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For which $f \in S(R^n)$, the Schwartz class, $\hat f \in D(R^n)$ ?

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Note that such functions must be real analytic, and take a look at the Paley-Wiener theorem. – Willie Wong Mar 9 '11 at 14:05
For instance, check Theorem 7.3.1 in the first volume of Lars Hormander's treatise – Piero D'Ancona Mar 9 '11 at 17:19
Just to expand Willie Wong's comment slightly: the Paley-Wiener Theorem exactly characterises the inverse Fourier transform of $L^2(-A,A)$. Since the Schwartz function $\widehat{f}$ is compactly supported if and only if it is in $L^2(-A,A)$ for some $A$, you immediately get the answer: $f$ extends to an entire function of exponential growth, which is $L^2$ on every horizontal line. See e.g. Wikipedia: I quite like the treatment in Koosis, Theory of $H_p$ spaces; many books on Hardy Spaces give it. – Zen Harper Mar 13 '11 at 12:02
Let me insist with Hormander's version :)) If you read through the book, you will see that the true spirit of the result is not exactly compact support vs. exponential growth, but, more precisely: if the support is contained in a closed convex set, then you can bound the growth of the F.transform in the dual directions, and viceversa – Piero D'Ancona Mar 29 '11 at 8:48

Gelfand-Shilov, Generalized Functions, Vol. 1 gives a rather extended discussion of this function class, as well as the corresponding class of distributions.

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Hmmm, Paley wiener provides a way through it. Since we have the result,

$\check f $ = $\hat \hat f$ (This result holds for $S(R^n)$,as periodicity of Fourier transform is 4)

Let $C$ ={ $ f \in S(R^n)$ | $\hat f \in D(R^n)$ }

Then we know that $f \in C$, $\check f $ is an entire function with exponential bound. Then $f$ will also be entire function with exponential bound.

Converse of Paley-Wiener Theorem will ensure that this is exactly the class. I hope I didn't make any mistake. :D

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