Let $g_n: [0,1]\to \mathbb{R}$ for all $n\in\mathbb{N}$ where
\begin{align*}
g_n(x)=
\begin{cases}
n^2x,\quad 0\leq x\leq \frac{1}{n}\\
\frac{1}{x},\quad \frac{1}{n}< x\leq 1.
\end{cases}
\end{align*}
I have proved that $\{g_n:n\in \mathbb{N}\}$ is not uniformly equicontinuous on $[0,1]$. Because there exists $\varepsilon_0=\frac{1}{n}$, such that for any $\delta>0$, there exists $n\in \mathbb{N}$ where $\frac{1}{n}<\delta$. By choosing $x=0$ and $y=\frac{1}{n}$, we have $|x-y|<\delta$, but $|g_n(x)-g_n(y)|=|g_n(0)-g_n(\frac{1}{n})|=n>\varepsilon_0$.
Then, I want to check, whether $\{g_n:n\in \mathbb{N}\}$ is equicontinuous on $[0,1]$?
From the proof that $\{g_n:n\in \mathbb{N}\}$ is not uniformly equicontinuous, I think it's not equicontinuous on $[0,1]$. Is my thought correct?
Thanks for any feedback.
Best Answer
EDIT: The below proof is invalid. The real solution is what OP wrote as a comment: the family cannot be equicontinuous, since it is not uniformly equicontinuous over a compact space. If the domain of the function were instead $(0, 1]$, then the below proof would be valid.
The family is in fact equicontinuous on $[0,1]$.
To see this, consider some $x_0 \in [0,1]$, and let $n_0 = 2\lceil 1/x_0\rceil$. Note that $g_n(x_0) = 1/x_0$ for all $n\geq n_0$. And in fact, we have that $g_n(x)=1/x$ for all $x$ sufficiently close to $x_0$ for all $n \geq n_0$, thanks to the slack from the factor of 2. Now, given some $\epsilon>0$, we can choose an appropriate $\delta>0$ to prove equicontinuity easily. With a small enough $\delta$, all $g_n(x)$ become identical for $n \geq n_0$ for $x$ near $x_0$. When including the finitely many values of $n <n_0$, we are left with a finite family of continuous functions, which is trivially equicontinuous.