Frostman's Theorem surprises me

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In my last post, I went into some detail about inner functions, Blaschke products, and singular inner functions.

The interesting thing about inner functions that I'll care about here is how they act when you compose them with an automorphism of the disk. This is called a Frostman shift:


Frostman shifts continue to be inner functions: inner functions have boundary value modulus 1 and Mobius transforms take the circle to the circle, so the composition also has boundary values modulus 1.

What's WILD to me is that for "most" mobius transforms, the Frostman shift of ANY inner function will be a Blaschke product! This surprises me because Blaschke products have zeros in the disk and singular inner functions are zero free in the disk. The proof that Garcia, Mashreghi, and Ross provide is measure theoretic and broadly speaking, it works by showing that the measure for the singular inner part of a Frostman shift is zero (so when integrated and exponentiated, the singular inner factor is just 1). 



I think it's helpful to kind of think through the proof backwards here, since the way it's presented, some integral estimates done at the beginning seem irrelevant. We want to show that the singular inner factor is 1, so we need to show that the integral is zero. This means we need to do some logarithms to get the integral out of the exponential.


Through some Fubini trickery and noting that (since \(B\) is an inner function and therefore bounded by 1) \(\log|B(r\zeta)|<0\), we get that the goal is to show



But the inner function we started with was a Frostman shift, so we have to show that for almost any one we choose, this holds. Since "almost any" means integrals, we need to have a variable to integrate over. This is done by fixing a \(\rho\in [0,1)\) and letting \(\xi\) vary around the circle of radius \(\rho\). We'll want to show that the integral above is zero for Lebesgue almost every \(\xi\):


To actually show this, you need to do some sneakiness with bounding certain logarithms and then use a bunch of theorems from grad measure theory (DCT, Fatou's lemma). So when you all go look up the actual proof, don't be confused like I was when the first two thirds of the proof are doing integrals and logarithms!

One last thing I want to talk about, because the terminology is rad: indestructible Blaschke products.

So Frostman's theorem says that you can take any inner function, Mobius transform it, and get a Blaschke product most of the time. But only for ALMOST every Mobius transform; not all of them. So it makes sense to ask "are there conditions on Blaschke products that mean they'll NEVER become not-Blaschke-products under a Mobius transformation?" These are called indestructible Blaschke products and there are still open questions!

Here's a link to a survey paper that goes over these more if you're interested! https://scholarship.richmond.edu/mathcs-faculty-publications/83/

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