Rational functions are exactly those with logarithmic characteristic

The identity function $z\mapsto z$ is entire and has no poles, so [F1, F10] give its characteristic $T(r,z)=\frac{1}{2}\log (1+r^2)=\log r+O(1)$. If $f$ is rational of degree $d\ge 1$, applying [F3] to the composition of $f$ with the identity function gives $T(r,f)=dT(r,z)+O(1)=d\log r+O(1)$, proving the forward implication and the degree formula.

Suppose $T(r,f)=O(\log r)$, and choose $A\ge 0$, $r_0\ge 1$ so $T(r,f)\le A\log r$ for every $r\ge r_0$; by [F1, F10], the integrand defining $m(r,\mathrm{\infty };f)$ is nonnegative, hence $N(r,\mathrm{\infty };f)\le T(r,f)\le A\log r$ for $r\ge r_0$.

Let $n_0=n(0,\mathrm{\infty };f)$, finite by [F5], and suppose there are infinitely many poles; since each bounded-disc count is finite by [F5], $n(k,\mathrm{\infty };f)$ is unbounded over positive integers $k$. Choose an integer $M>\max \{A,n_0\}$ and an integer $s>r_0$, and let $R_M$ be the least integer $k\ge s$ with $n(k,\mathrm{\infty };f)\ge M$; it exists by unboundedness and well-ordering. For $R\ge R_M$, closed-disc monotonicity gives $n(t,\mathrm{\infty };f)\ge M$ on $[R_M,R]$ and $n(t,\mathrm{\infty };f)\ge n_0$ everywhere, so [F2] yields $N(R,\mathrm{\infty };f)\ge n_0\log R+(M-n_0)\log (R\mathrm{/}{R}_M)=M\log R-(M-n_0)\log R_M$. This contradicts step 1.2 as $R\to \mathrm{\infty }$ because $M>A$, proving that $f$ has finitely many poles.

List the finite poles as $p_1,\dots ,p_s$ with orders $m_1,\dots ,m_s$, and set $q(z)={\prod }_{j=1}^s(z-p_j{)}^{m_j}$, taking $q=1$ for the empty pole set; at each $p_j$, [F6] gives the exact pole order, so the corresponding zero of $q$ cancels it and $g=qf$ extends holomorphically there, making $g$ entire.

Put $D={\sum }_j{m}_j$ and $C_q={\prod }_j(1+\mathrm{\mid }{p}_j\mathrm{\mid }{)}^{m_j}$, with $D=0$, $C_q=1$ for the empty product; for $r\ge 1$ and $\mathrm{\mid }z\mathrm{\mid }=r$, $\mathrm{\mid }q(z)\mathrm{\mid }\le C_q{r}^D$, so [F1, F10] and $\frac{1}{2}\log (1+\mathrm{\mid }w{\mathrm{\mid }}^2)\le {\log }^{+}\mathrm{\mid }w\mathrm{\mid }+\frac{1}{2}\log 2$ give $T(r,q)=O(\log r)$. The product law [F4] applied to $g=qf$ and step 1.2 give $T(r,g)=O(\log r)$. Define $T_0(r,g)=m_0(r,g)+N(r,\mathrm{\infty };g)$ as in [F7]; since $g$ is entire its pole count vanishes, and [F1, F10] give $T_0(r,g)\le T(r,g)=O(\log r)$.

If $g$ is constant then $f=g\mathrm{/}q$ is rational; otherwise [F7] with $R=2r$ gives ${\log }^{+}M(r,g)\le 3T_0(2r,g)=O(\log r)$, hence $M(r,g)\le C_1{r}^B$ for some $C_1>0$, $B\ge 0$ and all large $r$. Write $g(z)={\sum }_{n\ge 0}{a}_n{z}^n$; for every integer $n>B$, [F8] gives $\mathrm{\mid }{a}_n\mathrm{\mid }\le M(r,g)\mathrm{/}{r}^n\le C_1{r}^{B-n}\to 0$, so $a_n=0$. Thus only finitely many coefficients are nonzero, [F9] makes $g$ a polynomial, and $f=g\mathrm{/}q$ is rational.