Convex Hull of Precompact Subset is Precompact 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.
 A: 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 $F+U\supset K$.
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
subset $F’$ of the set $\operatorname{conv} F$ such that $\bigcup \{z+V: x\in F’\}\supset F$. Finally we obtain that $\operatorname{conv} K\subset \operatorname{conv} F+V\subset F’+V+V\subset F'+U$.
