- #1

joshmccraney

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## Homework Statement

Let ##V## be a Banach space. Let ##f:V\to V## and ##g:V\to V## be two ##q##-contracting maps, ##q\in(0,1)##. Assume they are uniformly close to each other. Show the distance between fixed points of ##f,g## is at most ##\epsilon/(1-q)##.

## Homework Equations

Definitions:

Uniformly close implies ##\forall \, \epsilon>0, v\in V## we have ##\| f(v) - g(v) \| < \epsilon##.

A map ##f## is ##q##-contracting if ##\exists \, q \in (0,1):\forall x,y\in\mathbb R^n, \| f(x) - f(y) \| \leq q\| x-y\|##.

## The Attempt at a Solution

My idea is to iterate one map starting at the fixed point of the other map and show after sufficient amount of iterations, you will be within ##\epsilon## of the other fixed point. Since we are proving the distance between fixed points, the Banach Fixed Point Theorem seems relevant: $$\| x-x_n \| \leq \frac{q^n}{1-q}\| x_1-x_0 \|$$ where ##x## is the fixed point and ##x_n## is the ##n##th iteration.

So trying to put all this together, if ##x_f## is the fixed point of ##f##, then consider

$$\| g(x_f) - f(x_f) \| = \| g(x_f) - x_f \| \implies\\

\| g(x_f) - x_f \| < \epsilon

$$

But this can't be right. Can someone help me smooth it out?