# Uniform Convergence and the Uniform Metric

1. Dec 25, 2012

### jmjlt88

Let X be a set, and let fn : X---> R be a sequence of functions. Let ρ be the uniform metric on the space RX. Show that the sequence (fn) converges uniformly to the function f:X--> R if and only if the sequence (fn) converges to f as elements of the metric space (RX, ρ). [Note: the ρ's should have a bar over them.]

I have a question concerning one direction of our implication.

[My attempt at the solution.]

Since I do not know how to make a d with a bar over it, I'll use ∂ to denote the standard bounded metric relative to d.

Proof:

Assume the sequence (fn) converges to f as elements of the metric space (RX, ρ).

Then, for any ε>0, there is a positive integer N such that ρ(fn,f)<ε for all n≥N.

That is, for n ≥ N, the sup{∂(fn(x),f(x))| x in X} < ε.

If ε≤1, our metrics are the same.

Hence for ε≤1, there exisits an integer N such that for n ≥ N, the sup{∂(fn(x),f(x))| x in X} < ε.

But since ∂(fn(x),f(x)) = d(fn(x),f(x)) for n ≥ N, d(fn(x),f(x)) ≤ sup{∂(fn(x),f(x))| x in X} < ε for all n≥N and x in X, which satisfies our condition for uniform convergence.

My question is about ε. Is this the right approach?

2. Dec 25, 2012

### jbunniii

To display a bar, you can use the TeX command \bar. For example, to typeset $\bar{\rho}$, use \bar{\rho}.

Just to make sure I understand the question correctly, let me paraphrase it with proper typesetting: by standard bounded metric relative to $d$, you mean
$$\bar{d}(x,y) = \min\{d(x,y), 1\}$$
and that the following definitions are in use
$$\rho(f,g) = \sup\{d(f(x),g(x)) : x \in X\}$$
$$\bar{\rho}(f,g) = \sup\{\bar{d}(f(x),g(x)) : x \in X\}$$
and that the goal is to show that
$$\bar{\rho}(f_n,f) \rightarrow 0 \implies \rho(f_n,f) \rightarrow 0$$
Assuming I interpreted the question correctly, what you have done is fine for $\epsilon \leq 1$. For $\epsilon > 1$, you can simply use the $N$ that works for $\epsilon = 1$.