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

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

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