Application de Stone-Weierstrass

Solution 1:

Edit: I only achieved the part that C.R. took for granted in his argument. Since it's already written, I leave it here for the sake of completeness.

Let $p_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^{k}}{k!} x^{k}$ be the $n$th Taylor polynomial of $e^{-x}$. Then by the Lagrange remainder theorem we have $e^{-x} = p_{n}(x) + \frac{(-1)^{n}}{(n+1)!} e^{-\xi} x^{n+1}$ for some $\xi \in [0,x]$.

Therefore $|e^{-2x} - e^{-x} p_{n}(x)| \leq \frac{e^{-\xi}}{(n+1)!} e^{-x} x^{n+1}$ and we want to show that the right hand side converges to zero uniformly in $x$. As $e^{-\xi} \leq 1$, it suffices to find the maximum of $e^{-x}x^{n+1}$. Differentiation yields that the maximum is located at $x = n+1$ and its value is $\frac{(n+1)^{n+1}}{e^{n+1}}$. On the other hand, Stirling's formula tells us that $n! = \sqrt{2\pi n} \left(\frac{n}{e}\right)^{n} e^{r_{n}}$ with $|\frac{1}{12n} - r_{n}| \leq \frac{1}{120n^2}$. Putting these things together we get \[ |e^{-2x} - e^{-x} p_{n}(x)| \leq \frac{1}{\sqrt{2\pi (n+1)}} e^{-r_{n+1}} \xrightarrow{n\to\infty} 0 \] as we wanted.