Denoted by \(\mathcal {P}\), the well-known Carathéodory’s class of analytic functions in \(\mathbb {D}\), normalized with P(0)=1, and having positive real part in \(\mathbb {D}\), that is ReP(z)>0 for all \(z\in \mathbb {D}\) (see [7]).
To prove our results, we need the following two lemmas.
Lemma 1
[8, Lemma 3] If \(p(z)=1+c_{1}z+c_{2}z^{2}+\dots \in \mathcal {P}\), and α is a complex number, then
$$\max\left|c_{2}-\alpha c_{1}^{2}\right|=2\max\{1;\left|2\alpha-1\right|\}. $$
Lemma 2
[9, Lemma 1] If \(p(z)=1+c_{1}z+c_{2}z^{2}+\dots \in \mathcal {P}\), then
$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq\left\{ \begin{array}{lll} -4\alpha+2, & \text{if} & \alpha\leq 0, \\ 2, & \text{if} & 0\leq\alpha\leq 1, \\ 4\alpha-2, & \text{if} & \alpha\geq 1. \end{array} \right. $$
When α<0 or α>1, the equality holds if and only if \(p(z)= \frac {1+z}{1-z}\) or one of its rotations.
If 0<α<1, then the equality holds if and only if \(p(z)=\frac {1+z^{2} }{1-z^{2}}\) or one of its rotations.
If α=0, the equality holds if and only if:
$$p(z)=\left(\frac{1}{2}+\frac{\lambda}{2}\right)\frac{1+z}{1-z}+\left(\frac{1 }{2}-\frac{\lambda}{2}\right)\frac{1-z}{1+z}, \quad\text{with}\quad 0\leq\lambda\leq 1, $$
or one of its rotations.
If α=1, the equality holds if and only if:
$$\frac{1}{p(z)}=\left(\frac{1}{2}+\frac{\lambda}{2}\right)\frac{1+z}{1-z} +\left(\frac{1}{2}-\frac{\lambda}{2}\right)\frac{1-z}{1+z}, \quad\text{with} \quad 0\leq\lambda\leq 1. $$
Like it was mentioned in [9, pages 162–163], although the above upper bound is sharp, it can be improved as follows when 0<α<1:
$$ \left|c_{2}-\alpha c_{1}^{2}\right|+\alpha\left|c_{1}\right|^{2}\leq 2,\quad \text{if}\quad 0<\alpha\leq\frac{1}{2}, $$
(7)
and
$$ \left|c_{2}-\alpha c_{1}^{2}\right|+(1-\alpha)\left|c_{1}\right|^{2}\leq 2,\quad\text{if}\quad\frac{1}{2}\leq\alpha<1. $$
(8)
Theorem 1
If the function f given by (1) belongs to the class \( \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)\), with Ψ(z)=1+B1z+B2z2+… satisfying the conditions of Definition 1, and μ is a complex number, then:
$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\frac{|\zeta|B_{1}}{2(1-\gamma)\psi_{3}} \cdot \max\left\{1;\left|\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma} -\frac{2\mu\zeta B_{1}\psi_{3}}{(1-\gamma)\psi_{2}^{2}} \right|\right\}, $$
where ψk,k∈{2,3}, are given by (6).
Proof
If \(f\in \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)\), then there exists a Schwarz function w, that is w is analytic in \(\mathbb {D}\), with w(0)=0 and \(\left |w(z)\right |<1, z\in \mathbb {D}\), such that:
$$ 1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z) \right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}f(z)+\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z)\right)^{\prime}}-1\right]=\Psi(w(z)),\;z\in \mathbb{D}. $$
(9)
Since w is a Schwarz function, it follows that the function p1 defined by:
$$ p_{1}(z):=\frac{1+w(z)}{1-w(z)}=1+c_{1}z+c_{2}z^{2}+\dots,\;z\in\mathbb{D}, $$
(10)
belongs to \(\mathcal {P}\). Defining the function p by:
$$ p(z):=1+\frac{1}{\zeta}\left[\frac{z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z) \right)^{\prime}} {(1-\gamma)\mathcal{N}_{\nu,q}^{\lambda}f(z)+\gamma z\left(\mathcal{N}_{\nu,q}^{\lambda}f(z)\right)^{\prime}}-1\right]= 1+d_{1}z+d_{2}z^{2}+\dots,\;z\in\mathbb{D}, $$
(11)
in view of (9) and (10), we have:
$$ p(z)=\Psi\left(\frac{p_{1}(z)-1}{p_{1}(z)+1}\right),\;z\in\mathbb{D}. $$
(12)
From (10), we easily get:
$$\frac{p_{1}(z)-1}{p_{1}(z)+1}=\frac{1}{2}\left[c_{1}z+\left(c_{2}-\frac{ c_{1}^{2}}{2}\right)z^{2}+\left(c_{3}+\frac{c_{1}^{3}}{4} -c_{1}c_{2} \right)z^{3}+\dots\right],\;z\in\mathbb{D}; $$
therefore,
$$\Psi\left(\frac{p_{1}(z)-1}{p_{1}(z)+1}\right)=1+\frac{1}{2}B_{1}c_{1}z + \left[\frac{1}{2}B_{1}\left(c_{2}-\frac{c_{1}^{2}}{2}\right)+\frac{1}{4} B_{2}c_{1}^{2}\right]z^{2}+\dots,\;z\in\mathbb{D}, $$
and from (12), we obtain:
$$ d_{1}=\frac{1}{2}B_{1}c_{1}\qquad\text{and}\qquad d_{2}=\frac{1}{2} B_{1}\left(c_{2}-\frac{c_{1}^{2}}{2}\right)+\frac{1}{4}B_{2}c_{1}^{2}. $$
(13)
On the other hand, from (11), according to (5), it follows that
$$ d_{1}=\frac{(1-\gamma)a_{2}\psi_{2}}{\zeta}\qquad\text{and}\qquad d_{2}= \frac{2(1-\gamma)a_{3}\psi_{3}}{\zeta}-\frac{(1-\gamma)(1+\gamma)a_{2}^{2} \psi_{2}^{2}}{\zeta}, $$
(14)
and combining (13) with (14), we have:
$$ a_{2}=\frac{\zeta B_{1}c_{1}}{2(1-\gamma)\psi_{2}}, $$
(15)
and
$$a_{3}=\frac{\zeta B_{1}}{4(1-\gamma)\psi_{3}}\left[c_{2}-\frac{c_{1}^{2}}{2} +\frac{1}{2}\frac{B_{2}}{B_{1}}c_{1}^{2}+\frac{\zeta B_{1}(1+\gamma) c_{1}^{2}}{2(1-\gamma)}\right]. $$
Therefore,
$$ a_{3}-\mu a_{2}^{2}=\frac{\zeta B_{1}}{4(1-\gamma)\psi_{3}} \left(c_{2}-\alpha c_{1}^{2}\right), $$
(16)
where
$$ \alpha=\frac{1}{2}\left[1-\frac{B_{2}}{B_{1}}-\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma}+\frac{2\mu\zeta B_{1}\psi_{3}} {(1-\gamma)\psi_{2}^{2}}\right], $$
(17)
and from Lemma 1, our result follows immediately. □
Putting q→1− in Theorem 1, we obtain the next corollary:
Corollary 1
If the function f given by (1) belongs to the class \(\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)\), with Ψ(z)=1+B1z+B2z2+… satisfying the conditions of Definition 1, and μ is a complex number, then
$$\begin{array}{*{20}l} \left|a_{3}-\mu a_{2}^{2}\right|\leq \\ \frac{8|\zeta|B_{1}(\lambda+1)_{2}(\nu+1)_{2}}{3(1-\gamma)}\cdot \max\left\{1;\left|\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma) (\lambda+2)(\nu+2)} \right|\right\}. \end{array} $$
Using a similar proof like for Theorem 1 combined with Lemma 2, we can obtain the following theorem:
Theorem 2
If the function f given by (1) belongs to the class \( \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)\), with Ψ(z)=1+B1z+B2z2+… satisfying the conditions of Definition 1 and \(\mu,B_{2}\in \mathbb {R}\), and ζ>0, then
$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\left\{ \begin{array}{lll} \frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}\left[\frac{B_{2}}{B_{1}}+\frac{ \zeta B_{1}(1+\gamma)}{1-\gamma}- \frac{2\mu\zeta B_{1}\psi_{3}}{ (1-\gamma)\psi_{2}^{2}}\right], & \text{if} & \mu\leq\sigma_{1}, \\[1em] \frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}, & \text{if} & \sigma_{1}\leq\mu \leq\sigma_{2}, \\[1em] \frac{-\zeta B_{1}}{2(1-\gamma)\psi_{3}}\left[\frac{B_{2}}{B_{1}} +\frac{ \zeta B_{1}(1+\gamma)}{1-\gamma}-\frac{2\mu\zeta B_{1}\psi_{3}}{ (1-\gamma)\psi_{2}^{2}}\right], & \text{if} & \mu\geq\sigma_{2}, \end{array} \right. $$
with
$$ \sigma_{1}=\frac{(1-\gamma)\psi_{2}^{2}}{2\zeta B_{1}\psi_{3}}\left[-1+ \frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$
(18)
and
$$ \sigma_{2}=\frac{(1-\gamma)\psi_{2}^{2}}{2\zeta B_{1}\psi_{3}}\left[1+\frac{ B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}\right], $$
(19)
where ψk,k∈{2,3}, are given by (6).
Proof
With the same proof like those of Theorem 1, we obtain the equalities (16) and (17) hold.
(i) According to the first part of Lemma 2, we have:
$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq-4\alpha+2,\;\text{if}\;\alpha\leq0. $$
Using (17), simple computation shows that the inequality α≤0 is equivalent to μ≤σ1, and from (16) combined with the inequality \(\left |c_{2}-\alpha c_{1}^{2}\right |\leq -4\alpha +2\), the first of our theorem is proved.
(ii) The second part of Lemma 2 shows that:
$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq2,\;\text{if}\;0\leq\alpha\leq1. $$
From (17), it is easy to check that the inequality 0≤α≤1 is equivalent to σ1≤μ≤σ2. From the relation (16), the inequality \(\left |c_{2}-\alpha c_{1}^{2}\right |\leq 2\) proves the second part of our result.
(iii) Finally, form the third part of Lemma 2, we have:
$$\left|c_{2}-\alpha c_{1}^{2}\right|\leq4\alpha-2,\;\text{if}\;\alpha\geq1. $$
The relation (17) shows immediately that α≥1 is equivalent to μ≥σ2, while (16) combined with the inequality \(\left |c_{2}-\alpha c_{1}^{2}\right |\leq 4\alpha -2\) proves the last part of our result. □
Taking q→1− in Theorem 2, we get the next special case:
Corollary 2
If the function f given by (1) belongs to the class \(\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)\), with Ψ(z)=1+B1z+B2z2+… satisfying the conditions of Definition 1 and \(\mu,B_{2}\in \mathbb {R}\), and ζ>0, then
$$\left|a_{3}-\mu a_{2}^{2}\right|\leq\left\{ \begin{array}{l} \frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}} {3(1-\gamma)}\left[\frac{ B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma)(\lambda+2)(\nu+2)}\right], \\[1em] \hfill\text{if}\quad\mu\leq\eta_{1}, \\[1em] \frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}} {3(1-\gamma)},\hfill\text{if} \quad \eta_{1}\leq\mu\leq\eta_{2}, \\[1em] \frac{-8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}} {3(1-\gamma)}\left[\frac{ B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{1-\gamma} -\frac{3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma)(\lambda+2)(\nu+2)}\right], \\[1em] \hfill\text{if}\quad\mu\geq\eta_{2}, \end{array} \right. $$
with
$$ \eta_{1}=\frac{2(1-\gamma)(\lambda+2)(\nu+2)}{3\zeta B_{1}(\lambda+1)(\nu+1)} \left[-1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)} \right], $$
(20)
and
$$ \eta_{2}=\frac{2(1-\gamma)(\lambda+2)(\nu+2)}{3\zeta B_{1}(\lambda+1)(\nu+1)} \left[1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)}\right]. $$
(21)
With a similar proof like for Theorem 1 and using the inequalities (7) and (8), we obtained the next result.
Theorem 3
If the function f given by (1) belongs to the class \( \mathcal {M}_{\nu,q}^{\lambda,\gamma }(\zeta ;\Psi)\), with Ψ(z)=1+B1z+B2z2+… satisfying the conditions of Definition 1 and \(\mu,B_{2}\in \mathbb {R}\), and ζ>0, then the next inequalities hold:
(i) for σ1<μ≤σ3, we have
$$ \left|a_{3}-\mu a_{2}^{2}\right|+\frac{(1-\gamma)\psi_{2}^{2}} {2\zeta B_{1}\psi_{3}}\left[1-\frac{B_{2}}{B_{1}}-\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma}+\frac{2\mu\zeta B_{1}\psi_{3}} {(1-\gamma)\psi_{2}^{2}}\right] \left|a_{2}\right|^{2}\leq\frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}; $$
(22)
(ii) for σ3≤μ≤σ2, we have
$$ \left|a_{3}\!-\mu a_{2}^{2}\right|+\frac{(1-\gamma)\psi_{2}^{2}} {2\zeta B_{1}\psi_{3}}\left[1+\frac{B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{ 1-\gamma}-\frac{2\mu\zeta B_{1}\psi_{3}} {(1-\gamma)\psi_{2}^{2}}\right] \left|a_{2}\right|^{2}\leq\frac{\zeta B_{1}}{2(1-\gamma)\psi_{3}}, $$
(23)
where σ1 and σ2 are defined by (18) and (19), respectively,
$$\sigma_{3}=\frac{(1-\gamma)\psi_{2}^{2}}{2\zeta B_{1}\psi_{3}} \left[\frac{ B_{2}}{B_{1}}+\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)}\right], $$
and ψk,k∈{2,3}, are given by (6).
Proof
With the same computations like in the proof of Theorem 1, we obtain the relations (16) and (17), while (15) is equivalent to:
$$ c_{1}=\frac{2(1-\gamma)\psi_{2}}{\zeta B_{1}}. $$
(24)
(i) To prove the first part of our theorem, we will use the inequality (7). Thus, according to (16), (17), and the above relation, it is easy to check that (7) could be written in the equivalent form (22), while the assumption \(0<\alpha \leq \frac {1 }{2}\) is equivalent to σ1<μ≤σ3.
(ii) For the proof of the second part of our result, we will use the inequality (8). From (16), (17), and (24), it follows that (8) could be written in the form (23), and the assumption \(\frac {1}{2}\leq \alpha <1\) is equivalent to σ3<μ≤σ2. □
Putting q→1− in Theorem 3, we obtain the following result:
Corollary 3
If the function f given by (1) belongs to the class \(\mathcal {G}_{\nu }^{\lambda,\gamma }(\zeta ;\Psi)\), with Ψ(z)=1+B1z+B2z2+… satisfying the conditions of Definition 1 and \(\mu,B_{2}\in \mathbb {R}\), and ζ>0, then the next inequalities hold:
(i) for η1<μ≤η3, we have
$$\begin{array}{*{20}l} \left|a_{3}-\mu a_{2}^{2}\right| \\ +\frac{2(1-\gamma)(\lambda+2)(\nu+2)} {3\zeta B_{1}(\lambda+1)(\nu+1)}\left[ 1-\frac{B_{2}}{B_{1}} -\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}+\frac{ 3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma) (\lambda+2)(\nu+2)}\right] \left|a_{2}\right|^{2} \\ \leq\frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}}{3(1-\gamma)}; \end{array} $$
(ii) for η3≤μ≤η2, we have
$$\begin{array}{*{20}l} \left|a_{3}-\mu a_{2}^{2}\right| \\ +\frac{2(1-\gamma)(\lambda+2)(\nu+2)}{3\zeta B_{1} (\lambda+1)(\nu+1)}\left[ 1+\frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{1-\gamma}-\frac{ 3\mu\zeta B_{1}(\lambda+1)(\nu+1)}{2(1-\gamma) (\lambda+2)(\nu+2)}\right] \left|a_{2}\right|^{2} \\ \leq\frac{8\zeta B_{1}(\lambda+1)_{2}(\nu+1)_{2}}{3(1-\gamma)}, \end{array} $$
where η1 and η2 are defined by (20) and (21), respectively, and
$$\eta_{3}=\frac{2(1-\gamma)(\lambda+2)(\nu+2)} {3\zeta B_{1}(\lambda+1)(\nu+1) }\left[\frac{B_{2}}{B_{1}} +\frac{\zeta B_{1}(1+\gamma)}{(1-\gamma)}\right]. $$