# Wiener's Tauberian Theorem proof

Discussion in 'Math Research' started by David C. Ullrich, May 13, 2007.

1. ### David C. UllrichGuest

Writing up some notes on WTT for a student I stumbled
on a proof that seems to me to be infinitely superior
to the proofs I've seen. On the one hand I can't believe
it's not well known, since it's so simple; on the
other hand if it's well known I don't understand why
I've never seen it before. On the third hand, I don't
know the literature on Banach algebras very well,
hence this post - if anyone knows that this is
well known I'd appreciate hearing about it.

Let's call the following the fundamental theorem
of commutative Banach algebras:

ftcba: If A is a commutative Banach algebra with
identity and J is a proper ideal then J lies
in the kernel of some (non-zero) complex homomorphism
of A.

And now let L^1 = L^1(R) (actually everything here
works just as well on an arbitrary locally compact
abelian group), and let f^ denote the Fourier transform
of f.

WTT: If J is an ideal in L^1 and for every y there
exists g in J with g^(y) <> 0 then J is dense in L^1.
In particular, if f in L^1 and f^ has compact support
then f is in J.

Let's say "f is in J_E" if there exists g in J
with f^ = g^ on E, and "f is in J near y" if
f is in J_V for some neighborhood V of y. The
usual proof of WTT proceeds in two steps:

(i) For every y, f is in J near y.

This is proved by an argument that's sort of
inspired by the proof of ftcba, instead of as
a corollary of ftcba itself. Then, using a
partition of unity or something analogous,
we show that (i) implies (ii):

(ii) If K is compact then f is in J_K,

and we're done.

Not that the proof is all that hard, but it
turns out that we can prove (ii) directly,
without going through (i), and more to the
point we can prove (ii) as a simple application
of ftcba itself!

Let sigma(A) denote the space of all (non-zero)
complex homorphisms of A.

Lemma. Suppose A is a commutative Banach algebra
and I is a closed ideal in A. Then A/I is a
Banach algebra, and sigma(A/I) is "equal to"
the space of all phi in sigma(A) such that
phi vanishes on I.

Proposition. Suppose E is a closed subset of R
and let I be the set of all f in L^1 such that
f^ vanishes on E. Then sigma(L^1/I) = E (ie
if phi is in sigma(L^1/I) then there exists
y in E with phi(f+I) = f^(y)). If E is
compact then L^1/I has an identity.

Pf: The first statement follows from the lemma.
Suppose E is compact. If j in L^1 is such that
j^ = 1 on E then j+I is an identity in L^1/I. QED.

Pf of WTT: Suppose J is as in the statement of WTT
and f in L^1 such that f^ has compact support K.
Let I be the set of g with g^=0 on K. Let
q : L^1 -> L^1/I be the quotient map q(g) = g + I.

Now q(J) is an ideal in L^1/I, and the lemma shows
that g(J) is not annihilated by any complex
homorphism of L^1/I. Hence ftcba shows that
q(J) = L^1/I. In particular there exists j in
J such that j^ = 1 on K; hence f = f*j. QED

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

David C. Ullrich

David C. Ullrich, May 13, 2007

2. ### David C. UllrichGuest

She(?) says she intended to post the same comments
to sci.math.research, but she's not certain whether
she did it right.

Since the post has in any case not appeared yet,
and since I'm going to be away for the next week
and a half, I thought that I'd post a reply to
you have any reply to _this_ post I won't be able

____________
The only "regularity" I know of in this context is the
definition that an ideal I in A is regular if A/I has
a unit. I don't quite see what (i) has to do with that,
exactly - maybe you're using the word in a different
sense.

But that doesn't matter. The important point, if there
is one, is that no, I'm simply _not_ using (i) in my
proof! Honest - it seems to me you're misunderstanding
something, exactly what I'm not sure. Briefly, the
"as f is in I, f^(E)=0" is exactly backwards - in fact
f is in I _because_ f^(E) = 0.

In detail: Let's recall what (i) stated, including
context explaining the notation and hypotheses:

Now about the proof of the proposition:

(Note that the I in the Proposition is not the
same as the J in the statement of WTT...)

You're exactly right when you say I need to prove
the following:

(iii) If t is not in E then there exists f in I
with f^(t) <> 0.

Here's the proof of (iii): Let phi be a function in
C^2_c(R) (twice continuously differentiable, with
compact support) such that g(t) <> 0 while g(E) = 0.
There exists f in L^1 such that g = f^. QED.

Yes, I omitted the proof of (iii) in my post.
That was because the proof is entirely trivial;
it certainly does not use (i). (Offhand I don't
see how it _could_ use (i), since (i) is a statement
in any case the trivial proof of (iii) does _not_
use (i), explicitly or implicitly.)

In fact it's very easy to give a proof of (iii)
that's valid in the context of locally compact
abelian groups, using just the Plancherel theorem.
Say G is an LCA group with dual group Gamma.
Say E is a closed subset of Gamma, let I be the
set of f in L^1(G) with f^(E) = 0, and now
consider

(iii') If t is in Gamma but t is not in E then there
exists I in E with f^(t) <> 0.

Proof: Let U and V be open sets of finite measure
such that t is in U + V but U + V is disjoint
from E. Let phi and psi be the characteristic
functions of U and V. The Plancherel theorem shows
that there exists f in L^1(G) with f^ = phi*psi
(the convolution of phi and psi), and it follows
Yes, I use that fact - that part of the proof is exactly
the same as in the standard proof. And that fact is also
very trivial. For R it follows using an explicit approximate
identity. For G: The Plancherel theorem shows that the
set of f in L^2(G) such that f^ has compact support is
dense in L^2(G). Say f is in L^1(G). Choose g in L^2(G)
with f = gg. Choose g_n in L^2(G) such that g_n^ has
compact support ang g_n -> g in L^2. Then (g_n g_n)^
has compact support and g_n g_n -> f in L^1. QED.
I can't look up that proof for a little while (see above).
It sounds to me like it's the same as the standard proof
that I know.

In any case: Does the proof in Loomis involve an
_application_ of what I called the fundamental
theorem for commutative Banach algebras? If not
then it doesn't seem like the same proof to
me - my whole point is that we can get WTT by
_applying_ that theorem, instead of by using
arguments that are like the proof of the theorem.

Or: Look at the proof of (i) in Loomis. It's
nowhere near as trivial as the proof of (iii)
above, right?

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

David C. Ullrich

David C. Ullrich, May 19, 2007

3. ### Julia KuznetsovaGuest

Sorry, there is indeed a misunderstanding. I had in mind not your (i),
property
which is called regularity: for any y in R and its neighborhood V
there exists f in L^1(R) such that f^(y)<>0 and support of f^ is
contained in V.

It is another meaning of the word. Regularity in this sense is
specific to
harmonic analysis and not used in general theory of Banach algebras.
The definition I use you can find in Loomis 25A (cited book) or in
Hewitt&Ross, Abstract harmonic analysis, vol. II, 39.1.

What you prove below --- (iii) --- follows at once from regularity,
and although it is not formally equivalent one proves regularity in
the same
way.
Well, one often splits proof of a theorem into trivial steps.
But every such step requires proof.
Yes, in fact it does, see Loomis 25C.

Let's return now to your condition (i). It isn't necessary for WTT
but for spectral synthesis, i.e. to prove that if I is a closed ideal
in L^1
and the set of common zeroes of functions {f^: f in I} equals E
then I is equal to the ideal of all functions with f^ vanishing on E.
Some extra conditions are required on E. If E is the empty set
the theorem reduces itself to WTT but it is stronger than WTT.
Of course it's less trivial but it is another result.
Couldn't you tell where have you seen the WTT reasoned
from it? In the books I know it is used for the purpose above
and WTT is proved without it.

Julia Kuznetsova, May 19, 2007
4. ### David C. UllrichGuest

Well, yesterday you said that _I_ had used (i) in proving
WTT. I take it you're no longer claiming that. I don't have
the two books in front of me, but my recollection is that
the proofs in Helson's book and also in Rudin "Functional
Analysis" use (i), or something equivalent to it.

It turns out that the library is open on Saturday during
the summer - I didn't think it was. So I now have Loomis
in front of me. There are actually _two_ proofs of WTT
in section 25. One of them is essentially the proof
I knew of, using (i), or rather something clearly
equivalent to (i), and the other is more or less
exactly the proof I "discovered" here, using the
fact that that quotient algebra is a Banach algebra
with identity:

WTT is the Corollary in section 25D. It follows from
Theorem 25D. The proof of that theorem uses Lemma 25C.
And Loomis gives two proofs of 25C: First he points out
that 25C is immediate from preceding results and
Lemma 25B. Lemma 25B is the "partition of unity, or
anlogous device" that I referred to in my post as
a part of the proof I knew - Lemma 25B follows by
that trick from Lemma 25A, and 25A is the same as
what I called (i).

So one of the proofs of WTT in Loomis does use (i)
(the statement of 25A is not identical to (i),
but it's close enough that that proof is definitely
in the class of proofs I was referring to as proofs
that proceed by first using (i).)

Then Loomis says that in view of the importance of
25C he will give a different proof. And the second
proof of 25C he gives is exactly the proof I thought
I might have discovered, using the fact that a
certain quotient algebra has an identity.

whether that argument was well-known - thanks.

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

David C. Ullrich

David C. Ullrich, May 19, 2007
5. ### David C. UllrichGuest

I just realized I lied when I said just now that 25A in
Loomis was the same as (i). Of course it's not, because
the f obtained in 25A is just an element of A, not some
ideal.

Otoh I'll add Rudin "Fourier Analysis on Groups" to
the list of places where I _have_ seen WTT obtained
from (i).

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

David C. Ullrich

David C. Ullrich, May 19, 2007
6. ### Julia KuznetsovaGuest

I'm not. I noticed you skipped regularity -- and was mistaken
to say it was your (i). I should have been more attentive.
I have Rudin, although I don't use it usually. It is exactly
as you say there: the proof goes through (i). I believe it's
the same in Helson.
BUT: it's not a proof of WTT but of a theorem of Ditkin on
spectral synthesis. If you need WTT only you can skip this step.
In Hewitt&Ross they also give separate proofs for Ditkin and WTT.

I hope now the issue is clear for both of us.

With best regards,
Julia Kuznetsova

Julia Kuznetsova, May 20, 2007
7. ### Julia KuznetsovaGuest

It's perhaps worth adding that the proof under discussion is
originally due to
G.E. Shilov (1947, in Trudy Mat. Inst. Steklov 21).

Julia Kuznetsova, Jun 12, 2007