Rudin's book "Principles of Mathematical Analysis" gives the following definition for equicontinuity:
A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\epsilon >0$ there exists a $\delta>0$ such that:
$$ \vert f(x) – f(y) \vert < \epsilon$$
whenever $d(x,y) < \delta, x \in E, y \in E$ and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$
Based on that definition I would think that if one considers a family $\mathscr{F}$ of functions $f$ such that each $f$ is uniformly continuous on a set $E$ then $\mathscr{F}$ would be equicontinuous on $E$, but it seems that this is not true. Example $7.21$ of the book considers the sequence $f_n:[0,1] \to \mathbb{R}$ given by:
$$ f_n(x) = \dfrac{x^2}{x^2+(1-nx)^2}$$
which is a sequence of uniformly continuous functions (since each $f_n$ is continuous and is defined on a compact set), but the book says that the sequence is not equicontinuous. Why is that? Each $f_n$ is uniformly continuous so shouldn't it satisfy the definition of equicontinuity?
Best Answer
The uniform continuity gives you a $\delta_n$ such that $d(x,y)<\delta_n$ implies $|f_n(x)-f_n(y)|<\epsilon$. There is no reason why the same $\delta$ will work for all $n$ (which is what equicontinuity means).
Here you have $f_n(0)=0$ for all $n$, but $f_n(1/n)=1$ for all $n$. So for large $n$, the $f_n$ will stay away from $0$ for longer. This can be made into a formal statement that equicontinuity fails.