UDC 517.54

MATHEMATICS

I. M. MILIN

HEYMAN’S REGULARITY THEOREM FOR THE COEFFICIENTS OF UNIVALENT FUNCTIONS

(Presented by Academician M. A. Lavrent’ev on 2 X 1969)

Let \(S\) be the class of functions \(f(z)=z+c_2z^2+\cdots\), regular and univalent in the unit disk, and let \(\Sigma\) be the class of functions \(F(z)=z+a_0+a_1z^{-1}+\cdots\), meromorphic and univalent in the domain \(|z|>1\). Denote the expansions:

\[ \ln \frac{f(z)}{z}=\sum_{k=1}^{\infty}2\gamma_k z^k,\qquad f(z)\in S; \tag{1} \]

\[ \ln \frac{z-t}{F(z)-F(t)}=\sum_{k=1}^{\infty} A_k(t)z^{-k},\qquad F(z)\in\Sigma; \tag{2} \]

\[ (1-z)^{-h}=\sum_{k=0}^{\infty}d_k(h)z^k,\qquad h>0. \tag{3} \]

In addition, by the symbol \(\{\psi(z)\}_n\) we shall denote the coefficient of \(z^n\) in the expansion of the function \(\psi(z)\) about \(z=0\), and by \(M(r,f)\) the maximum of \(|f(z)|\) on the circle \(|z|=r,\ 0<r<1\).

It is known \((^1)\) that for every function \(f(z)\in S\) there exists

\[ \lim_{r\to 1}(1-r)^2M(r,f)=\alpha,\quad 0\leq \alpha\leq 1, \]

and, for \(\alpha>0\), a radius of greatest growth of the function, i.e. such a \(\varphi_0\in[0,2\pi]\) that

\[ \lim_{r\to 1}(1-r)^2|f(re^{i\varphi_0})|=\alpha,\qquad 0<\alpha\leq 1. \tag{4} \]

Using (4) and the “area inequality” \((^2)\):

\[ \sum_{n=1}^{\infty} n|A_n(z)|^2\leq \ln \frac{1}{1-r^2},\qquad |z|=\frac{1}{r}>1, \]

I. E. Bazilevich \((^3)\) proved for \(f(z)\in S\) the inequality

\[ \sum_{k=1}^{\infty} k\left|\gamma_k-\frac{1}{k}e^{ik\varphi_0}\right|^2\leq \frac{1}{2}\ln\frac{1}{\alpha}. \tag{5} \]

It is proved below that conditions (4) and (5) are already sufficient for the regular growth of the Taylor coefficients even of non-univalent functions, i.e. the following holds.

Theorem. For a function \(f(z)=z+c_2z^2+\cdots\) satisfying conditions (4) and (5), for any \(\lambda>1/4\) the asymptotic equality

\[ \frac{\{(f(z)/z)^\lambda\}_n}{d_n(2\lambda)} \sim \alpha^\lambda \exp\{i[\lambda\arg f(re^{i\varphi_0})-(n+\lambda)\varphi_0]\} \qquad (n\to\infty), \tag{6} \]

holds, where \(r\) is connected with \(n\) by the relation

\[ 1-r=\theta/n,\qquad m<\theta<M \tag{7} \]

(\(m\) and \(M\) are positive constants).

The proof of the theorem is based on two lemmas.

Lemma 1. If the coefficients of the power series \(\displaystyle \sum_{k=1}^{\infty} A_k z^k\) satisfy—

satisfy the conditions:

\[ \text{a)}\quad \sum_{k=1}^{\infty} k |A_k|^2 < \infty; \]

\[ \text{b)}\quad \operatorname{Re}\sum_{k=1}^{n} A_k = O(1)\quad (n\to\infty) \tag{8} \]

and if the power series is formed

\[ \sum_{k=0}^{\infty} D_k z^k=\exp\left(\sum_{k=1}^{\infty} A_k z^k\right), \tag{9} \]

then for every \(h>1/2\) the following asymptotic equality holds:

\[ \sum_{k=0}^{n} D_k-\frac{1}{d_n(h)}\sum_{k=0}^{n} d_{n-k}(h)\overline{D}_k=o(1)\quad (n\to\infty). \tag{10} \]

Denoting by \(S_n^{(h)}\) the numerator of the Cesàro mean in (10), i.e.,

\[ S_n^{(h)}=\sum_{k=0}^{n} d_{n-k}(h)D_k = \left\{ \frac{1}{(1-z)^h}\exp\left(\sum_{k=1}^{\infty} A_k z^k\right) \right\}_n \quad (n=0,1,\ldots), \tag{11} \]

we write for the Cesàro means the obvious identity

\[ \frac{S_n^{(h)}}{d_n(h)}-\frac{S_n^{(h+1)}}{d_n(h+1)} = \sum_{k=0}^{n-1} \left[ \frac{d_k(h)}{d_n(h)}-\frac{d_k(h+1)}{d_n(h+1)} \right]D_{n-k} = \sum_{k=1}^{n} d_{n-k}(h)\cdot kD_k/(n+h)\times \]

\[ \times d_n(h) = \sum_{k=1}^{n} d_{n-k}(h)\sum_{\nu=1}^{k}D_{k-\nu}\nu A_\nu \big/ (n+h)d_n(h) = \sum_{k=1}^{n} S_{n-k}^{(h)}kA_k/(n+h)d_n(h). \tag{12} \]

Cauchy’s inequality, applied to the right-hand side of (12), gives the estimate

\[ \left| \frac{S_n^{(h)}}{d_n(h)}-\frac{S_n^{(h+1)}}{d_n(h+1)} \right| \le \frac{1}{(n+h)d_n(h)} \left( \sum_{k=0}^{n-1}|S_k^{(h)}|^2 \sum_{k=1}^{n} k^2|A_k|^2 \right)^{1/2}. \tag{13} \]

In order to estimate \(\sum_{k=0}^{n-1}|S_k^{(h)}|^2\), we note that, by the conditions of the lemma, the relation

\[ \sum_{k=1}^{n} k\left|\frac{h}{k}+A_k\right|^2 = h^2\sum_{k=1}^{n}\frac{1}{k} + 2h\operatorname{Re}\sum_{k=1}^{n}A_k + \sum_{k=1}^{n}k|A_k|^2 \le h^2\sum_{k=1}^{n}\frac{1}{k}+O(1) \]

holds, and, consequently, as \(n\to\infty\),

\[ \delta_n(h) = \max_{1\le \nu\le n} \left\{ \frac{1}{h^2} \sum_{k=1}^{\nu} k\left|\frac{h}{k}+A_k\right|^2 - \sum_{k=1}^{\nu}\frac{1}{k} \right\} = O(1). \tag{14} \]

But earlier, for the coefficients of a compound function of exponential form, inequality (4) was established:

\[ \sum_{k=0}^{n-1} \left| \left\{ \exp\left(\sum_{k=1}^{\infty}\left(\frac{h}{k}+A_k\right)z^k\right) \right\}_k \right|^2 \le \exp\bigl(h\delta_{n-1}(h)\bigr) \sum_{k=0}^{n-1} d_k^2(h), \]

using which, and taking into account (11) and (14), we shall have

\[ \sum_{k=0}^{n-1}|S_k^{(h)}|^2 = O(1)\sum_{k=0}^{n-1}d_k^2(h) \quad (n\to\infty),\quad h>0. \tag{15} \]

Simple calculations give, for the sum of squares of binomial coefficients when \(h>1/2\), the estimate

\[ \sum_{k=0}^{n-1}d_k^2(h)=O(1)n^{2h-1}\quad (n\to\infty), \]

which, together with (15), (13), and condition (8a), leads to the result

\[ S_n^{(h)}/d_n(h)-S_n^{(h+1)}/d_n(h+1)=o(1)\quad (n\to\infty),\qquad h>\frac12 . \tag{16} \]

In particular, for \(h=1\), it follows from (16) that

\[ \frac1n\sum_{k=1}^{n} kD_k=o(1)\quad (n\to\infty). \tag{17} \]

Now consider the difference

\[ \sum_{k=0}^{n} D_k-\frac{S_n^{(h)}}{d_n(h)} = \frac{1}{d_n(h)}\sum_{k=1}^{n}\frac1k\,[d_n(h)-d_{n-k}(h)]\,kD_k. \tag{18} \]

Since for \(h\ge 2\) the sequence of numbers
\[ \frac1k\,[d_n(h)-d_{n-k}(h)] \]
\((k=1,2,\ldots,n)\) is nondecreasing, by Abel’s inequality, from (18) and (17) we obtain

\[ \sum_{k=0}^{n} D_k-\frac{S_n^{(h)}}{d_n(h)} =o(1)\quad (n\to\infty),\qquad h\ge 2. \tag{19} \]

Equalities (16) and (19), considered together, prove the lemma completely.

Lemma 2. Under the conditions of Lemma 1, the following asymptotic equalities hold for \(h\ge \frac12\):

\[ \frac{1}{d_n(h)}\sum_{k=0}^{n} d_{n-k}(h)D_k \sim \sum_{k=0}^{\infty} D_k r^k \sim \exp\left(\sum_{k=1}^{n} A_k\right) \quad (n\to\infty), \tag{20} \]

where \(r\) is connected with \(n\) by formula (7).

Taking into account Lemma 1 and condition (8b), it suffices to prove the relations

\[ \sum_{k=0}^{n} D_k \sim \sum_{k=0}^{\infty} D_k r^k \sim \exp\left(\sum_{k=1}^{n} A_k\right) \quad (n\to\infty). \tag{21} \]

We first derive the second part of (21). To this end consider the difference

\[ \sum_{k=1}^{n} A_k-\sum_{k=1}^{\infty} A_k r^k = \sum_{k=1}^{n} A_k(1-r^k) - \sum_{k=n+1}^{\infty} A_k r^k . \]

Applying Cauchy’s inequality, we have

\[ \left| \sum_{k=1}^{n} A_k-\sum_{k=1}^{\infty} A_k r^k \right| \le \left[ \sum_{k=1}^{n} k|A_k|^2(1-r^k) \sum_{k=1}^{n}\frac1k(1-r^k) \right]^{1/2} + \]

\[ +\frac1n \left[ \sum_{k=n+1}^{\infty} k|A_k|^2\cdot \sum_{k=n+1}^{\infty} kr^{2k} \right]^{1/2} \le \left[ \sum_{k=1}^{n} k|A_k|^2(1-r^k) \right]^{1/2} [n(1-r)]^{1/2} + \]

\[ + \left[ \sum_{k=n+1}^{\infty} k|A_k|^2 \right]^{1/2} \frac{1}{n(1-r)}, \]

whence, taking (7) and (8a) into account, we conclude that

\[ \sum_{k=1}^{n} A_k-\sum_{k=1}^{\infty} A_k r^k=o(1)\quad (n\to\infty), \tag{22} \]

which, after exponentiation, gives the second part of (21).

Next, consider the other difference for \(0<r<1\):

\[ \sum_{k=0}^{n} D_k-\sum_{k=0}^{\infty} D_k r^k = \sum_{k=0}^{n} D_k(1-r^k) - \sum_{k=n+1}^{\infty} D_k r^k . \tag{23} \]

The first sum on the right-hand side of (23) is estimated with the aid of Abel’s inequality, namely
\[ \left|\sum_{k=0}^{n} D_k(1-r^k)\right| = \left|\sum_{k=1}^{n} kD_k\,\frac{1}{k}(1-r^k)\right| \leq (1-r)\max_{1\leq \nu \leq n}\left|\sum_{k=1}^{\nu} kD_k\right|. \tag{24} \]
From (24), (7), and (17) we obtain
\[ \sum_{k=0}^{n} D_k(1-r^k)=o(1)\qquad (n\to\infty). \tag{25} \]

The second sum can be estimated, also using Abel’s transformation:
\[ \left|\sum_{k=n+1}^{N} D_k r^k\right| \leq \left(\sum_{k=n+1}^{N}\frac{1}{k}r^k+\frac{2n+1}{n+1}r^{n+1}\right) \max_{n+1\leq k\leq N}\frac{1}{k} \left|\sum_{\nu=n+1}^{k}\nu D_\nu\right|. \tag{26} \]
If \(r\) is chosen according to (7), then the first factor in (26) is uniformly bounded with respect to \(n\) and \(N\), while the second factor, as \(n\to\infty\), tends to zero uniformly with respect to \(N\) by (17), and, consequently,
\[ \sum_{k=n+1}^{\infty}D_k r^k=o(1), \]
which, together with (25) and (86), gives the first part of (21). The lemma is proved.

Proof of the theorem. Without loss of generality one may assume that \(\varphi_0=0\), since otherwise, instead of \(f(z)\), it suffices to consider \(e^{-i\varphi_0}f(e^{i\varphi_0}z)\).

Starting from the identity
\[ \left(\frac{f(z)}{z}\right)^{\lambda} = \frac{1}{(1-z)^{2\lambda}} \left[\frac{f(z)}{z}(1-z)^2\right]^{\lambda} \tag{27} \]
and denoting
\[ \ln\left[\frac{f(z)}{z}(1-z)^2\right]^{\lambda} = \sum_{k=1}^{\infty} A_k z^k,\qquad A_k=2\lambda\left(\gamma_k-\frac{1}{k}\right), \]
\[ \left[\frac{f(z)}{z}(1-z)^2\right]^{\lambda} = \exp\left(\sum_{k=1}^{\infty} A_k z^k\right) = \sum_{k=0}^{\infty} D_k z^k, \tag{28} \]
we write the equality of the Taylor coefficients for the functions in (27):
\[ \left\{\left(\frac{f(z)}{z}\right)^{\lambda}\right\}_n = \sum_{k=0}^{n} d_{n-k}(h)D_k \qquad (n=0,1,\ldots),\qquad h=2\lambda. \tag{29} \]

It is not difficult to verify that the coefficients of the power series \(\sum_{k=1}^{\infty} A_k z^k\) satisfy conditions (8). Therefore, for \(h>1/2\) and for \(r\) chosen according to (7), by Lemma 2 we shall have:
\[ \frac{1}{d_n(h)}\sum_{k=0}^{n} d_{n-k}(h)D_k \sim \sum_{k=0}^{\infty} D_k r^k = \left[|f(r)|\,\frac{(1-r)^2}{r}\right]^{\lambda} \exp(i\lambda\arg f(r)), \]
which, together with (29) and (4), proves the theorem.

From the theorem just obtained there follows directly Hayman’s regularity theorem \({}^{1}\) for the coefficients of univalent functions in the case \(\alpha>0\) (proved by him for \(p\)-valent functions). If \(\alpha=0\), then the regularity theorem is proved more elementarily.

Received
26 IX 1969

REFERENCES

\({}^{1}\) W. K. Hayman, Multivalent Functions, IL, 1960.
\({}^{2}\) I. M. Milin, DAN, 154, No. 2 (1964).
\({}^{3}\) I. E. Bazilevich, Matem. sbornik, 74 (116), 1 (1967).
\({}^{4}\) I. M. Milin, DAN, 176, No. 5 (1967).