Why the following sequence of function does not converge uniformly at $[0, \infty)$

Why the following sequence of function does not converge uniformly at $[0, \infty)$ but converge uniformly for some $a>0, [a,\infty)$

$$f_n(x) := n^2x^2e^{-nx}$$

So I know the limit function $f$ is $f=0$. Hence $\lim _{n\rightarrow\infty} \left \| f_n - 0 \right \| = 0$.

Shouldn't this mean uniform convergence, but why is this not true if I include $0$ in my interval. Any help or insights to this is deeply appreciated. If I made a mistake in my working I would be very grateful if it can be pointed out.

Thank you.

$f_n(1/n) = 1/e$ for all $n$, so the sup-norm of $f_n$ on $[0,\infty)$ is at least $1/e$.

• Wow, a one sentence proof, but hits where it should ! Thank you! Oct 13 '16 at 3:41

all $f_n$ are positive.

the derivative of $f_n$ is

$f'_n(x)=n^2xe^{-nx}(2-nx)$

it reaches its maximum at

$x=\frac{2}{n}$

which is

$max(|f_n(x)-0|)=f_n(\frac{2}{n})=2e^{-1}$

this doesn't go to 0 when $n$ goes

to $+\infty$ and the convergence is not uniform on $[0,+\infty)$.

but on $[a,+\infty)$ with $a>0$, the maximum will be $f_n(a)$ for enough large $n$ since $f_n$ is decreasing on $[a,+\infty)$.

( $n>n_0$ with $\frac{2}{n_0}<a$).

this max goes to $0$ when $n$ tends to $+\infty$ and the convergence is uniform.