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I'm doing Ex 3.9 in Brezis's book of Functional Analysis.

Let $E$ be a Banach space; let $M \subset E$ be a linear subspace, and let $f \in E'$. Prove that there exists some $g \in M^{\perp}$ such that $$ \left\|g-f\right\| = \alpha :=\inf_{h \in M^{\perp}}\left\|h-f\right\| . $$

In my below proof, I don't use the the fact that $M$ is a linear subspace. It seems to me this hold for any subset $M$ of $E$. Could you have a check on my attempt?

I posted my proof separately so that I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

Analyst
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  • You can apply the "special case" to the subspace generated by $M$ which has the same annihilator. – Jochen Feb 22 '22 at 20:57
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    This is true for every weak-star closed subset of $E'$; see this for instance. Your problem is a special case of this, since $M^\bot $ is a weak-star closed (subspace) in $E'$. In fact, if $M' $ is a weak-star closed subspace of $E'$ then $M'=M^\bot$ for some subspace $M$ of $E$. – Evangelopoulos Foivos Feb 23 '22 at 09:37

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There is a sequence $(h_n)$ in $M^{\perp}$ such that $\|h_n -f\| \searrow \alpha$. It follows that $(h_n)$ is bounded by some $\beta \in \mathbb R_{>0}$. Let $B := (h_n)$. Then $B$ is a subset of the closed ball $\beta B_{E'}$ which is compact in weak$^\star$ topology $\sigma(E',E)$ by Banach–Alaoglu theorem. Then the weak$^\star$ closure $\overline{B}^\star$ is also compact in $\sigma(E',E)$. Also, the map $h \mapsto \|h-f\|$ is lower semi-continuous in $\sigma(E',E)$. Hence there is $g \in \overline{B}^\star$ such that $$ \|g-f\| = \inf_{h\in \overline{B}^\star} \|h-f\|. $$

It remains to show that $\overline{B}^\star \subseteq M^{\perp}$. It suffices to show that $M^{\perp}$ is closed in $\sigma(E',E)$. Let $(h_d)$ be a net in $M^{\perp}$ such that $h_d \overset{\ast}{\rightharpoonup} h\in E'$. This implies $0=\langle h_d, x \rangle \to \langle h, x \rangle$ for all $x\in M$. So $\langle h, x \rangle=0$ for all $x\in M$ and thus $h\in M^{\perp}$. This completes the proof.

Analyst
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  • I want to point out a small detail: $\overline{B}^\star$ is indeed weak-star compact but not weak-star sequentially compact (unless $E$ is separable). Meaning, there may exist sequences that are bounded and do not admit weak-star convergent subsequences. This problem is solved by replacing sequences by nets. So, in order to show that $||g-f||= \inf_{h\in \overline{B}^\star} ||h-f||$, you have to extract a weak-star convergent subnet (not subsequence) of $(h_n)$ and use the weak-star lower semi-continuity of the dual norm $||\cdot||_*$. – Evangelopoulos Foivos Feb 23 '22 at 13:36
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    @EvangelopoulosF. I use the property that l.s.c. function attains the minimum on a compact set. This holds for arbitrary real-valued function on a compact topological space. You can find a proof here or here. – Analyst Feb 23 '22 at 14:45