Is this property preserved under weak$^*$ convergence?

Question

Let $1 \le p < n$ and let $p^*$ be the Sobolev conjugate of $p$, i.e. $p^* = np/(n - p)$. Let $(\Omega_m)$ be an increasing sequence of bounded, convex and open sets such that $$ \lim_{m \to \infty} \sup \{r : B(0, r) \subset \Omega_m \} = \infty. $$ Assume $u_m \in W^{1, p}(\Omega_m)$, $0 \le u \le 1$ and $$ \lim_{m \to \infty} \frac{1}{|\Omega_m|} \int_{\Omega_m} |u_m|^{p^*} \, dx = 0, \quad \|\nabla u_m\|_{L^p(\Omega_m, \mathbb{R}^n)} \le C, $$ where $C$ does not depend on $m$. Since $(u_m)$ is uniformly bounded in $L^\infty(\mathbb{R}^n)$, we can assume without loss of generality that there exists $u \in \dot{W}^{1, p}(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n)$ such that $u_m \overset{\ast}\rightharpoonup u $ in $L^\infty(\mathbb{R}^n)$. Is it true that $$\lim_{m \to \infty} \frac{1}{|\Omega_m|} \int_{\Omega_m} |u|^{p^*} \, dx = 0 $$ as well?

EDIT: We actually have more than weak$^*$ convergence. By the Rellich-Kondrachov theorem and the uniform boundedness of $(u_m)$ in $L^\infty$, it is true that $u_m \to u$ in $L^q(K)$ for all compact $K$ and all $1 \le q < \infty$.

Answer 1

Let me do for balls $\Omega_m$ and I use your Edit. Note that $\bar u_m=\frac{1}{|\Omega_m|} \int_{\Omega_m} u_m \to 0$ by H"older inequality and your assumption. Next I use Poincarè-Wirtinger inequality $$ \|u_m-\bar u_m\|_{p^*} \leq C\|\nabla u_m\|_p $$ with a constant independent of the radius of the ball. This is where I need balls or, more general, dilated domains, since Sobolev inequality is scale invariant for these exponents. If you let $m \to \infty$, you get $\|u\|_{p^*} \leq C$ and then the result follows since $|\Omega_m| \to \infty$. Of course this generalizes to the case where the biggest ball $B_m$ satisfies $|B_m| \geq c|\Omega_m|$ but not to the rectangles you are interested in, so that this is only a partial answer.

EDIT Let me complete the proof in the general case. Let $B_m \subset \Omega_m$ be a ball or radius $r_m \to \infty$ and $\bar u_m$ the mean of $u_m$ over $B_m$. If $\bar u_m \to 0$ the argument above gives $u \in L^{p*}$ and the result holds.

If $u_m \to \ell$ (up to a subsequence) then, as above, $u-\ell \in L^{p^*}$ and then $\frac{1}{|\Omega_m|}\int_{\Omega_m}|v|^{p*} \to 0$ with $v=u$ and $v=u-l$. This yields $\ell=0$.