= Solution
Use the <Nevanlinna proximity function> $m(r,f)=(2\pi)^{-1}\int\log^+|f(re^{i\theta})|\,d\theta$ and the <Nevanlinna integrated counting function> $N(r,f)$ for <poles>, counting <multiplicity>. If the pole order at zero is $\nu_0$, then
$$
N(r,f)=\nu_0\log r+\sum_{0<|b|<r}\nu_b\log\frac r{|b|},\qquad T(r,f)=m(r,f)+N(r,f).
$$
Boundedness of the <Nevanlinna characteristic> refers to $r\uparrow1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when $\nu_0>0$.
The <Nevanlinna first main theorem> states, for every fixed finite value $a$ and every nonconstant <meromorphic function> on the disk,
$$
\boxed{N(r,1/(f-a))+m(r,1/(f-a))=T(r,f)+O(1).}
$$
For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while <Jensen's formula> gives $T(r,h)-T(r,1/h)=\log|c|$ when $h(z)=cz^\nu(1+O(z))$ near zero; these observations explain the first theorem.
The <Nevanlinna second main theorem in the unit disc> states that for $q\ge3$ distinct values $a_j$ on the <Riemann sphere>,
$$
\boxed{(q-2)T(r,f)\le\sum_{j=1}^q\overline N(r;a_j)+O\left(\log^+T(r,f)+\log\frac1{1-r}\right)}
$$
as $r\uparrow1$ outside an exceptional set $E$ satisfying $\int_Edr/(1-r)<\infty$. The <truncated Nevanlinna counting function> $\overline N$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in https://mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .
Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge0$, its <poles> satisfy the <Blaschke condition>, counting their orders: let $r\uparrow1$ in their integrated count and use $1-|b|\le-\log|b|$. Let $B$ be their <Blaschke product>, including the origin factor if necessary. Then $g=Bf$ extends to a <holomorphic function> on the whole disk, and $|B|\le1$ gives
$$
\frac1{2\pi}\int_0^{2\pi}\log^+|g(re^{i\theta})|\,d\theta\le m(r,f)\le C.
$$
Here the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge\nu_0\log r_0$ on $r\ge r_0$.
The nonnegative <subharmonic function> $\log^+|g|$ has a finite <harmonic majorant>. Indeed its <Poisson integrals> $H_R$ on expanding disks dominate it by the <maximum principle for subharmonic functions>, increase with $R$, and have $H_R(0)\le C$. The <Harnack inequality for harmonic functions> bounds them on every compact subdisk, so their increasing limit is a finite <harmonic function> $H\ge\log^+|g|$. Since the disk is simply connected, $H$ has a <harmonic conjugate>, giving a <holomorphic function> $A$ with $\operatorname{Re}A=H$. Consequently
$$
u=ge^{-A},\qquad v=Be^{-A},\qquad |u|\le1,\quad |v|\le1,\quad v\not\equiv0,\quad f=u/v.
$$
This proves the constructive half of the <quotient characterization of bounded characteristic>.
Conversely, if $f=u/v$ with $u,v$ bounded <holomorphic functions> and $v\not\equiv0$, each <pole> of $f$ is a zero of $v$ of at least the same order. The elementary inequality $\log^+|u/v|\le\log^+|u|+\log^+|1/v|$ and the <Nevanlinna first main theorem> give
$$
T(r,f)\le m(r,u)+T(r,1/v)+O(1)\le\log^+\|u\|_\infty+T(r,v)+O(1)=O(1).
$$
Any cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus \b[bounded characteristic is exactly a quotient of two bounded analytic functions].
An unbounded example with <bounded characteristic> is
$$
\boxed{g(z)=\frac1{1-z}.}
$$
It is a ratio of the bounded <holomorphic functions> $1$ and $1-z$. More explicitly, <Jensen's formula> gives the mean of $\log|1-re^{i\theta}|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le\log2$, although $g(r)\to\infty$.
Finally, set $Y=(1+|g|^2)^{p/2}\ge1$. Pointwise $\log^+|g|\le p^{-1}\log Y$. <Jensen's inequality> for the concave <logarithm> gives
$$
T(r,g)=m(r,g)\le\frac1p\int\log Y\,\frac{d\theta}{2\pi}\le\frac1p\log\int Y\,\frac{d\theta}{2\pi}<\frac{\log C}{p}.
$$
There is no pole term because $g$ is <holomorphic>. Hence \b[the stated integral bound implies bounded characteristic]. This argument in fact works for every $p>0$.
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