Wiener-Hopf factorization problem

Discussion in 'Math Research' started by Alan, Apr 18, 2004.

  1. Alan

    Alan Guest

    I would appreciate any suggestions on the following problem.
    As part of a WH factorization problem, I have a
    function G(z), with z a complex number, and q > 0 a real number,
    of the form

    G(z) = q + z^2 - (F(z) - 1).

    For my problem, F(z) is the fourier transform (f.t.) of a
    probability density on (-infinity, infinity). Specifically, I have

    F(z) = \int e^(I z x) [ c e^(eta x) 1(x < 0) + f(x) 1(x >0) ] dx,

    where c and eta are positive constants, independent of x or z.
    The integral is over the entire real x-axis.

    In other words, F(z) is the f.t. of the sum of two probability densities.
    One is an exponential density with support on (x < 0). The
    other is a completely arbitrary (real, integrable) density f(x) with support
    on (x > 0).
    Assume the entire density is normalized to one in the usual way: F(0) = 1,
    which means that G(0) = q > 0.

    My question (below) concerns the zeros of G(z)
    in the upper half z-plane Im z > 0.

    It's easy to show that G(z) is analytic in Im z > 0, except for a
    simple pole at z = i eta. It's also easy to show that G(z) has
    two purely imaginary zeros i a (q) and i b(q),
    such that 0 < a < eta < b. [To see it, draw a graph of G(i y), y real,
    and remember that eta > 0.]

    I can also prove that G(z) has no other zeros in the strip 0 < Im z < a(q).

    What I cannot prove directly yet, but strongly believe to be true, is that
    G(z) has *no other zeros* in Im z > 0 [In particular for Im z >= a(q)].

    [If f(x) = 0, the assertion is trivial because then G(z) is a
    ratio of polynomials.The numerator polynomial is a cubic,
    with 3 roots in known locations, two on the upper
    imaginary axis an one on the lower imaginary axis.]

    Any suggestions on establishing this last part, for non-trivial f(x),
    would be most welcome. If you can see an f(x) which provides a
    counter-example to the assertion, that would be very interesting,
    too, and most surprising.

    Alan, Apr 18, 2004
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