Re: proof of Wiener theorem ?

On Tue, 03 Oct 2006 01:55:01 -0400, Igor Khavkine <igor.kh@xxxxxxxxx>
wrote:

On 2006-10-02, David C Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
On 2 Oct 2006 06:21:25 -0700, "Fedor" <malabar_carotte@xxxxxxxxx>
wrote:


David C. Ullrich wrote:
On 1 Oct 2006 05:29:10 -0700, "Fedor" <malabar_carotte@xxxxxxxxx>
wrote:

[proof of WTT]

Thank you for this reply, I will try to understand this proof (yes, the
only proof I knew is this one in Rudin). Anyway, what is this
straightforward proof of Wiener theorem mentionned in this old post
http://groups.google.fr/group/sci.math/browse_thread/thread/6c540572272895c5/21a98561b122c879?lnk=gst&q=wiener+proof+operator&rnum=3#21a98561b122c879
?

Where Mattew P. Wiener said "Wiener's tauberian theorem is a
five-liner using operator theory". I'm not sure exactly what he
was referring to there, sorry.

I'm just taking a stab in the dark here, but could it be for the
following reason?

The Wiener tauberian theorem basically states that any function in L^1
with a nowhere-vanishing Fourier transform is cyclic under the action of
translation operators (replace the last part with OP's statement of the
theorem to get the definition of "cyclic"). But, since the group of
translation operators has no multiplicities when decomposed into a
direct integral of unitary irreducible representations, any vector that
does not lie in one of its proper invariant subspace is cyclic (in L^2,
which should not be hard to extend to L^1). Any function lying in a
proper invariant subspace of the translation operators should have its
Fourier transform vanish on a set of non-zero measure.

Not knowing any operator theory I'm not the person to say, but
as far as I can see that's a possibility.

I was going to suggest something else - again, not knowing
any operator theory I could be way off base (and for that
matter, it could be that if I knew some operator theory
I'd see that what I'm about to suggest is essentially
equivalent to what you just suggested):

Say f is in L^1(R). Then convolution by f defines a
bounded operator T on L^1; the spectrum of this operator
is the closure of the range of F(f), which is to say
the range of F(f) plus 0.

Given eps > 0, say phi is a continuous function in
the plane with phi(z) = 1/z for |z| > eps (or maybe
phi has compact support and phi(z) = 1/z for
eps < |z| < R.) Take phi to be infinitely differentiable
if it helps. If some version of the "symbolic calculus"
is suffices to show that there is such a thing as
phi(T), and that phi(T) is given by convolution with
an L^1 function (for example because it's in some
algebra generated by T or whatever) then WTT follows
immediately (because then if F(f) has no zero then
for any compact set K there exists g in L^1 with
F(g) = 1/F(f) on K.)

The above is possibly too terse to be useful, a 10-liner though.

Igor


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

David C. Ullrich
.