Hölder’s inequality via complex analysis
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 be a bounded analytic function in the strip . Furthermore, assume that extends to a continuous function on the boundary of and satisfies
for . Then, for , we have that
Proof: Let and consider the function (analytic in )
One easily checks that if or . Furthermore, since is uniformly bounded for , we must have that
for . We now claim that, for sufficiently large, for where . This follows by the previous remarks on the boundary of , and by the maximum modulus principle in its interior. This completes the proof.
We are now ready to give a complex variables proof of Hölder’s Inequality.
Theorem (Hölder’s Inequality)Let be a measure space, such that . If and then
Proof: By a standard limiting argument (preformed first with, say, fixed) it will suffice to assume that and are simple functions. If we let we may now rewrite Hölder’s inequality as
Indeed, . Using the fact that and are simple, we can define a function, , analytic in the strip by
It follows that , and that is bounded on the closure of . We record that and . Now, by the three lines lemma, we have that
Taking we recover Hölder’s inequality.
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.
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.
Thanks for inaugurating the comments!
I agree with your remarks. Of course, as you point out, you should be able to reduce the multilinear case to the bilinear case.
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).
There’s a typo in your definition of B. It should be 0<R(z)<1 rather than 0<R(z)<0.
Fixed. Thanks!
very interesting