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Newsgroups: comp.dsp, alt.engineering.electrical, sci.math, uk.radio.amateur
From: David C. Ullrich <ullr...@math.okstate.edu>
Date: Mon, 13 Dec 2004 15:51:33 -0600
Local: Mon, Dec 13 2004 10:51 pm
Subject: Re: Disappointed
On Mon, 13 Dec 2004 16:49:01 +0100, Han de Bruijn
<Han.deBru...@DTO.TUDelft.NL> wrote: _Some_ sort of growth condition is needed or none of >David C. Ullrich wrote: >> Let's see. Of course we need some technical hypotheses >As is demonstrated further on, this assumption is essential. the integrals exist. One could replace boundedness by exponential growth here. >> Using the fact that the gaussian has integral 1 we need Yes. (Also decided later that "delta" was a bad choice >> to show that >> int(-oo,+oo) (f(t) - f(T)).g_sigma(t-T) dt -> 0. >> Let's assume T = 0 just to save typing - we need to show >> int(-oo,+oo) (f(t) - f(0)).g_sigma(t) dt -> 0 >> as sigma -> 0. So it's more than enough to show that >> int(-oo,+oo) |f(t) - f(0)|.g_sigma(t) dt -> 0 >> Let epsilon > 0. Choose delta > 0 so that |f(t) - f(0)| >I suppose you mean |t| < delta instead of |t| < 0 . of variable name...) >Mind your typo's! This is SCI.MATH! Yes, sir. >And I would rather say _three_ integrals instead of two, As I said, it's a very standard argument in some circles. >but, anyway, I understand what you mean. >> The integral over |t| < delta is less than epsilon >This is correct, as far as I can see. >No more details needed. Thanx! ************************ >Han de Bruijn David C. Ullrich You must Sign in before you can post messages.
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