Algebras of convolution type operators with continuous data do not always contain all rank one operators

TL;DR

This paper investigates the structure of convolution operator algebras on Banach function spaces, revealing that for certain non-reflexive spaces like Lorentz spaces, these algebras do not include all rank one operators, contrary to previous assumptions.

Contribution

It demonstrates that in non-reflexive Banach function spaces, the algebra of convolution type operators may fail to contain all rank one operators, extending understanding beyond reflexive spaces.

Findings

Algebra does not always contain all rank one operators in non-reflexive spaces

Lorentz spaces $L^{p,1}( )$ with $1<p< $ are examples where the algebra lacks some rank one operators

Previous results hold only for reflexive spaces, not in general

Abstract

Let $X(\mathbb{R})$ be a separable Banach function space such that the Hardy-Littlewood maximal operator is bounded $X(\mathbb{R})$ and on its associate space $X'(\mathbb{R})$. The algebra $C_X(\dot{\mathbb{R}})$ of continuous Fourier multipliers on $X(\mathbb{R})$ is defined as the closure of the set of continuous functions of bounded variation on $\dot{\mathbb{R}}=\mathbb{R}\cup\{\infty\}$ with respect to the multiplier norm. It was proved by C. Fernandes, Yu. Karlovich and the first author \cite{FKK19} that if the space $X(\mathbb{R})$ is reflexive, then the ideal of compact operators is contained in the Banach algebra $\mathcal{A}_{X(\mathbb{R})}$ generated by all multiplication operators $aI$ by continuous functions $a\in C(\dot{\mathbb{R}})$ and by all Fourier convolution operators $W^0(b)$ with symbols $b\in C_X(\dot{\mathbb{R}})$. We show that there are separable and non-reflexive Banach function spaces $X(\mathbb{R})$ such that the algebra $\mathcal{A}_{X(\mathbb{R})}$ does not contain all rank one operators. In particular, this happens in the case of the Lorentz spaces $L^{p,1}(\mathbb{R})$ with $1<p<\infty$.