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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?

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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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Yep, that answers it. Thanks! –  Isaac Goldbring Nov 1 '13 at 21:37
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More generally: If $f:V\to V$ is an endomorphism and $P(X)=\prod_i (X-a_i)$ a polynomial with $P(f)=0$ and pairwise distince zeros $a_i$, then the projection onto the $a_i$-Eigenspace of $f$ is given by $\prod_{j\neq i} \frac{f-a_j}{a_i-a_j}$. –  Johannes Hahn Nov 1 '13 at 22:38

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