Differential sandwich theorems for p-valent analytic functions associated with a generalization of the integral operator

Norah Saud Almutairi; Awatef Shahen; Hanan Darwish

doi:10.1515/dema-2024-0051

Article Open Access

Differential sandwich theorems for p-valent analytic functions associated with a generalization of the integral operator

Norah Saud Almutairi , Awatef Shahen and Hanan Darwish

Published/Copyright: December 10, 2024

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From the journal Demonstratio Mathematica Volume 57 Issue 1

Abstract

In this study, subordination, superordination, and sandwich theorems are established for a class of p-valent analytic functions involving a generalized integral operator that has as a special case p-valent Sălăgean integral operator. Relevant connections of the new results with several well-known ones are given as a conclusion for this investigation.

Keywords: multivalent function; differintegral operator; differential subordination; differential superordination; sandwich theorem; p-valent Sălăgean integral operator

MSC 2010: 30C45

1 Introduction and definitions

Let ℋ be the class of functions analytic in the open unit disk

U ≔ { z : z ∈ C and ∣ z ∣ < 1 }

and ℋ [ a , p ] ( a ∈ C , p ∈ N ≔ 1 , 2 , 3 , … ) be the subclass of ℋ consisting of functions of the following form:

f ( z ) = a + a p z p + a p + 1 z p + 1 + … ( z ∈ U ) .

Suppose that f and g are in ℋ . We say that f is subordinate to g (or g is superordinate to f ), which can be written as

f ≺ g i n U o r f ( z ) ≺ g ( z ) ( z ∈ U ) ,

if there exists a function ω ∈ ℋ , satisfying the conditions of the Schwarz lemma (i.e., ω ( 0 ) = 0 and ∣ ω ( z ) ∣ < 1 ) such that

f ( z ) = g ( ω ( z ) ) ( z ∈ U ) .

It follows that

f ( z ) ≺ g ( z ) ( z ∈ U ) ⇒ f ( 0 ) = g ( 0 ) and f ( U ) ⊂ g ( U ) .

In particular, if g is univalent in U , then the reverse implication also holds (cf. [1]).

We recall here some more definitions and terminologies from the theory of differential subordination and differential superordination developed by Miller and Mocanu (cf. [1,2]).

Let ϕ ( r , s ; z ) : C 2 × U → C and L ( z ) be univalent in U . If p ∈ ℋ satisfies

(1) ϕ ( p ( z ) , z p ′ ( z ) ; z ) ≺ L ( z ) ( z ∈ U ) ,

then p ( z ) is called a solution of the first-order differential subordination (1). A univalent function q is called a dominant of the solutions of the differential subordination, or more precisely a dominant if p ≺ q , for all p satisfying (1). A dominant q ˜ that satisfies q ˜ ≺ q , for all dominant q of (1) is called the best dominant of (1). The best dominant is unique up to rotations of U .

Similarly, let φ ( r , s ; z ) : C 2 × U → C and L ∈ ℋ . Let p ∈ ℋ be such that p ( z ) and φ ( p ( z ) , z p ′ ( z ) ; z ) are univalent in U . If p ( z ) satisfies

(2) L ( z ) ≺ φ ( p ( z ) , z p ′ ( z ) ; z ) ( z ∈ U ) ,

then p ( z ) is called a solution of the first-order differential superordination (2).

An analytic function q is called a subordinant of the solutions of the differential superordination, or more precisely a subordinant, if q ≺ p , for all p satisfying (2). A univalent subordinant q ˜ that satisfies q ≺ q ˜ , for all subordinants q of (2) is said to be the best subordinant. The best subordinant is unique up to rotations of U . The well-known monograph of Miller and Mocanu [1] and the more recent work of Bulboacă [3] provide detailed expositions on the theory of differential subordination and superordination.

Miller and Mocanu [1,2] obtained sufficient conditions on certain broad class of functions L 1 , q 1 , L 2 , q 2 , φ 1 and φ 2 for which the following implications hold true:

φ 1 ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) ; z ) ≺ L 1 ( z ) ⇒ p ( z ) ≺ q 1 ( z ) ( z ∈ U )

and

L 2 ( z ) ≺ φ 2 ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) ; z ) ⇒ q 2 ( z ) ≺ p ( z ) ( z ∈ U ) .

Bulboacă [4,5], Ali et al. [6], and Shanmugam et al. [7,8] found adequate conditions on the normalized analytic function f in a series of follow-up studies such that sandwich subordinations of the following kind are true:

q 1 ( z ) ≺ z f ′ ( z ) f ( z ) ≺ q 2 ( z ) ( z ∈ U ) ,

where q 1 , q 2 are univalent in U and I is a suitable operator. Refer [9–18] for sandwich results from more recent studies.

Studies with intriguing results were recently inspired by p-valent analytic classes of functions. Recent publications have provided information on the following topics: the properties of p-valent analytic functions related to cosine and exponential functions [19], results of subordination and superordination obtained by applying operators on p-valent analytic functions [20,21], and the introduction of new classes through the application of operators on p-valent analytic functions [22].

The following studies, also recently published, served as further inspiration and motivation for the study’s findings. Two new classes of p-valent functions were introduced using generalized differential operators [23,24]. Geometric features of a newly developed operator involving p-valent functions were studied and a subclass of multivalent functions was introduced in [25]. A new generalized integral operator is presented and analyzed considering numerous subordination and coefficient properties in [26].

In view of the recent investigation listed above, the subclass ℋ p of ℋ [ 0 , p ] consists of functions of the following form:

(3) f ( z ) = z p + ∑ k = p + 1 ∞ a k z k ( z ∈ U ) ,

which will be investigated using a new generalized integral operator [27] defined for p ∈ N , n ∈ N 0 = N ∪ { 0 } , λ > 0 and f ∈ ℋ p , defined as follows:

I p , λ 0 f ( z ) = f ( z ) I p , λ 1 f ( z ) = p λ z p − p λ ∫ 0 z t p λ − p − 1 f ( t ) d t = z p + ∑ k = p + 1 ∞ p p + λ ( k − p ) a k z k I p , λ 2 f ( z ) = p λ z p − p λ ∫ 0 z t p λ − p − 1 I p , λ 1 f ( t ) d t = z p + ∑ k = p + 1 ∞ p p + λ ( k − p ) 2 a k z k

and (in general)

(4) I p , λ n f ( z ) = p λ z p − p λ ∫ 0 z t p λ − p − 1 I p , λ n − 1 f ( t ) d t = z p + ∑ k = p + 1 ∞ p p + λ ( k − p ) n a k z k = I p , λ 1 z p 1 − z ⁎ I p , λ 1 z p 1 − z ⁎ … ⁎ I p , λ 1 z p 1 − z ⁎ f ( z ) ︸ n -times ,

then from (4), we can easily deduce that

(5) λ p z ( I p , λ n f ( z ) ) ′ = I p , λ n − 1 f ( z ) − ( 1 − λ ) I p , λ n f ( z ) ( p , n ∈ N ; λ > 0 ) .

We note that

( i ) I 1 , λ n f ( z ) = I λ − n f ( z ) ( see [ 28 ] ) = f ( z ) ∈ ℋ : I λ − n f ( z ) = z + ∑ k = 2 ∞ [ 1 + λ ( k − 1 ) ] − n a k z k ( n ∈ N 0 ) , ( i i ) I 1 , 1 n f ( z ) = I n f ( z ) ( see [ 28 ] ) = f ( z ) ∈ ℋ : I n f ( z ) = z + ∑ k = 2 ∞ k − n a k z k ( n ∈ N 0 ) .

Also, we note that I p , 1 n f ( z ) = I p n f ( z ) , where I p n is p-valent Sălăgean integral operator

(6) I p n f ( z ) = f ( z ) ∈ ℋ p : I p n f ( z ) = z p + ∑ k = p + 1 ∞ p k n a k z k ( p ∈ N , n ∈ N 0 ) .

In the sequel to earlier investigations, in the present study, we find interesting sufficient conditions on the functions f ∈ ℋ p and q 1 , q 2 ∈ ℋ such that sandwich relation of the form [29]

q 1 ( z ) ≺ I p , λ n f ( z ) z p ≺ q 2 ( z )

q 1 ( z ) ≺ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ≺ q 2 ( z )

holds. For particular values of the parameters λ and p , our results obtained here include several classical as well as recent results. We will derive several subordination results, superordination results, and sandwich results involving the operator I p , λ n .

2 Preliminaries

To establish our results, we need the following:

Definition 1

([2], Definition 2, p. 817; also see [1], Definition 2.2b, p. 21) Let Q be the set of functions f that are analytic and injective on U ¯ \ E ( f ) , where

E ( f ) ≔ { ζ : ζ ∈ ∂ U and lim z → ζ f ( z ) = ∞ }

and such that f ′ ( ζ ) ≠ 0 for ζ ∈ ∂ U \ E ( f ) .

Lemma 1

([1], Theorem 3.4h, p. 132) Let q be univalent in the open unit disk U and θ and ϕ be analytic in a domain D containing q ( U ) with ϕ ( w ) ≠ 0 when w ∈ q ( U ) . Set Φ ( z ) = z q ′ ( z ) ϕ ( q ( z ) ) and L ( z ) = θ ( q ( z ) ) + Φ ( z ) . Suppose that

Φ is starlike in U
and
ℜ z L ′ ( z ) Φ ( z ) > 0 ( z ∈ U ) .

If p ∈ ℋ [ q ( 0 ) , n ] for some n ∈ N with p ( U ) ⊂ D and

(7) θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) ≺ θ ( q ( z ) ) + z q ′ ( z ) ϕ ( q ( z ) ) ,

then p ≺ q and q is the best dominant.

Lemma 2

[7] Let q be univalent convex in the open unit disk U and ψ , γ ∈ C with ℜ 1 + z q ″ ( z ) q ′ ( z ) > max { 0 , − ℜ ( ψ \ γ ) } . If p ( z ) is analytic and

ψ ( p ( z ) ) + γ z p ′ ( z ) ≺ ψ ( q ( z ) ) + γ z q ′ ( z ) ,

then p ≺ q and q is the best dominant.

Lemma 3

[30] Let q be univalent in the open unit disk U and θ and ϕ be analytic in a domain D containing q ( U ) . Set Φ ( z ) = z q ′ ( z ) ϕ ( q ( z ) ) . Suppose that

Φ is univalent starlike in U
and
ℜ θ ′ ( q ( z ) ) ϕ ( q ( z ) ) > 0 ( z ∈ U ) .

If p ∈ ℋ [ q ( 0 ) , 1 ] ∩ Q with p ( U ) ⊆ D ; θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) is univalent in U and

θ ( q ( z ) ) + z q ′ ( z ) ϕ ( q ( z ) ) ≺ θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) ( z ∈ U ) ,

then q ≺ p and q is the best dominant.

Lemma 4

([2], Theorem 8, p. 822) Let q be univalent convex in the open unit disk U and γ ∈ C , with ℜ ( γ ) > 0 . If p ∈ ℋ [ q ( 0 ) , 1 ] ∩ Q , p ( z ) + γ z p ′ ( z ) is univalent in U and

q ( z ) + γ z q ′ ( z ) ≺ p ( z ) + γ z p ′ ( z ) ( z ∈ U ) ,

then q ≺ p and q is the best subordinant.

3 Subordination and superordination results

We state and prove the following subordination and superordination results.

Theorem 1

Let q ∈ ℋ be a convex univalent function in U with q ( 0 ) = 1 . Let the function f ∈ ℋ p satisfy the following subordination condition:

(8) τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p ≺ q ( z ) + τ λ p z q ′ ( z ) ( z ∈ U ; p , n ∈ N , τ , λ > 0 ) ,

where I p , λ n is defined by (4). Then,

(9) I p , λ n f ( z ) z p ≺ q ( z ) ( z ∈ U )

and q is the best dominant.

Proof

Let the function p be defined by

(10) p ( z ) = I p , λ n f ( z ) z p ( z ∈ U )

p ′ ( z ) = z p ( I p , λ n f ( z ) ) ′ − p z p − 1 ( I p , λ n f ( z ) ) ( z P ) 2 z p + 1 p ′ ( z ) = z ( I p , λ n f ( z ) ) ′ − p ( I p , λ n f ( z ) ) z p + 1 p ′ ( z ) + p ( I p , λ n f ( z ) ) = z ( I p , λ n f ( z ) ) ′ ,

which, upon differentiation followed by multiplication by z , gives

(11) z p + 1 p ′ ( z ) + p z p p ( z ) = z ( I p , λ n f ( z ) ) ′ .

By using (5) we obtain the following, after a routine simplification:

z p + 1 p ′ ( z ) + p z p p ( z ) = p λ ( I p , λ n − 1 f ( z ) − ( 1 − λ ) I p , λ n f ( z ) ) z p + 1 p ′ ( z ) + p z p p ( z ) + p λ ( 1 − λ ) I p , λ n f ( z ) = p λ ( I p , λ n − 1 f ( z ) ) z p + 1 p ′ ( z ) + p + p λ ( 1 − λ ) z p p ( z ) = p λ ( I p , λ n − 1 f ( z ) ) z p + 1 p ′ ( z ) + p + p λ − p z p p ( z ) = p λ ( I p , λ n − 1 f ( z ) ) , λ p z p ′ ( z ) + p ( z ) = I p , λ n − 1 f ( z ) z p .

This further gives that

λ p z p ′ ( z ) = I p , λ n − 1 f ( z ) z p − p ( z ) λ p z p ′ ( z ) = I p , λ n − 1 f ( z ) − z p p ( z ) z P τ λ p z p ′ ( z ) = τ ( I p , λ n − 1 f ( z ) − I p , λ n f ( z ) ) z P p ( z ) + τ λ p z p ′ ( z ) = τ ( I p , λ n − 1 f ( z ) − I p , λ n f ( z ) ) z P + I p λ n f ( z ) z p p ( z ) + τ λ p z p ′ ( z ) = τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p .

Therefore, in the light of the hypothesis (9), we have

p ( z ) + τ λ p z p ′ ( z ) ≺ q ( z ) + τ λ p z q ′ ( z ) .

Now, an application of Lemma 2 with

γ = τ λ p and ψ = 1

gives the assertion in (10). This completes the proof of Theorem 1.□

Theorem 2

Let q ∈ ℋ be a univalent convex function in U with q ( 0 ) = 1 . Also, let the function f ∈ ℋ p , be such that

I p , λ n f ( z ) z p ∈ H [ 1 , 1 ] ∩ Q

and for τ > 0 , the function τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p be univalent in U , where I p , λ n is defined by (4). If

(12) q ( z ) + τ λ p z q ′ ( z ) ≺ τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p ( z ∈ U ; p , n ∈ N , τ , λ > 0 ) ,

then

q ( z ) ≺ I p , λ n f ( z ) z p ( z ∈ U )

and q is the best subordinant.

Proof

As in the proof of our Theorem 1, let the function p ( z ) be defined by (10). Then,

τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p = p ( z ) + τ λ p z p ′ ( z ) .

Therefore, the hypothesis (12) is equivalent to

q ( z ) + τ λ p z q ′ ( z ) ≺ p ( z ) + τ λ p z p ′ ( z ) .

Now, an application of Lemma 4 yields

q ( z ) ≺ p ( z ) = I p , λ n f ( z ) z p ,

and q is the best subordinant. The proof of Theorem 2 is completed.□

Theorem 3

Let the function q ∈ ℋ be nonzero univalent in U with q ( 0 ) = 1 and

(13) ℜ 1 + z q ″ ( z ) q ′ ( z ) − z q ′ ( z ) q ( z ) > 0 ( z ∈ U ) .

Let 0 ≤ ρ ≤ 1 , λ , p ∈ N , p > 0 , and η ∈ C . If f ∈ ℋ p satisfies the following:

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p ≠ 0 ( z ∈ U )

and

(14) η ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p ≺ z q ′ ( z ) q ( z ) ,

then

(15) ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ≺ q ( z )

and q is the best dominant in (15).

Proof

Let the function p ( z ) be defined on U by

(16) p ( z ) = ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η .

Then, p is analytic in U . The logarithmic differentiation of (16) yields

(17) z p ′ ( z ) p ( z ) = η ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p .

In order to apply Lemma 1, we set

θ ( z ) = 1 , ϕ ( w ) = 1 w ( w ∈ C \ { 0 } ) ,

Φ ( z ) = z q ′ ( z ) ϕ ( q ( z ) ) = z q ′ ( z ) q ( z ) ( z ∈ U ) ,

and

L ( z ) = θ ( q ( z ) ) + Φ ( z ) = 1 + z q ′ ( z ) q ( z ) .

By making use of hypothesis (13), we see that Φ ( z ) is univalent starlike in U. Since L ( z ) = 1 + Φ ( z ) , we further obtain that

ℜ z L ′ ( z ) Φ ( z ) > 0 .

By a routine calculation using (16) and (17), we have

θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) = 1 + η ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p .

Therefore, hypothesis (14) is equivalently written as follows:

θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) ≺ 1 + z q ′ ( z ) q ( z ) = θ ( q ( z ) ) + z q ′ ( z ) ϕ ( q ( z ) ) .

We see that condition (7) is also satisfied. Now, by an application of Lemma 1, we have

p ( z ) ≺ q ( z ) .

We, thus, obtain the assertions in (15). This completes the proof of Theorem 3.□

Theorem 4

Let q ∈ ℋ be a univalent mapping of U into the right half plane with q ( 0 ) = 1 and

(18) ℜ 1 + z q ″ ( z ) q ′ ( z ) − z q ′ ( z ) q ( z ) > 0 ( z ∈ U ) .

Let 0 ≤ ρ ≤ 1 , λ , p ∈ N , p > 0 , and η ∈ C . Suppose that the function f ∈ ℋ p satisfies the following:

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p ≠ 0 ( z ∈ U ) .

Set

(19) Δ ( z ) = ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η + η ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p ( z ∈ U ) .

(20) Δ ( z ) ≺ q ( z ) + z q ′ ( z ) q ( z ) ,

then

(21) ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ≺ q ( z )

and q is the best dominant in (21).

Proof

We follow the lines of proof of Theorem 3. Let the function p ( z ) be defined as in (16). We set

θ ( w ) = w , ϕ ( w ) = 1 w ( w ∈ C \ { 0 } ) ,

Φ ( z ) = z q ′ ( z ) ϕ ( q ( z ) ) = z q ′ ( z ) q ( z ) ( z ∈ U ) ,

and

L ( z ) = θ ( q ( z ) ) + Φ ( z ) = q ( z ) + Φ ( z ) .

In this case,

ℜ z L ′ ( z ) Φ ( z ) = ℜ q ( z ) + 1 + z q ″ ( z ) q ′ ( z ) − z q ′ ( z ) q ( z ) > 0 ( z ∈ U ) .

By making use of (17), hypothesis (20) can be equivalently written as

θ ( p ( z ) ) + z p ′ ( z ) ϕ ( z ) ≺ θ ( q ( z ) ) + z q ′ ( z ) ϕ ( q ( z ) ) .

Therefore, by applying Lemma 1, we obtain

p ( z ) ≺ q ( z ) ( z ∈ U ) .

We obtain the assertion in (21). The proof of Theorem 4 is completed.□

Theorem 5

Let q ∈ ℋ be a univalent mapping of U into the right half plane with q ( 0 ) = 1 and satisfy

(22) ℜ 1 + z q ″ ( z ) q ′ ( z ) − z q ′ ( z ) q ( z ) > 0 ( z ∈ U ) .

Let 0 ≤ ρ ≤ 1 and η ∈ C . Let the function f ∈ ℋ p be such that

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ∈ ℋ [ 1 , 1 ] ∩ Q .

Suppose that the function Δ ( z ) is also univalent in U , where Δ ( z ) is defined by (19). If

(23) q ( z ) + z q ′ ( z ) q ( z ) ≺ Δ ( z ) ,

then

(24) q ( z ) ≺ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η

and q is the best subordinant in (24).

Proof

In order to apply Lemma 3, we set

θ ( w ) = w , ϕ ( w ) = 1 w ( w ∈ C \ { 0 } )

and

Φ ( z ) = z q ′ ( z ) ϕ ( q ( z ) ) = z q ′ ( z ) q ( z ) ( z ∈ U ) .

We first observe that Φ is starlike in U . Furthermore,

ℜ θ ′ ( q ( z ) ) ϕ ( z ) = ℜ { q ( z ) } > 0 ( z ∈ U ) .

Now, let the function p be defined on U as in (16). By a routine calculation using (17), we have

θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) = Δ ( z ) .

Hence, condition (23) is equivalent to the following:

θ ( q ( z ) ) + z q ′ ( z ) ϕ ( q ( z ) ) ≺ θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) .

Therefore, by using Lemma 3, we have

q ( z ) ≺ p ( z ) ( z ∈ U ) ,

and q is the best subordinant. This is precisely the assertion of (24). The proof of Theorem 5 is completed.□

Theorem 6

Let 0 ≤ ρ ≤ 1 and α , η ∈ C . Let the function q ∈ ℋ be univalent in U and

(25) ℜ 1 + z q ″ ( z ) q ′ ( z ) > max { 0 , − ℜ ( α ) } .

Suppose that f ∈ ℋ p satisfies the following:

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p ≠ 0 ( z ∈ U ) .

Set

(26) Ω ( z ) = ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η α + η ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p ( z ∈ U )

(27) Ω ( z ) ≺ α q ( z ) + z q ′ ( z ) ,

then

(28) ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ≺ q ( z )

and q is the best dominant.

Proof

The proof of this theorem is similar to the proof of Theorem 4. Therefore, we sketch only the main steps. Let the function p ( z ) be defined on U by (16). By using (17), we write:

z p ′ ( z ) p ( z ) = η ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p

(29) z p ′ ( z ) = η p ( z ) ( 1 − ρ ) z ( I p , λ n − 1 f ( z ) ) ′ + ρ z ( I p , λ n f ( z ) ) ′ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) − p .

In this case setting,

θ ( w ) = α w , ϕ ( w ) = 1 ( w ∈ C ) ,

Φ ( z ) = z q ′ ( z ) ϕ ( q ( z ) ) = z q ′ ( z ) ,

and

L ( z ) = θ ( q ( z ) ) + Φ ( z ) = α q ( z ) + z q ′ ( z ) ,

we see that, by (25), Φ is starlike in U and

ℜ z L ′ ( z ) Φ ( z ) = R α + 1 + z q ″ ( z ) q ′ ( z ) > 0 .

Furthermore, by substituting the expression for p ( z ) from (16) and the expression for z p ′ ( z ) from (29), we have

θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) = α p ( z ) + z p ′ ( z ) = Ω ( z ) ,

where Ω ( z ) is defined by (26). The hypothesis (27) is now equivalently written as

θ ( p ( z ) ) + z p ′ ( z ) ϕ ( p ( z ) ) ≺ θ ( q ( z ) ) + z q ′ ϕ ( q ( z ) ) .

An application of Lemma 1 yields

p ( z ) ≺ q ( z ) .

This last statement gives the assertion in (28). The proof of Theorem 6 is completed.□

Theorem 7

Let 0 ≤ ρ ≤ 1 , η ∈ C , α ∈ C \ { 0 } , and ℜ ( α ) > 0 . Let the function q be univalent convex in U with q ( 0 ) = 1 . Suppose that the function f ∈ ℋ p is such that

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p ≠ 0 ( z ∈ U )

and

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ∈ ℋ [ 1 , 1 ] ∩ Q .

If Ω ( z ) defined by (26) is univalent and satisfies the following:

(30) α q ( z ) + z q ′ ( z ) ≺ Ω ( z ) ,

then

(31) q ( z ) ≺ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η .

The function q is the best subordinant in (31).

Proof

Let the function p ( z ) be defined as in (17). Then, by making use of (18), we write

Ω ( z ) = α p ( z ) + z p ′ ( z ) .

The hypothesis (31) is now equivalently written as

q ( z ) + 1 α z q ′ ( z ) ≺ p ( z ) + 1 α z p ′ ( z ) .

Therefore, an application of Lemma 4 with γ = 1 α yields (32). The proof of Theorem 7 is completed.□

4 Sandwich theorems

By combining Theorem 1 with Theorem 2, we obtain the following differential sandwich theorem:

Theorem 8

Let the functions q 1 and q 2 be univalent convex in U with q 1 ( 0 ) = q 2 ( 0 ) = 1 . Let f ∈ ℋ p be such that

I p , λ n f ( z ) z p ∈ ℋ [ 1 , 1 ] ∩ Q

and for τ > 0 , the function

τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p

is univalent in U , where I p , λ n is defined by (4). If

q 1 ( z ) + τ λ p z q 1 ′ ( z ) ≺ τ I p , λ n − 1 f ( z ) z P + ( 1 − τ ) I p , λ n f ( z ) z p ≺ q 2 ( z ) + τ λ p z q 2 ′ ( z ) ,

then

(32) q 1 ( z ) ≺ τ I p , λ n f ( z ) z p ≺ q 2 ( z ) .

The functions q 1 and q 2 are, respectively, the best subordinant and the best dominant in (32).

By combining Theorems 4 and 5, we obtain following.

Theorem 9

Let the functions q 1 , q 2 ∈ ℋ be univalent mappings of U into the right half plane and further satisfy the following conditions:

q 1 ( 0 ) = q 2 ( 0 ) = 1

and

ℜ 1 + z q j ″ ( z ) q j ′ ( z ) − z q j ′ ( z ) q j ( z ) > 0 ( j = 1 , 2 ; z ∈ U ) .

Let 0 ≤ ρ ≤ 1 and η ∈ C . Let f ∈ ℋ p be such that the following conditions hold true:

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p ≠ 0 ( z ∈ U )

and

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ∈ ℋ [ 1 , 1 ] ∩ Q .

Let the function Δ ( z ) be defined on U as in (19). If

q 1 ( z ) + z q 1 ′ ( z ) q 1 ( z ) ≺ Δ ( z ) ≺ q 2 + z q 2 ′ ( z ) q 2 ( z ) ,

then

(33) q 1 ( z ) ≺ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ≺ q 2 ( z ) ,

where q 1 and q 2 are, respectively, the best subordinant and the best dominant in (33).

By combining Theorems 6 and 7, we obtain following.

Theorem 10

Let 0 ≤ ρ ≤ 1 , η ∈ C , and α ∈ C \ { 0 } with ℜ ( α ) > 0 . Let the functions q 1 and q 2 be univalent convex in U with q 1 ( 0 ) = q 2 ( 0 ) = 1 . Suppose that f ∈ ℋ p is such that

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p ≠ 0 ( z ∈ U )

and

( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ∈ ℋ [ 1 , 1 ] ∩ Q .

Let the function Ω ( z ) be defined by (27). If

α q 1 ( z ) + z q 1 ′ ( z ) ≺ Ω ( z ) ≺ α q 2 ( z ) + z q 2 ′ ( z ) ,

then

(34) q 1 ( z ) ≺ ( 1 − ρ ) I p , λ n − 1 f ( z ) + ρ I p , λ n f ( z ) z p η ≺ q 2 ( z ) ,

where q 1 and q 2 are, respectively, the best subordinant and the best dominant in (34).

5 Concluding remarks

By taking particular values for the parameters λ , p , n , and choosing different dominant functions q ( z ) in our results of Section 3, we obtain several interesting consequences. As the first example, let Ω k ( 0 , ≤ k < ∞ ) be the convex conic region in the w -plane defined by the following:

Ω k ≔ { w = u + i v ∈ C : u 2 > k 2 ( u − 1 ) 2 + k 2 v 2 , u > 0 } .

Also, let R k be the Riemann map of U onto Ω k satisfying R k ( 0 ) = 1 and R k ′ ( 0 ) > 0 . Let the function q k be defined by

(35) q k ( z ) = exp ∫ 0 z R k ( s ) − 1 s d s ( z ∈ U ) .

The region Ω k ; the functions R k ( z ) and q k ( z ) are widely discussed in the literature in the context of k-uniformly convex functions. (See e.g. [31], also see [32].) Moreover, we can readily verify that

ℜ 1 + z q k ″ ( z ) q k ′ ( z ) − z q k ′ ( z ) q k ( z ) = ℜ R k ′ ( z ) R k ( z ) − 1 > 1 2 ( z ∈ U ) .

Therefore, condition (13) is satisfied. Now, by choosing p = 1 , ρ = 1 , n = 1 , λ = 1 , η real, and q ( z ) = q k ( z ) in Theorem 3, where q k ( z ) is defined by (35), we obtain

If the function f ∈ ℋ 1 satisfies the following:

f ( z ) z ≠ 0 ( z ∈ U )

and

η z f ′ ( z ) f ( z ) − 1 ≺ ( R k ( z ) − 1 ) ( η ∈ C , z ∈ U ) ,

then

(36) f ( z ) z η ≺ q k ( z )

and q k ( z ) is the best dominant in (36).

For η = 1 , this result is due to Kanas and Wisniowska [33]. (Also see [32,34,35] for generalizations.)

In the second example, we choose q ( z ) = 1 + A z 1 + B z ( − 1 ≤ B < A ≤ 1 ) , ρ = 1 in Theorem 3, obtain the following:

If the function f ∈ ℋ p satisfies

η z ( I p , λ n f ( z ) ) ′ I p , λ n f ( z ) − p ≺ ( A − B ) z ( 1 + A z ) ( 1 + B z ) ,

then

η ( I p , λ n f ( z ) ) z p η ≺ 1 + A z ( 1 + B z ) ( z ∈ U )

and 1 + A z 1 + B z is the best dominant.

Similarly setting p = 1 , ρ = 1 , n = 1 , η real, and q ( z ) = ( 1 + B z ) η ( A − B ) ⁄ B , which is univalent if and only if ∣ ( η ( A − B ) ⁄ B ) − 1 ∣ ≤ 1 or ∣ ( η ( A − B ) ⁄ B ) + 1 ∣ ≤ 1 [36], Theorem 3 reduces to the following.

Let the real numbers A , B be such that − 1 ≤ B < A ≤ 1 and suppose that the real number η satisfies 1 ≤ η ( A − B ) B ≤ 2 . For f ∈ ℋ 1 , if

z f ′ ( z ) f ( z ) ≺ 1 + A z 1 + B z ( z ∈ U ) ,

then

f ( z ) z η ≺ ( 1 + B z ) η ( A − B ) ⁄ B

and ( 1 + B z ) η ( A − B ) ⁄ B is the best dominant.

By further specializing A = 1 − 2 α , ( 0 ≤ α < 1 ) , B = − 1 , and η = 1 , here, we obtain the following well-known result on univalent starlike functions (see [28], also see [38]):

If f ∈ ℋ 1 is univalent starlike of order α ( 0 ≤ α < 1 ) in U , then

(37) f ( z ) z ≺ 1 ( 1 − z ) 2 ( 1 − α )

and 1 ( 1 − z ) 2 ( 1 − α ) is the best dominant.

Again, setting ρ = 0 , p = 1 , η = 1 , and q ( z ) = 1 ( 1 − z ) 2 ( 1 − α ) ( 0 ≤ α < 1 ) in Theorem 3, we obtain the following well-known result for univalent convex functions (see [28], also see [37,38]).

If f ∈ ℋ 1 is univalent convex of order α ( 0 ≤ α < 1 ) in U , then

f ′ ( z ) ≺ 1 ( 1 − z ) 2 ( 1 − α )

and 1 ( 1 − z ) 2 ( 1 − α ) is the best dominant.

Particular cases of our Theorems 1, 4, and 6 also yield interesting consequences. However, we omit the details for the sake of brevity. Finally, we address the following problem:

For 0 ≤ α < 1 , let the function q be defined on U by

(38) q ( z ) = 1 2 − α 1 ( 1 − z ) ( 1 − 2 α ) − 1 ; α ≠ 1 2 , − log ( 1 − z ) ; α = 1 2 ,

A result analogous to (37) for univalent convex functions of order α ( 1 ⁄ 2 ≤ α < 1 ) is well known, i.e.,

f ( z ) z ≺ q ( z ) ( z ∈ U ) ,

where q ( z ) is defined by (38). However, a similar result in the range 0 ≤ α < 1 ⁄ 2 seems to be an open problem [39].

Acknowledgements

Norah Saud Almutairi would like to thank his father Saud Dhaifallah Almutairi for supporting this work.

Funding information: Authors state no funding involved.
Author contributions: Investigation: N.S.A.; supervision: N.S.A., A.S., and H.D.; writing – original draft: N.S.A; writing – review and editing: N.S.A. and H.D. All authors have read and agreed to the published version of the manuscript.
Conflict of interest: Authors state no conflict of interest.

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Received: 2024-01-14

Revised: 2024-05-02

Accepted: 2024-08-06

Published Online: 2024-12-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2024-0051

Keywords for this article

multivalent function; differintegral operator; differential subordination; differential superordination; sandwich theorem; p-valent Sălăgean integral operator

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