BOUNDEDNESS OF OPERATORS ON HARDY SPACES

Abstract

In [1], the author provided an example which shows that there is a linear functional bounded uniformly on all atoms in $H^1(\mathbb{R}^n)$, and it can not be extended to a bounded functional on $H^1(\mathbb{R}^n)$. In this note, we first give a new atomic decomposition, where the decomposition converges in $L^2(\mathbb{R}^n)$ rather than only in the distribution sense. Then using this decomposition, we prove that for $0 \lt p \leq 1$, $T$ is a linear operator which is bounded on $L^{2}(\mathbb{R}^n)$, then $T$ can be extended to a bounded operator from $H^{p}(\mathbb{R}^n)$ to $L^{p}(\mathbb{R})$ if and only if $T$ is bounded uniformly on all $(p,2)$-atoms in $L^{p}(\mathbb{R}^n)$. A similar result from $H^{p}(\mathbb{R}^n)$ to $H^{p}(\mathbb{R}^n)$ is also obtained. These results still hold for the product Hardy space and Hardy space on spaces of homogeneous type.