Jensen Theory and Nevanlinna's First Main Theorem: Examples and Counterexamples

The only zero in the radius-$R$ disc is $b$, with multiplicity two; there are no poles or boundary divisor points, and $f(0)=b^2\mathrm{\ne }0$.

By [F1], Poisson–Jensen therefore reads $2\log \mathrm{\mid }z-b\mathrm{\mid }=\frac{1}{2\pi }{\int }_0^{2\pi }\frac{R^2-\mathrm{\mid }z{\mathrm{\mid }}^2}{\mathrm{\mid }R{e}^{it}-z{\mathrm{\mid }}^2}2\log \mathrm{\mid }R{e}^{it}-b\mathrm{\mid }dt-2G_R(z,b)$ for $\mathrm{\mid }z\mathrm{\mid }<R$ with $z\mathrm{\ne }b$. The coefficient is exactly two because the local factor is squared.

At $z=0$, $G_R(0,b)=\log (R\mathrm{/}\mathrm{\mid }b\mathrm{\mid })$. Also $R{e}^{it}-b=R{e}^{it}(1-(b\mathrm{/}R)e^{-it})$, and the convergent logarithm series has zero angular mean, so the boundary mean of $\log \mathrm{\mid }f\mathrm{\mid }$ is $2\log R$. The formula becomes $2\log \mathrm{\mid }b\mathrm{\mid }=2\log R-2\log (R\mathrm{/}\mathrm{\mid }b\mathrm{\mid })$, displaying the stated centre correction.

Since $f(z)-a=c{z}^m$ with $c\mathrm{\ne }0$, its only $a$-point is $0$, of multiplicity $m$, and it has no poles. Thus $n(t,a;f)=m$ for every $t\ge 0$, while $n(t,\mathrm{\infty };f)=0$.

Substituting these counts into [F2] gives $N(r,a;f)=m\log r+{\int }_0^r(m-m)dt\mathrm{/}t=m\log r$ for every $r>0$. The central term is the whole regularized count.

For $0<\epsilon <r$, the unregularized integral from $\epsilon$ to $r$ is ${\int }_{\epsilon }^{r}n(t,a;f)dt\mathrm{/}t=m\log (r\mathrm{/}\epsilon )\to +\mathrm{\infty }$ as $\epsilon \downarrow 0$. Thus ${\int }_0^{r}n(t,a;f)dt\mathrm{/}t$ diverges for every $r>0$, even though the regularized $N(r,a;f)$ is finite.

On $\mathrm{\mid }z\mathrm{\mid }=r$, $\log \mathrm{\mid }f(z)-a\mathrm{\mid }=\log \mathrm{\mid }c\mathrm{\mid }+m\log r$, so its circular mean is $\log \mathrm{\mid }c\mathrm{\mid }+m\log r$. Since $N(r,\mathrm{\infty };f)=0$, [F3] gives $M_r\log \mathrm{\mid }f-a\mathrm{\mid }=\log \mathrm{\mid }c\mathrm{\mid }+N(r,a;f)=\log \mathrm{\mid }c\mathrm{\mid }+m\log r$, agreeing with the direct boundary calculation.

By [F1], the finite-target chordal identity averages to $m(r,a;f)=T(r,f)+\frac{1}{2}\log (1+\mathrm{\mid }a{\mathrm{\mid }}^2)-M_r\log \mathrm{\mid }f-a\mathrm{\mid }$, because $f$ has no poles and hence $T(r,f)=m(r,\mathrm{\infty };f)$. Using step 2.1 gives $m(r,a;f)+N(r,a;f)=T(r,f)+\frac{1}{2}\log (1+\mathrm{\mid }a{\mathrm{\mid }}^2)-\log \mathrm{\mid }c\mathrm{\mid }$. This computes the finite-target constant directly.

Here $f(z)-a=c{z}^m$, so $c_a=c$ and $k_a=m$ in [F3] and [F5]; the First Main Theorem constant is exactly $C(f,a)=\frac{1}{2}\log (1+\mathrm{\mid }a{\mathrm{\mid }}^2)-\log \mathrm{\mid }c\mathrm{\mid }$, agreeing with step 3.1. Since $f(0)-a=0$, $\log \mathrm{\mid }f(0)-a\mathrm{\mid }$ is not a finite logarithm and cannot replace the term $\log \mathrm{\mid }c\mathrm{\mid }$.

Put $m_0(r,h)=(2\pi {)}^{-1}{\int }_0^{2\pi }{\log }^{+}\mathrm{\mid }h(r{e}^{it})\mathrm{\mid }dt$ for an entire $h$. Since such $h$ has no poles, [F1] gives $T(r,h)=m(r,\mathrm{\infty };h)$. For every finite $w$, ${\log }^{+}\mathrm{\mid }w\mathrm{\mid }\le \frac{1}{2}\log (1+\mathrm{\mid }w{\mathrm{\mid }}^2)\le {\log }^{+}\mathrm{\mid }w\mathrm{\mid }+\frac{1}{2}\log 2.$ Thus $m_0(r,h)\le T(r,h)\le m_0(r,h)+\frac{1}{2}\log 2$. [F1, algebra] 1.2 For $h(z)=z^d$, the pole count is zero and $\mathrm{\mid }h(r{e}^{it})\mathrm{\mid }=r^d$ at every angle. Hence $T(r,z^d)=\frac{1}{2}\log (1+r^{2d})=d\log r+\frac{1}{2}\log (1+r^{-2d})=d\log r+O(1)\mathrm{.}$ [F1, algebra] 1.3 Set $E(z)=\exp (2iz)$. The linear polynomial $z\mapsto 2iz$ is entire by [F16], the complex exponential is entire by [F14], and [F15] shows $E$ is entire with $E^{\mathrm{\prime }}(0)={\exp }^{\mathrm{\prime }}(0)2i=2i\exp 0=2i\mathrm{\ne }0$, using [F13]. Thus $E$ is nonconstant. [F13, F14, F15, F16, algebra] 2.1 On $\mathrm{\mid }z\mathrm{\mid }=r$, Euler's formula in [F6] gives $z=r{e}^{it}=r(\cos t+i\sin t)$, so $\mathrm{\mid }\exp (\lambda r{e}^{it})\mathrm{\mid }=e^{\lambda r\cos t}(\lambda >0)\mathrm{.}$ The logarithmic positive part is therefore $(\lambda r\cos t{)}^{+}$. From [F7], cosine is strictly decreasing on $[0,\pi ]$ and strictly increasing on $[\pi ,2\pi ]$. By [F8] and [F9], it has values $1,0,-1,0,1$ at $0,\pi \mathrm{/}2,\pi ,3\pi \mathrm{/}2,2\pi$, respectively. These facts show cosine is positive on $[0,\pi \mathrm{/}2)\cup (3\pi \mathrm{/}2,2\pi ]$ and negative on $(\pi \mathrm{/}2,3\pi \mathrm{/}2)$. By [F11], [F12], and [F10], the cosine integrals below exist and are evaluated using the primitive $\sin t$: ${\int }_0^{2\pi }(\cos t{)}^{+}dt={\int }_0^{\pi \mathrm{/}2}\cos tdt+{\int }_{3\pi \mathrm{/}2}^{2\pi }\cos tdt=(\sin \frac{\pi }{2}-\sin 0)+(\sin 2\pi -\sin \frac{3\pi }{2})=2.$ Consequently $m_0(r,\exp (\lambda z))=\lambda r\mathrm{/}\pi$. Step 1.1 now gives $T(r,\exp (\lambda z))=\lambda r\mathrm{/}\pi +O(1)$, in particular the asserted formula for $\exp z$. [F6, F7, F8, F9, F10, F11, F12, step 1.1, algebra] 2.2 Let $R(w)=-i(w-1)\mathrm{/}(w+1)$. This is a degree-one rational map. By [F2], $R(E)$ is the meromorphic composition to which the characteristic law applies. The definitions in [F5] and the addition law [F4] give, wherever $\cos z\mathrm{\ne }0$, $\frac{\sin z}{\cos z}=-i\frac{\exp (2iz)-1}{\exp (2iz)+1}=-i\frac{E(z)-1}{E(z)+1}\mathrm{.}$ Indeed [F4] and [F13] give $\exp (iz)\exp (-iz)=1$, so multiplying the numerator and denominator in [F5] by $\exp (iz)$ is legitimate. They give $\cos z=\frac{\exp (-iz)(E(z)+1)}{2},\sin z=\frac{\exp (-iz)(E(z)-1)}{2i}\mathrm{.}$ Thus $\cos z=0$ exactly when $E(z)=-1$. Since $E$ is nonconstant by step 1.3, [F17] makes each zero of $E+1$ isolated, and [F18] factors it with a finite positive order there. At such a point $E-1=-2$, so $R(E)$ has a pole of that order, while the quotient has the same pole because its numerator is nonzero. At every other point the quotient equals $R(E)$. Hence $R(E)$ is precisely the meromorphic continuation of $\sin z\mathrm{/}\cos z$, the complex tangent used here. [F2, F4, F5, F13, F17, F18, step 1.3, algebra] 3.1 For $z=r{e}^{it}$, [F6] and the decomposition in step 2.1 give ${\log }^{+}\mathrm{\mid }E(r{e}^{it})\mathrm{\mid }=(-2r\sin t{)}^{+}\mathrm{.}$ By [F7], [F8], and [F9], sine increases from $0$ to $1$ on $[0,\pi \mathrm{/}2]$, decreases to $0$ on $[\pi \mathrm{/}2,\pi ]$, and its shift by $\pi$ changes sign. Thus it is positive on $(0,\pi )$ and negative on $(\pi ,2\pi )$. By [F11], [F12], and [F10], $-\sin t$ is integrable and has primitive $\cos t$. Hence ${\int }_0^{2\pi }(-\sin t{)}^{+}dt={\int }_{\pi }^{2\pi }-\sin tdt=\cos 2\pi -\cos \pi =2,$ and $m_0(r,E)=2r\mathrm{/}\pi$. Step 1.1 gives $T(r,E)=2r\mathrm{/}\pi +O(1)$. [F6, F7, F8, F9, F10, F11, F12, step 1.1, step 2.1, algebra] 4.1 The numerator and denominator of $R$ are coprime linear polynomials, so $R$ has degree one. By [F2], step 1.3, step 2.2, and step 3.1, $T(r,\tan z)=T(r,R(E(z)))=T(r,E)+O(1)=\frac{2r}{\pi }+O(1)\mathrm{.}$ [F2, step 1.3, step 2.2, step 3.1, algebra] 5.1 For each $d\ge 1$, step 1.2 has $T(r,z^d)=d\log r+O(1)>1$ eventually, so $\log T(r,z^d)\mathrm{/}\log r\to 0$. Steps 2.1 and 4.1 give positive linear growth for $\exp z$ and $\tan z$, so for either function $\log T(r,h)=\log r+O(1)$ and the ratio tends to $1$. The three functions are nonconstant (the monomial has $d\ge 1$, $\exp z$ has derivative $1$ at zero by [F14], and step 1.3 shows $E$ is nonconstant; the positive linear growth of $R(E)$ in step 4.1 also rules out a constant tangent). Thus [F3] applies; in each case the limsup and liminf agree with the computed limit.

The numerator and denominator of $M_a$ have determinant $-(1+\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{\ne }0$, so $M_a$ is a degree-one Möbius map; also $M_a(a)=\mathrm{\infty }$ and the limit at $w=\mathrm{\infty }$ is $\bar{a}$. The identity $M_a(w)=\bar{a}+(1+\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{/}(w-a)$ holds for finite $w\mathrm{\ne }a$.

For finite $w\mathrm{\ne }a$, $\mathrm{\mid }1+\bar{a}w{\mathrm{\mid }}^2+\mathrm{\mid }w-a{\mathrm{\mid }}^2=(1+\mathrm{\mid }a{\mathrm{\mid }}^2)(1+\mathrm{\mid }w{\mathrm{\mid }}^2)$, so [F1] gives $\delta (M_a(w),\mathrm{\infty })=\mathrm{\mid }w-a\mathrm{\mid }\mathrm{/}\sqrt{(1+\mathrm{\mid }a{\mathrm{\mid }}^2)(1+\mathrm{\mid }w{\mathrm{\mid }}^2)}=\delta (w,a)$. At $w=a$, both sides are zero since $M_a(a)=\mathrm{\infty }$; at $w=\mathrm{\infty }$, $\delta (M_a(\mathrm{\infty }),\mathrm{\infty })=1\mathrm{/}\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}=\delta (\mathrm{\infty },a)$. Thus the pointwise identity holds on the whole sphere, including both endpoints.

Let $F=M_a\circ f$. At a finite point $b$ with $f(b)=a$ of order $m$, [F4] gives $f(z)-a=(z-b{)}^{m}h(z)$ with $h(b)\mathrm{\ne }0$; the numerator $1+\bar{a}f(z)$ equals $1+\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{\ne }0$ at $b$, so $F$ has a pole of order $m$. At a pole $b$ of $f$ of order $m$, [F5] says $1\mathrm{/}f$ has a zero of order $m$; since $1\mathrm{/}(f-a)=(1\mathrm{/}f)\mathrm{/}(1-a\mathrm{/}f)$ and $1-a\mathrm{/}f$ is nonzero at $b$, step 1.1 shows $F=\bar{a}+(1+\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{/}(f-a)$ extends holomorphically and finitely there. At every other point $f$ is finite and different from $a$, so $F$ is finite and holomorphic. Hence the poles of $F$ are exactly the $a$-points of $f$ with the same multiplicities; the closed-disc counts and their integrated versions satisfy $N(r,\mathrm{\infty };F)=N(r,a;f)$ for every $r>0$.

By step 2.1, the integrands defining $m(r,\mathrm{\infty };F)$ and $m(r,a;f)$ are equal at every point of the circle, with the same logarithmic singularity at an $a$-point and the same finite value at a pole of $f$. Therefore $m(r,\mathrm{\infty };F)=m(r,a;f)$ for every $r>0$.

The map $M_a$ has degree one and is invertible, so $F$ is nonconstant; [F3] gives $T(r,F)=T(r,f)+O_{f,a}(1)$. Steps 2.2–3.1 also give $T(r,F)=m(r,a;f)+N(r,a;f)$, and [F2] identifies this sum exactly as $T(r,f)+C(f,a)$. This proves the asserted characteristic estimate and confirms the target count/proximity relation.

By [F3], the product defines an entire function $g$. On every compact $K$, for all large $n$ the series for $\mathrm{Log}(1+z\mathrm{/}n)-z\mathrm{/}n$ is bounded by $C_K{n}^{-2}$ uniformly on $K$, so the product tail is the exponential of a locally uniformly convergent sum and is nowhere zero. The finitely many factors then give simple zeros exactly at $0,-1,-2,\dots$; [F5] identifies these with the simple poles of the meromorphic continuation of $\mathrm{\Gamma }$, and $g$ has no poles.

Fix $r\ge 2$ and $\mathrm{\mid }z\mathrm{\mid }=r$. At a product zero the upper bound is immediate; otherwise $\log \mathrm{\mid }g(z)\mathrm{\mid }=\log r+\mathrm{\Re }(\gamma z)+{\sum }_{n\ge 1}(\log \mathrm{\mid }1+z\mathrm{/}n\mathrm{\mid }-\mathrm{\Re }z\mathrm{/}n)$. For $n\le 2r$, the summands are at most $\log (1+r\mathrm{/}n)+r\mathrm{/}n$, with total $O(r\log r)$. For $n>2r$, $w=z\mathrm{/}n$ satisfies $\mathrm{\mid }w\mathrm{\mid }<1\mathrm{/}2$, so $\log \mathrm{\mid }1+w\mathrm{\mid }-\mathrm{\Re }w=\mathrm{\Re }(\mathrm{Log}(1+w)-w)\le {\sum }_{k\ge 2}\mathrm{\mid }w{\mathrm{\mid }}^k\mathrm{/}k\le C\mathrm{\mid }w{\mathrm{\mid }}^2$; the tail is at most $C{r}^2{\sum }_{n>2r}{n}^{-2}=O(r)$. Since $\mathrm{\mid }\mathrm{\Re }(\gamma z)\mathrm{\mid }\le \mathrm{\mid }\gamma \mathrm{\mid }r$, this gives ${\log }^{+}\mathrm{\mid }g(z)\mathrm{\mid }\le Cr\log r$ uniformly on the circle, and hence $m_0(r,g)=O(r\log r)$.

Set $\delta =\pi \mathrm{/}4$ and $I=[2\pi \mathrm{/}3,3\pi \mathrm{/}4]$, a fixed arc in the closed sector $\mathrm{\mid }\arg z\mathrm{\mid }\le 3\pi \mathrm{/}4$. For $z=r{e}^{it}$ with $t\in I$, [F3] identifies $g$ with $1\mathrm{/}\mathrm{\Gamma }$ and [F4] applies uniformly: writing its error as $E(z)=O(r^{-1})$, $\log \mathrm{\mid }g(r{e}^{it})\mathrm{\mid }=-r\cos t\log r+rt\sin t+r\cos t+\frac{1}{2}\log r-\frac{1}{2}\log (2\pi )+O(r^{-1})$. Since $-\cos t\ge 1\mathrm{/}2$, $t\sin t\ge 0$, and $\cos t\ge -1$, this is at least $\frac{1}{4}r\log r>0$ for all sufficiently large $r$, uniformly on $I$. This closed arc meets the exact sector hypotheses in [F4].

Integrating the lower bound of step 1.3 over the arc of length $\pi \mathrm{/}12$ gives $m_0(r,g):=(2\pi {)}^{-1}{\int }_0^{2\pi }{\log }^{+}\mathrm{\mid }g(r{e}^{it})\mathrm{\mid }dt\ge (1\mathrm{/}96)r\log r$ for all sufficiently large $r$.

Since $g$ is entire, [F1] gives $T(r,g)=m(r,\mathrm{\infty };g)$. The pointwise inequality ${\log }^{+}\mathrm{\mid }w\mathrm{\mid }\le \frac{1}{2}\log (1+\mathrm{\mid }w{\mathrm{\mid }}^2)\le {\log }^{+}\mathrm{\mid }w\mathrm{\mid }+\frac{1}{2}\log 2$ gives $m_0(r,g)\le T(r,g)\le m_0(r,g)+\frac{1}{2}\log 2$. Steps 1.2 and 2.1 prove $T(r,g)=\mathrm{\Theta }(r\log r)$, so $T>1$ eventually. By [F2], $\log T(r,g)\mathrm{/}\log r=(\log r+\log \log r+O(1))\mathrm{/}\log r\to 1$, proving both order assertions.

Polynomial division gives $R(z)=z+1+2\mathrm{/}(z-1)$. The numerator equals $2$ at $z=1$, so this is the unique pole and it is simple; the numerator and denominator are coprime and their maximum degree is $2$. Also $R(0)=-1$, so $n(0,\mathrm{\infty };R)=0$; then $n(t,\mathrm{\infty };R)=0$ for $t<1$ and $n(t,\mathrm{\infty };R)=1$ for $t\ge 1$. By [F3, F4], $N(1,\mathrm{\infty };R)=0$ (the boundary pole has logarithmic weight $\log 1=0$), while for every $r>1$, $N(r,\mathrm{\infty };R)={\int }_1^{r}dt\mathrm{/}t=\log r$.

For $\mathrm{\mid }z\mathrm{\mid }=r\ge 4$, the decomposition gives $\mathrm{\mid }R(z)\mathrm{\mid }\ge r-1-2\mathrm{/}(r-1)\ge r\mathrm{/}2$ and $\mathrm{\mid }R(z)\mathrm{\mid }\le r+1+2\mathrm{/}(r-1)\le 2r$. Hence [F2] bounds the pointwise proximity by $\log r-\log 2\le \frac{1}{2}\log (1+\mathrm{\mid }R(z){\mathrm{\mid }}^2)\le \log r+\frac{1}{2}\log (4+r^{-2})=\log r+O(1)$ uniformly on the circle. Averaging gives $m(r,\mathrm{\infty };R)=\log r+O(1)$.

By [F1] and steps 1.1–1.2, $T(r,R)=2\log r+O(1)$. Also, [F1, F2] give $T(r,z)=\frac{1}{2}\log (1+r^2)=\log r+O(1)$ because $z$ has no poles and $\mathrm{\mid }z\mathrm{\mid }=r$ on the averaging circle. Since $R$ has degree two, [F5] applied to the identity map gives the same $T(r,R)=2T(r,z)+O(1)$; this agrees with the exact rational degree law [F6].