On the gap between two classes of analytic functions (Q1262968)

The author considers functions analytic and non-zero in the unit disc U with Taylor expansion \(f(z)=1+a_ 1z+... \). Then for \(\alpha\geq 0\), \(\beta\geq 0\), f belongs to the Kaplan class K(\(\alpha\),\(\beta)\) if \[ - \alpha \pi \leq \int^{\theta_ 2}_{\theta_ 1}\{Re(zf'(z)/f(z))- (\alpha -\beta)\}d\theta \leq b\pi, \] where \(z=re^{i\theta}\), \(0<r<1\), \(\theta_ 1<\theta_ 2<\theta_ 1+2\pi\). Alternatively f belongs to the class T(\(\alpha\),\(\beta)\) if the Hadamard product \[ f(z)*\{(1+uz)^{[\alpha]}(1+vz)^{\alpha -[\alpha]}(1-z)^{-\beta})\} \] is not zero in U for \(| u| \leq 1\), \(| v| \leq 1\). It is proved that if \(\beta\geq 1\) and \(\gamma\) is real then \(1+\gamma z+z^ 2\) is in the dual of T(1,\(\beta)\), but not in K(1,\(\beta)\) if and only if 2 cos(\(\pi\) /(1\(+\beta))\leq | \gamma | \leq 2(1-\beta^{- 2})^{1/2}\).