# Pointwise and uniform convergence of $f_n(x)=xe^{-nx}$

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 $$0 for all $$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 $$\displaystyle{\lim_{y\rightarrow +\infty}ye^{-y}=0}$$ ?

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.
• Could you explain to me how it follows that the convergence is uniform knowing that $\lim_{n\to\infty}\max f_n=0$ ? Commented Jan 13, 2021 at 20:02
• 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$$\bigl|f_n(x)-0\bigr|<\varepsilon.$$ Commented Jan 13, 2021 at 20:07
• 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? Commented Jan 13, 2021 at 20:27