I proved that $f_n(x)$ converges uniformly to $0$ on $[0,1]$ on the previous problem. In that case I could easily show that $$\forall \epsilon>0, |f_n(x) - f(x)| = |\frac{x}{n^2} - 0| < \epsilon \quad\forall n > N = \frac{1} {\sqrt{\epsilon}}$$ where I could find $N$ by manipulating inequality $\frac{x}{n^2} < \epsilon$ since I knew $x=1$ at maximum.
With the interval $[0, \infty)$, I can't use that same argument because $x \ne 1$ at maximum? How would I go about this? I am assuming $f_n$ converges to 0 uniformly in this new interval as well.
But it doesn't converge uniformly on that interval. Note that there is no $N\in\mathbb N$ such that the set$$\left\{\bigl\lvert f_n(x)-0\bigr\rvert\,\middle|\,x\in[0,\infty)\right\}$$is bounded. In particular, t is not true that there is some $N\in\mathbb N$ such that$$n\geqslant N\implies\bigl(\forall x\in[0,\infty)\bigr):\bigl\lvert f_n(x)-0\bigr\rvert<1.$$