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 $→∞$ 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 $x=1/$? 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 $m\ne n$.