Coefficient estimates for some classes of functions associated with \(q\)-function theory

Theorem 2.4. Let $f \in \mathcal{S}_q^*$ be of the form (1.1) and $\mu$ be any complex number. Then $$|a_3 - \mu a_2^2| \le \max \left\{ \left| 2(1 - 2\mu) \left( \frac{\ln q}{q - 1} \right)^2 + 2 \left( \frac{\ln q}{q^2 - 1} \right) \right|, 2 \left( \frac{\ln q}{q^2 - 1} \right) \right\}.$$ Equality occurs for the functions (2.1) $$F_1(z) := z \left\{ \exp\left[\sum_{n=1}^{\infty} \frac{2\ln q}{q^n - 1} z^n\right] \right\}$$ and (2.2) $$F_2(z) := z \left\{ \exp \left[ \sum_{n=1}^{\infty} \frac{2 \ln q}{q^{2n} - 1} z^{2n} \right] \right\}.$$ The next result is the estimation of second order Hankel determinant for the class $\mathcal{S}_q^*$

Theorem 2.5. Let $f \in \mathcal{S}_q^*$ be of the form (1.1). Then $$|H_2(2)| = |a_2 a_4 - a_3^2| \le 4 \left(\frac{\ln q}{q^2 - 1}\right)^2.$$ Equality occurs for the function $F_2(z)$ defined in (2.2). Remark 2.6. For $q \to 1$ , Theorem 2.4 gives the Fekete-Szegö problem for the class $S^*$ [12, Theorem 1]. Remark 2.7. For $q \to 1$ , Theorem 2.5 gives the Hankel determinant for the class $S^*$ [11, Theorem 3.1]. Now we present the Herglotz representation of functions belonging to the class $C_q(\alpha)$ :

Theorem 2.8. Let $f \in \mathcal{A}$ . Then $f \in \mathcal{C}_q(\alpha)$ , $0 \le \alpha < 1$ , if and only if there exists a probability measure $\mu$ supported on the unit circle such that $$\frac{z(D_q f)'(z)}{(D_q f)(z)} = \int_{|\sigma|=1} \sigma z F'_{q,\alpha}(\sigma z) d\mu(\sigma)$$ where (2.3) $$F_{q,\alpha}(z) = \sum_{n=1}^{\infty} \frac{(-2)\left(\ln\frac{q}{1-\alpha(1-q)}\right)}{1-q^n} z^n, \quad z \in \mathbb{D}.$$ Remark 2.9. It is clear that when $q \to 1$ , $$F'_{q,\alpha}(z) \to 2(1-\alpha)/(1-z)$$ and $z(D_q f)'(z)/(D_q f)(z) \to zf''(z)/f'(z)$ . Hence, when q approaches to 1, Theorem 2.8 leads to the Herglotz Representation of convex functions of order $\alpha$ (see for instance [8, pp. 172, Problem 3]). The Bieberbach type problem for the class $C_q(\alpha)$ , $0 \le \alpha < 1$ , is stated below:

Proposition 3.1. Let $f \in C_q(\alpha)$ , $0 \le \alpha < 1$ . Then there exists a unique function $g \in S_q^*(\alpha)$ , $0 \le \alpha < 1$ , such that $$(3.1) g(z) = z(D_q f)(z)$$ holds. Similarly, for a given function $g \in \mathcal{S}_q^*(\alpha)$ there exists a unique function $f \in \mathcal{C}_q(\alpha)$ satisfying (3.1). Next result is a characterization for a function to be in the class $C_q(\alpha)$ .

Theorem 3.2. Let $f \in A$ . Then $f \in C_q(\alpha)$ , $0 \le \alpha < 1$ , if and only if $$\left| q \frac{(D_q f)(qz)}{(D_q f)(z)} - \alpha q \right| \le 1 - \alpha, \quad z \in \mathbb{D}.$$

Corollary 3.3. The class $C_q(\alpha)$ satisfies the inclusion relation $$\bigcap_{q$$

Lemma 3.4. [1, Lemma 2.4] If $h \in B_q$ then the infinite product $\prod_{n=0}^{\infty} \{((1-\alpha)h(zq^n) + \alpha q)/q\}$ converges uniformly on compact subsets of $\mathbb{D}$ .

Lemma 3.5. If $h \in B_q^0$ then the infinite product $\prod_{n=0}^{\infty} \{((1-\alpha)h(zq^n) + \alpha q)/q\}$ converges uniformly on compact subsets of $\mathbb{D}$ to a nonzero function in $\mathcal{H}(\mathbb{D})$ with no zeros. Furthermore, the function f satisfying the relation (3.2) $$z(D_q f)(z) = \frac{z}{\prod_{n=0}^{\infty} \{ ((1-\alpha)h(zq^n) + \alpha q)/q \}}$$ belongs to $$C_q(\alpha)$$ and $h(z) = \left(q \frac{(D_q f)(qz)}{(D_q f)(z)} - \alpha q\right) / (1 - \alpha)$ .

Lemma 3.6. [9, Lemma 2.4] A function $g \in B_q^0$ if and only if it has the representation $$(3.4) g(z) = \exp\{(\ln q)p(z)\},$$ where p(z) belongs to the class $\mathcal{P}$ .

Theorem 3.7. The mapping $\rho: \mathcal{C}_q(\alpha) \to B_q^0$ defined by $$\rho(f)(z) = \left(q \frac{(D_q f)(qz)}{(D_q f)(z)} - \alpha q\right) / (1 - \alpha)$$ is a bijection.

Lemma 4.1. [9, Theorem 1.13] The mapping $\rho: \mathcal{S}_q^* \to B_q^0$ defined by $$\rho(f)(z) = \frac{f(qz)}{f(z)}$$ is a bijection.

Lemma 4.2. [9, Theorem 1.15] Let $f \in \mathcal{A}$ . Then $f \in \mathcal{S}_q^*$ if and only if there exists a probability measure $\mu$ supported on the unit circle such that $$\frac{zf'(z)}{f(z)} = 1 + \int_{|\sigma|=1} \sigma z F_q'(\sigma z) d\mu(\sigma)$$ where (4.1) $$F_q(z) = \sum_{n=1}^{\infty} \frac{2 \ln q}{q^n - 1} z^n, \quad z \in \mathbb{D}.$$

Lemma 4.3. [15, pp. 254-256] Let the function $p \in \mathcal{P}$ and be given by the power series (3.3). Then $$2p_2 = p_1^2 + x(4 - p_1^2),$$ $$4p_3 = p_1^3 + 2(4 - p_1^2)p_1x - p_1(4 - p_1^2)x^2 + 2(4 - p_1^2)(1 - |x|^2)z,$$ for some x and z satisfying $|x| \le 1$ , $|z| \le 1$ , and $p_1 \in [0,2]$ .

Lemma 4.4. [17, Lemma 1] Let the function $p \in \mathcal{P}$ and be given by the power series (3.3). Then for any real number $\lambda$ , $$|p_2 - \lambda p_1^2| \le 2 \max\{1, |2\lambda - 1|\}$$ and the result is sharp. Proof of Theorem 2.4. Let $f \in \mathcal{S}_q^*$ . Then by Lemma 4.1, there exist a function $g \in B_q^0$ such that g(z) = f(qz)/f(z). Since $g \in B_q^0$ , by Lemma 3.6, g(z) has the representation (3.4). That is, $$\frac{f(qz)}{f(z)} = \exp\{(\ln q)p(z)\}.$$ Define the function $\phi(z) = \text{Log}\{f(z)/z\}$ and set $$\phi(z) = \operatorname{Log} \frac{f(z)}{z} = \sum_{n=1}^{\infty} \phi_n z^n.$$ On solving, we get $$\ln q + \phi(qz) = \phi(z) + (\ln q)p(z).$$ This implies $$\phi_n = p_n \left( \frac{\ln q}{q^n - 1} \right).$$ So, f(z) can be written as (4.3) $$f(z) = z \exp\left[\sum_{n=1}^{\infty} \phi_n z^n\right],$$ where $\phi_n$ is defined in (4.2) and f(z) has the form (1.1). Equating the coefficients of both sides in (4.3) and using the value of $\phi_n$ given in (4.2), we obtain $$(4.4) a_2 = \phi_1 = p_1 \left(\frac{\ln q}{q-1}\right), a_3 = \phi_2 + \frac{\phi_1^2}{2} = p_2 \left(\frac{\ln q}{q^2-1}\right) + \frac{p_1^2}{2} \left(\frac{\ln q}{q-1}\right)^2.$$ Thus, $$|a_{3} - \mu a_{2}^{2}| = \left| p_{2} \left( \frac{\ln q}{q^{2} - 1} \right) + \frac{p_{1}^{2}}{2} \left( \frac{\ln q}{q - 1} \right)^{2} - \mu p_{1}^{2} \left( \frac{\ln q}{q - 1} \right)^{2} \right|$$ $$= \left( \frac{\ln q}{q^{2} - 1} \right) \left| p_{2} - (2\mu - 1) \frac{\left( \frac{\ln q}{q - 1} \right)^{2}}{\left( \frac{2 \ln q}{q^{2} - 1} \right)} p_{1}^{2} \right|$$ $$\leq \max \left\{ \left| 2(1 - 2\mu) \left( \frac{\ln q}{q - 1} \right)^{2} + 2 \left( \frac{\ln q}{q^{2} - 1} \right) \right|, 2 \left( \frac{\ln q}{q^{2} - 1} \right) \right\},$$ where the last inequality follows from Lemma 4.4. It now remains to prove the sharpness part. This can easily be shown by the definition of $\mathcal{S}_q$ that the functions $F_1$ and $F_2$ defined in the statement of Theorem 2.4 belong to the class $\mathcal{S}_q$ . One can also see that $F_1 \in \mathcal{S}_q^*$ as a special case to Lemma 4.2, when the measure has a unit mass. The functions $F_1$ and $F_2$ show the sharpness of the result. This completes the proof of the theorem. We now pose the following conjecture on Fekete-Szegö problem for $\mathcal{S}_q^*(\alpha)$ . Conjecture 4.5. Let $f \in \mathcal{S}_q^*(\alpha)$ , $0 \le \alpha < 1$ , be of the form (1.1) and $\mu$ be any complex number. Then $$|a_3 - \mu a_2^2| \le \max \left\{ \left| 2(1 - 2\mu) \left( \frac{\ln \frac{q}{1 - \alpha(1 - q)}}{q - 1} \right)^2 + 2 \left( \frac{\ln \frac{q}{1 - \alpha(1 - q)}}{q^2 - 1} \right) \right|, 2 \left( \frac{\ln \frac{q}{1 - \alpha(1 - q)}}{q^2 - 1} \right) \right\}.$$ Equality occurs for the functions (4.5) $$F_1(z) := z \left\{ \exp \left[ \sum_{n=1}^{\infty} \frac{2 \ln \frac{q}{1 - \alpha(1 - q)}}{q^n - 1} z^n \right] \right\}$$ and (4.6) $$F_2(z) := z \left\{ \exp \left[ \sum_{n=1}^{\infty} \frac{2 \ln \frac{q}{1 - \alpha(1 - q)}}{q^{2n} - 1} z^{2n} \right] \right\}.$$ Proof of Theorem 2.5. Given that $f \in \mathcal{S}_q^*$ having the form (1.1). In (4.4), we already obtained the values of $a_2$ and $a_3$ . In the similar way one can find the value of $a_4$ . Indeed, $$a_4 = \phi_3 + \phi_1 \phi_2 + \frac{\phi_1^3}{6} = p_3 \left( \frac{\ln q}{q^3 - 1} \right) + p_1 p_2 \left( \frac{\ln q}{q - 1} \right) \left( \frac{\ln q}{q^2 - 1} \right) + \frac{p_1^3}{6} \left( \frac{\ln q}{q - 1} \right)^3.$$ Hence, $$|a_2 a_4 - a_3^2| = \left| -\frac{p_1^4}{12} \left( \frac{\ln q}{q - 1} \right)^4 + p_1 p_3 \left( \frac{\ln q}{q - 1} \right) \left( \frac{\ln q}{q^3 - 1} \right) - p_2^2 \left( \frac{\ln q}{q^2 - 1} \right)^2 \right|.$$ Suppose now that $p_1 = c$ and $0 \le c \le 2$ . Using Lemma 4.3, we obtain $$\begin{aligned} |a_2a_4 - a_3^2| &= \left| -\frac{c^4}{12} \left[ \left( \frac{\ln q}{q-1} \right)^4 - 3 \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) + 3 \left( \frac{\ln q}{q^2-1} \right)^2 \right] \right. \\ &\quad + \frac{c^2}{2} (4 - c^2) x \left[ \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) - \left( \frac{\ln q}{q^2-1} \right)^2 \right] \\ &\quad + \frac{(4-c^2)(1-|x|^2)cz}{2} \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) \\ &\quad - \left[ \frac{c^2}{4} (4-c^2) \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) + \frac{(4-c^2)^2}{4} \left( \frac{\ln q}{q^2-1} \right)^2 \right] x^2 \right| \\ &\leq \frac{c^4}{12} \left| \left( \frac{\ln q}{q-1} \right)^4 - 3 \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) + 3 \left( \frac{\ln q}{q^2-1} \right)^2 \right| + \frac{(4-c^2)c}{2} \left( \frac{\ln q}{q-1} \right) \\ &\quad \left( \frac{\ln q}{q^3-1} \right) + \frac{c^2}{2} (4-c^2) \left[ \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) - \left( \frac{\ln q}{q^2-1} \right)^2 \right] \rho \\ &\quad + \left( \frac{4-c^2}{4} \right) \left[ (4-c^2) \left( \frac{\ln q}{q^2-1} \right)^2 + c(c-2) \left( \frac{\ln q}{q-1} \right) \left( \frac{\ln q}{q^3-1} \right) \right] \rho^2 \\ &= F(\rho) \end{aligned}$$ with $\rho = |x| \le 1$ . Furthermore, $$F'(\rho) \ge 0.$$ This implies that F is an increasing function of $\rho$ and thus the upper bound for $|a_2a_4 - a_3^2|$ corresponds to $\rho = 1$ . Hence, $$|a_2a_4 - a_3^2| \le F(1) = G(c)$$ (say). We can see that $$\left(\frac{\ln q}{q-1}\right)^4 - 3\left(\frac{\ln q}{q-1}\right)\left(\frac{\ln q}{q^3-1}\right) + 3\left(\frac{\ln q}{q^2-1}\right)^2 > 0, \text{ for } 0 < q < 1.$$ Now, a simple calculation gives that $$G(c) = \frac{c^4}{12} \left[ \left( \frac{\ln q}{q - 1} \right)^4 - 12 \left( \frac{\ln q}{q - 1} \right) \left( \frac{\ln q}{q^3 - 1} \right) + 12 \left( \frac{\ln q}{q^2 - 1} \right)^2 \right] + c^2 \left[ 3 \left( \frac{\ln q}{q - 1} \right) \left( \frac{\ln q}{q^3 - 1} \right) - 4 \left( \frac{\ln q}{q^2 - 1} \right)^2 \right] + 4 \left( \frac{\ln q}{q^2 - 1} \right)^2.$$ The expression G'(c) = 0 gives either c = 0 or $$c^{2} = \frac{6\left[4\left(\frac{\ln q}{q^{2}-1}\right)^{2} - 3\left(\frac{\ln q}{q-1}\right)\left(\frac{\ln q}{q^{3}-1}\right)\right]}{\left(\frac{\ln q}{q-1}\right)^{4} - 12\left(\frac{\ln q}{q-1}\right)\left(\frac{\ln q}{q^{3}-1}\right) + 12\left(\frac{\ln q}{q^{2}-1}\right)^{2}}.$$ We can verify that G''(c) is negative for c = 0 and positive for other values of c. Hence the maximum of G(c) occurs at c = 0. Thus, we obtain $$|a_2a_4 - a_3^2| \le 4\left(\frac{\ln q}{q^2 - 1}\right)^2.$$ The function $F_2$ defined in the statement of the theorem shows the sharpness of the result. This completes the proof of the theorem. We pose one more conjecture which is about the Hankel Determinant for the class $\mathcal{S}_q^*(\alpha)$ . Conjecture 4.6. Let $f \in \mathcal{S}_q^*(\alpha)$ , $0 \le \alpha < 1$ , be of the form (1.1). Then $$|a_2 a_4 - a_3^2| \le 4 \left( \frac{\ln \frac{q}{1 - \alpha(1 - q)}}{q^2 - 1} \right)^2.$$ Equality occurs for the function $F_2$ defined in (4.6). Remark 4.7. Here we remark that the proofs of Conjectures 4.5 and 4.6 will follow in the similar manner as the proofs of Theorem 2.4 and Theorem 2.5, respectively. However, the conjectures are all about to find the extremal functions which we believe to be (4.5) and (4.6). Proof of Theorem 2.8. Let $f \in C_q(\alpha)$ , $0 \le \alpha < 1$ . By definition of $C_q(\alpha)$ , $z(D_q f)(z) \in S_q^*(\alpha)$ . Then by [1, Theorem 1.1], we have $$z \frac{(z(D_q f)(z))'(z)}{z(D_q f)(z)} = 1 + \int_{|\sigma|=1} \sigma z F'_{q,\alpha}(\sigma z) d\mu(\sigma)$$ or, $$1 + \frac{z(D_q f)'(z)}{(D_q f)(z)} = 1 + \int_{|\sigma|=1} \sigma z F'_{q,\alpha}(\sigma z) d\mu(\sigma),$$ where $F_{q,\alpha}$ is defined in (2.3). Hence the proof is complete. Proof of Theorem 2.10. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{C}_q(\alpha)$ . By definition of $\mathcal{C}_q(\alpha)$ , $z(D_q f)(z) = z + \sum_{n=2}^{\infty} (1 - q^n)/(1 - q)a_n z^n \in \mathcal{S}_q^*(\alpha)$ . Then by [1, Theorem 1.3], we have $$\left| \frac{1 - q^n}{1 - q} a_n \right| \le c_n.$$ Next, we show that equality holds for the function $E_q \in \mathcal{C}_q(\alpha)$ . As a special case to Theorem 2.8, when the measure has a unit mass, it is clear that $E_q \in \mathcal{C}_q(\alpha)$ . Let $E_q(z) = z + \sum_{n=2}^{\infty} b_n z^n$ . From this representation of $E_q$ and the definition of $D_q f$ , we get (4.7) $$z(D_q E_q)(z) = z + \sum_{n=2}^{\infty} b_n (1 - q^n) / (1 - q) z^n.$$ Since Eq(z) = Iq{exp[Fq,α(z)]}, z(DqEq)(z) = z{exp[Fq,α(z)]} and since cⁿ is the n-th coefficient of the function z exp[Fq,α(z)], we have (4.8) $$z(D_q E_q)(z) = z + \sum_{n=2}^{\infty} c_n z^n.$$ By comparing [\(4.7\)](#page-10-1) and [\(4.8\)](#page-11-20) we get, bⁿ = cn(1 − q)/(1 − q n ). i.e. $$E_q(z) = z + \sum_{n=2}^{\infty} \left(\frac{1-q}{1-q^n}\right) c_n z^n.$$ This completes the proof of our theorem.

Definition 2.1. [1, Definition 1.1] A function $f \in \mathcal{A}$ is said to be in the class $\mathcal{S}_q^*(\alpha)$ , $0 \le \alpha < 1$ , if $$\left| \frac{z(D_q f)(z)}{f(z)} - \alpha \frac{1}{1 - q} \right| \le \frac{1}{1 - q}, \quad z \in \mathbb{D}.$$ Note that the choice $\alpha=0$ gives the definition of the class of q-starlike functions, denoted by $\mathcal{S}_q$ , (see [9, Definition 1.3]). Indeed, a function $f\in\mathcal{A}$ is said to belong to $\mathcal{S}_q$ if $|(z(D_qf)(z))/f(z)-1/(1-q)|\leq 1/(1-q), z\in\mathbb{D}$ . By using the idea of the well-known Alexander's theorem [5, Theorem 2.12], Baricz and Swaminathan in [3] defined a q-analog of convex functions, denoted by $\mathcal{C}_q$ , in the following way.

Definition 2.2. [3, Definition 3.1] A function $f \in \mathcal{A}$ is said to belong to $\mathcal{C}_q$ if and only if $z(D_q f)(z) \in \mathcal{S}_q^*$ . We call the functions of the class $C_q$ as q-convex functions. The class $C_q$ is non-empty as shown in [3, Theorem 3.2]. Note that as $q \to 1$ , the classes $S_q$ and $C_q$ reduce to $S$ and C respectively. It is natural to define the q-convex functions of order $\alpha$ , $0 \le \alpha < 1$ , denoted by $C_q(\alpha)$ , in the following way:

Definition 2.3. A function $f \in \mathcal{A}$ is said to be in the class $C_q(\alpha)$ , $0 \le \alpha < 1$ , if and only if $z(D_q f)(z) \in \mathcal{S}_q^*(\alpha)$ . We can see that as $q \to 1$ , the class $C_q(\alpha)$ reduces to the class of convex functions of order $\alpha$ , $C(\alpha)$ (for definition of $C(\alpha)$ see [8]). Bieberbach type problem is estimated for the classes $\mathcal{S}_q$ and $\mathcal{S}_q^(\alpha)$ in the articles [9] and [1] respectively. But the Fekete-Szegö problem and the Hankel determinant were not considered there. In this article, we first discuss these two problems for the class $\mathcal{S}_q$ and posed two conjectures on the Fekete-Szegö problem and Hankel determinant for the class $\mathcal{S}_q^(\alpha)$ . Since the Bieberbach type problem for the class $\mathcal{C}_q(\alpha)$ , $0 \le \alpha < 1$ , is not available in literature, here we obtain the Bieberbach type problem for the class $\mathcal{C}_q(\alpha)$ for $0 \le \alpha < 1$ . In addition, we find the Herglotz representation formula for functions belonging to the class $\mathcal{C}_q(\alpha)$ . One can also think of Hankel determinant, Fekete-Szegö problems for $\mathcal{C}_q(\alpha)$ as well. The concept of q-integral is useful in this setting. Thomae was a pupil of Heine who introduced, the so-called, q-integral [21] $$\int_0^1 f(t) d_q t = (1 - q) \sum_{n=0}^{\infty} q^n f(q^n),$$ provided the q-series converges. In 1910, Jackson defined the general q-integral [10] (see also [7,21]) in the following manner: $$\int_{a}^{b} f(t) d_{q}t := \int_{0}^{b} f(t) d_{q}t - \int_{0}^{a} f(t) d_{q}t,$$ where $$I_q(f(x)) := \int_0^x f(t) d_q t = x(1-q) \sum_{n=0}^\infty q^n f(xq^n),$$ provided the q-series converges. Observe that $$D_q I_q f(x) = f(x)$$ and $I_q D_q f(x) = f(x) - f(0)$ , where the second equality holds if f is continuous at x = 0. For more background on q-integrals, we refer to [7]. Now, we state our main results. The Fekete-Szegö problem for the class $\mathcal{S}_q^*$ is obtained as follows: