3.9 Let $E$ be a Banach space; let $M\subset E$ be a linear subspace, and let $f_{0} \in E^{\star}$. Prove that there exists some $g_{0} \in M^{\perp}$ such that $$ \inf_{g\in M^{\perp}}\lVert f_{0}-g\rVert=\lVert f_{0}-g_{0}\rVert. $$ Two methods are suggested:
- Use Theorem 1.12.
- Use the weak$^{\star}$ topology $\sigma(E^{\star},E)$.
I'm trying to solve this problem using method number 2. I've already shown that $M^\bot$ is closed in the weak* topology, and I'm aware of the fact that $B_{E^*}$ is compact in that same topology.
Any ideas on how to proceed on this one?
Obs:
$$ M^\bot = \left\{ g\in E^*\ ;\ g(v)=0\ ,\ v\in M \right\} $$
There's a net $(g_i)_i$ in $M^\perp$ such that
$$ \lim_{i} \|f_0-g_i\| = \inf_{g \in M^\perp} \|f_0 - g\|. $$ Because $\|g_i\| \leq \|f_0\| + \|f_0-g_i\|$, $(g_i)_i$ is a bounded net. Then by the Banach-Alaoglu theorem, there exists a convergent subnet $(g_{f(j)})_j$ in the weak* topology. Let $g_0$ be its weak* limit. You already proved that $M^\perp$ is weak* closed, so $g_0 \in M^\perp$. Then we immediately have $$ \|f_0-g_0\| \geq \inf_{g \in M^\perp} \|f_0-g\|. $$ So it's sufficient to prove that $\lvert f_0(x) -g_0(x) \rvert \leq \inf_{g \in M^\perp} \|f_0-g\| \|x\|$ for all $x \in E$. This is easy to show, since for any $x \in E$ we have that $$ \lvert f_0(x) -g_0(x) \rvert = \lim_{j} \lvert f_0(x) -g_{f(j)}(x)\rvert \leq \lim_{j} \|f_0 -g_{f(j)}\| \|x\|. $$