Showing $f_n(x)=nxe^{-nx^2}$ converges nonuniformly

Question

If $f_n(x)=nxe^{-nx^2}$ and $\lim_{n\to \infty}f_n(x)=f(x)$ show that the sequence converges nonuniformly on the interval $0\le x \le1$ and that $$ \int_0^1f(x)dx\ne \lim_{n\to \infty}\int_0^1 f_n(x)dx \quad(*) $$

I'm not quite sure where to start on showing that it isn't uniformly convergent. But for the second part I started to show that if $f_n$ was uniformly convergent then the equality $(*)$ would hold.

It seems to be the case that $$\lim_{n\to \infty}f_n=0=f(x)$$ since we have an exponential in the denominator.

So I tried to solve the $RHS$ of $(*)$:

$$\lim_{n\to \infty}\int_0^1 \frac{nx}{e^{-nx^2}}dx=\lim_{n \to \infty} \frac{2}{n}e^{-n}=0 $$

So I'm stuck on showing the second part as well since I get $0$ for both sides.

Answer 1

For every fixed $x$, $f_n(x)\to0$ hence $f=0$, but $f_n(1/\sqrt{n})=\sqrt{n}/\mathrm e$ hence the convergence $f_n\to f$ is not uniform.

Furthermore, $f_n(x)=\sqrt{n}f_1(\sqrt{n}x)$ hence $$ \int_0^1f_n=\int_0^{\sqrt{n}}f_1\to\int_0^{+\infty}f_1=\left[-\frac12\mathrm e^{-x^2}\right]_0^{+\infty}=\frac12\ne0=\int_0^1f. $$

Answer 2

$f_n$ converges nonuniformly to 0, because the maximum of $xe^{-nx^2}$ on $[0, 1]$ is $e^{-1/2}/\sqrt{2n}$ (indeed, $f_n'(x) = ne^{-nx^2}(1-2nx^2)$, which is $0$ when $x = \sqrt{1/(2n)}$). So the maximum of $f_n$ is in $O(\sqrt n)$.