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

by P. Fazioli   Last Updated September 21, 2018 22:20 PM

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.

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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 $$|g(y)| \le 1$$.