Question

If $f\left(x\right)ϵ[1,2]$ when $xϵR$ and for a fixed positive real number $p$,
$f(x+p)=1+\sqrt{2f\left(x\right)-|f\left(x\right){|}^2}$ for all $xϵR$ then prove that $f\left(x\right)$ is a periodic function.

Answer 1

$|f\left(x\right){|}^2=\left(f\right(x){)}^2$

So, $f(x+p)=1+\sqrt{2f\left(x\right)-\left(f\right(x){)}^2}$

$\Rightarrow f(x+2p)=1+\sqrt{2f(x+p)-\left(f\right(x+p){)}^2}=1+\sqrt{2(1+\sqrt{2f\left(x\right)-\left(f\right(x){)}^2})-(1+\sqrt{2f\left(x\right)-\left(f\right(x){)}^2}{)}^2}$

$\Rightarrow f(x+2p)=1+\sqrt{2+2\sqrt{2f\left(x\right)-\left(f\right(x){)}^2}-(1+(\sqrt{2f\left(x\right)-\left(f\right(x){)}^2}{)}^2+2\sqrt{2f\left(x\right)-\left(f\right(x){)}^2})}$

$\Rightarrow f(x+2p)=1+\sqrt{1-2f\left(x\right)+\left(f\right(x{)}^2)}=1+\sqrt{\left(f\right(x)-1{)}^2}=1+|f\left(x\right)-1|$

As $f\left(x\right)\in [1,2]$, $|f\left(x\right)-1|=f\left(x\right)-1$

ie, $f(x+2p)=f\left(x\right)$

So, $f\left(x\right)$ is a periodic function with a period of $2p$