Certain Subclasses of Multivalent Analytic Functions

Lemma 1. Let f ∈ A(p) defined by (2) satisfy

Then where

Proof. For f ∈ A(p) defined by (2), the function f_p,k(z) in (8) can be expressed as

with In view of (1) and (9), we see that

Let inequality (6) hold. Then from (10) and (12) we deduce that

Hence, by the maximum modulus theorem, we arrive at (7).

Definition 2. A function f ∈ A(p) defined by (2) is said to be in the class H_p,k(λ, A, B) if and only if it satisfies the coefficient inequality (6).

Definition 3. A function f ∈ A(p) defined by (2) is said to be in the class Q_p,k(λ, A, B) if and only if it satisfies

It is obvious from Definitions 2 and 3 that

If we write

then it is easy to verify that Thus we obtain the following inclusion relations: Therefore, by Lemma 1, we see that each function in the classes H_p,k(λ, A, B) and Q_p,k(λ, A, B) is starlike with respect to k-symmetric points. Analytic functions which are starlike with respect to symmetric points and related functions have been extensively investigated in [1–6]. Recently, several authors have obtained many important properties and characteristics of multivalent analytic functions (see, e.g., [7–11]).

The main object of this paper is to present some distortion inequalities of functions in the classes H_p,k(λ, A, B) and Q_p,k(λ, A, B) which we have introduced here. In particular some results of inclusion relation and convolution of functions in these classes are also given. Further we derive several interesting results of the partial sums of functions in these classes.

Theorem 4. Let p/k ∉ N and suppose that either

• (a)

  1 − B ≥ p(1 − A) and 0 ≤ λ ≤ 1, or

• (b)

  1 − B < p(1 − A) and 0 ≤ λ ≤ (1 − B)/p(1 − A).

•  (i)

  If f ∈ H_p,k(λ, A, B), then, for z ∈ U,

The bounds in (19) are best possible for the function f defined by

•  (ii)

  If f ∈ Q_p,k(λ, A, B), then, for z ∈ U,

The bounds in (21) are best possible for the function f defined by

Proof. Let p/k ∉ N. For n ≥ 2p  (n ∈ N) and (n − p)/k ∉ N, we have δ_n,p,k = δ_2p,p,k = 0, and so

For n ≥ 2p  (n ∈ N) and (n − p)/k ∈ N, we have δ_n,p,k = 1 and If either (a) or (b) is satisfied, then

•  (i)

  If

then it follows from (23) to (25) that Hence we have for z ∈ U.

•  (ii)

  If

then (23) to (25) yield This leads to (21). The proof of the theorem is complete.

Theorem 5. Let

•  (i)

  If f(z) = z^p + a_2pz^2p + ⋯∈H_p,k(λ, A, B), then, for z ∈ U,

Equalities in (33) are attained, for example, by

•  (ii)

  If f(z) = z^p + a_2pz^2p + ⋯∈Q_p,k(λ, A, B), then, for z ∈ U,

Equalities in (36) are attained, for example, by

Proof. Note that (1 − B)/p(1 − A) < λ ≤ 1 implies that

•  (i)

  For f(z) = z^p + a_2pz^2p + ⋯∈H_p,k(λ, A, B), it follows from (23), (24), and (38) that

From this we can get (33).

•  (ii)

  For f(z) = z^p + a_2pz^2p + ⋯∈Q_p,k(λ, A, B), from (23), (24), and (38) we deduce that

Hence we have (36). The proof of the theorem is complete.

Theorem 6. Let p/k ∈ N.

• (i)

  If f ∈ H_p,k(λ, A, B), then, for z ∈ U,

  $$ ()

The bounds in (41) are sharp for the function f defined by

•  (ii)

    If f ∈ Q_p,k(λ, A, B), then, for z ∈ U,

The bounds in (43) are sharp for the function f defined by

Proof. Let p/k ∈ N. For n ≥ 2p  (n ∈ N) and (n − p)/k ∈ N, we have n = 2p + k(l − 1)  (l ∈ N), δ_n,p,k = δ_2p,p,k = 1, and

For n ≥ 2p  (n ∈ N) and (n − p)/k ∉ N, we have δ_n,p,k = δ_2p+1,p,k = 0, and so

• (i)

    If

then it follows from (45) and (46) that Hence we have for z ∈ U.

• (ii)

    If

  $$ ()

then (45) and (46) yield This leads to (43). Thus we complete the proof.

Theorem 7. If −1 ≤ D < 0, then

where

Proof. Since B < A ≤ 1 and −1 ≤ D < 0, we see that

Let f ∈ Q_p,k(λ, A, B). In order to prove that f ∈ H_p,k(λ, C(D), D), we need only to find the smallest C  (D < C ≤ 1) such that

for all n ≥ 2p, that is, that

For n ≥ 2p and (n − p)/k ∈ N, (56) is equivalent to

Noting (1), a simple calculation shows that (∂φ(λ, x))/∂x < 0 for all real x ≥ 2p and 0 ≤ λ ≤ 1, and so the function φ(λ, n) is decreasing in n  (n ≥ 2p). Therefore For n ≥ 2p and (n − p)/k ∉ N, (56) becomes

Consequently, by taking

it follows from (55) to (60) that f ∈ H_p,k(λ, C(D), D). The proof is complete.

Remark 8. If we take D = B in Theorem 7, then from (1) we have C(D) = (A + B)/2 < A. This shows that

Theorem 9. Let f ∈ H_p,k(λ, A, B). Then

where

Proof. For f ∈ H_p,k(λ, A, B), from Lemma 1 we have (7), which is equivalent to

or to Obviously If we put then, for p/k ∈ N, and, for p/k ∉ N,

Now, making use of (65) to (69), we arrive at

for z ∈ U,   σ ∈ C, and |σ | = 1. This gives the desired result (62). The proof of the theorem is complete.

Corollary 10. Let f ∈ Q_p,k(λ, A, B). Then

where h_σ(z) is the same as in Theorem 9.

Proof. Since f ∈ Q_p,k(λ, A, B) if and only if

it follows from Theorem 9 that Thus we complete the proof.

Theorem 11. Let f ∈ H_p,k(λ, A, B) and let either

• (a)

  1 − B ≥ p(1 − A) and 0 ≤ λ ≤ 1, or

• (b)

  1 − B < p(1 − A) and 0 ≤ λ ≤ (1 − B)/p(1 − A).

Then, for m ∈ N, we have The bounds in (75) and (76) are best possible for each m.

Proof. If either (a) or (b) is satisfied, then, for n ≥ 2p,

Let f ∈ H_p,k(λ, A, B). Then it follows from (77) that

If we put

for z ∈ U and m ∈ N∖{1}, then p₁(0) = 1 and we deduce from (78) that This implies that Re{p₁(z)} > 0 for z ∈ U, and so (75) holds true for m ∈ N∖{1}.

Similarly, by setting

it follows from (78) that Hence we have (76) for m ∈ N∖{1}.

For m = 1, replacing (78) by

and proceeding as the above, we see that (75) and (76) are also true.

Furthermore, taking the function f defined by

we have s_m(z) = z^p, Thus the proof of Theorem 11 is completed.

Remark 12. Replacing H_p,k(λ, A, B) by Q_p,k(λ, A, B), it follows from Theorem 11 that inequalities (75) and (76) are true. In Theorem 13 we improve the bounds in (75) and (76) for f ∈ Q_p,k(λ, A, B).

Theorem 13. Let f ∈ Q_p,k(λ, A, B) and let either (a) or (b) in Theorem 11 be satisfied. Then, for m ∈ N, one has

The bounds in (86) are sharp for the function f defined by

Proof. In view of the assumptions of Theorem 13, it follows from (77) that

If we put then (88) leads to Re  (p_j(z)) > 0  (z ∈ U;   m ∈ N;   j = 1,2). Hence we have (86). Sharpness can be verified easily.