Problem
Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that its derivative $f'$ is a continuous function. Moreover, assume that for all $x\in\mathbb{R}$, $$0\leqslant \vert f'(x)\vert\leqslant \frac{1}{2}$$Define a sequence of real numbers $\{a_n\}_{n\in\mathbb{N}}$ by :$$a_1=1~~\text{and}~~a_{n+1}=f(a_n)~\text{for all}~n\in\mathbb{N}$$Prove that there exists a positive real number $M$ such that for all $n\in\mathbb{N}$, $$\vert a_n\vert \leqslant M$$
My Proof
Notice that $$|a_{n}-a_{n-1}|=|f(a_{n-1})-f(a_{n-2})|=|f'(\xi)||a_{n-1}-a_{n-2}|\leq \frac{1}{2}|a_{n-1}-a_{n-2}|,$$where $n=3,4,\cdots$, and $a_{n-1} \gtrless \xi \gtrless a_{n-2}.$ Hence, $$|a_{n}-a_{n-1}|\leq \frac{1}{2}|a_{n-1}-a_{n-2}|\leq \frac{1}{2^2}|a_{n-2}-a_{n-3}|\leq \cdots \leq \frac{1}{2^{n-2}}|a_2-a_1|.$$Therefore,\begin{align*}|a_n-a_1|&=|(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\cdots+(a_2-a_1)|\\&\leq |a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+\cdots+|a_2-a_1|\\&\leq \left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^{n-2}}\right)|a_2-a_1|\\&\leq 2|a_2-a_1|.\end{align*}Hence, $$|a_n|\leq 2|a_2-a_1|+|a_1|$$for $n=3,4,\cdots$. But we can verify that it also holds for $n=1,2$. Now, take $M=2|a_2-a_1|+|a_1|$. We obtain $$|a_n|\leq M$$ for $n=1,2,\cdots$.This is exactly what we want to prove.