# Does uniform continuity on a compact subset imply equicontinuity?

I gave a proof here some time ago but was wondering about the following:

Suppose that $$f_n:K\subseteq \Bbb{R}\to\Bbb{R}$$ is continuous for each $$n\in \Bbb{N}$$, then $$\{f_n\}$$ is uniformly continuous. So, for every $$\epsilon > 0$$ and $$n\in \Bbb{N}$$, there exists $$\delta_n>0$$ such that $$\forall\,x,y\in K, \;|x − y| < \delta_n,$$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}.$$ Taking $$\delta=\min\{\delta_n:\;n\in \Bbb{N}\}$$ (which might not be true), then we have the definition given here, that "for every $$\epsilon > 0$$, there exists $$\delta>0$$ such that $$\forall\,x,y\in K, \;|x − y| < \delta,$$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}.$$ which implies that $$\{f_n\}$$ is equicontinuous.

QUESTION: I'm I right? If not, can you please provide a counter-example?

• What is the min of an infinite set of numbers? What is $\min\{1/n : n \in \mathbb{N}\}$? Your proof is flawed. – RRL Jan 5 at 8:04
• @RRL: Thanks for that! But how do I get an example? – Omojola Micheal Jan 5 at 8:07

To construct a sequence $$(f_n)$$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.
An example is $$f_n(x) = x^n$$ with $$K = [0,1]$$. To see that these are not equicontinuous, suppose that there exists $$\delta > 0$$ such that if $$|x - 1| < \delta$$ then $$|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$$ for every $$n$$. However, since $$x^n \to 0$$ when $$1- \delta < x < 1$$, with $$x$$ fixed we can find $$n$$ such that $$|x^n - 1| > 1/4$$ a contradiction.
• So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right? – Omojola Micheal Jan 5 at 8:25
• If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family $\{f_n\}$ may not be equicontinuous. – RRL Jan 5 at 8:27
• So, is it right to say that $f_n:K\subseteq \Bbb{R}\to\Bbb{R}$ is continuous for each $n\in \Bbb{N}$, then $\{f_n\}$ is uniformly continuous. So, for every $\epsilon > 0$ and $n\in \Bbb{N}$, there exists $\delta_n>0$ such that $\forall\,x,y\in K, \;|x − y| < \delta_n,$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}?$$ – Omojola Micheal Jan 5 at 8:27