# Finite Intersection of uniformly convex Banach spaces

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.

• Why is $X\cap Y$ a Banach space? You need some kind of compatibility between $X$ and $Y$ else it shouldn't work. Commented Sep 11, 2020 at 20:10
• Ok. I understand now. Yes, there should be compatibility between $X$ and $Y$. Commented Sep 12, 2020 at 9:00
• 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 $L^p(\Omega)$. Commented Sep 12, 2020 at 9:02
• 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). Commented Sep 13, 2020 at 5:56
• Thanks. But here $X$ and $Y$ are not finite dimensional space. @dmw64 Commented Sep 13, 2020 at 8:38

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.
• 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 $W_{0}^{s,p}(\Omega)$. Commented Sep 12, 2020 at 20:31