Bohr operator on analytic functions

Abstract

For $f(z)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and a fixed $z$ in the unit disk, $|z|=r,$ the Bohr operator ${\mathcal{M}}_r$ is given by \[\mathcal{M}_r (f) = \sum_{n=0}^{\infty} |a_n| |z^n| = \sum_{n=0}^{\infty} |a_n| r^n.\] This papers develops normed theoretic approaches on ${\mathcal{M}}_r$. Using earlier results of Bohr and Rogosinski, the following results are readily established: if $f(z)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ is subordinate (or quasi-subordinate) to $h(z)={\sum }_{n=0}^{\infty }{b}_n{z}^n$ in the unit disk, then \[\mathcal{M}_{r}(f) \leq \mathcal{M}_{r}(h), \quad 0 \leq r \leq 1/3,\] that is, \[\sum_{n=0}^{\infty} \ | a_{n}\ | |z|^{n} \leq \sum_{n=0}^{\infty} \ | b_{n}\ |t |z|^{n}, \quad 0 \leq |z| \leq 1/3. \] Further, each $k$-th section $s_k(f)=a_0+a_{1}z+\cdots +a_k{z}^k$ satisfies \[\ | s_k(f)\ | \leq \mathcal{M}_r \ ( s_k(h)\ ), \quad 0 \leq r \leq 1/2,\] and \[\mathcal{M}_{r}\ ( s_{k}(f) \ ) \leq \mathcal{M}_{r}(s_{k}(h)), \quad 0 \leq r \leq 1/3.\] A von Neumann-type inequality is also obtained for the class consisting of Schwarz functions in the unit disk.

8 pages

References in corpus (2)

• On the Bohr inequality
• Bohr--Rogosinski radius for analytic functions