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Let $E$ a normed linear space and $H$ the closed hyperplane $H=\ker f$, where $f\in L(E,\mathbb{R})$, $f\not\equiv 0$. Show that if $a\in E$ then $$d(a,H)=\frac{|f(a)|}{||f||}$$ And the problem have a second part: Let $E$ the space of all $x_n\to 0$. With $\displaystyle||(x_n)||=\sup_{n\in\mathbb{N}}|x_n|$. Let $H$ the closed hyperplane $H=f^{-1}(\{0\})$, with $$f(x)=\sum_{n=0}^{\infty}\frac{x_n}{2^n}$$ if $x=(x_n)$. Show that if $a\notin H$ then there is no $b\in H$ such that $d(a,H)=||a-b||$.

PD: i'm stuck in the two parts, any help. I try to write the definition $d(a,H)=\inf \{||a-x||:x\in H\}$ but no have more ideas!

Luis Valerin
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  • $L(E,\mathbb{R})$ is the space of functions linear and continuous. – Luis Valerin Jul 02 '14 at 11:45
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    See this for an elementary proof of the first part. – David Mitra Jul 02 '14 at 23:31
  • @DavidMitra thanks +1!! Any idea with the second part? I tried by contradiction but :( – Luis Valerin Jul 02 '14 at 23:39
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    $f$ has norm $1$. Show that this norm is not attained (that is, there is no non-zero $y$ with $f(y)=\Vert y\Vert$). Now if $d(a,H)=\Vert a-b\Vert$, then by your first problem we'd have $\Vert a-b\Vert=|f(a)|=|f(a-b)|$. – David Mitra Jul 03 '14 at 00:08
  • Oops, $f$ has norm $2$, actually. But the argument I wrote previously still works: Show for any $y\ne0$ that $|f(y)|<2\Vert y\Vert$. Use this to derive a contradiction assuming $\Vert a-b\Vert=d(a,H)$. – David Mitra Jul 03 '14 at 00:37
  • @David how to prove that $f$ has norm $2$? – Roiner Segura Cubero Jul 03 '14 at 00:42
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    $|f(y)|<\sum\limits_{i=0}^\infty {\Vert y\Vert \over 2^i}= 2 \Vert y\Vert$ (the inequality is strict since $y_i\rightarrow 0$). So $\Vert f\Vert\le 2$. To see $\Vert f\Vert=2$, consider $f(x_n)$ for $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$. – David Mitra Jul 03 '14 at 01:00

1 Answers1

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Here is a hint for the first part. I'll assume that $a \notin H$. By a corollary of the Hahn-Banach Theorem, there exists $x^* \in L(E,\mathbb{R}) = E^*$ such that:

  1. $H \subset \ker x^*$;
  2. $x^*(a) = d(a,H)$;
  3. $\|x^*\|=1$.

By a well-known result, condition 1. implies that $x^*$ is linearly dependent on $f$, i.e. for some $k \in \mathbb{R}$ we have $x^*=k f$. Now condition 3 implies that $k = \pm 1/\|f\|$, and condition 2 yields $$d(a,H) = x^*(a) = k f(a) = \pm \frac{f(a)}{\|f\|}.$$ It is now easy to conclude.

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