Exercise about weak topology

Question

Let $(E,||\cdot||)$ be a Banach space, $M\subset E$ a linear subspace and $f_0 \in E^*$. Prove that there exists $g_0 \in M^\perp$ such that $$ \inf_{g \in M^\perp} ||f_0-g|| = ||f_0-g_0|| $$ HINT: Note that $B_{E^*}$ is compact for $\sigma(E^*, E)$ and $M^\perp$ is closed in $\sigma(E^*,E) $.

I tried following the hint and I showed that $M^\perp$ is closed, since $M^\perp= \bigcap_{x\in M} J_x^{-1}(\{0\})$ is an intersection of closed sets; I also know that $B_{E^*}$ is compact. Anyway, I do not know how to proceed (I thought of using sequential compactness of the closed unit ball in $E^*$, but it seems that the hypothesis that $E^*$ is separable is necessary to have it). Any help is really appreciated.

Do you know that compactness implies the existence of a weak* convergent subnet? — gerw, Mar 25 '19 at 12:22
Actually, I do not know what nets are...I'll read something about, thank you! — mark, Mar 25 '19 at 13:10

score 0 · Accepted Answer · answered Mar 25 '19 at 12:22

There exists a sequence $\{g_n\}\subset M^{\perp}$ such that $\|f_0-g_n\|$ converges to LHS. By compactness the sequence $\{g_n\}$ has a subnet $\{g_n'\}$ converging to some $g_0 \in M^{\perp}$. For any $x \in E$ we have $g_{n'}(x) \to g_0(x)$ and $|f_0(x)-g_{n'}(x)|\leq \|x\| \|f_0-g_{n'}\| \to $ (LHS) $\|x\|$. So $|f_0(x)-g_0(x)| \leq $ (LHS) $\|x\|$. Taking sup over $x$ in the unit ball of $E$ we get RHS $\leq $ LHS. The reverse inequality is obvious.

Exercise about weak topology

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