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In a Banach space, is the convex hull of finite set compact?

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closed as too localized by George Lowther, Matthew Daws, Andreas Blass, Bill Johnson, Alain Valette Nov 4 '11 at 15:26

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The convex hull of a set of $n$ points is the image of the $n$-simplex in $\mathbb R^n$ under a continuous function, so yes. – Mariano Suárez-Alvarez Apr 26 '10 at 5:46
Mariano: do you want to write this as an answer (community wiki if you wish) so I can vote it up? I think it's slightly more to the point than Pete's answer (no offence, Pete!) which works fine but I think is slightly over-elaborate. To be fair, the same idea underlies both. – Yemon Choi Apr 26 '10 at 8:05
I am also not keen on questions which give little to no indication of (a) why the questioner wants to know (b) what they've tried doing (c) what level they are at. – Yemon Choi Apr 26 '10 at 8:06
@Yemon: I agree so much that I deleted my answer. – Pete L. Clark Apr 26 '10 at 16:58

Suppose $X$ is your Banach space and let $\{x_1,\dots,x_n\}$ be a finite subset of $X$. Let $$S=\{(t_1,\dots,t_n)\in\mathbb R^n:t_1,\dots,t_n\geq0,\\,t_1+\cdots+t_n=1\}$$ be the standard simplex in $\mathbb R^n$. The map $$\phi:(t_1,\dots,t_n)\in S\mapsto t_1x_1+\cdots+t_nx_n\in X$$ is evindently continuous and its image is $\mathrm{conv}\{x_1,\dots,x_n\}$. Since $S$ is compact, so is $\mathrm{conv}\{x_1,\dots,x_n\}$.

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Just to make perfectly clear what is used in the proof: $\phi$ is continuous because $X$ is a topological vector space and $S$ is compact as $X$ is a separated space. – Torsten Ekedahl Apr 26 '10 at 15:46
$S$ is compact because it is closed and bounded in $\mathbb R^n$, rather! :) – Mariano Suárez-Alvarez Apr 26 '10 at 15:48
Sorry, misread I thought $S$ was the image. – Torsten Ekedahl Apr 26 '10 at 16:48

Of course yes: the n points lie in the finite-dimentional linear subspace generated by themselves (remember that any norm, when restricted to a finite-dimensional linear subspace, gives rise to the same topology on that space).

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Actually, the convex hull of a sequence of points $(x_n)$ is (relatively) compact when $x_n\rightarrow 0$, and this easily gives a positive answer to your question (but is somewhat overkill). In fact, a closed convex set K in a Banach space is compact if and only if it's contained in the closed convex hull of a sequence $(x_n)$ with $x_n\rightarrow 0$. See, for example, Lindenstrauss and Tazfriri, vol I, Proposition 1.e.2.

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Actually, the statement in the question holds for all topological vector spaces, not just Banach spaces. The generalisation here does not apply so widely. I think you need local convexity. – George Lowther Nov 4 '11 at 12:21

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