Abstract
Full Text
Mathematics
I. V. OSTROVSKII
A GENERALIZATION OF M. G. KREIN’S THEOREM
(Presented by Academician S. N. Bernstein on 27 IV 1957)
In this note we consider meromorphic functions of the form
\[ f(z)=\sum \frac{A_k}{z-h_k}, \tag{1} \]
where: 1) \(\sum |\operatorname{Im} h_k^{-1}|<\infty\), 2) \(\sum |A_k|<\infty\).
M. G. Krein showed \({}^{1}\) that for such a function, if one additionally assumes that it has no zeros,
\[ T_f(r)<C_f(r+1), \]
where \(C_f\) is some constant, and \(T_f(r)\) is the Nevanlinna characteristic of the meromorphic function. Here we remove the assumption that there are no zeros and establish the following theorem.
Theorem 1. For a function of the form (1) the following asymptotic inequality holds\(^*\)
\[ T_f(r)\leqslant N_f(\theta r,0)+C_{f,\theta}r, \tag{2} \]
in which \(\theta\) is an arbitrary number greater than one; \(C_{f,\theta}\) is some constant.
Proof. \(1^\circ\). By Nevanlinna’s first theorem,
\[ T_f(\rho)=N_f(\rho,0)+m_f(\rho,0)+O(1). \]
Multiplying both sides of this relation by \(\rho\), integrating with respect to \(\rho\) from \(r\) to \(\theta r\), and finally using the monotonicity of \(T_f\) and \(N_f\), we obtain
\[ T_f(r)\leqslant N_f(\theta r,0)+ \frac{2}{(\theta^2-1)^2r^2}\int_r^{\theta r} m_f(\rho,0)\rho\,d\rho+O(1). \tag{3} \]
We need to show that
\[ \int_r^{\theta r} m_f(\rho,0)\rho\,d\rho<C'_{f,\theta}r^3. \]
For this we shall use the inequality
\[ \int_r^{\theta r} m_f(\rho,0)\rho\,d\rho = \int_{r\leqslant |z|\leqslant \theta r}\ln_-|f(z)|\,dx\,dy \leqslant (1+\theta^2r^2)\int_{|z|\leqslant \theta r} \frac{\ln_-|f(z)|}{1+x^2+y^2}\,dx\,dy \leqslant \]
\[ \leqslant (1+\theta^2r^2) \int_{-\theta r}^{\theta r}\int_{-\infty}^{\infty} \frac{\ln_-|f(z)|}{1+x^2+y^2}\,dx\,dy = (1+\theta^2r^2) \left\{ \int_{-1}^{1}\int_{-\infty}^{\infty} + \int_{1}^{\theta r}\int_{-\infty}^{\infty} + \int_{-\theta r}^{-1}\int_{-\infty}^{\infty} \right\}. \tag{4} \]
\[ \text{* This proposition was put forward as a hypothesis by B. Ya. Levin (oral communication).} \]
Let us estimate the integrals in the brackets.
2°. Assuming \(|h_k|>1\) \((k=0,\ \pm 1,\ \pm 2,\ldots)\) (which entails no loss of generality), we draw around each point \(h_k\) a circle \((C_k)\) of radius \(|\operatorname{Im} h_k|^{-1}\). For nonreal \(z\) lying outside the circles \((C_k)\), we have
\[ |f(z)|\leq \sum_{|z-h_k|<y/2}\frac{|A_k|}{|z-h_k|} + \sum_{|z-h_k|\geq y/2}\frac{|A_k|}{|z-h_k|} \leq \frac{C_1}{|y|}(|z|^2+1). \]
Putting \(g(z)=f(z)\chi(\{h_k\},z)\), where
\[ \chi(\{h_k\},z)= \prod_{\operatorname{Im}h_k>0} \left(1-\frac{z}{h_k}\right) \left(1-\frac{z}{\overline{h}_k}\right)^{-1}, \]
we obtain, by the maximum-modulus principle, the estimate for \(g(z)\) when \(\operatorname{Im}z>0\):
\[ |g(z)|\leq C_2 |y|^{-1}(|z|^2+1). \]
Hence the inequality follows
\[ \int_{-\infty}^{\infty} \frac{\ln_+|g(t+ih)|}{1+t^2}\,dt < C_3\ln\frac{1}{h}+C_4 \qquad (0<h\leq 1). \tag{5} \]
We shall next need the following fact \((^2)\). If \(F(z)\) is holomorphic for \(\operatorname{Im}z>0\), continuous for \(\operatorname{Im}z\geq 0\), and
\[ 1)\quad \frac{\ln|F(z)|}{|z|}\leq C_F<\infty, \]
and
\[ 2)\quad \int_{-\infty}^{\infty} \frac{\ln|F(z)|}{1+t^2}\,dt \quad \text{converges,} \]
then
\[ \ln|F(z)|= \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{\ln|F(t)|}{(x-t)^2+y^2}\,dt + \ln|\chi(\{b_k\},z)|+k_F y, \tag{6} \]
where \(b_k\) are the zeros of \(F(z)\), and \(k_F\) is a constant.
Applying (6) to \(g(z+ih)\) and then putting \(z=i\), we obtain
\[ \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\ln_-|g(t+ih)|}{1+t^2}\,dt = \]
\[ = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\ln_+|g(t+ih)|}{1+t^2}\,dt + k_g + \ln|\chi(\{a_k-ih\},z)| - \ln|g((h+1)i)|, \]
where \(a_k\) are the zeros of \(f(z)\). Hence, in view of (5), it follows that
\[ \int_{-\infty}^{\infty} \frac{\ln_-|g(t+ih)|}{1+t^2}\,dt < C_5\ln\frac{1}{h}+C_6 \qquad (0<h\leq 1), \]
and, since \(|f(t+ih)|\geq |g(t+ih)|\) for \(h>0\), then
\[ \int_{-\infty}^{\infty} \frac{\ln_-|f(t+ih)|}{1+t^2}\,dt < C_5\ln\frac{1}{h}+C_6 \qquad (0<h\leq 1). \]
Integrating the last inequality with respect to \(h\) from \(0\) to \(1\), we shall have
\[ \int_0^1\int_{-\infty}^{\infty} \frac{\ln_-|f(t+ih)|}{1+t^2}\,dh\,dt < C_7. \]
Estimating in the same way the integral from \(-1\) to \(0\), we obtain
\[ \int_{-1}^{1}\int_{-\infty}^{\infty} \frac{\ln_-|f(t+ih)|}{1+t^2}\,dh\,dt<C_8 . \tag{7} \]
\(3^\circ\). Let us estimate the two other integrals in (4). Using (6), we have for \(y>1\)
\[ \ln_-|f(x+iy)|\leq \ln_-|g(x+iy)|\leq \frac{y-\tfrac12}{\pi}\int_{-\infty}^{\infty} \frac{\ln_-|g(t+i/2)|}{(x-t)^2+(y-\tfrac12)^2}\,dt + \]
\[ +|k|(y-\tfrac12)+\ln_-|\chi(\{a_k-i/2\},\,z-i/2)|. \]
Dividing this inequality by \(1+x^2+(y-\tfrac12)^2\) and integrating with respect to \(x\), we obtain
\[ \int_{-\infty}^{\infty} \frac{\ln_-|f(x+iy)|}{1+x^2+(y-\tfrac12)^2}\,dx \leq \frac{y-\tfrac12}{\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{\ln_-|g(t+i/2)|}{(x-t)^2+(y-\tfrac12)^2} \frac{dx\,dt}{1+x^2+(y-\tfrac12)^2} + \]
\[ +|k|+ \int_{-\infty}^{\infty} \frac{\ln_-|\chi(\{a_k-i/2\},\,z-i/2)|}{1+x^2+(y-\tfrac12)^2}\,dx = I_1+|k|+I_2 . \tag{8} \]
Further, it is easy to see that
\[ I_1= \frac{y-\tfrac12}{\pi} \int_{-\infty}^{\infty} \ln_-|g(t+i/2)| \left\{ \int_{-\infty}^{\infty} \frac{dx}{[(x-t)^2+(y-\tfrac12)^2][1+x^2+(y-\tfrac12)^2]} \right\}dt \leq \]
\[ \leq \frac{y-\tfrac12}{\pi} \int_{-\infty}^{\infty} \ln_-|g(t+i/2)| \frac{2\pi}{(y-\tfrac12)(t^2+1)}\,dt <C_9 . \tag{9} \]
Introducing \(a_k'=a_k-i/2\), \(z'=z-i/2\), and denoting these quantities again by \(a_k\) and \(z\), consider
\[ I_2=\int_{-\infty}^{\infty} \frac{\ln_-|\chi(\{a_k\},z)|}{1+x^2+y^2}\,dx \qquad \left( a_k=\alpha_k+i\beta_k,\ \beta_k>0,\ \sum \frac{\beta_k}{\alpha_k^2+\beta_k^2}<\infty \right). \]
From the identity
\[ \int_{-\infty}^{\infty} \frac{\ln |(x-a)^2+(y-b)^2|}{1+x^2+y^2}\,dx = \]
\[ = \begin{cases} \displaystyle \frac{\pi}{\sqrt{1+y^2}} \ln\!\left[a^2+\left(y-b+\sqrt{1+y^2}\right)^2\right], & (y-b>0),\\[1.2em] \displaystyle \frac{\pi}{\sqrt{1+y^2}} \ln\!\left[a^2+\left(-y+b+\sqrt{1+y^2}\right)^2\right], & (y-b\leq 0), \end{cases} \]
it follows that
\[ \frac{2\sqrt{1+y^2}}{\pi}I_2 = \frac{\sqrt{1+y^2}}{\pi} \sum \int_{-\infty}^{\infty} \ln \frac{(\alpha_k-x)^2+(\beta_k+y)^2}{(\alpha_k-x)^2+(\beta_k-y)^2} \frac{dx}{1+x^2+y^2} = \]
\[ = \sum_{y<\beta_k} \ln \frac{\alpha_k^2+(y+\beta_k+\sqrt{1+y^2})^2} {\alpha_k^2+(-y+\beta_k+\sqrt{1+y^2})^2} + \sum_{y>\beta_k} \ln \frac{\alpha_k^2+(y+\beta_k+\sqrt{1+y^2})^2} {\alpha_k^2+(y-\beta_k+\sqrt{1+y^2})^2} \leq \]
\[ \leq 4y\sum_{y<\beta_k} \frac{\beta_k+\sqrt{1+y^2}} {\alpha_k^2+(-y+\beta_k+\sqrt{1+y^2})^2} + \]
\[ +4\left(y+\sqrt{1+y^2}\right) \sum_{y<\beta_k} \frac{\beta_k}{\alpha_k^2+(y-\beta_k+\sqrt{1+y^2})^2} \leq \]
\[ \leq 8y\sum \frac{\beta_k}{\alpha_k^2+\beta_k^2} +4\bigl(y+\sqrt{1+y^2}\bigr)\sum \frac{\beta_k}{\alpha_k^2+\beta_k^2} \leq C_{10}(y+1), \]
whence \(I_2\leq C_{11}\). Substituting the last inequality from (9) into (8), and then integrating from \(1\) to \(\theta r\), we obtain
\[ \int_1^{\theta r}\int_{-\infty}^{\infty} \frac{\ln_-|f(z)|}{1+x^2+y^2}\,dy\,dx\leq C_{12}r; \qquad \int_{-\theta r}^{-1}\int_{-\infty}^{\infty} \frac{\ln_-|f(z)|}{1+x^2+y^2}\,dy\,dx\leq C_{13}r \tag{10} \]
(the validity of (10) is obtained by applying the arguments of this item to \(f(z)\) for \(\operatorname{Im} z\leq -1\)).
\(4^\circ\). Substituting (7) and (10) into (4), we obtain
\[ \int_r^{\theta r} m_f(\rho,0)\,\rho\,d\rho<C'_{f,\theta}r^3. \tag{11} \]
From this inequality, in view of (3), (2) follows. Theorem 1 is proved.
\(5^\circ\). Inequality (11) makes it possible to establish the following theorem:
Theorem 2. If for a function of the form (1) \(\displaystyle \lim_{r\to\infty}T_f(r)r^{-1}=\infty\), then for every \(a\), including \(a=\infty\), \(\delta_f(a)=0\).
This proposition is adjacent to the following result of M. V. Keldysh \((^3)\):
A meromorphic function of finite order \(f(z)=\displaystyle \sum \frac{A_k}{z-h_k}\) has no nonzero deficient values, and if \(\displaystyle \sum |A_k|<\infty\) and \(\displaystyle \sum A_k\neq 0\), then also \(\delta_f(0)=0\).
From Theorem 2 (and the remark, see below) it follows, in particular, that if \(\sum |\operatorname{Im} h_k^{-1}|<\infty\), then for the function \(f(z)\) the equality \(\delta_f(0)=0\) holds also without these additional assumptions.
Proof of Theorem 2. In view of (11) we have
\[ \frac{C'_{\theta,f}r}{T_f(r)}\geq \frac{\displaystyle \int_r^{\theta r} m_f(\rho,0)\rho\,d\rho}{r^2T_f(r)} = \frac{m_f(\theta_r r,0)\theta_r r}{r^2T_f(r)}(\theta-1)r > \frac{m_f(\theta_r r,0)}{T_f(\theta_r r)}(\theta-1), \]
where \(1<\theta_r<\theta\), whence
\[ \delta_f(0)=\lim_{\rho\to\infty}\frac{m_f(\rho,0)}{T_f(\rho)}=0. \]
The equality \(\delta_f(a)=0\) for \(a\neq 0,\infty\) follows from consideration of the function \(f_1=(f-a)z^{-1}\), which is represented in the form (1), and \(m_{f_1}(r,0)\geq m_f(r,a)\). The equality \(\delta_f(\infty)=0\) was proved by M. V. Keldysh \((^3)\) also for functions of infinite order.
Remark. The assertions of Theorems 1 and 2 remain valid if the condition \(\sum |A_k|<\infty\) is dropped, for, if the latter is not fulfilled for \(f(z)\), then, by virtue of the absolute convergence of the series (1), it is fulfilled for \(f(z)z^{-1}\).
I am deeply grateful to B. Ya. Levin, whose advice I used in carrying out the present work.
Kharkov State University
named after A. M. Gorky
Received
7 I 1957
REFERENCES
- M. G. Krein, Izv. AN SSSR, ser. matem., 11, No. 4, 309 (1947).
- B. Ya. Levin, Distribution of zeros of entire functions, 1956, p. 311.
- M. V. Keldysh, DAN, 94, No. 3, 377 (1954).