MATHEMATICS

Academician of the Academy of Sciences of the Armenian SSR M. M. Dzhrbashyan

ON A CERTAIN PROPERTY OF BLASCHKE FUNCTIONS

1°. In the author’s paper \((^1)\) the class \(A_\alpha(-1<\alpha<+\infty)\) of functions \(f(z)\), analytic in the disk \(|z|<1\), was defined and studied, for which the integrals

\[ m_\alpha(r;f)\equiv m_\alpha(r;f(z))= \frac{r^{-\alpha}}{2\pi}\int_0^{2\pi}D_{(+)}^{-\alpha}\log |f(re^{i\theta})|\,d\theta \tag{1} \]

are bounded as \(r\to 1-0\). Here \(D^{-\alpha}\) \((-1\le \alpha<+\infty)\) is the operator of integration (for \(0<\alpha<+\infty\)) or differentiation (for \(-1<\alpha<0\)) in the Riemann–Liouville sense with initial point at zero, and

\[ D_{(+)}^{-\alpha}\varphi(r)=\max\{D^{-\alpha}\varphi(r),0\},\qquad D^{-\alpha}\varphi(r)\big|_{\alpha=0}=\varphi(r). \tag{2} \]

The classes \(A_\alpha(-1<\alpha<+\infty)\) expand monotonically as the parameter \(\alpha\) increases, and therefore, in particular, the strict inclusion

\[ A_\alpha\subset A_0\qquad (-1<\alpha<0), \tag{3} \]

holds, where, by virtue of (2), \(A_0=A\) is the well-known class of A. Ostrovskii.

It is also known that if a function \(f(z)\not\equiv 0\) belongs to the class \(A_\alpha\) and \(\{z_k\}_1^\infty\) \((0<|z_k|\le |z_{k+1}|<1)\) is the sequence of its zeros, different from \(z=0\), then the quantity

\[ \sigma_\alpha\{z_k\}=\sum_{k=1}^{\infty}(1-|z_k|)^{1+\alpha} \tag{4} \]

is always finite. On the other hand, if \(\sigma_0\{z_k\}<+\infty\), then the Blaschke function

\[ B(z)=\prod_{k=1}^{\infty}\frac{z_k-z}{1-\overline{z}_k z}\,\frac{|z_k|}{z_k}\qquad (|z|<1) \tag{5} \]

with zeros at the points of the sequence \(\{z_k\}_1^\infty\) is analytic and bounded in modulus by one in the disk \(|z|<1\), and therefore, obviously, belongs to the class \(A=A_0\).

2°. In the present note an answer is given to the question: when can one assert that a function \(B(z)\in A_0\) belongs to a given class \(A_\alpha\) \((-1<\alpha<0)\).

Theorem 1. In order that the function \(B(z)\) belong to the class \(A_\alpha(-1<\alpha<0)\), it is necessary and sufficient that the condition \(\sigma_\alpha\{z_k\}<+\infty\) be fulfilled.

Proof. a) Since, by virtue of the property of the class \(A_\alpha\), the necessity is obvious, we need to establish the sufficiency of the condition \(\sigma_\alpha\{z_k\}<+\infty\) of the theorem. b) Let us denote

\[ A_0(z;r)=\frac{\zeta-z}{1-\overline{\zeta}z}\,\frac{|\zeta|}{\zeta}\qquad (0<|\zeta|<1) \tag{6} \]

and prove the formula

\[ \Omega_\alpha(re^{i\varphi};\zeta)\equiv r^{-\alpha}D^{-\alpha}\log|A_0(re^{i\varphi};\zeta)| = -\frac{1}{\Gamma(1+\alpha)}\log\frac{1}{|\zeta|} + \]

\[ +\frac{1-|\zeta|^2}{\Gamma(1-\alpha)} \int_0^1 \frac{x(1+|\zeta|^2)-|\zeta|r^{-1}(1+r^2x^2)\cos(\varphi-\arg\zeta)} {|x-\zeta/re^{i\varphi}|^2\,|1-x\overline{\zeta}re^{i\varphi}|^2} (1-x)^\alpha\,dx. \tag{7} \]

For this purpose, first of all, we note that the formula

\[ r^{-\alpha}D^{-\alpha}\log\left|1-\frac{re^{i\varphi}}{\zeta}\right| =\operatorname{Re}\frac{1}{\Gamma(1+\alpha)} \int_0^1 \frac{(1-x)^\alpha}{x-\zeta/re^{i\varphi}}\,dx \tag{8} \]

is valid for all \(r\) \((0\le r\le 1)\), \(\varphi\) \((0\le \varphi\le 2\pi)\), and \(\zeta\) \((0<|\zeta|<1)\), where in the case \(\varphi=\arg \zeta\), \(|\zeta|\le r\le 1\), the integral should be understood in the sense of the principal value (see (1), Lemma 9.3). Further, since, by definition, for \(-1<\alpha<0\)

\[ r^{-\alpha}D^{-\alpha}\log|1-\bar{\zeta}re^{i\varphi}| = r^{-\alpha}\frac{d}{dr}D^{-(1+\alpha)} \log|1-\bar{\zeta}re^{i\varphi}|, \]

and, as is easy to see,

\[ D^{-(1+\alpha)}\log|1-\bar{\zeta}re^{i\varphi}| = -\operatorname{Re}\frac{1}{\Gamma(2+\alpha)} \int_0^r \frac{(r-t)^{\alpha+1}}{1-\bar{\zeta}te^{i\varphi}}\,\bar{\zeta}e^{i\varphi}\,dt, \]

we shall have

\[ r^{-\alpha}D^{-\alpha}\log|1-\bar{\zeta}re^{i\varphi}| = -\operatorname{Re}\frac{1}{\Gamma(1+\alpha)} \int_0^1 \frac{(1-x)^\alpha}{1-\bar{\zeta}xre^{i\varphi}}\, \bar{\zeta}re^{i\varphi}\,dx. \tag{9} \]

Finally, taking into account the elementary formula

\[ r^{-\alpha}D^{-\alpha}\{\log|\zeta|\} = -\frac{1}{\Gamma(1+\alpha)}\log\frac{1}{|\zeta|}, \]

from (6), (8), and (9) we obtain

\[ \Omega_\alpha(re^{i\varphi};\zeta) = -\frac{1}{\Gamma(1+\alpha)}\log\frac{1}{|\zeta|} + \]

\[ +\frac{1}{\Gamma(1+\alpha)} \operatorname{Re}\int_0^1 \left\{ \frac{1}{x-\zeta/re^{i\varphi}} + \frac{\bar{\zeta}re^{i\varphi}}{1-x\bar{\zeta}re^{i\varphi}} \right\}(1-x)^\alpha\,dx, \]

whence formula (7) follows immediately. Introduce the following notation:

\[ I_\alpha^{(1)}(re^{i\varphi};\zeta) = 2\int_0^1 \frac{|\zeta|r^{-1}(1+r^2x^2)\sin^2((\varphi-\arg\zeta)/2)} {|x-\zeta/re^{i\varphi}|^2\,|1-x\bar{\zeta}re^{i\varphi}|^2} (1-x)^\alpha\,dx, \tag{10} \]

\[ I_\alpha^{(2)}(re^{i\varphi};\zeta) = \int_0^1 \frac{x(1+|\zeta|^2)-|\zeta|r^{-1}(1+r^2x^2)} {|x-\zeta/re^{i\varphi}|^2\,|1-x\bar{\zeta}re^{i\varphi}|^2} (1-x)^\alpha\,dx, \tag{11} \]

with the aid of which we write formula (7) in the form

\[ \Omega_\alpha(re^{i\varphi};\zeta) = -\frac{1}{\Gamma(1+\alpha)}\log\frac{1}{|\zeta|} + \frac{(1-|\zeta|)^2}{\Gamma(1+\alpha)} \left\{ I_\alpha^{(1)}(re^{i\varphi};\zeta) + I_\alpha^{(2)}(re^{i\varphi};\zeta) \right\}. \tag{7'} \]

b) We pass to estimating the integrals \(I_\alpha^{(k)}(re^{i\varphi};\zeta)\) \((k=1,2)\), assuming that \(0\le r<1\), \(0<|\zeta|<1\). From (10), for \(\varphi\ne \arg\zeta\), we obtain

\[ \left|I_\alpha^{(1)}(re^{i\varphi};\zeta)\right| \le \frac{|\zeta|(\varphi-\arg\zeta)^2}{r\delta^2(\varphi;\zeta)} \int_0^1 \frac{(1-x)^\alpha}{|1-x\bar{\zeta}re^{i\varphi}|^2}\,dx, \]

where

\[ \delta(\varphi;\zeta) = \min_{0\le x\le 1}\left|x-\frac{\zeta}{re^{i\varphi}}\right| \ge \begin{cases} (|\zeta|/r)|\sin(\varphi-\arg\zeta)|, & |\varphi-\arg\zeta|<\pi/2,\\ |\zeta|/r, & \pi/2\le |\varphi-\arg\zeta|\le \pi. \end{cases} \]

Therefore, in general, for any \(r\) \((0<r<1)\), \(\zeta\) \((0<|\zeta|<1)\), and \(\varphi\) \((0\le \varphi\le 2\pi)\) the estimate

\[ \left|I_\alpha^{(1)}(re^{i\varphi};\zeta)\right| \le \frac{\pi^2}{|\zeta|} \int_0^1 \frac{(1-x)^\alpha}{|1-x\bar{\zeta}re^{i\varphi}|^2}\,dx \tag{12} \]

holds.

Let us now note that

\[ x(1+|\zeta|^2)-|\zeta|r^{-1}(1+r^2x^2) =(1-x|\zeta|r)(x-|\zeta|/r), \]

and also that

\[ |1-x\bar\zeta r e^{i\varphi}|\ge 1-x|\zeta|r\ge 1-x|\zeta| \qquad (0\le x\le 1,\ 0\le r\le 1). \]

Then from (11) it follows that

\[ \left| I_\alpha^{(2)}(re^{i\varphi};\zeta)\right| \le \int_0^1 \frac{|x-|\zeta|/r|} {|x-(\zeta/r)e^{-i\varphi}|^2(1-|\zeta|x)} (1-x)^\alpha\,dx \tag{13} \]

for any \(\varphi\ne \arg\zeta\). We next estimate the integrals

\[ U_k(r;\zeta)\equiv \frac1{2\pi}\int_0^{2\pi} \left| I_\alpha^{(k)}(re^{i\varphi};\zeta)\right|\,d\varphi, \]

using Poisson’s integral formula

\[ \frac1{2\pi}\int_0^{2\pi} \frac{1-\omega^2}{|1-\omega e^{i(\varphi-\psi)}|^2}\,d\varphi =1 \qquad (0\le \omega<1,\ 0\le \psi\le 2\pi). \tag{14} \]

First, integrating inequality (12) with respect to \(\varphi\), by virtue of (14), we obtain

\[ U_1(r;\zeta) \le \frac{\pi^2}{|\zeta|} \int_0^1 \frac{(1-x)^\alpha}{1-x^2|\zeta|^2r^2}\,dx \le \frac{\pi^2}{|\zeta|} \int_0^1 \frac{(1-x)^\alpha}{1-|\zeta|x}\,dx \qquad (0<r<1), \tag{15} \]

where the interchange of the order of integration is, of course, permissible.

Proceeding analogously with inequality (13), and observing that, according to (14),

\[ \frac1{2\pi}\int_0^{2\pi} \frac{d\varphi}{|x-(\zeta/r)e^{-i\varphi}|^2} = \frac{1}{|x^2-|\zeta|^2/r^2|} \le \frac1{|\zeta|}\frac1{|x-|\zeta|/r|}, \qquad x\ne |\zeta|/r, \]

we obtain the estimate

\[ U_2(r;\zeta) \le \frac1{|\zeta|} \int_0^1 \frac{(1-x)^\alpha}{1-|\zeta|x}\,dx, \tag{16} \]

where here too the interchange of the order of integration is permissible, for example by Fubini’s theorem, since in the end on the right we obtain a finite quantity.

From identity (7)—(7′), in view of estimates (15) and (16), we arrive at the inequality

\[ m_\alpha(r;A_0(z;\zeta)) = \frac1{2\pi}\int_0^{2\pi} \left|\Omega_\alpha(re^{i\varphi};\zeta)\right|_{(+)}\,d\varphi \le \]

\[ \le \frac1{2\pi}\int_0^{2\pi} \left|\Omega_\alpha(re^{i\varphi};\zeta)\right|\,d\varphi \le \frac1{\Gamma(1+\alpha)}\log\frac1{|\zeta|} + \]

\[ + \frac{2(1+\pi^2)}{\Gamma(1+\alpha)|\zeta|(1-|\zeta|)} \int_0^1 \frac{(1-x)^\alpha}{1-|\zeta|x}\,dx \qquad (0<r<1). \tag{17} \]

Let us now estimate the integrals

\[ V_\alpha(\zeta) = \int_0^1 \frac{(1-x)^\alpha}{1-|\zeta|x}\,dx = \sum_{k=0}^{\infty} \frac{\Gamma(1+\alpha)\Gamma(1+k)} {\Gamma(2+\alpha+k)} |\zeta|^k \qquad (0<|\zeta|<1). \tag{18} \]

Since, by Stirling’s formula, as \(k\to\infty\),

\[ \frac{\Gamma^2(1+k)} {\Gamma(2+\alpha+k)\Gamma(-\alpha+k)} \sim 1, \]

then there exists a constant \(C_\alpha>0\), independent of \(k\geqslant 0\), such that

\[ \frac{\Gamma(1+\alpha)\Gamma(1+k)}{\Gamma(2+\alpha+k)} \leqslant C_\alpha \frac{\Gamma(-\alpha+k)}{\Gamma(-\alpha)\Gamma(1+k)} \qquad (k=0,1,2,\ldots). \]

Hence, and from (18), it follows that

\[ V_\alpha(\xi)\leqslant C_\alpha \sum_{k=0}^{\infty} \frac{\Gamma(-\alpha+k)}{\Gamma(-\alpha)\Gamma(1+k)} |\xi|^k = C_\alpha(1-|\xi|)^\alpha \qquad (0<|\xi|<1). \tag{19} \]

From (17) and (19) it follows that, for \(0<r<1,\ |z_1|\leqslant|\zeta|<1\),

\[ m_\alpha(r; A_0(z;\zeta))\leqslant C(\alpha;|z_1|)(1-|\zeta|)^{1+\alpha}, \tag{20} \]

where \(C(\alpha;|z_1|)\) does not depend on \(|\zeta|\).

Finally, from (20) we obtain the inequality

\[ m_\alpha(r;B)\leqslant \sum_{k=1}^{\infty} m_\alpha(r;A_0(z;z_k)) \leqslant C(\alpha;|z_1|) \sum_{k=1}^{\infty} (1-|z_k|)^{1+\alpha} \qquad (0<r<1), \]

i.e. our assertion \(B(z)\in A_\alpha\).

Let us note that Blaschke functions for which \(\sigma_\alpha\{z_k\}<+\infty\) \((-1<\alpha<0)\) were first considered by Frostman \({}^{(2)}\), who established for such functions a subtle characteristic of the exceptional set \(E_\alpha\subset[0,2\pi]\) of measure zero at the points of which the limit \(|B(re^{i\theta})|\) as \(r\to 1-0\) may fail to exist.

\(3^\circ\). Earlier we constructed \({}^{(1)}\) another important example of a function of the class \(A_\alpha\) \((-1<\alpha<0)\), vanishing on the sequence \(\{z_k\}_1^\infty\), \(\sigma_\alpha\{z_k\}<+\infty\). This function is constructed in the form of the product

\[ B_\alpha(z;z_k)= \prod_{k=1}^{\infty} \left(1-\frac{z}{z_k}\right)e^{-W_\alpha(z;z_k)}, \tag{21} \]

where

\[ W_\alpha(z;\zeta)= \frac{1}{2\pi}\int_{0}^{2\pi} S_\alpha(e^{-i\theta}z)V_\alpha(e^{i\theta};\zeta)\,d\theta, \tag{22} \]

\[ S_\alpha(z)= \frac{1}{\Gamma(1+\alpha)} \left\{ \frac{2}{(1-z)^{1+\alpha}}-1 \right\}, \qquad V_\alpha(re^{i\theta};\xi)= r^{-\alpha}D^{-\alpha}\log\left|1-\frac{re^{i\theta}}{\xi}\right|. \tag{23} \]

Theorem 2. Under the condition \(\sigma_\alpha\{z_k\}<+\infty\) \((-1<\alpha<0)\), the representation

\[ B_\alpha(z;z_k)= B(z)\exp\left\{ \frac{1}{2\pi}\int_{0}^{2\pi} S_\alpha(e^{-i\theta}z)\,d\omega(\theta) \right\} \tag{24} \]

holds, where \(\omega(\theta)\) is a certain function of bounded variation on \([0;2\pi]\).

Proof. According to Theorem 1, \(B(z)\in A_\alpha\), and since we also have \(B_\alpha(z;z_k)\in A_\alpha\), the function analytic in the disk \(|z|<1\),

\[ B_\alpha(z;z_k)B^{-1}(z)\ne 0\qquad (|z|<1), \]

will also belong to the class \(A_\alpha\). But then representation (24) follows directly from the general representation of functions of the class \(A_\alpha\) (see \({}^{(1)}\), Theorem 9.13).

Remark. The property \(\operatorname{Re} S_\alpha(z)\geqslant 0\) \((|z|<1)\) for \(-1<\alpha<0\) of the kernel \(S_\alpha\), as well as the recently established inequality \({}^{(3)}\)

\[ |B_\alpha(z;z_k)|\leqslant |B(z)|\qquad (|z|<1), \]

give some grounds for conjecturing that Theorem 2 remains valid in a stronger form, namely with the additional assertion that in representation (24) the function \(\omega(\theta)\) is nonincreasing.

Institute of Mathematics and Mechanics
Academy of Sciences of the Armenian SSR

Received
17 IV 1967

References

\({}^{1}\) M. M. Dzhrbashyan, Integral transformations and representations of functions in the complex domain, “Nauka,” 1966.
\({}^{2}\) D. Frostman, Fysiogr. Sällsk. Lund Förh., 12 (1942).
\({}^{3}\) M. M. Dzhrbashyan, V. S. Zakharyan, DAN, 173, No. 6 (1967).