Could you please tell me what does it mean $u_n \rightarrow u$ weak-star in $L^\infty(0,T;L^2(\Omega))$ ?

  • 1
    $\begingroup$ $L^\infty(0,T;L^2(\Omega))$ should be the dual space of something. There is a natural candiate. // You didn't say what $\Omega$ is. Presumably an open subset of $\mathbb{R}^n$? $\endgroup$ – Martin Feb 23 '13 at 12:26
  • $\begingroup$ Please explain a bit more where you got lost, in order to be able to help. $\endgroup$ – vonbrand Feb 23 '13 at 12:29
  • $\begingroup$ yes, $\Omega$ is domain in $R^n$ $\endgroup$ – Igor Feb 23 '13 at 19:31
  • $\begingroup$ $L^{\infty}(0,T;L^2(\Omega))$ is dual space to $L^1(0,T;L^2(\Omega))$ $\endgroup$ – Igor Feb 23 '13 at 19:34

General definition. If $X$ is a normed space with dual $X^*$, we say that a sequence $u_n$ in $X^*$ converges to $u$ in weak* topology if for every $v\in X$ we have $\langle u_n, v\rangle\to \langle u, v\rangle$. So, the question reduces to identifying $X$ and $X^*$, and the pairing of $X^*$ on $X$.

Given that $X=L^1(0,T;L^2(\Omega))$ and $X^*=L^{\infty}(0,T;L^2(\Omega))$, the presumed pairing is $$\langle u,v\rangle=\int_0^T \int_\Omega u(x,t)\,v(x,t)\,dx \,dt$$ Or with a conjugate $\overline{v(x,t)}$ if we are dealing with complex spaces.

Thus: for every $v\in L^1(0,T;L^2(\Omega))$ $$\int_0^T \int_\Omega u_n(x,t)\,v(x,t)\,dx \,dt\to \int_0^T \int_\Omega u(x,t)\,v(x,t)\,dx \,dt$$


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.