Let $a \in \R,$ and let $k(a)$ be the largest constant such that $sup\vert cos(na)-cos(nb)\vert \textless{ } k(a)$ for $b\in \R$ implies that $b \in \pm a+2\pi\Z. $ We show that if a cosine sequence $(C(n))\_{ n\in \Z}$ 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 $n\in \Z.$ Since ${ \sqrt 5\over 2} \le k(a) \le { 8\over 3\sqrt 3}$ for every $a \in \R,$ 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 ${ \sqrt 5\over 2}$ is optimal. We also describe the set of all real numbers $a \in [0,\pi]$ satisfying $k(a)\le { 3\over 2}.$
Categories: math.FA