Brezis Exercise 3.9

Question

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:

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

Answer 1

Using nets is very good, but Brezis does not mention nets in the book, and possibly most students have not learned about nets before. We can prove by using the weak* topology directly.

First we show $M^\perp$ is closed. Let $f\in E^*\setminus M^\perp$. Choose any $x\in M$ 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 $$\|f_0-g_0\|\le \inf_{g\in M^\perp}\|f_0-g\|.$$

Answer 2

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 $$ \lvert f_0(x) -g_0(x) \rvert = \lim_{j} \lvert f_0(x) -g_{f(j)}(x)\rvert \leq \lim_{j} \|f_0 -g_{f(j)}\| \|x\|. $$