From: "G. A. Edgar" Subject: Re: defining Dirac's delta, or rather, "mathematical rigor" Date: Mon, 03 Jul 2000 08:27:00 -0400 Newsgroups: sci.math Summary: [missing] > > > > Don't forget the Mikusinski operators... generalized functions > > rigorously defined using abstract algebra... > > do you have any reference? Mikusi\'nski, Jan. Operational calculus. Vol. 1. Translated from the Polish by Janina Smólska. Second edition. International Series of Monographs in Pure and Applied Mathematics, 109. Pergamon Press, Oxford-Elmsford, NY; PWN---Polish Scientific Publishers, Warsaw, 1983. 318 pp. ISBN: 0-08-025071-8 A nice idea. In the simplest form... Take continuous functions on [0,infinity). With addition and convolution it is an integral domain (without unit). The "field of fractions" construction yields generalized functions (including the Dirac delta). So: functions --> generalized functions is just like integers --> rationals. -- Gerald A. Edgar edgar@math.ohio-state.edu ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: defining Dirac's delta, or rather, "mathematical rigor" Date: 3 Jul 2000 13:50:15 -0400 Newsgroups: sci.math In article <20000703133205.03428.00001305@ng-fk1.aol.com>, Penny314 wrote: :dear edgar, : Sounds just like my post only taken further than my quick idea. One :must be careful about this since the convolution of the function y=1 with :itself is not defined. :The convolution of continuous functions on the line may not be finite. : Anyway , I will get and read that reference . : Thanks : pennysmith : :>A nice idea. In the simplest form... Take continuous functions :>on [0,infinity). With addition and convolution :>it is an integral domain (without unit). The "field of fractions" :>construction yields generalized functions (including the Dirac : :etc. This depends on the playground, and that has been mentioned. Convolutions associated with Laplace Transform are the ones treated by Mikusinski: all functions are assumed to be zero for negative values of the variable (so, only restrictions to [0, infinity) are considered). In this case, the convolution with constant 1 on [0, infinity) is integration with zero initial condition at 0. An important property of this convolution is that it makes the continuous functions on [0, infinity) an "integral domain" (a commutative cancellation ring) - in plain terms, if * means Laplace convolution then from f*g = 0 on an interval [0, a) there follows that either f=0 or g=0 on the interval [0, a). (The "almost everywhere" clause can be dropped as long as we start with continuous functions.) This is known as Titchmarsh's Theorem; the proofs I have seen are quite advanced. And this cancellation property makes the quotient field possible. Cheers, ZVK(Slavek). ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: defining Dirac's delta, or rather, "mathematical rigor" Date: 5 Jul 2000 19:12:22 -0400 Newsgroups: sci.math In article <8jqjon$k71@mcmail.cis.mcmaster.ca>, Zdislav V. Kovarik wrote: [...] :Convolutions associated with Laplace Transform are the ones treated by :Mikusinski: all functions are assumed to be zero for negative values of :the variable (so, only restrictions to [0, infinity) are considered). It should be mentioned here that this convolution (denoted (*) to distinguish it from pointwise multiplication): (f (*) g)(x) = integral[0 to x] f(x-t) * g(t) dt makes sense even for functions which are not Laplace transformable, such as x |-> exp(x^2). The convolution has been "divorced" from Laplace Transformation. Titchmarsh's Theorem still holds, and we have a larger field than the one built from Laplace transformable functions. (The proof is not hard.) Cheers, ZVK(Slavek). ============================================================================== From: "G. A. Edgar" Subject: Re: defining Dirac's delta, or rather, "mathematical rigor" Date: Tue, 04 Jul 2000 11:43:24 -0400 Newsgroups: sci.math In article <20000703133205.03428.00001305@ng-fk1.aol.com>, Penny314 wrote: > dear edgar, > Sounds just like my post only taken further than my quick idea. One must > be > careful about this since the convolution of the function y=1 with itself is > not > defined. Notice I said functions defined on [0,infinity). So the convolution of f(x) = 1 with itself IS defined... you get f*f(x) = x. > >A nice idea. In the simplest form... Take continuous functions > >on [0,infinity). With addition and convolution > >it is an integral domain (without unit). The "field of fractions" > >construction yields generalized functions (including the Dirac -- Gerald A. Edgar edgar@math.ohio-state.edu