# Some Connections between Class𝒰- and α-Convex Functions

Authors: Edmond Aliaga; Nikola Tuneski
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The class𝒰
(
λ
,
μ
) of normalized analytic functions that satisfy |
(
z
/
f
(
z
)
)
1
+
μ
·
f
′
(
z
)
−
1
|
<
λ for all z in the open unit disk is studied and sufficient conditions for an α-convex function to be in 𝒰
(
λ
,
μ
) are given.


---

## Body

## 1. Introduction

Let𝒜 denote the class of functions f
(
z
) which are analytic in the unit disk 𝔻
=
{
z
:
|
z
|
<
1
} with the normalization f
(
0
)
=
0 and f
′
(
0
)
=
1.For a functionf
(
z
)
∈
𝒜, we say that f is starlike of order α, 0
≤
α
<
1, if and only if
(1)
Re
{
z
f
′
(
z
)
f
(
z
)
}
>
α
,
z
∈
𝔻
.
We denote by 𝒮
*
(
α
) the class of all such functions. Also, we denote by 𝒦
(
α
) the class of convex functions of order α, 0
≤
α
<
1, that is, the class of functions f
(
z
)
∈
𝒜 for which
(2)
Re
{
1
+
z
f
′′
(
z
)
f
′
(
z
)
}
>
α
,
z
∈
𝔻
.
For α
=
0, we have the classes of 𝒮
* and 𝒦 of starlike and convex functions, respectively. All of the above classes are subclasses of the class of univalent functions in 𝔻 and even more, 𝒦
⊂
𝒮
*. For details see [1].Further, forf
(
z
)
∈
𝒜 and μ
∈
ℂ, let us define the operator
(3)
U
(
f
,
μ
;
z
)
=
(
z
f
(
z
)
)
1
+
μ
·
f
′
(
z
)
and the class
(4)
𝒰
(
λ
,
μ
)
=
{
f
∈
𝒜
:
z
f
(
z
)
≠
0
,
|
U
(
f
,
μ
;
z
)
-
1
|
<
λ
,
z
∈
𝔻
}
.
This class and its special cases 𝒰
(
λ
)
≡
𝒰
(
λ
,
1
) and 𝒰
≡
𝒰
(
1
)
=
𝒰
(
1,1
) are widely studied in the past decades ([2–12]). It is known [2, 12] that functions in 𝒰
(
λ
) are univalent if 0
<
λ
≤
1, but not necessarily univalent if λ
>
1. Further, Fournier and Ponnusamy [3] proved that assuming Re
μ
<
1 the following equivalency holds:
(5)
𝒰
(
λ
,
μ
)
⊂
𝒮
*
⟺
0
<
λ
≤
|
1
-
μ
|
(
1
-
μ
)
2
+
μ
2
;
that is, in general case, 𝒰
(
λ
,
μ
) is not a subset of 𝒮
*. In particular,
(6)
𝒰
(
1
,
μ
)
⊂
𝒮
*
⟺
μ
=
0
;
that is, 𝒰
⊈
𝒮
*, which can be also verified by the function
(7)
f
(
z
)
=
z
1
+
(
1
/
2
)
z
+
(
1
/
2
)
z
3
∈
𝒰
∖
𝒮
*
.
Finally, let us consider the classes
(8)
ℳ
(
α
,
γ
)
=
{
f
∈
𝒜
:
Re
J
(
f
,
α
;
z
)
>
γ
,
z
∈
𝔻
}
,
ℳ
′
(
α
,
β
)
=
{
f
∈
𝒜
:
|
J
(
f
,
α
;
z
)
-
1
|
<
β
,
z
∈
𝔻
}
,
where
(9)
J
(
f
,
α
;
z
)
≡
(
1
-
α
)
z
f
′
(
z
)
f
(
z
)
+
α
(
1
+
z
f
′
′
(
z
)
f
′
(
z
)
)
,
α, γ
∈
ℝ, and β
>
0. These classes make a “bridge” between the classes of starlike and convex functions (of some order). The class ℳ
(
α
,
γ
) is in fact the class of α-convex functions of order γ and ℳ
′
(
α
,
β
), in the case when 0
<
β
≤
1 is a subclass of ℳ
(
α
,
γ
). Further, α-convex functions of some order are also starlike ([13], page 10). Therefore, it gives rise to the question (which is studied in this paper) of finding the sufficient conditions for
(10)
f
∈
𝒰
(
λ
,
-
1
α
)
∩
ℳ
(
α
,
γ
)
,
f
∈
ℳ
′
(
α
,
β
)
⟹
f
∈
𝒰
(
λ
,
-
1
α
)
.Letf
(
z
) and g
(
z
) be analytic in the unit disk. We say that f
(
z
) is subordinate to g
(
z
), and we write f
(
z
)
≺
g
(
z
); if g
(
z
) is univalent in 𝔻, then f
(
0
)
=
g
(
0
) and f
(
𝔻
)
⊆
g
(
𝔻
). Further, we use the method of differential subordination introduced by Miller and Mocanu [14]. In fact, if ϕ
:
ℂ
2
→
ℂ (ℂ is the complex plane) is analytic in domain D
⊂
ℂ, if h
(
z
) is univalent in 𝔻, and if p
(
z
) is analytic in 𝔻 with (
p
(
z
)
,
z
p
′
(
z
)
)
∈
D, when z
∈
𝔻, then we say that p
(
z
) satisfies a first-order differential subordination if
(11)
ϕ
(
p
(
z
)
,
z
p
′
(
z
)
)
≺
h
(
z
)
.
The univalent function q
(
z
) is called a dominant of the differential subordination (11) if p
(
z
)
≺
q
(
z
) for all p
(
z
) satisfying (11). If q
~
(
z
) is a dominant of (11) and q
~
(
z
)
≺
q
(
z
) for all dominants of (11), then we say that q
~
(
z
) is the best dominant of the differential subordination (11).We will make use of the following lemma.Lemma 1 (see [15]).
Letq be univalent in the unit disk 𝔻, and let θ
(
w
) and ϕ
(
w
) be analytic in a domain D containing q
(
𝔻
), with ϕ
(
w
)
≠
0 when w
∈
q
(
𝔻
). Set Q
(
z
)
=
z
q
′
(
z
)
ϕ
(
q
(
z
)
), h
(
z
)
=
θ
(
q
(
z
)
)
+
Q
(
z
) and suppose that (i)
Qis starlike in the unit disk 𝔻;
(ii)
Re
(
z
h
′
(
z
)
)
/
Q
(
z
)
=
Re
[
(
θ
′
(
q
(
z
)
)
/
ϕ
(
q
(
z
)
)
)
+
(
z
Q
′
(
z
)
/
Q
(
z
)
)
]
>
0, z
∈
𝔻. If p is analytic in 𝔻, with p
(
0
)
=
q
(
0
), p
(
𝔻
)
⊆
D, and
(12)
θ
(
p
(
z
)
)
+
z
p
′
(
z
)
ϕ
(
p
(
z
)
)
≺
θ
(
q
(
z
)
)
+
z
q
′
(
z
)
ϕ
(
q
(
z
)
)
=
h
(
z
)
,
then p
(
z
)
≺
q
(
z
), and q is the best dominant of (12).


## 2. Main Results and Consequences

Now we will prove the following theorem that will further lead to connections between class𝒰
(
λ
,
μ
) and classes ℳ
′
(
α
,
β
) and ℳ
(
α
,
γ
).Theorem 2.
Letf
∈
𝒜, 0
<
λ
≤
1 and α
≠
0. If f
′
(
z
)
≠
0 for all z
∈
𝔻, and if
(13)
J
(
f
,
α
;
z
)
≺
1
+
α
λ
z
1
+
λ
z
≡
h
(
z
)
,
then
(14)
U
(
f
,
-
1
α
;
z
)
≺
1
+
λ
z
;
that is, f
∈
𝒰
(
λ
,
-
1
/
α
), and 1
+
λ
z is the best dominant of (13).Proof.
Letp
(
z
)
=
U
(
f
,
-
1
/
α
;
z
)
=
(
z
/
f
(
z
)
)
1
-
1
/
α
·
f
′
(
z
), q
(
z
)
=
1
+
λ
z, θ
(
ω
)
=
1, and ϕ
(
ω
)
=
α
/
ω, where ω
∈
(
𝔻
). Then q
(
z
) is univalent in 𝔻
,
θ
(
ω
) and ϕ
(
ω
) are analytic in domain D
=
ℂ
∖
{
0
} which contains q
(
𝔻
)
=
{
1
+
z
:
|
z
|
<
λ
} (q
(
z
)
=
1
+
λ
z and 𝔻 is the unit disk), and ϕ
(
ω
)
≠
0 when ω
∈
q
(
𝔻
). On the other hand, let
(15)
Q
(
z
)
=
z
q
′
(
z
)
ϕ
(
q
(
z
)
)
=
z
λ
α
q
(
z
)
=
α
λ
z
1
+
λ
z
=
h
(
z
)
-
1
.
Then
(16)
z
Q
′
(
z
)
Q
(
z
)
=
z
h
′
(
z
)
Q
(
z
)
=
1
1
+
λ
z
,
Re
{
z
Q
′
(
z
)
Q
(
z
)
}
=
Re
{
z
h
′
(
z
)
Q
(
z
)
}
=
Re
1
q
(
z
)
>
1
1
+
λ
>
0
,
11111111111111
z
∈
𝔻
.
The last inequality holds since Re
q
(
z
)
=
1
+
λ
·
Re
z
<
1
+
λ for all z
∈
𝔻 and q
(
𝔻
) does not contain the origin. So, conditions (i) and (ii) from Lemma 1 are satisfied.
Further,p is analytic in 𝔻 and p
(
0
)
=
q
(
0
)
=
1. Also, p
(
z
)
≠
0 for all z
∈
𝔻; that is, p
(
𝔻
)
⊆
D, since f
′
(
z
)
≠
0 for all z
∈
𝔻 (condition of the theorem); z
/
f
(
z
)
=
1
≠
0 for z
=
0 (because f
∈
𝒜) and f
(
z
) has no poles on 𝔻. Hence from Lemma 1 and the fact that
(17)
J
(
f
,
α
;
z
)
=
1
+
α
·
z
p
′
(
z
)
p
(
z
)
=
θ
(
p
(
z
)
)
+
z
p
′
(
z
)
ϕ
(
p
(
z
)
)
≺
α
λ
z
1
+
λ
z
=
θ
(
q
(
z
)
)
+
z
q
′
(
z
)
ϕ
(
q
(
z
)
)
,
we receive that p
(
z
)
≺
q
(
z
), that is, relation (14), and we also receive that q
(
z
) is the best dominant of (13).Applying the definition of subordination to Theorem2, we receive the following.Corollary 3.
Letf
∈
𝒜, 0
<
λ
≤
1, and α
≠
0. Also, let f
(
z
)
/
z
≠
0 and f
′
(
z
)
≠
0 for all z
∈
𝔻. If
(18)
|
J
(
f
,
α
;
z
)
-
1
|
<
λ
|
α
|
1
+
λ
,
z
∈
𝔻
,
then f
∈
𝒰
(
λ
,
-
1
/
α
); that is,
(19)
|
U
(
f
,
-
1
α
;
z
)
-
1
|
<
λ
,
z
∈
𝔻
.
This result is sharp; that is, the constant λ in inequality (19) cannot be replaced by a smaller one such that the implication still holds.Proof.
First, let us note that the functionh defined by expression (13) is univalent in the unit disk such that
(20)
min
⁡
⁡
{
|
h
(
z
)
-
1
|
:
|
z
|
=
1
}
=
λ
|
α
|
1
+
λ
.
So, the disk {
w
:
|
w
-
1
|
<
λ
|
α
|
/
(
1
+
λ
)
} is contained in h
(
𝔻
) which, having in mind the definition of subordination, means that inequality (18) implies subordination (13). Further, from Theorem 2 follows subordination (14), which is equivalent to the inequality (19). Even more, Theorem 2 says that q
(
z
)
=
1
+
λ
z is the best dominant of (13).
In order to prove the sharpness of the result let us assume the opposite; that is, there existsλ
*, 0
<
λ
*
<
λ, such that inequality (18) implies
(21)
|
U
(
f
,
-
1
α
;
z
)
-
1
|
<
λ
*
,
z
∈
𝔻
;
that is,
(22)
U
(
f
,
-
1
α
;
z
)
≺
1
+
λ
*
z
≡
q
*
(
z
)
.
On the other hand, inequality (18) implies subordination (13) with best dominant q
(
z
), meaning that q
(
z
)
≺
q
*
(
z
). This is a contradiction to the assumption λ
*
<
λ which proves the sharpness of the result.Previous corollary can be written in the following, equivalent, form that gives conditions for inclusion of the classℳ
′
(
α
,
β
) into the class 𝒰
(
λ
,
-
1
/
α
).Corollary 4.
Letf
∈
𝒜, 0
<
λ
≤
1, α
≠
0, and β
=
λ
|
α
|
/
(
1
+
λ
). Also, let f
(
z
)
/
z
≠
0 and f
′
(
z
)
≠
0 for all z
∈
𝔻. Then
(23)
f
∈
ℳ
′
(
α
,
β
)
⟹
f
∈
𝒰
(
λ
,
-
1
α
)
.
The constant λ, for the class 𝒰
(
λ
,
-
1
/
α
), cannot be replaced by a smaller one such that the inclusion still holds.Next result gives the connection between classesℳ
(
α
,
γ
) and 𝒰
(
λ
,
-
1
/
α
).Corollary 5.
Letf
∈
𝒜, 0
<
λ
≤
1, α
≠
0, and
(24)
γ
:
=
{
1
+
α
-
α
1
-
λ
,
i
f
α
>
0
,
λ
≠
1
,
1
+
α
-
α
1
+
λ
,
i
f
α
<
0
.
Also, let f
(
z
)
/
z
≠
0 and f
′
(
z
)
≠
0 for all z
∈
𝔻. If subordination (13) holds, then
(25)
f
∈
𝒰
(
λ
,
-
1
α
)
∩
ℳ
(
α
,
γ
)
.Proof.
It is easy to check that all conditions of Theorem2 are fulfilled; hence, f
∈
𝒰
(
λ
,
-
1
/
α
).
It remains to verify thatf
∈
ℳ
(
α
,
γ
); that is, subordination (13) implies
(26)
Re
J
(
f
,
α
;
z
)
>
γ
,
z
∈
𝔻
.
Having in mind the definition of subordination and the fact that h
(
z
)
=
1
+
(
α
λ
z
)
/
(
1
+
λ
z
) is univalent, it is enough to show that Re
h
(
z
)
≥
γ for all z
∈
𝔻. The last is true because
(27)
inf
{
Re
h
(
z
)
:
z
∈
𝔻
}
=
min
⁡
{
Re
h
(
z
)
:
|
z
|
=
1
}
=
min
⁡
{
h
(
1
)
,
h
(
-
1
)
}
=
γ
.Remark 6.
The caseα
>
0 and λ
=
1 is not covered by the previous corollary since then inf
⁡
{
Re
h
(
z
)
:
z
∈
𝔻
}
=
-
∞.


## 3. Examples

Now we will apply the results from the previous section on specific functionsf
∈
𝒜 and receive interesting conclusions.Example 1.
Let0
<
λ
≤
1, α
≠
0, and a
∈
ℝ. Consider the following.(i)
Let(28)
γ
:
=
{
1
+
α
-
α
1
-
λ
,
if
α
>
0
,
λ
≠
1
1
+
α
-
α
1
+
λ
,
if
α
<
0
.
Ifa
≠
0, a
≠
-
1, λ
≤
|
1
+
a
|, a
α
>
0, and |
a
|
>
|
α
|, then
(29)
f
(
z
)
=
z
·
(
1
+
λ
1
+
a
z
)
a
∈
𝒰
(
λ
,
-
1
α
)
∩
ℳ
(
α
,
γ
)
.
(ii)
If|
a
|
<
1 and one of the following two sets of conditions holds:
(30)
α
>
0
,
|
a
|
(
1
+
λ
)
(
1
-
|
a
|
+
α
)
<
λ
α
(
1
-
|
a
|
)
or
(31)
α
<
0
,
|
a
|
(
1
+
λ
)
(
1
+
|
a
|
+
α
)
<
λ
|
α
|
(
1
+
|
a
|
)
,
then
(32)
f
(
z
)
=
z
·
e
a
z
∈
𝒰
(
λ
,
-
1
α
)
.In both cases, power is taken by its principal value.Proof.
(i) For the functionf
(
z
)
=
z
(
1
+
(
λ
/
(
1
+
a
)
)
z
)
a, we have f
(
0
)
=
0 and
(33)
f
′
(
z
)
=
(
1
+
λ
1
+
a
z
)
a
·
(
1
+
α
λ
z
1
+
α
+
λ
z
)
⟹
f
′
(
0
)
=
1
.
Conditionλ
≤
|
1
+
a
| guarantees that 1
+
α
+
λ
z
≠
0 for all z
∈
𝔻; hence, f is an analytic function and f
∈
𝒜. For the function f, it is easy to verify that
(34)
J
(
f
,
α
;
z
)
≡
(
1
-
α
)
z
f
′
(
z
)
f
(
z
)
+
α
(
1
+
z
f
′
′
(
z
)
f
′
(
z
)
)
=
1
+
a
λ
z
1
+
λ
z
.
Further, from the definition of subordination, we have thata
z
≺
α
z when |
a
|
>
|
α
| and a
α
>
0; that is, both are positive or both negative. Therefore,
(35)
J
(
f
,
α
;
z
)
=
1
+
a
λ
z
1
+
λ
z
≺
1
+
α
λ
z
1
+
λ
z
.
So, all conditions of Corollary5 bring us to the conclusion that f
∈
𝒰
(
λ
,
-
1
/
α
)
∩
ℳ
(
α
,
γ
).
(ii) It is easy to verify thatf
∈
𝒜 and that
(36)
J
(
f
,
α
;
z
)
≡
(
1
-
α
)
z
f
′
(
z
)
f
(
z
)
+
α
(
1
+
z
f
′′
(
z
)
f
′
(
z
)
)
=
1
+
a
z
(
1
+
α
1
+
a
z
)
.
Therefore,(37)
Δ
≡
sup
⁡
{
|
J
(
f
,
α
;
z
)
-
1
|
:
z
∈
𝔻
}
=
max
⁡
{
|
J
(
f
,
α
;
z
)
-
1
|
:
|
z
|
=
1
}
=
|
a
|
·
max
⁡
{
1
+
α
1
+
a
z
:
z
∈
𝔻
}
=
|
a
|
·
{
1
+
α
1
-
|
a
|
,
if
α
>
0
,
1
+
α
1
+
|
a
|
,
if
α
<
0
.
Further, if one of the conditions (30) or (31) holds, then Δ
≤
(
λ
|
α
|
)
/
(
1
+
λ
); that is, we receive inequality (18). Finally, we have shown that all conditions of Corollary 3 are fulfilled, which leads to f
(
z
)
∈
𝒰
(
λ
,
-
1
/
α
).The following example exhibits some concrete conclusions that can be obtained from the results of the previous section by specifying the values ofλ and/or α.Example 2.
Letf
∈
𝒜, 0
<
λ
≤
1 and let f
(
z
)
/
z
≠
0 and f
′
(
z
)
≠
0 for all z
∈
𝔻. Consider the following: (i)
f
∈
ℳ
′
(
α
,
|
α
|
/
2
)
⇒
f
∈
𝒰
(
1
,
-
1
/
α
) (λ
=
1 in Corollary 4);
(ii)
f
∈
𝒰
(
1
,
-
1
/
α
)
∩
ℳ
(
α
,
1
+
α
/
2
) (λ
=
1 and α
<
0 in Corollary 5).



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*Source: 692327-2014-03-13.xml*