Let Ap be the class of functions f(z) of the form

f ( z ) = z p + ∑ n = p + 1 ∞ a n z n ( p = 1 , 2 , 3 , … )

which are analytic in the open unit disk U = { z ∈ ℂ : | z | < 1 } and satisfy

z p f ( z ) = 1 + ∑ n = p + 1 ∞ b n z n - p ≠ 0 ( z ∈ U ) .

For f ( z ) ∈ A p , we say that f(z) belongs to the class Up(λ) if it satisfies

z2zp-1 f(z) - 1 z′≦ λ(z ∈ U)

for some real number λ > 0.

Let us consider a function f_δ(z) given by

f δ ( z ) = z p ( 1 - z ) δ ( δ ∈ ℝ ) .

Then, we can write that

f δ ( z ) = z p 1 + ∑ n = 1 ∞ a n z n

with

a n = ( - 1 ) n δ n

and

z 2 z p - 1 f δ ( z ) - 1 z ′ = ∑ n = 1 ∞ ( n - 1 ) a n z n < ∑ n = 1 ∞ ( n - 1 ) | a n | .

Thus, if δ = 2, then

z 2 z p - 1 f 2 ( z ) - 1 z ′ < 1 .

This shows that f 2 ( z ) ∈ U p ( λ ) for λ ≧ 1.

If δ = 3, then we have that

z 2 z p - 1 f 3 ( z ) - 1 z ′ < 5

Which shows that f ( z ) ∈ U p ( λ ) for λ ≧ 5.

Further, if δ = 4, then

z 2 z p - 1 f 4 ( z ) - 1 z ′ < 1 1

which shows that f(z) ∈Up(λ) for λ ≧ 11.

If p = 1, then f ( z ) ∈ U 1 ( λ ) is defined by

z 2 1 f ( z ) - 1 z ′ ≦ λ ( z ∈ U )

for some real number λ > 0. Note that (1.5) is equivalent to

f ′ ( z ) z f ( z ) 2 - 1 ≦ λ ( z ∈ U ) .

Therefore, this class U 1 ( λ ) was considered by Obradović and Ponnusamy 1. Further-more, this class was extended as the class U ( β 1 , β 2 ; λ ) by Shimoda et al. 2.

Let Sp*(α) denotes the subclass of Ap consisting of f(z) which satisfy

R e z f ′ ( z ) f ( z ) > α ( z ∈ U )

for some real α (0 ≦ α < p).

A function f(z) ∈Sp*(α) is said to be p-valently starlike of order α in U (cf. Robertson 3).

For f(z) ∈Ap , we consider the sufficient condition for f(z) to be in the class Up(λ) .

Lemma 1 If f(z) ∈Ap satisfies

∑ n = p + 2 ∞ ( n - p - 1 ) | b n | ≦ λ ,

then f(z) ∈U1(λ) .

Proof We note that

z 2 z p - 1 f ( z ) - 1 z ′ = ∑ n = p + 1 ∞ ( n - p - 1 ) b n z n - p < ∑ n = p + 1 ∞ ( n - p - 1 ) | b n | .

Therefore, if

∑ n = p + 1 ∞ ( n - p - 1 ) | b n | = ∑ n = p + 2 ∞ ( n - p - 1 ) | b n | ≦ λ ,

then f(z) ∈Up(λ) .

Example 1 If we consider a function f(z) ∈Ap given by

z p f ( z ) = 1 + b p + 1 z + ∑ n = p + 2 ∞ λ e i φ ( n - p ) ( n - p - 1 ) 2 z n - p ≠ 0 ( z ∈ U )

with

b n = λ e i φ ( n - p ) ( n - p - 1 ) 2 ( λ > 0 , φ ∈ ℝ )

for n ≧ p + 2, then we see that

∑ n = p + 2 ∞ ( n - p - 1 ) | b n | = ∑ n = p + 2 ∞ λ e i φ ( n - p ) ( n - p - 1 ) < λ ∑ n = p + 2 ∞ 1 n - p - 1 - 1 n - p = λ .

Thus, this function f(z) satisfies the inequality (2.1). Also, we see that

| z 2 ( z p − 1 f ( x ) − 1 z ) ′ | = | ∑ n = p + 2 ∞ λ e i φ n − p − 1 ) ( n − p ) z n − p |                 < λ ∑ n = p + 2 ∞ ( 1 n − p − 1 − 1 n − p ) = λ .

Therefore, we say that f(z) ∈Up(λ) .

Next, we discuss the necessary condition for the class Sp*(α) .

Lemma 2 If f(z) ∈Sp*(α) satisfies

z p f ( z ) = 1 + ∑ n = p + 1 ∞ b n z n - p ≠ 0 ( z ∈ U )

with b_n = |b_n| e^i(n-p)θ (n = p + 1, p + 2, p + 3,...), then

∑ n = p + 1 ∞ ( n + α - 2 p ) | b n | ≦ p - α .

Proof Let us define the function F(z) by

F ( z ) = z p f ( z ) = 1 + ∑ n = p + 1 ∞ b n z n - p .

It follows that

R e z f ′ ( z ) f ( z ) = R e p - z F ′ ( z ) F ( z ) = R e p - ∑ n = p + 1 ∞ ( n - 2 p ) b n z n - p 1 + ∑ n = p + 1 ∞ b n z n - p = R e p - ∑ n = p + 1 ∞ ( n - 2 p ) | b n | e i ( n - p ) θ z n - p 1 + ∑ n = p + 1 ∞ | b n | e i ( n - p ) θ z n - p > α

for z ∈ U . Letting z = |z| e^-iθ, we have that

p - ∑ n = p + 1 ∞ ( n - 2 p ) | b n | | z | n - p 1 + ∑ n = p + 1 ∞ | b n | | z | n - p > α ( z ∈ U ) .

If we take |z| → 1^-, we obtain that

p - ∑ n = p + 1 ∞ ( n - 2 p ) | b n | 1 + ∑ n = p + 1 ∞ | b n | ≧ α

which implies that

∑ n=p+1∞(n + α - 2p)|b n|≦ p - α.

Remark 1 If we take p = 1 in Lemmas 1 and 2, then we have that

(i) f ( z ) ∈ A 1 , ∑ n = 2 ∞ ( n - 2 ) | b n | ≦ λ ⇒ f ( z ) ∈ U 1 ( λ )

and

(ii) f ( z ) ∈ S * ( α ) , | b n | = | b n | e i ( n - 1 ) θ ⇒ ∑ n = 2 ∞ ( n + α - 2 ) | b n | ≦ 1 - α .

Our main result for the radius problem is contained in

Theorem 1 Let f(z) ∈Sp*(α) (p - 1 ≦ α < p) with

zp f(z) = 1 + ∑ n=p+1∞b nzn-p≠0(z ∈ U).

and b_n = | b_n| e^i(n-p)θ (n = p + 1, p + 2, p + 3, ...). If δ ∈ ℂ ( | δ | < 1 ) , then 1 δ p f ( δ z ) belongs to the class Up(λ) for 0 < | δ | ≦ | δ 0 ( λ ) | , where |δ₀(λ)| is the smallest positive root of the equation

| δ | 2 1 - α - ( 1 - | δ | 2 ) λ = 0 ,

that is,

| δ 0 ( λ ) | = λ λ + 1 - α .

Proof Since

f ( δ z ) = δ p z p + ∑ n = p + 1 ∞ a n δ n z n ,

we have that

z p 1 δ p f ( δ z ) = 1 + ∑ n = p + 1 ∞ b n δ n - p z n - p .

In view of Lemma 1, we have to show that

∑ n = p + 2 ∞ ( n - p - 1 ) | b n | | δ | n - p ≦ λ .

Note that f(z) ∈Sp*(α) satisfies

| b n | ≦ p - α n + α - 2 p < 1 ( p - 1 ≦ α < p ) .

Applying Cauchy-Schwarz inequality, we obtain that

∑ n = p + 2 ∞ ( n - p - 1 ) | b n | | δ | n - p ≦ ∑ n = p + 2 ∞ ( n - p - 1 ) | b n | 2 1 2 ∑ n = p + 2 ∞ ( n - p - 1 ) | δ | 2 ( n - p ) 1 2 ≦ ∑ n = p + 2 ∞ ( n - p - 1 ) | δ | 2 ( n - p ) 1 2 p - α .

Let |δ|² = x. Then, we have that

∑ n = p + 2 ∞ ( n - p - 1 ) x n - p = x 2 ∑ n = p + 2 ∞ ( n - p - 1 ) x n - p - 2 = x 2 ∑ n = p + 2 ∞ x n - p - 1 ′ = x 2 ∑ n = 1 ∞ x n - 1 = x 2 ( 1 - x ) 2 .

This gives us that

∑ n = p + 2 ∞ ( n - p - 1 ) | b n | | δ | n - p ≦ | δ | 2 p - α 1 - | δ | 2 .

Let us define the function h(|δ|) by

h ( | δ | ) = | δ | 2 p - α - ( 1 - | δ | 2 ) λ .

Then, h (|δ|) satisfies h (0) = -λ < 0 and h ( 1 ) = p - α > 0 . Indeed, we have that h (|δ₀(λ) |) = 0 for

0 < | δ 0 ( λ ) | = λ λ + p - α < 1 .

This completes the proof of the theorem.

Corollary 1 Let f ( z ) ∈ S 1 * ( α ) ( 0 ≦ α < 1 ) with

z f ( z ) = 1 + ∑ n = 2 ∞ b n z n - 1 ≠ 0 ( z ∈ U )

and b_n = |b_n| e^i(n-1)θ (n = 2, 3, 4,...). If δ ∈ ℂ (|δ| < 1), then 1 δ f ( δ z ) belongs to the class U1(λ) for 0 < |δ|≦|δ0(λ)| , where |δ₀(λ)| is the smallest positive root of the equation

|δ|21 - α - (1 -|δ|2)λ = 0,

that is,

|δ0(λ)| = λ λ + 1 - α.

Remark 2 In view of (3.2), we define the function g(λ) by

g ( λ ) = | δ 0 ( λ ) | = λ λ + p - α .

Then, we have that

g ′ ( λ ) = 1 2 p - α λ ( λ + p - α ) 3 > 0

for λ > 0. Therefore, |δ₀(λ)| given by (3.2) is increasing for λ > 0.

Remark 3 If we put α = p - 1 2 in Theorem 1, then

| δ 0 ( λ ) | = 2 λ 2 λ + 2 .

Therefore, if we consider λ = 1 2 , then we see that

δ 0 1 2 = 1 1 + 2 = 0 . 6 4 3 5 9 …

and if we make λ = 5, then we have that

| δ 0 ( 5 ) | = 1 0 1 0 + 2 = 0 . 9 3 6 0 0 …

The authors declare that they have no competing interests.

QF carried out the main part of this article. All authors read and approved the final manuscript.