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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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    I am not really familiar with the latter of these two spaces. What is wrong with using the “obvious” duality between these two spaces?2012-08-24
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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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    Thank you for your answer. I just can't see, why such a v exists. Can you give me a hint?2012-08-24
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    By definition of the norm on the dual (as a supremum). Please do not make new answers. Your answers should go into comments.2012-08-24
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    This is nice. There is a small gap though; you are choosing $v(t)$ arbitrarily for each $t$, and so it could turn out that $t \mapsto v(t)$ is not measurable, in which case $v$ is not an element of $L^\infty H$.2012-08-25
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    @Nate: Thanks for your comment! Please check if the update makes sense.2012-08-25
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    Yes, looks good now.2012-08-25
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    @ timur: Sorry I was not familiar with that platform. I read some more theory in order to comprehend, what you have written. Did you use Petti's theorem, which assumes a separable "target space" to conclude that indeed $v \to v(t)$ is measurable? I also read that we do not use separability in order to approximate u via simple functions. Last thing concerns your estimate: $T_u v = \int \langle u(t) v(t) \rangle dt = \int \langle u - u_n,v \rangle dt + \int \langle u_n,v\rangle dt $. The problem is that the first term is estimated in the wrong direction, isn't it?2012-08-25
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    @Markus: For the last question, you have $a+b\geq a-|b|$. I don't know separability is really necessary, but the claim (it is actually the definition) you mentioned gives you simple functions converging pointwise only. We need $L^1$ convergence.2012-08-25
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    @timur: the problem is, I can't see that $t \to v(t)$ is indeed measurable uwhen using simple functions2012-08-25
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    Sorry I was not precise, if we choose $u$ to be simple.2012-08-25
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    @Markus: That is fine because when $u$ is simple, $v$ can be chosen to be simple.2012-08-25
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    Many thanks to you!2012-08-26