Chebyshev centres of a bounded closed convex set in a strictly convex Banach space 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 $C(K) > 0$ ? 
 A: 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.
