Abstract
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MATHEMATICS
L. V. KRESNYAKOVA
ON REGULAR FUNCTIONS WITH BOUNDED MEAN MODULUS
(Presented by Academician V. I. Smirnov on 28 V 1962)
Let us denote by \(H_\delta\) \((\delta > 0)\) the class of functions
\[ f(\zeta)=\sum_{n=0}^{\infty} c_n \zeta^n, \tag{1} \]
regular in the disk \(|\zeta|<1\) and satisfying, for \(0<r<1\), the condition
\[ \frac{1}{2\pi}\int_0^{2\pi} |f(re^{i\theta})|^\delta\,d\theta \leq 1. \tag{2} \]
It is easy to see that \(H_{\delta_1}\subset H_{\delta_2}\) if \(\delta_1>\delta_2\).
Theorem 1. If \(f(z)\in H_\delta\), \(\delta>1\), then for \(n=1,2,\ldots\), with \(\zeta=z\), \(|z|=r<1\), the estimate holds
\[ |f^{(n)}(z)| \leq \frac{n!}{(1-r^2)^{n+1/\delta}} F^{\frac{\delta-1}{\delta}} \left( -\frac{\delta(n-1)+2}{2(\delta-1)}, -\frac{\delta(n-1)+2}{2(\delta-1)}, 1;\, r^2 \right), \tag{3} \]
where \(F(\alpha,\beta,\gamma;z)\) is the hypergeometric series. The order of this estimate as \(r\to 1\) is sharp.
Proof. We have
\[ f^{(n)}(z)=\frac{n!}{2\pi i}\int_{|\zeta|=1} \frac{f(\zeta)\,d\zeta}{(\zeta-z)^{n+1}} \qquad (n=1,2,\ldots). \]
Using Hölder’s inequality, putting \(\frac{1}{\delta}+\frac{1}{\delta_1}=1\), as in paper \((^1)\), we shall have
\[ |f^{(n)}(z)| \leq \frac{n!}{2\pi} \left\{\int_{|\zeta|=1}|f(\zeta)|^\delta\,|d\zeta|\right\}^{1/\delta} \left\{\int_0^{2\pi} \frac{d\theta}{|1-re^{i\theta}|^{\delta_1(n+1)}} \right\}^{1/\delta_1} = \]
\[ = 2\pi F\left( \frac{\delta_1(n+1)}{2}, \frac{\delta_1(n+1)}{2}, 1;\, r^2 \right). \]
Hence, applying the formula
\[ F(\alpha,\beta,\gamma;z) = (1-z)^{\gamma-\alpha-\beta} F(\gamma-\alpha,\gamma-\beta,\gamma;z), \]
we obtain estimate (3).
An example of a function
\[ f(\zeta)= \left[ \frac{1-r^2}{(1-z\zeta)^2} \right]^{1/\delta} \in H_\delta, \qquad r=|z|,\quad |\zeta|<1, \tag{4} \]
shows that the order of estimate (3) as \(r\to 1\) is sharp.
Remark. When \(\delta(n-1)/2(\delta-1)\) is equal to \(m\), a positive integer, formula (3) gives
\[ |f^{(n)}(z)| \leq \frac{n!}{(1-r^2)^{n+1/\delta}} \left[ 1+\binom{m}{1}^{2}r^2+\cdots+\binom{m}{m}^{2}r^{2m} \right]^{(\delta-1)/\delta}. \]
This estimate is sharp and was previously found in paper \((^2)\), p. 301.
Letting \(\delta\to\infty\) in (3), we obtain the estimate for the class of bounded functions previously found by Szasz \((^3)\).
Theorem 2. If \(f(\zeta)=c_m\zeta^m+c_{m+1}\zeta^{m+1}+\cdots \in H_1\) \((m=0,1,2,\ldots;\ c_m\ne0)\), then for \(\zeta=z,\ |z|=r<1\),
\[ |f'(z)|\le \frac{r^{m-1}\left[m+(2-m)r^2\right]}{2(1-r^2)^2} \quad \text{when} \quad |f(z)|\le \frac{r^m}{2(1-r^2)}, \tag{5} \]
\[ |f'(z)|\le \frac{r^{m-1}\left[m(1-r^2)+2r^2+ \sqrt{m^2(1-r^2)^2+4m(1-r^2)r^2+4r^4+4r^2}\right]} {2(1-r^2)^2} \]
\[ \text{when} \quad |f(z)|\ge \frac{r^m}{2(1-r^2)} . \tag{6} \]
The estimate is sharp for every \(|z|=r\), and equality is attained for the function
\[ f(\zeta)=\zeta^m \left( \sqrt{\frac{\sqrt{4r^2+k^2}+k}{2\sqrt{4r^2+k^2}}} + \sqrt{\frac{\sqrt{4r^2+k^2}-k}{2\sqrt{4r^2+k^2}}} \,\frac{\zeta-z}{1-\bar z\zeta} \right)^2 \frac{1-r^2}{(1-\bar z\zeta)^2} \in H_1, \]
where \(k=m(1-r^2)+2r^2,\ r=|z|,\ |\zeta|<1\).
Proof. For a function \(a_0+a_1z+\cdots\in H_1\), G. M. Goluzin proved ([4], p. 72) that for \(n>2k\)
\[ |a_n|\le \begin{cases} 1, & \text{when } |a_k|\le 1/2,\\[4pt] 2\sqrt{|a_k|(1-|a_k|)}, & \text{when } |a_k|\ge 1/2. \end{cases} \tag{7} \]
Put \(\varphi(z)=f(z)/z^m=c_m+c_{m+1}z+\cdots\). If \(f(z)\in H_1\), then also \(\varphi(z)\in H_1\). Consider the function
\[ g(\zeta)=\varphi\left(\frac{\zeta+z}{1+\bar z\zeta}\right) \frac{1-|z|^2}{(1+\bar z\zeta)^2}\in H_1 \quad (|\zeta|<1). \]
Applying (7) to the function \(g(\zeta)\) for \(k=0\) and \(n=1\), we estimate \(|f'(z)|\) as a function of \(|f(z)|\), whence the estimates (5) and (6) follow.
For \(m=0\), (6) gives the estimate of G. M. Goluzin ([4], p. 34), and for \(m=1\) the estimate of Macintyre–Rogosinski ([2], p. 317).
Theorem 3. If \(f(\zeta)\in H_\delta,\ f(\zeta)\ne0\) for \(|\zeta|<1\), and \(0<\delta\le1\), then for \(\zeta=z,\ |z|=r<1\),
\[ |f'(z)|\le \frac{2}{\delta(2-\delta)} \frac{r(1-\delta)+\sqrt{2\delta-\delta^2+r^2}}{(1-r^2)^{1+1/\delta}} \left\{ \frac{2-\delta+r^2+r\sqrt{2\delta-\delta^2+r^2}}{2(1+r^2)} \right\}^{1/\delta}. \tag{8} \]
The estimate is sharp, and equality is attained for the function
\[ f(\zeta)= \left(a_0+a_1\frac{\zeta-z}{1-\bar z\zeta}\right)^{2/\delta} \frac{(1-r^2)^{1/\delta}}{(1-\bar z\zeta)^{2/\delta}} \in H_\delta, \]
where
\[ a_0= \sqrt{\frac{2-\delta+r^2+r\sqrt{2\delta-\delta^2+r^2}}{2(1+r^2)}}; \qquad a_1= \sqrt{\frac{\delta+r^2-r\sqrt{2\delta-\delta^2+r^2}}{2(1+r^2)}}. \]
Proof. The function \(f_1(\zeta)=[f(\zeta)]^{\delta/2}\in H_2\). From the inequality
\[ |f_1(0)|^2+|f_1'(0)|^2\le 1 \]
we find
\[ |f'(0)|\le \frac{2}{\delta} \sqrt{|f(0)|^{2-\delta}-|f(0)|^2}. \tag{9} \]
Applying (9) to the function
\[ g(\zeta)= f\left(\frac{\zeta+z}{1+\bar z\zeta}\right) \left[ \frac{1-r^2}{(1+\bar z\zeta)^2} \right]^{1/2} \in H_\delta, \qquad r=|\zeta|<1, \]
we obtain
\[ |f'(z)|\le \frac{2}{\delta(1-r^2)} \left\{ r|f(z)|+ \sqrt{\frac{|f(z)|^{2-\delta}}{1-r^2}-|f(z)|^2} \right\}. \]
Determining the greatest value of the right-hand side as a function of \(|f(z)|\) on the interval \([0,(1-r^2)^{-1/\delta}]\), we obtain the estimate (8).
Using the results of G. M. Goluzin ((\(^{4}\)), pp. 34 and 45), one can prove the following two theorems.
Theorem 4. If \(f(z)\in H_\delta\) \((\delta>0)\), then
\[ |f(z)|\leq \frac{1}{(1-r^2)^{1/\delta}},\qquad r=|z|<1. \tag{10} \]
The estimate is sharp, and equality is attained for the function (4).
This theorem is known (\(^{5}\)). In what follows we use its result.
Theorem 5. For any integer \(n\geq 1\) and \(\alpha=e^{2\pi i/n}\), for a function \(f(\zeta)\in H_\delta\) \((\delta>0)\), for \(|z|=r<1\), we have the estimate
\[ \sum_{k=1}^{n}|f(\alpha^k z)|\leq \frac{n}{(1-r^2)^{1/\delta}} \qquad \text{for } \delta\geq 1, \tag{11} \]
\[ \sum_{k=1}^{n}|f(\alpha^k z)|\leq \left(\frac{n}{1-r^{2n}}\right)^{1/\delta} \qquad \text{for } \delta\leq 1. \tag{12} \]
The estimate (11) is sharp, and equality is attained for the function
\[ f(\zeta)=\left[\frac{1-r^{2n}}{(1-z^n\zeta^n)^2}\right]^{1/\delta}, \qquad r=|z|. \]
The estimate (12) is asymptotically sharp, as is shown by the example
\[ f(\zeta)=\left[\frac{1-r^2}{(1-\bar z\zeta)^2}\right]^{1/\delta}. \]
In the following theorem an estimate is given for the mean value of the modulus of a function, improving the estimate contained in the book of I. I. Privalov ((\(^{6}\)), p. 84).
Theorem 6. If \(f(\zeta)\in H_\delta\), \(0<\delta<1\), then
\[ \frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|\,d\theta \leq \frac{1}{(1-r^2)^{1/\delta-1}}, \qquad 0\leq r<1. \tag{13} \]
Proof. Applying \(k\) times the Schwarz–Bunyakovsky inequality, we shall have
\[ \frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|\,d\theta \leq \frac{1}{2\pi}(2\pi)^{1/2} \left(\int_{0}^{2\pi}|f(re^{i\theta})|^{2-\delta}\,d\theta\right)^{1/2} \leq \cdots \]
\[ \cdots \leq \frac{1}{2\pi}(2\pi)^{\frac12+\frac14+\cdots+\frac{1}{2^k}} \left(\int_{0}^{2\pi}|f(re^{i\theta})|^{2^k-(2^k-1)\delta}\,d\theta\right)^{2^{-k}} . \]
Letting \(k\to\infty\), and using the relation
\[ \lim_{\lambda\to\infty} \left(\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|^\lambda\,d\theta\right)^{1/\lambda} = \max_{0\leq\theta\leq\pi}|f(re^{i\theta})| \]
and inequality (10), we obtain the estimate (13).
Consider the subclass \(\widetilde H_1\) of functions \(f(z)\) of the class \(H_1\), representable in the form
\[ f(z)=\sum_{k=1}^{\infty}c_k z^k = z\left(\sum_{k=0}^{\infty}b_k z^k\right)^2 = zF^2(z), \tag{14} \]
where the functions \(F(z)\) are regular in the disk, \(|c_0|\ne 0\) and fixed.
Theorem 7. The radii of univalence \(R_1\) and of starlikeness \(\rho_1\) for the class \(H_1\) coincide and are equal to the root contained between 0 and 1 of the equation
\[ \frac{9r^2-2r^4+r^6}{1+6r^2+r^4}=|c_1|. \tag{15} \]
The estimate is sharp, and equality is attained for the function
\[ f(z)=z\left[b_0-\frac{b_1}{3}\,\frac{3z-\rho_1 z^2}{(1-\rho_1 z)^2}\right]\in \widetilde H_1, \tag{16} \]
where \(b_0,b_1>0\), \(b_0^2=|c_1|\), and
\[ b_1^2=9(1-b_0^2)\,\frac{(1-\rho_1^2)^3}{9-2\rho_1^2+\rho_1^4}. \tag{17} \]
Proof. As is known, the disk \(|z|<r\) is mapped by the function \(w=f(z)\) onto a domain starlike with respect to the point \(w=0\), if for all \(z\) in this disk the inequality
\[ \operatorname{Re}\left\{z\frac{f'(z)}{f(z)}\right\}\geq 0 \tag{18} \]
is satisfied.
Condition (18) is a consequence of the inequality
\[ \left|\frac{zf'(z)}{f(z)}-1\right|\leq 1, \]
or, on the basis of (14), a consequence of the inequality
\[ \sum_{k=1}^{\infty}(2k+1)|b_k|r^k\leq |b_0|. \tag{19} \]
Using the Cauchy–Schwarz inequality and noting that \(\sum_{k=0}^{\infty}|b_k|\leq 1\), we obtain
\[ \sum_{k=1}^{\infty}(2k+1)|b_k|r^k \leq \left[\frac{9r^2-2r^4+r^6}{(1-r^2)^3}\right]^{1/2} \sqrt{1-|b_0|^2}. \]
Denote the root of equation (15) by \(\rho_1\). The function \(f(z)\in \widetilde H_1\) will also be univalent in the disk \(|z|<\rho_1\).
The function (16), provided condition (17) is fulfilled, belongs to the class \(\widetilde H_1\), and its derivative for \(z=\rho_1\) vanishes. Hence, for this function \(R_1=\rho_1\), and \(\rho_1\) is the radius of starlikeness.
The following theorem is proved analogously:
Theorem 8. The radius of convexity \(r_1\) for the class of functions \(f(z)\in H_2\), \(f(0)=0\), \(|c_1|\ne 0\) fixed, is equal to the root contained between 0 and 1 of the equation
\[ \frac{r^2(16+r^2-11r^4-5r^6+r^8)}{(1-r^2)^5} = \frac{|c_1|^2}{1-|c_1|^2}. \tag{20} \]
The estimate is sharp, and equality is attained for the function
\[ f(z)=c_1z-\frac{c_2}{4}\, \frac{4z^2-2r_1z^3+r_1^2z^4}{(1-r_1z)^3}, \tag{21} \]
where
\[ c_1+\frac{c_2^2}{16}\, \frac{16+r_1^2+11r_1^4-5r_1^6+r_1^8}{(1-r_1^2)^5} =1,\qquad c_1>0,\quad c_2>0. \]
Gorky State University
named after N. I. Lobachevsky
Received
24 V 1962
References
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- A. I. Macintyre, W. W. Rogosinsky, Acta Math., 82, 275 (1950).
- O. Szász, Math. Zs., 8, 303 (1920).
- G. M. Goluzin, Tr. Mat. inst. im. V. A. Steklova AN SSSR, 18 (1946).
- Takenaka, Tohoku Math. J., 27, 21 (1926).
- I. I. Privalov, Boundary Properties of Analytic Functions, 1950.