Let $f_n:[0, +\infty)\rightarrow \mathbb{R}$, $f_n(x)=xe^{-nx}$.

Show that $(f_n)_n$ converges pointwise and calculate $$f(x):=\lim_{n\rightarrow +\infty}f_n(x)\ \ \ \forall x\in [0,+\infty)$$ Does $f_n$ converge also uniformly to $f$ on $[0, +\infty)$ ?


I have done the following:

We have that \begin{equation*}f(x)=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}xe^{-nx}=0 \text{ since } x\geq 0\end{equation*} So the sequence of functions converges pointwise on $[0, +\infty)$ to $0$.

Is that the way to show that $(f_n)_n$ converges pointwise? Or do we have to prove that in an otherway because after that question it is asked to calculate the limit?

As for the uniform convergence:

Let $\epsilon> 0$ arbitrary.

It holds that $\displaystyle{\lim_{y\rightarrow +\infty}ye^{-y}=0}$.

So there is a $M>0$ such that $0<ye^{-y}<\epsilon$ for all $y>M$.

For each $n\in \mathbb{N}$ with $n>M$ we get $\frac{1}{n}<\frac{1}{M}$.

We have that \begin{equation*}|f_n(x)|=\left |xe^{-nx}\right |=\left |\frac{1}{n}\cdot nxe^{-nx}\right |=\left |\frac{1}{n}\right |\cdot \left |nxe^{-nx}\right |\ \overset{y:=nx}{=}\ \left |\frac{1}{n}\right |\cdot \left |ye^{-y}\right |<\frac{1}{M}\cdot \epsilon\end{equation*} So we get also uniform convergence on $[0, +\infty)$ to $0$.

Is everythig correct?

Or do we not need to use $\displaystyle{\lim_{y\rightarrow +\infty}ye^{-y}=0}$ ?


1 Answer 1


Your proof of pointwise convergence is fine.

Concerning uniform convergence, note that $f_n'(x)=e^{-nx}(1-nx)$ and that it follows from this that $f_n$ is increasing on $\left[0,\frac1n\right]$ and decreasing on $\left[\frac1n,\infty\right)$. Therefore, $\max f_n=f_n\left(\frac1n\right)=\frac1{ne}$. Since $\lim_{n\to\infty}\max f_n=0$, the convergence is uniform.

  • $\begingroup$ Could you explain to me how it follows that the convergence is uniform knowing that $\lim_{n\to\infty}\max f_n=0$ ? $\endgroup$
    – Mary Star
    Jan 13, 2021 at 20:02
  • 1
    $\begingroup$ Take $\varepsilon>0$. There is some $N\in\Bbb N$ such that$$n\geqslant N\implies\max f_n<\varepsilon.$$But this is the same thing as asserting that$$n\geqslant N\implies\max|f_n-0|<\varepsilon,$$from which it follows that, if $n\geqslant N$ and if $x\in[0,\infty)$, then$$\bigl|f_n(x)-0\bigr|<\varepsilon.$$ $\endgroup$ Jan 13, 2021 at 20:07
  • $\begingroup$ The last part follows because $\max|f_n-0|<\varepsilon$ and this is the maximum as for the variable $x$ and so $|f_n-0|<\max_x|f_n-0|<\varepsilon$, right? $\endgroup$
    – Mary Star
    Jan 13, 2021 at 20:27
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    $\begingroup$ Yes, that is correct. $\endgroup$ Jan 13, 2021 at 21:58
  • $\begingroup$ Great!! Thank you very much!! :-) $\endgroup$
    – Mary Star
    Jan 13, 2021 at 22:08

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