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From: sullivan@harvard.ARPA (John M. Sullivan)
Newsgroups: net.math
Subject: Re: Interview Q revisited
Message-ID: <9@harvard.ARPA>
Date: Thu, 4-Apr-85 18:12:06 EST
Article-I.D.: harvard.9
Posted: Thu Apr  4 18:12:06 1985
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Organization: Harvard University
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Using integration by parts shows
$\int \delta' f + \int \delta f' = (\delta f) |^\infty_{-\infty} = 0.$
(I hope those of you who don't have TeX can read this notation anyway.)

Thus the desired constant is -1.

I don't see that distributions are that different from linear
functionals.  Viewing $\delta$ as a measure has the disadvantage of
making it hard to define things like its derivative.

-- 
John M. Sullivan
sullivan@harvard