Lebesgue-Bochner space Does there exist any isometric imbedding of $L^1(a,b;H^*)$ into the dual space of $L^{\infty}(a,b;H)$ where $H$ is a separable Hilbert space and $H^*$ denotes its dual?
 A: The obvious duality defines an isometric embedding. Given $u\in L^1H^*$, let 
$T_u:L^\infty H\to \mathbb{R}$ be defined by 
$$
T_uv=\int_a^b\langle u(t),v(t)\rangle \mathrm{d}t.
$$
We have
$$
|T_uv|\leq\|u\|_{L^1H^*}\|v\|_{L^\infty H},
$$
so $T_u\in (L^\infty H)^*$ and $\|T_u\|\leq\|u\|_{L^1H^*}$.
Moreover, for any $\delta>0$ and for almost every $t$ there exists $v(t)\in H$ with $\|v(t)\|_H=1$ such that
$$
\langle u(t),v(t)\rangle \geq\|u(t)\|_{H^*}-\delta.
$$
If $t\mapsto v(t)$ is measurable, this would show that
$$
\|T_u\|\geq\|u\|_{L^1H^*}-\delta|b-a|,
$$
establishing the claim.
However, as Nate pointed out in the comments, we did not make sure that $t\mapsto v(t)$ is measurable. This is obvious if $u$ is simple. By measurability, there is a sequence of simple functions $u_n$ converging to $u$ pointwise almost everywhere. Moreover, by making use of separability and the fact that $u$ is integrable, we can arrange that $u_n\to u$ in $L^1H^*$.
Now we choose $n$ so large that $\|u_n-u\|_{L^1H^*}\leq\delta$, hence
$$
\|T_u\|\geq\|u\|_{L^1H^*}-\delta|b-a|-2\delta.
$$
