Cauchy sequence of functions converges uniformly

Question

Given a Cauchy sequence of functions $\{f_n(x)\}$, it is obvious that it converges pointwise to some function $f(x)$. To prove uniform convergence, the standard argument goes as follows:

Let $\varepsilon>0$. Choose $N$ such that for all $m,n \geq N$, $|f_n(x)-f_m(x)|<\varepsilon$. Fix $n$ and take $m \to \infty$. Since $f_m(x) \to f(x)$, it follows that $|f_n(x)-f(x)| \leq \varepsilon$ for all $n \geq N$ and $x \in E$.

I don't find the "take the limit of $m$ to infinity" part rigorous enough because $f_m(x) \to f(x)$ is only pointwise and depends on $x$. How can we derive a conclusion of for all $x \in E$ then? What am I missing here? It also sounds a bit like heuristics and not really Analysis. How to make the argument formal?

Answer 1

First you pick $N$ such that for all $n,m>N$ and for all $x$ we have (here we're using that $(f_n)_{n\in \mathbb{N}})$ is a Cauchy sequence with respect to the supremum norm) $$ \vert f_n(x)-f_m(x) \vert < \varepsilon/2.$$ Now fix some $x$ and pick $m(x)>n>N$ such that $$ \vert f_{m(x)} (x) - f(x)\vert < \varepsilon/2$$ this you can do as $f_n(x) \rightarrow f(x)$. By the triangle inequality we get $$ \vert f_n(x)- f(x)\vert \leq \vert f_n(x)- f_{m(x)} (x) \vert + \vert f_{m(x)}(x) - f(x)\vert < \varepsilon/2 + \varepsilon/2 $$ Hence we get for every $\varepsilon >0$ some $N$ such that for all $n>N$ $$ \sup_{x\in E}\ \vert f_n(x) -f(x) \vert < \varepsilon.$$ Thus, $(f_n)_{n\in \mathbb{N}}$ converges uniformly to $f$.