Is $f_n$ uniformly convergent on $(-\infty,\infty)$

Question

Let $f_n(x)=\frac{2n^2x}{1+n^4x^4}$ where $n\in\mathbb{N}$ and $x\in(-\infty,\infty)$.

I've already shown that this function is pointwise convergent to the limit function $f(x)=0$. My question is whether it is uniformly convergent anywhere on $(-\infty,\infty)$.

Answer 1

We have uniform convergence on $[a,\infty)$ and $(-\infty, -a]$ for any $a > 0$.

Note that

$$\left|\frac{2n^2x}{1+n^4x^4}-0 \right| = \frac{2n^2|x|}{1 + n^4||x|^4} \leqslant \frac{2n^2|x|}{n^4||x|a^3} = \frac{2}{n^2a^3} \underset{n \to \infty}\longrightarrow 0$$

Convergence is not uniform on any set $S$ with $0$ as an interior or boundary point -- and, thus, $(-\infty,\infty)$ -- since

$$\sup_{x \in S}\left|\frac{2n^2x}{1+n^4x^4}-0 \right| \geqslant \frac{2n^2\cdot \frac{1}{n}}{1 + n^4 \cdot \left(\frac{1}{n} \right)^4} = n \underset{n \to \infty}\longrightarrow \infty $$

Answer 2

Note that $f_n(\frac{1}{n})=n$, which is very far from the limit function. So for every $n$ we have:

$\sup \{x\in\mathbb{R}: |f_n(x)-f(x)|\}\geq f_n(\frac{1}{n})-f(\frac{1}{n})=n$

So in particular the sequence of supremums doesn't converge to zero. There is no uniform convergence.