Proving that a sequence of function does not uniformly converge

Question

I am trying to prove that the sequence of functions $f_n (t) = n^2 t e^{-nt}$ does not converge uniformly on $\mathbb{R}_{\geq 0}$. I was able to show that it converges point-wise to the zero function. To prove that it doesn't converge uniformly, I need to show that $$ \exists x, \exists \epsilon, \forall N, \exists n > N, |f_n (t)| \geq \epsilon. $$ I found that $f_n (t)$ has a maximum of $\frac{n}{e}$ at $t = \frac{1}{n}$, but I'm stumped because I can't choose an $x$ that depends on $n$. Is it preferable to proceed by contradiction?

Answer 1

Your negation is not correct as written, it seems to me. Uniform convergence, once we know that the limit is $0$, is $\def\e{\varepsilon}$

$$\tag1 \forall \e>0,\ \exists n_0,\ \forall n\geq n_0,\ \forall x,\ |f_n(x)|<\e. $$

Hence the negation is

$$\tag2 \exists \e>0,\ \forall n_0,\ \exists n\geq n_0,\ \exists x,\ |f_n(x)|\geq\e. $$

So we need to find an $\e>0$ and, given an arbitrarily large $n$, an $x_n$ such that $|f_n(x_n)|\geq\e$. Since the function is uniformly continuous away from $0$ (being bounded, continuous, with limit zero at infinity) the problem has to occur at $0$. If we take $x_n=\frac1n$, then $$ f_n(x_n)=n^2\frac1n\,e^{-n\,\frac1n}=ne^{-1}. $$ So if we take $\e=e^{-1}$, given any $n_0\in\mathbb N$ we can take $n=n_0+1$, and $x_n=\frac1n$, to get $f_n(x_n)\geq e^{-1}$.