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Let $f$ be differentiable on $\mathbb R$ with $a=\sup \{|f'(x)| :x\in \mathbb R\}<1$ Select $S_0\in \mathbb R$ and define $S_n=f(S_{n-1})$ for $n \geq 1.$ Prove that $\{S_n\}$ is a convergent sequence.

My attempt

I know that I might have to show that $S_n$ is cauchy since it's convergent. By definition $S_n$ is cauchy if there exists a natural number $N$ such that $m,n>N$ implies $|X_m - X_n| < \epsilon$.

I was given a hint that we have to use the IVT to show that it is cauchy. The professor wrote on the board:

By IVT,

$$|S_{n+1}-S_n|=|f(S_n)-f(S_{n-1})|$$

I understand this step but I don't understand the next few steps

$$ = f'(c)|S_n - S_{n-1}|<a|S_n-S_{n-1}| \\ < a^2 |S_{n-1}-S_{n-2}| \\ a^n|S_1 - S_0|$$

Nigel Overmars
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P_Drach
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1 Answers1

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It's not actually an application of IVT, rather MVT.

Since $f$ is differentiable on all of $\mathbb{R}$, then $$ \frac{f(b)-f(a)}{b-a}= f'(c),\qquad c\in[a, b] $$ (with $a\neq b$). Hence

$$ 0 \le \frac{|f(S_n)-f(S_{n-1})|}{|S_n-S_{n-1}|} = |f'(c)| < a $$

Rearranging gives the form your professor wrote. The rest follows by induction on $n$.

user3002473
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