Blaschke factors and Blaschke products

For $a\in \mathbb{D}$, let ${\phi }_a(z)=\frac{a-z}{1-\bar{a}z}$ be the published Blaschke factor of The unit disc, the upper half-plane, and Blaschke factors, a biholomorphic self-map of $\mathbb{D}$ by Blaschke factors are automorphisms of the disc. Define the normalized Blaschke factor $b_a(z):=\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}{\phi }_a(z)(a\mathrm{\ne }0),b_0(z):=z\mathrm{.}$ Since $\mathrm{\mid }\bar{a}\mathrm{/}\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }=1$, each $b_a$ is holomorphic on an open neighbourhood of the closed disc (the denominator $1-\bar{a}z$ does not vanish there), $b_a$ is zero-free on $\mathbb{D}$ except for the simple zero at $a$ (the zero of ${\phi }_a$), and $b_a(0)=\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}a=\mathrm{\mid }a\mathrm{\mid }\ge 0,\mathrm{\mid }{b}_a(z)\mathrm{\mid }\le 1(z\in \mathbb{D}),$ the last inequality because ${\phi }_a$ maps $\mathbb{D}$ into itself. On the unit circle one has $\mathrm{\mid }{b}_a(\zeta )\mathrm{\mid }=1$ for every $\zeta \in \mathbb{T}$ (identified with the unit circle): for $\mathrm{\mid }\zeta \mathrm{\mid }=1$, $\mathrm{\mid }1-\bar{a}\zeta \mathrm{\mid }=\mathrm{\mid }\zeta \mathrm{\mid }\cdot \mathrm{\mid }\bar{\zeta }-\bar{a}\mathrm{\mid }=\mathrm{\mid }a-\zeta \mathrm{\mid }$ by Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive and Real and imaginary parts, complex conjugation, and modulus, so $\mathrm{\mid }{\phi }_a(\zeta )\mathrm{\mid }=1$.

Blaschke sequences and products. A sequence $(a_n{)}_{n\ge 1}$ in $\mathbb{D}$ is a Blaschke sequence if ${\sum }_n(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }$; its Blaschke product is the holomorphic product $B(z):={\prod }_{n\ge 1}{b}_{a_n}(z),$ defined as the locally uniform limit of the partial products $B_N:={\prod }_{n\le N}{b}_{a_n}$; for the empty sequence set $B:=1$. This definition is meaningful because the product converges normally on $\mathbb{D}$ in the sense of Normal convergence of holomorphic products: a Blaschke sequence satisfies $\mathrm{\mid }{a}_n\mathrm{\mid }\to 1$, so for every compact $K\subseteq \mathbb{D}$ only finitely many $a_n$ lie in $K$, and for every $a\in \mathbb{D}\setminus \{0\}$ and $\mathrm{\mid }z\mathrm{\mid }\le r<1$ the identity $1-b_a(z)=(1-\mathrm{\mid }a\mathrm{\mid })\frac{1+\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}z}{1-\bar{a}z},\mathrm{\mid }1-b_a(z)\mathrm{\mid }\le (1-\mathrm{\mid }a\mathrm{\mid })\frac{1+r}{1-r}$ holds: the identity is a direct computation from $b_a=\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}\frac{a-z}{1-\bar{a}z}$ using $\mathrm{\mid }a{\mathrm{\mid }}^2=a\bar{a}$, and the estimate uses $\mathrm{\mid }1+\frac{\bar{a}}{\mathrm{\mid }a\mathrm{\mid }}z\mathrm{\mid }\le 1+r$ and $\mathrm{\mid }1-\bar{a}z\mathrm{\mid }\ge 1-r$. For $a=0$, $b_0(z)=z$ and $\mathrm{\mid }1-b_0(z)\mathrm{\mid }\le 1+r=(1-\mathrm{\mid }0\mathrm{\mid })(1+r)\le (1-\mathrm{\mid }0\mathrm{\mid })(1+r)\mathrm{/}(1-r)$, so the same estimate holds without dividing by $\mathrm{\mid }a\mathrm{\mid }$. Hence ${\sum }_n{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le r}\mathrm{\mid }1-b_{a_n}(z)\mathrm{\mid }\le \frac{1+r}{1-r}{\sum }_n(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }\mathrm{.}$ The published normal-convergence theorem therefore applies: the partial products converge locally uniformly to a holomorphic $B$ that has exactly the zeros $a_n$ with the multiplicity with which they occur, and satisfies $\mathrm{\mid }B(z)\mathrm{\mid }\le 1$ on $\mathbb{D}$ (Normally convergent products define holomorphic functions with the expected zeros). The limit does not depend on the enumeration of the sequence: the normal-convergence criterion is a condition on the set of factors, and for two enumerations and finite initial segments $A,B$ containing a common block $\{1,\dots ,N\}$ the quotient ${\prod }_{n\in A}{b}_{a_n}\mathrm{/}{\prod }_{n\in B}{b}_{a_n}$ deviates from $1$ by at most a constant times the tail sum ${\sum }_{n>N}{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le r}\mathrm{\mid }1-b_{a_n}(z)\mathrm{\mid }$, which tends to $0$; hence the two partial-product sequences have the same locally uniform limit. The Blaschke sequence is reproduced by The zero set of a Hardy function satisfies the Blaschke condition from the zero set of a Hardy function. The boundary-modulus property $\mathrm{\mid }{B}^{\ast }\mathrm{\mid }=1$ almost everywhere is not part of this definition; it is proved in Boundary values and zeros of a Blaschke product ↗.