Language around Pointwise convergence v. Uniform convergence I've heard uniform convergence described loosely as convergence "occurring at the same rate at every point." Is that really accurate? If the absolute difference between a member of a sequence of functions and the function to which that sequence converges pointwise can be represented by a function of the point with a finite maximum, then isn't it true that the sequence of functions uniformly convergent AND the rate of convergence varies with the point?
 A: Given an $\epsilon>0$ and $x$, we can define $N_x(\epsilon)$ as the smallest $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n\geq N.$
In some sense, $N_x(\epsilon)$ is a measure of the rate convergence at $x.$ 
Uniform convergence means that for any $\epsilon>0,$ $N(\epsilon)=\sup_{x} N_x(\epsilon)<+\infty.$
So this doesn't strictly mean they converge at the same speed, but that there is some speed of convergence which is slower than (or equal to) all of them.
A: Perhaps it would be more accurate to say "convergence bounded by some rate at every point." You are right that the convergence rate may be faster in some places than others. For instance, the function $f_n(x)=x^n$ on $[0,1/2]$ converges to $f \equiv 0$ uniformly, since $\sup_{x \in [0,1/2]}|f_n(x)-f(x)|=|f_n(x)| \le 2^{-n}$. But at $x=1/4$, the rate of convergence is the faster $4^{-n}$.
The point of the definition of uniform convergence is to exclude cases where the convergence at some points is arbitrarily slow.
A: It looks like an acceptable translation in English of "for all $\varepsilon$ there is $n$ such that, for all $m>n$ and for all $x$, $\lvert f(x)-f_m(x)\rvert<\varepsilon$", though it might be misinterpreted in a set-up where a theory of "rates of convergence" has been introduced. If the reader understands it like this, it's ok. Otherwise he'd be misled. A possible misintepretation might be like this: if we say that a sequence $a_n$ "converges to $a$ like $n^{-2}$" if $\limsup_{n\to\infty} n^2\lvert a-a_n\rvert<+\infty$, then there are sequences of functions $f_n$ such that $f_n(x)$ converges to $f(x)$ like $n^{-2}$ for all $x$, but $f_n$ does not converge uniformly to $f$; on the other hand, your phrasing strongly hints towards the former.
What the formulation would like you to think is that there is a function $\varepsilon(n)$ such that $\lim_{n\to\infty} \varepsilon(n)\to 0$ and such that for all $x$, $\lvert f(x)-f_n(x)\rvert<\varepsilon(n)$. In this sense, there is a "global" rate of convergence. But globality is the important part: you want a single bound holding for all $x$.
A: Let's call the sequence $f_n$ and the limit $f$. If $f_n\to f$ uniformly, it's true that $f_n(x)-f(x)$ will not generally be constant in $x$ for any given $n$, but we have a uniform (in $x$) bound on the difference $|f_n(x)-f(x)|$ that goes to zero as $n\to \infty$. That upper bound is what people mean by a uniform rate of convergence. The convergence at some points $x$ may be faster, but the convergence is at least as fast as the uniform rate at every point.
A: Its not totally true to say that the points converge "at the same rate." First, "the same rate" isn't really a mathematical statement. Second, a better descriptor, in my opinion, is that the points converge in some similar fashion.
The definition of uniform convergence is that, for a given error bound $\epsilon$, there is eventually an $f_N$ for which all of the $f_n$'s beyond this point are within the error bound.
Once you understand this, it is almost hard to think of a function which is not uniformly convergent. In a sense, a sequence has to behave badly for this to happen. Take $\{1/(nx)\}$ converging to $0$ on $(0,1]$, for example.
