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 $\sup_{x\in [\epsilon,\infty)}f_n(x) = f_n(\epsilon)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.