From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Q: distributions and scalar products Date: 11 Mar 1999 12:21:16 -0500 Newsgroups: sci.math.research In article <7c6rsq\$rmq\$1@platane.wanadoo.fr>, Eric Chopin wrote: >Hi, > >My main question is : if a distribution is a >sum of a dirac distribution and of a smooth function, >how can I "extract" the smooth part of this distribution >(I mean, through the action of a kind of operator on distributions). > An indefinite integral of such a distribution is a function f of locally bounded variation, and such a function has a derivative almost everywhere; it also has a unique (modulo constants) decomposition into an absolutely continuous function f_a, a jump function f_j (derivative equal zero except for a countable set of jump discontinuities), and a continuous singular function f_s (whose derivative is zero almost everywhere). The function f_a is obtained as an indefinite integral of the derivative of f. This derivative should be the "smooth" part of your required decomposition. (In general, it is just locally integrable.) >A related question is: >Does it exist now some scalar product for >distributions (more precisely I am interested in >tempered distribution)? > Yes, in many ways if you are willing to respect some restrictions. Every time you introduce a scalar product < , > on the space of your test functions (with no attempt to make the space complete), you also create a space of distributions continuous with respect to the induced norm. Then use Riesz representation theorem to represent a bounded distribution F as a scalar product with a function f uniquely defined by F; f will belong to the completion of the space of test functions: F(h) = for every test function h . Then you can extend the scalar product to the completion, and the "upstairs" scalar product of F and G will be the "downstairs" scalar product of the representatives of F and G. (It looks great in abstract form, but e.g. in case of Sobolev type scalar products the representation leads to solving a differential equation :-( ) There are cases when a scalar product (and the induced norm) is so demanding that the domain of the norm turns out to be trivial. If I recall correctly, take w(x) = exp(x^2) and let T be the Fourier transform. If you require that a test function h be admissible if both w*h and w*T(h) are square integrable (with the obvious definition of the norm) then h=0 everywhere. (I heard it in connection with "uncertainty principles" in signal processing.) Good luck, ZVK(Slavek).