My question concerns the following problem
Let $x_n(t) = t^n$ for $n\ge 1$. Is $\mathrm{span}(x_n; n\ge 1)$ dense in $L^1([0,1])$?
By Weierstrass' approximation theorem, it suffices to check whether the constant function $1$ is contained in the $L^1$-closure of the linear span of the $x_n$.
I think this is not the case, but I haven't found a proof so far.
Some simple observations: I know that if $1$ is contained in the $L^1$-closure, then we can find a sequence of polynomials $p_n\in \mathrm{span}(x_n; n\ge 1)$ such that $p_n\to 1$ in $L^1$, almost everywhere and almost uniformly. Furthermore the sequence $q_n(x) = \int_0^x p_n(t) \, dt$ will converge uniformly to $x$ on $[0,1]$.
I don't see how this helps, though.