# Tag Info

14

Given a square-integrable, positive semi-definite function $f$, with its Fourier transform $\hat{f}$, then the function $$F=f^2+\hat{f}\star\hat{f},$$ with $\star$ the convolution, is its own Fourier transform: $\hat{F}=F$. If we require that $F$ is a probability density (absolutely integrable and positive semi-definite), then any $F$ with $\hat{F}=F$ is ...

10

I am somewhat confused that, despite saying many true and useful things, no one has said directly that $H^{-1}$ on the circle can be characterized as the set of distributions $\theta$ such that $\sum_{n\in \mathbb Z}(1+n^2)^{-1}\cdot |\theta(x\to e^{inx})|^2<\infty$. (All distributions on the circle are compactly supported, so can be applied to the ...

6

There is a whole chapter in Titchmarsh, Fourier transform, describing all eigenfunctions in great detail. He calls them "self-reciprocal" functions. A more difficult question is about the eigenvectors of discrete Fourier transform, but this one you do not ask. Edit. Since you are asking now about discrete transform too, it has 4 eigenspaces corresponding to ...

6

If $1\leq p<2$ then $\mathscr{F}: L^p \to L^{p'}$ is not surjective. I had this as a homework problem a week back. The reason is the bounded inverse theorem: $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, ...

4

I understand that you are dealing with semi-classical symbols $$b(x,\xi, h)=a(x,h\xi), \quad \vert\partial_x^\alpha\partial_\xi^\beta b\vert\le C_{\alpha\beta} h^{\vert \beta\vert} m(x),\quad 0<h\le 1.$$ According to Hörmander's terminology, we have $$b\in S(m, g),\quad g_{x,\xi}(t,\tau)={\vert dt\vert^2} +h^2{\vert d\tau\vert^2}.$$ If $b\ge 0$, then ...

4

For any even function, $\hat{f}(x) + f(x)$ is a fixed point of the Fourier transform.

4

Note first of all that $H^{-1}$ will contain objects that are not functions, such as the Dirac $\delta$ concentrated at a point. (Think that $2\delta_0=\frac{d^2}{dx^2}|x|$.) The correct definition is that $H^{-1}$ is the completion of $C^\infty(S^1)$ with respect to the norm $\newcommand{\ii}{\boldsymbol{i}}$ $$\Vert ... 4 Your sum diverges for most x\in\mathbb{R}. The innermost sum is$$\sum\limits_{\underset{(p,q)=1} {p=1}}^{q-1} f(n) \; e^{2 i \pi n (x+ \frac{p}{q}) } = f(n)e^{2 i \pi nx}\sum\limits_{\underset{(p,q)=1} {p=1}}^{q-1}e^{2 i \pi n\frac{p}{q} } = f(n)e^{2 i \pi nx}\sum_{d\mid (n,q)}d\mu\left(\frac{q}{d}\right),$$by the well-known formula for Ramanujan's sum. ... 2 This kernel is called the Bessel potential. It is smooth away from 0, and in your case, this kernel is locally integrable. 1 In Section 8, the notes say that the set of Hecke characters is \widehat{J/\mathbb{Q}^\times} or \widehat{J^0/\mathbb{Q}^\times}, which can be identified with (\mathbb{Q}^\times)^\perp, but I do not see any reason to think that the Hecke characters should be \widehat J/(\mathbb{Q}^\times)^\perp. Indeed, \widehat{J^0/\mathbb{Q}^\times} is discrete, ... 1 This function G(x) is not periodic, even if the terms of the sum are the same. The fact to add a rational number \frac{c}{d} changes the order of summation of the terms and finally the sum defining G(x+\frac{c}{d}) exists but is different from G(x). Suppose G(x) is \mathbb{Q} periodic then as G(x) is also equal to:$$ G(x)=\sum\limits_{q ...

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