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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable fucntion with m = sup{|$f'(x)$|| $x$ $\in \mathbb{R}$} < $1$. And let $a_0 \in \mathbb{R}$ and define $a_{n+1} = f(a_n)$ with $n \in \mathbb{N}$. Proof $a_n$ is a cauchy sequence.

We know that every converging sequence in the real numbers is a cauchy sequence. So I think it is easier to show that this sequence is convergent, but I can't seem to find a good way... Because $|f'(x)| < 1$, I think we can conclude that $|f(x) - f(a)| < |x - a|$ or that $|f(a_n) - f(a_{n+1})| < |a_n - a_{n+1}|$ which is the same as $|a_{n+1} - a_{n+2}| < |a_n - a_{n+1}|$. But this is not enough to conclude the sequence converges, because for example 1 + 1/2 + 1/3 + 1/4...

I think showing $a_n$ is bounded doesn't help either because $|a_{n+1} - a_{n+2}| < |a_n - a_{n+1}|$ doesn't imply that $a_n$ is monotonic increasing/decreasing. As you can see, i'm lost here. Am i on the right track or is there something i'm missing?

D.G. van de Schepop
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    You are using $|f'(x)|<1$ when you received a much more generous gift: $|f'(x)|\leq m<1$. Try that idea again but take advantage of the stronger condition. – B. S. Thomson Apr 08 '22 at 22:31

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You have$$|a_3-a_2|=\bigl|f(a_2)-f(a_1)\bigr|\leqslant m|a_2-a_1|.$$Then, you have$$|a_4-a_3|=\bigl|f(a_3)-f(a_2)\bigr|\leqslant m|a_3-a_2|\leqslant m^2|a_2-a_1|.$$After this, you have$$|a_5-a_4|=\bigl|f(a_4)-f(a_3)\bigr|\leqslant m|a_4-a_3|\leqslant m^3|a_2-a_1|.$$And so on. More generally,$$|a_{n+1}-a_n|\leqslant m^{n-1}|a_2-a_1|.$$Since $m<1$, it's not hard now to prove that the sequence $(a_n)_{n\in\Bbb N}$ is a Cauchy sequence.

José Carlos Santos
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