Limit of sequence $a_0=-1$, $a_{n+1}=f(a_n)$

Question

I have to prove convergence and find its limit of sequnce $a_0=-1$, $a_{n+1}=f(a_n)$ for $n \ge 0$ where $f: \mathbb{R} \rightarrow [0; \infty)$ is continous and $f(0)=0$ and $\forall_{x>0}{f(x)<x}$. If $f(-1) = 0$ then $a_n = 0$ for $n >0$ and limit is $0$. When $f(-1) > 0 $ sequnce is decreasing and its infimum is $0$ because of function's set of values so limit is also $0$. Am I right?

Answer 1

I would say yes. Let $a =\lim_{n\to \infty} a_n$. Since $f$ is continous we have $$a= \lim_{n\to \infty} a_{n+1} = \lim_{n\to \infty} f(a_n) = f(\lim_{n\to \infty} a_n) = f(a)$$

So $a$ is fixed point for $f$. If $a>0$ then $f(a)<a$ a contradiction, thus $a=0$.