> [!theoremb] Theorem > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], and $p, q$ be [[Hölder's Inequality|Hölder conjugates]]. If $p \in (1, \infty)$ or $\mu$ is $\sigma$-finite and $p = \infty$, then for any [[Bounded Linear Functional|bounded linear functional]] $\phi \in (L^p)^*$ there exists $g \in L^q$ such that > $ > \angles{f, \phi} = \angles{f, g} \quad \forall f \in L^p > $ > thus the [[Isometry of Lq Representation|isometry]] $L^p \to (L^q)^*$ is an isomorphism. # Finite > [!theorem] > > If $\mu$ is finite and $\phi \in (L^p)^*$, then there exists a [[Complex Measure|complex measure]] $\nu$ on $(X, \cm)$ such that > $ > \angles{f, \nu} = \angles{f, \phi} \quad \forall f \in L^p > $ > and $\nu \ll \mu$. > > *Proof*. Let $\seq{E_n} \subset \cm$ be a disjoint family of sets, $E = \bigsqcup_{n \in \nat}E_n$, and $\sigma: \nat \to \nat$ be a bijection, then $\one_E = \sum_{n = 1}^\infty \one_{E_{\sigma(n)}}$. Moreover, > $ > \begin{align*} > \norm{\one_E - \sum_{k = 1}^\infty \one_{E_{\sigma(k)}}}_p &= \norm{\sum_{k = n + 1}^\infty \one_{E_{\sigma(k)}}}_p \\ > &\le \mu\paren{\bigsqcup_{k > n}E_{\sigma(k)}}^{1/p} \\ > &= \braks{\sum_{k > n}\mu(E_{\sigma(k)})}^{1/p} \to 0 > \end{align*} > $ > as $n \to \infty$, so the series converges unconditionally. Define $\nu(E) = \phi(\one_{E})$, then by the continuity of $\phi$, > $ > \begin{align*} > \nu(E) = \angles{\one_E, \phi} &= \limv{n}\angles{\sum_{k = 1}^n\one_{E_{\sigma(k)}}, \phi} \\ > &= \angles{\limv{n}\sum_{k = 1}^n\one_{E_{\sigma(k)}}, \phi} \\ > &= \sum_{k = 1}^\infty \angles{\one_{E_{\sigma(k)}}, \phi} = \sum_{k = 1}^\infty \nu(E_{\sigma(k)}) > \end{align*} > $ > as the sum converges unconditionally, it converges absolutely. Thus $\nu$ is a complex measure. If $E$ is $\mu$-null, then $\nu(E) = \phi(\one_E) = \phi(0) = 0$, so $\nu \ll \mu$. > > Since $\nu$ and $\phi$ agree on indicator functions and simple functions, and [[Simple Function|simple functions]] are [[Dense|dense]] in [$L^p$](Lp%20Space), they are equal by the [[Linear Extension Theorem|linear extension theorem]]. > [!theorem] > > If $\mu$ is finite and $\phi \in (L^p)^*$, then there exists $g \in L^q$ such that > $ > \angles{f, g} = \angles{f, \phi} \quad \forall f \in L^p > $ > *Proof*. Let $\nu$ be the complex measure corresponding to $\phi$, then by the [[Lebesgue-Radon-Nikodym Theorem]], there exists $g = \frac{d\nu}{d\mu}$ such that $\angles{f, g} = \angles{f, \phi}$ for all $f \in \Sigma$. From here, since $\phi$ is a [[Functional on Lp with Representation|functional on simple functions with representation]] $g$, $g \in L^q$ with $\norm{g}_q = \norm{\phi}$. > > By the linear extension theorem, $\angles{f, g} = \angles{f, \phi}$ for all $f \in L^p$. # $\sigma$-Finite > [!theorem] > > If $\mu$ is $\sigma$-finite and $\phi \in (L^p)^*$, then there exists $g \in L^q$ such that > $ > \angles{f, g} = \angles{f, \phi} \quad \forall f \in L^p > $ > *Proof*. Let $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \nat$. Identify $L^p(E_n) \subset L^p(X)$ and $L^q(E_n) \subset L^q(X)$ as functions in $L^p$/$L^q$ that vanish outside of $E_n$. > > For each $E_n$, let $g_n \in L^q(E_n)$ such that $\angles{f, g_n} = \angles{f, \phi}$ for all $f \in L^p(E_n)$. > > Let $x \in X$, then there exists $n \in \nat$ such that $x \in E_n$. Define $g(x) = g_n(x)$. The uniqueness of the Radon-Nikodym derivative means that this function is [[Almost Everywhere|a.e.]] well-defined, where $g|_{E_n} = g_n$ a.e. > > By the the [[Monotone Convergence Theorem for Sequences|monotone convergence theorem]], $\norm{g_n}_q \upto \norm{g}_q$. As $\norm{g_n}_q \le \norm{\phi}$ for all $n \in \nat$, $\norm{g}_q < \infty$ with $g \in L^q$. > > Let $f \in L^p$, then by continuity and the [[Dominated Convergence Theorem]], > $ > \angles{f, \phi} = \limv{n}\angles{f \cdot \one_{E_n}, \phi} = \limv{n}\angles{f \cdot \one_{E_n}, g} = \angles{f, g} > $ # Arbitrary > [!theorem] > > Let $p > 1$ and $\mu$ be arbitrary, then for any $\phi \in (L^p)^*$, there exists $g \in L^q$ such that > $ > \angles{f, g} = \angles{f, \phi} \quad \forall f \in L^p > $ > *Proof*. Let $E$ be a $\sigma$-finite set, then there exists $g_E \in L^q(E)$ such that $\angles{f, g_E} = \angles{f, \phi}$ for all $f \in L^p(E)$. By uniqueness of Radon-Nikodym, if $F \supset E$, then $g_F|_E = g_E$ [[Almost Everywhere|a.e.]] > > Let $M = \sup_{E} \norm{g_E}_q$ over all $\sigma$-finite sets, and let $\seq{E_n}$ be a sequence of $\sigma$-finite sets such that $\norm{g_{E_n}} \upto M$ and $E_n \upto E = \bigcup_{n \in \nat}E_N$, then $M = \norm{g_E}_q \ge \norm{g_{E_n}}_q$ for each $n \in \nat$. Let $A \supset E$ be a $\sigma$-finite set, then > $ > \norm{g_A}_q^q = \norm{g_E}_q^q + \norm{g_{A\setminus E}}_q^q \le M^q = \norm{g_E}^q_q > $ > so $\norm{g_{A \setminus E}}_q = 0$, and $g_A = g_E$ [[Almost Everywhere|a.e.]] > > Let $f \in L^p$, and $A = E \cup \supp{f}$, then $A$ is $\sigma$-finite and > $ > \angles{f, \phi} = \angles{f, g_A} = \angles{f, g_E} > $ > so $g_E$ is the desired representation.