$K$ is weakly-compact $\Longleftrightarrow$ $\Pi(K)$ is weak*-compact Let $X$ be a Banach space and $K\subset X$.
$\displaystyle \Pi:X \longrightarrow X$** canonical injection
$\Pi(x)(f)=f(x)$
How can we prove that:
$K$ is weakly-compact $\Longleftrightarrow$  $\Pi(K)$ is weak*-compact
Any hints would be appreciated.
 A: I'll assume the base field is $\mathbb{C}$, but this could be $\mathbb{R}$ without changing anything.
The weak*-topology on $X^{**}$ is the weakest topology that makes the point evaluations $e_{x^*}:x^{**}\longmapsto x^{**}(x^*)$ continuous. It follows that a map $\phi:Y\longrightarrow (X^{**},w^*)$ is continuous if and only if $y\longmapsto \phi(y)(x^*)$ is continuous from $Y$ to $\mathbb{C}$ for every $x^*\in X^*$. 
The weak topology on $X$ is the weakest topology that makes the functionals of the topological dual $X^*$ continuous. It follows that a map $\theta: Z\longrightarrow (X,w)$ is continuous if and only if $x^* \circ \theta:Z\longrightarrow \mathbb{C}$ is continuous for every $x^*\in X^*$. 
Now for every $x^*\in X^*$, the map $x\longmapsto \Pi(x)(x^*)=x^*(x)$ is just the functional $x^*\in X^*$, which is continuous on $(X,w)$. Therefore $\Pi:(X,w)\longrightarrow (X^{**},w^*)$ is continuous.
Conversely, for every $x^*\in X^*$, the map $\Pi(x)\longmapsto (x^*\circ \Pi^{-1})(\Pi(x))=x^*(x)=\Pi(x)(x^*)$ is continuous on $(\Pi(X),w^*)$ as the restriction of $e_{x^{*}}$. Hence $\Pi^{-1}:(\Pi(X),w^*)\longrightarrow (X,w)$ is continuous.

Conclusion:  the isometric embedding $\Pi:X\longrightarrow X^{**}$, which is obviously a norm-norm homeomorphism onto its range, is also a $w-w^*$ homeomorphism onto its range. In particular, since the continuous image of a compact is compact, $K$ is $w$-compact if and only if $\Pi(K)$ is $w^*$-compact. Note that since both topologies are Hausdorff, such compacts are Hausdorff as well.

Note: we never used that $X$ is a Banach space. This is true for any normed vector space. And even for a topological vector space, where we only lose the isometry of $\Pi$.
