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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!

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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 $W^{-1,q}$.

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  • $\begingroup$ 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

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