Let $(X,d)$ be a complete metric linear space whose balls are convex. Let $Y\subseteq X$ be a bounded, closed and convex subset that verifies the following property: for all $y_0\in Y$, the distance function $y\rightarrow d(y_0,y)$ attains its sup in $Y$. Does $Y$ have extreme points?

I was trying to adapt the classical Krein-Milman's proof: the family of non-empty closed faces is not empty (because, given $y_0$, the set of maxima of $d(y_0,\cdot)$ is an extremal non-empty set and one can get a face by taking a maximal and convex superset of $\bar y$, being $\bar y$ a point that maximizes the distance from $y_0$). It is inductive by completeness, boundedness and convexity of the faces. The last step is the hardest one and I'm not sure it is true: given a face $F$ with two different points $x$ and $y$, can I say that $d(x,\cdot)$ attains its sup in $F$?

Or.. are there counterexamples?


2 Answers 2


There is a counter-example.

Note that in any normed space, its unit ball satisfies your supremum-attaining property. Indeed, for any $x_0\in X$ the supremum of $d(x_0,\cdot)$ on the ball is attained at the point $-x_0/\|x_0\|$ if $x_0\ne 0$, and at any point of the sphere if $x_0=0$.

It remains to construct a Banach space whose unit ball has no extreme points. Let $X$ be the space of all sequences $(x_i)_{i=1}^\infty$ such that $x_i\to 0$ as $i\to\infty$, equipped with the $\ell_\infty$ norm $\|x\|=\sup|x_i|$. It is a closed subspace of $\ell_\infty$ and hence complete. But clearly its unit ball has no extreme points.


For a counterexample, let $X=c_0$ and let $Y$ be its unit ball.


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