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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?

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    @Markus: your accounts have been merged. In the future, please do not use answers to make comments. You can comment on answers to your own questions, but to make sure you keep logging into the same account you should register.2012-08-25

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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. $

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    Many thanks to you!2012-08-26