Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $S$.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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 $r>0$.)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.