0
$\begingroup$

I'm having bit trouble in understanding weak convergences in Bochner space. I have following question for some general measurable space $\Omega$:

Let $\{x_n\}$ be a bounded sequence in $L^2((0,T)\times \Omega ; H^1(\mathbb{R}^d))$, then we can extract a weakly convergent sub sequence $\{x_{n_k}\}$ st $x_{n_k} \rightharpoonup x $ in $L^2((0,T)\times \Omega ; H^1(\mathbb{R}^d))$. Can we conclude for a.e. $t$, $x_{n_k}(t) \rightharpoonup x(t) $ in $L^2(\Omega ; H^1(\mathbb{R}^d))$?

Improve this question
$\endgroup$
2
$\begingroup$

(a) is not even true in ordinary Lebesgue spaces so it's certainly not true in Bochner. The standard counterexample is something like $x_n(t) = e^{2\pi i n t}$ in $L^2([0,1])$. It converges weakly to 0 but $x_n(t)$ diverges for almost every $t$.

In your specific context, you could take instead $x_n(t) = e^{2 \pi i n t} \phi$ where $\phi $ is some fixed nonzero element of $L^2(\Omega; H^1(\mathbb{R}^d))$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Would this example also work for part b) ? $\endgroup$ – MathAnimal Sep 24 '18 at 4:54
  • $\begingroup$ Hello sir, could you please elaborate on this. $\endgroup$ – MathAnimal Sep 25 '18 at 14:04
  • $\begingroup$ @MathAnimal: Sorry, my idea was wrong. Maybe someone else will answer. $\endgroup$ – Nate Eldredge Sep 25 '18 at 14:33
  • $\begingroup$ Hi, shall I post this as a different question, this post seems to have saturated. $\endgroup$ – MathAnimal Sep 25 '18 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.