Suppose that $f$ and $g$ satisfy the equation $f(x+y)+f(x-y)=2f(x)g(y)$, $x$,$y$ $\in \mathbb{R}$.
Show that If $f$ is not identically zero and $|f(x)| \le 1$ for $x$ $\in \mathbb{R}$, then also $|g(x)| \le 1$ for $x \in \mathbb{R}$.
This is how I proceeded:
Putting $x=0$ in the functional equation We have $f(y)+f(-y)=2f(0)g(y)$. Taking modulus on both sides $|f(y)+f(-y)|=2|f(0)|g(y)|$ which gives $2|f(0)||g(y)|\le |f(y)|+|f(-y)|\le 2$
$\implies |f(0)||g(y)|\le 1, \forall y \in \mathbb{R}$.
All I need to show is that $f(0) \ge 1$. Putting both $x=y=0$ in the functional equation we have $2f(0)=2f(0)g(0)\implies 2f(0)(1-g(0))=0.$
Now only putting $y=0$ in the equation we have $2f(x)(1-g(0))=0$. Since $f(x)$ is not identically $0$, $g(0)=1$. I am unable to show that $f(0) \ge 1$.
Thanks for the help!!
Let $M=\sup_{\Bbb{R}}|f(t)|$. Since $f$ is not identically zero we conclude that $M>0$.
Consider $\epsilon\in(0,M)$. There is a real $x_\epsilon$ such that $|f(x_\epsilon)|>M-\epsilon$. Now, for a real $y\in\Bbb{R}$ we have $$ 2|g(y)|\cdot |f(x_\epsilon)|=|f(x_\epsilon+y)+f(x_\epsilon-y)|\leq |f(x_\epsilon+y)|+|f(x_\epsilon-y)|\leq 2M $$ Thus $$ 2(M-\epsilon)|g(y)|<2 |g(y)|\cdot |f(x_\epsilon)|\leq 2M $$ or $$ \forall\,y\in\Bbb{R},\quad|g(y)|\leq \frac{M}{M-\epsilon} $$ But $\epsilon\in(0,M)$ is arbitrary, letting $\epsilon$ tend to $0$, we conclude that $|g(y)|\leq1$ for every $y\in\Bbb{R}$.$\qquad\square$