> [!theorem] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]] and $f$ be a [[Measurable Function|measurable function]]. Then > $ > \norm{f}_\infty = \inf\bracs{a \ge 0: \mu(\bracs{x \in X: \abs{f(x)} > a}) = 0} > $ > is the **essential supremum** of $f$, where the [[Supremum and Infimum|infimum]] is taken to be $\infty$ if the set is empty. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space, then define > $ > L^{\infty}(X, \cm, \mu) = \bracs{f: X \to \complex: \norm{f}_\infty < \infty} > $ > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space. > - [[Hölder's Inequality]]: If $f, g$ are measurable functions on $X$, then $\norm{fg}_1 \le \norm{f}_1 \cdot \norm{g}_\infty$. > - $\norm{ \cdot }_{\infty}$ is a [[Normed Vector Space|norm]] on $L^{\infty}$. > - $L^{\infty}$ is a [[Banach Space|Banach space]]. > - The [[Simple Function|simple functions]] are [[Dense|dense]] in $L^{\infty}$. > > *Proof*. > > ### Hölder's Inequality > > If $f, g$ are measurable functions on $X$, then $\norm{fg}_1 \le \norm{f}_1 \cdot \norm{g}_\infty$. > > *Proof*. > $ > \begin{align*} > \norm{fg}_1 &= \int \abs{fg} \\ > &= \int \abs{f} \cdot \abs{g} \\ > &\le \int \abs{f} \cdot \norm{g}_\infty \\ > &= \norm{g}_\infty \int \abs{f} = \norm{g}_\infty \norm{f}_1 > \end{align*} > $ > > ### Norm > > $\norm{ \cdot }_{\infty}$ is a [[Normed Vector Space|norm]] on $L^{\infty}$. > > *Proof*. $\norm{f}_{\infty} = 0$ if and only if $f = 0$ [[Almost Everywhere|a.e.]], > $ > \begin{align*} > \norm{\lambda f}_\infty &= \norm{f}_\infty = \inf\bracs{a \ge 0: \mu(\bracs{x \in X: \abs{\lambda f(x)} > a}) = 0} \\ > &= \inf\bracs{a \ge 0: \mu(\bracs{x \in X: \abs{\lambda}\abs{ f(x)} > a}) = 0} \\ > &= \inf\bracs{\abs{\lambda}a \ge 0: \mu(\bracs{x \in X: \abs{\lambda}\abs{ f(x)} > \abs{\lambda}a}) = 0} \\ > &= \abs{\lambda}\inf\bracs{a \ge 0: \mu(\bracs{x \in X: \abs{ f(x)} > a}) = 0} \\ > &= \abs{\lambda} \cdot \norm{f}_\infty > \end{align*} > $ > Lastly, since $\abs{f + g} \le \norm{f}_\infty + \norm{g}_\infty$ a.e., $\norm{f + g}_\infty \le \norm{f}_\infty + \norm{g}_\infty$. > > ### Completeness > > Let $\seq{f_n} \subset L^{\infty}$ be a [[Sequence|sequence]] of functions such that $\sum_{n \in \nat}\norm{f_n}_\infty < \infty$, then $\sum_{n = 1}^{\infty}f_n \in L^{\infty}$. > > *Proof*. Since > $ > \abs{\sum_{n = 1}^{\infty}f_n(x)} \le \sum_{n = 1}^{\infty}\abs{f_n(x)} \le \sum_{n = 1}^{\infty}\norm{f}_\infty \quad \text{a.e.} > $ > we have $\norm{\sum_{n = 1}^{\infty}f_n}_{\infty} < \infty$. > > > ### Simple Functions > > The simple functions are dense in $L^{\infty}$. > > *Proof*. Let $f \in L^\infty$, then $f$ is bounded by a constant a.e., therefore there exists a sequence of simple functions converging to $f$ a.e. uniformly.