# A map on the unit ball

Let $B$ be the unit ball of $\ell^2(\mathbb{N})$, i.e. $B=\lbrace x\in \ell^2(\mathbb{N}): \|x\|\le 1\rbrace.$ For each $x=(x_1,x_2,\cdots)\in B$, let $$f(x)=(1-\|x\|,x_1,x_2,\cdots).$$ Define $T:B\to 2^B$ by $$T(x)=B(f(x),r(x))\cap B, \mbox{ where }r(x)=\frac{1}{2}(\|x-f(x)\|).$$ Is it true that $$D(Tx,Ty)\le \|x-y\| \mbox{ for all }x,y\in B?$$ Here $B(y,r)$ denotes the closed ball with radius $r$ centered at $y$, and $D$ 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 $S$.

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It's wrong: Take $x=0$ and $y=e_1=(1,0,0,\dots)$. Then $f(x)=e_1$ and $f(y)=e_2=(0,1,0,\dots)$, so $r(x) = 1/2$ and $r(y) = \sqrt2/2$. Now it is not hard to see that $p = (-\sqrt7/4,3/4,0,\dots)$ is the point in $Ty = B(f(y),r(y))\cap B$ farthest away from $f(x)$. From this one deduces that $$D(Tx,Ty) = \sqrt{(1+\sqrt7/4)^2+(3/4)^2} - 1/2 = \sqrt{2+\sqrt7/2} - 1/2 > 1 = \|x-y\|.$$ (As $Ty$ is the larger chunk, geometric considerations show that $Ty \subseteq N_r(Tx)$ implies $Tx \subseteq N_r(Ty)$, for all $r>0$.)
Let $p=(-\sqrt{7}/4, 3/4,0,0,\cdots)\in T(y)$. According to my calculation the minimum distance $d(p, Tx)$ from $p$ to $Tx$ is $$\sqrt{(1+\sqrt{7}/4)^2+9/16}-1/2\approx 1.32287…$$ So $D(Tx, Ty)\ge 1.32287>\|x-y\|$ (but I am not able to show $D(Tx, Ty)\ge\|f(x)-f(y)\|$). This is enough to show that $T$ is not nonexpansive. Thus my previous question [Multi-valued Nonexpansive Maps][1] is still open. [1]:math.stackexchange.com/questions/12444/… – TCL Dec 18 '10 at 16:06
@TCL: I'm not exactly sure what you're doing with the $p$. Have a close look at your definition of $D(A,B)$: It's not the minimum distance of $A$ and $B$ but the Hausdorff distance. There you've got $D(B(x,s),B(y,r))=\|x−y\|+|r−s|≥\|x−y\|$. – Hendrik Vogt Dec 18 '10 at 16:25
Don't forget that the definition of $T$ includes an intersection with B(0,1). – TCL Dec 18 '10 at 16:41
@TCL: You're quite right, sorry; thanks for pointing that out. As you see from my corrected answer, now I do understand what you're doing with the $p$ :-). – Hendrik Vogt Dec 18 '10 at 17:40