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Arif,marifmaths@yahoo.com Received 6 January 2009; Revised 6 March 2009; Accepted 19 March 2009 Recommended by Narendra Kumar Govil We establish a relation between the functions of bound

Volume 2009, Article ID 813687, 12 pages

doi:10.1155/2009/813687

Research Article

On Bounded Boundary and

Bounded Radius Rotations

K I Noor, W Ul-Haq, M Arif, and S Mustafa

Department of Mathematics, COMSATS Institute of Information Technology, 44000 Islamabad, Pakistan

Correspondence should be addressed to M Arif,marifmaths@yahoo.com

Received 6 January 2009; Revised 6 March 2009; Accepted 19 March 2009

Recommended by Narendra Kumar Govil

We establish a relation between the functions of bounded boundary and bounded radius rotations

by using three different techniques A well-known result is observed as a special case from our main result An interesting application of our work is also being investigated

Copyrightq 2009 K I Noor et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Let A be the class of functions f of the form

f z  z ∞

n2

which are analytic in the unit disc E  {z : |z| < 1} We say that f ∈ A is subordinate to g ∈ A, written as f ≺ g, if there exists a Schwarz function wz, which by definition is analytic in

E with w 0  0 and |wz| < 1 z ∈ E, such that fz  gwz In particular, when g is univalent, then the above subordination is equivalent to f0  g0 and fE ⊆ gE.

For any two analytic functions

f z ∞

n0

a n z n , g z ∞

n0

the convolutionHadamard product of f and g is defined by



f ∗ gz ∞

n0

We denote by S∗α, Cα, 0 ≤ α < 1, the classes of starlike and convex functions of order

α, respectively, defined by

S∗α 



f ∈ A: Re zfz

f z > α, z ∈ E



,

C α f ∈ A: zfz ∈ S∗α, z ∈ E.

1.4

For α 0, we have the well-known classes of starlike and convex univalent functions denoted

by S∗and C, respectively.

Let P k α be the class of functions pz analytic in the unit disc E satisfying the properties p0  1 and

2π

0

p z − α

where z  re iθ , k ≥ 2, and 0 ≤ α < 1 For α  0, we obtain the class P kintroduced in1 Also,

for p ∈ P k α, we can write pz  1 − αq1z  α, q1∈ P k We can also write, for p ∈ P k α ,

p z  1 2π

2π

0

1 1 − 2αze −it

where μt is a function with bounded variation on 0, 2π such that

2π

0

dμ t  2π,

2π

0

For1.6 together with 1.7, see 2 Since μt has a bounded variation on 0, 2π, we may write μt  At − Bt, where At and Bt are two non-negative increasing functions

on0, 2π satisfying 1.7 Thus, if we set At  k/4  1/2μ1t and Bt  k/4 −

1/2μ2t, then 1.6 becomes

p z 

k

4 1 2

1

2π

2π

0

1 1 − 2αze −it

1− ze −it dμ1t

−

k

4 −1 2

1

2π

2π

0

1 1 − 2αze −it

1− ze −it dμ2t.

1.8

Now, using Herglotz-Stieltjes formula for the class Pα and 1.8, we obtain

p z 

k

4  1 2

p1z −

k

4 −1 2

where Pα is the class of functions with real part greater than α and p i ∈ Pα, for i  1, 2.

We define the following classes:

R k α 



f: f ∈ A and zfz

f z ∈ P k α, 0 ≤ α < 1



,

V k α  f: f ∈ A and



zfz

fz ∈ P k α, 0 ≤ α < 1



.

1.10

We note that

For α  0, we obtain the well-known classes R k and V kof analytic functions with bounded radius and bounded boundary rotations, respectively These classes are studied by Noor3 5

in more details Also it can easily be seen that R2α  S∗α and V2α  Cα.

Goel6 proved that f ∈ Cα implies that f ∈ S∗β, where

β  βα 

⎧

⎪

⎪

4α 1 − 2α

4− 22α1 , α /1

2, 1

2,

1.12

and this result is sharp

In this paper, we prove the result of Goel6 for the classes V k α and R k α by using

three different methods The first one is the same as done by Goel 6, while the second and

third are the convolution and subordination techniques

2 Preliminary Results

We need the following results to obtain our results

Lemma 2.1 Let f ∈ Vk α Then there exist s1, s2∈ S∗α such that

fz  s1z/z k/41/2

Proof It can easily be shown that f ∈ V k α if and only if there exists g ∈ V ksuch that

From Brannan7 representation form for functions with bounded boundary rotations, we have

gz 

g1z

z

⎛⎝k

4

⎞

⎠

⎛

⎝1 2

⎞

⎠

g

2z

z

⎛⎝k

4

⎞

⎠−

⎛

⎝1 2

⎞

⎠

Now, it is shown in8 that for s i ∈ S∗α, we can write

s i z  z



g i z

z

1−α

Using2.3 together with 2.4 in 2.2, we obtain the required result

Lemma 2.2 see 9 Let u  u1  iu2, v  v1 iv2, and Ψu, v be a complex-valued function

satisfying the conditions:

i Ψu, v is continuous in a domain D ⊂ C2,

ii 1, 0 ∈ D and Re Ψ1, 0 > 0,

iii Re Ψiu2, v1 ≤ 0, whenever iu2, v1 ∈ D and v1≤ −1/21  u2

2.

If h z  1  c1z  · · · is a function analytic in E such that hz, zhz ∈ D and

ReΨhz, zhz > 0 for z ∈ E, then Re hz > 0 in E.

Lemma 2.3 Let β > 0, β  γ > 0, and α ∈ α0, 1 , with

α0 max



β − γ − 1

−γ

β



If



h z  zhz

βh z  γ



≺ 1 1 − 2αz

then

h z ≺ Qz ≺ 1 1 − 2αz

where

βG z−

γ

β ,

G z 

1

0



1− z

1− tz

2β1−α

t β γ−1 dt 2F1



2β1 − α, 1, β  γ  1; z/z − 1



2.8

2F1denotes Gauss hypergeometric function From2.7, one can deduce the sharp result that h ∈ Pβ,

with

β  βα, β, γ

This result is a special case of the one given in [ 10 , page 113].

3 Main Results

By using the same method as that of Goel6, we prove the following result We include all the details for the sake of completeness

3.1 First Method

Theorem 3.1 Let f ∈ Vk α Then f ∈ R k β, where β  βα is given by 1.12 This result is

sharp.

Proof Since f ∈ V k α, we useLemma 2.1, with relation1.11 to have

1zf”z

fz 

k

4 1 2

zs

1z

s1z −

k

4 −1 2

zs

2z

s2z

k

4 1 2

zf

1z

f1z −

k

4 − 1 2

zf

2z

f2z ,

3.1

where s i ∈ S∗α and f i ∈ Cα, i  1, 2.

Therefore, from2.4, we have

zfz

f z 

k

4 1 2

zg

1z/z1−α

z

0



g1

φ

/φ1−α

dφ−

k

4 −1 2

zg

2z/z1−α

z

0



g2

φ

/φ1−α

that is,

zfz

f z 

k

4 1 2

⎡

⎣ z

0



z φ

1−αg

1



φ

g1z

1−α

dφ z

⎤

⎦

−1

−

k

4 −1 2

⎡

⎣ z

0



z φ

1−αg

2



φ

g2z

1−α

dφ z

⎤

⎦

−1

,

3.3

where we integrate along the straight line segment0, z, z ∈ E.

zfz

f z 

k

4 1 2

p1z −

k

4 1 2

and using3.3, we have

p i z 

⎡

⎣ z

0



z φ

1−αg

i



φ

g i z

1−α

dφ z

⎤

⎦

−1

where p i0  1 and hence by 11 we have

p i z −1 r2

1− r2

2r

Therefore,

min

f i ∈Cαmin

|z|rRe

p i z min

f i ∈Cαmin

|z|r p i z 3.7

Let z  re iθ and φ  Re iθ , 0 < R < r < 1 For fixed z and φ, we have from2.4

g i

φ

g i z

R r

1 r

1 R

2

Now, using3.8, we have, for a fixed z ∈ E, |z|  r,

z

0



z φ

1−αg

i



φ

g i z

1−α

dφ z

r

0

1 r

1 R

21−α

dR

Let

T r 

r

0

1 r

1 R

21−α

dR

with R  rt, 0 < t < 1, we have

T r 

1

0

1 r

1 rt

21−α

By differentiating we note that

Tr  21 − α

1

0

1 − t

1  rt2

1 r

1 rt

1−2α

dt > 0, 3.12

and therefore T r is a monotone increasing function of r and hence

max

0≤r≤1T r  T1  221−α 1

0

dt

1  t21−α



⎧

⎪

⎪



2− 41−α

2α − 1 , if α /

1 2

2 .

3.13

By letting

β α  min

⎡

⎣ z

0



z φ

1−αg

i



φ

g i z

1−α

dφ z

⎤

⎦

−1

for all g i z ∈ S∗, we obtain the required result from3.7, 3.13, and 3.14

Sharpness can be shown by the function f0 ∈ V k α given by



zf0z

f0z 

k

4 1 2

1− 1 − 2αz

1 z

−

k

4 −1 2

1 1 − 2αz

1− z

It is easy to check that f0 ∈ R k β, where β is the exact value given by 1.12

3.2 Second Method

Theorem 3.2 Let f ∈ Vk α Then f ∈ R k β, where

β 1 4

"

Proof Let

zfz

f z 



1− βp z  β

1− β k

4 1 2

p1z −

k

4 − 1 2

p2z



 β

3.17

p z is analytic in E with p0  1 Then



zfz

fz 



1− βp z  β 



1− βzpz



that is,

1

1− α



zfz

fz − α

1− α



1− βp z  β − α 



1− βzpz



1− βp z  β





β − α

1− α 



1− β

1− α

p z 



1/

1− βzpz

p z β/

1− β

.

3.19

Since f ∈ V k α, it implies that



β − α

1− α 



1− β

1− α

p z 



1/

1− βzpz

p z β/

1− β

We define

ϕ a,b z  1

1 b

z

1 − z a b

1 b

z

with a  1/1 − β, b  β/1 − β By using 3.17 with convolution techniques, see 5, we have that

ϕ a,b z

z ∗ pz 

k

4 1 2



ϕ a,b z

z ∗ p1z



−

k

4 −1 2



ϕ a,b z

z ∗ p2z



3.22

implies

p z  azpz

p z  b 

k

4  1 2

p1z  azp1z

p1z  b

−

k

4 −1 2

p2z  azp2z

p2z  b

Thus, from3.20 and 3.23, we have



β − α

1− α 



1− β

1− α

p i z  azpi z

p i z  b

We now form the functionalΨu, v by choosing u  p i z, v  zp

i z in 3.24 and note that the first two conditions ofLemma 2.2are clearly satisfied We check conditioniii as follows:

Re

ψ iu2, v1 1

1− α



β − α Re

%

v1

iu2β/

1− β

&

1− α



β − α v1



β/

1− β

u2

2β/ 1 − β2

1− α



β − α−1

2



1 u2 2



β/

1− β

u2

2β/ 1 − β2

 2



β − α'u2

2β/ 1 − β2(

−1 u2

2



β/

1− β

2'

u22β/ 1 − β2(

1 − α



"

2

β − α'β2/

1− β2(

−β/

1− β$2β − 2α −β/

1− βu22

2'

u2

2β/ 1 − β2(

1 − α

 A  Bu22

2C , 2C > 0,

3.25

where

A β

1 − β2



2

β − αβ−1− β,

B 1

1− β



2

β − α1− β− β,

C  1 − α

%

u22

1− β

2&

> 0.

3.26

The right-hand side of3.25 is negative if A ≤ 0 and B ≤ 0 From A ≤ 0, we have

β  βα  14"2α − 1 #4α2− 4α  9 $, 3.27

and from B ≤ 0, it follows that 0 ≤ β < 1.

Since all the conditions ofLemma 2.2are satisfied, it follows that p i ∈ P in E for i  1, 2 and consequently p ∈ P k and hence f ∈ R k β, where β is given by 3.16 The case k  2 is

discussed in12

3.3 Third Method

Theorem 3.3 Let f ∈ Vk α Then f ∈ R k β, where

β  β1α, 1, 0 

⎧

⎪

⎪

2α− 1

2− 221−α, if α /1

2, 1

2 ln 2, if α 1

2 .

3.28

Proof Let

zfz

f z  pz 

k

4 1 2

zs

1z

s1z −

k

4 − 1 2

zs

2z

and let

zsi z

Then p, p i are analytic in E with p0  1, p i 0  1, i  1, 2.

Logarithmic differentiation yields



zfz

fz  pz  zp p zz 

k

4 1 2

zs

1z

s1z −

k

4 − 1 2

zs

2z

s2z



k

4  1 2

%

p1z  zp1z

p1z

&

−

k

4 − 1 2

%

p2z  zp2z

p2z

&

.

3.31

Since f ∈ V k α, it follows that zs

i/si ∈ Pα, z ∈ E, or s i ∈ Cα for z ∈ E Consequently,

%

p i z  zpi z

p i z

&

where zsi z/s i z  p i z, i  1, 2 We useLemma 2.3with γ  0, β  1 > 0, α ∈ 0, 1, and

h  p iin3.32, to have p i ∈ Pβ, where β is given in 3.28 and this estimate is best possible,

extremal function Q is given by

Q z 

⎧

⎪

⎨

⎪

⎩

1 − 2αz

1 − z"1− 1 − z1−2α$, if α /12,

z

z − 1 log1 − z , if α

1

2,

3.33

see10 MacGregor 13 conjectured the exact value given by 3.28 Thus s i ∈ S∗β and consequently f ∈ R k β, where the exact value of β is given by 3.28

3.4 Application of Theorem 3.3

Theorem 3.4 Let g and h belong to Vk α Then Fz, defined by

F z 

z

0

g t

t

μ

h t

t

η

is in the class V k δ, where 0 ≤ μ < η ≤ 1, δ  δα  1 − μ  η1 − β, and βα is given by

1.12.

Proof From3.34, we can easily write

zFz

Fz  μ

zgz

g z  η

zhz

h z  1 −



Since g and h belong to V k α, then, byTheorem 3.3, zgz/gz and zhz/hz belong to

P k β, where β  βα is given by 1.12 Using

zgz

g z 



1− βq1z  β, q1∈ P k ,

zhz

h z 



1− βq2z  β, q2∈ P k ,

3.36

in3.35, we have

1

1− δ

zFz

Fz − δ

μ  η q1z  η

Now by using the well-known fact that the class P kis a convex set together with3.37, we obtain the required result

For α  0, μ  0, and η  1, we have the following interesting corollary.

Corollary 3.5 Let f belongs to Vk 0 Then Fz, defined by

F z 

z

0

f t

t dt

 Alexander’s integral operator

is in the class V k 1/2.

Acknowledgments

The authors are grateful to Dr S M Junaid Zaidi, Rector, CIIT, for providing excellent research facilities and the referee for his/her useful suggestions on the earlier version of this paper W Ul-Haq and M Arif greatly acknowledge the financial assistance by the HEC, Packistan, in the form of scholarship under indigenous Ph.D fellowship

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1 B Pinchuk, “Functions of bounded boundary rotation,” Israel Journal of Mathematics,...