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96 views

Derivative bounds for self convolution of the spherical measure in $R^d$

While reading this article on near $L^1$ estimates for the spherical lacunary maximal function, I came across the estimate $$ |\partial^{\gamma} (\widetilde{\sigma} \ast \sigma)(x)| \lesssim |x|^{-(1 +...
Zygmund's user avatar
0 votes
1 answer
161 views

Does convolution with $(1+|x|)^{-n}$ define an operator $L^p(\mathbb R^n) \to L^p(\mathbb R^n)$

Suppose that $f : \mathbb R^n \to \mathbb R^n$ is a locally integrable function. I am interested in the integral $$ x \to \int_{\mathbb R^n} ( 1 + |y| )^{-n} f(x-y) \;dy $$ If the decay of the ...
AlpinistKitten's user avatar
2 votes
0 answers
116 views

A technical question concerning convolution product

Let $v\in L^p(\Bbb R^d)$, $1\leq p<\infty$ be nonzero function, i.e., $v\not\equiv 0$. Define $$u(x)= |v|*\phi(x)= \int_{\Bbb R^d} |v(y)|\phi(x-y)d y$$ with $\phi(x)= ce^{-|x|^2}$ and $c>0$ so ...
Guy Fsone's user avatar
  • 1,101
3 votes
1 answer
626 views

Can we show that the characteristic function of an infinitely divisible probability measure has no zeros

Let $E$ be a normed $\mathbb R$-vector space, $\mu$ be a probability measure on $\mathcal B(E)$ and $\varphi_\mu$ denote the characteristic function$^1$ of $\mu$. Assume $\mu$ is infinitely divisible, ...
0xbadf00d's user avatar
  • 167
0 votes
1 answer
134 views

How can we show this estimate for the convolution of two probability measures?

Let $(\delta_k)_{k\in\mathbb N}\subseteq(0,\infty)$ be nonincreasing with $\delta_k\xrightarrow{k\to\infty}0$ and $(\varepsilon_k)_{k\in\mathbb N}\subseteq(0,\infty)$ with $\sum_{k\in\mathbb N}\...
0xbadf00d's user avatar
  • 167
2 votes
2 answers
322 views

If $(\exp(\mu_n))_{n\in\mathbb N}$ is weakly convergent, is the normalized sequence convergent as well?

Let $E$ be a metric space and $\mathcal M(E)$ denoote the space of finite signed measures on $\mathcal B(E)$ equipped with the total variation norm $\left\|\;\cdot\;\right\|$. I would like to know ...
0xbadf00d's user avatar
  • 167
3 votes
2 answers
278 views

The disintegration of the convolution of two probability measures

Let $G$ be a topological group with all the topological conditions in order that some form of the disintegration theorem be applicable (for instance, take $G$ metrizable). Let $N$ be normal and closed,...
Alex M.'s user avatar
  • 5,407
5 votes
4 answers
824 views

Self convolutions of singular continuous measure

Let $\mu$ be a finite measure on $\mathbb{R}$. Define the measures $(\mu_n)_{n\geq 1}$ by $\mu_{n+1}=\mu\ast \mu_n$ and $\mu_1=\mu$ Is there a singular (with respect to the Lebesgue measure) ...
Chris's user avatar
  • 321
3 votes
0 answers
741 views

Justification of the convolution operation of $L^1(G)$ functions where $G$ is a LCA group (measurability)

Suppose $G$ is a locally compact abelian Hausdorff group (LCA), and $\lambda$ is the Haar measure on it. We all know the convolution of two $L^1(\lambda)$ functions $f$ and $g$ on $G$ is defined as $$...
Hua Wang's user avatar
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