Bohr operator on analytic functions (1912.11787v1)

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 $s_k(f) = a_0 + a_1 z + \cdots + a_kz^k$ 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.