Separated-radius Poisson–Jensen derivative bound

(Reduction to centre value $1$) Let $k\in \mathbb{Z}$ be the signed order of $f$ at $0$ (negative for a pole), and write $f(z)=c{z}^{k}h(z)$ with $c\mathrm{\ne }0$, $h$ meromorphic on $\mathbb{C}$, and $h(0)=1$. The local zero and pole factorizations justify this form; when $k=0$, take $c=f(0)$. Then $f^{\mathrm{\prime }}\mathrm{/}f=k\mathrm{/}z+h^{\mathrm{\prime }}\mathrm{/}h$. The proximity sum inequality gives, for $r\ge 2$, $m_0(r,f^{\mathrm{\prime }}\mathrm{/}f)\le m_0(r,h^{\mathrm{\prime }}\mathrm{/}h)+{\log }^{+}(\mathrm{\mid }k\mathrm{\mid }\mathrm{/}r)+\log 2\le m_0(r,h^{\mathrm{\prime }}\mathrm{/}h)+{\log }^{+}\mathrm{\mid }k\mathrm{\mid }+\log 2,$ where the $k\mathrm{/}z$ term is zero when $k=0$. If $h$ is constant then $h\equiv 1$, so $f^{\mathrm{\prime }}\mathrm{/}f=k\mathrm{/}z$ and $m_0(r,f^{\mathrm{\prime }}\mathrm{/}f)\le {\log }^{+}\mathrm{\mid }k\mathrm{\mid }$ for $r\ge 2$, proving the stated bound directly with a constant depending on $f$. In all subsequent steps assume $h$ is nonconstant. The characteristic laws and the direct rational-map estimate $T(R,c^{-1}{z}^{-k})=\mathrm{\mid }k\mathrm{\mid }\log R+O_f(1)$ for $R\ge 2$ give $T(R,h)\le T(R,f)+\mathrm{\mid }k\mathrm{\mid }\log R+O_f(1)\mathrm{.}$

(Vanishing centre constants) For $h$ with $h(0)=1$, [F3] gives $C(h,0)=\frac{1}{2}\log 1-\log 1=0$ and $C(h,\mathrm{\infty })=0$, hence $m(s,0;h)+N(s,0;h)=T(s,h)=m(s,\mathrm{\infty };h)+N(s,\mathrm{\infty };h)$. For every $w$ one has ${\log }^{+}\mathrm{\mid }w\mathrm{\mid }\le \frac{1}{2}\log (1+\mathrm{\mid }w{\mathrm{\mid }}^2)$ and ${\log }^{+}\frac{1}{\mathrm{\mid }w\mathrm{\mid }}\le \frac{1}{2}\log (1+\mathrm{\mid }w{\mathrm{\mid }}^{-2})$, so taking angular means gives $\frac{1}{2\pi }{\int }_0^{2\pi }\mid \log \mathrm{\mid }h(s{e}^{it})\mathrm{\mid }\mid dt\le m(s,0;h)+m(s,\mathrm{\infty };h)\le 2T(s,h)\mathrm{.}$

(Differentiated Poisson–Jensen) Let $0<r<s<R$ with $\mathrm{\mid }z\mathrm{\mid }=s$ carrying no zero or pole of $h$; [F1] applies on $\mathrm{\mid }z\mathrm{\mid }<s$. Differentiation in $z$ of [F1], whose boundary kernel and Green kernels are smooth for $\mathrm{\mid }z\mathrm{\mid }<s$ and whose divisor sum is finite, gives for $\mathrm{\mid }z\mathrm{\mid }<s$ $\frac{h^{\mathrm{\prime }}(z)}{h(z)}=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }h(s{e}^{it})\mathrm{\mid }\frac{2s{e}^{it}}{(s{e}^{it}-z{)}^2}dt+{\sum }_{h(c)=0}\frac{s^2-\mathrm{\mid }c{\mathrm{\mid }}^2}{(s^2-\bar{c}z)(z-c)}-{\sum }_{p\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{e}}\frac{s^2-\mathrm{\mid }p{\mathrm{\mid }}^2}{(s^2-\bar{p}z)(z-p)},$ the divisor sums running over the zeros and poles in $\mathrm{\mid }z\mathrm{\mid }<s$ with multiplicity; at a radius $s$ meeting the divisor, take regular $s_j\downarrow s$ and pass to the limit using the continuity of [F5].

(Angular integral of one kernel) For any $c$ and every $\phi$ one has $\mathrm{\mid }r{e}^{i\phi }-c\mathrm{\mid }\ge r\mathrm{\mid }\sin \phi \mathrm{\mid }$ after rotating $c$ to $\mathrm{\mid }c\mathrm{\mid }$: indeed $r^2-2r\mathrm{\mid }c\mathrm{\mid }\cos \phi +\mathrm{\mid }c{\mathrm{\mid }}^2-r^2{\sin }^2\phi =(r\cos \phi -\mathrm{\mid }c\mathrm{\mid }{)}^2\ge 0$. Hence, using $\sin \phi \ge 2\phi \mathrm{/}\pi$ on $[0,\pi \mathrm{/}2]$, $\frac{1}{2\pi }{\int }_0^{2\pi }\frac{d\phi }{\mathrm{\mid }r{e}^{i\phi }-c{\mathrm{\mid }}^{\alpha }}\le \frac{1}{2\pi r^{\alpha }}{\int }_0^{2\pi }\frac{d\phi }{\mathrm{\mid }\sin \phi {\mathrm{\mid }}^{\alpha }}\le \frac{2}{(1-\alpha )r^{\alpha }}\mathrm{.}$

(Kernel bounds) For $\mathrm{\mid }z\mathrm{\mid }=r<s$: $\mid \frac{2s{e}^{it}}{(s{e}^{it}-z{)}^2}\mid \le \frac{2s}{(s-r{)}^2}$; and for a divisor point $c$ with $\mathrm{\mid }c\mathrm{\mid }<s$, using $\mathrm{\mid }{s}^2-\bar{c}z\mathrm{\mid }\ge s^2-\mathrm{\mid }c\mathrm{\mid }r$, $\mid \frac{s^2-\mathrm{\mid }c{\mathrm{\mid }}^2}{(s^2-\bar{c}z)(z-c)}\mid =\frac{(s+\mathrm{\mid }c\mathrm{\mid })(s-\mathrm{\mid }c\mathrm{\mid })}{\mathrm{\mid }{s}^2-\bar{c}z\mathrm{\mid }\mathrm{\mid }z-c\mathrm{\mid }}\le \frac{s+\mathrm{\mid }c\mathrm{\mid }}{s}\cdot \frac{1}{\mathrm{\mid }z-c\mathrm{\mid }}\le \frac{2}{\mathrm{\mid }z-c\mathrm{\mid }},$ because $\frac{(s-\mathrm{\mid }c\mathrm{\mid })s}{s^2-r\mathrm{\mid }c\mathrm{\mid }}\le 1$ for $0\le \mathrm{\mid }c\mathrm{\mid }<s$ and $r<s$.

(Pointwise bound) Combining steps 1.3 and 2.1 with step 1.2 at $\mathrm{\mid }z\mathrm{\mid }=r$, $\mid \frac{h^{\mathrm{\prime }}(z)}{h(z)}\mid \le \frac{2s}{(s-r{)}^2}\cdot 2T(s,h)+2{\sum }_{\mathrm{\mid }c\mathrm{\mid }<s}\frac{1}{\mathrm{\mid }z-c\mathrm{\mid }}=\frac{4sT(s,h)}{(s-r{)}^2}+2\mathrm{\Sigma }(z),$ where the last sum extends over all zeros and poles of $h$ in $\mathrm{\mid }z\mathrm{\mid }<s$ (each repeated according to multiplicity) and is finite by [F5].

($\alpha$-power mean) Fix $\alpha \in (0,1)$. By $(x+y{)}^{\alpha }\le x^{\alpha }+y^{\alpha }$ for $x,y\ge 0$, $\frac{1}{2\pi }{\int }_0^{2\pi }\mid \frac{h^{\mathrm{\prime }}(r{e}^{i\phi })}{h(r{e}^{i\phi })}{\mid }^{\alpha }d\phi \le (\frac{4sT(s,h)}{(s-r{)}^2}{)}^{\alpha }+\frac{2^{\alpha }}{2\pi }{\int }_0^{2\pi }\mathrm{\Sigma }(r{e}^{i\phi }{)}^{\alpha }d\phi ,$ and $\mathrm{\Sigma }(r{e}^{i\phi }{)}^{\alpha }\le {\sum }_{\mathrm{\mid }c\mathrm{\mid }<s}\mathrm{\mid }r{e}^{i\phi }-c{\mathrm{\mid }}^{-\alpha }$ again by the subadditivity for exponent $\alpha <1$.

(Counting the divisor) Let $n(s;0,\mathrm{\infty }):=n(s,0;h)+n(s,\mathrm{\infty };h)$. From [F5], $N(R,a;h)-N(s,a;h)\ge n(s,a;h)\log (R\mathrm{/}s)$ for $a=0,\mathrm{\infty }$; by [F3], $N(R,a;h)\le T(R,h)+C(h,a)$, and $\log \frac{R}{s}\ge \frac{R-s}{R}$ for $s<R$. With $s:=\frac{R+r}{2}$ this gives $n(s,a;h)\le \frac{2R(T(R,h)+C(h,a))}{R-r}$, hence $n(s;0,\mathrm{\infty })\le \frac{4RT(R,h)}{R-r}+C_h^{\mathrm{\prime }}$ for a constant $C_h^{\mathrm{\prime }}\ge 0$.

(Proximity bound) On normalized angular measure put $X(\phi ):=\mathrm{\mid }{h}^{\mathrm{\prime }}(r{e}^{i\phi })\mathrm{/}h(r{e}^{i\phi }){\mathrm{\mid }}^{\alpha }$, defined arbitrarily at its measure-zero singularities. Step 4.1 gives $X\in L^1$, so $\log (1+X)\in L^1$. For $u\ge 0$, ${\log }^{+}u\le {\alpha }^{-1}\log (1+u^{\alpha })$; applying this pointwise and [F8] to $X$ yields $m_0(r,h^{\mathrm{\prime }}\mathrm{/}h)\le \frac{1}{\alpha }\log (1+\frac{1}{2\pi }{\int }_0^{2\pi }X(\phi )d\phi )\mathrm{.}$ For $s=\frac{R+r}{2}$, steps 4.1 and 4.2 bound the integral mean by $A^{\alpha }+B$, where $A:=\frac{16RT(R,h)}{(R-r{)}^2},B:=\frac{2^{\alpha +1}}{(1-\alpha )r^{\alpha }}(\frac{4RT(R,h)}{R-r}+C_h^{\mathrm{\prime }})\mathrm{.}$ Thus $m_0(r,h^{\mathrm{\prime }}\mathrm{/}h)\le {\alpha }^{-1}\log (1+A^{\alpha }+B)$. Since $r\ge 2$, $\log (1+A^{\alpha }+B)\le \log 3+{\log }^{+}A+{\log }^{+}B,$ while ${\log }^{+}A\le {\log }^{+}T(R,h)+\log R+2{\log }^{+}\frac{1}{R-r}+\log 16$ and ${\log }^{+}B\le C_{\alpha }+C_h+{\log }^{+}T(R,h)+\log R+{\log }^{+}\frac{1}{R-r}\mathrm{.}$ Absorbing fixed terms into $C_{h,\alpha }$ and enlarging $C_{\alpha }$ proves the required bound.

(Undoing the normalization) By step 1.1, $T(R,h)\le T(R,f)+O_f(\log R)$ and ${\log }^{+}T(R,h)\le {\log }^{+}T(R,f)+\log R+O_f(1)$ for $R\ge 2$ after enlarging the fixed constants, by continuity on any remaining compact radius interval; absorbing the constants depending on $f$ (including ${\log }^{+}\mathrm{\mid }k\mathrm{\mid }$ and the fixed normalization terms) into $C_{f,\alpha }$ and the numerical factors into $C_{\alpha }$ yields $m_0(r,f^{\mathrm{\prime }}\mathrm{/}f)\le C_{f,\alpha }+C_{\alpha }({\log }^{+}T(R,f)+\log R+{\log }^{+}\frac{1}{R-r})$ for all $2\le r<R$.