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Fractal Fract 2019, 3(1), 5; https://doi.org/10.3390/fractalfract3010005

Article
On q-Uniformly Mocanu Functions
*
Author to whom correspondence should be addressed.
Received: 28 January 2019 / Accepted: 10 February 2019 / Published: 11 February 2019

Abstract

:
Let f be analytic in open unit disc $E = { z : | z | < 1 }$ with $f ( 0 ) = 0$ and $f ′ ( 0 ) = 1$. The q-derivative of f is defined by: $D q f ( z ) = f ( z ) − f ( q z ) ( 1 − q ) z , q ∈ ( 0 , 1 ) , z ∈ B − { 0 } ,$ where $B$ is a q-geometric subset of $C$. Using operator $D q$, q-analogue class $k − U M q ( α , β )$, k-uniformly Mocanu functions are defined as: For $k = 0$ and $q → 1 −$, $k −$ reduces to $M ( α )$ of Mocanu functions. Subordination is used to investigate many important properties of these functions. Several interesting results are derived as special cases.
Keywords:
q-calculus; q-starlike; uniformly convex; subordination; Mocanu functions; q-Ruscheweyh derivative

1. Introduction

Let A denote the class of functions f that are analytic in the open unit disc E and are also normalized by the conditions $f ( 0 ) = 0$, $f ′ ( 0 ) = 1$. Let $f , g ∈ A .$ f is said to be subordinate to g (written as $f ≺ g$), if there exists a Schwartz function $w ( z )$ such that $f ( z ) = g ( w ( z ) ) .$
q-calculus is ordinary calculus without a limit, and it has been used recently by many researchers in the field of geometric function theory. q-derivatives and q-integrals play an important and significant role in the study of quantum groups and q-deformed super-algebras, the study of fractal and multi-fractal measures and in chaotic dynamical systems. The name q-calculus also appears in other contexts; see [1,2]. The most sophisticated tool that derives functions in non-integer order is the long-known fractional calculus; see [1,2,3,4].
We recall here some basic concepts from q-calculus for which we refer to [5,6,7,8,9,10,11,12,13,14,15,16] and the references therein.
A subset $β ⊂ C$ is called q-geometric, if $z q ∈ β ,$ whenever $z ∈ B$, and it contains all the geometric sequences ${ z q m } 0 ∞$.
The q-derivative $D q$ of a function $f ∈ A$ is defined by:
$D q f ( z ) = f ( z ) − f ( q z ) ( 1 − q ) z , ( z ∈ B − { 0 } )$
and $D q f ( 0 ) = f ′ ( 0 ) .$
Under this definition, we have the following rules for q-derivative $D q .$
(i).$D q z m = 1 − q m 1 − q z m − 1 = [ n , q ] z m − 1 ,$ where $[ m , q ] = 1 − q m 1 − q .$
Let $f ( z )$ and $g ( z )$ be defined on a q-geometric set $B ⊂ C$ such that q-derivatives of $f ( z )$ and $g ( z )$ exist for all $z ∈ B$. Then, for $a , b$ complex numbers, we have:
(ii).$D q ( a f ( z ) ± b g ( z ) ) = a D q f ( z ) ± b D q g ( z ) .$
(iii).$D q ( f ( z ) . g ( z ) ) = g ( z ) D q f ( z ) + f ( q z ) D q g ( z ) .$
(iv).$D q ( f ( z ) g ( z ) ) = g ( z ) D q f ( z ) − f ( z ) D q g ( z ) g ( z ) g ( q z ) , g ( z ) . g ( q z ) ≠ 0 .$
(v).$D q ( log f ( z ) ) = D q f ( z ) f ( z ) .$
Let $P ( z )$ be the class of functions $p ( z ) = 1 + c 1 z + . . . ,$ analytic in E and satisfying:
$| p ( z ) − 1 1 − q | ≤ 1 1 − q , ( z ∈ E , q ∈ ( 0 , 1 ) ) .$
It is known  that $p ∈ P ( q )$ implies that $p ( z ) ≺ 1 + z 1 − q z$, where ≺ denotes subordination, and from this, it easily follows that $R e p ( z ) > 0 , z ∈ E .$
Now, we have:
Definition 1.
[4,5] Let $f ∈ A$. Then, it is said to belong to the class $S q ∗ ( α )$ of q-starlike functions of order α, $0 ≤ α ≤ 1 ,$ if and only if,
$1 1 − α z D q f ( z ) f ( z ) − α ≺ 1 + z 1 − q z .$
We can write (3) as:
$| z D q f ( z ) f ( z ) − 1 − α q 1 − q | ≤ 1 − α 1 − q .$
By taking $a = 1 − α 1 − q$, $b = 1 − α 1 − q$ in 4, it has been shown in  that $f ∈ S q ∗ ( α ) ,$ if and only if,
$z D q f ( z ) f ( z ) ≺ 1 + A z 1 + B z , − 1 < B < 0 ≤ A ≤ 1 ,$
where $A = 1 − ( 1 + q ) α$ and $B = − q$.
As a special case, we note that:
$lim q → 1 − S q ∗ ( α ) = S ∗ ( α ) with A = 1 − 2 α ,$
which is the class of starlike functions denoted as $S ∗ ( α )$.
Furthermore, for $α = 0$, we obtain the class $S q ∗$ of q-starlike functions introduced and studied in .
Definition 2.
Let $f ∈ A$ and $k ≥ 0 , 0 ≤ α , β ≤ 1$, $q ∈ ( 0 , 1 )$. Then,
$f ∈ k − U M q ( α , β ) ,$
if and only if, for $z ∈ E ,$
$R e ( 1 − α ) z D q f ( z ) f ( z ) + α D q ( z D q f ( z ) ) D q f ( z ) > k | ( 1 − β ) z D q f ( z ) f ( z ) + β D q ( z D q f ( z ) ) D q f ( z ) − 1 | .$
Selecting special values of parameters $α , β$ and k and letting $q → 1 −$, we obtain a number of known classes of analytic functions; see [5,9,18,19,20,21]. We list some of these as follows:
(i)
Choosing $k = 0$, we get $lim q → 1 − M q ( α ) = M ( α )$, the class of $α$-convex functions; see .
(ii)
For $β = 0$, $k = 1$, and $q → 1 −$, we have the class $M N$; see .
(iii)
Choosing $β = 0 , q → 1 −$, we get the class $k − M N$ introduced in [18,19].
(iv)
$k − U M q ( 0 , 0 ) = k − s T , k − U M q ( 1 , 1 ) = k − U C V q$.
Throughout this paper, we shall assume that $q ∈ ( 0 , 1 ) , 0 ≤ α < 1$ and $z ∈ E ,$ unless otherwise mentioned.

2. Preliminary Results

Lemma 1.
. Let $ϕ ( z )$ be analytic with $ϕ ( 0 ) = 0$. If $| ϕ ( z 0 ) |$ attains its maximum value on the circle $| z | = r$ at a point $z 0 ∈ E$, then we have:
$z 0 D q ϕ ( z 0 = m ϕ ( z 0 ) , m ≥ 1 , r e a l n u m b e r .$
Lemma 2.
. Let $α ≥ 0$ and $0 ≤ r < 1$. Let $p ( z )$ be analytic in E with $p ( 0 ) = 1$.
If:
$p ( z ) + α z p ′ ( z ) p ( z ) ≺ 1 + ( 1 − 2 r ) z 1 − z ,$
then:
$p ( z ) ≺ 1 + ( 1 − 2 § ) z 1 − z ,$
where:
$δ = 1 4 ( 2 r − α ) + ( 2 r − α ) 2 + 8 α .$

3. Main Results

Theorem 1.
Let $p ( z )$ be analytic in E with $p ( 0 ) = 1 .$ Let, for $k > 1 + q q$,
$R e 1 + z D q p ( z ) p ( z ) > k | z D q p ( z ) p ( z ) | , z ∈ E .$
Then, $p ( z )$ is subordinate to $1 1 − q z$, that is, $p ( z ) ≺ 1 1 − q z$ in $E .$
Proof.
Let $p ( z ) = 1 1 − q ϕ ( z )$. It can easily be seen that $ϕ ( z )$ is analytic in E and $ϕ ( 0 ) = 0$. We shall show that $| ϕ ( z ) | < 1$ for all $z ∈ E .$ We suppose on the contrary that there exists a $z 0 ∈ E$ such that $| ϕ ( z 0 ) | = 1 .$
Then:
$R e 1 + z ∘ D q p ( z ∘ ) p ( z ∘ ) − k | z ∘ D q p ( z ∘ ) p ( z ∘ ) | = R e 1 + q z ∘ D q ϕ ( z ∘ ) 1 − q ϕ ( z ∘ ) − k | q z ∘ D q ϕ ( z ∘ ) 1 − q ϕ ( z ∘ ) | .$
Now, by Lemma 1, $z ∘ D q ϕ ( z ∘ ) = m ϕ ( z ∘ ) = m e i θ , m ≥ 1 ,$ and we use it in (6) for:
$R e 1 + m q e i θ 1 − q e i θ > k | m q e i θ 1 − q e i θ | > k | q e i θ 1 − q e i θ | .$
From (6), (7), and choosing $θ = π$, we have:
$R e 1 + z ∘ D q p ( z ∘ ) p ( z ∘ ) − k | z ∘ D q p ( z ∘ ) p ( z ∘ ) | = 1 − m q 1 + q − k q 1 + q < 0 f o r k > 1 + q q .$
This is a contradiction, and hence, $| ϕ ( z ) | < 1$ for all $z ∈ E$. This proves that:
$p ( z ) ≺ 1 1 − q z i n E .$
□
We apply Theorem 1 to have the following results.
Corollary 1.
Let $p ( z ) = f ′ ( z )$, $q → 1 −$, and $k > 2$. Then, from Theorem 1, it follows that:
$R e 1 + z f ″ ( z ) z f ′ ( z ) > k | z f ″ ( z ) z f ′ ( z ) | ,$
which implies $f ∈ k − U C V$, and so, $R e f ′ ( z ) > 1 2$ in E.
Corollary 2.
For $k > q + 1 q$, let $f ∈ k − S T q .$ Then, $f ( z ) z ≺ 1 1 − q z$ in E.
The proof is immediate when we take $p ( z ) = f ( z ) z$ in Theorem 1.
As a special case, when $q → 1 − , k > 2 , f ∈ k − S T$ implies $R e f ( z ) z > 1 2$ in $E .$
Using a similar technique, we can prove the following results.
Theorem 2.
Let $k ≥ 0 , α , β ∈ [ 0 , 1 ] , q k < 1$, and let $p ( z )$ be analytic in E with $p ( 0 ) = 1 .$
If:
$R e p ( z ) + α z D q p ( z ) p ( z ) − 1 − k q 1 + q > k | p ( z ) + β z D q p ( z ) p ( z ) − 1 | ,$
then $p ( z ) ≺ 1 1 − q z , z ∈ E .$
We can easily deduce some special cases of Theorem 2 as given below.
Corollary 3.
Let $β = 0 , p ( z ) = z D q f ( z ) f ( z )$ in (8). Then:
$R e ( 1 − α ) z D q f ( z ) f ( z ) + α D q ( z D q f ( z ) ) D q f ( z ) − 1 − k q 1 + q − k | z D q f ( z ) f ( z ) − 1 | > 0$
implies:
$f ∈ S q ∗ ( 1 1 + q ) , z ∈ E .$
As a special case of this corollary, we observe that $U S T ⊂ S ∗ ( 1 2 ) ,$ when we choose $k = 1 , α = 0$, and let $q → 1 − .$
Corollary 4.
Let $q → 1 −$ and $p ( z ) = f ′ ( z ) .$ Then:
$R e f ′ ( z ) + α ( z f ′ ( z ) ) ′ f ′ ( z ) − ( α + 1 − k 2 ) > k | f ′ ( z ) + β ( z f ′ ( z ) ) ′ f ′ ( z ) − ( 1 + β ) | = k | ( 1 + β ) − f ′ ( z ) − β ( z f ′ ( z ) ) ′ f ′ ( z ) | ≥ R e ( 1 + β ) − f ′ ( z ) − β ( z f ′ ( z ) ) ′ f ′ ( z ) .$
This gives us:
$R e f ′ ( z ) + ( α + β 1 + k ) ( z f ′ ( z ) ) ′ f ′ ( z ) ≥ k ( 1 + β ) + ( α + γ ) 1 + k = η , ( γ = 1 − k 2 ) .$
Now, using Lemma 2 together with Theorem 2 when $q → 1 − ,$ we obtain the result that:
$R e f ′ ( z ) > δ = 1 4 [ ( 2 η − ρ ) + ( 2 η − ρ ) 2 + 8 ρ ] , ρ = α + β 1 + k .$
Corollary 5.
In (8), if we take $β = 0 , α = 1 , k = 1$ and $p ( z ) = z D q f ( z ) f ( z )$, then:
$R e D q ( z D q f ( z ) ) D q ( f ( z ) ) − 1 − q 1 + q > | z D q f ( z ) f ( z ) − 1 | .$
implies
$f ∈ S q ∗ ( 1 1 + q ) i n E .$
Furthermore, with $β = 1 , α = 1 , k = 1$ and $p ( z ) = z D q f ( z ) f ( z )$ in (8), it follows that:
$R e D q ( z D q f ( z ) ) D q f ( z ) − 1 − q 1 + q > | D q ( z D q f ( z ) ) D q f ( z ) − 1 |$
implies $f ∈ S q ∗ ( 1 1 + q ) .$
Next, we prove the following:
Theorem 3.
Let $p ( z )$ be analytic in E with $p ( 0 ) = 1 .$ Let:
$R e p ( z ) + ( z D q p ( z ) ) λ p ( z ) + c − r − k | p ( z ) + z D q p ( z ) λ p ( z ) + c − 1 | > 0 ,$
where $r = 1 1 + q , λ$, and c are positive real. Then, $p ( z ) ≺ 1 1 − q z$ in $E .$
Proof.
We shall follow the same procedure to prove this result as was used in Theorem 1. Let $p ( z ) = 1 1 − q ϕ ( z )$. Clearly, $ϕ ( 0 ) = 0$, and $ϕ ( z )$ is analytic. We prove that $ϕ ( z )$ is a Schwartz function, that is $| ϕ ( z ) | < 1 , ∀ z ∈ E$. Suppose on the contrary that there exists $z ∘ ∈ E$ such that $| ϕ ( z ∘ ) | = 1 = | e i θ | , 0 ≤ θ ≤ 2 π$.
Now, with some computations, we have:
$p ( z ) + z D q p ( z ) λ p ( z ) + c = 1 1 − q ϕ ( z ) + ( q λ ) z D q ϕ ( z ) 1 − q ϕ ( z ) − ( q λ ) c z D q ϕ ( z ) ( λ + c ) − q c ϕ ( z ) .$
We apply Lemma 1 to have $z ∘ D q ϕ ( z ∘ ) = m ϕ ( z ∘ ) , m ≥ 1$, and note that:
$R e q λ z ∘ D q ϕ ( z ∘ ) 1 − q ϕ ( z ∘ ) = R e m q λ ϕ ( z ∘ ) 1 − q ϕ ( z ∘ ) = R e m q λ e i θ 1 − q e i θ = m q λ ( c o s θ − q ) | 1 − q e i θ | 2 ,$
$R e q λ c z ∘ D q ϕ ( z ∘ ) ( λ + c ) − q c ϕ ( z ∘ ) = q λ c m ( λ + c ) c o s θ − q 2 c 2 m λ | ( λ + c ) − q c e i θ | 2 ,$
and:
$| 1 1 − q e i θ + { q λ m e i θ ( 1 − q e i θ ) } − { q λ c m e i θ ( λ + c ) − q c e i θ − 1 } | θ = π = | 1 − q 1 + q − m q λ 1 + q + q c m λ ( λ + c ) + q c |$
Using (10), (11), (12), and (13), we get a contradiction to the given hypothesis (9), when we assume $| ϕ ( z ∘ ) | = 1$ for some $z ∘ ∈ E$. Hence $| ϕ ( z ) | < 1$ for all $z ∈ E$ and:
$p ( z ) ≺ 1 1 − q z , z ∈ E .$
This completes the proof. □
In order to develop some applications of Theorem 3, we need the following.
Let the operator $D q n : A → A$ be defined as:
$D q n f ( z ) = F n + 1 , q ( z ) ∗ f ( z ) = z + ∑ m = 2 ∞ [ m + n − 1 , q ] ! [ n , q ] ! [ m − 1 , q ] ! a m z n ,$
where:
$f ( z ) = z + ∑ m = 2 ∞ a m z m ,$
and:
$F n + 1 , q ( z ) = z + ∑ [ m + n − 1 , q ] ! [ n , q ] ! [ m − 1 , q ] ! z m .$
This series is absolutely convergent in E, and * denotes convolution. The operator $D q n$ is called the q-Ruscheweyh derivative of order n; see .
It can easily be seen that $D q ∘ f ( z ) = f ( z )$ and $D q ′ f ( z ) = z D q f ( z ) .$
The relation (14) can be expressed as:
$D q n f ( z ) = z D q n ( z n − 1 f ( z ) ) [ n , q ] ! , n ∈ N .$
Furthermore,
$l i m q → 1 D q n f ( z ) = z ( 1 − z ) n + 1 ∗ f ( z ) = D n f ( z ) ,$
which is called the Ruscheweyh derivative of order n; see .
Let $f ∈ A$. Then, f is said to belong to the class $S q ∗ ( n , α ) ,$ if and only if, $D q n f ∈ S q ∗ ( α ) , z ∈ E .$
The following identity can easily be obtained:
$z D q ( D q n f ( z ) ) = 1 + [ n , q ] q n D q n + 1 f ( z ) − [ n , q ] q n D q n f ( z )$
We now take $p ( z ) = z D q ( D q n f ( z ) ) D q n f ( z )$ in relation (9) of Theorem 3 to have:
Theorem 4.
Let $D q n f = F n$ denote the q-Ruscheweyh derivative of order n for $f ∈ A .$ Let:
$R e z D q F n + 1 ( z ) F n + 1 ( z ) − 1 1 + q > k | z D q F n + 1 ( z ) F n + 1 ( z ) − 1 | , k ≥ 0 .$
Then:
$z D q F n ( z ) F n ( z ) ≺ 1 1 − q z , z ∈ E .$
That is, $f ∈ S q ∗ ( n , α ) , α = 1 1 + q .$
Proof.
Let p be analytic in E with $p ( 0 ) = 0$, and let:
$p ( z ) = z D q ( D q n f ( z ) ) D q n f ( z ) = z D q F n ( z ) F n ( z ) .$
Using identity (15) and some computation, we have:
$R e p ( z ) + z D q p ( z ) p ( z ) + n − 1 1 + q − k | p ( z ) + z D q p ( z ) p ( z ) − 1 | > 0 .$
Now, the required result follows immediately from Theorem 3. □
Corollary 6.
In Theorem 4, we take $k = 0$. Then, it gives us:
$S q ∗ ( n + 1 , α ) ⊂ S q ∗ ( n , α ) ⊂ . . . ⊂ S q ∗ ( α ) , α = 1 1 + q .$
When $q → 1 − , 1 1 + q → 1 2$, and we have:
$S q ∗ ( n + 1 , 1 2 ) ⊂ S ∗ ( n , 1 2 ) ⊂ . . . ⊂ S ∗ ( 1 2 ) .$
Corollary 7.
Let $f ∈ A$, and let:
$R e z D q f ( z ) f ( z ) − 1 1 + q > k | z D q f ( z ) f ( z ) − 1 | , k ≥ 0 .$
Define:
$L B ( f ) = F c ( z ) = [ c + 1 ] q z c ∫ 0 z t c − 1 f ( t ) d q t , c ∈ N ∘ .$
Then:
$z D q F c ( z ) F c ( z ) ≺ 1 1 − q z , z ∈ E .$
Proof.
The integral operator $L B : A → A$ defined in (16) is known as the q-Bernardi integral operator $L B ( f ) = F c .$ When $q → 1 − ,$ (16) reduces to the well-known Bernardi operator; see .
Let,
$z D q F c ( z ) F c ( z ) = p ( z ) .$
Then, from (16), (17), (18), and some computations, this leads us to:
$R e z D q f ( z ) f ( z ) − 1 1 + q − k | z D q f ( z ) f ( z ) − 1 | = R e p ( z ) + z D q p ( z ) p ( z ) + q c [ c , q ] − k | p ( z ) + z D q p ( z ) p ( z ) + q c [ c , q ] − 1 | > 0 , z ∈ E .$
We now apply Theorem 3, and it follows that:
$p ( z ) = z D q F c ( z ) F c ( z ) ≺ q 1 − q z in E .$
That is, $F c ∈ S q ∗ ( 1 1 + q ) .$ □
As a special case, when $q → 1 − ,$ then $f ∈ K − U S T ( 1 2 )$, and then, $F c ,$ defined by 17, belongs to $S ∗ ( 1 2 )$ in $E .$

4. Conclusions

In this paper, we have used q-calculus, conic domains, and subordination to define and study some new subclasses involving Mocanu functions. Some interesting inclusion and subordination properties of these new classes have been derived. The q-analogue of the Ruscheweyh derivative has been used to obtain a new subordination result for q-Mocanu functions. Some special cases have been discussed as applications of our main results. The technique and ideas of this paper may stimulate further research in this dynamic field.

Author Contributions

Conceptualization, K.I.N.; formal analysis, K.I.N.; investigation, R.S.B. and K.I.N.; methodology, R.S.B. and K.I.N.; supervision, K.I.N.; validation, R.S.B. and K.I.N.; writing, original draft, R.S.B. and K.I.N.; writing, review and editing, K.I.N.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

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