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

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 $$f$$

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

• The "i.e." in the first sentence is only an implication in one direction. As a trivial example, if $u_n$ takes the values $1$ and $-1$ on two crazy complementary sets, then $u_n^2 = 1 \in W^{1,2}$, but the weak derivative of $u_n$ may not even be a function, let alone an $L^4$ function. – Nate Eldredge Jul 1 at 3:10

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

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

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

• What you show is true, but you only show convergence for a subsequence. It is not necessary that the whole sequence converges weakly, as can be seen by considering $u_n = (-1)^n$. – PhoemueX Dec 27 '18 at 17:00