Uniformly convergent implies equicontinuous I'm trying to prove that if I have a sequence of continuously differentiable functions $f_n$ that converge uniformly on $[a,b]$, then $\{f_n\}$ is equicontinuous for all $x_0 \in [a, b]$.
My idea is to use uniform convergence to deal with the "tail" and then use continuity to deal with the finitely many $f_n$'s left. But I'm having trouble writing it down.
 A: There is no need to assume differentiability. Let $\epsilon>0$ and $x_0$ be given. Uniform convergence implies that the sequence is uniformly Cauchy. Choose $n_0$ such that $|f_n(x)-f_{n_0}(x)|<\epsilon$ for all $x$ and all $n\ge n_0$. Choose $\delta_k \ (k\in \{1,\ldots,n_0\})$ such that $|x-x_0|<\delta_k$ implies $|f_k(x)-f_k(x_0)|<\epsilon$. Put $\delta=\min_k{\delta_k}$.
Now, if $n\ge n_0$ and $|x-x_0|<\delta$, the triangle inequality yields
$|f_n(x)-f_n(x_0)|\le |f_n(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(x_0)|+|f_{n_0}(x_0)-f_n(x_0)|<3\epsilon$.
If $k\le n_0$, then $|x-x_0|<\delta\le\delta_k$ implies $|f_k(x)-f_k(x_0)|<\epsilon$ by choice of $\delta_k$ above.
A: If you have a sequence of continuous functions $f_n \to f$ uniformly, it suffices to assume they are continuous. The sequence, along with its limit, is compact in the uniform norm.  By Arzela-Ascoli, it is equicontinous.
A: See this

$$ |f_n(x)-f_n(y)| \leq |f'_n(\eta)||x-y|,\quad \eta \in (x,y) $$

by mean value theorem which implies 

$$|f_n(x)-f_n(y)| \leq M |x-y|   $$

since $|f'_n(x)|$ is continuous on $[a,b]$. I think there is nothing left for you to do except some details. 
