Tag Info

New answers tagged

0

And any tempered distribution can be written as a finite sum of derivatives of continuous moderate growth functions.


1

Sorry to be a little technical, but there's some confusion concerning integration in some of the answers here. Once this is cleared up, it seems the convolution theorems are really at the heart of the matter in explaining the transformation of the differing integrations to relatively simple multiplications/divisions in the reciprocal Fourier or Laplace ...


3

Convolution extends to $f\in \mathscr{S}'$, $g\in \mathscr{S}$, so also to your case since $L^p$ spaces are contained in $\mathscr{S}'$. Also, the formula for the Fourier transform is valid, but has to be understood as an equality in $\mathscr{S}'$. See Reed and Simon, Vol.2 Theorem IX.4 for further reference.


2

This is an answer as to why $MI_{\omega_0,\omega_1}(t)$ of $f_\theta$ is the same for all $\theta$, which is one of the motivating questions. First, let $f_\theta=f\cos\theta+f_h\sin\theta$, where $f_h=\mathcal{H}f$ is the Hilbert transform of $f$. Then let $$ z^\theta_t(\omega)=(\chi_{(0,\omega)}\,\hat f_\theta)^\vee, $$ where $\chi_{(0,\omega)}$ is the ...


0

I would like to prove for the case of jump discontinuity of the function itself. (rather than that of one of its derivatives). Let $t_0>0$ be a point where $f$ jumps. The curve $(X_{t_0}(s),Y_{t_0}(s))$ is asymptotic to the straight line $(\frac{(f(t_0^+)+f(t_0^-)}{2},s)$ which is parallel to the y-axis. For any other $t\ne t_0$, the curve ...


1

Here is one approach you can use. The differences $a_k$ provide you with bounds on the difference between the characteristic functions of the two distributions. This may be easy or not depending on your example. Then you can apply known bounds on the difference between two distributions based on the difference between their characteristic functions, for ...


3

Here's a very simple example which illustrates one of the difficulties of this approach. Consider $t=1/2$ and random variables $X$ which is $1/2$ with probability $1$ and $Y_\epsilon$ which is $1/2 + \epsilon$ with probability $1$, where $\epsilon > 0$. We have $P(X \le 1/2) = 1$ while $P(Y_\epsilon \le 1/2) = 0$. On the other hand, $E X^k = (1/2)^k$ ...


1

Apparently such functions have been studied before, in cryptography. This condition is called the Strict Avalanche Criterion, and a guy called Daniel K. Biss proved a lower bound of $2^{2^{n-o(1)}}$. ...



Top 50 recent answers are included