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Let $\Omega$ be a smooth bounded domain in $\mathbb{R}^N$, $X=W_{0}^{1,p}(\Omega)$ and $Y=W_{0}^{s,p}(\Omega)$ be the classical and fractional Sobolev spaces. Both are uniformly convex Banach spaces. Let $Z=X\cap Y$ be the space under the norm $$ \|z\|_Z=\|z\|_X+\|z\|_Y. $$ Then it can be easily seen that $Z$ is a Banach space.

But I am unable to predict whether $Z$ is uniformly convex under the above norm, which would also give the reflexivity since uniform convex spaces are reflexive.

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  • $\begingroup$ Why is $X\cap Y$ a Banach space? You need some kind of compatibility between $X$ and $Y$ else it shouldn't work. $\endgroup$
    – s.harp
    Commented Sep 11, 2020 at 20:10
  • $\begingroup$ Ok. I understand now. Yes, there should be compatibility between $X$ and $Y$. $\endgroup$
    – Mathlover
    Commented Sep 12, 2020 at 9:00
  • $\begingroup$ Indeed, I have edited the question a little with two kinds of Sobolev spaces. Now $Z$ is a Banach space, since both $X$ and $Y$ are compactly embedded in $L^p(\Omega)$. $\endgroup$
    – Mathlover
    Commented Sep 12, 2020 at 9:02
  • $\begingroup$ Is you main target the uniform convexity? Or maybe only the reflexivity? For reflexivity you could notice the all norm on a finite dimensional space are equivalent. So on $\mathbb{R}^2$ you could use $||.||_p$ instead of $||.||_1$. So switching to $|| (||z||_X,||z||_Y) ||_p$ would solve the question of uniform convexity and reflexivity (but using an equivalent norm). $\endgroup$
    – dmw64
    Commented Sep 13, 2020 at 5:56
  • $\begingroup$ Thanks. But here $X$ and $Y$ are not finite dimensional space. @dmw64 $\endgroup$
    – Mathlover
    Commented Sep 13, 2020 at 8:38

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In fact, it is enough to notice that, for any $s\geq 0$ we have $$ X\cap Y = W^{k,p}_0(\Omega), \quad \hbox{where}\quad k=\max\{1,s\}. $$ Then, since $W^{k,p}_0$ is uniformly convex (as you said on your post), we deduce the uniform convexity of the intersection.

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  • $\begingroup$ Thank you very much. But how this intersection become $W_{0}^{k,p}(\Omega)$? Indeed, I have assumed $0<s<1$ so the intersection should be $W_{0}^{1,p}(\Omega)$. Can you kindly help how does it follow? Indeed, you mean to say that any $u\in W_{0}^{1,p}(\Omega)$ belong to $W_{0}^{s,p}(\Omega)$. $\endgroup$
    – Mathlover
    Commented Sep 12, 2020 at 20:31

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