# Sequential Banach-Alaoglu theorem for a Bochner space

I have a question concerning the sequential weak-$*$ compactness in a Bochner space. Let $H$ be some non-separable Hilbert space. Consider the Bochner space $L^{\infty}(0,T;H)$. It is known that this space is isometrically isomorph to the dual of $L^1(0,T;H)$. Consequently, the Banach-Alaoglu theorem applies and yields the weak-$*$ compactness of a unit ball in $L^{\infty}(0,T;H)$. Now, since $L^1(0,T;H)$ is not separable (as $H$ is not), one cannot use the sequential version of the Banach-Alaoglu theorem in this case. But it seems that the conclusion must still be true in this case, i.e., a ball in $L^{\infty}(0,T;H)$ should be sequentially weak-$*$ compact. Here is my argumentation. Consider $L^2(0,T;H)$. This space is reflexive, so that Eberlein–Šmulian theorem applies. This result ensures in particular that each $L^{\infty}(0,T;H)$-bounded sequence $x_n$ has a subsequence $x_{n_k}$ which weakly converges to some $x\in L^2(0,T;H)$. But $L^2(0,T;H)$ is dense in $L^1(0,T;H)$, and the sequence is uniformly bounded in $(L^1(0,T;H))^*$. Therefore, due to the Banach-Steinhaus theorem the limit $x$ should actually belong to $(L^1(0,T;H))^*$ as well. Is this a correct argumentation? If yes, could somebody please recommend a suitable citation? I searched, e.g., in the book on vector measures by Diestel and Uhl, but could not find such a result.

Thanks a lot in advance!