Let $\Omega$ be an open bounded set in $R^n$ with smooth boundary. Suppose that $u_n$ is a sequence in $H^1_0(\Omega)$ which converges weakly to $u$ in the sense that for all $y \in H^1_0(\Omega)$

$$\int_\Omega \nabla u_n\cdot \nabla y \to \int_\Omega \nabla u\cdot \nabla y$$

Then, why is it true that $u_n$ converges to $u$ weakly in $L^2(\Omega)$, that is,

$$\int_\Omega u_n\cdot y \to \int_\Omega u\cdot y$$ for all $y \in L^2(\Omega)$?

  • 1
    $\begingroup$ The embedding $H_0^1(\Omega) \hookrightarrow L^2(\Omega)$ is linear and continuous hence weakly continuous. $\endgroup$ – daw Mar 21 '17 at 8:54

Let $f \in L^2(\Omega)$ be given. Then, there is a unique $w \in H_0^1(\Omega)$ with $$\int_\Omega \nabla w \cdot \nabla y \, \mathrm{d}x = \int_\Omega f \cdot y\, \mathrm{d}x$$ for all $y \in H_0^1(\Omega)$. (Note that $w$ is essentially the Riesz representative of the functional $y \mapsto \int_\Omega f \cdot y \, \mathrm{d}x$.)

Then, $$ \int_\Omega u_n \cdot f \, \mathrm{d}x = \int_\Omega \nabla u_n \cdot \nabla w\,\mathrm{d}x \to \int_\Omega \nabla u \cdot \nabla w\,\mathrm{d}x = \int_\Omega u \cdot f \, \mathrm{d}x.$$

  • $\begingroup$ Could you elaborate more on your answer. I don't see how $w$ is the Riesz representative since your Hilbert space is $H^1_0(\Omega)$ not $\dot H^1_0(\Omega)$, the homogeneous Sobolev space. $\endgroup$ – Jacky Chong Mar 22 '17 at 1:22
  • $\begingroup$ $u \mapsto (\int_\Omega |\nabla u|^2 \, \mathrm{d}x)^{1/2}$ is a norm on $H_0^1(\Omega)$ and induces the scalar product $(u,v) = \int_\Omega \nabla u \cdot \nabla v \, \mathrm{d}x$. $\endgroup$ – gerw Mar 22 '17 at 7:52
  • $\begingroup$ Thanks. I forgot about the equivalence of norm by Poincare inequality. $\endgroup$ – Jacky Chong Mar 22 '17 at 23:37

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