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The Berezin range of a bounded operator $T$ acting on a reproducing kernelHilbert space $\mathcal{H}$ is the set $\text{Ber}(T)$ := $\{\langleT\hat{k}_{x},\hat{k}_{x} \rangle_{\mathcal{H}} : x \in X\}$, where$\hat{k}_{x}$ is the normalized reproducing kernel for $\mathcal{H}$ at $x \inX$. In general, the Berezin range of an operator is not convex. In this paper,we discuss the convexity of range of the Berezin transforms. We characterizethe convexity of the Berezin range for a class of composition operators actingon the Hardy space and the Bergman space of the unit disk. Also for so-calledsuperquadratic functions, we prove the Berezin set mapping theorem for positiveself-adjoint operators $A$ on the reproducing kernel Hilbert space$\mathcal{H}(\Omega)$, namely we prove that$f(\mathrm{Ber}(\Phi(A)))=\mathrm{Ber}(\Phi(f(A)))$, where$\Phi:\mathcal{B}%\left( \mathcal{H}\left( \Omega\right) \right)\mathcal{\rightarrow}\mathcal{B}\left( \mathcal{K(}Q\mathcal{)}\right) $ is anormalized positive linear map.

Composition operators, convexity of their Berezin range and related questions