> [!theorem] Lemma > > Let $a, b \ge 0$ and $\lambda \in (0, 1)$, then > $ > a^\lambda b^{1 - \lambda} \le \lambda a + (1 - \lambda) b > $ > with equality if and only if $a = b$. > > *Proof*. If $b = 0$, then the inequality holds directly. Otherwise, divide both sides by $b$, and set $t = \frac{a}{b}$, then the problem becomes > $ > t^\lambda \le \lambda t + (1 - \lambda) > $ > Since $t^\lambda - \lambda t$ is strictly increasing for $t > 1$ and decreasing for $t < 1$ (by checking the [[Derivative|derivative]]), its maximum value $1 - \lambda$ occurs at $t = 1$, in which case $t^\lambda - \lambda t \le (1 - \lambda)$ for all $t \in \real$, and $\frac{a}{b}$, with equality when $a/b = 1$. > [!theoremb] Theorem > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], $p, q \in (1, \infty)$ such that $\frac{1}{p} + \frac{1}{q} = 1$ ($q = p/(p - 1)$), and $f, g$ be [[Measurable Function|measurable functions]] on $X$, then > $ > \norm{fg}_1 \le \norm{f}_p\norm{g}_q > $ > the $L^1$ norm of the product is bounded by the product of conjugate [[Lp Space|Lp norms]]. > > Moreover, if $f \in L^p$ and $g \in L^q$, then $fg \in L^1$, and the equality holds if and only if there exists $(\alpha, \beta) \ne (0, 0)$ such that $\alpha \abs{f}^p = \beta \abs{g}^q$ [[Almost Everywhere|a.e.]] > > ### Edge Cases > > If $\norm{f}_p \in \bracs{0, \infty}$ or $\norm{g}_q \in \bracs{0, \infty}$, then the inequality holds. > > *Proof*. If $\norm{f}_p = 0$ or $\norm{g}_q = 0$, then > $ > 0 = \norm{0}_1 = \norm{fg}_1 = \norm{f}_p \norm{g}_q = 0 > $ > If $\norm{f}_p = \infty$ or $\norm{g}_q = \infty$ with $\norm{f}_p, \norm{g}_q \ne 0$, then $\norm{f}_p\norm{g}_q = \infty$ and the inequality holds no matter the value of $\norm{fg}_1$. > > > ### Norm Reduction > > If $\norm{fg}_1 \le \norm{f}_p\norm{g}_q$, then $\norm{\alpha \beta fg}_1 \le \norm{\alpha f}_p\norm{\beta g}_q$ for all $\alpha, \beta \in \complex$. > > If $\norm{fg}_1 = \norm{f}_p\norm{g}_q$, then $\norm{\alpha \beta fg}_1 \le \norm{\alpha f}_p\norm{\beta g}_q$. > > This means that it's sufficient to prove the in/equality for functions of norm $1$. > > *Proof*. Since $\norm{\alpha\beta fg}_1 = \abs{\alpha} \abs{\beta} \norm{fg}_1$ and > $ > \norm{\alpha f}_p \norm{\beta g}_1 = \abs{\alpha}\abs{\beta}\norm{f}_p\norm{f_q} > $ > Multiplying the functions also multiply both sides of the in/equality. > > > ### Inequality > > Let $f \in L^p$ and $g \in L^q$ such that $\norm{f}_p = \norm{g}_q = 1$, then $\norm{fg}_1 \le \norm{f}_p \norm{f}_q$. > > *Proof*. Apply the lemma with $a = \abs{f(x)}^p$, $b = \abs{g(x)}^q$, and $\lambda = 1/p$, then > $ > \begin{align*} > a^\lambda b^{1 - \lambda} &\le \lambda a + (1 - \lambda) b \\ > \abs{f(x)} \cdot \abs{g(x)} &\le \frac{\abs{f(x)}^p}{p} + \frac{\abs{g(x)}^q}{q} \\ > \int\abs{fg} &\le \frac{1}{p}\int\abs{f}^p + \frac{1}{q}\int\abs{g}^q \\ > \norm{fg}_1 &\le \frac{1}{p}\norm{f}_p^p + \frac{1}{q}\norm{f}_q^q \\ > &= 1 = \norm{f}_p\norm{g}_q > \end{align*} > $ > > ### Equality > > Let $f \in L^p, g \in L^q$. $\norm{fg}_1 = \norm{f}_p\norm{g}_q$ is and only if $\alpha \abs{f}^p = \beta\abs{g}^q$ a.e. for some $(\alpha, \beta) \ne (0, 0)$. > > *Proof*. If $f$ or $g$ is equal to $0$ a.e., then the equality holds directly. If the equality holds with $\norm{f}_p = \norm{g}_q = 0$, then both functions are $0$ a.e. > > Suppose that $\norm{f}_p = \norm{g}_q = 1$ and $\alpha \abs{f}^p = \beta\abs{g}^q$ a.e., then we can assume that $\abs{f}^p = C\abs{g}^q$ for some $C = \alpha/\beta$. In which case, > $ > \norm{f}_p^p = \int\abs{f}^p = C\int\abs{g}^q = C\norm{g}_q > $ > we have $C = 1$, meaning that > $ > \abs{fg} = \abs{f^{p/p}g^{q/q}} = \abs{f^{p/p}f^{p/q}} = \abs{f^{p \cdot(1/p + 1/q)}} = \abs{f^p} > $ > and $\int\abs{fg}_1 = \int\abs{f^p} = \int \abs{f}^p = 1$, giving us the equality. > > If $\int\abs{fg}_1 = 1$, then > $ > \abs{f(x)} \cdot \abs{g(x)} = \frac{\abs{f(x)}^p}{p} + \frac{\abs{g(x)}^q}{q} \quad \text{a.e.} > $ > which only happens when $\abs{f}^p = \abs{g}^q$ a.e. > [!theorem] > > Let $\seqf{p_j}$ and $\seqf{f_j}$ such that $f_j \in L^{p_j}$ for each $j$ and $\sum_{j = 1}^n \frac{1}{p_j} = \frac{1}{r}$, then > $ > \norm{\prod_{j = 1}^nf_j}_r \le \prod_{j = 1}^n \norm{f_j}_{p_j} > $ > *Proof*. If $n = 2$, then $\abs{f}^r \in L^{p/r}, \abs{g}^r \in L^{q/r}$, where $r/p + r/q = 1$, so > $ > \begin{align*} > \norm{fg}_r^r &\le \norm{\abs{f}^r}_{p/r} \cdot \norm{\abs{g}^r}_{q/r} \\ > &= \norm{f}_{p}^r \cdot \norm{g}_{q}^r \\ > \norm{fg}_r &\le \norm{f}_p \norm{g}_q > \end{align*} > $ > Now suppose that the result holds for $n$. By Hölder's inequality, > $ > \begin{align*} > \int \abs{\prod_{j = 1}^{n + 1} f_j} &\le \norm{f_{n + 1}}_{p_{n + 1}} \cdot \norm{\prod_{j = 1}^n f_j}_{\paren{\sum_{j = 1}^n\frac{1}{p_j}}^{-1}} \\ > &\le \norm{f_{n + 1}}_{p_{n + 1}} \cdot \prod_{j = 1}^n\norm{f_j}_{p_j} > \end{align*} > $ > giving us Hölder's inequality for $n + 1$ functions. Now let $r$ such that $1/r = \sum_{j \in [n]}1/p_j$. Since each $f_j \in L^{p_j/r}$ and $\sum_{j \in [n]}r/p_j = 1$, > $ > \norm{\prod_{j = 1}^n f_j}_r^r \le \prod_{j = 1}^n \norm{\abs{f_j}^r}_{p_j/r} = \prod_{j = 1}^n \norm{{f_j}}_{p_j}^r > $ > taking the $r$-th root on both sides yields the desired result.