Is $W_0^{1,p}(\Omega)\cap L^q(\Omega)$ Uniformly Convex?

Question

Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $p<q$ with $p,q\in (1,\infty)$. Is $W_0^{1,p}(\Omega)\cap L^q(\Omega)$ Uniformly Convex with respect to the norm: $\|u\|=\|u\|_{1,p}+\|u\|_q$?

Note: Observe that by an similar argument like this, we have that $W_0^{1,p}(\Omega)\cap L^q(\Omega)$ is Banach with respect to the norm $\|\cdot\|$

Answer 1

Let's equip $W^{1,p}_0\cap L^q$ with the norm $\|\cdot\|_*$ such that $$\|u\|_*^2=\|\nabla u\|_{L^p}^2+\| u\|_{L^q}^2 \tag1$$ Then the map $u\mapsto (\nabla u,u)$ is an isometric embedding of $W^{1,p}_0\cap L^q$ into the direct sum space $L^p\oplus L^q$ with the $2$-norm. By the latter I mean the norm $\sqrt{\|x\|_X^2+\|y\|_Y^2}$ on the direct sum $X\oplus Y$ of two Banach spaces.

Theorem 5.2.25 in An introduction to Banach space theory by Megginson asserts that $X\oplus Y$ is uniformly convex (with the above norm) if and only if $X$ and $Y$ are both uniformly convex. This answers the question.

If one takes for granted the equivalence of

• uniform convexity of $X$
• uniform smoothness of $X^*$
• uniform continuity of the duality map $X^*\to X$ on bounded subsets.

then the statement is transparent: $(X\oplus Y)^*$ is isometric to $X^*\oplus Y^*$ (only because we use the $2$-norm), and the duality map on the product is the coordinate-wise combination of duality maps.