On convolution and q-calculus

We consider the convolution operator dζf(z)=1zf(z)∗z(1-ζz)(1-z)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \mathrm{d}_\zeta f(z) = \frac{1}{z}\left\{ f(z)*\frac{z}{(1-\zeta z)(1-z)}\right\} \end{aligned}$$\end{document}on the class of analytic functions f(z)=z+a2z2+⋯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f(z)=z+a_2z^2+\cdots $$\end{document}, |z|<1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$|z|<1$$\end{document}, in the complex plane, where ζ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\zeta $$\end{document} is complex, |ζ|≤1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$|\zeta |\le 1$$\end{document}. For ζ=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\zeta =1$$\end{document}, the operator becomes the derivative f′(z)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f'(z)$$\end{document}, while for real ζ=q\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\zeta =q$$\end{document}, 0<q<1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$0<q<1$$\end{document} we obtain the Jackson’s q-derivative dqf(z)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{d}_qf(z)$$\end{document}.

Let the function f ∈ H be univalent in the unit disc D with the normalization f (0) = 0. Then f maps D onto a starlike domain with respect to w 0 = 0 if and only if Re{J ST ( f ; z)} > 0 for all z ∈ D. (1.2) Such function f is said to be starlike in D with respect to w 0 = 0 (or briefly starlike).
Recall that a set E ⊂ C is said to be starlike with respect to a point w 0 ∈ E if and only if the linear segment joining w 0 to every other point w ∈ E lies entirely in E, while a set E is said to be convex if and only if it is starlike with respect to each of its points, that is, if and only if the linear segment joining any two points of E lies entirely in E.
is sufficient for starlikeness of f . In this paper we shall consider certain sufficient conditions for starlikeness of order 1/2. The class S * (α) of starlike functions of order α < 1 may be defined as The class S * (α) and the class K(α) of convex functions of order α < 1 were introduced by Robertson in [10]. It is known the old Strohhäcker result [16] that Robertson [11] proved that if f ∈ A with f (z)/z = 0 and if there exists a k, 0 < k ≤ 2, such that then f (z) ∈ S * (2/(2+k)). In [8], it was proved that for then f (z) ∈ S * . Several more complicated sufficient conditions for starlikeness and for convexity are collected in the book [7], Chap. 5. Recall also, that if f ∈ A satisfies for some g ∈ S * and some α ∈ (−π/2, π/2), then f is said to be close-to-convex in D and denoted by f ∈ C. An univalent function f ∈ A belongs to C if and only if the complement E of the image-region F = { f (z) : |z| < 1} is the union of rays that are disjoint (except that the origin of one ray may lie on another one of the rays).
On the other hand, if f ∈ A satisfies for some g ∈ S * and some β ∈ [0, 1], then f is said to be a Bazilevic function of type β and denoted by f ∈ B(β). Jackson in [5,6] introduced and studied the q-derivative, 0 < q < 1, as and [n] q a n z n ,

Making use of q-derivative, Argawal and Sahoo in [1] introduced the class S
If q → 1 − , the class S * q (α) reduces to the class S * (α). If α = 0, the class S * q (α) coincides with the class S * q (0) = S * q , which was first introduced by Ismail et al. in [3] and was considered in [2,9,12,14]. Let [n] q a n z n , z ∈ D, Proof It is easy to check that for each q ∈ C, |q| ≤ 1, q = 1, the function is convex univalent in D, see also [4,Th.17,p.170]. Hence, is starlike univalent in D. With a little more effort, we can find that zh q (z) is starlike of order 1−q 2(1+q) . On the other hand, we have that where * denotes the Hadamard product, or convolution, of power series. Hence, zd q f (z) is a convolution of f (z) with a starlike function of order 1−q 2(1+q) . Because of the famous result [13] that K * S * (α) = S * (α), we finally obtain that the function in (2.1) is in the class S * 1−q 2(1+q) .
The known Alexander theorem says that Therefore, Theorem 2.1 is an equivalent of Alexander theorem It is known that A question is: Is it true that In terms of the convolution, this problem we may write as: exists there a q, 0 < q < 1, that for given f (z) = z + a 2 z 2 + · · · ∈ S * , we have z + ∞ n=2 a n [n] q z n ∈ K ? (2.5) The answer on the question (2.5) is: no. Namely, for the starlike function nz n , z ∈ D, a n = n and the function in (2.5) becomes which is not in the class K because it has the coefficients n/[n] q greater than 1, otherwise than in K.

-derivative operator
It is easy to check that if q → 1 − the function zh q (z) (2.3) becomes the well-known Koebe function (3.1) For each f ∈ A, we can express its derivative in terms of the Koebe function as It is a natural to consider a generalization of (3.2) for ζ ∈ C, |ζ | ≤ 1: .
For ζ = 1, we have the derivative f , while for ζ = q, 0 < q < 1 we obtain the Jackson's q-derivative of f , namely d q f , which is defined in (1.7). Therefore, for we have [n] ζ a n z n , (3.4) where For these reasons, we can look on q-derivative d q f as a special case of the convolution operator (3.4).

Corollary 3.1 If f is in the class K of convex univalent functions, then for all
[n] ζ a n z n , z ∈ D, (3.5) is in the class S * of starlike univalent functions.
Proof The proof runs in the same way as the proof of Theorem 2.1, because the function is starlike for all complex ζ , |ζ | < 1.

Theorem 3.2 If f is in the class K of convex univalent functions, then for all ζ , |ζ | ≤ 1, we have
, z ∈ D, (3.6) or Proof It is known [13], [15, p.10], that if f ∈ K, then for all z, v, and w ∈ D, we have If we put w = ζ z and v = ζ 2 z in (3.8), then we obtain Trivial calculations give This establishes (3.6) and (3.7). Re

Corollary 3.3 If f is in the class K of convex univalent functions, then for all q,
, z ∈ D, (3.9) or , z ∈ D. (3.10) Remark For ζ = 1, condition (4.1) becomes (4.2) and the class S * (1, α) becomes the well-known class of starlike functions of order α. For ζ = 1, condition (4.1) becomes It is a natural question whether is in the class of ζ -starlike functions S * (ζ, α) defined in (4.1), for some α.
We answer this question in the following theorem.
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