Take the 2-minute tour ×

I recently ran across a description of uniform convergence that claimed: "The difference between pointwise convergence and uniform convergence is that with pointwise convergence, the convergence of $f_n$ to $f$ can vary in speed at each point of $f$. With uniform convergence, the speed of convergence is roughly the same all across the domain."

Is this necessarily true? Taken strictly, all uniform convergence says is that, given an $\epsilon$, I can find an $N(\epsilon)$ such that $\forall n > N$, $|f_n(x) - f(x)|<\epsilon$ for all $x$ on the relevant domain. Meaning, if I go out far enough in the sequence, the entire function will be contained within an $\epsilon$-tube centered around $f$.

This doesn't seem to say anything about speed of convergence at different points: I can conceive of the sequence of functions converging very quickly at some points while "lagging behind" at others (but nonetheless eventually getting there), so that those laggers just require I take a larger $N$ for a given $\epsilon$. More precisely, while I may be able to find different $N$'s for different $x$, I can simply take the largest of these as my $N(\epsilon)$ (supposing it exists). I'm specifically thinking in contrast to situations like $f_n(x) = x^n, x \in [0,1]$, where we have a "stuck point" at $x = 1$ that screws everything up; or, similarly, in a Fourier series with the Gibbs phenomenon present, where the overshoot never in fact decays, it just moves outward toward the endpoints. (In both of these cases, there is no largest $N$ to take as my $N(\epsilon)$ independent of $x$, and the best we can do is pointwise convergence.)

So, I'm trying to construct examples where we have uniform convergence, but the speed of convergence still depends on x. My first thought was $f_n(x) = \frac{\sin^n(\pi x)}{n}$. Since $|\sin(x)| \leq 1$ , this clearly converges uniformly to $f(x) = 0$ (just let $N$ = $1/\epsilon$) -- but the closer $x$ is to an integer, the faster the convergence, with $x = \frac{k\pi}{2}, k$ odd, taking the longest to converge and in fact preventing any lower N from working.

Is this correct? Does this example work? Are there other, more interesting ones?

What's more: is there some extra condition I can add that will guarantee the same speed of convergence everywhere?

share|improve this question
Well, you wrote "roughly" the same speed, which i would take as: speed varies along the domain but is bounded. If you don't have "slowest" points, you don't have uniform convergence. –  wisefool Dec 24 '12 at 1:09
Point taken. But isn't it possible to have some points converge as slowly as I like, while others converge as quickly as I like, and still have uniform convergence? I guess what I'm getting at is: anything can happen inside the epsilon-tube, so long as the tube is shrinking at some rate. Right? But then uniform convergence doesn't seem as strong as it once did to me. –  AndrewG Dec 24 '12 at 1:23
Uniform convergence is related to the convergence in the supremum norm. So, it is related to the maximum and the minimum of the difference; as long as these two numbers go to zero, you have uniform convergence. If you set $v(x, m)=\min\{n : \forall k>n, |f_k(x)-f(x)|<1/m\}$, then you have uniform convergence if there exists $N(m)$ such that $v(x,m)<N(m)$ for every $x$ (i.e. $\|v(x,m)\|_\infty< N(m)$). In this sense, the speed is roughly constant with respect to $x$, as a bounded function is roughly constant. Obviously, the bounds can be as big as you wish, as long as they are finite. –  wisefool Dec 24 '12 at 13:45

1 Answer 1

up vote 3 down vote accepted

The most extreme example I can construct is the sequence of functions $f_n:\mathbb{R}\to\mathbb{R}$ defined by $$f_n(x)=\begin{cases}\tfrac{1}{n} & \text{ if }x=0,\\\\ 0 & \text{ if }x\neq 0 \end{cases}$$ which converges uniformly to the zero function but where $x=0$ is the only laggard.

share|improve this answer
Oh, that's even simpler! Awesome. –  AndrewG Dec 23 '12 at 23:06

Your Answer


