If $K$ is $w$−compact and convex, $f\in X^\ast \implies f$ attains its maximum on $K$

Question

Let $X$ be a real Banach space

If $K\subset X$ is weakly compact and convex, then for a given $f\in X^\ast$ (dual space) we can always

find $k\in K$ such that

$$\displaystyle \sup_{x\in K}{f(x)}=f(k)$$

Any hints would be appreciated.

Answer 1

By definition of the weak topology, $f$ is weakly continuous, so $f(K)$ is a compact subset of $\mathbb{R}$.

Convexity of $K$ is not needed, but it ensures that $f(K)$ is actually an interval.