A problem on weak convergence: weak convergence of a sequence implies the weak convergence of the square of this sequence?

Question

Suppose that $\{u_m\}_m\subset H^1(\mathbb{R}^3)$ and $u_m\rightharpoonup u$ weakly in $H^1(\mathbb{R}^3)$. From the classic results, I know that there exists a subsequence such that $u_m\rightharpoonup u$ weakly in $L^2(\mathbb{R}^3)$. My question is whether there exists another subsequence such that

$u_m^2\rightharpoonup u^2$ weakly in $L^2(\mathbb{R}^3)$.

I know that $u_m^2\rightharpoonup v$ in $L^2(\mathbb{R}^3)$ for some $v$ but I don't know how to prove that $v=u^2$.

Thanks for any help!

Answer 1

Let $u_m$ be weakly convergent as in your post. Take $K$ to be a compact cube $\mathbb R^3$. Then the restrictions $u_m|_K$ converge weakly as well: $u_m|_K \rightharpoonup u$ in $H^1(K)$ and $u_m^2|_K \rightharpoonup v$ in $L^2(K)$. By compact embeddings, a subsequence of $(u_m|_K)$ converges strongly in $L^4(K)$ to $u$, which implies that the corresponding subsequence of $(u_m^2|_K)$ converges strongly to $u^2$. This shows $u^2=v$ a.e. on $K$. We can cover $\mathbb R^3$ be countably such cubes, and it follows $u^2=v$ a.e. on $\mathbb R^3$.