Re: fourier transform, scalar product

The World Wide Wade <aderamey.addw@xxxxxxxxxxx> writes:

In article <47d6c6f7$0$4792$4fafbaef@xxxxxxxxxxxxxxxxxxx>,
Joubert <luckyguy227223@xxxxxxxxxxx> wrote:

Prove that fourier transforms don't change the scalar product of
functions
of X where X:={f in L1(R): Ff in L1(R)} where the mentioned scalar
product
is: Integral(Ff)(Conj(Fg)) over R. What I must prove is that Integral(Ff
(Conj(Fg)) over R = Integral((f)Conj(g)) over R. With Ff I mean the
fourier
transform of f. With Conj, I mean the complex conjugate. Looks like it
should be a consequence of the duality lemma.

Both f and g are equal to their inverse Fourier transforms; apply
Fubini.

??? Neither f nor g is likely to be its own inverse Fourier transform.
I think you mean f (or g) is the inverse Fourier transform of its Fourier
transform.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.