Özet:

The Berezin transform $\widetilde{A}$ and the Berezin number of an operator $A$ on the reproducing kernel Hilbert space over some set $\Omega$ with normalized reproducing kernel $\widehat{k}_{\lambda}$ are defined, respectively, by $\widetilde{A}(\lambda)=\left\langle {A}\widehat{k}_{\lambda },\widehat{k}_{\lambda}\right\rangle ,\ \lambda\in\Omega$ and $\mathrm{ber}% (A):=\sup_{\lambda\in\Omega}\left\vert \widetilde{A}{(\lambda)}\right\vert .$ A straightforward comparison between these characteristics yields the inequalities $\mathrm{ber}\left( A\right) \leq\frac{1}{2}\left( \left\Vert A\right\Vert _{\mathrm{ber}}+\left\Vert A^{2}\right\Vert _{\mathrm{ber}}% ^{1/2}\right) $. In this paper, we study further inequalities relating them. Namely, we obtained some refinement of Berezin number inequalities involving convex functions. In particular, for $A\in\mathcal{B}\left( \mathcal{H}% \right) $ and $r\geq1$ we show that \[ \mathrm{ber}^{2r}\left( A\right) \leq\frac{1}{4}\left( \left\Vert A^{\ast }A+AA^{\ast}\right\Vert _{\mathrm{ber}}^{r}+\left\Vert A^{\ast}A-AA^{\ast }\right\Vert _{\mathrm{ber}}^{r}\right) +\frac{1}{2}\mathrm{ber}^{r}\left( A^{2}\right) . \]

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