Wiener's Tauberian Theorem proof

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

  1. 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
    #1
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  2. I receieved an email about this from Julia Kuznetsova.
    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
    the email instead:
    Thanks for your comments, and please note that if
    you have any reply to _this_ post I won't be able
    to reply to it for about 10 days (see above).

    ____________
    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
    about J while (iii) is a statement about I. But
    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
    #2
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  3. Sorry, there is indeed a misunderstanding. I had in mind not your (i),
    it's another thing I'll return to later. I meant the following
    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
    #3
  4. 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.

    So this does answer my original question about
    whether that argument was well-known - thanks.

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

    David C. Ullrich
     
    David C. Ullrich, May 19, 2007
    #4
  5. 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
    #5
  6. 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.
    Your (i) is Loomis 25E.

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

    With best regards,
    Julia Kuznetsova
     
    Julia Kuznetsova, May 20, 2007
    #6
  7. 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
    #7
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