# Question about limits of weakly convergent sequence in $H^1_0(\Omega)$

Let $H = H_{0}^{1}(\Omega)$ where $\Omega$ is a bounded domain in $R^N$ whose boundary $\partial\Omega$ is a smooth manifold. We know that the embedding $$H\hookrightarrow L^s(\Omega)$$ is compact for each $s\in [1, 2^*)$ and continuous for each $s\in [1, 2^*]$

Suppose that $\{u_n\}$ is a sequence in $H$ such that $u_n\rightharpoonup u$. Then $\{u_n\}$ is bounded, so by the compact embedding there exists a subsequence $\{u_{n_k}\}$ such that $u_{n_k}\rightarrow u_0$ in $L^s(\Omega)$ for $s\in [1, 2^*)$

How do we know that $u_0 = u$?

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Can you define what you mean exactly by $u_n \hookrightarrow u$? Is that weak convergence? – user1736 Dec 12 '11 at 6:25
For any continuous and linear functional $f$ on $H$, $u_n\rightharpoonup u$ in $H$ means that $f(u_n)\rightarrow f(u)$. – Jiahua Jin Dec 12 '11 at 6:49
$H\hookrightarrow L^s(\Omega)$, i.e. there is a identity operate $I$ and a positive constant $M$ such that $$\|u\|_L\leq M\|u\|_H$$ for any $u\in H$. – Jiahua Jin Dec 12 '11 at 7:06

We have $2^*=\frac{2N}{N-2}>\frac{2(N-2)}{N-2}=2\geq 1$ so $u_{n_k}\to u_0$ in $L^2$. Since the sequence $\{u_{n_k}\}$ is bounded in $L^2$, we can extract a converging subsequence $\{u_{\psi(k)}\}$ in $L^2$ to a function $v$ (taking again a subsequence we can assume it converges almost everywh. Using test functions and weak convergence, we can see that in fact $v=u$.

We show the following result

If $\{f_n\}\subset H^1_0(\Omega)$ is a sequence which converges weakly to $f$ in $H^1_0(\Omega)$, then this sequence converges weakly to $f$ in $L^2(\Omega)$.

Since $\{ f_n\}$ and $\{ \nabla f_n\}$ are bounded in $L^2$, we can extract converging subsequences $\{f_{\psi(n)}\}$ and $\{\nabla f_{\psi(n)}\}$, which converges respectively to $g$ and $h$. But for $\varphi\in\mathcal D(\Omega)$ and $1\leq i\leq N$ $$\int_{\Omega}gD_i\varphi dx=\lim_k\int_{\Omega}f_{\psi(k)}D_i\varphi dx =-\lim_k\int_{\Omega}D_if_{\psi(k)}\varphi dx =-\int_{\Omega}h_i\varphi dx,$$ so $h=\nabla g$ and $f_{\psi(k)}$ converges to $g$ weakly in $H^1_0(\Omega)$, so $f=g$.

So the sequence $\{u_n\}$ admit a subsequence which converges weakly to $u$ and $u_0$, which implies that $\langle w,u-u_0\rangle=0$ for each $w\in H$, so $u=u_0$.

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Maybe my question is not clear:$u_n\rightharpoonup u$ in $H$ implies subsequential $u_{n_k}\rightharpoonup u$ in $L^t(\Omega)$? – Jiahua Jin Dec 12 '11 at 14:33
Where is $t$ supposed to be? – Davide Giraudo Dec 12 '11 at 16:06
We known that $u\in L^t(\Omega)$, but $\nabla u$ isnot sure in $L^t$ from the definition of the space $L^t$! $t\in [1,2^*)$, where $2^*=\frac{2N}{N-2}$. – Jiahua Jin Dec 13 '11 at 0:51

If $t$ is the conjugate exponent to $s$ (so that $\frac{1}{s} + \frac{1}{t} = 1$), and $g \in L^t(\Omega)$, note that $$f \mapsto \int_\Omega fg\,dx \quad (*)$$ defines a continuous linear functional on $H$. This is because (*) defines a continuous linear functional on $L^s$, and the inclusion $H \hookrightarrow L^s$ is continuous.

Then for any $g \in L^t$, we have $\int u_{n_k} g\,dx \to \int u g \,dx$ by weak convergence. On the other hand,by Hölder's inequality we have $\int u_{n_k} g \,dx \to \int u_0 g \,dx$. It follows that $u = u_0$ almost everywhere.

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