# Homework Help: Cauchy Sequence

1. Dec 14, 2011

### manooba

Let f : [a,b] → [a,b] satisfy

|f(x)-f(y)| ≤ λ|x-y|

where 0<λ<1. Prove f is continuous. Choose any Xo ε [a,b] and for n ≥ 1 define X_n+1 = f(Xn). Prove that the sequence (Xn) is convergent and that its limit L is a 'fixed point' of f, namely f(L)=L

2. Dec 14, 2011

The continuity part seems pretty straightforward, unless I'm mistaken. Let c be an arbitraty point in [a, b]. Let ε > 0 be given. Let δ = ε. Them for all x such that |x - c| < δ, we have: |f(x) - f(c)| <= ...

3. Dec 14, 2011

### manooba

reckon you can help me further please i am really struggling

4. Dec 14, 2011

### micromass

No. You must make your own efforts on homework problems. We are not here to solve the questions for you. We will help you as soon as you make an attempt.

5. Dec 14, 2011

### Yuqing

Well, you initially begin with
$$\left|x_2 - x_1\right|=\left|f(x_1) - f(x_0)\right|\le \lambda\left|x_1 - x_0\right|$$
if you take one more step down the sequence, you can see that
$$\left|x_3 - x_2\right|=\left|f(x_2) - f(x_1)\right|\le \lambda\left|x_2 - x_1\right|\le \lambda^2\left|x_1 - x_0\right|$$
I don't think it's difficult to see how this generalizes.

Can you see how this implies the sequence is Cauchy? What do we know about Cauchy sequences? Finally, what do we know about the limit of a sequence in a continuous function?

6. Dec 14, 2011

And for the first part, can you continue where I wrote "..."? What does your function satisfy, by defintion?

7. Dec 14, 2011

### manooba

satisfy's |f(x)-f(y)| ≤ λ|x-y) where 0<λ<1

Last edited: Dec 14, 2011
8. Dec 14, 2011

OK. Now, we're looking at all x such that |x - c| < δ, right? So, apply your function property to these x's...

9. Dec 14, 2011

### manooba

ok i get that, so what do i plug in to the x's?

10. Dec 14, 2011