I searched but couldn't find. Are the family of continuous functions $C^0(I,[0,1])$ equicontinuous for the finite interval $I\subset\mathbb{R}$?
To claim this, I guess for every $\epsilon>0$ there must exist $\delta>0$ s.t for all $f\in\mathcal{F}$ and $x,y\in I$, $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$.
It seems to me like it is true. But I cannot prove it. I think for all $x$ and $y$ I have $\max(|x-y|)<I$, the domain is bounded. Since $f\in[0,1]$, co-domain is also bounded. I can choose $x=a$ and $y=a-k\delta$, ($-1<k<1$) such that $|x-y|<\delta$. Now I consider $|f(a)-f(a-k\delta)|<\epsilon$. This should hold for all $\epsilon$ but I don't see why it must hold.
If the claim is not true, can we add a constraint such as all $f\in\mathcal{F}$ are increasing functions s.t. the claim becomes correct?
Thank you very much.
Consider the family $f_n(x) = (1+\sin(nx))/2$. This is not equicontinuous. To prove it, take $\varepsilon=1/2$, pick an arbitrary $\delta > 0$ and consider $n > 2\pi/\delta$.
Assuming increasing is not enough, either. For this, note that the Arzela-Ascoli theorem is an equivalence: a sequence of continuous functions on a compact set has a uniformly convergent subsequence if and only if it is uniformly bounded and equicontinuous. So now consider $g_n(x) = x^n$ with $I=[0,1]$. No subsequence converges uniformly; if it did, it would have to converge to the pointwise limit, but this pointwise limit is not continuous, and uniform limits of continuous functions are continuous. Since this sequence is clearly uniformly bounded we conclude that it is not equicontinuous.
If you prefer, here's a direct proof of the second result. Take $\varepsilon=1/2$. The maximal $\delta$ so that $|g_n(1)-g_n(x)| \leq 1/2$ for all $x \in [1-\delta,1]$ satisfies $(1-\delta)^n = 1/2$. (Why?) Thus $(1-\delta)=(1/2)^{1/n}$ and $\delta=1-(1/2)^{1/n}$. This may be made arbitrarily small by taking arbitrarily large $n$ (how large?), so there is no $\delta$ that works for every $n$.