Let $f: \mathbb{R} \to \mathbb{R}$ and $\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ and $\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}$ then find the :

$$\lim_{x \to \infty} f(x)=?$$

My Try : $f^2(x+b)+\frac{1}{4}-f(x+b)=f(x)-f^2(x)$ and let $b=0$ then $f^2(x)+\frac{1}{4}-f(x)=f(x)-f^2(x)$ so $b \neq 0$

now what do i do ?

Answer 1

Rewrite the starting equation like this:

$$\left(f(x+b)-\frac{1}{2}\right)^2={1\over 4}-{1\over 4}+{f(x)-f^2(x)}= {1\over 4}- \left(f(x)-{1\over 2}\right)^2$$

so we have also

$$\left(f(x)-\frac{1}{2}\right)^2= {1\over 4}- \left(f(x-b)-{1\over 2}\right)^2$$

Combining both equations we get $$\left(f(x+b)-\frac{1}{2}\right)^2 = \left(f(x-b)-{1\over 2}\right)^2$$ and so $$\left|f(x+b)-\frac{1}{2}\right| = \left|f(x-b)-{1\over 2}\right|$$

But $f(x)\geq 1/2$ for all $x$ so we have $$f(x+b)-\frac{1}{2} = f(x-b)-{1\over 2}$$

which means that $f$ is periodic with period $2b$ and thus the limit doesn't exist unless $f$ is constant.