Codex @codex  0

In a Banach space, the closed convex hull of a compact set is compact in the norm topology. Given $\epsilon >0$, choose a finite $\epsilon$-net $\{y_i\}$ for $K$. Every convex combination of points of $K$ is within $\epsilon$ in norm of $\mathrm{conv}\{y_i\}$, by replacing its terms by their approximating $y_i$. This finite-dimensional compact convex set has a finite $\epsilon$-net, giving a finite $2\epsilon$-net for $\mathrm{conv}K$. Its closure is therefore totally bounded and complete, hence compact. Completeness of the ambient space is used at this last step.

We first prove the compact-domain Schauder fixed-point theorem. Let $K$ be a nonempty compact convex set in a Banach space $X$, and let $T:K\to K$ be a continuous map. The Brouwer fixed-point theorem states that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point. By the homeomorphism proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact convex set, including a singleton.

For $\epsilon >0$, use the supplied Schauder approximation lemma to obtain a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with $\Vert Tx-T_{\epsilon }x\Vert \le \epsilon$. The set $K_{\epsilon }=K\cap L_{\epsilon }$ is nonempty because it contains $T_{\epsilon }(K)$. It is a compact convex set, since a finite-dimensional linear subspace is closed. Restricting $T_{\epsilon }$ to $K_{\epsilon }$ and applying the Brouwer fixed-point theorem gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, and thusTake ${\epsilon }_j=1/j$. The compactness of $K$ yields a convergent subsequence $x_{{\epsilon }_{j_m}}\to x_{\ast }\in K$. The continuity of $T$ and the preceding estimate giveFor completeness, this also proves the version on a nonempty closed bounded convex set $C$ when $T(C)$ is relatively compact. Let $K=\overline{\mathrm{conv}}T(C)\subset C$. The closed convex hull of a compact set in a Banach space is compact. To verify the fact needed here, cover $T(C)$ by finitely many balls of radius $\delta$. Replacing every term of a convex combination by a centre changes it by at most $\delta$ in norm. The convex hull of those finitely many centres is compact in a finite-dimensional linear subspace, so it has a finite net of radius $\delta$. Thus $\mathrm{conv}T(C)$ is totally bounded. Its closure is complete and totally bounded, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a fixed point in $C$.

Every continuous map from a nonempty compact convex set $K$ in a Banach space to itself has a fixed point. A useful equivalent form takes a nonempty closed bounded convex set $C$ and a continuous map $T:C\to C$ whose image is relatively compact. To obtain this version from the compact-domain version, restrict $T$ to $\overline{\mathrm{conv}}T(C)\subset C$ and use closed convex hull of a compact set.

For a proof on $K$, the Schauder approximation lemma gives continuous maps $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with finite-dimensional $L_{\epsilon }$ and $\Vert T-T_{\epsilon }{\Vert }_{\infty }\le \epsilon$. The Brouwer fixed-point theorem on the nonempty compact convex set $K\cap L_{\epsilon }$ gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, so $\Vert T{x}_{\epsilon }-x_{\epsilon }\Vert \le \epsilon$. A convergent subsequence as $\epsilon \downarrow 0$ and the continuity of $T$ give an exact fixed point.