# Why does $(f_{n})_{n}$ equicontinuous and $f_{n}$ uniformly continuous not imply $(f_{n})_{n}$ uniformly equicontinuous

Why does $$(f_{n})_{n}$$ equicontinuous and $$f_{n}$$ uniformly continuous not imply $$(f_{n})_{n}$$ uniformly equicontinuous. I have already seen the example of the case $$(f_{n})_{n}$$ where $$f_{n}(x)=\arctan{(nx)}$$ where $$x \in ]0,\infty[$$. But I fail to understand why it is so?

My logic: Since $$(f_{n})_{n}$$ equicontinuous, for any $$x$$ and $$\epsilon > 0$$, there exists a $$\delta_{\operatorname{equi}} >0$$ so that for any $$y: \vert x-y\vert<\delta_{\operatorname{equi}}\Rightarrow$$ $$\vert f_{n}(x)-f_{n}(y)\vert<\epsilon$$ for all $$n \in \mathbb N$$. And further, since for each respective $$n \in \mathbb N$$, $$f_{n}$$ is uniformly continuous, i.e. there exists $$\delta_{\operatorname{unif}}>0$$ so that for $$\vert x-y\vert<\delta_{\operatorname{unif}}\Rightarrow$$ $$\vert f_{n}(x)-f_{n}(y)\vert<\epsilon$$, couldn't we get uniform equicontinuity by taking the minimum, namely $$\delta:=\min\limits_{n \in \mathbb N} \{\delta_{\operatorname{unif_{n}}},\delta_{\operatorname{equi}}\}$$. I can see that my idea is wrong but I would just like clarity

• How do you guarantee $\delta>0?$ BTW, $\delta_{\text{equi}}$ depends on $x$. – saulspatz Jul 12 '19 at 15:41

## 2 Answers

An infinite set that is bounded below does not need to attain its minimum. Your expression $$\delta:=\min\limits_{n \in \mathbb N} \{\delta_{\operatorname{unif_{n}}},\delta_{\operatorname{equi}}\}$$ is not well-defined. For example, says we have $$\delta_{\operatorname{unif_{n}}}=\frac 1n$$, then $$\left\{\delta_{\operatorname{equi}},1, \frac 12, \frac 13, \dots \right\}$$ has $$0$$ as its infimum so you'd need $$\delta=0$$ here.

You could also consider a sequence of "tent" functions, f_n, defined on (0, 2] by letting f_n be the function that increases linearly from 0 to 1 on the interval (0, 1/n] and then descends down to 0 linearly on [1/n, 2/n]. Clearly they are uniformly continuous and they are pointwise equicontinuous because for any fixed x in (0, 2] all tents will eventually be to the left of x so you only have to worry about finitely many of them. However, if the f_n s are defined on a compact metric space, say $$C$$, you can show that pointwise equicontinuity implies uniform equicontinuity. Indeed, fix $$\epsilon>0$$ and for each $$x\in C$$ let $$\delta_x>0$$ be such that for each $$n$$, $$d(x, y)<\delta_x \rightarrow d(f_n(x), f_n(y))<\epsilon$$. By compactness $$\{B(x; \delta_x) : x\in X\}$$ has a finite subcover, say $$B(x_1; delta_{x_1}), B(x_2; delta_{x_2}), \cdots, B(x_1; delta_{x_1})$$. Define a continuous function, $$C \rightarrow (0,\infty), x \mapsto \sup\{r : B(x;r) \text{ is contained in one of the } B(x_i; delta_{x_i})\}$$. It attains a minimum value, say $$\delta>0$$ and this is the delta you need to prove uniform equicontinuity.