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If $m(\xi)$ satisfies $$D^{\alpha}m(\xi)\leq \frac{C}{(1+|\xi|)^{|\alpha|+1}}$$ then is $m$ a Fourier transform of a $L^{1}$ function? (Note that the Bernstein theorem can't be applied here, since $m(\xi)$ may not be in $H^{s}$, where $s>\frac{n}{2}$.)

Generally, are there some simple ways to make sure that a given function belongs to $\mathcal{F}L^{1}$?

Kuba Helsztyński
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Tomas
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1 Answers1

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There is a well known class of functions known as Schwartz class. In this class, the functions have the property that they and their derivatives tends to zero as $|x|\rightarrow \infty $, faster than any positive power of $x^{-1}$, or in other words, suppose that for each positive integer $N$ and $n$, $$ \lim_{|x|\rightarrow \infty} x^N g^{(n)}(x) = 0\,. $$

Also, these kind of functions are known as good functions. For example, $ x^m {\rm e}^{-x^2} $ is a good function. One of the properties of these functions is

$$ |f(x)| < C \frac{1}{(1+|x|)^m} $$ for any $m \in N \,.$

This space of functions plays an important rule in Fourier analysis, since the Fourier transform of a good function is well defined (you can use the above property to show this) and it is a good function.

Mhenni Benghorbal
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    Right,$\mathcal{FS}$ belonngs to any $L^{p}$,$1\leq p \leq \infty$,and it's well known that $\mathcal{F}L^{1}\subset C_{0}$,so I'm very interested of how to identify a fuction(together with it's derivatives) of some decay at $\infty$ is the fourier transform of a integral function(see the example above) – Tomas Aug 28 '12 at 09:00