$\newcommand{\loc}{\mathrm{loc}}$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.
Best Answer
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 $\Omega$.