Mugur Alexandru Acu  ULB...
Embed Size (px)
Transcript of Mugur Alexandru Acu  ULB...

Mugur Alexandru Acu
SUBCLASSES OF αCONVEX FUNCTIONS
”Lucian Blaga” University Publishing House
2008

2

Preface
The concept of αconvex functions was introduced in
1969, by the great romanian mathematician Petru T.
Mocanu, with the aim of making a continuously con
nection between the notions of starlike functions and
convex functions. Taking account of the importance de
rived from this connection, the study of variously sub
classes of αconvex functions become a pursuit for many
mathematicians from all over the world.
The present book contain results of the author (with
complete proofs), and connected results of other math
ematicians (without proofs for efficiency reasons), re
garding some subclasses of αconvex convex functions,
and it is addressed to researchers in the field of Geo
3

metric Functions Theory, students in mathematics and
other researchers or students in connected fields such is
engineering (fluids mechanics).
4

Contents
Preface 3
1 Preliminaries 7
1.1 Univalent functions . . . . . . . . . . . . 7
1.2 Starlike functions . . . . . . . . . . . . . 15
1.3 Convex functions . . . . . . . . . . . . . 21
1.4 αconvex functions . . . . . . . . . . . . 26
1.5 Differential subordinations.
Admissible functions method . . . . . . . 31
1.6 BriotBouquet differential subordinations 42
2 Uniformly starlike and
uniformly convex functions 45
5

2.1 Uniformly starlike functions . . . . . . . 45
2.2 Uniformly convex functions . . . . . . . 49
3 Subclasses of αconvex functions 66
3.1 The subclasses UM(α) and UMα . . . . 66
3.2 The subclass UDn,α(β, γ) . . . . . . . . . 68
3.3 The subclasses UMα(q) and UDn,α(q) . . 75
3.4 The subclass Mλ,α(q) . . . . . . . . . . . 83
3.5 The subclass MLn,α(q) . . . . . . . . . . 94
3.6 The subclass MLβ,α(q) . . . . . . . . . . 110
Bibliography 122
6

Chapter 1
Preliminaries
1.1 Univalent functions
Definition 1.1.1 A holomorphic (or meromorphic) func
tion which is injective in a domain D, is called univalent
in D.
We denote with Hu(D) the set of all univalent func
tions in a domain D. In the case D = U
= {z ∈ C : z < 1}, we will denote with Hu(U) theclass of holomorphic and univalent in U . The class of
all holomorphic functions in a domain D will be denoted
with H(D).
7

Examples
1.1.1) If f ∈ Hu(D), g ∈ Hu(E) and f(D) ⊂ Ethen g ◦ f ∈ Hu(D).
1.1.2) The Koebe function f(z) = z(1−z)2 , z ∈ Uis univalent in U .
Theorem 1.1.1 [25] If f ∈ Hu(D), then f ′(z) 6= 0 forall z ∈ D.
We remark that for the function f(z) = ez we have
f ′(z) 6= 0 for all z ∈ C, but ez = ez+2πi show to us thatthis function it is not univalent. From Theorem 1.1.1 we
deduce that the univalent functions are also conformal
mappings.
We denote with
H [a, n] = {f ∈ H(U) : f(z) = a + anzn + ....} ,
A = {f ∈ H(U); f(0) = f ′(0)− 1 = 0}(1.1)
and with
S = {f ∈ A; f it is univalent .}(1.2)8

We remark that a function f ∈ A will have the following series expansion in the unit disk U :
f(z) = z + a2z2 + ... + anz
n + ... =(1.3)
= z +∞∑
j=2
ajzj , z ∈ U,
and S = A ∩ Hu(U) = {f ∈ Hu(U); f with the seriesexpansion (1.3) }.
We can use the unit disc and the above normalization
conditions, because them are not restrictions, such it is
easy to see from the next Theorem:
Theorem 1.1.2 (Riemann′s Theorem)[25] Let D ⊆C, D 6= C a simpleconnected domain, w0 ∈ D and α ∈(−π, π). Then will exist a unique function ϕ ∈ Hu(D)such that ϕ(U) = D, ϕ(0) = w0 and arg ϕ
′(0) = α.
To study in the same time with the class S the mero
morphic and univalent functions, will be considered the
class∑
of the meromorphic and univalent in U− = C\Ufunctions, having ∞ unique pole and the Laurent series
9

expansion:
ϕ(ζ) = ζ + α0 +α1ζ
+α2ζ2
+ ... +αnζn
+ ..., ζ > 1.
A function ϕ from∑
will verify the normalization
conditions ϕ(∞) = ∞ and ϕ′(∞) = 1. We will alsodenote by
E(ϕ) = C\ϕ(U−).
This set it is a continuum in C and contain at least one
point. The coefficient α0 from the above series expansion
is given by
α0 =1
2π
2π∫
0
ϕ(peiθ)dθ, p > 1.
We will also use in this book the following notation
∑0
= {ϕ ∈∑
; ϕ(ζ) 6= 0, ζ ∈ U−}.
Remark 1.1.1 Let f ∈ S, f(z) = z + a2z2 + .... Then,the function
g(z) = f
(1
z
)=
1
z−1 + a2z−2 + ...=
z
1 + a2z−1 + ...
10

= z − a2 + a3z
+ ... ∈∑
and g(z) 6= 0 for all z ∈ U−, because f ∈ S has no poles.
Conversely, if g ∈ ∑,
g(z) = z + b0 +b1z
+ ...
and c ∈ C∞\g(U−), then the function
f(z) =1
g(1
z
)− c
=z
1 + (b0 − c)z + ... = z + (c− b0)z2 + ... ∈ S
This mean that we have a bijection between S and∑
0.
Theorem 1.1.3 (Gronwall  Bieberbach) [24], [16]
Let g be a functions with the Laurent series
g(z) = z +∞∑
n=0
bnz−n, z ∈ U−.(1.4)
Then g ∈ ∑, then the area
E(g) = π
(1−
∞∑n=1
nbn2)≥ 0,
11

and thus∞∑
n=1
nbn2 ≤ 1.
The equality take place for the function
gθ(z) = z +eiθ
z, θ ∈ R.
The above Theorem it′s the starting point for the next
results.
Consequence 1.1.1 Let g ∈ ∑ having the form (1.4).Then b1 ≤ 1, and the equality take place if and only ifg(z) = z + b0 + e
2iθ/z, where b0 ∈ C and θ ∈ R.
Consequence 1.1.2 Let f ∈ S having the form (1.3).Then
a3 − a22 ≤ 1.
More, if f it′s a odd functions, then a3 ≤ 1, and a3 =1 if and only if
f(z) =z
1 + e2iθz2, θ ∈ R.
Theorem 1.1.4 [17] If f ∈ S and
f(z) = z + a2z2 + ..., z ∈ U,
12

then a2 ≤ 2 and a2 = 2 if and only if f = Kθ; where
Kθ(z) =z
(1− eiθz)2 = z +∞∑
n=2
ne(n−1)iθzn, z ∈ U,
θ ∈ R.
It′s easy to see that the domain Kθ(U) is the complex
plane except a radii with the origin in the point −14e−iθ.From the Theorem 1.1.4 it is easy to obtain the fol
lowing result:
Theorem 1.1.5 (KoebeBieberbach)[17]
If f ∈ S and w0 6∈ f(U), then w0 ≥ 1/4 andw0 = 1/4 if and only if f = Kθ, where θ is give byw0 = −e−iθ/4.
The Theorem (1.1.5) has the following geometric in
terpretation: the disk U(0; 1/4) it′s the disk, centered in
the origin, with the maximum radii such that to be cov
ered by the image f(U) of the unit disk for all functions
f ∈ S :U(0; 1/4) =
⋂
f∈Sf(U).
13

We call U(0; 1/4) the Koebe disk of the class S, and
1/4 will be named the Koebe constant of the class S.
Theorem 1.1.6 (Covering and distortion
Theorem)[17] If z ∈ U is a fixed point and r = z,then for all f ∈ S the following inequalities holds:
r
(1 + r)2≤ f(z) ≤ r
(1− r)2(1.5)1− r
(1 + r)3≤ f ′(z) ≤ 1 + r
(1− r)3(1.6)1− r1 + r
≤∣∣∣∣zf ′(z)f(z)
∣∣∣∣ ≤1 + r
1− r .(1.7)
The above inequalities are sharp and the equalities
holds if and only if f = Kθ, for a proper value of the
parameter θ.
Remark 1.1.2 Let r1 =r
(1+r)2 and r2 =r
(1−r)2 . Then
the geometric interpretations of the inequalities (1.5)
are:
U(0, r1) =⋂
f∈Sf(U(0, r))
U(0, r2) =⋃
f∈Sf(U(0, r))
14

From (1.5) it is easy to see that S is a compact class
of analytic functions.
Theorem 1.1.7 (Bieberbach)[17] If f ∈ S and f(z) =z+a2z
2+ ..., z ∈ U , then an ≤ n, n ≥ 2 . The extremalfunctions are Kθ, θ ∈ R.
The above Theorem was proved, by using the Loewner
(see [45]) method, in 1984 by the mathematician Louis
de Branges (see [20]).
1.2 Starlike functions
Let f a analytic function in U , f(0) = 0 and
f(z) 6= 0, z 6= 0. We will denote by Cr the image of thecircle {z ∈ C : z = r}, 0 < r < 1, thro′ the function f .
Definition 1.2.1 We say that Cr it is a starlike curve
in respect to the origin (or briefly, starlike) if the angle
ϕ = ϕ(r, θ) = arg f(reiθ) between the radius of f(z), z =
reiθ and the real positive axis, is a increasing function
15

on θ, when θ increase from 0 to 2π. This condition can
be express by the following inequality
∂ϕ
∂θ=
∂
∂θarg f(z) > 0, z = reiθ, θ ∈ (0, 2π)(1.8)
We say that f it is starlike function onto the circle
{z = r} if Cr it is a starlike curve.
Because f(z) 6= 0, for all z 6= 0 we obtain
Logf(z) = log f(z)+ i arg f(z),
where z = reiθ. By differentiating with respect to θ and
using the following equality
∂z
∂θ=
∂reiθ
∂θ= rieiθ = iz,
we obtain
izf ′(z)f(z)
=∂
∂θlog f(z)+ i ∂
∂θarg f(z)
From the above we obtain
∂
∂θarg f(z) = Re
zf ′(z)f(z)
, z = reiθ(1.9)
The condition (1.8) can be write in the following form
Rezf ′(z)f(z)
> 0, for z = r(1.10)
16

which express the necessary and sufficiently condition
such that a function f to be starlike onto the circle {z ∈C : z = r}.
Because zf′(z)
f(z) is a harmonic function and the above
condition hold for z = r, we can conclude that theabove condition will hold also for z ≤ r. From theabove, we conclude that from f starlike function onto
the circle {z ∈ C : z = r}, we obtain that f willbe starlike onto every circle {z ∈ C : z = r′}, where0 < r′ < r.
Definition 1.2.2 We define the radii of starlikeness for
the function f by
r∗(f) = sup{
r; Rezf ′(z)f(z)
> 0, z ≤ r}
.(1.11)
If r∗(f) ≥ 1 we will say that f is a starlike function ontothe unit disk U (or briefly, starlike)
Remark 1.2.1 1) The equality Rezf ′(z)f(z)
= 0 can not
hold for z ∈ U , because the function f will reduce toa constant.
17

2) The condition Rezf ′(z)f(z)
> 0, z ∈ U , do not assurethat the function f will be univalent in the unit disk.
If we impose the additional condition f ′(0) 6= 0, thenthe condition Re
zf ′(z)f(z)
> 0 will assure that the func
tion f will be univalent in the unit disk and f(U)
it is a starlike domain (with respect to the origin),
namely, the segment [0, w] ∈ f(U) for all w ∈ f(U).
Theorem 1.2.1 [43] Let f be a analytic function in U ,
with f(0) = 0. Then f is univalent in U , and f(U) is a
starlike domain (with respect to the origin) if and only
if f ′(0) 6= 0 and
Rezf ′(z)f(z)
> 0, for all z ∈ U(1.12)
Definition 1.2.3 Let denote by S∗ the class of func
tions analytic in U, with f(0) = 0, f ′(0) = 1 and which
are starlike (with respect to the origin) in U , namely
S∗ = {f ∈ H(U); f(0) = f ′(0)− 1 = 0,
Rezf ′(z)f(z)
> 0, z ∈ U}.
18

Remark 1.2.2 By using the subordination, the class S∗
may be defined: if f(z) = z + a2z2 + ..., z ∈ U , then
f ∈ S∗ if and only if zf′(z)
f(z)≺ 1 + z
1− z , z ∈ U .
Because the Koebe function Kθ(z) =z
(1 + eiθz)2, θ ∈
R is starlike (for a proper value of the parameter θ), we
conclude that the distortion theorem for the class S hold
also for the class S∗:
Theorem 1.2.2 [43] If f ∈ S∗, then the following inequalities are sharp:
r
(1 + r)2≤ f(z) ≤ r
(1− r)2(1.13)1− r
(1 + r)3≤ f ′(z) ≤ 1 + r
(1− r)3(1.14)1− r1 + r
≤∣∣∣∣zf ′(z)f(z)
∣∣∣∣ ≤1 + r
1− r(1.15)
where z ∈ U, z = r, and the extremal function is theKoebe function f = Kθ (for a proper value of the pa
rameter θ).
From the above theorem, we conclude that the class
S∗ is a compact set.
19

Let
M [a, b] = {µ : [a, b] → R+,(1.16)
where µ it is a increasing function
onto [a, b] ,
b∫
a
dµ(t) = µ(b)− µ(a) = 1}
Theorem 1.2.3 [43] The function f(z) = z + a2z2 +
..., z ∈ U belong to the class S∗ if and only if there exista function µ ∈ M [0, 2π] such that
f(z) = z exp
−2
2π∫
0
log(1− ze−it)dµ(t) , z ∈ U.
Two important subclasses of the class S∗ are the sub
class of starlike functions of order α(0 ≤ α < 1), denotedby S∗(α) and the subclass of strongly starlike of order
α(0 < α ≤ 1), denoted by S∗[α].
Definition 1.2.4 The function f ∈ A is called starlikeof order α, 0 ≤ α < 1, if the following inequality hold
Rezf ′(z)f(z)
> α, z ∈ U.(1.17)
We denote this class by S∗(α).
20

Definition 1.2.5 The function f ∈ A is calledstrongly starlike of order α, 0 < α ≤ 1 if the followinginequality hold
∣∣∣∣argzf ′(z)f(z)
∣∣∣∣ < απ
2, z ∈ U.(1.18)
We denote this class by S∗[α].
It is easy to see that S∗(0) = S∗ and S∗[1] = S∗.
1.3 Convex functions
Let f a analytic function in U , with f ′(z) 6= 0, for all0 < z < 1. Let Cr be the image of the circle {z ∈ C :z = r}, 0 < r < 1, by using the function f .
Definition 1.3.1 The curve Cr is called convex if the
angle
ψ(r, θ) =π
2+ arg zf ′(z), z = reiθ
between the tangent, into the point f(z), to the curve
Cr and the real positive axis, is a increasing function on
θ ∈ [0, 2π] .
21

Definition 1.3.2 The function f is called convex onto
the circle z = r if Cr is a convex curve.The function f is convex onto the circle {z ∈ C :
z = r} if and only if
Re
{1 +
zf ′′(z)f ′(z)
}> 0, z = r.(1.19)
From the above we obtain that for f convex onto the
circle
{z ∈ C : z = r}, we have f convex onto every circle{z ∈ C : z = r′}, where 0 < r′ < r.
Definition 1.3.3 We define the radii of convexity for
the function f by
rc(f) =(1.20)
sup
{r; Re
{zf ′′(z)f ′(z)
+ 1
}> 0, z ≤ r
}.
If rc(f) ≥ 1, we will say that the function f is convexin the unit disk U (or briefly, convex), and f will verify
the condition
Re
{1 +
zf ′′(z)f ′(z)
}> 0, z < 1.(1.21)
22

The condition(1.21) imply f ′(z) 6= 0, for all0 < z < 1.
Remark 1.3.1 The condition Re
{1 +
zf ′′(z)f ′(z)
}> 0, z ∈
U do not assure that the function f is univalent in the
unit disk U (for example the function f(z) = z2 verify
the above condition, but it is not univalent in U).
Theorem 1.3.1 [43] Let f be a analytic function in U .
the function f is univalent in U , and f(U) is a convex
domain, if and only if f ′(0) 6= 0 and
Re
{1 +
zf ′′(z)f ′(z)
}> 0, for all z ∈ U(1.22)
Definition 1.3.4 We will denote by Sc (or by K) the
class of all analytic functions in U, with f(0) = 0,
f ′(0) = 1 and which are convex in U , namely
Sc = {f ∈ H(U); f(0) = f ′(0)− 1 = 0,
Re
{1 +
zf ′′(z)f ′(z)
}> 0, z ∈ U}
and Sc ⊂ S.23

The connection between the classes S∗ and Sc is es
tablish by the following theorem:
Theorem 1.3.2 [13] Let f ∈ A and g(z) = zf ′(z).Then f ∈ Sc if and only if g ∈ S∗.
Let consider the integral operator IA : A → A, f =IA(g), g ∈ A, where
f(z) =
z∫
0
g(t)
tdt, z ∈ U.(1.23)
The integral operator IA is called Alexander′s opera
tor. By using this operator, the above theorem can be
express in the following form: Sc = IA(S∗), and IA is a
bijection between the classes S∗ and Sc.
Between the classes S∗ and Sc can also be establish
the following connection:
Theorem 1.3.3 (Marx and Strohhäcker)[37], [52]
If f ∈ A, then
Re
{1 +
zf ′′(z)f ′(z)
}> 0, z ∈ U(1.24)
24

⇒ Rezf′(z)
f(z)>
1
2, z ∈ U
We conclude that Sc ⊂ S∗(1/2).
Theorem 1.3.4 [43] If f(z) = z + a2z2 + a3z
3 + ...
belong to the class Sc, then an ≤ 1, for all n ≥ 2. Theequality hold if and only if the function f have the form
f(z) = z1 + eiτz
, τ ∈ R, z ∈ U .
Theorem 1.3.5 [43] If f ∈ Sc, then the following inequalities are sharp:
r
1 + r≤ f(z) ≤ r
1− r(1.25)
1
(1 + r)2≤ f ′(z) 1
(1− r)2(1.26)
where z ∈ U, z = r. The equalities holds for the function f(z) =
z
1 + eiτz, τ ∈ R, z ∈ U , where τ is properly
choose.
From (1.25) we conclude that the class Sc is a compact
set.
25

Remark 1.3.2 Letting r → 1 in (1.25) we find that theKoebe constant for the class Sc is 1/2.
Definition 1.3.5 We say that the function f ∈ A isconvex of order α, 0 ≤ α, < 1, if the following inequalityhold
Re
{1 +
zf ′′(z)f ′(z)
}> α, z ∈ U(1.27)
We denote by Sc(α) the class of all this functions.
1.4 αconvex functions
Let f be a analytic function in U , with f(0) = 0,
f(z)f ′(z)z 6= 0, and let α be a fixed real number.Let χ(r, θ) = (1− α)ϕ(r, θ) + αψ(r, θ).
Definition 1.4.1 The curve Cr is called αconvex if the
function χ(r, θ) is a increasing function on the parame
ter θ, where θ ∈ [0, 2π], namely∂χ(r, θ)
∂θ=(1.28)
(1− α)∂ϕ(r, θ)∂θ
+ α∂ψ(r, θ)
∂θ> 0,
26

where θ ∈ [0, 2π].
The function f is called αconvex onto the circle
{z ∈ C; z = r} if Cr is a αconvex curve.The function f is αconvex onto the circle
{z ∈ C; z = r} if and only if
ReJ(α, f ; z) > 0, z = r,(1.29)
where
J(α, f ; z) =(1.30)
(1− α)zf′(z)
f(z)+ α
(1 +
zf ′′(z)f ′(z)
).
Taking into the account the properties of the har
monic functions, from f is a αconvex function onto the
circle {z ∈ C; z = r}, we conclude that F is a αconvex function onto every circle {z ∈ C; z = r′},where 0 < r′ < r.
Definition 1.4.2 We define the radii of αconvexity for
the function f by
rα(f) = sup{r; ReJ(α, f ; z) > 0, z ≤ r}.27

If rα(f) ≥ 1 we say that the function f is αconvex inthe unit disk U , and f will verify the following condition
ReJ(α, f ; z) > 0, z ∈ U.(1.31)
Definition 1.4.3 Let denote by Mα the class of analytic
functions in U , with f(0) = 0, f ′(0) = 1 and which are
αconvex in U , namely
Mα = {f ∈ H(U), f(0) = f ′(0)− 1 = 0,
ReJ(α, f ; z) > 0, z ∈ U}
We remark that M0 = S∗ and M1 = Sc.
Remark 1.4.1 1. By taking p(z) = zf′(z)
f(z) we obtain
J(α, f ; z) = p(z) + αzp′(z)p(z)
,
and thus (1.31) can be write in the following form
Re
[p(z) + α
zp′(z)p(z)
]> 0, z ∈ U(1.32)
or
p(z) + αzp′(z)p(z)
≺ 1 + z1− z .(1.33)
28

2. If the condition (1.32) hold, then p(z) =zf ′(z)f(z)
is
analytic in U and p(z) 6= 0, z ∈ U . We conclude that the condition
f(z)f ′(z)z
6= 0, z ∈ U willalso hold.
Theorem 1.4.1 [43] For α, β ∈ R with0 ≤ β/α ≤ 1, we have Mα ⊂ Mβ.
Corollarly 1.4.1 For all α ∈ [0, 1], we haveSc ⊂ Mα ⊂ S∗.
Theorem 1.4.2 [43] If α > 0, then f ∈ Mα if and onlyif F ∈ S∗, where
F (z) = f(z) =
[zf ′(z)f(z)
]α.
From the above theorem it is easy to obtain the fol
lowing result:
Theorem 1.4.3 [43] If α > 0, then f ∈ Mα if and onlyif there exist a function F ∈ S∗ such that
f(z) =
1
α
z∫
0
F 1/α(ζ)
ζdζ
α
, z ∈ U.(1.34)
29

Definition 1.4.4 A function f ∈ Mα is called αconvexof order γ, 0 ≤ γ < 1, if
ReJ(α, f ; z) > γ, z ∈ U.(1.35)
We denote by Mα(γ) the class of all this functions.
Concerning the radii of αconvexity for the class S, in
1972 V.V. Cernikov give the following result:
Theorem 1.4.4 [19] If coth 1 − 1 = 0.313... ≤ α ≤ 1,then rα(S) = 1 + α−
√α(α + 2) .
In 1974 S.S. Miller, P.T. Mocanu and M.O. Reade,
prove in [42] that the above result hold also for α > 1.
Theorem 1.4.5 [42] The radii of αconvexity for the
class S∗ is
rα(S∗) =
1 + α−√
α(α + 2), α ≥ 0√2−√−α2 +
√−α, −3 < α < 0
−(1 + α)−√
α(α + 2), α ≤ −3.
30

1.5 Differential subordinations.
Admissible functions method
Definition 1.5.1 Let f and g be analytic functions in
U . We say that the function f is subordinate to the
function g, if there exist a function w, which is analytic
in U, w(0) = 0, w(z) < 1, z ∈ U , such that
f(z) = g(w(z)), (∀)z ∈ U.
We denote by ≺ the subordination relation.
Theorem 1.5.1 [43] Let f be a analytic function in U
and g be a analytic and univalent function in U . Then
f ≺ g if and only if f(0) = g(0) and f(U) ⊆ g(U).
1.4.2 Subordination′s Principle Let f be a an
alytic function in U and g be a analytic and univalent
function in U . Then f(0) = g(0) and f(U) ⊆ g(U) implies f(Ur) ⊆ g(Ur), r ∈ (0, 1]. The equality f(Ur) =g(Ur) hold for r < 1 if and only if f(U) = g(U), or
f(z) = g(eiθz), z ∈ U, θ ∈ R.31

The differential subordinations method (also called
admissible functions method) was introduced by the P.T.
Mocanu and S.S. Miller in [38] and [39].
Let Ω , ∆ ∈ C, p be a analytic function in U, withp(0) = a, and let ψ(r, s, t; z) : C3 × U → C.
Let consider the following implication:
(1.36)
{ψ(p(z), zp′(z), z2p′′(z); z) z ∈ U} ⊂ Ω ⇒ p(U) ⊂ ∆.
Concerning the above implication we can consider the
following three problems:
• Problem 1. We know the sets Ω and ∆, and wewant to find conditions on the function ψ such that
the implication (1.36) hold. A function ψ, which
is a solution of the above problem, it is called a
admissible function.
• Problem 2. We know the set Ω and the functionψ, and we want to find the set ∆ such that the
32

implication (1.36) hold. If it is possible, we want to
find the smallest ∆, which is a solution of the above
problem.
• Problem 3. We know the set ∆ and the functionψ, and we want to find the set Ω such that the impli
cation (1.36) hold. If it is possible, we want to find
the greatest set Ω such that the implication (1.36)
hold.
If ∆ is a simpleconnected domain, with a ∈ ∆ and∆ 6= C, then there exist a conformal mapping q fromU to ∆ and such that q(0) = a. In this conditions the
implication (1.36) can be write in the following form:
(1.37)
{ψ(p(z), zp′(z), z2p′′(z); z)  z ∈ U} ⊂ Ω ⇒p(z) ≺ q(z).
If Ω it is also a simple connected domain and Ω 6= C,then there exist a conformal mapping h from U to Ω
and such that h(0) = ψ(a, 0, 0; 0).
33

More, if ψ(p(z), zp′(z), z2p′′(z); z) is a analytic func
tion in U , then the implication (1.36) can be write in
the following form:
ψ(p(z), zp′(z), z2p′′(z); z) ≺ h(z) ⇒(1.38)
p(z) ≺ q(z).
From the above we derive the following definitions:
Definition 1.5.2 Let ψ : C3 × U → C and h be a univalent function in U . If p is a analytic function in U
which satisfy the following differential subordination:
ψ(p(z), zp′(z), z2p′′(z); z) ≺ h(z),(1.39)
then p is called a solution of the differential subordina
tion.
Definition 1.5.3 A univalent function q is called a dom
inant of the differential subordination (1.39) if p ≺ q, forevery p which satisfy (1.39).
Definition 1.5.4 A dominant q̃ which satisfy q̃ ≺ q forevery dominant q of the differential subordination (1.39)
34

is called the best dominant of the differential subordina
tion (1.39). We remark that the best dominant it is
unique except a rotation of U .
If Ω and ∆ are sample connected domains, the above
three problems can be write in the following forms:
Problem 1′. Let consider the univalent functions h
and q. We want to find the class of admissible functions
ψ[h, q] such that the differential subordination (1.38)
hold.
Problem 2′. Let consider the differential subordi
nation (1.38). We want to find the dominant q of this
subordination. If it is possible, we want to find the best
dominant.
Problem 3′. Let consider the class ψ and the domi
nant q. We want to find the greater class of functions h
such that the differential subordination (1.38) hold.
Fundamental lemmas:
For z0 = r0eiθ0, 0 < r0 < 1, we will denote by Ur0 =
35

{z ∈ C; z < r0} the disk with the center into the originand with the radii r0, U r0 =
{z ∈ C; z ≤ r0}.
Lemma 1.5.1 Let z0 ∈ U , r0 = z0 and f(z) = anzn++an+1z
n+1 + ... a continuous function into U r0 and an
alytic onto Ur0 ∪ {z0} with f(z) 6≡ 0 and n ≥ 1. If
f(z0) = max{f(z); z ∈ U r0}(1.40)
then there exist a number m ≥ n ≥ 1 such that(i)
z0f′(z0)
f(z0)= m, and
(ii) Re
{1 +
z0f′′(z0)
f ′(z0)
}≥ m.
A particular version (z0 = f(z0) = 1) of the first item
of the above lemma was considered in 1925 like a open
problem by K. Loewner. The form presented above, was
considered in 1971 by I.S. Jack.
Definition 1.5.5 Let Q be the class of all analytical and
injective functions q defined onto U\E(q), where
E(q) =
{ζ ∈ ∂U ; lim
z→ζq(z) = ∞
},
36

and q′(ζ) 6= 0 for every ζ ∈ ∂U\E(q).
Lemma 1.5.2 Let q ∈ Q cu q(0) = a and p(z) = a +pnz
n + ... be a analytic function in U , with p(z) 6≡ a andn ≥ 1. If there exist z0 ∈ U and ζ0 ∈ ∂U\E(q) such thatp(z0) = q(ζ0) and p(Ur0) ⊂ q(U), where r0 = z0, thenthere exist a number m ≥ n such that:
(i) z0p′(z0) = mζ0q′(ζ0) and
(ii) Re
{z0p
′′(z0)p′(z0)
+ 1
}≥ mRe
{ζ0q
′′(ζ0)q′(ζ0)
+ 1
}.
Lemma 1.5.3 Let q ∈ Q, with q(0) = a, and let p(z) =a + pnz
n + ... be a analytic function in U with p(z) 6≡ aand n ≥ 1. If p 6≺ q then there exist z0 = r0eiθ0 ∈ U andζ0 ∈ ∂U\E(q) and a number m ≥ n ≥ 1 such that
(i) p(Ur0) ⊂ q(U),(ii) p(z0) = q(ζ0),
(iii) z0p′(z0) = mζ0q′(ζ0) and
(iv) Re
{z0p
′′(z0)p′(z0)
+ 1
}≥ Re
{ζ0q
′′(ζ0)q′(ζ0)
+ 1
}.
Definition 1.5.6 Let Ω be a set from C, q ∈ Q and n ∈N. The class of admissible functions ψn[Ω, q] contain all
37

the functions ψ : C3×U → C which satisfy the followingadmissibility condition care
ψ(r, s, t; z) 6∈ Ω for r = q(ζ), s = mζq′(ζ),Re
[ts + 1
] ≥ mRe[
ζq′′(ζ)q′(ζ) + 1
],
z ∈ U, ζ ∈ ∂U\E(q) and m ≥ n.(1.41)
We will also use the following notation ψ1[Ω, q] =
ψ[Ω, q].
For Ω a sample connected domain, Ω 6= C and h bea conformal mapping from U to Ω, we will denote the
class of admissible functions by ψn[h, q].
If ψ : C2 × U → C, then the admissibility condition(1.41) become
ψ(q(ζ),mζq′(ζ); z) 6∈ Ω,
where z ∈ U, ζ ∈ ∂U\E(q) and m ≥ n.If Ω ⊂ Ω̃, then ψn(Ω̃, q) ⊂ ψn(Ω, q) and
ψn[Ω, q] ⊂ ψn+1[Ω, q].
Theorem 1.5.2 [43] Let ψ ∈ ψn[Ω, q], with q(0) = a.If p(z) = a+pnz
n+... is a analytic function in U , which
38

satisfy the following condition
ψ(p(z), zp′(z), z2p′′(z); z) ∈ Ω, z ∈ U ,(1.42)
then p ≺ q.
Corollarly 1.5.1 Let Ω ⊂ C and q a univalent functionin U with q(0) = a. Also, let ψ ∈ ψn[Ω, qρ], ρ ∈ (0, 1),where qρ(z) = q(ρz).
If p(z) = a+pnzn + ... is a analytic function in U , which
satisfy the following condition
ψ(p(z), zp′(z), z2p′′(z); z) ∈ Ω, z ∈ U ,
then p ≺ q.
Let consider Ω 6= C a sample connected domain.
Theorem 1.5.3 [43] Let ψ ∈ ψn[h, q] with q(0) = aand ψ(a, 0, 0; 0) = = h(0). If p(z) = a + pnz
n + ... and
ψ(p(z), zp′(z), z2p′′(z); z) is analytic in U and satisfy the
following condition
ψ(p(z), zp′(z), z2p′′(z); z) ≺ h(z) ,(1.43)
then p ≺ q.39

Corollarly 1.5.2 Let h , q be two univalent functions
in U and q(0) = a. Also, let ψ : C3 × U → C, withψ(a, 0, 0; 0) = h(0), which satisfy one from the following
conditions:
(i) ψ ∈ ψn[h, qρ], for a number ρ ∈ (0, 1)(ii) there exist a number ρ0 ∈ (0, 1) such that ψ ∈
ψn[hρ, qρ] for every ρ ∈ (ρ0, 1), where qρ(z) = q(ρz) andhρ(z) = h(ρz)
If p(z) = a + pnzn + ..., ψ(p(z), zp′(z), z2p′′(z); z) is
analytic in U and
ψ(p(z), zp′(z), z2p′′(z); z) ≺ h(z),
then p ≺ q.
The following theorems will refer to the best dominant
of the differential subordination (1.43):
Theorem 1.5.4 [43] Let h be a univalent function in
U and ψ : C3×U → C. Let assume that the differentialequation
ψ(p(z), zp′, z2p′′(z); z) = h(z)(1.44)
40

has a solution q which satisfy one from the following
conditions:
(i) q ∈ Q and ψ ∈ ψ[h, q] or
(ii) q is univalent in U and ψ ∈ ψ[h, qρ], for a numberρ ∈ (0, 1) or
(iii) q is univalent in U and there exist a number
ρ0 ∈ (0, 1) such that ψ ∈ ψn[hρ, qρ] for every ρ ∈ (ρ0, 1).
If p(z) = q(0) + p1z + ..., ψ(p(z), zp′, z2p′′(z); z) is
analytic in U , and p satisfy the condition (1.43), then
p ≺ q and q is the best dominant.
Theorem 1.5.5 [43] Let ψ ∈ ψn[Ω, q] and f a analyticfunction in U with f(U) ⊂ Ω. If the differential equation
ψ(p(z), zp′, z2p′′(z); z) = f(z)
has a solution p(z) which is analytic in U , with p(0) =
q(0), then p ≺ q.
41

1.6 BriotBouquet differential subordinations
Definition 1.6.1 Let β and γ be two complex numbers,
h a univalent function in U and p(z) = h(0)+p1z + ... a
analytic function in U which satisfy the subordination:
p(z) +zp′(z)
βp(z) + γ≺ h(z).(1.45)
This first order differential subordination is called a
BriotBouquet differential subordination.
By using the differential subordinations method, in
[40] and [41], are obtained many usefully result regarding
the BriotBouquet differential subordinations or regard
ing generalizations of BriotBouquet differential subor
dinations.
Theorem 1.6.1 [40], [41] Let h be a convex function
in U such that Re[βh(z) + γ] > 0, z ∈ U . If p is aanalytic function in U , with p(0) = h(0), and p satisfy
the BriotBouquet differential subordination (1.45), then
p ≺ h.
42

Theorem 1.6.2 [40], [41] Let h be a convex function
in U and assume that the differential equation
q(z) +zq′(z)
βq(z) + γ= h(z), q(0) = h(0)(1.46)
has a univalent solution which satisfy the subordination
q ≺ h. If p is a analytic function in U and satisfythe BriotBouquet differential subordination (1.45), then
p ≺ q ≺ h and q is the best dominant.
Remark 1.6.1 By using the previously result, the proof
of the Marx and Strohäcker Theorem′s (1.3.3) become a
sample verification (with h(z) =
(1 + z)/(1− z), q(z) = 1/(1− z), β = 1 and γ = 0).
Theorem 1.6.3 [40], [41] Let h be a convex function
in U such that Re[βh(z) + γ] > 0, z ∈ U and assumethat the differential equation
q(z) +zq′(z)
βq(z) + γ= h(z), q(0) = h(0)
has a univalent solution q. If p is a analytic function
in U which satisfy the BriotBouquet differential sub
43

ordination (1.45), then p ≺ q ≺ h and q is the bestdominant.
Theorem 1.6.4 [40], [41] Let q be a convex function
in U and j : U → C with Re[j(z)] > 0.If p ∈ H(U), which satisfy the subordinationp(z) + j(z) · zp′(z) ≺ q(z), then p ≺ q.
44

Chapter 2
Uniformly starlike and
uniformly convex functions
2.1 Uniformly starlike functions
The notion of uniformly starlike function was introduced
in 1991 by A.W. Goodman (see [22]) and was inspired
by the following open problem:
Pinchuk′s problem. Let f ∈ S∗ and let γ be acircle contained in U with the center ζ also in U . It is
f(γ) a closed starlike curve with respect to f(ζ)?
Because the above problem has a negative answer,
45

A.W. Goodman consider a more ”strongly” condition in
the definition of the uniformly starlike functions:
Definition 2.1.1 A function f is called uniformly star
like in U if f ∈ S∗ and for any circular arc γ from U ,with the center ζ also in U , the arc f(γ) is starlike with
respect to f(ζ). We denote by US∗ the class of all this
functions.
In [22] Goodman prove that for the arc γ = z(t), we
have the arc f(γ) starlike with respect to ω0, if and only
if
Im
[f ′(z)
f(z)− ω0 ·dz
dt
]≥ 0,(2.1)
for z onto γ. For γ = Cr = {z ∈ C; z = r} and ω0 = 0it is easy to see that we obtain the condition (1.10).
For z = ζ + reit we have z′(t) = i(z − ζ), and weobtain:
Theorem 2.1.1 [22] Let f ∈ S. Then f ∈ US∗ if andonly if
Ref(z)− f(ζ)(z − ζ)f ′(z) > 0,(2.2)
46

for any (z, ζ) from U × U .
Theorem 2.1.2 [22] The function
f1(z) =z
1− Az = z +∞∑
n=2
An−1zn, z ∈ U(2.3)
belong to the class US∗ if and only if A ≤ 1√2 '0, 7071.
Theorem 2.1.3 [22] If
f2(z) = z + Azn, n ≥ 1, z ∈ U(2.4)
and A ≤ √2/(2n), then f2 belong to the class US∗.
Remark 2.1.1 If f ∈ US∗, then for ζ = −z we obtain
Re2zf ′(z)
f(z)− f(−z) ≥ 0, z ∈ U.(2.5)
A function f ∈ A which verify the condition (2.5) itis called a starlike function with respect to symmetric
points. This class of functions was introduced by Sak
aguci in [49].
Theorem 2.1.4 [22] If f ∈ US∗ and z = r < 1, then:r
1 + 2r≤ f(z) ≤ −r + 2 ln 1
1− r .(2.6)
47

Theorem 2.1.5 [22] Let f ∈ S, f(z) = z +∞∑
n=2anz
n. If
∞∑n=2
nan ≤√
2/2,(2.7)
then f ∈ US∗.
Definition 2.1.2 A function f ∈ S is called uniformlystarlike of order α, α ∈ [0, 1) if
Ref(z)− f(ζ)(z − ζ)f ′(z) ≥ α,(2.8)
for any (z, ζ) from U × U . We denote by US∗(α) theclass of all this functions. We remark that US∗(0) =
US∗.
We also remark that the uniformly starlikeness of or
der α do not imply the starlikeness of order α. Indeed, if
we consider ζ = 0 in the uniformly starlikeness of order
α, with α ∈ (0, 1], it follow that Re f(z)zf ′(z) ≥ α, z ∈ U ,or equivalently, zf
′(z)f(z) take all values in the disc centered
in 12α and with the radius12α . From the above do not
follow that Rezf′(z)
f(z) ≥ α, z ∈ U .
48

Theorem 2.1.6 [32] Let f1(z) =z
1−Az =
z +∞∑
n=2An−1zn, z ∈ U and α ∈ [0, 1).
If A ≤ 1− α√2(α2 + 1)
,(2.9)
then f1 ∈ US∗(α).
Theorem 2.1.7 [32] Let f ∈ S, f(z) = z +∞∑
n=2anz
n
and α ∈ [0, 1). If∞∑
n=2
√2(α2 + 1)
1− α nan ≤ 1,(2.10)
then f ∈ US∗(α).
2.2 Uniformly convex functions
The notion of uniformly convex function was introduced
by A.W. Goodman in 1991 (see [23]) through analogy
with the notion of uniformly starlike function.
Definition 2.2.1 A function f is called uniformly con
vex in U if f ∈ Sc and for any circular arc γ from U ,with the center ζ also in U , the arc f(γ) is convex. We
denote by UCV or USc the class of all this functions.
49

For Γ(t) = f(γ) and γ = z(t), then f(γ) is convex if
and only if
Im
[z′′(t)z′(t)
+f ′′(z)f ′(z)
z′(t)]≥ 0,(2.11)
for any z onto Γ.
We remark that for γ = Cr = {z ∈ C; z = r} weobtain the condition (1.19).
If for the circular arc γ with the center ζ we consider
z = ζ + reit, then z′(t) = i(z− ζ), z′′(t) = −(z− ζ), andfrom (2.11) we obtain:
Theorem 2.2.1 [23] Let f ∈ S. Then f ∈ USc if andonly if
1 + Re
[f ′′(z)f ′(z)
(z − ζ)]≥ 0,(2.12)
for any (z, ζ) from U × U .
Theorem 2.2.2 [23] The function
f1(z) =z
1− Az = z +∞∑
n=2
An−1zn, z ∈ U
belong to the class USc if and only if A ≤ 1/3.
50

Theorem 2.2.3 [23] The function
f2(z) = z − Az2, z ∈ U
belong to the class USc if and only if A ≤ 1/6.
Theorem 2.2.4 [23] Let f ∈ S, f(z) = z +∞∑
n=2anz
n. If
∞∑n=2
n(n− 1)an ≤ 13,(2.13)
then f ∈ USc and the constant 1/3 can not be replacedwith a greater one.
Theorem 2.2.5 [23] If f ∈ USc, f(z) = z +∞∑
n=2
anzn,
then
an ≤ 1n
, n ≥ 2 .
F. Ronning introduce in [46] the class SP which is
important because it can be used to translate the results
obtained from this class, directly to the class USc.
Definition 2.2.2 SP = {F ∈ S∗F (z) = zf ′(z),f ∈ USc}.
51

Ma and Minda (see [31]) ,and independently, Ronning
(see [46]) gives a characterization, which use only one
variable, for the uniformly convex functions:
Theorem 2.2.6 [31], [46] Let f ∈ S. Thenf ∈ USc if and only if:
Re
{1 +
zf ′′(z)f ′(z)
}>
∣∣∣∣zf ′′(z)f ′(z)
∣∣∣∣ , z ∈ U(2.14)
For g(z) = zf ′(z) we obtain:
Corollarly 2.2.1 [46] A function g ∈ S belong to theclass SP if and only if
Rezg′(z)g(z)
>
∣∣∣∣zg′(z)g(z)
− 1∣∣∣∣ , z ∈ U.(2.15)
From the geometric interpretation of the relation (2.15),
we deduce that the class SP is the class of all the func
tion g ∈ S for which zg′(z)/g(z), z ∈ U take all thevalues into the parabolic domain
Ω = {ω : ω − 1 < Reω} =(2.16)
{ω = u + iv; v2 < 2u− 1}.
52

Theorem 2.2.7 [46] g(z) = z+anzn belong to the class
SP if and only if
an ≤ 12n− 1 .(2.17)
Corollarly 2.2.2 f(z) = z + bnzn belong to the class
USc if and only if
bn ≤ 1n(2n− 1) .(2.18)
Definition 2.2.3 [47] We will denote by SP (α, β),
α > 0, β ∈ [0, 1) the class of all the functions f ∈ Swhich verify the condition:
(2.19)∣∣∣∣zf ′(z)f(z)
− (α + β)∣∣∣∣ ≤ Re
zf ′(z)f(z)
+ α− β, z ∈ U.
Geometric interpretation: f ∈ SP (α, β) if and only ifzf ′(z)/f(z), z ∈ U , take all the values into the parabolicdomain
(2.20)
Ωα,β = {ω : ω − (α + β) ≤ Reω + α− β} =53

{ω = u + iv : v2 ≤ 4α(u− β)}.
Stankiewicz and Wisniowska introduce in [51], rela
tive to a hyperbolic domain, the following new class of
functions:
Definition 2.2.4 We say that the function f ∈ S belongto the class SH(α) if
∣∣∣∣zf ′(z)f(z)
− 2α(√
2− 1)∣∣∣∣ < Re
{√2zf ′(z)f(z)
}+
2α(√
2− 1)
, z ∈ U , α > 0 .
Remark 2.2.1 For the function f ∈ SH(α) the expression
zf ′(z)f(z)
take all values into the interior of the posi
tive branch of the hyperbola v2 = 4αu + u2 , u > 0, and
the function Hα, with Hα(0) = 1 and H′α(0) > 0, which
is univalent and maps U into the above domain, is given
by
Hα(z) = (1 + 2α)
√1 + bz
1− bz − 2α
where
b = b(α) =1 + 4α− 4α2
(1 + 2α)2.
54

Definition 2.2.5 We say that a function f ∈ S is uniformly convex of type α, α ≥ 0 if:
Re
{1 +
zf ′′(z)f ′(z)
}≥ α
∣∣∣∣zf ′′(z)f ′(z)
∣∣∣∣ , z ∈ U.(2.21)
We denote by USc(α) (or k − UCV ) the class of allthis functions.
Remark 2.2.2 The class USc(α) was introduced by Kanas
and Wisniowska in [27] by using the following definition:
Let 0 ≤ k < ∞. We say that a function f ∈ S iskuniformly convex in U if the image of any circle arc
γ contained in U , with the center ζ (ζ ≤ k), is convex.We denote by k − UCV the class of all this functions.
We remark that USc(1) = USc and USc(0) = Sc.
By this remark we obtain a continuously connection be
tween convexity and uniformly convexity.
The geometric interpretation of the definition 2.2.5:
f ∈ USc(α) if and only if 1 + zf ′′(z)/f ′(z) take all thevalues into the domain Dα, where Dα is:
55

i) the elliptic region:(u− α2α2−1
)2(
αα2−1
)2 +v2(1√
α2−1
)2 < 1, for α > 1
ii) the parabolic region:
v2 < 2u− 1, for α = 1
iii) the hyperbolic region:(u + α
2
1−α2)2
(α
1−α2)2 −
v2(1√
1−α2)2 > 1, and u > 0,
for 0 < α < 1
iv) the halfplane u > 0, for α = 0.
56

We also remark that USc(α) ⊂ Sc ( αα+1).
Theorem 2.2.8 [27] Let f ∈ S, f(z) = z+∞∑
j=2ajz
j and
α ≥ 0. If∞∑
j=2
j(j − 1)aj ≤ 1α + 2
(2.22)
then f ∈ USc(α). The constant 1/(α + 2) can not bereplaced be a greater one.
Remark 2.2.3 Inspired by the class USc(α) Kanas and
Wisniowska introduce, in [29], the class α−ST by usingthe following definition:
α− ST := {f ∈ S : f(z) = zg′(z) , g ∈ USc(α)} ,
α ≥ 0 , z ∈ U.
In [53] the authors introduced the class of uniformly
convex of order γ functions by using the following defi
nition:
Definition 2.2.6 We say that a function f ∈ S is uniformly convex of order γ ∈ [−1, 1) if
(2.23)
57

Re
{1 +
zf ′′(z)f ′(z)
}≥
∣∣∣∣zf ′′(z)f ′(z)
∣∣∣∣ + γ, z ∈ U.
We denote by USc[γ] the class of all this functions.
The following subclasses are introduce by using the
Sălăgean differential operator (see [50]):
Dn : A → A , n ∈ N and D0f(z) = f(z)(2.24)
D1f(z) = Df(z) = zf ′(z) , Dnf(z) = D(Dn−1f(z)
)
In 1999 I. Magdaş (see [33]), and independently, S.
Kanas and T. Yaguchi (see [30]) introduce the class of
nuniformly starlike of type α functions:
Definition 2.2.7 We say that a function f ∈ A is nuniformly starlike of type α, α ≥ 0 and n ∈ N if:
Re
{Dn+1f(z)
Dnf(z)
}≥ α
∣∣∣∣Dn+1f(z)
Dnf(z)− 1
∣∣∣∣ ,(2.25)
for all z ∈ U .we denote by USn(α) the class of all this functions.
We remark that US0(1) = SP, US1(1) = USc, US1(α) =
USc(α), where USc(α) is the class defined by (2.21).
58

Geometric interpretation of the relation (2.25): f ∈USn(α) if and only if D
n+1f(z)/Dnf(z) take all values in
the domain Dα, where Dα is a elliptic region for α > 1,
a parabolic region for α = 1, a hyperbolic region for
0 < α < 1, respectively the halfplan u > 0 for α = 0
(see also the Definition (2.2.5)).
From the above we remark that ReDn+1f(z)
Dnf(z) >α
α+1 .
We have USn(α) ⊂ Sn(
αα+1 , 1
) ⊂ S∗, and so we conclude that the functions from USn(α) are univalent.
Remark 2.2.4 In [30], S. Kanas and T. Yaguchi, the
above mentioned functions are denominate (k, n)uniformly
convex functions and the class of all functions is denoted
by (k, n)− UCV .In the same paper, the authors introduced also the
class (k, n)− ST by the following definition:For f ∈ S, k ∈ [0,∞) and n ∈ N, we say that f
belong to the class (k, n)− ST if
Re
(Dnf(z)
f(z)
)> k
∣∣∣∣Dnf(z)
f(z)− 1
∣∣∣∣ , z ∈ U .
59

I. Magdaş introduce in [34] the uniformly convex of
type α and order γ functions and the nuniformly star
like of order γ and type α functions:
Definition 2.2.8 We say that a function f ∈ A is uniformly convex of type α and order γ, α ≥ 0, γ ∈ [−1, 1),α + γ ≥ 0 if:
Re
{1 +
zf ′′(z)f ′(z)
}≥ α
∣∣∣∣zf ′′(z)f ′(z)
∣∣∣∣ + γ,(2.26)
for all z ∈ U .
We denote by USc(α, γ) the class of all this functions.
We remark that USc(α, 0) = USc(α) and USc(1, γ) =
USc[γ].
Geometric interpretation of the relation (2.26):
f ∈ USc(α, γ) if and only if 1 + zf ′′(z)f ′(z) take all values inthe domain Dα,γ, where Dα,γ is:
i) a elliptic region:(u− α2−γα2−1
)2[
α(1−γ)α2−1
]2 +v2(
1−γ√α2−1
)2 < 1, for α > 1;
60

ii) a parabolic region:
v2 < 2(1− γ)u− (1− γ2), for α = 1;
iii) a hyperbolic region:(u− γ−α21−α2
)2[
α(1−γ)1−α2
]2 +v2(
1−γ√1−α2
)2 > 1 and u > 0, for 0 < α < 1;
iv) the halfplane u > γ, for α = 0
We have Re{
1 + zf′′(z)
f ′(z)
}> α+γα+1 .
We also remark that USc(α, γ) ⊂ Sc (α+γα+1).
Definition 2.2.9 We say that a function f ∈ A is nuniformly starlike of order γ and type α, where
61

α ≥ 0, γ ∈ [−1, 1), α + γ ≥ 0 and n ∈ N if
ReDn+1f(z)
Dnf(z)≥ α
∣∣∣∣Dn+1f(z)
Dnf(z)− 1
∣∣∣∣ + γ,(2.27)
for all z ∈ U .
We denote by USn(α, γ) the class of all this functions.
We remark that
US1(α, γ) = USc(α, γ), US0(α, γ) = S
∗(γ),
US0(1, γ) = SP
(1− γ
2,
1 + γ
2
)and USn(α, 0) = USn(α).
Geometric interpretation of the relation (2.27):
f ∈ USn(α, γ) if and only if Dn+1f(z)/Dnf(z) take allvalues in the domain Dα,γ, where Dα,γ was defined at
the geometric interpretation of the definition of the class
USc(α, γ).
We remember that for the functions f ∈ USn(α, γ)we have
Re{Dn+1f(z)/Dnf(z)
}> (α + γ)/(α + 1),
and thus
USn(α, γ) ⊂ Sn(
α + γ
1 + α
)⊂ S∗.
62

This mean that the functions from USn(α, γ) are univa
lent.
Definition 2.2.10 Let f, g ∈ A; f(z) = z+∞∑
j=2ajz
j, z ∈
U and g(z) = z+∞∑
j=2bjz
j, z ∈ U . We will denote by f ∗gthe convolution (or Hadamard) product of the functions
f and g, defined by
(f ∗ g)(z) = z +∞∑
j=2
ajbjzj, z ∈ U.(2.28)
Definition 2.2.11 [48] We define the Ruscheweyh op
erator Rn : A → A, n ∈ N , z ∈ U , by:
Rnf(z) =z
(1− z)n+1 ∗ f(z) =z(zn−1f(z))(n)
n!.(2.29)
Remark 2.2.5 1. If f ∈ A, f(z) = z+∞∑
j=2ajz
j, z ∈ U ,then
Rnf(z) = z +∞∑
j=2
Cnn+j−1ajzj, z ∈ U.(2.30)
2. We remark that the inequality
ReRn+1f(z)
Rnf(z)>
1
2, z ∈ U(2.31)
63

become for n = 1 the convexity condition.
We will denote by Kn the class of all functions f ∈ Awhich satisfy (2.31).
By using the Ruscheweyh operator, in [35], a new
subclass of o uniformly convex functions is defined by:
Definition 2.2.12 Let n ∈ N. We say that the functionf ∈ A belong to the class UKn(δ), δ ∈ [−1, 1), if:
ReRn+1f(z)
Rnf(z)≥
∣∣∣∣Rn+1f(z)
Rnf(z)− 1
∣∣∣∣ + δ,(2.32)
for all z ∈ U .
Geometric interpretation: f ∈ UKn(δ), if and only ifRn+1f(z)/Rnf(z) take all values in the domain
Ω 1−δ2 ,
1+δ2
not= Ωδ bounded by the parabola:
v2 = 2(1− δ)u− (1− δ2).(2.33)
The corresponding Carathéodory function is
Qδ(z) = 1 +2(1− δ)
π2
(log
1 +√
z
1−√z)2
, z ∈ U.(2.34)
64

We remark that the function Qδ is convex and satisfy
ReQδ(z) >1 + δ
2. We conclude that, f ∈ UKn(δ) if and
only if Rn+1f(z)Rnf(z) ≺ Qδ(z).
We remark that for n = 0 we have UK0(δ) =
SP(1−δ
2 ,1+δ2
), and for n = 1 and δ = 1/2, we have
UK1(1/2) = USc.
65

Chapter 3
Subclasses of αconvex
functions
3.1 The subclasses UM(α) and UMα
In [26] S. Kanas define the following subclass of αuniformly
convex functions:
Definition 3.1.1 Let α ∈ [0, 1]. We say that a univalent function f is called αuniformly convex if the image
of every circle arc Γz contained in U and with the center
ζ ∈ U , is a αconvex curve (see Definition 1.4.1) withrespect to f(ζ). We denote by UM(α) the class of all
66

this functions. We remark that UM(α) ⊂ Mα, whereMα is the class of αconvex functions (see section 1.4)
Theorem 3.1.1 [26] Let α ∈ [0, 1] and f be a univalentfunction. Then, f is a αuniformly convex function if
and only if
Re
{(1− α)(z − ζ)f
′(z)f(z)− f(ζ)
+α
(1 +
(z − ζ)f ′′(z)f ′(z)
)}> 0 , z, ζ ∈ U .
Theorem 3.1.2 [26] If f is a αuniformly convex func
tion and 0 ≤ β < α, then f is also a βuniformly convexfunction, or briefly UM(α) ⊂ UM(β).
In [36] I. Magdaş introduce the following subclass of
αuniformly convex functions:
Definition 3.1.2 Let f ∈ A. We say that f is αuniform convex function, α ∈ [0, 1] if
Re
{(1− α)zf
′(z)f(z)
+ α
(1 +
zf ′′(z)f ′(z)
)}
≥∣∣∣∣(1− α)
(zf ′(z)f(z)
− 1)
+ αzf ′′(z)f ′(z)
∣∣∣∣ , z ∈ U.
67

We denote this class with UMα.
Remark 3.1.1 Geometric interpretation: f ∈ UMα ifand only if
J(α, f ; z) = (1− α)zf′(z)
f(z)+ α
(1 +
zf ′′(z)f ′(z)
)
takes all values in the parabolic region Ω = {w : w−1 ≤Re w} = {w = u + iv ; v2 ≤ 2u − 1}. We have UM0 =SP (see definition 2.2.2). Also, we have UMα ⊂ Mα,where Mα is the class of αconvex functions.
3.2 The subclass UDn,α(β, γ)
The results included in this section are obtained in [1].
Definition 3.2.1 Let α ∈ [0, 1] and n ∈ N. We saythat f ∈ A is in the class UDn,α(β, γ) , β ≥ 0 , γ ∈[−1, 1) , β + γ ≥ 0, if
Re
[(1− α)D
n+1f(z)
Dnf(z)+ α
Dn+2f(z)
Dn+1f(z)
]
≥ β∣∣∣∣(1− α)
Dn+1f(z)
Dnf(z)+ α
Dn+2f(z)
Dn+1f(z)− 1
∣∣∣∣ + γ ,
68

where Dn is the differential operator defined by (2.24).
Remark 3.2.1 We have UDn,0(β, γ) = USn(β, γ) ⊂ S∗
,UD0,α(1, 0) = UMα and UD0,1(β, γ) = USc(β, γ) ⊂
Sc(
β + γ
β + 1
), where USn(β, γ) is the class given in the
definition 2.2.9, UMα is the class of αuniformly convex
functions defined in the previously section, USc(β, γ) is
the class of the uniformly convex functions of type β and
order γ (see definition 2.2.8) and Sc(δ) is the class of
convex functions of order δ (see definition1.3.5).
Remark 3.2.2 Geometric interpretation:
f ∈ UDn,α(β, γ) if and only if
Jn(α, f ; z) = (1− α)Dn+1f(z)
Dnf(z)+ α
Dn+2f(z)
Dn+1f(z)
takes all values in the convex domain Dβ,γ, where Dβ,γ is
defined at the geometric interpretation of the definition
2.2.8.
Theorem 3.2.1 For all α, α′ ∈ [0, 1] with α < α′, wehave UDn,α′(β, γ) ⊂ UDn,α(β, γ).
69

Proof. From f ∈ UDn,α′(β, γ) we have
Re
[(1− α′)D
n+1f(z)
Dnf(z)+ α′
Dn+2f(z)
Dn+1f(z)
]
≥ β∣∣∣∣(1− α′)
Dn+1f(z)
Dnf(z)+ α′
Dn+2f(z)
Dn+1f(z)− 1
∣∣∣∣ + γ .
With the notationsDn+1f(z)
Dnf(z)= p(z), where p(z) =
1 + p1z + · · ·, we have
zp′(z) = z
(Dn+1f(z)
)′ ·Dnf(z)−Dn+1f(z) · (Dnf(z))′(Dnf(z))2
=Dn+2f(z)
Dnf(z)−
(Dn+1f(z)
Dnf(z)
)2,
zp′(z)p(z)
=Dn+2f(z)
Dn+1f(z)− D
n+1f(z)
Dnf(z),
and thus we obtain
Jn(α′, f ; z) = p(z) + α′ · zp
′(z)p(z)
.
Now we have that p(z) + α′ · zp′(z)
p(z)takes all values in
the convex domain Dβ,γ which is included in right half
plane.
If we consider h ∈ Hu(U), with h(0) = 1, which mapsthe unit disc U into the convex domain Dβ,γ, we have
70

Reh(z) > 0 and from hypothesis α′ > 0. From here
follows that Re1
α′· h(z) > 0. In this conditions from
Theorem 1.6.1 , with δ = 0 we obtain p(z) ≺ h(z), orp(z) take all values in Dβ,γ.
If we consider the function g : [0, α′] → C,g(u) = p(z) + u · zp
′(z)p(z)
, with g(0) = p(z) ∈ Dβ,γ andg(α′) ∈ Dβ,γ. Since the geometric image of g(α) is on thesegment obtained by the union of the geometric image
of g(0) and g(α′), we have g(α) ∈ Dβ,γ, or
p(z) + α · zp′(z)
p(z)∈ Dβ,γ .
Thus Jn(α, f ; z) takes all values in Dβ,γ, or f ∈ UDn,α(β, γ).
Remark 3.2.3 From Theorem 3.2.1 we have UDn,α(β, γ)
⊂ UDn,0(β, γ) for all α ∈ [0, 1], and from Remark 3.2.1we obtain that the functions from the class UDn,α(β, γ)
are univalent.
71

Let consider the integral operator La : A → A definedas (see [44]):
f(z) = LaF (z) =1 + a
za
z∫
0
F (t)·ta−1dt, a ∈ C, Re a ≥ 0.
(3.1)
Remark 3.2.4 If we take a = 1, 2, 3, ... in the above
definition, we obtain the Bernardi integral operator (see
[15]).
Theorem 3.2.2 If F (z) ∈ UDn,α(β, γ) then f(z) =La(F )(z) ∈ USn(β, γ), where La is the integral operator defined by (3.1) and the class USn(β, γ) is given in
the definition 2.2.9.
Proof. From (3.1) we have
(1 + a)F (z) = af(z) + zf ′(z)
By means of the application of the linear operator Dn+1
we obtain
(1 + a)Dn+1F (z) = aDn+1f(z) + Dn+1(zf ′(z))
72

or
(1 + a)Dn+1F (z) = aDn+1f(z) + Dn+2f(z) .
Thus:
Dn+1F (z)
DnF (z)=
Dn+2f(z) + aDn+1f(z)
Dn+1f(z) + aDnf(z)
=
Dn+2f(z)
Dn+1f(z)· D
n+1f(z)
Dnf(z)+ a · D
n+1f(z)
Dnf(z)
Dn+1f(z)
Dnf(z)+ a
.
With the notationDn+1f(z)
Dnf(z)= p(z) where p(z) = 1+
p1z + ..., we have:
zp′(z) = z ·(
Dn+1f(z)
Dnf(z)
)′
=z(Dn+1f(z)
)′ ·Dnf(z)−Dn+1f(z) · z (Dnf(z))′(Dnf(z))2
=Dn+2f(z) ·Dnf(z)− (Dn+1f(z))2
(Dnf(z))2
and
1
p(z)· zp′(z) = D
n+2f(z)
Dn+1f(z)− D
n+1f(z)
Dnf(z)=
Dn+2f(z)
Dn+1f(z)− p(z) .
73

It follows:
Dn+2f(z)
Dn+1f(z)= p(z) +
1
p(z)· zp′(z) .
Thus we obtain:
Dn+1F (z)
DnF (z)=
p(z) ·(zp′(z) · 1p(z) + p(z)
)+ a · p(z)
p(z) + a
= p(z) +1
p(z) + a· zp′(z) .
If we denoteDn+1F (z)
DnF (z)= q(z), with q(0) = 1, and
we consider h ∈ Hu(U), with h(0) = 1, which maps theunit disc U into the convex domain Dβ,γ, we have from
F (z) ∈ UDn,α(β, γ) (see Remark 3.2.2):
q(z) + α · zq′(z)
q(z)≺ h(z) .
From Theorem 1.6.1 , with δ = 0 we obtain q(z) ≺ h(z),or
p(z) +1
p(z) + a· zp′(z) ≺ h(z) .
Using the hypothesis and the construction of the func
tion h(z) we obtain from Theorem 1.6.1 p(z) ≺ h(z) or74

f(z) ∈ USn(β, γ) (see the geometric interpretation ofthe definition 2.2.9).
Remark 3.2.5 From Theorem 3.2.2 with α = 0 we ob
tain the Theorem 3.1 from [7] which assert that the in
tegral operator La, defined by (3.1), preserve the class
USn(β, γ).
3.3 The subclasses UMα(q) and UDn,α(q)
In the beginning of this section we will recall some re
sults due to D. Blezu (see [18]):
Definition 3.3.1 The function f ∈ A is nstarlike withrespect to convex domain included in right half plane D if
the differential expressionDn+1f(z)
Dnf(z)takes values in the
domain D, where Dn is the differential operator defined
by (2.24).
If we consider q(z) an univalent function with
q(0) = 1, Re q(z) > 0, q′(0) > 0 which maps the unit
75

disc U into the convex domain D we have:
Dn+1f(z)
Dnf(z)≺ q(z).
We note by S∗n(q) the set of all these functions.
The following results are obtained in [2].
Let q(z) be an univalent function with q(0) = 1,
q′(0) > 0, which maps the unit disc U into a convex
domain included in right half plane D.
Definition 3.3.2 Let f ∈ A and α ∈ [0, 1]. We say thatf is αuniform convex function with respect to D, if
J(α, f ; z) = (1− α)zf′(z)
f(z)+ α
(1 +
zf ′′(z)f ′(z)
)≺ q(z).
We denote this class with UMα(q).
Remark 3.3.1 Geometric interpretation: f ∈ UMα(q)if and only if J(α, f ; z) take all values in the convex
domain included in right half plan D.
Remark 3.3.2 We have UMα(q) ⊂ Mα, where Mα isthe well know class of αconvex function. If we take
D = Ω (see Remark 3.1.1) we obtain the class UMα.
76

Remark 3.3.3 From the above definition it easily re
sults that q1(z) ≺ q2(z) implies UMα(q1) ⊂ UMα(q2).
Theorem 3.3.1 For all α, α′ ∈ [0, 1] with α < α′ wehave UMα′(q) ⊂ UMα(q).
Proof. From f ∈ UMα′(q) we have
J(α′, f ; z) = (1− α)zf′(z)
f(z)+ α
(1 +
zf ′′(z)f ′(z)
)≺ q(z),
(3.2)
where q(z) is univalent in U with q(0) = 1, q′(0) > 0, and
maps the unit disc U into the convex domain included
in right half plane D.
With notationzf ′(z)f(z)
= p(z), where p(z) = 1+p1z+...
we have:
J(α′, f ; z) = p(z) + α′ · zp′(z)
p(z).
From (3.2) we have p(z) + α′ · zp′(z)
p(z)≺ q(z) with
p(0) = q(0) and Re q(z) > 0, z ∈ U .In this conditions from Theorem 1.6.1, with δ = 0, we
obtain p(z) ≺ q(z), or p(z) take all values in D.77

If we consider the function g : [0, α′] → C,g(u) = p(z) + u · zp
′(z)p(z)
, with g(0) = p(z) ∈ D andg(α′) = J(α′, f ; z) ∈ D. Since the geometric imageof g(α) is on the segment obtained by the union of the
geometric image of g(0) and g(α′), we have g(α) ∈ D orp(z) + α
zp′(z)p(z)
∈ D.Thus J(α, f ; z) take all values in D, or
J(α, f ; z) ≺ q(z). This means f ∈ UMα(q).
Theorem 3.3.2 If F (z) ∈ UMα(q) thenf(z) = La(F )(z) ∈ S∗0(q), where La is the integral operator defined by (3.1) and α ∈ [0, 1].
Proof. From (3.1) we have
(1 + a)F (z) = af(z) + zf ′(z).
With notationzf ′(z)f(z)
= p(z), where p(z) = 1+p1z+...
we have
zF ′(z)F (z)
= p(z) +zp′(z)
p(z) + a.
78

If we denotezF ′(z)F (z)
= h(z), with h(0) = 1, we have
from F (z) ∈ UMα(q) (see Definition 3.3.2):
h(z) + α · zh′(z)
h(z)≺ q(z),
where q(z) is univalent un U with q(0) = 1, q′(z) > 0 and
maps the unit disc U into the convex domain included
in right half plane D.
From Theorem 1.6.1 we obtain h(z) ≺ q(z) orp(z) +
zp′(z)p(z) + a
≺ q(z).Using the hypothesis and the construction of the func
tion q(z) we obtain from Theorem 1.6.1zf ′(z)f(z)
= p(z) ≺ q(z) or f(z) ∈ S∗0(q) ⊂ S∗.
Definition 3.3.3 Let f ∈ A, α ∈ [0, 1] and n ∈ N.We say that f is α − nuniformly convex function withrespect to D if
Jn(α, f ; z) = (1− α)Dn+1f(z)
Dnf(z)+ α
Dn+2f(z)
Dn+1f(z)≺ q(z),
where Dn is the differential operator defined by (2.24).
We denote this class with UDn,α(q).
79

Remark 3.3.4 Geometric interpretation: f ∈ UDn,α(q)if and only if Jn(α, f ; z) take all values in the convex do
main included in right half plane D.
Remark 3.3.5 We have UD0,α(q) = UMα(q) and if in
the above definition we consider D = Dβ,γ (see Remark
3.2.2) we obtain the class UDn,α(β, γ).
Remark 3.3.6 It is easy to see that q1(z) ≺ q2(z) implies UDn,α(q1) ⊂ UDn,α(q2).
Theorem 3.3.3 For all α, α′ ∈ [0, 1] with α < α′ wehave UDn,α′(q) ⊂ UDn,α(q).
Proof. From f ∈ UDn,α′(q) we have:
Jn(α′, f ; z) = (1−α)D
n+1f(z)
Dnf(z)+ α
Dn+2f(z)
Dn+1f(z)≺ q(z),
(3.3)
where q(z) is univalent in U with q(0) = 1, q′(0) > 0, and
maps the unit disc U into the convex domain included
in right half plane D.
80

With notationDn+1f(z)
Dnf(z)= p(z), where
p(z) = 1 + p1z + ... we have
Jn(α′, f ; z) = p(z) + α′ · zp
′(z)p(z)
.
From (3.3) we have p(z) + α′ · zp′(z)
p(z)≺ q(z) with
p(0) = q(0) and Re q(z) > 0, z ∈ U . In this conditionfrom Theorem 1.6.1 we obtain p(z) ≺ q(z), or p(z) takeall values in D.
If we consider the function g : [0, α′] → C,g(u) = p(z) + u · zp
′(z)p(z)
, with g(0) = p(z) ∈ D andg(α′) = Jn(α′, f ; z) ∈ D, it easy to see thatg(α) = p(z) + α
zp′(z)p(z)
∈ D.Thus we have Jn(α, f ; z) ≺ q(z) or f ∈ UDn,α(q).
Theorem 3.3.4 If F (z) ∈ UDn,α(q) thenf(z) = La(F )(z) ∈ S∗n(q), where La is the integral operator defined by (3.1).
Proof. From (3.1) we have
(1 + a)F (z) = af(z) + zf ′(z). By means of the applica
81

tion of the linear operator Dn+1 we obtain:
(1 + a)Dn+1F (z) = aDn+1f(z) + Dn+1(zf ′(z))
or
(1 + a)Dn+1F (z) = aDn+1f(z) + Dn+2f(z).
With notationDn+1f(z)
Dnf(z)= p(z), where
p(z) = 1 + p1z + ..., we have:
Dn+1F (z)
DnF (z)= p(z) +
1
p(z) + a· zp′(z).
If we denoteDn+1F (z)
DnF (z)= h(z), with h(0) = 1, we
have from F ∈ UDn,α(q):
h(z) + αzh′(z)h(x)
≺ q(z),
where q(z) is univalent in U with q(0) = 1, q′(0) > 0, and
maps the unit disc U into the convex domain included
in right half plane D.
From Theorem 1.6.1 we obtain h(z) ≺ q(z) orp(z) +
zp′(z)p(z) + a
≺ q(z).Using the hypothesis we obtain from Theorem 1.6.1
p(z) ≺ q(z) or f(z) ∈ S∗n(q).82

Remark 3.3.7 If we consider D = Dβ,γ in Theorem
3.3.3 and Theorem 3.3.4 we obtain the main results from
the previously section and if we take D = Dβ,γ and α = 0
in Theorem 3.3.4 we obtain the Theorem 3.1 from [7].
3.4 The subclass Mλ,α(q)
For the main results from this section we need to recall
here the following definitions and theorems:
Definition 3.4.1 [3] Let λ ∈ R , λ ≥ 0 andf(z) = z +
∞∑
j=2
ajzj. We define the generalized Sălăgean
operator by Dλ : A → A
Dλf(z) = z +∞∑
j=2
jλajzj .
Remark 3.4.1 [3] It is easy to observe that the general
ized Sălăgean operator defined above is a linear operator.
Also, we observe that for λ ∈ N we obtain the Sălăgeandifferential operator.
83

Definition 3.4.2 [3] Let q(z) ∈ Hu(U), with q(0) = 1and q(U) = D, where D is a convex domain contained in
the right half plane. We say that a function f(z) ∈ A isa λqstarlike function if
Dλ+1f(z)
Dλf(z)≺ q(z). We denote
this class by S∗λ(q).
Definition 3.4.3 [8] Let q(z) ∈ Hu(U), with q(0) = 1and q(U) = D, where D is a convex domain contained
in the right half plane. We say that a function f(z) ∈ Ais a λqconvex function if
Dλ+2f(z)
Dλ+1f(z)≺ q(z). We denote
this class by Scλ(q).
The main results of this section are obtained in [4].
Definition 3.4.4 Let α ∈ [0, 1], q(z) ∈ Hu(U), withq(0) = 1 and q(U) = D, where D is a convex domain
contained in the right half plane. We say that a function
f(z) ∈ A is a λqαconvex function if
Jλ(α, f ; z) = (1− α)Dλ+1f(z)
Dλf(z)+ α
Dλ+2f(z)
Dλ+1f(z)≺ q(z) .
We denote this class with Mλ,α(q).
84

Remark 3.4.2 Geometric interpretation: f(z) ∈ Mλ,α(q)if and only if Jλ(α, f ; z) take all values in the convex do
main D contained in the right halfplane.
Remark 3.4.3 It is easy to observe that if we choose
different function q(z) we obtain variously classes of
αconvex functions, such as (for example), for λ = 0,
the class of αconvex functions, the class of αuniform
convex functions with respect to a convex domain (see
the previously section), and, for λ = n ∈ N, the classUDn,α(β, γ), β ≥ 0, γ ∈ [−1, 1), β + γ ≥ 0 (see the section 3.2), the class of αnuniformly convex functions
with respect to a convex domain (see the previously sec
tion).
Remark 3.4.4 We have Mλ,0(q) = S∗λ(q) and
Mλ,1(q) = Scλ(q).
Remark 3.4.5 For q1(z) ≺ q2(z) we haveMλ,α(q1) ⊂ Mλ,α(q2) .From the above we obtain Mλ,α(q) ⊂ Mλ,α
(1 + z
1− z)
.
85

Theorem 3.4.1 Let λ ∈ R , λ ≥ 0.For all α, α′ ∈ [0, 1], with α < α′ we haveMλ,α′(q) ⊂ Mλ,α(q) .
Proof. From f(z) ∈ Mλ,α′(q) we have
Jλ(α′, f ; z) = (1− α′)D
λ+1f(z)
Dλf(z)+ α′
Dλ+2f(z)
Dλ+1f(z)≺ q(z) ,
(3.4)
where q(z) is univalent in U with q(0) = 1 and maps the
unit disc U into the convex domain D contained in the
right halfplane.
With notation
p(z) =Dλ+1f(z)
Dλf(z),
where
p(z) = 1 + p1z + . . . ,
we have
p(z) + α′ · zp′(z)
p(z)
=Dλ+1f(z)
Dλf(z)+ α′
Dλf(z)
Dλ+1f(z)
86

·z(Dλ+1f(z)
)′Dλf(z)−Dλ+1f(z) (Dλf(z))′
(Dλf(z))2
=Dλ+1f(z)
Dλf(z)+ α′
Dλf(z)
Dλ+1f(z)
(z(Dλ+1f(z)
)′Dλf(z)
−Dλ+1f(z)
Dλf(z)· z
(Dλf(z)
)′Dλf(z)
)
=Dλ+1f(z)
Dλf(z)+ α′ · D
λf(z)
Dλ+1f(z)
z
(z +
∞∑j=2
jλ+1ajzj
)′
Dλf(z)
−Dλ+1f(z)
Dλf(z)·z
(z +
∞∑
j=2
jλajzj
)′
Dλf(z)
=Dλ+1f(z)
Dλf(z)+ α′ · D
λf(z)
Dλ+1f(z)
z
(1 +
∞∑j=2
j(jλ+1aj)zj−1
)
Dλf(z)
87

−Dλ+1f(z)
Dλf(z)·z
(1 +
∞∑j=2
j(jλaj)zj−1
)
Dλf(z)
=Dλ+1f(z)
Dλf(z)+ α′ · D
λf(z)
Dλ+1f(z)
z +∞∑
j=2
jλ+2ajzj
Dλf(z)
−Dλ+1f(z)
Dλf(z)·z +
∞∑j=2
jλ+1ajzj
Dλf(z)
=Dλ+1f(z)
Dλf(z)+ α′ · D
λf(z)
Dλ+1f(z)
(Dλ+2f(z)
Dλf(z)− D
λ+1f(z)
Dλf(z)
·Dλ+1f(z)
Dλf(z)
)
=Dλ+1f(z)
Dλf(z)+ α′ · D
λ+2f(z)
Dλ+1f(z)− α′ · D
λ+1f(z)
Dλf(z)
=Dλ+1f(z)
Dλf(z)(1− α′) + α′ · D
λ+2f(z)
Dλ+1f(z)= Jλ(α
′, f ; z)
88

From (3.4) we have
p(z) +zp′(z)1
α′· p(z)
≺ q(z) ,
with p(0) = q(0), Re q(z) > 0 , z ∈ U , and α′ > 0. Inthis conditions from Theorem 1.6.1 we obtain
p(z) ≺ q(z) or p(z) take all values in D.If we consider the function g : [0, α′] → C,
g(u) = p(z) + u · zp′(z)
p(z),
with g(0) = p(z) ∈ D and g(α′) = Jλ(α′, f ; z) ∈ D, iteasy to see that
g(α) = p(z) + α · zp′(z)
p(z)∈ D , 0 ≤ α < α′ .
Thus we have
Jλ(α, f ; z) ≺ q(z)
or
f(z) ∈ Mλ,α(q) .
Remark 3.4.6 From the above theorem we have, for ev
ery α ∈ [0, 1], that Mλ,α(q) ⊂ S∗λ(q).89

Remark 3.4.7 If we consider λ = 0 we obtain the The
orem 3.3.1 from the section 3.3. Also, for λ = n ∈ N,we obtain the Theorem 3.3.3 from the previously section.
Remark 3.4.8 If we consider D = Dβ,γ (see the geo
metric interpretation of the definition 2.2.8) in the above
theorem we obtain the Theorem 3.2.1 from the section
3.2.
Theorem 3.4.2 If F (z) ∈ Mλ,α(q) thenf(z) = LaF (z) ∈ S∗λ(q), where La is the integral operatordefined by (3.1).
Proof. From (3.1) we have
(1 + a)F (z) = af(z) + zf ′(z)
and, by using the linear operator Dλ+1, we obtain
(1 + a)Dλ+1F (z) = aDλ+1f(z) + Dλ+1 (zf ′(z))
= aDλ+1f(z) + Dλ+1
(z +
∞∑
j=2
jajzj
)
= aDλ+1f(z) + z +∞∑
j=2
jλ+1(jaj)zj
90

= aDλ+1f(z) + Dλ+2f(z)
or
(1 + a)Dλ+1F (z) = aDλ+1f(z) + Dλ+2f(z) .
Similarly, we obtain
(1 + a)DλF (z) = aDλf(z) + Dλ+1f(z) .
Then
Dλ+1F (z)
DλF (z)=
Dλ+2f(z)
Dλ+1f(z)· D
λ+1f(z)
Dλf(z)+ a · D
λ+1f(z)
Dλf(z)
Dλ+1f(z)
Dλf(z)+ a
.
With notation
Dλ+1f(z)
Dλf(z)= p(z) , p(0) = 1 ,
we obtain
Dλ+1F (z)
DλF (z)=
Dλ+2f(z)
Dλ+1f(z)· p(z) + a · p(z)
p(z) + a(3.5)
We have
Dλ+2f(z)
Dλ+1f(z)=
Dλ+2f(z)
Dλf(z)· D
λf(z)
Dλ+1f(z)=
1
p(z)· D
λ+2f(z)
Dλf(z)(3.6)
91

Also, we have
Dλ+2f(z)
Dλf(z)=
z +∞∑
j=2
jλ+2ajzj
z +∞∑
j=2
jλajzj
and
zp′(z) =z(Dλ+1f(z)
)′Dλf(z)
− Dλ+1f(z)
Dλf(z)· z
(Dλf(z)
)′Dλf(z)
=
=
z +∞∑
j=2
jλ+2ajzj
z +∞∑
j=2
jλajzj
− p(z) ·z +
∞∑j=2
jλ+1ajzj
z +∞∑
j=2
jλajzj
=
=Dλ+2f(z)
Dλf(z)− p(z) · D
λ+1f(z)
Dλf(z).
Thus
zp′(z) =Dλ+2f(z)
Dλf(z)− p(z)2
orDλ+2f(z)
Dλf(z)= p(z)2 + zp′(z) .
From (3.6) we obtain
Dλ+2f(z)
Dλ+1f(z)=
1
p(z)
[p(z)2 + zp′(z)
]= p(z) +
zp′(z)p(z)
.
92

From (3.5) we obtain
Dλ+1F (z)
DλF (z)=
(p(z) +
zp′(z)p(z)
)· p(z) + a · p(z)
p(z) + a
= p(z) +zp′(z)
p(z) + a
If we denote
Dλ+1F (z)
DλF (z)= h(z) , with h(0) = 1 ,
we have from F (z) ∈ Mλ,α(q) (see the proof of the abovetheorem):
Jλ(α, F ; z) = h(z) + α · zh′(z)
h(z)≺ q(z) .
Using the hypothesis we obtain from Theorem 1.6.1
h(z) ≺ q(z)
or
p(z) +zp′(z)
p(z) + a≺ q(z) .
By using the Theorem 1.6.1 and the hypothesis we have
p(z) ≺ q(z)93

orDλ+1f(z)
Dλf(z)≺ q(z) .
This means that f(z) ∈ S∗λ(q) .
Remark 3.4.9 If we consider λ = 0 we obtain the The
orem 3.3.2 from previously section. Also, for λ = n ∈ N,we obtain the Theorem 3.3.4 from the section 3.3.
Remark 3.4.10 If we consider D = Dβ,γ (see remark
3.4.8) in the above theorem we obtain the Theorem 3.2.2
from the section 3.2.
3.5 The subclass MLn,α(q)
In the first part of this section we will introduce some
usefully definitions and remarks:
Definition 3.5.1 [14] Let n ∈ N and λ ≥ 0. We denotewith Dnλ the operator defined by
Dnλ : A → A ,
D0λf(z) = f(z) , D1λf(z) = (1−λ)f(z)+λzf ′(z) = Dλf(z) ,
94

Dnλf(z) = Dλ(Dn−1λ f(z)
).
Remark 3.5.1 [14] We observe that Dnλ is a linear op
erator and for f(z) = z +∞∑
j=2
ajzj we have
Dnλf(z) = z +∞∑
j=2
(1 + (j − 1)λ)n ajzj .
Also, it is easy to observe that if we consider λ = 1 in
the above definition we obtain the Sălăgean differential
operator (see (2.24)).
Definition 3.5.2 [9] Let q(z) ∈ Hu(U), withq(0) = 1 and q(U) = D, where D is a convex domain
contained in the right half plane, n ∈ N and λ ≥ 0. Wesay that a function f(z) ∈ A is in the class SL∗n(q) ifDn+1λ f(z)
Dnλf(z)≺ q(z) , z ∈ U .
Remark 3.5.2 Geometric interpretation: f(z) ∈ SL∗n(q)if and only if
Dn+1λ f(z)
Dnλf(z)take all values in the convex do
main D contained in the right halfplane.
95

Definition 3.5.3 [10] Let q(z) ∈ Hu(U), with q(0) =1 and q(U) = D, where D is a convex domain con
tained in the right half plane, n ∈ N and λ ≥ 0. Wesay that a function f(z) ∈ A is in the class SLcn(q) ifDn+2λ f(z)
Dn+1λ f(z)≺ q(z) , z ∈ U .
Remark 3.5.3 Geometric interpretation: f(z) ∈ SLcn(q)if and only if
Dn+2λ f(z)
Dn+1λ f(z)take all values in the convex do
main D contained in the right halfplane.
The main results of this section are obtained in [5].
Definition 3.5.4 Let q(z) ∈ Hu(U), with q(0) = 1,q(U) = D, where D is a convex domain contained in
the right half plane, n ∈ N, λ ≥ 0 and α ∈ [0, 1]. Wesay that a function f(z) ∈ A is in the class MLn,α(q) if
Jn,λ(α, f ; z) = (1− α)Dn+1λ f(z)
Dnλf(z)+ α
Dn+2λ f(z)
Dn+1λ f(z)≺ q(z)
, z ∈ U .
Remark 3.5.4 Geometric interpretation:
f(z) ∈ MLn,α(q) if and only if Jn,λ(α, f : z) take all96

values in the convex domain D contained in the right
halfplane.
Remark 3.5.5 It is easy to observe that if we choose
different function q(z) we obtain variously classes of α
convex functions, such as (for example), for λ = 1 and
n = 0, the class of αconvex functions, the class of α
uniform convex functions with respect to a convex do
main (see the section 3.3), and, for λ = 1, the class
UDn,α(β, γ), β ≥ 0, γ ∈ [−1, 1), β + γ ≥ 0 (see the section 3.2), the class of αnuniformly convex functions
with respect to a convex domain (see the section 3.3).
Remark 3.5.6 We have MLn,0(q) = SL∗n(q) and
MLn,1(q) = SLcn(q).
Remark 3.5.7 For q1(z) ≺ q2(z) we haveMLn,α(q1) ⊂ MLn,α(q2) . From the above we obtainMLn,α(q) ⊂ MLn,α
(1 + z
1− z)
.
Theorem 3.5.1 For all α, α′ ∈ [0, 1], with α < α′, wehave MLn,α′(q) ⊂ MLn,α(q) .
97

Proof. From f(z) ∈ MLn,α′(q) we have
Jn,λ(α′, f ; z) = (1− α′)D
n+1λ f(z)
Dnλf(z)+ α′
Dn+2λ f(z)
Dn+1λ f(z)≺ q(z) ,
(3.7)
where q(z) is univalent in U with q(0) = 1 and maps the
unit disc U into the convex domain D contained in the
right halfplane.
With notation
p(z) =Dn+1λ f(z)
Dnλf(z),
where
p(z) = 1 + p1z + . . . and f(z) = z +∞∑
j=2
ajzj
we have
p(z) + α′ · λ · zp′(z)
p(z)
=Dn+1λ f(z)
Dnλf(z)+ α′λ
Dnλf(z)
Dn+1λ f(z)
·z(Dn+1λ f(z)
)′Dnλf(z)−Dn+1λ f(z) (Dnλf(z))′
(Dnλf(z))2
=Dn+1λ f(z)
Dnλf(z)+ α′λ
Dnλf(z)
Dn+1λ f(z)
(z
(Dn+1λ f(z)
)′Dnλf(z)
98

−Dn+1λ f(z)
Dnλf(z)· z (D
nλf(z))
′
Dnλf(z)
)
=Dn+1λ f(z)
Dnλf(z)+ α′λ
· Dnλf(z)
Dn+1λ f(z)
z
(z +
∞∑j=2
(1 + (j − 1)λ)n+1 ajzj)′
Dnλf(z)
−Dn+1λ f(z)
Dnλf(z)·z
(z +
∞∑
j=2
(1 + (j − 1)λ)n ajzj)′
Dnλf(z)
=Dn+1λ f(z)
Dnλf(z)+ α′λ
· Dnλf(z)
Dn+1λ f(z)
z
(1 +
∞∑j=2
j (1 + (j − 1)λ)n+1 ajzj−1)
Dnλf(z)
99

−Dn+1λ f(z)
Dnλf(z)·z
(1 +
∞∑j=2
j (1 + (j − 1)λ)n ajzj−1)
Dnλf(z)
or
p(z) + α′ · λ · zp′(z)
p(z)=
Dn+1λ f(z)
Dnλf(z)+ α′λ(3.8)
· Dnλf(z)
Dn+1λ f(z)
z +∞∑
j=2
j (1 + (j − 1)λ)n+1 ajzj
Dnλf(z)
−Dn+1λ f(z)
Dnλf(z)·z +
∞∑j=2
j (1 + (j − 1)λ)n ajzj
Dnλf(z)
We have
z +∞∑
j=2
j (1 + (j − 1)λ)n+1 ajzj
= z +∞∑
j=2
((j − 1) + 1) (1 + (j − 1)λ)n+1 ajzj
100

= z +∞∑
j=2
(1 + (j − 1)λ)n+1 ajzj
+∞∑
j=2
(j − 1) (1 + (j − 1)λ)n+1 ajzj
= z + Dn+1λ f(z)− z +∞∑
j=2
(j − 1) (1 + (j − 1)λ)n+1 ajzj
= Dn+1λ f(z) +1
λ
∞∑j=2
((j − 1)λ) (1 + (j − 1)λ)n+1 ajzj
= Dn+1λ f(z)
+1
λ
∞∑j=2
(1 + (j − 1)λ− 1) (1 + (j − 1)λ)n+1 ajzj
= Dn+1λ f(z)−1
λ
∞∑j=2
(1 + (j − 1)λ)n+1 ajzj
+1
λ
∞∑j=2
(1 + (j − 1)λ)n+2 ajzj
= Dn+1λ f(z)−1
λ
(Dn+1λ f(z)− z
)
+1
λ
(Dn+2λ f(z)− z
)
= Dn+1λ f(z)−1
λDn+1λ f(z) +
z
λ+
1
λDn+2λ f(z)−
z
λ
101

=λ− 1
λDn+1λ f(z) +
1
λDn+2λ f(z)
=1
λ
((λ− 1)Dn+1λ f(z) + Dn+2λ f(z)
).
Similarly we have
z +∞∑
j=2
j (1 + (j − 1)λ)n ajzj
=1
λ
((λ− 1)Dnλf(z) + Dn+1λ f(z)
).
From (3.8) we obtain
p(z) + α′ · λ · zp′(z)
p(z)
=Dn+1λ f(z)
Dnλf(z)+ α′λ
Dnλf(z)
Dn+1λ f(z)
1
λ·(
(λ− 1)Dn+1λ f(z)
Dnλf(z)
+Dn+2λ f(z)
Dnλf(z)− D
n+1λ f(z)
Dnλf(z)(λ− 1)−
(Dn+1λ f(z)
Dnλf(z)
)2)
=Dn+1λ f(z)
Dnλf(z)+ α′
Dn+2λ f(z)
Dn+1λ f(z)− α′D
n+1λ f(z)
Dnλf(z)
=Dn+1λ f(z)
Dnλf(z)(1− α′) + α′D
n+2λ f(z)
Dn+1λ f(z)= Jn,λ(α
′, f ; z)
From (3.7) we have
p(z) +zp′(z)1
α′λ· p(z)
≺ q(z) ,
102

with p(0) = q(0), Re q(z) > 0 , z ∈ U , α′ > 0 andλ ≥ 0. In this conditions from Theorem 1.6.1 we obtainp(z) ≺ q(z) or p(z) take all values in D.
If we consider the function g : [0, α′] → C,
g(u) = p(z) + u · λzp′(z)
p(z),
with g(0) = p(z) ∈ D and g(α′) = Jn,λ(α′, f ; z) ∈ D, iteasy to see that
g(α) = p(z) + α · λzp′(z)
p(z)∈ D , 0 ≤ α < α′ .
Thus we have
Jn,λ(α, f ; z) ≺ q(z)
or
f(z) ∈ MLn,α(q) .
From the above theorem we have
Corollarly 3.5.1 For every n ∈ N and α ∈ [0, 1], we103

have
MLn,α(q) ⊂ MLn,0(q) = SL∗n(q) .
Remark 3.5.8 If we consider λ = 1 and n = 0 we
obtain the Theorem 3.3.1 from the section 3.3. Also, for
λ = 1 and n ∈ N, we obtain the Theorem 3.3.3 from thesame section.
Remark 3.5.9 If we consider λ = 1 and D = Dβ,γ (see
the geometric interpretation of the definition 2.2.8) in
the above theorem we obtain the Theorem 3.2.1 from the
section 3.2.
Theorem 3.5.2 Let n ∈ N, α ∈ [0, 1] and λ ≥ 1 . IfF (z) ∈ MLn,α(q) then f(z) = LaF (z) ∈ SL∗n(q), whereLa is the integral operator defined by (2.24).
Proof. From (2.24) we have
(1 + a)F (z) = af(z) + zf ′(z)
104

and, by using the linear operator Dn+1λ and if we consider
f(z) =∞∑
j=2
ajzj, we obtain
(1 + a)Dn+1λ F (z) = aDn+1λ f(z) + D
n+1λ
(z +
∞∑
j=2
jajzj
)
= aDn+1λ f(z) + z +∞∑
j=2
(1 + (j − 1)λ)n+1 jajzj
We have (see the proof of the above theorem)
z +∞∑
j=2
j (1 + (j − 1)λ)n+1 ajzj(3.9)
=1
λ
((λ− 1)Dn+1λ f(z) + Dn+2λ f(z)
)
Thus
(1 + a)Dn+1λ F (z) = aDn+1λ f(z)
+1
λ
((λ− 1)Dn+1λ f(z) + Dn+2λ f(z)
)
=
(a +
λ− 1λ
)Dn+1λ f(z) +
1
λDn+2λ f(z)
or
λ(1+a)Dn+1λ F (z) = ((a + 1)λ− 1) Dn+1λ f(z)+Dn+2λ f(z) .105

Similarly, we obtain
λ(1 + a)DnλF (z) = ((a + 1)λ− 1) Dnλf(z) + Dn+1λ f(z) .
ThenDn+1λ F (z)
DnλF (z)
=
Dn+2λ f(z)
Dn+1λ f(z)· D
n+1λ f(z)
Dnλf(z)+ ((a + 1)λ− 1) · D
n+1λ f(z)
Dnλf(z)
Dn+1λ f(z)
Dnλf(z)+ ((a + 1)λ− 1)
.
With notation
Dn+1λ f(z)
Dnλf(z)= p(z) , p(0) = 1 ,
we obtainDn+1λ F (z)
DnλF (z)(3.10)
=
Dn+2λ f(z)
Dn+1λ f(z)· p(z) + ((a + 1)λ− 1) · p(z)
p(z) + ((a + 1)λ− 1)Also, we obtain
Dn+2λ f(z)
Dn+1λ f(z)=
Dn+2λ f(z)
Dnλf(z)· D
nλf(z)
Dn+1λ f(z)=
1
p(z)· D
n+2λ f(z)
Dnλf(z)(3.11)
106

We have
Dn+2λ f(z)
Dnλf(z)=
z +∞∑
j=2
(1 + (j − 1)λ)n+2 ajzj
z +∞∑
j=2
(1 + (j − 1)λ)n ajzj
and
zp′(z) =z(Dn+1λ f(z)
)′Dnλf(z)
− Dn+1λ f(z)
Dnλf(z)· z (D
nλf(z))
′
Dnλf(z)
=
z
(1 +
∞∑j=2
(1 + (j − 1)λ)n+1 jajzj−1)
Dnλf(z)
−p(z) ·z
(1 +
∞∑
j=2
(1 + (j − 1)λ)n jajzj−1)
Dnλf(z)
or
zp′(z) =
z +∞∑
j=2
j (1 + (j − 1)λ)n+1 ajzj
Dnλf(z)(3.12)
−p(z) ·z +
∞∑
j=2
j (1 + (j − 1)λ)n ajzj
Dnλf(z).
107

By using (3.9) and (3.12) we obtain
zp′(z) =1
λ
((λ− 1)Dn+1λ f(z) + Dn+2λ f(z)
Dnλf(z)
−p(z)(λ− 1)Dnλf(z) + D
n+1λ f(z)
Dnλf(z)
)
=1
λ
((λ− 1)p(z) + D
n+2λ f(z)
Dnλf(z)− p(z) ((λ− 1) + p(z))
)
=1
λ
(Dn+2λ f(z)
Dnλf(z)− p(z)2
)
Thus
λzp′(z) =Dn+2λ f(z)
Dnλf(z)− p(z)2
orDn+2λ f(z)
Dnλf(z)= p(z)2 + λzp′(z) .
From (3.11) we obtain
Dn+2λ f(z)
Dn+1λ f(z)=
1
p(z)
(p(z)2 + λzp′(z)
).
Then, from (3.10), we obtain
Dn+1λ F (z)
DnλF (z)=
p(z)2 + λzp′(z) + ((a + 1)λ− 1) p(z)p(z) + ((a + 1)λ− 1)
= p(z) + λzp′(z)
p(z) + ((a + 1)λ− 1) ,
108

where a ∈ C, Rea ≥ 0 and λ ≥ 1 .If we denote
Dn+1λ F (z)
DnλF (z)= h(z), with h(0) = 1, we
have from F (z) ∈ MLn,α(q) (see the proof of the aboveTheorem):
Jn,λ(α, F ; z) = h(z) + α · λ · zh′(z)
h(z)≺ q(z)
Using the hypothesis, from Theorem 1.6.1, we obtain
h(z) ≺ q(z)
or
p(z) + λzp′(z)
p(z) + ((a + 1)λ− 1) ≺ q(z) .
By using the Theorem 1.6.1 and the hypothesis we
have
p(z) ≺ q(z)
orDn+1λ f(z)
Dnλf(z)≺ q(z) .
This means f(z) = LaF (z) ∈ SL∗n(q) .
Remark 3.5.10 If we consider λ = 1 and n = 0 we
obtain the Theorem 3.3.2 from the section 3.3. Also, for
109

λ = 1 and n ∈ N, we obtain the Theorem 3.3.4 from thesame section.
Remark 3.5.11 If we consider λ = 1 and D = Dβ,γ
(see remark 3.4.6) in the above theorem we obtain the
Theorem 3.2.2 from the section 3.2.
3.6 The subclass MLβ,α(q)
For the main results of this section we will need the
following definitions and theorems:
Definition 3.6.1 [11] Let β, λ ∈ R, β ≥ 0, λ ≥ 0 andf(z) = z+
∞∑j=2
ajzj. We denote by Dβλ the linear operator
defined by
Dβλ : A → A ,
Dβλf(z) = z +∞∑
j=2
(1 + (j − 1)λ)β ajzj .
Definition 3.6.2 [11] Let q(z) ∈ Hu(U), with q(0) = 1and q(U) = D, where D is a convex domain contained
in the right half plane, β, λ ∈ R, β ≥ 0 and λ ≥ 0. We110

say that a function f(z) ∈ A is in the class SL∗β(q) if
Dβ+1λ f(z)
Dβλf(z)≺ q(z) , z ∈ U .
Theorem 3.6.1 [11] Let β, λ ∈ R, β ≥ 0 and λ ≥ 1 . IfF (z) ∈ SL∗β(q) then f(z) = LaF (z) ∈ SL∗β(q), where Lai