New subclass of the class of close-to-convex harmonic mappings defined by a third-order differential inequality

Lemma 1. Suppose $\mathfrak{s}$ and $\mathfrak{t}$ are analytic in $\mathcal{U}$ with $|\mathfrak{t}'(0)| < |\mathfrak{s}'(0)|$ and $F_{\epsilon} = \mathfrak{s} + \epsilon \mathfrak{t}$ is close-to-convex for each $\epsilon$ ( $|\epsilon| = 1$ ), then $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}}$ is close-to-convex in $\mathcal{U}$ .

Theorem 2. The harmonic mapping $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ if and only if $F_{\epsilon} = \mathfrak{s} + \epsilon \mathfrak{t} \in \mathcal{R}(\gamma, \delta, \lambda)$ for each $\epsilon (|\epsilon| = 1)$ . Proof. Suppose $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ . For each $|\epsilon| = 1$ , $$\operatorname{Re}\left\{\gamma F_{\epsilon}'(z) + \delta z F_{\epsilon}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} F_{\epsilon}'''(z)\right\}$$ $$= \operatorname{Re}\left\{\gamma \mathfrak{s}'(z) + \delta z \mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{s}'''(z)\right\}$$ $$+ \epsilon \left(\gamma \mathfrak{t}'(z) + \delta z \mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{t}'''(z)\right)\right\}$$ $$> \operatorname{Re}\left\{\gamma \mathfrak{s}'(z) + \delta z \mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{s}'''(z)\right\}$$ $$- \left|\gamma \mathfrak{t}'(z) + \delta z \mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{t}'''(z)\right|$$ $$> \lambda \qquad (z \in \mathcal{U}).$$ Thus, $F_{\epsilon} \in \mathcal{R}(\gamma, \delta, \lambda)$ for each $\epsilon(|\epsilon| = 1)$ . Conversely, let $F_{\epsilon} = \mathfrak{s} + \epsilon \mathfrak{t} \in \mathcal{R}(\gamma, \delta, \lambda)$ then $$\operatorname{Re}\left\{\gamma\mathfrak{s}'(z) + \delta z\mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right)z^{2}\mathfrak{s}'''(z)\right\}$$ $$> \operatorname{Re}\left\{-\epsilon\left(\gamma\mathfrak{t}'(z) + \delta z\mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right)z^{2}\mathfrak{t}'''(z)\right)\right\} + \lambda \ (z \in \mathcal{U}).$$ With appropriate choice of $\epsilon(|\epsilon|=1)$ , it follows that $$\operatorname{Re}\left\{\gamma\mathfrak{s}'(z) + \delta z\mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right)z^{2}\mathfrak{s}'''(z) - \lambda\right\}$$ $$> \left|\gamma\mathfrak{t}'(z) + \delta z\mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right)z^{2}\mathfrak{t}'''(z)\right| \quad (z \in \mathcal{U}),$$ and hence $\mathfrak{f} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ .

Lemma 3. (Jack-Miller-Mocanu Lemma [12, 13]) Let w defined by $w(z) = c_n z^n + c_{n+1} z^{n+1} + ...$ be analytic in $\mathcal{U}$ , with $c_n \neq 0$ , and let $z_0 \neq 0$ , $z_0 = r_0 e^{i\theta_0} (0 < r_0 < 1)$ be a point of $\mathcal{U}$ such that $$|w(z_0)| = \max_{|z| \le |z_0|} |w(z)|$$ then there is a real number $k, k \ge n \ge 1$ , such that $$\frac{z_0 w'(z_0)}{w(z_0)} = k \quad and \quad Re\left\{1 + \frac{z_0 w''(z_0)}{w'(z_0)}\right\} \ge k.$$

Lemma 4. If $F \in \mathcal{R}(\gamma, \delta, \lambda)$ then $Re\{F'(z)\} > 0$ , and hence F is close-to-convex in $\mathcal{U}$ . Proof. Suppose $F \in \mathcal{R}(\gamma, \delta, \lambda)$ and $\frac{2\gamma F'(z) + 2\delta z F''(z) + (\delta - \gamma)z^2 F'''(z) - 2\lambda}{2(\gamma - \lambda)} =: \Psi(z)$ . Then $\text{Re}\{\Psi(z)\} > 0$ for $z \in \mathcal{U}$ . Consider an analytic function w in $\mathcal{U}$ with w(0) = 0 and $$F'(z) = \frac{1 + w(z)}{1 - w(z)}, \quad w(z) \neq 1.$$ We need to prove that |w(z)| < 1 for all z ∈ U. Then we have $$\Psi(z) = \frac{2\gamma F'(z) + 2\delta z F''(z) + (\delta - \gamma) z^{2} F'''(z) - 2\lambda}{2 (\gamma - \lambda)} = \frac{\gamma}{\gamma - \lambda} \frac{1 + w(z)}{1 - w(z)} + \frac{2\delta}{\gamma - \lambda} \frac{zw'(z)}{(1 - w(z))^{2}} + \frac{\delta - \gamma}{\gamma - \lambda} \frac{z^{2} \left[w''(z) (1 - w(z)) + 2 (w'(z))^{2}\right]}{(1 - w(z))^{3}} - \frac{\lambda}{\gamma - \lambda} = \frac{1}{\gamma - \lambda} \left(\gamma \frac{1 + w(z)}{1 - w(z)} + 2\delta \frac{zw'(z)}{(1 - w(z))^{2}} \right. + (\delta - \gamma) \frac{zw'(z)}{(1 - w(z))^{2}} \frac{zw''(z)}{w'(z)} + 2 (\delta - \gamma) \frac{(zw'(z))^{2}}{(1 - w(z))^{3}} - \lambda\right).$$ Since w is analytic in U and w(0) = 0, if there are z⁰ ∈ U such that $$\max_{|z| \le |z_0|} |w(z)| = |w(z_0)| = 1,$$ then by Lemma 3, we can write $$w(z_0) = e^{i\theta}, \quad z_0 w'(z_0) = k w(z_0) = k e^{i\theta}, \quad (k \ge 1, \ 0 < \theta < 2\pi).$$ and $$\operatorname{Re}\left\{\frac{z_0w''(z_0)}{w'(z_0)}\right\} \ge k - 1.$$ For such a point z⁰ ∈ U, we obtain $$\operatorname{Re}\{\Psi(z_{0})\} = \frac{1}{\gamma - \lambda} \operatorname{Re}\left(\gamma \frac{1 + w(z_{0})}{1 - w(z_{0})} + 2\delta \frac{z_{0}w'(z_{0})}{(1 - w(z_{0}))^{2}} + (\delta - \gamma) \frac{z_{0}w'(z_{0})}{(1 - w(z_{0}))^{2}} \frac{z_{0}w''(z_{0})}{w'(z_{0})} + 2(\delta - \gamma) \frac{(z_{0}w'(z_{0}))^{2}}{(1 - w(z_{0}))^{3}} - \lambda\right) \\ = \frac{1}{\gamma - \lambda} \left[ -\frac{\delta k}{1 - \cos \theta} - \frac{(\delta - \gamma) k}{2(1 - \cos \theta)} \operatorname{Re}\left\{ \frac{zw''(z_{0})}{w'(z_{0})} \right\} + \frac{(\delta - \gamma) k^{2}}{2(1 - \cos \theta)} - \lambda \right] \\ \leq \frac{1}{\gamma - \lambda} \left[ -\frac{\delta k}{1 - \cos \theta} + \frac{(\delta - \gamma) k}{2(1 - \cos \theta)} (1 - k) + \frac{(\delta - \gamma) k^{2}}{2(1 - \cos \theta)} - \lambda \right] \\ = -\frac{1}{\gamma - \lambda} \left[ \frac{(\delta + \gamma) k}{2(1 - \cos \theta)} + \lambda \right] < 0,$$ which contradicts our assumption. Hence, there is no z⁰ ∈ U such that |w(z0)| = 1, which means that |w(z)| < 1 for all z ∈ U. Therefore, we obtain that Re{F ′ (z)} > 0. Theorem 5 The functions in the class R⁰ ^H(γ, δ, λ) are close-to-convex in U. Proof. Referring to Lemma 4, we derive that functions F^ǫ = s + ǫt ∈ R(γ, δ, λ) are close-to-convex in U for each ǫ(|ǫ| = 1). Now in view of Lemma 1 and Theorem 2, we obtain that functions in R⁰ ^H (γ, δ, λ) are close-to-convex in U.

Theorem 6. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ then for $m \geq 2$ , $$|b_m| \le \frac{2(\gamma - \lambda)}{m^2 \left[2\gamma + (\delta - \gamma)(m - 1)\right]}.$$ (4) The result is sharp and equality holds for the function $f(z) = z + \frac{2(\gamma - \lambda)}{m^2[2\gamma + (\delta - \gamma)(m - 1)]} \bar{z}^m$ . Proof. Suppose that $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Using the series representation of $\mathfrak{t}(z)$ , we derive $$r^{m-1}m^{2}\left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right]|b_{m}|$$ $$\leq \frac{1}{2\pi} \int_{0}^{2\pi} \left|\gamma \mathfrak{t}'(re^{i\theta}) + \delta re^{i\theta}\mathfrak{t}''(re^{i\theta}) + \left(\frac{\delta - \gamma}{2}\right)r^{2}e^{2i\theta}\mathfrak{t}'''(re^{i\theta})\right|d\theta$$ $$< \frac{1}{2\pi} \int_{0}^{2\pi} \operatorname{Re}\left\{\gamma \mathfrak{s}'(re^{i\theta}) + \delta re^{i\theta}\mathfrak{s}''(re^{i\theta}) + \left(\frac{\delta - \gamma}{2}\right)r^{2}e^{2i\theta}\mathfrak{s}'''(re^{i\theta}) - \lambda\right\}d\theta$$ $$= \frac{1}{2\pi} \int_{0}^{2\pi} \operatorname{Re}\left\{\gamma - \lambda + \sum_{m=2}^{\infty} m^{2}\left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right]a_{m}r^{m-1}e^{i(m-1)\theta}\right\}d\theta$$ $$= \gamma - \lambda.$$ Allowing $r \to 1^-$ gives the desired bound. Moreover, it is easy to verify that the equality holds for the function $\mathfrak{f}(z) = z + \frac{2(\gamma - \lambda)}{m^2[2\gamma + (\delta - \gamma)(m - 1)]}\bar{z}^m$ .

Theorem 7. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ . Then for $m \geq 2$ , we have (i) $$|a_m| + |b_m| \le \frac{4(\gamma - \lambda)}{m^2 [2\gamma + (\delta - \gamma)(m - 1)]},$$ (ii) $||a_m| - |b_m|| \le \frac{4(\gamma - \lambda)}{m^2 [2\gamma + (\delta - \gamma)(m - 1)]},$ (iii) $|a_m| \le \frac{4(\gamma - \lambda)}{m^2 [2\gamma + (\delta - \gamma)(m - 1)]}.$ All these results are sharp and all equalities hold for the function $\mathfrak{f}(z) = z + \sum_{m=2}^{\infty} \frac{4(\gamma-\lambda)}{m^2[2\gamma+(\delta-\gamma)(m-1)]} z^m$ . Proof. (i) Suppose that $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ , then from Theorem 2, $F_{\epsilon} = \mathfrak{s} + \epsilon \mathfrak{t} \in \mathcal{R}(\gamma, \delta, \lambda)$ for $\epsilon$ ( $|\epsilon| = 1$ ). Thus for each $|\epsilon| = 1$ , we have <span id="page-5-0"></span>Re $$\left\{ \gamma(\mathfrak{s} + \epsilon \mathfrak{t})' + \delta z(\mathfrak{s} + \epsilon \mathfrak{t})'' + \left( \frac{\delta - \gamma}{2} \right) z^2(\mathfrak{s} + \epsilon \mathfrak{t})''' \right\} > \lambda$$ for $z \in \mathcal{U}$ . This implies that there exists an analytic function p of the form $p(z) = 1 + \sum_{m=1}^{\infty} p_m z^m$ , with Re[p(z)] > 0 in $\mathcal{U}$ such that $$\gamma \mathfrak{s}'(z) + \delta z \mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^2 \mathfrak{s}'''(z) + \epsilon \left(\gamma \mathfrak{t}'(z) + \delta z \mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^2 \mathfrak{t}'''(z)\right)$$ $$= \lambda + (\gamma - \lambda) p(z). \tag{5}$$ Comparing coefficients on both sides of (5) we have $$m^2 \left[ \gamma + \frac{\delta - \gamma}{2} (m - 1) \right] (a_m + \epsilon b_m) = (\gamma - \lambda) p_{m-1} \text{ for } m \ge 2.$$ Since $|p_m| \leq 2$ for $m \geq 1$ , and $\epsilon(|\epsilon| = 1)$ is arbitrary, proof of (i) is complete. Proofs of (ii) and (iii) follows from (i). The function $\mathfrak{f}(z) = z + \sum_{m=2}^{\infty} \frac{4(\gamma - \lambda)}{m^2[2\gamma + (\delta - \gamma)(m-1)]} z^m$ , shows that all inequalities are sharp. The following result gives a sufficient condition for a function to be in the class $\mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Theorem 8 Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{H}^0$ with <span id="page-6-0"></span> $$\sum_{m=2}^{\infty} m^2 \left[ 2\gamma + (\delta - \gamma) (m - 1) \right] (|a_m| + |b_m|) \le 2 (\gamma - \lambda), \tag{6}$$ then $\mathfrak{f} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ . Proof. Suppose that $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{H}^0$ . Then using (6), $$\operatorname{Re}\left\{\gamma\mathfrak{s}'(z) + \delta z\mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right)z^{2}\mathfrak{s}'''(z) - \lambda\right\}$$ $$= \operatorname{Re}\left\{\gamma - \lambda + \sum_{m=2}^{\infty} m^{2} \left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right] a_{m}z^{m-1}\right\}$$ $$> \gamma - \lambda - \sum_{m=2}^{\infty} m^{2} \left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right] |a_{m}|$$ $$\geq \sum_{m=2}^{\infty} m^{2} \left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right] |b_{m}|$$ $$> \left|\sum_{m=2}^{\infty} m^{2} \left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right] b_{m}z^{m-1}\right|$$ $$= \left|\gamma\mathfrak{t}'(z) + \delta z\mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right)z^{2}\mathfrak{t}'''(z)\right|.$$ Hence, $\mathfrak{f} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Corollary 9 Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{H}^0$ satisfies the inequality (6), then f is stable harmonic close-to-convex in $\mathcal{U}$ .

Theorem 10. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Then $$|z| + 4(\gamma - \lambda) \sum_{m=2}^{\infty} \frac{(-1)^{m-1} |z|^m}{m^2 [2\gamma + (\delta - \gamma) (m-1)]} \le |\mathfrak{f}(z)|,$$ $$|\mathfrak{f}(z)| \le |z| + 4\left(\gamma - \lambda\right) \sum_{m=2}^{\infty} \frac{|z|^m}{m^2 \left[2\gamma + (\delta - \gamma)\left(m - 1\right)\right]}.$$ Inequalities are sharp for the function $f(z) = z + \sum_{m=2}^{\infty} \frac{4(\gamma - \lambda)}{m^2[2\gamma + (\delta - \gamma)(m-1)]} \overline{z}^m$ . Proof. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Then using Theorem 2, $F_{\epsilon} \in \mathcal{R}(\gamma, \delta, \lambda)$ and for each $|\epsilon| = 1$ we have $\text{Re}\{\psi(z)\} > \lambda$ where $$\psi(z) = \gamma F_{\epsilon}'(z) + \delta z F_{\epsilon}''(z) + \frac{\delta - \gamma}{2} z^2 F_{\epsilon}'''(z).$$ Then, we have $$\psi(z) = \left(\frac{\delta - \gamma}{2}\right) \left[\frac{2\gamma}{\delta - \gamma} F'_{\epsilon}(z) + \left(\frac{2\gamma}{\delta - \gamma} + 2\right) z F''_{\epsilon}(z) + z^{2} F'''_{\epsilon}(z)\right]$$ $$= \left(\frac{\delta - \gamma}{2}\right) \left[\frac{2\gamma}{\delta - \gamma} \left(z F'_{\epsilon}(z)\right)' + \left(z^{2} F''_{\epsilon}(z)\right)'\right]$$ $$= \left(\frac{\delta - \gamma}{2}\right) \left[\frac{2\gamma}{\delta - \gamma} \left(z F'_{\epsilon}(z)\right) + \left(z^{2} F''_{\epsilon}(z)\right)\right]'$$ $$= \left(\frac{\delta - \gamma}{2}\right) \left[z^{2 - \frac{2\gamma}{\delta - \gamma}} \left(z^{\frac{2\gamma}{\delta - \gamma}} F'_{\epsilon}(z)\right)'\right]'.$$ Then integrating from 0 to z gives $$\left(\frac{2}{\delta - \gamma}\right) z^{\frac{2\gamma}{\delta - \gamma} - 2} \int_{0}^{z} \psi(\omega) d\omega = \left(z^{\frac{2\gamma}{\delta - \gamma}} F_{\epsilon}'(z)\right)'.$$ Making the substitution $\omega = r^{\frac{\delta-\gamma}{2}}z$ in the above integral and integrating again, change of variables gives <span id="page-7-0"></span> $$F'_{\epsilon}(z) = \frac{1}{\gamma} \int_{0}^{1} \int_{0}^{1} \psi(v^{\frac{\delta - \gamma}{2\gamma}} uz) du dv.$$ (7) On the other hand, since $\operatorname{Re}\left\{\frac{\psi(z)-\lambda}{\gamma-\lambda}\right\} > 0$ then $\psi(z) \prec \frac{\gamma+(\gamma-2\lambda)z}{1-z}$ where $\prec$ denotes the subordination [5]. Let $$\phi(z) = \int_{0}^{1} \int_{0}^{1} \frac{dudv}{1 - uv^{\frac{\delta - \gamma}{2\gamma}} z} = 1 + \sum_{m=1}^{\infty} \frac{z^{m}}{(1 + m)\left(1 + \frac{\delta - \gamma}{2\gamma} m\right)}$$ and $$h(z) = \frac{1}{\gamma} \left( \frac{\gamma + (\gamma - 2\lambda)z}{1 - z} \right) = 1 + \sum_{m=1}^{\infty} \frac{2(\gamma - \lambda)}{\gamma} z^{m}.$$ Then, from (7) we have $$\begin{split} F_{\epsilon}'(z) & \prec & (\phi h)(z) \\ & = \left(1 + \sum_{m=1}^{\infty} \frac{z^m}{(1+m)\left(1 + \frac{\delta - \gamma}{2\gamma}m\right)}\right) \left(1 + \sum_{m=1}^{\infty} \frac{2\left(\gamma - \lambda\right)}{\gamma}z^m\right) \\ & = 1 + \sum_{m=1}^{\infty} \frac{4\left(\gamma - \lambda\right)}{m^2\left(\delta - \gamma\right) + m\left(\delta + \gamma\right) + 2\gamma}z^m. \end{split}$$ Since $$|F'_{\epsilon}(z)| = |\mathfrak{s}'(z) + \epsilon \mathfrak{t}'(z)|$$ $$\leq 1 + 4(\gamma - \lambda) \sum_{m=1}^{\infty} \frac{|z|^m}{m^2(\delta - \gamma) + m(\delta + \gamma) + 2\gamma}$$ and $$\begin{aligned} |F'_{\epsilon}(z)| &= |\mathfrak{s}'(z) + \epsilon \mathfrak{t}'(z)| \\ &\geq 1 + 4 \left(\gamma - \lambda\right) \sum_{m=1}^{\infty} \frac{(-1)^m |z|^m}{m^2 \left(\delta - \gamma\right) + m \left(\delta + \gamma\right) + 2\gamma}, \end{aligned}$$ in particular we have $$|\mathfrak{s}'(z)| + |\mathfrak{t}'(z)| \le 1 + 4(\gamma - \lambda) \sum_{m=1}^{\infty} \frac{|z|^m}{m^2(\delta - \gamma) + m(\delta + \gamma) + 2\gamma}$$ and $$|\mathfrak{s}'(z)| - |\mathfrak{t}'(z)| \ge 1 + 4(\gamma - \lambda) \sum_{m=1}^{\infty} \frac{(-1)^m |z|^m}{m^2 (\delta - \gamma) + m (\delta + \gamma) + 2\gamma}.$$ Let $\Gamma$ be the radial segment from 0 to z, then $$\begin{aligned} |\mathfrak{f}(z)| &= \left| \int\limits_{\Gamma} \frac{\partial \mathfrak{f}}{\partial \zeta} d\zeta + \frac{\partial \mathfrak{f}}{\partial \overline{\zeta}} d\overline{\zeta} \right| \leq \int\limits_{\Gamma} \left( |\mathfrak{s}'(\zeta)| + |\mathfrak{t}'(\zeta)| \right) |d\zeta| \\ &\leq \int\limits_{0}^{|z|} \left( 1 + 4 \left( \gamma - \lambda \right) \sum_{m=1}^{\infty} \frac{|\tau|^m}{m^2 \left( \delta - \gamma \right) + m \left( \delta + \gamma \right) + 2\gamma} \right) d\tau \\ &= |z| + 4 \left( \gamma - \lambda \right) \sum_{m=1}^{\infty} \frac{|z|^{m+1}}{(m+1) \left[ m^2 \left( \delta - \gamma \right) + m \left( \delta + \gamma \right) + 2\gamma \right]} \\ &= |z| + 4 \left( \gamma - \lambda \right) \sum_{m=2}^{\infty} \frac{|z|^m}{m \left[ (m-1)^2 \left( \delta - \gamma \right) + (m-1) \left( \delta + \gamma \right) + 2\gamma \right]} \\ &= |z| + 4 \left( \gamma - \lambda \right) \sum_{m=2}^{\infty} \frac{|z|^m}{m^2 \left[ 2\gamma + \left( \delta - \gamma \right) \left( m - 1 \right) \right]} \end{aligned}$$ and $$|\mathfrak{f}(z)| \geq \int_{\Gamma} (|\mathfrak{s}'(\zeta)| - |\mathfrak{t}'(\zeta)|) |d\zeta|$$ $$\geq \int_{0}^{|z|} \left( 1 + 4(\gamma - \lambda) \sum_{m=1}^{\infty} \frac{(-1)^{m} |\tau|^{m}}{m^{2} (\delta - \gamma) + m(\delta + \gamma) + 2\gamma} \right) d\tau$$ $$= |z| + 4(\gamma - \lambda) \sum_{m=2}^{\infty} \frac{(-1)^{m-1} |z|^{m}}{m^{2} [2\gamma + (\delta - \gamma)(m-1)]}.$$

Theorem 11. The class $\mathcal{R}_H^0(\gamma, \delta, \lambda)$ is closed under convex combinations. Proof. Suppose $\mathfrak{f}_i = \mathfrak{s}_i + \overline{\mathfrak{t}_i} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ for i = 1, 2, ..., n and $\sum_{i=1}^n \varrho_i = 1$ ( $0 \le \varrho_i \le 1$ ). The convex combination of functions $\mathfrak{f}_i$ (i = 1, 2, ..., n) may be written as $$f(z) = \sum_{i=1}^{n} \varrho_i f_i(z) = \mathfrak{s}(z) + \overline{\mathfrak{t}(z)},$$ where $$\mathfrak{s}(z) = \sum_{i=1}^{n} \varrho_{i} \mathfrak{s}_{i}(z)$$ and $\mathfrak{t}(z) = \sum_{i=1}^{n} \varrho_{i} \mathfrak{t}_{i}(z)$ . Then both $\mathfrak{s}$ and $\mathfrak{t}$ are analytic in $\mathcal{U}$ with $\mathfrak{s}(0) = \mathfrak{t}(0) = \mathfrak{s}'(0) - 1 = \mathfrak{t}'(0) = 0$ and $$\operatorname{Re}\{\gamma \mathfrak{s}'(z) + \delta z \mathfrak{s}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{s}'''(z) - \lambda\}$$ $$= \operatorname{Re}\left\{\sum_{i=1}^{n} \varrho_{i} \left(\gamma \mathfrak{s}'_{i}(z) + \delta z \mathfrak{s}''_{i}(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{s}'''_{i}(z) - \lambda\right)\right\}$$ $$> \sum_{i=1}^{n} \varrho_{i} \left|\gamma \mathfrak{t}'_{i}(z) + \delta z \mathfrak{t}''_{i}(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{t}'''_{i}(z)\right|$$ $$\geq \left|\gamma \mathfrak{t}'(z) + \delta z \mathfrak{t}''(z) + \left(\frac{\delta - \gamma}{2}\right) z^{2} \mathfrak{t}'''(z)\right|$$ showing that $\mathfrak{f} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . A sequence $\{c_m\}_{m=0}^{\infty}$ of non-negative real numbers is said to be a convex null sequence, if $c_m \to 0$ as $m \to \infty$ , and $c_0 - c_1 \ge c_1 - c_2 \ge c_2 - c_3 \ge ... \ge c_{m-1} - c_m \ge ... \ge 0$ . To prove results for convolution, we shall need the following Lemma 12 and Lemma 13.

Lemma 12. [7] If $\{c_m\}_{m=0}^{\infty}$ be a convex null sequence, then function $$q(z) = \frac{c_0}{2} + \sum_{m=1}^{\infty} c_m z^m$$ is analytic and $Re\{q(z)\} > 0$ in $\mathcal{U}$ .

Lemma 13. [20] Let the function p be analytic in $\mathcal{U}$ with p(0) = 1 and $Re\{p(z)\} > 1/2$ in $\mathcal{U}$ . Then for any analytic function F in $\mathcal{U}$ , the function p \* F takes values in the convex hull of the image of $\mathcal{U}$ under F.

Lemma 14. Let $$F \in \mathcal{R}(\gamma, \delta, \lambda)$$ , then $Re\left\{\frac{F(z)}{z}\right\} > \frac{1}{2}$ . Proof. Suppose F ∈ R(γ, δ, λ) be given by F(z) = z + P^∞ ^m=2 Amz ^m, then $$\operatorname{Re}\left\{\gamma + \sum_{m=2}^{\infty} m^2 \left[\gamma + \frac{\delta - \gamma}{2}(m-1)\right] A_m z^{m-1}\right\} > \lambda \quad (z \in \mathcal{U}),$$ which is equivalent to Re{p(z)} > 1 2 in U, where $$p(z) = 1 + \frac{1}{4(\gamma - \lambda)} \sum_{m=2}^{\infty} m^2 [2\gamma + (\delta - \gamma)(m - 1)] A_m z^{m-1}.$$ Now consider a sequence {cm} ∞ ^m=0 defined by $$c_0 = 1 \text{ and } c_{m-1} = \frac{4(\gamma - \lambda)}{m^2 [2\gamma + (\delta - \gamma)(m-1)]} \text{ for } m \ge 2.$$ It can be easily seen that the sequence {cm} ∞ ^m=0 is a convex null sequence. Using Lemma 12, this implies that the function $$q(z) = \frac{1}{2} + \sum_{m=2}^{\infty} \frac{4(\gamma - \lambda)}{m^2 \left[2\gamma + (\delta - \gamma)(m - 1)\right]} z^{m-1}$$ is analytic and Re{q(z)} > 0 in U. Writing $$\frac{F(z)}{z} = p(z) * \left(1 + \sum_{m=2}^{\infty} \frac{4(\gamma - \lambda)}{m^2 \left[2\gamma + (\delta - \gamma)(m - 1)\right]} z^{m-1}\right),$$ and making use of Lemma 13 gives that Re F(z) z > 1 2 for z ∈ U. Lemma 15 Let Fⁱ ∈ R(γ, δ, λ) for i = 1, 2. Then F¹ ∗ F² ∈ R(γ, δ, λ). Proof. Suppose F1(z) = z + P^∞ ^m=2 Amz ^m and F2(z) = z + P^∞ ^m=2 Bmz ^m. Then the convolution of F1(z) and F2(z) is defined by $$F(z) = (F_1 * F_2)(z) = z + \sum_{m=2}^{\infty} A_m B_m z^m.$$ Since F ′ (z) = F ′ 1 (z) ∗ F2(z) z , zF′′(z) = zF′′ 1 (z) ∗ F2(z) z and zF′′′(z) = zF′′′ 1 (z) ∗ F2(z) z then we have <span id="page-10-0"></span> $$\frac{2\gamma F'(z) + 2\delta z F''(z) + (\delta - \gamma) z^2 F'''(z) - 2\lambda}{2(\gamma - \lambda)}$$ $$= \left(\frac{2\gamma F_1'(z) + 2\delta z F_1''(z) + (\delta - \gamma) z^2 F_1'''(z) - 2\lambda}{2(\gamma - \lambda)}\right) * \frac{F_2(z)}{z}.$$ (8) Since F¹ ∈ R(γ, δ, λ), $$\operatorname{Re}\left\{\frac{2\gamma F_{1}'(z)+2\delta z F_{1}''(z)+\left(\delta-\gamma\right)z^{2} F_{1}'''\left(z\right)-2\lambda}{2\left(\gamma-\lambda\right)}\right\}>0\ \left(z\in\mathcal{U}\right)$$ and using Lemma 14, Re $\left\{\frac{F_2(z)}{z}\right\} > \frac{1}{2}$ in $\mathcal{U}$ . Now applying Lemma 13 to (8) yields Re $\left(\frac{2\gamma F'(z) + 2\delta z F''(z) + (\delta - \gamma)z^2 F'''(z) - 2\lambda}{2(\gamma - \lambda)}\right) > 0$ in $\mathcal{U}$ . Thus, $F = F_1 * F_2 \in \mathcal{R}(\gamma, \delta, \lambda)$ . Now using Lemma 15, we prove that the class $\mathcal{R}_H^0(\gamma, \delta, \lambda)$ is closed under convolutions of its members. We make use of the techniques and methodology introduced by Dorff [4] for convolution.

Theorem 16. Let $\mathfrak{f}_i \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ for i = 1, 2. Then $\mathfrak{f}_1 * \mathfrak{f}_2 \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Proof. Suppose $\mathfrak{f}_i = \mathfrak{s}_i + \overline{\mathfrak{t}_i} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ (i = 1, 2). Then the convolution of $\mathfrak{f}_1$ and $\mathfrak{f}_2$ is defined as $\mathfrak{f}_1 \mathfrak{f}_2 = \mathfrak{s}_1 \mathfrak{s}_2 + \overline{\mathfrak{t}_1} \overline{\mathfrak{t}_2}$ . In order to prove that $\mathfrak{f}_1 \mathfrak{f}_2 \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ , we need to prove that $F_{\epsilon} = \mathfrak{s}_1 \mathfrak{s}_2 + \epsilon(\mathfrak{t}_1 \mathfrak{t}_2) \in \mathcal{R}(\gamma, \delta, \lambda)$ for each $\epsilon(|\epsilon| = 1)$ . By Lemma 15, the class $\mathcal{R}(\gamma, \delta, \lambda)$ is closed under convolutions for each $\epsilon(|\epsilon| = 1)$ , $\mathfrak{s}_i + \epsilon \mathfrak{t}_i \in \mathcal{R}(\gamma, \delta, \lambda)$ for i = 1, 2. Then both $F_1$ and $F_2$ given by $$F_1 = (\mathfrak{s}_1 - \mathfrak{t}_1) * (\mathfrak{s}_2 - \epsilon \mathfrak{t}_2)$$ and $F_2 = (\mathfrak{s}_1 + \mathfrak{t}_1) * (\mathfrak{s}_2 + \epsilon \mathfrak{t}_2)$ , belong to $\mathcal{R}(\gamma, \delta, \lambda)$ . Since $\mathcal{R}(\gamma, \delta, \lambda)$ is closed under convex combinations, then the function $$F_{\epsilon} = \frac{1}{2}(F_1 + F_2) = \mathfrak{s}_1 \mathfrak{s}_2 + \epsilon(\mathfrak{t}_1 \mathfrak{t}_2)$$ belongs to $\mathcal{R}(\gamma, \delta, \lambda)$ . Hence $\mathcal{R}_H^0(\gamma, \delta, \lambda)$ is closed under convolution. Now we consider the Hadamard product of a harmonic function with an analytic function which is defined by Goodloe [10] as $$\mathfrak{f} \widetilde{} \varphi = \mathfrak{s} \varphi + \overline{\mathfrak{t} * \varphi}$$ where $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}}$ is harmonic function and $\varphi$ is an analytic function in $\mathcal{U}$ .

Theorem 17. Let $\mathfrak{f} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ and $\varphi \in \mathcal{A}$ be such that $\operatorname{Re}\left(\frac{\varphi(z)}{z}\right) > \frac{1}{2}$ for $z \in \mathcal{U}$ , then $\mathfrak{f} * \varphi \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Proof. Suppose that $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ , then $F_{\epsilon} = \mathfrak{s} + \epsilon \mathfrak{t} \in \mathcal{R}(\gamma, \delta, \lambda)$ for each $\epsilon(|\epsilon| = 1)$ . By Theorem 2, in order to show that $\widetilde{\mathfrak{f}} \varphi \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ , we need to show that $G = \mathfrak{s} \varphi + \epsilon(\mathfrak{t} \varphi) \in \mathcal{R}(\gamma, \delta, \lambda)$ for each $\epsilon(|\epsilon| = 1)$ . Write G as $G = F_{\epsilon} \varphi$ , and $$\frac{1}{2(\gamma - \lambda)} \left( 2\gamma G'(z) + 2\delta z G''(z) + (\delta - \gamma) z^2 G'''(z) - 2\lambda \right) = \frac{1}{2(\gamma - \lambda)} \left( 2\gamma F'_{\epsilon}(z) + 2\delta z F''_{\epsilon}(z) + (\delta - \gamma) z^2 F'''_{\epsilon}(z) - 2\lambda \right) * \frac{\varphi(z)}{z}.$$ Since $\operatorname{Re}\left(\frac{\varphi(z)}{z}\right) > \frac{1}{2}$ and $\operatorname{Re}\left\{2\gamma F_{\epsilon}'(z) + 2\delta z F_{\epsilon}''(z) + (\delta - \gamma) z^2 F_{\epsilon}'''(z) - 2\lambda\right\} > 0$ in $\mathcal{U}$ , Lemma 13 proves that $G \in \mathcal{R}(\gamma, \delta, \lambda)$ . Corollary 18 Let $\mathfrak{f} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ and $\varphi \in \mathcal{K}$ , then $f * \varphi \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ . Proof. Suppose $\varphi \in \mathcal{K}$ , then $\operatorname{Re}\left(\frac{\varphi(z)}{z}\right) > \frac{1}{2}$ for $z \in \mathcal{U}$ . As a corollary of Theorem 17, $f * \varphi \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ .

Lemma 19. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}}$ , where $\mathfrak{s}$ and $\mathfrak{t}$ are given by (1). Further, let $$\sum_{m=2}^{\infty} m^2 \left[ |a_m| + |b_m| \right] \le 1. \tag{9}$$ Then $\mathfrak{f}$ is harmonic univalent in $\mathcal{U}$ , and $\mathfrak{f} \in \mathcal{F}\mathcal{K}_H^0$ .

Lemma 20. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}}$ , where $\mathfrak{s}$ and $\mathfrak{t}$ are given by (1). Further, let $$\sum_{m=2}^{\infty} m \left[ |a_m| + |b_m| \right] \le 1. \tag{10}$$ Then $\mathfrak{f}$ is harmonic univalent in $\mathcal{U}$ , and $\mathfrak{f} \in \mathcal{FS}_H^{*,0}$ . The following identities are useful in the proof of the theorems: <span id="page-12-0"></span>(i) $$\sum_{m=2}^{\infty} mr^{m-1} = \frac{r(2-r)}{(1-r)^2},$$ (11) (ii) $$\sum_{m=2}^{\infty} m^2 r^{m-1} = \frac{r(4-3r+r^2)}{(1-r)^3}.$$ (12)

Theorem 21. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ . Then $\mathfrak{f}$ is fully convex in $|z| < r_c$ , where $r_c$ is the unique real root of pc(r) = 0 in (0,1), and where $$pc(r) = \left(-\delta - 2\gamma + \lambda\right)r^3 + \left(3\delta + 6\gamma - 3\lambda\right)r^2 + \left(-3\delta - 7\gamma + 4\lambda\right)r + \delta + \gamma.$$ Proof. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}_H^0(\gamma, \delta, \lambda)$ where $\mathfrak{s}(z) = z + \sum_{m=2}^{\infty} a_m z^m$ and $\mathfrak{t}(z) = \sum_{m=2}^{\infty} b_m z^m$ . For $r \in (0, 1)$ , it is sufficient to show that $\mathfrak{f}_r \in \mathcal{FK}_H^0$ where $$f_r(z) = \frac{f(rz)}{r} = z + \sum_{m=2}^{\infty} a_m r^{m-1} z^m + \sum_{m=2}^{\infty} b_m r^{m-1} z^m.$$ Consider this time, the sum <span id="page-12-1"></span> $$S = \sum_{m=2}^{\infty} m^2 (|a_m| + |b_m|) r^{m-1}.$$ (13) In view of Theorem 7 (i) and (12), (13) gives $$S \leq \sum_{m=2}^{\infty} m^2 \left( \frac{4 (\gamma - \lambda)}{m^2 [2\gamma + (\delta - \gamma) (m - 1)]} \right) r^{m-1}$$ $$\leq \frac{\gamma - \lambda}{\delta + \gamma} \sum_{m=2}^{\infty} m^2 r^{m-1}$$ $$= \frac{\gamma - \lambda}{\delta + \gamma} \frac{r (4 - 3r + r^2)}{(1 - r)^3} =: X_1.$$ Lemma 19 implies that in order to show that $\mathfrak{f}_r \in \mathcal{F}\mathcal{K}_H^0$ , it is sufficient to show that $X_1 \leq 1$ . Thus, we need to prove that $(-\delta - 2\gamma + \lambda) r^3 + (3\delta + 6\gamma - 3\lambda) r^2 + (-3\delta - 7\gamma + 4\lambda) r + \delta + \gamma \geq 0$ . Suppose $pc(r) := (-\delta - 2\gamma + \lambda) r^3 + (3\delta + 6\gamma - 3\lambda) r^2 + (-3\delta - 7\gamma + 4\lambda) r + \delta + \gamma$ , so that $X_1 \le 1$ whenever $pc(r) \ge 0$ . It is easy to observe that $pc(0) = \delta + \gamma > 0$ and $pc(1) = 2(\lambda - \gamma) < 0$ , and hence pc(r) has at least one root in (0, 1). To show that pc(r) has exactly one root in (0,1), it is sufficient to prove that pc(r) is monotonic function on (0,1). A simple computation shows that $$pc'(r) = (-6\gamma + 3\lambda - 3\delta) r^{2} + (12\gamma - 6\lambda + 6\delta) r - 3\delta - 7\gamma + 4\lambda$$ $$pc'(0) = -3\delta - 7\gamma + 4\lambda = -3(\gamma + \delta) - 4(\gamma - \lambda) < 0$$ $$pc'(1) = \lambda - \gamma < 0$$ $$pc''(r) = (-6\delta - 12\gamma + 6\lambda) r + 6\delta + 12\gamma - 6\lambda$$ $$= [-6(\gamma + \delta) - 6(\delta - \lambda)] r - [-6(\gamma + \delta) - 6(\delta - \lambda)]$$ $$= [-6(\gamma + \delta) - 6(\delta - \lambda)] (r - 1) > 0 \text{ for } r \in (0, 1).$$ Hence pc'(r) is a strictly monotonic increasing function on (0,1). Since pc'(1) < 0, we conclude that pc'(r) < 0 on (0,1). This shows that pc(r) is strictly monotonically decreasing on (0,1). Thus pc(r) = 0 has exactly one root in (0,1). Since pc(r) is strictly monotonically decreasing on (0,1) with pc(0) > 0 and $pc(r_c) = 0$ , it is easy to see that $pc(r) \ge 0$ for $0 < r \le r_c$ . Hence $\mathfrak{f}$ is fully convex in $|z| < r_c$ .

Theorem 22. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ . Then $\mathfrak{f}$ is fully starlike in $|z| < r_s$ , where $r_s$ is the unique real root of ps(r) = 0 in (0,1), and where $$ps(r) = (\delta + 2\gamma - \lambda) r^2 + (-2\delta - 4\gamma + 2\lambda) r + \delta + \gamma.$$ Proof. Let $\mathfrak{f} = \mathfrak{s} + \overline{\mathfrak{t}} \in \mathcal{R}^0_H(\gamma, \delta, \lambda)$ . For $r \in (0, 1)$ , let $$f_r(z) = \frac{f(rz)}{r} = z + \sum_{m=2}^{\infty} a_m r^{m-1} z^m + \sum_{m=2}^{\infty} b_m r^{m-1} z^m$$ Consider the sum <span id="page-13-0"></span> $$S = \sum_{m=2}^{\infty} m(|a_m| + |b_m|) r^{m-1}$$ (14) Using Theorem 7(i) and [\(11\)](#page-12-0), [\(14\)](#page-13-0) gives $$S \leq \sum_{m=2}^{\infty} m \left( \frac{4 (\gamma - \lambda)}{m^2 [2\gamma + (\delta - \gamma) (m - 1)]} \right) r^{m-1}$$ $$\leq \frac{\gamma - \lambda}{\delta + \gamma} \sum_{m=2}^{\infty} m r^{m-1}$$ $$= \frac{\gamma - \lambda}{\delta + \gamma} \frac{r (2 - r)}{(1 - r)^2} =: X_2.$$ In view of Lemma 19 in order to prove that f^r ∈ FSH^∗,⁰ , it is sufficient to show that X² ≤ 1. This implies that it suffices to show (δ + 2γ − λ) r ² + (−2δ − 4γ + 2λ) r + δ + γ ≥ 0. Suppose ps(r) = (δ + 2γ − λ) r ² + (−2δ − 4γ + 2λ) r + δ + γ, so that X² ≤ 1 whenever ps(r) ≥ 0. It is easy to observe that ps(0) = δ+γ > 0 and ps(1) = λ−γ < 0, and hence ps(r) has a real root in (0, 1). To show that ps(r) has exactly one root in (0, 1), it is sufficient to prove that ps(r) is monotonic function on (0, 1). A simple computation shows that $$ps'(r) = 2 (\delta + 2\gamma - \lambda) (r - 1) < 0, \text{ for } r \in (0, 1)$$ $ps'(0) = 2 (\lambda - \delta - 2\gamma),$ $ps'(1) = 0,$ $ps''(r) = 2 (\delta + 2\gamma - \lambda) > 0.$ Hence ps′ (r) is a strictly monotonically increasing function on (0, 1). Since ps′ (1) = 0, we conclude that ps′ (r) < 0 on (0, 1). This shows that ps(r) is strictly monotonically decreasing on (0, 1). Thus ps(r) has exactly one root in (0, 1). Since ps(r) is strictly monotonically decreasing on (0, 1) with ps(0) > 0 and ps(rs) = 0, it is easy to see that ps(r) ≥ 0 for 0 < r ≤ rs. Hence f is fully starlike in |z| < rs. Lemma 23 [\[19,](#page-16-3) Corollary 3.2] Let λ < 1, δ ≥ 1, and f ∈ R(1, δ, λ). If λ satisfies <span id="page-14-0"></span> $$7 - 3\delta = 4\lambda + 4(1 - \lambda) \sum_{m=1}^{\infty} \frac{2m(3 - \delta) + (\delta - 5)}{(m+1)(m(\delta - 1) + 2)},$$ (15) then f is convex in U. Theorem 24 Suppose f ∈ R⁰ ^H (1, δ, λ) with λ < 1, δ ≥ 1. If λ satisfies [\(15\)](#page-14-0), then f is fully convex in U. Proof. Let λ < 1, δ ≥ 1 and f ∈ R⁰ ^H(1, δ, λ). Then F^ǫ = s + ǫt ∈ R(1, δ, λ) for each ǫ(|ǫ| = 1). If λ satisfies [\(15\)](#page-14-0), then F^ǫ is convex in U. In view of [\[15,](#page-15-6) Corollary 2.4], it follows that f is fully convex in U. Problem 25 Find γ, δ and λ so that functions f ∈ R⁰ ^H(γ, δ, λ) are fully convex in U.