# Wiener-Hopf factorization problem

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

1. ### AlanGuest

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.

regards,
alan

Alan, Apr 18, 2004