Inner function

Remarks

An inner function satisfies $\abs{f(z)} \le 1$ on $\mathbb{D}$, because $f$ is the Poisson integral of $f^*$.

Example: let $S(z) = \exp\left(-\frac{1 + z}{1 - z}\right)$. For $z = r e^{it}$ one has $\operatorname{Re} \frac{1 + z}{1 - z} = \frac{1 - r^2}{\abs{1 - z}^2} > 0$, so $\abs{S(z)} < 1$ and $S$ is bounded. For $t \notin 2\pi\Z$ this real part tends to $0$ as $r \to 1$, so $\abs{S^*(e^{it})} = 1$ and $S$ is inner. It has no zeros, so it is not a Blaschke product.

Non-example: $f(z) = (1 + z)/2$ is bounded and holomorphic, but $\abs{f^*(e^{it})} = \abs{\cos(t/2)} < 1$ for $t \notin \pi\Z$.

Every inner function has the form $c B S_\mu$, where $\abs{c} = 1$ and $B$ is a Blaschke product. Here $S_\mu(z) = \exp\left(-\int \frac{e^{it} + z}{e^{it} - z} \, d\mu(t)\right)$, and $\mu$ is a finite positive Borel measure on the circle that is singular to Lebesgue measure. The factor $S_\mu$ is called a singular inner function.