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Newsgroups: comp.dsp, alt.engineering.electrical, sci.math, uk.radio.amateur
From: Han de Bruijn <Han.deBru...@DTO.TUDelft.NL>
Date: Mon, 13 Dec 2004 16:49:01 +0100
Local: Mon, Dec 13 2004 4:49 pm
Subject: Re: Disappointed
David C. Ullrich wrote: As is demonstrated further on, this assumption is essential. > Let's see. Of course we need some technical hypotheses > of f; let's assume that f is bounded and continuous. > Using the fact that the gaussian has integral 1 we need I suppose you mean |t| < delta instead of |t| < 0 . > 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)| Mind your typo's! This is SCI.MATH! And I would rather say _three_ integrals instead of two, > The integral over |t| < delta is less than epsilon This is correct, as far as I can see. > for all sigma, by our choice of delta and the fact > that the gaussian has integral 1. On the other > hand the fact that f is bounded and property (iii) > above shows that the integral over |t| > delta is > < epsilon if sigma is small enough. QED. (more details > if you want.) No more details needed. Thanx! Han de Bruijn You must Sign in before you can post messages.
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