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2

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 ...


2

Inventiones papers up to 1996 are publicly available here.


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} ...


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 ...


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 ...


0

Mathematica gives the following for the inverse Fourier transform: $$ \frac{\left| s\right| ^{-1-i \alpha } \left(\left| s\right| ^{1+i \alpha } \left(\alpha s \, _1F_2\left(\frac{i \alpha }{2}+\frac{1}{2};\frac{3}{2},\frac{i \alpha }{2}+\frac{3}{2};-\frac{s^2}{4}\right)+(1+i \alpha ) \, _1F_2\left(\frac{i \alpha }{2};\frac{1}{2},\frac{i \alpha ...


0

You should have a look at http://arxiv.org/abs/1110.4873 It appeared also peer reviewed journal but may not be freely accessible. There are more papers of Vega and collaborators on Hardy's uncertainty principle where you might find your answer. I have not clarified whether this answers your question exactly but it is at least very related and you might be ...


0

Unni addresses $L^p$ Paley Wiener type theorems for Hankel transforms in (MR0174941). In the introduction, he also lists references for $L^p$ Paley Wiener theorems for Fourier transforms.


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 = ...


4

Let us look at the definition of Besov spaces from [Bergh and Löfström, 1976]. Suppose $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ is a Schwartz class function satisfying the support of $\varphi$ is contained in $\{ \omega : 2^{-1} \leq |\omega| \leq 2 \}$ $\varphi(\omega)>0$ for $2^{-1} <|\omega| <2$ $\sum_{k\in\mathbb{Z} } ...


2

The property $F(0)=0$ follows from applying $F$ to the zero function, but the regularity is not as good as you ask for. As a partial counterexample, let $F(x)=|x|$ and $s=1$. Now $F(f)\in H^1$ for all $f\in H^1$ but $F\notin C^1$. Although $F$ is not classically differentiable, it is weakly differentiable, and for example $F\in W^{1,\infty}_{\text{loc}}$. I ...


0

This has essentially been cleared up by Christian Remling but let me be more precise since there still seems to be some confusion. We are dealing with two categories---measure spaces (say, finite) and Hilbert spaces, and the natural functor from the former to the latter which assigns to a probability space the corresponding $L^2$-space. Both spaces have ...


3

I think the suggested example is not a good fit for illustrating a tensor product decomposition, because $L^2$ functions on an interval are most naturally identified with states of a single particle in the interval (with some potential). The tensor product is then identified with the states of a pair of particles in two intervals, or a single particle in ...



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