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I'm looking at this proof of Schauder theorem and I am struggling with a passage. This is my problem:

Let $X$ be a Banach space, $K \subset X$ a convex, close and bounded set and $F:K \rightarrow K$ a compact operator.

$F$ compact and $K$ bounded imply that $ \:\overline{F(K)} \subset K$ is a compact set. So, fixed $\epsilon >0, $there exist $\{v_1,...,v_{n_\epsilon}\}$ such that $F(K) \subset\overline{F(K)} \subset \bigcup_{i=1}^{n_\epsilon}B(v_i, \epsilon)$.

I call $E_\epsilon$ the vector space generated by $v_1,...,v_{n_\epsilon}$ and for $u\in F(K)$ I consider

$G_\epsilon(u)=\sum_{k=1}^{n_\epsilon} \lambda_i(u) v_i$ with $\lambda_i\in (0,1)$ and such that $ \sum_{k=1}^{n_\epsilon} \lambda_i(u) =1 $ for every $u$.

Clearly $G_\epsilon\in E_\epsilon$ and then I have that $G_\epsilon(u)\in M:= co(v_1,...,v_n)$ (the convex hull of $ \{ v_1,...,v_n \} $).

$\textbf{The claim is this}$: if $u\in F(u)$ then $G_\epsilon(u) \in E_\epsilon \cap K$.

In order to prove that it is said: $G_\epsilon(u) \in E_\epsilon \cup F(K) \subset E_\epsilon \cup K$.

$\textbf{My problem:}$ The thing that I don't understand is why $G_\epsilon \in F(K)$.

I would have concluded in this way: $G\epsilon\in M$, which is the smallest convex set containg the elements $v_i\in K$. Since $K$ is convex I have directly that $G_\epsilon(u)\in M\subset K$. Is it correct?

Many thanks for the help.

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1 Answer 1

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I think we can argue as follows.

Let's fix $\epsilon > 0.$ Since $\overline{F(K)}$ is compact, there is a finite set $$ W_\epsilon := \{w_1,\ldots,w_{n_\epsilon}\} \subseteq \overline{F(K)} $$ such that $$ \overline{F(K)} \subseteq \bigcup_{j=1}^{n_\epsilon}B(w_j,\frac{\epsilon}{2}). $$ Moreover, for a fixed index $j\in\{1,\ldots,n_\epsilon\},$ since $w_j\in\overline{F(K)},$ there is a $v_j\in F(K)$ such that $$ \|v_j - w_j\| < \frac{\epsilon}{2}. $$ This implies $$ B(w_j,\frac{\epsilon}{2}) \subseteq B(v_j,\epsilon). $$ Thus, we have the finite set $$ V_\epsilon := \{v_1,\ldots,v_{n_\epsilon}\} \subseteq F(K) $$ with $$ \overline{F(K)} \subseteq \bigcup_{j=1}^{n_\epsilon}B(w_j,\frac{\epsilon}{2}) \subseteq \bigcup_{j=1}^{n_\epsilon}B(v_j,\epsilon). $$ If we use this particular set $V_\epsilon,$ we can further argue that $$ G_\epsilon(u) = \sum_{j=1}^{n_\epsilon} \lambda_j(u)v_j $$ is a convex combination of elements of $F(K),$ and since $F(K)$ is convex (since $K$ is convex, and $F$ is linear), we must have $$ G_\epsilon(u)\in F(K), $$ as desired.

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