Let $X=]0,\infty[: \{x \mapsto \arctan{(nx)}: n \in \mathbb N \}$

so $f_{n}: ]0,\infty[ \to \mathbb R, x \mapsto \arctan{(nx)}$ for all $n \in \mathbb N$

There is a statement, saying (w.r.t. $d_{\infty}$):

$1.$ $(f_{n})_{n}$ is equicontinuous

$2.$ $(f_{n})_{n}$ is not uniformly equicontinuous

$3.$ each function $f_{n}$ is uniformly continuous.

The difference between $1.$ and $2.$ is clear but I do not understand why $(f_{n})_{n}$ is equicontinuous because with every increasing $n$ my slope close to $0$ increases, so my chosen $\delta$ will need to get smaller and smaller.

And surely if $1.$ and $3.$ are true our sequence $(f_{n})_{n}$ necessarily has to be uniformly equicontinuous. Why is this not the case, any explanations are greatly appreciated.

  • $\begingroup$ For 1, notice that $0$ is not included in $X$ and at any point $x\neq 0$ the functions "become eventually flat." $\endgroup$
    – Pavel
    Jun 14, 2019 at 6:44

1 Answer 1


In equicontinuity, we have a $\delta$ dependent of $\epsilon$ and the point (say $x_0$) but not of $n$. In uniform equicontinuity, the dependence is only of $\epsilon$.

In the case of your sequence, as $$f_n'(x) = \frac{n}{n^2 x^2 + 1},$$ for any fixed $x_0$ we have $$x > x_0/2\implies |f_n'(x)|\le\frac 4{n x_0^2}\le\frac 4{x_0^2}.$$ The bound is independent of $n$ (equicontinuity is true), but dependent of $x_0$ (uniform equicontinuity is false and fails nearing $0$ where $f_n'(x)\approx n$).


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