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Let $\Omega$ be an opend bounded subset of $\mathbb{R}^n$ and let $p, q$ be two real numbers such that $p, q\geq 1$. Let $(w_n)_n\subset W_0^{1, p}(\Omega)$ and $(z_n)_n\subset W_0^{1, q}(\Omega)$ such that $\exists w\in W_0^{1, p}(\Omega)$ such that $$ w_n\longrightarrow w \quad \mbox{ in } L^{r}(\Omega) \quad \mbox{ for } \ 1\leq r < p^{\ast}$$ and $$ w_n\longrightarrow w \quad \mbox{ a.e. in } \Omega.$$ Moreover, fix $k\geq 1$ and consider $$\Omega_{n, k}:=\left\lbrace x\in\Omega \mid \vert w_n(x), z_n(x)\vert > k\right\rbrace.$$ I would like to show that the integral $$\int_{\Omega\setminus\Omega_{n, k}} F(x, w_n, z_n) w \vert\nabla z_n\vert^{q} dx$$ can be estimated with a positive constant, i.e. $\exists c\in\mathbb{R}$ such that $\displaystyle\int_{\Omega\setminus\Omega_{n, k}} F(x, w_n, z_n) w \vert\nabla z_n\vert^{q} dx\leq c$.

Here, I assume $F:\Omega\times\Omega\times\mathbb{R}\to\mathbb{R}$ such that $$ \sup_{\vert (u, v)\vert\leq t} \vert F(\cdot, u, v)\vert\in L^{\infty}(\Omega)$$ for any $t>0$.

Could anyone please help? Thank you in advance!

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  • $\begingroup$ Any assumptions on the sequence $z_n$? $\endgroup$
    – Chris
    Jun 27 '20 at 16:47
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Since $z_n \in W_{0}^{1, q}$, your only hope on estimating the integral of $|\nabla z_n|^q$ is to put it in $L^1$ and pull out an $L^\infty$ norm of the $F$ term: $$ \int_{\Omega \setminus \Omega_{n,k}} |F(x, w_n, z_n)||w||\nabla z_n|^q \, dx \leq \|{z_n}\|_{W^{1 , q}}\sup_{x \in \Omega \setminus \Omega_{n, k}} |F(x, w_n(x), z_n(x))||w(x)|. $$ So to get a uniform bound we'd first need to know that the $z_n$'s are uniformly bounded in $W^{1, q}$, or at least that their gradients are bounded uniformly in $L^q$. Meanwhile the supremum term is finite for each $k$, since $\Omega_{n,k}$ is the set where $w_n, z_n$ are greater than $k$, but this bound potentially depends on $k$. Therefore we'd probably also need some assumptions on $F$'s $L^\infty$ behavior.

If there are no such assumptions on $F$ or $z_n$, we can cook up a counterexample. Let $F \equiv 1$ and let $w_n \equiv w \in W_0^{1, q}$ be some smooth bump function. Then let $z_n$ be some sequence in $W_0^{1, q}$ such that $\sup_{x \in \Omega} |z_n| \leq C$ for all $n$, but such that $\|\nabla z_n\|_{L^q} \to \infty$ as $n \to \infty$ (for instance, on $\mathbb{R}$ we could let $z_n$ be something like $\sin(nx)$, perhaps a polygonal version of this). Then the integral above is always over all of $\Omega$ for $k > C$ and in fact it's equal to $$ \int_{\Omega} |w(x)||\nabla z_n|^q\, dx. $$ So as long as we choose, say, $w$ to be equal to 1 on some set of large enough measure, this integral will go to infinity as $n \to \infty$.

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