GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 4, 1998, 361-366

PROPERTIES OF THE SALAGEAN OPERATOR

LI JIAN LIN AND SHIGEYOSHI OWA

Abstract. The object of the present paper is to show the properties of the Salagean operator for analytic functions in the open unit disk. The main results obtained here extend and improve the earlier results obtained by several authors.

Introduction. Let A be the class of functions f(z) of the form f(z) = z + ∑_{k=2}^{∞} a_{k}z^{k} which are analytic in the open unit disk U = {z : |z| < 1}. For f(z) belonging to A, Salagean [1] has introduced the following operator called the Salagean operator:

D^{0}f(z) = f(z), D^{1}f(z) = Df(z) = zf′(z),
D^{n}f(z) = D(D^{n−1}f(z)) (n ∈ N = {1, 2, 3, … }).

Note that D^{n}f(z) = z + ∑_{k=2}^{∞} k^{n}a_{k}z^{k}, n ∈ N_{0} = {0} ∪ N.

A function f(z) ∈ A is said to belong to the class S_{n}(α) if it satisfies

Re \{\frac{D^{n+1}f(z)}{D^{n}f(z)}\} > α (z ∈ U)

for some α (0 ≤ α < 1) and n ∈ N_{0}.

For the class S_{n}(α), Owa, Shen, and Obradović [2] showed that if f(z) ∈ A satisfies

|\frac{D^{n+1}f(z)}{D^{n}f(z)} − 1|^{1−β} |\frac{D^{n+2}f(z)}{D^{n+1}f(z)} − 1|^{β} <

< (1 − α)^{1−2β}(1 − \frac{3}{2}α + α^{2})^{β} (z ∈ U)

for some α (0 ≤ α ≤ 1/2) and β (0 ≤ β ≤ 1), then f(z) ∈ S_{n}(α). Also, Uralegaddi [3] proved that if

f(z) = z + a_{m+1}z^{m+1} + a_{m+2}z^{m+2} + ⋯ ∈ S_{n}(0)

361 for some m ∈ N and n ∈ N, then Re \{\frac{D^{n}f(z)}{z}\}^{1/(n+1)} > \frac{1}{2} (z ∈ U).

1991 Mathematics Subject Classification. 30C45.

Key words and phrases. Analytic function, Salagean operator, Jack’s lemma.

1072-947X/98/0700-0361$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

In the present paper, we first show that the above conclusions hold under a much weaker hypothesis. We then sharpen the above-mentioned result of Uralegaddi and some related results. Our main tool is the well-known Jack’s lemma [4] (see also [5]):

Lemma. Let w(z) be a nonconstant analytic function in U with w(0) = 0. If |w(z)| attains its maximum value at a point z_{0} ∈ U on the circle |z| = r < 1, then z_{0}w′(z_{0}) = mw(z_{0}), where m is real and m ≥ 1.

Results and Proofs. Now we derive

Theorem 1. If f(z) ∈ A satisfies

|\frac{D^{n+1}f(z)}{D^{n}f(z)} − 1|^{γ} |\frac{D^{n+2}f(z)}{D^{n+1}f(z)} − 1|^{β} < M(α, β, γ) (z ∈ U)  (1)

for some α (0 ≤ α < 1), β (β ≥ 0), and γ ≥ 0 with β + γ > 0, then f(z) ∈ S_{n}(α), where n ∈ N_{0} and

M(α, β, γ) = \begin{cases}(1 − α)^{γ}(\frac{3}{2} − α)^{β} & (0 ≤ α ≤ 1/2), \\ 2^{β}(1 − α)^{β+γ} & (1/2 ≤ α < 1). \end{cases}

Proof. (i) Let 0 ≤ α ≤ 1/2. Then we define the function w(z) by

\frac{D^{n+1}f(z)}{D^{n}f(z)} = \frac{1 + (1 − 2α)w(z)}{1 − w(z)} (z ∈ U)  (2)

with w(z) ≠ 1. It follows from (2) that

|\frac{D^{n+1}f(z)}{D^{n}f(z)} − 1|^{γ} |\frac{D^{n+2}f(z)}{D^{n+1}f(z)} − 1|^{β} = |\frac{2(1 − α)w(z)}{1 − w(z)}|^{β+γ} ×

×|1 + \frac{zw′(z)}{w(z)(1 + (1 − 2α)w(z))}|^{β} < (1 − α)^{γ} (\frac{3}{2} − α)^{β}

for all z ∈ U. Suppose that there exists a point z_{0} ∈ U such that

max_{|z|≤|z_{0}|} |w(z)| = |w(z_{0})| = 1 (w(z_{0}) ≠ 1).

Then, by Lemma, we have w(z_{0}) = e^{iθ} and z_{0}w′(z_{0}) = mw(z_{0}) (m ≥ 1). Therefore we have

|\frac{D^{n+1}f(z_{0})}{D^{n}f(z_{0})} − 1|^{γ} |\frac{D^{n+2}f(z_{0})}{D^{n+1}f(z_{0})} − 1|^{β} = |2(1 − α)\frac{w(z_{0})}{1 − w(z_{0})}|^{β+γ} ×

×|1 + \frac{m}{1 + (1 − 2α)w(z_{0})}|^{β} = \frac{2^{β+γ}(1 − α)^{β+γ}}{|1 − e^{iθ}|^{β+γ}} |1 + \frac{m}{1 + (1 − 2α)e^{iθ}} |^{β} ≥

≥ (1 + α)^{β+γ} |1 + \frac{m}{2(1 − α)}|^{β} ≥ (1 − α)^{γ} (\frac{3}{2} − α)^{β}

which contradicts condition (1) for 0 ≤ α ≤ 1/2. This gives us |w(z)| < 1 for all z ∈ U, i.e., f(z) = S_{n}(α).

(ii) Let 1/2 ≤ α < 1. Define the function w(z) by

\frac{D^{n+1}f(z)}{D^{n}f(z)} = \frac{α}{α − (1 − α)w(z)} (z ∈ U).

Then condition (1) for 1/2 ≤ α < 1 leads us to

\frac{|zw′(z) + w(z)|^{β}|w(z)|^{γ}}{|α − (1 − α)w(z)|^{β+γ}} (z ∈ U).  (3)

If there exists a point z_{0} ∈ U such that max_{|z|≤|z_{0}|} |w(z)| = |w(z_{0})| = 1, then by

using the lemma we have

\frac{|z_{0}w′(z_{0}) + w(z_{0})|^{β}|w(z_{0})|^{γ}}{|α − (1 − α)w(z_{0})|^{β+γ}} = \frac{(m + 1)|^{β}}{|α − (1 − α)w(z_{0})|^{β+γ}} ≥ 2^{β},

which contradicts (2.7). Therefore we conclude that f(z) ∈ S_{n}(α).

Remark 1. Several cases of Theorem 1 with special values of the parameters n, α, β, and γ will improve some other interesting known results. For example, taking n = α = 0 and β = γ = 1 in Theorem 1, we have

|\frac{zf′′(z)}{f′(z)}(\frac{zf′′(z)}{f(z)} − 1)| < \frac{3}{2}, f(z) ∈ S^{∗} = S_{0}(0),

which improves the result by Obradović [6, p. 229].

Theorem 2. Let f(z) = z + a_{m+1}z^{m+1} + a_{m+2}z^{m+2} + ⋯ be analytic in U and satisfy

Re \{\frac{D^{n+1}f(z)}{D^{n}f(z)}\} > \frac{2 − m(n + 1)}{2} (z ∈ U)  (4)

for some m ∈ N and n ∈ N_{0}. Then

Re \{\frac{D^{n}f(z)}{z}\}^{1/(n+1)} > \frac{1}{2} (z ∈ U).  (5)

Proof. Define the function w(z) by

(\frac{D^{n}f(z)}{z})^{1/(n+1)} = \frac{1}{1 + w(z)} (z ∈ U).  (6)

Clearly, w^{(j)}(0) = 0 (j = 0, 1, 2, …, m − 1) and w(z) ≠ −1. Making use of the logarithmic differentiation of both sides in (6), we have

\frac{D^{n+1}f(z)}{D^{n}f(z)} = 1 − (n + 1)\frac{zw′(z)}{1 + w(z)}.

Suppose that |w(z)| ≮ 1 for some z ∈ U. Then by Lemma there exists a point z_{0} ∈ U such that z_{0}w′(z_{0}) = kw(z_{0}) (k ≥ m) with |w(z_{0})| = 1. This implies that

Re \{\frac{D^{n+1}f(z_{0})}{D^{n}f(z_{0})}\} = 1 − (n + 1)k Re \{\frac{w(z_{0})}{1 + w(z_{0})}\} ≤ \frac{2 − m(n + 1)}{2}

which contradicts condition (4). Thus we have |w(z)| < 1 for all z ∈ U. This completes the proof of the theorem.

Remark 2. It is worth mentioning that the lower bound 1/2 in (5) is the best possible one. The function

f_{0}(z) = ∑_{k=1}^{∞} \frac{Γ(k + n)}{Γ(n + 1)Γ(k)\{(k − 1)m + 1\}^{n}} (−1)^{k−1}z^{(k−1)m+1}  (7)

or its rotation is extremal in the following sense:

inf_{z∈U} Re \{\frac{D^{n+1}f_{0}(z)}{D^{n}f_{0}(z)}\} = \frac{2 − (n + 1)m}{2}, inf_{z∈U} Re \{\frac{D^{n}f_{0}(z)}{z}\} = \frac{1}{2}.

Indeed, for the function f_{0}(z) defined by (7), we have

D^{n}f_{0}(z) = ∑_{k=1}^{∞} \frac{Γ(k + n)}{Γ(n + 1)Γ(k)}(−1)^{k−1}z^{(k−1)m+1} = \frac{z}{(1 + z^{m})^{n+1}} ,

which yields

inf_{z∈U} Re \{\frac{D^{n}f_{0}(z)}{z}\}^{1/(n+1)} = inf_{z∈U} Re \{\frac{1}{1 + z^{m}} \} = \frac{1}{2},

inf_{z∈U} Re \{\frac{D^{n+1}f_{0}(z)}{D^{n}f_{0}(z)}\} = inf_{z∈U} Re \{1 − m(n + 1)\frac{z^{m}}{1 + z^{m}} \} = \frac{2 − (n + 1)m}{2}.

Theorem 3. If f(z) = z+a_{m+1}z^{m+1}+a_{m+2}z^{m+2}+⋯ ∈ Sn(α) for some α (0 ≤ α < 1), n ∈ N_{0}, and m ∈ N, then for any β (0 < β ≤ m/(2(1 − α))) we have the sharp estimate

Re \{\frac{D^{n}f(z)}{z}\}^{β} > 2^{2β(α−1)/m} (z ∈ U).  (8)

Proof. Define the function w(z) by

\frac{D^{n}f(z)}{z} = \frac{1}{(1 + w(z))^{2(1−α)/m}} (z ∈ U).

Then, clearly, w^{(j)}(0) = 0 (j = 0, 1, 2, …, m − 1) and w(z) ≠ −1. Using the same method as the proof of Theorem 2, one can show that |w(z)| < 1 for all z ∈ U. Thus

Re \{\frac{D^{n}f(z)}{z}\}^{β} = Re \{\frac{1}{(1 + w(z))^{2β(1−α)/m}} \} ≥

≥ \{ Re (\frac{1}{1 + w(z)})\}^{2β(1−α)/m} > 2^{2β(α−1)/m} (z ∈ U),

where we use the fact that Re{z^{λ}} ≥ {Re(z)}^{λ} for Re(z) > 0 and 0 ≤ λ ≤ 1.

That the rezult (8) is the best possible one can be seen by considering the function

f_{0}(z)=∑_{k=1}^{∞} \frac{Γ(k + 2(1 − α)/m − 1)}{Γ(2(1−α)/m)Γ(k)\{(k−1)m+1\}^{n}} (−1)^{k−1}z^{k−1}z^{(k−1)m+1}.

Since D^{n}F_{0}(z) = z(1 + z^{m})^{2(α−1)/m} for F_{0}(z), we have

inf_{z∈U} Re \{\frac{D^{n+1}F_{0}(z)}{D^{n}F_{0}(z)}\} = α,

inf_{z∈U} Re \{\frac{D^{n}F_{0}(z)}{z}\}^{β} = inf_{z∈U} Re \{\frac{1}{(1+z^{m})^{2β(1−α)/m}} \}=2^{2β(α−1)/m}.

Remark 3. Theorem 3 sharpens the result by Owa et al. [7] when m = 1. Furthermore, if we take α = 0 and β = 1/(n + 1) in Theorem 3, then we obtain

Corollary 1. If f(z) = z + a_{m+1}z^{m+1} + a_{m+2}z^{m+2} + ⋯ ∈ S_{n}(0) for some m ∈ N and n ∈ N_{0} with m(n + 1) ≥ 2, then we have the estimate

Re \{\frac{D^{n}f(z)}{z}\}^{1/(n+1)} > \frac{1}{4^{1/m(n+1)}} (z ∈ U).

This sharpens the above-mentioned result by Uralegaddi [3]. If we take n = α = 0 and β = m/2 in Theorem 3, then we have

Corollary 2. If f(z) = z + a_{m+1}z^{m+1} + a_{m+2}z^{m+2} + ⋯ ∈ S^{∗} with m ∈ N, then

Re \{\frac{f(z)}{z}\}^{m/2} > \frac{1}{2} (z ∈ U).

The result is sharp.

It should be pointed out that Corollary 2 shows that the result by Golusin [8, Theorem 5] (see also [5, Theorem 11]) is sharp.

Acknowledgements

The second author was supported, in part, by the Japanese Ministry of Education, Science and Culture under a grant-in-aid for general scientific research.

The authors would like to express their gratitude to the referee for valuable advices in the preparation of this paper.

References

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2. S. Owa, C. Y. Shen, and M. Obradović, Certain subclasses of analytic functions. Tamkang J. Math. 20(1989), 105–115.

3. B. A. Uralegaddi, Certain subclasses of analytic functions. New Trends in Geometric Function Theory and Applications (Madras, 1990), 159–161, World Scientific Publishing Company, Singapore, New Jersey, London, and Hong Kong, 1991.

4. I. S. Jack, Functions starlike and convex of order α. J. London Math. Soc. 3(1971), 469–474.

5. S. S. Miller and P. T. Mocanu, Second-order differential inequalities in the complex plane. J. Math. Anal. Appl. 65(1978), 289–305.

6. H. M. Srivastava and S. Owa (Editors), Current topics in analytic function theory. World Scientific Publishing Company, Singapore, New Jersey, London, and Hong Kong, 1992.

7. S. Owa, M. Obradović, and S. K. Lee, Notes on a certain subclass of analytic functions introduced by Salagean. Bull. Korean Math. Soc. 23(1986), 133–140.

8. G. M. Golusin, Some estimates for coefficients of univalent functions. (Russian) Mat. Sb. 2(1938), No. 3(45), 321–330.

(Received 22.02.1996; revised 17.04.1997)

Authors’ addresses:

Li Jian Lin Department of Applied Mathematics Northwestern Polytechnical University Xi An, Shaan Xi 710072 People’s Republic of China Shigeyoshi Owa Department of Mathematics Kinki University Higashi-Osaka, Osaka 577 Japan