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I would like to show elementary - using the canonical embedding - that the Sobolevspace $W^{1,4}(0,1)$ is reflexive.

Therefore I set $X=W^{1,4}(0,1)$ and now I have to show that the canonical embedding

$$ i\colon X\to X'', i(x)(x')=x'(x) $$

is bijective and isometric.


I think the canonical embedding is always injective and isometric. So I only have to show here, that it is surjective.

Am I right?

How can I show that?

Let $x''$ be in $X''$. Now I have to find a $x\in X$ with $i(x)=x''$, right?

But - how?

Greetings

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I have removed the tag (reflexive). According to the tag-wiki, this tag is for questions about reflexive relations. –  Martin Sleziak Aug 30 '13 at 9:14
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1 Answer

up vote 2 down vote accepted
  1. Use Clarkson's inequality to see that $L^p(\Omega)$ is reflexive for any $p\in (1,\infty)$. So is $(L^p(\Omega))^{N+1}$.
  2. Show that $W^{1,p}(\Omega)$ is a closed subset of $(L^p(\Omega))^{N+1}$. We can use sequential closeness.
  3. Use the non-trivial fact that a closed subspace of a reflexive space is reflexive.
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1. und 3. are okay. I can use that. But i did not come along with 2. How can I show that it is a closed subset? –  math12 Jan 24 '13 at 15:50
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I think it is better to use the space $(L^p)^{N+1}$ instead of $L^p$ –  Tomás Jan 24 '13 at 15:57
    
@Tomás Right, $L^p$ alone looks misleading. –  Davide Giraudo Jan 24 '13 at 16:11
    
Why $(L^p(\Omega))^{N+1}$? And how can i show that it is closed subset? –  math12 Jan 24 '13 at 16:15
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Since uniform convexity is a condition on all 2-dimensional subspaces, it passes to closed subspaces. Hence the Milman-Pettis theorem already used in 1 can be applied again --- although Milman-Pettis (combined with Clarkson's inequality) is probably way harder than the non-trivial fact in 3. –  Martin Jan 24 '13 at 17:43
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