For a Schwartz function $f\in\mathcal S(\mathbb R)$ it is known that
$$\exists F\in\mathcal S(\mathbb R): F'=f \iff \int\limits_{-\infty}^{+\infty}f(x) dx = 0 $$
in this case $F(x) = \int_{x}^{+\infty}f(t)dt$, see related question here.
Question: Does there exist $f\in\mathcal S(\mathbb R)$ with $f^{(-k)}\in\mathcal S(\mathbb R)$ for all $k\ge 0$ (my guess would be no)
I.e. for a given $f$ we can test if it has an antiderivate $F$ in the Schwartz space by simply checking if its mean is $0$. Then we can do the same with $F$. Does this process always stop at some point?
My work: It seems that all functions of the type $f(x) = p(x)e^{-\alpha x^2}$ with $p$ polynomial only have antiderivatives of finite order in $\mathcal S(\mathbb R)$. (I think I can prove this). As one can construct an orthonormal basis of $\mathcal S(\mathbb R)$ with such functions, e.g. the Hermite functions, I would guess that there is no function with antiderivates of arbitrary order in $\mathcal S(\mathbb R)$.
Remark: If there is such an $f$, then $(x^k \star f) = 0$ for all $k\ge 0$; then via Fourier transform, this would imply that in the sense of distributions: $$0 = \mathcal F[x^k\star f] = \mathcal F [x^k] \mathcal F[f] = \Big(\frac{i}{2\pi}\Big)^k\delta^{(k)}(w)\hat f(w)$$
You have pretty much answered your own question already: looking at the Fourier transforms, we want a $g\in\mathcal S$ such that $g(\xi)\xi^{-k}\in\mathcal S$ also for all $k\ge 1$. This is easy: our only potential problem occurs at $\xi =0$, so it suffices to take a $g\in\mathcal S$ with $0\notin\textrm{supp}(g)$.