Weakly * continuous definition

Question

What does it mean that $$t\to u(t,\cdot)$$ is waakly* continuous from $[0,T]$ to $L^\infty(\mathbb{R}^d)$? I guess that I have to see $L^\infty(\mathbb{R}^d)$ as the dual of $L^1(\mathbb{R}^d)$ and then consider the convergence with respect to the weak star topology on $L^\infty(\mathbb{R}^d)$. Is it right?

Answer 1

You are right. $t \mapsto u(t, \cdot)$ being weak$^*$ly continuous means, that for every $f \in L^1(\mathbb R^d)$, the map $$ t \mapsto \int_{\mathbb R^d} f(x)u(t,x)\, dx $$ is continuous.