Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $u_0 = u$?

share|improve this question
    
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

2 Answers 2

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 $u=u_0$.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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