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

Question

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

Answer 1

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.