Certain classes of regular univalent functions

Theorem 1. If $f(z) \in S^(\alpha)$ and $g(z) \in S^(\gamma)$ , then the function F(z) defined by $$F(z) = \frac{2}{g(z)} \int_0^z f(t) dt$$ is $\beta$ -starlike for $|z| < \sigma = \sigma(\alpha, \beta, \gamma)$ where $\sigma$ is given by $$\sigma[1 + 2\alpha - 2\gamma - \beta] = [(2 - \alpha - \gamma)^2 + (1 - \beta)(1 + 2\alpha - 2\gamma - \beta)]^{1/2} - (2 - \alpha - \gamma)$$ when $1+2\alpha-2\gamma-\beta\neq 0$ , $2\sigma(2-\alpha-\gamma)=1-\beta$ when $\beta=1+2\alpha-2\gamma$ .

Lemma 1. Suppose $p(z) = [1 + Dw(z)][1 + Bw(z)]^{-1}$ where w(z) is regular in E, w(0) = 0, |w(z)| < 1 for $z \in E$ and $-1 \le D < B \le 1$ ; then for any $C \ge B$ , we have on |z| = r < 1, $$\begin{split} \text{Re} \left[ Cp(z) + \frac{D}{p(z)} \right] - \left[ \frac{r^2 \, | \, Bp(z) - \, D \, |^2 - | \, 1 - \, p(z) \, |^2}{(1 - \, r^2) \, | \, p(z) \, |} \right] \ & \geq \begin{split} P_1(r) \, \, for \, \, R_0 \leq R_1 \, \, , \ P_2(r) \, \, for \, \, R_0 \geq R_1 \, \, , \end{split}$$ where $$\begin{align} P_{_1}\!(r) &= C\!\left(\!\frac{1+Dr}{1+Br}\right) + D\!\left(\!\frac{1+Br}{1+Dr}\!\right), \ \ P_{_2}\!(r) &= \frac{2}{(1-r^2)} \{\!(1+D)^{_{1/2}}\![1+C-(D(1+C)+B^2+C)r^2\ + D(B^2+C)r^4]^{_{1/2}} - (1-BDr^2)\!\}\;, \ \ R_{_0}^2 &= \frac{(1+D)(1-Dr^2)}{(1+C)-r^2(C+B^2)} \end{split}$$ and $$R_{\scriptscriptstyle 1} = \frac{1 \, + \, Dr}{1 \, + \, Br}$$ .

Lemma 2. If w(z) is regular in E and satisfies $|w(z)| \leq |z|$ for $z \in E$ and $p(z) = [1 + Dw(z)][1 + Bw(z)]^{-1}$ , then $$\begin{split} \operatorname{Re} \Big\{ & \frac{zw'(z)}{[1 + Bw(z)][1 + Dw(z)]} \Big\} \leq & -\frac{1}{(B - D)^2} \Big\{ \operatorname{Re} \Big[ \frac{D}{p(z)} + Bp(z) \Big] \\ & - \Big[ \frac{r^2 \, |Bp(z) - D|^2 - |1 - p(z)|^2}{(1 - r^2) \, |p(z)|} \Big] \\ & - (B + D) \Big\} \; . \end{split}$$

Theorem 2. Let $g(z) \in S^(\gamma)$ and $F(z) \in S^(\alpha)$ and define f(z) by $$F(z)g(z)=2\int_0^z f(t)dt,$$ or, equivalently, by $$f(z) = 2^{-1}(g(z)F(z))'$$ , then for |z| = r, $$\operatorname{Re}\left[\frac{zf'(z)}{f(z)}\right] \geq \begin{cases} P_1(r) & for & R_0 \leq R_1 \ P_2(r) & for & R_0 \geq R_2 \ \end{cases}$$ where $$\begin{align} P_{_1}\!(r) &= \frac{1}{(1-A)}\!\!\left\{\!\!\left[(3-2A)\!\!\left[\frac{1+Ar}{1+r}\right]\! + A\!\left[\frac{1+r}{1+Ar}\right]\! - 2\right\}\!, \right. \ P_{_2}\!(r) &= \frac{2}{1-A}\!\!\left\{\!\!\left[\frac{(1+A)(4-2A))(1-Ar^2)}{1-r^2}\right]^{\!1/2}\! - \!\left[\frac{7-Ar^2}{1-r^2}\right]\! - 1\!\right\}\!, \ R_{_0}^2 &= (1+A)(1-Ar^2)/(4-2A)(1-r^2)\;, \ R_{_1} &= (1+Ar)/(1+r)\;, \end{array}$$ and $$A = \alpha + \gamma - 1$$ . These bounds are sharp. Proof. $$2f(z) = g(z)F'(z) + F(z)g'(z)$$ . So $$\frac{2zf(z)}{g(z)F(z)} = \frac{zF'(z)}{F(z)} + \frac{zg'(z)}{g(z)}.$$ $F(z) \in S^(\alpha)$ and $g(z) \in S^(\gamma)$ imply that $zF'(z)/F(z) \in P(\alpha)$ and $zg'(z)/g(z) \in P(\gamma)$ . Thus $zf(z)/g(z)F(z) \in P((\alpha + \gamma)/2)$ and so, by a well-known representation formula, we have $$\frac{zf(z)}{g(z)F(z)} = \frac{1 + Aw(z)}{1 + w(z)}$$ where $A = \alpha + \gamma - 1$ and w(z) is regular in E and satisfies $|w(z)| \le |z|$ for $z \in E$ . Let $p(z) = [1 + Aw(z)][1 + w(z)]^{-1}$ , then we have $$(4) zf(z) = g(z)F(z)p(z).$$ Thus, $$\begin{align} \frac{zf'(z)}{f(z)} &= \frac{zg'(z)}{g(z)} + \frac{zF'(z)}{F(z)} + \frac{zp'(z)}{p(z)} - 1 \ &= 2p(z) + \frac{zp'(z)}{p(z)} - 1 \; , \end{gathered}$$ using (3) and (4). Now, $$\frac{zp'(z)}{p(z)} = \frac{-(1-A)zw'(z)}{[1+Aw(z)][(1+w(z)]}.$$ An application of Lemma 2 and then Lemma 1 with C=3-2A, B=1 and D=A gives immediately the inequality stated in Theorem 2. We now show that the bounds are sharp. Let us choose $g_1(z) \in S^(\gamma)$ and $F_1(z) \in S^(\alpha)$ defined by the formulas, $$\frac{zg_1'(z)}{g_1(z)} = \frac{1 + (2\gamma - 1)w_1(z)}{1 + w_1(z)} , \quad \frac{zF_1'(z)}{F_1(z)} = \frac{1 + (2\alpha - 1)w_1(z)}{1 + w_1(z)} .$$ The corresponding $f_1(z) = 2^{-1}(g_1(z)F_1(z))'$ satisfies $$p_{i}(z) = \frac{zf_{i}(z)}{g_{i}(z)F_{i}(z)} = \frac{1 + Aw_{i}(z)}{1 + w_{i}(z)} \; , \quad A = \alpha + \gamma - 1 \; .$$ From the proof of Lemma 1 it is evident that the bounds $P_1(r)$ and $P_2(r)$ will be attained only when R=a-d and $R=R_0$ respectively, where $R=|p(z)|=\operatorname{Re}\,p(z)$ . Since a-d=(1+Ar)/(1+r), the function $w_1(z)\equiv z$ will give $g_1(z),\, F_1(z)$ such that $|p_1(z)|=\operatorname{Re}\,p_1(z)=a-d$ at z=r and so the bound $P_1(r)$ will be attained by the corresponding function $zf_1'(z)/f_1(z)$ , at z=r, for all r satisfying $R_0\leqq R_1$ . When $R_0\geqq R_1=a-d$ , we choose $w_2(z)=z(z-q)/(1-qz)$ where q is determined by the condition that $p_2(z)=(1+Aw_2(z))/(1+w_2(z))$ satisfies $|p_2(z)|=R_0=\operatorname{Re}\,p_2(z)$ at z=r. Now we have $$a-d=R_{\scriptscriptstyle 1} \leqq R_{\scriptscriptstyle 0} \leqq a+d$$ , so $$\frac{1+Ar}{1+r} \leq \frac{1+AT}{1+T} \leq \frac{1-Ar}{1-r}$$ , where $T = w_2(r)$ and this in turn implies one of the three following equivalent conditions, $$\frac{r^2(r-q)^2}{(1-qr)^2}=\,T^2 \leqq r^2$$ , $q^2 \leqq 1$ , $\mid q\mid \leqq 1$ . So $w_2(z)$ is indeed subordinate to z in E and hence $p_2(z)$ belongs to $P((\alpha + \gamma)/2)$ . Clearly, $g_2(z) \in S^(\gamma)$ and $F_2(z) \in S^(\alpha)$ defined by the formulas $$\frac{zg_2'(z)}{g_2(z)} = \frac{1+(2\gamma-1)w_2(z)}{1+w_2(z)} \; , \quad \frac{zF_2'(z)}{F_2(z)} = \frac{1+(2\alpha-1)w_2(z)}{1+w_2(z)} \; ,$$ give rise to $f_2(z)$ which satisfies $|p_2(r)| = \text{Re } p_2(r) = R_0$ , where $$p_{\scriptscriptstyle 2}\!(z) = \frac{z f_{\scriptscriptstyle 2}\!(z)}{g_{\scriptscriptstyle 2}\!(z) F_{\scriptscriptstyle 2}\!(z)} = \frac{1 \, + \, A w_{\scriptscriptstyle 2}\!(z)}{1 \, + \, w_{\scriptscriptstyle 2}\!(z)}$$ , and so $zf_2'(z)/f_2(z)$ attains the bound $P_2(r)$ at z=r.

Theorem 3. If the family Q is defined by $$Q = \{f(z) = 2^{-1}(g(z)F(z))'; g(z) \in S^(\gamma), F(z) \in S^(\alpha) \text{ and } \alpha + \gamma \leq 1\},$$ then the radius of $\beta$ -starlikeness of the family Q is the least positive root of the equation $P_1(r) - \beta = 0$ , where $P_1(r)$ is as defined in Theorem 2.