MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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 $$f_1=f+Ff+F^2f+F^3f,$$ $$f_2=f+iFf-F^2f-iF^3f,$$ $$f_3=f-Ff+F^2f-F^3f,$$ $$f_4=f-iFf-F^2f+iF^3f.$$

share|cite|improve this answer
Yep, that answers it. Thanks! – Isaac Goldbring Nov 1 '13 at 21:37
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

Your Answer


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.