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Let $f: \Bbb{R} \rightarrow \Bbb{R}$ be a differentiable function , and suppose that there is a constant $A<1$ such that $|f'(t)|\le A$ for all real $t$. Define a sequence $\{x_n\}$ by $ $$$x_{n+1}=\frac{2x_n+3f(x_n)}{5}$$

Prove that the sequence $\{x_n\}$ is convergent and that its limit is the unique fixed point of $f$

I tried using the Banach fixed point theorem but it doesn't seem to apply here

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Here is a related problem. –  Mhenni Benghorbal Jan 15 '13 at 5:44
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1 Answer

up vote 4 down vote accepted

By the mean value theorem, for any $u, v \in \Bbb R$, we can find $t \in (u, v)$ so that: $$ \frac{f(u) - f(v)}{u - v} = f'(t) $$

Given that $\left|f'(t)\right| \le A < 1$, it follows that: $$ \left|f(u) - f(v)\right| \le A |u - v| $$

Hence, $f$ is a contraction.

Now define $g(t) = \dfrac{2t + 3f(t)}{5}$. By a simple calculation we find that: $$ \left|g(u) - g(v)\right| = \left|\dfrac{2(u-v) + 3\left(f(u) - f(v)\right)}{5}\right| \le \dfrac{2 + 3A}{5} |u - v| $$

Since $A < 1$, $\dfrac{2 + 3A}{5} < 1$ and $g$ is also a contraction.

Now apply the Banach fixed point theorem to $x_{n+1} = g(x_n)$ to arrive at the desired result.

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