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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?

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Hint: $f_n$ is decreasing on $[\epsilon,\infty)$ if $\frac 1n < \epsilon$. Hence $\sup_{x\in [\epsilon,\infty)}f_n(x) = f_n(\epsilon)$.

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  • $\begingroup$ 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 $$\text{sup}_{x \in [0,\epsilon ]} |f_n(x) - 0| =f_n(\frac{1}{n}) =\frac{1}{e} \nrightarrow 0$$ $\endgroup$ – Phoenix_10 Oct 31 '19 at 12:57
  • $\begingroup$ 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". $\endgroup$ – amsmath Oct 31 '19 at 15:19
  • $\begingroup$ Yes that it is what I mean. So then for $\epsilon>\frac{1}{n}$, $[0, \epsilon) $ is not compact and and $[\epsilon, \infty) $ ? $\endgroup$ – Phoenix_10 Oct 31 '19 at 19:36

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