# Weak convergence in $H^1_0(U)$ implies convergence in $L^2(U)$

Recall some definitions :

$H^1_0(U) =W_0^{1,2}(U)$ and $u\in W^{k,p}_0(U)$ i there exists $u_m\in C_c^\infty(U)$ s.t. $u_m\rightarrow u$ in $W^{k,p}(U)$

question For $1\leq p \leq \infty$, $W^{k,p}(U)$ is Banach. How can we prove that $H^1_0(U)$ is Hilbert ?

question 2 If $\{u_k\}$ is bounded in $H^1_0(U)$ then there exists a subsequence s.t. $u_{k_j}$ converges to $u$ in $L^2(U)$. How can we prove this ?

question 3 If $\{u_k\}$ is bounded in reflexive Banach then there exists a subsequence s.t. $u_{k_j}$ converges to $u$ weakly. How can we prove this ?

From your definition, $W_0^{k,p}(U)$ is the completion of the space $C^\infty_c(U)$ with the metric

$$||u||_{k,p} = \sum_{|I|\leq k} \bigg|\bigg|\frac{\partial^k u}{\partial x_I}\bigg|\bigg|_p\ .$$

$H^1_0(U)$ is a Hilbert space because $||\cdot||_{1,2}$ norm is induced by the inner product

$$\langle f, g\rangle_{W^{1, 2}} = \langle f, g\rangle_{L^2} + \sum_i \langle \frac{\partial f}{\partial x_i}, \frac{\partial f}{\partial x_i}\rangle_{L^2}$$

For the latter case, one need to use the Kondrachov compactness theorem, which says that the imbedding

$$W^{1,p}_0(U) \to L^q(U)$$

is compact for all $q< np/(n-p)$ if $p<n$. Thus you have a compcat embedding

$$H^1_0(U) \to L^2(U)\ .$$

Then every bounded sequence $\{u_m\}\in H^1_0(U)$ is weakly convergent in $H^1_0(U)$ to $u$. Under the compact operator it converges to $u$ strongly in $L^2(U)$.

• This is not right, the compactness of the operator gives you only a convergent subsequence. Nevertheless, any subsequence has a subsequence converging to the same $u$. – Jorge E. Cardona May 7 '17 at 2:29