A nonnegative even polynomial has a bounded parity complement

If $A=0$, choose $B=C=0$. Otherwise evenness makes $\mathrm{deg}A=2n$ for some $0\le n\le d$. By [F3], [F8], and repeated use of [F4], factor $A$ over $\mathbb{C}$ into its finite multiset of roots with multiplicity.

The root multiset, with multiplicities, is invariant under $\zeta \mapsto -\zeta$ because substituting $-x$ preserves the order of a zero, and under conjugation as follows: if $A=(x-\zeta {)}^{m}H$ with $H(\zeta )\mathrm{\ne }0$, conjugating the entire coefficient factorization gives $A=(x-\bar{\zeta }{)}^m\bar{H}$ and $\bar{H}(\bar{\zeta })=\overline{H(\zeta )}\mathrm{\ne }0$. Hence conjugate roots have exactly equal multiplicities, strengthening the root existence in [F6]. The multiplicity at zero is even because every nonzero coefficient of an even polynomial has even exponent. Every real root $s$ with $0<\mathrm{\mid }s\mathrm{\mid }<1$ has even multiplicity: write $A(x)=(x-s{)}^{m}h(x)$ with $h(s)\mathrm{\ne }0$. Division by $x-s$ gives $h(x)-h(s)=(x-s)k(x)$ for a polynomial $k$. On a bounded interval around $s$, $k$ is bounded (each of its finitely many monomials is bounded); choosing the interval small enough makes $\mathrm{\mid }h(x)-h(s)\mathrm{\mid }<\mathrm{\mid }h(s)\mathrm{\mid }$, so $h$ has the sign of $h(s)$ there by [F9]. If $m$ were odd, $(x-s{)}^{m}h(x)$ would then have opposite signs on the two sides of $s$, contrary to $A\ge 0$.

Pair the roots into the following classes, choosing one representative from each symmetric class. A zero root of multiplicity $2r$ contributes $x^r$. A real pair $\pm s$ with $0<s<1$ and multiplicity $m$ at each root contributes $(x^2-s^2{)}^{m\mathrm{/}2}$; this is a polynomial because $m$ is even. A real pair $\pm s$ with $s\ge 1$ and multiplicity $m$ contributes $R_s(x{)}^m$, where $R_s(x)=\sqrt{s^2-1}x+is\sqrt{1-x^2}$ and $R_s{R}_s^{\ast }=s^2-x^2$. A pure-imaginary pair $\pm ib$ with $b>0$ and multiplicity $m$ contributes $I_b(x{)}^m$, where $I_b(x)=\sqrt{b^2+1}x+ib\sqrt{1-x^2}$ and $I_b{I}_b^{\ast }=x^2+b^2$.

A nonreal quadruple $\pm a\pm ib$ with $a,b>0$ and common multiplicity $m$ contributes $Q_{a,b}(x{)}^m$. Put $u=a^2+b^2$ and $v=2(a^2+1)b^2+(a^2-1{)}^2+b^4$, and set $c=u+\sqrt{v}$. Since $\sqrt{v}\ge \mathrm{\mid }{a}^2-1\mathrm{\mid }$, the two cases $a^2\le 1$ and $a^2\ge 1$ give $c\ge 1$, so all displayed square roots are real by [F5]. Define $Q_{a,b}(x)=c{x}^2-u+i\sqrt{c^2-1}x\sqrt{1-x^2}$. For $x\in [-1,1]$, its quartic norm is $Q_{a,b}(x)\overline{Q_{a,b}(x)}=\mathrm{\mid }{Q}_{a,b}(x){\mathrm{\mid }}^2\ge 0$; because the quartic roots are the nonreal numbers $\pm a\pm ib$, it is strictly positive on the real interval. Direct expansion gives $Q_{a,b}{Q}_{a,b}^{\ast }=x^4+2(b^2-a^2)x^2+(a^2+b^2{)}^2$, the product of the four linear root factors.

Let $G(x)$ be the product of the nonnegative real factors associated with these root classes: $x^{2r}$ for zero roots, $(x^2-s^2{)}^m$ for interior real pairs, $(s^2-x^2{)}^m$ for exterior real pairs, $(x^2+b^2{)}^m$ for pure-imaginary pairs, and the quartics from step 1.4. Then $A=\kappa G$ for a nonzero real constant $\kappa$, because both sides have the same roots with the same multiplicities and the same degree. Enumerate reduced rationals in $(-1,1)$ by increasing positive denominator and then increasing numerator, and take the first $x_0$ outside the finite root set; this search terminates because a nonzero polynomial has finitely many roots by step 1.1. There $A(x_0)>0$ by the nonnegativity hypothesis and $A(x_0)\mathrm{\ne }0$, and $G(x_0)>0$ by its displayed factors, so $\kappa =A(x_0)\mathrm{/}G(x_0)>0$ by [F9]. Let $K=\sqrt{\kappa }$ by [F5].

Multiply the corresponding root-class factors defined above to form $W(x)=K\prod R(x)$, retaining the interior and zero factors as real factors. Repeated use of $(B+i\sqrt{1-x^2}C)(D+i\sqrt{1-x^2}E)=(BD-(1-x^2)CE)+i\sqrt{1-x^2}(BE+CD)$ shows that $W=B+i\sqrt{1-x^2}C$ for real polynomials $B,C$. Their factor identities give $W{W}^{\ast }=A$ for every $x\in [-1,1]$. Both $W{W}^{\ast }=B^2+(1-x^2)C^2$ and $A$ are polynomials; their difference vanishes at infinitely many real points, so it is the zero polynomial, since a nonzero polynomial has at most its degree many roots by repeated division by linear factors.

In each root class, the real part and the coefficient of $i\sqrt{1-x^2}$ have opposite parities; a real zero or interior factor has only a real part. The multiplication rule in step 3.1 preserves this parity relation. If $\mathrm{deg}A=2n$, the degrees of the factors give $\mathrm{deg}B\le n$ and $\mathrm{deg}C\le n-1$: every imaginary term contains at least one factor whose imaginary coefficient has degree at least one below that factor's real degree. When $n=d$, the pair has a unique parity profile $b$ determined by its nonzero component: $B$ has parity $b$ if $B\mathrm{\ne }0$, and otherwise $C$ has parity $1-b$. If $b\mathrm{\ne }d$, then $B$ has degree at most $d-1$, while $C$, whose parity is $d$, has degree at most $d-2$; both terms of $B^2+(1-x^2)C^2$ would then have degree at most $2d-2$, contradicting $\mathrm{deg}A=2d$. Thus $b=d$ and the required parities hold.

If $n<d$, let $b$ be the parity profile from step 4.1. If $b\mathrm{\ne }d$, multiply $W$ by $x+i\sqrt{1-x^2}$. Its conjugate product is $x^2+(1-x^2)=1$, and the multiplication rule changes the profile from $b$ to $1-b$. It raises the degrees by at most one, so the new bounds are $\mathrm{deg}B\le n+1\le d$ and $\mathrm{deg}C\le n\le d-1$. If $b=d$, no adjustment is needed. This also covers $B=0$ or $C=0$, since the profile is read from the nonzero component. Thus the resulting $B,C$ meet the required bounds and parities, completing the identity.