I am trying to show that $f_n(x) = \frac{nx}{1+n^2x^2}$ where $x\in \mathbb{R}$ does not converge uniformly.
I have made an attempt and would like to make sure that I am going about it in the right way (i.e. if my thought process is correct or I am missing something).
First note that $f_n(x) \rightarrow 0$ pointwise and so if it converges uniformly it should converge to $f(x) = 0$ as well. To show this is not the case, note that $f_n(\frac{1}{n}) = 1/2$.
Hence, given $\epsilon = \frac{1}{3}$ for all $N\in \mathbb{N}$ we can find $n\geq N$ and $x\in \mathbb{R}$ where $|f_n(x)-f(x)|\geq \epsilon$, indeed we can just take $n = N$ and $x = \frac{1}{n} \in \mathbb{R}$ and $|f_n(x)-f(x)| = |f_n(\frac{1}{n})| = \frac{1}{2}>\frac{1}{3} = \epsilon$.
This shows that $(f_n)$ does not converge uniformly.
Best Answer
The negation of uniform convergence on a set $A$ is, there exists a number $\epsilon>0$, such that for all $N$, there exists a number $n>N$ and an $x\in A$ with $|f_n(x)-f(x)|\ge \epsilon$.
If one finds a number $\epsilon>0$ such that for all $N$, there exists an $x\in A$ with $|f_n(x)-f(x)|\ge \epsilon$ for all $n>N$, then certainly this more than suffices as proof of uniform convergence on $A$.