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.

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