2
$\begingroup$

$\newcommand{\loc}{\mathrm{loc}}$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.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

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 $\Omega$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.