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
Best Answer
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.