# The Unapologetic Mathematician

## Cauchy’s Condition for Uniform Convergence

As I said at the end of the last post, uniform convergence has some things in common with convergence of numbers. And, in particular, Cauchy’s condition comes over.

Specifically, a sequence $f_n$ converges uniformly to a function $f$ if and only if for every $\epsilon>0$ there exists an $N$ so that $m>N$ and $n>N$ imply that $|f_m(x)-f_n(x)|<\epsilon$.

One direction is straightforward. Assume that $f_n$ converges uniformly to $f$. Given $\epsilon$ we can pick $N$ so that $n>N$ implies that $|f_n(x)-f(x)|<\frac{\epsilon}{2}$ for all $x$. Then if $m>N$ and $n>N$ we have

$|f_m(x)-f_n(x)|<|f_m(x)-f(x)|+|f(x)-f_n(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

In the other direction, if the Cauchy condition holds for the sequence of functions, then the Cauchy condition holds for the sequence of numbers we get by evaluating at each point $x$. So at least we know that the sequence of functions must converge pointwise. We set $f(x)=\lim\limits_{n\rightarrow\infty}f_n(x)$ to be this limit, and we’re left to show that the convergence is uniform.

Given an $\epsilon$ the Cauchy condition tells us that we have an $N$ so that $n>N$ implies that $|f_n(x)-f_{n+k}(x)|<\frac{\epsilon}{2}$ for every natural number $k$. Then taking the limit over $k$ we find

$|f_n(x)-f(x)|=\lim\limits_{k\rightarrow\infty}|f_n(x)-f_{n+k}(x)|\leq\frac{\epsilon}{2}<\epsilon$

Thus the convergence is uniform.

September 8, 2008