UDC 517.535.6

MATHEMATICS

A. A. GOL’DBERG

ON DEFECTIVE NONASYMPTOTIC VALUES OF MEROMORPHIC FUNCTIONS

(Presented by Academician S. N. Bernstein on 29 I 1966)

Let \(f(z)\) be a function meromorphic in the finite \(z\)-plane. Throughout we shall use the standard notation of Nevanlinna theory \((^{1})\). If \(a\) is a Picard exceptional value of \(f(z)\), then, by the well-known theorem of Iversen \((^{1}\), p. 293), \(a\) is an asymptotic value for the function \(f(z)\). It is also known \((^{2-5})\) that in this theorem the requirement that \(a\) be a Picard exceptional value cannot be replaced by the weaker condition that \(a\) be a deficient value, even if one restricts oneself to entire functions. However, from results of Edrei and Fuchs \((^{6})\) it follows that for entire functions of finite order, for which the value \(a\) has maximal deficiency \(\delta(a)=1\), \(a\) is necessarily an asymptotic value. The question naturally arises (see, for example, \((^{7})\)) whether this will also be true for meromorphic functions. Here we give a negative answer to this question. Namely, for any \(\rho,\ 1<\rho\leqslant\infty\), we construct a meromorphic function of order \(\rho\) for which \(\delta(\infty)=1\) and \(\infty\) is not an asymptotic value. Moreover, in our example the ratio \(T(r,f)/r^\rho\) for \(r>r_0\) is bounded above and below by positive constants if \(1<\rho<\infty\), and \(\ln r=o(\ln T(r,f))\) if \(\rho=\infty\), while \(N(r,\infty)\sim Cr^\beta,\ \rho/(2\rho-1)<\beta<1,\ 0<C<\infty\). Thus \(\infty\) is also a nonasymptotic Borel exceptional value*.

First let \(1<\rho<\infty\). Take on \([0,\infty)\) two continuously differentiable functions \(l_j(r)\), \(j=1,2\), possessing the following properties:

A. \(\alpha\leqslant l_j(r)\leqslant\rho\), where \(\alpha\) is some number such that \(\rho/(2\rho-1)<\alpha<1\).

B. \(|l'_j(r)|=O\{(r\ln r\ln_2 r\ln_3 r)^{-1}\}\), where \(\ln_i r\) denotes the \(i\)-th iterated logarithm.

C. There exist sequences of segments \([a_n^{(j)}, b_n^{(j)}]\), \(n=1,2,\ldots;\ j=1,2\), such that \(a_n^{(j)}\to\infty,\ b_n^{(j)}/a_n^{(j)}\to\infty\) as \(n\to\infty\), and \(l_j(r)\equiv\alpha\) for \(r\in[a_n^{(j)}, b_n^{(j)}]\).

D. \(l(r)=\max\{l_1(r),l_2(r)\}\equiv\rho\) for \(0\leq r<\infty\).

It is not difficult to verify that such functions \(l_j(r)\) can be constructed.

Everywhere below we have \(z=re^{i\varphi}\), \(|\varphi|\leqslant\pi\). Consider the functions analytic in \(|\varphi|<\pi\)

\[ \lambda_j(z)=z\int_0^\infty \frac{l_j(t)\,dt}{(t+z)^2}. \]

We shall show that in each angle \(|\varphi|\leqslant\pi-\delta\), \(\delta>0\), the following holds uniformly (cf. \((^{10})\), p. 53):

\[ \lambda_j(z)=l_j(r)+o\{(\ln r)^{-1}\}. \tag{1} \]

* A Borel exceptional value may have zero deficiency. Apparently this was first noted by Valiron \((^{8})\); see also \((^{9})\), p. 271.

We note that for \(|\varphi|\leq \pi-\delta,\ 0\leq t<\infty\), one has \(|t+re^{i\varphi}|\geq \sin\delta/2(t+r)\). Hence

\[ |\lambda_j(z)-l_j(r)|=\left|z\int_0^\infty \frac{l_j(t)-l_j(r)}{(t+z)^2}\,dt\right| \leq \frac{r}{\sin^2\delta/2}\int_0^\infty \frac{|l_j(t)-l_j(r)|}{(t+r)^2}\,dt . \tag{2} \]

We split the last integral into three parts: over \([0,r(\ln r)^{-2}]\), over \([r(\ln r)^{-2},r(\ln r)^2]\), and over \([r(\ln r)^2,\infty)\). In the first and third integrals we estimate \(|l_j(t)-l_j(r)|\leq \rho\), and in the second
\[ |l_j(t)-l_j(r)|=|\xi l_j'(\xi)|\,|\ln(t/r)|=O\{(\ln r\ln_3 r)^{-1}\}, \]
where \(\xi\) lies between \(t\) and \(r\) (condition B was used, and \(|\ln(t/r)|\leq 2\ln_2 r\)). Substituting in (2), we obtain (1). We note that (1) remains valid if condition A is replaced by the condition \(l_j(r)=o(\ln r)\). We also note that for \(|\varphi|\leq \pi-\delta\)

\[ z^{\lambda_j(z)}=(1+o(1))z^{l_j(r)} . \tag{3} \]

Let \(\theta\) be a constant such that \(1<\theta<2\rho\alpha/(\rho+\alpha)<2\). Denote by \(\Gamma_j\) the boundary of the domain
\[ D_j=\{r>1,\ |\varphi|<\pi\theta/2l_j(r)\}, \]
which is taken so that \(D_j\) lies to the right of \(\Gamma_j\). In the complement of \(\overline D_j\) we define an analytic function \(F_j(z)\) by an integral of Cauchy type (the convergence follows at once from (3)):

\[ F_j(z)=\frac{1}{2\pi i}\int_{\Gamma_j} \frac{e^{\zeta^{\lambda_j(\zeta)}}}{\zeta-z}\,d\zeta,\qquad z\in C\overline D_j, \tag{4} \]

and then analytically continue \(F_j(z)\) to the whole finite \(z\)-plane, replacing in (4) \(\Gamma_j\) by the boundary of the domain \(D_j\cap\{|z|>r_0\}\), \(r_0>1\). Applying the usual arguments in these questions (cf., for example, \((^{10})\), p. 54), for the entire function \(F_j(z)\) we find

\[ F_j(z)= \begin{cases} O(1), & \text{for } z\in C D_j,\\ \exp\{z^{\lambda_j(z)}\}+O(1), & \text{for } z\in D_j . \end{cases} \]

Denote by \(\Delta_1\) the domain \(D_1\), and by \(\Delta_2\) the domain symmetric to \(D_2\) with respect to the imaginary axis. By our choice of \(\theta\), one has
\[ \pi-\frac{\pi\theta}{2l_2(r)}>\frac{\pi\theta}{2l_1(r)}, \]
and therefore \(\Delta_1\) and \(\Delta_2\) do not intersect. Let
\[ \Delta_3=C(\Delta_1\cup\Delta_2). \]
Then for the function
\[ \Psi(z)\equiv F_1(z)+F_2(-z) \]
we have

\[ \Psi(z)= \begin{cases} \exp\{z^{\lambda_1(z)}\}+O(1), & \text{for } z\in\Delta_1,\\ \exp\{(-z)^{\lambda_2(-z)}\}+O(1), & \text{for } z\in\Delta_2,\\ O(1), & \text{for } z\in\Delta_3 . \end{cases} \]

Let \(\alpha<\beta<1\) and

\[ \Phi(z)=\prod_{n=1}^{\infty}\left[1-\left(\frac{z}{n^{1/\beta}}\right)^2\right]. \]

Using the known formulas \((^{11})\), it is easy to obtain asymptotic expressions for \(|\Phi(z)|\). If from \(|z|<\infty\) we exclude certain disks with centers at \(\pm n^{1/\beta}\) and radii \(r_n,\ n=1,2,\ldots\), such that
\[ \sum_{k=1}^{n} r_k=o(n^{1/\beta}), \]
then in the remaining domain \(G\), as \(r\to\infty\), uniformly \((|\varphi|\leq\pi)\),
\[ \ln|\Phi(re^{i\varphi})|=(1+o(1))A(\beta)\cos\beta(|\varphi|-\pi/2)r^\beta,\qquad A(\beta)=\pi\operatorname{cosec}\pi\beta/2 . \]
It is obvious that in the domain \(G\) the function \(\Phi(z)\) tends uniformly to \(\infty\). We shall show that the function
\[ f(z)=\Psi(z)/\Phi(z) \]
has the required properties. Indeed, taking into account (3) and the boundedness of the set \(\Delta_3\cap C G\),

we obtain

\[ |f(z)| = \begin{cases} \exp \{(1+o(1))[r^{l_1(r)}\cos l_1(r)\varphi - A(\beta)\cos\beta(|\varphi|-\pi/2)r^\beta]\}+o(1), & \text{for } z\in\Delta_1\cap G,\\ \exp \{(1+o(1))[r^{l_2(r)}\cos l_2(r)(\pi-|\varphi|)-A(\beta)\cos\beta(|\varphi|-\pi/2)r^\beta]\} \\ \qquad {}+o(1), & \text{for } z\in\Delta_2\cap G,\\ o(1), & \text{for } z\in\Delta_3 . \end{cases} \tag{5} \]

It is easy to see that \(n(r,\infty,f)=n(r,0,\Phi)+O(1)\sim 2r^\beta\), \(N(r,\infty,f)\sim (2/\beta)r^\beta\). Denote by \(I_1\) the intersection of \(G\) with the positive real ray, \(I_2=(0,\infty)\setminus I_1\). Obviously, \(\operatorname{mes}\{(0,r)\cap I_2\}=o(r)\). Taking condition \(\Gamma\) into account, from (5) we obtain that, as \(r\to\infty\), \(r\in I_1\), one has
\(m(r,\infty,f)=(1+o(1))K(r)r^\rho\), where \((\pi\rho)^{-1}\le K(r)\le 2(\pi\rho)^{-1}\). Consequently, for \(r\in I_1\) we have

\[ T(r,f)=m(r,\infty)+N(r,\infty)=(1+o(1))K(r)r^\rho . \tag{6} \]

Since the function \(T(r,f)\) is increasing, relation (6) is valid for all \(r>0\). Thus \(\delta(\infty)=1\), and \(\infty\) is a Borel exceptional value.

We shall now show that \(\infty\) is not an asymptotic value. For all sufficiently large \(n\), in the segments \([a_n^{(j)},b_n^{(j)}]\) there are points \(r_n^{(j)}\in I_1\). On the arc \(\{|z|=r_n^{(j)}, z\in\Delta_j\}\), \(j=1,2\), by virtue of (5) and condition B, one has \((r=r_n^{(j)})\):

\[ |f(re^{i\varphi})|=\exp\{(1+o(1))[O(r^\alpha)-A(\beta)\cos\beta(|\varphi|-\pi/2)r^\beta]\}+o(1)\le \]

\[ \le \exp\{-(1+o(1))\pi\operatorname{ctg}\frac{\pi\beta}{2}\,r^\beta\}+o(1)=o(1). \]

Since, moreover, \(|f(re^{i\varphi})|=o(1)\) for \(z\in\Delta_3\), it is clear that \(\infty\) cannot be an asymptotic value.

In the case \(\rho=\infty\) the example must be modified only slightly. Let \(1/2<\alpha<\beta<1\), \(1<\theta<2\alpha\). We choose the functions \(l_j(r)\) so that they satisfy conditions B and C, and also the following conditions:

A′. \(\alpha\le l_j(r)<\infty\).

\(\Gamma'\). \(l(r)\to\infty\), \(l(r)=o(\ln r)\) as \(r\to\infty\).

D. If \(l_1(r)<2\theta\alpha/(2\alpha-\theta)\), then \(l_2(r)>\theta\alpha\), and if \(l_2(r)<2\theta\alpha/(2\alpha-\theta)\), then \(l_1(r)>\theta\alpha\).

After this the previous arguments go through almost without change.

Remark. As is known \(\left(^{11}\right.\), Ch. I, §§ 10, 11), an entire function of finite order can have a finite Borel exceptional value only when the order \(\rho\) is a natural number and the type is mean or maximal. If the order of \(N(r,a)\) is less than the order of \(T(r,f)\), then \(a\) is an asymptotic value (stronger results in this direction are also known \(\left(^{12}\right)\)). But the type of \(N(r,a)\) may be lower than the type of the entire function without \(a\) being an asymptotic value. We indicate corresponding examples. Let \(\rho=1\). Take for \(f(z)\) the canonical product of genus 1 with zeros at points \(a_n\) on the real axis, \(|a_n|=n\), the signs of \(a_n\) being chosen so that
\(\limsup_{n\to\infty} A_n=+\infty\),

\[ \liminf_{n\to\infty} A_n=-\infty,\qquad \text{where } A_n=\sum_{k=1}^n a_k^{-1}. \]

It is clear that \(|f(iy)|\ge 1\), \(-\infty<y<+\infty\). Using known results \(\left(^{11}\right.\), Ch. I, § 11), one can show that there exist semicircles
\(\{|z|=R_n,\ \operatorname{Re} z\ge 0\}\) and
\(\{|z|=R_n',\ \operatorname{Re} z\le 0\}\), \(R_n\to\infty\), \(R_n'\to\infty\), on which \(|f(z)|\) tends uniformly to \(\infty\). Consequently, zero is not an asymptotic value. But \(N(r,0)\) has mean type, while the function \(f(z)\) is of maximal type,

i.e., \(0\) is a Borel exceptional value. If we take \(|a_n|=n\ln n\), then \(N(r,0)\) will be of minimal type, while the type of \(f(z)\), as before, is maximal. If, with \(|a_n|=n\ln n\), we choose the signs of \(a_n\) so that \(\lim\sup\limits_{n\to\infty} A_n=1\), \(\lim\inf\limits_{n\to\infty} A_n=-1\), then the type of \(N(r,0)\) is minimal, the type of \(T(r,f)\) is mean, and zero again will not be an asymptotic value. If \(\rho\) is an integer, \(\rho>1\), then we consider \(f(z^\rho)\), where \(f(z)\) are the functions indicated above.

Ivan Franko Lviv State University

Received
25 I 1966

CITED LITERATURE

\(^{1}\) R. Nevanlinna, Single-Valued Analytic Functions, Moscow–Leningrad, 1941.
\(^{2}\) O. Teichmüller, Deutsche Math., 4, 163 (1939).
\(^{3}\) M.-H. Schwartz, C. R., 212, 382 (1941).
\(^{4}\) W. K. Hayman, J. London Math. Soc., 28, 369 (1953).
\(^{5}\) A. A. Gol’dberg, Scientific Notes, Uzhgorod University, 18, 191 (1957).
\(^{6}\) A. Edrei, W. H. J. Fuchs, Comm. math. Helv., 33, 258 (1959).
\(^{7}\) U. K. Kheĭman, Collection of translations, Mathematics, 4, No. 4, 21 (1960).
\(^{8}\) G. Valiron, Rend. Circ. mat. Palermo, 57, No. 1, 71 (1933).
\(^{9}\) A. A. Gol’dberg, Supplement to G. Wittich’s book Recent Investigations on Single-Valued Analytic Functions, Moscow, 1960.
\(^{10}\) M. A. Evgrafov, Asymptotic Estimates and Entire Functions, Moscow, 1957.
\(^{11}\) B. Ya. Levin, Distribution of Zeros of Entire Functions, Moscow, 1956.
\(^{12}\) S. M. Shah, Compositio math., 9, 227 (1951).