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 $0\notin\textrm{supp}(g)$.

  • $\begingroup$ So the idea is to use the time integration formula $$ \mathcal F \Big[\int_{-\infty}^t f(x)dx \Big](w) = \frac{\hat f(w)}{w} + \hat f(0) \delta(w)$$ (with some additional constant factors depending on the normalization of $\mathcal F$) with $\hat f = g$ as you described above. I checked and this seems to work nicely! (I wasn't really aware of this formula) The only thing we need of course is not simply $0\notin\text{supp}(g)$, but rather $g^{(k)}(0) = 0 \implies \lim_{x\to 0} \frac{g(x)}{x^k} = 0$ for all $k\ge 0$ $\endgroup$ – Hyperplane Apr 8 '17 at 13:13
  • $\begingroup$ So one concrete example is given by $$f = \mathcal F[e^{-(\frac{1}{x^2} + x^2)}]$$ which looks similar to a $\text{sinc}$ function. Is it then correct to say that the qualitative difference between $f$ and functions of the type $p(x)e^{-x^2}$ is that the latter only features a finite amount of oscillations ($\sim$ number of isolated zeroes), and the number of oscillations decreases under integration. Thus a finite order antiderivative of it is either positive or negative for all $x$. This doesn't happen necessarily with $f$ since you have an infinite amount of oscillations to begin with. $\endgroup$ – Hyperplane Apr 8 '17 at 13:22
  • $\begingroup$ @Hyperplane: The support is defined as the closure of the set of points where $g\not=0$, so $0\notin\textrm{supp}(g)$ means that $g=0$ in a neighborhood of zero. But your example (a $g$ that vanishes at zero fast enough) works fine too. $\endgroup$ – user138530 Apr 8 '17 at 16:09

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