Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. Let $p \geq 1$ and suppose that $u_n \rightharpoonup u$ in $L^p(\Omega)$ and $u_n \to v$ in $H^{-1}(\Omega)$.

How to show that $u=v$?

I can do this for $p=2$ by using definition of weak convergence but for other $p$ I have no idea.


Both assumptions imply convergence in the sense of distributions: that is, for every smooth compactly supported function $\varphi$ we have $$ \int \varphi u_n\to \int \varphi u,\qquad \int \varphi u_n\to \int \varphi v \tag1$$ So, $$\int \varphi u = \int \varphi v\tag2$$ for every such $\varphi$. This means exactly that $u=v$ in the sense of distributions. But $u$ is an $L^p$ function, so $v$ is a distribution represented by an $L^p$ function. By virtue of (2) it's the same function. Indeed, the elements of $L^p$ can be identified with bounded linear functionals on $L^q$ ($1/p+1/q=1$) and if two such functionals agree on a dense subspace, they are equal.

Summary: both modes of convergence imply distributional convergence, and distributional limit is unique.

  • $\begingroup$ Thank you. Can you explain one thing: why does $u_n \to u$ in $H^{-1}$ imply convergence in distribution? I was thinking: $(u_n, f)_{H^{-1}} \to (u, f)_{H^{-1}}$ for all $f \in H^{-1}$ (the brackets means the inner product on $H^{-1}$), and then we use Riesz map $\langle u_n, F \rangle_{H^{-1}, H^1} \to \langle u, F \rangle_{H^{-1}, H^1}$ for all $F \in H^1$. I think we want to pick $F \in C_c^\infty$, but the angled bracket only turns into an integral when $u_n$ and $u$ are in $L^2$ which is not necessary. So I guess this is not the way to prove it converges in distribution? $\endgroup$
    – asda
    Jul 11 '14 at 9:44
  • $\begingroup$ @asda I was not thinking of inner product in $H^{-1}$, but rather of the definition of this space as the space of linear functionals on $H^1_0$. The way these functionals are evaluated is consistent with the evaluation of distributions on test functions. It's a matter of unwrapping the definition of this dual. $\endgroup$
    – user147263
    Jul 11 '14 at 12:22

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.