A note on Ruscheweyh type of integral operators for uniformly \(\alpha\)-convex functions (Q1607832)
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scientific article; zbMATH DE number 1780331
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | A note on Ruscheweyh type of integral operators for uniformly \(\alpha\)-convex functions |
scientific article; zbMATH DE number 1780331 |
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A note on Ruscheweyh type of integral operators for uniformly \(\alpha\)-convex functions (English)
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13 August 2002
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Let \(A\) denote the class of functions \(f(z)= z+ a_2z^2\cdots\) analytic in the unit disk \(U\), \(f\in U\) is uniformly \(\alpha\)-convex iff \[ \text{Re}\{(1- \alpha)(z-\zeta) f'(z)/(f(z)- f(\zeta))+ \alpha(1+ (z-\zeta) f''(z))/f'(z))\}> 0 \] for all \(z,\zeta\in U\) and \(0\leq\alpha\leq 1\). In this note the authors consider the integral operator \[ F(z)= (F_\alpha(z,\zeta)- F_\alpha(0,\zeta))/F_\alpha'(0,\zeta), \] where \(\alpha> 0\) and \[ F_\alpha(z,\zeta)= \Biggl\{(c+ 1/\alpha)/(z- \zeta)^c \int^z_\zeta (t-\zeta)^{c- 1} (f(t)- f(\zeta))^{1/\alpha} dt\Biggr\}^\alpha \] (\(z\in U\), \(\zeta\in U\) fixed and \(z=\zeta\)) and prove that \(F\) is a uniformly \(\alpha\)-convex function when \(f\) is a uniformly \(\alpha\)-convex function.
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\(\alpha\)-convex functions
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integral operator
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uniformly \(\alpha\)-convex functions
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0.9148007
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0.9118892
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0.8995214
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