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Let $W^{m,p}(\Omega)$ be a Sobolev space, where $\Omega\subset\mathbb{R}^{n}$ is a open set.

So the question is: For what $p$ and $m$ the space $W^{m,p}(\Omega)$ is Uniformly Convex.

Some reference or the answer would be appreciated.

ViktorStein
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Tomás
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  • I just found a reference: Theorem 4.2 in Marcellán, Quintana, and Rodríguez, Bull. Math. Sci. (2018) 8:233–256. The proof is essentially the same as given by @martini below. – Mage Jan 28 '20 at 00:09

1 Answers1

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Note, that the map ($N := |\{\alpha \in \mathbb N_{\ge 0}^n \mid |\alpha|\le m\}|$) \begin{align*} T\colon W^{m,p}(\Omega) &\to L^p(\Omega, \mathbb R^{N})\\ x &\mapsto (D^\alpha x)_{|\alpha| \le m} \end{align*} is a closed and isometric embedding. As $L^p(\Omega, \mathbb R^N)$ is uniformly convex for $1 < p < \infty$, so are its closed subspaces, and hence as $W^{m,p}(\Omega)$ is isometric to its image under $T$, so is $W^{m,p}(\Omega)$ for these $p$.

On the other side, uniformly convex spaces are reflexive and $W^{m,1}(\Omega)$ and $W^{m,\infty}(\Omega)$ aren't.

martini
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