TITLE: counter example of equicontinuous
QUESTION [2 upvotes]: Consider the functions on $[0,1]$: 
$f_n(x)=nx$, when $x$ is between $0$ and $1/n$
$f_n(x)=2-nx$, when $x$ is between $1/n$ and $2/n$
$f_n(x)=0$, otherwise
How to see it is not (uniformly) equicontinuous? 
I checked $0$. For $\epsilon$=$1/2$, no matter how small the neighborhood around $0$ we pick, we can choose a $N$ large enough s.t. $1/N$ < $\delta$. So for $n> N$ , we will have some $x$=$1/n$, which satisfies $|x- 0|$ < $1/N$< $\delta$, but makes $|f_n(x)-0|$= $1$, instead of smaller than $1/2$.
Is it correct?

REPLY [1 votes]: The idea is fine, but you may want to make the formalism a bit more rigorous:
Let $1 > \epsilon > 0$. We must show that $\forall\delta > 0 \exists x\in[0,1], N\in\mathbb N$ such that $|x|<\delta$ but $|f_N(x)| > \epsilon$.
For this pick $N = \lceil \frac1\delta \rceil
+1$ and $x = \frac1N < \delta$. Then $|f_N(x)| = 1 > \epsilon$ as claimed. Therefor $(f_n)_n$ is not equicontinuous.
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