I'm a physics grad student who encountered the mathematics of spectral measures and orthogonal polynomials on the unit circle in the study of quantum time of arrival statistics. My question in particular pertains to the math in this paper (sec. 3.1). Consider the Caratheodory function of $z\in \mathbb{C}$:
$$F(z) = \int \mu(du) \frac{u+z}{u-z}$$ where $\mu$ is the probability measure of a unitary $U$ on the unit circle $S^1 = \{u\in \mathbb{C}; |u|=1\}$ w.r.t. a vector $\phi$. That is to say $\mu(du) = \langle \phi |E(du)| \phi \rangle$ where $E(du)$ is the spectral measure of $U$ for some interval $du$ on $S^1$. Obviously $\int \mu(du) = 1$.
Now defining the Schur function as $$f(z) = \frac{1}{z} \frac{F(z)-1}{F(z)+1}$$
this paper claims that it's evident from Schwarz's lemma that $$|f(z)| <1$$ in the open unit disk, i.e when $z \in \mathbb{D}= \{z;|z|<1\}$. I'm failing to see how one arrives here since neither $F(z)$ nor $f(z)$ vanish at $0$ and hence don't even qualify in the class of functions the lemma deals with.
What I do see already:
For $z \in \mathbb{D}$
$$\Re{F(z)} = \int \mu(du) \frac{1-|z|^2}{|u-z|^2} = \frac{1-|zf(z)|^2}{|1-zf(z)|^2} > 0$$
This in turn gives $$|f(z)| < \frac{1}{|z|}\quad \forall z \in \mathbb{D}$$ This, of course, is not enough to show $|f(z)|<1$ in $\mathbb{D}$ since $\frac{1}{|z|}$ can take values much larger than $1$ inside the disk.
Any help will be much appreciated. Thanks in advance.
Apply Schwarz Lemma to $zf(z)$. You already know that this maps the unit disk to itself. It also vanishes at $0$. So $|zf(z)| \leq |z|$ and equality cannot hold unless $f$ is a constant.