# Uniform convergence

1. Mar 26, 2013

### Zondrina

1. The problem statement, all variables and given/known data

Suppose that $s_n(x)$ converges uniformly to $s(x)$ on $[b, ∞)$.

If $lim_{x→∞} s_n(x) = a_n$ for each n and $lim_{n→∞} a_n = a$ prove that :

$lim_{x→∞} s(x) = a$

2. Relevant equations

$\space ε/N$

3. The attempt at a solution

I see a quick way to do this one using first principle definitions.

$\forall ε>0, \exists N(ε) \space | \space n > N(ε) \Rightarrow |s_n(x) - s(x)| < ε/3, \space \forall x \in [b, ∞)$

$\forall ε>0, \exists N_1 \space | \space x > N_1 \Rightarrow |s_n(x) - a_n| < ε/3$

$\forall ε>0, \exists N_2 \space | \space n > N_2 \Rightarrow |a_n - a| < ε/3$

We want to prove :

$\forall ε>0, \exists N \space | \space x > N \Rightarrow |s(x) - a| < ε$

So :

$|s(x) - a| = |s(x) - s_n(x) + s_n(x) - a_n + a_n - a| ≤ |s_n(x) - s(x)| + |s_n(x) - a_n| + |a_n - a| < ε/3 + ε/3 + ε/3 = ε$

Does this look okay?

2. Mar 26, 2013

### Dick

Almost. You should have an N1, N2, N3 in the first three statements. Then you just want to say how your final N is related to those three N's.

3. Mar 26, 2013

### Zondrina

Ah, I see. I thought being explicit about $N(ε)$ would make the context clear.

So I should also say choosing $N = max\{N_1, N_2, N_3\}$?

4. Mar 26, 2013

### Dick

Yes, that's it. But now that I've stared at it for a bit there is a second problem. The $N_2$ is also going to depend on n. There's a bit more work to do. You have to choose an $N_2$ independent of n.

Last edited: Mar 26, 2013