$ \frac{1}{n}\sum_{i=1}^{n}{C_i}\text{ is weakly compact} $

Question

Let $X$ be a separable Banach space. Let $C_1,...,C_n$ are nonempty weakly compact convex subsets of $X$. Why

$$ \frac{1}{n}\sum_{i=1}^{n}{C_i}\text{ is weakly compact} $$ An idea please.

Answer 1

In general for a topological vector space $X$ and compact subsets $A$ and $B$, $A+B$ is the image of the compact set $A\times B$ under the continuous map $+: X\times X\rightarrow X$. Hence $A+B$ is compact. Multiplication by scalars also preserves compactness.