# Homework Help: Contraction Mapping Theorem Question

1. May 3, 2010

### nugget

1. The problem statement, all variables and given/known data

Consider the function g : [0,∞) → R defined by g(x) = x + e−2x.

Given |g(x2) − g(x1)| < |x2 − x1| for all x1, x2 ∈ [0,∞) with x1 ≠ x2.

Is g a contraction on [0,∞)? Why?

2. Relevant equations

I think we are intended to use the given equation and the CMT

CMT states that |f(x)-f(y)|=c|x-y|, where x,y∈R.

Which is similar to |g(x2) − g(x1)|<|x2 − x1|

3. The attempt at a solution

I really don't know where to begin with this; I'm not simply after the answer though, just a prod in the right direction would be great

2. May 3, 2010

### Tedjn

On my computer, the formula for g(x) displays an unreadable symbol before the e-2x. Also, I confess that I have only seen CMT from browsing my friend's analysis book, and it doesn't exactly say
.
Are they equivalent?

EDIT: The actual statement I read is at http://en.wikipedia.org/wiki/Banach_fixed_point_theorem

Last edited: May 3, 2010
3. May 3, 2010

### gauss^2

g is a contraction precisely when there is some positive real number c < 1 with

$$|g(y) - g(x)| < c |y-x|$$

for all x,y in the domain.

You have to either find a positive c < 1 that satisfies the inequality above for all $$x,y \in [0,\infty)$$ or show no positive c < 1 exists which satisfies the inequality. Note that

$$|g(y) - g(x)| < |y-x|$$

does not show g is a contraction. The constant 0 < c < 1 really is needed (check the proof of the contraction mapping theorem to see why if you're interested; it's a really cool piece of mathematics).

Last edited: May 3, 2010
4. May 3, 2010

### HallsofIvy

No, it doesn't. That is simply the definition of "contraction map". The CMT states that if f is a contraction map that take set S into itself, then it has exactly one fixed point in S.