Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
18  
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
4  
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
4  
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
1  
@Yemon: I agree so much that I deleted my answer. –  Pete L. Clark Apr 26 '10 at 16:58
add comment

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

3 Answers

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 $\mathrm{conv}\{x_1,\dots,x_n\}$.

share|improve this answer
    
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
2  
$S$ 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
add comment

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).

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.