1. Introduction and Motivation
The study of the geometric properties of some classes of analytic functions associated with some special functions in the unit disk in the complex plane has always attracted several researchers. One of the special functions for which the geometric properties have been studied widely is the Mittag-Leffler function [
1,
2,
3,
4]. Interested readers can find more information on the various geometric properties of certain analytic functions like the Wright function [
5], generalized Bessel function [
6,
7,
8], and Fox–Wright functions [
9] in the listed references. For the geometric behavior of other special functions, one can refer to [
10,
11,
12,
13,
14,
15] and the references cited therein.
The Mittag-Leffler function is closely related to the Barnes–Mittag-Leffler function. The Mittag-Leffler function play a crucial role in fractional calculus, approximation theory, and various branches of science and engineering. Our main goal of the present paper is to study several potentially geometric properties of the normalized form of the Barnes–Mittag-Leffler function. This paper is a continuation along some lines of the authors’ previous results.
Now, we recall some known definitions and results related to the context of Geometric Functions Theory. Let
denote the class of all analytic functions inside the unit disk
Assume that
denotes the collection of all functions
satisfying the normalization
such that
A function
is said to be a starlike function (with respect to the origin 0) in
if
is univalent in
and
is a starlike domain with respect to 0 in
. The analytic characterization of the class of starlike functions is given below [
16]:
Some geometric characterization of
k-starlike functions is given in [
17] and the references therein.
If
is a univalent function in
and
is a convex domain in
, then
is said to be a convex function in
. The class of convex functions can be described as follows:
However, an analytic function
is convex if and only if the function
is starlike.
An analytic function
in
is called close-to-convex in the open unit disk
if there exists a function
that is starlike in
such that
It can be easily verified that every starlike (and hence, convex) function is close-to-convex. It can be noted that every close-to-convex function in
is also univalent in
.
A function
in
is called uniformly convex (or uniformly starlike) in
if for every circular arc
contained in
with center
, the image arc
is convex or starlike with respect to
; for more details, see [
18]. This class of functions is denoted by UCV; the analytic description of the class of uniformly convex function is given as [
19]:
An analytic function
is said to be strongly starlike of order
if and only if
where
We denote that if
, then
is starlike (or strongly starlike of order 1).
An analytic function
is said to be strongly convex of order
if and only if
where
We denote that if
, then
is convex (or strongly convex of order 1) in
Let
and
be two analytic functions in
The function
is said to be subordinate to the function
, denoted by
if
where
is an analytic function in
such that
Hence, in view of the Lemma of Schwartz, we deduce that
if and only if
and
Now, we introduce the definitions of the Mittag-Leffler function
and its two-parameter version
, respectively, as defined by [
20,
21]:
and
For some of the properties of the Mittag-Leffler function, we refer the reader to [
22,
23] and the references cited therein.
One of the most important generalizations of the Mittag-Leffler function is the Barnes–Mittag-Leffler function
, which is defined [
24] as
It can be noted that
,
, and
.
By using the fact that
, we introduce the following normalization of the Barnes–Mittag-Leffler function:
where
In the present paper, we will restrict our attention to the conditions involving positive real-valued parameters
, and
and the argument
In this paper, we study some geometric properties (such as starlikness, uniformly starlike (convex), strongly starlike (convex), convexity, close-to-convexity) of a class of analytic function related to the Barnes–Mittag-Leffler function (consult (
8)).
At the end of this section, we offer some helpful lemmas that aid with the completion of the proofs of the major findings.
Lemma 1 ([
25])
. Assume that If the inequalityholds for all then f is starlike in Lemma 2 ([
26])
. Assume that and is satisfied for each ; then f is convex in . Lemma 3 ([
27])
. If F such that is convex univalent in and G with is analytic in such that in , then we get Lemma 4 (Ozaki [
28])
. If φ is of the form (1) such thator ifthen φ is close-to-convex with respect to the function Lemma 5 ([
29])
. Assume that the real sequence is positive and decreasing and satisfies If is a convex sequence, then 2. Main Results and Their Consequences
Our first major finding is asserted by the following theorem.
Theorem 1. Assume that one of the following sets of conditions holds true:
Suppose that , and , where is the abscissa of the minimum of the gamma function, and is the Riemann zeta function defined by Suppose that , and also, the following inequalityis valid. Then, the function is starlike in
Proof. Let
and assume that the conditions of
are valid; then we have
Since
and using the fact that the Gamma function
is increasing on
we deduce that
Hence, in view of the above inequality and (8) combined with the condition
, we obtain
Then, with the aid of Lemma 1, we conclude that the function
is starlike in
Finally, we assume that the hypotheses
hold true; then we have
where
In [
30], Proof of Theorem 2.8, the authors proved that the function
is decreasing on
if the inequality (
7) holds. By these observations and under the conditions of
, we deduce that
Again, by means of Lemma 1, we derive the desired result asserted by Theorem 1 under the hypotheses
□
Corollary 1. Assume that , and ; then the function is starlike in
Proof. Setting in the hypotheses of Theorem 1, we obtain the required result. □
Example 1. The function is starlike in
Remark 1. If we set in Theorem 1 under the conditions , we obtain that the normalized form of the Mittag-Leffler function defined byis starlike in Remark 2. In [4], Theorem 2.4, the authors obtained that the function is starlike in for and In view of Remark 1, we see that the function is starlike. However, Theorem 1 improves the corresponding results derived in [4]. Theorem 2. Let the parameters κ and ν satisfy the following inequality:If or if then the function is convex in Proof. For
we have
where
However, in view of the fact that the function (cf. [
30], Proof of Theorem 3.8)
is decreasing on
when the parameters
and
satisfy inequality (
13) and with the help of (
14), we find that
It is easy to obtain that the following inequalities of the function
read as follows:
and
However, combining the above inequalities with (
16), we readily derived the desired result by means of Lemma 2. □
Corollary 2. If then the function is convex in Furthermore, if , then the function is convex in
Proof. Firstly, specifying and in Theorem 2, we obtained the first stated result asserted by the above corollary. Secondly, taking and in the above theorem, we readily derived the second result in Corollary 2. □
Remark 3. If we set in the first set of conditions in Theorem 2, we deduce that the normalized form of the Mittag-Leffler function defined in (12) is convex in Remark 4. In [4], Theorem 2.4, Bansal and Prajapat proved that the function is convex in if and However, in view of the above remark, we can easily conclude that the function is convex in Hence, Theorem 2 improves Theorem 2.4 in [4]. Theorem 3. Let the parameter space be the same as in Theorem 2. Upon settingandthen the function is strongly starlike of order Proof. According to (
16) and (
17) (respectively, (
16) and (
18)), we get
where
Then, by using the above inequality, we obtain
This implies that
Now, we apply Lemma 3 for
, where
and
; we obtain
Therefore, we get
However, keeping (
19) and (
20) in mind, we have
and this completes the proof. □
Remark 5. If , then the function is starlike in
Theorem 4. Assume that and If , then the function is starlike in
Proof. According to the analytic description of starlike functions, to show that the function
is starlike in
, it suffices to prove that the following inequality
It suffices to establish that the following inequality
holds for all
In view of (8) and by using routine algebra, we have
where
is defined in (
15). Since the digamma function
is increasing on
, we deduce that the sequence
is decreasing for all
This, in turn, implies
Having (
22) and (
23) in mind, we obtain
Since
for all
, for
, we get
So by combining (
24) and (
25), we obtain that the inequality (
21) is valid under our assumption. This is what we intended to show. □
Corollary 3. Let If then the function is starlike in In particular, if then the function is starlike in
Proof. Firstly, we set , and secondly, we let and in the above theorem; we derive the desired results asserted by Corollary 3. □
Theorem 5. Suppose that Also, if the following inequality is valid:then the function is convex in . Proof. According to the analytic characterizations of a convex function, to show that the function
is convex in
, it is enough to prove that the function
is starlike in
For this, it suffices to prove the following inequality:
Again, by (8), we have
Moreover, in the proof of Theorem 3.1 [
30], under the relation (
26), the authors proved that the sequence
defined by
is decreasing, and consequently, for all
, we obtain
Elementary calculus gives
where
, and
Under our assumption, the sequence
is decreasing, and consequently, the sequence
is decreasing. This means that
Finally, by combining (
30) and (
32), we deduce that the inequality (
27) holds true under the conditions we imposed on the parameters. With this the proof, Theorem 5 is complete. □
If we set in Theorem 5, we compute the following result:
Corollary 4. If then the function is convex in .
If we let in Theorem 5, we derive the following result:
Corollary 5. If then the function is convex in .
Example 2. The functions and are convex in .
Theorem 6. Assume that the inequality (26) holds true. Also, suppose thatThen the function is uniformly convex in . Proof. Let
; then by (8), we get
where
is defined in (
29). By using the fact that the sequence
is decreasing, we obtain
Further, for
, we obtain
where
Hence, the sequence
is decreasing as the product of three decreasing sequences. This, in turn, implies that the following inequality holds true:
Finally, in virtue of the above inequality and (
34) and with the help of the analytic description given in (
2), we obtain the desired result. □
Taking in Theorem 6, we compute the following result:
Corollary 6. If then the function is uniformly convex in .
Next, we set in the above corollary; we obtain the following result:
Corollary 7. If then the function is uniformly convex in .
Theorem 7. Assume that of all the hypotheses of Theorem 5 hold true. Then the function is strongly convex in .
Proof. Let
; a short computation gives
Let us write the expression of the sequence
as follows:
Then the sequence
is decreasing as the product of three decreasing sequences. Thus, we get
Thus, we get
and consequently, we have
By applying Lemma 3 once more, where
and the functions
F and
G are defined by
we obtain
Having the above inequality and (
38) in mind, then for
we have
So the proof of Theorem 7 is completed. □
Theorem 8. Let the parameter ranges for , and be such that . If the following inequality is valid:then the function is close-to-convex with respect to the function Proof. Firstly, we note that the condition
implies
In addition, for
, we find that
However, in [
30], Proof of Theorem 4.1, it was proved that the sequence
defined by
is decreasing if (
39) is valid. From this observation and with the aid of (
40), we deduce that the sequence
is decreasing. Thus, finally, by applying Lemma 4, we readily establish the desired result asserted by Theorem 8. □
Specifying in Theorem 8, we compute the following corollary:
Corollary 8. For all the function is close-to-convex with respect to the function
Theorem 9. Let the parameter space be the same as in Theorem 8. Then, we get Proof. From (8), we have
Under Condition (
39), the sequence
is decreasing, and consequently, the sequence
is also decreasing. Therefore, the sequence
is decreasing. In addition, the inequality
implies
. Next, we show that the sequence
is also decreasing. So it suffices to prove that
Then, we get
Since the function
is increasing on
for each
, we have
We set
in our assumption; we obtain
and in view of the above inequality and (
41), we deduce that
Hence, the sequence
is decreasing. So Lemma 5 yields the asserted result. The proof of Theorem 9 is completed. □