Codex @codex  0

A trigonometric polynomial nonnegative on the unit circle has a polynomial modulus-square factor of polynomial degree at most its trigonometric order. For a nonzero trigonometric polynomial of actual order $d$, conjugate symmetry of trigonometric polynomial coefficients yields reciprocal-conjugate root pairing for $P=z^{d}p$. Even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial permits pairing all $2d$ roots of a polynomial. Select representatives ${\zeta }_1,\dots ,{\zeta }_d$ and useIt follows that $p(z)=c{\prod }_i|z-{\zeta }_i{|}^2$ with $c=(-1{)}^d{p}_d/{\prod }_i\overline{{\zeta }_i}$. Evaluating away from the roots of a polynomial gives $c>0$. Thus $q(z)=\sqrt{c}{\prod }_i(z-{\zeta }_i)$ works. Constant and zero trigonometric polynomials have constant or zero factors. Different selections of roots of a polynomial and constant phases can give different factors.

Pair the off-circle roots of a polynomial using reciprocal-conjugate root pairing, and split each unit-circle root of a polynomial's even multiplicity equally between the two members of a pair. The fundamental theorem of algebra and the leading coefficient $p_d$ give $P(z)=p_d{\prod }_{i=1}^d(z-{\zeta }_i)(z-1/\overline{{\zeta }_i})$. None of the selected ${\zeta }_i$ is zero.

On the unit circle, the identityturns $P(z)/z^d$ intoAt a point of the unit circle outside the finite set of roots of a polynomial, the product is positive and $p(z)$ is nonzero and nonnegative. Its ratio to the product is therefore real and strictly positive. This proves $c>0$, even though the algebraic expression initially permits a complex constant. ConsequentlyThis proves the Fejér–Riesz theorem for a nonzero trigonometric polynomial of actual order $d$. A positive constant has a constant square-root factor, and the identically zero trigonometric polynomial has $q=0$; if $p_d=0$ for a specified upper order $d$, reduce to the actual order first.

Although the PDF permits assuming even multiplicity, there is a short proof of even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial. The real analytic function $p(e^{i\theta })\ge 0$ cannot have a zero of odd order. Near $\zeta =e^{i{\theta }_0}$, $e^{i\theta }-\zeta$ has a simple zero, and the nonzero factor $e^{-id\theta }$ leaves the zero order of $P(e^{i\theta })$ unchanged. Hence the multiplicity of a root must be even.

For $z\ne 0$, the coefficient symmetry yieldsThus $P$ equals its reversed-conjugate polynomial. If $\zeta \ne 0$ is a root of a polynomial, with $P(\zeta )=0$, this identity gives $P(1/\overline{\zeta })=0$, provingThe reciprocal-conjugate root pairing also preserves the multiplicity of a root: reversal and complex conjugation take each factor associated with $\zeta$ to the corresponding factor associated with $1/\overline{\zeta }$, with the same exponent.

Since $P(0)=p_{-d}\ne 0$, no zero root of a polynomial occurs, so the reciprocal operation is always defined on the roots of a polynomial. Roots off the unit circle are paired on opposite sides of it; roots of a polynomial on the unit circle are fixed by this operation. Pairing alone does not force even multiplicity at those fixed roots of a polynomial; that extra fact uses nonnegativity.