부산대학교 도서관

Let $\Bbb{D}=\{z\in\Bbb{C}:|z|<1\}$ denote the unit disk in the complexplane $\Bbb{C}$. Also, let $\Cal{B}$ be the class of all analyticfunctions $f$ defined on $\Bbb{D}$ such that $|f(z)|\leq1$,$z\in\Bbb{D}$. In 1914, it was discovered by H. Bohr [Proc. LondonMath. Soc. (2) {\bf 13} (1914), 1--5; MR1577494] thatif $f(z)=\sum_{n=0}^{\infty} a_nz^n\in\Bbb{B}$, then$$\sum_{n=0}^{\infty} |a_nz^n|\leq 1$$for $|z|\leq 1/3$ and the constant $1/3$ cannot be improved. Theinequality is known as the Bohr inequality while the constant $1/3$ isknown as the Bohr radius of the class $\Cal{B}$. Recently, the conceptof Bohr radius has been extended to the study of integral operators,which are defined on certain analytic function classes. In this work,the authors manage to obtain two Bohr-type radii by considering the$\beta$-Cesàro operator $T_{\beta}$ and the Bernadi operator$L_{\gamma}$.\par In 2020, the Cesàro operator $T$ on the class $\Cal{B}$, as definedin [G.~H. Hardy and J.~E. Littlewood, Math. Z. {\bf 34} (1932),no.~1, 403--439; MR1545260],$$T[f](z)=\int_0^1\frac{f(tz)}{1-tz}\,dt=\sum_{n=0}^{\infty}\left(\frac{1}{n+1}\sum_{k=0}^n a_k\right)z^n,$$was first applied to the study of Bohr radii in [I.~R. Kayumov,D.~M. Khammatova and S. Ponnusamy, C. R. Math. Acad. Sci. Paris {\bf 358} (2020), no.~5, 615--620; MR4149861]. Motivated bythis work, the authors investigate the Bohr radius for a generalizedversion of Cesàro operator. They introduce the $\beta$-Cesàrooperator $T_{\beta}$ ($\beta>0$), which is defined on the class$\Cal{B}$ and given by$$T_{\beta}[f](z)=\int_0^1\frac{f(tz)}{(1-tz)^{\beta}}\,dt =\sum_{n=0}^{\infty} \left(\frac{1}{n+1}\sum_{k=0}^n\frac{\Gamma(n-k+\beta)}{\Gamma(n-k+1)\Gamma(\beta)}a_k\right)z^n,\quad z\in\Bbb{D},$$for $f(z)=\sum_{n=0}^{\infty} a_nz^n\in\Cal{B}$. It can be seen the$\beta$-Cesàro operator reduces to the classical Cesàro operatorwhen $\beta=1$. The authors show that the following Bohr-typeinequality$$\sum_{n=0}^{\infty} \left(\frac{1}{n+1}\sum_{k=0}^n\frac{\Gamma(n-k+\beta)}{\Gamma(n-k+1)\Gamma(\beta)}|a_k|\right)r^n\leq\frac{1}{r}\left[\frac{1-(1-r)^{1-\beta}}{1-\beta}\right]\tag1$$holds for $r\leq R(\beta)$, where $R(\beta)$ is the positive root ofthe equation$$\frac{3[1-(1-x)^{1-\beta}]}{1-\beta}-\frac{2[(1-x)^{-\beta}-1]}{\beta}=0.$$The Möbius transformation$$\phi_a(z)=\frac{z-a}{1-az}\in \Cal{B},\quad a\in[0,1),$$can be used to verify that the constant $R(\beta)$ cannot be improved.The proof makes use of Wiener's estimate ($|a_n|\leq 1-a^2$, $n\geq 1$)to find a non-negative function $\phi(a)$ which dominates the infinitesum on the left of inequality (1). The authors then show that $\phi(a)$is an increasing function in terms of $a$ and is dominated by thefunction on the right hand side of inequality (1).\par Let $\Cal{B}_0=\{f\in \Cal{B}:f(0)=0\}$ be the class of normalizedfunctions in $\Cal{B}$. If $g(z)=\sum_{n=1}^{\infty}b_nz^n\in\Cal{B}_0$, then there is another form of $\beta$-Cesàrooperator $C_{\beta}$ ($\beta>0$) defined on the class $\Cal{B}_0$ [S. Kumar and S.~K. Sahoo, Rocky Mountain J. Math. {\bf 50} (2020),no.~5, 1723--1746; MR4170682]:$$C_{\beta}[g](z)=\int_0^1\frac{g(tz)}{t(1-tz)^{\beta}}\,dt=\sum_{n=0}^{\infty} \left(\frac{1}{n+1}\sum_{k=0}^n\frac{\Gamma(n-k+\beta)}{\Gamma(n-k+1)\Gamma(\beta)}b_{k+1}\right)z^{n+1},\quad z\in\Bbb{D}.$$The authors remark that the Bohr radii for the operators $T_{\beta}$and $C_{\beta}$ are exactly the same where the corresponding Bohr-typeinequality is given by$$\sum_{n=0}^{\infty} \left(\frac{1}{n+1}\sum_{k=0}^n\frac{\Gamma(n-k+\beta)}{\Gamma(n-k+1)\Gamma(\beta)}|b_{k+1}|\right)r^{n+1}\leq\frac{1}{r}\left[\frac{1-(1-r)^{1-\beta}}{1-\beta}\right].$$\par Last but not least, the authors also obtain Bohr-type radii for theBernadi operator $L_{\gamma}$ [S.~S. Miller and P.~T. Mocanu, {\it Differential subordinations}, Monogr. Textbooks Pure Appl. Math., 225,Dekker, New York, 2000 (p. 11); MR1760285], which isdefined on the class $\Cal{B}$ and given by$$L_{\gamma}[f](z)\coloneq \sum_{n=m}^{\infty}\frac{a_n}{n+\gamma}z^n=\int_0^1f(zt) t^{\gamma-1}\,dt,$$for $f(z)=\sum_{n=m}^{\infty} a_nz^n\in\Cal{B}$ and $\gamma>-m$. Theauthors prove that the following Bohr-type inequality$$\sum_{n=m}^{\infty} \frac{|a_n|}{n+\gamma} r^n\leq\frac{1}{m+\gamma}r^m\tag2$$holds for $r\leq R(\gamma)$. The constant $R(\gamma)$ is a positiveroot of some equation and cannot be improved. In particular, if$\gamma=1$ and $m=0$, then the Bernadi operator $L_{\gamma}$ is reducedto the well-known Libera operator [see S.~S. Miller and P.~T.Mocanu, op. cit.; R. Parvatham, S. Ponnusamy and S.~K. Sahoo, HiroshimaMath. J. {\bf 38} (2008), no.~1, 19--29; MR2397377].The techniques applied in the proof of this result are very muchsimilar to the proof of Bohr-type radii for the Cesàro operator, andthe Möbius transformation $\phi_a$ can be used to show that$R(\gamma)$ cannot be improved.