5
$\begingroup$

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

$\endgroup$
  • 6
    $\begingroup$ 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.] $\endgroup$ – Daniel Fischer Feb 27 '16 at 20:35
1
$\begingroup$

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 $$\sup_{\alpha\in I}\sup\{d'(f_\alpha(x),f_\alpha(y)) : d(x,y)<\delta\}<\varepsilon.$$

$\endgroup$
15
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – jamesmartini May 20 '18 at 8:15
  • 1
    $\begingroup$ 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 '18 at 8:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.