Well-definedness and radius conventions for Nevanlinna quantities

For $a=\mathrm{\infty }$, [F1] identifies the counted points as poles; [F4] makes them isolated, so compactness gives finitely many poles in each bounded closed disc, each of finite order.

At a finite $a$-point $b$ of order $m_b$, [F2], [F3] and [F5] give $f-a=(z-b{)}^{m_b}h$ with $h(b)\mathrm{\ne }0$ and ${\psi }_a(z):=\log (1\mathrm{/}\delta (f(z),a))=-m_b\log \mathrm{\mid }z-b\mathrm{\mid }+q(z)$ for a continuous $q$ near $b$.

At any pole, [F6] makes $w=1\mathrm{/}f$ extend holomorphically with $w(b)=0$; for finite $a$, [F2] rewrites the chordal distance as $\delta (f,a)=\mathrm{\mid }1-aw\mathrm{\mid }\mathrm{/}(\sqrt{1+\mathrm{\mid }w{\mathrm{\mid }}^2}\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2})$, which has a positive limit, so ${\psi }_a$ extends continuously over the pole.

For $b\mathrm{\ne }0$ and $r<\mathrm{\mid }b\mathrm{\mid }$, factor $r{e}^{it}-b=-b(1-(r\mathrm{/}b)e^{it})$; for $r>\mathrm{\mid }b\mathrm{\mid }$, factor it as $r{e}^{it}(1-(b\mathrm{/}r)e^{-it})$. The uniformly convergent series $\log (1-\zeta )=-{\sum }_{n\ge 1}{\zeta }^n\mathrm{/}n$ for $\mathrm{\mid }\zeta \mathrm{\mid }<1$ has zero mean term by term, so the mean of $\log \mathrm{\mid }r{e}^{it}-b\mathrm{\mid }$ is $\log \mathrm{\mid }b\mathrm{\mid }$ or $\log r$, respectively; for $b=0$ it is $\log r$.

For $a=\mathrm{\infty }$ at a pole $b$ of order $m_b$, [F6] gives the reciprocal zero order $m_b$ from the leading Laurent term, and [F5] applied to the zero $w=1\mathrm{/}f$ from step 1.3 gives $f=(z-b{)}^{-m_b}h$ with $h$ holomorphic and nonzero; hence ${\psi }_{\mathrm{\infty }}+m_b\log \mathrm{\mid }z-b\mathrm{\mid }=\frac{1}{2}\log (\mathrm{\mid }z-b{\mathrm{\mid }}^{2m_b}+\mathrm{\mid }h{\mathrm{\mid }}^2)$ extends continuously.

For finite $a$, [F3] isolates the zeros of $f-a$ away from poles, and [F6] prevents such zeros from accumulating at a pole. The finitely many poles from step 1.1 have neighborhoods free of $a$-points; the remaining compact set contains only finitely many isolated zeros. Thus every bounded-disc finite-target count is finite.

If $r=\mathrm{\mid }b\mathrm{\mid }>0$, rotate to $b=\mathrm{\mid }b\mathrm{\mid }$. Then the mean is $\log \mathrm{\mid }b\mathrm{\mid }+(2\pi {)}^{-1}{\int }_0^{2\pi }\log (2\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid })dt=\log \mathrm{\mid }b\mathrm{\mid }$: with $J={\int }_0^{\pi \mathrm{/}2}\log (\sin x)dx$, symmetry and $\sin (2x)=2\sin x\cos x$ give $2J=-(\pi \mathrm{/}2)\log 2+J$, hence $J=-(\pi \mathrm{/}2)\log 2$ and the integral is zero. The endpoint singularities have finite absolute integral since ${\int }_0^{ϵ}\mathrm{\mid }\log x\mathrm{\mid }dx<\mathrm{\infty }$, so [F7] applies.

Integrating the step function in [F1] gives $N(r,a;f)=n(0,a;f)\log r+{\sum }_{0<\mathrm{\mid }b\mathrm{\mid }\le r}{m}_b\log (r\mathrm{/}\mathrm{\mid }b\mathrm{\mid })$, a finite sum by steps 1.1 and 2.2; each term is zero when first included at $r=\mathrm{\mid }b\mathrm{\mid }$, so $N$ is continuous.

Around any fixed $r_0>0$, take a compact annulus containing all nearby circles. Steps 1.1 and 2.2 give finitely many relevant divisor points there; by steps 1.2, 1.3 and 2.1, adding $m_b\log \mathrm{\mid }z-b\mathrm{\mid }$ at each singular divisor leaves a continuous function on the annulus, whose circular mean varies continuously with $r$.

Splitting the defining integral at $r_0$ gives $N(r,a;f)-N(r_0,a;f)=n(0,a;f)\log (r\mathrm{/}{r}_0)+{\int }_{r_0}^r(n(t,a;f)-n(0,a;f))dt\mathrm{/}t={\int }_{r_0}^{r}n(t,a;f)dt\mathrm{/}t$, also for $r<r_0$ as an oriented integral.

By steps 1.4 and 2.3, each removed logarithm has continuous mean $\log \max (r,\mathrm{\mid }b\mathrm{\mid })$, including at $r=\mathrm{\mid }b\mathrm{\mid }$. Combining those means with the continuous remainder from step 3.2 proves $m(r,a;f)$ finite and continuous at every radius. Only finite divisor lists are used, so no AC is needed.

For $f(z)=z+1\mathrm{/}z$ and $a=\mathrm{\infty }$, the reverse triangle inequality gives $\mathrm{\mid }f(r{e}^{it})\mathrm{\mid }\ge \mathrm{\mid }r-r^{-1}\mathrm{\mid }$. Thus [F2] gives $m(r,\mathrm{\infty };f)\ge \frac{1}{2}\log (1+(r-r^{-1}{)}^2)$. At $r=1$, $\mathrm{\mid }f(e^{it})\mathrm{\mid }=\mathrm{\mid }2\cos t\mathrm{\mid }\le 2$, so $m(1,\mathrm{\infty };f)\le \frac{1}{2}\log 5$. At both $r=1\mathrm{/}4$ and $r=4$, the lower bound is $\frac{1}{2}\log (241\mathrm{/}16)>\frac{1}{2}\log 5$. Consequently $m(1\mathrm{/}4,\mathrm{\infty };f)>m(1,\mathrm{\infty };f)$ and $m(4,\mathrm{\infty };f)>m(1,\mathrm{\infty };f)$, ruling out both nondecreasing and nonincreasing behaviour. Proximity alone need not be monotone.