A family of meromorphically multivalent functions which are starlike with respect to k-symmetric points

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Citation for this paper:

Liu, J.-L.; Srivastava, H. M.; Yuan, Y. (2017). A family of meromorphically

multivalent functions which are starlike with respect to k-symmetric points. Journal of Mathematical Inequalities, 11(3), 781-798. DOI: 10.7153/jmi-2017-11-61

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A family of meromorphically multivalent functions which are starlike with respect to k-symmetric points

Jin-Lin Liu, H. M. Srivastava and Yuan Yuan 2017

This article was originally published at: https://dx.doi.org/10.7153/jmi-2017-11-61

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Inequalities

Volume 11, Number 3 (2017), 781–798 doi:10.7153/jmi-2017-11-61

A FAMILY OF MEROMORPHICALLY MULTIVALENT FUNCTIONS WHICH ARE STARLIKE WITH RESPECT TO k –SYMMETRIC POINTS

JIN-LINLIU, H. M. SRIVASTAVA AND YUANYUAN (Communicated by S. Hencl)

Abstract. In this paper, two new subclassesRp,k(λ,A,B) and Tp,k(λ,A,B) of meromorphically

multivalent functions starlike with respect to k -symmetric points are studied. Distortion bounds, inclusion relations and convolution properties for each of these classes are obtained.

1. Introduction, definitions and preliminaries

Thoughout this paper, we assume that

N = {1,2,3,...}, k ∈ N \ {1}, −1 B < 0, B < A −B and λ 1. (1.1) For functions f and g analytic in the open unit disk

U = {z : z ∈ C and |z| < 1},

the function f is said to be subordinate to g, written f(z) ≺ g(z) (z ∈ U), if there exists an analytic function w in _{U, with w(0) = 0 and |w(z)| < 1, such that f (z) = g(w(z)).}

Let Σp denote the class of functions of the form: f_{(z) = z}−p₊

∑

∞

n=p

anzn (p ∈ N), (1.2)

which are analytic in the punctured open unit disk U0= U \ {0}.

A function f ∈ Σp is said to be meromorphically starlike with respect to k -symmetric points, if it satisfies

ℜ −z f_f (z) p,k(z) > 0, where f_p,k_{(z) =} 1 k k−1

∑

j=0ε jp k f(εkjz) and εk = exp ₂ πi k .

Mathematics subject classification (2010): Primary 30C45.

Keywords and phrases: Meromorphically multivalent functions, Hadamard product (or convolution), distortion bounds, inclusion relations, convolution properties, subordination properties, symmetric points.

c

, Zagreb

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Let fj(z) = z−p+ ∞

∑

n=p a_{n, j}zn∈ Σp ( j = 1,2).

Then the Hadamard product (or convolution) of f1 and f2 is defined by

( f1∗ f2)(z) = z−p+ ∞

∑

n=p

a_n,1a_n,2zn_{= ( f}₂_{∗ f}₁_)(z).

The following lemma will be required in our investigation. LEMMA. Let f ∈ Σp defined by (1.2) satisfy

∞

∑

n=p[λn(1 − B) + p(1 − A)δn,p,k]|an| p(A − B).

(1.3) Then p(1 −λ)z−p₋_{λz f}_(z) p fp,k(z) ≺ 1+ Az 1+ Bz (z ∈ U), (1.4) where fp,k(z) = 1_k k−1

∑

j=0ε jp k f(εkjz), εk= exp ₂ πi k (1.5) and δn,p,k= 0n+p_k /∈ N, 1n+p_k _{∈ N}_. (1.6)

Proof. For f ∈ Σp defined by (1.2), the function fp,k in (1.5) can be expressed as f_p,k(z) = z−p+

∑

∞ n=pδn,p,k anzn (1.7) with δn,p,k= 1_k k−1

∑

j=0ε j(n+p) k = 0 n+p_k _{/∈ N}_, 1 n+p_k ∈ N.

In view of (1.1) and (1.6), we see that

Apδn,p,k+ Bλn B(λn − pδn,p,k) 0 (n p). (1.8)

Let the inequality (1.3) hold true. Then from (1.7) and (1.8) we deduce that p(1−λ )z−p−λ z f(z) p fp,k(z) − 1 A− Bp(1−λ )z_{p f}−p_p,k−_(z)λ z f(z) = ∑∞n=p(λn + pδn,p,k)anzn+p p_{(A − B) + ∑}∞_n_=p(Apδn,p,k+ Bλn)anzn+p _p ∑∞n=p(λn + pδn,p,k)|an| (A − B) + ∑∞n=p(Apδn,p,k+ Bλn)|an| 1 (|z| = 1).

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Hence, by the Maximum Modulus Theorem, we arrive at (1.4). We now consider the following two subclasses of _Σp.

DEFINITION 1. A function f ∈ Σ_p defined by (1.2) is said to be in the class Rp,k(λ,A,B) if and only if it satisfies the coefficient inequality (1.3).

It follows from the Lemma that, if f _{∈ Rp,k}(λ,A,B), then the subordination rela-tion (1.4) holds true.

DEFINITION 2. A function f ∈ Σ_p defined by (1.2) is said to be in the class Tp,k(λ,A,B) if and only if it satisfies

∞

∑

n=pn[λn(1 − B) + p(1 − A)δn,p,k]|an| p

2_{(A − B).} _(1.9)

For f ∈ Σp defined by (1.2), we have

2z−p+z f_p(z) = z−p+

∑

∞

n=p

n panzn,

which implies that

f _{∈ T}_p,k(λ,A,B) if and only if 2z−p+z f_p(z) ∈ Rp,k(λ,A,B). (1.10)

If we write

αn= λn(1 − B) + p(1 − A)δ_p n,p,k

(A − B) and βn=

n

p αn (n p), (1.11)

then it is easy to verify that ∂βn ∂λ = n p ∂αn ∂λ > 0, ∂βn ∂A = n p ∂αn ∂A < 0 and ∂βn ∂B = n p ∂αn ∂B 0. Thus we have the following inclusion relations. If

1λ1λ, −1 B1 B < 0, B < A −B and A A1 −B1,

then

Tp,k(λ,A,B) ⊂ Rp,k(λ,A,B) ⊂ Rp,k(λ1,A1,B1) ⊂ Rp,k(1,1,−1). (1.12)

Therefore, by the Lemma, we see that each function in the classes Rp,k(λ,A,B) and

Tp,k(λ,A,B) is meromorphically starlike with respect to k-symmetric points. Mero-morphic (and analytic) functions which are starlike with respect to symmetric points and related functions have been extensively studied by several authors (see, e.g., [1, 2, 3, 6, 7, 8, 9] and [12] to [15]; see also the recent works [10] and [11]).

There are several papers which study the convolution properties of functions in dif-ferent function classes, and sometimes these questions might turn out to be very difficult (see, e.g., [5] and the references therein). Also, many authors investigate the distortion bounds of functions in various function classes (see, e.g., [4] and the references therein). In the present paper, we obtain distortion bounds, inclusion relations and convolution properties for each of the above-defined classes Rp,k(λ,A,B) and Tp,k(λ,A,B).

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2. Distortion bounds

THEOREM 1. Let 2p_k ∈ N and suppose that either

(a) 1− B p(1 − A) and λ 1 or (b) 1_{− B < p(1 − A) and} _λ p(1−A)₁_−B . Then, if we denote

C₁₌ A− B

λ(1 − B) + 1 − A,

we have the following:

(i) If f ∈ Rp,k(λ,A,B), then for z ∈ U0,

|z|−p−C1|z|p | f (z)| |z|−p+C1|z|p. (2.1)

(ii) If f ∈ Tp,k(λ,A,B), then for z ∈ U0,

p|z|−p−1−C1|z|p−1 | f(z)| p|z|−p−1+C1|z|p−1. (2.2)

The bounds in (2.1) and (2.2) are sharp.

Proof. Let 2p_k _{∈ N. For n p and} n+p_k _{∈ N, we have n = p+k(m−1) (m ∈ N),}

δn,p,k= 1, and so

λn(1 − B) + p(1 − A)δn,p,k p(A − B)

λ(1 − B) + 1 − A

A− B . (2.3)

For n p and n+p_k /∈ N, we have δn,p,k=δp+1,p,k= 0 and

λn(1 − B) + p(1 − A)δn,p,k p_{(A − B)}

λ(p + 1)(1 − B)

p_{(A − B)} . (2.4)

If either (a) or (b) is satisfied, then λ(p + 1)(1 − B) p_{(A − B)} λ(1 − B) + 1 − A A_{− B} . (2.5) (i) If f(z) = z−p+

∑

∞ n=p anzn∈ Rp,k(λ,A,B),

then it follows from (1.3) and (2.3) to (2.5) that λ(1 − B) + 1 − A A− B ∞

∑

n=p|an| 1. Hence we have | f (z)| |z|−p+ |z|p

∑

∞ n=p|an| |z| −p₊ A− B λ(1 − B) + 1 − A|z|p

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and | f (z)| |z|−p− |z|p

_∑

∞ n=p|an| |z| −p₋ A− B λ(1 − B) + 1 − A|z|p> 0 for z∈ U0. (ii) If f(z) = z−p+

∑

∞ n=panz n_{∈ T} p,k(λ,A,B),

then it follows from (1.9) and (2.3) to (2.5) that λ(1 − B) + 1 − A p(A − B) ∞

∑

n=p n_{|an| 1.} This leads to (2.2).

Furthermore, the bounds in (2.1) and (2.2) are sharp for the function

f_{(z) = z}−p₊ A− B

λ(1 − B) + 1 − Azp∈ Tp,k(λ,A,B) ⊂ Rp,k(λ,A,B). (2.6) THEOREM 2. Let 2p_k ∈ N and suppose that

(1 − B) < p(1 − A) and 1λ < p(1 − A)₁ − B , and let f_{(z) = z}−p₊

∑

∞ n=p anzn. Then, if we denote C₂= p(A − B) − p(λ(1 − B) + 1 − A)|ap| λ(p + 1)(1 − B) ,

|z|−p− |ap||z|p−C2|z|p+1 | f (z)| |z|−p+ |ap||z|p+C2|z|p+1. (2.7)

(ii) If f _{∈ T}_p,k(λ,A,B), then for z ∈ U0,

p|z|−p−1− |ap||z|p−1−C2|z|p | f(z)| p|z|−p−1+ |ap||z|p−1+C2|z|p. (2.8)

The bounds in (2.7) and (2.8) are sharp.

Proof. Note that 1λ < p(1−A)₁_−B implies that λ(1 − B) + 1 − A

A_{− B}

λ(p + 1)(1 − B)

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(i) For f(z) = z−p_{+ a}_p_zp_{+ ··· ∈ R}

p,k(λ,A,B), it follows from (2.3), (2.4) (used

in the proof of Theorem 1) and (2.9) that λ(1 − B) + 1 − A A− B |ap| + λ(p + 1)(1 − B) p(A − B) ∞

∑

n=p+1|an| 1.

From this we easily have (2.7).

The bounds in (2.7) are sharp for the function

f(z) = z−p+ p(A − B)

λ(p + 1)(1 − B)zp+1∈ Rp,k(λ,A,B). (2.10) (ii) For f_{(z) = z}−p_{+ a}pzp+ ··· ∈ Tp,k(λ,A,B), from (2.3), (2.4) and (2.9) we

deduce that λ(1 − B) + 1 − A A− B |ap| + λ(p + 1)(1 − B) p2_{(A − B)} ∞

∑

n=p+1 n_{|an| 1.} Hence we have (2.8).

The bounds in (2.8) are sharp for the function

f(z) = z−p+ p2(A − B)

λ(p + 1)2_{(1 − B)}zp+1∈ Tp,k(λ,A,B). (2.11)

THEOREM 3. Let 2p_k /∈ N. Then, if we denote C3= A− B

λ(1 − B),

|z|−p−C3|z|p | f (z)| |z|−p+C3|z|p. (2.12)

(ii) If f ∈ Tp,k(λ,A,B), then for z ∈ U0,

p|z|−p−1−C3|z|p−1 | f(z)| p|z|−p−1+C3|z|p−1. (2.13)

The bounds in _{(2.12) and (2.13) are sharp.}

Proof. Let 2p_k /∈ N. For n p and n+p_k /∈ N, we have δn,p,k=δp,p,k= 0 and

λn(1 − B) + p(1 − A)δn,p,k p(A − B)

λ(1 − B)

A− B . (2.14)

For n p and n+p_k ∈ N, we have

δn,p,k= 1, n = k _2p k + m − p > p (m ∈ N),

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and λn(1 − B) + p(1 − A)δn,p,k p(A − B) > λ(1 − B) + 1 − A A− B λ(1 − B) A− B , (2.15)

where [a] denotes the integer part of a given real number a. (i) If f(z) = z−p_{+ ∑}∞

n=panzn ∈ Rp,k(λ,A,B), then it follows from (2.14) and

(2.15) that λ(1 − B) A_{− B} ∞

∑

n=p|an| 1, which leads to (2.12). (ii) If f(z) = z−p_{+ ∑}∞

n=panzn∈ Tp,k(λ,A,B), then (2.14) and (2.15) give

λ(1 − B) p_{(A − B)} ∞

∑

n=pn|an| 1, which yields (2.13).

Furthermore, the function f defined by

f(z) = z−p+ _{λ(1 − B)}A− B zp∈ Tp,k(λ,A,B) ⊂ Rp,k(λ,A,B) (2.16)

shows that the bounds in (2.12) and (2.13) are best possible.

3. Inclusion relations

In this section, we generalize the above-mentioned inclusion relation (1.12)

Tp,k(λ,A,B) ⊂ Rp,k(λ,A,B) (3.1) as follows. THEOREM 4. If −1 D B, then Tp,k(λ,A,B) ⊂ Rp,k(λ,C(D),D), (3.2) where C_{(D) = D +}(1 − D)(A − B) 1− B . (3.3) The number C_{(D) cannot be decreased for each D.}

Proof. Since B_{< A −B and −1 D B < 0, we see that} D< C(D) D −2B₁(1 − D)

− B −D.

Let f ∈ Tp,k(λ,A,B). In order to prove that f ∈ Rp,k(λ,C(D),D), we only need

to find the smallest C (D < C −D) and show that it equals to C(D) such that λn(1 − D) + p(1 −C)δn,p,k

p_{(C − D)}

n[λn(1 − B) + p(1 − A)δn,p,k]

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for all n p, that is, that (λn + pδn,p,k)(1 − D) p(C − D) −δn,p,k n p (λn + pδn,p,k)(1 − B) p(A − B) −δn,p,k (n p). (3.5) For n p and n+p_k ∈ N, (3.5) is equivalent to

C D + _n_(1−B)1− D

p(A−B)− λ n+pn−p

=ϕ(n) (say). (3.6) Noting that (1.1), a simple calculation shows that ϕ(n) (n p,λ 1) is decreasing in

n. Therefore ϕ(n) ⎧ ⎨ ⎩ ϕ(p) 2p_k ∈ N, ϕk2p_k + 1− p 2p_k /∈ N, (3.7)

where [a] in (3.7) denotes the integer part of a given real number a. For n_{p and} n+p_k _{/∈ N, (3.5) becomes}

C D + 1_n_(1−B)− D p(A−B) =ψ(n) (say) (3.8) and ψ(n) ⎧ ⎨ ⎩ ψ(p + 1)2p_k ∈ N, ψ(p) 2p_k /∈ N. (3.9) Consequently, by taking C=ϕ(p) = ψ(p) = D +(1 − D)(A − B)₁ − B = C(D), (3.10)

it follows from (3.4) to (3.10) that f _{∈ Rp,k}(λ,C(D),D). Furthermore, for 2p_k ∈ N and D < C0< C(D), we have

λ(1 − D) + 1 −C0 C0− D · A_{− B} λ(1 − B) + 1 − A> λ(1 − D) + 1 −C(D) C(D) − D · A_{− B} λ(1 − B) + 1 − A= 1, which implies that the function f ∈ Tp,k(λ,A,B) defined by (2.6) is not in the class Rp,k(λ,C0,D). Also, for 2p_k /∈ N and D < C0< C(D), we have

λ(1 − D) C0− D · A_{− B} λ(1 − B) > λ(1 − D) C(D) − D· A_{− B} λ(1 − B)= 1,

which implies that the function f ∈ Tp,k(λ,A,B) defined by (2.16) is not in the class Rp,k(λ,C0,D). The proof of Theorem 4 is thus completed.

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4. Convolution properties

In this section, we assume that

− 1 Bj< 0 and Bj < Aj −Bj ( j = 1,2). (4.1)

Furthermore, we denote by λ1 the root in (1,+∞) of the equation:

h(λ) = aλ2+ bλ + c = 0, where ⎧ ⎨ ⎩ a= −(1 − B1)(1 − B2), b= (p − 1)(1 − B1)(1 − B2) − p[(1 − B1)(A2− B2) + (1 − B2)(A1− B1)], c_{= p[(1 − A}₁_{)(1 − A}₂_{) + (A}₁_{− B}₁_)(A₂_{− B}₂_)]. (4.2) We also denote A(B) = B + 1− B (λ + 1)∏2 j=1A1−Bj−Bjj − ∑ 2 j=1A1−Bj−Bjj + 2 λ +1 (4.3) and A(B) = B +_{λ(p + 1)}p(1 − B)

∏

2 j=1 Aj− Bj 1_{− Bj} . (4.4) THEOREM 5. Let fj∈ Rp,k(λ,Aj,Bj) ( j = 1,2) with 2p k ∈ N and − 1 B max{B1,B2}. Then we have the following:

(i) If p(1 − A1)(1 − A2) (1 − B1)(1 − B2) and λ 1, then

f1∗ f2∈ Rp,k(λ, A(B),B).

(ii) If p(1 − A1)(1 − A2) > (1 − B1)(1 − B2) and λ λ1, then

f1∗ f2∈ Rp,k(λ, A(B),B).

(iii) If p(1 − A1)(1 − A2) > (1 − B1)(1 − B2) and 1 λ < λ1, then

f₁_{∗ f}₂_{∈ R}_p,k_(λ,A(B),B).

In all cases (i)–(iii) the numbers A(B) and A(B) are optimal in the sense that they

cannot be decreased for each B.

Proof. Suppose that −1 B max{B1,B2} = Bj ( j = 1 or 2). It follows from

(4.1) and (4.4) that 1− B A(B) − B = λ(p + 1) p 2

∏

j=1 1− Bj Aj− Bj 1− Bj Aj− Bj − 1− Bj 2Bj − 1− B 2B >0,

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which implies that B< A(B) −B. Also, (4.1) and (4.3) give that 1− B A(B) − B = (λ + 1) 2

∏

_j =1 1− Bj Aj− Bj− 2

∑

j=1 1− Bj Aj− Bj + 2 λ + 1 = (λ + 1)

∏

2 j=1 1− Bj A_j_{− B}_j− 2

∏

j=1 1− Bj A_j_{− B}_j− 2

∏

j=1 1− Aj A_j_{− Bj} + 1 + 2 λ + 1 =λ

∏

2 j=1 1− Bj Aj− Bj + 2

∏

j=1 1− Aj Aj− Bj− λ − 1 λ + 1

∏

2 j=1 1− Bj Aj− Bj − 1− B 2B >0, which implies that B< A(B) −B.

Let 2p_k _{∈ N and} fj(z) = z−p+ ∞

∑

n=p a_{n, j}zn∈ Rp,k(λ,Aj,Bj) ( j = 1,2). Then ∞

∑

n=p 2

∏

j=1 λn(1 − Bj) + p(1 − Aj)δn,p,k p(Aj− Bj) |an,1an,2|

∏

2 j=1 ∞

∑

n=p λn(1 − Bj) + p(1 − Aj)δn,p,k p_(Aj− Bj) |an, j| 1. (4.5) Also, f1∗ f2∈ Rp,k(λ,A,B) if and only if

∞

∑

n=p

λn(1 − B) + p(1 − A)δn,p,k

p(A − B) |an,1an,2| 1. (4.6) In order to prove Theorem 5, it follows from (4.5) and (4.6) that we need only to find the smallest A such that

λn(1 − B) + p(1 − A)δn,p,k p(A − B) 2

∏

j=1 λn(1 − Bj) + p(1 − Aj)δn,p,k p(Aj− Bj) (n p). (4.7)

For n p and n+p_k ∈ N, (4.7) is equivalent to A B + _{(λ n+p)} 1− B p ∏2j=1A1j−B−Bjj − ∑ 2 j=1A1−Bj−Bjj + 2p λ n+p =ϕ1(n) (say). (4.8)

It can be verified that ϕ1(n) (n p,λ 1) is decreasing in n and so, in view of 2p k ∈ N, ϕ1(n) ϕ1(p) = B + 1− B (λ + 1)∏2 j=1A1j−B−Bjj − ∑ 2 j=1A1−Bj−Bjj + 2 λ +1 . (4.9)

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For n p and n+p_k /∈ N, (4.7) becomes A_{B +} 1− B λ n p ∏2j=1A1−Bj−Bjj =ψ1(n) (say) (4.10) and we have ψ1(n) ψ1(p + 1) = B + _{λ (p+1)}1− B p ∏2j=1A1j−B−Bjj . (4.11) Now (λ + 1)

∏

2 j=1 1_{− Bj} Aj− Bj− 2

∑

j=1 1_{− B}j Aj− Bj + 2 λ + 1− λ(p + 1) p 2

∏

j=1 1_{− Bj} Aj− Bj = _p h(λ) (λ + 1)(A1− B1)(A2− B2), (4.12) where h(λ) = (p − λ)(λ + 1)(1 − B1)(1 − B2) − p(λ + 1)[(1 − B1)(A2− B2)

+ (1 − B2)(A1− B1)] + 2p(A1− B1)(A2− B2)

= aλ2+ bλ + c, (4.13)

a_{= −(1 − B}₁_{)(1 − B}₂_),

b_{= (p − 1)(1 − B}₁_{)(1 − B}₂_{) − p[(1 − B}₁_)(A₂_{− B}₂_{) + (1 − B}₂_)(A₁_{− B}₁_)],

c= p(1−B1)(1−B2)+2p(A1−B1)(A2−B2)−p[(1−B1)(A2−B2)+(1−B2)(A1−B1)]

= p[(1 − A1)(1 − A2) + (A1− B1)(A2− B2)].

Note that a_{< 0, h(0) = c > 0 and}

h(1) = 2(p − 1)(1 − B1)(1 − B2) − 2p[(1 − B1)(A2− B2) + (1 − B2)(A1− B1)]

+ 2p(A1− B1)(A2− B2)

= 2[p(1 − A1)(1 − A2) − (1 − B1)(1 − B2)]. (4.14)

Therefore, if (i) or (ii) is satisfied, then it follows from (4.7) to (4.14) that h(λ) 0 for λ λ1, ψ1(p + 1) ϕ1(p) = A(B), and f1∗ f2∈ Rp,k(λ, A(B),B).

Furthermore, for B< A0< A(B), we have

λ(1−B)+1−A A0−B 2

∏

j=1 Aj−Bj λ(1−Bj)+1−Aj > λ(1−B)+1−A(B) A(B)−B 2

∏

j=1 Aj−Bj λ(1−Bj)+1−Aj = 1.

Hence the functions fj defined by fj(z) = z−p+ Aj− Bj

λ(1 − Bj) + 1 − Ajz p_{∈ R}

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show that f1∗ f2 /∈ Rp,k(λ,A0,B).

(iii) If p(1−A1)(1−A2) > (1−B1)(1−B2) and 1 λ < λ1, then we have h(λ) >

0, ϕ1(p) <ψ1(p+1) = A(B), and f1∗ f2∈ Rp,k(λ,A(B),B). Furthermore, the number

A_{(B) cannot be decreased as can be seen from the functions f}j(z) defined by fj(z) = z−p+ p(Aj− Bj)

λ(p + 1)(1 − Bj)z

p+1_{∈ Rp,k}_(λ,A_j,B

j) ( j = 1,2).

THEOREM 6. Let

f1∈ Rp,k(λ,A1,B1) and f2∈ Tp,k(λ,A2,B2)

with

2p

k ∈ N and − 1 B max{B1,B2}.

Also let A(B), A(B) and λ1 be given as in Theorem 5. Then we have the

follow-ing:

(i) If p_{(1 − A}₁_{)(1 − A}₂_{) (1 − B}₁_{)(1 − B}₂_{) and}λ 1, then

f1∗ f2∈ Tp,k(λ, A(B),B).

(ii) If p(1 − A1)(1 − A2) > (1 − B1)(1 − B2) andλ λ1, then

f1∗ f2∈ Tp,k(λ, A(B),B).

(iii) If p(1 − A1)(1 − A2) > (1 − B1)(1 − B2) and 1 λ < λ1, then

f1∗ f2∈ Tp,k(λ,A(B),B).

In all cases (i)–(iii) the numbers A(B) and A(B) are optimal in the sense that they

cannot be decreased for each B. Proof. Since [see Eq. (1.10)]

f1 ∈ Rp,k(λ,A1,B1), 2z−p+z f 2(z) p ∈ Rp,k(λ,A2,B2) and f₁_{(z) ∗} 2z−p₊z f2(z) p = 2z−p+ z( f1∗ f_p2)(z) (z ∈ U0),

an application of Theorem 5 yields the theorem.

Next, we denote by λ2 the root in (1,+∞) of the equation:

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where ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ a1= −(2p + 1)(1 − B1)(1 − B2), b1= (p2− 2p − 1)(1 − B1)(1 − B2) − p2[(1 − B1)(A2− B2) + (1 − B2)(A1− B1)], c₁_{= p}2_{(1 − B} 1)(1 − B2) − p2[(1 − B1)(A2− B2) + (1 − B2)(A1− B1)] +2p2_(A 1− B1)(A2− B2) = p2_{[(1 − A} 1)(1 − A2) + (A1− B1)(A2− B2)]. (4.15) We also denote A1(B) = B + p 2_{(1 − B)} λ(p + 1)2 2

∏

_j =1 Aj− Bj 1− Bj . (4.16) THEOREM 7. Let

f₁_{∈ R}_p,k(λ,A1,B1) and f2∈ Tp,k(λ,A2,B2)

with

2p

k ∈ N and − 1 B max{B1,B2}.

(i) If p2(1 − A1)(1 − A2) (2p + 1)(1 − B1)(1 − B2) andλ 1, then

f1∗ f2∈ Rp,k(λ, A(B),B). (ii) If p2_{(1 − A} 1)(1 − A2) > (2p + 1)(1 − B1)(1 − B2) andλ λ2, then f1∗ f2∈ Rp,k(λ, A(B),B). (iii) If p2_{(1 − A} 1)(1 − A2) > (2p + 1)(1 − B1)(1 − B2) and 1 λ < λ2, then f1∗ f2∈ Rp,k(λ,A1(B),B).

In all cases (i)–(iii) the numbers A(B) and A1(B) are optimal in the sense that

they cannot be decreased for each B. Proof. It can be verified that

1− B A1(B) − B = λ(p + 1)2 p2 2

∏

j=1 1_{− Bj} Aj− Bj > 2

∏

j=1 1_{− B}j Aj− Bj − 1− B 2B >0 and so B< A1(B) < −B.

In order to prove Theorem 7, we need only to find the smallest A such that λn(1 − B) + p(1 − A)δn,p,k p_{(A − B)} n p 2

∏

j=1 λn(1 − Bj) + p(1 − Aj)δn,p,k p_(Aj− Bj) (4.17) for all n p.

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For n p and n+p_k ∈ N, (4.17) is equivalent to A B + _n_{(λ n+p)} 1− B p2 ∏2j=1A1j−B−Bjj − n p∑2j=1A1j−B−Bjj + n+p λ n+p =ϕ2(n) (say). (4.18)

Defining the function g(λ,x) by

g(λ,x) = x(λx + p)_p₂

∏

2 j=1 1− Bj Aj− Bj− x p 2

∑

j=1 1− Bj Aj− Bj + x+ p λx + p (x p;λ 1), then ∂g(λ,x) ∂x = 2λx + p p2 2

∏

j=1 1_{− B}_j Aj− Bj− 1 p 2

∑

j=1 1_{− B}_j Aj− Bj − p(λ − 1) (λx + p)2 2λ + 1_p

∏

2 j=1 1− Bj Aj− Bj − 1 p 2

∏

_j =1 1− Bj Aj− Bj − 2

∏

_j =1 1− Aj Aj− Bj + 1 − _p λ − 1 (λ + 1)2 = 2_pλ

∏

2 j=1 1− Bj Aj− Bj + 1 p 2

∏

j=1 1− Aj Aj− Bj − 1 − _p_{(λ + 1)}λ − 1 ₂ 2_{p −}λ 1_{p −} _p_{(λ + 1)}λ − 1 ₂ > 0 (x p;λ 1),

which implies that ϕ2(n) defined by (4.18) is decreasing in n (n p). Hence, in view

of 2p_k ∈ N, we have ϕ2(n) ϕ2(p) = B + 1− B (λ + 1)∏2 j=1A1j−B−Bjj − ∑ 2 j=1A1−Bj−Bjj + 2 λ +1 .

For n p and n+p_k /∈ N, (4.17) reduces to A B + _{λ n}₂ 1− B

p2 ∏2j=1A1j−B−Bjj

=ψ2(n) (say)

and, in view of 2p_k ∈ N, we have

ψ2(n) ψ2(p + 1) = B + _{λ (p+1)}21− B p2 ∏2j=1A1−Bj−Bjj . Now (λ + 1)

∏

2 j=1 1− Bj Aj− Bj− 2

∑

j=1 1− Bj Aj− Bj + 2 λ + 1− λ(p + 1)2 p2 2

∏

j=1 1− Bj Aj− Bj = _p₂_{(λ + 1)(A}h1(λ) 1− B1)(A2− B2), (4.19)

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where h1(λ) = a1λ2+ b1λ + c1 and a1, b1, c1 are given by (4.15). Note that a1< 0,

h1(0) = c1> 0 and

h1(1) = [4p2−2(p+1)2](1−B1)(1−B2)−2p2[(1−B1)(A2−B2)+(1−B2)(A1−B1)]

+ 2p2(A1− B1)(A2− B2)

= 2[p2(1 − A1)(1 − A2) − (2p + 1)(1 − B1)(1 − B2)].

The remaining part of the proof of Theorem 7 is much akin to Theorem 5 and hence we omit it. The proof of the theorem is completed.

By applying Theorem 7, we can derive the following theorem immediately. THEOREM 8. Let

fj∈ Tp,k(λ,Aj,Bj) ( j = 1,2) with

2p

k ∈ N and − 1 B max{B1,B2}. Also let A(B), A1(B) and λ2 be given as in Theorem 7.

(i) If p2_{(1 − A} 1)(1 − A2) (2p + 1)(1 − B1)(1 − B2) andλ 1, then f1∗ f2∈ Tp,k(λ, A(B),B). (ii) If p2_{(1 − A} 1)(1 − A2) > (2p + 1)(1 − B1)(1 − B2) andλ λ2, then f1∗ f2∈ Tp,k(λ, A(B),B).

(iii) If p2(1 − A1)(1 − A2) > (2p + 1)(1 − B1)(1 − B2) and 1 λ < λ2, then

f1∗ f2∈ Tp,k(λ,A1(B),B).

they cannot be decreased for each B.

Finally, we denote by λ3 the root in (1,+∞) of the equation:

h2(λ) = a2λ2+ b2λ + c2= 0, where ⎧ ⎨ ⎩ a2= −(3p2+ 3p + 1)(1 − B1)(1 − B2), b₂_{= (p}3_−3p2_{−3p−1)(1−B} 1)(1−B2)−p3[(1−B1)(A2−B2)+(1−B2)(A1−B1)], c2= p3[(1 − A1)(1 − A2) + (A1− B1)(A2− B2)]. (4.20) We also denote A₂_{(B) = B +} p3(1 − B) λ(p + 1)3 2

∏

j=1 Aj− Bj 1− Bj . (4.21)

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THEOREM 9. Let fj∈ Tp,k(λ,Aj,Bj) ( j = 1,2) with 2p k ∈ N and − 1 B max{B1,B2}. (i) If p3_{(1 − A} 1)(1 − A2) (3p2+ 3p + 1)(1 − B1)(1 − B2) and λ 1, then f1∗ f2∈ Rp,k(λ, A(B),B). (ii) If p3_{(1 − A} 1)(1 − A2) > (3p2+ 3p + 1)(1 − B1)(1 − B2) and λ λ3, then f1∗ f2∈ Rp,k(λ, A(B),B). (iii) If p3_(1−A

1)(1−A2) > (3p2+3p+1)(1−B1)(1−B2) and 1 λ < λ3, then

f1∗ f2∈ Rp,k(λ,A2(B),B).

they cannot be decreased for each B.

Proof. It can be seen that B< A2(B) < −B. In order to prove Theorem 9, we need

only to find the smallest A such that λn(1 − B) + p(1 − A)δn,p,k p(A − B) n p _{2 2}

∏

j=1 λn(1 − Bj) + p(1 − Aj)δn,p,k p(Aj− Bj) (4.22) for all n p.

For n p and n+p_k ∈ N, (4.22) can be written as A_{B +} 1− B n2_{(λ n+p)} p3 ∏2j=1A1j−B−Bjj − n2 p2∑2j=1A1−Bj−Bjj + n2_+p2 p(λ n+p) =ϕ3(n) (say). (4.23)

Since ϕ3(n) (n p,λ 1) is decreasing in n and so

ϕ3(n) ϕ3(p) = B + 1− B (λ + 1)∏2 j=1A1−Bj−Bjj − ∑ 2 j=1A1j−B−Bjj + 2 λ +1 = A(B).

For n p and n+p_k /∈ N, (4.22) becomes A_{B +} 1− B

λ n3

p3 ∏2j=1A1j−B−Bjj

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and we have ψ3(n) ψ3(p + 1) = B + _{λ (p+1)}31− B p3 ∏2j=1A1−Bj−Bjj . Now (λ + 1)

∏

2 j=1 1− Bj Aj− Bj− 2

∑

j=1 1− Bj Aj− Bj + 2 λ + 1− λ(p + 1)3 p3 2

∏

j=1 1− Bj Aj− Bj = _p₃ h2(λ) (λ + 1)(A1− B1)(A2− B2),

where h2(λ) = a2λ2+ b2λ + c2 and a2,b2,c2 are given by (4.20).

We note that a2< 0, h2(0) = c2> 0 and

h2(1) = 2[p3(1 − A1)(1 − A2) − (3p2+ 3p + 1)(1 − B1)(1 − B2)].

The remaining part of the proof is similar to that of Theorem 5 and thus we omit it.

5. Concluding remarks and observations

In our present investigation, we have introduced and studied several properties of the two new subclasses Rp,k(λ,A,B) and Tp,k(λ,A,B) of meromorphically

mul-tivalent functions which are starlike with respect to k -symmetric points. Among the various properties derived in this paper for each of these classes are obtained, we in-clude distortion bounds, inclusion relations and convolution properties. Our results are motivated by a number of recent works (see, for example, [1] to [15]).

Acknowledgement. The authors would like to express sincere thanks to the referee

for careful reading and suggestions which helped us to improve the paper. This work is supported by National Natural Science Foundation of China(Grant No. 11571299) and Natural Science Foundation of Jiangsu Province(Grant No. BK20151304).

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[10] H. TANG, H. M. SRIVASTAVA, S.-H. LI ANDL.-N. MA, Third-order differential subordination and superordination results for meromorphically multivalent functions associated with the Liu-Srivastava operator, Abstr. Appl. Anal. 2014 (2014), Article ID 792175, 1–11.

[11] Z.-G. WANG, H. M. SRIVASTAVA ANDS.-M. YUAN, Some basic properties of certain subclasses of meromorphically starlike functions, J. Inequal. Appl. 2014 (2014), Article ID 2014:29, 1–12.

[12] Z.-G. WANG, C.-Y. GAO ANDS.-M. YUAN, On certain subclasses of close-to-convex and quasi-convex functions with respect to k -symmetric points, J. Math. Anal. Appl. 322 (2006), 97–106. [13] Z. WU, On classes of Sakaguchi functions and Hadamard products, Sci. Sinica Ser. A 30 (1987),

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(Received April 10, 2016) Jin-Lin Liu

Department of Mathematics Yangzhou University Yangzhou 225002, People’s Republic of China e-mail:jlliu@yzu.edu.cn H. M. Srivastava Department of Mathematics and Statistics University of Victoria Victoria, British Columbia V8W 3R4, Canada and Department of Medical Research China Medical University Hospital China Medical University Taichung 40402, Taiwan, Republic of China e-mail:harimsri@math.uvic.ca Yuan Yuan Department of Mathematics, Maanshan Teacher’s College Maanshan 243000, People’s Republic of China e-mail:47341653@qq.com

Journal of Mathematical Inequalities

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