# Does $L_{p}(\mu, X)^*=L_{q} (\mu, X^*)$ hold for $\sigma$-finite measure spaces?

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 $$\left [ \sum_{m=1}^n \frac{|H_m (f_m)|}{\|f_m\|_{L_{p}(\mu_m, X)}} \right ]^q \le \left [ \frac{|H (f)|}{\|f\|_{L_{p}(\mu, X)}} \right ]^q.$$

• Hint: check that $\|L\|^q = \sum \|L_n\|^q$. Nov 12, 2022 at 17:30
• @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? Nov 13, 2022 at 13:40
• Now show that $\|L\|^q \geq \sum_1^n \|L_k\|^q$ for any $n$. Nov 13, 2022 at 13:57
• 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 $\varphi:\Omega \to (0,\infty)$? Nov 13, 2022 at 14:38
• Yeah, looks fine. Good for you. Nov 20, 2022 at 17:02

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.