This is a result that I have known for a long time whose proof I feel I have never really understood. I will offer a specific version of it (as opposed to the most general formulation).
Let $A \subset \mathbb{R}$ and let $f_n : A \to \mathbb{R}$ be a sequence of continuous functions on $A$. Moreover, assume $f_n$ is uniformly Cauchy, i.e. for any $\epsilon > 0$ we have some $N := N(\epsilon)$, such that for all $m ,n \geq N$ :
$$
\sup_{x \in A} | f_n(x) – f_m(x) | < \epsilon
$$
Then, in fact, there exists a function $f:A \to \mathbb{R}$ for which $f_n \to f$ uniformly.
This is basically the result that $C(X)$ is complete where $X$ is say a compact Hausdorff space (although here we do not assume $A$ is necessarily compact, not that it matters.)
The proof I have seen proceeds like this:
For any fixed $x_0 \in A$ the sequence $f_n(x_0)$ is a Cauchy Sequence in $\mathbb{R}$ as:
$$
|f_n(x_0) – f_m(x_0)| \leq \sup_{x \in A} |f_n(x) – f_m(x)| < \epsilon
$$
for $n,m$ sufficiently large. Ergo, the pointwise limit of the $f_n$ exists for any $x \in A$ by the completeness of $\mathbb{R}$ and thus the obvious candidate function $f$ is given pointwise by $f(x) := \lim_{n \to \infty} f_n(x)$.
This is well and good, and all that needs to be shown now that $f_n$ converges uniformly to $f$, rather than (just) pointwise. To do this, we consider:
\begin{align*}
\sup_{x\in A} |f_n(x) – f(x)| &= \sup_{x \in A}|f_n(x) – f_m(x) + f_m(x) – f(x)| \\ &\leq \sup_{x \in A} |f_n(x) – f_m(x)| + \sup_{x\in A}|f_m(x) – f(x)|
\end{align*}
The first term:
$$
\sup_{x \in A} |f_n(x) – f_m(x)|
$$
is small for $n,m$ chosen sufficiently large. The second term, I am not sure what to do with. Most proofs I have seen write something to the effect of:
$$
|f_m(x) – f(x)| = \lim_{n \to \infty} \sup_{x\in A}|f_m(x) – f_n(x)|
$$
and then conclude that the right-hand term is small, but frankly with this sort of reasoning, I don't seem to understand why that couldn't have been done in the very first step. In particular, the sort of limit-taking being done here is in the pointwise sense, and so it doesn't seem to make sense to interchange the norm and the limit as limits only interchange with norms if the limit is being taken in the norm topology. However, I'm fairly certain I'm overthinking this.
I would appreciate some assistance with this question – frankly I am embarrassed to ask it, as someone who specializes in functional analysis, I am having difficulty proving such a simple fact.
Also we of course do not need the $f_n$ to be continuous – that is only required if we'd like the limit $f$ to be continuous.
Best Answer
For $\varepsilon>0$, take $N$ for which $m,n>N$ implies $|f_n(x)-f_m(x)|<\varepsilon$ for all $x\in A$. Then for any $x$, we can take $m$ such that $|f_m(x)-f(x)|<\varepsilon$ and we may assume $m>N$. Then
$$|f_n(x)-f(x)|\le|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<2\varepsilon.$$
The key fact here is that $m$ exists for any $x$ and can always be taken to be larger than $n$. This means that, with no mention of $m$, we always have $|f_n(x)-f(x)|<2\varepsilon$. This holds for all $x$, giving the uniformity.
I agree this is fairly confusing at first glance. It helps to think about just sequence of points. The same $N$ used in the Cauchy criterion to obtain $|x_n-x_m|<\varepsilon$ can be used in the usual definition to obtain $|x_n-x|\le\varepsilon$. This is true pointwise, and thus $|f_n(x)-f_m(x)|<\varepsilon$ implies $|f_n(x)-f(x)|\le\varepsilon$.