Characterization of weak compactness in a Banach space

Question

While studying the textbook of Fernando Albiac and Nigel J. Kalton (Topics in Banach Space Theory), I came across the following result:

A subset $A$ of a Banach space $X$ is relatively weakly compact if and only if it is norm-bounded and the $\sigma(X^{**},X^*)$-closure of $A$ in $X^{**}$ is contained in $A$.

I am relatively unexperienced with weak topologies, so I would appreciate help in comprehending how that result might be proven. I am especially interested in the "if"-direction.

A first idea to prove the "if" direction is, to apply the Banach-Alaoglu theorem in $X^{**}$ and use that the canonical map $i:X\to i(X) $ given by $i(x):=(X^*\ni f\mapsto f(x))$ is actually a homeomorphism with respect to $(X,\sigma(X,X^*))$ and $(X^{**},\sigma(X^{**},X^*))$.

If that might work, I would appreciate getting ideas on how to prove the $(X,\sigma(X,X^*))$, $(X^{**},\sigma(X^{**},X^*))$ - continuity of $i$.

Answer 1

If $x\in B_X$, then $\|i(x)(f)\|=|f(x)|\le \|f\|\|x\|\le \|f\|\Rightarrow \|i(x)\|\le 1\Rightarrow \sup_{x\in X}\|i(x)\|\le 1,$ so $i$ is bounded. It is clearly linear, so it is continuous.

Another way is to take a neighborhood $\mathcal N$ of $0\in X^{**}$ and show that there is a neighborhood $\mathcal M$ of $0$ in $X$ and show that $i(\mathcal M)\subseteq \mathcal N$. Without loss of generality, assume that $\mathcal N_{f^{*}}$ is a subbasis element; that is, that $\mathcal N_{f^{*}}=\{\phi\in X^{**}:|f_*\phi|=|\phi(f)|<\epsilon\}.$ Then, if $\mathcal M=\{x\in X:|f(x)|<\epsilon\},$ we have $|f_*(i(x))|=|i(x)(f)|=|f(x)|<\epsilon$

Answer 2

Working with nets make this almost trivial:

Suppose $(x_\gamma)$ is a net in $X$ that is weakly convergent to $x\in X$. If $f\in X^*$, then we have $$i(x_\gamma)(f)=f(x_\gamma)\to f(x)=i(x)(f),$$ so $(i(x_\gamma))$ is weak$^*$ convergent to $i(x)$, and thus $i$ is weak-weak$^*$ continuous.