Timeline for Tauberian measures on a locally compact abelian group
Current License: CC BY-SA 3.0
9 events
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| Sep 20, 2013 at 15:14 | comment | added | Asaf | @M. Gonzalez, usually the decomposition is written as $\mu=\mu_{a.c}+\mu_{sing}+\mu_{d}$, as (for example) the Cantor measure is not atomic, but not a.c. wrt to Lebesgue. Anyways, as the Tauberian theorem usual proof involves analysing a suitable summation kernel via harmonic analysis, it seems profitable to try to attack the problem you're facing via Fourier transform on $L^1(G)$ (I think that maybe Kauffman did work in those lines). Detecting atoms in measure via its Fourier transform can be done via the so-called Wiener's lemma. | |
| Sep 20, 2013 at 6:38 | comment | added | M.González | @Asaf, the decomposition I considered is $\mu=\mu_d +\mu_c$, discrete part plus continuous part. Then the continuous (no atoms) part can be decomposed in absolutely continuous plus singular continuous. | |
| Sep 19, 2013 at 21:03 | comment | added | Asaf | @M. Gonzalez, the Lebesgue decomposition statement you've written is false, as $\mu$ can be decomposed as a.c. part plus a singular part, but the singular need not be atomic. It can be for example a Cantor measure, or any measure of dimension less than $m$, but need not be discrete. | |
| Sep 17, 2013 at 7:21 | comment | added | M.González | CONTINUATION: It was proved by R. Doss [Studia Math. 45 (973), 111-117] that for every continuous measure $\mu$ there exists a non absolutely continuous measure $\mu$ such that $\mu *\nu$ is absolutely continuous. Therefore a measure satisfying the tauberian property has non-zero atomic part. | |
| Sep 17, 2013 at 7:14 | comment | added | M.González | Every $\mu\in M(G)$ can be decomposed as $\mu= \mu_d + \mu_c$ where $\mu_d$ is the discrete (atomic) part and $\mu_c$ is the continuous part (no atoms).\newline | |
| Sep 16, 2013 at 14:49 | comment | added | M.González | $(M(G),*)$ is a commutative Banach algebra. $\mu$ invertible means that there is $\nu$ such that $\mu * \nu =\delta_e$, where $e$ is the unit of $G$ ($\delta_e$ is the unit of $M(G)$), and $\mu_2$ a.c. wrt m means $A$ Borel subset of $G$ and $m(A) =0$ implies $\mu_1(A)=0$; equiv., there is $f_1 \in L_1(m)$ such that $d\mu_1 = f_1\cdot d m$. | |
| Sep 16, 2013 at 13:14 | comment | added | Asaf | What is the definition of invertible measure? and what do you mean by $\mu_{2}$ to be a.c? wrt to $m$? | |
| Sep 16, 2013 at 6:46 | history | edited | M.González | CC BY-SA 3.0 |
added complementary question.
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| Sep 15, 2013 at 18:09 | history | asked | M.González | CC BY-SA 3.0 |