# Hölder’s Inequality

Let ${(\Omega,\mu)}$ be a ${\sigma}$–finite measure space and ${f,g : \Omega \rightarrow \mathbb{C}}$ be measurable functions. Hölder’s inequality asserts for ${p,q \in [1, \infty]}$, ${p^{-1} + q^{-1} = 1}$ one has $\displaystyle \int |fg| \leq ||f||_{p} ||g||_{q}.$

One standard proof is to integrate the AM-GM inequality $\displaystyle \frac{|f(x)||g(x)|}{||f||_{p} ||g||_q} \leq \frac{|f(x)|^p}{p||f||_p^p} + \frac{|g(x)|}{q ||g||_q^q}.$

Here, I provide an alternative proof. First consider the case where ${g = \chi_E}$ is the characteristic function of a set ${E}$ of finite measure. Let $\displaystyle f_{

Then $\displaystyle \int |f| \chi_E = \int |f_{

By choosing the optimal value for ${M= \frac{||f||_p}{\mu(E)^{\frac{1}{p}}}}$, we obtain $\displaystyle \int |f| \chi_E \leq 2 \mu(E)^{1-\frac{1}{p}} ||f||_p.\ \ \ \ \ (1)$

Remark: Note that if ${|f|}$ is bounded away from ${\lambda}$, we may improve the above argument. For instance, suppose ${f(x) \notin [M^{-1} \lambda , M^{\frac{1}{p-1}} \lambda]}$ for some ${M\geq 1}$ and for all ${x \in \Omega}$. Then $\displaystyle \int |f| \chi_E \leq 2 M^{-1} \mu(E)^{1-\frac{1}{p}} ||f||_p.$

We can get rid of the constant 2 in (1) by using the tensor product trick. That is, apply (1) to $\displaystyle F: \Omega^n \rightarrow \mathbb{C} , \ \ \ \ \ \ \ F(x_1 , \ldots , x_n) = f(x_1) \cdots f(x_n).$

It follows that $\displaystyle \left(\int |f| \chi_E \right)^n = \int |F|\chi_{E^n} \leq \mu(E^n)^{1-\frac{1}{p}} ||F||_p = 2 \left(\mu(E)^{1-\frac{1}{p}} ||f||_p\right)^n.$

Taking ${n^{\rm th}}$ roots and letting ${n \rightarrow \infty}$ establishes Hölder’s inequality in the case ${g = \chi_E}$.

Remark: In many applications, (1) is enough. Nevertheless, extending to simple functions is an easy matter and Hölder’s inequality follows by the density of simple functions in ${L^p}$. ♠