# Question about the non-uniform convergence of $f(x)=2nxe^{-nx^2}$

I came across an analysis problem where I am asked to determine if $$f(x)=2nxe^{-nx^2}$$ converges to zero on the interval $$[0,1]$$ a) point wise and b) uniformly.

Part a was fine. I noted that $$|f_n -f|→0$$ as $$n→∞$$ since exponentials beat polynomials.

For b) I noted that $$f(1/\sqrt{n})=\sqrt[]{\frac 2e}\sqrt{n}$$ which goes to infinity of course, and so $$sup|f_n -f|$$ does not converge to $$0$$ as $$𝑛→∞$$ and so the convergence is not uniform. But since this supremum goes off to infinity doesn't this mean that the function doesn't converge to $$0$$ at $$x=1/𝑛$$? Wouldn't this contradict my part a answer which says the convergence is pointwise?

Thanks a lot!

No, there is no contradiction there. Here's a simpler example: define$$\begin{array}{rccc}f_n\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}n&\text{ if }x=\frac1n\\0&\text{ otherwise.}\end{cases}\end{array}$$Then, for each $$x\in\Bbb R$$, $$\lim_{n\to\infty}f_n(x)=0$$. But, but, if $$f$$ is the null function, $$\sup|f-f_n|=n$$ and therefore $$\lim_{n\to\infty}\sup|f-f_n|=\infty$$.
However, for each $$n\in\Bbb N$$, you still have $$\lim_{m\to\infty}f_m\left(\frac1n\right)=0$$, since $$f_m\left(\frac1n\right)=0$$ for every $$m\ne n$$.
• Thanks for your reply. But I don't see how $\lim_{n\to \infty}f_n (x)$ is $0$ for each $x$. Surely at $x=1/n$ we have $f_n (x)=n$ which goes to infinity as $n$ does?
• Why? Take $n=1$, for instance. And take $x=\frac1n=1$. Then $f_1(x)=1$, $f_2(x)=0$, $f_3(x)=0$, $f_4(x)=0$, and so on. So, $\lim_{n\to\infty}f_n(x)=0$. If you disagree, then please give me an example of a natural number $n$ such that $\lim_{m\to\infty}f_m\left(\frac1n\right)\ne0$. Apr 19 '20 at 15:43
• I see I see. I think my confusion stems from me forgetting that we are fixing x when talking about pointwise convergence. So with the function you gave the x value of the non-zero point gets closer to 0 when we increase $n$. And so if we fix $x$ we must have that $\lim_{n\to \infty}f_n (x)$ goes to 0 since there is at most only ever one potential value of n where $f_ n$ is non-zero and for any n greater than this $f_n$ will be 0. I understand where I have gone wrong now in the original question, thank you.