# Show reflexivity of Sobolevspace $W^{1,4}(0,1)$

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

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.
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