# Weak star convergence in $L^{\infty}(0,T;H^1(\Omega))$

Let $\Omega$ be a bounded Lipschitz domain in $\mathbb{R}^n$.

What is the meaning of $u_n \rightharpoonup^\star u$ in $L^{\infty}(0,T;H^1(\Omega))$?

Thank you!

Let $X$ be a Banach space and $X^\star$ its dual. Consider the weak star topology $\sigma (X^\star,X)$ in $X^\star$ (see Brezis - Functional Analysis book for reference).

We say that a sequence $u_n\in X^\star$ converge in the weak star topology to some $u\in X^\star$, denoted by $u_n \rightharpoonup^\star u$, if $u_n(x)\to u(x)$ for all $x\in X$.

For $u,v\in H^1(\Omega)$, let $\langle u,v\rangle=\int_ \Omega uv+\int_\Omega \nabla u\nabla v$.

One can prove that $$(L^1(0,T;H^1(\Omega)))^\star=L^\infty(0,T;H^1(\Omega)),$$

by noting that the function $$T:(L^1(0,T;H^1(\Omega)))^\star\to L^\infty(0,T;H^1(\Omega)),$$

defined by $T(f)=u$ is an isomorphism, where $u$ is such that $$f(v)=\int_0^T\langle u(t),v(t)\rangle dt,\ \forall\ v\in L^1(0,T;H^1(\Omega)),$$

therefore, by taking $X=L^1(0,T;H^1(\Omega))$, we have that, $u_n \rightharpoonup^\star u$ if and only if $$\int_0^T \langle u_n(t),v(t)\rangle dt\to \int_0^T \langle u(t),v(t)\rangle dt,\ \forall\ v\in L^1(0,T;H^1(\Omega)).$$

• Please check your answer. There seem to be some typos in the notion of the spaces. Moreover, you should specify the integrals in the last three lines. Also note that the question is about $H^1(\Omega)$ and not $H_0^1(\Omega)$. – gerw Sep 6 '16 at 6:55
• Thank you @gerw for your observations. I have fixed it accordingly. – Tomás Sep 6 '16 at 11:09
• I have fixed some other small typos. – gerw Sep 6 '16 at 11:44