Question

Let f and g be real valued functions such that f(x+y)+f(x-y)=2f(x).g(y)$\forall x,yϵR$. Prove that, if f(x) is not identically zero and $\left|f\left(x\right)\right|\le 1\forall xϵR$, then $\left|g\left(y\right)\right|\le 1\forall yϵR$.

Answer 1

Let maximum value of $f\left(x\right)$ be $M$.
$\Rightarrow max|f\left(x\right)|=M,$ where $0<M\le 1$ (i)

(Since, $f$ is not identically zero and $\left|f\left(x\right)\right|\le 1\forall xϵR$)
Now, $f(x+y)+f(x-y)=2f\left(x\right).g\left(y\right)$
$\Rightarrow$ $|2f\left(x\right)|.|g\left(y\right)|=|f(x+y)+f(x-y)|$
$\Rightarrow$ $2\left|f\left(x\right)\right|.\left|g\left(y\right)\right|\le \left|f(x+y)\right|+\left|f(x-y)\right|$ (as $|a+b|\le \left|a\right|+\left|b\right|$)
$\Rightarrow$ $2\left|f\left(x\right)\right|.\left|g\left(y\right)\right|\le M+M$ [Using Eq. (i), ie, $max|f\left(x\right)|=M$]
$\Rightarrow$ $\left|g\left(y\right)\right|\le 1$ for $yϵR$
Thus, if $f\left(x\right)$ is not identically zero and $\left|f\left(x\right)\right|\le 1\forall xϵR$
then, $\left|g\left(y\right)\right|\le 1$ for $yϵR$