, U, V ) is a solution of the Bezout equation with V (y) = U (1 ? y). Now, by uniqueness of the solution of the Bezout equation such that both U and V have degrees at most L + M ? 1, we see that this solution must satisfy V (y) = U (1 ? y) (since otherwise (V (1 ? y), U (1 ? y)) would provide a different solution). Hence we obtain a unique solution U = r of (44) of degree, Note that U is a solution of (44) if and only if

, Other solutions (U, V ) of the Bezout equation are obtained by taking U (y) = r(y)+s(1?y) q(y) with q any polynomial. Looking for such a solution of (44), we easily get that it is one if and only if q satisfies q(1 ? y) = ?q(y). The proof of Assertion

, 20) provided that q satisfies q(1 ? y) = ?q(y), which we assume in the following. As explained above, the factorization holds if and only if R is non-negative on the unit circle, or, equivalently, if r(y) + s(1 ? y) q(y) is non-negative for y in (0, 1). By antisymmetry of q around 1/2, this is equivalent to have

, Since s(y) > 0 for all y ? (0, 1) and using that (44) holds with U = r, we finally obtain that the claimed factorization holds if and only if

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