# Homework Help: Analysis + Cauchy's criterion

1. Feb 4, 2006

### Pearce_09

there is N so that
$$|S_n(x) - S_m(x)| \leq \epsilon$$ for ever x in I if n,m N

( prove by cauchy's criterion )

claim: $$lim S_n(x) = S(x)$$

$$|S_n(x) - S(x)| < \epsilon /2$$ if n$$\geq N$$

then,

$$|S_n(x) - S_m(x)| < |S_n(x) - S(x)| + |S(x) - S_m(x)|$$
< $$\epsilon /2 + \epsilon /2$$
< $$\epsilon$$

therefor the series converges pointwise to a funtion S(x)
im stuck here, I dont know where to go, to say that $$S_n(x)$$ converges uniformly on I.

2. Feb 4, 2006

### JasonRox

Have you tried using Triangle Inequalities?

3. Feb 4, 2006

### Pearce_09

no, im not sure how that will show uniform convergence??

4. Feb 4, 2006

### JasonRox

You are working with inequalities. It is your job it is to show that there is uniform convergence. The Triangle Inequality is a really good tool, especially with limits.

It's your job to deduce this fact.

First are you proving Cauchy's Criterion implies that the sequence is convergent?

5. Feb 4, 2006

### Pearce_09

yes i used Cauchy's Criterion to show that it converges pointwise

6. Feb 4, 2006

### JasonRox

Can you assume that a Cauchy sequence is bounded?

If not, try proving that first. It would be a great tool to use.

7. Feb 4, 2006

### Pearce_09

bounded eh, well ill try that.. ...

8. Feb 5, 2006

### matt grime

What is I, what are the S_n? One presumes I is a compact interval, probably [0,1], and that S_n are continuous functions.

What exactly are you trying to prove?

As far as I can tell what you wrote states that the S_n converge uniformly.

I find it impossible to deduce what you've been given and what you're asked to prove.

9. Feb 5, 2006

### JasonRox

He's trying to prove that if a sequence is a Cauchy Sequence then the sequence converges to some limit L.

It's definitely possible, but like you said, probably not with the stuff he's been given.

10. Feb 5, 2006

### Pearce_09

yes, thanks jasonrox...thats exactly what im trying to do..but unfortunetly I cant seem to do.

11. Feb 5, 2006

### JasonRox

If you haven't proved that it is bounded yet, ignore that. Just move on with the assumption that it is bounded, prove that it is later.

So, what do you know about bounded sequences?

12. Feb 5, 2006

### Pearce_09

well i know that every bounded sequence in the Reals has a convergent subsequence....Also every Cauchy sequence In the Reals conveges.
and some stuff about Reimman measurable/measure ... which wont help this problem

13. Feb 5, 2006

### matt grime

a sequence of what? Functions, we are to assume, I imagine, and continuous ones, probably. It would be nice for that to be stated.

that is impossible to do since it has not been stated in what space are looking at this Cauchy sequence.

14. Feb 5, 2006

### matt grime

Let me state what I think the question appears to be:

let S_n be a cauchy sequence in the sup norm on C([0,1]), prove that S_n converges to a continuous function.

15. Feb 5, 2006

### Pearce_09

yes, thats pretty much what im trying to prove.

16. Feb 5, 2006

### matt grime

Good, but what I wrote bears only passing relation to what you actually stated.