Some applications of differential subordination

Lemma 1. Let \$3yeC3he H(E) be convex univalent in E with h(0) = 1 and Re(B?z(3)+Y) > 0 , z e E and let q e H(E) with q(0) = 1 and q{z) -e h(s) , z e E . If p(z) = 1 + p z + ... is analytic in E , then $$p(z) + \frac{zp'(z)}{\beta q(z) + \gamma} < h(z) \Rightarrow p(z) < h(z)$$ The proof is essentially the same as that of Theorem A. However we give the details below for the sake of completeness. Proof Of Lemma 1. We first suppose that all functions under consideration are analytic in E , and show that if p(z) is not subordinate to h(z) , then there is a z in E such that $$(1) p(z_o) + \frac{z_o p'(z_o)}{\beta q(z_o) + \gamma} \not\in h(E)$$ which contradicts the hypothesis. If p(s) is not subordinate to h(z) , then by Lemma B in P. Eenigenburg, S.S. Miller, P.T. Mocanu and M.O. Reade [3], we conclude that there are ZQ e E , c e 3E and m , m > 1 such that o o , arg [z^'{ZQ)} = arg Uj> • ( KQ)\ and j?' (2O) I = m\toh'(?o) | > 0 . Hence we can write (2) $$p(z_o) + \frac{z_o p'(z_o)}{\beta q(z_o) + \gamma} = h(\zeta_o) + m \frac{\zeta_o h'(\zeta_o)}{\beta q(z_o) + \gamma}.$$ Now q(z) < h(z) =\* 3<7(2) + y < &h{z) + y and so Re(g?z(3)+Y) > 0 =» Re(g<7(2)+Y) > 0 , or equivalently $$\left|\arg(\beta q(z)+\gamma)\right| < \frac{\pi}{2} \text{ for } z \in E$$ . Hence |arg(617(2 )+Y)| < y an ^d C fc'(C ) is in the direction of the outer normal to the convex domain h(E) , so that the right-hand member of (2) is a complex number outside h{E) ; that is, (1) holds. We conclude that p(z) < h{z) provided all functions under consideration are analytic in E . To remove the restriction imposed, we need to replace p(z) by p (2) = p(p2) and h(z) by h (2) = h{pz) , 0 < p < 1 . All the hypotheses of the lemma are satisfied and we conclude that p (2) < h (2) for each p , 0 < p < 1 . By letting p -> 1 we obtain p(z) < h(z) . Let K (2) = where a is any real number. We consider the convolution of / e A with K . We need the following easily verified result: (3) $$z(K_{\alpha} f)'(z) = \alpha(K_{\alpha+1} f)(z) - (\alpha-1)(K_{\alpha} * f)(z) .$$ In the sequel h e H(E) is convex univalent in E and satisfies fe(O) = 1 and Re(h(z)) > 0 for 2 e E .$

Theorem 1. If f e S [h) , then f e 5fl(/2) /zoZds /or a > 1 (X \*/) (2) provided ^ 0 for z e E . ³ ziK \/)'((2) Proof. Let p(2) = —T-S — . Using (3), we get (Xa\/)(2) $$p(z) + (\alpha - 1) = \frac{\alpha(K_{\alpha + 1} f)(z)}{(K_{\alpha} f)(z)}.$$ Taking logarithmic derivatives and multiplying by 2 , we get $$\frac{zp'(z)}{p(z) + (a-1)} = \frac{z(K_{a+1} f)'(z)}{(K_{a+1} f)(z)} - p(z) ,$$ which yields (4) $$\frac{z(K_{a+1} f)'(z)}{(K_{a+1} f)(z)} = \frac{zp'(z)}{p(z) + (a-1)} + p(z) .$$ This means that if / e S +Ah) , then (5) $$\frac{zp'(z)}{p(z) + (a-1)} + p(z) < h(z) .$$ From Theorem A, it follows that for a 2 1 , p(z) < h(z) ; that is z(K \* f) '(2) tv . ^ . . < h{3) which means f e S (h) for all a > 1 . CL

Theorem 2. Suppose f e SQ(^) ; then F e Sa(h) provided (K \* F) (s) — ^ — f 0 /or 3 e ff . ² Proof. We have $$zF'(z)+rF(z)=(r+1)f(z)$$ ; and so $$(K_{\alpha}(zF'))(z)+r(K_{\alpha}F)(z)=(r+1)(K_{\alpha}*f)(z).$$ Using the fact (6) $$z(K_{a} F)'(z) = (K_{a} zF')(z) ,$$ we obtain (7) $$z(K_a F) (z) + r(K_a F) (z) = (r+1)(K_a f) (z) .$$ Let $$p(z) = \frac{z(K_a F) '(z)}{(K_a F) (z)}$$ . Then (7) yields $$p(z) + r = (r+1) \frac{(K_a f) (z)}{(K_a F) (z)}$$ taking logarithmic derivatives and multiplying by z , we get (8) $$\frac{zp'(z)}{p(z) + r} + p(z) = \frac{z(K_a f)'(z)}{(K_a f)(z)} .$$ Since $f \in S_a(h)$ , it follows that $\frac{zp!(z)}{p(z)+r}+p(z) \prec h(z)$ which implies that $p(z) \prec h(z)$ by Theorem A. Thus $$\frac{z(K_{\alpha} \star F)'(z)}{(K_{\alpha} \star F)(z)} = p(z) < h(z) ;$$ that is $F \in S_a(h)$ . REMARK 2. If $\alpha$ is a positive integer $\geq 1$ and $h(z) = \frac{1-z}{1+z}$ , then the above theorem reduces to Theorem 5 in St. Ruscheweyh [8]. REMARK 3. If $\alpha = 1$ , r = 1 and $h(z) = \frac{1-z}{1+z}$ , we deduce Theorem 1 of R.J. Libera (1965). REMARK 4. If $\alpha=1$ and $h(z)=\frac{1+Az}{1+Bz}$ , $-1\leq A < B \leq 1$ , we deduce Lemma 2 of R.M. Goel and B.S. Mehrok [4].

Theorem 3. If $f \in C_{a+1}(h)$ , then $f \in C_a(h)$ holds for $a \ge 1$ provided $\frac{(K_a * \phi)(z)}{z} \ne 0 \text{ for } z \in E.$ Proof. Let $p(z) = \frac{z(K_a f)'(z)}{(K_a \phi)(z)}$ . Using (3) and simplifying $$(\mathcal{K}_a^{\phantom{a}} \star \phi) (z) p(z) + (a-1) (\mathcal{K}_a^{\phantom{a}} \star f) (z) = a (\mathcal{K}_{a+1}^{\phantom{a}} \star f) (z) \ .$$ Differentiating and multiplying by z , we obtain (9) $$(K_a \phi)(z) z p'(z) + z p(z) (K_a \phi)'(z) + (a-1) z (K_a * f)'(z)$$ $$= z a (K_{a+1} * f)'(z) .$$ Let $q(z) = \frac{z(K_{\alpha} \phi)'(z)}{(K_{\alpha} \phi)(z)}$ . Then once again using (3), (10) $$q(z) + (\alpha - 1) = \frac{\alpha(K_{\alpha+1} \phi)(z)}{(K_{\alpha} \phi)(z)}.$$ From (9) and (10), we obtain (11) $$\frac{zp'(z)}{q(z) + (\alpha - 1)} + p(z) = \frac{z(K_{\alpha + 1} f)'(z)}{(K_{\alpha + 1} \phi)(z)} .$$ If $f \in \mathcal{C}_{\alpha+1}(h)$ , then $\frac{zp'(z)}{q(z)+(\alpha-1)}+p(z) < h(z)$ . Applying Lemma 1 for $a \geq 1$ , we get $\frac{z(K_{\alpha}f)'(z)}{(K_{\alpha}\phi)(z)}=p(z) < h(z)$ . Since $\phi \in S_{\alpha+1}(h)$ , $\phi$ also belongs to $S_{\alpha}(h)$ for $\alpha \geq 1$ by Theorem 1. Hence $f \in \mathcal{C}_{\alpha}(h)$ .

Theorem 4. Suppose $f \in C_{\alpha}(h)$ with respect to the function $\phi \in S_{\alpha}(h)$ . Define $\psi$ by $\psi(z) = (\phi h_r)(z)$ . Then $F \in C_{\alpha}(h)$ with respect to $\psi$ , provided $\frac{(K_{\alpha} \psi)(z)}{z} \neq 0$ for $z \in E$ . Proof. Since $f \in C_a(h)$ with respect to $\phi \in S_a(h)$ , we have $\frac{z(K_a f)'(z)}{(K_a \phi)(z)} < h(z) \text{. Since } \psi(z) = (\phi * h_p)(z) \text{, we have}$ $$\psi(z) = \frac{(r+1)}{z^r} \int_0^z t^{r-1} \phi(t) dt . \text{ By Theorem 2 } \psi \in S_a(h) \text{ provided}$$ $$\frac{(K_a * \psi)(z)}{z} \neq 0 \text{ for } z \in E . \text{ Also}$$ (12) $$z(K_{\alpha} \psi)'(z) + r(K_{\alpha} \psi)(z) = (r+1)(K_{\alpha} * \phi)(z).$$ Also we have (13) $$z(K_a F)'(z) + r(K_a F)(z) = (r+1)(K_a * f)(z).$$ Let $$p(z) = \frac{z(K_{\alpha} F)'(z)}{(K_{\alpha} \psi)(z)}$$ . Then $$p\left(z\right)\left(K_{\alpha}^{}\psi\right)\left(z\right) \;+\; r\left(K_{\alpha}^{}F\right)\left(z\right) \;=\; (r+1)\left(K_{\alpha}^{*}f\right)\left(z\right) \;\;;$$ differentiating with respect to z , multiplying by z and dividing by $(K_{\alpha} * \psi)(z)$ , we get $$zp'(z) + p(z)(q(z)+r) = (r+1)\frac{z(K_a f)'(z)}{(K_a \psi)(z)}$$ where $q(z) = \frac{z(K_a \psi)'(z)}{(K_a \psi)(z)} < h(z)$ . Hence $$\frac{zp'(z)}{q(z) + r} + p(z) = (r+1)\frac{z(K_a f)'(z)}{z(K_a \psi)'(z) + r(K_a * \psi)(z)}$$ $$= \frac{z(K_a f)'(z)}{(K_a \phi)(z)}$$ by (12). Since $f \in C_a(h)$ , $\frac{zp'(z)}{q(z)+r}+p(z) < h(z)$ . Once again by Lemma 1, p(z) < h(z) and hence $F \in C_a(h)$ . REMARK 6. For $\alpha=1$ , r=1 and $h(z)=\frac{1-z}{1-z}$ , this theorem reduces to Theorem 3 in R.J. Libera [6].

Theorem 5. If $f \in C_a^{\alpha}(h)$ , then $f \in C_a^{\alpha}(h) = C_a(h)$ , for $a \ge 1$ . Proof. Let $$p(z) = \frac{z(K_a f)'(z)}{(K_a \phi)(z)}$$ where $\phi \in S_a(h)$ . By (11) $$\frac{z\left(K_{\alpha+1}f\right)'(z)}{\left(K_{\alpha+1}\phi\right)(z)} = \frac{zp'(z)}{q(z) + (\alpha-1)} + p(z)$$ where $$q(z) = \frac{z(K_{\alpha} \phi)'(z)}{(K_{\alpha} \phi)(z)}$$ . Hence $$\begin{split} J_{\alpha}(\alpha;f;\phi) &= \alpha \Big( \frac{zp'(z)}{q(z) + (\alpha - 1)} + p(z) \Big) + (1 - \alpha)p(z) \\ &= \frac{\alpha zp'(z)}{q(z) + (\alpha - 1)} + p(z) \ . \end{split}$$ If $f \in C^\alpha_\alpha(h)$ , then $\frac{\alpha z p^+(z)}{q(z) + (\alpha - 1)} + p(z) \prec h(z)$ . By Lemma 1, $p(z) \prec h(z)$ which implies $f \in C_\alpha(h)$ .

Theorem 6. For $\alpha > \beta \ge 0$ $C^{\alpha}_{\sigma}(h) \subset C^{\beta}_{\sigma}(h)$ . Proof. The case $\beta=0$ was treated in Theorem 5. Hence we can assume that $\beta\neq 0$ . Suppose that $f\in C^\alpha_a(h)$ . Then $J_\alpha(\alpha;f;\phi)(z)\prec h(z)$ . Let $z_1$ be any arbitrary point in E. Then (14) $$J_{\sigma}(\alpha; f; \phi)(z_{1}) \in h(E) .$$ Also $\frac{z(K_{\alpha} f)'(z)}{(K_{\alpha} \phi)(z)} < h(z)$ by the previous theorem; so (15) $$\frac{z_1(K_\alpha f)'(z_1)}{(K_\alpha \phi)(z_1)} \in h(E) .$$ Also $$J_{\alpha}(\beta;f;\phi)(z) = (1-\frac{\beta}{\alpha}) \frac{z(K_{\alpha} f)'(z)}{(K_{\alpha} \phi)(z)} + \frac{\beta}{\alpha} J_{\alpha}(\alpha;f;\phi)(z)$$ . Since $\frac{\beta}{\alpha} < 1$ and h(E) is convex $J_{\alpha}(\beta; f; \phi) (z_1) \in h(E)$ by virtue of (14) and (15). It follows that $J_{\alpha}(\beta; f; \phi) (z) \prec h(z)$ . That is, $f \in \mathcal{C}^{\beta}_{\alpha}(h)$ . REMARK 7. If $\alpha = 1$ and $h(z) = \frac{1-z}{1+z}$ , Theorem 5 and Theorem 6 reduce to Theorem 1 and Theorem 2 respectively of Pram Nath Chichra [2]. REMARK 8. If $$a = 1$$ and $h(z) = \frac{1 + (2^{\rho} - 1)z}{1 + z}$ , $0 \le \rho \le 1$ , Theorem 5 reduces to Theorem 1 of V.A. Zmorovich and V.A. Pokhilevich [9].

Theorem 7. If $f \in S_a^{\alpha}(h)$ then $f \in S_a^{O}(h) = S_a(h)$ , for $a \ge 1$ . Proof. Let $$p(z) = \frac{z(K_a f)'(z)}{(K_a f)(z)}$$ . Then using (4) we have $$\begin{split} J_{\alpha}(\alpha,f) &= \alpha \Big( \frac{zp'(z)}{p(z) + (\alpha-1)} \Big) + (1-\alpha)p(z) \\ &= \frac{\alpha zp'(z)}{p(z) + (\alpha-1)} + p(z) \end{split} .$$ Since f e S (h) , a P ) — + p(z) < h(z) and Theorem A implies that \*2 p (3 J + (<2— 1] p(s) -e ^(s) for a > 1 . Hence f e S (h) .

Theorem 8. For a > B > 0 , S\*(h) c 53 (^) . Proof. The proof of this theorem is similar to that of Theorem 6, and hence is omitted. REMARK 9. If a is an integer > 1 and h{.z) = } ~ f in J. + 3 Theorem 7, then we ge t Theorem 1 of Al-flmiri [J] . REMARK 10 . If a = 1 and 7z(3) = 2 then Theorem 7 reduce s to 1 + 3 the well-known result that all a-convex functions are starlike by S.S. Miller, P.T. Mocanu and M.O. Reade [7].

Definition 1. Let S (h) denote the class of functions f e A such z(Ka\f)'(z) (Ka\f){z) that IK \ f) (z) < ^ ³ ) where ^ 0 for 2 e E . REMARK 1. If a = 1 and h{z) = , " Z , then (K \/) (2) = f(z) , and 5 (h) = S , the class of starlike univalent functions.

Definition 2. Let $$f(z) = z + \sum_{j=2}^{\infty} a_j z^j$$ . Define $$h_{r}(z) = \sum_{j=1}^{\infty} \left(\frac{r+1}{r+j}\right) z^{j} \quad \text{for Re } r > 0 \quad \text{and} \quad F(z) = (f \star h_{r})(z) = \sum_{j=1}^{\infty} \left(\frac{r+1}{r+j}\right) a_{j} z^{j}$$ Then $$F(z) = \frac{(r+1)}{z^r} \int_{0}^{z} t^{r-1} f(t) dt$$ .

Definition 3. Let $C_{\alpha}(h)$ (h as above) denote the class of functions $f \in A$ such that $\frac{z(K_{\alpha} \star f) \cdot (z)}{(K_{\alpha} \star \phi) \cdot (z)} \prec h(z)$ for some $\phi \in S_{\alpha}(h)$ . REMARK 5. If a=1 and $h(z)=\frac{1-z}{1+z}$ , then $C_a(h)=C$ the class of close-to-convex functions introduced by W. Kaplan [5].

Definition 4. Let $\ {\it C}^{\alpha}_{\alpha}(h)$ , $\alpha > 0$ denote the class of functions $f \in A$ such that $$J_{\alpha}(\alpha;f;\phi) \ = \ \frac{z\left(K_{\alpha+1} \star f\right) \cdot (z)}{\left(K_{\alpha+1} \star \phi\right) \left(z\right)} \ + \ \left(1 - \alpha\right) \frac{z\left(K_{\alpha} \star f\right) \cdot (z)}{\left(K_{\alpha} \star \phi\right) \left(z\right)} \ \ < \ h(z)$$ for some $\phi \in S_a(h)$ satisfying $\frac{(K_{a+1} \star \phi)(z)}{z} \neq 0$ for $z \in E$ .

Definition 5. Let $S^{\alpha}_{\alpha}(h)$ , $\alpha\geq 0$ denote the class of functions $f\in A$ such that $$J_{\alpha}(\alpha, f) = \alpha \frac{z(K_{\alpha+1} f)'(z)}{(K_{\alpha+1} f)(z)} + (1-\alpha) \frac{z(K_{\alpha} f)'(z)}{(K_{\alpha} f)(z)} < h(z)$$ with $$\frac{(K_{a+1} * f)(z)}{z} \neq 0$$ and $\frac{(K_a * f)(z)}{z} \neq 0$ for $z \in E$ .