Does Weak Convergence in $W^{1,2}$ imply weak convergence in $W^{1,4}$

Question

Say I have a sequence $u_{n} \in W^{1,4}(\mathbb{T}^2)$, i.e $u^{2}_{n} \in W^{1,2}(\mathbb{T}^2)$.

If $u^{2}_{n}$ converges weakly to $v$ in $W^{1,2}(\mathbb{T}^2)$, does $u_{n}$ converge weakly to 'something' in $ W^{1,4}(\mathbb{T}^2).$

Comments : since $ W^{1,2}(\mathbb{T}^2)$ is a Hilbert space weak convergence can be characterised by the inner product. The space $ W^{1,4}(\mathbb{T}^2)$ is not a Hilbert space so for weak convergence (of $u_{n}$ to $u$) we need to show that for all linear bounded functionals $f$

$f(u_{n})\to f(u)$

Answer 1

Yes, this is true, at least for some subsequence.

Let $u_n \in W^{1,4}$ such that

$$u_n^2 \longrightarrow v \quad \text{ weakly in } W^{1,2}$$

for some function $v \in W^{1,2}$. Now, remember that weak convergence implies uniform boundedness, i.e.

$$\sup_n \|u_n^2\|_{W^{1,2}} < C$$

and therefore, also $\|u_n\|_{W^{1,4}}<C$. By the theorem of Banach-Alaoglu there is a function $w \in W^{1,4}$ such that

$$u_{n_k} \longrightarrow w \quad \text{ weakly in } W^{1,4},$$

for some subsequence $\{u_{n_k}\}_{k \in \mathbb{N}}$.