Understanding Uniform Convergence and Cauchy's Criterion for Series

[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 don't know where to go, to say that S_n(x) converges uniformly on I.

 

[JasonRox]

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)| &lt; \epsilon /2 if n\geq N

then,

|S_n(x) - S_m(x)| &lt; |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 don't know where to go, to say that S_n(x) converges uniformly on I.

Have you tried using Triangle Inequalities?

 

[Pearce_09]

no, I am not sure how that will show uniform convergence??

 

[JasonRox]

no, I am not sure how that will show uniform convergence??

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?

 

[Pearce_09]

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

 

[JasonRox]

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

Can you assume that a Cauchy sequence is bounded?

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

 

[Pearce_09]

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

 

[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.

 

[JasonRox]

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.

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.

 

[Pearce_09]

yes, thanks jasonrox...thats exactly what I am trying to do..but unfortunately I can't seem to do.

 

[JasonRox]

yes, thanks jasonrox...thats exactly what I am trying to do..but unfortunately I can't seem to do.

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?

 

[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 won't help this problem

 

[matt grime]

He's trying to prove that if a sequence

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

is a Cauchy Sequence then the sequence converges to some limit L.

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

 

[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.

 

[Pearce_09]

yes, that's pretty much what I am trying to prove.

 

[matt grime]

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