Weak convergence in Sobolev space $H_1$ implies weak convergence in $L^2$

Question

I am trying to prove this statement

Let $D \subset \mathbb{R}^m$. If $\{v_n\}$ is a weakly converging sequence in Sobolev spapce $H_1(D)$, then $\{v_n\}$ weakly converges in $L^2(D)$.

This is the exercise 10.12 from the book An Introduction to Partial Differential Equations by Y.Pinchover and J. Rubinstein. I followed the solution manual.

This is my attempt: Denote the weak limit of $\{v_n\}$ with $v$. Because $\{v_n\}$ is a weakly convergent sequence, that means that it is bounded so there exists a $M \geq 0$ so that \begin{align*} &||v_n||_{H_1(D)} \leq M \quad \text{for every } n = 1, 2 \dots \\ \implies &||v_n||^2_{H_1(D)} = \int_D ((\nabla v_n)^2 + v_n^2) d{\vec{x}} \leq M^2 \quad \text{for every } n = 1, 2 \dots\\ \implies &\int_D (\nabla v_n)^2 d{\vec{x}} = ||\nabla v_n ||_{L^2(D)}^2 \leq M^2 \\ \implies &\int_D v_n^2 d{\vec{x}} = || v_n||_{L^2(D)}^2 \leq M^2. \end{align*} That means that $\{v_n\}$ and $\{\nabla v_n\}$ are also bounded in $L^2(D)$.

Since the sequence $\{\nabla v_n\} = \{(\frac{\partial v_n}{\partial x_1}, \frac{\partial v_n}{\partial x_2}, \dots, \frac{\partial v_n}{\partial x_m})\}_{n=1}^{\infty}$ is bounded in $L^2(D)$, the sequences $\{\frac{\partial v_n}{\partial x_i}\}_{n=1}^{\infty}$ are also bounded in $L^2(D)$ for every $i = 1, 2, \dots, m$. This means that the sequences $\{v_n\}$ and $\{\frac{\partial v_n}{\partial x_i}\}$ contain a weakly convergent subsequence in $L^2(D)$. Denote the weak limit of $\{v_{n_k}\}_k$ with $\tilde{v}$ and the weak limit of $\{\frac{\partial v_{n_k}}{\partial x_i}\}$ with $\tilde{v_i}$ for every $i = 1, 2, \dots, m$. This means that for every $u \in {L^2(D)}$ and $i = 1, 2, \dots, m$ $ \langle \tilde{v}_i, u\rangle_{L^2(D)} =\lim\limits_{k \to \infty} \langle \frac{\partial v_{n_k}}{\partial x_i}, u\rangle_{L^2(D)}$. If we take $u$, such that $u_{|\partial D} = 0$ then \begin{align*} \langle \tilde{v}_i, u\rangle _{L^2(D)} &= \lim\limits_{k \to \infty} \int\limits_D \frac{\partial v_{n_k}}{\partial x_i} u d{\vec{x}}\\ &= -\lim\limits_{k \to \infty} v_{n_k} \frac{\partial u}{\partial x_i} d{x} \\ &= - \int\limits_D \tilde{v} \frac{\partial u}{\partial x_i} d{x} \\ &= \int\limits_D \frac{\partial \tilde{v}}{\partial x_i} u d{x} \\ & = \langle \frac{\partial \tilde{v}}{\partial x_i} ,u\rangle_{L^2(D)}. \end{align*} This means $\tilde{v_i} = \frac{\partial \tilde{v}}{\partial x_i}$ for every $i = 1, 2, \dots, m$.

I do not quite know what to do with this. - I think I am missing the final conclusion, I also do not understand why I even needed to show $\tilde{v_i} = \frac{\partial \tilde{v}}{\partial x_i}$. Did I also prove that $\{\nabla v_n\}$ is weakly convergent in $L^2(D)$?

Answer 1

It is unnecessary to make things this knid complicated. Anyway, I'll finish the proof you presented in OP, and then I'll give a more elegant proof.

$\tilde{v_i} = \frac{\partial \tilde{v}}{\partial x_i}$ implies that $\frac{\partial \tilde{v}}{\partial x_i}\in L^2$, thus $\tilde{v}\in H_1$ and then we know that $v_{n_k}\rightharpoonup\tilde v$ in $H_1$. Since we also have $v_n\rightharpoonup v$ in $H_1$, it follows that $\tilde v=v$ a.e.

Now, we have proved that $$\text{if}\ \ v_n\rightharpoonup v \ \ \text{in}\ \ H_1, \text{then there exists a subsequence} \ \ \{v_{n_k}\}\ \ \text{such that }\ v_{n_k}\rightharpoonup v \ \text{ in}\ \ L^2(D).$$

This will imply that $v_n\rightharpoonup v$ in $L^2$. Indeed, fix any $u\in L^2(D)$, we need to show that $$\lim_{n\to \infty}\langle v_n,u\rangle_{L^2}=\langle v,u\rangle_{L^2}.\tag{1}$$ We use the fact that $\lim_{n\to\infty}a_n=A$ if and only if every subsequence of $\{a_n\}$ has a sub-subsequence converging to $A$, see here for the proof of this result. Now, for a subsequence $\langle v_{n_k},u\rangle_{L^2}$ of $\langle v_n,u\rangle_{L^2}$, since $v_{n_k}\rightharpoonup v$ in $H_1$, by the third paragraph, there is a sub-subsequence $v_{n_{k_j}}$ of $v_{n_k}$ converging weakly to $v$ in $L^2$, thus $$\lim_{j\to \infty}\langle v_{n_{k_j}},u\rangle_{L^2}=\langle v,u\rangle_{L^2}.$$ Now, $(1)$ follows and the proof is complete.

About $\{\nabla v_n\}$: the same trick implies that $\nabla v_n\rightharpoonup\nabla v$ in $L^2$.

Another method. For each $u\in L^2$, the linear functional $$T_u: H_1\to\mathbb R, w\mapsto \langle w,u\rangle_{L^2}$$ is continuous since $|T_u(w)|\leq \|w\|_{L^2}\|u\|_{L^2}\leq \|u\|_{L^2}\|w\|_{H_1}$ for all $w\in H_1$. Hence $T_u(v_n)\to T_u(v)$, i.e. $$\lim_{n\to \infty}\langle v_n,u\rangle_{L^2}=\langle v,u\rangle_{L^2}.$$ This proves the weak convergence of $\{v_n\}$ in $L^2(D)$.