Dual of Lp is Lq: Detail for Surjectivity?

Question

I am trying to prove $(L^p(X,\mu))^* = L^q(X,\mu)$ (where $p,q$ are conjugate pairs and $1\le p< \infty$) from first principles, and I am stuck on a detail. Let $\Phi: L^q(X,\mu) \to (L^p(X,\mu))^*$ be the map given by

$\Phi(g)(f) := \int_X fg\ d \mu$

I want to prove that $\Phi$ is surjective. Here is my attempt when $X$ is a compact Hausdorff space. Let $F\in (L^p)^*$. Then by Riesz representation theorem for bounded linear functionals, there exists a measure $\nu$ such that

$\Phi(g)(f) = \int_X f \ d \nu$

I want to now apply the Radon-Nikodym theorem to $\nu$ to obtain $g = \frac{d \nu}{d \mu}$. We know that this $g$ is in $L^q$ since Holder's inequality says $||F||_{(L^p)^*} = ||g||_q$. I have two questions with my attempted proof:

My Question.

1. How (if at all) can we get that $\nu << \mu$ to satisfy the condition for Radon-Nikodym?
2. How can one generalize this when $X$ is an arbitrary measure space? (According to Bass's book, Riesz representation theorem only applies when the space is compact Hausdorff.)

Note: there are many online sources on proof of the above duality theorem. However, I did not find a proof which used the above approach specifically. (If there is such a resource, then that would be good enough of an answer for me.)

Answer 1

The Riesz representation theorem extends to a measure space $(X, \mathcal{M}, \mu)$, indeed the identification for the dual still holds:

• $1<p< \infty$ for a general measure $\mu$ (see Gariepy and Ziemer Theorem 6.43 and Folland Theorem 6.15)
• $p=1$ provided $\mu$ is $\sigma-$finite (see Rudin Theorem 6.16)

Let us recall (and to define notation):

Riesz Representation Theorem: Let $1 \leqslant p<\infty$ and denote $$ p^{\prime}:=\left\{\begin{array}{l} \frac{p}{p-1} \quad \text { if } 1<p<\infty \\ \infty \text { if } p=1 \end{array} \quad(\text { conjugate exponent of } p)\right. $$ Then the mapping $T: L^{p^{\prime}}(\Omega) \rightarrow\left(L^{p}(\Omega)\right)^{\prime}$, defined by $$ \langle T(u), f\rangle_{\left(L^{p}(\Omega)\right)^{\prime} \times L^{p}(\Omega)}:=\int_{\Omega} u f d x \quad \forall f \in L^{p}(\Omega) $$ is an isometric isomorphism and we will mean this representation by means of the identification $$ L^{p^{\prime}}(\Omega) \equiv\left(L^{p}(\Omega)\right)^{\prime} \quad \text { if } 1 \leqslant p<\infty $$

First we prove that $T$ is an isometry.

Then, (what you need, i.e. surjectivity): Suppose first that $|\Omega|<\infty$ (then you can relax it) and let us prove that $T$ is onto, that is, $\forall \phi \in\left(L^{p}(\Omega)\right)^{\prime} \exists u \in L^{p^{\prime}}(\Omega)$ such that $$ T(u)=\phi \quad \Longleftrightarrow \quad\langle T(u), f\rangle_{\left(L^{p}(\Omega)\right)^{\prime} \times L^{p}(\Omega)}=\phi(f) \quad \forall f \in L^{p}(\Omega) $$ Because of $|\Omega|<\infty, \chi_{E} \in L^{p}(\Omega)$ whenever $E \in \mathcal{M}:=\mathcal{M}_{n} \cap \Omega, p \in[1, \infty) .$ Define the set function $\nu: \mathcal{M} \rightarrow \mathbb{R}$ $$ \nu(E):=\phi\left(\chi_{E}\right) \quad E \in \mathcal{M} $$ Show that $\begin{array}{ll}\text { (A) } & \nu \text { is a } \sigma \text { -finite, signed measure; }\end{array}$ (B) $ \nu<<\mathcal{L}^{n} \quad \text { on } \mathcal{M}. $ Indeed $|\nu(E)|<\infty$ for each $E \in \mathcal{M},$ thus $\nu$ is $\sigma$ -finite. Suppose $\left(E_{h}\right)_{h} \subset \mathcal{M}$ is a disjoint sequence, we have to prove that $\nu\left(\cup_{h=1}^{\infty} E_{h}\right)=\sum_{h=1}^{\infty} \nu\left(E_{h}\right) .$ Set $E:=\cup_{h=1}^{\infty} E_{h}$. Then for any positive integer $m$ $$ \begin{aligned} \left|\nu(E)-\sum_{h=1}^{m} \nu\left(E_{h}\right)\right| &=\left|\phi\left(\chi_{E}\right)-\sum_{h=1}^{m} \phi\left(\chi_{E_{h}}\right)\right|=\left|\phi\left(\chi_{E}\right)-\phi\left(\sum_{h=1}^{m} \chi_{E_{h}}\right)\right| \\ &=\left|\phi\left(\chi_{E}-\sum_{h=1}^{m} \chi_{E_{h}}\right)\right|=\left|\phi\left(\sum_{h=m+1}^{\infty} \chi_{E_{h}}\right)\right| \\ & \leqslant\|\phi\|_{\left(L^{p}(\Omega)\right)^{\prime}}\left\|\sum_{h=m+1}^{\infty} \chi_{E_{h}}\right\|_{L^{p}(\Omega)}=\|\phi\|_{\left(L^{p}(\Omega)\right)^{\prime}}\left|\cup_{h=m+1}^{\infty} E_{h}\right|^{\frac{1}{p}} \end{aligned} $$ and $\left|\cup_{h=m+1}^{\infty} E_{h}\right|=\sum_{h=m+1}^{\infty}\left|E_{h}\right| \rightarrow 0$ as $m \rightarrow \infty$ since $|E|<\infty .$ Thus $$ \nu(E)=\sum_{h=1}^{\infty} \nu\left(E_{h}\right) $$ and since the same result holds for any rearrangement of the sequence $\left(E_{h}\right)_{h},$ the series converges asbolutely and (A) follows. Moreover, since $$ |\nu(E)| \leqslant\|\phi\|_{\left(L^{p}(\Omega)\right)^{\prime}}|E|^{\frac{1}{p}} \quad \forall E \in \mathcal{M} $$ (B) follows, too (and then you keep going, proving surjectivity). Warning: Note that if $p=\infty$ in the previous estimate, then (B) need not hold!

I honestly would not use the Riesz representation theorem you mentioned, but the one I stated above.