THOMAS, D.K.

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TitleAn inequality for analytic functions(Topics in UnivalentFunctions and Its Applications)Author(s)THOMAS, D.K.Citation数理解析研究所講究録 (1990), 714: 22-24Issue Date1990-03URL http://hdl.handle.net/2433/101739RightType Departmental Bulletin PaperTextversionpublisherKyoto University22An inequality for analytic functionsD.K.THOMAS ( )INTRODUCTIONDenote by $A$ the class of functions which are analytic in the unit disc$D=\{z : |z|<1\}$ and are normalised so that $f(O)=0$ and $f’(O)=1$ .In [3], Obradovi\v{c} showed that if $f\in A$ and satisfies ${\rm Re} f(z)/z>\alpha$ for$\alpha<1$ , then${\rm Re} \frac{a+1}{z^{a+1}}\int_{0}^{z}t^{a-1}f(t)dt>\alpha+\frac{1-\alpha}{3+2a}$for $a>-1$ and $z\in D$ . This result, (which is not sharp) was then usedby Odradovi\v{c} to establish certain non-sharp lower bound estimates forthe real parts of some integral operators of functions in various classesof univalent functions.In this note, we prove the sharp version of Obradivi\v{c}’s result and givea generalisation. The method is quite elementary. Other applications ofthe method have been given in [1] and [2].RESULTSLet $f\in A$ and $z\in D$ . For $n=1,2\ldots$ , and $a>-1$ , define$I_{n}(z)= \frac{a+1}{z^{a+1}}\int_{0}^{z}t^{a}I_{n-1}(t)dt$ ,wltere $I_{0}(z)=f(z)/z$ .We prove:THEOREM. Let $f\in A$ $al1d\alpha<1$ . Then for $7?\geq 0$ and $z\in D$ , th $e$$i_{l1}eqnality{\rm Re} f(z)/z>\alpha$ implies${\rm Re} I_{71}(z)\geq\gamma_{?l}(?)>\gamma_{n}(1)$ ,Typeset by $A_{J}w^{s}- IE^{X}$714 1990 2-2423$wl_{l}ere$A$0< \gamma_{n}(’)=1+2(a+1)^{n}(1-\alpha)\sum_{j=1}^{\infty}\frac{(-?\cdot)^{j}}{(j+a+1)^{n}}<1$. (1)Equality is at tain $ed$ vvhen $\frac{f(z)}{z}=\alpha+(1-\alpha)\frac{1-z}{1+z}$ .PROOF: The case $n=0$ is trivial. Suppose that $n=1$ , then since${\rm Re} f(z)/z>\alpha$ , we have, with $z=re^{i\theta}$ ,$\frac{\partial}{\partial r}\int o^{z}t^{a-1}f(t)dt=z^{a^{\backslash }}\frac{f(z)}{z}e^{i\theta}=z^{a}e^{i\theta}[\alpha+(1-\alpha)h(z)]$ ,where ${\rm Re} h(z)>0$ for $z\in D$ .Thus integrating and noting that ${\rm Re} h(z) \geq\frac{1-\rho}{1+\rho}$ for $0\leq\rho<1$ , itfollows that if $a>-1$ ,${\rm Re} \frac{a+1}{z^{a+1}}\int_{0}^{z}t^{a-1}f(t)dt\geq\frac{a,+1}{7^{a+1}}\int_{0}^{r}p^{a}[\alpha+(1-\alpha)(\frac{1-\rho}{1+\rho})]d\rho$ ,$= \frac{a+1}{r^{a+1}}\int_{0}^{r}p^{a}(1+2(1-\alpha)\sum_{j=1}^{\infty}(-\rho)^{j})d\rho$ ,$=1+2(a+1)(1- \alpha)\sum_{j=1}^{\infty}\frac{(-r)^{j}}{j+a+1}$ ,whiclt proves the theorem in the case $n=1$ .Ncxt note tliat writing $t=\rho e^{i0}$ , we have${\rm Re} I_{?\iota+1}(z)={\rm Re} \frac{a+1}{z^{a+1}}\int_{0}^{z}t^{a}I_{n}(t)dt$ ,$= \frac{a.+1}{7^{a+1}}\int_{0}^{r}p^{a}{\rm Re} I_{?l}(\rho e^{i\theta})dp$ ,$\geq\frac{a+1}{\uparrow^{\backslash }a+1}.1_{0}^{r}(\rho^{a}+2(a+1)^{n}(1-\alpha)\sum_{j=1}^{\infty}\frac{(-1)^{j}p^{j+a}}{(j+a+1)^{7l}})d\rho$ ,$=\gamma_{?\iota+1}(r)$ ,wltere the inequality follows by induction.Rearranging terms in the infinite series given in (1), shows tliat $0<$$\gamma_{n}(\uparrow)<1$ and the proof is complete.224REFERENCES..1. S.A.Halixn and D.K. $T1_{1OlI1}as,$ A note on Bazilevi\v{c} functions, to appear.2. S.A.Halim and D.K.Thomas, Sharp estimates for some integral operators ofconvex functions of order alpha, to appear.3. M.Obradovi\v{c}, On certain inequalities for some regular functions in the unitdisc, Internat. J. Math. and Math. Sci. 8 (1985), 677-681.1980 Mathematics subject classifications: Primary $30C45$3

An inequality for analytic functions(Topics in Univalent Functions and Its Applications)

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An inequality for analytic functions(Topics in Univalent Functions and Its Applications)

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