# Brezis Exercise 3.9

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:

1. Use Theorem 1.12.
2. 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:

$$M^\bot = \left\{ g\in E^*\ ;\ g(v)=0\ ,\ v\in M \right\}$$

First we show $$M^\perp$$ is closed. Let $$f\in E^*\setminus M^\perp$$. Choose any $$x\in E$$ such that $$\langle f,x\rangle\ne 0$$. Since $$x(\cdot)=\langle\cdot,x\rangle$$ is continuous on $$E^*$$ in weak* topology, $$x^{-1}(I)$$ is weak* open for any open interval $$I\subset \mathbb{R}$$. In particular, choosing $$I=(\langle f,x\rangle - \frac{1}{2}|\langle f,x\rangle|,\langle f,x\rangle + \frac{1}{2}|\langle f,x\rangle|)$$, $$x^{-1}(I)$$ is a weak* open neighborhood of $$f$$ contained in $$E^*\setminus M^\perp$$.
Now let $$g_n$$ be a sequence in $$M^\perp$$ such that $$\lim_{n\to \infty}\|f_0-g_n\|=\inf_{g\in M^\perp}\|f_0-g\|.$$ It's easy to show $$g_n$$ is bounded, i.e. there exists $$r>0$$ such that $$g_n\in rB_{E^*}=\{f\in E^*:\|f\|\le r\}$$ for all $$n$$. Let $$E_k=\{g_n:n\ge k\}$$ for $$k\in\mathbb{N}$$, and let $$\overline{E_k}$$ be the weak* closure of $$E_k$$. Since $$rB_{E^*}\cap M^\perp$$ is a weak* compact set containing all $$E_k$$, $$\overline{E_1}\supset\overline{E_2}\supset\cdots$$ is a decreasing sequence of nonempty weak* compact sets. Hence there exists $$g_0\in\cap_k \overline{E_k}$$. Obviously $$g_0\in M^\perp$$, and hence $$\|f_0-g_0\|\ge \inf_{g\in M^\perp}\|f_0-g\|.$$ It remains to show the reverse inequality. For fixed $$y\in B_E$$ (the closed unit ball in $$E$$), we have \begin{align}|\langle f_0-g_0,y\rangle|&\le |\langle f_0-g_n,y\rangle|+|\langle g_n-g_0,y\rangle|\\ & \le \|f_0-g_n\|+|\langle g_n-g_0,y\rangle|.\tag{1}\end{align} Given $$\varepsilon>0$$, let $$N\in\mathbb{N}$$ be such that $$\|f_0-g_n\|<\inf_{g\in M^\perp}\|f_0-g\|+\varepsilon\quad\forall\,n\ge N.$$ Since $$g_0\in \overline{E_N}$$, every weak* neighborhood of $$g_0$$ has a nonempty intersection with $$E_N$$. In particular, there is $$N_1\ge N$$ such that $$g_{N_1}\in y^{-1}(\langle g_0,y\rangle-\varepsilon,\langle g_0,y\rangle+\varepsilon)=\{g\in E^*:|\langle g-g_0,y\rangle|<\varepsilon\}.$$ Thus, taking $$n=N_1$$ in $$(1)$$ we obtain $$|\langle f_0-g_0,y\rangle|< \inf_{g\in M^\perp}\|f_0-g\| + 2\varepsilon.$$ Since $$\varepsilon>0$$ and $$y\in B_E$$ are arbitrary, we get $$\|f_0-g_0\|\le \inf_{g\in M^\perp}\|f_0-g\|.$$
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 $$\lvert f_0(x) -g_0(x) \rvert = \lim_{j} \lvert f_0(x) -g_{f(j)}(x)\rvert \leq \lim_{j} \|f_0 -g_{f(j)}\| \|x\|.$$