In a normed space $E$, is the convex envelope of any compact set also compact?

Question

In a normed space $E$;

One can show that, if $E$ is finte-dimensional, the convex envelope of any compact set is also compact. $(*$)

Is (*) true if $E$ is no longer supposed finite-dimensional ?

The answer is negative in both cases. See here: https://math.stackexchange.com/questions/340324/is-the-convex-hull-of-closed-set-in-rn-is-closed — uniquesolution, Dec 03 '17 at 07:29

score 1 · Accepted Answer · answered Dec 03 '17 at 07:42

The convex hull of a compact set in an infinite-dimensional Banach space is not necessarily closed, and thus not necessarily compact. For example, in a Hilbert space with orthonormal basis $\{u_n:\; n \in \mathbb N\}$ consider the compact set $S$ consisting of $0$ and $u_n/n$ for positive integers $n$. Each point in the convex hull of this set is a linear combination of only finitely many $u_n$. However, e.g. $\sum_{n=1}^\infty 2^{-n} u_n/n$ is in the closure of the convex hull.

In a normed space $E$, is the convex envelope of any compact set also compact?

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