In Evans’s book “Measure theory and fine properties of functions”, the author proved that any bounded sequence of functions in $L^p$ admits a weak convergence subsequence for $1<p<\infty$, and then the author said that
“The $L^p$ weak compactness theorem fails for $L^1$, since its dual space $L^\infty$ is not separable”
I don’t know what would happen if it’s dual space is separable. By what I have known from functional analysis, a reflexive Banach space is weak compact. However, it seems that the separable dual space cannot imply the reflexivity.
Best Answer
$c_0^{*}=\ell^{1}$ is separable. But the closed unit ball of $c_0$ is not weakly compact. [Look at $(1,1,...,1,0,0...)$].