parmenides51 wrote: Let $f$ and $g$ be real-valued functions defined for all real numbers $x$ and $a$, and $s$, $m$ be some positive constants, such that $f$, $g$ satisfy the equations 1. $f(x + a) + f(x - a) = \frac{2f(x)g(a)}{s}$ 2. $|f(x)|\le m$ for all $x$, $a$. Prove that if $|f|$ is not identically zero, and attains a maximum value, then $|g(a)| \le s$ for all $a$. Let $x$ be such that $|f(x)|$ is maximal. Note that $|f(x)|> 0$. Hence, we have for all $a \in \mathbb{R}$ that $$\frac{|g(a)|}{s} = \frac{|f(x+a)+f(x-a)|}{2 |f(x)|} \le \frac{|f(x+a)|+|f(x-a)|}{2 |f(x)|} \le \frac{2|f(x)|}{2|f(x)|}=1,$$as desired.