I am very confused. Thanks in advance.
Our definition is that:
Uniformly Equicontinuous: $\forall \epsilon>0,\exists\delta>0 \ such \ that \ |s-t|< \delta \ and \ n \in \mathbb{N} \ then \ |f_n(t)-f_n(s)|<\epsilon$
Uniformly continuous: $\forall \epsilon>0,\exists\delta>0 \ such \ that \ \forall s,t \in [a,b], \ |s-t|< \delta \ and \ n \in \mathbb{N} \ then \ |f_n(t)-f_n(s)|<\epsilon$
Best Answer
The family of function $(f_n)$ defined on $[a,b]$ is said :
Uniformly equicontinuous : $\forall \epsilon>0,\exists\delta>0,\forall n \in \mathbb{N}, \forall s,t \in [a,b], \ |s-t|< \delta \ \Rightarrow \ |f_n(t)-f_n(s)|<\epsilon$
Uniformly continuous : $\forall \epsilon>0,\forall n \in \mathbb{N},\exists\delta>0, \forall s,t \in [a,b], \ |s-t|< \delta \ \Rightarrow \ |f_n(t)-f_n(s)|<\epsilon$
Look at the place of $\forall n \in \mathbb{N}$. In the first case, you have the same $\delta$ for the whole family of functions. While in the second case, the $\delta$ may depend on the function you are considering. One can remark that uniform equicontinuity implies uniform continuity. So uniform equicontinuity is a more strong condition.