Let $f$ and $g$ be real-valued functions defined for all real values of $x$ and $y$, and satisfying the equation $f(x + y) + f(x − y) = 2f(x)g(y)$ for all $x$, $y$.
Is it true that if $f(x)$ is not identically zero, and if $|f(x)| ≤ 1$ for all $x$, then $|g(y)| ≤ 1$ for all $y$?
Take $x_n$ such that $ |f(x_n)| \geq \sup |f(x)| - \epsilon$.
Then set $x=x_n$ and obtain that
\begin{align*} |g(y)|& =| \frac{\frac{ f(x_n+y) +f(x_n -y) }{f(x_n)}}{2}|\\ &\leq \frac{|f(x_n + y)|}{2|f(x_n)|}+\frac{|f(x_n - y)|}{2|f(x_n)|}\\ &\leq \frac{\sup |f(x)|}{2(\sup |f(x)| - \epsilon)}+\frac{\sup |f(x)|}{2(\sup |f(x)| - \epsilon)}\\ &=\frac{\sup |f(x)|}{(\sup |f(x)| - \epsilon)} \end{align*} Now taking $\epsilon \rightarrow 0$ yields the result.
