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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?

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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
up vote 3 down vote accepted

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?

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My savior! Thanks. – Jonas Teuwen Nov 15 '10 at 21:25

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