Volume 8 (2007), Issue 3, Article 109, 8 pp.
AN APPLICATION OF SUBORDINATION ON HARMONIC FUNCTION
H.A. AL-KHARSANI
D EPARTMENT OF M ATHEMATICS
G IRLS C OLLEGE
P.O. B OX 838, DAMMAM , S AUDI A RABIA
hakh73@hotmail.com
Received 28 June, 2006; accepted 15 May, 2007
Communicated by A. Sofo
A BSTRACT. The purpose of this paper is to obtain sufficient bound estimates for harmonic func∗
tions belonging to the classes SH
[A, B], KH [A, B] defined by subordination, and we give some
convolution conditions. Finally, we examine the closure properties of the operator Dn on these
classes under the generalized Bernardi integral operator.
Key words and phrases: Harmonic, Univalent, Subordination, Convex, Starlike, Close-to-convex.
2000 Mathematics Subject Classification. Primary 30C45; Secondary 58E20.
1. I NTRODUCTION
A continuous function f = u + iv is a complex-valued harmonic function in a complex
domain C if both u and v are real harmonic in C. In any simply connected domain D ⊂ C,
we can write f = h + g, where h and g are analytic in D. We call h the analytic part and g
the co-analytic part of f . A necessary and sufficient condition for f to be locally univalent and
orientation-preserving in D is that |g 0 (z)| < |h0 (z)| in D [2].
We denote by SH the family of functions f = h + g which are harmonic univalent and
orientation-preserving in the open disk U = {z : |z| < 1} so that f = h + g is normalized by
f (0) = h(0) = fz (0) − 1 = 0. Therefore, for f = h + g ∈ SH , we can express the analytic
functions h and g by the following power series expansion:
(1.1)
h(z) = z +
∞
X
m=2
m
am z ,
g(z) =
∞
X
bm z m .
m=1
Note that the family SH of orientation-preserving, normalized harmonic univalent functions
reduces to the class S of normalized analytic univalent functions if the co-analytic part of f =
h + g is identically zero.
∗
Let K, S ∗ , C, KH , SH
and CH denote the respective subclasses of S and SH where the images
of f (u) are convex, starlike and close-to-convex.
176-06
2
H.A. A L -K HARSANI
A function f (z) is subordinate to F (z) in the disk U if there exists an analytic function w(z)
with w(0) = 0 and |w(z)| < 1 such that f (z) = F (w(z)) for |z| < 1. This is written as
f (z) ≺ F (z).
Let K[A, B], S ∗ [A, B] denote the subclasses of S defined as follows:
1 + Az
zf 0 (z)
∗
S [A, B] = f ∈ S,
≺
, −1 ≤ B < A ≤ 1 ,
f (z)
1 + Bz
1 + Az
(zf 0 (z))0
K[A, B] = f ∈ S,
≺
, −1 ≤ B < A ≤ 1 .
f 0 (z)
1 + Bz
We now introduce the following subclasses of harmonic functions in terms of subordination.
Let f = h + g ∈ SH such that
h(z) − g(z)
,
1 − b1
h(z) − eiθ g(z)
(1.3)
ψ(z) =
, 0 ≤ θ < 2π,
1 − eiθ b1
∗
and let −1 ≤ B < A ≤ 1, then we can construct the classes KH [A, B], SH
[A, B] using
subordination as follows:
(zψ 0 (z))0
1 + Az
KH [A, B] = f ∈ SH ,
≺
,
ψ 0 (z)
1 + Bz
zϕ0 (z)
1 + Az
∗
SH [A, B] = f ∈ SH ,
≺
.
ϕ(z)
1 + Bz
P
m
Let Dn denote the n-th Ruscheweh derivative of a power series t(z) = z + ∞
m=2 tm z which
is given by
z
Dn t =
∗ t(z)
(1 − z)n+1
∞
X
=z+
C(n, m)tm z m ,
(1.2)
ϕ(z) =
m=2
where
(n + 1)m−1
(n + 1)(n + 2) · · · (n + m − 1)
=
.
(m − 1)!
(m − 1)!
In [5], the operator Dn was defined on the class of harmonic functions SH as follows:
C(n, m) =
Dn f = Dn h + Dn g.
The purpose of this paper is to obtain sufficient bound estimates for harmonic functions be∗
longing to the classes SH
[A, B], KH [A, B], and we give some convolution conditions. Finally,
we examine the closure properties of the operator Dn on the above classes under the generalized
Bernardi integral operator.
2. P RELIMINARY R ESULTS
Cluni and Sheil-Small [2] proved the following results:
Lemma 2.1. If h, g are analytic in U with |h0 (0)| > |g 0 (0)| and h + g is close-to-convex for
each , || = 1, then f = h + g is harmonic close-to-convex.
Lemma 2.2. If f = h + g is locally univalent in U and h + g is convex for some , || ≤ 1,
then f is univalent close-to-convex.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
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A N A PPLICATION OF S UBORDINATION ON H ARMONIC F UNCTION
3
A domain D is called convex in the direction γ (0 ≤ γ < π) if every line parallel to the line
through 0 and eiγ has a connected intersection with D. Such a domain is close-to-convex. The
convex domains are those that are convex in every direction.
We will make use of the following result which may be found in [2]:
Lemma 2.3. A function f = h + g is harmonic convex if and only if the analytic functions
h(z) − eiγ g(z), 0 ≤ γ < 2π, are convex in the direction γ2 and f is suitably normalized.
∗
Necessary and sufficient conditions were found in [2, 1] and [4] for functions to be in KH , SH
∗
and CH . We now give some sufficient conditions for functions in the classes SH
[A, B] and
KH [A, B], but first we need the following results:
P∞
m
Lemma 2.4
([7]).
If
q(z)
=
z
+
is analytic
m=2 Cm z
P∞
P∞in U , 2then q maps onto a starlike
domain if m=2 m|Cm | ≤ 1 and onto convex domains if m=2 m |Cm | ≤ 1.
Lemma 2.5 ([4]). If f = h + g with
∞
X
m|am | +
m=2
∞
X
m|bm | ≤ 1,
m=1
then f ∈ CH . The result is sharp.
Lemma 2.6 ([4]). If f = h + g with
∞
X
2
m |am | +
m=2
∞
X
m2 |bm | ≤ 1,
m=1
then f ∈ KH . The result is sharp.
Lemma 2.7 ([6]). A function f (z) ∈ S is in S ∗ [A, B] if
∞
X
{m(1 + A) − (1 + B)} |am | ≤ A − B,
m=2
where −1 ≤ B < A ≤ 1.
Lemma 2.8 ([6]). A function f (z) ∈ S is in K[A, B] if
∞
X
m {m(1 + A) − (1 + B)} |am | ≤ A − B,
m=2
where −1 ≤ B < A ≤ 1.
Lemma 2.9 ([3]). Let h be convex univalent in U with h(0) = 1 and Re(λh(z)+µ) > 0 (λ, µ ∈
C). If p is analytic in U with p(0) = 1, then
p(z) +
zp0 (z)
≺ h(z)
λp(z) + µ
(z ∈ U )
implies
p(z) ≺ h(z)
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
(z ∈ U ).
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4
H.A. A L -K HARSANI
3. M AIN R ESULTS
Theorem 3.1. If
∞
X
(3.1)
{m(1 + A) − (1 + B)} |am | +
m=2
∞
X
{m(1 + A) − (1 + B)} |bm | ≤ A − B,
m=1
∗
then f ∈ SH
[A, B]. The result is sharp.
∗
Proof. From the definition of SH
[A, B], we need only to prove that ϕ(z) ∈ S ∗ [A, B], where
φ(z) is given by (1.2) such that
∞ X
am − b m
φ(z) = z +
zm.
1
−
b
1
m=2
Using Lemma 2.7, we have
∞
∞
X
{m(1 + A) − (1 + B)} am − bm X {m(1 + A) − (1 + B)} |am | + |bm |
1 − b1 ≤
A−B
A−B
1 − |b1 |
m=2
m=2
≤1
if and only if (3.1) holds and hence we have the result.
The harmonic function
f (z) = z +
∞
X
1
xm z m
(A
−
B){m(1
+
A)
−
(1
+
B)}
m=2
+
∞
X
1
ymzm
(A − B){m(1 + A) − (1 + B)}
m=1
where
∞
X
|xm | +
m=2
∞
X
!
|ym | = A − B − 1
m=1
shows that the coefficient bound given by (3.1) is sharp.
Corollary 3.2. If A = 1, B = −1, then we have the coefficient bound given in [1] with a
different approach.
Theorem 3.3. If f = h + g with
∞
X
{m(1 + A) − (1 + B)}|am |C(n, m)
m=2
+
∞
X
{m(1 + A) − (1 + B)}|bm |C(n, m) ≤ A − B,
m=1
∗
then Dn f = H + G ∈ SH
[A, B]. The function
f (z) = z +
(1 + δ)(A − B)
zm,
{m(1 + A) − (1 + B)}C(n, m)
δ>0
shows that the result is sharp.
Corollary 3.4. If A = 1, B = −1, then we have the coefficient bound given in Theorem 3.1,
α = 0 [5] with a different approach.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
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A N A PPLICATION OF S UBORDINATION ON H ARMONIC F UNCTION
5
Theorem 3.5. If
(3.2)
∞
X
m{m(1 + A) − (1 + B)}|am | +
m=2
∞
X
m{m(1 + A) − (1 + B)}|bm | ≤ A − B,
m=1
then f ∈ KH [A, B]. The result is sharp.
Proof. From the definition of the class KH [A, B] and the coefficient bound of K[A, B] given in
Lemma 2.8, we have the result. The function
f (z) = z +
(1 + δ)(A − B)
zm,
m{m(1 + A) − (1 + B)}
δ>0
shows that the upper bound in (3.2) cannot be improved.
Theorem 3.6. If f = h + g with
∞
X
m{m(1 + A) − (1 + B)}C(n, m)|am |
m=2
+
∞
X
m{m(1 + A) − (1 + B)}C(n, m)|bm | ≤ A − B,
m=1
then Dn f ∈ KH [A, B]. The function
f =z+
(1 + δ)(A − B)
zm,
m{m(1 + A) − (1 + B)}C(n, m)
δ>0
shows that the result is sharp.
Corollary 3.7. If n = 0, A = 1, B = −1, we have Theorem 3 in [4] and if A = 1, B = −1, we
have Theorem 2 in [5].
In the next two theorems, we give necessary and sufficient convolution conditions for func∗
tions in SH
[A, B] and KH [A, B].
∗
Theorem 3.8. Let f = h + g ∈ SH . Then f ∈ SH
[A, B] if
!
!
(−1−Aξ) 2
2
z
z
z + (ξ−A)
ξ
z
−
A−B
A−B
6= 0,
h(z) ∗
+ B g(z)
(1 − z)2
(1 − z)2
Proof. Let S(z) =
h(z)−g(z)
,
1−b1
|ξ| = 1, 0 < |z| < 1.
then S ∈ S ∗ [A, B] if and only if
zS 0
1 + Az
≺
S
1 + Bz
or
zS 0 (z)
1 + Aeiθ
6=
,
S(z)
1 + Beiθ
0 ≤ θ < 2π, z ∈ U.
It follows that
1 + Aeiθ
0
zS (z) − S(z)
6= 0.
1 + Beiθ
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
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6
H.A. A L -K HARSANI
Since zS 0 (z) = S(z) ∗
(3.3)
z
,
(1−z)2
the above inequality is equivalent to
z
1 + Aeiθ z
0 6= S(z) ∗
−
(1 − z)2 1 + Beiθ 1 − z



(−e−iθ −A) 2


z + (A−B) z
1
 , 1 − b1 = λeit
= it S(z) ∗ 
λe 
(−e−iθ − B)(1 − z)2 
(
!
−iθ
z + (−eA−B−A) z 2
1
= it h(z) ∗
− g(z)
λe
(−e−iθ − B)(1 − z)2
z + (−e−iθ − A)z 2
2
iθ
∗
(1 − z) (−B − e )
(A − B)(e−iθ /B)
(
!
−iθ
z + (−eA−B−A) z 2
1
=
h(z) ∗
λ
(1 − z)2 eit
!)
−iθ −A)eiθ
Beiθ z + B(−eA−B
z2
−g(z) ∗
.
eit (−B − eiθ )(1 − z)2
Now, if z1 − z2 6= 0 and |z1 | =
6 |z2 |, then z1 − z 2 6= 0, || = 1, i.e.,
"
!#
−iθ
z + (−eA−B−A) z 2
1
=
h(z) ∗
λ(−B − e−iθ )
(1 − z)2 eit


(−1−Aeiθ )B 2
+iθ
Be z +
z
A−B

− g(z) ∗  it
e (−B − eiθ )(1 − z)2
"
!#
−iθ
z + (−eA−B−A) z 2
1
h(z) ∗
=
λ(−B − e−iθ )
(1 − z)2 eit
−iθ )
− g(z) ∗
(−B)(−e−iθ z + B(−1−Ae
A−B
(1 − z)2 e−it
z2
!
.
Since arg(1 − b1 ) = t 6= π, we obtain the result and the proof is thus completed.
Corollary 3.9. If A = 1, B − 1 and = 1, then we have Theorem 2.6 in [1] with a different
approach.
Theorem 3.10. Let f = h + g ∈ SH . Then f ∈ KH [A, B] if and only if
"
#
"
#
−2+(A+B)ξ 2
2
z + 2ξ−A−B
z
ξz
−
z
A−B
A−B
h(z) ∗
+ g(z) ∗
6= 0
(1 − z)3
(1 − z)3
|| = 1, |ξ| = 1,
0 < |z| < 1
iγ
g(z)
Proof. Let ψ(z) = h(z)−e
, 0 ≤ γ < 2π and 1 − eiγ b1 = λ eit , then from (1.3) and (3.3),
1−eiγ b1
0
∗
zψ (z) ∈ SH [A, B] if and only if
"
#
(−e−iθ −A) 2
z
z
+
A−B
zψ 0 (z) ∗
6= 0
(−e−iθ − B)(1 − z)2
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
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A N A PPLICATION OF S UBORDINATION ON H ARMONIC F UNCTION
7
i.e.,
)#
)
(
(−e−iθ −A) 2
(−e−iθ −A) 2
z
+
z
z
+
z
1
A−B
A−B
zh0 ∗
− zg 0 ∗
.
it
iθ
2
−iθ
λe
(−e − B)(1 − z)
(−e − B)(1 − z)2

(
)0
(
)0 
(−e−iθ −A) 2
(−e−iθ −A) 2
z
+
z
+
z
z
1 
A−B
A−B

h(z) ∗
− g(z) ∗
it
2
−iθ
2
λe
(1 − z) (−e − B)
(1 − z) (−e−iθ − B)
!
!#
"
−iθ −A−B
−iθ −A−B
z + −2e A−B
z + −2e A−B
z2
z2
1
− g(z) ∗
h(z) ∗
(1 − z)3 (−e−iθ − B)
(1 − z)3 (−e−iθ − B)
λeit
"
!
!#
−iθ −A−B
−iθ −A−B
z + −2e A−B
z2
z + −2e A−B
z2
1
− g(z) ∗
h(z) ∗
−iθ
λ
eit (1 − z)3 (−e−iθ − B)
eit (1 − z)3 (−B − e−iθ ) e B



iθ
−2e−iθ −A−B 2
iθ z + −2B−(A+B)Be z 2
z
+
z
Be
1
A−B
A−B

h(z) ∗ it
− g(z) ∗  it
3
−iθ
3
λ
e (1 − z) (−e − B)
e (1 − z) (−B − eiθ )
"
!#
−iθ
−iθ −A−B
2
z + −2e A−B
(−B)(−e−iθ )z + −2B−(A+B)Be
z
1
A−B
h(z) ∗ it
− g(z) ∗
λ
e (1 − z)3 (−e−iθ − B)
e−it (−B − e−iθ )(1 − z)3
"
!#
−2+(A+B)(−e−iθ ) 2
−2e−iθ −A−B
−iθ
z
+
(−e
)z
−
z
1
A−B
A−B
h(z) ∗ it
+ Bg(z) ∗
,
3
−iθ
−it
−iθ
λ
e (1 − z) (e − B)
e (−B − e )(1 − z)3
"
0 6=
=
=
=
=
=
=
(
and we have the result.
Corollary 3.11. If A = 1, B = −1, = −1, then we have Theorem 2.7 of [1].
Theorem 3.12. If f = h + g ∈ SH with
∞
∞
X
X
(3.4)
mC(n, m)|am | +
mC(n, m)|bm | ≤ 1,
m=2
m=1
n
then D f = H + G ∈ CH . The result is sharp.
Proof. The result follows immediately. Using Lemma 2.5, the function
1+δ
f (z) = z +
zm, δ > 0
mC(n, m)
shows that the upper bound in (3.4) cannot be improved.
P∞
Theorem 3.13. If f = h + g is locally univalent with m=2 m2 C(n, m)|am | ≤ 1, then Dn f ∈
CH .
Proof. Take = 0 in Lemma 2.2 and apply Lemma 2.4.
Corollary 3.14. Dn f = H + G ∈ CH if |G0 (z)| ≤
1
2
and
P∞
m=2
m2 C(n, m)|am | ≤ 1.
Proof. The function Dn f is locally univalent if |H 0 (z)| > |G0 (z)| for z ∈ U . Since
∞
∞
X
X
2
mC(n, m)|am | ≤
m2 C(n, m)|am | ≤ 1,
m=2
m=2
we have
0
|H (z)| > 1 −
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
∞
X
1
m|am |C(n, m)| ≥ .
2
m=2
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8
H.A. A L -K HARSANI
Corollary 3.15. If h(z) ∈ K and w(z) is analytic with |w(z)| < 1, then
Z z
n
f (z) = D h(z) +
w(t)(Dn h(t))0 dt ∈ CH .
0
∗
Theorem 3.16. Let f = h + g ∈ SH . If Dn+1 f ∈ R, then Dn f ∈ R, where R can be SH
[A, B]
or KH [A, B] or CH .
h
i
h−g
∗
∗
Proof. We can prove the result when R ≡ SH
[A, B]. If Dn+1 f ∈ SH
[A, B], then Dn+1 1−b
∈
1
S ∗ [A, B] and |Dn+1 h| > |Dn+1 g|. Using Lemma 2.9, we have
n h−g
D
∈ S ∗ [A, B].
1 − b1
Since
0 z
1
z
n+1
0 ∗
h
=
z
∗
h
|D h| = z
,
(1 − z)n+1
z (1 − z)n+1
∗
this implies |Dn h| > |Dn g|, or Dn (h) + Dn g ∈ SH
[A, B] and we have the result.
R
z c−1
Theorem 3.17. Let f = h + g ∈ SH and let Fc (f ) = 1+c
t f (t)dt. If Dn f ∈ R, then
zc
0
∗
Dn Fc (f ) ∈ R, where R can be SH
[A, B] or KH [A, B] or CH .
h−g
∗
Proof. If Dn f ∈ SH
[A, B], then Dn 1−b
∈ S ∗ [A, B]. Using Lemma 2.9, we have Dn Fc (f ) ∈
1
∈ S ∗ [A, B] or Dn Fc (h) − Dn Fc (g) ∈ S ∗ [A, B]. Since
S ∗ [A, B]. That is, Dn Fc (h−g)
1−b1
∗
|Dn Fc (n)| > |Dn Fc (g)|, then Dn Fc (f ) ∈ SH
[A, B].
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J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 109, 8 pp.
http://jipam.vu.edu.au/