$\newcommand{\loc}{\mathrm{loc}}$Let $\Omega$ be a bounded open set (smooth as we wish if necessary) in $\mathbb{R}^n$, $(\omega\_k)$ a sequence of open subsets whose closure is contained in $\Omega$ and whose union covers $\Omega$. Let $(u\_k)\_{k\in\mathbb{N}}$ a sequence in $H^1(\Omega)$ and assume the uniform bound
$$
\| u\_k \|\_{H^1(\omega\_k)} \leq M
$$
for every $k\in\mathbb{N}$, with the positive constant $M>0$ independent of $k$.


By a diagonal argument, we can infer the existence of a subsequence (not relabeled here) $(u\_k)\_{k\in\mathbb{N}}$ which weakly converge in $H^1\_{\loc}(\Omega)$ to an element $u\in H^1(\Omega)$. Therefore
$$
(\*)\qquad \qquad \int\_\omega \nabla u\_k \cdot \psi \to \int\_\omega \nabla u \cdot \psi
$$
for every open set $\omega$ whose closure is contained in $\Omega$, and every $\psi\in H^1(\Omega,\mathbb{R}^n)$.


**Question.** Given that both $(u\_k)\_{k\in\mathbb{N}}$ and the weak limit $u$ belong to $H^1(\Omega)$, is this sufficient to infer that we have weak convergence in $H^1$ instead of $H^1\_{\loc}$? In other words, is it true that $(\*)$ can be replaced by$$
(\*\*)\qquad \qquad \int\_\Omega \nabla u\_k \cdot \psi \to \int\_\Omega \nabla u \cdot \psi
$$
for every $\psi\in H^1(\Omega,\mathbb{R}^n)$?.


**Update** I realized that the way it is formulated is trivial. I was trying to minimize the number of formulas to write and I came up with this formulation that I was thinking equivalent to my real question but it is not. Nevertheless I am going to accept the answer because it answers to the question I wrongly asked.



Weak convergence in $H^1_{\mathrm{loc}}$
The sequence $(u\_k)$ need not converge in $H^1(\Omega)$ under the given hypotheses. Indeed, recall that a weakly convergent sequence is bounded. Therefore, a sequence whose terms have $\lVert u\_k \rVert\_{1;\Omega} \to \infty$ but $\lVert u\_k \rVert\_{1;\omega\_k} \leq M$ would not converge weakly in $\Omega$.


