# 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.

### 6 Responses

1. K said, on October 13, 2009 at 11:35 am

This is nice. But to get a multi-linear version [without the reduction of the multilinear case to the bilinear case], it appears you would need a higher dimensional version of the three lines lemma.

• Mark Lewko said, on October 15, 2009 at 1:48 am

I agree with your remarks. Of course, as you point out, you should be able to reduce the multilinear case to the bilinear case.

2. George Lowther said, on October 31, 2009 at 10:53 pm

Interesting proof, which I haven’t seen before.
The basic idea seems to be that since your function phi(z) is analytic, it must be bounded by the values on the lines Re(z)=0 and Re(z)=1. Also, on each line Re(z)=constant it is maximised at the point Im(z)=0, which gives you the bounds on Re(z)=0 and Re(z)=1. Using almost the same argument, phi(z) is log-convex on [0,1].

However, log-convexity of phi(z) follows directly from the fact that it is an integral over exponential functions of z, which also proves the Holder inequality (and a bit more efficiently imo).

3. George Lowther said, on October 31, 2009 at 10:55 pm

There’s a typo in your definition of B. It should be 0<R(z)<1 rather than 0<R(z)<0.

4. Marcelo Fernandes said, on June 5, 2012 at 1:14 am

very interesting