> [!theorem] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], $p \in [1, \infty)$, and $f, g \in L^p$, then > $ > \norm{f + g}_p \le \norm{f}_p + \norm{g}_p > $ > the [$L^p$](Lp%20Space) norm satisfies the [[Triangle Inequality|triangle inequality]]. > > *Proof*. If $p = 1$ or $f + g = 0$ [[Almost Everywhere|a.e.]], then the inequality holds directly. Suppose that $f + g \ne 0$ a.e., then > $ > \abs{f + g}^p \le (\abs{f} + \abs{g})\abs{f + g}^{p - 1} > $ > Since the conjugate exponent of $p$ is $p/(p - 1)$, by [[Hölder's Inequality]], > $ > \begin{align*} > \int \abs{f + g}^p &\le \int \abs{f} \cdot \abs{f + g}^{p - 1} + \int \abs{g} \cdot \abs{f + g}^{p - 1} \\ > &\le \norm{f}_p \cdot \norm{\abs{f + g}^{p - 1}}_q + \norm{g}_p \cdot \norm{\abs{f + g}^{p - 1}}_q \\ > &= \paren{\norm{f}_p + \norm{g}_p} \cdot \braks{\int \abs{f + g}^{q(p - 1)}}^{1/q}\\ > &= \paren{\norm{f}_p + \norm{g}_p} \cdot \braks{\int \abs{f + g}^{p}}^{1/q} > \end{align*} > $ > where dividing both sides by the integral on the right yields > $ > \begin{align*} > \int \abs{f + g}^p &\le \paren{\norm{f}_p + \norm{g}_p} \cdot \braks{\int \abs{f + g}^{p}}^{1/q} \\ > \braks{\int \abs{f + g}^p}^{1 - 1/q}&\le \norm{f}_p + \norm{g}_p\\ > \braks{\int \abs{f + g}^p}^{1/p} &\le \norm{f}_p + \norm{g}_p \\ > \norm{f + g}_p &\le \norm{f}_p + \norm{g}_p > \end{align*} > $ > [!theorem] > > Let $(X, \cm, \mu)$ and $(Y, \cn, \mu)$ be $\sigma$-finite measure spaces. Let $f$ be a $(\cm \otimes \cn)$-[[Measurable Function|measurable function]] on $X \times Y$. Firstly, if $f \ge 0$ and $p \in [1, \infty)$, then > $ > \braks{\int \paren{\int f(x, y)d\nu(y)}^pd\mu(x)}^{1/p} \le \int\norm{f(\cdot, y)}_p d\nu(y) > $ > Now, if the following conditions are satisfied: > 1. $p \in [0, \infty]$. > 2. $f(\cdot, y) \in L^p(\mu)$ $\nu$-a.e. > 3. $y \mapsto \norm{f(\cdot, y)}_p$ is in $L^1(\nu)$. > > then > > 1. $f(x, \cdot) \in L^1(\nu)$ $\mu$-a.e. > 2. $x \mapsto \int f(x, y)d\nu(y)$ is in $L^p(\mu)$. > > and > $ > \norm{\int f(\cdot, y)d\nu(y)}_p \le \int \norm{f(\cdot, y)}_p d\nu(y) > $ > > *Proof*. If $p = 1$, then [[Fubini-Tonelli Theorem|Tonelli's theorem]] yields the desired result. Now suppose that $p \in (1, \infty)$. Let $q$ be its Hölder conjugate, and $g \in L^q(\mu) \cap L^+$. By [[Hölder's Inequality]], > $ > \begin{align*} > \int \braks{\int f(x, y)d\nu(y)} \cdot g(x)d\mu(x) &= \int \int f(x, y)g(x)d\mu(x)d\nu(y) \\ > &\le \norm{g}_q \int \norm{f(\cdot, y)}_p d\nu(y) > \end{align*} > $ > Since the integral acts as a [[Bounded Linear Map|bounded linear map]], from the [[Functional on Lp with Representation]] we know that the mapping $x \mapsto \braks{\int f(x, y)d\nu(y)}$ is in $L^p$, with $L^p$ norm less than the integral on the right.