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El teorema de punto fijo de Schauder

The proof of Schauder's fixed point theorem we saw in class was, sadly, incomplete, as the approximating functions defined on the finite dimensional convex sets were not well defined. Here is a correct proof of the theorem.

Theorem (Schauder). Let $latex V$ be a compact convex subset of the Banach space $latex X$ and $latex f:V\to V$ continuous. Then $latex f$ has a fixed point.

Proof. For a given $latex n\in\Z_+$, let $latex x_1, \ldots, x_k\in V$ be such that

$latex \displaystyle V\subset\bigcup_i^k B_{1/n}(x_i)$.

Such $latex x_i$ exist because $latex V$ is compact. Define, for each $latex i$, the functions $latex \lambda_i:V\to\R$ by

$latex \lambda_i(x) = \begin{cases}1/n-||x_i - x||& x\in B_{1/n}(x_i)\\0 & \text{otherwise.}\end{cases}$

The functions $latex \lambda_i$ are continuous and $latex \sum_i \lambda_i(x)\not=0$ for every $latex x\in V$. Thus, if we define

$latex \pi_n(x) = \dfrac{\sum \lambda_i(x)x_i}{\sum\lambda_i(x)}$,

$latex \pi_n$ maps $latex V$ into the convex hull $latex V_n$ of $latex x_1, \ldots, x_k$, because $latex \pi_n(x)$ is a convex combinations of the $latex x_i$. $latex V_n\subset V$ because $latex V$ is convex.

Moreover, as $latex \lambda_i(x)=0$ if $latex ||x_i-x||\ge1/n$, we have

$latex \lambda_i(x)||x_i - x|| \le \lambda_i(x)\dfrac{1}{n}$

for all $latex i$ and all $latex x\in V$, and thus

$latex ||\pi_n(x) - x|| = \Big|\Big|\dfrac{\sum\lambda_i(x)(x_i-x)}{\sum\lambda_i(x)}\Big|\Big| \le \dfrac{1}{n}.$

Now, set $latex f_n = \pi_n\circ f|_{V_n}$. Then $latex f_n:V_n\to V_n$ is continuous on the finite dimensional compact convex set $latex V_n$ so, by Brower's fixed point theorem, it has a fixed point, say $latex y_n$. Since $latex V$ is compact, $latex y_n$ has a convergent subsequence. So passing to such subsequence we can assume $latex y_n$ converges, and let's say $latex y_n\to y_0$.

We prove $latex f(y_0) = y_0$.

Let $latex \e>0$ be given. Since $latex f$ is continuous, there exists $latex \delta>0$ such that $latex \delta<\e/3$ and, if $latex ||y_0-x||<\delta$ then $latex ||f(y_0) - f(x)||<\e/3$.

Now, let $latex N>3/\e$ such that, if $latex n\ge N$, then $latex ||y_n - y_0||<\delta$.


$latex ||f(y_0) - y_0||\le ||f(y_0) - f(y_N)|| + ||f(y_N) - f_N(y_N)||+ ||f_N(y_N) - y_0|| \\ \hspace*{.9in} < \dfrac{\e}{3} + \dfrac{1}{N} + \dfrac{\e}{3} < \e.$

Since $latex \e>0$ is arbitrary, $latex f(y_0) = y_0$. $latex \Box$