Boundary Values and Inner Functions

Someday, I'll do a post that doesn't draw heavily from "Introduction to Model Spaces and their Operators" by Garcia, Mashreghi, and Ross, but today is not that day. I wish I had studied this book as a grad student: it's so well written!

Anyway, let's talk about inner functions!

First, we need to establish the world in which we live. Today, this is going to be the Banach space of analytic functions bounded on the unit disk, \(H^\infty(\mathbb{D})\).


In a complex analysis class, you probably did the exercise where you express the Poisson kernel in a couple of different ways: as a "difference quotient," as the real part of a rational function analytic on the disk, as a geometric series, and in terms of the modulus and argument of the complex variable \(z=re^{i\theta}\):


Since the Poisson kernel is the real part of an analytic function, it's harmonic, and taking derivatives, this lets us see that the Poisson integral of a measure \(\mu\) is (complex) harmonic on the circle. (There's usually also a discussion about moments of a measure, but I'm not going to get into that here.


So if you have a function on the circle, you can integrate against the Poisson kernel to get a function on the disk. It can also be shown that all sufficiently nice functions on the disk are the Poisson integral against some measure on the circle.


Fortunately for us, our \(H^\infty\) functions are nice! Because these functions are bounded in the sup norm, they're bounded in the total variation norm, and they're analytic, so they're automatically harmonic. This means that each of our \(H^\infty\) functions is the Poisson integral of some measure on the circle.


To go the other direction (function on the disk to function on the circle), we think about radial limits: given an analytic function \(f(z)\) on the (open) disk, consider \(f(r\zeta)\) where \(\zeta\in\mathbb{T}\), and let \(r\to1\). This seems straightforward, but how do we know these limits exist???

Thankfully the fact that our functions are Poisson integrals comes to our rescue! There's a theorem of Fatou that says that Poisson integrals have finite radial limits Lebesgue-almost-everywhere on the circle. This theorem is proved with "hard analysis:" actual integral estimates involving epsilons and deltas! I'm not going to go through it here, but it's an excellent exercise for people studying measure theory!

(Another place to find more discussion about this is Chapter 3 of Agler and McCarthy "Pick Interpolation and Hilbert Function Spaces." They discuss \(H^p\) spaces for cases where \(p\neq\infty\). Good things continue to hold!)

Ok, so why did I go through all of this? Two reasons:
    1. I think it's good for me to have had to write all that out. A big part of this blog is about me building character.
    2. I need you to believe me when I say that we can talk about boundary values, at least almost everywhere!

We care about boundary values, because when we define inner functions, it's in terms of their boundary values!

A function \(u\in H^\infty\) is inner if \(|u(\zeta)|=1\) almost everywhere on \(\mathbb{T}\).

(Note: the maximum modulus principle then guarantees that our function $u$ is modulus-bounded by 1 on the disk.)

It can be shown that every inner function can be factored as \(u(z)=z^k B(z) s_{\mu}(z)\) where \(B\) is a Blaschke product and \(s_{\mu}\) is a singular inner function. Let's talk about what those are, because it's kind of the point of this post.

Blaschke products are of the form


Finite Blaschke products are interesting and useful especially in the context of model spaces, but as long as the zeros \(a_n\) satisfy the Blaschke condition, the product will converge. (I vaguely remember doing this in my grad complex analysis homework…)

That these actually are inner functions can be shown by noting that each factor is inner (in fact each factor is a Mobius transformation!) and that this guarantees that finite products are inner and also that \(B\) is bounded by 1 in the sup norm of the unit disk. Through some convergence theorems and measure theory, one can then show:


Singular inner functions are defined as the exponential of an integral against a measure singular with respect to Lebesgue measure. A classic example sets the measure \(\mu\) to be a point mass at z=1.


It's somewhat easier to see that these are inner because when we take the modulus, the integral becomes a Poisson integral, and the measure we're Poisson integrating is SINGULAR with respect to the Lebesgue measure. This means that the (Radon-Nikodym) derivative of $\mu$ is zero Lebesgue almost everywhere. Fatou's theorem then says that the radial limits of the Poisson integral will be zero almost everywhere, so when we exponentiate, we get that \(s_{\mu}\) is inner.

There are lots of interesting things to ask about inner functions that I'm not going to get into here. We've said that they have modulus 1 almost everywhere on the boundary, but WHERE? Also, inner functions have angular derivatives in the sense of Julia-Caratheodory in predictable ways. (Garcia, Mashreghi, and Ross have some discussion of this.)

In my next post though, I'll talk about Frostman's theorem and how inner functions behave when you compose them with a Mobius Transform!