On neighborhoods of functions associated with conic domains
Nihat Ya˘gmur
Abstract
Letk−ST[A, B], k≥0,−1≤B < A≤1 be the class of normalized analytic functions defined in the open unit disk satisfying
<
(B−1)zff0(z)(z) −(A−1) (B+ 1)zff0(z)(z) −(A+ 1)
> k
(B−1)zff(z)0(z)−(A−1) (B+ 1)zff(z)0(z)−(A+ 1)−1
.
and letk−U CV[A, B], k≥0,−1≤B < A≤1 be the corresponding class satisfying
<
(B−1)(zf0(z))0
f0(z) −(A−1) (B+ 1)(zff00(z)(z))0 −(A+ 1)
> k
(B−1)(zf0(z))0
f0(z) −(A−1) (B+ 1)(zff00(z)(z))0 −(A+ 1) −1
.
For an appropriate δ > 0, the δ neighborhood of a function f ∈ k− U CV[A, B] is shown to consist of functions in the classk−ST[A, B].
1 Introduction
LetAdenote the family of functionsf of the form f(z) =z+
∞
X
n=2
anzn (1)
Key Words: Analytic function,k−starlike functions,k−uniformly convex functions.
2010 Mathematics Subject Classification: 30C45.
Received: 8 May, 2014.
Accepted: 30 June, 2014.
291
which are analytic in the open unit disk U = {z:|z|<1}. The classes S∗ and C are the well-known classes of starlike and convex univalent functions respectively; for details, see[4].
Noor and Malik introduced and studied the classk−U CV[A, B] and the corresponding classk−ST[A, B] in [10] as following:
A function f(z)∈Ais said to be in the classk−ST[A, B], k ≥0,−1≤ B < A≤1,if and only if,
<
(B−1)zff(z)0(z)−(A−1) (B+ 1)zff(z)0(z)−(A+ 1)
> k
(B−1)zff(z)0(z)−(A−1) (B+ 1)zff(z)0(z)−(A+ 1)−1
. (2)
A function f(z)∈ Ais said to be in the classk−U CV[A, B], k≥0,−1 ≤ B < A≤1,if and only if,
<
(B−1)(zf0(z))0
f0(z) −(A−1) (B+ 1)(zff00(z)(z))0 −(A+ 1)
> k
(B−1)(zf0(z))0
f0(z) −(A−1) (B+ 1)(zff00(z)(z))0 −(A+ 1) −1
. (3)
It can be easily seen that
f(z)∈k−U CV[A, B]⇐⇒zf0(z)∈k−ST[A, B]. (4) Special cases.
i. k−ST[1,−1] =k−ST, k−U CV[1,−1] =k−U CV,the well-known classes of k−starlike and k−uniformly convex functions respectively, intro- duced by Kanas and Wisniowska [6, 7].
ii. k−ST[1−2α,−1] =SD(k, α), k−U CV[1−2α,−1] =KD(k, α),the classes introduced by Shams et al. in [17].
iii. 0−ST[A, B] = S∗[A, B], 0−U CV[A, B] = C[A, B], the well-known classes of Janowski starlike and Janowski convex functions respectively, intro- duced by Janowski [5].
Geometrically, if a functionf(z)∈k−ST[A, B] then (B−1)
zf0(z) f(z) −(A−1) (B+1)zff(z)0(z)−(A+1) = wtakes all values from the domain Ωk, k≥0, as
Ωk={w:<w > k|w−1|} (5) or equivalently
Ωk=n
u+iv:u > kp
(u−1)2+v2o
. (6)
The domain Ωk represents the right half plane for k= 0, a hyperbola for 0< k <1,a parabola fork= 1 and an ellipse fork >1, for more details see [10].
Given δ ≥ 0, Ruscheweyh [16] defined the δ−neighborhood Nδ(f) of a functionf(z)∈Aby
Nδ(f) = (
g(z) :g(z) =z+
∞
X
n=2
bnzn,and
∞
X
n=2
n|an−bn| ≤δ )
. (7) Ruscheweyh [16] proved among other results thatN1/4(f)⊂S∗forf ∈C.
Sheil-Small and Silvia [18] introduced more general notions of neighborhood of an analytic function. These included noncoefficient neighbourhoods as well.
Problems related to the neighborhoods and some other properties of analytic functions were considered by many authors, for example, see [1, 2, 3, 8, 9, 11, 12, 13, 14].
In this paper, the neighborhoodNδ(f) for functionsf(z)∈k−U CV[A, B]
is investigated. It is shown that all functions g ∈ Nδ(f) are in the class k−ST[A, B] for certainδ >0.
2 Main Results
In order to obtain the main results, a characterization of the classk−ST[A, B]
in terms of the functions in another classk−ST0[A, B] is needed. Forz∈U, a functionHtis said to be in the classk−ST0[A, B] if the functionHtis of the form
Ht(z) = 1
(B−A)
1−kt± q
t2−(kt−1)2i
×
z
B−1−(B+ 1)
kt± q
t2−(kt−1)2i
(1−z)2
− z
A−1−(A+ 1) (kt± q
t2−(kt−1)2i)
1−z
(8)
where
t≥k+11 , if 0≤k≤1,
1
k+1 ≤ t≤k−11 , if k >1. (9)
Recall that for any two functionsf(z) andg(z) given by f(z) =z+
∞
X
n=2
anzn, g(z) =z+
∞
X
n=2
bnzn
the Hadamard product (or convolution) off andg is defined by (f∗g)(z) =z+
∞
X
n=2
anbnzn=(g∗f)(z) (10) Lemma 2.1. A functionf is in the classk−ST[A, B]if and only if
1
z(f∗Ht)(z)6= 0, z∈U (11) for allHt∈k−ST0[A, B].
Proof. Let f ∈ k − ST[A, B]. Then the image of U under
w = u + iv = (B−1)
zf0(z) f(z) −(A−1)
(B+1)zff(z)0(z)−(A+1) lies in the conic type regions Ωk =n
u+iv:u > kp
(u−1)2+v2o
so that forz∈U (B−1)zff(z)0(z)−(A−1)
(B+ 1)zff(z)0(z)−(A+ 1) 6=kt± q
t2−(kt−1)2i (12) where
t≥k+11 , if 0≤k≤1,
1
k+1 ≤ t≤k−11 , if k >1. (13) Thusf ∈k−ST[A, B] if and only if
B−1−(B+ 1) (kt± q
t2−(kt−1)2i)
zf0(z)
−
A−1−(A+ 1) (kt± q
t2−(kt−1)2i)
f(z)
/
z(B−A)
1−kt± q
t2−(kt−1)2i
6= 0 (14) or equivalently
1
z(f ∗Ht)(z)6= 0, z∈U for allHt∈k−ST0[A, B].
Lemma 2.2. If
Ht(z) =z+
∞
X
n=2
hn(t)zn ∈k−ST0[A, B], z∈U (15) then
|hn(t)|< 2n(t+ 1)
|B−A|t (16)
wheret is given in (13).
Proof. Writing Ht(z) =z+
∞
P
n=2
hn(t)zn,and comparing coefficients ofzn in (8) we obtain
hn(t) = [n(B−1)−(A−1)−(n(B+ 1)−(A+ 1))kt
±(n(B+ 1)−(A+ 1)) q
t2−(kt−1)2i]
/
(B−A)
1−kt± q
t2−(kt−1)2i
(17) Thus
|hn(t)|2 = [|n(B−1)−(A−1)−(n(B+ 1)−(A+ 1))kt
±(n(B+ 1)−(A+ 1)) q
t2−(kt−1)2i 2] /
(B−A)
1−kt± q
t2−(kt−1)2i
2
= h
{n(B−1)−(A−1)−[n(B+ 1)−(A+ 1)]kt}2 +(n(B+ 1)−(A+ 1))2
t2−(kt−1)2i /h
(B−A)2
(1−kt)2+t2−(kt−1)2i
= 4(n−1) [(nB−A) (kt−1) + (n−1)kt] + [nB−A+n−1]2t2
(B−A)2t2 .
If−1≤B < A≤1, n≥2 thennB−A < n−1,and from (13),kt−1< t.
So
|hn(t)|2≤ 4(n−1)2(t+ 1)2
(B−A)2t2 < 4n2(t+ 1)2
(B−A)2t2 (18)
and
|hn(t)|<2n(t+ 1)
|B−A|t.
Lemma 2.3. For a functionf ∈Aandε∈C,define the functionFε by
Fε=f(z) +εz
1 +ε . (19)
If for every ε, |ε| < δ we have Fε ∈ k−ST[A, B], then for every Ht ∈ k−ST0[A, B]
1
z(f∗Ht)(z)
≥δ, z∈U. (20) Proof. If Fε ∈ k−ST[A, B] for every ε, |ε| < δ, then by Lemma 2.1 for all Ht∈k−ST0[A, B]
1
z(Fε∗Ht)(z)6= 0, z∈U or equivalently
(f∗Ht)(z) +εz
(1 +ε)z 6= 0. (21)
Since|ε|< δ,it is easily follows that 1
z(f ∗Ht)(z)
≥δ.
Theorem 2.1. Let f ∈A,ε∈C andδ >0. If for every ε,|ε|< δ,we have Fε∈k−ST[A, B] then
Nδ0(f)⊂k−ST[A, B] (22)
for
δ0 =|B−A|t
2(t+ 1) δ (23)
where−1≤B < A≤1 and
t≥k+11 , if 0≤k≤1,
1
k+1 ≤ t≤k−11 , if k >1.
Proof. Letg(z) =z+
∞
P
n=2
bnzn∈Nδ0(f).For anyHt∈k−ST0[A, B],
1
z(g∗Ht)(z)
= 1
z(f ∗Ht)(z) +1
z((g−f)∗Ht) (z)
≥ 1
z(f ∗Ht)(z)
− 1
z((g−f)∗Ht) (z)
From lemma 2.3, 1
z(g∗Ht)(z)
≥ δ−
∞
X
n=2
(bn−an)hn(t)zn z
≥ δ−
∞
X
n=2
|(bn−an)| |hn(t)|
≥ δ− 2(t+ 1)
|B−A|t
∞
X
n=2
n|(bn−an)|.
Using Lemma 2.2 and noting thatg∈Nδ0(f) and whence
∞
P
n=2
n|(bn−an)|< δ0,
thus
1
z(g∗Ht)(z)
≥δ− 2(t+ 1)
|B−A|tδ0= 0 Therefore
1z(g∗Ht)(z)
6= 0 inUfor allHt∈k−ST0[A, B] if δ0 =|B−A|t
2(t+ 1) δ
By Lemma 2.1g∈k−ST[A, B].This proves thatNδ0(f)⊂k−ST[A, B].
Lemma 2.4. ([15])Ifφis a convex univalent function withφ(0) = 0 =φ0(0)−
1 in Uand g is starlike univalent in U, then for each analytic function F in U,
(φ∗F g)(z)
(φ∗g)(z) ⊂coF(U), (z∈U) (24) wherecostands for the closed convex hull.
Lemma 2.5. If φ(z)∈C, f(z)∈k−ST[A, B] andg(z) = (B+ 1)zf0(z)− (A+ 1)f(z)∈S∗ then(φ∗f)(z)∈k−ST[A, B].
Proof. AssumeF(z) = (B−1)
zf0(z) f(z) −(A−1)
(B+1)zff(z)0(z)−(A+1).Then forG(z) = (φ∗f)(z), we have zG0(z) =φ(z)∗zf0(z).
Hence
(B−1)zGG(z)0(z)−(A−1) (B+ 1)zGG(z)0(z)−(A+ 1) =
(B−1)φ(z)∗zf(φ∗f)(z)0(z)−(A−1) (B+ 1)φ(z)∗zf(φ∗f)(z)0(z)−(A+ 1)
= φ(z)∗[(B−1)zf0(z)−(A−1)f(z)]
φ(z)∗[(B+ 1)zf0(z)−(A+ 1)f(z)]
=
φ(z)∗
(B−1)zff(z)0(z)−(A−1) (B+1)zff(z)0(z)−(A+1)
[(B+ 1)zf0(z)−(A+ 1)f(z)]
φ(z)∗[(B+ 1)zf0(z)−(A+ 1)f(z)]
= φ(z)∗(F g)(z)
φ(z)∗g(z) . (25)
By Lemma 2.4 the image ofUunder φ(z)∗(F g)(z)
φ(z)∗g(z) is a subset of the convex hull ofF(U).That means (φ∗f)(z)∈k−ST[A, B].
Theorem 2.2. Iff ∈k−U CV[A, B]andg(z) = (B+1)zf0(z)−(A+1)f(z)∈ C thenFε= f(z)+εz1+ε ∈k−ST[A, B]for|ε|< 14.
Proof. Letf(z) =z+
∞
P
n=2
anzn∈k−U CV[A, B].Then Fε = f(z) +εz
1 +ε
=
z(1 +ε) +
∞
P
n=2
anzn 1 +ε
=
f(z)∗
z(1 +ε) +
∞
P
n=2
zn
1 +ε
= f(z)∗
z−1+εε z2
1−z =f(z)∗h(z) whereh(z) =z−
ε 1+εz2 1−z .Now, zh0(z)
h(z) = 1−2ρz+ρz2
(1−ρz) (1−z), where ρ= ε 1 +ε. Hence|ρ|= 1−|ε||ε| <13 gives|ε|< 14. Thus
<
zh0(z) h(z)
≥ 1−2|ρ| |z| − |ρ| |z|2
(1− |ρ| |z|) (1 +|z|) >0. (26)
This inequality holds for all |ρ|< 13 which is true for |ε|< 14. Thereforehis starlike inUand so
z
Z
0
h(t)
t dt=z+
∞
X
n=2
cn
nzn=h(z)∗log 1
1−z
(27)
is convex for|ε|< 14
(f∗h) (z) =zf0(z)∗
h(z)∗log 1
1−z
(28)
f(z)∈k−U CV[A, B] =⇒zf0(z)∈k−ST[A, B] and h(z)∗log 1
1−z
∈C.
Alsog(z)∈C implies thatzg0(z)∈S∗.By Lemma 2.5, we have zf0(z)∗
h(z)∗log 1
1−z
∈k−ST[A, B].
Thus
(f∗h) (z) =f(z) +εz
1 +ε ∈k−ST[A, B].
Theorem 2.3. Let f ∈k−U CV[A, B] then
Nδ0(f)⊂k−ST[A, B] (29)
for
δ0= |B−A|t
8(t+ 1) (30)
where−1≤B < A≤1 and
t≥k+11 , if 0≤k≤1,
1
k+1 ≤ t≤k−11 , if k >1.
Proof. Let f ∈ k−U CV[A, B] then we get the result by taking δ = 14 in Theorem 2.1 and using Theorem 2.2.
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Nihat Ya˘gmur,
Department of Mathematics, Faculty of Science and Art, Erzincan University, 24000, Erzincan, Turkey.
Email: [email protected]