TITLE: Uniform Convergence: The two definitions
QUESTION [11 upvotes]: I know of the two equivalent definitions for uniform convergence. Namely:
$f_n(x)$ converges uniformly to $f(x)$ if either:
a)$$ \forall \epsilon \exists N \in \mathbb N s.t. \forall x \in D \ \forall n \geq N:
|f_n(x) - f| < \epsilon $$
or
b) $$\lim_{n \rightarrow \infty} \sup_{x\in D} |f_n(x) - f(x)| = 0 $$
While I see why in the case of a) the same "speed of convergence" is guaranteed since one epsilon is chosen for all $x$, I unfortunately cannot make sense of b) on an intuitive level. So I see that b) states that the biggest difference between the sequences of the functions and the limit functions must become zero for n going to infinity. But how does that also guarantee this convergence process is of "uniform speed" for all x?

REPLY [1 votes]: It isn't necessarily that the convergence takes place at a uniform speed, so much as that there is a speed of convergence such that all pointwise convergence takes place at at least that speed. Put another way, there's a sequence $c_n>0$ such that $c_n\to 0$ and such that for $n$ big enough, $f_n$ stays within the vertical window between $f+c_n$ and $f-c_n$.
For an example where we don't have a "uniform speed", take $f$ to be the constant $0$ function on $[a,b]$ for some $-1<a<0<b<1$, and let each $f_n$ be defined on $[a,b]$ by $$f_n(x)=\begin{cases}0 & x<0\\ x^n & \text{otherwise.}\end{cases}$$ Then we have uniform convergence of the $f_n$ to $f$ on $[a,b]$, but the convergence was immediate in $[a,0]$, and not in $(0,b]$.
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