# There is N so that$$|S_n(x) - S_m(x)| \leq \epsilon$$ for

1. Feb 3, 2006

### Pearce_09

there is N so that
$$|S_n(x) - S_m(x)| \leq \epsilon$$ for ever x in I if n,m $$\geq$$ 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)
..... and im not sure how to show that this converging uniformly on I

Last edited: Feb 3, 2006
2. Feb 3, 2006

### TD

How did you define uniform convergence?

I have that a sequence of functions $s_n \left( x \right)$ converges uniformely to the function $\ell :\left( {a,b} \right) \to \mathbb{R}$ if:

$$\forall \varepsilon > 0,\exists n > N \Rightarrow \forall x \in \left( {a,b} \right):\left| {s_n \left( x \right) - \ell \left( x \right)} \right| < \varepsilon$$

If I understand, you need to show that a sequence of functions $s_n \left( x \right)$ converges uniformely over that interval iff (I believe you only have to prove it one way, using this to prove uniform convergence) it satisfies Cauchy's criterion, being:

$$\forall \varepsilon > 0,\exists N:m,n > N \Rightarrow \forall x \in \left( {a,b} \right):\left| {s_n \left( x \right) - s_m \left( x \right)} \right| < \varepsilon$$ (*)

So assume that $\left( {s_n } \right)$ satisfies (*). Then for all x in I, the numerical sequence $\left( {s_n \left( x \right)} \right)$ is a Cauchy-sequence and is therefore convergent, call the limit l(x). So we already have pointwise convergence to l(x), the only thing you need to show now is that the convergence is uniform. Now use the assumption of Cauchy's criterion.