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Suppose $X$ is a strictly convex Banach space. Does there exist a bounded closed convex set $K$ in $X$ such that the set of all Chebyshev centers $C(K)$ of $K$ is a proper subset of $K$ with diameter of $C(K) > 0$ ?

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    $\begingroup$ As a general rule, titles more to the point are preferred (it make it easier to reach potentially interested people). Also, you can use more tags, and more precise (banach-spaces, convexity,..). I suggest to change the title accordingly. $\endgroup$ – Pietro Majer Sep 5 '14 at 12:21
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My purpose is to show that there exists $X$ for which the mentioned phenomenon occurs. Of course it does not occur for all strictly convex spaces because it does not occur for uniformly convex spaces.

I think that one can work out an example of such space $X$ and set $K$ on the following lines. We consider the space $\ell_2$ with the following equivalent norm: We pick a tending to $\infty$ increasing sequence $\{p_i\}$ of numbers $>1$. For $a=\{a_i\}_{i=0}^\infty$ we let $$||a||_s=\sup_{i\ge 1}\left(|a_0|^{p_i}+|a_i|^{p_i}\right)^{1/p_i}+\left(\sum_{i=1}^\infty|a_i|^2\right)^{1/2}.$$ This norm is strictly convex because the supremum is always attained.

Denote by $\{e_i\}_{i=0}^\infty$ the unit vector basis of this space. Let set $K$ be defined as the closed convex hull of $\left\{\pm \frac34e_0,\pm b_ie_i, i\ge 1\right\}$ where $\{b_i\}$ are chosen in such a way that $$\left\|\frac14e_0\pm b_ie_i\right\|_s=1.$$ Then $|b_i|\uparrow \frac12$. This implies that $\hbox{diam}K=2$. Therefore each element of the line segment joining $\frac14e_0$ and $-\frac14e_0$ is in the Chebyshev center of $K$. It seems straighforward to verify that actually $C(K)$ is equal to this line segment.

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