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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 ...


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 ...


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} } ...


3

Inventiones papers up to 1996 are publicly available here.


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 ...


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 ...


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 = ...



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