GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 5, 1995, 535-545

PROPERTIES OF CERTAIN INTEGRAL OPERATORS

SHIGEYOSHI OWA

Abstract. Two integral operators P^{α} and Q_{β}^{α} for analytic functions in the open unit disk are introduced. The object of the present paper is to derive some properties of integral operators P^{α} and Q_{β}^{α}.

1. Introduction

Let A be the class of functions of the form

f(z) = z + ∑_{n=2}^{∞} a_{n}z^{n}  (1.1)

which are analytic in the open unit disk U = {z : |z| < 1}. Recently, Jung, Kim, and Srivastava [1] have introduced the following one-parameter families of integral operators:

P^{α}f = P^{α}f(z) = \frac{2^{α}}{zΓ(α)} ∫_{0}^{z} ( log \frac{z}{t} )^{α−1}f(t)dt (α > 0),  (1.2)

Q_{β}^{α}f = Q_{β}^{α}f(z) = (\frac{α + β}{β} )\frac{α}{z^{β}} ∫_{0}^{z} (1 − \frac{t}{z} )^{α−1}t^{β−1}f(t)dt  (1.3)

(α > 0, β > −1)

and

J_{α}f = J_{α}f(z) = \frac{α + 1}{z^{α}} ∫_{0}^{z} t^{α−1}f(t)dt (α > −1),  (1.4)

535

1072-947X/95/900-0535$7.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1995 Plenum Publishing Corporation

1991 Mathematics Subject Classification. 30C45.

Key words and phrases. Analytic functions, logarithmic differentiation, integral operator.

where Γ(α) is the familiar Gamma function, and (in general)

[\begin{matrix}α \\ β \end{matrix}] = \frac{Γ(α + 1)}{Γ(α − β + 1)Γ(β + 1)} = [\begin{matrix}α \\ α − β \end{matrix}] .  (1.5)

For α ∈ ℕ = {1, 2, 3, … }, the operators P^{α}, Q_{1}^{α}, and J_{α} were considered by Bernardi ([2], [3]). Further, for a real number α > −1, the operator J_{α} was used by Owa and Srivastava [4], and by Srivastava and Owa ([8], [6]).

Remark 1. For f(z) ∈ A given by (1.1), Jung, Kim, and Srivastava [1] have shown that

P^{α}f(z) = z + ∑_{n=2}^{∞} (\frac{2}{n + 1})^{α}a_{n}z^{n} (α > 0),  (1.6)

Q_{β}^{α}f(z) = z + \frac{Γ(α + β + 1)}{Γ(β + 1)} ∑_{n=2}^{∞} \frac{Γ(β + n)}{Γ(α + β + n)}a_{n}z^{n}  (1.7)
(α > 0, β > −1)

and

J_{α}f(z) = z + ∑_{n=2}^{∞} (\frac{α + 1}{α + n})a_{n}z^{n} (α > −1).  (1.8)

By virtue of (1.6) and (1.8), we see that

J_{α}f(z) = Q_{α}^{1}f(z) (α > −1).  (1.9)

2. An Application of the Miller–Mocanu Lemma

To derive some properties of operators, we have to recall here the following lemma due to Miller and Mocanu [7].

Lemma 1. Let w(u, v) be a complex valued function,

w : D → ℂ, D ⊂ ℂ × ℂ (ℂ is the complex plane),

and let u = u_{1} + iu_{2} and v = v_{1} + iv_{2}. Suppose that the function w(u, v) satisfies the following conditions:

(i) w(u, v) is continuous in D; (ii) (1, 0) ∈ D and Re{w(1, 0)} > 0; (iii) Re{w(iu_{2}, v_{1})} ≤ 0 for all (iu_{2}, v_{1}) ∈ D and such that v_{1} ≤ −(1 +

u_{2}^{2})/2. Let p(z) be regular in U and p(z) = 1 + p_{1}z + p_{2}z^{2} + … such that

(p(z), zp′(z)) ∈ D for all z ∈ U. If Re{w(p(z), zp′(z))} > 0 (z ∈ U), then Re{p(z)} > 0 (z ∈ U).

Applying the above lemma, we derive

PROPERTIES OF CERTAIN INTEGRAL OPERATORS 537

Theorem 1. If f(z) ∈ A satisfies

Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > β (α > 2; z ∈ U)  (2.1)

for some β (β < 1), then

Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{4β − 1 + \sqrt{16β^{2} − 8β + 17}}{8} (z ∈ U).  (2.2)

Proof. Noting that

z(P^{α}f(z))′ = 2P^{α−1}F (z) − P^{α}f(z) (α > 1),  (2.3)

we have

\frac{z(P^{α}f(z))′}{P^{α}f(z)} = 2\frac{P^{α−1}f(z)}{P^{α}f(z)} − 1 (α > 1).  (2.4)

Define the function p(z) by

\frac{P^{α−1}f(z)}{P^{α}f(z)} = γ + (1 − γ)p(z)  (2.5)

with

γ = \frac{4β − 1 + \sqrt{16β^{2} − 8β + 17}}{8}.  (2.6)

Then p(z) = 1 + p_{1}z + p_{2}z^{2} + … is analytic in the open unit disk U. Since

\frac{z(P^{α−1}f(z))′}{P^{α−1}f(z)} − \frac{z(P^{α}f(z))′}{P^{α}f(z)} = \frac{(1 − γ)zp′(z)}{γ + (1 − γ)p(z)},  (2.7)

or

\frac{P^{α−2}f(z)}{P^{α−1}f(z)} = \frac{P^{α−1}f(z)}{P^{α}f(z)} + \frac{(1 − γ)zp′(z)}{2\{γ + (1 − γ)p(z)\}},  (2.8)

we have

Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)} − β\} =

= Re \{γ + (1 − γ)p(z) + \frac{(1 − γ)zp′(z)}{2\{γ + (1 − γ)p(z)\}} − β\} > 0.  (2.9)

Therefore, if we define the function w(u, v) by

w(u, v) = γ − β + (1 − γ)u + \frac{(1 − γ)v}{2\{γ + (1 − γ)u\}},  (2.10)

then we see that (i) w(u, v) is continuous in D = (C − \{\frac{γ}{γ−1}\}) × C; (ii) (1, 0) ∈ D and Re{w(1, 0)} = 1 − β > 0; (iii) for all (iu_{2}, v_{1}) ∈ D and such that v_{1} ≤ −(1 + u_{2}^{2})/2,

Re\{w(iu_{2}, v_{1})\} = γ − β + \frac{γ(1 − γ)v_{1}}{2\{γ^{2} + (1 − γ)^{2}u_{2}^{2}\}} ≤

≤ γ − β − \frac{γ(1 − γ)(1 + u_{2}^{2})}{4\{γ^{2} + (1 − γ)^{2}u_{2}^{2}\}} =

= −\frac{(1 − γ)(γ + 4(1 − γ)β)}{4\{γ^{2} + (1 − γ)^{2}u_{2}^{2}\}}u_{2}^{2} ≤ 0.

This implies that the function w(u, v) satisfies the conditions in Lemma 1. Thus, applying Lemma 1, we conclude that

Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > γ =

= \frac{4β − 1 + \sqrt{16β^{2} − 8β + 17}}{8} (z ∈ U).  (2.11)

Taking the special values for β in Theorem 1, we have

Corollary 1. Let f(z) be in the class A. Then

(i) Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > −\frac{1}{2} (z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{1}{4} (z ∈ U),

(ii) Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > −\frac{1}{4} (z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{\sqrt{5} − 1}{4} (z ∈ U),

(iii) Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > 0 (z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{\sqrt{17} − 1}{8} (z ∈ U),

(iv) Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > \frac{1}{4} (z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{1}{2} (z ∈ U),

and

(v) Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > \frac{1}{2} (z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{\sqrt{17} + 1}{8} (z ∈ U).

PROPERTIES OF CERTAIN INTEGRAL OPERATORS 539

Next, we have Theorem 2. If f(z) ∈ A satisfies

Re \{\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)}\} > γ (α > 2, β > −1; z ∈ U)  (2.12)

for some γ ((α + β − 3)/2(α + β − 1) ≤ γ < 1), then

Re \{\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)}\} > δ (z ∈ U),  (2.13)

where

δ = \frac{1 + 2γ(α + β − 1) + \sqrt{(1 + 2γ(α + β − 1))^{2} + 8(α + β)}}{4(α + β)}. (2.14)

Proof. By the definition of Q_{β}^{α}f(z), we know that

z(Q_{β}^{α}f(z))′ = (α + β)Q_{β}^{α−1}f(z) − (α + β − 1)Q_{β}^{α}f(z)  (2.15)
(α > 1, β > −1),

so that

\frac{z(Q_{β}^{α}f(z))′}{Q_{β}^{α}f(z)} = (α + β)\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)} − (α + β − 1).  (2.16)

We define the function p(z) by

\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)} = δ + (1 − δ)p(z).  (2.17)

Then p(z) = 1 + p_{1}z + p_{2}z^{2} + … is analytic in U. Making use of the logarithmic differentiations in both sides of (2.17), we have

\frac{z(Q_{β}^{α−1}f(z))′}{Q_{β}^{α−1}f(z)} = \frac{z(Q_{β}^{α}f(z))′}{Q_{β}^{α}f(z)} + \frac{(1 − δ)zp′(z)}{δ + (1 − δ)p(z)}.  (2.18)

Applying (2.15) to (2.18), we obtain that

\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)} = \frac{1}{α + β − 1}\{(α + β)\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)} − 1 +

+\frac{(1 − δ)zp′(z)}{δ + (1 − δ)p(z)}\} = \frac{1}{α + β − 1}\{δ(α + β) − 1 +

+(α + β)(1 − δ)p(z) + \frac{(1 − δ)zp′(z)}{δ + (1 − δ)p(z)}\},  (2.19)

that is, that

Re \{\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)} − γ\} = \frac{1}{α + β − 1} Re \{δ(α + β) − 1 +

+(α + β)(1 − δ)p(z) + \frac{(1 − δ)zp′(z)}{δ + (1 − δ)p(z)}\} − γ > 0.  (2.20)

Now, we let

w(u, v) = \frac{1}{α + β − 1}\{δ(α + β) − 1 +

+(α + β)(1 − δ)u + \frac{(1 − δ)v}{δ + (1 − δ)u}\} − γ  (2.21)

with u = u_{1} + iu_{2} and v = v_{1} + iv_{2}. Then w(u, v) satisfies that (i) w(u, v) is continuous in D = (ℂ − \{\frac{δ}{δ−1}\}) × ℂ; (ii) (1, 0) ∈ D and Re{w(1, 0)} = 1 − γ > 0; (iii) for all (iu_{2}, v_{1}) ∈ D and such that v_{1} ≤ −(1 + u_{2}^{2})/2,

Re\{w(iu_{2}, v_{1})\} = \frac{1}{α + β − 1}\{δ(α + β) − 1 +

+\frac{δ(1 − δ)v_{1}}{δ^{2} + (1 − δ)^{2}u_{2}^{2}} \} − γ ≤ \frac{1}{α + β − 1}\{δ(α + β) − 1 −

−γ(α + β − 1) − \frac{δ(1 − δ)(1 + u_{2}^{2})}{2\{δ^{2} + (1 − δ)^{2}u_{2}^{2}\}}\} ≤ 0.

Thus, the function w(u, v) satisfies the conditions in Lemma 1. This shows that Re{p(z)} > 0, or

Re \{\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)}\} > δ (z ∈ U).  (2.22)

If we take γ = (α + β − 3)/2(α + β − 1) in Theorem 2, then we have Corollary 2. If f(z) ∈ A satisfies

Re \{\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)}\} > \frac{α + β − 3}{2(α + β − 1)}

(α > 2, β > −1; z ∈ U),

then

Re \{\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)}\} > \frac{1}{2} (z ∈ U).

Further, letting α = 2 − β and γ = 1/2 in Theorem 2, we have Corollary 3. If f(z) ∈ A satisfies

Re \{\frac{Q_{β}^{−β}f(z)}{Q_{β}^{1−β}f(z)}\} > \frac{1}{2} (β > −1; z ∈ U),

then

Re \{\frac{Q_{β}^{1−β}f(z)}{Q_{β}^{2−β}f(z)}\} > \frac{1 + \sqrt{5}}{4} (z ∈ U).

3. An Application of Jack’s Lemma

We need the following lemma due to Jack [8], (also, due to Miller and Mocanu [7]).

Lemma 2. Let w(z) be analytic in U with w(0) = 0. If |w(z)| attains its maximum value on the circle |z| = r < 1 at a point z_{0} ∈ U, then we can write

z_{0}w′(z_{0}) = kw(z_{0}),  (3.1)

where k is a real number and k ≥ 1.

Applying Lemma 2 for the operator P^{α}, we have

Theorem 3. If f(z) ∈ A satisfies

Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > β (α > 2, β ≤ 1/4; z ∈ U),  (3.2)

then

Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > γ (z ∈ U),  (3.3)

where

γ = \frac{3 + 4β + \sqrt{16β^{2} − 40β + 9}}{8}.  (3.4)

Proof. Defining the function w(u, v) by

\frac{P^{α−1}f(z)}{P^{α}f(z)} = \frac{1 − (1 − 2γ)w(z)}{1 + w(z)},  (3.5)

we see that w(z) is analytic in U and w(0) = 0. It follows from (3.5) that

\frac{z(P^{α−1}f(z))′}{P^{α−1}f(z)} = \frac{z(P^{α}f(z))′}{P^{α}f(z)} −

−\frac{(1 − 2γ)zw′(z)}{1 − (1 − 2γ)w(z)} − \frac{zw′(z)}{1 + w(z)}.  (3.6)

542 SHIGEYOSHI OWA

Using (2.3), we obtain that

\frac{P^{α−2}f(z)}{P^{α−1}f(z)} = \frac{1 − (1 − 2γ)w(z)}{1 + w(z)} −

−\frac{1}{2} \frac{zw′(z)}{w(z)}(\frac{(1 − 2γ)w(z)}{1 − (1 − 2γ)w(z)} + \frac{w(z)}{1 + w(z)}).  (3.7)

If we 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 Lemma 2 gives us that

z_{0}w′(z_{0}) = kw(z_{0}) (k ≥ 1).

Therefore, letting w(z_{0}) = e^{iθ} (θ ≠ π), we have

Re \{\frac{P^{α−2}f(z_{0})}{P^{α−1}f(z_{0})}\} = Re \{\frac{1 − (1 − 2γ)w(z_{0})}{1 + w(z_{0})} −

−\frac{k}{2} (\frac{(1 − 2γ)w(z_{0})}{1 − (1 − 2γ)w(z_{0})} + \frac{w(z_{0})}{1 + w(z_{0})})\} =

= Re \{\frac{1 − (1 − 2γ)e^{iθ}}{1 + e^{iθ}} − \frac{k}{2} (\frac{(1 − 2γ)e^{iθ}}{1 − (1 − 2γ)e^{iθ}} + \frac{e^{iθ}}{1 + e^{iθ}} )\} =

= γ − \frac{k}{2} (\frac{(1 − 2γ)(cos θ − (1 − 2γ))}{1 + (1 − 2γ)^{2} − 2(1 − 2γ) cos θ} + \frac{1}{2}) ≤

≤ γ − \frac{k}{2} (\frac{1}{2} − \frac{1 − 2γ}{2(1 − γ)}) ≤ \frac{γ(3 − 4γ)}{4(1 − γ)} = β.  (3.8)

This contradicts our condition (3.2). Therefore, we have

|w(z)| = | \frac{\frac{P^{α−1}f(z)}{P^{α}f(z)} − 1}{\frac{P^{α−1}f(z)}{P^{α}f(z)} + (1 − 2γ)}| < 1 (z ∈ U),  (3.9)

which implies (3.3).

Taking β = 0 and β = 1/4 in Theorem 3, we have Corollary 4. Let f(z) be in the class A. Then

Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > 0 (α > 2; z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{3}{4} (z ∈ U)

PROPERTIES OF CERTAIN INTEGRAL OPERATORS 543

and

Re \{\frac{P^{α−2}f(z)}{P^{α−1}f(z)}\} > −\frac{1}{4} (α > 2; z ∈ U)

⟹ Re \{\frac{P^{α−1}f(z)}{P^{α}f(z)}\} > \frac{1}{2} (z ∈ U).

Finally, we prove

Theorem 4. If f(z) ∈ A satisfies

Re \{\frac{Q^{α−2}f(z)}{Q^{α−1}f(z)}\} > γ (α > 2; β > −1; z ∈ U)  (3.10)

for some γ (γ < 1), then

Re \{\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)}\} > δ (z ∈ U),  (3.11)

where δ (0 ≤ δ < 1) is the smallest positive root of the equation

2(α + β)δ^{2} − {2(α + β)(γ + 1) −

−(2γ − 1)}δ + 2{(α + β − 1)γ + 1} = 0.  (3.12)

Proof. Defining the function w(z) by

\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)} = \frac{1 − (1 − 2δ)w(z)}{1 + w(z)},  (3.13)

we obtain that

\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)} = \frac{1}{α + β − 1}\{(α + β)\frac{1 − (1 − 2δ)w(z)}{1 + w(z)} − 1 −

−\frac{zw′(z)}{w(z)}(\frac{(1 − 2δ)w(z)}{1 − (1 − 2δ)w(z)} + \frac{w(z)}{1 + w(z)})\}.  (3.14)

Therefore, supposing that there exists a point z_{0} ∈ U such that

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

we see that

Re \{\frac{Q_{β}^{α−2}f(z_{0})}{Q_{β}^{α−1}f(z_{0})}\} ≤

≤ \frac{1}{α + β − 1}\{(α + β)δ − 1 − \frac{δ}{2(1 − δ)}\} = γ.  (3.15)

Making γ = 0 in Theorem 4, we have Corollary 5. If f(z) ∈ A satisfies

Re \{\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)}\} > 0 (α > 2; β > −1; z ∈ U),  (3.16)

then

Re \{\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)}\} > \frac{−1}{α + β − 1} (z ∈ U).  (3.17)

Letting γ = 1/2 in Theorem 4, we have

Corollary 6. If f(z) ∈ A satisfies

Re \{\frac{Q_{β}^{α−2}f(z)}{Q_{β}^{α−1}f(z)}\} > \frac{1}{2} (α > 2; β > −1; z ∈ U),  (3.18)

then

Re \{\frac{Q_{β}^{α−1}f(z)}{Q_{β}^{α}f(z)}\} > \frac{α + β − 3}{α + β − 1} (z ∈ U).  (3.19)

Further, if α = 3 − β, we have

Re \{\frac{Q_{β}^{1−β}f(z)}{Q_{β}^{2−β}f(z)}\} > \frac{1}{2} (β > −1; z ∈ U)

⟹ Re \{\frac{Q_{β}^{2−β}f(z)}{Q_{β}^{3−β}f(z)}\} > 0 (z ∈ U).

Acknowledgement

This research was supported in part by the Japanese Ministry of Education, Science and Culture under a Grant-in-Aid for General Scientific Research (No. 04640204).

References

1. S. D. Bernardi, Convex and starlike univalent functions. Trans. Amer. Math. Soc. 135 (1969), 429-446.

2. S. D. Bernardi, The radius of univalence of certain analytic functions. Proc. Amer. Math. Soc. 24 (1970), 312-318.

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

4. I. B. Jung, Y. C. Kim, and H. M. Srivastava, The Hardy space of analytic functions associated with certain one-parameter families of integral operators, to appear.

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. S. Owa and H. M. Srivastava, Some applications of the generalized Libera operator. Proc. Japan Acad. 62 (1986), 125-128.

7. H. M. Srivastava and S. Owa, New characterization of certain starlike and convex generalized hypergeometric functions. J. Natl. Acad. Math. India 3 (1985), 198-202.

8. H. M. Srivastava and S. Owa, A certain one-parameter additive family of operators defined on analytic functions. J. Anal. Appl. 118 (1986), 80-87.

(Received 3.06.94)

Author’s address: Department of Mathematics Kinki University Higashi-Osaka, Osaka 577 Japan