Why isn't $nxe^{-nx}$ uniformly convergent for $x \geq 0$?

Question

Why isn't $nxe^{-nx}$ uniformly convergent for $x \geq 0$?

The definition of uniform convergence states that $f_n$ is uniformly convergent if $$\lim_{n\to\infty} ||f_n-f|| = 0.$$

I apply the definition with $f = 0$ and get $$\lim_{n\to\infty}||\frac{nx}{e^{nx}} - 0 || = 0,$$ since $e^{nx}$ grows much faster than $nx$ regardless of $x$ (except when $x = 0$, but if that's the case everything is $0$).

Where am I wrong?

Answer 1

Note that$$(\forall n\in\mathbb{N}):f_n\left(\frac1n\right)=\frac1e.$$Therefore, your sequence does not converge uniformly to the null function (although it does converge pointwise).

Answer 2

For any given $x$ it's true that $nxe^{-nx}$ tends to zero as $n\to\infty$. However, the function $f_n$ as a whole, when viewed over all $x\ge0$, will always have a bump somewhere, which keeps the norm $||f_n-f||$ from getting small. The quickest way to see this: $nxe^{-nx}$ is simply the function $xe^{-x}$ but with a compressed $x$-axis. So each $f_n$ has the same maximum value.