Functions satisfying $f(x+y)+f(x-y)=2f(x)g(y)$

Question

Given $f,g:\mathbb{R}\longrightarrow\mathbb{R}$ such that $f$ is not the zero function, and $\forall x\in\mathbb{R},\; |f(x)|\leq 1$, and $\forall x,y\in\mathbb{R},\; f(x+y)+f(x-y)=2f(x)g(y)$, can we claim $\forall y\in\mathbb{R},\; |g(y)|\leq 1$ ?... as it would be in the special case $f=\sin$ and $g=\cos$. Many thanks. Edit: title edited, I had forgotten the factor $2$.

Answer 1

Let $M$ be the supremum of $|f(x)|$ on $\Bbb R$. For $0 < \epsilon < M$ choose $x \in \Bbb R$ such that $|f(x)| > M - \epsilon$. Then $$ |g(y)| = \frac{|f(x+y) + f(x-y)|}{2|f(x)|} \le \frac{M}{M - \epsilon} \, . $$ With $\epsilon \to 0$ it follows that $|g(y)| \le 1$.

The bound for $|g|$ is best possible, as the examples $f(x) = 1$, $g(y) = 1$, or $f(x) = \sin(x)$, $g(y) = \cos(y)$ show.

It is also necessary to require that $f$ is bounded, otherwise $f(x) = e^x$, $g(y) = (e^y + e^{-y})/2$ would be a counter-example.