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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

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    $\begingroup$ How do you guarantee $\delta>0?$ BTW, $\delta_{\text{equi}}$ depends on $x$. $\endgroup$
    – saulspatz
    Jul 12, 2019 at 15:41

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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.

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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.

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