ON THE INVERSE POWER INEQUALITY FOR THE BEREZIN NUMBER OF OPERATORS

The Berezin symbol (A) over tilde of operator A acting on the reproducing kernel Hilbert space H =H(Omega) over some set Omega is defined by (A) over tilde(lambda) = < A (K) over cap (H),(lambda),(K) over cap (H),(lambda)>, lambda is an element of Omega, where (K) over cap (H),(lambda) = k(H),(lambda)/parallel to k(H),(lambda)parallel to(H) is the normalized reproducing kernel of H. The Berezin number of operator A is the following number: ber(A): = sup{vertical bar(A) over tilde(lambda)vertical bar: lambda is an element of Omega}. Clearly, ber(A) <= w(A), where w(A) = sup{vertical bar >< Ax,x >vertical bar: x is an element of H, parallel to x parallel to(H) = 1} is the numerical radius of A. The power inequality for the numerical radius of Hilbert space operator A is the following: w(A(n)) <= (w(A))(n), for all n >= 1. Since ber(A) <= w(A), the following question naturally arises: is it true that ber(A(n)) <= (ber(A))(n) for any operator A and any integer n > 1 ? Although we do not solve this question, in this paper, by using some Hardy type inequality, we prove the inverse power inequality for ber(A) for positive operators on H(Omega); namely, we prove that (ber(A))(n) <= C(n,m)ber(A(n)) for any positive operator A on H(Omega), where C(n,m) > 1 is the constant depending only on n and its conjugate m, where 1/n + 1/m = 1.