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Let $C$ be a nonempty bounded closed convex subset, say the unit ball, of $\ell^2(\mathbb{N})$. Let $T: C\to 2^C$ be a map such that $T(x)$ is nonempty closed for each $x$, and that $$D(Tx,Ty)\le \|x-y\|$$ for all $x,y\in C$, where $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r>0: N_r(A)\supset B, N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the $r$-neighborhood of $S$.

The question is whether $T$ has a fixed point, i.e. a point $x\in C$ such that $x\in Tx$.

The answer is yes if $T$ is a contraction, i.e. replacing the $\le \|x-y\|$ by $\le \lambda \|x-y\|$ for some $0\le\lambda<1$ in the definition.; or if $Tx$ is compact for each $x$.

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Let $C$ be a nonempty bounded closed convex subset, say the unit ball, of $\ell^2(\mathbb{N})$. Let $T: C\to 2^C$ be a map such that $T(x)$ is nonempty closed for each $x$, and that $$D(Tx,Ty)\le \|x-y\|$$ for all $x,y\in C$, where $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r>0: N_r(A)\supset B, N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the $r$-neighborhood of $S$.

The question is whether $T$ has a fixed point, i.e. a point $x\in C$ such that $x\in Tx$.

The answer is yes if $T$ is a contraction, i.e. replacing the $\le \|x-y\|$ by $\le \lambda \|x-y\|$ for some $0\le\lambda<1$ in the definition; or if each $Tx$ is weakly closed.

2 added 33 characters in body

Let $C$ be a nonempty bounded closed convex subset, say the unit ball, of $\ell^2(\mathbb{N})$. Let $T: C\to 2^C$ be a map such that $T(x)$ is nonempty closed for each $x$, and that $$D(Tx,Ty)\le \|x-y\|$$ for all $x,y\in C$, where $D$ is the Hausdorff metric defined by $$D(A,B)=\inf\lbrace r>0: N_r(A)\supset B, N_r(B)\supset A\rbrace,$$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the $r$-neighborhood of $S$.

The question is whether $T$ has a fixed point, i.e. a point $x\in C$ such that $x\in Tx$.

The answer is yes if $T$ is a contraction, i.e. replacing the $\le \|x-y\|$ by $\le \lambda \|x-y\|$ for some $0\le\lambda<1$ in the definition; or if each $Tx$ is weakly closed.

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