# On which interval is $f_n = nx e^{-nx}$ is the pointwise limit of $f_n$ uniform convergent?

Given is that $$f_n(x) = nx e^{-nx}$$ on $$[0,\infty)$$
I computed that the pointwise limit is $$f(x)= 0$$ for all $$x \in[0,\infty)$$
I would like to compute the interval in $$[0,\infty)$$ such on which $$f_n(x)$$ is uniform convergent.
So I began as follows:
$$\text{sup}_{x \in [0,\infty)} |f_n(x) - 0| \\ = f_n(\frac{1}{n}) \\ =\frac{1}{e}$$ So $$f_n(x)$$ is not uniformly convergent on the whole of $$[0,\infty)$$ since $$\text{sup}_{x \in [0,\infty)} |f_n(x) - 0| =\frac{1}{e} \nrightarrow 0$$
On which intervals is it then uniformly convergent and on which is it not and why?

Hint: $$f_n$$ is decreasing on $$[\epsilon,\infty)$$ if $$\frac 1n < \epsilon$$. Hence $$\sup_{x\in [\epsilon,\infty)}f_n(x) = f_n(\epsilon)$$.
• Thanks, that helps. So then on $[0,\epsilon)$ it not uniform convergent and on $[\epsilon,\infty)$ it is right? Is this correct what I say: $f_n$ is uniform convergent on $[0,\epsilon]$ since $$\text{sup}_{x \in [0,\epsilon ]} |f_n(x) - 0| =f_n(\frac{1}{n}) =\frac{1}{e} \nrightarrow 0$$ – Phoenix_10 Oct 31 '19 at 12:57
• You mean it is not uniformly convergent on $[0,\epsilon]$, right? Your reasoning for this is of course correct. Your sequence is uniformly convergent on every compact subset of $(0,\infty)$, but not on $(0,\infty)$. This is called "locally uniform convergence". – amsmath Oct 31 '19 at 15:19
• Yes that it is what I mean. So then for $\epsilon>\frac{1}{n}$, $[0, \epsilon)$ is not compact and and $[\epsilon, \infty)$ ? – Phoenix_10 Oct 31 '19 at 19:36