4
$\begingroup$

I'm trying to prove that, if $K$ is a precompact (I've also heard the phrase totally bounded used for this) subset of a Banach Space $X$, then its convex hull is also precompact.

I've come across a similar statement with Hilbert spaces that suggested I fix some $x\in X$ and define a bounded conjugate linear form on $X$ by $x\mapsto B(x,y)$ and then use the Riesz Representation Theorem. I'm not sure if this can be generalized to a proof for general Banach spaces. Any help would be appreciated.

Thank you.

$\endgroup$
5
$\begingroup$

A subset $K$ of an abelian topological group $X$ is called precompact provided for each neighborhood $U$ of the zero of the group $X$ there exists a finite subset $F$ of the space $X$ such that $F+U\supset K$.

It seems the claim holds for each locally convex topological vector space $X$. Indeed, let $U$ be an arbitrary neighborhood of the zero of the group $X$. Since the space $X$ is locally convex, there exists a convex open neighborhood $V$ of the zero of the space $X$ such that $V+V\subset U$. Since the set $K$ is precompact, there exists a finite subset $F$ of the space $X$ such that $F+V\supset K$.

Let $x$ be a point of the convex hull $\operatorname{conv} K$ of the set $K$. Then there exist a natural number $n$, non-negative real numbers $\lambda_1, \dots, \lambda_n$ with sum $1$, and points $x_1,\dots,x_n$ of the set $K$ such that $x=\sum_{i=1}^n \lambda_ix_i$. Since $F+V\supset K$, for each $i$ from $1$ to $n$ there exist points $f_i\in F$ and $y_i\in V$ such that $x_i=f_i+y_i$.

Then $x=\sum_{i=1}^n \lambda_i (f_i+y_i)= \sum_{i=1}^n \lambda_i f_i+\sum_{i=1}^n \lambda_i y_i$. The second summand is contained in the set $V$, because the set $V$ is convex. The first summand is contained in the convex hull $\operatorname{conv} F$ of the finite set $F$.

The set $\operatorname{conv} F$ is compact as a continuous image of a compact set $\Lambda_n\times F^n$, where $\Lambda_n=\{\lambda\in \mathbb R^n: \sum_{i=1}^n \lambda_i=1$ and $ \lambda_i\ge 0$ for every $i\}$ is an $n-1$-dimensional simplex. Since $\{z+V: z\in \operatorname{conv} F \}$ is an open cover of the compact set $\operatorname{conv} F$, there exists a finite a finite subset $F’$ of the set $\operatorname{conv} F$ such that $\bigcup \{z+V: x\in F’\}\supset F$. Finally we obtain that $\operatorname{conv} K\subset \operatorname{conv} F+V\subset F’+V+V\subset F'+U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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