Real analysis solution

Question

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 solution: by integrating as we get $$-x/2 +f(0) \leq f(x) \leq x/2 +f(0)$$ WLOG assume $f(0)=0$, then , then $$ -1/4 \leq f(1)/2 \leq 1/4$$ and $$ -1/8 \leq f(f(1))/2 \leq 1/8$$ and so on... Inductively, we get $$-1/2^n \leq f^n (1)/2 \leq 1/2^n$$. Thus we can set $M=1/2$, which will be the bound. My concerns regarding this proof:

Is taking WLOG $f(0)=0$ correct here?
The other solutions I saw took the bound to be $M=2|a_2-a_1|+|a_1|$, not what I took. Considering my concerns, is this correct?

You would need to compare your answer, with $f(x)=g(x)-g(0)$ to the more general function $g(x)$. — Michael Burr, Dec 27 '23 at 12:21
Right, so you it looks like you want to do this without quoting the Banach fixed-point theorem, which would be the easiest way to go. — Adam Rubinson, Dec 27 '23 at 15:04

score 0 · Answer 1 · answered Dec 27 '23 at 12:16

0

Hint: By Mean Value Theorem $|a_{n+1}-a_n|\leq \frac 1 2 |a_{n}-a_{n-1}|$. This gives $|a_{n+1}-a_n|\leq \frac 1 {2^{n-1}}|a_2-a_1|$. This in turn gives $$|a_{n+1}-a_1|$$ $$\le |a_{n+1}-a_n|+|a_n-a_{n-1}|+...+|a_2-a_1|$$ $$\le (\frac 1 2+\frac 1 {2^{2}}+...+\frac 1{2^{n-1}}) |a_2-a_1|$$$$\le |a_2-a_1|.$$

answered Dec 27 '23 at 12:16

geetha290krm

36,632

Is it sufficient to assume $f'(x)<1$, or is the best we can do $f'(x)\leq a < 1$? – Carlyle Dec 27 '23 at 13:08

Real analysis solution

1 Answers1