I'm trying to prove an identity for Lemetski operators and I'm having a problem in the case $n = 2$.
For a bounded $\Omega \subseteq \mathbb{R}^2$ I have two functions $u \in L^1(\Omega)$ and $v \in H_0^1(\Omega)$ and I want to prove that $uv \in L^1(\Omega)$. By the Sobolev imbedding theorem I know that $v \in L^p(\Omega), \forall p \in [1, \infty)$ but since the dimension is 2 the $L^{\infty}$ inclusion is false.
So, I have an unbounded function that is integrable when raised to any power, and a regular integrable function, do you have an idea of how I can prove that their product is integrable? Or can you help me find a counterexample?
Thank you very much for your time!
Kavi Rama Murthy was right in his deleted answer. The statement is false, in the sense that there are $u\in L^1(\Omega)$ and $v\in H^1_0(\Omega)$ such that $uv\notin L^1(\Omega)$.
To prove this, let us consider a function $v\in H^1_0(\Omega)$ such that $v\notin L^\infty(\Omega)$. (For example, $v(x)=\log\log(1+\tfrac1{|x|})$, if $\Omega$ is the unit disk in $\mathbb R^2$). We claim that there must be $u\in L^1(\Omega)$ such that $uv\notin L^1(\Omega)$. Indeed, if this were not the case, then a standard argument based on the uniform boundedness principle would imply that the linear functional $$ T_v(u)=\int_\Omega uv\, dx $$ would be continuous on $L^1(\Omega)$, that is, $T_v\in (L^1(\Omega))^\star$. But we know that this is only possible if $v\in L^\infty$; to express this, we usually say that “the dual of $L^1(\Omega)$ is $L^\infty(\Omega)$”.