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Let $\Omega$ be a bounded open set of $\mathbb{R}^N$, define $\|u\|_1 = \Big(\int |\nabla u|^p \Big)^\frac{1}{p}$ as a norm on $W_0^{1,p}(\Omega)$ for $2\leq p<\infty$, where $$\int |\nabla u|^p = \int \big(\sum_i u_{x_i}^2\big)^{\frac{p}{2}},$$ I know this is equivalent with another standard norm $$\|u\|_2 = \Big(\int \sum_i |u_{x_i}|^p \Big)^\frac{1}{p},$$ which makes $\big(W_0^{1,p}(\Omega), \|\cdot\|_1\big)$ a Banach space.

However is there any way to show that $\big(W_0^{1,p}(\Omega), \|\cdot\|_1\big)$ is also uniformly convex?

Since even if two norms are equivalent, uniform convexity with one norm does not imply the other. For example $|\cdot|_{\max}$ and the euclidean norm in $\mathbb{R}^2$.

I am working on this right now, and I think Clarkson's inequality will be helpful, for any $f,g \in L^p$, $p\geq 2$, we have $$\left\|\frac{f+g}{2}\right\|^p + \left\|\frac{f-g}{2}\right\|^p \leq \frac{1}{2}\big(\|u\|^p + \|v\|^p\big)$$

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  • $\begingroup$ I think you need to show that Clarkson's inequality is also true for vector-valued $f,g$. This means that vector-valued $L^p$ is uniformly convex. And since the map $u\mapsto \nabla u$ is an isometric embedding, the Sobolev space is uniformly convex too. $\endgroup$ – user147263 Dec 4 '14 at 6:50
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I think it is. Let $1<p<\infty$ be given and notice that the mapping $T$ from $W^{1,p}(\Omega)\to L^p(\Omega, R^{N+1})$ via $$ T[u]\to (u,\nabla u) $$ is isomorphic and closed. Together with the fact that $L^p(\Omega, R^{M})$ is uniformly convex for any $M\geq 1$, here we are interested in the case $M=N+1$, hence we know that $W^{1,p}$ is uniformly convex.

Moreover, via the same argument, we have $W^{k,p}$ is uniformly convex as well.

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