# Lewko's blog

## Hölder’s inequality via complex analysis

Posted in expository, math.CA by Mark Lewko on October 13, 2009

In this post I will give a complex variables proof of Hölder’s inequality due to Rubel. The argument is very similar to Thorin’s proof of the Riesz-Thorin interpolation theorem. I imagine that there is a multilinear form of Riesz-Thorin that provides a common generalization of the two arguments, however we won’t explore this here. We start by establishing the well-known three lines lemma.

Lemma (three lines lemma) Let ${\phi(z)}$ be a bounded analytic function in the strip ${B=\{z : 0<\Re(z)<1\}}$. Furthermore, assume that $\phi(z)$ extends to a continuous function on the boundary of $B$ and satisfies

$\displaystyle |\phi(it)|\leq M_{0} \hspace{1cm}\text{and}\hspace{1cm} |\phi(1+it)|\leq M_{1}$

for $t\in \mathbb{R}$. Then, for $0\leq \sigma \leq 1$, we have that

$\displaystyle|\phi(\sigma+it)|\leq M^{1-\sigma}_{0}M^{\sigma}_{1}.$

Proof: Let $\epsilon>0$ and consider the function (analytic in $B$)

$\displaystyle \phi_{\epsilon}(z)=\phi(z)M_{0}^{z-1}M_{1}^{-z} e^{\epsilon z(z-1)}.$

One easily checks that ${\phi_{\epsilon}(z)\leq 1}$ if ${\Re(z)=0}$ or ${\Re(z)=1}$. Furthermore, since ${|\phi(z)|}$ is uniformly bounded for ${z \in \overline{B}}$, we must have that

$\displaystyle \lim_{\Im{z}\rightarrow\infty}|\phi_{\epsilon}(z)|=0$

for ${ 0 < \Re z < 1}$. We now claim that, for ${A}$ sufficiently large, ${|\phi_{\epsilon}(z)|\leq 1}$ for ${z \in B_{A}}$ where ${B_{A}=\{z : 0\leq\Re(z)\leq 1, -A \leq \Im(z) \leq A\}}$. This follows by the previous remarks on the boundary of ${B_{A}}$, and by the maximum modulus principle in its interior. This completes the proof. $\Box$

We are now ready to give a complex variables proof of Hölder’s Inequality.

Theorem (Hölder’s Inequality)Let ${(X, \mathcal{M}, \mu)}$ be a measure space, ${p,q\in [1,\infty]}$ such that ${\frac{1}{p}+\frac{1}{q}=1}$. If ${f \in L^p(X)}$ and ${g \in L^q(X)}$ then

$\displaystyle ||fg||_{L^1} \leq ||f||_{L^p}||g||_{L^q}.$

Proof: By a standard limiting argument (preformed first with, say, ${g}$ fixed) it will suffice to assume that ${f}$ and ${g}$ are simple functions. If we let ${z=1/q}$ we may now rewrite Hölder’s inequality as

$\displaystyle \int_{X}|f|^{p(1-z)}|g|^{qz}d\mu \leq ||f||_{L^p}^{p(1-z)}||g||_{L^q}^{qz}$

Indeed, ${p(1-z)=p/p=1}$. Using the fact that ${f}$ and ${g}$ are simple, we can define a function, ${\phi(z)}$, analytic in the strip ${B}$ by

$\phi(z)=\int_{X}|f|^{p(1-z)}|g|^{qz}d\mu=\sum_{n=1}^{N}\sum_{m=1}^{M}a_n b_m e^{\lambda_n p(1-z)}e^{\kappa_m q z}.$

It follows that ${|\phi(\sigma+it)| \leq |\phi(\sigma)|}$, and that ${\phi(z)}$ is bounded on the closure of ${B}$. We record that ${|\phi(it)| \leq |\phi(0)| = \int_{X}|f|^{p}d\mu}$ and ${|\phi(1+it)| \leq \phi(1) = \int_{X}|g|^{q}d\mu }$. Now, by the three lines lemma, we have that

$\displaystyle \int_{X}|f|^{p(1-\sigma)}|g|^{q\sigma}d\mu=|\phi(\sigma)|\leq (\int_{X}|f|^{p}d\mu)^{1-\sigma}(\int_{X}|g|^{q}d\mu)^{\sigma}.$

Taking ${\sigma=1/q}$ we recover Hölder’s inequality. $\Box$

Updated 10/13/2009: typos corrected

Updated 10/14/2009: the statement of the three lines lemma was truncated in the original post

Update 10/31/2009: typo in definition of B.