functions associated with some hyperbola
Irina Dorca and Daniel V. Breaz
Abstract. In this paper we study several relevant properties of certain subclasses of functions with negative coefficients associated with some hyperbola.
M.S.C. 2010: 30C45.
Key words: β-starlike functions; β-convex functions; hyperbola; Libera-Pascu inte- gral operator; Briot-Bouquet differential subordination; generalized S˘al˘agean opera- tor.
1 Introduction
Let H(U ) be the set of functions which are regular in the unit disk U , A = {f ∈ H(U ) | f (0) = f0(0) − 1 = 0}
and S = {f ∈ A | f is univalent in U }.
In [13], the subfamily T of S is introduced as following:
(1.1) T = {f (z) | f (z) = z − X∞ j=2
ajzj, aj≥ 0, j = 2, 3, ..., z ∈ U } .
We recall here the definitions of the well - known classes of starlike functions and convex functions
S∗=
½
f ∈ A | Rezf0(z)
f (z) > 0 , z ∈ U
¾ ,
Snc(α) =
½
f ∈ A | ReDn+2f (z)
Dn+1f (z) > α, z ∈ U
¾ .
We further consider the Libera-Pascu integral operator, La: A → A :
(1.2) f (z) = LaF (z) = 1 + a za
Zz
0
F (t) · ta−1dt , a ∈ C , Re a ≥ 0.
D
ifferential Geometry - Dynamical Systems, Vol.14, 2012, pp. 46-54.° Balkan Society of Geometers, Geometry Balkan Press 2012.c
For a = 1 we obtain the Libera integral operator, for a = 0 we obtain the Alexan- der integral operator and if a = 1, 2, 3, ... we obtain the Bernardi integral operator.
Mathematicians like P. T. Mocanu (see [11]), E. Drˇaghici (see [7]) and D. Breaz (see [6]) studied and generalized the Libera-Pascu integral operator.
Let Dn: A → A, n ∈ N be the S˘al˘agean differential operator ([12]) defined as:
D0f (z) = f (z), D1f (z) = Df (z) = zf0(z) , Dnf (z) = D(Dn−1f (z)).
Let β, λ ∈ R, β ≥ 0, λ ≥ 0 and f (z) = z + X∞ j=2
ajzj. We consider as well the linear operator Dλβ: A → A defined by ([5])
Dβλf (z) = z + X∞ j=2
[(1 + (j − 1)λ)β]ajzj.
From [3] we have:
Definition 1.1. Let f ∈ T , f (z) = z −P∞
j=2
ajzj, aj ≥ 0, j = 2, 3, ... and z ∈ U .
a) If ReDλβ+1f (z)
Dβλf (z) > α, α ∈ [0, 1), λ ≥ 0, β ≥ 0, z ∈ U , then we say that f is in the class T?Lβ(α).
b) If ReDλβ+2f (z)
Dλβ+1f (z) > α, α ∈ [0, 1), λ ≥ 0, β ≥ 0, z ∈ U , then we say that f is in the class TcLβ(α).
We have the following results:
Theorem 1.1. Let α ∈ [0, 1), λ ≥ 0 and β ≥ 0. The function f ∈ T of the form (1) is in the class T?Lβ(α) iff
(1.3)
X∞ j=2
[(1 + (j − 1)λ)β(1 + (j − 1)λ − α)]aj < 1 − α.
Theorem 1.2 ([4]). Let α ∈ [0, 1), λ ≥ 0 and β ≥ 0. The function f ∈ T of the form (1) is in the class TcLβ(α) iff
(1.4)
X∞ j=2
[(1 + (j − 1)λ)β+1(1 + (j − 1)λ − α)]aj< 1 − α.
From [1] we know the following:
Definition 1.2. Let f ∈ S and α > 0 . If
¯¯
¯¯Dn+1f (z)
Dnf (z) − 2α(√ 2 − 1)
¯¯
¯¯ < Re
½√
2Dn+1f (z) Dnf (z)
¾
+ 2α(√
2 − 1) , z ∈ U, then we say that the function f is in the class SHn(α) , n ∈ N.
Theorem 1.3. If F (z) ∈ SHn(α) , α > 0 , n ∈ N and f (z) = LaF (z) , where La is the integral operator defined by (2), then f (z) ∈ SHn(α) , α > 0 , n ∈ N .
Theorem 1.4. Let n ∈ N and α > 0 . If f (z) ∈ SHn+1(α) , then f (z) ∈ SHn(α) . Theorem 1.5. Let h convex in U and Re[βh(z) + γ] > 0, z ∈ U. If p ∈ H(U ) with p(0) = h(0) and p satisfied the Briot-Bouquet differential subordination
(1.5) p(z) + zp0(z)
βp(z) + γ ≺ h(z) , then p(z) ≺ h(z) .
From [2] we know the following:
Definition 1.3. A function f ∈ A is said to be in the class CV H(α) if it satisfies
¯¯
¯¯
¯ zf00(z)
f0(z) − 2α³√
2 − 1
´ + 1
¯¯
¯¯
¯< Re (√
2zf00(z) f0(z)
)
+ 2α³√
2 − 1
´
for some α (α > 0) and for all z ∈ U .
Definition 1.4. Let f ∈ A and α > 0 . We say that the function f is in the class CV Hn(α) , n ∈ N , if
¯¯
¯¯Dn+2f (z)
Dn+1f (z)− 2α(√ 2 − 1)
¯¯
¯¯ < Re
½√
2Dn+2f (z) Dn+1f (z)
¾
+ 2α(√
2 − 1) , z ∈ U .
Theorem 1.6. If F (z) ∈ CV Hn(α) , α > 0 , n ∈ N and f (z) = LaF (z) , where La
is the integral operator defined by (2), then f (z) ∈ CV Hn(α) , α > 0 , n ∈ N . Theorem 1.7. Let n ∈ N and α > 0 . If f (z) ∈ CV Hn+1(α) , then f (z) ∈ CV Hn(α) .
In [14] we have the following:
Definition 1.5. A function f ∈ S is said to be in the class SH(α) if it satisfies
¯¯
¯¯
¯ zf0(z)
f (z) − 2α³√
2 − 1´¯
¯¯
¯¯< Re (√
2zf0(z) f (z)
)
+ 2α³√
2 − 1´
for some α (α > 0) and for all z ∈ U .
Theorem 1.8. Let f ∈ SH(α) and f (z) = z + b2z2+ b3z3+ ... . Then
(1.6) |b2| ≤1 + 4α
1 + 2α, |b3| ≤ (1 + 4α)(3 + 16α + 24α2) 4(1 + 2α)3 .
The next theorem is the result of the so called ”admissible functions method” due to P.T. Mocanu and S.S. Miller (see [8], [9] , [10]).
2 Results
The purpose of this paper is to define subclasses of β-starlike functions and of β-convex functions with negative coefficients associated with some hyperbola. Estimations for the coefficients of the series expansion and relevant properties of these classes are also developed.
Definition 2.1. Let f ∈ T?Lβ(α), f (z) = z−P∞
j=2
ajzj, aj≥ 0, j ≥ 2, α ∈ [0, 1), λ ≥ 0 and α1> 0. We say that the function f is in the class T?HLβ(α; α1), β ≥ 0, if
¯¯
¯¯
¯
Dβ+1λ f (z)
Dλβf (z) − 2α1· (√ 2 − 1)
¯¯
¯¯
¯< Re (√
2 · Dβ+1λ f (z) Dλβf (z)
)
+ 2α1· (√
2 − 1), z ∈ U.
Remark 2.2. Geometric interpretation : If the functions pα1 are analytic and uni- valent, with the properties pα1(0) = 1, p0α1(0) > 0 and pα1(U ) = Ω(α1) , then f ∈ T?HLβ(α; α1) if and only if Dλβ+1f (z)
Dβλf (z) ≺ pα1(z), where the symbol ” ≺ ” denotes the subordination in U . We have pα1(z) = (1 + 2α1)
r1 + bz
1 − z − 2α1, b = b(α1) = 1 + 4α1− 4α21
(1 + 2α1)2 and the branch of the square root√
w is chosen so that Im√
w ≥ 0. If
we consider pα1(z) = 1 − C1z − ..., we have C1= 1 + 4α1
1 + 2α1.
Theorem 2.1. Let f ∈ T?HLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and f (z) = z − P∞
j=2
ajzj, aj≥ 0, j ≥ 2. Then
|a2| ≤ 1
(1 + λ)β(1 + λ − α)· 1 + 4α1
1 + 2α1,
|a3| ≤ 1
(1 + 2λ)β(1 + 2λ − α) ·(1 + 4α1)(3 + 16α1+ 24α21) 4(1 + 2α1)3 . Proof. If we consider by Dβλf (z) = g(z) , g(z) = z−P∞
j=2
bjzj, we have f ∈ T?HLβ(α; α1) if and only if g ∈ T?HLβ(α; α1). We take into account the series expansions from above and we obtain:
|aj| ≤ 1
[(1 + (j − 1)λ)β(1 + (j − 1)λ − α)]|bj|, j ≥ 2.
Using the estimations from Theorem 1.8, the proof is complete. ¤ Theorem 2.2. If F (z) ∈ T?HLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and f (z) = LaF (z), where Lais the integral operator defined by (1.2), then f (z) ∈ T?HLβ(α; α1).
Proof. By differentiating (1.2), we obtain (1 + a)F (z) = af (z) + zf0(z). We apply the linear operator Dβλ and we obtain
(1 + a)Dβ+1λ F (z) = aDλβ+1f (z) + Dβ+1λ (zf0(z)) or
(1 + a)Dβ+1λ F (z) =
·
a +λ − 1 λ
¸
· Dβ+1λ f (z) + 1
λ· Dβ+2λ f (z) . We apply, in a similar way as before, the linear operator Dβλ and we obtain
(1 + a)DλβF (z) =
·
a +λ − 1 λ
¸
· Dβλf (z) + 1
λ· Dβ+1λ f (z) . Thus,
Dλβ+1F (z) DβλF (z) =
Dβ+2λ f (z) +
·
a +λ − 1 λ
¸
· Dβ+1λ f (z)
Dλβ+1f (z) +
·
a +λ − 1 λ
¸
· Dβλf (z)
(2.1) =
Dλβ+2f (z)
Dλβ+1f (z)·Dβ+1λ f (z)
Dλβf (z) + [λ(a + 1) − 1] ·Dλβ+1f (z) Dβλf (z) Dβ+1λ f (z)
Dλβf (z) + [λ(a + 1) − 1]
.
With the notation Dβ+1λ f (z)
Dλβf (z) = p(z), where p(z) = 1 − p1(z) − ... , we have
zp0(z) = z · Ã
Dβ+1λ f (z) Dλβf (z)
!0
= z(Dβ+1λ f (z))0· Dλβf (z) − Dβ+1λ f (z) · z(Dβλf (z))0 (Dλβf (z))2
=Dβ+2λ f (z) · Dβλf (z) − (Dβ+1λ f (z))2 (Dλβf (z))2
and
1
p(z)· λzp0(z) = Dβ+2λ f (z)
Dβ+1λ f (z)−Dβ+1λ f (z)
Dβλf (z) =Dβ+2λ f (z)
Dβ+1λ f (z) − p(z) . From above, we have that
Dβ+2λ f (z)
Dβ+1λ f (z) = p(z) + 1
p(z)· λzp0(z) .
Thus, from (2.1) we deduce the following
(2.2) Dβ+1λ F (z) DβλF (z) =
p(z) · µ
λzp0(z) ·1 z + p(z)
¶
+ [λ(a + 1) − 1] · p(z) p(z) + [λ(a + 1) − 1]
= p(z) + 1
p(z) + [λ(a + 1) − 1]· λzp0(z) . We have Dβ+1λ F (z)
DλβF (z) ≺ pα1(z) and thus, using (2.2), we notice that
p(z) + 1
p(z) + [λ(a + 1) − 1]· λzp0(z) ≺ pα1(z) .
From the hypothesis, we see that Re(pα1(z) + a) > 0, z ∈ U. Furthermore, from Theorem 1.5, we obtain p(z) ≺ pα1(z) or Dβ+1λ f (z)
Dλβf (z) ≺ pα1(z). This means that
f (z) = LaF (z) ∈ T?HLβ(α; α1). ¤
Theorem 2.3. Let a ∈ C, Re a ≥ 0, α ∈ [0, 1), α1 > 0 and β ≥ 0. If F (z) ∈ T?HLβ(α; α1), F (z) = z − P∞
j=2
ajzj, aj ≥ 0, j ≥ 2, f (z) = LaF (z), f (z) = z − P∞
j=2
bjzj, where La is the integral operator defined by (1.2), λ ≥ 0, β ≥ 0, then
|b2| ≤
¯¯
¯¯a + 1 a + 2
¯¯
¯¯ · 1
(1 + λ)β(1 + λ − α)· 1 + 4α1
1 + 2α1
,
|b3| ≤
¯¯
¯¯a + 1 a + 3
¯¯
¯¯ · 1
(1 + 2λ)β(1 + 2λ − α)·(1 + 4α1)(3 + 16α1+ 24α21) 4(1 + 2α1)3 .
Proof. From f (z) = LaF (z) we have that (1 + a)F (z) = af (z) + zf0(z). Using the series of expansions from above, we obtain that
(1 + a)z − X∞ j=2
(1 + a)ajzj = az − X∞ j=2
(a + j)bjzj+ z
and thus, bj(a + j) = (1 + a)aj, j ≥ 2 . Furthermore, we have that |bj| ≤
¯¯
¯¯a + 1 a + j
¯¯
¯¯ ·
|aj|, j ≥ 2 . Using the estimations from Theorem 2.1, we obtain the needed results. ¤ For a = (j − 1)λ − α, for any j ≥ 2, and if La is the integral operator Libera, we obtain the following result:
Corollary 2.4. Let α ∈ [0, 1), α1> 0 and β ≥ 0. If F (z) ∈ T?HLβ(α; α1), F (z) = z − P∞
j=2
ajzj, aj ≥ 0, j ≥ 2, and f (z) = LF (z), f (z) = z − P∞
j=2
bjzj, where L is the
Libera integral operator defined by
L(F (z)) = 1 + (j − 1)λ − α z(j−1)λ−α
Zz
0
F (t) · t(j−1)λ−α−1dt,
then
|b2| ≤ 1
(1 + λ)β · 1 + 4α1
(1 + 2α1)(2 + λ − α), (j = 2) ,
|b3| ≤ 1
(1 + 2λ)β · (1 + 4α1)(3 + 16α1+ 24α21)
4(1 + 2α1)3(3 + 2λ − α) , (j = 3) .
Theorem 2.5. Let β ≥ 0, α ∈ [0, 1) and α1 > 0. If f (z) ∈ T?HLβ+1(α; α1), then f (z) ∈ T?HLβ(α; α1).
Proof. If Dλβ+1f (z)
Dβλf (z) = p(z), we have (see the proof of Theorem 3.2):
Dλβ+2f (z)
Dλβ+1f (z) = p(z) + 1
p(z)· λzp0(z) .
From f (z) ∈ T?HLβ+1(α; α1) we obtain p(z) + 1
p(z)· λzp0(z) ≺ pα1(z) (see Remark 2.2). By considering the functions pα1(z), we have that Repα1(z) > 0 and from Theorem 1.5 we obtain p(z) ≺ pα1(z) or f (z) ∈ T?HLβ(α; α1) . ¤ Remark 2.3. From the above theorem we obtain that T?HLβ(α; α1) ⊂ T?HL0(α; α1) = T?HL(α; α1) ⊂ T∗L f or β ≥ 0, α ∈ [0, 1), and α1> 0.
Remark 2.4. We obtain similar results for β-convex functions with negative coef- ficients called TcHLβ(α; α1), where β ≥ 0, α ∈ [0, 1), α1 > 0 . In this case we make use of Theorem 2.2 instead of Theorem 2.1 .
Definition 2.5. Let f ∈ TcLβ(α), f (z) = z−P∞
j=2
aizj, aj ≥ 0, j ≥ 2, α ∈ [0, 1), λ ≥ 0 and α1> 0. We say that the function f is in the class TcHLβ(α; α1), β ≥ 0, if
¯¯
¯¯
¯
Dβ+2λ f (z)
Dβ+1λ f (z)− 2α1· (√ 2 − 1)
¯¯
¯¯
¯< Re (√
2 · Dβ+2λ f (z) Dβ+1λ f (z)
)
+ 2α1· (√
2 − 1), z ∈ U.
Theorem 2.6. Let f ∈ TcHLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and f (z) = z − P∞
j=2
ajzj, with aj≥ 0, j ≥ 2. Then,
|a2| ≤ 1
(1 + λ)β+1(1 + λ − α)·1 + 4α1
1 + 2α1,
|a3| ≤ 1
(1 + 2λ)β+1(1 + 2λ − α)· (1 + 4α1)(3 + 16α1+ 24α21) 4(1 + 2α1)3 .
Theorem 2.7. If F (z) ∈ TcHLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and f (z) = LaF (z), where La is the integral operator defined by (1.2), then f (z) ∈ TcHLβ(α; α1).
Theorem 2.8. Let a ∈ C, Re a ≥ 0, α ∈ [0, 1), α1 > 0 and β ≥ 0. If F (z) ∈ TcHLβ(α; α1), F (z) = z − P∞
j=2
ajzj, aj ≥ 0, j ≥ 2, f (z) = LaF (z), f (z) = z − P∞
j=2
bjzj, where La is the integral operator defined by (1.2), λ ≥ 0, β ≥ 0, then
|b2| ≤
¯¯
¯¯a + 1 a + 2
¯¯
¯¯ · 1
(1 + λ)β+1(1 + λ − α)·1 + 4α1
1 + 2α1,
|b3| ≤
¯¯
¯¯a + 1 a + 3
¯¯
¯¯ · 1
(1 + 2λ)β+1(1 + 2λ − α) ·(1 + 4α1)(3 + 16α1+ 24α21) 4(1 + 2α1)3 . Corollary 2.9. Let α ∈ [0, 1), α1 > 0, β ≥ 0. If F (z) ∈ TcHLβ(α; α1), F (z) = z − P∞
j=2
ajzj, aj ≥ 0, j ≥ 2, and f (z) = LF (z), f (z) = z − P∞
j=2
bjzj, where L is the Libera integral operator defined by
L(F (z)) = 1 + (j − 1)λ − α z(j−1)λ−α
Zz
0
F (t) · t(j−1)λ−α−1dt,
then
|b2| ≤ 1
(1 + λ)β+1 · 1 + 4α1
(1 + 2α1)(2 + λ − α), (j = 2) ,
|b3| ≤ 1
(1 + 2λ)β+1 ·(1 + 4α1)(3 + 16α1+ 24α21)
4(1 + 2α1)3(3 + 2λ − α) , (j = 3) .
Theorem 2.10. Let β ≥ 0, α ∈ [0, 1) and α1 > 0. If f (z) ∈ TcHLβ+1(α; α1) then f (z) ∈ TcHLβ(α; α1).
Remark 2.6. From the above theorem we see that TcHLβ(α; α1) ⊂ TcHL0(α; α1) = TcHL(α; α1) ⊂ TcL f or β ≥ 0, α ∈ [0, 1), α1> 0.
Acknowledgment. This work was partially supported by the strategic project POSDRU 107/1.5/S/77265, inside POSDRU Romania 2007-2013 co-financed by the European Social Fund-Investing in People.
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Author’s address:
Irina Dorca University of Pite¸sti, Department of Mathematics, Arge¸s, Romania.
E-mail: [email protected] Daniel V. Breaz
University ”1st December 1918” of Alba Iulia, Department of Mathematics, Alba Iulia, Romania.
E-mail: [email protected]