MATHEMATICS

A. A. GOL'DBERG

AN ESTIMATE OF THE SUM OF DEFECTS OF A MEROMORPHIC FUNCTION OF ORDER LESS THAN ONE

(Presented by Academician M. A. Lavrent'ev on 7 XII 1956)

Let \(\delta(a)\) denote the Nevanlinna defect at the point \(a\) of a function \(f(z)\), meromorphic in the finite \(z\)-plane \(\left({}^{1}\right)\). R. Nevanlinna showed that for meromorphic functions of nonintegral order \(\rho\), for any \(a \ne b\), \(\delta(a)+\delta(b)\) is less than a certain constant, which is less than \(2\) and depends only on \(\rho\), and gave estimates \(\left({}^{2}\right)\). For the case \(0 \le \rho < 1\), Shah \(\left({}^{3}\right)\), refining Nevanlinna’s estimates, showed that \(\delta(a)+\delta(b) \le 1+\rho\).

The purpose of the present note is to prove the inequality

\[ \delta(a)+\delta(b)\le \begin{cases} 1, & 0 \le \rho \le 1/3,\\[4pt] 2-2^\rho \sqrt{\pi}\,\Gamma\!\left(1-\dfrac{\rho}{2}\right)\Gamma^{-1}\!\left(\dfrac{1-\rho}{2}\right), & 1/3<\rho<1, \end{cases} \tag{1} \]

which, as is not difficult to verify, is a sharper estimate than Shah’s estimate. The estimate (1) is probably not the best possible (for \(1/3<\rho<1\)), but, in any case, it does not deviate very greatly from it, since for every entire function of order \(0 \le \rho < 1/2\), \(\delta(\infty)+\delta(0)=1\), and for the entire function

\[ \prod_{n=1}^{\infty} (1-zn^{-1/\rho}) \]

of order \(1/2 \le \rho < 1\), \(\delta(\infty)+\delta(0)=2-\sin \pi\rho\). The graphs of the corresponding functions are shown in Fig. 1.

Fig. 1

Let us have a meromorphic function \(w=f(z)\) of order \(\rho\), \(0 \le \rho < 1\). Without loss of generality, we may restrict ourselves to estimating \(\delta(0)+\delta(\infty)\) and assume that \(f(0)=1\). Then

\[ f(z)=\prod_{\nu}\left(1-za_\nu^{-1}\right)\cdot \prod_{\mu}\left(1-zb_\mu^{-1}\right)^{-1}, \]

where the series \(\sum_{\nu}|a_\nu|^{-\lambda}\), \(\sum_{\mu}|b_\mu|^{-\lambda}\) and the integrals

\[ \int_{0}^{\infty} N(r,0) r^{-\lambda-1}\,dr, \]

\[ \int_{0}^{\infty} N(r,\infty) r^{-\lambda-1}\,dr \]

converge for all \(\lambda\), \(\rho<\lambda\le 1\); from this will follow the convergence of all the series and integrals that we shall encounter below.

\[ m(r,0)=\frac{1}{2\pi}\int_0^{2\pi}\ln^+\left|\frac{1}{f(re^{i\varphi})}\right|\,d\varphi \leq \sum_\nu \frac{1}{2\pi}\int_0^{2\pi}\ln^+\left|\frac{a_\nu}{re^{i\varphi}-a_\nu}\right|\,d\varphi+ \]
\[ +\sum_\mu \frac{1}{2\pi}\int_0^{2\pi}\ln^+\left|\frac{re^{i\varphi}-b_\mu}{b_\mu}\right|\,d\varphi =\sum_\nu u_\nu(r)+\sum_\mu U_\mu(r). \tag{2} \]

Denote by \(c^\nu\) (\(C^\mu\)) the disk \(|z-a_\nu|<|a_\nu|\) (\(|z-b_\mu|>|b_\mu|\)); by \(\gamma_r^\nu\) (\(\Gamma_r^\mu\)) the arcs of the circle \(|z|=r\) lying in \(c^\nu\) (\(C^\mu\)); and by \(\omega_\nu(r)\) (\(\Omega_\mu(r)\)) the radian measure of the arc of the circle \(|z-a_\nu|=|a_\nu|\) (\(|z-b_\mu|=|b_\mu|\)) lying inside the disk \(|z|<r\).

A simple calculation gives that \(\omega_\nu(r)=4\arcsin(2^{-1}|a_\nu|^{-1}r)\) for \(0\leq r\leq 2|a_\nu|\); \(\Omega_\mu(r)=4\arcsin(2^{-1}|b_\mu|^{-1}r)\) for \(0\leq r\leq 2|b_\mu|\); \(\Omega_\mu(r)=2\pi\) for \(2|b_\mu|\leq r<\infty\).

Further,
\[ \frac{du_\nu(r)}{d\ln r} =\frac{1}{2\pi}\int_{\gamma_r^\nu}\frac{\partial}{\partial r}\ln\left|\frac{a_\nu}{re^{i\varphi}-a_\nu}\right|\,r\,d\varphi =-\frac{1}{2\pi}\int_{\gamma_r^\nu} d\arg(z-a_\nu). \]

Consequently,
\[ \frac{du_\nu}{d\ln r}=\frac{\omega_\nu(r)}{2\pi} \quad \text{for } 0\leq r<|a_\nu|; \]
\[ \frac{du_\nu}{d\ln r}=\frac{\omega_\nu(r)-2\pi}{2\pi}=0 \quad \text{for } |a_\nu|<r\leq 2|a_\nu|; \]
\[ \frac{du_\nu}{d\ln r} \quad \text{for } 2|a_\nu|<r<\infty. \]

\[ u_\nu(r)= \begin{cases} \displaystyle \frac{2}{\pi}\int_0^r \arcsin\frac{r}{2|a_\nu|}\,\frac{dr}{r} -\ln^+\frac{r}{|a_\nu|}, & 0\leq r\leq 2|a_\nu|,\\[1.2em] 0, & 2|a_\nu|\leq r<\infty. \end{cases} \tag{3} \]

\[ u_\nu(r)+\ln^+\frac{r}{|a_\nu|} = \begin{cases} \displaystyle \frac{2}{\pi}\int_0^r \arcsin\frac{r}{2|a_\nu|}\,\frac{dr}{r}, & 0\leq r\leq 2|a_\nu|,\\[1.2em] \displaystyle \ln\frac{r}{|a_\nu|}, & 2|a_\nu|\leq r<\infty. \end{cases} \tag{4} \]

Analogously we obtain
\[ \frac{dU_\mu(r)}{d\ln r} =\frac{1}{2\pi}\int_{\Gamma_r^\mu}\frac{\partial}{\partial r}\ln\left|\frac{re^{i\varphi}-b_\mu}{b_\mu}\right|\,r\,d\varphi =\frac{1}{2\pi}\int_{\Gamma_r^\mu}d\arg(z-b_\mu) =\frac{\Omega_\mu(r)}{2\pi}, \]

\[ U_\mu(r)= \begin{cases} \displaystyle \frac{2}{\pi}\int_0^r \arcsin\frac{r}{2|b_\mu|}\,\frac{dr}{r}, & 0\leq r\leq 2|b_\mu|,\\[1.2em] \displaystyle \ln\frac{r}{|b_\mu|}, & 2|b_\mu|\leq r<\infty. \end{cases} \tag{5} \]

Introduce the function \(\chi(x)=\dfrac{2}{\pi}\int_0^x \arcsin\dfrac{x}{2}\,\dfrac{dx}{x}\) for \(0\leq x\leq 2\); \(\chi(x)=\ln x\) for \(2<x<\infty\). It is not difficult to establish the continuity of \(\chi(x)\) for \(0\leq x<\infty\).

We also note that, since \(u_\nu(r)>0\) for \(0<r<2|a_\nu|\), it follows from (3) that

\[ \frac{2}{\pi}\int_0^x \arcsin \frac{x}{2}\,\frac{dx}{x}>\ln^+x,\qquad 0<x<2. \tag{6} \]

From (2), (4), (5) we obtain

\[ T(r,f)=m(r,0)+N(r,0)=m(r,0)+\sum_\nu \ln^+\frac{r}{|a_\nu|}\leq \]

\[ \leq \sum_\nu\left(u_\nu(r)+\ln^+\frac{r}{|a_\nu|}\right)+\sum_\mu U_\mu(r) =\sum_\nu \chi\left(\frac{r}{|a_\nu|}\right)+\sum_\mu \chi\left(\frac{r}{|b_\mu|}\right). \tag{7} \]

We shall now prove the following auxiliary inequality: for \(0<a<\infty\), \(0<\lambda<1\),

\[ \varkappa(\lambda)=\Gamma\left(\frac{1-\lambda}{2}\right)2^{-\lambda}\pi^{-1/2}\Gamma^{-1}\left(1-\frac{\lambda}{2}\right) \]

\[ \int_a^\infty \chi(x)x^{-\lambda-1}\,dx < \varkappa(\lambda)\int_a^\infty (\ln^+x)x^{-\lambda-1}\,dx. \tag{8} \]

Similar auxiliary inequalities for obtaining various estimates have been used by many authors (see, for example, \((^{4-6})\)). By a simple, though somewhat cumbersome, calculation one verifies that

\[ \int_0^\infty \chi(x)x^{-\lambda-1}\,dx = \varkappa(\lambda)\int_0^\infty (\ln^+x)x^{-\lambda-1}\,dx. \tag{9} \]

From (6), (9), and the definition of \(\chi(x)\) it follows that \(\varkappa(\lambda)>1\). Let

\[ \Phi(a)=\int_a^\infty [\varkappa(\lambda)\ln^+x-\chi(x)]x^{-\lambda-1}\,dx =\int_a^\infty [-\psi(x)]x^{-\lambda-1}\,dx. \]

\[ \Phi'(a)=\psi(a)a^{-\lambda-1}. \]
Obviously, \(\psi(a)>0\) for \(0<a\leq 1\). Further,
\[ \psi'(a)=2(\pi a)^{-1}\arcsin(a/2)-a^{-1}\varkappa(\lambda)<0 \]
for \(1<a\leq2\), and
\[ \psi'(a)=[1-\varkappa(\lambda)]a^{-1}<0 \]
for \(2\leq a<\infty\). Consequently, for \(1<a<\infty\), \(\psi\) is strictly monotonically decreasing, and since \(\psi(1)>0\) and \(\psi(2)=[1-\varkappa(\lambda)]\ln 2<0\), there exists a unique point \(a_0\), \(\psi(a_0)=0\), \(1<a_0<2\), such that \(\psi(a)>0\) for \(0<a<a_0\) and \(\psi(a)<0\) for \(a_0<a<\infty\). Hence \(\Phi(a)\) is strictly monotonically increasing for \(0<a<a_0\) and strictly monotonically decreasing for \(a_0<a<\infty\). But \(\Phi(a)\to0\) as \(a\to\infty\), and \(\Phi(0)=0\) by virtue of (9); therefore \(\Phi(a)>0\) for all \(a>0\), and this is equivalent to (8).

From (8), for \(\rho<\lambda<1\), it follows that

\[ \int_0^\infty \left[\sum_\nu \chi\left(\frac{r}{|a_\nu|}\right)+\sum_\mu \chi\left(\frac{r}{|b_\mu|}\right)\right]r^{-\lambda-1}\,dr< \]

\[ <\varkappa(\lambda)\int_a^\infty\left(\sum_\nu \ln^+\frac{r}{|a_\nu|} +\sum_\mu \ln^+\frac{r}{|b_\mu|}\right)r^{-\lambda-1}\,dr = \]

\[ =\varkappa(\lambda)\int_a^\infty [N(r,0)+N(r,\infty)]r^{-\lambda-1}\,dr. \]

Since in (10) \(a\) can be chosen arbitrarily large, there exists a sequence \(r_k \to \infty\) such that

\[ \sum_{\nu}\chi\left(\frac{r_k}{|a_\nu|}\right)+\sum_{\mu}\chi\left(\frac{r_k}{|b_\mu|}\right)<\varkappa(\lambda)\,[N(r_k,0)+N(r_k,\infty)]. \]

By (7),

\[ T(r_k,f)<\varkappa(\lambda)[N(r_k,0)+N(r_k,\infty)]; \]

\[ \overline{\lim}_{r\to\infty}[N(r,0)+N(r,\infty)]/T(r,f)\geq \varkappa^{-1}(\lambda). \]

Since \(\varkappa(\lambda)\) is a continuous function, letting \(\lambda\to\rho\) we obtain

\[ \overline{\lim}_{r\to\infty}[N(r,0)+N(r,\infty)]/T(r,f)\geq \varkappa^{-1}(\rho). \]

Hence it follows that \(\delta(0)+\delta(\infty)\leq 2-\varkappa^{-1}(\rho)\), \(0\leq \rho<1\). For \(1/3<\rho<1\) this is our desired estimate (1). For \(0\leq \rho\leq 1/3\), estimate (1) follows from assertion (7): if for \(0\leq \rho<1/2\) \(\delta(a)>1-\cos\pi\rho\), then \(a\) is the only deficient value of \(f(z)\); indeed, for \(0\leq \rho\leq 1/3\), \(\max[1,2(1-\cos\pi\rho)]=1\).

Uzhgorod
State University

Received
8 IX 1956

CITED LITERATURE

\(^{1}\) R. Nevanlinna, Univalent Analytic Functions, 1941.
\(^{2}\) R. Nevanlinna, Le théorème de Picard—Borel et la théorie des fonctions méromorphes, 1929.
\(^{3}\) S. M. Shah, Math. Student, 12, 67 (1944).
\(^{4}\) E. Titchmarsh, Theory of Functions, 1951, pp. 8.74.
\(^{5}\) G. Valiron, Mathematica, 11, 264 (1935).
\(^{6}\) O. Teichmüller, Deutsche Math., 4, 163 (1939).
\(^{7}\) A. A. Gol’dberg, DAN, 98, 893 (1954).