---
tag: T8032P33B7
kind: definition
name: Inner function
msc: [30J05, 30H10]
defines: NN8XH5RJYB
uses: [8G5ABXRK85, 0DG0Z72CQD, Q2N917JQ6A]
sources:
  - url: https://en.wikipedia.org/wiki/Hardy_space
---

## Definition

Let $\mathbb{D}$ be the open [[unit disc|8G5ABXRK85]] and let $f \colon \mathbb{D} \to \C$ be a bounded [[holomorphic function|0DG0Z72CQD]].
By [[Fatou's theorem|Q2N917JQ6A]], the radial limit $f^*(e^{it}) = \lim_{r \to 1^-} f(r e^{it})$ exists for [[almost every|RARTMA0ZEC]] $t \in \R$.
The function $f$ is an *inner function* if $\abs{f^*(e^{it})} = 1$ for almost every $t$.

## 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|9N6R5ADJXT]].

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.