---
tag: C2082GXJ3Q
kind: definition
name: Blaschke product
msc: [30J10, 30J05]
defines: 9N6R5ADJXT
uses: [8G5ABXRK85, 0DG0Z72CQD, 0B346DDAM5]
sources:
  - url: https://en.wikipedia.org/wiki/Blaschke_product
---

## Definition

Let $\mathbb{D} = \set{z \in \C : \abs{z} < 1}$ be the [[unit disc|8G5ABXRK85]].
Let $m \ge 0$ be an integer.
Let $(a_n)$ be a finite or infinite sequence of points of $\mathbb{D} \setminus \set{0}$, with repetitions allowed, such that
$$\sum_n (1 - \abs{a_n}) < \infty.$$
The *Blaschke product* with zeros $0$ (of order $m$) and $(a_n)$ is the function $B \colon \mathbb{D} \to \C$ given by
$$B(z) = z^m \prod_n \frac{\abs{a_n}}{a_n} \, \frac{a_n - z}{1 - \overline{a_n} z}.$$
It is a *finite Blaschke product* when the sequence $(a_n)$ is finite.

## Remarks

Under the condition above, the product converges uniformly on compact subsets of $\mathbb{D}$.
So $B$ is [[holomorphic|0DG0Z72CQD]] with $\abs{B} \le 1$, and its zeros are exactly $0$ and the $a_n$, with multiplicity.
Moreover the radial limit of $B$ has modulus $1$ at almost every point of the circle, so $B$ is an [[inner function|NN8XH5RJYB]].
The factor $\abs{a_n}/a_n$ makes the $n$-th factor positive at $z = 0$; some authors also allow a constant factor of modulus $1$.

Example: for $a \in \mathbb{D} \setminus \set{0}$, the single factor $B(z) = \frac{\abs{a}}{a} \frac{a - z}{1 - \bar a z}$ is a finite Blaschke product.
For $\abs{z} = 1$ one has $\abs{1 - \bar a z} = \abs{\bar z - \bar a} = \abs{a - z}$, so $\abs{B(z)} = 1$ on the circle.

Non-example: the points $a_n = 1 - 1/n$ for $n \ge 2$ fail the summability condition, since $\sum 1/n = \infty$.
For them the partial products tend to $0$ at every point of $\mathbb{D}$, so no Blaschke product has these zeros.
In fact no bounded holomorphic function other than $0$ vanishes at all of them.