An application of certain fractional operator
Let \(f\) be holomorphic in the open unit disc \(E\) such that \(f(0)= f'(0)- 1=0\). The fractional integral of order \(\lambda\) is defined by \[ D^{-\lambda}_z f(z)= {1\over \Gamma(\lambda)} \int^z_0 {f(\xi)\over(z- \xi)^{1-\lambda}} d\xi,\quad \lambda>0. \] The fractional derivative of order \(\lambda\) is defined by \[ D^\lambda_z f(z)= {1\over\Gamma(1-\lambda)} {d\over dz} \int^z_0 {f(\xi)\over(z- \xi)^\lambda} d\xi,\quad 0\leq \lambda<1. \] The fractional derivative of order \((n+\lambda)\) is defined by \[ D^{n+ \lambda}_z f(z)= {d^n\over dz^n} D^\lambda_z f(z),\quad 0\leq\lambda<1,\quad n\in\mathbb{N}_0. \] The fractional operator \(J^\lambda_z f(z)\) is defined by \[ J^\delta_z f(z)= \Gamma(2-\delta) z^\delta D^\delta_z f(z),\quad\delta< 1. \] The authors establish coefficient criteria for \(J^\delta_zf\) to be in the class \(S^*(\alpha)\) and in the class \(K(\alpha)\). They also find convolution conditions for concerning \(h(z)\) for \(f\) so that \(J^\delta_z f(z)\in S^*(\alpha)\). A similar result has been proved so that \(J^\delta_z f(z)\in K(\alpha)\). They show for the same \(h\) also \(K\) if \(f\in S^*(\alpha)\), \(J^\delta_z f(z)\in S^*(\alpha)\). A similar result when \(f\in K(\alpha)\) is also received.
scientific article; zbMATH DE number 1174970