The other day someone reminded me of something I'd thought about some years ago. As back then it took me a little while to see why there was any problem; this time I got much farther on a solution than I did back then.
This is supposed to be a question. Question: Is the stuff below (especially the result at the very end) actually true? Question: Reference?
Background If $\Omega\subset\Bbb C$ is open the notation $H^\infty(\Omega)$ denotes the bounded holomorphic functions in $\Omega$. If $\Bbb D$ is the unit disk it's very well known that $f\in H^\infty(\Bbb D)$ has radial limits, in fact "non-tangential" limits, at almost every boundary point. A function $f\in H^\infty(\Bbb D)$ is an inner function if the boundary values have modulus $1$ almost everywhere. If $f\in H^\infty(\Bbb D)$ and the zeroes of $f$ are $(a_j)$ then $$\sum(1-|a_j|)<\infty.$$Conversely, this condition implies that $(a_j)$ is the zero set of some inner function (a Blaschke product).
The Problem
Now fix $R\in(0,1)$ and let $A$ be the annulus $R<|z|<1$. Define "inner function" in the obvious way. What are the zero sets of the inner functions in $A$?
Obvious Conjecture (OC)
The obvious conjecture is that $(a_j)$ is the zero set of an inner function in $A$ if and only if $$\sum d(a_j,\partial A)<\infty.$$Seems like an obvious conjecture to me anyway. The zero set of any (non-trivial) $f\in H^\infty(A)$ satisfies this condition; this follows in any of various ways from the result in the disk.
Conversely, given $(a_j)$ with $\sum d(a_j,\partial A)<\infty$, separate the $a_j$ into two classes, those closer to $|z|=1$ and those closer to $|z|=R$. Let $B_1$ be the Blachke product formed from the $a_j$ in the first class, and let $B_2$ be the Blaschke product with zeroes at $R/a_j$ for $a_j$ in the second class. Let $$f(z)=B_1(z)B_2(R/z).$$For minute I thought that was an inner function. Of course it's not. But it is a bounded holomorphic function, so we've established this:
Theorem The sequence $(a_j)$ is the zero set of some $f\in H^\infty(A)$ if and only if $\sum d(a_j,\partial A)<\infty$.
To get an inner function with the right zero set I thought I'd try constructing some things analogous to Blaschke products. So given $a\in A$ I want to construct or say something about an inner function with just one zero, at $a$; then I could hope that a product of such things converged.
Getting an inner function with zero set $\{a\}$ is very simple. Just say $$f(z)=(z-a)e^{u(z)+iv(z)},$$where $u$ is the solution to the Dirichlet problem with boundary data $-\log|z-a|$ and $v$ is a harmonic conjugate of $u$. For a few hours some years ago and a few hours the other day I thought that did it. Oops, harmonic functions need not have harmonic conjugates. Too much time in the disk...
So this raises the question of which harmonic functions in $A$ have harmonic conjugates. And then come to think of it, $e^{u+iv}$ can be single-valued even if $v$ is not; the actual question of interest is which harmonic functions in $A$ are equal to $\log|f|$ for some non-vanishing holomorphic function $f$. I was surprised that both these questions have simple answers.
In general let $u^\#$ be the radialization or radial part of $u$: $$u^\#(z)=\frac1{2\pi}\int_0^{2\pi}u(e^{it}z)\,dt.$$
Calculus $$r\frac d{dr}u^\#(r)=\frac1{2\pi i}\int_{|z|=r}2\frac{\partial u}{\partial z}\,dz.$$(Express $\partial u/\partial z$ in polar coordinates...)
Now note that if $u$ is harmonic in $A$ then there exist $a,b\in\Bbb R$ with $$u^\#(r)=a+b\log(r).$$We will say $b(u)=b$ below.
Theorem Suppose $u$ is harmonic in $A$. (i) The function $u$ has a harmonic conjugate if and only if $b(u)=0$. (ii) There exists a nonvanishing holomorphic $f$ with $u=\log|f|$ if and only if $b(u)\in\Bbb Z$.
Proof (i) One direction is just Cauchy's Theorem. For the other direction, suppose $b(u)=0$. Note that $\partial u/\partial z$ is holomorphic. The calculus above shows that $\int_\gamma\partial u/\partial z=0$ for any closed curve $\gamma$. Hence $\partial u/\partial z$ has an antiderivative: There exists a holomorphic $f$ with $2\partial u/\partial z=f'=\partial f/\partial z$. This says that $2u-\overline f$ is holomorphic. Hence $2u-(\overline f+f)$ is holomorphic and real-valued, hence constant.
(ii) Suppose that $u=\log|f|$. Then $2\partial u/\partial z=f'/f$. Since $\frac1{2\pi i}\int_{|z|=r}f'/f\in\Bbb Z$ the calculus above shows that $u^\#=c+n\log(r)$, $n\in\Bbb Z$. Conversely, suppose $b(u)=n\in\Bbb Z$. Let $v(z)=u(z)-n\log|z|$. Part (i) shows that $v$ has a harmonic conjugate $w$, and hence $v=\log\left|e^{v+iw}\right|$. So $u=\log\left|z^ne^{v+iw}\right|$. QED.
We can now determine which finite subsets of $A$ are the zero sets of inner functions. Given $a_1,\dots,a_n\in A$, let $$p(z)=\prod_{j=1}^n(z-a_j),$$and let $u$ be the solution to the Dirichlet problem with boundary data $-\log|p|$. There exists an inner function with zero set the same as $p$ if and only if there exists $f$ with $u=\log|f|$. This happens if and only if $b(u)\in\Bbb Z$. One can calulate $u^\#(1)$ and $u^\#(R)$ explicitly, and it turns out that
Theorem Suppose $a_1,\dots,a_n\in A$. There exists an inner function with precisely these zeroes if and only if $$\frac1{\log(R)}\sum_{j=1}^n\log|a_j|\in\Bbb Z.\quad(*)$$That seems interesting enough to justify reading this far, not that anybody will.
Corollary Inner functions are not as fundamental in $A$ as they are in $\Bbb D$.
Corollary $n=1$ is impossible: There is no inner function in $A$ with exactly one zero.
But $n=2$ is possible. Very curious.
That's as far as I got way back then. What about inner functions with infinitely many zeroes? Condition ($*$) makes no sense.
Say $R<R'<1$. An analysis like the above, with that $B_1(z)B_2(R/z)$ thing in place of $p$, proves this:
Theorem. Suppose $a_1,\dots\in A$. There exists an inner function with zero set $(a_j)$ if and only if $\sum d(a_j,\partial A)<\infty$ and $$\frac1{\log(R)}\left(\sum_{|a_j|>R'}\log|a_j|-\sum_{|a_j|\le R'}\log\left|\frac{R}{a_j}\right|\right)\in\Bbb Z.$$I'll spare you the details. Heh.
