Pointwise equicontinuity

Question

Is $\Phi:=\{(t \mapsto t^n):n \in \mathbb{N}\}\subseteq C([0,1),\mathbb{R})$ where $C([0,1),\mathbb{R})$ describes the set of all continuous functions from $[0,1) \to \mathbb{R}$ pointwise equicontinuous?

The family $\Phi$ is equicontinuous at a point $y$ if for every $\epsilon > 0 $, there exists a $\delta > 0$ such that $|f(y)- f(x)|<\epsilon$ for all $f \in \Phi$ and all $x$ such that $|y-x| < \delta$

The family is pointwise equicontinuous if it is equicontinuous at each point $y$.

I tried the following (which is most likely wrong):

Let $\epsilon>0$ and let $x \le y$

$|x^n-y^n|=|x-y|\sum_{j=0}^{n-1}x^{n-1-j}y^j \le |x-y|\sum_{j=0}^{n-1}(y+\delta)^{n-1-j}(y+\delta)^j=|x-y|\sum_{j=0}^{n-1}(y+\delta)^{n-1}=|x-y|n(y + \delta)^{n-1}$

Now $n(y + \delta)^{n-1} < M$ for large enough $n$. So we can pick $\delta=\epsilon/M$

Answer 1

You are on the right track but a few details are missing. You want $n(y+\delta)^{n-1} <M$for $|x-y|< \delta$, for $n$ sufficiently large, say $n \geq m$. This $m$ should depend only on $x$, not on $y$. So choose $\delta $ such that $x +2\delta <1$ and choose $m$ corresponding to $y=x+2\delta$. This $m$ will then work for all $y$. Secondly, you want the inequality for all $n$, not for $n$ sufficiently large. So you will have to take the minimum of finite number of $\delta$'s to complete the proof.

Answer 2

Of course $t^n$ is the canonical example of a sequence that converges to $0$ pointwise on $[0,1)$ but not uniformly; hence it's interesting to note that you can get the pointwise equicontinuity from uniform convergence:

Fix $x\in[0,1)$. Choose $\delta>0$ so that $$K=[x-\delta,x+\delta]\cap[0,1]\subset[0,1)$$.

Now $$t^n\le(x+\delta)^n\quad(t\in K);$$since $x+\delta<1$ this shows that $t^n\to0$ uniformly on $K$, hence $(t_n)$ is equicontinuous at $x$.