A zero-sqrt(5)/ 2 law for cosine families

Abstract

Let $a\in \mathbb{R},$ and let $k(a)$ be the largest constant such that $sup\vert cos(na)-cos(nb)\vert \textless{} k(a)$ for $b\in \mathbb{R}$ implies that $b\in \pm a+2\pi \mathbb{Z}.$ We show that if a cosine sequence $(C(n))\_n\in \mathbb{Z}$ with values in a Banach algebra $A$ satisfies $sup\_{n\ge 1}\Vert C(n) -cos(na).1\_A\Vert \textless{} k(a),$ then $C(n)=cos(na)$ for $n\in \mathbb{Z}.$ Since $\frac{\sqrt{5}}{2}\le k(a)\le \frac{8}{3\sqrt{3}}$ for every $a\in \mathbb{R},$ this shows that if some cosine family $(C(g))\_g\in G$ over an abelian group $G$ in a Banach algebra satisfies $sup\_{g\in G}\Vert C(g)-c(g)\Vert \textless{} {\sqrt 5\over 2}$ for some scalar cosine family $(c(g))\_g\in G,$ then $C(g)=c(g)$ for $g\in G,$ and the constant $\frac{\sqrt{5}}{2}$ is optimal. We also describe the set of all real numbers $a\in [0,\pi ]$ satisfying $k(a)\le \frac{3}{2}.$