Let $f, g\colon\mathbb{R}\rightarrow\mathbb{R}$ functions so that for all $\,x, y\in\mathbb{R}$, $f(x+y)+f(x-y)=2f(x)g(y).$ Prove that if $f$ is not constant zero function and $|f(x)|\leqslant 1$ for all $x\in\mathbb{R}$ then $|g(x)|\leqslant 1$ for all $x\in\mathbb{R}$.
Exercise of functions of a real variable
-1
$\begingroup$
functions
inequality
functional-equations
1 Answers
1
The condition $|f|\leqslant 1$ can be removed by a homogeneity argument (we actually just need $f$ bounded). We have for all $x,y\in\Bbb R$ that $2|f(x)|\cdot |g(y)|\leqslant 2\sup_{t\in\Bbb R}|f(t)|,$ and taking the supremum over $x\in\Bbb R$, this gives $\sup_{t\in\Bbb R}|f(t)|\cdot |g(y)|\leqslant \sup_{t\in\Bbb R}|f(t)|.$ We can divide by $\sup_{t\in\Bbb R}|f(t)|$ as $f$ does identically vanish.