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.