# compare N(f,a,r) with T(f,r)

I'm reading William Cherry and Zhuan Ye's book 'Nevanlinna's theory of value distribution, the second main theorem and its error terms'. In Section 1.12, they explains why $$N$$ and $$T$$ is used in Nevanlinna theory instead of $$n$$ and $$A$$, where $$A(f,r)=\int_{D(t)}f^\ast\omega$$. Then they gave some results on the comparison of $$n(f,a,r)$$ and $$A(f,r)$$. For example,

Gol'dberg in 1978 constructed an entire function $$f$$ such that for every $$a\in \mathbb{C}$$ such that $$\limsup_{r\to \infty}\frac{n(f,a,r)}{A(f,r)}=\infty.$$

In another direction, Hayman and Stewart in 1954 formulate a theorem

Let $$f$$ be a non-constant meromorphic function on $$\mathbb{C}$$ and set $$n(f,r)=\sup_{a\in \mathbb{P}^1}n(f,a,r)$$. Then $$1\le\liminf_{r\to \infty}\frac{n(f,r)}{A(f,r)}\le e.$$

My question is: what can we say about $$\frac{N(f,a,r)}{T(f,r)}$$?

Firstly, by FMT we know $$\frac{N(f,a,r)}{T(f,r)}\le 1$$.

I also checked some elementaty functions. For examples, exponential function $$f(z)=e^z$$. By a simple calculation, $$N(f,\infty,r)=0,N(f,a,r)=\frac{r}{\pi}+O(\log r),T(f,r)=\frac{r}{\pi}$$. So $$\frac{N(f,\infty,r)}{T(f,r)}=0,\frac{N(f,0,r)}{T(f,r)}=1.$$ When does this ratio nonzero for a general meromorphic function?

I'm aware that the purpose of Nevanlinna theory is to give the upper bound and lower bound of $$N(f,a,r)$$ by $$T(f,r)$$. But I'm still interested in the value of the ratio by taking $$r\to \infty$$.

Any reply or reference is appreciated.

• @user64494 The value distribution theory has many similarities with Roth’s theorem on diophantine approximation. In fact, Vojta has established a dictionary between Nevanlinna theory and number theory. Maybe this introduction will be helpful. I want to know when the ratio is rational, raised from some calculation. – Wei Xia Aug 1 '18 at 8:40
• Thank you. However, I don't see this ratio in the dictionary. – user64494 Aug 1 '18 at 10:10
• The answers to these questions are contained in any book on Nevanlinna theory, for example, Hayman, Meromorphic functions. – Alexandre Eremenko Aug 2 '18 at 1:04
• @user64494 Thanks. I just know defect $\delta(f,a)$ is defined to be $1-\limsup_{r\to\infty}\frac{N(f,a,r)}{T(f,r)}$. It measures to what extent $f$ does not take on the value $a$ with the expected frequency. Using defect, we can reformulate a weaker version of SMT. And the defect relation plays the role of the Fundamental Theorem of Algebra in some sense. Chapter 4 of Hayman's Meromorphic functions provides a broader study of defects. – Wei Xia Aug 2 '18 at 6:09

Chapter 4 of Hayman's Meromorphic functions and Fuchs'The development of the theory of deficient values since Nevanlinna provide fruitful results on the deficient value theory. I will quote some facts I'm interested here.

The deficiency of the meromorphic function $f(z)$ with respect of $\infty$ is $$\delta(\infty,f)=\liminf_{r\to\infty}\frac{m(r,f)}{T(r,f)}=1-\limsup_{r\to\infty}\frac{N(r,f)}{T(r,f)}.$$ For $a\in \mathbb{C}$, $$\delta(a,f)=\delta(\infty,\frac{1}{f-a}).$$ A value $c\in\hat{\mathbb{C}}$ is called deficient if $\delta(c,f)>0$. If $f$ never takes on the value $a$, or takes on the value $a$ very infrequently, then $\delta(a,f)=1$. In general, $\delta(a,f)$ measures the tendency of $f$ to omit the value $a$.(In another word,$1-\delta(a,f)$ measures the tendency of $f$ to take on the value $a$.)

From FMT, $0\le \delta(c,f)\le 1, c\in\hat{\mathbb{C}}$.

From SMT, there is at most a countable set of $c$ with $\delta(c,f)>0$ and $S(f):=\sum_{c\in\hat{\mathbb{C}}}\delta(c,f)\le 2$.(Defect Relation). This is a generalizaion of Picard's theorem.

Next, define the order $\rho$ and the lower order $\lambda$ of $f$ by

$\rho=\limsup_{r\to\infty} \frac{\log T(f,r)}{\log r}$ and $\lambda=\liminf_{r\to\infty} \frac{\log T(f,r)}{\log r}$.

In order to further distinguish the growth of $T(f,r)$, define the type $K$ of $f$ by $$K=\limsup_{r\to \infty}\frac{T(f,r)}{r^\rho}.$$ The order and type of $f$ is the 'appropriate' value such that $|f(z)|\le A\cdot e^{K|z|^\rho}$.

A value, $c=\infty$, say, is deficient if $\log|f(re^{i\theta})|$ is very large compared to $T$ in a very small set on all large circles $|z|=$const. This can be the case, if $f(z)$ is of infinite order.

Set of deficient values of a meromorphic function of order $\rho$ can be any countable set if $\rho>0$ and consists of at most one point if $\rho=0$.

If $f$ is an entire function of finite order $\rho$ and $S(f):=\sum_{c\in\hat{\mathbb{C}}}\delta(c,f)=2$,then

1. $\rho$ is a positive integer.
2. Each deficiency is an integral multiple of $\frac{1}{\rho}$. ln particular $f(z)$ can not have more than $\rho$ finite deficient values.

Conversely, the existence of deficient values have implications for the growth properties of a function. An entire function with a finite deficient value must have lower order$\ge 2$.

In addition, the problem of finding functions with assigned deficiencies attracted is called inverse Nevanlinna problem. Apart from its intrinsic interest this problem was of great importance, because the attempts to solve it led to many advances and new questions in the theory of Riemann surfaces. It has a definitive solution for any prescribed $\{a_j\}_{j=1}^N$ and $\{\delta_j\}_{j=1}^N$ satisfying $0<\delta_j\le 1$ and $\sum_{j=1}^N\delta_j\le 2$.