Tag Info

New answers tagged

5

Observe that if $n$ and $m$ are relatively prime, $\gamma_{nm}\leq \gamma_n \gamma_m$ because you can combine functions on $\mathbb Z/n$ and $\mathbb Z/m$ using the Chinese remainder theorem. So if we show $\gamma_n \leq 1- \epsilon$ for some $\epsilon$ and all sufficiently large $n$, then we can make $\gamma_n$ arbitrarily small. Probably $x(t) = \sin( ...


3

I believe no such constant exists. $\newcommand{\cF}{{\mathcal F}}\newcommand{\FA}{{\rm A}}$ Let $G$ be a LCA group with Pontryagin dual $\Gamma$, let $\cF_\Gamma: L^1(\Gamma)\to C_0(G)$ be the Fourier transform for $\Gamma$, and let $\FA(G)$ denote the image $\cF_\Gamma(L^1(\Gamma))$ equipped with the norm pushed forward from $L^1(\Gamma)$. Then if I ...


1

You can approximate this using a fast numerical method (note this answer assumes you are performing max-convolution on an equivalent transformed problem on the ring $(\times, \max)$ rather than $(+, \min)$-- you can convert from this one to the one in the question using logarithms and negation if necessary): If $M[m]$ is the exact result at index $m$ (i.e., ...


0

The answer to (1) is yes, due to the Jordan criterion for convergence of the Fourier series. The answer to (2) is no, since by the Riemann-Lebesgue lemma $\gamma$ cannot be periodic (or even almost periodic) and be the (inverse) Fourier transform of an $L^1$ function.


1

Regarding the classic Littlewood-Paley, there is an analogue. You can look it up e.g. in "Littlewood-Paley and Multiplier Theory" by R. E. Edwards, G. I. Gaudry. Rubio de Francia's result most probably can be transplanted as well. Gillespie and Torrea (in Transference of a Littlewood-Paley-Rubio inequality and dimension free estimates), mentioned that they ...


4

The relationship has actually motivated several studies: Eisenstein series and the Riemann zeta function (1981) Moments of the Riemann zeta function and Eisenstein series part I and part II (2004) The Riemann hypothesis for certain integrals of Eisenstein series (2004) Some of its limitations are discussed in an answer to this MO question.


2

The identity $\hat{\psi}(t)=(it)^{-1}\eta(t)$ you quoted is indeed hold except for the typo that $\eta\in C^{\infty}$ instead of Schwartz functions. Although the function on the right hand side is not integrable as you have noticed, itself can be the Fourier transform of a $L^1$ function. The more proper setting here is to use the Fourier transform of ...



Top 50 recent answers are included