> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]] and $p, q$ be conjugate exponents with $q \in [1, \infty]$, then this induces a [[Bounded Linear Functional|bounded linear functional]] > $ > L^q \to (L^p)^* \quad g \mapsto \angles{\cdot, g} = \phi_g > $ > by [[Hölder's Inequality]]. > [!theoremb] Theorem > > If $q \in [1, \infty)$ or $q = \infty$ and $\mu$ is [[Semifinite Measure|semifinite]], then the mapping is an isometry. > [!theorem] > > If $g = 0$, then $\norm{\phi_g} = 0$. > [!theorem] > > If $q \in (1, \infty)$ and $g \ne 0$, then > $ > \norm{g}_q = \norm{\phi_g} > $ > *Proof*. If $\beta \in \complex$ with $\alpha \beta = \abs{\beta}^q$, then > $ > \alpha = \beta^{-1}\abs{\beta}^q = \ol{\sgn(\beta)} \cdot \abs{\beta}^{q - 1} > $ > where $\abs{\alpha}^p = \abs{\beta}^q$. Let > $ > f = \ol{\sgn(g)} \cdot \abs{g}^{q - 1} \quad \abs{f}^p = \abs{g}^q > $ > then > $ > \norm{f}_p = \braks{\int\abs{g}^q}^{1/p} = \norm{g}_q^{q/p} = \norm{g}_q^{q - 1} > $ > Let $\wh f = f/\norm{f}_p = f/\norm{g}_q^{q - 1}$, then $||{\wh f}||_p = 1$ and > $ > \angles{\wh f, g} = \frac{1}{\norm{g}_q^{q - 1}}\angles{f, g} = \frac{\norm{g}_q^q}{\norm{g}_q^{q - 1}} = \norm{g}_q > $ > so $\norm{g}_q \le \norm{\phi_g} \le \norm{g}_q$. > [!theorem] > > If $q = 1$ and $g \ne 0$, then $\norm{g}_1 = \norm{\phi_g}$. > > *Proof*. Let $f = \ol{\sgn(g)}$, then $\norm{f}_\infty = 1$ and > $ > \angles{f, g} = \int \abs{g} = \norm{g}_1 > $ > [!theorem] > > If $\mu$ is semifinite, $q = \infty$, and $g \ne 0$, then $\norm{g}_\infty = \norm{\phi_g}$. > > *Proof*. Let $\eps \in (0, \norm{g}_\infty)$ and $E = \bracs{x: \abs{g(x)} \ge \norm{g}_\infty - \eps}$. Then $E$ has positive measure. Let $f = \ol{\sgn{g}}$ such that $fg = \abs{g}$. > > Let $F \subset E$ such that $0 < \mu(F) < \infty$, then $\norm{f \cdot \one_F}_1 = \mu(F)$. Let $\wh f = \one_F \cdot f/\mu(F)$, then $||\wh f||_1 = 1$. In this case, > $ > \begin{align*} > \angles{\wh f, g} &= \frac{1}{\mu(F)}\int_F \abs{g} \\ > &\ge \frac{1}{\mu(F)}\int_F (\norm{g}_\infty - \eps) = \norm{g} - \eps > \end{align*} > $ > as $\eps$ was arbitrary, $\norm{\phi_g} = \norm{g}_\infty$.