TITLE: eigenfunctions of the Fourier transform in the Schwartz space
QUESTION [4 upvotes]: Recall that $L^2(\mathbb R)$ decomposes into the direct sum of the eigenspaces of the Fourier transform corresponding to its four eigenvalues, namely the four fourth roots of unity.  If $f\in L^2(\mathbb R)$ actually belongs to the  Schwartz space  of functions (namely $\|x^mf^{(n)}(x)\|_\infty$ is finite for all $m,n\in \mathbb N$), do the orthogonal projections of $f$ onto the four eigenspaces also lie in the Schwartz space?

REPLY [7 votes]: Let F denote the Fourier transform, and let f be a given function. Then consider the decomposition
$$f=(f_1+f_2+f_3+f_4)/4,$$
where
$$f_1=f+Ff+F^2f+F^3f,$$
$$f_2=f+iFf-F^2f-iF^3f,$$
$$f_3=f-Ff+F^2f-F^3f,$$
$$f_4=f-iFf-F^2f+iF^3f.$$
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