On a Class of Functions whose Derivatives Map the Unit Disc into a Half Plane

MALAYSIAN MATHEMATICAL

SCIENCES SOCIETY

On a Class of Functions whose Derivatives Map the Unit Disc into a Half Plane

DAUD MOHAMAD

Universiti Teknologi MARA, Kampus Bukit Sekilau, 25200 Kuantan, Pahang, Malaysia

Abstract. Let G(α,δ) denote the class of functions f, f(0) = f(0) − 1 = 0 for which

α ′( )>δ

Re eⁱ f z in D={z: z <1} where α ≤ π and cos α−δ > 0. We discuss some basic properties of the class including representation theorem, extremals and argument of G(α,δ).

1. Introduction

We denote G(α,δ) the class of normalized analytic functions f in the unit disc D where L

L + + +

+

= z a z a_nzⁿ

z

f( ) ₂ ²

satisfying Re e^iαf′(z) > δ where α ≤ π and cos α − δ > 0.

Many of the classes G(α,δ) have been studied by several researchers such as MacGregors [3] for G(0,0), Goel and Mehrok [1] for G(α,δ)(δ ≥ 0)and Silverman and Silvia [4] for G(α,0). Writing

, ) ( cos

sin ) ) (

( e f z i z D

z p

i

− ∈

−

′ −

= α δ

δ

α α

(1) clearly f ∈G(α,δ) if and only if p∈P, the class of functions with positive real parts.

Solving (1) for f'(z) yields

) ( ) sin ) ( ( )

(z e Ap z i z D

f′ = ^−iα + α +δ ∈ (2) where A=cosα−δ.

2. Representation theorem

We obtain the representation theorem for G(α,δ), sharing the same approach through Herglotz Representation Theorem for functions in P.

Theorem 2.1. Let f ∈g(α,δ). Then for some probability measure μ on the unit circle X,

[

]

)

(z e e z e Ax xz d x

f

X

i i

i^{α α} δ ^α μ

∫

= ^{− − −} (3)

Conversely, if f is given by the above equation, then f ∈G(α,δ). Proof. For some probability measure μ on the circle X,

).

( 1

) 1 (

d x

xz z xz

p P

p^{∈ ⇔ =}

∫

Using (2), we have

) ( 1 sin

) 1

( i d x

xz A xz e z

f ^iα α δ ⎥⎦⎤ μ

⎢⎣⎡ + +

−

= +

′ ⁻

∫

and so

ψ ψ μ

δ

ψ ψ μ

ψ δ

ψ μ δ ψ α

ψ

α α α

α α

α

d x x d

A e e

e

d x x d

x e e

d x d x i

A x e

z f

i i

X i z

X

i i

z

X i

) ( 1

) 2 2 (

) ( 1

) 2 (

1

) ( ) sin 1 (

) 1 (

z

0 0

2 0

∫ ∫

∫ ∫

∫ ∫

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+ −

−

−

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−

−

= +

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ⎟⎟⎠ + +

⎜⎜ ⎞

⎝

⎛

−

= +

− −

−

−

−

−

(4)

and the desired representation theorem is obtained by reversing the order of integration and integrating with respect to ψ .

We note that the extreme points of G(α,δ) are the unit point masses ) 1 ( log 2

) 2 (

)

(z e e z e Ax xz

f_x = − ^{−iα −iα} − δ − ^−iα −

with x = 1 and the derivatives of the extreme points for G(α,δ)are the point masses

. 1 , 1

) 2 (

) 1 (

2

− =

−

= + ^{− −} x

xz

xz e z e

f

i i

x

α

α δ

3. Extremal properties

Following Silverman and Silvia [4], we now obtain a coefficient bound for functions in )

, (α δ

g and distortion theorems for the derivatives of these functions.

Theorem 3.1. If f∈G(α,δ), then a_n ≤ 2A/n, n=2,3,4,_L and equality is attained for each n when f is an extreme point of G(α,δ).

Proof. Using (4) and since 1 (1⁻xψ) ⁼

∑

. ) ( 2

) (

2

∫ ∑

=

−

+ −

=

X

n

n n i

n x z d x A e z z

f ^α μ

Now, let ∑^∞

=

+

=

2

) (

n n nz a z z

f . Then 2 ₁ ( )

x d n x

A a e

X n i

n ^{= −α}

∫

Our further result will be based on the following theorem.

Theorem 3.2. Let f∈G(α,δ). Then f′ maps z ≤ r into the disc D_r with center )

1 /(

) 2 ( ) 2

(e e A r²

e ^{i i} − + ⁱ −

− ^{−α −α} δ ^−α and radius 2Ar/(1 − r²).

Proof. If a and b are complex numbers with b <1, and if 0 < r <1, the range of the function (1 + arω) (1 + brω)(|ω| ≤ 1) is the disc with center and radius

2 2 2

1 1

r b

r b a

−

− ,

2 2

1 b r

r b a

−

−

respectively. By taking a=(e^−i2α− 2δ e^−iα)xr and b=xr where x =1, we see that

1

) 2 (

1 ²

xz

xz e

e ^{i i}

−

−

+ ^{− α} δ ^−α

maps z ≤ r onto D_r. By convexity, any linear combination of functions of this form also maps D onto D_r. Since for some probability measure μ, we have

, ) ( 1

) 2 (

) 1 (

2

x xz d

xz e z e

f

X

i

i^α δ ^α μ

∫

′ = ^{− −}

the stated result now follows.

Theorem 3.3. If f ∈G(α,δ), then

2 2

2 2

1

2 ) 1 ) ( 2 ( 1 ) ( Re 1

2 ) 1 ) ( 2 ( 1

r

rA A

A z r

f r

rA A

A r

−

+

− +

≤ +

≤ ′

−

−

− +

+ δ δ

(5) and

2

2 2

2

1

) ) ( 1 1 ( 2 ) ( Im 1

) ) ( 1 1 ( 2

r A r

z Ar f r

A r

Ar

− +

−

≤ +

≤ ′

−

+

− +

− δ δ

.

All bounds are sharp for any extreme point f of G(α,δ). Proof. By Theorem 3.2, we can write

2

2 1

2 1

) 2 2 (

) (

r Ar r

A e e

e z

f

i i

i

≤ −

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+ −

−

−

′ − ^{−α −α} δ ^−α (6) so that

2 2

2 1

2 1

) 2 2 (

) ( Re 1

2

r Ar r

A e e

e z f r

Ar _{i i} ⁱ

≤ −

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

− −

−

′ +

− ≤

− _{−α −α} δ ^−α

and also

2 2

2 1

2 1 ) 2 2 (

) ( Im 1

2

r Ar r

A e e

e z f r

Ar _{i i} ⁱ

≤ −

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

− −

−

′ +

− ≤

− _{−α −α} δ ^−α .

The results are obtained by simplifying the above inequalities.

We note that if f ∈G(α,δ), then since f₀′(0)=1, we have Ref′(z) > 0 for ρ

<

z and some ρ in (0,1]. However if

, ) ( , 1

) 2 (

) 1 (

2

D z z

z e z e

f

i i

o ∈

−

−

= + ^{− α} δ ^−α

then the left side of inequality (5) is sharp so that

) 1 ( ) 1 )(cos (cos

2 2 ) 1 ) ( 2 ( 1 ) ( Re ) 1

( −r² f₀′ −r = + r² A A+δ − − rA → α−δ α − r→

and the last expression is negative if α ≠ 0. This shows that ρ≠1 in general, and it is natural to ask for the best possible value of ρ. We answer this question in the following application of Theorem 3.2

Theorem 3.4. Let f ∈g(α,δ) and put ρ = 1(A + 1−A(2δ + A)). Then 1

0 < ρ ≤ and Re f′(z) ≥ 0 fo r z <ρ . If ρ ≤ r ≤ 1, then Ref₀′(z) < 0 for some z on z < r.

Proof. Let f ∈G(α,δ) and define ρ as above. Obviously ρ>0 since A>0,and 0

cos 1

) 2 (

1−A δ + A = + δ²− α ≥ . The inequality ρ ≤ 1 is equivalent to 1

) 2 (

1− + ≥

+ A A

A δ and this is obviously true if A ≥ 1. If A < 1, it is true if and only if 1 − A(2δ + A) ≥ (1 − A)², and thus reduces to the trivially true inequality cos α ≤ 1. So in both cases, ρ ≤1.

Now, put σ(x) = (2A(A + δ) − 1)x²− 2x + 1 for real values of x. From (5), we have (1− r²)Ref′(z) ≥ σ(r) (0 ≤ z = r < 1) with equality for each r when

fo

f = and z is a suitable value on z = r. To prove the theorem, it is sufficient to show that σ(x) is positive on [0,ρ) and non-positive on [ρ,1].

If 2A(A + δ) = 1, so that σ(x) is linear in x, then ρ=1/(2A) and it is clear that )

σ(x is positive on [0,ρ)and non-positive on [ρ,1]. When 2A(A + δ ≠) 1, σ(x) is quadratic and has zeros

. ) 2 ( 1 1 1 ) ( 2

) 2 ( 1

A A

A A A

A A

x A

+

= −

− +

+

−

= ±

δ δ δ

m (7) One of the zeros is ρ. Let the other zero be μ, If 2A(A + δ) < 1, then μρ < 0 and (7) shows that μ< 0 and ρ > 0. Since σ is concave, σ(x) is positive on [0,ρ)and

non-positive on [ρ,1]. If 2A(A + δ)>1, then μ,ρ>0 since μρ>0, μ+ρ>0.

Also ρ <μ by (5). In this case σ is convex so σ(x) is positive on [0,ρ) and non-positive on [ρ,μ]. In particular, since σ(1) = 2A(cosα−1) ≤ 0,σ(x) is non-positive on [ρ,1]. This completes the proof.

We next obtain a distortion theorem for G(α,δ). Theorem 3.5. If f∈G(α,δ), then

1 2

) 2 ( ) (

r r Ar

C z

f′ ≤ + −

where

1 1

1 ) 4

( _{2 2}

2 ⎟⎟ +

⎠

⎞

⎜⎜

⎝

⎛ +

−

= − δ

r A r r Ar

C (8)

and the bound is sharp for any extreme point f of G(α,δ).

Proof. Let ₂

1 ) 2 2 ( )

( r

A e e

e r

i i

i

+ −

−

−

=

Γ ^{−α −α} δ ^−α . By using (6) we have

1 2

) 2 ( ) (

r r Ar z

f′ ≤ Γ + −

1 2

) 2 (

r r Ar

C + −

= as required.

4. Argument of f′(z)

We see that if δ ≥0, then f′ is non-zero throughout D, and has continuous argument.

But if δ <0, and if f_o is any extreme function of G(α,δ), then at some point of D, f0′ has a zero and hence no argument. So to obtain result for argument of f′, we restrict the values of z considered in the case δ <0. We will also use the following property for argument: for a given α in [−π,π] and as x varies in some interval [0, c], so that e^iα+ x ≠ 0, φ_α(x) is the continuous argument of e^iα+x, for which

α

φ_α(0)= . We have

0 cos if 2 ,

/

0 cos if , cos

tan sin

0 cos if cos ,

tan sin

)

( ₁

1

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

= +

<

⎟ +

⎠

⎜ ⎞

⎝

⎛ + +

>

⎟ +

⎠

⎜ ⎞

⎝

⎛ +

= ₋

−

π α

α α

π α

α α α

φ_α

x x x

x x

x

when 0<α <π, and similar formulae for the case −π <α <0, α = 0, ±π. Theorem 4.1. Let f ∈G(α,δ), and put x(r)=2Ar² (1−r²) (0≤r<1). Let

⎪⎪

⎩

⎪⎪⎨

⎧

<

−

≥

= , 0.

4 1

1

0 , 1

δ δ

δ

A r_o

Then, for 0 < z = r < r_o, and for suitable determination of argument

) ( ) 1 ( sin 2 )) ( ( )

(

arg 2

1

r C r r Ar

x z

f′ + α −φ_α ≤ ⁻ − (9)

where φ_α(x) is defined on [0,x(r_o)) as above and C(r) is given by (8).

Proof. We restrict the value of z = r by the condition

2

2 1

2 2 1

2

r e Ar

r

A i

> −

−

− +

−α

δ

to ensure that f′(z) ≠ 0. Squaring both sides and simplifying, we have 0

1 4 1

4

2 − + >

− δ δ

A r

A .

The inequality holds for all r in [0,1) if δ ≥ 0 and for 0 ≤ r < 1 1−4δA if δ <0. This establishes the restriction on z . By using (6) and Theorem 3.5, we deduce that

) ( ) 1 ( sin 2 ) ( arg ) (

arg 2

1

r C r r Ar

z

f′ − Γ ≤ ⁻ − (10)

and also

. 1

arg 2

1 ) 2 2 (

arg ) ( arg

2 2

2

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ + − +

−

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+ −

−

−

=

Γ ^{− − −}

r e Ar

r A e e

e r

i

i i

i

α α α α

α

δ

Put x(r)=2Ar² /(1−r²), then argΓ(r) = −α + φ_α(x(r)) and the desired result follows using (10).

We obtain another result for argument of G(α,δ), features arg(f′(z) + k) for some real k that satisfy f′(z) + k ≠ 0 for z∈D and for all f ∈G(α,δ). When

2 π /

α = , such a choice is impossible, for if f_o is an extreme function in ,

) , (α δ

G then f₀′(z)+k maps D onto either Imw >δ or Imw<−δ and since δ <0 both these half planes contain 0. If α ≠ π/2, any choice of k with kcosα + δ >0 ensures that f₀′(z) + k ≠ 0 for z∈D, f ∈G(α,δ).

In the statement of the following theorem, for a given α∈[−π,π], and as x varies in some interval [0,c), so that (k+1)e^iα +x≠0, ψ_α(α) is the continuous argument of

x e k+1) ^iα +

( for which ψ_α(0) is principal.

Theorem 4.2. Let f ∈G(α,δ), where α ≠ π/2. Put x(r)=2A/(1−r²)(0 ≤ r<1) and let k be a real number such that kcosα + δ > 0. Then

) ( ) 1 ( sin 2 )) ( ( )

) ( ( arg

1 2 1

r C r r Ar

x k

z

f′ + + α − ψ_α ≤ ⁻ −

where ψ_α(α) is defined on [0,∞) as above, and

2 2

2 2

1 cos ( 1)

1 1

) 4

( ⎟⎟ + +

⎠

⎞

⎜⎜

⎝

⎛ + +

−

= − k k

r A r

r Ar

C α δ . (11)

Proof. Let α ≠ π/2, and let k satisfy kcosα + δ >0. We have, using (6),

1 2

) 2 ) ( ( )

(

r k Ar

r k z

f′ + − Γ + ≤ −

where

1 . 1 2 1

) 2 2 ( )

( ₂

2 2

α α α

α ⁱ δ ^{i i}

i e

r Ar r

A e e

e

r ⁻

− −

−

+ −

− = +

−

−

= Γ

Hence

) ( ) 1 ( sin 2 ) ) ( arg(

) ) ( arg(

1 2 1

r C r k Ar

r k

z

f′ + − Γ + ≤ ⁻ − (12)

where C₁(r) = Γ(r) + k and is written as in (11). Now

)) ( ( 1

2 2 arg )

) (

arg( ₂ ke xr

r e A

k

r α δ ^{iα iα}⎥⎦⎤ = −α +ψ_α

⎢⎣⎡ +

+ −

− +

−

= +

Γ ⁻

and the proof is complete by using (12).

References

1. R.M. Goel and B.S. Mehrok, A class of univalent functions, J. Austral. Maths Soc. (Series A) 35 (1983), 1-17.

2. A.W. Goodman, Univalent Functions, Vol. I and II, Mariner Publishing Co. Inc. Tampa, Florida, 1983.

3. T.H. MacGregor, Functions whose derivative has a positive real part, Trans. Amer. Math. Soc.

104 (1962), 532-537.

4. H. Silverman and E.M. Silvia, On α-close to convex function, Publ. Math. Debrecen, 49 (1996), 532-537.