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 $n \leq N$?

  • 4
    $\begingroup$ Possible duplicate of Show that f is uniformly continuous and that $f_n$ is equicontinuous $\endgroup$ – iamvegan Sep 21 '16 at 1:18
  • $\begingroup$ The proof in that post omitted the case when n<= N. $\endgroup$ – user1559897 Sep 21 '16 at 1:20
  • $\begingroup$ You should look at the second solution $\endgroup$ – iamvegan Sep 21 '16 at 1:23
  • $\begingroup$ 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" $\endgroup$ – 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 $n$.


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