Suppose $B_1$ is a unit ball in $\mathbb{R}^n$ and $u_n$ is a sequence of function in $H^1(B_1)$. It is well-known that if $|u_n|_{H^1(B_1)}$ is uniformly bounded, then $u_n$ converges weakly to $u_0$, up to a subsequence.
Now suppose we have a sequence of function $u_{n,m}$ in $H^1(B_1)$. If $|u_{n,m}|_{H^1(B_1)}$ is uniformly bounded in both $n$ and $m$, I believe that we also have the fact that $u_{n,m}$ is weakly convergent up to some subsequences. Here we do not relabelled the subsequence.
I would like to ask that is the weak convergence uniform in $n$ and $m$? Is it true that
$$ \lim_{n \to \infty}\lim_{m \to \infty} u_{n,m} = \lim_{m \to \infty}\lim_{n \to \infty} u_{n,m}? $$ The above limit is taken as weak sense in $H^1$. Thank you so much!
Even if you replace $H^1$ by $\mathbb R$, this does not seem to be true. We pick $$ a_{n,m} = \begin{cases}42 & \text{for } n > m, \\ 23 & \text{for } m \ge n.\end{cases}$$ Note that we do not need to select subsequences, since we have convergence for $n \to \infty$ and for $m \to \infty$ separately. Similarly, the iterated limits exist, but are not equal.