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I'm reading Theorem 1 at page 98 of Vector Measures by Joseph Diestel, John Jerry Uhl. Here we use the Bochner integral.

Theorem 1 Let $(\Omega, \Sigma, \mu)$ be a $\sigma$-finite measure space, $1 \leq p<\infty$, and $(X, |\cdot|)$ be a Banach space. Then $L_{p}(\mu, X)^*=L_{q} (\mu, X^*)$ where $p^{-1}+q^{-1}=1$, if and only if $X^{*}$ has the Radon-Nikodým property with respect to $\mu$.

The isometric isomorphism given in the book is $\varphi: L_{p}(\mu, X)^* \to L_{q} (\mu, X^*)$ such that $$ H (f) = \int_\Omega \langle \varphi(H), f \rangle \mathrm d \mu \quad \forall H \in L_{p}(\mu, X)^*, \forall f \in L_{p}(\mu, X). $$

I feel that below result holds, i.e.,

Corollary: Let $N \in \Sigma$ and $K \in L_{p}(\mu, X)^*$ such that $K(f1_N)=0 \quad \forall f \in L_{p}(\mu, X)$. Then $\varphi (K)1_N = 0$, i.e., $\varphi(K)=0$ on $N$.

Could you provide me with some hints? Does it hold in case of other isometric isomorphisms?

Thank you so much for your elaboration.

My attempt: We need to show that $$ \int_N |\varphi(K)|_{X^*} \mathrm d \mu =0. $$

We have $$ K(f1_N) = \int \langle \varphi(K)1_N, f \rangle \mathrm d \mu = 0 \quad \forall f \in L_{p}(\mu, X). $$

It suffices to prove that if $T \in L_{q} (\mu, X^*)$ such that $$ \int \langle T, f \rangle \mathrm d \mu = 0 \quad \forall f \in L_{p}(\mu, X), $$ then $T=0$ $\mu$-a.e.

I posted this question on MSE, but have not received any answer so far. So I post it here.

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It is always true that $L_q(\mu,X^*)\hookrightarrow L_p(\mu,X)^*$ isometrically (without the assumption that $X$ has the Radon-Nikodym property). We need the Radon-Nikodym property to guarantee that this isometric embedding is a surjection.

From this it follows that for any $T\in L_q(\mu,X^*)$, $$\|T\|=\sup\Bigl\{\Bigl|\int \langle T,f\rangle d\mu\Bigr| : f\in B_{L_p(\mu,X)}\Bigr\}.$$ If $\int \langle T,f\rangle d\mu=0$ for all $f\in L_p(\mu,X)$, then it follows that $\|T\|=0$, and $T$ is the zero vector. From this it follows that $T=0$ $\mu$ a.e.

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  • $\begingroup$ My answer gets you to the conclusion that you want, but it doesn't really explain why. To see why the result is true, you will want to look at the proof that the inclusion of $L_q(\mu,X^*)$ into $L_p(\mu,X)^*$ is an isometric embedding. The fact that it's norm at most $1$ follows from H\:{o}lder. To see why it's isometric, I'd recommend looking first at simple functions. $\endgroup$
    – user469053
    Commented Nov 19, 2022 at 17:32
  • $\begingroup$ Thank you so much for your enlightening answer. It's not so complicated but I could not recognize... Can we extend this corollary to all other isometric isomorphisms? $\endgroup$
    – Akira
    Commented Nov 19, 2022 at 22:48
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    $\begingroup$ Are you asking this: If $\varphi:L_p(\mu,X)^*\to L_q(\mu,X^*)$ is an isometric isomorphism, $N\in \Sigma$, and $K\in L_p(\mu,X)^*$ is such that $K(f1_N)=0$ for all $f\in L_p(\mu,X)$, then $\varphi(K)=0$ a.e. on $N$? The answer to this would be no. Consider $\Omega=[0,1]$, $\mu$ Lebesgue measure. Let $\psi:L_p^*\to L_q$ be the usual isometric isomorphism and let $\varphi K (x) = \psi K(1-x)$. Then any function which vanishes on $[0,1/2]$ would satisfy $K(f 1_{[1/2,1]})=0$ for all $f\in L_p$. $\endgroup$
    – user469053
    Commented Nov 20, 2022 at 13:41
  • $\begingroup$ Thank you so much for your response. That's exactly what I'm asking for. Now I understand why the canonical isometric isomorphism is beautiful... $\endgroup$
    – Akira
    Commented Nov 20, 2022 at 13:44

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