Some Properties of Certain Multivalent Analytic Functions Involving the Cătas Operator

Definition 1.1 (see [1].)Let f(z) ∈ A_p(n). For p, n ∈ N,  δ, λ ≥ 0, ℓ ≥ 0, define the multiplier transformations I_p(δ, λ, ℓ) on A_p(n) by the following infinite series:

It is easily verified from (1.7), that

It should be remarked that the class of multiplier transforms I_p(δ, λ, ℓ) is a generalization of several other linear operators considered, in earlier investigations (see [2–12]).

If f(z) is given by (1.1), then we have

where In particular, we set

For two functions f(z) and g(z), analytic in U, we say that the function f(z) is subordinate to g(z) in U, and write f(z)≺g(z)  (z ∈ U) if there exists a Schwarz function w(z), which is analytic in U with

such that Indeed, it is known that Furthermore, if the function g is univalent in U, then we have the following equivalence: By making use of the linear operator I_p(δ, λ, ℓ) and the above-mentioned principle of subordination between analytic functions, we introduce and investigate the following subclass of the class A_p(n) of p-valent analytic functions.

Definition 1.2. A function f(z) ∈ A_p(n) is said to be in the class $$ if it satisfies the following subordination condition:

where (and throughout this paper unless otherwise mentioned) the parameters α, β, p, n, λ, δ, ℓ, A, and B are constrained as follows:

Definition 2.1 (see [16].)Denote by Q the set of all functions f(z) that are analytic and injective on $$, where

and such that f^′(ε) ≠ 0 for ε ∈ ∂U − E(f).

Lemma 2.2 (see [17].)Let the function h be analytic and univalent (convex) in U with h(0) = 1. Suppose also that the function k given by

is analytic in U. If then and χ(z) is the best dominant of (2.3).

Lemma 2.3 (see [18].)Let q(z) be a convex univalent function in U and let σ, η ∈ ℂ with

If the function p is analytic in U and then p(z)≺q(z) and q(z) is the best dominant.

Lemma 2.4 (see [16].)Let q be convex univalent in U and k ∈ C. Further assume that $$ if

and p(z) + kzp^′(z) is univalent in U, then implies that q(z)≺p(z) and q(z) is the best subdominant.

Lemma 2.5 (Jach’s Lemma [19]). Let w(z) be a noncostant analytic function in U with w(0) = 0. If |w| attains its maximum value on the circle |z| = r < 1 at z₀, then

where k ≥ 1 is a real number.

Lemma 2.6 (see [20].)Let F be analytic and convex in U. If f(z), g(z) ∈ A and f(z), g(z)≺F(z); then

Lemma 2.7 (see [21], [22].)Let k, v ∈ ℂ. Suppose also that m is convex and univalent in U with

If u is analytic in U with u(0) = 1, then the following subordination: implies that

Lemma 2.8 (see [23].)Let $$ analytic in U and $$ be analytic and convex in U. If f(z)≺g(z), then |a_k| ≤ |b₁|,     (k ∈ N).

Lemma 2.9 (see [24].)Let δ ≠ 0,  δ ∈ R,  v/δ > 0,0 ≤ ρ < 1,  p ∈ H[1, n], and p(z)≺1 + kz  (k : = vM/(nδ + v)), where

If q(z) ∈ H[1, n] satisfies the following subordination condition: then

Theorem 3.1. Let $$ with Re(α) > 0. Then

Proof. Define the function P(z) by

Then P(z) is analytic in U with P(0) = 1. By taking the derivatives in the both sides in equality (3.2) and using (1.8), we get An application of Lemma 2.2 to (3.3) yields where The proof of Theorem 3.1 is thus completed.

Theorem 3.2. Let q(z) be univalent in U, 0 ≠ α ∈ ℂ. Suppose also that q(z) satisfies

If f(z) ∈ A_p(n) satisfying the following subordination: then and q(z) is the best dominant.

Proof. Let the function P(z) be defined by (3.2). We know that (3.3) holds true. Combining (3.3) and (3.7), we find that

Corollary 3.3. Let α ∈ ℂ and −1 ≤ B < A ≤ 1. Suppose also that (1 + Az)/(1 + Bz) satisfies the condition (3.6). If f(z) ∈ A_p(n) satisfies the following subordination:

then and (1 + Az)/(1 + Bz) is the best dominant.

If f(z) is subordinate to F(z), then F(z) is superordinate to f(z). We now derive the following superordination result for the class $$.

Theorem 3.4. Let q(z) be convex univalent in U, α ∈ ℂ, with Re(α) > 0. Also let

be univalent in U. If then and q(z) is the best subdominant.

Proof. Let the function P(z) be defined by (3.2). Then

An application of Lemma 2.4 yields the assertion of Theorem 3.4.

Corollary 3.5. Let q(z) be convex univalent in U and −1 ≤ B < A ≤ 1, α ∈ ℂ with Re(α) > 0. Also let

be univalent in U. If then and (1 + Az)/(1 + Bz) is best subdominant.

Corollary 3.6. Let q₁(z) be convex univalent and let q₂(z) be univalent in U, α ∈ ℂ, Re(α) > 0. Let q₂(z) satisfies (3.6). If

is univalent in U, also then and q₁(z),  q₂(z) are, respectively, the best subordinate, and dominant.

Theorem 3.7. Let f(z) ∈ A_p(n),  ξ ∈ ℂ∖{0}, and 0 ≤ γ < 1. Also let the function φ be defined by

If φ satisfies one of the following conditions: or then

Proof. We define the function ϕ(z) by

It is easy to see that the function ϕ is analytic in U with ϕ(0) = 0.

Differentiating both sides of (3.26) with respect to z logarithmically, we get

We now consider the function φ defined by Assume that there exists a point z₀ ∈ U such that by Lemma 2.5, we know that If follows from (3.28) and (3.30) that But the inequalities in (3.31) and (3.32) contradict, respectively, the inequalities in (3.23) and (3.24). Therefore, we can conclude that which implies that We thus complete the proof of Theorem 3.7.

Corollary 3.8. Let f(z) ∈ A,  δ = 0,  p = λ = ℓ = ξ = 1, and γ = 0. Also let the function φ be defined by (3.22). If φ satisfies one of the following conditions:

then f(z) ∈ ß(β).

Theorem 3.9. Let Re(α) > 0,  β > 0, and $$. Then $$ for |z| < R(α, β, ℓ, λ, p), where

The bound R(α, β, ℓ, λ, p) is the best possible.

Proof. Suppose that

where h is analytic and has a positive real part in U. By taking the derivatives in the both sides in equality (3.37) and using (1.9), we get By making use of the following well-known estimate (see [25]): in (3.38), we obtain that for r < R(α, β, ℓ, λ, p), where R(α, β, ℓ, λ, p) is given by (3.36).

To show that the bound R(α, β, ℓ, λ, p) is the best possible, we consider the function f(z) ∈ A_p(n) defined by

By noting that for z = R(α, β, ℓ, λ, p), we conclude that the bound is the best possible. Theorem 3.9 is thus proved.

Theorem 3.10. Let f(z)∈$$ with Re(α) > 0. Then

where w(z) is analytic in U with w(0) = 0 and |w(z)| < 1  (z ∈ U).

Proof. Suppose that $$. It follows from (3.1) that

where w(z) is analytic in U with w(0) = 0 and |w(z)| < 1  (z ∈ U).

By virtue of (3.44), we easily find that

Combining (1.10), (1.16), and (3.45), we have The assertion (3.43) of Theorem 3.10 can now easily be derived from (3.46).

Theorem 3.11. Let $$ with Re(α) > 0. Then

Proof. Suppose that $$ with Re(α) > 0. We know that (3.1) holds true, which implies that

It is easy to see that the condition (3.48) can be written as follows: Combining (1.9), (1.10), and (3.49), we easily get the convolution property (3.47) asserted by Theorem 3.11.

Theorem 3.12. Let α₂ ≥ α₁ ≥ 0 and −1 ≤ B₁ ≤ B₂ < A₂ < A₁ ≤ 1. Then

Proof. Suppose that $$. We know that

Since −1 ≤ B₁ ≤ B₂ < A₂ ≤ A₁ ≤ 1, we easily find that that is, $$. Thus the assertion of Theorem 3.12 holds for α₂ = α₁ ≥ 0.

If α₂ > α₁ ≥ 0, by Theorem 3.1 and (3.52), we know that $$, that is,

At the same time, we have Moreover, since 0 ≤ (α₁/α₂) < 1 and h₁(z): = (1 + A₁z)/(1 + B₁z), is analytic and convex in U. Combining (3.52)–(3.54) and Lemma 2.6, we find that that is, $$, which implies that the assertion (3.50) of Theorem 3.12 holds.

Let ρ denote the class of functions of the following form:

which are analytic and convex in U and satisfy the following condition: By making use of the principle of subordination between analytic functions, we introduce the subclasses $$ and C_p,n(μ, ϕ) of the class A_p(n): Next, by using the operator defined by (1.7), we define the following two subclasses S_p,n(δ, λ, ℓ; μ; ϕ) and C_p,n(δ, λ, ℓ; μ; ϕ) of the class A_p(n): Clearly, we know that

We now derive some inclusion relationships for the classes S_p,n(δ, λ, ℓ; μ; ϕ) and C_p,n(δ, λ, ℓ; μ; ϕ), by similarly applying the method of proof of Proposition 1 obtained by Cho et al. [26] and Wang et al. [27].

Theorem 3.13. Let 0 ≤ μ < p, λ > −p, and ϕ ∈ ρ with

Then

Theorem 3.14. Let 0 ≤ μ < p,  λ > −p and ϕ ∈ ρ with (3.61) holds. Then

Proof. By virtue of (3.60) and Theorem 3.13, we observe that

From (3.64), we conclude that the assertion of Theorem 3.14 holds true.

Corollary 3.15. Let 0 ≤ μ < p,  λ > −p, and −1 ≤ B < A ≤ 1. Then

Theorem 3.16. Let $$ with Re(α) > 0 and −1 ≤ B < A ≤ 1. Then

The extremal function of (3.66) is defined by

Proof. Let $$ with Re(α) > 0. From Theorem 3.1, we know that (3.1) holds, which implies that

Combining both equations of (3.68), we get (3.66). By noting that the function I_p(δ, λ, ℓ)F_α,β,A,B(z) defined by (3.67) belongs to the class $$, we obtain that the equality (3.66) is sharp. The proof of Theorem 3.16 is evidently completed.

Corollary 3.17. Let $$ with Re(α) > 0 and −1 ≤ A < B ≤ 1. Then

The extremal function of (3.69) is defined by (3.67).

Corollary 3.18. Let $$ with Re(α) > 0 and−1 ≤ B < A ≤ 1. Then for |z| = r < 1, we have

The extremal function of (3.70) is defined by (3.67).

Corollary 3.19. Let $$ with Re(α) > 0 and−1 ≤ A < B ≤ 1. Then for |z| = r < 1, we have

The extremal function of (3.71) is defined by (3.67).

Corollary 3.20. Let $$ with Re(α) > 0 and −1 ≤ B < A ≤ 1. Then

Corollary 3.21. Let $$ with Re(α) > 0 and −1 ≤ A < B ≤ 1. Then

Theorem 3.22. Let

Then

The inequality (3.76) is sharp, with extremal function defined by (3.67).

Proof. Combining (1.16) and (3.75), we obtain

An application of Lemma 2.8 to (3.77) yields Thus, from (3.78), we easily arrive at (3.76) asserted by Theorem 3.22.

Theorem 3.23. Let 0 ≠ α ∈ ℝ,  β ∈ ℝ,  β/α > 0,  λ > −p and 0 ≤ ρ < 1. If $$ with

then

Proof. Suppose that $$. By definition, we have

Let the function P(z) be defined by (3.2). We then find from (3.1) and (3.81) that We now suppose that Then q(z) ∈ H[1, n]. It follows from (3.81) and (3.83) that An application of Lemma 2.9 to (3.84) yields Combining (3.83) and (3.85), we find that The assertion of Theorem 3.23 can now easily be derived from (1.8) and (3.86).

Theorem 3.24. Let $$ with β > 0, A > 0, Re(α) > 0, and |α|(n + Re(p + ℓ)β/λα) > A((p + ℓ)β/λ). Then

Proof. Let the function P(z) be defined by (3.2). It follows from (3.3) that

where is analytic in U with |w(z)| < 1  (z ∈ U). From (3.88), we easily get It follows from (3.90) that We next find from (3.90) and (3.91) that

Now from (3.88) we get

and from (3.90) we get from (3.93) and (3.94), we get

Thus, from (3.2) and (3.95), we easily arrive at the assertion of Theorem 3.24.