Weak $H^1$ convergence and uniform convergence on finite intervals Suppose $\{u_k\}$ is weakly convergent to $u$ in $H^1(\mathbb{R})$, does that imply that $u_k$ uniformly converges to $u$ in finite interval? 
On the other hand, if $\{u_k\}$ is weakly convergent to $u$ in $H^1(\mathbb{R})$, $u_k$ uniformly converges to $v$ in finite interval $[a,b]$, does $u=v$ in $[a,b]$?
 A: Let's start with the second question. The answer is yes, by the usual "common denominator" type of convergence, distributional convergence. Namely, both $u_k\to u$ uniformly and $u_k\to v$ weakly in $H^1$ imply that for every $\varphi\in C_c^\infty(\mathbb{R})$ we have 
$$
\int u_k \varphi\to\int u\,\varphi,\qquad \int u_k \varphi\to\int v\,\varphi
$$
So, $\int (u-v)\varphi=0$ for every such $\varphi$, which implies $u=v$ a.e. In other words, the distributional limit is unique. 
Now the first question can be dealt with as follows, avoiding the Rellich-Kondrachov theorem. A weakly convergent sequence $(u_k)$ is bounded. Being bounded in $H^1([a,b])$, the sequence is not only bounded in uniform norm, but also equicontinuous, since
$$
|u(x)-u(y)| \le \int_{[x,y]}|u'| \le \sqrt{|x-y|}\|u\|_{H^1}
$$
By Arzela-Ascoli, it has a uniformly convergent subsequence. By the previous discussion, the uniform limit is also the weak limit of $u_k$. It remains to recall a general fact about metric spaces: if every subsequence of $(u_k)$ has a subsubsequence converging to $u$, then $u_k\to u$. 
