English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

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 papersdevelops normed theoretic approaches on $\mathcal{M}_r$. Using earlier resultsof 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\[\mathcal{M}_{r}(f) \leq \mathcal{M}_{r}(h), \quad 0 \leq r \leq 1/3,\] thatis, \[\sum_{n=0}^{\infty} \ | a_{n}\ | |z|^{n} \leq \sum_{n=0}^{\infty} \ |b_{n}\ |t |z|^{n}, \quad 0 \leq |z| \leq 1/3. \] 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 vonNeumann-type inequality is also obtained for the class consisting of Schwarzfunctions in the unit disk.

Bohr operator on analytic functions