Consider the sequence of functions $f_n(x) = nxe^{-nx}$. We notice that if $x < 0$ the sequence diverges to $-\infty$ while if $x \geq 0$ the sequence converges to $0$. This is the analysis of the pointwise convergence of $f_n$. The case $x >0$ can be made rigorous noticing that $e^{y} > 1 + y+ \frac{y^2}{2}$. And using this to conclude that
$$0 < f_n(x) < \dfrac{2}{nx}.$$
In that sense we notice that by the archemedian property of $\Bbb R$ there is $n_0\in \Bbb N$ such that $n_0 > 2/\epsilon x$. Thus if $n \geq n_0$ we have $n > 2/\epsilon x$ and hence $1/n < \epsilon x/2$ and then
$$|f_n(x)| < \dfrac{2}{nx} < \dfrac{\epsilon x}{2}\dfrac{2}{x} = \epsilon.$$
We have that $n_0$ obviously depends on $x$ so it appears the convergence is not uniform.
Anyway I don't know how to prove this. My first instict is that we should use the supremum norm and consider $\sup \{|f_n(x)| : x\in (0,\infty)\}$ and we should try to prove this converges to zero, but I'm quite unsure if that's the way. Nor I know how to compute that supremum.
So how can I decide whether the convergence is uniform or not and how can I indeed prove it?
Best Answer
In the topology of uniform convergence, a sequence of functions $g_n$ tends to a function $g$ if and only if $\sup_{x} \lvert g_n(x) - g(x) \rvert \to 0$ as $n \to \infty$.
Thus to see if the convergence to zero is uniform, we need only check the maximum of each $f_n$. If the sequence of maxima goes to zero then the convergence is uniform; otherwise it isn't. Here we see $$f_n'(x) = ne^{-nx} -n^2xe^{-nx} = ne^{-nx}(1 - nx).$$ Then $f'(x) = 0$ iff $x = 1/n$. It is straight-forward to check that these points are maxima. At these points, we see $$f_n(1/n) = 1/e.$$ Thus $\sup_x \lvert f_n(x) - 0 \rvert \not \to 0$ so the convergence is not uniform.