I am currently trying to figure out the following. If I consider the Sobolev space $W^{1,p}_0$ is it possible to show that the operator given by $$\langle Au,v\rangle=\int u^3 v dx$$ is strongly (weak to strong) continuous for all $p$? We may assume we are on a bounded Lipschitz domain. It is possible using the Sobolev embedding theorem but that will then restrict our choices of $p$ based on the dimension. I was told it is possible using the reverse dominated convergence theorem. Does anybody have any ideas?
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$\begingroup$ I don't see how this is even well defined, in general. The exponents in the Sobolev embedding theorem are sharp, are they not? So take $p=2$ and dimension $d=7$, then there is a $u \in W^{1,p}_0 \setminus L^3$. If we take $v=1$ (with appropriate cutoff) then the integral is not finite. $\endgroup$ – Nate Eldredge Jan 20 at 3:23

$\begingroup$ Maybe I misunderstand the question. What space(s) are the domain and codomain of $A$? Where is $v$ supposed to live? $\endgroup$ – Nate Eldredge Jan 20 at 3:24

$\begingroup$ Concretely, if your domain is the unit ball in $\mathbb{R}^7$ and $p=2$, take $u(x) = x^{2.4}$. Let $v$ be any smooth function which is $1$ on some neighborhood of the origin. $\endgroup$ – Nate Eldredge Jan 20 at 3:58