On an integral operator for convex univalent functions

Lemma 2.1. Let a,m and M be real numbers such that 0 < a < 1 and \m-l\ < M < m . If (2.1) $$t = 1-\alpha + \alpha m \quad \text{and} \quad T = \alpha M,$$ then $K(t,T) \subseteq K(m,M)$ . Proof. It suffices to show that $$(2.2) m-M \le t-T \text{ and } t+T \le m+M.$$ We need consider only the case when 0 < a < 1 . Suppose m-M > t-T . Then, m-M > 1-a. + a(m-M) , which implies that m-M > 1 . But this is contrary to the assumption \m-l | < M . Next, suppose t+T > m+M . Then, l-a+a(m+M) > rrH-M , which implies rrH-M < 1 . This is also contrary to \m-l\ < M . Therefore the inequalities (2.2) hold. LEMMA 2.2 Let a,m and M be real numbers such that $$(2.3) max \left\{ -\frac{1-(m-M)}{m+M-1}, -\frac{m+M-1}{1-(m-M)} \right\} \le \alpha < 0$$ and \m-l\ < M < m . If $$(2.4) \widetilde{t} = 1 - \alpha + \alpha m \quad and \quad \widetilde{T} = -\alpha M,$$ then KC£,T) £ K(m,M) . Proof. We need to show that $$(2.5) m-M \le t-T and t+T \le m+M.$$ Suppose m-M > t-T . Then, m-M > l-a+a(m+M) , which implies that a < -(l-(m-M))/(m+M-l) . Further, suppose t+T > m+M . Then, l-a+a(m-M) > m+M , which implies a < -(m+M-1)/(l-(m-M)) . Now if $$\alpha < \max \left\{-\frac{1-(m-M)}{m+M-1}, -\frac{m+M-1}{1-(m-M)}\right\}$$ then at least one of the inequalities in (2.5) does not hold. Whence the disc centred at t and having radius T is not contained in the disc centred at m and having radius M . Therefore K(t,T) c^ K(m,M) if a satisfies (2.3). It is noticeable that the truth of |t-1\ <T < t requires a S -1/m+M-l) , which automatically holds since $$\alpha \ge \max \left\{ -\frac{1 - (m - M)}{m + M - 1}, -\frac{m + M - 1}{1 - (m - M)} \right\}$$ $$\ge -(1 - (m - M))/(m + M - 1)$$ $$\ge -1/(m + M - 1).$$ NOTE. From now on t,T and t,T will be as in (2.1) and (2.4) respectively. 3. Jakubowski Convex Integral Operators. In this section we obtain the main results of this paper.

Theorem 3.1. If $f \in K(m,M)$ , then the function F, defined by (3.1) $$F(z) = \int_{0}^{z} \{f'(u)\}^{\alpha} du,$$ also belongs to K(m,M), provided $$(3.2) - \phi \leq \alpha \leq 1$$ where $\phi = (M - |m-1|)/M + |m-1|)$ . The result is sharp. Proof. From (3.1) we have $$F'(z) = \{f'(z)\}^{\alpha}.$$ Logarithmic differentiation yields $$1 + z \; \frac{f''(z)}{f'(z)} = \alpha \; \{1 + z \; \frac{f''(z)}{f'(z)}\} \; + \; (1 - \alpha) \; .$$ Therefore (3.3) $$1+z \frac{f''(z)}{f'(z)} - (1-\alpha+\alpha m) = \alpha \{1+z \frac{f''(z)}{f'(z)} - m\}.$$ Now consider two cases, namely $\alpha > \theta$ and $\alpha < \theta$ . Case I. When $\alpha > 0$ , it follows from (3.3) that $$\left| 1+z \frac{f''(z)}{f'(z)} - (1-\alpha+\alpha m) \right| = \alpha \left| 1+z \frac{f''(z)}{f'(z)} - m \right|$$ $$< \alpha M, \text{ since } f \in K(m,M).$$ Thus $F \in K(t,T)$ , and hence, by Lemma 2.1, $F \in K(m,M)$ provided $$(3.4) 0 < \alpha \le 1.$$ Case II. When $\alpha < \theta$ , from (3.3) we get $$\left|1+z \frac{f''(z)}{f'(z)} - (1-\alpha+\alpha m)\right| = -\alpha \left|1+z \frac{f''(z)}{f'(z)} - m\right|$$ $$< -\alpha M.$$ Thus $F \in K(\tilde{t}, \tilde{T})$ , and hence, by Lemma 2.2, $F \in K(m, M)$ provided (3.5) $$\max \left\{ -\frac{1-(m-M)}{m+M-1} , -\frac{m+M-1}{1-(m-M)} \right\} \le \alpha < 0.$$ It is now easy to compute that $$\max \left\{ -\frac{1-(m-M)}{m+M-1} , -\frac{m+M-1}{1-(m-M)} \right\} = \begin{cases} -\frac{1-(m-M)}{m+M-1} , & \text{when } m > 1 \\ -1 , & \text{when } m = 1 \\ -\frac{m+M-1}{1-(M-M)} , & \text{when } m < 1 \end{cases}.$$ The expressions on the right hand side are easily seen to be equal to - $\phi$ . Hence, combining (3.4), (3.5) and the trivial case a = 0 , we conclude that F e K(rrijM) if a satisfies (3.2) . To establish the sharpness of the result we take $$f(z) = \begin{cases} \int_{0}^{z} (1-bu)^{-(a+b)/b} du, & \text{when } m \neq 1, \\ \int_{0}^{z} e^{Mu} du, & \text{when } m = 1, \end{cases}$$ where (3.6) $$a = \frac{M^2 - m^2 + m}{M} \text{ and } b = \frac{m-1}{M}$$ It is easy to see that 1+zf '(z)/f(z) maps U onto the disc centred at m and having radius M . It follows then from (3.3) that 1+zF''(z)/F'(z) maps U onto the disc centred at t and having radius T , when a > 0 . Now if a > 1 , none of the inequalities in (2.2) holds, and consequently the disc centred at t and having radius T is not contained in the disc centred at m and having radius M . Therefore F / K(m,M) if a > 1 . Similarly we can show that F / K(m,M) if a < - <j> . Hence the result is sharp. In the particular case when m = M and m -\•<\• 3 K(m,M) is equal to the well known class K of convex functions. Hence the following result of Kim and Merkes [ 3 follows from Theorem 3.1.

Corollary 3.1. If f e K , then the function F defined by (3.1) also belongs to K , provided 0 ^ a S 1 . The result is sharp with the 2 -2 extremal function f(z) = / (1-u) du . Very recently, Pandey and Bhargava [5] have shown that if f e K(m,U)} then the function F defined by (3.1) also belongs to K(m,M) provided a is a complex number such that |a| < (l-b)/2 , where b is given by (3.6). We show below that their result is incorrect: Let us consider the function (3.7) $$f(z) = \int_{0}^{z} (1-bu)^{-(a+b)/b} du$$ with m = .7 and M = .6 , so that a = .95 and b = -.5. Evidently, / e K(.7,.6) and (l-b)/2 = .75 . Now, by the abovementioned result of Pandey and Bhargava, F e K(.7, .6) provided |ct| £ .75 . But, if we take a = -.75 , then from (3.1)and (3.7) we are led to $$\left| 1+z \frac{F''(z)}{F'(z)} - .7 \right| = \left| -.75 \left\{ \frac{1+.95z}{1+.5z} \right\} + 1.05 \right| ;$$ and at z = -.8 , the right hand side of this equation is equal to .75 . Thus we see that $$1+z \frac{F''(z)}{F'(z)} - .7 > .6$$ at some point in the unit disc U . Therefore F / K(,7t.6) . Hence the result of Pandey and Bhargava [5] is incorrect. We now proceed to improve and correct the result of Pandey and Bhargava [5] . First we prove:

Theorem 3.2. If f e K(m,M) , then the function F defined by (3.1) also belongs to K(m,M) provided a is a complex number satisfying (3.8) $$\left|\alpha\right| < 1$$ , $\frac{\left|1-\alpha\right|}{1-\left|\alpha\right|} \le \frac{M}{\left|m-1\right|} = \frac{1+\phi}{1-\phi}$ , when $m \ne 1$ and $$|\alpha| \leq 1 \qquad \text{, when } m = 1 \text{,}$$ where <f> = (M-\m-l\ )/(M+\m-l\ ) . Proof. We have (as in Theorem 3.1) $$1+z \frac{F''(z)}{F'(z)} = \alpha \left\{ 1+z \frac{f''(z)}{f'(z)} - m \right\} + (1-\alpha)$$ Therefore, when m / 1 , we obtain $$\begin{vmatrix} 1+z & \frac{f''(z)}{f'(z)} & -m \end{vmatrix} = \begin{vmatrix} \alpha \{1+z & \frac{f''(z)}{f'(z)} & -m \} & -(1-\alpha)(m-1) \end{vmatrix}$$ $$\leq |\alpha| \begin{vmatrix} 1+z & \frac{f''(z)}{f'(z)} & -m \end{vmatrix} + |1-\alpha| |m-1|$$ $$\leq |\alpha| M+|1-\alpha| |m-1|$$ $$\leq M,$$ provided $$|\alpha| < 1$$ and $\frac{|1-\alpha|}{|1-|\alpha|} \le \frac{M}{|m-1|} = \frac{1+\phi}{1-\phi}$ Further, when m=l , we obtain $$\begin{vmatrix} 1+z & \frac{f''(z)}{f'(z)} - 1 \end{vmatrix} = \left| \alpha \left\{ 1+z & \frac{f''(z)}{f'(z)} - 1 \right\} \right|,$$ $$= \left| \alpha \right| \left| 1+z & \frac{f''(z)}{f'(z)} - 1 \right|,$$ $$< \left| \alpha \right| M,$$ $$\leq M,$$ provided |a| 5 1 . This completes the proof of the theorem. Using the above theorem we now establish the following corollary which improves and corrects the result of Pandey and Bhargava [5], It is worth noting that the technique employed by us is different; and the result is sharp also.

Corollary 3.2. If / e K(m,M) , then the function F defined by (3.1) also belongs to K(m,M) provided a is a complex number satisfying (3.10) ] cc j < (J> . The result is sharp in the sense that the region given by (3.10) cannot be extended into any larger disc centred at the origin on the a-plane. Proof. When m=l , the inequality (3.10) is identical with the inequality (3.9). We need therefore consider only the case when m ^ 1 . In this case, from (3.8), we have $$\frac{|1-\alpha|}{|1-\alpha|} \le \frac{1+\phi}{|1-\phi|}.$$ This inequality is implied by $$\frac{1+|\alpha|}{1-|\alpha|} \le \frac{1+\phi}{1-\phi}$$ which is equivalent to $$|\alpha| \leq \phi$$ Hence, by Theorem 3.2,F e K(m,M) if lett £ <j>. To show the sharpness of the inequality (3.10), let us assume that F e K(m,M) for all values of a lying in a larger disc | ex ] ^ §+s , for some s > 0 . Then, in particular, F e K(m,M) for all values of o lying in the annulus <(> < |o| i <j>+s . From this it follows that when a is real and negative, then F e K(m,M) even if -(§+s) < a < -4 . But this is contrary to the sharpness of a > -<(> in (3.2). Therefore the region given by (3.10) cannot be extended into any larger disc centred at the origin on the a-plane. Hence the result is sharp. Remark. We could not establish the sharpness of Theorem 3.2 when m ^ 1 . However the result is sharp for m=l (as shown in Corollary 3.2) . Therefore, it would be interesting to establish the sharpness of the inequality (3.8).