# How should I prove a uniformly continuous sequence of real-valued continuous function is equicontinuous?

How should I prove a uniformly continuous sequence of real-valued continuous function is equicontinuous? how should I handle case when $n \leq N$?

This is what I have so far.

Given $\epsilon>0$, there exists N s.t. for all $n\geq N$ we have $|f_n-f|<\epsilon$.

Whenever $\rho(x,y)<\epsilon$, $|f_n(x)-f_n(y)|\leq|f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_(y)|$.

However, how should I handle case when $n \leq N$?

• Possible duplicate of Show that f is uniformly continuous and that $f_n$ is equicontinuous – iamvegan Sep 21 '16 at 1:18
• The proof in that post omitted the case when n<= N. – user1559897 Sep 21 '16 at 1:20
• You should look at the second solution – iamvegan Sep 21 '16 at 1:23
• the second answer specifically mentioned "I'll leave it to you that this δδ will be good enough to show equicontinuity for all (fn)n∈N" – user1559897 Sep 21 '16 at 1:28

Since $f_n$ is uniformly convergent, for $n \leq N$ define $\delta_n >0$ such that $$|x-y|<\delta_n \quad \Rightarrow \quad |f_n(x)-f(y)|<\epsilon$$ Now define $$\delta := \min \{ \delta_1 , \cdots , \delta_N \}.$$ Using this $\delta$ and the inequality you wrote in your post, you can write $$|x-y|<\delta \quad \Rightarrow \quad |f_n(x)-f(y)|<\epsilon$$ for all $n$.