Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f_n$ be bounded uniformly in the $H^1$ norm, so we have (weak convergence) $$f_n \rightharpoonup f \qquad \text{in} \qquad H^1(\Omega\times [0,T]).$$ Then by compact embedding, we have strong convergence $$f_n \to f \qquad \text{in}\qquad L^2(\Omega \times [0,T]).$$

How can I show that $$\frac{d}{dt}\int_{\Omega(t)}f_n\phi \to \frac{d}{dt}\int_{\Omega(t)}f\phi$$ for $\phi \in H^1$?

How to approach this problem? What conditions do I need? Obviously the statement is true if the derivatives weren't there but I'm not sure what to do. I can't take the absolute value of the difference and put the absolute value into the integral either.

share|improve this question
    
$\Omega(t)=?{}{}$ –  Giuseppe Negro Jun 14 '12 at 12:11
    
@GiuseppeNegro For fixed $t$, $\Omega(t)$ is a $C^2$ hypersurface in $\mathbb{R}^n$. –  blahb Jun 14 '12 at 12:13
1  
Without actually thinking much about the details of your question, I'll just mention that integration by parts usually works well in these cases. Let $\chi \in C^\infty_c((0,T))$ be arbitrary, and try to show that $$\int_0^T \int_{\Omega(t)} \chi' f_n \phi \to \int_0^T \int_{\Omega(t)} \chi' f \phi.$$ –  Nate Eldredge Jun 14 '12 at 14:23
    
Thanks @NateEldredge. Showing what you wrote above is fine I think (take the difference and use Cauchy-Schwarz etc.) I take it: $$\int_0^T \chi g_n \to \int_0^T \chi g \qquad \text{for all $\chi$}$$ implies $$g_n \to g \qquad \text{a.e.}$$ is what you were hinting at? –  blahb Jun 14 '12 at 17:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.