# Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$

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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I am curious: what is what? – Włodzimierz Holsztyński May 16 '13 at 18:07
@Wlodzimierz I added some details. These are Bochner spaces. – lollypop May 16 '13 at 18:09
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? – András Bátkai May 16 '13 at 19:09
@Andras you are right. But I am interested in more general $X$, but your answers caters for this. Thanks. – lollypop May 17 '13 at 19:23
@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.) – lollypop May 18 '13 at 14:29

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