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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 $V$ be a compact convex subset of the Banach space $X$ and $f:V\to V$ continuous. Then $f$ has a fixed point.

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

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

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

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

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

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

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

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

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

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

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

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

We prove $f(y_0) = y_0$.

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

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

Therefore

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

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