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?


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


2 Answers 2


Using net is very good, but Brezis do not mention net in the book, and possibly most students do not learn it 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 $x\in E$ 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 $$ \lvert f_0(x) -g_0(x) \rvert = \lim_{j} \lvert f_0(x) -g_{f(j)}(x)\rvert \leq \lim_{j} \|f_0 -g_{f(j)}\| \|x\|. $$

  • $\begingroup$ I made an error in assuming that Banach-Alaoglu implies sequential compactness. Fortunately, you can just replace every sequence with a net and the proof can be salvaged. $\endgroup$
    – Demophilus
    Nov 15, 2017 at 13:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .