Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer 1

up vote 2 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?

share|improve this answer
    
My savior! Thanks. –  Jonas Teuwen Nov 15 '10 at 21:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.