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How is the proof that $$[L^2(0,T;X)]' = L^2(0,T;X')$$ looking like, where $X$ is a Hilbert space? I am asking for the proof that the dual space of $L^2(0,T;X)$ is the space $L^2(0,T;X^*)$.

Is the proof much different to the $L^p(a,b)$ case?

I can't find any easy to understand proofs.

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  • $\begingroup$ I am curious: what is what? $\endgroup$ May 16, 2013 at 18:07
  • $\begingroup$ @Wlodzimierz I added some details. These are Bochner spaces. $\endgroup$
    – lollypop
    May 16, 2013 at 18:09
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    $\begingroup$ I am a bit confused. If $X$ is a Hilbert space, then $L^2(0,T;X)$ is a Hilbert space (complete + norm comes from a scalar product). Hence if you also identify $X$ with its dual (as you do with $L^2$), then the statement follows. Maybe this is not what you ask? $\endgroup$ May 16, 2013 at 19:09
  • $\begingroup$ @Andras you are right. But I am interested in more general $X$, but your answers caters for this. Thanks. $\endgroup$
    – lollypop
    May 17, 2013 at 19:23
  • $\begingroup$ @AndrasBatkai Is it as simple as: $L^2(0,T;X)' = L^2(0,T;X) = L^2(0,T;X')$? Where the first equality is by Riesz representation theorem (RRT) for the Bochner space (which is Hilbert), and the second is by RRT for $X$ which is also Hilbert. Would this be a good proof?? (The longer way to do this requires us to show that the map is isometric, which seems difficult to do even in this case.) $\endgroup$
    – lollypop
    May 18, 2013 at 14:29

1 Answer 1

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To give you a reference: Diestel-Uhl, Vector measures, page 98, Chapter 4, Theorem 1:

$$L^p(\mu,X)^\ast = L^q(\mu,X^\ast)$$

if and only if $X^\ast$ has the Radon-Nikodym property with respect to $\mu$.

Here $\mu$ is a finite measure and $p$ and $q$ as usual. Proof can be read there.

ADDED: It should be noted that $L^q(\mu,X^\ast) \subset L^p(\mu,X)^\ast $ always holds, without any condition on the Banach space $X$. The proof of this is quite the same as in the scalar case.

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    $\begingroup$ You should also mention that reflexive Banach spaces (hence, Hilbert spaces) possesses the Radon-Nikodym property. $\endgroup$
    – gerw
    May 17, 2013 at 19:03

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