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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!

Mohsen Shahriari
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Ignacio Riego Hunt
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  • I thought a little bit about this. If this is false, a counterexample should come from the function $$v(x)=\log\log\left(1+\frac1{|x|}\right), $$ which is in $H^1_0(D^2)$ (here $D^2$ is the unit disk), but $v\notin L^\infty(D^2)$. – Giuseppe Negro Oct 05 '20 at 10:14
  • For example, is it true that $$u(x)=\frac{1}{\lvert x \rvert^2 \log \log( 1+\tfrac1{\lvert x \rvert} )}$$ is in $L^1(D^2)$? If this is the case, that’s a counterexample, because $uv\notin L^1(D^2)$. – Giuseppe Negro Oct 05 '20 at 11:00

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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 $L^\infty(\Omega)$”.

Giuseppe Negro
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  • Thanks! I was having trouble with the idea of a linear functional T on $L^1$ that is not continuous but is defined in all points, but now I see that that cannot happen, since I could use the uniform boundedness principle to construct a norm and it would therefore be continous. – Ignacio Riego Hunt Oct 07 '20 at 14:16
  • Glad it helped. We discussed here something a bit similar, and arrived at the conclusion that “every linear or multilinear operator that is everywhere defined must be bounded”. This is not a precise statement but a guiding principle, which applies also to the case at hand. By the way, if you found this answer useful, you could upvote and/or accept it, thanks. – Giuseppe Negro Oct 07 '20 at 16:06