# 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 -

1. […] we’ve got Cauchy’s condition: a series converges uniformly if for every there is an so that and both greater than zero […]

Pingback by Uniform Convergence of Series « The Unapologetic Mathematician | September 9, 2008 | Reply

2. In the Cauchy –> Uniform convergence direction, could you clarify the step “taking the limit over k”? until this point you have only established pointwise convergence. But in this step it feels like you’ve presupposed uniform convergence in saying f_n+k –> f(x).

I know you’re not doing anything wrong, because Rudin has the same proof. I’m just confused and would appreciate the help

Comment by Anirudh | February 21, 2011 | Reply

• I believe they forgot to mention that the conditions for the Cauchy criterion here also include that for each epsilon the N chosen must hold for all x in the domain of the sequence of functions; from there you can see that is indeed true.

Comment by Armitage | January 6, 2013 | Reply

• If one assumes that for any x f_{n+k}(x) -> f(x), one is only presupposing pointwise convergence, which follows from the normal Cauchy’s condition for the real or complex numbers.

To see the difference between pointwise and uniform convergence, it might be recommended to have a look at the the function series sin(x)^n (For a graph of this see https://en.wikipedia.org/wiki/File:Drini-nonuniformconvergence.png). If you now take any x in the open interval (0, PI/2), you will find that sin(x)^n -> 0, because sin(x) PI/2. Therefore, the function converges on the open interval (0, PI/2) pointwise, but not uniformly.

• If one assumes that for any x f_{n+k}(x) -> f(x), one is only presupposing pointwise convergence, which follows from the normal Cauchy’s condition for the real or complex numbers.

To see the difference between pointwise and uniform convergence, it might be recommended to have a look at the the function series sin(x)^n (For a graph of this see https://en.wikipedia.org/wiki/File:Drini-nonuniformconvergence.png). If you now take any x in the open interval (0, PI/2), you will find that sin(x)^n -> 0, because sin(x) 1 – epsilon for all arbitrarily small positive epsilon, because sin(x) -> 1 if x -> PI/2. Therefore, the function converges on the open interval (0, PI/2) pointwise, but not uniformly.