Can you tell me if the following is right:
When arguing why we have that the characters of the torus are complete for $L^2$ we in particular need to show that the closure of their linear hull is $L^2$. We can show this by showing that the characters are an algebra containing 1, separating points. (If we have that we can apply Stone-Weierstrass.)
What I don't understand about this: Stone-Weierstrass gives me density with respect to $\|\cdot\|_\infty$. But here I want density with respect to $\|\cdot\|_{L^2}$.

@experimentX: If you are interested, there is another method. Since in laplace transform we do not care for values of $x<0$ we can put $|sin(x)|$ as following
\begin{equation}
|sin(x)| = sin(x) + u(t-\pi) \left( sin(t- \pi) + u(t-2\pi) \left( sin(t-2\pi) + ... \right) \right)
\end{equation}

Take the series, evaluate the nth term and then add them up. The method is relatively self-contained since you don't have to explicitly evaluate integrals.

This questions seems interesting, if there is already no such question on the site, we might consider putting it up. What do you say?