Given $\Omega\in \mathbb R^N$ open bounded smooth boundary, assume $u_n$, $u\in L^q(\Omega,\mathbb R^d)$ for some $d\in\mathbb N$ and $1<q<\frac{N}{N-1}$. We also assume that $u_n\to u$ weakly star in $\mathcal M(\Omega;\mathbb R^d)$ where the space $\mathcal M(\Omega;\mathbb R^d)$ denotes all finite Redon measures.

Now, I want to conclude that $u_n\to u$ strongly in space $W^{-1,q}(\Omega,\mathbb R^d)$, where $W^{-1,q}(\Omega,\mathbb R^d)$ defined as the dual space of $W^{1,q'}_0(\Omega,\mathbb R^d)$, and I also want to conclude that $\nabla u_n\to \nabla u$ strongly in $W^{-2,q}(\Omega,\mathbb R^d)$. How may I conclude this result? Is there kind of Sobolev embedding I can use for negative Sobolev space?

Thank you!


Let $E:W^{1,q'}_0(\Omega)\to L^{q'}(\Omega)$ denote the compact Sobolev embedding operator. Let $I:L^q(\Omega) \to (L^{q'}(\Omega))^*$ be the usual isomorphism between these two spaces. Since $E$ is compact, $E^*$ is compact, and the operator $E^*I:L^q(\Omega) \to (W^{1,q'}_0(\Omega))^*$ is compact, as well. Hence $E^*Iu_k$ converges strongly in $(W^{1,q'}_0(\Omega))^*$.

It remains to check that this convoluted construction does the right thing: Let $v\in W^{1,q'}_0(\Omega)$ be given. Then $$ (E^*Iu)(v) = (Iu_k)(Ev) = \int_\Omega u(x) (Ev)(x) \ dx= \int_\Omega u(x) v(x) \ dx, $$ thus $E^*I$ maps the function $u$ to the functional $v\mapsto \int_\Omega u v$, which is the standard identification of a a function in $L^q$ with a functional in $(W^{1,q'})^*$. In this sense, $u_k$ converges strongly to $u$ in $W^{-1,q}$.

| cite | improve this answer | |
  • $\begingroup$ Thank you! I will think the part of $\nabla u_n\to \nabla u$ works in the similar way? $\endgroup$ – spatially Jun 2 '15 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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