Re: must specify free variable for consistent operator on L^2(R)?

On Fri, 15 Jun 2007 02:05:53 -0000, Dan Greenhoe <dgreenhoe@xxxxxxxxx>
wrote:

On Jun 13, 9:24 pm, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> >

[(Fx)(t)](w) = int_t x(t) e^{-iwt} dt

No! That's wrong, even though you see things written
that way in various books (not in books written by
careful mathematicians). The point is that _functions_
have Fourier transforms, _numbers_ do not have Fourier
transforms. Here x is a function and x(t) is _not_ a
function, x(t) is a _number_ (namely the value of
the function x evaluated at t.) So that definition
should be written

(Fx)(w) = etc.

Here x is a function, Fx is another function, and
then (Fx)(w) is the value of that other function at w,
defined to be that integral.

Got it --- thanks ^__^
And I think this kind of notation more easily adapts to operators on
arbitrary vector spaces as in
Fx
where F is an operator F:X-->Y on vector spaces X and Y
and x \in X.
Does that sound right or like someone pretending to be a real
mathematician?

Um, yes.

Say we've defined the Fourier transform Fx of a function x as
above. And now we want to talk about the Fourier transform
of the function x(2t). You can't write

(Fx(2t))(w) = int x(2t) exp(-iwt) dt

Sad news :(

You could do this:
"If y(t) = x(2t) then (Fy)(w) = etc",

I try to avoid introducing new variables _if_ possible --- too much
clutter in math proofs sometimes.

Or you could invent a dilation operator D, or maybe
D_2, defined by
(D_2x)(t) = x(2t).
(FD_2x)(w) = int x(2t) exp(-iwt) dt.

I like this one better (my humble opinion).

Here FD_2x must mean F(D_2x), since the other grouping,
(FD_2)x makes no sense.

My understanding was that this is true by definition (of operator
multiplication). That is, if A and B are operators and x is a vector
then
(AB)x = A(Bx) by definition.

Right - I didn't know that we were familiar with that
standard bit of notation.

Thank you again very much for your help and for spending so much time
in providing that help. I greatly appreciate it.

Dan Greenhoe


************************

David C. Ullrich
.