Volume 10 (2009), Issue 3, Article 83, 9 pp.
CERTAIN SUBCLASSES OF p-VALENTLY CLOSE-TO-CONVEX FUNCTIONS
OH SANG KWON
D EPARTMENT OF M ATHEMATICS
K YUNGSUNG U NIVERSITY
B USAN 608-736, KOREA
oskwon@ks.ac.kr
Received 28 May, 2007; accepted 14 October, 2007
Communicated by G. Kohr
A BSTRACT. The object of the present paper is to drive some properties of certain class Kn,p (A, B)
of multivalent analytic functions in the open unit disk E.
Key words and phrases: p-valently starlike functions of order α, p-valently close-to-convex functions of order α, subordination, hypergeometric series.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let Ap be the class of functions of the form
(1.1)
p
f (z) = z +
∞
X
ap+k z p+k
k=1
which are analytic in the open unit disk E = {z ∈ C : |z| < 1}. A function f ∈ Ap is said to
be p-valently starlike of order α of it satisfies the condition
0 zf (z)
Re
> α (0 ≤ α < p, z ∈ E).
f (z)
We denote by Sp∗ (α).
On the other hand, a function f ∈ Ap is said to be p-valently close-to-convex functions of
order α if it satisfies the condition
0 zf (z)
Re
> α (0 ≤ α < p, z ∈ E),
g(z)
for some starlike function g(z). We denote by Cp (α).
This research was supported by Kyungsung University Research Grants in 2006.
174-07
2
O H S ANG K WON
For f ∈ Ap given by (1.1), the generalized Bernardi integral operator Fc is defined by
Z
c+p z
Fc (z) =
f (t)tc−1 dt
zc 0
∞
X
c+p
p
(1.2)
=z +
ap+k z p+k (c + p > 0, z ∈ E).
c+p+k
k=1
For an analytic function g, defined in E by
g(z) = z p +
∞
X
bp+k z p+k ,
k=1
η
Flett [3] defined the multiplier transform I for a real number η by
∞
X
η
I g(z) =
(p + k + 1)−η bp+k z p+k (z ∈ E).
k=0
η
Clearly, the function I g is analytic in E and
I η (I µ g(z)) = I η+µ g(z)
for all real numbers η and µ.
For any integer n, J. Patel and P. Sahoo [5] also defined the operator Dn , for an analytic
function f given by (1.1), by
−n
∞ X
p+k+1
n
p
ap+k z p+k
D f (z) = z +
1
+
p
k=1
"
#
−n
∞ X
k
+
1
+
p
(1.3)
= f (z) ∗ z p−1 z +
z k+1
(z ∈ E),
1
+
p
k=1
where ∗ stands for the Hadamard product or convolution.
It follows from (1.3) that
z(Dn f (z))0n−1 f (z) − Dn f (z).
(1.4)
We also have
zf 0 (z) + f (z)
.
p+1
If f and g are analytic functions in E, then we say that f is subordinate to g, written f < g
or f (z) < g(z), if there is a function w analytic in E, with w(0) = 0, |w(z)| < 1 for z ∈ E,
such that f (z) = g(w(z)), for z ∈ U . If g is univalent then f < g if and only if f (0) = g(0)
and f (E) ⊂ g(E).
Making use of the operator notation Dn , we introduce a subclass of Ap as follows:
D0 f (z) = f (z) and D−1 f (z) =
Definition 1.1. For any integer n and −1 ≤ B < A ≤ 1, a function f ∈ Ap is said to be in the
class Kn,p (A, B) if
z(Dn f (z))0
p(1 + Az)
<
,
p
z
1 + Bz
(1.5)
where < denotes subordination.
For convenience, we write
Kn,p
2α
1−
, −1 = Kn,p (α),
p
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
http://jipam.vu.edu.au/
C ERTAIN S UBCLASSES OF p-VALENTLY C LOSE - TO - CONVEX F UNCTIONS
3
where Kn,p (α) denote the class of functions f ∈ Ap satisfying the inequality
z(Dn f (z))0
> α (0 ≤ α < p, z ∈ E).
Re
zp
We also note that K0,p (α) ≡ Cp (α) is the class of p-valently close-to-convex functions of
order α.
In this present paper, we derive some properties of a certain class Kn,p (A, B) by using differential subordination.
2. P RELIMINARIES AND M AIN R ESULTS
In our present investigation of the general class Kn,p (A, B), we shall require the following
lemmas.
Lemma 2.1 ([4]). If the function p(z) = 1 + c1 z + c2 z 2 + · · · is analytic in E, h(z) is convex
in E with h(0) = 1, and γ is complex number such that Re γ > 0. Then the Briot-Bouquet
differential subordination
zp0 (z)
< h(z)
p(z) +
γ
implies
Z z
γ
tγ−1 h(t)dt < h(z) (z ∈ E)
p(z) < q(z) = γ
z 0
and q(z) is the best dominant.
For complex numbers a, b and c 6= 0, −1, −2, . . . , the hypergeometric series
(2.1)
2 F1 (a, b; c; z)
=1+
ab
a(a + 1)b(b + 1) 2
z+
z + ···
c
2!c(c + 1)
represents an analytic function in E. It is well known by [1] that
Lemma 2.2. Let a, b and c be real c 6= 0, −1, −2, . . . and c > b > 0. Then
Z 1
Γ(b)Γ(c − b)
tb−1 (1 − t)c−b−1 (1 − tz)−a dt =
2 F1 (a, b; c; z),
Γ(c)
0
z
−a
(2.2)
2 F1 (a, b; c; z) = (1 − z) 2 F1 a, c − b; c;
z−1
and
(2.3)
2 F1 (a, b; c; z)
= 2 F1 (b, a; c; z).
Lemma 2.3 ([6]). Let φ(z) be convex and g(z) is starlike in E. Then for F analytic in E with
g
F (0) = 1, φ∗F
(E) is contained in the convex hull of F (E).
φ∗g
Lemma 2.4 ([2]). Let φ(z) = 1 +
∞
P
ck z k and φ(z) <
k=1
1+Az
.
1+Bz
Then
|ck | ≤ (A − B).
Theorem 2.5. Let n be any integer and −1 ≤ B < A ≤ 1. If f ∈ Kn,p (A, B), then
(2.4)
z(Dn+1 f (z))0
p(1 + Az)
<
q(z)
<
zp
1 + Bz
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
(z ∈ E),
http://jipam.vu.edu.au/
4
O H S ANG K WON
where
(2.5)


2 F1 (1, p + 1; p + 2; −Bz)



p+1
+ p+2
Az 2 F1 (1, p + 2; p + 3; −Bz), B 6= 0;
q(z) =



 1 + p+1 Az,
B = 0,
p+2
and q(z) is the best dominant of (2.4). Furthermore, f ∈ Kn+1,p (ρ(p, A, B)), where

p2 F1 (1, p + 1; p + 2; B)




− p(p+1)
A2 F1 (1, p + 2; p + 3; B), B 6= 0;
p+2
(2.6)
ρ(p, A, B) =




1 − p+1
A,
B = 0.
p+2
Proof. Let
p(z) =
(2.7)
z(Dn+1 f (z))0
,
pz p
where p(z) is analytic function with p(0) = 1.
Using the identity (1.4) in (2.7) and differentiating the resulting equation, we get
(2.8)
z(Dn f (z))0
zp0 (z)
1 + Az
=
p(z)
+
<
(≡ h(z)).
p
pz
p+1
1 + Bz
Thus, by using Lemma 2.1 (for γ = p + 1), we deduce that
Z z p
t (1 + At)
−(p+1)
p(z) < (p + 1)z
dt(≡ q(z))
1 + Bt
0
Z 1 p
s (1 + Asz)
= (p + 1)
ds
1 + Bsz
0
Z 1
Z 1
sp+1
sp
(2.9)
= (p + 1)
ds + (p + 1)Az
ds.
0 1 + Bsz
0 1 + Bsz
By using (2.2) in (2.9), we obtain


2 F1 (1, p + 1; p + 2; −Bz)



+ p+1
Az 2 F1 (1, p + 2; p + 3; −Bz), B 6= 0;
p+2
p(z) < q(z) =



 1 + p+1 Az,
B = 0.
p+2
Thus, this proves (2.5).
Now, we show that
(2.10)
Re q(z) ≥ q(−r)
(|z| = r < 1).
Since −1 ≤ B < A ≤ 1, the function (1 + Az)/(1 + Bz) is convex(univalent) in E and
1 + Az
1 − Ar
Re
≥
> 0 (|z| = r < 1).
1 + Bz
1 − Br
Setting
g(s.z) =
1 + Asz
1 + Bsz
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
(0 ≤ s ≤ 1,
z ∈ E)
http://jipam.vu.edu.au/
C ERTAIN S UBCLASSES OF p-VALENTLY C LOSE - TO - CONVEX F UNCTIONS
5
and dµ(s) = (p + 1)sp ds, which is a positive measure on [0, 1], we obtain from (2.9) that
Z 1
q(z) =
g(s, z)dµ(s) (z ∈ E).
0
Therefore, we have
1
Z
Z
1
Re g(s, z)dµ(s) ≥
Re q(z) =
0
0
1 − Asr
dµ(s)
1 − Bsr
which proves the inequality (2.10).
Now, using (2.10) in (2.9) and letting r → 1− , we obtain
z(Dn+1 f (z))0
Re
> ρ(p, A, B),
zp
where

p2 F1 (1, p + 1; p + 2; B)




A2 F1 (1, p + 2; p + 3; B), B 6= 0
− p(p+1)
p+2
ρ(p, A, B) =




p − p(p+1)
A,
B = 0.
p+2
This proves the assertion of Theorem 2.5. The result is best possible because of the best dominant property of q(z).
Putting A = 1 −
2α
p
and B = −1 in Theorem 2.5, we have the following:
Corollary 2.6. For any integer n and 0 ≤ α < p, we have
Kn,p (α) ⊂ Kn+1,p (ρ(p, α)),
where
(2.11)
ρ(p, α) = p · 2 F1 (1, p + 1; p + 2; −1) −
p(p + 1)
(1 − 2α)2 F1 (1, p + 2; p + 3; −1).
p+2
The result is best possible.
Taking p = 1 in Corollary 2.6, we have the following:
Corollary 2.7. For any integer n and 0 ≤ α < 1, we have
Kn (δ) ⊂ Kn+1 (δ(α)),
where
(2.12)
δ(α) = 1 + 4(1 − 2α)
∞
X
k=1
1
(−1)k .
k+2
Theorem 2.8. For any integer √
n and 0 ≤ α < p, if f (z) ∈ Kn+1,p (α), then f ∈ Kn,p (α) for
|z| < R(p), where R(p) =
−1+
1+(p+1)2
.
p+1
The result is best possible.
Proof. Since f (z) ∈ Kn+1,p (α), we have
z(Dn+1 f (z))0
= α + (p − α)w(z), (0 ≤ α < p),
zp
where w(z) = 1 + w1 z + w2 z + · · · is analytic and has a positive real part in E. Making use of
logarithmic differentiation and using identity (1.4) in (2.13), we get
z(Dn f (z))0
zw0 (z)
(2.14)
− α = (p − α) w(z) +
.
zp
p+1
(2.13)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
http://jipam.vu.edu.au/
6
O H S ANG K WON
Now, using the well-known (by [5])
|zw0 (z)|
2r
1−r
≤
and
Re
w(z)
≥
(|z| = r < 1),
Re w(z)
1 − r2
1+r
in (2.14), we get
1 Re zw0 (z)
z(Dn f (z))0
Re
− α = (p − α) Re w(z) 1 +
zp
p + 1 Re w(z)
1 |zw0 (z)|
≥ (p − α) Re w(z) 1 −
p + 1 Re w(z)
1−r
1
2r
≥ (p − α)
1−
.
1+r
p + 1 1 − r2
It is√easily seen that the right-hand side of the above expression is positive if |z| < R(p) =
−1+ 1+(p+1)2
. Hence f ∈ Kn,p (α) for |z| < R(p).
p+1
To show that the bound R(p) is best possible, we consider the function f ∈ Ap defined by
1−z
z(Dn+1 f (z))0
= α + (p − α)
p
z
1+z
(z ∈ E).
Noting that
z(Dn f (z))0
1−z
1
−2z
− α = (p − α) ·
1+
zp
1+z
p + 1 (p + 1)(1 − z 2 )
1 − z (p + 1) − (p + 1)z 2 − 2z
= (p − α) ·
1+z
(p + 1) − (p + 1)z 2
=0
for z =
−1+
√
1+(p+1)2
,
p+1
we complete the proof of Theorem 2.8.
Putting n = −1, p = 1 and 0 ≤ α < 1 in Theorem 2.8, we have the following:
Corollary 2.9. If Re f 0 (z) > α, then Re{zf 00 (z) + 2f 0 (z)} > α for |z| <
√
−1+ 5
.
2
Theorem 2.10.
(a) If f ∈ Kn,p (A, B), then the function Fc defined by (1.2) belongs to Kn,p (A, B).
(b) f ∈ Kn,p (A, B) implies that Fc ∈ Kn,p (η(p, , c, A, B)) where

p2 F1 (1, p + c; p + c + 1; B)




− p(p+c)
A F (1, p + c + 1; p + c + 2; B), B 6= 0
η(p, c, A, B) =
p+c+1 2 1



 p − p(p+c) A,
B = 0.
p+c+1
Proof. Let
z(Dn Fc (z))0
,
pz p
where φ(z) is an analytic function with φ(0) = 1. Using the identity
(2.15)
(2.16)
φ(z) =
z(Dn Fc (z))0n f (z) − cDn Fc (z)
in (2.15) and differentiating the resulting equation, we get
z(Dn f (z))0
zφ0 (z)
=
φ(z)
+
.
pz p
p+c
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
http://jipam.vu.edu.au/
C ERTAIN S UBCLASSES OF p-VALENTLY C LOSE - TO - CONVEX F UNCTIONS
7
Since f ∈ Kn,p (A, B),
zφ0 (z)
1 + Az
<
.
p+c
1 + Bz
By Lemma 2.1, we obtain Fc (z) ∈ Kn,p (A, B). We deduce that
1 + Az
,
(2.17)
φ(z) < q(z) <
1 + Bz
where q(z) is given by (2.5) and is the best dominant of (2.17).
This proves part (a) of the theorem. Proceeding as in Theorem 2.10, part (b) follows.
φ(z) +
Putting A = 1 −
2α
p
and B = −1 in Theorem 2.8, we have the following:
Corollary 2.11. If f ∈ Kn,p (A, B) for 0 ≤ α < p, then Fc ∈ Kn,p H(p, c, α), where
H(p, c, α) = p · 2 F1 (1, p + c; p + c + 1; −1)
p+c
−
(p − 2α)2 F1 (1, p + c; p + c + 1; −1).
p+c+1
Setting c = p = 1 in Theorem 2.10, we get the following result.
Corollary 2.12. If f ∈ Kn,p (α) for 0 ≤ α < 1, then the function
Z
2 z
G(z) =
f (t)dt
z 0
belongs to the class Kn (δ(α)), where δ(α) is given by (2.12).
Theorem 2.13. For any integer n and 0 ≤ α < p and c > −p, if F
(α) then the
√c ∈ Kn,p
2
−1+ 1+(p+c)
function f defined by (1.1) belongs to Kn,p (α) for |z| < R(p, c) =
. The result is
p+c
best possible.
Proof. Since Fc ∈ Kn,p (α), we write
z(Dn Fc )0
= α + (p − α)w(z),
zp
where w(z) is analytic, w(0) = 1 and Re w(z) > 0 in E. Using (2.16) in (2.18) and differentiating the resulting equation, we obtain
z(Dn f (z))0
zw0 (z)
(2.19)
Re
− α = (p − α) Re w(z) +
.
zp
p+c
Now, by following the line of proof of Theorem 2.8, we get the assertion of Theorem 2.13. (2.18)
Theorem 2.14. Let f ∈ Kn,p (A, B) and φ(z) ∈ Ap convex in E. Then
(f ∗ φ(z))(z) ∈ Kn,p (A, B).
Proof. Since f (z) ∈ Kn,p (A, B),
z(Dn f (z))0
1 + Az
<
.
p
pz
1 + Bz
Now
z(Dn (f ∗ φ)(z))0
φ(z) ∗ z(Dn f )0
=
pz p ∗ φ(z)
φ(z) ∗ pz p
(2.20)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
=
z(Dn f (z))0
pz p
pz p
.
φ(z) ∗ pz p
φ(z) ∗
http://jipam.vu.edu.au/
8
O H S ANG K WON
Then applying Lemma 2.3, we deduce that
z(Dn f (z))0
pz p
pz p
φ(z) ∗ pz p
φ(z) ∗
<
1 + Az
.
1 + Bz
Hence (f ∗ φ(z))(z) ∈ Kn,p (A, B).
Theorem 2.15. Let a function f (z) defined by (1.1) be in the class Kn,p (A, B). Then
|ap+k | ≤
(2.21)
p(A − B)(p + k + 1)n
(1 + p)n (p + k)
for
k = 1, 2, . . . .
The result is sharp.
Proof. Since f (z) ∈ Kn,p (A, B), we have
z(Dn f (z))0
1 + Az
≡ φ(z) and φ(z) <
.
p
pz
1 + Bz
Hence
0p
n
z(D f (z)) φ(z) and φ(z) = 1 +
(2.22)
∞
X
ck z k .
k=1
From (2.22), we have
!0
n
∞ X
1
+
p
z(Dn f (z))0 = z z p +
ap+k z p+k
p
+
k
+
1
k=1
n
∞ X
1+p
p
= pz +
(p + k)ap+k z p+k
p
+
k
+
1
k=1
!
∞
X
= pz p 1 +
ck z k .
k=1
Therefore
(2.23)
By using Lemma 2.4 in (2.23),
1+p
p+k+1
1+p
p+k+1
n
n
(p + k)ap+k = pck .
(p + k)|ap+k |
p
= |ck | ≤ A − B.
Hence
p(A − B)(p + k + 1)n
.
(1 + p)n (p + k)
The equality sign in (2.21) holds for the function f given by
|ap+k | ≤
(2.24)
Hence
pz p−1 + p(A − B − 1)z p
(D f (z)) =
.
1−z
n
0
z(Dn f (z))0
1 + (A − B − 1)z
1 + Az
=
<
p
pz
1−z
1 + Bz
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
for k = 1, 2, . . . .
http://jipam.vu.edu.au/
C ERTAIN S UBCLASSES OF p-VALENTLY C LOSE - TO - CONVEX F UNCTIONS
9
The function f (z) defined in (2.24) has the power series representation in E,
∞
X
p(A − B)(p + k + 1)n p+k
p
f (z) = z +
z .
n (p + k)
(1
+
p)
k=1
R EFERENCES
[1] M. ABRAMOWITZ AND I.A. STEGUN, Hand Book of Mathematical Functions, Dover Publ. Inc.,
New York, (1971).
[2] V. ANH, k-fold symmetric starlike univalent function, Bull. Austrial Math. Soc., 32 (1985), 419–436.
[3] T.M. FLETT, The dual of an inequality of Hardy and Littlewood and some related inequalities, J.
Math. Anal. Appl., 38 (1972), 746–765.
[4] S.S. MILLER AND P.T. MOCANU, Differential subordinations and univalent functions, Michigan
Math. J., 28 (1981), 157–171.
[5] J. PATEL AND P. SAHOO, Certain subclasses of multivalent analytic functions, Indian J. Pure. Appl.
Math., 34(3) (2003), 487–500.
[6] St. RUSCHEWEYH AND T. SHEIL-SMALL, Hadamard products of schlicht functions and the
Polya-Schoenberg conjecture, Comment Math. Helv., 48 (1973), 119–135.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 83, 9 pp.
http://jipam.vu.edu.au/