Conclude $u\frac{\partial v}{\partial x_i}\in L^2(\Lambda)$ for all $u,v\in H_0^1(\Lambda)$ from the Sobolev embedding theorem

Question

Let

• $d\in\left\{2,3\right\}$
• $\Lambda\subseteq\mathbb R^d$ be open
• $u,v\in H_0^1(\Lambda)$
• $i\in\left\{1,\ldots,d\right\}$

I want that $$u\frac{\partial v}{\partial x_i}\in L^2(\Lambda)\tag 1\;.$$ Which further assumption on $\Lambda$ do I need?

By Hölder's inequality, $$\left\|u\frac{\partial v}{\partial x_i}\right\|_{L^2(\Lambda)}\le\left\|\left|u\right|^2\right\|_{L^p(\Lambda)}\left\|\left|\frac{\partial v}{\partial x_i}\right|^2\right\|_{L^q(\Lambda)}\tag 2$$ for all $p,q\in[1,\infty]$ with $1/p+1/q=1/2$ and hence (for $p=q=2$) $$\left\|u\frac{\partial v}{\partial x_i}\right\|_{L^2(\Lambda)}\le\left\|u\right\|_{L^4(\Lambda)}\left\|\frac{\partial v}{\partial x_i}\right\|_{L^4(\Lambda)}\tag 3\;.$$

So, it would be sufficient, if each element of $H_0^1(\Lambda)$ would admit a representative in $W^{1,\:4}(\Lambda)$.

The following is according to Theorem 5.6.3 and Theorem 5.6.6 of Partial Differential Equations by Lawrence C. Evans: If $\Lambda$ is bounded and $d>2$, then $$\left\|w\right\|_{L^6(\Lambda)}\le C_1\left\|\nabla w\right\|_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }w\in H_0^1(\Lambda)\tag 4$$ for some $C_1\ge 0$. If $\Lambda$ is bounded with $\partial\Lambda\in C^1$ and $d>2$, then $$\left\|w\right\|_{L^6(\Lambda)}\le C_2\left\|u\right\|_{H^1(\Lambda)}\;\;\;\text{for all }w\in H^1(\Lambda)\tag 5$$ for some $C_2\ge 0$.

However, the corresponding statements for $d=2$ are useless and even in the case $d=3$ I cannot conclude $(1)$ from $(4)$ or $(5)$. So, is there any chance for $(1)$?

Answer 1

Which further assumption on Λ do I need?

You need $\Lambda=\emptyset$.

The statement

For all $u,v\in H^1_0(\Lambda)$ it holds $u\frac{\partial v}{\partial x_i}\in L^2(\Lambda)$

will never be true in space dimensions larger than one. The weak derivative $\frac{\partial v}{\partial x_i}$ is only $L^2(\Lambda)$ and not better in general. In order that the claim is true, one would need $u\in L^\infty(\Lambda)$, which does not hold for $d\ge2$.

In order to construct an explicit counterexample, take a function $u\in L^2(\Lambda)\setminus L^4(\Lambda)$ (such functions exist), and define $v(x):=\int_{-\infty}^{x_i} u(x_1,\dots,x_{i-1}, s, x_{i+1},\dots,x_d)ds$.