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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 $f$

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

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  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Jul 1 at 3:10
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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 $\{u_{n_k}\}_{k \in \mathbb{N}}$.

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  • $\begingroup$ 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 $u_n = (-1)^n$. $\endgroup$ – PhoemueX Dec 27 '18 at 17:00

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