# Uniform Convergence

1. Sep 3, 2010

### Enjoicube

1. The problem statement, all variables and given/known data
Alright, here is the problem. Given a compact metric space $$X$$, and a sequence of functions fn which are continuous and $$f_{n}:X->R$$ (reals), also $$f_n->f$$ (where f is an arbitrary function $$f:X->R$$). Also, given any convergent sequence in $$X$$ $$x_{n}->x$$, $$f_{n}(x_{n})->f(x)$$. The problem is to show that fn converges uniformly to f.

3. The attempt at a solution
Alright, I can prove this relatively easily if I can prove that f (the limit function) is continuous. However, I don't know if this is possible, does anyone see a way to do this? Only little hints if you see a way.

Last edited: Sep 4, 2010
2. Sep 5, 2010

### Enjoicube

Nevermind continuity, I don't think the continuity of f directly matters in this situation. However, maybe a start on the problem would be to assume that f is not uniformly continuous, and then, picking an epsilon e>0, for each n, pick one x in X such that |fn(x)-f(x)|>=e, and so create a sequence of points in X, which then must have a convergent subsequence due to the compactness of X. From there on, I am stumped, even if this is the correct way to solve the problem.

3. Sep 5, 2010

### Office_Shredder

Staff Emeritus
I would think proof by contradiction is a good way to go. If the fn do not converge to f uniformly, then for all $$\epsilon$$, there exists an infinite subsequence of functions $$f_{n_k}$$ such that for each such function, there is a point $$x_{n_k}$$ with $$|f(x_{n_k})-f_{n_k}(x_{n_k})|>\epsilon$$

Relabel these sequences as $$f_k$$ and $$x_k$$ for convenience. $$x_k$$ has a convergent subsequence which we will label $$y_k$$. Now let $$f_k$$ just be the functions corresponding to the points $$y_k$$

What can you say about $$|f(y_k)-f_k(y_k)|$$

4. Sep 5, 2010

### Enjoicube

Haha, took a walk and that's exactly what came to me. I think I had a proof of a similar theorem stuck in my head and wanted to follow that technique.