On starlikeness of certain integral transforms

Theorem 1. If $\delta = (2 \ln 2 - 1)(3 - 2 \ln 2)/[3 - (2 \ln 2 - 1)(3 - 2 \ln 2)] = 0.262...$ and $f \in R(-\delta)$ then the function F defined by (1) is starlike in U. Proof. From (1) we deduce (6) $$zF''(z) + F'(z) = f'(z), \quad z \in U.$$ Let P(z) = F'(z) and Q(z) = F(z)/z. Since Re $f'(z) > -\delta$ in U, by using (4) and (6) we find that Re $F'(z) > -\delta + (1+\delta)(2\ln 2 - 1)$ for $z \in U$ . Again by using (4) this in turn implies Re $[Q(z)] > 2\delta > 0$ , $z \in U$ . Now if we set p(z)=zF'(z)/F(z) then p is analytic in U, p(0)=1 and f'(z)=m(p(z),zp'(z);z), where $m(u,v;z)=(u^2+v)Q(z)$ . To prove $F\in S^*$ , it is enough to show $\operatorname{Re} p(z)>0$ in U. Since $\operatorname{Re} f'(z)>-\delta$ in U, by (6), we get $\operatorname{Re} m(p(z),zp'(z);z)>-\delta$ in U. Now for all real $x,y\leq -(1+x^2)/2$ and $z\in U$ , we have $\operatorname{Re} m(ix,y;z)=(y-x^2)\operatorname{Re} Q(z)\leq -(1+3x^2)\operatorname{Re}(Q(z)/2)\leq -\delta$ and so applying Lemma A with $\Omega=\{w\in\mathbb{C}:\operatorname{Re} w>-\delta\}$ , we get $\operatorname{Re} p(z)>0$ in U. Hence the theorem. For the proof of Theorem 2 we prove the following lemmas. Theorem 2 further improves Theorem 1.

Lemma 1. If $g \in R(\beta)$ then G defined by (2) belongs to $R(\beta + (1-\beta)(3-4\ln 2))$ ( $\beta < 1$ ). Proof. From (2) we deduce $$zG'(z) + G(z) = 2g(z),$$ (8) $$zG''(z) + 2G'(z) = 2g'(z).$$ Since $g \in R(\delta)$ , by using (5), we obtain $$G'(z) \prec \beta + (1 - \beta)L(z), \quad z \in U$$ and so $\operatorname{Re} G'(z) > \beta + (1 - \beta)(3 - 4 \ln 2)$ , $z \in U$ . Here L(z) is as defined earlier. This proves Lemma 1.

Lemma 2. Let $M=2(2\ln 2-1)(1-\ln 2),\ \theta=0.911621907,\ N=\tan\theta,\ a=4(1+M)^2-\frac{4}{3}N^2(2\ln 2-1)^4-4,\ b=-4(1-2M)(1+M)-\frac{8}{3}(2\ln 2-1)^4N^2,\ c=(1-2M)^2-\frac{2}{3}(2\ln 2-1)^2N$ and $\varrho=(-b-(b^2-4ac)^{1/2})/(2a)$ . Suppose that Q is a complex function with Q(0)=1 satisfying (9) $$Q(U) \subset E_1 \cap E_2 \cap E_3 \quad \text{where}$$ $$E_1 = \{ w \in \mathbb{C} : \text{Re } w > 1 - 2M(1 + \varrho) \},$$ $$E_2 = \{ w \in \mathbb{C} : |\text{arg } (w - (1 - 2(2 \ln 2 - 1)(\varrho + 1)))| < \theta \},$$ $$E_3 = \{ w \in \mathbb{C} : |\text{Im } w| < 2(2 \ln 2 - 1)(\varrho + 1)\pi \}.$$ If p is analytic in U with p(0) = 1 and if Re $$Q(z)[zp'(z) + p^2(z) + p(z)] > -2\varrho$$ , $z \in U$ , then $\operatorname{Re} p(z) > 0$ in U. Throughout the paper M, $\theta$ , N, a, b, c, $\varrho$ , and E's are all as defined above. Proof of Lemma 2. If we let $m(u,v;z)=Q(z)(v+u^2+u)/2$ then for all $x, y \leq -(1+x^2)/2$ and $z \in U$ , we have $$\operatorname{Re} m(ix, y; z) = [(y - x^{2}) \operatorname{Re} Q(z) - x \operatorname{Im} Q(z)]/2$$ $$< -[3x^{2} \operatorname{Re} Q(z) + 2x \operatorname{Im} Q(z) + \operatorname{Re} Q(z)]/4.$$ Thus $\operatorname{Re} m(ix, y; z) \leq -\varrho$ if Q satisfies (10) $$[(X - 2\varrho)^2/(2\varrho)^2] - [Y^2/(3(2\varrho)^2)] \ge 1$$ where $X = \operatorname{Re} Q(z)$ and $Y = \operatorname{Im} Q(z)$ . Since Q satisfies (9), to prove (10) it is enough to show that the point $(X_0, Y_0)$ with $X_0 = 1 - 2M(\varrho + 1)$ and $Y_0 = 2(2 \ln 2 - 1)^2(\varrho + 1) \tan \theta$ lies on the hyperbola $[(X - 2\varrho)^2/(2\varrho)^2] - [Y^2/(3(2\varrho)^2)] = 1$ . Thus by substituting this value in this hyperbola, we get, by a simple calculation, $$(12) a\varrho^2 + b\varrho + c = 0.$$ Hence by hypothesis, we deduce that $\operatorname{Re} m(ix, y; z) \leq -\varrho$ for all $z \in U$ . Now by Lemma A, with $\Omega = \{w \in \mathbb{C} : \operatorname{Re} w > -\varrho\}$ , we obtain $\operatorname{Re} p(z) > 0$ in U. This completes the proof of Lemma 2. Remark 1. If we let $\varrho'$ and $\varrho''$ be the roots of the quadratic equation (12) then the approximate calculations show that $$\varrho' = (-b - (b^2 - 4ac)^{1/2})/(2a) \approx 0.09032572...,$$ $$\varrho'' = (-b + (b^2 - 4ac)^{1/2})/(2a) \approx 1.2113303378...$$ (Here $a \approx 2.071919132...$ , $b \approx -2.701014071$ , $c \approx 0.227066802...$ and $b^2 - 4ac = \frac{80}{3} \tan^2 \theta (2 \ln 2 - 1)^4 + 16(1 - 2M)^2 \approx [2.326718893...]^2$ .)

Theorem 2. Let $\varrho$ be as defined in Lemma 2, i.e., $\varrho \approx 0.09032572...$ and $g \in R(-\varrho)$ . Then the Libera transform G defined by (2) is in $S^*$ . Proof. Since $g \in R(-\varrho)$ , by using Lemma 1 we obtain $G \in R(\beta)$ with (13) $$\beta = -\rho + (1+\rho)(3-4\ln 2) = 1 - 2(2\ln 2 - 1)(\rho + 1).$$ Now using (4) and the fact that $G \in R(\beta)$ we get $$(G(z)/z) \prec \beta + (1-\beta)l(z), \quad z \in U,$$ where $l(z) = -1 - (2/z) \log(1-z)$ . By (13), a simple calculation yields $\beta + (1-\beta)(2 \ln 2 - 1) = 1 - 2M(1+\varrho)$ . This, from (14) and the observation made earlier, shows that the complex function Q defined by Q(z) = G(z)/z satisfies (9). If we set p(z) = zG'(z)/G(z), by using (2) we obtain (15) $$zG''(z) + 2G'(z) = 2g'(z).$$ Since Re $g'(z) > -\rho$ in U, by using (15) we easily get $$\text{Re}\{Q(z)[zp'(z) + p^2(z) + p(z)]\} > -2\varrho, \quad z \in U,$$ and by Lemma 2 we deduce $\operatorname{Re} p(z) > 0$ in U, which shows that $G \in S^*$ . Hence the theorem. The following theorem can be proved along similar lines and so we omit its

Theorem 3. If $h \in A$ satisfies $\operatorname{Re}\{h'(z)h(z)/z\} > -\varrho$ in U then the function H defined by $H(z) = \int_0^z (h(t)/t) dt$ is starlike in U. Remark 2. In [6], the author showed that for $f \in A$ and $1/6 \le \beta < 1$ , $\text{Re}[h'(z)h(z)/z] > \beta((3\beta - 1)/2)$ implies $\text{Re}(f(z)/z) > \beta$ in U. R e m a r k 3. For α ≥ 0 and β < 1, let R(α, β) be the class of functions f in A satisfying Re[f 0 (z) + αzf⁰⁰(z)] > β for z in U. From a result of Ponnusamy and Karunakaran [5], we have R(α, β) ⊂ R(α 0 , β + (α − α 0 )(1 − β)/(2 + α)) for all α > α⁰ ≥ 0. This relation and Theorem 2 show that $$R(\alpha, (-2\varrho(2+\alpha)+1-2\alpha)/5) \subset S^*$$ for all $\alpha \ge 1/2$ . As an immediate consequence of the above observation, we have

Theorem 4. If f ∈ A satisfies Re[f 0 (z)] > (−2%(2 + α) + 1 − 2α)/5, z ∈ U, for α ≥ 1/2, then the function F defined by $$F(z) = \alpha z^{1-1/\alpha} \int_{0}^{z} f(t)t^{1/\alpha - 2} dt$$ is in S ∗ . Corollary. If f ∈ A satisfies Re f 0 (z) > −(6% + 1)/5 ≈ 0.3083908 . . . for z in U, then the function F defined by (1) is starlike in U. The above corollary improves Theorem 1. R e m a r k 4. For g defined by g(z) = z(2 + z)/2(1 − z) (and hence g satisfies Re[zg⁰ (z)/g(z)] > −1/2 in U) it is well known that the corresponding Libera transform G is starlike in U. On the other hand, a simple calculation shows that g ∈ R(−1/8). Hence the natural problem which arises is to find the best possible % 0 (> %) such that g ∈ R(−% 0 ) implies G is starlike in U. I would like to thank Prof. Dr. K.-J. Wirths for encouraging me by sending a copy of [3].