Upper bound for the second Hankel determinant of certain subclass of analytic and bi-univalent functions

Lemma 1.1. ([4]) Let P be the class of all analytic functions p(z) of the form $$p(z) = 1 + p_1 z + p_2 z^2 + \dots = 1 + \sum_{n=1}^{\infty} p_n z^n$$ (1.8) satisfying $\operatorname{Re}(p(z)) > 0$ , $z \in U$ and p(0) = 1. Then, $|p_n| \le 2$ , for every n = 1, 2, 3, .... This inequality is sharp for each n. Moreover, $$2p_2 = p_1^2 + (4 - p_1^2)x,$$ $$4p_3 = p_1^3 + 2(4 - p_1^2)p_1x - (4 - p_1^2)p_1x^2 + 2(4 - p_1^2)(1 - |x|^2)z,$$

Theorem 2.1. Let the function f(z) given by (1.1) be in the class $Q_{\Sigma}(G, \beta, t)$ , $\beta \in [0,1], t \in (\frac{1}{2},1)$ , where the function G is an analytic function given by (1.4). Then, $$\left|a_{2}a_{4}-a_{3}^{2}\right| \leq \begin{cases} H(t,2-), & \text{if } \Delta(\beta,t) \geq 0 \text{ and } c(\beta,t) \geq 0, \\ \max\left\{\frac{4t^{2}}{\left(1+2\beta\right)^{2}}, H(t,2-)\right\}, & \text{if } \Delta(\beta,t) > 0 \text{ and } c(\beta,t) < 0, \\ \frac{4t^{2}}{\left(1+2\beta\right)^{2}}, & \text{if } \Delta(\beta,t) \leq 0 \text{ and } c(\beta,t) \leq 0, \\ \max\left\{H(t,\tau_{0}), H(t,2-)\right\}, & \text{if } \Delta(\beta,t) < 0 \text{ and } c(\beta,t) > 0, \end{cases}$$ where $$H(t,2-) = \frac{8t^{2} \left| \left(2t^{2} - 1\right)\left(1+\beta\right)^{3} - 2t^{2}\left(1+3\beta\right) \right|}{\left(1+\beta\right)^{4} \left(1+3\beta\right)},$$ $$H(t,\tau_{0}) = \frac{4t^{2}}{\left(1+2\beta\right)^{2}} - \frac{c^{2}(\beta,t)}{4\left(1+\beta\right)^{4} \left(1+2\beta\right)^{2} \left(1+3\beta\right)}, \quad p_{0} = \sqrt{\frac{-2c(\beta,t)}{\Delta(\beta,t)}},$$ $$\Delta(\beta,t) = 16t^{2} \left| \left(2t^{2} - 1\right)\left(1+\beta\right)^{3} - 2t^{2}\left(1+3\beta\right) \left| \left(1+2\beta\right)^{2} - 8t\left[t^{2} - \left(4t^{2} - 1\right)\left(1+\beta\right)\left(1+2\beta\right)\right]\left(1+\beta\right)^{2} \left(1+2\beta\right)\left(1+3\beta\right) - 8t^{2}\left(1+\beta\right)^{3}\beta^{2},$$ $$c(\beta,t) = 8t\left[\left(5t^{2} - 1\right)\left(1+3\beta\right) + 2\left(4t^{2} - 1\right)\beta^{2}\right]\left(1+\beta\right)^{2}\left(1+2\beta\right) - 8t^{2}\left[\left(1+\beta\right)^{2} + \left(2+\beta\right)\beta\right]\left(1+\beta\right)^{3}.$$ Proof. Let $f \in Q_{\Sigma}(G, \beta, t)$ , $\beta \in [0,1]$ , $t \in \left(\frac{1}{2}, 1\right)$ and $g = f^{-1}$ . Then, according to Definition 1.1, there are analytic functions $\omega: U \to U$ , $\varpi: D_{r_0} \to D_{r_0}$ with $\omega(0) = 0 = \varpi(0)$ , $|\omega(z)| < 1$ , $|\varpi(w)| < 1$ satisfying the following conditions $$\left(1-\beta\right)\frac{f(z)}{z} + \beta f'(z) = G(t,\omega(z)), \ z \in U$$ (2.1) and $$(1-\beta)\frac{g(w)}{w} + \beta g'(w) = G(t,\varpi(w)), \ w \in D_{r_0}.$$ (2.2) Let also the functions $p, q \in P$ be define as follows $$p(z) := \frac{1 + \omega(z)}{1 - \omega(z)} = 1 + p_1 z + p_2 z^2 + \dots = 1 + \sum_{n=1}^{\infty} p_n z^n$$ and $$q(w) := \frac{1 + \varpi(w)}{1 - \varpi(w)} = 1 + q_1 w + q_2 w^2 + \dots = 1 + \sum_{n=1}^{\infty} q_n w^n$$ . It follows that $$\omega(z) := \frac{p(z) - 1}{p(z) + 1} = \frac{1}{2} \left[ p_1 z + \left( p_2 - \frac{p_1^2}{2} \right) z^2 + \left( p_3 - p_1 p_2 + \frac{p_1^3}{4} \right) z^3 + \cdots \right]$$ (2.3) and $$\varpi(w) := \frac{q(w) - 1}{q(w) + 1} = \frac{1}{2} \left[ q_1 w + \left( q_2 - \frac{q_1^2}{2} \right) w^2 + \left( q_3 - q_1 q_2 + \frac{q_1^3}{4} \right) w^3 + \cdots \right]. \tag{2.4}$$ From (2.3) and (2.4), considering (1.4), we can easily show that $$G(t,\omega(z)) = 1 + \frac{U_1(t)}{2} p_1 z + \left[ \frac{U_1(t)}{2} \left( p_2 - \frac{p_1^2}{2} \right) + \frac{U_2(t)}{4} p_1^2 \right] z^2 + \left[ \frac{U_1(t)}{2} \left( p_3 - p_1 p_2 + \frac{p_1^3}{4} \right) + \frac{U_2(t)}{2} p_1 \left( p_2 - \frac{p_1^2}{2} \right) + \frac{U_3(t)}{8} p_1^3 \right] z^3 + \cdots$$ $$(2.5)$$ and $$G(t,\varpi(w)) = 1 + \frac{U_1(t)}{2}q_1w + \left[\frac{U_1(t)}{2}\left(q_2 - \frac{q_1^2}{2}\right) + \frac{U_2(t)}{4}q_1^2\right]w^2 + \left[\frac{U_1(t)}{2}\left(q_3 - q_1q_2 + \frac{q_1^3}{4}\right) + \frac{U_2(t)}{2}q_1\left(q_2 - \frac{q_1^2}{2}\right) + \frac{U_3(t)}{8}q_1^3\right]w^3 + \cdots\right]$$ (2.6) From (2.1), (2.5) and (2.2), (2.6), we can easily obtain that $$(1+\beta)a_2 = \frac{U_1(t)}{2}p_1, \tag{2.7}$$ $$(1+2\beta)a_3 = \frac{U_1(t)}{2} \left(p_2 - \frac{p_1^2}{2}\right) + \frac{U_2(t)}{4}p_1^2, \tag{2.8}$$ $$(1+3\beta)a_4 = \frac{U_1(t)}{2} \left(p_3 - p_1p_2 + \frac{p_1^3}{4}\right) + \frac{U_2(t)}{2}p_1 \left(p_2 - \frac{p_1^2}{2}\right) + \frac{U_3(t)}{8}p_1^3$$ (2.9) and $$-(1+\beta)a_2 = \frac{U_1(t)}{2}q_1, \tag{2.10}$$ $$(1+2\beta)(2a_2^2-a_3) = \frac{U_1(t)}{2} \left(q_2 - \frac{q_1^2}{2}\right) + \frac{U_2(t)}{4}q_1^2, \qquad (2.11)$$ $$-\left(1+3\beta\right)\left(5a_{2}^{3}-5a_{2}a_{3}+a_{4}\right)=\frac{U_{1}(t)}{2}\left(q_{3}-q_{1}q_{2}+\frac{q_{1}^{3}}{4}\right)+\frac{U_{2}(t)}{2}q_{1}\left(q_{2}-\frac{q_{1}^{2}}{2}\right)+\frac{U_{3}(t)}{8}q_{1}^{3}. \quad (2.12)$$ From (2.7) and (2.10), we obtain that $$\frac{U_1(t)}{2(1+\beta)}p_1 = a_2 = -\frac{U_1(t)}{2(1+\beta)}q_1. \tag{2.13}$$ Subtracting (2.11) from (2.8) and considering (2.13), we can easily obtain that $$a_{3} = a_{2}^{2} + \frac{U_{1}(t)}{4(1+2\beta)} \left(p_{2} - q_{2}\right) = \frac{U_{1}^{2}(t)}{4(1+\beta)^{2}} + \frac{U_{1}(t)}{4(1+2\beta)} \left(p_{2} - q_{2}\right). \tag{2.14}$$ On the other hand, subtracting (2.12) from (2.9) and considering (2.13) and (2.14), we get $$a_{4} = \frac{5U_{1}^{2}(t)p_{1}(p_{2}-q_{2})}{6(1+\beta)(1+2\beta)} + \frac{U_{1}(t)(p_{3}-q_{3})}{4(1+3\beta)} + \frac{U_{2}(t)-U_{1}(t)}{4(1+3\beta)}(p_{2}+q_{2}) + \frac{U_{1}(t)-2U_{2}(t)+U_{3}(t)}{8(1+3\beta)}p_{1}^{3}.$$ (2.15) Thus, from (2.13), (2.14) and (2.15), we can easily establish that $$a_{2}a_{4} - a_{3}^{2} = \frac{U_{1}^{3}(t)p_{1}^{2}(p_{2} - q_{2})}{32(1+\beta)^{2}(1+2\beta)} + \frac{U_{1}^{2}(t)p_{1}(p_{3} - q_{3})}{8(1+\beta)(1+3\beta)} + \frac{\left[U_{2}(t) - U_{1}(t)\right]U_{1}(t)}{8(1+\beta)(1+3\beta)}p_{1}^{2}(p_{2} + q_{2}) - \frac{U_{1}^{2}(t)(p_{2} - q_{2})^{2}}{16(1+2\beta)^{2}} + \frac{U_{1}(t)p_{1}^{4}}{16(1+\beta)^{4}(1+3\beta)} \left\{ \left[U_{1}(t) - 2U_{2}(t) + U_{3}(t)\right](1+\beta)^{3} - U_{1}^{3}(t)(1+3\beta) \right\}.$$ (2.16) According to Lemma 1.1, we have $$2p_2 = p_1^2 + (4 - p_1^2)x$$ and $2q_2 = q_1^2 + (4 - q_1^2)y$ (2.17) and $$4p_{3} = p_{1}^{3} + 2(4 - p_{1}^{2})p_{1}x - (4 - p_{1}^{2})p_{1}x^{2} + 2(4 - p_{1}^{2})(1 - |x|^{2})z$$ $$4q_{3} = q_{1}^{3} + 2(4 - q_{1}^{2})q_{1}y - (4 - q_{1}^{2})q_{1}y^{2} + 2(4 - q_{1}^{2})(1 - |y|^{2})w,$$ (2.18) for some x, y, z, w with $|x| \le 1$ , $|y| \le 1$ , $|z| \le 1$ , $|w| \le 1$ . Since (see (2.13)) $p_1 = -q_1$ , from (2.17) and (2.18), we get $$p_2 - q_2 = \frac{4 - p_1^2}{2} (x - y), \ p_2 + q_2 = p_1^2 + \frac{4 - p_1^2}{2} (x + y)$$ (2.19) and $$p_{3} - q_{3} = \frac{p_{1}^{3}}{2} + \frac{\left(4 - p_{1}^{2}\right)p_{1}}{2}(x + y) - \frac{\left(4 - p_{1}^{2}\right)p_{1}}{4}(x^{2} + y^{2}) + \frac{4 - p_{1}^{2}}{2} \left[\left(1 - |x|^{2}\right)z - \left(1 - |y|^{2}\right)w\right].$$ (2.20) According to Lemma 1.1, we may assume without any restriction that $\tau \in [0,2]$ , where $\tau = |p_1|$ . Thus, substituting the expressions (2.19) and (2.20) in (2.16) and using triangle inequality, letting $|x| = \xi$ , $|y| = \eta$ , we can easily obtain that $$|a_2 a_4 - a_3^2| \le c_1(t,\tau) (\xi + \eta)^2 + c_2(t,\tau) (\xi^2 + \eta^2) + c_3(t,\tau) (\xi + \eta) + c_4(t,\tau) := F(\xi,\eta), \quad (2.21)$$ where $$\begin{split} c_1(t,\tau) &= \frac{U_1^2(t) \left(4-\tau^2\right)^2}{64 \left(1+2\beta\right)^2} \geq 0, \ c_2(t,\tau) = \frac{U_1^2(t) \tau \left(\tau-2\right) \left(4-\tau^2\right)}{32 \left(1+\beta\right) \left(1+3\beta\right)} \leq 0, \\ c_3(t,\tau) &= \frac{U_1^3(t) \tau^2 \left(4-\tau^2\right)}{64 \left(1+\beta\right)^2 \left(1+2\beta\right)} + \frac{U_1(t) U_2(t) \tau^2 \left(4-\tau^2\right)}{16 \left(1+\beta\right) \left(1+3\beta\right)} \geq 0, \\ c_4(t,\tau) &= \frac{U_1(t) \left| \left(1+\beta\right)^3 U_3(t) - \left(1+3\beta\right) U_1^3(t) \right|}{16 \left(1+\beta\right)^4 \left(1+3\beta\right)} \tau^4 + \frac{U_1^2(t) \tau \left(4-\tau^2\right)}{8 \left(1+\beta\right) \left(1+3\beta\right)} \geq 0, t \in \left(\frac{1}{2},1\right), \tau \in \left[0,2\right]. \end{split}$$ Now, we need to maximize the function $F(\xi,\eta)$ on the closed square $\Omega = \{(\xi,\eta) \colon \xi,\eta \in [0,1]\}$ for $\tau \in [0,2]$ . Since the coefficients of the function $F(\xi,\eta)$ is dependent to variable $\tau$ for fixed value of t, we must investigate the maximum of $F(\xi,\eta)$ respect to $\tau$ taking into account these cases $\tau = 0$ , $\tau = 2$ and $\tau \in (0,2)$ . Let $\tau = 0$ . Then, we write $$F(\xi,\eta) = c_1(t,0) = \frac{U_1^2(t)}{4(1+2\beta)^2} (\xi + \eta)^2.$$ It is clear that the maximum of the function $F(\xi, \eta)$ occurs at $(\xi, \tau) = (1, 1)$ , and $$\max \left\{ F(\xi, \eta) : \xi, \eta \in [0, 1] \right\} = F(1, 1) = \frac{U_1^2(t)}{\left(1 + 2\beta\right)^2}$$ (2.22) Now, let $\tau = 2$ . In this case, $F(\xi, \eta)$ is a constant function (respect to $\tau$ ) as follows: $$F(\xi,\eta) = c_4(t,2) = \frac{U_1(t) \left| \left( 1 + \beta \right)^3 U_3(t) - \left( 1 + 3\beta \right) U_1^3(t) \right|}{\left( 1 + \beta \right)^4 \left( 1 + 3\beta \right)}.$$ (2.23) In the case $\tau \in (0,2)$ , we will examine the maximum of the function $F(\xi,\eta)$ taking into account the sing of $\Lambda(\xi,\eta) = F_{\xi\xi}(\xi,\eta)F_{\eta\eta}(\xi,\eta) - \left[F_{\xi\eta}(\xi,\eta)\right]^2$ . By simple computation, we can easily see that $$\Lambda(\xi,\eta) = 4c_2(t,\tau) \Big[ 2c_1(t,\tau) + c_2(t,\tau) \Big].$$ Since $c_2(t,\tau) < 0$ for all $t \in \left(\frac{1}{2},1\right), \tau \in \left(0,2\right)$ and $$2c_1(t,\tau) + c_2(t,\tau) = \frac{U_1^2(t)(4-\tau^2)(2-\tau)}{32(1+\beta)(1+2\beta)^2(1+3\beta)}\varphi(\tau),$$ where $\varphi(\tau) = 2(1+\beta)(1+3\beta) - \beta^2 \tau$ since $2-\beta^2 \tau > 0$ , $\varphi(\tau) > 6\beta^2 + 8\beta^2 > 0$ ; that is $2c_1(t,\tau) + c_2(t,\tau) > 0$ for all $\tau \in (0,2)$ and $\beta \in [0,1]$ , we conclude that $\Lambda(\xi,\eta) < 0$ for all $(\xi,\eta) \in \Omega$ . Consequently, the function $F(\xi,\eta)$ cannot have a local maximum in $\Omega$ . Therefore, we must investigate the maximum of the function $F(\xi,\eta)$ on the boundary of the square $\Omega$ . Let $$\partial \Omega = \left\{ \left(0, \eta\right) : \eta \in \left[0, 1\right] \right\} \cup \left\{ \left(\xi, 0\right) : \xi \in \left[0, 1\right] \right\} \cup \left\{ \left(1, \eta\right) : \eta \in \left[0, 1\right] \right\} \cup \left\{ \left(\xi, 1\right) : \xi \in \left[0, 1\right] \right\}.$$ We can easily show that the maximum of the function $F(\xi,\eta)$ on the boundary $\partial\Omega$ of the square $\Omega$ occurs at $(\xi,\eta)=(1,1)$ , and $$\max \left\{ F\left(\xi, \eta\right) : \left(\xi, \eta\right) \in \partial\Omega \right\} = F\left(1, 1\right) = 4c_1(t, \tau) + 2\left[c_2(t, \tau) + c_3(t, \tau)\right] + c_4(t, \tau), t \in \left(\frac{1}{2}, 1\right), \tau \in (0, 2).$$ (2.24) Now, let us define the function $H:(0,2) \to \mathbb{R}$ as follows: $$H(t,\tau) = 4c_1(t,\tau) + 2[c_2(t,\tau) + c_3(t,\tau)] + c_4(t,\tau)$$ (2.25) for fixed value of t. Substituting the value $c_i(t,\tau)$ , j=1,2,3,4 in the (2.25), we obtain $$H(t,\tau) = \frac{U_1^2(t)}{(1+2\beta)^2} + \frac{\Delta(\beta,t)\tau^4 + 4c(\beta,t)\tau^2}{32(1+\beta)^4(1+2\beta)^2(1+3\beta)},$$ where $$\begin{split} \Delta(\beta,t) &= 2U_{1}(t) \Big| \Big(1+\beta\Big)^{3} U_{3}(t) - \Big(1+3\beta\Big) U_{1}^{3}(t) \Big| \Big(1+2\beta\Big)^{2} - \\ U_{1}(t) \Big[ U_{1}^{2}(t) - 4U_{2}(t) \Big(1+\beta\Big) \Big(1+2\beta\Big) \Big] \Big(1+\beta\Big)^{2} \Big(1+2\beta\Big) \Big(1+3\beta\Big) - 2U_{1}^{2}(t)\beta^{2} \Big(1+\beta\Big)^{3} , \\ c(\beta,t) &= U_{1}(t) \Big[ U_{1}^{2}(t) \Big(1+3\beta\Big) + 4U_{2}(t) \Big(1+\beta\Big) \Big(1+2\beta\Big) \Big] \Big(1+\beta\Big)^{2} \Big(1+2\beta\Big) - \\ 2U_{1}^{2}(t) \Big[ \Big(1+\beta\Big)^{2} + \Big(2+\beta\Big)\beta\Big] \Big(1+\beta\Big)^{3} . \end{split}$$ Now, we must investigate the maximum (respect to $\tau$ ) of the function $H(t,\tau)$ in the interval (0,2) for fixed value of t. By simple computation, we can easily show that $$H'(t,\tau) = \frac{\Delta(\beta,t)\tau^2 + 2c(\beta,t)}{8(1+\beta)^4(1+2\beta)^2(1+3\beta)}\tau.$$ We will examine the sign of the function $H'(t,\tau)$ depending on the different cases of the signs of $\Delta(t,\tau)$ and $c(t,\tau)$ as follows. (i) Let $\Delta(\beta,t) \ge 0$ and $c(\beta,t) \ge 0$ , then $H'(t,\tau) \ge 0$ , so $H(t,\tau)$ is an increasing function. Therefore, $$\max \left\{ H(t,\tau) : \tau \in (0,2) \right\} = H(t,2-) = \frac{U_1(t) \left| (1+\beta)^3 U_3(t) - (1+3\beta) U_1^3(t) \right|}{(1+\beta)^4 (1+3\beta)}. \tag{2.26}$$ That is, $$\max\left\{\max\left\{F\left(\xi,\eta\right)\colon\xi,\eta\in\left[0,1\right]\right\}\colon\tau\in\left(0,2\right)\right\}=H(t,2-)\;.$$ (u) Let $\Delta(\beta,t) > 0$ and $c(\beta,t) < 0$ , then $\tau_0 = \sqrt{\frac{-2c(\beta,t)}{\Delta(\beta,t)}}$ is a critical point of the function $H(t,\tau)$ . We assume that $\tau_0 \in (0,2)$ . Since $H''(t,\tau_0) > 0$ , $\tau_0$ is a local minimum point of the function $H(t,\tau)$ . That is, the function $H(t,\tau)$ cannot have a local maximum. (uu) Let $\Delta(\beta,t) \leq 0$ and $c(\beta,t) \leq 0$ , then $H'(t,\tau) \leq 0$ . Thus, $H(t,\tau)$ is an decreasing function on the interval (0,2). Therefore, $$\max \left\{ H(t,\tau) : \tau \in (0,2) \right\} = H(t,0+) = 4c_1(t,0) = \frac{U_1^2(t)}{\left(1+2\beta\right)^2}. \tag{2.27}$$ (iv) Let $\Delta(\beta,t) < 0$ and $c(\beta,t) > 0$ , then $\tau_0$ is a critical point of the function $H(t,\tau)$ . We assume that $\tau_0 \in (0,2)$ . Since $H''(t,\tau_0) < 0$ , $\tau_0$ is a local maximum point of the function $H(t,\tau)$ and maximum value occurs at $\tau = \tau_0$ . Therefore, $$\max \{H(t,\tau) : \tau \in (0,2)\} = H(t,\tau_0), \qquad (2.28)$$ where $$H(t,\tau_0) = \frac{4t^2}{(1+2\beta)^2} - \frac{c^2(\beta,t)}{4(1+\beta)^4(1+2\beta)^2(1+3\beta)}.$$ Thus, from (2.22)-(2.28), the proof of Theorem 2.1 is completed. In the special cases from Theorem 2.1, we arrive at the following results.

Corollary 2.1. Let the function f(z) given by (1.1) be in the class $Q_{\Sigma}(G,0,t) = N_{\Sigma}(G,t)$ , $t \in \left(\frac{1}{2},1\right)$ , where the function G is an analytic function given by (1.4). Then, $$\left| a_2 a_4 - a_3^2 \right| \le 8t^2.$$

Corollary 2.2. Let the function f(z) given by (1.1) be in the class $Q_{\Sigma}(G,1,t) = \Re_{\Sigma}(G,t)$ , $t \in \left(\frac{1}{2},1\right)$ , where the function G is an analytic function given by (1.4). Then, $$|a_2a_4-a_3^2| \le t^2(1-t^2)$$ . From the Corollary 2.2, we can easily arrive at the following result.

Corollary 2.3. Let the function f(z) given by (1.1) be in the class $Q_{\Sigma}(G,1,\frac{\sqrt{2}}{2}) = \Re_{\Sigma}(G)$ , where the function G is an analytic function given by (1.4). Then, $$\left|a_2a_4-a_3^2\right| \leq \frac{1}{4}$$ .

Definition 1.1. A function $f \in \Sigma$ given by (1.1) is said to be in the class $Q_{\Sigma}(G, \beta, t), \ \beta \in [0,1], t \in \left(\frac{1}{2}, 1\right)$ , where G is an analytic function given by (1.4), if the following conditions are satisfied $$(1-\beta)\frac{f(z)}{z} + \beta f'(z) \prec G(t,z), \ z \in U$$ (1.6) and $$(1-\beta)\frac{g(w)}{w} + \beta g'(w) \prec G(t, w), \ w \in D_{r_0}$$ (1.7) where $g = f^{-1}$ . Remark 1.1. Taking $\beta = 1$ , we have $Q_{\Sigma}(G, 1, t) = \Re_{\Sigma}(G, t)$ , $t \in \left(\frac{1}{2}, 1\right)$ ; that is, $$f \in \mathfrak{R}_{\Sigma}(G,t) \Leftrightarrow f'(z) \prec G(t,z), \ z \in U \ \ and \ \ g'(w) \prec G(t,w), \ w \in D_{r_0},$$ where $g = f^{-1}$ . Remark 1.2. Taking $\beta = 0$ , we have $Q_{\Sigma}(G, 0, t) = N_{\Sigma}(G, t)$ , $t \in \left(\frac{1}{2}, 1\right)$ ; that is, $$f \in N_{\Sigma}(G,t) \Leftrightarrow \frac{f(z)}{z} \prec G(t,z), \ z \in U \ \ and \ \ \frac{g(w)}{w} \prec G(t,w), \ w \in D_{r_0},$$ where $g = f^{-1}$ . The object of this paper is to determine the second Hankel determinant for the function class $Q_{\Sigma}(G,\beta,t)$ and its special classes, and is to give upper bound estimate for $|H_2(2)|$ . In the study, the bound estimates for the initial coefficients of the functions belonging to this class are also obtained. For the functions belonging to class $Q_{\Sigma}(G,\beta,t)$ , Fekete-Szöge inequality is also obtained. In order to prove our main results, we shall need the following lemma.