# Tag Info

## New answers tagged fourier-analysis

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, ...

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. ...

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: $$G(x)=\sum\limits_{q ... 0 Alternatively, the self-adjoint operator -\Delta on the domain H^2\subseteq L^2 has spectrum [0,\infty) in any dimension. Since -1 is not in the spectrum, (-\Delta+1)^{-1} is bounded on L^2 and thus for any f\in L^2, there is a unique u\in H^2 such that (1-\Delta)u=f. 6 If 1\leq p<2 then \mathscr{F}: L^p \to L^{p'} is not surjective. I had this as a homework problem a week back. The reason is the bounded inverse theorem: \mathscr{F}: L^p \to L^{p'} is injective, (by fourier inversion on the dense subspace of schwarz functions). If the map were surjective then there would be an inverse that would be continuous, ... 2 This kernel is called the Bessel potential. It is smooth away from 0, and in your case, this kernel is locally integrable. 4 I understand that you are dealing with semi-classical symbols$$ 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}. $$If b\ge 0, then ... 6 There is a whole chapter in Titchmarsh, Fourier transform, describing all eigenfunctions in great detail. He calls them "self-reciprocal" functions. A more difficult question is about the eigenvectors of discrete Fourier transform, but this one you do not ask. Edit. Since you are asking now about discrete transform too, it has 4 eigenspaces corresponding to ... 4 For any even function, \hat{f}(x) + f(x) is a fixed point of the Fourier transform. 14 Given a square-integrable, positive semi-definite function f, with its Fourier transform \hat{f}, then the function$$F=f^2+\hat{f}\star\hat{f},$$with \star the convolution, is its own Fourier transform: \hat{F}=F. If we require that F is a probability density (absolutely integrable and positive semi-definite), then any F with \hat{F}=F is ... 10 I am somewhat confused that, despite saying many true and useful things, no one has said directly that H^{-1} on the circle can be characterized as the set of distributions \theta such that \sum_{n\in \mathbb Z}(1+n^2)^{-1}\cdot |\theta(x\to e^{inx})|^2<\infty. (All distributions on the circle are compactly supported, so can be applied to the ... 4 Note first of all that H^{-1} will contain objects that are not functions, such as the Dirac \delta concentrated at a point. (Think that 2\delta_0=\frac{d^2}{dx^2}|x|.) The correct definition is that H^{-1} is the completion of C^\infty(S^1) with respect to the norm \newcommand{\ii}{\boldsymbol{i}}$$\Vert ...

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