# The Unapologetic Mathematician

## Uniform Convergence

Today we’ll give the answer to the problem of pointwise convergence. It’s analogous to the notion of uniform continuity in a metric space. In that case we noted that things became nicer if we could choose our $\delta$ the same for every point, and something like that will happen here.

To reiterate: we say that a sequence $f_n$ converges pointwise to a function $f$ if for every $x$, and for every $\epsilon$, there is an $N$ so that $n>N$ implies that $|f_n(x)-f(x)|<\epsilon$. Just like we did for uniform continuity we’re going to move around the quantifiers so that $N$ can depend only on $\epsilon$, not on $x$.

We say that a sequence of functions converges uniformly to a function $f$ if for every $\epsilon$ there is an $N$ so that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<\epsilon$. In pointwise convergence, the value at each point does converge to the value of the limiting function, but the rates can vary widely enough to make it impossible to control convergence at two different parts of the domain simultaneously. But in uniform convergence we have “uniform” control of the convergence over the entire domain.

So let’s see how we can use this to show that the limiting function $f$ is continuous if each function $f_n$ in the sequence is. Uniform convergence tells us that for every $\epsilon$ there is an $N$ so that $n>N$ implies that $|f_n(x)-f(x)|<\frac{\epsilon}{3}$ for every $x$. But since $f_n$ is continuous at $x_0$ there is some $\delta$ so that $|x-x_0|<\delta$ implies that $|f_n(x)-f_n(x_0)|<\frac{\epsilon}{3}$.

And now we can use this $\delta$ to show the continuity of $f$. For if $|x-x_0|<\delta$, we find \begin{aligned}|f(x)-f(x_0)|<|f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|\\<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon\end{aligned}

The essential point here is that we were able to keep control of the convergence of the sequence both at the point of interest $x_0$, and at all points $x$ in the $\delta$-wide neighborhood.

Uniform convergence isn’t the only way to be assured of continuity in the limit, but it’s surely one of the most convenient. One thing that’s especially nice about uniform convergence is the way that we can control the separation of sequence terms from the limiting function by a single number $\epsilon$ instead of a whole function of them.

That is, instead of fixing an $\epsilon$, fix an $N$ and consider how far sequence terms can be from the limit. Take the maximum $\max\limits_{n>N}|f_n(x)-f(x)|$

This depends on $x$, but if the convergence is uniform we can keep it down below some constant function. For pointwise convergence that isn’t uniform, no matter how big we pick the $N$ there will still be some differences that are “large” compared to others.

In this way, uniform convergence is more like convergence of numbers than pointwise convergence of functions. Uniform convergence just isn’t as floppy as pointwise convergence can be.

September 5, 2008 -

## 11 Comments »

1. hullo john,

nice blog you have there.

shouldn’t it read |f_n(x) – f_n(x_0)| < epsilon/3 in the last line of the fourth paragraph? Comment by rustam | September 7, 2008 | Reply

2. Thanks. And yes, you’re right. Thanks for catching that. Comment by John Armstrong | September 7, 2008 | Reply

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10. In your last frase:
“For pointwise convergence that isn’t uniform, no matter how big we pick the N there will still be arbitrarily large differences”

I think you meant this: no matter how big we pick the N the differences won’t be arbitrary small. I don’t think that equals that the differeces will be arbitrary large, consider for example $f:[0,1] \to \mathbb{R}$, $f(x) = x^n$. The differences are all in $[-1,1]$. Comment by ghi | July 21, 2015 | Reply

• I think you’re right; I was a bit sloppy with my language there. I’ll clean that up, thanks. Comment by John Armstrong | July 21, 2015 | Reply