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: $\forall \epsilon>0,\exists\delta>0 \ such \ that \ \forall s,t \in [a,b], \ |s-t|< \delta \ and \ n \in \mathbb{N} \ then \ |f_n(t)-f_n(s)|<\epsilon$

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    $\begingroup$ Uniform continuity is a property of a single function. Uniform equicontinuity is a property of a family of functions. [Uniform equicontinuity implies that every member of the family is uniformly continuous, but it's more than just that.] $\endgroup$ Feb 27, 2016 at 20:35

2 Answers 2


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.

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    $\begingroup$ Hi Nico, do you not think it would be a good idea to label $\delta$ as $\delta_n$ for the second case to avoid such a possible confusion/ambiguity? $\endgroup$
    – user311475
    May 20, 2018 at 8:15
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    $\begingroup$ Hi. In this answer or in general ? In this answer I think it is better to let it that way. Indeed, this emphasizes the fact that the place in which you put things does really matter. Making a difference would be nice if the confusion could occur in a proof for example. @jamesmartini $\endgroup$
    – nicomezi
    May 20, 2018 at 8:41

If $E,E'$ are metric spaces then a function $f:E\to E'$ is uniformly continuous if for each $\varepsilon>0$ there exists $\delta>0$ such that $$\sup\{d'(f(x),f(y)) : d(x,y)<\delta\}<\varepsilon.$$ A family of functions $(f_\alpha)_{\alpha\in I}$ from $E$ to $E'$ is uniformly equicontinuous if for each $\varepsilon>0$ there exists $\delta>0$ such that $$\sup_{\alpha\in I}\sup\{d'(f_\alpha(x),f_\alpha(y)) : d(x,y)<\delta\}<\varepsilon.$$


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