# Cauchy sequence of functions converges uniformly

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?

• What is the norm for your functional space? Sup norm? Mar 30, 2019 at 9:18

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 $$f$$.