---- $\mathbb{F}$ denotes $\mathbb{R}$ or $\mathbb{C}$. > [!theorem] Theorem. ([[Hölder's inequality]]) > Suppose $(X, \Sigma, \mu)$ is a [[measure|measure space]], $1 \leq p \leq \infty$, and $f,h:X \to \mathbb{F}$ are [[measurable function|measurable functions]]. [[Lp-norm|Then]] $\|f h\|_{1} \leq \|f\|_{p} \|h\|_{p'},$ where $p'$ denotes [[dual exponent]] of $p$. ^theorem > [!specialization] > In the special case $p=2=p'$ one recovers the [[Cauchy-Schwarz Inequality]]. ^specialization > [!proposition] Corollary. (Hölder's inequality for $m$ functions) > An inductive argument shows that if $p_{1},p_{2},\dots p_{m}$ are such that $\frac{1}{p_{1}}+\dots + \frac{1}{p_{m}}=1$, then $\|f_{1} \cdots f_{m}\|_{1} \leq \|f_{1}\|_{p_{1}} \cdots \|f_{1}\|_{p_{m}}$ for all $f_{1} \in L^{p_{1}}(\mu), \dots, f_{m} \in L^{p_{m}}$. ^proposition > [!proof]- Proof. ([[Hölder's inequality]]) > First consider the case $1<p<\infty$. Consider the subcase where $\|f\|_{p}=\|h\|_{p'}=1$. [[Young's inequality]] implies $|f(x)h(x)| \leq \frac{|f(x)|^{p}}{p}+ \frac{|h(x)|^{p'}}{p'}$ > for all $x \in X$. [[integral|Integrating both sides]] with respect to $\mu$ shows that[^1] $\|fh\|_{1} \leq 1=\|f\|_{p} \|h\|_{p'}$, completing the proof in this special case. > > Now in general: If $\|f\|_{p}=0$ or $\|h\|_{p}=0$ then $\|fh\|_{1}=0$[^2] and the desired inequality holds. Similarly, if $\|f\|_{p}=\infty$ or $\|h\|_{p}=\infty$ then it holds. Thus we assume $0<\|f\|_{p} <\infty$ and $0<\|h\|_{p}<\infty$. Now define [[measurable function|measurable functions]] $f_{1},h_{1}:X \to \mathbb{F}$ as the normalizations $f_{1}=\frac{f}{\|f\|_{p}}\text{ and }h_{1}=\frac{h}{\|h\|_{p'}}.$ > Then $\|f\|_{1}=1=\|h\|_{1}$, so our work in the special case above applies to give $\underbrace{ \|f_{1}h_{1}\| }_{ =\frac{\|f h\|_{1}}{\|f\|_{p} \|h\|_{p\prime}} } \leq 1,$ > the result follows. > > The cases $p=1$ and $p=\infty$ are part of the pset. > [^1]: $\int \frac{|f(x)|^{p}}{p} \, d\mu=\frac{\overbrace{ \|f\|_{p}^{p} }^{ =1 }}{p}=\frac{1}{p}$. Similarly, $\int \frac{|h(x)|^{p'}}{p'} \, d\mu=\frac{1}{p'}$. Now $\frac{1}{p}+\frac{1}{p'}=1$ [[dual exponent|by definition]]. [^2]: Explicating: if $\|f\|_{p}=0$, then $\|f\|_{p}^{p}=0$, that is, $\int |f|^{p} \, d\mu=0$. A positive function which [[integral|integrates]] to zero is zero [[almost-everywhere]], so $|f|^{p}=0 \text{ a.e.}$ In turn, $|f|=0 \text{ a.e.}$, so $f=0 \text{ a.e.}$ It follows that $fh$ (and therefore $|fh|$) is zero almost everywhere, so $\|fh\|_{1}=0$. ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```