Absolutely convergent Fourier series. An improvement of the Beurling--Helson theorem

TL;DR

This paper improves the Beurling--Helson theorem by showing that if the Fourier coefficients of a continuous function on the circle decay slightly faster than a certain logarithmic rate, then the function must be linear.

Contribution

It establishes a new, weaker growth condition under which a continuous function's phase must be linear, refining previous results in harmonic analysis.

Findings

Proves linearity of phase under a new logarithmic decay condition

Extends the Beurling--Helson theorem with a weaker assumption

Shows the decay rate involving iterated logarithms implies linearity

Abstract

We consider the space $A(\mathbb T)$ of all continuous functions $f$ on the circle $\mathbb T$ such that the sequence of Fourier coefficients $\hat{f}=\{\hat{f}(k), ~k \in \mathbb Z\}$ belongs to $l^1(\mathbb Z)$. The norm on $A(\mathbb T)$ is defined by $\|f\|_{A(\mathbb T)}=\|\hat{f}\|_{l^1(\mathbb Z)}$. According to the known Beurling--Helson theorem, if $\phi : \mathbb T\rightarrow\mathbb T$ is a continuous mapping such that $\|e^{in\phi}\|_{A(\mathbb T)}=O(1), ~n\in\mathbb Z,$ then $\phi$ is linear. It was conjectured by Kahane that the same conclusion about $\phi$ is true under the assumption that $\|e^{in\phi}\|_{A(\mathbb T)}=o(\log |n|)$. We show that if $\|e^{in\phi}\|_{A(\mathbb T)}=o((\log\log |n|/\log\log\log |n|)^{1/12})$ then $\phi$ is linear.