Disappointed

Typo for 'a > 0', sorry.

Typo, sorry. 's -> 0' was supposed to be 'sigma -> 0'

(I was using s instead of sigma in a first draft, decided that was a bad idea because s already has a traditional meaning here, but I missed that one when I changed all the s's to sigma's.)

So all together, (iii) is this:

(iii) If a > 0 is fixed then int_{|t-T| > a} g_sigma(t-T) dt -> 0 as sigma -> 0.

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David C. Ullrich

As I mentioned before, the document contains mathematical typesetting that is difficult to reproduce in a text-only format.

And posting a link to a web site that gives more information IS a standard and accepted practice on Usenet when the information is a) too large or detailed to be quoted directly b) in a format that cannot be adequately represented in text-only or c) copyrighted and cannot be legally/ethically quoted. Obviously b) applies here.

This sure sounds like another attempt to avoid understanding, which is not consistent behavior for a truth-seeker such as yourself.

Uh, you mean the gist of a Usenet discussion is not to read what you are replying to? That actually explains a lot about your postings.

Another example of Bean's inconsistency was his request for exactly this kind of thing only a few weeks ago. He was asking in usenet for urls that pointed to a 1924 paper that he was interested in.

Ah, yes, here it is, 14 Nov 2004 in comp.dsp:

"Airy R. Bean" wrote: "In pursuit of this objective, I am having difficulty in finding on the web, an original transcript of Nyquist's

1924 paper, "Certain Factors Affecting Telegraph Speed", or a retyping of it without others' commentaries. Can anyone provide a URL?"

So, how was he going to make that work? Did he, or did he not, intend to 'follow web leads from usenet'? What happened between then and now? Another ISP failure?

No. Enough is enough.

Han de Bruijn

(i) g_sigma >= 0.

(ii) int g_sigma = 1.

OK. Got that. But I don't see (immediately) how to prove the diaphragm property (Dutch: "diafragma eigenschap") with the above knowledge i.e.:

int(-oo,+oo) f(t).g_sigma(t-T) dt = f(T)

Han de Bruijn

Well of course you can't prove that because it's not true - by now I know that you mean the limit of that as sigma -> 0, but the fact that you mean that doesn't make the notation correct. How to prove that

int(-oo,+oo) f(t).g_sigma(t-T) dt -> f(T)

as sigma -> 0 is a standard thing in harmonic analysis - look for "approximate identity" arguments.

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 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 that

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)| < epsilon whenever |t| < 0. Write the last integral as the sum of two integrals, one where |t| < delta and one where |t| > delta.

The integral over |t| < delta is less than epsilon 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.)

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David C. Ullrich

As is demonstrated further on, this assumption is essential.

I suppose you mean |t| < delta instead of |t| < 0 . Mind your typo's! This is SCI.MATH!

And I would rather say _three_ integrals instead of two, but, anyway, I understand what you mean.

This is correct, as far as I can see. No more details needed. Thanx!

Han de Bruijn

_Some_ sort of growth condition is needed or none of the integrals exist. One could replace boundedness by exponential growth here.

Yes. (Also decided later that "delta" was a bad choice of variable name...)

Yes, sir.

As I said, it's a very standard argument in some circles.

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David C. Ullrich

Maybe. But it's not very standard that your vicious circles intersect with mine. ;-)

Han de Bruijn