> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], $p, q$ be [[Hölder's Inequality|Hölder conjugates]], and define $\Sigma$ as the subspace of [[Simple Function|simple functions]] that vanish outside of a set of finite measure. If $\phi$ is a [[Bounded Linear Functional|bounded linear functional]] on $\Sigma$, and there exists $g: X \to \complex$ [[Measurable Function|measurable]] such that $\abs{fg} \in L^1$ for all $f \in \Sigma$ and > $ > \angles{f, \phi} = \angles{f, g} \quad \forall f \in \Sigma > $ > then $\phi$ is a **functional on simple functions with representation** $g$. > > By the [[Linear Extension Theorem]], $\phi$ and $\angles{\cdot, g}$ admit the same extension to $L^p$, so $\phi$ is a **functional on $L^p$ with representation $g$**. > [!theorem] > > Let $\phi$ be a functional on simple functions with representation $g$, then $\phi$ uniquely extends to $L^p(X, \cm, \mu)$ with $\norm{\phi} = \norm{\phi|_\Sigma}$. Moreover, $\abs{fg} \in L^1$ for all $f \in L^p$ with > $ > \angles{f, \phi} = \angles{f, g} \quad \forall f \in L^p > $ > *Proof*. Since the simple functions that vanish outside of a set of finite measure are [[Dense|dense]] in $L^p$, by the [[Linear Extension Theorem]], $\phi$ uniquely extends to $L^p$. If $\seq{\psi_n}$ is a family of simple functions that vanish outside of sets of finite measure, with $\abs{\psi_n} \le \abs{f}$ and $\phi_n \to \phi_f$ pointwise, then by the [[Dominated Convergence Theorem]], > $ > \angles{f, \phi} = \limv{n}\angles{\psi_n, \phi} = \limv{n}\angles{\psi_n, g} = \angles{f, g} > $ > [!theorem] > > If $\mu$ is [[Semifinite Measure|semifinite]], $q < \infty$ ($p > 1$), then for any $\eps > 0$, the set $E = \bracs{x \in X: \abs{g(x)} > \eps}$ has finite measure. Thus $\bracs{x \in X: g(x) \ne 0}$ is $\sigma$-finite. > > *Proof*. Let $F \subset E$ with $0 < \mu(F) < \infty$, and > $ > f = \one_E \cdot \frac{\sgn{(g)}}{\mu(F)^{1/p}} \quad \norm{f}_p = 1 > $ > then > $ > \begin{align*} > \norm{\phi} \ge \abs{\phi(f)} &\ge \int fg = \frac{\abs{g} \cdot \one_{F}}{\mu(F)^{1/p}} = \frac{1}{\mu(F)^{1/p}}\int_F \abs{g} \\ > &\ge \frac{\eps \mu(F)}{\mu(F)^{1/p}} = \eps\mu(F)^{1/q} > \end{align*} > $ > Since this uniformly bounds the subsets of finite measure, $\mu(E)^{1/q}$ is bounded by $\norm{\phi}/\eps$ as well. > [!theorem] > > If $\bracs{x \in X: g(x) \ne 0}$ is $\sigma$-finite and $q < \infty$, then $g \in L^q$ with $\norm{g}_q = \norm{\phi_g}$. > > *Proof*. Let $\seq{E_n}$ be a sequence of measurable sets such that $\mu(E_n) < \infty$ for all $n \in \nat$ and $E_n \upto \bracs{x \in X: g(x) \ne 0}$. Let $g_n = g \cdot \one_{E_n}$, then $g_n \upto g$ pointwise, $\abs{g_n} \le \abs{g}$, and $g_n$ vanishes outside of $E_n$. > > Let > $ > f_n = \frac{\abs{g_n}^{q - 1} \cdot \ol{\sgn(g)}}{\norm{g_n}_q^{q - 1}} \quad f_n g_n = \frac{\abs{g_n}^{q}}{\norm{g_n}_q^{q - 1}} \quad \norm{f_n}_p = 1 > $ > then > $ > \angles{f_n, g_n} = \frac{\norm{g_n}^q}{\norm{g_n}_q^{q - 1}} = \norm{g_n}_q > $ > So by [[Fatou's Lemma]], > $ > \begin{align*} > \norm{g}_q^q &= \int \liminf_{n \to \infty} \abs{g_n}^q \le \liminf_{n \to \infty}\int \abs{g_n}^q \\ > \norm{g}_q &\le \liminf_{n \to \infty}\norm{g_n}_q = \liminf_{n \to \infty}\angles{f_n, g_n} \\ > &\le \liminf_{n \to \infty}\angles{f_n, g} \le \norm{\phi_g} > \end{align*} > $ > On the other hand, $\norm{\phi_g} \le \norm{g}_q$, so the $\norm{\phi_g} = \norm{g}_q$. > [!theorem] > > If $\bracs{x \in X: g(x) \ne 0}$ is $\sigma$-finite, $\mu$ is semi-finite, and $q = \infty$, then $g \in L^\infty$ with $\norm{g}_q = \norm{\phi_g}$. > > *Proof*. Let $\eps > 0$ and $E = \bracs{x \in X: \abs{g(x)} \ge \norm{\phi_g} + \eps}$. If $\mu(E) = 0$, then $\norm{g}_\infty \le \norm{\phi_g}$. > > Suppose that $\mu(E) > 0$, and let $F \subset E$ with $0 < \mu(F) < \infty$. Let > $ > f = \frac{\one_F \cdot \ol{\sgn(g)}}{\mu(F)} \quad \norm{f}_p = 1 > $ > so > $ > \begin{align*} > \angles{f, g} = \frac{1}{\mu(F)}\int_F g \cdot \ol{\sgn(g)} \ge \norm{\phi_g} + \eps > \end{align*} > $ > which contradicts the fact that $\angles{f, g} \le \norm{\phi_g}$.