Blaschke factorization of a Nevanlinna-class function

The zero sequence satisfies the Blaschke condition. For every $0<r<1$, monotonicity of the integral and $\log \mathrm{\mid }f\mathrm{\mid }\le {\log }^{+}\mathrm{\mid }f\mathrm{\mid }\le h_0$ give ${\int }_{\mathbb{T}}\log \mathrm{\mid }f(r\zeta )\mathrm{\mid }dm(\zeta )\le {\int }_{\mathbb{T}}{h}_0(r\zeta )dm(\zeta )=h_0(0)<+\mathrm{\infty }$ by the mean value property; hence the liminf hypothesis of [L2] holds and ${\sum }_n(1-\mathrm{\mid }{a}_n\mathrm{\mid })<+\mathrm{\infty }$, so $(a_n)$ is a Blaschke sequence and $B$ is a well-defined Blaschke product with $\mathrm{\mid }B\mathrm{\mid }\le 1$.

The logarithm of one factor. Let $a\in \mathbb{D}$ and $0<r<1$ with $r\mathrm{\ne }\mathrm{\mid }a\mathrm{\mid }$. Since $\mathrm{\mid }{b}_a(r\zeta )\mathrm{\mid }=\mathrm{\mid }a-r\zeta \mathrm{\mid }\mathrm{/}\mathrm{\mid }1-\bar{a}r\zeta \mathrm{\mid }$ and ${\int }_{\mathbb{T}}\log \mathrm{\mid }1-\bar{a}r\zeta \mathrm{\mid }dm(\zeta )=0$ by the mean value property of the zero-free harmonic function $\log \mathrm{\mid }1-\bar{a}rw\mathrm{\mid }$ on a neighbourhood of $\bar{\mathbb{D}}$, [L5] gives ${\int }_{\mathbb{T}}\log \frac{1}{\mathrm{\mid }{b}_a(r\zeta )\mathrm{\mid }}dm(\zeta )=\log \frac{1}{r}-{\int }_{\mathbb{T}}\log \mathrm{\mid }\zeta -\frac{a}{r}\mathrm{\mid }dm(\zeta )=\log \frac{1}{r}-{\log }^{+}\frac{\mathrm{\mid }a\mathrm{\mid }}{r}=\{\begin{array}{ll}\log \frac{1}{r},& \mathrm{\mid }a\mathrm{\mid }<r,\\ \log \frac{1}{\mathrm{\mid }a\mathrm{\mid }},& \mathrm{\mid }a\mathrm{\mid }>r\mathrm{.}\end{array}$

The quotient. Put $g:=f\mathrm{/}B$, holomorphic on the complement of the zero set of $B$. At a point $a$ occurring $m\ge 1$ times in the zero sequence, $B$ has a zero of order $m$ and $f$ a zero of order at least $m$ (the sequence lists all zeros with multiplicity), so $g$ is bounded near $a$ and extends holomorphically there by [L4]; hence $g$ extends holomorphically to all of $\mathbb{D}$. Since the zeros of $B$ are exactly the $a_n$ and $g$ is holomorphic at those points with $g\mathrm{\ne }0$ there — the multiplicity of the numerator's zero is exactly exhausted when the sequence is repeated with multiplicity — $g$ has no zero in $\mathbb{D}$; and $\mathrm{\mid }g\mathrm{\mid }=\mathrm{\mid }f\mathrm{\mid }\mathrm{/}\mathrm{\mid }B\mathrm{\mid }\ge \mathrm{\mid }f\mathrm{\mid }$ off that zero set because $\mathrm{\mid }B\mathrm{\mid }\le 1$, with the inequality extending to the zeros by continuity.

The mean of $-\log \mathrm{\mid }{B}_r\mathrm{\mid }$ at non-exceptional radii. Fix $0<r<1$ with $r\mathrm{\notin }\{\mathrm{\mid }{a}_n\mathrm{\mid }\}$. The partial sums ${\sum }_{n\le N}-\log \mathrm{\mid }{b}_{a_n}(r\zeta )\mathrm{\mid }$ are nonnegative and increase to $-\log \mathrm{\mid }B(r\zeta )\mathrm{\mid }$ (the product converges and each factor has modulus $\le 1$), so the monotone convergence theorem [L6] and step 1.2 give ${\int }_{\mathbb{T}}\log \frac{1}{\mathrm{\mid }B(r\zeta )\mathrm{\mid }}dm(\zeta )={\sum }_n{\int }_{\mathbb{T}}\log \frac{1}{\mathrm{\mid }{b}_{a_n}(r\zeta )\mathrm{\mid }}dm(\zeta )=N(r)\log \frac{1}{r}+{\sum }_{\mathrm{\mid }{a}_n\mathrm{\mid }>r}\log \frac{1}{\mathrm{\mid }{a}_n\mathrm{\mid }}=:R(r),$ where $N(r)<+\mathrm{\infty }$ because $(a_n)$ has no accumulation point in $\{\mathrm{\mid }z\mathrm{\mid }<r\}$ and the sum is finite: only finitely many zeros have $\mathrm{\mid }{a}_n\mathrm{\mid }<1\mathrm{/}2$, while the remaining terms obey $\log \frac{1}{\mathrm{\mid }{a}_n\mathrm{\mid }}\le 2(1-\mathrm{\mid }{a}_n\mathrm{\mid })$ by [L7].

Bound at non-exceptional radii. For $r\ge 1\mathrm{/}2$, $R(r)\le 2N(r)(1-r)+2S\le 2S+2S=4S$, because $N(r)(1-r)\le {\sum }_{\mathrm{\mid }{a}_n\mathrm{\mid }<r}(1-\mathrm{\mid }{a}_n\mathrm{\mid })\le S$ and by [L7].

Exceptional radii. Let $r\in [1\mathrm{/}2,1)$ be arbitrary and choose a sequence $r_k\downarrow r$ with $r<r_k<1$ and $r_k\mathrm{\notin }\{\mathrm{\mid }{a}_n\mathrm{\mid }\}$ for all $k$ (the exceptional set is countable and has no accumulation point below $1$). The functions $\zeta \mapsto -\log \mathrm{\mid }B(r_k\zeta )\mathrm{\mid }$ are nonnegative and converge pointwise $m$-almost everywhere to $-\log \mathrm{\mid }B(r\zeta )\mathrm{\mid }$: for $\zeta$ outside the finite set where $B(r\zeta )=0$, continuity of $B$ gives $B(r_k\zeta )\to B(r\zeta )\mathrm{\ne }0$. Fatou's inequality [L6] and step 3.1 therefore give ${\int }_{\mathbb{T}}\log \frac{1}{\mathrm{\mid }B(r\zeta )\mathrm{\mid }}dm(\zeta )\le {\mathrm{lim\hspace{0.17em}inf}}_k{\int }_{\mathbb{T}}\log \frac{1}{\mathrm{\mid }B(r_k\zeta )\mathrm{\mid }}dm(\zeta )\le 4S\mathrm{.}$

The quotient lies in $N(\mathbb{D})$. Since $g=f\mathrm{/}B$ and $\mathrm{\mid }B\mathrm{\mid }\le 1$, [L7] gives ${\log }^{+}\mathrm{\mid }g\mathrm{\mid }\le {\log }^{+}\mathrm{\mid }f\mathrm{\mid }+\log (1\mathrm{/}\mathrm{\mid }B\mathrm{\mid })$. For $1\mathrm{/}2\le r<1$, steps 2.2, 3.1 and 4.1 and the majorant bound of step 1.1 give ${\int }_{\mathbb{T}}{\log }^{+}\mathrm{\mid }g(r\zeta )\mathrm{\mid }dm(\zeta )\le h_0(0)+4S;$ for $0\le r\le 1\mathrm{/}2$ the holomorphic function $g$ is continuous on the compact disc $\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}2$, so ${\int }_{\mathbb{T}}{\log }^{+}\mathrm{\mid }g(r\zeta )\mathrm{\mid }dm(\zeta )\le {\log }^{+}({\sup }_{\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}2}\mathrm{\mid }g(z)\mathrm{\mid })<+\mathrm{\infty }$. Hence ${\sup }_{0<r<1}{\int }_{\mathbb{T}}{\log }^{+}\mathrm{\mid }g(r\zeta )\mathrm{\mid }dm(\zeta )<+\mathrm{\infty }$, and the sup-mean criterion [L1] shows that ${\log }^{+}\mathrm{\mid }g\mathrm{\mid }$ has a harmonic majorant, that is, $g\in N(\mathbb{D})$.

Assembly. Step 1.1 produces the Blaschke sequence and the product $B$; step 2.1 produces the holomorphic zero-free extension $g=f\mathrm{/}B$ with $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }$; and steps 1.2–3.1 verify the sup-mean criterion for $g$, giving $g\in N(\mathbb{D})$.