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Let $B$ be a refexive Banach space. I want to show that $$(L^2(0,T;B))^* = L^2(0,T;B^*)$$ and that the dual pairing is $$\langle F,f \rangle_{L^2(0,T;B^*), L^2(0,T;B)} = \int_0^T \langle F(t), f(t) \rangle_{B^*,B}.$$ Can anyone help me with either part? Thanks.

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  • $\begingroup$ When following the proof of $(L^2)^*=L^2$ for scalar functions, where do you get stuck? $\endgroup$ – user53153 Feb 20 '13 at 16:38
  • $\begingroup$ @5pm Well in that case I just use Riesz representation theorem. Unless there is another way (maybe echoing proof for $L^p$) I don't know.. $\endgroup$ – soup Feb 23 '13 at 10:50
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No proof in the generality that is desired is going to be very brief. But all the essential ideas come from the usual, scalar case. First we need the following:

Definition. A Banach space $V$ is said to have the Radon-Nikodym property with respect to a measure space $(X, \Sigma, \mu)$ if, for every $V$-valued vector measure of bounded variation $\nu$ which is absolutely continuous with respect to $\mu$, there exists a Bochner integrable $f \in L^1(X,V)$ such that $\nu(E) = \int_E f \, d\mu$ for every $E \in \Sigma$.

In other words, spaces with the Radon-Nikodym property are precisely those in which the generalization of the Radon-Nikodym theorem for scalar functions holds. Then the key result that one needs to prove is due to Phillips:

Reflexive Banach spaces have the Radon-Nikodym property.

A proof of this requires a decent amount of machinery and a proof of this fact and other related facts is the subject of Chapter 2 of Vector Measures by Diestel and Uhl. Now, with this fact under our belt, we have the following result:

Theorem. Let $(X, \Sigma, \mu)$ be a finite measure space, and let $V$ be a Banach space such that $V'$ has the Radon-Nikodym property. Then, for $p \in [1, \infty)$, $L^p(X,V)' \cong L^q(X,V')$ isometrically, where $1/p + 1/q = 1$.

Proof. Suppose $g \in L^q(X,V')$. Then, for $f \in L^p(X,V)$, the map $\Lambda_g(f) = \int_X \langle f, g \rangle \, d\mu$ is clearly a continuous linear functional, so $L^q(X,V') \subseteq L^p(X,V)'$. To prove the reverse inclusion, let $\Lambda \in L^p(X,V)'$. For $h \in L^{\infty}(X,V)$, $|\Lambda(h)| \le C \|h\|_p \le C \|h\|_{\infty}$, and hence $\Lambda$ is also a continuous functional on $L^{\infty}(X,V)$. Now, $\Lambda$ induces a $V'$-valued vector measure absolutely continuous with respect with $\mu$ given by $$ \langle \nu(E), x \rangle = \Lambda(x 1_E)$$ for each $x \in V$. Then, as $V'$ has the Radon-Nikodym property, then there exists $g \in L^1(X,V')$ such that $\nu(E) = \int_E g \, d\mu$. Then, by approximating each $f \in L^p(X,V)$ by simple functions, we have that $\Lambda(f) = \int_X \langle f, g\rangle \, d\mu$. Finally, it is straightforward to show that $g$ is in fact also in $L^q$.

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  • $\begingroup$ Thanks! Does it need to be isometric for us to identify the two together? Do you know of a simple way to show it's isometric when $X$ is a Hilbert space? $\endgroup$ – michael_faber May 20 '13 at 9:36
  • $\begingroup$ Yes, I should clarify that by Banach space equality, I just mean isometrically isomorphic. It seems likely there is a simpler proof for the Hilbert space case, but unfortunately I do not know it. $\endgroup$ – Christopher A. Wong May 20 '13 at 9:51
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    $\begingroup$ For the Hilbert space case it should be possible to prove that $L^2(X,H)$ is a Hilbert space, and then the result follows simply from the Riesz representation theorem. $\endgroup$ – Christopher A. Wong May 20 '13 at 10:13
  • $\begingroup$ Oops, in your notation I meant if both $X=[0,T]$ and $H$ are Hilbert. It's definitely true that $L^2(X,H)$ is Hilbert. I tried doing it in detail but couldn't show it. Let me see again. $\endgroup$ – michael_faber May 20 '13 at 10:29
  • $\begingroup$ If $X$ is a Hilbert space, what measure are you using? $\endgroup$ – Christopher A. Wong May 20 '13 at 10:31

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