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A Compact Counterexample to Su\'arez's Berezin Approximation Question

Sep 2026 · 0 citations · 6 references
Mathematics

Abstract

Let $A^2(\mathbb D)$ be the unweighted Bergman space and write $Q_m(S)=T_{B_m(S)}$ for the map induced by the $m$th higher-order Berezin transform. Su\'arez asked in 2005 whether $Q_m(S)$ converges to $S$ in operator norm for every $S$ in the full Bergman Toeplitz algebra. We answer this question negatively in a strong form: there is a compact operator $S$ such that \[ \sup_{m\geq 0}\|Q_m(S)\|=\infty. \] In particular, along a strictly increasing sequence $(m_n)$ one has \[ \|Q_{m_n}(S)-S\|\longrightarrow\infty. \] The obstruction is a moving matrix edge. The proof uses a moving family of rank-one test operators, each supported in the $m$th matrix column. Near the corresponding edge, these test operators are sent to weighted Hankel matrices, and the limiting coefficients form an explicit Pascal kernel. We prove a standalone Pascal--Hankel theorem showing that the associated weighted Hankel transformation fails to map $\ell^2$ boundedly into trace class. Finite-section convergence, trace duality, and the Uniform Boundedness Principle then transfer this instability to the exact Bergman maps.

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