As the title implies, let $\{x^n/n \}_n\subseteq C[0,1]$, where $x\in X$ and $X$ is a metric space. I want to show that this family is equicontinuous at each $x\in[0,1]$. By the definition then, we need to show that for any $\epsilon>0$, there is some $\delta>0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $f$ in the family, and for all $x\in [0,1]$. My approach is as follows.
Consider\begin{equation*} \left|\frac{x^n}{n}-\frac{y^n}{n} \right|=\frac{|x^n-y^n|}{n}\end{equation*}Since $x^n/n$ is continuous on $[0,1]$ and differentiable on $(0,1)$, by the Mean Value Theorem there is a $z\in(0,1)$ such that\begin{equation*} z^{n-1} = \frac{\frac{x^n}{n}-\frac{y^n}{n}}{x-y}\implies \frac{x^n}{n}-\frac{y^n}{n}=(x-y)z^{n-1}\end{equation*}Since $z\in(0,1)$, $z^{n-1}\leq 1$, so\begin{equation*} \frac{|x^n-y^n|}{n}\leq |x-y|\implies |x^n-y^n|\leq n|x-y|\end{equation*}for all $x\in [0,1]$. We have then that\begin{equation*} \left|\frac{x^n}{n}-\frac{y^n}{n} \right|\leq |x-y|<\epsilon\end{equation*}Choose $\delta=\epsilon$. Then\begin{equation*} |x-y|<\delta\implies \left|\frac{x^n}{n}-\frac{y^n}{n} \right|<|x-y|<\delta = \epsilon \end{equation*}So the sequence is equicontinuous for every $x\in[0,1]$.
Something about this doesn't feel correct to me. First of all, it feels as I didn't do anything differently than if we were showing a sequence $x^n/n$ is continuous; that is, I don't really understand how, if I even have, shown that for all$x\in [0,1]$, equicontinuity holds. It is possible I just don't understand the actual meaning of the definition well. Also, my professor defined the 'less than $\epsilon$' part of the definition with $||f(x)-f(y)||_\infty<\epsilon$...for all $x\in X$ and $f\in \mathcal{F}$. I did not use the infinity norm here because many of the examples I have seen do not use it in proving equicontinuity. Any pointers would be appreciated.