$|a_{n+1} -a_n| \le M\cdot |a_n -a_{n-1}|$. Prove that $a_n$ is convergent when $M=\frac{1}{2}$, and could be divergent when $M=1$

Question

Let $f$ be a function with a bounded derivative in $R$.

Meaning that there exists $M >0 $ , in a way that $|f'(x)|\le M$ to all $x$.

Also, let $(a_n)$ be the sequence defined as: $a_1 = 1, a_{n+1} = f(a_n)$ for every $n \in N$.

1. Prove that $|a_{n+1} -a_n| \le M\cdot |a_n -a_{n-1}|$ for every $n$
2. Prove that $(a_n)$ is convergent when $M=\frac{1}{2}$
3. Show by giving an example, that when $M=1$, it is possible that $(a_n)$ is divergent

My idea:

1. for every $n$, if $a_n = a_{n+1}$, than the statement is trivial. for every $n$ where $a_n \neq a_{n+1}$: $f$ is continuous over $R$, therefore it is continuous over the bounded interval $[a_n, a_{n+1}]$. According to Lagrange's mean value theorem, there exists $c\in (a_n, a_{n+1})$, in a way that: $$M \geq |f'(c)|=|\frac{f(a_{n})-f(a_{n-1})}{a_{n}-a_{n-1}}| = \frac{|f(a_{n})-f(a_{n-1})|}{|a_{n}-a_{n-1}|} = \frac{|a_{n+1}-a_n|}{|a_n-a_{n-1}|} \implies M\cdot|a_n -a_{n-1}|\geq |a_{n+1}|-|a_n|$$

2. Knowing that $M=\frac{1}{2}$, I get that for every $n$, $|a_{n+1} -a_n| \le \frac{1}{2}\cdot |a_n -a_{n-1}|$. Now, let $(t_n)$ be the sequence defined like this: $t_n = a_n-a_{n-1}$. I get that: $0 \leq \frac{|t_{n+1}|}{|t_n|}\leq \frac{1}{2}$, therefore, $\lim_{n\to\infty}t_n=0$. I'm stuck now. Is this a good direction?

3. I'm thinking maybe $f(x) = \sin x$ but don't know how to prove it

I need help with tasks 2,3

Answer 1

1. seems ok!

2. If $|a_{n+1}-a_n|\leq\frac{1}{2}|a_{n}-a_{n-1}|$, then $|a_{n+m}-a_{n+m-1}|\leq(\frac{1}{2})^m|a_{n}-a_{n-1}|$, and thus for all (arbitrarily large) $m$ we have that $|a_{n+m}-a_n|\leq |a_{n+m}-a_{n+m-1}|+\dots+|a_{n+1}-a_{n}|\leq ((\frac{1}{2})^m+\dots+\frac{1}{2})|a_{n}-a_{n-1}|$, and the last term is obviously no more than $|a_{n}-a_{n-1}|\leq (\frac{1}{2})^{n-1}|a_1-a_0|$, which converges to $0$ for $n\to\infty$. Then, $a_n$ is bounded, and thus (by Bolzano-Weirstrass's Theorem) must have a subsequence converging to some $a$; and the whole $a_n$ must then converge to $a$, since for any arbitrarily small $\epsilon>0$ there exists some $m$ such that $|a_m-a|\leq \epsilon$ and $|a_n -a_m|\leq \epsilon$ for all $n>m$, so that $|a_n-a|\leq 2\epsilon$ for all $n>m$.

3. just take $a_{n+1}=a_n+1$.