Cj67

Welcome!

Hello, Cj67, and welcome to Wikipedia! I am CTSWyneken. Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:

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Again, welcome! And if you have any questions, please do not hesitate to ask. --CTSWyneken 23:59, 29 May 2006 (UTC)^(talk)

Speed of gravity

Welcome to to the organized chaos that is Wikipedia. Kindly be advised that your edits to speed of gravity have been reverted for the reasons described on talk:speed of gravity. Kindly feel free to read the prior discussions and to weigh in on the issue of mentioning TVF in that article. You will not be permitted to unilaterally remove that content, but I am happy to listen to your reasons for wanting it removed. Given a good case that myself and another editor can comfortably get behind and defend, that mention will become history. --EMS | Talk 03:14, 3 June 2006 (UTC)

Annihilation theorem on open set

Hi,

so this is the place you said right?

His proof, which was a very very hasty sketch went along these lines :

let f be $\in L_{1}^{loc}(\Omega )$

he said, consider the functions

$\rho _{\epsilon }=e^{\frac {\epsilon ^{2}}{||x||^{2}-\epsilon ^{2})}}$ now convolve :

$f*\rho _{\epsilon }$

if $\epsilon$ goes to zero, it is known that $\rho _{\epsilon }$ becomes an approximation of unity under convolving that should give the result

But is this really okay? Doesn't convolving require the function to be defined everywhere at first?

Thanks, Evilbu 18:18, 11 June 2006 (UTC)

OK, a couple of things. First, $\rho _{\epsilon }$ needs to be defined a bit more carefully, but no big deal, you can show it converges (in the sense of measures or distributions) to the Dirac. But, this begs the question a bit, because this means convolving with continuous functions works. Proving that it works with $L_{1}^{loc}(\Omega )$ functions requires the kind of approximation that I was describing earlier. And, you don't need things to be defined everywhere, since convolving is defined in terms of integration, which doesn't see sets of measure zero. (Cj67 21:12, 11 June 2006 (UTC))

R^n is an l^p space

Lebesgue spaces can be defined on any measure space. A Lebesgue space on a measure space S with counting measure is the space $\ell ^{p}(S)$. In the case that S = N, the set of natural numbers, the standard space $\ell ^{p}$ is obtained. In the case that S = n, the set with n elements, the space Rⁿ with its p-norm is obtained. -lethe ^{talk +} 09:59, 19 June 2006 (UTC)

From this, you agree that R^n is not l^p, though it is L^p for a certain measure. (Cj67 15:26, 19 June 2006 (UTC))

No, I don't quite agree. $\ell ^{p}$ is, by definition, $L^{p}$ with the counting measure. Since $\mathbb {R} ^{n}$ is $L^{p}(n)$ where $n$ is the set with n elements and its counting measure, we conclude that according to the definition, $\mathbb {R} ^{n}$ is an $\ell ^{p}$ space. In particular, it is $\ell ^{p}(n)$. It is the space of sequences of length n. Rⁿ with its p-norm is indeed an $\ell ^{p}$ space, though it is not the standard $\ell ^{p}$ space of sequences of countable length. R^n is not standard l^p, but it is some l^p space. Your wording makes it sound like you think there is only one $\ell ^{p}$ space, but there are many. -lethe ^{talk +} 15:34, 19 June 2006 (UTC)

Edit summary

Hi. Just a reminder, it is good if you use an edit summary when you contribute, it helps others understand what you change. Thanks. You can reply here if you have comments. Oleg Alexandrov (talk) 20:51, 24 June 2006 (UTC)

The preview button is a good thing too. :) Oleg Alexandrov (talk) 20:51, 24 June 2006 (UTC)

Yes, I am aware of both. I guess I made a mistake on some page, so I should have previewed? On what page? (Cj67 20:57, 24 June 2006 (UTC))

Never mind about the preview. You were making a lot of edits at Sobolev space but only later I realized it was in different sections. The edit summary comment is valid though. :) Oleg Alexandrov (talk) 20:58, 24 June 2006 (UTC)

Sobolev space

I am sure you know about these things more than me, but are you sure about the phrase:

$W_{loc}^{1,1}$ is the space of absolutely continuous functions

Why is "loc" there? Oleg Alexandrov (talk) 20:54, 24 June 2006 (UTC)

E.g., the constant function 1 is not $W^{1,1}(R)$, but it is $W_{loc}^{1,1}$. But, you are right, loc isn't correct either. (Cj67 20:58, 24 June 2006 (UTC))

OK, I leave it up to you to sort that out. :) Oleg Alexandrov (talk) 21:06, 24 June 2006 (UTC)

CfD

Check this out: [1] bunix 02:01, 14 October 2006 (UTC)

culminated in

Yeah, didn't you see South Park? We're all atheists now. Hail Science! — coelacan talk — 03:30, 12 December 2006 (UTC)

Please more productive edit summaries

Dear Cj,

Please provide edit summaries on Religion that describe the neutral reason for the edit and not ones like "remove silly fluff". I reverted your recent change because "way of life" however ambiguous you may think it sounds is a very common comparative descriptor of religion. Perhaps there are other reasons for making the change you made, but calling this "silly fluff" is wholly unproductive.PelleSmith 21:01, 13 March 2007 (UTC)

ArbCom elections are now open!

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