Fourier transform of a special Schwartz function

In Classical Fourier Analysis by Loukas Grafakos we have in Proposition 2.3.25 the following definition for $\mathcal{S}_\infty(\mathbf{R}^n)$, namely that these are all the Schwartz functions $\phi$ such that for all multi-indices $\alpha$ we have that

$$\int_{\mathbf{R}^n} x^\alpha \phi(x) \, dx = 0.$$

Now I'm trying to find non-trivial functions in this space. I know that the Fourier transform maps the Schwartz-functions to itself, so I note that the requirement is actually that the Fourier transform evaluated in $0$ of $x^\alpha \phi(x)$ is $0$. So, $x^\alpha \mapsto i^\alpha d/dx^\alpha$ in the Fourier domain, so we actually want a function $\phi$ such that (let's take $n = 1$):

$$\left . \frac{d}{dx^\alpha} \widehat{\phi}(x) \right |_{x = 0} = 0.$$

For all $n \geq 0$ An obvious candidate is $f(x) = \text{exp}(-1/x^2)$ for $x > 0$ and $0$ otherwise.

However, if I now compute the Fourier transform (or the inverse) of this function (with Maple) I get another function say $g$, but if I plot the real part of $g$ it is not smooth (it has a cusp in 0), how is this possible? Further the integral which I want to be zero is only zero for odd $n$, for even $n$ it is complex! What goes wrong?

• Jonas, you can also regard this as saying that $\phi\in\mathcal{S}_\infty$ if $\phi$ is orthogonal to all polynomials. It is a moment problem - maybe you should take a look at Paul Koosis book "The logarithmic integral" (I learned a lot from this book). – AD. Nov 15 '10 at 22:32
• @AD: Thanks, I will look at it. – Jonas Teuwen Nov 16 '10 at 17:53

As $x\to\infty$, $\exp(-1/x^2)\to1$ so this isn't in the Schwartz space. Why not multiply this by $\exp(-x)$ to get it decaying to $0$ at infinity rapidly enough?
In his book "Modern Fourier Analysis", Grafakos is concerning radial Schwartzfunctions $\Psi$ with support in the annulus $1 - \frac{1}{7} \leq |\xi| \leq 2$, which also are constant $1$ in the smaller annulus $1 \leq |\xi| \leq 2 - \frac{2}{7}$ and satisfy$$\sum_{j \in \mathbb{Z}} \Psi(2^{-j} \, \xi) = 1$$ for all $\xi \in \mathbb{R}^n\setminus\{0\}$. Such functions can be constructed from the standard mollifier. You can imagine these functions as a partition of unity of the euclidean space. However, the function $\Phi = \hat{\Psi}$ lies in $\mathscr{S}_\infty$. To see this, choose $\alpha \in \mathbb{N}^n$ and consider the function $f(x) = x^\alpha \, \Phi(x)$. Then we have $$\int x^\alpha \, \Phi(x) = \langle 1, f\rangle = \, \langle \delta_0, f^{\vee}\rangle = c \langle \delta_0, \partial^\alpha \Psi \rangle = 0,$$ since $\Psi$ has compact support away from the origin. But $\alpha$ was arbitrary, so we just found a non trivial example of a function which has vanishing moments of all orders.