Question

Let $f$ and $g$ be real-valued functions such that
$f(x+y)+f(x-y)=2f\left(x\right)\cdot g\left(y\right)\forall x,yϵR$
if $f$ is not identically zero and $f\left|\right(x\left)\right|\le 1,\forall xϵR,$ then $\left|g\right(y\left)\right|\le 1,\forall yϵR$.
If true enter 1 else enter 0

• A. True
• B. False

Answer 1

The correct option is A True

Let $max\left|f\right(x\left)\right|=M$ where $0<M\le 1$.
(Since, f is not identically zero and $\left|f\right(x\left)\right|\le 1\forall x\in R$)
Now, $f(x+y)+f(x-y)=2f\left(x\right).g\left(y\right)$
$\Rightarrow \left|2f\right(x).g(y\left)\right|=\left|f\right(x+y)+f(x-y\left)\right|$
$\Rightarrow \left|2f\right(x\left)\right|.\left|g\right(y\left)\right|\le \left|f\right(x+y\left)\right|+\left|f\right(x-y\left)\right|\le M+M$
$\Rightarrow 2\left|f\right(x\left)\right|\left|g\right(y\left)\right|\le 2M$
$\Rightarrow \left|g\right(y\left)\right|\le 1$ for $y\in R$