14.29.4 Prescribing the determinantCB8BE6BADB1A

On $U$, one has $t=\sigma (r)$. We first construct two triples of coefficients, positive on the middle band, with product $f_*(\sigma )(\sigma ')^2$. At each endpoint of the band, exactly one of the coefficients of $P$ and $Q$ will vanish.

Put $d_*(r)=f_*(\sigma (r))(\sigma '(r))^2$. Since $a=\sigma (-1)$, $b=\sigma (1)$, define the positive smooth divided differences

and $H_*(r)=k_-(r)k_+(r)\sigma '(r)>0$. The fundamental theorem of calculus gives

These formulas show that $H_*$ is smooth and positive also at the endpoints, without any division by a vanishing function. The symmetry $\sigma (-r)=1-\sigma (r)$ makes $d_*$, and hence $H_*$, even.

Choose the function

All its derivatives vanish at $r=\pm 1$: each derivative is a finite sum of polynomials in $r$ times powers of $(1-r^2)^{-1}$, multiplied by the exponential, and the exponential factor dominates every such power. On $|r|<1$, $q=-2rp/(1-r^2)^2$. Define

Both radicands are strictly positive everywhere, so these formulas define smooth functions on $\mathbb R$, including at the endpoints. They satisfy

The coefficients $x_ b,y_ b$ are nonnegative, positive away from the endpoints, with values $(1,0)$ at $-1$ and $(0,1)$ at $1$. On the open band,

so $x_ b>p$, $y_ b>p$. Consequently both triples

are strictly positive there.

We now verify that the two triples have the same product. Set $D_*=x_ by_ b-p^2$, $E_*=p(x_ b-y_ b)=-rp$. Equations (14.29.8) and the formula for $q$ give

The two products in (14.29.9) therefore equal

For $|r|\geq 1$, $p=q=0$ and the same identity reads $x_ by_ bz_ b=d_*$. Thus both triples have exactly the prescribed product, and at the endpoints exactly one of their entries vanishes.