TITLE: question related to equicontinuous set
QUESTION [1 upvotes]: I have come across one problem. I have a set,  $T=\{f | f:[a, b]\rightarrow \mathbb{R}\}$ where each $f$ is a piecewise continuous function on $[a, b]$. Can we talk about equicontinuity of the ser $T?$

REPLY [1 votes]: (1).Yes. This comes up in the proofs of existence of solutions of certain integral or differential equations. Let $F$ be a family of continuous functions from $[a,b]$ to $\mathbb R.$ For $x\in [a,b]$ we say that $F$ is equicontinuous at $x$ iff $$\forall r>0 \;\exists s>0 \;
\forall f\in F \;(y\in [a,b]\land |y-x|<s\implies |f(y)-f(x)|<r).$$
(2).A continuous $f:[a,b]\to \mathbb R$ is uniformly continuous. That is $$\forall r>0\;\exists s>0\; ((x,y\in [a,b]\land |x-y|<s)\implies |f(x)-f(y)|<r).$$
If the $F,$ above, is equicontinuous at every $x\in [a,b]$ then we also have $$\forall r>0\; \exists s>0\; \forall f\in F \;((x,y\in [a,b]\land |x-y|<s)\implies |f(x)-f(y)|<r).$$ To prove this, by contradiction, suppose not. Then take $r_0>0$ such that for each $n\in \mathbb N$ we have some $f_n\in F$ and $x_n,y_n \in [a,b]$ with $$(0<|x_n-y_n|<1/n)\; \land \;(|f_n(x_n)-f_n(y_n)|\geq r_0).$$ Now let $(x_{n_j})_j$ be a convergent subsequence of $(x_n)_n,$ converging to $x.$ Since $|x_{n_j}-y_{n_j}|<1/n_j,$ the sequence $(y_{n_j})_j$ also converges to $x.$ 
Now $F$ is equicontinuous at $x,$ so take $s>0$ such that $$\forall f\in F ((y\in [a,b]\land |y-x|<s)\implies |f(y)-f(x)|<r_0/2.$$  But there exists $j$ such that $\{x_{n_j},y_{n_j}\}\subset (x-s,x+s)$. (In fact this holds for all but finitely many values of $j$.) For such $j$ we have  $$r_0\leq |f_{n_j}(x_{n_j})-f_{n_j}(y_{n_j})|\leq |f_{n_j}(x_{n_j})-f_{n_j}(x)|+ |f_{n_j}(x)-f_{n_j}(y_{n_j})|<r_0/2+r_0/2=r_0,$$ implying $r_0<r_0.$
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