Timeline for Can singular measures be viewed as vanishing distributions? (Answer No!)
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Jul 14, 2011 at 21:55 | vote | accept | Anand | ||
| Jul 14, 2011 at 21:54 | answer | added | Anand | timeline score: 0 | |
| Jul 14, 2011 at 21:53 | history | edited | Anand | CC BY-SA 3.0 |
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| Jul 14, 2011 at 21:51 | comment | added | Anand | @André Henriques, thank you for your suggestion. I will do it.:-) | |
| Jul 14, 2011 at 19:30 | comment | added | André Henriques | @Anand: I think that the correct protocol is to type the answer to your question in the answer box, and then click on the check mark to accept your own answer. (Moderators correct me if I'm wrong) | |
| Jul 14, 2011 at 18:53 | history | edited | Anand | CC BY-SA 3.0 |
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| Jul 14, 2011 at 18:45 | comment | added | Anand | @Gerald Edgar, thank you for the simple example. :-) @S. Carnahan, I think I have understood my question. Thanks! @Mariano, I will revise my post, thanks. :-) | |
| Jul 14, 2011 at 12:42 | comment | added | Gerald Edgar | The simplest example of a singular measure is the so-called "delta function". The derivative of the unit step function. But of course $\int_R \psi(x)\delta(dx) = \psi(0)$. | |
| Jul 14, 2011 at 7:25 | comment | added | S. Carnahan♦ | Do you have a question in your revised post? | |
| Jul 14, 2011 at 7:23 | comment | added | Mariano Suárez-Álvarez | Anand: please add all the information that you provided in the comments to the question itself. It is much easier for everyone of it is complete. | |
| Jul 14, 2011 at 7:15 | history | edited | Anand | CC BY-SA 3.0 |
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| Jul 14, 2011 at 7:13 | comment | added | Anand | @Andreas Blass and Willie Wong, thank you very much. I am clear now.:-) | |
| Jul 13, 2011 at 11:20 | comment | added | Willie Wong | Anand, you should edit your question and replace it by the comment you gave (second in the thread right now). As it stands, what you wrote about the function $f$ is not true. The Cantor function is a singular continuous function, and clearly if $\psi = 1$ on $[1/3,2/3]$ and is non-negative elsewhere, $\int f\psi dx > 1/6$. | |
| Jul 12, 2011 at 22:43 | comment | added | Andreas Blass | In view of your answer to André Henriques's comment, I suppose the $f$ and $dx$ in the integrand in your question actually refer to the singular measure that acts as the "derivative" of some increasing, continuous function $g$. In that case, the answer to your question is no, the integral need not vanish. Suppose, for example, that $g$ is constant outside some interval $[a,b]$ (so the measure $f\,dx$ concentrates on $[a,b]$) while $\psi$ is identically 1 on $[a,b]$. Then the integral will be $g(b)-g(a)$. (Apologies if I misunderstood the question.) | |
| Jul 12, 2011 at 21:33 | history | edited | Anand | CC BY-SA 3.0 |
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| Jul 12, 2011 at 21:32 | comment | added | Anand | I am sorry for the lack of explanation. $\mathcal{D}(R)$ is the space of test functions $C_c^\infty(R)$ (smooth functions with compact support) with the usual topology. | |
| Jul 12, 2011 at 21:30 | comment | added | André Henriques | You should also explain what $\mathcal D(R)$ is... it's not clear from the context. | |
| Jul 12, 2011 at 21:29 | comment | added | Anand | Yes, it is. :-) | |
| Jul 12, 2011 at 21:28 | comment | added | André Henriques | After reading your other post, I think that I now understand what you mean by "singular". You're probably thinking of an increasing functions on $\mathbb R$, and you want the derivative of that function to be a measure that is singular w.r.t Lebesgue measure. | |
| Jul 12, 2011 at 21:27 | comment | added | Anand | Yes, it is better to put in this way: let $\mu$ be a distribution, singular with respect to Lebesgue's measure. For any test function $\psi\in\mathcal{D}(R)$, do we have $\int_R \psi\mu(d x)=0$? | |
| Jul 12, 2011 at 21:12 | comment | added | André Henriques | A non-zero continuous function can be viewed as a distribution. That distribution will then also be non-zero... What do you mean by "singular"? | |
| Jul 12, 2011 at 21:04 | history | edited | Anand | CC BY-SA 3.0 |
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| Jul 12, 2011 at 20:38 | history | edited | Anand | CC BY-SA 3.0 |
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| Jul 12, 2011 at 20:29 | history | asked | Anand | CC BY-SA 3.0 |