On a Univalent Class Involving Differential Subordinations with Applications

Corollary 3. Let $q,q(z) \neq 0$ be a univalent function in U. If $$\frac{zf'(z)}{\Phi(f(z))} \neq 0, z \in U$$ and $$(1-\alpha)\frac{zf'(z)}{\Phi(f(z))} + \alpha(1+\frac{zf''(z)}{f'(z)} - \frac{z\Phi(f(z))}{\Phi'(f(z))})$$ $$\prec (1-\alpha)q(z) + \alpha \frac{zq'(z)}{q(z)},$$ (6) then $\frac{zf'(z)}{\Phi(f(z))} \prec q$ and q is the best dominant Eq. 6. Finally, by assuming: $$\mu = 1 - \alpha, \nu = \alpha, g(z) = p(z) = \frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)}$$ in Theorem 1 we obtain the following result: Corollary 4: Let $q,q(z) \neq 0$ be a univalent function in U and $\frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)} \neq 0$ be analytic in U satisfy (2). If the subordination Eq. 7 $$\frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)} \{ (1 - \alpha) \frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)}$$ $$+\alpha(1+\frac{z(f(z)\Phi(z))''}{(f(z)\Phi(z))'}-\frac{z(f(z)\Psi(z))'}{(f(z)\Psi(z))})\}$$ $$\prec g(z)[(1-\alpha)q(z) + \alpha \frac{zq'(z)}{q(z)}],$$ (7) holds then $\frac{z(f(z)\Phi(z))'}{f(z)\Psi(z)} \prec q$ and q is the best dominant. Note that Corollary 4, gives sufficient conditions for functions $f \in A$ to be in the class $H(\alpha, 1, 1, \Phi(z); \Psi(z))$ . An application of Theorem 1, next result shows the sufficient conditions for functions $f \in A$ to be in the class $H(\alpha,\lambda,\delta,\Phi(z);\Psi(z))$ . By assuming $\mu:=\alpha$ and $\nu:=1-\alpha$ , we have the following result: Theorem 2: Let $f \in A$ and $q,q(z) \neq 0$ be a univalent function in U. Assume that $p(z) := \frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)} \neq 0$ is analytic in U satisfies (2-3) for some g If $\lambda, \delta \in R$ then $f \in H(\alpha, \lambda, \delta, \Phi(z); \Psi(z))$ . Sandwich Theorem: By employing the concept of the superordination (Lemma 2), we pose the sandwich theorem containing functions $f \in A$ . Theorem 3: Let q(z) be convex univalent in the unit disk U. Suppose that g ia an analytic in the unit disk such that $$\nu g(z) \frac{zq'(z)}{q(z)}$$ is starlike univalent in $U$ and $$\frac{\mu}{\nu}\Re\{q(z)q'(z)\}>0 \text{ for } Z\in U.$$ If $$p(z) \in H[q(0),1] \cap Q$$ , with $p(U) \subseteq D$ and $g(z)[\mu p(z) + \nu \frac{zp'(z)}{p(z)}]$ is univalent in U and $$g(z)[\mu q(z) + \nu \frac{zq'(z)}{q(z)}] \prec g(z)[\mu p(z) + \nu \frac{zp'(z)}{p(z)}]$$ then $q(z) \prec p(z)$ and q(z) is the best subordinant. Proof: Define the functions $\theta$ and $\phi$ as follows: $$\vartheta(w(z)) \coloneqq \mu w(z) g(z) \quad \text{and} \quad \phi(w(z)) \coloneqq \frac{\nu g(z)}{w(z)}.$$ Obviously, the functions $\vartheta$ and $\phi$ are analytic in domain $D=C\setminus\{0\}$ and $\phi(w)\neq 0$ in D hence the assumptions of Lemma 2 are satisfied. Combining Theorem 1 and Theorem 3 we get the following sandwich theorem: Theorem 4: Let $q_1(z), q_2 \neq 0$ be convex and univalent in U respectively. Suppose that g ia an analytic in U such that Eq. 8: $$\nu g(z) \frac{z q_l{}'(z)}{q_l(z)}$$ is starlike univalent in $U$ and $$\frac{\mu}{\nu} \Re\{q_1(z)q_1'(z)\} > 0 \text{ for } z \in U \text{ and}$$ $$\Re\{1+\frac{zq_2{''}(z)}{q_2{'}(z)}-\frac{zq_2{'}(z)}{q_2(z)}\}$$ $$> \max\{0, (\frac{\mu}{\nu}) \Re(q_2(z)[1 + \frac{g'(z)}{g(z)}(\frac{q_2(z)}{q_2'(z)} + \frac{\nu z}{\mu q_2(z)})])\} \,. \tag{8}$$ If $p(z) \neq 0 \in H[q(0),1] \cap Q$ , With: $$p(U) \subseteq D$$ And: $$g(z)[\mu p(z) + \nu \frac{zp'(z)}{p(z)}]$$ is univalent in U and $$g(z)[\mu q_1(z) + \nu \frac{zq_1'(z)}{q_1(z)}] \prec g(z)[\mu p(z) + \nu \frac{zp'(z)}{p(z)}]$$ $$\prec g(z)[\mu q_2(z) + \nu \frac{zq_2'(z)}{q_2(z)}]$$ Then: $$q_1(z) \prec p(z) \prec q_2(z), \quad (z \in U)$$ and $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively. By letting $$p(z) := \frac{zf'(z)}{f(z)}$$ in Theorem 4, we have Corollary 5: Let the conditions of Theorem 4 on the functions $q_1$ and $q_2$ hold. If for $f \in A$ $$\frac{zf'(z)}{f(z)} \neq 0 \in H[q(0),1] \cap Q,$$ With: $$(\frac{zf'}{f})(U) \subseteq D$$ And: $$g(z)[(\mu - \nu)\frac{zf'(z)}{f(z)} + \nu(1 + \frac{zf''(z)}{f'(z)})]$$ Is univalent in U and: $$g(z)[\mu q_1(z) + \nu \frac{zq_1'(z)}{q_1(z)}]$$ $$\prec g(z)[(\mu - \nu)\frac{zf'(z)}{f(z)} + \nu(1 + \frac{zf''(z)}{f'(z)})]$$ $$\prec g(z)[\mu q_2(z) + \nu \frac{zq_2'(z)}{q_2(z)}]$$ Then Eq. 9: $$q_1(z) \prec \frac{zf'(z)}{f(z)} \prec q_2(z), \quad (z \in U)$$ (9) And $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively. Note that (Ali et al., 2004) have used the results of (Bulboac, 2002) and obtained sufficient conditions for certain normalized analytic functions f(z) to satisfy (9). By assuming $$p(z) := \frac{f(z)}{zf'(z)}$$ in Theorem 4, we obtain Corollary 6: Let the conditions of Theorem 4 on the functions $q_1$ and $q_2$ hold. If for $f \in A$ $\frac{f(z)}{zf'(z)} \neq 0 \in Hq(0), 1] \cap Q,$ With: $$(\frac{f}{zf'})(U) \subseteq D$$ And: $$g(z)[\mu \frac{f(z)}{zf'(z)} + \nu(\frac{zf'(z)}{f(z)} - 1 - \frac{zf''(z)}{f'(z)})]$$ is univalent in U and $$g(z)[\mu q_1(z) + \nu \frac{zq_1'(z)}{q_1(z)}]$$ $$\prec g(z)[\mu \frac{f(z)}{zf'(z)} + \nu(\frac{zf'(z)}{f(z)} - 1 - \frac{zf''(z)}{f'(z)})]$$ $$\prec g(z)[\mu q_2(z) + \nu \frac{zq_2'(z)}{q_2(z)}]$$ Then Eq. 10: $$q_1(z) \prec \frac{f(z)}{zf'(z)} \prec q_2(z), \quad (z \in U)$$ (10) and $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively. Note that (Shanmugam et al., 2006) posed sufficient conditions for certain normalized analytic functions f(z) to satisfy (10). Again by considering $p(z) := \frac{z^2 f'(z)}{f^2(z)}$ in Theorem 4, we find Corollary 7: Let the conditions of Theorem 4 on the functions $q_1$ and $q_2$ hold. If for $f \in A$ . $$\frac{z^2 f'(z)}{f^2(z)} \neq 0 \in H[q(0), 1] \cap Q,$$ With: $$\frac{z^2f'}{f^2}(U) \subseteq D$$ Δnd $$g(z)\left[\mu \frac{z^2 f'(z)}{f^2(z)} + \nu \left(\frac{z f''(z)}{f'(z)} + 2 - 2\frac{z f'(z)}{f(z)}\right)\right]$$ is univalent in U and $$g(z)[\mu q_1(z) + \nu \frac{zq_1'(z)}{q_1(z)}]$$ $$\prec g(z))[\mu \frac{z^2f'(z)}{f^2(z)} + \nu(\frac{zf''(z)}{f'(z)} + 2 - 2\frac{zf'(z)}{f(z)})]$$ $$\prec g(z)[\mu q_2(z) + \nu \frac{zq_2'(z)}{q_2(z)}]$$ Then Eq. 11: $$q_1(z) < \frac{z^2 f'(z)}{f^2(z)} < q_2(z), \quad (z \in U)$$ (11) And $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively. Note that (Shanmugam et al., 2006) estimated sufficient conditions for certain normalized analytic functions f (z) to satisfy (11). Furthermore, by letting $p(z) := \frac{z(f g)'(z)}{\Phi(f g)(z)}$ in Theorem 4, we pose Corollary 8: Let the conditions of Theorem 4 on the functions $q_1$ and $q_2$ hold. If for $f \in A$ $$\frac{z(f g)'(z)}{\Phi(f g)(z)} \neq 0 \in H[q(0),1] \cap Q,$$ With: $$(\frac{z(f g)'}{\Phi(f g)})(U) \subseteq D$$ And: $$g(z)[\mu \frac{z(f g)'(z)}{\Phi(f g)(z)} - \nu(\frac{zf''(z)}{f'(z)} + 1 - \frac{z\Phi'(f g)(z)}{\Phi(f g)(z)})]$$ is univalent in U and: $$g(z)[\mu q_1(z) + \nu \frac{zq_1'(z)}{q_1(z)}]$$ $$\prec g(z)[\mu \frac{z(f g)'(z)}{\Phi(f g)(z)} - \nu(\frac{zf''(z)}{f'(z)} + 1 - \frac{z\Phi'(f g)(z)}{\Phi(f g)(z)})]$$ $$\prec g(z)[\mu q_2(z) + \nu \frac{zq_2'(z)}{q_2(z)}]$$ Then Eq. 12: $$q_1(z) < \frac{z(f g)'(z)}{\Phi(f g)(z)} < q_2(z), \quad (z \in U)$$ (12) and $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively. Note that (Shanmugam et al., 2007) posed sufficient conditions for certain normalized analytic functions f(z) to satisfy (12). Finally, by setting $$p(z) := (\frac{H_m^l[\alpha_1]f(z)}{z})^{\delta}$$ , where $f \in A$ and $H_m^l[\alpha_1]$ is the Dziok-Srivastava linear operator (Dziok and Srivastava, 2003), in Theorem 4, we have. Corollary 9: Let the conditions of Theorem 4 on the functions $q_1$ and $q_2$ hold. If for $f \in A$ , $$\left(\frac{H_{\mathrm{m}}^{1}[\alpha_{1}]f(z)}{z}\right)^{\delta}\neq0\in\mathrm{H}[q(0),1]\cap\mathrm{Q},$$ With: $$((\frac{H_m^l[\alpha_1]f}{z})^{\delta})(U) \subseteq D$$ And: $$g(z)[\mu(\frac{H_m^l[\alpha_1]f(z)}{z})^{\delta} - \nu \delta z(\frac{z}{H_m^l[\alpha_1]f(z)} - 1)]$$ is univalent in U and: $$g(z)[\mu q_1(z) + \nu \frac{zq_1'(z)}{q_1(z)}]$$ $$\prec g(z)[\mu(\frac{H_m^l[\alpha_1]f(z)}{z})^{\delta} - \nu \delta z(\frac{z}{H_m^l[\alpha_1]f(z)} - 1)]$$ $$\prec g(z)[\mu q_2(z) + \nu \frac{zq_2'(z)}{q_2(z)}]$$ Then Eq. 13. $$q_1(z) \prec \left(\frac{H_m^1[\alpha_1]f(z)}{z}\right)^{\delta} \prec q_2(z), \quad (z \in U)$$ (13) and $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively. Note that (Murugusundaramoorthy and Magesh, 2006) introduced sufficient conditions for certain normalized analytic functions f(z) to satisfy (13). Corollary 10: Let the assumptions of Theorem 4 on the function: $$p(z) := \frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)} \{ (1 - \alpha) \frac{z(f(z) \Phi(z))'}{f(z) \Psi(z)}$$ $$+\alpha(1+\frac{\lambda z(f(z)\Phi(z))''}{(f(z)\Phi(z))'}-\frac{\delta z(f(z)\Psi(z))'}{(f(z)\Psi(z))})\}$$ holds. Then: $$q_l(z) \! \prec \! \frac{z(f(z) \ast \Phi(z))'}{f(z) \ast \Psi(z)} \{ (1 \! - \! \alpha) \frac{z(f(z) \ast \Phi(z))'}{f(z) \ast \Psi(z)}$$ $$+\alpha(1+\frac{\lambda z(f(z)\Phi(z))''}{(f(z)\Phi(z))'}-\frac{\delta z(f(z)\Psi(z))'}{(f(z)\Psi(z))})\}\!\prec\!q_2(z),$$ and $q_1(z), q_2(z)$ are the best subordinant and the best dominant respectively.