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Let $E$ a finite dimensional real normed vector space and let $K\subset E$, such that $K\neq\emptyset$ and $K$ compact.

It is well known that there exists a closed ball $B$, of minimal radius, such that $K\subset B$ (see below for a proof).

Question Does this result remain true in an infinite dimensional space ?

Proof for the finite dimensional case

We denote by $\bar{B}(x,r)$ the closed ball with center at $x$ ($x\in E$) and radius $r$ ($r>0$).

Since $K$ is bounded, the set ${\cal R}=\{r\in[0,+\infty[;\,\exists x\in E;\,K\subset\bar{B}(x,r)\}$ is not empty and we can consider $\rho=\inf({\cal R})$.

For every $n\in\mathbb{N}^\star$, there exist $\rho_n\in[\rho,\rho+\frac{1}{n}[$ and $x_n\in E$ such that $K\subset\bar{B}(x_n,\rho_n)$

Fix $a\in K$. For all $n\ge1$, we have :

$$\Vert x_n\Vert-\Vert a\Vert\le\Vert x_n-a\Vert\le\rho_n\le\rho+\frac{1}{n}\le\rho+1$$

So, the sequence $(x_n)$ is bounded. Since $\dim(E)<\infty$, there exists a convergente subsequence $(x_{\varphi(n)})$.

Put $c=\lim_{n\to\infty}x_{\varphi(n)}$. Taking the limit ($n\to\infty$) in $\Vert x_{\varphi(n)}-a\Vert\le\rho_n$, we get $\Vert c-a\Vert\le\rho$. Hence $K\subset\bar{B}(c,\rho)$, which completes the proof.

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    If $E$ is Banach (i.e. complete) and reflexive, the same argument can be repeated using the fact that bounded sets are relatively compact. This time $x_n$ converges *weakly* up to subsequences, and using lower semicontinuity of the norm under weak convergence we conclude – 2017-01-17
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    Interesting remark. Thank you. – 2017-01-17
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    Why don't you consider balls $\overline{B}(x, r)$ with $x\in K$ instead of $x\in X$? In this way the sequence $(x_n)$ will be inside the compact set $K$. – 2017-01-17
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    I can't do that because the $x_n$ don't have any reason to belong to $K$. For example : if $K$ is a square (boundary only) in the euclidean plane ... – 2017-01-17
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    I see... If $X$ is a Banach space, without loss of generality you may also assume that $K$ is convex, but this doesn't help at all: By Konyagin's theorem (1988), "For every nonreflexive space, there exists an equivalent norm on it and a three point set $K$ without a Chebyshev center". For a partial positive answer, which is slightly stronger than Del's, "In every dual Banach space, every bounded set has a Chebyshev center." I couldn't find Konyagin's proof, only a paper by Vesely that strengthens it: https://cmuc.karlin.mff.cuni.cz/pdf/cmuc0101/vesely.pdf – 2017-01-18
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    @treedetective Do you know a simple counterexample (if there is) to the main question if $X$ is not complete? – 2017-01-20
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    @Del I can't think of any... But for (nonreflexive) Banach spaces the simplest example is by Garkavi (1964). You can find it here, page 114: https://books.google.gr/books?id=bkNZDcvfyDwC&printsec=frontcover&hl=el#v=onepage&q&f=false I have the impression that all these counterexamples take more or less advantage of the existence of non norm attaining functionals in $X^*$, either by explicitly finding one, or, in the general case, by James' characterization of reflexivity. – 2017-01-24

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