4
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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 $\prod_{j\neq i} \frac{f-a_j}{a_i-a_j}$. $\endgroup$ – Johannes Hahn Nov 1 '13 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.