# Convergent Sequences and Functions

1. Feb 7, 2012

### zbr

Hello all, I am having trouble with a convergent series problem.

The problem statement:

Let f:ℝ→ℝ be a function such that there exists a constant 0<c<1 for which:
|f(x)-f(y)| ≤c|x-y|
for every x,y in ℝ. Prove that there exists a unique a in ℝ such that f(a) = a.

There is a provided hint:
Consider a sequence {x(n)} defined recurrently by x(n+1) = f(x(n)). Prove that it converges and its limit a satisfies f(a) = a.

My confusion is how to show that the recursive function converges and satisfies the required limit. Should I use a specific function f?

I am fairly certain that we need to show that this sequence (defined in the hint section) is Cauchy. This is what I have so far:

Consider the sequence:
x1 = f(0)
x2 = f(x1)
x3 = f(x2)
...
x(n+1) = f(xn)

Now let us take |xn-xm|:
|xn-xm| = (f(x(n-1)) - f(x(m-1))) < c|x(n-1) - x(m-1)|

However, here I don't know how to pick N sch that for every e>0:
c|x(n-1) - x(m-1)|<e for every n,m>= N (i.e. the definition of a cauchy sequence)

Thanks in advance for all help!

Last edited: Feb 7, 2012
2. Feb 7, 2012

### Dick

$| x_2 - x_1 | < c | x_1 - x_0 |$. $| x_3 - x_2 | < c^2 | x_1 - x_0|$. $| x_4 - x_3 | < c^3 | x_1 - x_0 |$. Etc etc. Take that for a hint and see what you can do with it. Once you've show it's Cauchy and defines a fixed point prove there aren't two fixed points.

Last edited: Feb 7, 2012