t(u)$. As $g_t(\mathring K)$ is open, $u\in\partial K$ (since otherwise, $v\notin\partial g_t(\mathring K)$ which contradicts the construction). Therefore, $\left|u\right|=1$, and since $g_t$ is almost a retraction, $g_t(u)=u$. Now,

$\displaystyle v=g_t(u) = u \quad\implies\quad v\in\partial K.$

But by construction, $v$ is a point between $z\in K$ and $y\in g_t(\mathring K)$; however, $y\notin\partial K$, since $g_t(\mathring K)$ is open. Due to the convexity of $K$, we have no choice but $z\in\partial K$, and by retraction again, $g_t(z)=z$. In particular, $z\in g_t(\partial K)$.

We have shown that if $z\notin g_t(\mathring K)$, then $z\in g_t(\partial K)$. In particular, $z\in g_t(K)$ for any $z\in K$. $g_t$ is surjective. q.e.d.

Lemma (The Integral Application): The real-valued function

$\displaystyle V(t)=\int_K\det g_t'(x)dx$

is a polynomial and satisfies $V(1)>0$.

Proof: We have already seen in the previous lemma that $\det g_t'(x)>0$ on $x\in\mathring{\hat K}$ for $t<\varepsilon$. This fact allows us to apply the transformation formula to the integral:

$\displaystyle V(t) = \int_{g_t(K)}1dx.$

As $g_t$ is surjective, provided $t$ is this small, $g_t(K) = K$, and therefore

$\displaystyle V(t) = \int_K1dx = \mu(K).$

In particular, this no longer depends on $t$, which implies $V(t)>0$ for any $t<\varepsilon$.

By the Leibniz representation of the determinant, $\det g_t'(x)$ is a polynomial in $t$, and therefore, so is $V(t)$. The identity theorem shows that $V$ is constant altogether: in particular $V(1)=V(0)>0$. q.e.d.

Now we can readily conclude the proof of Brouwer’s Fixed Point Theorem, and we do it in a rather unexpected way. After the construction of $g_t$, we had found $\left|g_1(x)\right|=1$ for all $x\in V$. Let us write this in its components and take a partial derivative ($j=1,\ldots,p$)

\displaystyle \begin{aligned} &&1 &= \sum_{k=1}^p\bigl(g_{1,k}(x)\bigr)^2\\ &\implies& 0 &= \frac\partial{\partial x_j}\sum_{k=1}^p\bigl(g_{1,k}(x)\bigr)^2 = \sum_{k=1}^p2\frac{\partial g_{1,k}(x)}{\partial x_j}g_{1,k}(x) \end{aligned}

This last line is a homogeneous system of linear equations, that we might also write like this

$\displaystyle \begin{pmatrix}\frac{\partial g_{1,1}(x)}{\partial x_1}&\cdots &\frac{\partial g_{1,p}(x)}{\partial x_1}\\ \ldots&&\ldots\\ \frac{\partial g_{1,1}(x)}{\partial x_p}&\cdots&\frac{\partial g_{1,p}(x)}{\partial x_p}\end{pmatrix} \begin{pmatrix}\xi_1\\\ldots\\\xi_p\end{pmatrix} = 0,$

and our computation has shown that the vector $\bigl(g_{1,1}(x),\ldots,g_{1,p}(x)\bigr)$ is a solution. But the vector $0$ is a solution as well. These solutions are different because of $\left|g_1(x)\right| = 1$. If a system of linear equations has two different solutions, it must be singular (it is not injective), and the determinant of the linear system vanishes:

$\displaystyle 0 = \det \begin{pmatrix}\frac{\partial g_{1,1}(x)}{\partial x_1}&\cdots &\frac{\partial g_{1,p}(x)}{\partial x_1}\\ \ldots&&\ldots\\ \frac{\partial g_{1,1}(x)}{\partial x_p}&\cdots&\frac{\partial g_{1,p}(x)}{\partial x_p}\end{pmatrix} = \det g_1'(x).$

This means

$\displaystyle 0 = \int_K\det g_1'(x)dx = V(1) > 0$.

A contradiction, which stems from the basic assumption that Brouwer’s Fixed Point Theorem were wrong. The Theorem is thus proved. q.e.d.

Let us make some concluding remarks. Our proof made vivid use of the fact that if there is a retraction, Brouwer’s Theorem must be wrong (this is where we got our contradiction in the end: the retraction cannot exist). The proof may also be started the other way round. If we had proved Brouwer’s Theorem without reference to retractions (this is how Elstrodt does it), you can conclude that there is no retraction from $K$ to $\partial K$ as follows: if there was a retraction $g:K\to\partial K$, we could consider the mapping $-g$. It is, in particular, a mapping of $K$ to itself, but it does not have any fixed point – a contradiction to Brouwer’s Theorem.

Brouwer’s Theorem, as we have stated it here, is not yet ready to drink. For many applications, the set $K$ is too much of a restriction. It turns out, however, that the hardest work has been done. Some little approximation argument (which in the end amounts to continuous projections) allows to formulate, for instance:

• Let $C\subset\mathbb{R}^p$ be convex, compact and $C\neq\emptyset$. Let $f:C\to C$ be continuous. Then $f$ has a fixed point.
• Let $E$ be a normed space, $K\subset E$ convex and $\emptyset\neq C\subset K$ compact. Let $f:K\to C$ be continuous. Then $f$ has a fixed point.
• Let $E$ be a normed space, $K\subset E$ convex and $K\neq\emptyset$. Let $f:K\to K$ be continuous. Let either $K$ be compact or $K$ bounded and $f(K)$ relatively compact. Then $f$ has a fixed point.

The last two statements are called Schauder’s Fixed Point Theorems, which may often be applied in functional analysis, or are famously used for proofs of Peano’s Theorem in differential equations. But at the core of all of them is Brouwer’s Theorem. This seems like a good place to end.

# What does the determinant have to do with transformed measures?

Let us consider transformations of the space $\mathbb{R}^p$. How does Lebesgue measure change by this transformation? And how do integrals change? The general case is answered by Jacobi’s formula for integration by substitution. We will start out slowly and only look at how the measure of sets is transformed by linear mappings.

It is folklore in the basic courses on linear algebra, when the determinant of a matrix is introduced, to convey the notion of size of the parallelogram spanned by the column vectors of the matrix. The following theorem shows why this folklore is true; this of course is based on the axiomatic description of the determinant which encodes the notion of size already. But coming from pure axiomatic reasoning, we can connect the axioms of determinant theory to their actual meaning in measure theory.

First, remember the definition of the pushforward measure. Let $X$ and $Y$ be measurable spaces, and $f:X\to Y$ a measurable mapping (i.e. it maps Borel-measurable sets to Borel-measurable sets; we shall not deal with finer details of measure theory here). Let $\mu$ be a measure on $X$. Then we define a measure on $Y$ in the natural – what would $\mu$ do? – way:

$\displaystyle \mu_f(A) := \mu\bigl(f^{-1}(A)\bigr).$

In what follows, $X = Y = \mathbb{R}^p$ and $\mu = \lambda$ the Lebesgue measure.

Theorem (Transformation of Measure): Let $f$ be a bijective linear mapping and let $A\subset\mathbb{R}^p$ be a measurable set. Then, the pushforward measure satisfies:

$\displaystyle \lambda_f(A) = \left|\det f\right|^{-1}\lambda(A)\qquad\text{for any Borel set }A\in\mathcal{B}^p.$

Lemma (The Translation-Lemma): Lebesgue measure is invariant under translations.

Proof: Let $c,d\in\mathbb{R}^p$ with $c\leq d$ component-wise. Let $t_a$ be the shift by the vector $a\in\mathbb{R}^p$, i.e. $t_a(v) = v+a$ and $t_a(A) = \{x+a\in\mathbb{R}^p\colon x\in A\}$. Then,

$\displaystyle t_a^{-1}\bigl((c,d]\bigr) = (c-a,d-a],$

where the interval is meant as the cartesian product of the component-intervals. For the Lebesgue-measure, we get

$\displaystyle \lambda_{t_a}\bigl((c,d]\bigr) = \lambda\bigl((c-a,d-a]\bigr) = \prod_{i=1}^p\bigl((d-a)-(c-a)\bigr) = \prod_{i=1}^p(d-c) = \lambda\bigl((c,d]\bigr).$

The measures $\lambda$, $\lambda_{t_a}$ hence agree on the rectangles (open on their left-hand sides), i.e. on a semi-ring generating the $\sigma$-algebra $\mathcal{B}^p$. With the usual arguments (which might as well involve $\cap$-stable Dynkin-systems, for instance), we find that the measures agree on the whole $\mathcal{B}^p$.

q.e.d.

Lemma (The constant-multiple-Lemma): Let $\mu$ be a translation-invariant measure on $\mathcal{B}^p$, and $\alpha:=\mu\bigl([0,1]^p\bigr) < \infty$. Then $\mu(A) = \alpha\lambda(A)$ for any $A\in\mathcal(B)^p$.

Note that the Lemma only holds for finite measures on $[0,1]^p$. For instance, the counting measure is translation-invariant, but it is not a multiple of Lebesgue measure.

Proof: We divide the set $(0,1]^p$ via a rectangular grid of sidelengths $\frac1{n_i}$, $i=1,\ldots,p$:

$\displaystyle (0,1]^p = \bigcup_{\stackrel{k_j=0,\ldots,n_j-1}{j=1,\ldots,p}}\left(\times_{i=1}^p\left(0,\frac1{n_i}\right] + \left(\frac{k_1}{n_1},\ldots,\frac{k_p}{n_p}\right)\right).$

On the right-hand side there are $\prod_{i=1}^pn_i$ sets which have the same measure (by translation-invariance). Hence,

$\displaystyle \mu\bigl((0,1]^p\bigr) = \mu\left(\bigcup\cdots\right) = n_1\cdots n_p \cdot \mu\left(\times_{i=1}^p \left(0,\frac1{n_i}\right]\right).$

Here, we distinguish three cases.

Case 1: $\mu\bigl((0,1]^p\bigr) = 1$. Then,

$\displaystyle\mu\left(\times_{i=1}^p (0,\frac1{n_i}]\right) = \prod_{i=1}^p\frac1{n_i} = \lambda\left(\times_{i=1}^p (0,\frac1{n_i}]\right).$

By choosing appropriate grids and further translations, we get that $\mu(I) = \lambda(I)$ for any rectangle $I$ with rational bounds. Via the usual arguments, $\mu=\lambda$ on the whole of $\mathcal{B}^p$.

Case 2: $\mu\bigl((0,1]^p\bigr) \neq 1$ and $>0$. By assumption, however, this measure is finite. Setting $\gamma = \mu\bigl((0,1]^p\bigr)$, we can look at the measure $\gamma^{-1}\mu$, which of course has $\gamma^{-1}\mu\bigl((0,1]^p\bigr) = 1$. By Case 1, $\gamma^{-1}\mu = \lambda$.

Case 3: $\mu\bigl((0,1]^p\bigr) = 0$. Then, using translation invariance again,

$\displaystyle \mu(\mathbb{R}^p) = \mu\bigl(\bigcup_{z\in\mathbb{Z}^p}((0,1]^p+z)\bigr) = \sum_{z\in\mathbb{Z}^p}\mu\bigl((0,1]^p\bigr) = 0.$

Again, we get $\mu(A) = 0$ for all $A\in\mathcal{B}^p$.

That means, in all cases, $\mu$ is equal to a constant multiple of $\lambda$, the constant being the measure of $(0,1]^p$. That is not quite what we intended, as we wish the constant multiple to be the measure of the compact set $[0,1]^p$.

Remember our setting $\alpha:=\mu\bigl([0,1]^p\bigr)$ and $\gamma := \mu\bigl((0,1]^p\bigr)$. Let $A\in\mathcal{B}^p$. We distinguish another two cases:

Case $a)$ $\alpha = 0$. By monotony, $\gamma = 0$ and Case 3 applies: $\mu(A) = 0 = \alpha\lambda(A)$.

Case $b)$ $\alpha > 0$. By monotony and translation-invariance,

$\displaystyle \alpha \leq \mu\bigl((-1,1]^p\bigr) = 2^p\gamma,$

meaning $\gamma\geq\frac{\alpha}{2^p}$. Therefore, as $\alpha>0$, we get $\gamma>0$, and by Case 1, $\frac1\gamma\mu(A) = \lambda(A)$. In particular,

$\displaystyle \frac\alpha\gamma = \frac1\gamma\mu\bigl([0,1]^p\bigr) = \lambda\bigl([0,1]^p\bigr) = 1,$

and so, $\alpha = \gamma$, meaning $\mu(A) = \gamma\lambda(A) = \alpha\lambda(A)$.

q.e.d.

Proof (of the Theorem on Transformation of Measure). We will first show that the measure $\lambda_f$ is invariant under translations.

We find, using the Translation-Lemma in $(\ast)$, and the linearity of $f$ before that,

\displaystyle \begin{aligned} \lambda_{t_a\circ f}(A) = \lambda_f(A-a) &\stackrel{\hphantom{(\ast)}}{=} \lambda\bigl(f^{-1}(A-a)\bigr) \\ &\stackrel{\hphantom{(\ast)}}{=} \lambda\bigl(f^{-1}(A) - f^{-1}(a)\bigr) \\ &\stackrel{(\ast)}{=} \lambda\bigl(f^{-1}(A)\bigr) \\ &\stackrel{\hphantom{(\ast)}}{=} \lambda_f(A), \end{aligned}

which means that $\lambda_f$ is indeed invariant under translations.

As $[0,1]^p$ is compact, so is $f^{-1}\bigl([0,1]^p\bigr)$ – remember that continuous images of compact sets are compact (here, the continuous mapping is $f^{-1}$). In particular, $f^{-1}\bigl([0,1]^p\bigr)$ is bounded, and thus has finite Lebesgue measure.

We set $c(f) := \lambda_f\bigl([0,1]^p)\bigr)$. By the constant-multiple-Lemma, $\lambda_f$ is a multiple of Lebesgue measure: we must have

$\displaystyle \lambda_f(A) = c(f)\lambda(A)\text{ for all }A\in\mathcal{B}^p.\qquad (\spadesuit)$

We only have left to prove that $c(f) = \left|\det f\right|^{-1}$. To do this, there may be two directions to follow. We first give the way that is laid out in Elstrodt’s book (which we are basically following in this whole post). Later, we shall give the more folklore-way of concluding this proof.

We consider more and more general fashions of the invertible linear mapping $f$.

Step 1: Let $f$ be orthogonal. Then, for the unit ball $B_1(0)$,

$\displaystyle c(f)\lambda\bigl(B_1(0)\bigr) \stackrel{(\spadesuit)}{=} \lambda_f\bigl(B_1(0)\bigr) = \lambda\left(f^{-1}\bigl(B_1(0)\bigr)\right) = \lambda\bigl(B_1(0)\bigr).$

This means, that $c(f) = 1 = \left|\det(f)\right|$.

This step shows for the first time how the properties of a determinant encode the notion of size already: we have only used the basic lemmas on orthogonal matrices (they leave distances unchanged and hence the ball $B_1(0)$ doesn’t transform; besides, their inverse is their adjoint) and on determinants (they don’t react to orthogonal matrices because of their multiplicative property and because they don’t care for adjoints).

Step 2: Let $f$ have a representation as a diagonal matrix (using the standard basis of $\mathbb{R}^p$). Let us assume w.l.o.g. that $f = \mathrm{diag}(d_1,\ldots,d_p)$ with $d_i>0$. The case of $d_i<0$ is only notationally cumbersome. We get

$\displaystyle c(f)\lambda_f\bigl([0,1]^p\bigr) = \lambda\left(f^{-1}\bigl([0,1]^p\bigr)\right) = \lambda\bigl(\times_{i=1}^p[0,d_i^{-1}]\bigr) = \prod_{i=1}^pd_i^{-1} = \left|\det f\right|^{-1}.$

Again, the basic lemmas on determinants already make use of the notion of size without actually saying so. Here, it is the computation of the determinant by multiplication of the diagonal.

Step 3: Let $f$ be linear and invertible, and let $f^\ast$ be its adjoint. Then $f^\ast f$ is non-negative definite (since for $x\in\mathbb{R}^p$, $x^\ast(f^\ast f)x = (fx)^\ast(fx) = \left\|fx\right\|^2\geq0$). By the Principal Axis Theorem, there is some orthogonal matrix $v$ and some diagonal matrix with non-negative entries $d$ with $f^\ast f = vd^2v^\ast$. As $f$ was invertible, no entry of $d$ may vanish here (since then, its determinant would vanish and in particular, $f$ would no longer be invertible). Now, we set

$\displaystyle w:=d^{-1}v^\ast f^\ast,$

which is orthogonal because of

$\displaystyle ww^\ast = d^{-1} v^\ast f^\ast fv d^{-1} = d^{-1}v^\ast (vd^2v^\ast)v d^{-1} = d^{-1}d^2d^{-1}v^\ast v v^\ast v = \mathrm{id}.$

As $f = w^\ast dv$, we see from Step 1

\displaystyle \begin{aligned} c(f) = \lambda_f\bigl([0,1]^p\bigr) &= \lambda\left(v^{-1}d^{-1}w\bigl([0,1]^p\bigr)\right) \\ &= \lambda\left(d^{-1}w\bigl([0,1]^p\bigr)\right)\\ &= \left|\det d\right|^{-1}\lambda\left(w\bigl([0,1]^p\bigr)\right) \\ &= \left|\det d\right|^{-1}\lambda\bigl([0,1]^p\bigr) \\ &= \left|\det f\right|^{-1}\lambda\bigl([0,1]^p\bigr) \\ &= \left|\det f\right|^{-1}, \end{aligned}

by the multiplicative property of determinants again ($\det f = \det d$).

q.e.d.(Theorem)

As an encore, we show another way to conclude in the Theorem, once all the Lemmas are shown and applied. This is the more folklore way alluded to in the proof, making use of the fact that any invertible matrix is the product of elementary matrices (and, of course, making use of the multiplicative property of determinants). Hence, we only consider those.

Because Step 2 of the proof already dealt with diagonal matrices, we only have to look at shear-matrices like $E_{ij}(r) := \bigl(\delta_{kl}+r\delta_{ik}\delta_{jl}\bigr)_{k,l=1,\ldots,p}$. They are the identity matrix with the (off-diagonal) entry $r$ in row $i$ and column $j$. One readily finds $\bigl(E_{ij}(r)\bigr)^{-1} = E_{ij}(-r)$, and $\det E_{ij}(r) = 1$. Any vector $v\in[0,1]^p$ is mapped to

$\displaystyle E_{ij}(v_1,\ldots,v_i,\ldots, v_p)^t = (v_1,\ldots,v_i+rv_j,\ldots,v_p)^t.$

This gives

\displaystyle \begin{aligned}\lambda_{E_{ij}(r)}\bigl([0,1]^p\bigr) &= \lambda\left(E_{ij}(-r)\bigl([0,1]^p\bigr)\right) \\ &= \lambda\left(\left\{x\in\mathbb{R}^p\colon x=(v_1,\ldots,v_i-rv_j,\ldots,v_p), v_k\in[0,1]\right\}\right). \end{aligned}

This is a parallelogram that may be covered by $n$ rectangles as follows: we fix the dimension $i$ and one other dimension to set a rectangle of height $\frac1n$, width $1+\frac rn$ (all other dimension-widths = 1; see the image for an illustration). Implicitly, we have demanded that $p\geq2$ here; but $p=1$ is uninteresting for the proof, as there are too few invertible linear mappings in $\mathbb{R}^1$.

By monotony, this yields

$\lambda_{E_{ij}(r)}\bigl([0,1]^p\bigr) \leq n\frac1n\left(1+\frac{r}{n}\right) = 1+\frac rn\xrightarrow{n\to\infty}1.$

On the other hand, this parallelogram itself covers the rectangles of width $1-\frac rn$, and a similar computation shows that in the limit $\lambda_{E_{ij}(r)}\bigl([0,1]^p\bigr)\geq1$.

In particular: $\lambda_{E_{ij}(r)}\bigl([0,1]^p\bigr) = 1 = \left|\det E_{ij}(r)\right|^{-1}$.

q.e.d. (Theorem encore)

Proving the multidimensional transformation formula for integration by substitution is considerably more difficult than in one dimension, where it basically amounts to reading the chain rule reversedly. Let us state the formula here first:

Theorem (The Transformation Formula, Jacobi): Let $U,V\subset \mathbb{R}^p$ be open sets and let $\Phi:U\to V$ be a $\mathcal{C}^1$-diffeomorphism (i.e. $\Phi^{-1}$ exists and both $\Phi$ and $\Phi^{-1}$ are $\mathcal{C}^1$-functions). Let $f:V\to\mathbb{R}$ be measurable. Then, $f\circ\Phi:U\to\mathbb{R}$ is measurable and

$\displaystyle\int_V f(t)dt = \int_U f\bigl(\Phi(s)\bigr)\left|\det\Phi'(s)\right|ds.$

At the core of the proof is the Theorem on Transformation of Measure that we have proved above. The idea is to approximate $\Phi$ by linear mappings, which locally transform the Lebesgue measure underlying the integral and yield the determinant in each point as correction factor. The technical difficulty is to show that this approximation does no harm for the evaluation of the integral.

We will need a lemma first, which carries most of the weight of the proof.

The Preparatory Lemma: Let $U,V\subset \mathbb{R}^p$ be open sets and let $\Phi:U\to V$ be a $\mathcal{C}^1$-diffeomorphism. If $X\subset U$ is a Borel set, then so is $\Phi(X)\subset V$, and

$\displaystyle \lambda\bigl(\Phi(X)\bigr)\leq\int_X\left|\det\Phi'(s)\right|ds.$

Proof: Without loss of generality, we can assume that $\Phi$, $\Phi'$ and $(\Phi')^{-1}$ are defined on a compact set $K\supset U$. We consider, for instance, the sets

$\displaystyle U_k:=\left\{ x\in U\colon \left|x\right|\frac1k\right\}.$

The $U_k$ are open and bounded, $\overline U_k$ is hence compact, and there is a chain $U_k\subset\overline U_k\subset U_{k+1}\subset\cdots$ for all $k$, with $U=\bigcup_kU_k$. To each $U_k$ there is, hence, a compact superset on which $\Phi$, $\Phi'$ and $(\Phi')^{-1}$ are defined. Now, if we can prove the statement of the Preparatory Lemma on $X_k := X\cap U_k$, it will also be true on $X=\lim_kX_k$ by the monotone convergence theorem.

As we can consider all relevant functions to be defined on compact sets, and as they are continuous (and even more) by assumption, they are readily found to be uniformly continuous and bounded.

It is obvious that $\Phi(X)$ will be a Borel set, as $\Phi^{-1}$ is continuous.

Let us prove that the Preparatory Lemma holds for rectangles $I$ with rational endpoints, being contained in $U$.

There is some $r>0$ such that for any $a\in I$, $B_r(a)\subset U$. By continuity, there is a finite constant $M$ with

$\displaystyle M:=\sup_{t\in I}\left\|\bigl(\Phi'(t)\bigr)^{-1}\right\|,$

and by uniform continuity, $r$ may be chosen small enough such that, for any $\varepsilon>0$, even

$\displaystyle \sup_{x\in B_r(a)}\left\|\Phi'(x)-\Phi'(a)\right\|\leq\frac{\varepsilon}{M\sqrt p} \text{ for every }a\in I.$

With this $r$, we may now sub-divide our rectangle $I$ into disjoint cubes $I_k$ of side-length $d$ such that $d<\frac{r}{\sqrt p}$. In what follows, we shall sometimes need to consider the closure $\overline I_k$ for some of the estimates, but we shall not make the proper distinction for reasons of legibility.

For any given $b\in I_k$, every other point $c$ of $I_k$ may at most have distance $d$ in each of its components, which ensures

$\displaystyle\left\|b-c\right\|^2 \leq \sum_{i=1}^pd^2 = pd^2 < r^2.$

This, in turn, means, $I_k\subset B_r(b)$ (and $B_r(b)\subset U$ has been clear because of the construction of $r$).

Now, in every of the cubes $I_k$, we choose the point $a_k\in I_k$ with

$\displaystyle\left|\det\Phi'(a_k)\right| = \min_{t\in I_k}\left|\det\Phi'(t)\right|,$

and we define the linear mapping

$\displaystyle \Phi_k:=\Phi'(a_k)$.

Remember that for convex sets $A$, differentiable mappings $h:A\to\mathbb{R}^p$, and points $x,y\in A$, the mean value theorem shows

$\displaystyle \left\|h(x)-h(y)\right\|\leq\left\|x-y\right\|\sup_{\lambda\in[0,1]}\left\|h'\bigl(x+\lambda(y-x)\bigr)\right\|.$

Let $a\in I_k$ be a given point in one of the cubes. We apply the mean value theorem to the mapping $h(x):=\Phi(x)-\Phi_k(x)$, which is certainly differentiable, to $y:=a_k$, and to the convex set $A:=B_r(a)$:

\displaystyle \begin{aligned} \left\|h(x)-h(y)\right\|&\leq\left\|x-y\right\|\sup_{\lambda\in[0,1]}\left\|h'\bigl(x+\lambda(y-x)\bigr)\right\|\\ \left\|\Phi(x)-\Phi_k(x)-\Phi(a_k)+\Phi_k(a_k)\right\| & \leq \left\|x-a_k\right\|\sup_{\lambda\in[0,1]}\left\|\Phi'\bigl(x+\lambda(x-a_k)\bigr)-\Phi'(a_k)\right\|\\ \left\|\Phi(x)-\Phi(a_k)-\Phi_k(x-a_k)\right\| &< \left\|x-a_k\right\| \frac{\varepsilon}{M\sqrt p}\qquad (\clubsuit). \end{aligned}

Note that as $a_k\in I_k\subset B_r(a)$, $x+\lambda(x-a_k)\in B_r(a)$ by convexity, and hence the upper estimate of uniform continuity is applicable. Note beyond that, that $\Phi_k$ is the linear mapping $\Phi'(a_k)$ and the derivative of a linear mapping is the linear mapping itself.

Now, $\left\|x-a_k\right\|< d\sqrt p$, as both points are contained in $I_k$, and hence $(\clubsuit)$ shows

\displaystyle \begin{aligned} \Phi(I_k) &\subset \Phi(a_k)+\Phi_k(I_k-a_k)+B_{\frac{\varepsilon}{M\sqrt p}d\sqrt p}(0) \\ &\subset \Phi(a_k)+\Phi_k(I_k-a_k)+B_{\frac{d\varepsilon}{M}}(0). \end{aligned}

By continuity (and hence boundedness) of $\Phi'$, we also have

$\displaystyle \left\|(\Phi_k)^{-1}(x)\right\|\leq\left\|\bigl(\Phi'(a_k)\bigr)^{-1}\right\|\left\|x\right\|\leq M \left\|x\right\|$,

which means $B_{\frac{d\varepsilon}{M}}(0) = \Phi_k\left(\Phi_k^{-1}\bigl(B_{\frac{d\varepsilon}{M}}(0)\bigr)\right) \subset \Phi_k\bigl(B_{d\varepsilon}(0)\bigr)$.

Hence:

$\displaystyle \Phi(I_k) \subset \Phi(a_k) + \Phi_k\bigl(I_k-a_k+B_{d\varepsilon}(0)\bigr).$

Why all this work? We want to bound the measure of the set $\Phi(I_k)$, and we can get it now: the shift $\Phi(a_k)$ is unimportant by translation invariance. And the set $I_k-a_k+B_{d\varepsilon}(0)$ is contained in a cube of side-length $d+2d\varepsilon$. As promised, we have approximated the mapping $\Phi$ by a linear mapping $\Phi_k$ on a small set, and the transformed set has become only slightly bigger. By the Theorem on Transformation of Measure, this shows

\displaystyle \begin{aligned} \lambda\bigl(\Phi(I_k)\bigr) &\leq \lambda\left(\Phi_k\bigl(I_k-a_k+Blat_{d\varepsilon}(0)\bigr)\right) \\ &=\left|\det\Phi_k\right|\lambda\bigl(I_k-a_k+B_{d\varepsilon}(0)\bigr)\\ &\leq \left|\det\Phi_k\right|d^p(1+2\varepsilon)^p \\ &= \left|\det\Phi_k\right|(1+2\varepsilon)^p\lambda(I_k). \end{aligned}

Summing over all the cubes $I_k$ of which the rectangle $I$ was comprised, (remember that $\Phi$ is a diffeomorphism and disjoint sets are kept disjoint; besides, $a_k$ has been chosen to be the point in $I_k$ of smallest determinant for $\Phi'$)

\displaystyle \begin{aligned} \lambda\bigl(\Phi(I)\bigr) &\leq (1+2\varepsilon)^p\sum_{k=1}^n\left|\det \Phi_k\right|\lambda(I_k) \\ &= (1+2\varepsilon)^p\sum_{k=1}^n\left|\det \Phi'(a_k)\right|\lambda(I_k)\\ &= (1+2\varepsilon)^p\sum_{k=1}^n\int_{I_k}\left|\det\Phi'(a_k)\right|ds\\ &\leq (1+2\varepsilon)^p\int_I\left|\det\Phi'(s)\right|ds. \end{aligned}

Taking $\varepsilon\to0$ yields to smaller subdivisions $I_k$ and in the limit to the conclusion. The Preparatory Lemma holds for rectangles.

Now, let $X\subset U$ be any Borel set, and let $\varepsilon>0$. We cover $X$ by disjoint (rational) rectangles $R_k\subset U$, such that $\lambda\bigl(\bigcup R_k \setminus X\bigr)<\varepsilon$. Then,

\displaystyle \begin{aligned} \lambda\bigl(\Phi(X)\bigr) &\leq \sum_{k=1}^\infty \lambda\bigl(\Phi(R_k)\bigr)\\ &\leq\sum_{k=1}^\infty\int_{R_k}\left|\det \Phi'(s)\right|ds\\ &= \int_{\bigcup R_k}\left| \det\Phi'(s)\right| ds\\ &= \int_X\left| \det\Phi'(s)\right| ds + \int_{\bigcup R_k\setminus X}\left|\det\Phi'(s)\right|ds\\ &\leq \int_X\left| \det\Phi'(s)\right| ds + M\lambda\left(\bigcup R_k\setminus X\right)\\ &\leq \int_X\left| \det\Phi'(s)\right| ds + M\varepsilon. \end{aligned}

If we let $\varepsilon\to0$, we see $\lambda\bigl(\Phi(X)\bigr)\leq\int_X\bigl|\det\Phi'(s)\bigr|ds$.

q.e.d. (The Preparatory Lemma)

We didn’t use the full generality that may be possible here: we already focused ourselves on the Borel sets, instead of the larger class of Lebesgue-measurable sets. We shall skip the technical details that are linked to this topic, and switch immediately to the

Proof of Jacobi’s Transformation Formula: We can focus on non-negative functions $f$ without loss of generality (take the positive and the negative part separately, if needed). By the Preparatory Lemma, we already have

\displaystyle\begin{aligned} \int_{\Phi(U)}\mathbf{1}_{\Phi(X)}(s)ds &= \int_{V}\mathbf{1}_{\Phi(X)}(s)ds\\ &= \int_{\Phi(X)}ds\\ &= \lambda\bigl(\Phi(X)\bigr)\\ &\leq \int_X\left|\det\Phi'(s)\right|ds\\ &= \int_U\mathbf{1}_X(s)\left|\det\Phi'(s)\right|ds\\ &= \int_U\mathbf{1}_{\Phi(X)}\bigl(\Phi(s)\bigr)\left|\det\Phi'(s)\right|ds, \end{aligned}

which proves the inequality

$\displaystyle \int_{\Phi(U)}f(t)dt \leq \int_U f\bigl(\Phi(s)\bigr)\left|\det\Phi'(s)\right|ds,$

for indicator functions $f = \mathbf{1}_{\Phi(X)}$. By usual arguments (linearity of the integral, monotone convergence), this also holds for any measurable function $f$. To prove the Transformation Formula completely, we apply this inequality to the transformation $\Phi^{-1}$ and the function $g(s):=f\bigl(\Phi(s)\bigr)\left|\det\Phi'(s)\right|$:

\displaystyle \begin{aligned} \int_Uf\bigl(\Phi(s)\bigr)\left|\det\Phi'(s)\right|ds &= \int_{\Phi^{-1}(V)}g(t)dt\\ &\leq \int_Vg\bigl(\Phi^{-1}(t)\bigr)\left|\det(\Phi^{-1})'(t)\right|dt\\ &=\int_{\Phi(U)}f\Bigl(\Phi\bigl(\Phi^{-1}(t)\bigr)\Bigr)\left|\det\Phi'\bigl(\Phi^{-1}(t)\bigr)\right|\left|\det(\Phi^{-1})'(t)\right|dt\\ &=\int_{\Phi(U)}f(t)dt, \end{aligned}

since the chain rule yields $\Phi'\bigl(\Phi^{-1}\bigr)(\Phi^{-1})' = \bigl(\Phi(\Phi^{-1})\bigr)' = \mathrm{id}$. This means that the reverse inequality also holds. The Theorem is proved.

q.e.d. (Theorem)

There may be other, yet more intricate proofs of this Theorem. We shall not give any other of them here, but the rather mysterious looking way in which the determinant pops up in the transformation formula is not the only way to look at it. There is a proof by induction, given in Heuser’s book, where the determinant just appears from the inductive step. However, there is little geometric intuition in this proof, and it is by no means easier than what we did above (as it make heavy use of the theorem on implicit functions). Similar things may be said about the rather functional analytic proof in Königsberger’s book (who concludes the transformation formula by step functions converging in the $L^1$-norm, he found the determinant pretty much in the same way that we did).

Let us harvest a little of the hard work we did on the Transformation Formula. The most common example is the integral of the standard normal distribution, which amounts to the evaluation of

$\displaystyle \int_{-\infty}^\infty e^{-\frac12x^2}dx.$

This can happen via the transformation to polar coordinates:

$\Phi:(0,\infty)\times(0,2\pi)\to\mathbb{R}^2,\qquad (r,\varphi)\mapsto (r\cos \varphi, r\sin\varphi).$

For this transformation, which is surjective on all of $\mathbb{R}^2$ except for a set of measure $0$, we find

$\Phi'(r,\varphi) = \begin{pmatrix}\cos\varphi&-r\sin\varphi\\\sin\varphi&\hphantom{-}r\cos\varphi\end{pmatrix},\qquad \det\Phi'(r,\varphi) = r.$

From the Transformation Formula we now get

\displaystyle \begin{aligned} \left(\int_{-\infty}^\infty e^{-\frac12x^2}dx\right)^2 &= \int_{\mathbb{R}^2}\exp\left(-\frac12x^2-\frac12y^2\right)dxdy\\ &= \int_{\Phi\left((0,\infty)\times(0,2\pi)\right)}\exp\left(-\frac12x^2-\frac12y^2\right)dxdy\\ &= \int_{(0,\infty)\times(0,2\pi)}\exp\left(-\frac12r^2\cos^2(\varphi)-\frac12r^2\sin^2(\varphi)\right)\left|\det\Phi'(r,\varphi)\right|drd\varphi\\ &= \int_{(0,\infty)\times(0,2\pi)}\exp\left(-\frac12r^2\right)rdrd\varphi\\ &= \int_0^\infty\exp\left(-\frac12r^2\right)rdr\int_0^{2\pi}d\varphi \\ &= 2\pi \left[-\exp\left(-\frac12r^2\right)\right]_0^\infty\\ &= 2\pi \left(1-0\right)\\ &= 2\pi. \end{aligned}

In particular, $\int_{-\infty}^\infty e^{-\frac12x^2}dx=\sqrt{2\pi}$. One of the very basic results in probability theory.

Another little gem that follows from the Transformation Formula are the Fresnel integrals

$\displaystyle \int_{0}^\infty \cos(x^2)dx = \int_{0}^\infty\sin(x^2)dx = \sqrt{\frac{\pi}{8}}.$

They follow from the same basic trick given above for the standard normal density, but as other methods for deriving this result involve even trickier uses of similarly hard techniques (the Residue Theorem, for instance, as given in Remmert’s book), we shall give the proof of this here:

Consider

$\displaystyle F(t)=\int_{0}^\infty e^{-tx^2}\cos(x^2)dx\qquad\text{and}\qquad\int_{0}^\infty e^{-tx^2}\sin(x^2)dx.$

Then, the trigonometric identity $\cos a+b = \cos a \cos b - \sin a\sin b$ tells us

\displaystyle \begin{aligned} \bigl(F(t)\bigr)^2 - \big(G(t)\bigr)^2 &= \int_0^\infty\int_0^\infty e^{-t(x^2+y^2)}\cos(x^2)\cos(y^2)dxdy - \int_0^\infty\int_0^\infty e^{-t(x^2+y^2)}\sin(x^2)\sin(y^2)dxdy\\ &= \int_0^\infty\int_0^\infty e^{-t(x^2+y^2)}\bigl(\cos(x^2+y^2)+\sin(x^2)\sin(y^2) - \sin(x^2)\sin(y^2)\bigr)dxdy\\ &= \int_0^\infty\int_0^{\frac\pi2}e^{-tr^2}\cos(r^2)rd\varphi dr\\ &= \frac\pi2 \int_0^\infty e^{-tu}\cos u\frac12du. \end{aligned}

This integral can be evaluated by parts to show

$\displaystyle \int_0^\infty e^{-tu}\cos udu\left(1+\frac1{t^2}\right) = \frac1t,$

which means

$\displaystyle \bigl(F(t)\bigr)^2 - \bigl(G(t)\bigr)^2 = \frac\pi4\int_0^\infty\int_0^\infty e^{-tu}\cos udu = \frac\pi4\frac t{t^2+1}.$

Then we consider the product $F(t)G(t)$ and use the identity $\sin(a+b) = \cos a\sin b + \cos b\sin a$, as well as the symmetry of the integrand and integration by parts, to get

\displaystyle \begin{aligned} F(t)G(t) &= \int_0^\infty\int_0^\infty e^{-t(x^2+y^2)}\bigl(\cos(x^2)\sin(y^2)\bigr)dxdy\\ &=2\int_0^\infty\int_0^ye^{-t(x^2+y^2)}\sin(x^2+y^2)dxdy\\ &=2\int_0^\infty\int_0^{\frac\pi8}e^{-tr^2}\sin(r^2)rd\varphi dr\\ &=\frac\pi4\int_0^\infty e^{-tr^2}\sin(r^2)r dr\\ &=\frac\pi4\int_0^\infty e^{-tu}\sin u\frac12du\\ &=\frac\pi8\frac1{1+t^2}. \end{aligned}

We thus find by the dominated convergence theorem

\displaystyle \begin{aligned} \left(\int_0^\infty\cos x^2dx\right)^2-\left(\int_0^\infty\sin x^2dx\right)^2 &= \left(\int_0^\infty\lim_{t\downarrow0}e^{-tx}\cos x^2dx\right)^2-\left(\int_0^\infty\lim_{t\downarrow0}e^{-tx}\sin x^2dx\right)^2 \\ &=\lim_{t\downarrow0}\left(\bigl(F(t)\bigr)^2-\bigl(G(t)\bigr)^2\right)\\ &=\lim_{t\downarrow0}\frac\pi4\frac{t}{t^2+1}\\ &=0, \end{aligned}

and

\displaystyle \begin{aligned} \left(\int_0^\infty\cos x^2dx\right)^2 &= \left(\int_0^\infty\cos x^2dx\right)\left(\int_0^\infty\sin x^2dx\right)\\ &=\left(\int_0^\infty\lim_{t\downarrow0}e^{-tx}\cos x^2dx\right)\left(\int_0^\infty\lim_{t\downarrow0}e^{-tx}\sin x^2dx\right)\\ &=\lim_{t\downarrow0}F(t)G(t)\\ &=\lim_{t\downarrow0}\frac\pi8\frac1{1+t^2}\\ &=\frac\pi8. \end{aligned}

One can easily find the bound that both integrals must be positive and from the first computation, we get

$\int_0^\infty\cos x^2dx = \int_0^\infty\sin x^2dx,$

from the second computation follows that the integrals have value $\sqrt{\frac\pi8}$.

q.e.d. (Fresnel integrals)

Even Brouwer’s Fixed Point Theorem may be concluded from the Transformation Formula (amongst a bunch of other theorems, none of which is actually as deep as this one though). This is worthy of a seperate text, mind you.

In the past few weeks I have encountered a youtube channel that gives several nice ideas about art and about some artistic ideas work. Apparently, youtube (which is to mean: Google) knows enough about my musical preferences to present a video called “Lord of the Rings: How music elevates story”, which de-constructs the musical leitmotifs in the score of Peter Jackson’s Lord of the Rings trilogy:

In particular, I had recognized the different themes in the score by myself, but the way they were intertwined had stayed hidden from my conscious thinking. A very nice clip that most certainly made my day – and in particular, I found out that the nerdwriter-channel had plenty of other very insightful ideas to discover: understanding Picasso or Edvard Munch (he didn’t just paint The Scream, you know, even though it is his very most famous piece, and rightly so), looking deeper into Bob Dylan’s lyrics, analyzing the speeches and tweets of President Donald Trump, something on how great directors work, as in Sherlock, or Saving Private Ryan, even about the Beatles’ cover of Sgt. Pepper’s Lonely Hearts Club Band (which would tell me little new, but I wouldn’t expect that in a non-specialist video). A very nice channel, challenging me to think deeply about many different topics, both old and new to me.

# Johann Kepler and how planets move

Kepler’s Laws of planetary motion follow very smoothly from Newton’s Law of Gravitation. Very little tough mathematics is needed for the proofs, it can actually be done with ordinary differential calculus and some knowledge on path integration.

Of course, from a historical point of view, these laws appeared reversed. Kepler had neither Newton’s Law at his disposal, nor had he sufficient use of the calculus machinery that we have today. Instead, his way of coming up with his Laws included years of hard work on the astronomical tables compiled by Tycho Brahe; Kepler himself was unable to make astronomical observations himself, even with his self-invented telescope, as he was ill-sighted for all his life. Later, Newton could rely on Kepler’s results to find inspiration for his Law of Gravitation: indeed, unless he could relate his results to Kepler’s laws, he knew his results to be incomplete. Together with his many other achievements (for instance, Kepler was the first to state Simpson’s rule, which is accordingly called “Kepler’s barrel rule” sometimes) this makes him one of the most interesting minds of the early modern era. His interest in planet’s orbits was rooted in astrology and the construction of horoscopes; his observations and deductions led him to drop both Copernicus’ thought that the orbits were circles and, later, that there were no platonic solids involved. Both Copernicus and Kepler had revolutionized the thoughts on space itself, Copernicus by showing how much easier the orbits can be described when Earth is allowed to move itself (no more epicycles and the like), but Kepler made it even easier by dropping circles altogether.

To consider how hard the problem of finding the planet’s orbits is, consider that we have plenty of observational data of the planets, however the data contain two unknowns: the orbit of Earth and the orbit of the planet. On top of that, our observations only tell us about the angle under which the planet is observed, we don’t learn anything about the distances involved (unless we have Kepler’s third law at our disposal). In a very insightful talk for a wider audience, Terry Tao has explained some of Kepler’s ideas on this, especially how Kepler dealt with the orbit of Mars which had been for several reasons the most tricky one of the orbits in the models that preceded Kepler. Tao mentions that Einstein valued the finding of Kepler’s Laws one of the most shining moments in the history of human curiosity. “And if Einstein calls you a genius, you are really doing well.

From these Laws and from what Newton and his successors achieved, many things can be inferred that are impossible to measure directly. For instance the mass of the Sun and all planets can be computed from here, once the gravitational constant is known (which is tricky to pinpoint, actually). Voltaire is quoted with the sentence, regarding Newton’s achievements but this would fit to Kepler as well, that the insights gained “semblaient n’être pas faites pour l’esprit humain.

To give just a tiny bit of contrast, we mention that Kepler also had erroneous thoughts that show how deeply he was still rooted in ancient ideas of harmonics and aesthetics. For instance, Kepler tried to prove why the Solar system had exactly six planets (or to rephrase a little more accurately to his thinking: why God had found pleasure in creating exactly six planets). For some time, he believed that the reason was related to the fact that there are exactly five platonic solids which define the structure of the six orbits around the sun. Those were ideas also related to the integer harmonies of a vibrating string, as the planets were supposed to move in a harmonical way themselves. Of course, in these days the observations were limited up to Saturn, as the outer planets (and dwarf planets) cannot be found by eyesight or the telescopes at Kepler’s disposal; all such ideas were doomed to be incomplete. However, his quest for harmonics in the Solar system led him in the end to his Third Law. On another account, Kepler was mistaken in the deduction of his First Law, since he lacked the deep knowledge about integration that would be developed decades later; luckily, his mistake cancels out with another mistake later on: “Es ist schon atemberaubend, wie sich bei Keplers Rechnungen letztlich doch alles fügt.” (“It is stunning how everything in Kepler’s computations adds up in the end”; have a look at Sonar’s highly readable and interesting book on the history of calculus for this).

In what follows, we shall show how the Kepler’s Laws can be proved, assuming Newton’s Law of Gravitation, in a purely mathematical fashion. There will be no heuristics from physics or from astronomy, only the axiomatic mathematical deduction that mostly works without any intuition from the applications (though we will look at motivations for why some definitions are made the way they are).

As a nice aside, we can look at the mathematical descriptions of the conic sections on which the first Law relies. But here again, there’s no connection to why these curve are called this way.

Let us state Kepler’s Laws here first.

Kepler’s First Law of Planetary Motion: Planets orbit in ellipses, with the Sun as one of the foci.

Kepler’s Second Law of Planetary Motion: A planet sweeps out equal areas in equal times.

Kepler’s Third Law of Planetary Motion: The square of the period of an orbit is proportional to the cube of its semi-major axis.

Let us prove this, by following the account of Königsberger’s book on calculus. Many calculus books deal with Kepler’s Laws in a similar axiomatical fashion, yet we stick to this account as it appears to be the neatest one without conjuring up too much of physics.

We shall give a couple of technical lemmas first.

The Triangle-Lemma: The triangle marked by the points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ has area $\displaystyle \frac12\left|\mathrm{det}\begin{pmatrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{pmatrix}\right|$.

Proof: The triangle together with the coordinate axes marks the parallelograms:

$P_{13}$ with the points $(x_1,y_1)$, $(x_3, y_3)$, $(x_3,0)$, $(x_1,0)$;

$P_{23}$ with the points $(x_2,y_2)$, $(x_3, y_3)$, $(x_3,0)$, $(x_2,0)$;

$P_{12}$ with the points $(x_1,y_1)$, $(x_2, y_2)$, $(x_2,0)$, $(x_1,0)$.

Thus, the area of the triangle is:

$F_{T} = P_{13}+P_{23}-P_{12}$.

The sign represents the situation given in the figure. For other triangles, another permutation of the signs may be necessary, but there will always be exactly one negative sign. Other permutations of the sign only represent a re-numbering of the points and therefore a change of sign in the determinant given in the statement. As we put absolute values to our statement, we avoid any difficulties of this kind.

As each of the paralellograms has two of their points on the $x$-axis, we find

\begin{aligned} F_{T} &= \frac{y_1+y_3}{2}(x_3-x_1)+\frac{y_2+y_3}{2}(x_2-x_3)-\frac{y_1+y_2}{2}(x_2-x_1)\\ &= \frac12\left(x_1\bigl((y_1+y_2)-(y_1+y_3)\bigr) + x_2\bigl((y_2+y_3)-(y_1+y_2)\bigr) + x_3\bigl((y_1+y_3)-(y_2+y_3)\bigr)\right)\\ &= \frac12\left(x_1\bigl((y_1-y_3) + (y_2 - y_1)\bigr) + x_2(y_3-y_1) + x_3(y_1-y_2)\right)\\ &= \frac12\left((x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1)\right)\\ &= \frac12\mathrm{det}\begin{pmatrix}x_2-x_1&y_2-y_1\\x_3-x_1&y_3-y_1\end{pmatrix}\\ &= \frac12\mathrm{det}\begin{pmatrix}1&x_1&y_1\\0&x_2-x_1&y_2-y_1\\0&x_3-x_1&y_3-y_1\end{pmatrix}\\ &= \frac12\mathrm{det}\begin{pmatrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{pmatrix}. \end{aligned}

Q.e.d.

Lemma (Leibniz’ sector formula): Let $\gamma:[a,b]\to\mathbb{R}^2$ be a continuously differentiable path, $\gamma(t) = \begin{pmatrix}x(t)\\y(t)\end{pmatrix}$. Then the line segment from $0$ to the points of the path sweeps the area $\displaystyle\frac12\int_a^b(x\dot y - y\dot x)dt$.

Note that we have used Newton’s notation for derivatives. One might also write the integral as $\displaystyle\int_a^b\bigl(x(t)y'(t)-y(t)x'(t)\bigr)dt$.

Proof: Let us clarify first, what we understand by “sweeping” line segments. Consider the path $\gamma$ given in the image.

As this path is not closed (it’s not a contour), it doesn’t contain an area. But if we take the origin into account, we can define an area that is related to where the path is:

Now, pick a partition of $[a,b]$, such as $\{a=t_0,t_1,\ldots,t_{n-1},b=t_n\}$ and make a polygon of the partition and the origin – the corresponding triangles form an area that approximates the area bounded by $\gamma$, as above.

As the partition gets finer, we expect that the polygon-area converges to the $\gamma$-area. And this is where the definition originates, of the area that is swept by a path:

For any $\varepsilon>0$ there shall be $\delta>0$, such that for every partition $\{t_0,\ldots,t_n\}$ of $[a,b]$ that is finer than $\delta$, we get

$\displaystyle \left|\sum_{i=0}^{n-1}F_{T_i} - \frac12\int_a^b(x\dot y-y\dot x)dt\right| < \varepsilon.\qquad(\ast)$

Here, $F(T_i)$ is the area of the triangle bounded by $\gamma_{t_i}$, $\gamma_{t_{i+1}}$ and the origin. By the Triangle-Lemma, its area is $\displaystyle\frac12\mathrm{det}\begin{pmatrix}1&0&0\\1&x_i&y_i\\1&x_{i+1}&y_{i+1}\end{pmatrix}$.

Because the orientation of the $T_i$ might be of importance, $F(T_i)$ may keep its sign in what follows (Imagine, for instance, a path that traverses a line segment once from left to right and once from right to left; in total, no area is covered).

Now, let us prove that $(\ast)$ is true.

As $\dot x$ and $\dot y$ are continuous, choose $L=\max\left(\max_{[a,b]}\left|\dot x(t)\right|, \max_{[a,b]}\left|\dot y(t)\right|\right)$ and take $\delta = \frac{\varepsilon}{2L^2(b-a)}$. Take a partition $\{t_0,\ldots,t_n\}$ with $t_0=a$, $t_0=b$ and $\left|t_{k+1}-t_k\right| < \delta$ for $k=0,\ldots,n-1$. Then, for any such $k$,

\begin{aligned} F(T_k) &= \frac12\mathrm{det}\begin{pmatrix}1&0&0\\1&x_k&y_k\\1&x_{k+1}&y_{k+1}\end{pmatrix}\\ &=\frac12(x_ky_{k+1}-x_{k+1}y_k)\\ &= \frac12\Bigl(x_k\bigl(y_k+(y_{k+1}-y_k)\bigr) - y_k\bigl(x_k+(x_{k+1}-x_k)\bigr)\Bigr)\\ &=\frac12\bigl(x_k(y_{k+1}-y_k)-y_k(x_{k+1}-x_k)\bigr)\\ &=\frac12\left(x_k\int_{t_k}^{t_{k+1}}\dot y(t)dt - y_k\int_{t_k}^{t_{k+1}}\dot x(t)dt\right)\\ &=\frac12\int_{t_k}^{t_{k+1}}\bigl(x_k\dot y(t)-y_k\dot x(t)\bigr)dt. \end{aligned}

This yields, using the mean value theorem,

\begin{aligned} \left|2F(T_k)-\int_{t_k}^{t_{k+1}}(x\dot y-y\dot x)dt\right| &= \left|\int_{t_k}^{t_{k+1}}(x_k\dot y-y_k\dot x-x\dot y+y\dot x)dt\right|\\ &\leq \left|\int_{t_k}^{t_{k+1}}(x_k-x)\dot ydt\right| + \left|\int_{t_k}^{t_{k+1}}(y-y_k)\dot xdt\right|\\ &\leq\int_{t_k}^{t_{k+1}}\max_{[a,b]}\left|\dot x(t)\right|\left|{t_k-t}\right|\left|\dot y(t)\right|dt + \\ &\hphantom{=}+\int_{t_k}^{t_{k+1}}\max_{[a,b]}\left|\dot y(t)\right|\left|{t_k-t}\right|\left|\dot x(t)\right|dt\\ &\leq L^2\left|t_{k+1}-t_k\right|(t_{k+1}-t_k) + L^2\left|t_{k+1}-t_k\right|(t_{k+1}-t_k)\\ &< 2L^2\delta(t_{k+1}-t_k). \end{aligned}

We conclude

$\displaystyle\left|2\sum_{k=0}^{n-1}F(T_k) - \int_a^b(x\dot y - y\dot x)dt\right| < 2L^2\delta(b-a) = \varepsilon.$

Q.e.d.

One might as well prove this by applying Green’s Theorem, but in this case it just gets less elementary.

The $\times$-Lemma: Let $a,b\in\mathbb{R}^3$. We set

$\displaystyle a\times b := \begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$.

Obviously, this is linear both in $a$ and in $b$. We have

$(i)\quad\displaystyle \left\langle (a\times b),c\right\rangle = \mathrm{det}(a,b,c).$

$(ii)\quad a\times b$ is orthogonal to $a$ and to $b$.

$(iii)\quad\displaystyle (a\times b)\times c = -\left\langle b,c\right\rangle a + \left\langle a,c\right\rangle b$ (Grassmann’s equation).

$(iv)\quad\displaystyle (a\times b)^{\cdot} = \dot a\times b+a\times\dot b$.

Proof: For $(i)$:

\begin{aligned} \left\langle (a\times b),c\right\rangle &= \left\langle\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix},\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}\right\rangle\\ &=c_1(a_2b_3-a_3b_2)+c_2(a_3b_1-a_1b_3)+c_3(a_1b_2-a_2b_1)\\ &=c_1\mathrm{det}\begin{pmatrix}a_2&b_2\\a_3&b_3\end{pmatrix}-c_2\mathrm{det}\begin{pmatrix}a_1&b_1\\a_3&b_3\end{pmatrix}+c_3\mathrm{det}\begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\\ &=\mathrm{det}\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix}\\ &=\mathrm{det}(a,b,c). \end{aligned}

For $(ii)$: $\left\langle a\times b, a\right\rangle = \mathrm{det}(a,b,a)=0$ and $\left\langle a\times b,b\right\rangle = \mathrm{det}(a,b,b)=0$.

For $(iii)$:

\begin{aligned} (a\times b)\times c &=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}\times\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}\\ &=\begin{pmatrix}(a_3b_1-a_1b_3)c_3-(a_1b_2-a_2b_1)c_2\\(a_1b_2-a_2b_1)c_1-(a_2b_3-a_3b_2)c_3\\(a_2b_3-a_3b_2)c_2-(a_3b_1-a_1b_3)c_1\end{pmatrix}\\ &=\begin{pmatrix}a_3b_1c_3-a_1b_3c_3-a_1b_2c_2+a_2b_1c_2\\a_1b_2c_1-a_2b_1c_1-a_2b_3c_3+a_3b_2c_3\\a_2b_3c_2-a_3b_2c_2-a_3b_1c_1+a_1b_3c_1\end{pmatrix}\\ &=\begin{pmatrix}a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_1b_2c_2-a_1b_3c_3\\a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_2b_1c_1-a_2b_2c_2-a_2b_3c_3\\a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_3b_1c_1-a_3b_2c_2-a_3b_3c_3\end{pmatrix}\\ &=(a_1c_1+a_2c_2+a_3c_3)\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} - (b_1c_1+b_2c_2+b_3c_3)\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\\ &=\left\langle a,c\right\rangle b - \left\langle b,c\right\rangle a. \end{aligned}

For $(iv)$:

\begin{aligned} \bigl(a(t)\times b(t)\bigr)^\cdot &= \begin{pmatrix}a_2(t)b_3(t)-a_3(t)b_2(t)\\a_3(t)b_1(t)-a_1(t)b_3(t)\\a_1(t)b_2(t)-a_2(t)b_1(t)\end{pmatrix}^\cdot\\ &=\begin{pmatrix}\dot a_2b_3+a_2\dot b_3-\dot a_3b_2-a_3\dot b_2\\\dot a_3b_1+a_3\dot b_1-\dot a_1b_3-a_1\dot b_3\\\dot a_1b_2+a_1\dot b_2-\dot a_2b_1-a_2\dot b_1\end{pmatrix}\\ &=\begin{pmatrix}\dot a_2b_3-\dot a_3b_2\\\dot a_3b_1-\dot a_1b_3\\\dot a_1b_2-\dot a_2b_1\end{pmatrix}+\begin{pmatrix}a_2\dot b_3-a_3\dot b_2\\a_3\dot b_1-a_1\dot b_3\\ a_1\dot b_2-a_2\dot b_1\end{pmatrix} \\ &=\dot a\times b + a\times \dot b \end{aligned}

Q.e.d.

Now, let us look at conic sections and define them mathematically. We will not be interested in what these things have to do with cones – as stated at the beginning: pure mathematics here.

We are going to work in $\mathbb{R}^2$ here. Let $F$ be a point (the so-called focal point) and $l$ be a line (the so-called directrix), and the distance of $F$ and $l$ shall be some $p>0$. We are looking for all those points in $\mathbb{R}^2$ for which the distance to $F$ and the distance to $l$ are proportional – formally: For any point $(\xi,\eta)^t\in\mathbb{R}^2$, we set

$r=\mathrm{dist}\bigl(F,(\xi,\eta)^t\bigr)\qquad\text{and}\qquad d = \mathrm{dist}\bigl(l, (\xi,\eta)^t\bigr),$

and for $\varepsilon > 0$ we demand

$\displaystyle \frac{r}{d}=\varepsilon.$

For simplicity, we will put $F$ into the origin of our coordinate system, and $l$ parallel to one of the axes, as in the figure. In particular, $r^2 = \xi^2+\eta^2$ and $d = p+\xi$. Our equation for the interesting points thus becomes:

\begin{aligned} && r^2&=\varepsilon^2d^2\\ &\iff& \xi^2+\eta^2 &= \varepsilon^2p^2+2\varepsilon^2p\xi+\varepsilon^2\xi^2\\ &\iff&\xi^2(1-\varepsilon^2) &= \varepsilon^2p^2-\eta^2+2\varepsilon^2p\xi. \end{aligned}

Let us distinguish the following cases:

Case $\varepsilon = 1$. Then we set $x := \xi+\frac p2$, $y := \eta$ and find

\begin{aligned} &&\xi^2 (1-\varepsilon^2) &= \varepsilon^2p^2-\eta^2 + 2\varepsilon^2 p \xi \\ &\iff& 0 &= p^2-y^2+2p\left(x-\frac p2\right)\\ &\iff& 0 &= p^2-y^2+2px-p^2\\ &\iff& y^2 &= 2px. \end{aligned}

We see that the interesting points lie on a parabola which is open to the right (by choosing other coordinate systems, of course, any other parabola will appear; in some way, this is its normal form).

Case $\varepsilon < 1$. Here we set $x:=\xi-\frac{p\varepsilon^2}{1-\varepsilon^2}$, $y:=\eta$. Then we get

\begin{aligned} &&\xi^2 (1-\varepsilon^2) &= \varepsilon^2p^2-\eta^2 + 2\varepsilon^2 p \xi \\ &\iff &\left(x+\frac{p\varepsilon^2}{1-\varepsilon^2}\right)^2(1-\varepsilon^2) &= \varepsilon^2p^2-y^2+2\varepsilon^2p\left(x+\frac{p\varepsilon^2}{1-\varepsilon^2}\right)\\ &\iff& x^2(1-\varepsilon^2)+2xp\varepsilon^2 + \frac{p^2\varepsilon^4}{1-\varepsilon^2} &= \varepsilon^2p^2-y^2+2\varepsilon^2px+\frac{2\varepsilon^4p^2}{1-\varepsilon^2}\\ &\iff& x^2(1-\varepsilon^2) + y^2 &= \varepsilon^2p^2+\frac{\varepsilon^4p^2}{1-\varepsilon^2}\\ &\iff& x^2(1-\varepsilon^2) + y^2 &= \frac{(1-\varepsilon^2)\varepsilon^2p^2+\varepsilon^4p^2}{1-\varepsilon^2}\\ &\iff& x^2(1-\varepsilon^2) + y^2 &= \frac{\varepsilon^2p^2}{1-\varepsilon^2}\\ &\iff& \frac{x^2(1-\varepsilon^2)^2}{\varepsilon^2p^2} + \frac{y^2(1-\varepsilon^2)}{\varepsilon^2p^2} &= 1\\ &\iff& \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1, \end{aligned}

with $a = \frac{\varepsilon p}{1-\varepsilon^2}$ and $b = \frac{\varepsilon p}{\sqrt{1-\varepsilon^2}}$.

We have found that the interesting points lie on an ellipse.

Case $\varepsilon > 1$. This is exactly the same as the case $\varepsilon<1$, except for the last step. We mustn’t set $b$ as before, since $1-\varepsilon^2<0$ and we cannot get a real square root of this. Thus, we use $b = \frac{\varepsilon p}{\sqrt{\varepsilon^2-1}}$ and the resulting negative sign is placed in the final equation:

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1.$

This is a hyperbola.

To conclude this part, we give the general representation of conic sections in polar coordinates. From the figure given above, we see $d = p+r\cos\varphi$ and so

$\displaystyle r = \varepsilon d = \varepsilon p + \varepsilon r\cos\varphi,$

which means

$\displaystyle r = \frac{\varepsilon p}{1-\varepsilon\cos\varphi}.$

That yields the polar coordinates (only depending on parameters and on the variable $\varphi$:

$\displaystyle re^{i\varphi} = \frac{\varepsilon p}{1-\varepsilon\cos\varphi}e^{i\varphi}.$

Now, let us turn to our base for the proofs of Kepler’s Laws: Newton’s Law of Gravitation. Let $m$ be the mass of a planet, $M$ the mass of the Sun, $\gamma$ a real constant (the gravitational constant), and let $x(t)$ be a path (the planet’s orbit). By Newton’s Law we have the differential equation

$\displaystyle m\ddot x = -\gamma Mm\frac{x}{\left\|x\right\|^3}.$

On the left-hand side, there’s the definition of force as mass multiplied by acceleration. On the right-hand side is Newton’s Law stating the gravitational force between Sun and planet.

We define the vector-valued functions of $t$ (note that $x$ depends on $t$):

$\displaystyle J = x\times m\dot x\qquad\text{and}\qquad A=\frac1{\gamma Mm}J\times\dot x+\frac{x}{\left\|x\right\|}.$

($J$ is the angular momentum, $A$ is an axis; but our math doesn’t care for either of those names or intentions).

The $A$$J$-Lemma: As functions of $t$, $A$ and $J$ are constant.

Proof: Let us look at $J$ first. Using the fact that by definition $a\times a = 0$, and Newton’s Law of Gravitation, we get

\begin{aligned} \dot J \underset{\times\text{-Lemma}}{\overset{(iv)}{=}} (x\times m\dot x)^\cdot &= \dot x \times m\dot x + x\times m\ddot x\\ &= \dot x \times m\dot x + x\times\left(-\gamma Mm\frac{x}{\left\|x\right\|^3}\right)\\ &= m(\dot x\times \dot x) - \frac{\gamma Mm}{\left\|x\right\|^3}(x\times x)\\ &= 0 - 0 = 0. \end{aligned}

Now, for $A$, we will need a side-result first.

\begin{aligned} \frac{d}{dt}\frac{1}{\left\|x\right\|} &= \frac{d}{dt}\bigl(x_1^2(t)+x_2^2(t)+x_3^2(t)\bigr)^{-1/2}\\ &= -\frac12\bigl(x_1^2(t)+x_2^2(t)+x_3^2(t)\bigr)^{-3/2}\bigl(2x_1(t)\dot x_1(t)+2x_2(t)\dot x_2(t)+2x_3(t)\dot x_3(t)\bigr)\\ &= -\frac{\left\langle x,\dot x\right\rangle}{\left\|x\right\|^3}. \end{aligned}

This yields

\begin{aligned} \dot A &\underset{\hphantom{\times\text{-Lemma}}}{=} \left(\frac{1}{\gamma Mm}J\times \dot x + \frac{x}{\left\|x\right\|}\right)^\cdot \\ &\underset{\times\text{-Lemma}}{\overset{(iv)}{=}} \frac{1}{\gamma Mm}\bigl(\dot J\times\dot x + J\times\ddot x\bigr) + \left(\frac{1}{\left\|x\right\|}\dot x-\frac{\left\langle x,\dot x\right\rangle}{\left\|x\right\|^3}x\right)\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} \frac1{\gamma Mm}\left(0+J\times\left(-\gamma M\frac{x}{\left\|x\right\|^3}\right)\right) + \left(\frac{1}{\left\|x\right\|}\dot x-\frac{\left\langle x,\dot x\right\rangle}{\left\|x\right\|^3}x\right)\\ &\underset{\hphantom{\times\text{-Lemma}}}{=}-\frac1m\left((x\times m\dot x)\times \frac{x}{\left\|x\right\|^3}\right) + \left(\frac{1}{\left\|x\right\|}\dot x-\frac{\left\langle x,\dot x\right\rangle}{\left\|x\right\|^3}x\right)\\ &\underset{\hphantom{\times\text{-Lemma}}}{=}-\left((x\times \dot x)\times \frac{x}{\left\|x\right\|^3}\right) + \left(\frac{1}{\left\|x\right\|}\dot x-\frac{\left\langle x,\dot x\right\rangle}{\left\|x\right\|^3}x\right)\\ &\underset{\times\text{-Lemma}}{\overset{(iii)}{=}} -\left\langle x,\frac{x}{\left\|x\right\|^3}\right\rangle \dot x + \left\langle \dot x,\frac{x}{\left\|x\right\|^3}\right\rangle x + \frac{1}{\left\|x\right\|}\dot x-\frac{\left\langle x,\dot x\right\rangle}{\left\|x\right\|^3}x \\ &\underset{\hphantom{\times\text{-Lemma}}}{=} -\frac1{\left\|x\right\|^3}\left\langle x,x\right\rangle \dot x + \frac1{\left\|x\right\|^3}\left\langle \dot x, x\right\rangle x + \frac1{\left\|x\right\|}\dot x-\frac1{\left\|x\right\|^3}\left\langle x,\dot x\right\rangle x\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} -\frac1{\left\|x\right\|^3}\left\|x\right\|^2\dot x + \frac1{\left\|x\right\|}\dot x\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} 0. \end{aligned}

Q.e.d.

Now, we have all ingredients to prove Kepler’s Laws. We conclude axiomatically by assuming that planetary motion is governed by Newton’s Law of Gravitation (the differential equation given above).

Theorem (Kepler’s First Law of Planetary Motion): Planets orbit in ellipses, with the Sun as one of the foci.

Proof: Let $x(t)$ denote the orbit of a planet around the Sun. By the $A$$J$-Lemma, $J$ is constant. By definition of $J = x\times m\dot x$ and by $(ii)$ of the $\times$-Lemma, both $x$ and $\dot x$ are orthogonal to $J$; the orbit is therefore located in a two-dimensional plane. Let us introduce polar coordinates in this plane, with the Sun in the origin, and with axis $A$ (this works, as $A$ is located in the same plane as well: $A\bot J$ by the definition of $A$).

Now let $\varphi(t)$ be the angle of $x(t)$ and $A$, set $\varepsilon := \left\|A\right\|$. This means

$\displaystyle \cos\varphi(t) = \frac{\left\langle x(t),A\right\rangle}{\left\|x\right\|\left\|A\right\|},$

and so

$\displaystyle \left\langle A,x(t)\right\rangle = \varepsilon \left\|x\right\|\cos\varphi(t).$

By definition of $A$, we have

\begin{aligned} \left\langle A,x(t)\right\rangle &\underset{\hphantom{\times\text{-Lemma}}}{=} \left\langle\frac{1}{\gamma Mm}J\times\dot x + \frac{x}{\left\|x\right\|}, x(t)\right\rangle\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} \frac{1}{\gamma Mm}\left\langle J\times \dot x, x\right\rangle + \frac{1}{\left\|x\right\|}\left\langle x,x\right\rangle\\ &\underset{\times\text{-Lemma}}{\overset{(i)}{=}} \frac{1}{\gamma Mm}\mathrm{det}(J,\dot x,x)+\frac{\left\| x\right\|^2}{\left\|x\right\|}\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} (-1)^3\frac{1}{\gamma Mm}\mathrm{det}(x,\dot x,J) + \left\|x(t)\right\|\\ &\underset{\times\text{-Lemma}}{\overset{(i)}{=}} -\frac1{\gamma Mm}\left\langle x\times \dot x,J\right\rangle + \left\|x(t)\right\|\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} -\frac1{\gamma Mm^2}\left\langle x\times m\dot x, J\right\rangle + \left\|x(t)\right\|\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} -\frac1{\gamma Mm^2}\left\langle J,J\right\rangle + \left\|x(t)\right\|\\ &\underset{\hphantom{\times\text{-Lemma}}}{=} const. + \left\|x(t)\right\|. \end{aligned}

Now, if $A=0$, then we have found

$\displaystyle \left\|x(t)\right\| = \frac1{\gamma Mm^2}\left\|J\right\|^2,$

which means that the planet moves on a circular orbit.

If $A\neq0$, then we conclude

\begin{aligned} &&\varepsilon \left\|x(t)\right\|\cos\varphi(t) &= -\frac1{\gamma Mm^2}\left\|J\right\|^2+\left\|x(t)\right\|\\ &\implies&\varepsilon\left\|x(t)\right\|\cos\varphi(t)&=-\frac1{\gamma Mm^2}\left\|J\right\|^2+\left\|x(t)\right\|\\ &\implies&\frac1{\gamma Mm^2}\left\|J\right\|^2&=\bigl(1-\varepsilon\cos\varphi(t)\bigr)\left\|x(t)\right\|\\ &\implies&\left\|x(t)\right\|&=\frac{\frac{\left\|J\right\|^2}{\gamma Mm^2}}{1-\varepsilon\cos\varphi(t)} = \frac{\varepsilon\frac{\left\|J\right\|^2}{\gamma Mm^2\left\|A\right\|}}{1-\varepsilon\cos\varphi(t)} = \frac{\varepsilon p}{1-\varepsilon\cos\phi(t)}, \end{aligned}

with $p$ defined in the obvious fashion to make the last equation work.

Therefore, the planet moves on a conic section, with focus in the Sun. As the planet’s orbits are bounded, we have proved that it must follow an ellipse. Q.e.d.

Theorem (Kepler’s Second Law of Planetary Motion): A planet sweeps out equal areas in equal times.

Proof: We use cartesian coordinates in $\mathbb{R}^3$, such that $e_1$ is parallel to $A$ and $e_3$ is parallel to $J$. Then $x(t)$ is the plane $\mathrm{span}(e_1,e_2)$ and the Sun is in $(0,0)$. In particular, $x_3(t) = 0$ for all $t$ by the proof of the First Law. Then,

$\displaystyle \frac1m J = x\times \dot x = \begin{pmatrix}x_1\\x_2\\0\end{pmatrix}\times \begin{pmatrix}\dot x_1\\\dot x_2\\ 0\end{pmatrix} = \begin{pmatrix}0\\0\\x_1\dot x_2-x_2\dot x_1\end{pmatrix}.$

By Leibniz’ sector formula, the line segment between times $t_1$ and $t_2$ sweeps the area

$\displaystyle \frac12\left|\int_{t_1}^{t_2}(x_1\dot x_2-x_2\dot x_1)dt\right| = \frac1{2m}\left\|J\right\|(t_2-t_1).$

This area only depends on the difference of times, as stated. Q.e.d.

Theorem (Kepler’s Third Law of Planetary Motion): The square of the period of an orbit is proportional to the cube of its semi-major axis.

Proof: By Leibniz’ sector formula (used similarly to the proof of the Second Law), the area contained in the planet’s entire orbit is

$\displaystyle \frac12\left|\int_0^T (x_1\dot x_2-x_2\dot x_1)dt\right| = \frac1{2m}\left\|J\right\| T,$

where $T$ is the time taken for a full orbit around the Sun. By the First Law, this orbit is an ellipse, the area of which may be computed as follows: The cartesian coordinates of an ellipse are

$\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}a\cos t\\b\sin t\end{pmatrix},$

with $a$ and $b$ real constants (the larger one is called the semi-major axis). This is actually an ellipse because of

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{a^2\cos^2 t}{a^2}+\frac{b^2\sin^2t}{b^2} = 1.$

From the notations about normal forms of conic sections, we find $b = \frac{\varepsilon p}{\sqrt{1-\varepsilon2}} = a\sqrt{1-\varepsilon^2}$, which implies that $a>b$ (as $\varepsilon > 0$ for any conic section). Now, the area of the ellipse is, by Leibniz’ sector formula again,

\begin{aligned} \frac12\left|\int_0^{2\pi} (x_1\dot x_2-x_2\dot x_1)dt\right| &= \frac12\left|\int_0^{2\pi}\bigl(a \cos t \cdot b \cos t-b\sin t \cdot a (-\sin t)\bigr)dt\right| \\ &= \frac12 ab\int_0^{2\pi}dt \\ &= \pi ab. \end{aligned}

Both representations of the area covered by the orbit now yield

$\displaystyle T \frac1{2m}\left\|J\right\| = \pi a^2\sqrt{1-\varepsilon^2},$

and so, using the definition of $p$ obtained in the proof of the First Law,

\begin{aligned} \displaystyle T^2 &= \frac{4m^2}{\left\|J\right\|^2}\pi^2a^4(1-\varepsilon^2) \\ &= \frac{4\pi^2 m^2a^3}{\left\|J\right\|^2}a(1-\varepsilon^2) \\ &= \frac{4\pi^2m^2a^3}{\left\|J\right\|^2}\frac{\varepsilon p}{1-\varepsilon^2}(1-\varepsilon^2) \\ &= \frac{4\pi^2m^2a^3}{\left\|J\right\|^2}\varepsilon\frac{\left\|J\right\|^2}{\gamma Mm^2\left\|A\right\|} \\ &= \frac{4\pi^2}{\gamma M}a^3. \end{aligned}

The constant $\frac{4\pi^2}{\gamma M}$ is identical for any planet travelling around the Sun, and thus is constant. Q.e.d.

Let us conclude with a brief remark of how beautiful and elegant those Laws are – made my day.

# Loreena McKennitt’s music and lyrics

In the winter, when the dark evenings are long and I enjoy a cup of tea while looking out of my window to the stars, it is the time for me to listen to Loreena McKennitt’s music. This is by no means “easy listening” but it relaxes me a lot. It is music that I can sink into perfectly, especially when it’s dark outside and I can dream away.

Loreena McKennitt is a Canadian soprano singer and songwriter. Her music is inspired by mystic melodies, by celtic folk songs and middle eastern influences. She accordingly uses many unusual instruments: besides her singing, she performs many different instruments herself, such as a harp, an accordion and the keyboard. Many of her songs have been written by herself, but she also re-arranges classic folk music and even Elizabethan (resp. Shakespearian) songs.

My enjoyment of this music stems from her clear soprano voice, which has a marvelous, dreamy and still variable sound to it. Here is a very special singer who can transport her love for music to her listeners. I once had the distinct pleasure to listen to her in a live performance in an open-air concert – an evening that most certainly made my day.

I find it tricky to point out a few special songs here, as I can’t find any failures in her nine studio albums at all. They may all be recommended, and each has a certain feel to it, and I can’t put my finger on it – but different moods inspire me to listen to different albums. In particular, I refrain myself from posting links to any videos of her recordings – I wish to avoid cherry picking here.

My first encounters with her music came from the album “The Visit”, which contains the most beautiful rendition of the poem “The Lady of Shalott” – a poem known to every fan of Miss Marple’s detective stories, of course (“The mirror crack’d from side to side”) – which made me spend some time with English poetry for the first time. But then there is also the joyously sounding “All souls night”, the sad “Bonnie Portmore” and the ancient “Greensleeves” which apparently was only recorded in a short break in one take and still sounds wonderful on the album.

After I had been hooked, the album “Elemental” caught my attention with the harmonic duet “Carrighfergus”, a classic Irish song which is performed by a male singer and Loreena McKennitt rather contents herself with dreamy backing vocals. “Blacksmith” and “Stolen Child” are similar but the lyrics are performed by herself marvelously. Very notable is “She moved through the fair” for being entirely a cappella – there is not a hint of accompanying instrumentation to her crystal-clear voice. I am hard-pressed to think of any musician who would show this much courage on a recording (especially since this recording is about 30 years old now and couldn’t have been digitally enhanced).

“Parallel Dreams” was an album that I bought on a vacation trip many years ago, when there was too few information to know which albums were available – when I saw this during my trip to France, I bought it on the spot, the price certainly higher than what I’d have paid at home, but I wouldn’t regret that. With “Annachie Gordon” she performs a traditional Scottish song telling a sad love-story. And “Dicken’s Dublin” is a most beautiful and yet heart-breaking Christmas Song, speaking of the hope of a homeless child to find a home, and being voiced-over by a child narrating the story of the birth of Jesus. What beauty lies in this composition.

For “The Mask and the Mirror”, she wrote about her travels in the booklet, and how she was inspired for her various songs on this album. Here appear references to oriental and middle eastern music for the first time. “The dark Night of the Soul” has been her encore in her live performance that I attended, and the evening only got complete with this song. Never before have I witnessed a crowd that was so focused and so mesmerized by a single person performing on the stage.

“The Book of Secrets” appears to be one of her most famous albums, and rightfully so. With “Skellig”, “Dante’s Prayer” and “The Highwayman”, she again takes classical pieces uses this influence and produces wonderful art. I do not wish to make this whole post tedious by enumerating all the marvelous songs that can be found on “An Ancient Muse” and on “The Wind that shakes the Barley”, but let it be said that those most recent albums are by no means any worse than their predecessors. Wonderful music all over the place.

The sad thing is, that since 2010 there has not been any new music from her. But as she still travels the world and performs her music on stage, I have not given up hope that we’ll have new music eventually. And wouldn’t that be beautiful?

# On numerical integration and convergence issues

The methods for numerical integration are of extreme practical interest for any mathematician. In any number of applications one is obliged to find the value of some definite integral. One of the most trivial examples should be the values of probability distributions, a little less obvious example are solutions to differential equations.

Especially since the advent of cheap but high-performing computer power those possibilities have risen immensely. Things that were unattainable some decades ago are used thoroughly by today’s standards. This is why I found it very necessary to make myself familiar with how one can find definite integrals numerically (basically that’s the same motivation for why I wanted to know about numerical algorithms for finding eigenvalues – and there is a nice connection between the two fields, as we shall see).

At first, I had wanted to compile some of the basic ideas for numerical integration here. However, I dived deeper down into some certain aspect of the theory and decided to cut that one short to present something a little less known on convergence of quadrature methods. But to begin with, a little review of what every mathematician should learn in his basic course on numerical analysis.

The numerical integration is sometimes called quadrature, as one tries to find the value of a given shape (the “area under the curve”) and therefore find a square with the same area – similar to what one does in “squaring the circle” which is called “quadraturam circuli” in latin. It’s just that we don’t have to be restricted to ruler and compass anymore.

Basics

Of course, the natural way of numerical integration is to evaluate the function $f$ (to fix notation here) at certain points and make a weighted average of the values. This is related to the Riemann-definition of integrability, where you use sums of the form $\sum_{i=1}^nf(\xi_i)(x_{i}-x_{i-1})$ for $\xi_i\in(x_{i-1},x_{i})$ and let the grid $x_i$ get finer. This is a starting point for what we can achieve, but the first question is, how can we choose the $\xi_i$ smartly?

Let’s not delve into this too deeply, but the short answer from Riemann-integration is: do as you please (as long as $f$ is continuous). In the end, it will converge. The most common first choice is the appropriately-called midpoint-rule (here’s a picture). If you don’t know anything further about $f$, why not. But one will soon find the trapezoidal rule (picture) and there is no general rule which one of them will be better in any instance. Let’s just write those down for further reference:

Midpoint-rule: $\int_a^b f(t)dt \approx (b-a)f(\frac{a+b}2)$,

Trapezoidal rule: $\int_a^b f(t)dt \approx \frac{b-a}{2}\bigl(f(a)+f(b)\bigr)$.

A certain mixture of those two rules can be found in

Simpson’s rule: $\int_a^b f(t)dt \approx \frac{b-a}{6}(f(a)+4f(\frac{a+b}2)+f(b)\bigr)$.

All these rules are fit to integrate polynomials up to a certain degree exactly (which is why this feature is called the degree of exactness for the quadrature rule): the midpoint rule is exact for linear polynomials, the trapezoidal rule makes it up to degree 2 and Simpson’s rule can integrate cubic polynomials exactly.

Each of these rules can be constructed by interpolation of $f$ with low-degree polynomials: one chooses the unique polynomial $p$ for which $p(t)=f(t)$ in those $t$ that come up in the formulae given above and choose constants (“weights”) such that $\int_a^b p(t)dt = \sum_{i=1}^n w_ip(t_i)$. Note, that this will not be a Riemann-sum anymore, in general. You will find that for $p$ being of degree $n=1$, 2 or 3, you will end up with the three formulae given above.

Now, if you wish to be a little better in your interpolation, you can subdivide the interval $[a,b]$ a little further and use any of those rules on the smaller intervals (interpolating repeatedly) to get the compound rules:

– Midpoint-rule: $\int_a^b f(t)dt \approx \frac{b-a}{n}\sum_{i=1}^nf(\frac{x_{i}-x_{i-1}}2)$ with $x_k = a+k\frac{b-a}{n}$,

– Trapezoidal rule: $\int_a^b f(t)dt \approx \frac{b-a}{n}\sum_{i=1}^n\frac{f(x_{i-1})+f(x_i)}{2} = \frac12f(a)+\frac{b-a}{n}\sum_{i=1}^{n-1}f(x_i)+\frac12f(b)$ with $x_k = a+k\frac{b-a}{n}$,

– Simpsons rule: $\int_a^b f(t)dt \approx \frac{b-a}{6n}\bigl(f(a) + 2\sum_{i=1}^n f(x_i) + 4\sum_{i=1}^nf(\frac{x_{i-1}+x_{i}}{2}) + f(b)\bigr)$ with $x_k = a+k\frac{b-a}{n}$.

Note, that we have a constant grid-width here. This doesn’t have to be the case in general, as we shall see. Besides, note that in Simpson’s rule we have three evaluations of the $f$ in each of the sub-intervals, and only the boundary points $x_k$ are used twice (once as the left-hand point, once as the right-hand point). One might also define it slightly differently, by having the sub-intervals overlap and each evaluation of $f$ is used twice (once as the midpoint, the other time as a boundary point), except for $f(a)$ and $f(b)$.

Now, any basic text on numerical analysis and Wikipedia will tell you everything there is to know about the error that you get from these quadrature formulae. Let’s not go there, except for some small remarks. First, each of these rules has an error that will diminish with more sub-divisions of $[a,b]$. The error will depend on how smooth $f$ is, in terms of the norm $\left\|f^{(k)}\right\|_\infty$ for some $k\geq1$. Therefore, to know about the error, you will need to conjure some information on the derivatives of $f$, which will rarely be possible in applications. However, if $f$ is “only” continuous, you may find yourself in a bad position for error estimation: the quadrature rule might be just looking at the unusual points to evaluate $f$, which are not typical of the overall behaviour:

(here’s an example going with the picture: take $f$ to be some positive constant $c$, except for a tiny area around the evaluation, where it shall vanish; then $\int_a^bf(t)dt \approx c(b-a) \neq 0 = \sum_{i=1}^n w_if(x_i)$ – though not impossible, it’s harder to do this the smoother $f$ will be, and the derivatives of $f$ will blow up, giving a worse estimate). This effect has been called “resonance” by Pólya.

Apart from ad hoc methods for the error estimation there is the method of Peano kernels which can be applied to bound quadrature formulae (and other linear functionals). Let there just be mentioned that this method requires rather little effort to yield error bounds. I wouldn’t quite call it elegant, possibly since I lack the background in functional analysis, but it gets the job nicely done.

Next natural question: which $n$ should we use? Short answer: the more the better, though many evaluations of $f$ are “expensive” (especially in applications from the sciences and economics where you might even be unable to get more measurements). Apart from the compound rules, which are used quite often in practice, one might think that interpolation using polynomials of higher degree should work out better and give finer error estimates. The quadrature stemming from this idea is called Newton-Cotes-formula (of which our previous rules are special cases). In principle, this is sound reasoning, but here’s the showstopper: interpolating polynomials need not be similar to the function that they interpolate, and hence the quadrature runs wild. The most famous example is Runge’s phenomenon: for $f(x)=\frac{1}{1+x^2}$ the sequence of polynomials interpolating on an equidistant grid does not converge pointwise, which can be explained by the singularity at $x = \pm i$ that stays invisible while working on the real numbers only. In particular, an interpolating polynomial of sufficiently high degree won’t tell us anything useful about the behaviour of $f$.

For numerical applications, the next problem is that the weights in Newton-Cotes-formulae become negative when $n\geq 9$ (that means, if you use a polynomial of higher degree to interpolate $f$ on $[a,b]$). In effect, you will encounter annihilation in your computation from subtracting rather large numbers – there won’t be any guarantee for proper results anymore, even though in theory your method might work fine on your given function. Quadrature formulae with positive weights are called for.

Before the discussion of convergence of quadrature formulae, we will still look at what methods are best possible in terms of the degree of exactness. It’s the Gaussian integration (many practical things are named after Gauß, aren’t they?) which allows for exact integration of polynomials up to degree $2n-1$ from $n$ evaluations.

But first, why will we be content with a degree of exactness $2n-1$, why not go further? Well, there is always at least one polynomial of degree $2n$ that can’t be exactly integrated by a quadrature formula on $n$ evaluations. Let’s fix some quadrature formula, it will look like $\sum_{i=1}^nw_if(x_i)$ with weights $w_i$ and evaluations in $x_i$, and let’s look at the polynomial $\omega^2(x) = \prod_{i=1}^n(x-x_i)^2$ which has its zeroes at the points of evaluation of the quadrature formula. Then, we find

$\displaystyle \sum_{i=1}^nw_i\omega^2(x_i) = 0 < \int_a^b \omega^2(t)dt.$

Hence, there is always a polynomial of degree $2n$ that can’t be integrated exactly, and therefore $2n-1$ is the best we could hope to achieve. Now, how do we get there? Up to now, the quadrature formulae had used some $x_i$ (mostly because those $x_i$ were the natural thing to do or because we can’t always chose the $x_i$ at liberty, if they stem from experimental measurements, say) and the $w_i$ were computed accordingly to make the polynomials up to degree $n$ be integrated exactly. But we can of course choose the $x_i$ wisely as well: they can act as a degree of freedom, and then we’ll have $2n$ clever choices to make – this seems to indicate that we can achieve the degree of exactness $2n-1$. But can we? And how?

Let’s assume that there is some quadrature formula which has this degree of exactness, and let’s look at what it will do to the polynomials $\omega p$, where $p$ is any polynomial of degree less than $n$. Restricting ourselves to the interval $[a,b] = [-1,1]$ for simplicity, we should find

$\displaystyle 0 = \sum_{i=1}^nw_i\omega(x_i)p(x_i) = \int_{-1}^1\omega(t)p(t)dt.$

That shows that $p$ would be orthogonal to $\omega$, no matter what $p$ actually was (as the $L^2$-space is a Hilbert space, of course, with the integration as its scalar product). From the abstract theory of Hilbert spaces we can know that there are orthonormal bases of polynomials. For our setting here, we are led to use the Legendre polynomials as $\omega$; there are other orthonormal polynomials which will then use another weight function inside the integral – later more on that.

We have seen that the zeroes of the Legendre polynomials should be used to evaluate $f$. But how to find those? Here’s the pretty thing I alluded to above. There’s a recursion formula for the Legendre polynomials $P_n$ which can be shown by induction:

$\displaystyle \beta_{n+1}P_{n+1}(x) = xP_n(x) - \beta_nP_{n-1}(x)$, with $\displaystyle P_1\equiv\frac1{\sqrt2}$ and $\displaystyle P_2(x) = \frac32x$,

and using $\beta_n = \frac{n}{\sqrt{(2n-1)(2n+1)}}$. This recursion may be re-written as a vector-matrix-shape:

$\displaystyle\begin{bmatrix}0&\beta_1&0\\\beta_1&0&\beta_2&0\\0&\beta_2&0&\beta_3&0\\&&\ddots&\ddots&\ddots\\&&&\beta_{n-2}&0&\beta_{n-1}\\&&&0&\beta_{n-1}&0\end{bmatrix}\begin{bmatrix}P_0(x)\\P_1(x)\\P_2(x)\\\vdots\\P_{n-2}(x)\\P_{n-1}(x)\end{bmatrix}= x\begin{bmatrix}P_0(x)\\P_1(x)\\P_2(x)\\\vdots\\P_{n-2}(x)\\P_{n-1}(x)\end{bmatrix}-\begin{bmatrix}0\\0\\0\\\vdots\\0\\\beta_nP_n(x)\end{bmatrix}$

The last vector drops out for the zeroes of $P_n$, and then we have an eigenvalue-problem. In particular: find the eigenvalues of the matrix on the left-hand-side to find the zeroes of $P_n$. That’s some rather unexpected connection – I found it marvellous when I first saw this.

Computing eigenvalues is a tricky thing to do, as everybody knows, and getting eigenvectors is even harder. It turns out that you’ll need the eigenvectors to compute the proper weights $w_i$ for the quadrature formula, and many intelligent people have come up with ingenious algorithms to do so, such that we can have the proper values for the maximum degree of exactness that we have yearned for.

This shall suffice here. We can understand that these so-called Gauss-Legendre-formulae can’t be expressed in a closed form, but they yield satisfying results in practice. Of course, when there was no cheap computer power available, those formulae were more of an academic exercise, as you couldn’t easily evaluate any given function $f$ in those crude points $x_i$ (even though these $x_i$ came in tables) and you can’t re-use those evaluations when you take other values for $n$.

A numerical advantage is that the weights will be positive for the Gaussian quadrature, which is easy to see, as a matter of fact: with the Lagrange-polynomials $\ell_{ni}$, that are used for explicit interpolation with polynomials and that have $\ell_{ni}(x_j) = \delta_{ij}$ for the evaluation points, we find

$\displaystyle w_i = \sum_{j=1}^nw_j\ell_{ni}^2(x_j) = \int_{-1}^1\ell_{ni}^2(t)dt > 0.$

As an aside: there are some variants of Gauss-Legendre that may come in useful in some applications. For instance there is the Gauss-Chebyshev-formula which uses the weight function $\frac1{\sqrt{1-x^2}}$ in the integration (not in the quadrature – weights and weight functions are not related directly). This means, if your function $f$ is of the shape $f(x) = \frac{g(x)}{\sqrt{1-x^2}}$, use Gauss-Chebyshev on the function $g$, which may be easier or more stable; in that case, other $x_i$ and $w_i$ are necessary, of course. Similar things hold for Gauss-Laguerre (weight $e^{-x}$, the integral running on $[a,b)=[0,\infty)$, which makes it possible to integrate on infinite intervals, being a non-trivial task at a first glance) and Gauss-Hermite (weight $e^{-x^2}$, $(a,b)=(-\infty,\infty)$). Other variants have certain fixed points of evaluation, which certainly reduces the degree of exactness, but which might help in solving differential equations (when you always evaluate in the point $x_1=a$, you save one evaluation as this is your starting-point of your multi-step method; Gauss-Radau does this, for instance, and similarly Gauss-Lobatto with $x_1=a$ and $x_n=b$ fixed).

Obvious linear transformations will enable us to use different values for $a$ and $b$ than the standard ones for which the various orthonormal polynomials are constructed. Let’s ignore this here.

Those were pretty much the basic facts of what any mathematician should know or at least should have heard once, without much of a proof stated here. Now for a little more specialized material. Who guarantees that the quadrature gets better with increasing numbers of evaluation? Evaluations are expensive, generally speaking, and maybe it doesn’t even do any good (as was the case with the Newton-Cotes-method – or does it)? It seems that Stieltjes first came up with this question and gave partial answers in 1884, while Pólya answered it rather exhaustively a couple of decades later in 1933. We rely on Pólya’s original article “Über die Konvergenz von Quadraturverfahren” (Math. Zeitschr. 37, 1933, 264-286)

Convergence

A quadrature method is, for all intents and purposes that are to come, a triangular scheme of weights $w_{ij}$ and evaluation points $x_{ij}$ like

$\displaystyle \begin{matrix}w_{11}\hphantom{,}\\w_{21}, &w_{22}\hphantom{,}\\w_{31},&w_{32},&w_{33}\\\vdots\\w_{n1},&w_{n2},&\ldots&w_{nn}&\\\vdots&\end{matrix}\hspace{2em}$ and $\hspace{2em}\displaystyle \begin{matrix}x_{11}\hphantom{,}\\x_{21}, &x_{22}\hphantom{,}\\x_{31},&x_{32},&x_{33}\\\vdots\\x_{n1},&x_{n2},&\ldots&x_{nn}&\\\vdots&\end{matrix}$

For any $n$, we need $a\leq x_{n1}. Of course we shall combine those triangular schemes like this

$\displaystyle Q_n[f] := \sum_{i=1}^nw_{ni}f(x_{ni})$

and hold up our hope that $\lim_{n\to\infty} Q_n[f] = \int_a^bf(t)dt$. If so, the quadrature method converges for the funtion $f$. Of course, $f$ needs to be at least Riemann-integrable (We are unable to focus on Lebesgue-integrable functions here, as no quadrature method can converge for any Lebesgue-integrable function: there are only countably many evaluation points on which $f$ may be altered at will without effect on its integral).

It is clear that the Newton-Cotes-formulae fit in this definition. We can choose the $x_{ij}$ at liberty and the $w_{ij}$ follow from the demand that in stage $n$ polynomials up to degree $n-1$ are integrated exactly. In particular, the Newton-Cotes-formulae converge for any polynomial by definition.

But there’s the question, what do the $x_{ij}$ and $w_{ij}$ need to satisfy in order for the quadrature method to converge on a given set of functions (any polynomial, any continuous or even any integrable function)? If you switch to a larger set, what are the additional necessary assumptions?

For polynomials, the case is quite easy. We look for $\lim_{n\to\infty}\sum_{i=1}^nw_{ni}x_{ni}^k = \frac{1}{k+1}(b^{k+1}-a^{k+1})$ for any $k\geq0$, as every polynomial is a linear combination of the $x^k$ and both integration and quadrature are linear functionals. No big deal, and for $k=0$ in particular we find the important assumption $\lim_{n\to\infty}\sum_{i=1}^nw_{ni} = b-a$.

Pólya’s result that we shall examine in a little more detail is this

Theorem: The quadrature method converges for every continuous function, if and only if it converges for every polynomial and there is a constant $\Gamma$ such that $\sup_n\sum_{i=1}^n\left|w_{ni}\right| < \Gamma$.

One direction of the proof is easy. Let $\sup_n\sum_{i=1}^n\left|w_{ni}\right| < \Gamma$ and let $f$ be any continuous function. We show that the method converges for $f$. We know by Weierstrass’ Approximation Theorem that there are uniformly close polynomials for $f$, in the sense that for every $\varepsilon>0$ we can get a polynomial $p$ with $\left\|f(t) - p(t)\right\|_\infty<\varepsilon$. Thus,

\displaystyle \begin{aligned} \left|\int_a^bf(t)dt - Q_n[f]\right| &\leq \int_a^b\left|f(t)-p(t)\right|dt + \left|\int_a^bp(t)dt - Q_n[p]\right| + \sum_{i=1}^n\left|w_{ni}\right|\left|p(x_{ni})-f(x_{ni})\right| \\ &\leq \varepsilon(b-a) + \varepsilon + \varepsilon\Gamma. \end{aligned}

Here, we used that $Q_n$ converges for the polynomial $p$ and that for $n$ sufficiently large $\left\|f-p\right\|_\infty<\varepsilon$. This shows that $Q_n[f]$ is convergent.

The other direction of the proof is a little harder. We assume that there is no constant $\Gamma$ such that $\sup_n\sum_{i=1}^n\left|w_{ni}\right| < \Gamma$ – therefore $\limsup_n\sum_{i=1}^n\left|w_{ni}\right| = \infty$. We will construct a continuous function $f$ for which $Q_n[f]$ diverges.

This will involve the “resonance” that I alluded to earlier: choose a function that reacts as badly as possible on the evaluation points. For fixed $k$ take $g_k$ such that $\left\|g_k\right\|_\infty = 1$ and $Q_k[g_k] = \sum_{i=1}^k\left|w_{ki}\right|$.

This ensures that $Q_k[g_k]$ cannot possibly take a larger value on any function that itself is bounded by $1$. Such $g_k$ may be constructed, for instance, via $g_k(x_{kj}) = \mathrm{sgn} w_{kj}$ and linear connections between the evaluation points.

Now, one of two possible cases may occur: either there is some $k$ with $\limsup_n\left|Q_n[g_k]\right|=\infty$, or that expression stays bounded uniformly in $k$. In the first case, we have already found the continuous function which makes the quadrature method diverge and we are done. So, let us look at the second case only.

Let us inductively construct a subsequence of the $g_k$ as follows. Let $g_{k_1} = g_1$. If we have got $g_{k_{m-1}}$, set

$\displaystyle M_{m-1}:= \sup_nQ_n\left[\sum_{j=1}^{m-1}g_{k_j}3^{-j}\right].$

As we have already decided we’re in the second case, $M_{m-1}$ will be finite. Now, we wish to choose the next part $g_{k_m}$ of the subsequence, and we ensure both $k_m>k_{m-1}$ and (as $\limsup_n\sum_{i=1}^n\left|w_{ni}\right| = \infty$, we may find some index to make this sum of weights as large as we like)

$\displaystyle \sum_{i=1}^{k_m}\left|w_{k_mi}\right| > 3^m 2(M_{m-1} + m).$

With the subsequence $g_{k_m}$ at our disposal, let’s define the function

$\displaystyle f(x):=\sum_{j=1}^\infty 3^{-j}g_{k_j}(x).$

This is a normally convergent series, as each of the $g_{k_j}$ is bounded by $1$ and thus for sufficiently large $n$:

$\displaystyle\left\|\sum_{j=n}^\infty 3^{-j}g_{k_j}(x)\right\|_\infty\leq\sum_{j=n}^\infty 3^{-j} < C3^{-n}<\varepsilon.$

As a uniformly convergent limit of continuous functions, $f$ is continuous. Next, we invoke linearity of quadrature methods again and have a closer look at what the $Q_n$ do to $f$:

\displaystyle \begin{aligned} Q_{k_m}[f] &= Q_{k_m}\left[\sum_{j=1}^{m-1}g_{k_j}3^{-j}\right] + 3^{-m}Q_{k_m}\left[g_{k_m}\right] + Q_{k_m}\left[\sum_{j=m+1}^\infty g_{k_j}3^{-j}\right]\\ & > -M_{m-1} + 3^{-m}\sum_{j=1}^{k_m}\left|w_{k_mj}\right| - \sum_{\ell=1}^{k_m}\left|w_{k_m\ell}\right|\sum_{j=m+1}^\infty 1\cdot 3^{-j}, \end{aligned}

by the definition of $M_{m-1}$ and by the construction of $g_{k_m}$. Then,

\displaystyle \begin{aligned} Q_{k_m}[f] & = -M_{m-1} + 3^{-m}\sum_{j=1}^{k_m}\left|w_{k_mj}\right| - 3^{-m-1}\sum_{\ell=1}^{k_m}\left|w_{k_m\ell}\right|\cdot\frac1{1-\frac13}\\ & = -M_{m-1} + \frac123^{-m}\sum_{j=1}^{k_m}\left|w_{k_mj}\right|\\ & > m, \end{aligned}

which may explain the strange bound that we used in the construction of the subsequence.

Letting $m\to\infty$, we see that the quadrature method diverges for $f$. The entire construction hinges on the assumption that $\limsup_n\sum_{i=1}^n\left|w_{ni}\right| = \infty$; in particular, the condition given in the theorem is necessary for the convergence of the quadrature method.

The theorem is proved. q.e.d.

The condition for the convergence on Riemann-integrable functions is rather technical and we won’t dwell on this for too long. The basic ideas of resonance are similar, and in applications there is often good reason to assume that your function is continuous (as opposed to differentiable which you usually can’t know). We’ll content ourselves with the remark that quadrature methods converge on any Riemann-integrable function if and only if they converge for every continuous function and $\limsup_n\sum_{i\in I_n}\left|w_{ni}\right|=0$ where $I_n$ is a sequence of intervals which contains the $x_{ni}$ and the total length of the intervals converges to $0$.

A little more interesting is the following counter-example: There is an analytic function for which the Newton-Cotes-formulae on equidistant grids diverge.

Analytic functions are about the most well-behaved functions you may imagine, and even there the Newton-Cotes-method won’t do much good (as long as they’re computed on equidistant grids). Of course, by the theorem we proved, we can already see why: the weights are not bounded and the basic problem is what happened in Runge’s counterexample; the interpolating polynomials don’t have anything to do with the function $f$ that we wish to integrate. As Ouspensky put it:

La formule de Cotes perd tout valeur pratique tant que le nombre des ordonnées devient considérable.

However, the problem is not inherent in Newton-Cotes: if you choose the points of evaluation wisely (as zeroes of Legendre polynomials, or the like, usually clustering them around the endpoints of the interval), the sequence will converge by the theorem that we have proved: the quadrature is then Gaussian, its weights are positive and integrate the constant function $1$ properly (which even gives $\sum_{i=1}^n w_{ni} = b-a$ and so $b-a =: \Gamma <\infty$) – hence, it must converge for any continuous (and in particular: analytic) function. The message is: avoid negative weights in quadrature methods and choose wisely whatever you can choose; not only will you avoid trouble in numerical computations, but you may also keep nice convergence properties.

Let us at least sketch the proofs of this counter-example.

Let us switch the notation very slightly and fix it like this: take $x_{ni} = \frac in$ for $i=0,\ldots,n$, as we try to approximate $\int_0^1f(t)dt$. Then, $Q_n[f] = \sum_{i=0}^nw_{ni}f(\frac in)$.

We’ll need two preliminaries. First: Asymptotics on the quadrature weights. Second: An estimate for the interpolating polynomial.

First preliminary: Asymptotics on the quadrature weights. As we use Newton-Cotes, the idea is to interpolate and then integrate the corresponding polynomial. The interpolating polynomial can be expressed via the Lagrange polynomials $\ell_{ni}(x) := \prod_{j=1, j\neq i}^n\frac{x-x_{nj}}{x_{ni}-x_{nj}}$ and our idea narrows down to

$\displaystyle \int_0^1f(t)dt\approx \int_0^1p(t)dt = \int_0^1\sum_{i=0}^nf(x_{ni})\ell_{ni}(t)dt = \sum_{i=0}^nf(x_{ni})\int_0^1\ell_{ni}(t)dt.$

On the other hand, $\int_0^1f(t)dt\approx Q_n[f] = \sum_{i=0}^nf(x_{ni})w_{ni}$. Hence, the weights $w_{ni}$ must equal $\int_0^1\ell_{ni}(t)dt$, or to write it down for once

$\displaystyle w_{nk} = \int_0^1\frac{nt\cdot (nt-1)\cdots (nt-k+1)\cdot (nt-k-1)\cdots (nt-n)}{k(k-1)\cdots 1\cdot(-1)\cdots(k-n)}dt.$

The really troublesome part is an asymptotic formula for the growth of these $w_{nk}$. Pólya uses all kinds of neat ideas from the theory of the gamma function, far-sighted variable transformations and a lot of computation. This has been done by Ouspensky first (“Sur les valeurs asymptotiques des coefficients de Cotes“; Bull. Amer. Math. Soc., 31, 1925, 145-156), who concluded with the quote given above, that the weights for Newton-Cotes get larger and larger in absolute value and therefore become increasingly useless. The result (which was at least a little generalized and simplified by Pólya) goes like this

$\displaystyle w_{nk} = -\frac{1}{n(\log n)^2}\binom{n}{k}\left(\frac{(-1)^k}{k}+\frac{(-1)^{n-k}}{n-k}\right)(1+\eta_{nk})$

with $\lim_n\sup_k\eta_{nk}=0$. In particular, a close inspection shows the special case

$\displaystyle w_{n,\frac n2} > 2^n n^{-3}$

for even $n$. Let’s keep this guy in mind.

Second preliminary: An estimate for the interpolating polynomial. As we wish to say something about analytic functions, we can invoke the pretty results of complex analysis. This can tell us about a representation of the interpolation polynomial. In what follows, $\omega$ is, again, the polynomial with zeroes in the points of evaluation.  Of course, in the complex plane interpolation works with the Lagrange polynomials as well:

$\displaystyle p(x) = \sum_{i=0}^nf(x_{ni})\prod_{j=0, j\neq i}^n\frac{x-x_{nj}}{x_{ni}-x_{nj}}.$

Some close inspections show:

\displaystyle \begin{aligned} p(x) &= \sum_{i=0}^nf(x_{ni})\frac{\prod_{j=0, j\neq i}^n(x-x_{nj})}{\prod_{j=0, j\neq i}^n(x_{ni}-x_{nj})}\\ &= \sum_{i=0}^nf(x_{ni})\frac{\prod_{j=0}^n(x-x_{nj})}{(x-x_{ni})\sum_{k=0}^n\prod_{\ell=0, \ell\neq k}^n(x_{ni}-x_{n\ell})}\\ &= \sum_{i=0}^n\frac{\omega(x)f(x_{ni})}{(x-x_{ni})\omega'(x_{ni})}. \end{aligned}

For the sake of seeing how to invoke the residue theorem, we set $F(t) := \frac{\omega(x)-\omega(t)}{x-t}f(t)$ and $G(t):=\omega(t)$. This leads to

$\displaystyle p(x) = \sum_{i=0}^n\frac{F(x_{ni})}{G'(x_{ni})} = \sum_{i=1}^n\mathrm{res}_{x_{ni}}\left(\frac{F}{G}\right).$

The last equation followed from a little divine intuition – something we can’t avoid in this argument; in order to drop this intuition, we would have needed to start with an eerie expression that would have come from nowhere. So, how does it follow? The function $\frac1G$ has simple poles in the $x_{ni}$, while $F$ does not (the simple singularities in the $x_{ni}$ cancel out in the numerator to leave us with a polynomial without further zeroes in the $x_{ni}$). The Laurent expansion of $\frac FG$ starts with the $a_{-1}(x-x_{ni})^{-1}+\cdots$, $a_{-1}$ being the residue. Multiplying with $x-x_{ni}$, using the Taylor expansion of $G$ and taking the limit gives

\displaystyle \begin{aligned} \mathrm{res}_{x_{ni}}\left(\frac{F}{G}\right) = \lim_{x\to x_{ni}}(x-x_{ni})\frac{F(x)}{G(x)} &= \lim_{x\to x_{ni}}(x-x_{ni})\frac{F(x)}{(x-x_{ni})G'(x_{ni})+\cdots} \\ &= \lim_{x\to x_{ni}}\frac{F(x)}{G'(x_{ni})+\cdots} = \frac{F(x_{ni})}{G'(x_{ni})}. \end{aligned}

The argument looks slightly butchered as we read it the wrong way around – but it works nonetheless. Now comes the residue theorem into play, applying it on any reasonable domain $D$ that contains the $x_{ni}$ on its interior:

\displaystyle \begin{aligned} p(x) = \sum_{i=1}^n\mathrm{res}_{x_{ni}}\left(\frac{F(\zeta)}{G(\zeta)}\right)&=\sum_{i=1}^n\mathrm{res}_{x_{ni}}\left(\frac{\bigl(\omega(x)-\omega(\zeta)\bigr)f(\zeta)}{(x-\zeta)\omega(\zeta)}\right)\\ &= \frac1{2\pi i}\int_{\partial D}\frac{\omega(x)-\omega(\zeta)}{(x-\zeta)\omega(\zeta)}f(\zeta)d\zeta \end{aligned}

Re-writing this last expression with the help of Cauchy’s integral formula shows

\displaystyle \begin{aligned}p(x) &= \frac1{2\pi i}\int_{\partial D}\frac{\omega(x)f(\zeta)}{(x-\zeta)\omega(\zeta)}d\zeta - \frac1{2\pi i}\int_{\partial D}\frac{f(\zeta)}{x-\zeta}d\zeta \\ &= \frac1{2\pi i}\int_{\partial D}\frac{\omega(x)f(\zeta)}{(x-\zeta)\omega(\zeta)}d\zeta + f(x). \end{aligned}

Now, consider a certain domain $D$ in the complex plane

(half-circles of radius $1$ around $z=0$ and $z=1$, connected by straight lines), in which $f$ is bounded by some constant $M$. Let $f$ be analytic in this domain $D$ and being interpolated by the polynomial $p$ in the points $x_{n0},\ldots,x_{nn}$ which are located in the interval $[0,1]$. For $\zeta$ on the boundary of $D$, we are far away from the interval $[0,1]$ and all $x_{ni}$ in the sense that $\left|\omega(\zeta)\right|\geq1$ and $\left|x-\zeta\right|\geq1$, while for $x$ in $[0,1]$ we have $\left|\omega(x)\right|<1$. Hence,

$\displaystyle \left|p(x)-f(x)\right|\leq \left|\frac1{2\pi i}\int_{\partial D}\frac{\omega(x)f(\zeta)}{(x-\zeta)\omega(\zeta)}d\zeta\right| \leq \frac1{2\pi}(2+2\pi)M.$

Therefore

$\displaystyle \left|\int_0^1p(t)dt\right|\leq\left|\int_0^1p(t)-f(t)dt\right|+\left|\int_0^1f(t)dt\right|\leq \frac{\pi+1}{\pi}M+M\leq 3M.$

Let’s keep this one in mind as well.

From all of these preliminaries, let us state the promised counter-example: consider the function

$\displaystyle f(x) := \sum_{k=4}^\infty a^{k!}\frac{\sin(k! \pi x)}{-\cos(\pi x)},$

with $a\in(\frac12, 1)$ and where the denominator and the starting index $k=4$ are chosen to ensure that the singularity in $x=\frac12$ is removable by a positive constant. This function is actually analytic as the uniform limit of analytic functions (the pretty complex analysis again; and here we need $a<1$ for the summands to actually diminish quickly enough). Besides, it is periodic and one can use this to find its maximum value on the horizontal bounds of the domain: by complex analysis (yet again!) the maximum must be located on the boundary, and if it were on the vertical bounds, we could (by periodicity) shift the domain without losing any information and then contradictively find the maximum on the inside of the domain. Therefore, any summand of which $f$ is constructed may be bounded inside the domain by

$\displaystyle \left|a^{k!}\frac{\sin(k! \pi x)}{-\cos(\pi x)}\right| \leq Ca^{k!}\exp(k!\pi \mathrm{Im}(x)) \leq Ca^{k!}\exp(k!\pi),$

where the constant $C$ helps to encode the minimum of the cosine on $\mathrm{Im}z=1$ (it won’t vanish anywhere near this line).

Now that we have bounded the analytic summand of our function $f$, from our second preliminary, we get

$\displaystyle \left|\int_0^1 p(t)dt\right| < 3Ca^{k!}\exp(k!\pi)$

and we shall consume the 3 into the constant $C$ without further notice. On top of that, we are about to apply a Newton-Cotes-method; the integral of the interpolating polynomial is equal to what all the quadratures $Q_n[f]$ do:

$\displaystyle \left|\sum_{i=0}^nw_{ni}a^{k!}\frac{\sin(k! \pi \frac in)}{-\cos(\pi \frac in)}\right| = \left|\int_0^1 p(t)dt\right| < Ca^{k!}\exp(k!\pi).$

Now we see

\displaystyle \begin{aligned}Q_{n!}[f] &= Q_{n!}\left[\sum_{k=n}^\infty a^{k!}\frac{\sin(k!\pi x)}{-\cos(\pi x)}\right]+Q_{n!}\left[\sum_{k=4}^{n-1} a^{k!}\frac{\sin(k!\pi x)}{-\cos(\pi x)}\right]\\ &\geq Q_{n!}\left[\sum_{k=n}^\infty a^{k!}\frac{\sin(k!\pi x)}{-\cos(\pi x)}\right]-\sum_{k=4}^{n-1}Ca^{k!}\exp(k!\pi)\\ &= w_{n!, \frac{n!}{2}}\sum_{k=n}^\infty k! a^{k!} -\sum_{k=4}^{n-1}Ca^{k!}\exp(k!\pi), \end{aligned}

since in case $k\geq n$, $\sin(k!\pi x)=0$ for any relevant $x_{n!,i}=\frac i{n!}$ and all terms in $Q_{n!}$ vanish except for $x_{n!,\frac12n!}=\frac12$ where the denominator vanishes as well and the singularity is removable by taking the value $k!$.

This is the time to take our first preliminary out of the drawer:

\displaystyle \begin{aligned} Q_{n!}[f] &> \frac{2^{n!}}{(n!)^3} n! a^{n!} - (n-4)C(ae^\pi)^{(n-1)!}\\ &= \frac{(2a)^{n!}}{(n!)^2}\left(1-\bigl(n!\bigr)^2(n-4)C\left(\frac{ae^\pi}{(2a)^n}\right)^{(n-1)!}\right)\end{aligned}.

As $a>\frac12$, the term in the large brackets converges to 1 ($n!$ grows slower than $(2a)^{n!}$ does), and the entire term diverges. In particular, the Newton-Cotes-method diverges for $f$. That’s what we wanted to show. Oof. q.e.d.

So, we have found necessary and sufficient conditions for quadrature methods to converge and we dealt with a striking counter-example for a plausible method on a reasonably nice class of functions to make us dismiss this plausible method. However, whenever we spoke about convergence, we didn’t deal with the speed of convergence – and we couldn’t have. As Davis and Rabinovitz have rather un-nervedly put it: Nothing positive can be said about the degree of convergence. Lipow and Stenger (“How slowly can quadrature Formulas converge?“; Math. Comp. 26, 1972, 917-922) have shown that all quadrature methods will converge arbitrarily slowly on the space of continuous functions: for every sequence $a_n$ converging to $0$, there is a continuous function $f$ such that $\int_0^1f(t)dt - Q_n[f] \geq \varepsilon_n$.

Before closing this chapter, let’s say that in practical applications it’s hard to know whether your approximation to the integral is sufficient. The convergence theorems don’t speak about the speed of convergence, the error estimates deal with sup-norms of derivatives. Most applications deal with adaptive methods (which we haven’t covered here) to have a closer look where the function might become erratic, and in the end the methods are used with a little hope in the back of the user’s minds that stable methods with a high degree of exactness will do sufficiently well. There can be no certainty, however. But still, it’s some pretty mathematics to spend your time with.

# How does it feel? To be on your own, with no direction home, like a complete unknown, like a rolling stone?

Every October, I get mildly interested in who is going to be Nobel Prize Laureate this year. I don’t get totally excited, since most of the time I don’t understand enough about the physics, the chemistry and (Lord help me) the medicine. I can classify the importance of the Nobel Peace Prize reasonably, and I couldn’t care less about the Non-Nobel Prize for economics. Besides, I was never fond of the Nobel Prize for Literature, since it seems a much more random prize to award, as it is the only one for artists, totally ignoring musicians, sculpturists and whatnot. In particular, if you are not closely familiar with world’s modern literature, you can’t understand the first thing about the Laureates. It depends heavily on where you live, if you are to know the winners and to estimate whether the award is rightful or not. This seems different for the science prizes, as I can at least estimate how important the respective field of research is, and it is quite different for the peace prize, as everyone with an understanding of present-day politics can estimate the importance of the awardee.

This year was different. The Nobel Prize for Literature went to Bob Dylan. This surprised me for two reasons: I knew the winner beforehand, and the winner is not a writer in the classical sense of the word. As far as I know this is the first time that a singer/songwriter wins the Nobel Prize, and this is a fine decision. As a side note, much of the literature of the ancient times used to be presented in the shape of music (since songs and rhymes are easier to memorize, an important thing when you’re without much opportunity for written records), just think of the Iliad and the Odyssey. Today, those are only referred to by their lyrics, as the music has vanished from mankind’s memory, but they are considered classical literature nonetheless. So, a good thing to award the prize for literature to a songwriter.

Now, I haven’t had much time yet to dig deeply into Bob Dylan’s discography, something I urgently need to do in the weeks to come. I had been aware that he was quite an influential writer whose songs have been covered numerous times, like Blowin’ in the Wind and Like a Rolling Stone. But I had not been aware how many songs were originally his: Mr Tambourine Man, Knockin’ on Heaven’s Door, Times are a-chaingin’, It ain’t me Babe, … and that list doesn’t start to be exhaustive. In fact, Dylan seems to be the most-covered musician in the past 100 years; I had believed this had been the Beatles, but they “only” have the most-covered song Yesterday (which is not the strongest Beatles-song by far, actually, but that doesn’t matter here). Of those many covers, most are better-known to be interpreted by other singers though they were written by Bob Dylan in the first place. Which is the phenomenon that happened to me, in fact.

One reason for this is Dylan’s voice which can’t actually be called melodic. In fact, he’s not much of a fantastic singer, as far as my taste is concerned. But his arrangements and his lyrics have been an inspiration for the entire modern popular music ever since the 1960’s. There is a virtually un-ending list of movies that feature a Dylan song in their score, which is why I know so very much of his work – though rather unaware.

Another fact that amazed me was part of the video clip for Homesick Subterranean Blues that was shown in the news after the Nobel Prize was announced: it was the inspiration to the beautiful video clip for Nur ein Wort by Wir sind Helden with their lead-singer Judith Holofernes. I had always given them the credit for the idea with the lyrics cards that they drop as the lyrics come – but again, it was Dylan’s idea. Wir sind Helden evolved this further to use little video tricks of slow-motion, playing backwards and gestures. But the basic is purely due to Dylan. Now, that is influencing new generations of artists. Mesmerizing.

I guess there’s much more down the rabbit hole. There’s a lot to discover in the oeuvre of Bob Dylan. And the few things that I have discovered already tell me he’s a worthy Laureate of the Nobel Prize for Literature.

# Beyond

I have rather stopped watching movies in the cinema in recent years. This is partly due to my lack of leisure time, but for the most part it’s due to the lack of actual good and interesting movies. It appears to me that the overwhelming part of movies in the theatres are by now either re-makes (such as Spiderman) or sequels and mixtures of already well-known ingredients. I would consider Inception and Interstellar to be the last truly original movies in the past years (and, yes, most probably I’ve overlooked something – but considering the fact that I could overlook that would prove my point).

Paying respect to the franchise that I have spent so much time with, I watched Star Trek Beyond in a cinema near me the other day. Since I’ve only seen it once and there is not yet much access to background matrial, I can’t really say that many deep things about it. The punchline would be: it was entertaining.

Note: This text contains heavy spoilers. Don’t read it unless you have seen the movie, or unless you wish to be spoiled. You have been warned.

Star Trek Beyond tries very hard to be a modern Star Trek movie, being released in the 50th anniversary year of the franchise. One can tell from the optics, from the lines, the various references to the old movies and even to Enterprise. They also hid several small in-jokes in there (some more hidden, some less), such as the seat belts which are only present in the old-era ship. They really tried this time, and I believe they nearly succeeded. Beyond does not feel like classical Star Trek, it’s a modern action movie – but then again, you couldn’t make classical Star Trek episodes on the big screen these days. You couldn’t even make classical Star Trek episodes on the big screen in the old days (I don’t believe, Darmok would work in the cinema; and neither Wrath of Khan nor First Contact are the beloved Star Trek from TV: they are action flicks of their respective time as well).

The credit goes, as far as I can see, to Simon Pegg, writer of the movie and actor of Scotty. He did a fine job of trying to translate Star Trek to modern-day cinema. In the end, we have got a very entertaining movie with a proper balance of action and comedy, a tiny bit of character development and some very nice sequences of the entire crew working together. On top of that, there are interesting new aliens (in the first place of course Jaylah, but I also liked Ensign Syl). Jaylah was a really interesting character – if only Scotty had stopped calling her “Lassie”.

The fact that the crew works together to achieve their goals had been lost in the franchise. Both Next Generation and Deep Space 9 had been ensemble-shows. Each character was important in their own right, and they all got their moments to shine (not all of those moments were well-used, sadly, but nonetheless). Towards the final seasons of Voyager and for most of Enterprise, this had been dropped altogether. Problems were dealt by Janeway, Seven of Nine and the Doctor (only briefly assisted by the rest of the crew, dropping Chaktotay and Harry nearly completely) or by Archer, T’Pol and Trip. In Beyond, the crew works together, which is, of course, some sort of the message of the entire movie: the villain does not believe in teamwork, so he’s beaten by a strong team. This makes the movie like sort of an ensemble show again, down to the point where they complete each other’s sentences, and the even split the classical Star Trek monologue so that each of them has a couple of words in it.

I really liked the scenes of McCoy and Spock, who pay homage to the old series, but who also push the boundaries of the chemistry between these characters. Those two are the buddies of this movie, not Kirk and Spock. And still, Uhura gets her time with Spock again.

Completing the ensemble-topic, I have really come to appreciate the cast, especially the actors of Spock, Uhura and McCoy. They do a really fine job of not impersonating the well-known actors but of giving their characters a new level of personality (which comes from the different experiences they have had in the alternate reality).

I totally loved the looks of the starbase, a huge city in all three dimensions of space. The really put some effort in this one, as well for the looks of the old starship that had stranded on the planet. Also, the small attacking ships have a great and scary look to them – well done, there.

Many of the action-ideas are nicely executed, let me just name the capture of the three rogue ships by letting them smash in the starship, or the Sulu’s really awesome lift-off sequence of the ship by letting it fall down a cliff (only slightly cheesy be having it strike a mountain during its rise).

Now to the rather weak parts of the movie: there’s too much fast-paced action for my taste. Especially towards the destruction sequences of the Enterprise, I had lost track of what was going on. Too many quick cuts, too many close-up shots. I wasn’t even aware of what I should be frightened of, anymore. Action is well and good, and it has its place, but the thing is: I need to understand what’s going on. In particular, in 3d this is quite tough to deal with (as an aside: I’m not fond of 3d anyway and I think it sucks that there are too very little instances to watch the movie in 2d).

Then, the villain. I’ve even forgotten his name, and I didn’t really get his motivation. That’s sad, because they tried to construct a motivation that is tied into the Enterprise-stories of the Xindi and the Romulan War. They also tried to elaborate on this but – hey, couldn’t you do a little better than that? It’s already tough to understand why the villain is as bloodthirsty as he is, why he actually has a grudge against the Federation (he was a maco-soldier in duty to Starfleet, after all – what did he fight for, then? Even if they screwed him over, why not avoid the Federation completely?), why he wants to destroy the starbase, and what did he need that artifact for? The whole storyline doesn’t really fit. The revealing plot-twist isn’t explained either – why did Uhura notice that particular guy in that old movie (is it a log?) to be their villain? It couldn’t have been the looks, but was it the voice? No explanation. What is it that this artifact does – the death of Ensign Syl is shown, but what is going on there?

Then, the deus ex machina planned in advance: there was no actual reason for McCoy and Spock to steer the enemy ship. They didn’t do any good there (because music would save the day, huh?). But on the other hand, we needed a rescue option for Kirk 20 minutes later. In the meantime, they had plenty of time to return to their crewmates, and they didn’t have anything else to do either. But, they stay away in their enemy ship to save the day in the nick of time. Seriously?

Then, the music. Oh, how I loathed the first trailer of this movie when it came out (there’s a reason why I don’t even embed it here). It looked like some random action-packed over-the-top funny movie. When I re-watch the trailer now, it strangely fits, but there’s this music-topic. Rock music saves the day to stop the attack on the starbase. That might have been a nice idea, but it’s executed so strangely and with a ridiculously large explosion… my dear.

Then, is there a reason why this movie is called Beyond? There’s not so much of undiscovered countries, or of breaking frontiers. Even though the villain is talking about the frontier that pushes back – they deal with an unknown part of space. That’s what they always do, that’s their job on this mission. I actually expected a deeper mystery of some sort.

To conclude: too bad that they had to cling to their clichés so much. Too many fist-fights, in the end a fight with the villain to the death, even a ridiculous motorcycle sequence… possibly necessary for success to the box office, but it wouldn’t have been necessary for a proper movie. Especially since I am going to forget most of my impressions of this movie soon. It’s not like memorable movies that make me think even today. But it was fine entertainment and it will keep the Star Trek universe alive.