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Banach space as Banach algebra

  1. Sep 11, 2014 #1
    I find, in Kolmogorov-Fomin's Элементы теории функций и функционального анализа, at the end of § 5 of chapter IV, several statement on the spectral radius and the non-emptyness of the spectrum of a linear operator ina Banach space, which are left without proof.
    Nevertheless, in Tikhomirov's appendix, the same properties are prooven for non-commutative unitary Banach algebras.
    I wonder whether all Banach spaces can be provided with the structure of a unitary (not necessarily commutative) Banach algebras...
    ##\infty## thanks!
     
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  3. Sep 11, 2014 #2

    mathman

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    I haven't looked at it in detail, but I doubt it. For example, how would you make Hilbert space into an algebra?
     
  4. Sep 11, 2014 #3

    mathwonk

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    I haven't read either of your sources, but the connection between banach spaces and banach algebras seems to be that the space of continuous linear maps on a banach space is a banach algebra. maybe that suffices for your purpose.
     
  5. Sep 12, 2014 #4

    mathwonk

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    If that was not clear, the continuous linear operator T on the Banach space B is itself a member of the Banach algebra of operators, and thus the spectrum of T is non empty.
     
  6. Sep 12, 2014 #5

    WWGD

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    Maybe Davide is thinking of a sort of reverse situation. Given a Banach algebra B_A can we always find a Banach space B so that B_A is the algebra of continuous linear maps on B?
     
  7. Sep 12, 2014 #6

    mathwonk

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    but if you read his question, :

    "I find, in Kolmogorov-Fomin's Элементы теории функций и функционального анализа, at the end of § 5 of chapter IV, several statement on the spectral radius and the non-emptyness of the spectrum of a linear operator ina Banach space, which are left without proof.
    Nevertheless, in Tikhomirov's appendix, the same properties are prooven for non-commutative unitary Banach algebras."

    it sounds as if he just wants to know the non emptiness of the spectrum of a linear operator. Or am I missing something?
     
  8. Sep 12, 2014 #7

    WWGD

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    Maybe you're right, mathwonk, but the title says Banach spaces as Banach algebras; can you clarify for us, Davide?
     
    Last edited: Sep 12, 2014
  9. Sep 13, 2014 #8
    I was wondering whether a Banach space $B$ can be considered a Banach unitary, not necessarily commutative, algebra by defining some canonical multiplication between the vectors of $B$...
     
  10. Sep 13, 2014 #9

    dextercioby

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    Well, no, basically you're extending the Banach space which is itself a topological vector space into an algebra by adding a multiplication between the vectors. Banach space + vector multiplication =/= Banach space.

    A Banach algebra is thus an enhancement of a Banach space, a richer mathematical notion.
     
  11. Sep 14, 2014 #10
    I think OP's question is clear. A Banach algebra is ##\langle \mathbb A, ||\cdot||,+, *\rangle##, where ##\langle \mathbb A, ||\cdot||,+\rangle## is a Banach space and ##*: \mathbb A^2\to \mathbb A## is a binary operation satisfying some properties.

    The question is: given a Banach space ##\langle \mathbb A, ||\cdot||,+\rangle##, must there always exist some ##*## such that ##\langle \mathbb A, ||\cdot||,+,*\rangle## is a Banach algebra?

    A rephrasing of the question is as follows.
    Given any Banach algebra, we can get a Banach space by just forgetting about multiplication. If I tell you a Banach space was built this way, does that give you any information about what kind of Banach algebra you're looking at?
     
  12. Sep 14, 2014 #11
    To answer your question: I have no idea.
     
  13. Sep 15, 2014 #12
    Exactly what I meant.
    I thank any past, presend and future poster in this thread!
     
  14. Sep 15, 2014 #13

    mathwonk

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    i am puzzled since the question as clarified has absolutely nothing to do with the non emptiness of the spectrum, which apparently motivated it.
     
  15. Sep 17, 2014 #14
    @mathwonk: Tikhomirov's appendix, which is about Banach's algebras, proves those statements in the case of Banach spaces with multiplication as unitary non-commutative algebras. Kolmogorov-Fomin's text states them without a proof for Banach spaces (without multiplication). I haven't reached those proofs yet. I will check whether those proofs can be valid for Banach spaces without assuming them to be unitary algebras and I'll let you know. Thank you again!
     
  16. Sep 24, 2014 #15
    That is the case. Cfr. p. 519 here, corollary 2 and theorem 2, for those knowing Russian.
    Thank you all!!!
     
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