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I am attempting problem 3.9 in Brezis' book:

Let $E$ be a Banach space; let $M\subseteq E$ be a linear subspace, and let $f_0\in E^\star$. Prove that there exists some $g_0\in M^\perp$ such that $$ \inf_{g\in M^\perp}\{\Vert f_0-g\Vert \}=\Vert f_0-g_0\Vert $$ I have attempted the following solution:

Let $\varphi : E^\star\to \mathbb{R}:g\mapsto \Vert f_0 - g\Vert$. Note that $\varphi$ is weak$^*$ly l.s.c. In addition, $M^\perp$ is always closed. Let $\varepsilon > 0$ and $$ r_0\colon=\inf_{\overline{B_{E^\star}(f_0, r)}\cap M^\perp\neq \varnothing}\{r>0\} $$ $$ B\colon= \overline{B_{E^\star}(f_0, r_0+\varepsilon)} $$ Then $B\cap M^\perp$ is a closed subspace of $B$ and therefore compact in the weak$^*$ topology. Thus, by lower semicontinuity, $\varphi$ attains its infimum on $B\cap M^\perp$.

I am stuck on trying to prove that $\varphi$ attaining its infimum on $B$ for every $\varepsilon$ implies that the global infimum is attained. Specifically, is it true that the global infimum of $\varphi$ lies in $B$ for every $\varepsilon$? Any hints or suggestions? Thanks!

Ray
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  • Welcome to Mathematics Stack Exchange. I think this should answer your questions. Can you please check if this is indeed the case? – Dean Miller Jul 02 '25 at 19:11
  • @DeanMiller Thank you for the welcome. Can you please elaborate how I can reach the desired conclusion from the linked thread? I am new to this so it's hard for me to see the connection here. – Ray Jul 02 '25 at 19:50
  • No problem. Actually, now I see you are asking something more specific than the other questions. – Dean Miller Jul 02 '25 at 20:08
  • You may also find this and this useful for alternative solutions to the problem later on. – Dean Miller Jul 02 '25 at 20:26

1 Answers1

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Here is something that should help. If $g\in E^{*}$ and $\|g\| > 2\|f_{0}\|$, then by the triangle inequality we have $\|f_{0} - g\| \geq \|g\| - \|f_{0}\| > \|f_{0}\|$. From this it can be deduced that \begin{equation} \inf\{\|f_{0} - g\| : g\in M^{\perp}\} = \inf\{\|f_{0} - g\| : g\in M^{\perp} \text{ and } \|g\| \leq 2\|f_{0}\| \} . \end{equation} Now the set \begin{equation} \{\|f_{0} - g\| : g\in M^{\perp} \text{ and } \|g\| \leq 2\|f_{0}\| \} \end{equation} is the image of the lower semicontinuous function $\varphi$ under the set $\{g\in M^{\perp} : \|g\| \leq 2\|f_{0}\|\}$. See if you can use these observations and fill in the details to deduce that $\varphi$ attains its infimum on $M^{\perp}$.

See here for some complete solutions to this problem.

Dean Miller
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