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Math Help - [SOLVED] convolution operator, functional analysis, linear algebra

  1. #1
    michaelangelo
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    Question [SOLVED] convolution operator, functional analysis, linear algebra

    hi everyone, I have some questions, I hope someone can help..

    Let {a_n, n from 0 to infinity} denotes a sequence of complex numbers that is summable (i.e. sum of |a_n| for all n, is finite)

    and f is the continuous function defined on the unit circle by

    f(z)= SUM (from minus infinity to plus infinity) {a_n.z^n} where z=e^it where 0<= t <= 2pi


    1) Consider hilbert space H = l^2 (Z). how can we show that the convolution operator A defined by,

    (A g)_n = SUM (k from minus infinity to plus infinity) {a_(n-k).g_k}
    = SUM (j from minus infinity to plus infinity) {a_j.g_(n-j)}

    satisfies ||A|| <= (sum of |a_n| for n from zero to infinity)


    2) How can we show A is normal?

    3) what is the spectrum of A in concrete terms

    4) Assuming {a_n} is not trivial sequence, how can we deduce that A has no eigenvalues (i.e. no point spectrum).. when the operator A is invertible in terms of funciton f, and what is ||A|| exactly?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    The Fourier transform converts convolution to pointwise multiplication. By Parseval's theorem, it also transforms the sequence space \ell^2(\mathbb{Z}) to the space L^2(\mathbb{T}) of square-integrable functions on the unit circle. Under this isometric isomorphism, the convolution operator A gets mapped to the operator M_f of multiplication by f. The adjoint of M_f is the multiplication operator M_{\bar{f}} corresponding to the complex conjugate of f. Since multiplication operators commute with each other, it is clear that M_f commutes with its adjoint, hence so does A (so A is normal).

    The spectrum of a multiplication operator is the range of the function, so \text{sp}(A) = \text{sp}(M_f) = \{f(z):z\in\mathbb{T}\}.

    The norm of a multiplication operator is the sup norm of the function, so \|A\| = \|M_f\| = \sup\{|f(z)|:z\in\mathbb{T}\}.
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  3. #3
    michaelangelo
    Guest
    thanks for the reply Opalg..
    I need to clear a few points in my mind, may be someone can help..
    for the fourth question , thanks to Opalg, we know what is the norm of A, but for the remaining part what should we do?

    there was a hint saying that using third question, we assume there is a function f in L^2(unit circle) such that Af=\lambda f, for some complex number \lambda, and then use some measure theory plus a little inspiration to show that this leads to a contradiction (because lebesgue measure on the unit circle has no atoms)

    can someone explain the details in this hint??
    thanks
    Last edited by michaelangelo; December 25th 2007 at 03:28 AM.
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