Hilbert matrix norms on weighted Bergman spaces: even exponents and a counterexample to the beta formula

Abstract

Let $A_α^p$ be the weighted Bergman space on the unit disk, where $α>-1$. For $f(z)=\sum_{k=0}^{\infty}a_k z^k\in A_α^p$, consider the Hilbert matrix operator $\mathcal{H}f(z)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\frac{a_k}{n+k+1}\right)z^n =\int_0^1\frac{f(t)}{1-tz}\,dt$. For even exponents $p=2m$, we prove that $\|\mathcal{H}\|_{A_α^{2m}\to A_α^{2m}}=B(a,1-a)$, where $a=(α+2)/(2m)$, whenever $0<a\leq m/(2m-1)$. For $p=2,4,6,8,10$, the same formula holds throughout the admissible range. We also show that the beta-function norm formula does not hold for all admissible parameters. Set $a_0=800001/1000000$ and $α_p=a_0p-2$. Then, for every real $p\geq 1100000$, $\|\mathcal{H}\|_{A_{α_p}^p\to A_{α_p}^p}>B(a_0,1-a_0)$. The counterexample is based on the fixed function $f_0(z)=(1-z^2)^{-4/5}=\sum_{k=0}^{\infty}\frac{(4/5)_k}{k!}z^{2k}$. A rigorous interval estimate at $p=1100000$, together with monotonicity in $p$, yields the result on the entire half-line. In particular, the formula fails for every even exponent $p=2m$ with $m\geq 550000$.

Disclosure

“X (4/5)k 2k f0 (z) = (1 − z ) = z , k! k=0 was constructed with the assistance of OpenAI Codex. The authors independently verified the argument and take full responsibility for the final manuscript. References [1] NIST Digital Library of Mathematical Functions, Chapters 5 and 15,”

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8

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2026-08-23T19:06:38.305639Z

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