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).