Nevanlinna's Second Main Theorem and Defects: Examples and Counterexamples

By [F2] the function $f$ is entire, so as a meromorphic sphere map it has no poles: $f(z)\mathrm{\ne }\mathrm{\infty }$ for every $z$, i.e. $f$ omits $\mathrm{\infty }$; and $\mathrm{\mid }f(z)\mathrm{\mid }=e^{\mathrm{Re}z}>0$, so $f$ omits $0$.

Let $a\in \mathbb{C}\setminus \{0\}$ and let $b$ be a logarithm of $a$, so $e^b=a$ by [F4]. For $z\in \mathbb{C}$, by [F3], $e^z=a$ holds if and only if $e^{z-b}=1$, if and only if $z-b\in 2\pi i\mathbb{Z}$, if and only if $z=b+2\pi ik$ for some $k\in \mathbb{Z}$. Hence the preimage of every nonzero finite value is exactly this arithmetic progression in $b$.

At a point $z=b+2\pi ik$ the derivative is $f^{\mathrm{\prime }}(z)=e^z=a\mathrm{\ne }0$ by [F2] and step 1.2; therefore $f-a$ has a simple zero there, that is, the value $a$ is attained only with multiplicity one.

Combining steps 1.1, 1.2 and 2.1: the two sphere values $0$ and $\mathrm{\infty }$ are omitted, and no other sphere value is omitted. In the notation of [F1], for $a=0,\mathrm{\infty }$ the counting function $N(r,a;f)$ vanishes identically and the whole characteristic sits in the proximity term, $m(r,a;f)=T(r,f)+C(f,a)$; the corresponding deficiency-one computation is carried out in the companion example on this page devoted to deficiencies of elementary functions.

The argument is choice-free: the only choices made are the fixed logarithm $b$ of $a$ and the integer enumeration of the progression, both of which are data of the example. Countable Choice is carried only because the surrounding Nevanlinna quantities and their exceptional-set interface are stated under it.

(Exponential: characteristic) By [F1] and [F6], for the entire function $e^z$ one has $m_0(r,e^z)\le T(r,e^z)\le m_0(r,e^z)+\frac{1}{2}\log 2$. For $z=r{e}^{it}$ the modulus formula [F5] gives ${\log }^{+}\mathrm{\mid }{e}^z\mathrm{\mid }=\max (0,r\cos t)=r{\cos }^{+}t$, and ${\int }_0^{2\pi }{\cos }^{+}tdt={\int }_{-\pi \mathrm{/}2}^{\pi \mathrm{/}2}\cos tdt=2$, so $m_0(r,e^z)=\frac{r}{2\pi }\cdot 2=\frac{r}{\pi }$. Hence $T(r,e^z)=\frac{r}{\pi }+O(1)$.

(Exponential: $a$-point counts) By [F5] the kernel of the exponential is $2\pi i\mathbb{Z}$ and every nonzero value is attained, so for fixed $a\mathrm{\ne }0$ and a logarithm $b$ of $a$ the $a$-points of $e^z$ are exactly the points $b+2\pi ik$, $k\in \mathbb{Z}$, and they are simple because $(e^z{)}^{\mathrm{\prime }}=e^z\mathrm{\ne }0$ there. The number of $k\in \mathbb{Z}$ with $\mathrm{\mid }b+2\pi ik\mathrm{\mid }\le t$ is $\frac{t}{\pi }+O(1)$ for large $t$: writing $b=u+iv$, the condition is $\mathrm{\mid }v+2\pi k\mathrm{\mid }\le \sqrt{t^2-u^2}$, a nonempty integer interval of length $\frac{t}{\pi }+O(1)$. Therefore $n(t,a;e^z)=\frac{t}{\pi }+O(1)$ and $N(r,a;e^z)=\bar{N}(r,a;e^z)=\frac{r}{\pi }+O(\log r)$ with $N_1(r,a;e^z)=0$.

(Sine: characteristic) By [F7] sine is entire, so [F6] gives $m_0(r,\sin )\le T(r,\sin )\le m_0(r,\sin )+\frac{1}{2}\log 2$. For $z=r{e}^{it}$ the definition [F7] and the modulus formula [F5] give $\mathrm{\mid }\sin z\mathrm{\mid }\le \frac{1}{2}(e^{\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}+e^{-\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }})\le e^{\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}=e^{r\mathrm{\mid }\sin t\mathrm{\mid }}$, so ${\log }^{+}\mathrm{\mid }\sin z\mathrm{\mid }\le r\mathrm{\mid }\sin t\mathrm{\mid }$ and $m_0(r,\sin )\le \frac{r}{2\pi }{\int }_0^{2\pi }\mathrm{\mid }\sin t\mathrm{\mid }dt=\frac{2r}{\pi }$ by [F7]. Conversely, $\mathrm{\mid }\sin z\mathrm{\mid }\ge \frac{1}{2}(e^{\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }}-e^{-\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }})\ge \frac{1}{4}{e}^{r\mathrm{\mid }\sin t\mathrm{\mid }}$ whenever $r\mathrm{\mid }\sin t\mathrm{\mid }\ge 1$, so ${\log }^{+}\mathrm{\mid }\sin z\mathrm{\mid }\ge r\mathrm{\mid }\sin t\mathrm{\mid }-\log 4$ on that set, and since the integrand is nonnegative everywhere, $m_0(r,\sin )\ge \frac{1}{2\pi }{\int }_0^{2\pi }(r\mathrm{\mid }\sin t\mathrm{\mid }-\log 4)dt=\frac{2r}{\pi }-\log 4$. Hence $T(r,\sin )=\frac{2r}{\pi }+O(1)$.

(Sine: reduction to a quadratic) Fix $a\in \mathbb{C}$; putting $w=e^{iz}$, one has $\sin z=a$ if and only if $w^2-2iaw-1=0$, because $e^{iz}\mathrm{\ne }0$ and multiplying $\frac{w-w^{-1}}{2i}=a$ by $2iw$ is reversible. If $q(w)=w^2-2iaw-1$ has a double root $w$, then $w=ia$ and $w^2=-1$, forcing $a^2=1$; so for $a\mathrm{\ne }\pm 1$ the polynomial $q$ has two distinct roots $w_1,w_2$, and $q(0)=-1$ shows both are nonzero. By [F7] a root $w_1\mathrm{\ne }0$ exists, and $w_2:=-1\mathrm{/}{w}_1$ is a second root, since $w_1^2-2ia{w}_1-1=0$ gives $q(-1\mathrm{/}{w}_1)=w_1^{-2}(1+2ia{w}_1-w_1^2)=0$.

(Sine: simplicity away from $\pm 1$) At a solution of $\sin z=a$ with $w=e^{iz}$ one has $\cos z=\frac{w+w^{-1}}{2}=\frac{w^2+1}{2w}$ by [F7], which vanishes exactly when $w=\pm i$ by [F7]; then $\sin z=\frac{w-w^{-1}}{2i}$ equals $1$ for $w=i$ and $-1$ for $w=-i$. Hence for $a\mathrm{\ne }\pm 1$ no $a$-point has $\cos z=0$, and since ${\sin }^{\mathrm{\prime }}=\cos$ by [F7] every $a$-point is simple.

(Exponential: deficiencies) Since $e^z$ is entire, $N(r,\mathrm{\infty };e^z)=0$ and [F3] gives $\delta (\mathrm{\infty },e^z)=1$; since $e^z\mathrm{\ne }0$ for all $z$ the value $0$ is omitted and [F3] gives $\delta (0,e^z)=1$; for $0\mathrm{\ne }a\in \mathbb{C}$ step 1.2 gives $\frac{N(r,a;e^z)}{T(r,e^z)}\to 1$, hence $\delta (a,e^z)=1-{\mathrm{lim\hspace{0.17em}sup}}_r\frac{N(r,a;e^z)}{T(r,e^z)}=0$.

(Exponential: ramification indices) For finite $a\mathrm{\ne }0$ all $a$-points of $e^z$ are simple by step 1.2, so $N_1(r,a;e^z)=0$ and $\epsilon (a,e^z)=0$; for $a=0$ the counting function is identically zero, so $\epsilon (0,e^z)=0$; and $N_1(r,\mathrm{\infty };e^z)=0$ because $e^z$ has no poles, so $\epsilon (\mathrm{\infty },e^z)=0$.

(Sine: the $a$-point progressions) With $w_j\mathrm{\ne }0$ as in step 1.4, fix $b_j$ with $e^{b_j}=w_j$ by [F5]; then $e^{iz}=w_j$ if and only if $iz-b_j\in 2\pi i\mathbb{Z}$, that is, $z\in -i{b}_j+2\pi \mathbb{Z}$ by [F5]. Hence for $a\mathrm{\ne }\pm 1$ the solutions of $\sin z=a$ are exactly the two disjoint arithmetic progressions $-i{b}_1+2\pi \mathbb{Z}$ and $-i{b}_2+2\pi \mathbb{Z}$.

(Sine: the values $\pm 1$) For $a=1$ the quadratic of step 1.4 is $q(w)=(w-i{)}^2$, so $\sin z=1$ if and only if $e^{iz}=i$, that is, $z\in \frac{\pi }{2}+2\pi \mathbb{Z}$; at these points $\cos z=0$ by [F7] and ${\sin }^{\mathrm{\prime }\mathrm{\prime }}=-\sin =-1\mathrm{\ne }0$ by [F7], so each is a double point: $n(t,1;\sin )=\frac{2t}{\pi }+O(1)$ with multiplicity and $\bar{n}(t,1;\sin )=\frac{t}{\pi }+O(1)$, whence $N(r,1;\sin )=\frac{2r}{\pi }+O(\log r)$, $\bar{N}(r,1;\sin )=\frac{r}{\pi }+O(\log r)$ and $N_1(r,1;\sin )=\frac{r}{\pi }+O(\log r)$. The same computation with $-i$ and the progression $-\frac{\pi }{2}+2\pi \mathbb{Z}$ gives the analogous statements for $a=-1$.

(Sine: counting for $a\mathrm{\ne }\pm 1$) For a fixed $c\in \mathbb{C}$, the number of $k\in \mathbb{Z}$ with $\mathrm{\mid }c+2\pi ik\mathrm{\mid }\le t$ is $\frac{t}{\pi }+O(1)$ as $t\to \mathrm{\infty }$: the condition is $\mathrm{\mid }2\pi k+\mathrm{Im}c\mathrm{\mid }\le \sqrt{t^2-(\mathrm{Re}c{)}^2}$, an integer interval of length $\frac{t}{\pi }+O(1)$. The two disjoint progressions of step 2.3 therefore give $n(t,a;\sin )=\frac{2t}{\pi }+O(1)$, and by step 1.5 all $a$-points are simple, so $N(r,a;\sin )=\bar{N}(r,a;\sin )=\frac{2r}{\pi }+O(\log r)$ and $N_1(r,a;\sin )=0$.

(Sine: deficiencies and ramification indices) Since sine is entire, $N(r,\mathrm{\infty };\sin )=0$, so $\delta (\mathrm{\infty },\sin )=1$ by [F3]. For every finite $a$, steps 3.1 and 2.4 give $\frac{N(r,a;\sin )}{T(r,\sin )}\to 1$ because $T(r,\sin )=\frac{2r}{\pi }+O(1)$ by step 1.3, so $\delta (a,\sin )=1-{\mathrm{lim\hspace{0.17em}sup}}_r\frac{N}{T}=0$. For $a\mathrm{\notin }\{1,-1\}$ step 3.1 gives $N_1(r,a;\sin )=0$, so $\epsilon (a,\sin )=0$, including $a=0$; and $\epsilon (\mathrm{\infty },\sin )=0$ because the $N_1$-count at $\mathrm{\infty }$ vanishes for an entire function; while for $a=\pm 1$ step 2.4 gives $\frac{N_1(r,a;\sin )}{T(r,\sin )}\to \frac{r\mathrm{/}\pi }{2r\mathrm{/}\pi }=\frac{1}{2}$, so $\epsilon (1,\sin )=\epsilon (-1,\sin )=\frac{1}{2}$.

(Sine: combined defect sum) By step 4.1 the only nonzero terms of the total deficiency sum of [F3] are $\delta (\mathrm{\infty },\sin )=1$, $\epsilon (1,\sin )=\frac{1}{2}$ and $\epsilon (-1,\sin )=\frac{1}{2}$; the supremum of finite subsums is therefore $1+\frac{1}{2}+\frac{1}{2}=2$, so the combined defect sum of sine equals $2$, in agreement with the general upper bound of [F4].

(Exponential: combined defect sum) For $e^z$ steps 2.1 and 2.2 give the only nonzero terms $\delta (0,e^z)=\delta (\mathrm{\infty },e^z)=1$, so its combined defect sum is $2$ as well, again with equality in the bound of [F4].

The function $f(z)=z^d$ has exactly one zero, at $z=0$, of order $d$. Hence for every $t\ge 0$: $n(t,0;f)=d$, $\bar{n}(t,0;f)=1$ and $n_1(t,0;f)=d-1$.

Centre regularisation of the first count: $n(0,0;f)=d$, so $N(r,0;f)=d\log r+{\int }_0^r\frac{d-d}{t}dt=d\log r$ for every $r>0$.

For the truncated counts the centre values are $1$ and $d-1$ respectively, so $\bar{N}(r,0;f)=1\cdot \log r+{\int }_0^r\frac{1-1}{t}dt=\log r$ and $N_1(r,0;f)=(d-1)\log r$.

Consistency: $d\log r=\log r+(d-1)\log r$, in agreement with $N=\bar{N}+N_1$; and $N(r,0;f)=d\log r>\log r=\bar{N}(r,0;f)$ because $d\ge 2$ and $\log r>0$.

The rational degree of $z^d$ is $d$, so [F3] gives $T(r,f)=d\log r+O(1)$; thus $T=N(r,0;f)+O(1)$ while the truncated count is smaller by $(d-1)\log r$. The truncated count omits the weight $(d-1)\log r$ of the multiple zero. Replacing the full count by the truncated count in the First Main Theorem therefore changes its equality by this unbounded term. In a Second Main Theorem inequality with truncated counts on the right, replacing them by the larger full counts preserves the inequality but gives a weaker bound.

Set $r_n:=2^{n-1}$ and define integers ${\lambda }_1:=2$, ${\lambda }_{n+1}:=4n{\lambda }_n{2}^{n{\lambda }_n}+n+1$ for $n\ge 1$. Then ${\lambda }_n$ is strictly increasing with ${\lambda }_n>n$ and ${\lambda }_n\ge 2^n$, and ${\lambda }_{n+1}\ge 4n{\lambda }_n{2}^{n{\lambda }_n}\ge 16n{\lambda }_n$ for every $n$.

Consequences for $\nu \ge 2$: $\log {\lambda }_{\nu }\ge \log (4(\nu -1){\lambda }_{\nu -1})+(\nu -1){\lambda }_{\nu -1}\log 2\ge (\nu -1){\lambda }_{\nu -1}\log 2$, and ${\lambda }_{\nu -1}\to \mathrm{\infty }$, so $\log {\lambda }_{\nu }\mathrm{/}\nu \to \mathrm{\infty }$ and $\log {\lambda }_{\nu }\mathrm{/}{\lambda }_{\nu -1}\to \mathrm{\infty }$. Hence $\nu =o(\log {\lambda }_{\nu })$, $\log \nu =o(\log {\lambda }_{\nu })$ and ${\lambda }_{\nu -1}=o(\log {\lambda }_{\nu })$.

Define coefficients $a_j:=r_n^{-{\lambda }_n}$ when $j={\lambda }_n$ for some $n$, and $a_j:=0$ otherwise; strict increase of ${\lambda }_n$ makes this unambiguous. On $\mathrm{\mid }z\mathrm{\mid }\le R$, all but finitely many nonzero terms of ${\sum }_{j\ge 0}{a}_j{z}^j$ are bounded by $2^{-{\lambda }_n}$, and ${\sum }_n{2}^{-{\lambda }_n}<\mathrm{\infty }$. Thus [F2] gives absolute uniform convergence on every such disc; the power series has infinite radius and its nonzero terms, in increasing degree order, are precisely $f(z):={\sum }_{n\ge 1}(z\mathrm{/}{r}_n{)}^{{\lambda }_n}$. By [F3], $f$ is entire and $f^{\mathrm{\prime }}(z)={\sum }_{n\ge 1}\frac{{\lambda }_n}{r_n}(z\mathrm{/}{r}_n{)}^{{\lambda }_n-1}$; this is the differentiated power series with zero coefficients omitted. Its coefficient of $z^2$ is $r_1^{-2}=1$, so $f$ is nonconstant.

(Upper bound on lacunary circles) For $\nu \ge 2$ and $\mathrm{\mid }z\mathrm{\mid }=r_{\nu }$, the $n$-th term of $f$ has modulus $2^{(\nu -n){\lambda }_n}$ for $n<\nu$, modulus $1$ for $n=\nu$, and modulus $2^{-(n-\nu ){\lambda }_n}$ for $n>\nu$. In the first block each modulus is at most $2^{{\lambda }_{\nu -1}}$: for $n\le \nu -2$ one has $(\nu -n){\lambda }_n\le (\nu -1){\lambda }_{\nu -2}\le {\lambda }_{\nu -1}$ by step 1.1, while $n=\nu -1$ gives exponent ${\lambda }_{\nu -1}$ exactly. Hence the first block is at most $(\nu -1)2^{{\lambda }_{\nu -1}}$, and the tail is at most ${\sum }_{m\ge 1}{2}^{-m}=1$. So $\mathrm{\mid }f(z)\mathrm{\mid }\le (\nu -1)2^{{\lambda }_{\nu -1}}+2$ and $m_0(r_{\nu },f)\le \log ((\nu -1)2^{{\lambda }_{\nu -1}}+2)\le {\lambda }_{\nu -1}\log 2+\log (2(\nu -1))$.

(Lower bound for the derivative) For $\nu \ge 2$ and $\mathrm{\mid }z\mathrm{\mid }=r_{\nu }$: the $\nu$-th term of $f^{\mathrm{\prime }}$ has modulus ${\lambda }_{\nu }\mathrm{/}{r}_{\nu }$; the earlier terms satisfy ${\sum }_{n<\nu }\frac{{\lambda }_n}{r_n}(\frac{r_{\nu }}{r_n}{)}^{{\lambda }_n-1}=\frac{1}{r_{\nu }}{\sum }_{n<\nu }{\lambda }_n{2}^{(\nu -n){\lambda }_n}\le \frac{(\nu -1){\lambda }_{\nu -1}{2}^{(\nu -1){\lambda }_{\nu -1}}}{r_{\nu }}\le \frac{{\lambda }_{\nu }}{4r_{\nu }}$ by step 1.1; and the later terms satisfy ${\sum }_{n>\nu }\frac{{\lambda }_n}{r_n}(\frac{r_{\nu }}{r_n}{)}^{{\lambda }_n-1}=\frac{1}{r_{\nu }}{\sum }_{m\ge 1}{\lambda }_{\nu +m}{2}^{-m{\lambda }_{\nu +m}}\le \frac{2}{r_{\nu }}$ because ${\lambda }_{\nu +m}\ge 2$ makes each summand at most $2^{-m}$. Hence $\mathrm{\mid }{f}^{\mathrm{\prime }}(z)\mathrm{\mid }\ge \frac{1}{r_{\nu }}({\lambda }_{\nu }-\frac{{\lambda }_{\nu }}{4}-2)\ge \frac{{\lambda }_{\nu }}{2r_{\nu }}$ for all $\nu$, so $m_0(r_{\nu },f^{\mathrm{\prime }})\ge \log {\lambda }_{\nu }-\log 2-(\nu -1)\log 2\ge \frac{1}{2}\log {\lambda }_{\nu }$ for all large $\nu$ by step 2.1.

(Standard proximity of the quotient) For finite complex $u,v\mathrm{\ne }0$ one has ${\log }^{+}\mathrm{\mid }u\mathrm{/}v\mathrm{\mid }\ge {\log }^{+}\mathrm{\mid }u\mathrm{\mid }-{\log }^{+}\mathrm{\mid }v\mathrm{\mid }$; taking angular means gives $m_0(r_{\nu },f^{\mathrm{\prime }}\mathrm{/}f)\ge m_0(r_{\nu },f^{\mathrm{\prime }})-m_0(r_{\nu },f)\ge \frac{1}{2}\log {\lambda }_{\nu }-{\lambda }_{\nu -1}\log 2-\log (2(\nu -1))\ge \frac{1}{3}\log {\lambda }_{\nu }$ for all large $\nu$, by step 2.1.

(Comparison scale) By step 3.1 and [F1], ${\log }^{+}T(r_{\nu },f)+\log r_{\nu }=O({\lambda }_{\nu -1}+\nu +\log \nu )=o(\log {\lambda }_{\nu })$ by step 2.1.

(Infinite order) At $\mathrm{\mid }z\mathrm{\mid }=r_{\nu +1}=2r_{\nu }$ the $\nu$-th term of $f$ equals $2^{{\lambda }_{\nu }}$, the terms with $n<\nu$ sum to at most $2^{{\lambda }_{\nu }-1}$ by step 1.1 (each of the $\nu -1$ terms has exponent $(\nu +1-n){\lambda }_n\le 2{\lambda }_{\nu -1}$, and $(\nu -1)2^{2{\lambda }_{\nu -1}}\le 2^{{\lambda }_{\nu }-1}$ because ${\lambda }_{\nu }-1\ge 2{\lambda }_{\nu -1}+{\log }_2(\nu -1)$ whenever ${\lambda }_{\nu }\ge 16(\nu -1){\lambda }_{\nu -1}$ and ${\lambda }_{\nu -1}\ge 2^{\nu -1}$), and the terms with $n>\nu$ sum to at most $2$ as in step 3.1. Hence $\mathrm{\mid }f\mathrm{\mid }\ge 2^{{\lambda }_{\nu }-1}-2\ge 2^{{\lambda }_{\nu }-2}$ on that circle, so $m_0(r_{\nu +1},f)\ge ({\lambda }_{\nu }-2)\log 2$ and [F1] gives $T(r_{\nu +1},f)\ge m_0(r_{\nu +1},f)$. By step 2.1, $\log {\lambda }_{\nu }\mathrm{/}\nu \to \mathrm{\infty }$; since $\log r_{\nu +1}=\nu \log 2$, this gives $\log T(r_{\nu +1},f)\mathrm{/}\log r_{\nu +1}\to \mathrm{\infty }$, so $f$ has infinite order.

Combining steps 4.1 and 4.2, $\frac{m_0(r_{\nu },f^{\mathrm{\prime }}\mathrm{/}f)}{{\log }^{+}T(r_{\nu },f)+\log r_{\nu }}\ge \frac{\log {\lambda }_{\nu }\mathrm{/}3}{o(\log {\lambda }_{\nu })}\to \mathrm{\infty }$ along $r_{\nu }\to \mathrm{\infty }$; hence $m_0(r_{\nu },f^{\mathrm{\prime }}\mathrm{/}f)$ is not $O({\log }^{+}T(r_{\nu },f)+\log r_{\nu })$.