Hardy-Hilbert's inequality and power inequalities for Berezin numbers of operators (Q2830468)

The main goal of this article is to answer the question: Does there exists a constant \(C=C(n)>1\) such that the power inequality for the Berezin number NEWLINE\[CARRIAGE_RETURNNEWLINE\operatorname{ber}(A^n)\leq C (\operatorname{ber}(A))^n, \quad n>1,\tag{1}CARRIAGE_RETURNNEWLINE\]NEWLINE holds?NEWLINENEWLINEThe authors prove here that (1) holds for \(n=2\) for selfadjoint operators whose spectra are contained in a real interval \(I\).NEWLINENEWLINEThe main result depends mainly on the following lemma:NEWLINENEWLINELet \(f,g\) be two non-negative continuous functions on an interval \(J\subseteq (0,\infty)\). If \(p,q>1\) such that \(\frac{1}{p}+\frac{1}{q}=1\), then NEWLINE\[CARRIAGE_RETURNNEWLINE\begin{aligned} \frac{1}{2}\widetilde{f( A)g(A)}(\lambda) + \frac{1}{3}\widetilde{f( A )}( \lambda)\widetilde{g( B)}( \mu) + \frac{1}{3}\widetilde{f( A)}( \mu )\widetilde{g( B)}( \lambda) + \frac{1}{4}\widetilde{f(B )g( B )}( \mu) \\ \leq \pi \csc\left( \frac{\pi }{p} \right)\left[(f^p ( A) + f^p( B ))^{\frac{1}{p}}(g^q( A) + g^q ( B) )^{\frac{1}{q}}\right]^{\raisebox{-0.9ex}{\~{}}} ( \lambda)\end{aligned}CARRIAGE_RETURNNEWLINE\]NEWLINE for any selfadjoint operators \(A,B \in \mathcal{B}( \mathscr{H})\) with spectra contained in \(J\) and all \(\lambda, \mu \in \Omega\).NEWLINENEWLINEThe proof of this result follows by applying 1.1 under the assumption that \(a_n=0=b_n\) for all \(n\geq3\); i.e., NEWLINE\[CARRIAGE_RETURNNEWLINE\begin{aligned} \frac{{a_1 b_1 }}{2} + \frac{{a_1 b_2 }}{3} + \frac{{a_2 b_1 }}{3} + \frac{{a_2 b_2 }}{4} \leq \pi \csc \left( {\frac{\pi }{p}} \right)\left( {a_1^p + a_2^p } \right)^{\frac{1}{p}} \left( {b_1^q + b_2^q } \right)^{\frac{1}{q}}. \end{aligned}CARRIAGE_RETURNNEWLINE\]NEWLINE Since \(f\) and \(g\) are continuous positive functions on \(J\), by setting \(a_1=f(x)\), \(a_2=f(y)\), \(b_1=g(x)\) and \(b_2=g(y)\), \(x,y\in J\), the result follows by applying the continuous functional calculus and employing the Berezin symbol \(\widetilde{A}( \lambda ) = \langle {A\widehat{k}_\lambda ,\widehat{k}_\lambda } \rangle \), where \(\widehat{k}_\lambda\) is the normalized reproducing kernel of \(\mathscr{H}\).NEWLINENEWLINELet \(f\) be a non-negative continuous function defined on \(J\). For any positive operator \(A\in \mathcal{B}( \mathscr{H})\) whose spectrum is contained in \(J\), NEWLINE\[CARRIAGE_RETURNNEWLINE\begin{aligned} (\operatorname{ber}(f(A)))^2 \leq \left(3\pi \csc\left(\frac{\pi}{p}\right)-\frac{9}{8}\right)\operatorname{ber}(f^2(A)).\end{aligned}CARRIAGE_RETURNNEWLINE\]NEWLINE In particular, NEWLINE\[CARRIAGE_RETURNNEWLINE\begin{aligned} (\operatorname{ber}(f(A)))^2 \leq \left(3\pi -\frac{9}{8}\right)\operatorname{ber}(f^2(A)).\end{aligned}\tag{2}CARRIAGE_RETURNNEWLINE\]NEWLINE The proof follows by replacing \(B\) by \(A\), \(\mu\) by \(\lambda\) in the above lemma, and using the fact that \(\frac{1}{p}+\frac{1}{q}=1\); a~little manipulation gives the required result.NEWLINENEWLINEThe main result in [loc.\,cit.] yields a sharper estimate than (2), as follows:NEWLINENEWLINELet \(f\) be non-negative continuous function defined on \(J\). For any positive operator \(A\in \mathcal{B}( \mathscr{H})\) whose spectrum is contained in \(J\), NEWLINE\[CARRIAGE_RETURNNEWLINE\begin{aligned} (\operatorname{ber}(f(A)))^2 \leq \left( \frac{{16\pi ^2 - 77}}{{72}}\right)\operatorname{ber}(f^2(A)).\end{aligned}CARRIAGE_RETURNNEWLINE\]NEWLINE The proof of this result follows by applying 1.2 for \(p=2 \); indeed, NEWLINE\[CARRIAGE_RETURNNEWLINE\begin{aligned} \left( {\frac{{a_1 }}{2} + \frac{{a_2 }}{3}} \right)^2 + \left( {\frac{{a_1 }}{3} + \frac{{a_2 }}{4}} \right)^2 \leq \pi ^2 \left( {a_1^2 + a_2^2 } \right).\end{aligned}CARRIAGE_RETURNNEWLINE\]NEWLINE Since \(f\) is a continuous positive function on \(J\), by setting \(a_1=(x)\) and \(a_2=f(y)\), \(x,y\in J\), the result follows by applying the continuous functional calculus and employing the Berezin symbol \(\widetilde{A}( \lambda ) = \langle {A\widehat{k}_\lambda ,\widehat{k}_\lambda } \rangle \), where \(\widehat{k}_\lambda\) is the normalized reproducing kernel of \(\mathscr{H}\).