# Tag Info

## Hot answers tagged fourier-analysis

14

Here is a full characterization. Theorem. The function $e^{-|x|^a}$ is positive definite for $0\le a \le 2$ and is not positive definite for $a>2$. Thus, its Fourier transform is positive and non positive for the said ranges. Proof. The claim $a=0$ is trivial. Assume $0<a<2$. Then, it can be shown that \begin{equation*} -|x|^a = ...

13

no, it is not positive everywhere; notice that the integral only depends on $|x|$, so we may orient the vector $x$ along the $x_1$ axis; going to hyperspherical coordinates we have $$\Phi_n(x)=\frac{2\pi^{(n-1)/2}}{\Gamma[\tfrac{1}{2}(n-1)]}\int_0^\infty dr \int_0^\pi d\phi\; e^{-r^n}\cos\bigl(|x|r\cos\phi\bigr)r^{n-1}(\sin\phi)^{n-2}$$ I checked that it ...

5

I think I can prove a polynomial bound using complex analysis. This really seems like it shouldn't work but it seems to. The bound is that for all $\alpha_1, \dots, \alpha_n \in \mathbb R$ $$\sup_{t \in [A,B]}\left| \sum_{i=1}^n e( \alpha_i t) \right| \geq n^{1-\frac{\pi}{4 \arctan {\sqrt{\frac{B}{A}}}-\pi}}$$ So in the special case $[.9A,A]$ this is ...

3

For hypersurfaces there is a decent chance that the methods can be adapted to handle $L^q_t L^p_x$ type norms, but not $L^p_x L^q_t$ type norms. This is because the parabolic rescalings used in the Bourgain-Demeter argument (see Proposition 4.1 of http://arxiv.org/pdf/1403.5335.pdf ) react well with the former type of norm (after writing the parabolic ...

2

If I understand correctly, what you're doing amounts to: Starting with initial data $\psi(x,0)$ that is represented as a truncated Fourier series. Computing the exact (up to roundoff errors) time evolution of this initial condition on a periodic domain, using discrete Fourier transforms. This is often referred to as a Fourier spectral method. Note that ...

1

Great question. I've often used this heuristic but never thought about whether it had a rigorous meaning. Let me do this in one dimension; the generalization to higher dimensions is straightforward. My first comment is that the Fourier transform between $l^2(\frac{1}{2\pi r}\mathbb{Z})$ and $L^2(r\mathbb{T})$ genuinely sits inside of the Fourier transform ...

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