# Uniform Convergence on an Equicontinuous Set

1. Jul 10, 2008

### Kreizhn

1. The problem statement, all variables and given/known data
Let (X,d) be a compact metric space, (f_n) be an equicontinuous sequences of functions in C(X, $\mathbb{R}$)such that, for every fixed x in X, $(f_n(x)) \to 0$.

Show that (f_n) converges uniformly to the zero function

2. Relevant equations

3. The attempt at a solution
First things first, I started by noting that since X is compact, then C(X,R) (the space of all continuous functions from X to R) is equivalent to C_b(X,R), the set of all continuous bounded functions from X to R. Now R complete implies that F_b(X,R) is complete (set of bounded functions), and C_b(X,R) is a closed subset of F_b(X,R) and so is also complete. Thus it suffices to show that (f_n) is Cauchy with respect to the distance induced by the uniform norm. That is

$$\rho(f,g) = \sup \{ |f(x) - g(x) | | x \in X \}$$

Let $\epsilon > 0$. Then (f_n) uniformly continuous implies $\exists \delta >0$ such that $\forall x,y \text{ satisfying } d(x,y) < \delta, |f(x)-f(y)| < \epsilon$. Since $(f_n(x)) \to 0, \exists n_x$ for each fixed x, such that $|f_n(x)| < \epsilon$.

Since X is compact, let $x_1, \ldots, x_p$ be a delta-net for X. Then $\forall x \in X, \exists x_j \text{ such that } d(x,x_j) < \delta$ and so $|f_n(x) - f_n(x_j)| < \epsilon$ for all n.

I'm not sure if I ended up doing some of this for no reason, but I can't quite figure out where to go from here.

2. Jul 11, 2008

### Kreizhn

Anyone?

3. Jul 11, 2008

### Dick

You know f_n converges to zero. That means that for every e>0 there is a N(x) such that for all n>N(x) |f_n(x)|<e. You only have to show that it's uniform. If f_n is continuous that means that there is a d_n(x) such that if |x-x'|<d_n(x) then |f_n(x)-f_n(x')|<e. Equicontinuity let's you eliminate the dependence of d on n. Now you want to eliminate the dependence on x. Use compactness.