I'm reading Theorem 1 at page 98 of Vector Measures by Joseph Diestel, John Jerry Uhl.

THEOREM 1. Let $(\Omega, \Sigma, \mu)$ be a finite measure space, $1 \leq p<\infty$, and $X$ 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 $\mu$.

I would like to extend it to $\sigma$-finite measure space. However, I'm stuck at proving that $\varphi$ is an isometry. Could you elaborate on how to prove it?

Thank you so much!

My attempt: Let $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space. There is a countable measurable partition $(\Omega_n)$ of $\Omega$ such that $\mu(\Omega_n) < \infty$. Let $\mu_n(A) := \mu(A \cap \Omega_n)$ for all $n$. Then

  • $\mu_n$ is concentrated on $\Omega_n$,
  • $(\Omega, \Sigma, \mu_n)$ is a finite measure space, and
  • $\mu = \sum \mu_n$.

By Theorem 1, for each $n$ there is an isometric isomorphism $$ \varphi_n : L_{p}(\mu_n, X)^* \to L_{q} (\mu_n, X^*). $$

For $H \in L_{p}(\mu, X)^*$, we define $H_n \in L_{p}(\mu_n, X)^*$ by $$ H_n (f) := H (f 1_{\Omega_n}) \quad \forall f \in L_{p}(\mu_n, X). $$

Notice that $\varphi_n (H_n)$ is just an equivalence class of $L_{q} (\mu_n, X^*)$. If $g$ is a representative of $\varphi_n (H_n)$, then $g$ can take any value of $X^*$ on $\Omega \setminus \Omega_n$ and thus $\|g\|_{L_{q} (\mu_m, X^*)}$ can be $+\infty$ for some $m \neq n$. To avoid this situation, we define $$ \varphi : L_{p}(\mu, X)^* \to L_{q} (\mu, X^*), H \mapsto \sum_n \varphi_n (H_n) 1_{\Omega_n}. $$

It's straightforward to verify $\varphi$ is an isomorphism. Let's prove that it is an isometry. We have $$ \begin{align} \| \varphi (H) \|_{ L_{q} (\mu, X^*)}^q &= \int \bigg \| \sum_n \varphi_n (H_n) 1_{\Omega_n} \bigg \|_{X^{*}}^q \mathrm d \mu \\ &= \sum_m \int \bigg \| \sum_n \varphi_n (H_n) 1_{\Omega_n} \bigg \|_{X^{*}}^q \mathrm d \mu_m \\ &= \sum_m \int \big \| \varphi_m (H_m) \big \|_{X^{*}}^q \mathrm d \mu_m \\ &= \sum_m \| \varphi_m (H_m) \|_{ L_{q} (\mu_m, X^*)}^q \\ &= \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*}. \end{align} $$

So it suffices to prove that $$ \|H\|^q_{L_{p}(\mu, X)^*} = \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*}. $$

By Hölder's inequality, $$ \begin{align} \left [ \frac{|H(f)|}{\|f\|_{L_{p}(\mu, X)}} \right ]^q = \frac{\big |\sum_m H_m(f) \big |^q}{\left [\sum_m \|f\|^p_{L_{p}(\mu_m, X)} \right]^{q/p}} \le \sum_m \left [ \frac{|H_m (f)|}{\|f\|_{L_{p}(\mu_m, X)}} \right ]^q \quad \forall f \in L_{p}(\mu, X). \end{align} $$

As such, $$ \|H\|^q_{L_{p}(\mu, X)^*} \le \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*}. $$

Fix $\varepsilon>0$. Pick $f_m \in L_{p}(\mu_m, X)$ such that $$ \left [ \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q > \|H_m\|^q_{L_{p}(\mu_m, X)^*} - \varepsilon 2^{-m}. $$

WLOG, we can assume $$ H_m (f_m) \ge 0 \quad \forall m \in \mathbb N^*. $$

Then $$ \sum_m \left [ \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q > \sum_m \|H_m\|^q_{L_{p}(\mu_m, X)^*} - \varepsilon. $$

It remains to prove that $$ \sum_{m=1}^n \left [ \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q \le \|H\|^q_{L_{p}(\mu, X)^*}. $$

It suffices to prove that there is $f \in L_{p}(\mu, X)$ such that $$ \left [ \sum_{m=1}^n \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q \le \left [ \frac{|H (f)|}{\|f\|_{L_{p}(\mu, X)}} \right ]^q. $$

  • 3
    $\begingroup$ Hint: check that $\|L\|^q = \sum \|L_n\|^q$. $\endgroup$
    – Nik Weaver
    Nov 12, 2022 at 17:30
  • $\begingroup$ @NikWeaver I'm able to prove that $\|L\|^q \le \sum \|L_n\|^q$. However, I failed to prove the reverse inequality. Could you have a check on my update? $\endgroup$
    – Akira
    Nov 13, 2022 at 13:40
  • 1
    $\begingroup$ Now show that $\|L\|^q \geq \sum_1^n \|L_k\|^q$ for any $n$. $\endgroup$
    – Nik Weaver
    Nov 13, 2022 at 13:57
  • 1
    $\begingroup$ Did you try the simple fact that every $\sigma$-finite measure $\mu$ is of the form $\mu= \varphi\cdot \nu$ for a finite measure $\nu$ and a density $\varphi:\Omega \to (0,\infty)$? $\endgroup$ Nov 13, 2022 at 14:38
  • 1
    $\begingroup$ Yeah, looks fine. Good for you. $\endgroup$
    – Nik Weaver
    Nov 20, 2022 at 17:02

1 Answer 1


Below is my formalization of @Nik's hints to finish the proof.

Let's prove that $$ \sum_{m=1}^M \|H_m\|^q_{L_{p}(\mu_m, X)^*} \le \|H\|^q_{L_{p}(\mu, X)^*} \quad \forall M \in \mathbb N^*. $$

Let $\Omega' := \bigcup_{m=1}^M \Omega_m$. We define a measure $\mu'$ on $\Omega$ by $$ \mu' (B) := \mu(B \cap \Omega') \quad \forall B \in \Sigma. $$

Let $\varphi' : L_{p}(\mu', X)^* \to L_{q} (\mu', X^*)$ be the canonical isometric isomorphism, i.e., $$ K (f) = \int_\Omega \langle \varphi'(K), f \rangle \mathrm d \mu' \quad \forall K \in L_{p}(\mu', X)^*, \forall f \in L_{p}(\mu', X). $$

Then we have

Lemma Let $N \in \Sigma$ and $K \in L_{p}(\mu', X)^*$ such that $K(f1_N)=0$ for all $f \in L_{p}(\mu', X)$. Then $\varphi'(K)=0$ on $N$.

We define $H' ,H'_m \in L_{p}(\mu', X)^*$ by $$ H' (f) := H (f 1_{\Omega'}) \quad H'_m (f) := H (f 1_{\Omega_m}) \quad \forall f \in L_{p}(\mu', X). $$

Then $H' = \sum_{m=1}^M H'_m$. Also, $$ \| H'\|_{L_{p}(\mu', X)^*} \le \|H\|_{L_{p}(\mu, X)^*} \quad \text{and} \quad \|H'_m\|_{L_{p}(\mu', X)^*} = \|H_m\|_{L_{p}(\mu_m, X)^*}. $$

By our Lemma, the supports of $\{\varphi'(H'_m)\}_{m=1}^M$ are pairwise disjoint and thus justifies $(\star)$ below. We have $$ \begin{align} \sum_{m=1}^M \|H_m\|^q_{L_{p}(\mu_m, X)^*} &= \sum_{m=1}^M \|H'_m\|^q_{L_{p}(\mu', X)^*} \\ &= \sum_{m=1}^M \|\varphi'(H'_m) \|^q_{L_{q}(\mu', X^*)} \\ &\overset{(\star)}{=} \bigg \| \sum_{m=1}^M \varphi' (H'_m) \bigg \|^q_{L_{q}(\mu', X^*)} \\ &= \bigg \| \varphi'\bigg (\sum_{m=1}^MH'_m \bigg) \bigg \|^q_{L_{q}(\mu', X^*)} \\ &= \bigg \|\sum_{m=1}^MH'_m \bigg\|^q_{L_{p}(\mu', X)^*} \\ &= \| H'\|^q_{L_{p}(\mu', X)^*} \\ &\le \|H\|^q_{L_{p}(\mu, X)^*}. \end{align} $$

This completes the proof.


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