Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 20, 2006
SUBORDINATION RESULTS FOR THE FAMILY OF UNIFORMLY CONVEX
p−VALENT FUNCTIONS
H.A. AL-KHARSANI AND S.S. AL-HAJIRY
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE
G IRLS C OLLEGE , DAMMAM
S AUDI A RABIA .
ssmh1@hotmail.com
Received 16 May, 2005; accepted 02 October, 2005
Communicated by G. Kohr
A BSTRACT. The object of the present paper is to introduce a class of p−valent uniformly functions U CVp . We deduce a criteria for functions to lie in the class U CVp and derive several
interesting properties such as distortion inequalities and coefficients estimates. We confirm our
results using the Mathematica program by drawing diagrams of extremal functions of this class.
Key words and phrases: p−valent, Uniformly convex functions, Subordination.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Denote by A(p, n) the class of normalized functions
(1.1)
p
f (z) = z +
∞
X
ak+p−1 z k+p−1
k=2
regular in the unit disk D = {z : |z| < 1} and p ∈ N, consider also its subclasses C(p), S ∗ (p)
consisting of p−valent convex and starlike functions respectively, where C(1) ≡ C, S ∗ (1) ≡
S ∗ , the classes of univalent convex and starlike functions .
It is well known that for any f ∈ C, not only f (D) but the images of all circles centered at
0 and lying in D are convex arcs. B. Pinchuk posed a question whether this property is still
valid for circles centered at other points of D. A.W. Goodman [1] gave a negative answer to
this question and introduced the class U CV of univalent uniformly convex functions, f ∈ C
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
We express our thanks to Dr.M.K.Aouf for his helpful comments.
151-05
2
H.A. A L -K HARSANI AND S.S. A L -H AJIRY
4
1
2
0
0
-1
2
3
-2
4
5
6
such that any circular arc γ lying in D, having the center ζ ∈ D is carried by f into a convex
arc. A.W.Goodman [1] stated the criterion
f 00 (z)
> 0, ∀z, ζ ∈ D ⇐⇒ f ∈ U CV.
(1.2)
Re 1 + (z − ζ) 0
f (z)
Later F. Ronning (and independently W. Ma and D. Minda) [7] obtained a more suitable
form of the criterion , namely
00 zf (z) zf 00 (z)
, ∀z ∈ D ⇐⇒ f ∈ U CV.
(1.3)
Re 1 + 0
> 0
f (z)
f (z) This criterion was used to find some sharp coefficients estimates and distortion theorems for
functions in the class U CV .
2. T HE C LASS P ARp
We now introduce a subfamily P ARp of P . Let
υ2
(2.1)
Ω = w = µ + iυ :
< 2µ − p
p
(2.2)
= {w : Re w > |w − p|}.
Note that Ω is the interior of a parabola in the right half-plane which is symmetric about the
real axis and has vertex at (p/2, 0). The following diagram shows Ω when p = 3:
Let
P ARp = {h ∈ p : h(D) ⊆ Ω}.
o
n
π √
ν2
2
√
Example 2.1. It is known that z = − tan 2 2p w maps w = µ + iν : p < p − 2µ
√
conformally onto D. Hence, z = − tan2 2√π2p p − w maps Ω conformally onto D. Let
w = Q(z) be the inverse function. Then Q(z) is a Riemann mapping function from D to Ω
(2.3)
J. Inequal. Pure and Appl. Math., 7(1) Art. 20, 2006
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S UBORDINATION R ESULTS
3
which satisfies Q(0) = p; more explicitly,
√ 2 X
∞
2p
1+ z
√
=
(2.4)
Q(z) = p + 2 log
Bn z n
π
1− z
n=0
16p
184p 3
8p
z + 2 z2 +
z + ···
2
π
3π
45π 2
Obviously, Q(z) belongs to the class P ARp . Geometrically, P ARp consists of those holomorphic functions h(z) (h(0) = p) defined on D which are subordinate to Q(z), written
h(z) ≺ Q(z).
The analytic characterization of the class P ARp is shown in the following relation:
(2.5)
(2.6)
=p+
h(z) ∈ P ARp ⇔ Re{h(z)} ≥ |h(z) − p|
such that h(z) is a p−valent analytic function on D.
Now, we can derive the following definition.
00
(z)
Definition 2.1. Let f (z) ∈ A(p, n). Then f (z) ∈ U CVp if f (z) ∈ C(p) and 1+z ff 0 (z)
∈ P ARp .
3. C HARACTERIZATION OF U CVp
We present the nessesary and sufficient condition to belong to the class U CVp in the following
theorem:
Theorem 3.1. Let f (z) ∈ A(p, n). Then
00 00
f (z)
f (z)
(3.1)
f (z) ∈ U CVp ⇔ 1 + Re z 0
≥ z 0
− (p − 1) ,
f (z)
f (z)
z ∈ D.
00
(z)
Proof. Let f (z) ∈ U CVp and h(z) = 1 + z ff 0 (z)
. Then h(z) ∈ P ARp , that is, Re{h(z)} ≥
|h(z) − p|. Then
00
f (z)
f 00 (z)
Re 1 + z 0
≥ z 0
− (p − 1) .
f (z)
f (z)
Example 3.1. We now specify a holomorphic function K(z) in D by
1+z
(3.2)
K 00 (z)
= Q(z),
K 0 (z)
where Q(z) is the conformal mapping onto Ω given in Example 2.1. Then it is clear from
Theorem 3.1 that K(z) is in U CVp .
Let
∞
X
Ak z k+p−1 .
(3.3)
K(z) = z p +
k=2
From the relationship between the functions Q(z) and K(z), we obtain
(3.4)
(p + n − 1)(n − 1)An =
n−1
X
(k + p − 1)Ak Bn−k .
k=1
Since all the coefficients Bn are positive, it follows that all of the coefficients An are also
positive. In particular,
(3.5)
J. Inequal. Pure and Appl. Math., 7(1) Art. 20, 2006
A2 =
8p2
,
π 2 (p + 1)
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4
H.A. A L -K HARSANI AND S.S. A L -H AJIRY
and
(3.6)
p2
A3 =
2(p + 2)
k 0 (z)
log p−1 =
z
z
16
64p
+
3π 2
π4
.
Note that
(3.7)
Z
0
Q(ς) − p
dς.
ς
By computing some coefficients of K(z) when p = 3, we can obtain the following diagram
1
0.5
0
-0.5
-1
-2
2
0
-2
-1
0
1
2
4. S UBORDINATION T HEOREM
AND
C ONSEQUENCES
In this section, we first derive some subordination results from Theorem 4.1; as corollaries
we obtain sharp distortion, growth, covering and rotation theorems from the family U CVp .
00
00
(z)
(z)
Theorem 4.1. Assume that f (z) ∈ U CVp . Then 1 + z ff 0 (z)
≺ 1 + zK
and
K 0 (z)
00
f 0 (z)
z p−1
≺
K 0 (z)
.
z p−1
00
(z)
(z)
Proof. Let f (z) ∈ U CVp . Then h(z) = 1 + z ff 0 (z)
≺ 1+zK
is the same as h(z) ≺ Q(z).
K 0 (z)
Note that Q(z) − p is a convex univalent function in D. By using a result of Goluzin, we may
conclude that
Z z
Z z
f 0 (z)
h(ς) − 1
Q(ς) − p
K 0 (z)
(4.1)
log p−1 =
dς ≺
dς = log p−1 .
z
ς
ς
z
0
0
Equivalently,
f 0 (z)
z p−1
≺
K 0 (z)
.
z p−1
Corollary 4.2 (Distortion Theorem). Assume f (z) ∈ U CVp and |z| = r < 1. Then K 0 (−r) ≤
|f 0 (z)| ≤ K 0 (r).
Equality holds for some z 6= 0 if and only if f (z) is a rotation of K(z).
Proof. Since Q(z) − p is convex univalent in D, it follows that log K 0 (z) is also convex
univalent in D. In fact, the power series for log K 0 (z) has positive coefficients, so the image
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S UBORDINATION R ESULTS
5
of D under this convex function is symmetric about the real axis. As log
subordination principle shows that
(4.2)
0
K 0 (−r) = e{ log K (−r)} = e
min Re{log K 0 (z)}
≤e
0
(z)
≺ log Kzp−1
, the
|z|=r
{Re log K 0 (z)}
f 0 (z)
z p−1
= |f 0 (z)| ≤ e
max Re{log K 0 (z)}
|z|=r
0
= e{log K (r)} = K 0 (r).
Note that for |z0 | = r, either
Re{log f 0 (z0 )} = min Re{log K 0 (z)}
|z|=r
or
Re{log f 0 (z0 )} = max Re{log K 0 (z)}
|z|=r
for some z0 6= 0 if and only if log f 0 (z) = log K 0 (eiθ z) for some θ ∈ R.
Theorem 4.3. Let f (z) ∈ U CVp . Then
14p
(4.3)
|f 0 (z)| ≤ z p−1 e π2 ς(3) = z p−1 Lp
for |z| < 1. (L ≈ 5.502, ς(t) is the Riemann Zeta function.)
0
(z)
Proof. Let φ(z) = zgg(z)
, where g(z) = zf 0 (z). Then φ(z) ≺ Q(z) which means that φ(z) ≺
√ 1+ z
√
p + π2p2 log 1−
. Moreover ,
z
g(z)
log p =
z
Z
0
z
φ(s) − p
s
ds
and therefore, if z = reiθ and |z| = 1,
Z r
g(z) dt
log p =
<e(φ(teiθ ) − p)
z
t
0
√ Z r
2p
1
1+ t
√ dt
≤ 2
log
π 0 t
1− t
√ Z
2p 1 1
1+ t
√ dt
≤ 2
log
π 0 t
1− t
2p
= 2 (7ς(3)),
π
where
√ Z 1
1
1+ t
√ dt = 7ς(3)
log
[8].
1− t
0 t
Then we find that
0 zf (z) 2p
2 (7ς(3))
.
zp ≤ e π
J. Inequal. Pure and Appl. Math., 7(1) Art. 20, 2006
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6
H.A. A L -K HARSANI AND S.S. A L -H AJIRY
The following diagram shows the boundary of K(z)’s dervative when p = 2 in a circle has
the radius (5.5)2 :
20
1
0.5
0
-0.5
-1
0
-20
0
-20
20
Corollary 4.4 (Growth Theorem). Let f (z) ∈ U CVp and |z| = r < 1. Then −K(−r) ≤
|f (z)| ≤ K(r).
Equality holds for some z 6= 0 if and only if f (z) is a rotation of K(z).
Corollary 4.5 (Covering Theorem). Suppose f (z) ∈ U CVp . Then either f (z) is a rotation of
K(z) or {w : |w| ≤ −K(−1)} ⊆ f (D).
Corollary 4.6 (Rotation Theorem). Let f (z) ∈ U CVp and |z0 | = r < 1. Then
|Arg{f 0 (z0 )}| ≤ max Arg{K 0 (z).
(4.4)
|z|=r
Equality holds for some z 6= 0 if and only if f (z) is a rotation of K(z).
P
k+p−1
Theorem 4.7. Let f (z) = z p + ∞
and f (z) ∈ U CVp , and let An+p−1 =
k=2 ak+p−1 z
max |an+p−1 |. Then
f (z)∈U CVp
Ap+1 =
(4.5)
8p2
.
π 2 (p + 1)
The result is sharp. Further, we get
(4.6)
An+p−1
Proof. Let f (z) = z p +
n Y
8p2
8p
≤
1+
.
(n + p − 1)(n − 1)π 2 k=3
(k − 2)π 2
P∞
ak+p−1 z k+p−1 and f (z) ∈ U CVp , and define
∞
X
zf 00 (z)
φ(z) = 1 + 0
=p+
ck z k+p−1 .
f (z)
k=2
k=2
Then φ(z) ≺ Q(z). Q(z) is univalent in D and Q(D) is a convex region, so Rogosinski’s
theorem applies.
8p
16p
184p 3
Q(z) = p + 2 z + 2 z 2 +
z + ··· ,
π
3π
45π 2
so we have |cn | ≤ |B1 | = π8p2 := B. Now, from the relationship between functions f (z) and
Q(z), we obtain
(n + p − 1)(n − 1)an+p−1 =
n−1
X
(k + p − 1)ak+p−1 cn−k .
k=1
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S UBORDINATION R ESULTS
From this we get |ap+1 | =
pB
(p+1)
=
8p2
.
π 2 (p+1)
7
If we choose f (z) to be that function for which
zf 00 (z)
,
f 0 (z)
Q(z) = 1 +
then f (z) ∈ U CVp with ap+1 =
sharp. Now, when we put |c1 | = B, then
8p2
,
π 2 (p+1)
which shows that this result is
pap c2 + (p + 1)ap+1 c1
2(p + 2)
pB(1 + Bp)
.
|ap+2 | ≤
2(p + 2)
ap+2 =
When n = 3
pap c3 + (p + 1)ap+1 c2 + (p + 2)ap+2 c1
3(p + 3)
1 pB(1 + Bp)(2 + Bp)
|ap+3 | ≤
2
3(p + 3)
1
Bp
=
pB(1 + Bp) 1 +
.
3(p + 3)
2
ap+3 =
We now proceed by induction. Assume we have
Bp
Bp
1
pB(1 + Bp) 1 +
··· 1 +
|ap+n−1 | ≤
(n − 1)(p + n − 1)
2
n−2
n
Y
pB
Bp
=
1+
.
(n − 1)(p + n − 1) k=3
k−2
Corollary
4.8. Let f (z) = z p +
1
O n2 .
P∞
k=2
ak+p−1 z k+p−1 and f (z) ∈ U CVp . Then |ap+n−1 | =
5. G ENERAL P ROPERTIES OF F UNCTIONS IN U CVp
P
k+p−1
Theorem 5.1. Let f (z) = z p + ∞
and f (z) ∈ U CVp . Then f (z) is a
k=2 ak+p−1 z
p−valently convex function of order β in |z| < r1 = r1 (p, β), where r1 (p, β) is the largest
value of r for which
(5.1)
rk−1 ≤
(p − β)(k − 1)
Q (k + p − β − 1)B kj=3 1 +
pB
j−2
,
(k ∈ N − {1}, 0 ≤ β < p).
Proof. It is sufficient to show that for f (z) ∈ U CVp ,
00
zf
(z)
1 +
≤ p − β, |z| < r1 (p, β),
−
p
f 0 (z)
0 ≤ β < p,
where r1 (p, β) is the largest value of r for which the inequality (5.1) holds true. Observe that
P∞
00
k=2 (k + p − 1)(k − 1)ak+p−1 z k−1 zf
(z)
1 +
.
P
− p = k−1 f 0 (z)
p+ ∞
(k
+
p
−
1)a
z
k+p−1
k=2
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8
H.A. A L -K HARSANI AND S.S. A L -H AJIRY
Then we have 1 +
zf 00 (z)
f 0 (z)
− p ≤ p − β if and only if
P∞
(k + p − 1)(k − 1) |ak+p−1 | rk−1
k=2P
k−1
p− ∞
k=2 (k + p − 1) |ak+p−1 | r
∞
X
≤p−β
(k + p − 1)(k + p − 1 − β) |ak+p−1 | rk−1 ≤ p2 − pβ.
⇒
k=2
Then from Theorem 4.7 since f (z) ∈ U CVp , we have
k Y
pβ
Bp
|ak+p−1 | ≤
1+
(k + p − 1)(k − 1) j=3
j−2
and we may set
k Y
pβ
Bp
|ak+p−1 | =
1+
ck+p−1 , ck+p−1 ≥ 0,
(k + p − 1)(k − 1) j=3
j−2
)
(
∞
X
k ∈ N − {1},
ck+p−1 ≤ 1 .
k=1
Now, for each fixed r, we choose a positive integer k0 = k0 (r) for which
(k + p − 1 − β) k−1
r
(k − 1)
is maximal. Then
∞
X
(k + p − 1)(k + p − β − 1) |ak+p−1 | r
k=2
k−1
k (k0 + p − β − 1) k0 −1 Y
Bp
≤
r
1+
.
(k0 − 1)
j
−
2
j=3
Consequently, the function f (z) is a p−valently convex function of order β in |z| < r1 =
r1 (p, β) provided that
k (k0 + p − β − 1) k0 −1 Y
Bp
r
1+
≤ p(p − β).
(k0 − 1)
j
−
2
j=3
We find the value r0 = r0 (p, β) and the corresponding integer k0 (r0 ) so that
k (k0 + p − β − 1) k0 −1 Y
Bp
r
1+
= p(p − β), (0 ≤ β < p).
(k0 − 1)
j−2
j=3
Then this value r0 is the radius of p−valent convexity of order β for functions f (z) ∈ U CVp .
Theorem 5.2. h(z) = z p + bn+p−1 z n+p−1 is in U CVp if and only if
r≤
p2
,
(p + n − 1)(p + 2n − 2)
where |bn+p−1 | = r and bn+p−1 z n−1 = reiθ .
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S UBORDINATION R ESULTS
9
00
(z)
. Then h(z) ∈ U CVp if and only if w(z) ∈ P ARp which means
Proof. Let w(z) = 1 + zhh0 (z)
that Re{w(z)} ≥ |w(z) − p| . On the other side we have
zh00 (z)
zh00 (z)
Re 1 + 0
≥ 1 + 0
− p ,
h (z)
h (z)
then
zh00 (z)
p + (n + p − 1)nreiθ
Re 1 + 0
= Re (p − 1) +
h (z)
p + (n + p − 1)reiθ
p3 + p(n + p − 1)(n + 2p − 1)r cos θ + (n + p − 1)3 r2
=
.
|p + (n + p − 1)reiθ |2
The right-hand side is seen to have a minimum for θ = π and this minimal value is
p3 + p(n + p − 1)(n + 2p − 1)r + (n + p − 1)3 r2
.
|p + (n + p − 1)reiθ |2
Now, by computation we see that
00
zh
(z)
1 +
= (n + p − 1)(n − 1)r .
−
p
|p + (n + p − 1)reiθ |
0
h (z)
Then
p3 + p(n + p − 1)(n + 2p − 1)r + (n + p − 1)3 r2
,
(n + p − 1)(n − 1)r ≤
p − (n + p − 1)r
which leads to
(n + p − 1)(n − 1)r ≤ p2 − (n + p − 1)2 r.
Hence,
p2
.
r≤
(n + p − 1)(2n + p − 2)
Theorem 5.3. Let f (z) ∈ U CV, then (f (z))p ∈ U CVp .
Proof. Let w(z) = (f (z))p , then
1+z
w00 (z)
f 00 (z)
f 0 (z)
=
1
+
z
+
(p
−
1)z
.
w0 (z)
f 0 (z)
f (z)
Then we find
00
w (z)
w00 (z)
Re 1 + z 0
− z 0
− (p − 1)
w (z)
w (z)
f 00 (z)
f 0 (z)
= Re 1 + z 0
+ (p − 1)z
f (z)
f (z)
00
0
f (z)
f
(z)
− z 0
+ (p − 1)z
− (p − 1) .
f (z)
f (z)
Since f (z) ∈ U CV , therefore we have
00
0
f (z)
f 00 (z)
f 0 (z)
f
(z)
Re 1 + z 0
+ (p − 1)z
− z 0
+ (p − 1)z
− (p − 1)
f (z)
f (z)
f (z)
f (z)
0 0
f (z)
f (z)
≥ (p − 1)Re z
− z
− 1
f (z)
f (z)
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10
H.A. A L -K HARSANI
AND
S.S. A L -H AJIRY
f (z) ∈ U CV , then f (z) ∈ SP [7] which means that
0 0
f (z)
f (z)
Re z
− z
− 1 ≥ 0.
f (z)
f (z)
Then
00
w (z)
w00 (z)
Re 1 + z 0
− z 0
− (p − 1) ≥ 0.
w (z)
w (z)
The following diagram shows the extermal function k(z) of the class U CV when (k(z))p , p =
2:
The following diagram shows the extermal function K(z) of the class U CVp when p = 2:
And the following diagram shows that (k(z))p ≺ K(z):
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S UBORDINATION R ESULTS
11
5.1. Remarks. Taking p = 1 in Theorem 3.1, we obtain the corresponding Theorem 1 of [7].
Taking p = 1 in Theorem 4.1, we obtain the corresponding Theorem 3 of [3].
Taking p = 1 in inequality (4.3), we obtain Theorem 6 of [7], and in inequalities (4.5), (4.6),
we obtain Theorem 5 of [7].
Taking p = 1 in Theorem 5.2, we obtain Theorem 2 of [4].
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Math., 105(1-2) (1999), 327–336.
[3] W.C. MA AND D. MINDA, Uniformly convex functions, Ann. Polon. Math., 57(2) (1992), 165–175.
[4] S. OWA, On uniformly convex functions, Math. Japonica, 48(3) (1998), 377–384.
[5] M.S. ROBERTSON, On the theory of univalent functions, Ann. Math., 2(37) (1936), 347–408.
[6] F. RONNING, A survey on uniformly convex and uniformly starlike functions, Ann. Univ. Mariae
Curie-Sklodowska Sect. A, 47 (1993), 123–134.
[7] F. RONNING, Uniformly convex functions and a corresponding class of starlike functions, Proc.
Amer. Math. Soc., 118(1) (1993), 189–196.
[8] F. RONNING, On uniform starlikeness and related properties of univalent functions, Complex Variables Theory Appl., 24(3-4) (1994), 233–239.
J. Inequal. Pure and Appl. Math., 7(1) Art. 20, 2006
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