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Let be $T \in \mathbb{L}(E,F)$ (linear and bounded), where $E, F$ are Banach spaces. And let be $E^*$, the dual space of $E$ .

If we define: $Ker(T)^\perp=\{\gamma \in E^* : \gamma(x)=0, \forall x \in ker(T)\}$. Prove

$d(x, Ker(T))=max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}$.

I can prove $d=d(x, Ker(T))\geq max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}$. Let be $\epsilon>0$, then there is $w \in ker(T)$ such as $||x-w||<d+\epsilon$. Let be $\phi \in Ker(T)^\perp, ||\phi||=1$, then $|\phi(x)|=|\phi(x-w)|\leq||\phi||||x-w||<d+\epsilon$ then $max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}\leq d +\epsilon$, for all $\epsilon>0$, then we have $d\geq max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}.$

I can't prove $d\leq max\{|\phi(x)|: \phi \in Ker(T)^\perp, ||\phi||=1\}.$ Thanks a lot for your help.

Maxi
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  • Do you remember the Hahn-Banach extension theorem? – Daniel Fischer Jul 14 '14 at 15:43
  • Yes, I can use that because of H-B theorem we have $||x||=max{|\phi(x)|, \phi\in E^*, ||\phi||=1}$ but I don't know how to continue. – Maxi Jul 14 '14 at 15:49
  • Fix $x_0\in E\setminus \ker T$. You are looking for a $\phi\in E^\ast$ with $\lVert\phi\rVert = 1$, $\ker T \subset \ker\phi$, and $\phi(x_0) = d(x_0,\ker T)$. Doesn't that remind you of something? – Daniel Fischer Jul 14 '14 at 15:52
  • If I Fix $x_0 \in E\setminus KerT$ then by H-B theorem there is a $\phi \in E^*$ such as $||\phi||=1$ and $\phi(x_0)=||x_0||\geq d$, this is it? – Maxi Jul 14 '14 at 15:58
  • No, in general, if $\lVert\phi\rVert = 1$ and $\phi(x_0) = \lVert x_0\rVert$, then $\phi \notin (\ker T)^\perp$. You need something a little different. – Daniel Fischer Jul 14 '14 at 16:03
  • Ok, now I think I understand. If $x_0 \in kerT$ then is trivial. If I fix $x_0 \in E \setminus kerT$ then $d(x_0,kerT)>0$ and using H-B theorem I can prove, there is a $\phi \in E^*$ such as $||\phi||=1$, $\phi(x_0)=d(x_0, KerT)$ and $\phi(y)=0, \forall y \in kerT$ (so $\phi \in kerT^\perp$). Then we have $d(x_0,kerT)≤|\phi(x_0)|≤max{|\phi(x_0)|, \phi \in kerT^\perp, ||\phi||=1}.$ Is it correct? – Maxi Jul 14 '14 at 16:23
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    Right. And since you already have the inequality in the other direction, equality. – Daniel Fischer Jul 14 '14 at 16:25
  • possible duplicate of Distance between point and linear Space – Norbert Jul 15 '14 at 06:10

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