## 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 HÃ¶rmander'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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