Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From Wiki

A set $A$ of continuous functions between two uniform spaces $X$ and $Y$ is uniformly equicontinuous if for every element $W$ of the uniformity on $Y$, the set $\{ (u,v) ∈ X × X: \forall f ∈ A. (f(u),f(v)) ∈ W \}$ is a member of the uniformity on $X$.

  1. I was wondering if it is equivalent to

    A set $A$ of continuous functions between two uniform spaces $X$ and $Y$ is uniformly equicontinuous if for every element $W$ of the uniformity on $Y$, there is a member $V$ of the uniformity on $X$ such that $\forall (u,v) ∈ V, \forall f ∈ A. (f(u),f(v)) ∈ W $.

    This is because in the special case when the uniform spaces are metric spaces, for a family $F$ of continuous functions between two metric spaces:

    The family $F$ is uniformly equicontinuous if for every $ε > 0$, there exists a $δ > 0$ such that $d(f(x_1), f(x_2)) < ε$ for all $f ∈ F$ and all $x_1, x_2 ∈ X$ such that $d(x_1, x_2) < δ$.

    Also this is because I think that the relation between equicontinuity and uniform equicontinuity should be similar to the relation between continuity and uniform continuity.

  2. Does the definition require the set of functions to be continuous? Doesn't the definition itself, when without requiring continuity, imply uniform continuity and therefore continuity?

Thanks and regards!

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted
  1. Yes, the two are equivalent. This follows from clause (2) of this definition of a uniformity. Let $U=\{\langle u,v\rangle\in X\times X:\forall f\in A(\langle f(u),f(v)\rangle\in W\}$. If $U$ belongs to the uniformity on $X$, then $U$ itself is a member of the uniformity on $X$ such that for each $\langle u,v\rangle\in U$, $\langle f(u),f(v)\rangle\in W$. Conversely, if there is a member $V$ of the uniformity on $X$ such that for each $\langle u,v\rangle\in V$, $\langle f(u),f(v)\rangle\in W$, then $U\supseteq V$, so $U$ is in the uniformity on $X$.

  2. Yes, uniform equicontinuity of a family of functions implies that the functions are uniformly continuous. It is not necessary in the definition to specify that the members of $A$ are continuous, but it does no real harm.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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