## Hot answers tagged fourier-analysis

11

A very important role for the undetermined constants in indefinite integrals, in fact perhaps their first really essential role, is in solving basic differential equations. The constants of integration are used to find the solution satisfying the initial conditions for the differential equation. Changing the initial conditions will change the solution, which ...

6

This does not follow. There are compactly supported, finite, purely singular measures $\mu$ whose Fourier transform has power decay: $|\widehat{\mu}(x)|\lesssim (1+|x|)^{-\beta}$; for more background, you can search for Fourier dimension, for example on this site.
This function $g=\widehat{\mu}$ satisfies all your conditions, but its Fourier transform $\mu$ ...

5

If $f$ is an $L^2$ function, with Fourier transform $\hat{f}$, then the identity you're trying to prove is
$$\sum_{m = 0}^\infty \binom{-1}{m} \sum_{k = 0}^m \binom{m}{k} (-1)^{m - k} (-i\xi)^k \hat{f}(\xi) = \frac{1}{-i\xi} \hat{f}(\xi).$$
In other words, you want to show that
$$\sum_{m = 0}^\infty \binom{-1}{m} \sum_{k = 0}^m \binom{m}{k} (-1)^{m - k} ...

5

First, the paper is Bruedern and Wooley!
Next ... by Cauchy's inequality, one has
$$|\Phi(\alpha)|^2\le 6H\Psi(\alpha),$$
say, where
$$\Psi(\alpha)=\sum_{h\le 6H}\left| \sum_{P<x\le 2P}e(\alpha h(3x^2+3xh+h^2))\right|^2.$$
Thus the mean value in question is bounded above by $6H$ times
$$\int_0^1\Psi(\alpha)|\Phi(\alpha)|^2\, d\alpha ,$$
and by ...

4

Regarding smoothness: $f$ is smooth everywhere outside of 0, the singularity at zero can also be described.
More precisely $f$ is the inverse Fourier transform of $(x_+)^{-1+i\alpha}-1_{[-1,1]}\cdot (x_+)^{-1+i\alpha}$, where $1_{[-1,1]}$ is the indicator function of the interval $[-1,1]$.
Notice that both expressions are well defined distributions. The ...

3

Let me note $\phi_k(D)$ the Fourier multiplier $\phi_k(\xi)$, i.e.
$
\text{Fourier}\bigl(\phi_k(D)u\bigr)(\xi)=\phi_k(\xi)\hat u(\xi).
$
$\bullet$ The answer to (1) is yes since
$$
\Vert{u}\Vert_{L^1}=\Vert{\sum_{k\ge 0}\phi_k(D)u}\Vert_{L^1}\le
\sum_{k\ge 0}\Vert{\phi_k(D)u}\Vert_{L^1} =\Vert{u}\Vert_{B^0_{1,1}}.
$$
$\bullet$ The answer to (2) is no: take ...

2

Corduneanu, Almost Periodic Oscillations and Waves. (2009, Springer)

2

What do you mean "a new", of what years?
Classics are always new-the books of Bohr, Levitan and Zhikov, Besikovich himself.
There is also a book of Corduneanu with standard name "almost periodic function" of 1989. Look also for Diagana 2013, Albrecht Böttcher et al 2002, Gaston M. N’Guerekata 2001, Wolfgang Schwarz 1994, S. Zaidman 1985 ...

2

You should compare this construction to the dyadic (or Littlewood-Paley) decomposition.
The answer to (i) is simply that the volume of the cubes $Q_k$ stays $1$ while, e.g. the volume of the annuli in the Littlewood-Paley decomposition grows like $O(2^{nk})$.
The answer to (ii) is also very easy and follows from the property $\sum_{k\in \mathbb Z^{n}} ...

1

I suspect that you meant
\begin{equation*}
\widehat{f}(n)
=
\frac{1}{2\pi}
\int_{-\pi}^{\pi}
f(e^{i\theta})
e^{-in\theta}
d\theta
\end{equation*}
in which case
\begin{equation*}
F(x)
=
f(e^{ix})
=
\sum_{n\in\mathbb{Z}}
\widehat{f}(n)
e^{inx}.
\end{equation*}
The measure $\mu$ is then a series of shifted Diracs
\begin{equation*}
\mu
=
...

Only top voted, non community-wiki answers of a minimum length are eligible