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Let $E$ be a Banach space and $M\subset E$ a linear subspace, let $f_{0} \in E^*$.

Prove that: $\exists \, g_{0}\in M^\perp$ such that : $\inf_{g\in M^\perp} \|f_0−g\|=\|f_0 - g_0\|$

where:

$M^{\perp}:= \{f\in E^{*} \, : \, f(x)=0 \ ∀x\in M\}$.

I considered $a:=\inf_{g\in M^\perp}\|f_0−g\|$ and I tried the definition of the infinimum and arrived to a sequence $g_{n}$ in the closed ball $B(f_0,a+1)$ which is weak* compact by Banach-Alaoglu-Bourbaki theorem. I couldn't get any further.

Brian Moehring
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Jack Tell
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    As written $g_0 \in M \subseteq E$ and $f_0 \in E^*$. What does $|f_0 - g_0|$ even mean? These elements are in different spaces. Do you want something like $g_0 \in M^\perp$? – Rhys Steele Feb 12 '20 at 16:30
  • Please edit & repair the question. It is meaningless presently. – copper.hat Feb 12 '20 at 16:31
  • I appologize, I just edited the question. – Jack Tell Feb 12 '20 at 16:35
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    Well now $|f_0 - g| \geq 0$ for all $g$ and $|f_0 - f_0| = 0$ so clearly $g_0 = f_0$ works. I think you still need to reformulate something. – Rhys Steele Feb 12 '20 at 16:37
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    You have a sequence $g_n\in B(f_0,a+1) \cap M^\perp$ with $|f_0-g_n|\to a$. By compactness of the ball assume that this sequence converges to some $g_0$. Check that $M^\perp$ is weak* closed to see that $g_0\in M^\perp$. Now let $x_m\in E$ have norm $1$ and $\langle f_0-g_0, x_m\rangle \to |f_0-g_0|$, so for any $\epsilon$ you have an $m$: $$|f_0-g_0|-\epsilon ≤ |\langle f_0-g_0, x_m\rangle| = \lim_n |\langle f_0-g_n , x_m\rangle| ≤ \liminf_n|f_0-g_n| = a$$ giving that $|f_0-g_0|≤a$ but since $g_0\in M^\perp$ you also have $|f_0-g_0|≥a$. – s.harp Feb 12 '20 at 16:39
  • @s.harp Couldn't you post exactly that comment as an answer? – Rhys Steele Feb 12 '20 at 16:45

1 Answers1

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Let $g_n\in M^\perp$ with $\|f_0-g_n\|\to\inf_{g\in M^\perp}\|f_0-g\|$, by definition this sequence is bounded so by Banach-Alaoglu it will have a weak* convergent subsequence, by relabeling we then assume that $g_n$ is weak* convergent and let $g_0$ be the weak* limit of $g_n$.

$M^\perp$ is weak* closed, as follows directly from applying definitions. As such $g_0\in M^\perp$. Remembering that $\|f_0-g_0\|:=\sup_{x\in E, \|x\|≤1}|(f_0-g_0)(x)|$ you may find a sequence $x_m\in E$ of norm $1$ vectors so that $(f_0-g_0)(x_m)\to \|f_0- g_0\|$, in particular for any $\epsilon>0$ you have an $m$ with: $$\|f_0-g_0\|-\epsilon ≤ |(f_0-g_0)(x_m)| = |\lim_n (f_0-g_n)(x_m)| ≤\liminf_n \|f_0- g_n\|\,\|x_m\| = \inf_{g\in M^\perp}\|f_0-g\|$$ giving $\|f_0-g_0\|≤\inf_{g\in M^\perp}\|f_0-g\|$. But since $g_0\in M^\perp$ you also have $\|f_0-g_0\|≥\inf_{g\in M^\perp}\|f_0-g\|$, so $$\|f_0-g_0\|=\inf_{g\in M^\perp}\|f_0-g\|.$$

s.harp
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