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

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.

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

## 1 Answer

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 $$f_n'(x)\approx n$$).