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The following problem is from Functional analysis written by Brezis.

$\mathbf{3.9}$ Let $E$ be a Banach space; let $M\subset E$ be a linear subspace, and let $f_0\in E^*$. Prove that there exists some $g_0\in M^{\perp}$ such that \begin{align*} \inf_{g\in M^\perp}\|f_0-g\|=\|f_0-g_0\|. \end{align*}

Two methods are suggested :

  1. Use Theorem 1.12

  2. Use the weak$^*$ topology $\sigma(E^*,E).$

My attempt is following.

Observe that \begin{align*} M^\perp=\{ f\in E^* : \left<f,x \right>=0\hspace{5mm}\forall x\in M \}=\bigcap_{x\in M} \varphi_x^{-1}(\{0\}) \end{align*} where $\varphi_x \in E^{**}$ is evaluation map defined by $\varphi_x(f)=f(x)=\left<f,x \right>$. Thus, $M^\perp$ is the intersection of closed sets so is closed. Thus, there exists $g_0\in M^\perp$ such that \begin{align*} \inf_{g\in M^\perp} \|f_0-g\|=\|f_0-g_0\| \end{align*}

Now, as you can see I have not used any fact about weak* topology. I only used the fact that $M^\perp$ is closed and so the minimizer should be in there. I feel like I am clearly missing something based on what Brezis suggested. Could you let me know where I am missing? Thanks in advance!

Apparently, it has been asked several time in Brezis Exercise 3.9 and in Exercise about weak topology, and in A Consequence of Banach Alaoglu Bourbaki theorem and so on. But none of them seem clear about my concern.

Byeong-Ho Bahn
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    Why does the closedness of $M^\perp$ automatically imply that the minimizer exists? – mechanodroid Apr 28 '20 at 14:29
  • @mechanodroid By the definition of infimum, there exists a sequence ${g_n}\subset M^\perp$ such that $|f_0-g_n | \rightarrow \inf|f_0-g||$. Since $M^\perp$ is closed, any limit point of ${g_n}$ should be in $M^\perp$. – Byeong-Ho Bahn Apr 28 '20 at 14:32
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    I think you need Banach-Alaoglu to conclude that a limit point of $(g_n)_n$ exists, like in this answer. – mechanodroid Apr 28 '20 at 14:35
  • @mechanodroid Thank you for the link. But can I ask, in this proof, how Banach Alaoglu says there is a weak* convergent subsequence? All the theorem says is that the closed ball is compact. And compact implies any sequence has convergent subsequence if and only if the space is metrizable. – Byeong-Ho Bahn Apr 28 '20 at 14:41
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    I believe you also need Eberlein-Smulian theorem. – mechanodroid Apr 28 '20 at 15:25
  • @mechanodroid Thank you! But still I don't know how it works. For example, Eberlein-Smulian theorem is for weak convergence not weak* convergent we are talking about. Also, if you see the application section of your link, the article applied it with Alaoglu theorem to find the weakly convergent subsequence from the fact that the space is reflexive which is far from our assumption. – Byeong-Ho Bahn Apr 28 '20 at 15:35
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    You are right, it seems false (e.g. unit ball of $(\ell^\infty)^$ is not weak$^$ sequentially compact). However, this answer uses nets so it circumvents the issue. – mechanodroid Apr 28 '20 at 15:49
  • @mechanodroid Thank you so much! You are right! I get it. It is helpful! – Byeong-Ho Bahn Apr 28 '20 at 16:21

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