Let $(E,\|\cdot\|)$ be a normed space und let $E'$ be the topological dual space of $E$ provided with the norm $\|f\|:=\sup\{|f(x)|\colon\|x\|\leq 1\}$

Show, that for $F\subseteq E$ linear subspace provided with the restricted norm and is $g\in F'$, then exists $f\in E'$ such that $f_{|F}=g$ and $\|f\|=\|g\|$.

I want to solve this task. I think I can use the theorem of Hahn-Banach.

Hahn-Banach states:

Let $E$ be a $\mathbb{K}$-vectorspace, $p: E\to [0,\infty)$ a seminorm, $F\subseteq E$ a linear subspace and $g: F\to\mathbb{K}$ linear (hence $g\in F'$), with $|g|\leq p$ on $F$. Then exists a linear function $\hat{g}:E\to\mathbb{K}$ with $\hat{g}_{|F}=g$ and $|\hat{g}(x)|\leq p(x)$ for all $x\in E$.

To use this theorem it is missing, that $|g|\leq p$. So I have to show, that $\|g(x)\|_F\leq \color{red}{\|x\|}$ (where $\|\cdot\|_F$ notes the restriced norm on $F$.) Is that correct?

If I succeed we get the desired $f\in E'$ with $f_{|F}=g$ and $\|f\|\leq \color{red}{\|g\|}$ for all $x\in E$.

Then I need to show, that $\|f\|=\|g\|$, hence it would be missing, that $\|f\|\geq\|g\|$.

Am I right? I marked $\color{red}{red}$ where I think I made a mistake.

I appreciate any kind of help. Thanks in advance.

  • $\begingroup$ Take $p(x)=||g|| ||x||$. The inequality $||f|| \geq ||g||$ follows immediately from the fact that f extends g. $\endgroup$ – Kabo Murphy Nov 22 '17 at 8:02
  • $\begingroup$ Why can I take $p(x)=\|g\|\cdot\|x\|$? Isn't $p$ "fixed"? Can you tell me, if I made somewhere above a mistake, on what to show? $\endgroup$ – Cornman Nov 22 '17 at 8:12
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    $\begingroup$ Hahn Banach Theorem only requires p to be a semi-norm. When you apply it you can choose the p that works for you. $\endgroup$ – Kabo Murphy Nov 24 '17 at 9:13

Notice that your approach works if $\|g\|_F = 1$.

To make it work in general, define a seminorm $p : E \to [0, +\infty)$ as $p(x) = \|x\|\cdot\|g\|_F$, for $x \in E$.

Notice that $p$ is furthermore a norm on $E$ if $g \ne 0$.

Hahn-Banach theorem for the vector space $E$ equipped with the seminorm $p$ gives you the existence of a linear map $\hat{g} : E \to \mathbb{K}$ such that $\hat{g}$ extends $g$, and that $|\hat{g}| \le p$. This condition means:

$$|\hat{g}(x)| \le p(x) \le \|x\| \cdot \|g\|_F$$

So, $\hat{g}$ is bounded (hence $\hat{g} \in E'$) and $\|\hat{g}\|\le \|g\|_F$.

As noted in the comments, the reverse inequality follows from the fact that the supremum of a subset is less than or equal to the supremum of the entire set and that $\hat{g}(x) = g(x)$ for $x \in F$.

$$\|g\|_F = \sup\{|g(x)| : x \in F, \|x\| = 1\} \le \sup\{|\hat{g}(x)| : x \in E, \|x\| = 1\} = \|\hat{g}\|$$

We conclude $\|\hat{g}\| = \|g\|_F$, so $\hat{g}$ is the required linear functional.

  • $\begingroup$ Why can we simply define a seminorm $p$ and show that it works with this? $\endgroup$ – Cornman Nov 23 '17 at 1:28
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    $\begingroup$ @Cornman The Hahn-Banach theorem for seminorms you cited works for any seminorm. If you are asking why did we define $p$ as $p = \|g\|_F\|\cdot\|$, its because we have to be able to bound $|g(x)| \le p(x)$. Since $|g(x)| \le \|g\|_F\|x\|$, we see that it is sufficient to take $ \|g\|_F\|x\| \le p(x)$. On the other hand, we also wish that $|\hat{g}(x)| \le p(x)$ implies $|\hat{g}(x)| \le \|g\|_F\|x\|$ (so that $\|\hat{g}\| \le \|g\|_F$), so it is sufficient to take $p(x) \le \|g\|_F\|x\|$. We conclude that the argument works with $p(x) = \|g\|_F\|x\|$. $\endgroup$ – mechanodroid Nov 23 '17 at 12:31

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