RACSAM https://doi.org/10.1007/s13398-018-0517-9 ORIGINAL PAPER

Integral transforms and partial sums of certain p-valent starlike functions Jin-Lin Liu1

· Rekha Srivastava2 · Yi-Hui Xu3

Received: 9 January 2018 / Accepted: 13 March 2018 © Springer-Verlag Italia S.r.l., part of Springer Nature 2018

Abstract In this paper, two new subclasses of the class of p-valent starlike functions in the open unit disk U are introduced and investigated systematically. Several such properties as (for example) inclusion relations, integral transforms, partial sums and distortion bounds for each of these subclasses are obtained. Keywords Analytic function · p-Valent starlike function · Principle of subordination · Inclusion relation · Integral transform · Partial sum · Distortion bound Mathematics Subject Classification Primary 30C45; Secondary 30C80

1 Introduction A function f which is analytic in a domain D ⊂ C is called p-valent in D if, for every complex number w, the equation f (z) = w has at most p roots in D and there exists a complex number w0 such that the equation f (z) = w0 has exactly p roots in D. Let A( p) denote the class of functions of the form: f (z) = z p +

∞ 

an z n ( p ∈ N),

(1.1)

n= p+1

B

Jin-Lin Liu [email protected] Rekha Srivastava [email protected] Yi-Hui Xu [email protected]

1

Department of Mathematics,Yangzhou University, Yangzhou 225002, People’s Republic of China

2

Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada

3

Department of Mathematics, Suqian College, Suqian 223800, People’s Republic of China

J.-L. Liu et al.

which are analytic in the open unit disk U = {z : z ∈ C and |z| < 1}, N being the set of positive integers. We denote by S ∗p the well-known class of p-valent starlike functions in U. It is defined as follows:      z f (z) > 0 (z ∈ U) . S ∗p = f : f ∈ A( p) and  f (z) For functions f and g, analytic in U, we say that f is subordinate to g in U and write f (z) ≺ g(z) (z ∈ U), if there exists an analytic function w in U such that   |w(z)|  |z| and f (z) = g w(z) (z ∈ U). Let ∞ 

f j (z) = z p +

an, j z n ∈ A( p) ( j = 1, 2).

n= p+1

Then the Hadamard product (or convolution) of f 1 and f 2 is defined by ( f 1 ∗ f 2 )(z) = z p +

∞ 

an,1 an,2 z n = ( f 2 ∗ f 1 )(z).

n= p+1

Throughout the present paper, we assume that − 1  B < A  1,

B  0, 0  λ  1 and k ∈ N\{1}.

(1.2)

The following lemma will be required in our investigation. Lemma Let f ∈ A( p) defined by (1.1) satisfy: ∞  

 n(1 − B) − p(1 − A) 1 − λ(1 − δn, p,k ) |an |  p(A − B).

(1.3)

n= p+1

Then

1 + Az z f  (z) ≺p (z ∈ U), (1 − λ) f (z) + λ f p,k (z) 1 + Bz

where δn, p,k =

⎧ ⎪ ⎪ 0 ⎪ ⎨ ⎪ ⎪ ⎪ ⎩1

(1.4)



 n−p ∈ /N k   n−p ∈N k

(1.5)

for n  p + 1,   k−1 2πi 1  − jp  j  εk f εk z and εk = exp . f p,k (z) = k k j=0

(1.6)

Integral transforms and partial sums. . .

Proof The function f p,k in (1.6) can be expressed as follows: ∞ 

f p,k (z) = z p +

δn, p,k an z n ,

n= p+1

where δn, p,k =

1 k

k−1 

j (n− p)

εk

=

j=0

⎧ ⎪ ⎪ ⎨0 ⎪ ⎪ ⎩1



 n−p ∈ /N ,  k  n−p ∈N k

for n  p + 1. By applying (1.2) and (1.5), we can see that   p A 1 − λ(1 − δn, p,k ) − n B  − p Bλ(1 − δn, p,k )  0 (n  p + 1). Let the inequality (1.3) hold true. Then we have     ∞ z f  (z)    [n − p + pλ(1 − δn, p,k )]an z n− p  (1−λ) f (z)+λ f p,k (z) − p   n= p+1   =  

    p(A − B) + ∞ z f  (z) n− p   p A − B (1−λ) f (z)+λ f p,k (z)  n= p+1 p A 1 − λ(1 − δn, p,k ) − n B an z ∞ n= p+1 [n − p + pλ(1 − δn, p,k )]|an |  

  p(A − B) − ∞ n= p+1 p A 1 − λ(1 − δn, p,k ) − n B |an |  1.

Hence, by the Maximum Modulus Theorem, we arrive at (1.4). The proof of the Lemma is completed.

Now we define the following two new subclasses of A( p). Definition 1 A function f ∈ A( p) defined by (1.1) is said to be in the class G p,k (λ, A, B) if and only if it satisfies the coefficient inequality (1.3). Definition 2 A function f ∈ A( p) defined by (1.1) is said to be in the class T p,k (λ, A, B) if and only if it satisfies the following condition: ∞ 

 

n n(1 − B) − p(1 − A) 1 − λ(1 − δn, p,k ) |an |  p 2 (A − B).

(1.7)

n= p+1

It is obvious that f ∈ T p,k (λ, A, B) ⇐⇒

z f  (z) ∈ G p,k (λ, A, B). p

If we write αn =

(1.8)

  n(1 − B) − p(1 − A) 1 − λ(1 − δn, p,k ) n and βn = αn (n  p + 1), (1.9) p(A − B) p

then it is easy to verify that ∂βn n ∂αn =  0, ∂λ p ∂λ

∂βn n ∂αn = < 0 and ∂A p ∂A

∂βn n ∂αn =  0. ∂B p ∂B

Thus, we have the following inclusion relations: If 0  λ0  λ  1, −1  B0  B < A  A0  1 and B  0,

J.-L. Liu et al.

then T p,k (λ, A, B) ⊂ G p,k (λ, A, B) ⊆ G p,k (λ0 , A0 , B0 )

⊆ G p,k (0, 1, −1) ⊆ S ∗p      z f (z) > 0 (z ∈ U) . = f : f ∈ A( p) and  f (z) Therefore, by the Lemma, we see that each function in the classes G p,k (λ, A, B) and

T p,k (λ, A, B) is p-valent starlike in U.

Multivalent analytic (and meromorphic) functions, which are starlike, have been extensively studied by several authors (see, e.g., [1–11]; see also the related recent works [12–18] and the references cited therein). In the present paper, we investigate several such properties as (for example) inclusion relations, integral transforms, partial sums and distortion bounds for each of the above-defined classes G p,k (λ, A, B) and T p,k (λ, A, B).

2 Inclusion relations In this section, we generalize the above-mentioned inclusion relation: T p,k (λ, A, B) ⊂ G p,k (λ, A, B).

Theorem 1 If −1  D  0, then



(2.1) 

T p,k (λ, A, B) ⊂ G p,k λ, C(D), D ,

where C(D) = D +

(2.2)

p(1 − D)(A − B)(1 + λp) ( p + 1)(1 − B)(1 + λp) + p(1 − λ)(A − B)

(2.3)

and the number C(D) cannot be decreased for each D. Proof It is clear that D < C(D) < 1. Let f ∈ T p,k (λ, A, B). In order to prove that f ∈ G p,k λ, C(D), D , we need only to find the smallest C (D < C  1) such that   

 n n(1 − B) − p(1 − A) 1 − λ(1 − δn, p,k ) n(1 − D) − p(1 − C) 1 − λ(1 − δn, p,k )  p(C − D) p 2 (A − B) (2.4) for all n  p + 1, that is, that   n− p+λp(1−δn, p,k ) (1 − D) p   C  D+  . (2.5)    (1−B) n− p+λp(1−δ ) n, p,k n + 1 − λ(1 − δn, p,k ) − 1 − λ(1 − δn, p,k ) p p(A−B) For n  p + 1 and

n− p k

∈ / N, (2.5) becomes

C  D+

1− D n(1−B) p(A−B)

+

(1−λ)(n− p) n− p(1−λ)

=: ϕ(λ, n).

(2.6)

< 0 for all real x  p + 1 and 0  λ  1, and so A simple calculation shows that ∂ϕ(λ,x) ∂x ϕ(λ, n) is decreasing in n. Hence we have ϕ(λ, n)  ϕ(λ, p + 1) = D +

1− D ( p+1)(1−B) p(A−B)

+

1−λ 1+λp

.

(2.7)

Integral transforms and partial sums. . .

For n  p + 1 and

n− p k

∈ N, (2.5) reduces to C  D+

1− D n(1−B) p(A−B)

+1

= ϕ(0, n),

(2.8)

and we thus find that ϕ(0, n)  ϕ(0, p + k) = D +

1− D ( p+k)(1−B) p(A−B)

+1

.

(2.9)

It is obvious that ϕ(0, p + k) < ϕ(λ, p +  1). Thus, by  taking C = ϕ(λ, p + 1) = C(D), it follows from (2.4) to (2.9) that f ∈ G p,k λ, C(D), D . Furthermore, for D < C0 < C(D), we have p2 (A − B) ( p + 1)(1 − D) − p(1 − C0 )(1 − λ) · p(C0 − D) ( p + 1)[( p + 1)(1 − B) − p(1 − A)(1 − λ)]   ( p + 1)(1 − D) − p 1 − C(D) (1 − λ) p2 (A − B)   > · ( p + 1)[( p + 1)(1 − B) − p(1 − A)(1 − λ)] p C(D) − D = 1,

which implies that the function f (z) = z p +

p 2 (A − B) z p+1 ∈ T p,k (λ, A, B) (2.10) ( p + 1)[( p + 1)(1 − B) − p(1 − A)(1 − λ)]

is not in the class G p,k (λ, C0 , D). The proof of Theorem 1 is completed.



Putting D = B, Theorem 1 reduces to the following corollary. Corollary 1 It is asserted that T p,k (λ, A, B) ⊂ G p,k (λ, C(B), B), where C(B) = B +

p(1 − B)(A − B)(1 + λp) ∈ (B, A) ( p + 1)(1 − B)(1 + λp) + p(1 − λ)(A − B)

cannot be decreased for each B. Remark Corollary 1 refines the inclusion relation (2.1).

3 Integral transforms In this section, we consider several properties of integral transforms for the classes G p,k (λ, A, B) and T p,k (λ, A, B). Theorem 2 Let f ∈ G p,k (λ, A, B) and  μ + p z μ−1 t f (t)dt (μ > − p). Iμ (z) = zμ 0

(3.1)

Then Iμ ∈ G p,k (λ, C1 (D), D), where −1  D  0 and C1 (D) = D +

(μ + p)(1 + λp)(A − B)(1 − D) . (1 + λp)(μ + p + 1)(1 − B) + p(1 − λ)(A − B)

The number C1 (D) cannot be decreased for each D.

(3.2)

J.-L. Liu et al.

Proof Clearly D < C1 (D) < 1. For ∞ 

f (z) = z p +

an z n ∈ G p,k (λ, A, B),

n= p+1

it follows from (3.1) that Iμ (z) = z p +

∞  μ+ p an z n . μ+n

(3.3)

n= p+1

In order to prove that Iμ ∈ G p,k (λ, C1 (D), D), we need only to find the smallest C (D < C  1) such that   n(1 − D) − p(1 − C) 1 − λ(1 − δn, p,k ) μ + p · p(C − D) μ+n   n(1 − B) − p(1 − A) 1 − λ(1 − δn, p,k )  (3.4) p(A − B) for all n  p + 1. For n  p + 1 and

n− p k

∈ / N, (3.4) becomes

C  D+

1− D (μ+n)(1−B) (μ+ p)(A−B)

+

p(1−λ)(n− p) 



:= ψ(λ, n).

(3.5)

(μ+ p) n− p(1−λ)

It is easy to show that ψ(λ, n) (n  p + 1, 0  λ  1) is a decreasing function of n and so ψ(λ, n)  ψ(λ, p + 1) = D + For n  p + 1 and

n− p k

1− D (μ+ p+1)(1−B) (μ+ p)(A−B)

+

p(1−λ) (μ+ p)(1+λp)

.

(3.6)

∈ N, Eq. (3.4) reduces to

C  D+

1− D (μ+n)(1−B) (μ+ p)(A−B)

+

and we have ψ(0, n)  ψ(0, p + k) = D +

p μ+ p

= ψ(0, n),

(3.7)

1− D (μ+ p+k)(1−B) (μ+ p)(A−B)

+

p μ+ p

.

(3.8)

It is obvious that ψ(0, p + k) < ψ(λ, p + 1). Therefore, by taking C = ψ(λ, p + 1) = C1 (D), it follows from (3.4) to (3.8) that Iμ ∈ G p,k (λ, C1 (D), D). Finally, the number C1 (D) is the best possible for the function f (z) = z p +

p(A − B) z p+1 ∈ G p,k (λ, A, B). ( p + 1)(1 − B) − p(1 − λ)(1 − A)

The proof of Theorem 2 is completed.



Theorem 3 Let Iμ (μ > − p) andC1 (D) (−1   D  0) be the same as in Theorem 2. If f ∈ T p,k (λ, A, B), then Iμ ∈ T p,k λ, C1 (D), D . The number C1 (D) cannot be decreased for each D.

Integral transforms and partial sums. . .

Proof From (3.3) we have ⎛

⎞ ∞  μ + p n z ⎠ ∗ f (z) Iμ (z) = ⎝z + μ+n p

n= p+1

and so z Iμ (z) p

⎞    ∞  z f (z) μ + p = ⎝z p + zn ⎠ ∗ . μ+n p ⎛

(3.9)

n= p+1

In view of (3.9) and (1.8), an application of Theorem 2 yields Theorem 3. The proof is completed.

Theorem 4 Let f ∈ T p,k  (λ, A, B) and Iμ (μ > − p) be the same as in Theorem 2. Then Iμ ∈ G p,k λ, C2 (D), D , where −1  D  0 and C2 (D) = D +

p(μ + p)(1 + λp)(A − B)(1 − D) . ( p + 1)[(1 + λp)(μ + p + 1)(1 − B) + p(1 − λ)(A − B)] + p(1 − λ)(μ + p)(A − B)

(3.10) The number C2 (D) cannot be decreased for each D. Proof We note that D < C2 (D) < 1. Let f ∈ T p,k (λ, A, B). Then it follows from Theorem 3 and Corollary 1 that f ∈ T p,k (λ, A, B) ⇒ Iμ ∈ T p,k (λ, C1 (D), D) ⇒ Iμ ∈ G p,k (λ, C2 (D), D), where C1 (D) and C2 (D) are given by (3.2) and (3.10), respectively. Furthermore, for the function f (z) = z p +

p 2 (A − B) z p+1 ∈ T p,k (λ, A, B) ( p + 1)[( p + 1)(1 − B) − p(1 − λ)(1 − A)]

and D < C0 < C2 (D), we have Iμ (z) = z p +

p 2 (A − B)(μ + p) z p+1 ( p + 1)(μ + p + 1)[( p + 1)(1 − B) − p(1 − λ)(1 − A)]

and ( p + 1)(1 − D) − p(1 − λ)(1 − C0 ) p(C0 − D) p 2 (A − B)(μ + p) · ( p + 1)(μ + p + 1)[( p + 1)(1 − B) − p(1 − λ)(1 − A)]   ( p + 1)(1 − D) − p(1 − λ) 1 − C2 (D) > C2 (D) − D p(A − B)(μ + p) · ( p + 1)(μ + p + 1)[( p + 1)(1 − B) − p(1 − λ)(1 − A)] = 1. / G p,k (λ, C0 , D) and the proof of Theorem 4 is completed. Hence Iμ ∈



J.-L. Liu et al.

4 Partial sums In this section, we let f ∈ A( p) be given by (1.1) and define the partial sums s1 (z) and sm (z) by p+m−1  s1 (z) = z p and sm (z) = z p + an z n (m ∈ N\{1}). (4.1) n= p+1

For simplicity, we use the notation αn (n  p + 1) defined by (1.9). Theorem 5 Let f ∈ G p,k (λ, A, B). Also let either (a) p(1 − A)  1 − B and 0  λ  1; or 1−B (b) p(1 − A) > 1 − B and 0  λ  p(1−A) . Then, for m ∈ N,   

and 

f (z) sm (z)

sm (z) f (z)

 >1−

 >

1

(z ∈ U)

(4.2)

α p+m (z ∈ U). 1 + α p+m

(4.3)

α p+m

The bounds in (4.2) and (4.3) are sharp for each m. Proof If either (a) or (b) is satisfied, then (for n  p + 1)   n(1 − B) − p(1 − A) 1 − λ(1 − δn, p,k ) ( p + 1)(1 − B) − p(1 − A)  αn = p(A − B) p(A − B) p(A − B) + 1 − B >1 (4.4) = p(A − B) and 1 − B + λp(1 − A)(δn, p,k − δn+1, p,k ) p(A − B) 1 − B − λp(1 − A)  αn +  αn . p(A − B)

αn+1 = αn +

(4.5)

Let f ∈ G p,k (λ, A, B). Then it follows from (4.4) and (4.5) that p+m−1  n= p+1

|an | + α p+m

∞  n= p+m

∞ 

|an | 

αn |an |  1 (m ∈ N\{1}).

n= p+1

If we put  g1 (z) = 1 + α p+m

f (z) −1 sm (z)



(4.6)

Integral transforms and partial sums. . .

for z ∈ U and m ∈ N\{1}, then g1 (0) = 1 and we deduce from (4.6) that        n− p α p+m ∞ a z   g1 (z) − 1   n= p+m n  =     p+m−1    g (z) + 1   n− p  1  2 1 + n= p+1 an z n− p + α p+m ∞ n= p+m an z  α p+m ∞ n= p+m |an |   p+m−1  2 − 2 n= p+1 |an | − α p+m ∞ n= p+m |an |  1 (z ∈ U; m ∈ N\{1}).  This shows that  g1 (z) > 0 (z ∈ U ), and so (4.2) holds true for m ∈ N\{1}. Similarly, by taking 

g2 (z) = (1 + α p+m )

sm (z) − α p+m , f (z)

we can see from (4.6) that        n− p −(1 + α p+m ) ∞ a z   g2 (z) − 1   n n= p+m  =     p+m−1    g (z) + 1   ∞ 2  2 1 + n= p+1 an z n− p + (1 − α p+m ) n= p+m an z n− p   (1 + α p+m ) ∞ n= p+m |an |   p+m−1  2 − 2 n= p+1 |an | − (α p+m − 1) ∞ n= p+m |an |  1 (z ∈ U; m ∈ N\{1}). Hence we have (4.3) for m ∈ N\{1}. For m = 1, upon replacing (4.6) by α p+1

∞ 

|an | 

n= p+1

∞ 

αn |an |  1

n= p+1

and proceeding as the above, we see that (4.2) and (4.3) are also true. Finally, by taking the function f defined by f (z) = z p +

z p+m ∈ G p,k (λ, A, B), α p+m

we have sm (z) = z p ,     f (z) zm 1  = 1+ →1− sm (z) α p+m α p+m and

 

sm (z) f (z)



 =

α p+m m z + α p+m

The proof of Theorem 5 is completed.

 →

 as z → exp

α p+m 1 + α p+m

πi m



as z → 1.



Theorem 6 Let f ∈ T p,k (λ, A, B) and let either (a) or (b) in Theorem 5 be satisfied. Then, for m ∈ N,   p f (z) >1− (z ∈ U) (4.7)  sm (z) ( p + m)α p+m

J.-L. Liu et al.



and

sm (z)  f (z)



( p + m)α p+m (z ∈ U). p + ( p + m)α p+m

>

(4.8)

The bounds in (4.7) and (4.8) are sharp for the function f (z) given by f (z) = z p +

pz p+m ∈ T p,k (λ, A, B). ( p + m)α p+m

(4.9)

Proof In view of the assumptions of Theorem 6, it follows from (4.4) and (4.5) that p+m−1 

|an | +

n= p+1

( p + m)α p+m p

and

∞  n= p+m ∞ 

α p+1

∞  n αn |an |  1 (m ∈ N\{1}) p

|an | 

(4.10)

n= p+1

∞  n αn |an |  1. p

|an | 

n= p+1

(4.11)

n= p+1

If we put g1 (z) = 1 + and

( p + m)α p+m p



( p + m)α p+m g2 (z) = 1 + p





f (z) −1 sm (z)



sm (z) ( p + m)α p+m − , f (z) p

then (4.10) and (4.11) lead to    g j (z) > 0 (z ∈ U; j = 1, 2).



This completes the proof of Theorem 6.

Theorem 7 Let f ∈ T p,k (λ, A, B) and let either (a) or (b) in Theorem 5 be satisfied. Then, for m ∈ N,    f (z) 1   (z ∈ U) (4.12) >1− sm (z) α p+m 

and

s  (z)  m f (z)

 >

α p+m (z ∈ U). 1 + α p+m

(4.13)

The bounds in (4.12) and (4.13) are sharp. Proof By virtue of the assumptions of Theorem 7, it follows from (4.4) and (4.5) that p+m−1 α p+m 1  n|an | + p p n= p+1

and

∞ 

n|an | 

n= p+m

∞  n αn |an |  1 (m ∈ N\{1}) p

∞ ∞  α p+1  n n|an |  αn |an |  1. p p n= p+1

(4.14)

n= p+1

n= p+1

(4.15)

Integral transforms and partial sums. . .

By taking the functions

 g1 (z) = 1 + α p+m

f  (z) −1 sm (z)



and  s  (z)  g2 (z) = 1 + α p+m m − α p+m , f (z) we deduce from (4.14) and (4.15) that    g j (z) > 0 (z ∈ U; j = 1, 2), which shows that (4.12) and (4.13) hold true. Finally, the bounds in (4.12) and (4.13) are the best possible for the function f (z) defined by (4.9). The proof of Theorem 7 is completed.



5 Distortion bounds In this section, we shall derive several distortion theorems for the classes G p,k (λ, A, B) and T p,k (λ, A, B). Theorem 8 Suppose that either (a) p(1 − A) ≤ (k − 1)(1 − B) and 0 ≤ λ ≤ 1; or (b) p(1 − A) > (k − 1)(1 − B) and 0 ≤ λ ≤ (k−1)(1−B) p(1−A) . (i) If f ∈ G p,k (λ, A, B), then p(A − B) |z| p+1 ≤ | f (z)| ( p + 1)(1 − B) − p(1 − A)(1 − λ) p(A − B) ≤ |z| p + |z| p+1 ( p + 1)(1 − B) − p(1 − A)(1 − λ)

|z| p −

(5.1)

for z ∈ U. The bounds in (5.1) are sharp for the function f (z) = z p −

p(A − B) z p+1 ∈ G p,k (λ, A, B). ( p + 1)(1 − B) − p(1 − A)(1 − λ)

(ii) If f ∈ T p,k (λ, A, B), then   p(A − B) p |z| p−1 − |z| p ≤ | f  (z)| ( p + 1)(1 − B) − p(1 − A)(1 − λ)   p(A − B) ≤ p |z| p−1 + |z| p ( p + 1)(1 − B) − p(1 − A)(1 − λ)

(5.2)

(5.3)

for z ∈ U. The bounds in (5.3) are sharp for the function f (z) = z p −

p 2 (A − B) z p+1 ∈ T p,k (λ, A, B). ( p + 1)[( p + 1)(1 − B) − p(1 − A)(1 − λ)] (5.4)

Proof For n ≥ p + 1 and

n− p k

∈ N, we have n = p + mk (m ∈ N), δn, p,k = 1 and

n(1 − B) − p(1 − A)(1 − λ(1 − δn, p,k )) ≥ ( p + k)(1 − B) − p(1 − A) ≥ 0.

(5.5)

J.-L. Liu et al.

For n ≥ p + 1 and

n− p k

∈ / N, we have δn, p,k = 0, δ p+m, p,k = 0 (1 ≤ m ≤ k − 1) and

n(1 − B) − p(1 − A)(1 − λ(1 − δn, p,k )) ≥ ( p + 1)(1 − B) − p(1 − A)(1 − λ) ≥ 0. (5.6) If either (a) or (b) is satisfied, then ( p + k)(1 − B) − p(1 − A) ≥ ( p + 1)(1 − B) − p(1 − A)(1 − λ).

(5.7)

(i) If ∞ 

f (z) = z p +

an z n ∈ G p,k (λ, A, B),

n= p+1

then it follows from (5.5) to (5.7) that [( p + 1)(1 − B) − p(1 − A)(1 − λ)]

∞ 

|an |

n= p+1

≤

∞ 

[n(1 − B) − p(1 − A)(1 − λ(1 − δn, p,k ))]|an | ≤ p(A − B).

n= p+1

Hence we obtain | f (z)| ≤ |z| p + |z| p+1

∞ 

|an | ≤ |z| p +

n= p+1

p(A − B) |z| p+1 ( p + 1)(1 − B) − p(1 − A)(1 − λ)

and ∞ 

| f (z)| ≥ |z| p − |z| p+1

|an | ≥ |z| p −

n= p+1

p(A − B) |z| p+1 ≥ 0 ( p + 1)(1 − B) − p(1 − A)(1 − λ)

for z ∈ U. (ii) If f (z) = z p +

∞ 

an z n ∈ T p,k (λ, A, B),

n= p+1

then (5.5) to (5.7) yield [( p + 1)(1 − B) − p(1 − A)(1 − λ)]

∞ 

n|an | ≤ p 2 (A − B).

n= p+1

From this we easily have (5.3). The proof of Theorem 8 is completed. The following theorem will give the remaining part of Theorem 8.



Integral transforms and partial sums. . .

Theorem 9 Let p(1 − A) > (k − 1)(1 − B) and (i) If f (z) = z p +

∞

n= p+1 an z p+k−1 

| f (z)| ≥ |z| p −

n

(k − 1)(1 − B) < λ ≤ 1. p(1 − A)

(5.8)

∈ G p,k (λ, A, B), then for z ∈ U,

|an ||z|n

n= p+1

−

p(A − B) −

 p+k−1

n= p+1 [n(1 −

B) − p(1 − A)(1 − λ)]|an |

( p + k)(1 − B) − p(1 − A)

|z| p+k

(5.9)

and p+k−1 

| f (z)| ≤ |z| p +

|an ||z|n

n= p+1

+

 p+k−1

n= p+1 [n(1 −

p(A − B) −

B) − p(1 − A)(1 − λ)]|an |

( p + k)(1 − B) − p(1 − A)

|z| p+k .

(5.10)

Equalities in (5.9) and (5.10) are attained, for example, by the function f (z) = z p − (ii) If f (z) = z p +

p(A − B) z p+k ∈ G p,k (λ, A, B). ( p + k)(1 − B) − p(1 − A)

∞

n= p+1 an z

| f  (z)| ≥ p|z| p−1 −

p+k−1 

n

(5.11)

∈ T p,k (λ, A, B), then for z ∈ U,

n|an ||z|n−1

n= p+1

−

p 2 (A

− B) −

 p+k−1

n= p+1 n[n(1 −

B) − p(1 − A)(1 − λ)]|an |

( p + k)(1 − B) − p(1 − A)

|z| p+k−1

(5.12)

|z| p+k−1 .

(5.13)

and | f  (z)| ≤ p|z| p−1 +

p+k−1 

n|an ||z|n−1

n= p+1

+

p 2 (A − B) −

 p+k−1

n= p+1 n[n(1 −

B) − p(1 − A)(1 − λ)]|an |

( p + k)(1 − B) − p(1 − A)

Equalities in (5.12) and (5.13) are attained, for example, by the function f (z) = z p −

p 2 (A − B) z p+k ∈ T p,k (λ, A, B). ( p + k)[( p + k)(1 − B) − p(1 − A)]

(5.14)

J.-L. Liu et al.

 n Proof (i) If f (z) = z p + ∞ n= p+1 an z ∈ G p,k (λ, A, B), then from (5.5), (5.6) and (5.8) that ∞  p(A − B) ≥ [n(1 − B) − p(1 − A)(1 − λ(1 − δn, p,k ))]|an | n= p+1

≥

p+k−1 

[n(1 − B) − p(1 − A)(1 − λ)]|an | + [( p + k)(1 − B) − p(1 − A)]

n= p+1

×

∞ 

|an |.

n= p+k

From this we easily have  (5.9) and (5.10). n (ii) If f (z) = z p + ∞ n= p+1 an z ∈ T p,k (λ, A, B), then we have ∞ 

p 2 (A − B) ≥

n[n(1 − B) − p(1 − A)(1 − λ(1 − δn, p,k ))]|an |

n= p+1

≥

p+k−1 

n[n(1 − B) − p(1 − A)(1 − λ)]|an | + [( p + k)(1 − B) − p(1 − A)]

n= p+1

×

∞ 

n|an |.

n= p+k

This leads to (5.12) and (5.13). The proof of Theorem 9 is completed.



Acknowledgements The authors would like to express sincere thanks to the referees for careful reading and suggestions which helped us to improve the paper. This work was supported by the National Natural Science Foundation of the People’s Republic of China (Grant no. 11571299), the Natural Science Foundation of the Jiangsu Province of the People’s Republic of China (Grant no. BK20151304) and the Natural Science Foundation of Jiangsu Gaoxiao (Grant no. 17KJB110019).

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