Weighted rational minimizers satisfy the defect-sensitive alternation criterion

First suppose $E_r>0$ and that $r$ has $N+2$ alternating weighted extrema. If another admissible $s=p^{\mathrm{\prime }}\mathrm{/}{q}^{\mathrm{\prime }}$ had strictly smaller weighted error, write $e_i=w(x_i)(f(x_i)-r(x_i))=\sigma (-1{)}^i{E}_r$ and $e_i^{\mathrm{\prime }}=w(x_i)(f(x_i)-s(x_i))$, so $\mathrm{\mid }{e}_i^{\mathrm{\prime }}\mathrm{\mid }<E_r$. Then $\sigma (-1{)}^{i}w(x_i)(s(x_i)-r(x_i))=E_r-\sigma (-1{)}^i{e}_i^{\mathrm{\prime }}>0$. Since $w(x_i)q(x_i)q^{\mathrm{\prime }}(x_i)>0$, the numerator $p^{\mathrm{\prime }}q-p{q}^{\mathrm{\prime }}$ has nonzero alternating signs at those nodes. It lies in ${\mathcal{P}}_N$ by [F3], contradicting [F6]. Thus no rational candidate has smaller weighted error and $r$ is best.

Now suppose $r$ is best with $E:=E_r>0$, and put $e=w(f-r)$. It is continuous by [F8], and [F10] makes $\mathrm{\mid }e\mathrm{\mid }$ attain $E$ on compact $I$, so its active-extrema set is nonempty and compact. If it had no $N+2$ alternating points, [F4] would give a polynomial $g\in {\mathcal{P}}_N$ with $e(x)g(x)>0$ on that set. By [F3], choose $u,v$ with $\mathrm{deg}u\le m$, $\mathrm{deg}v\le n$, and $g=uq-pv$.

If the best error is zero, then $w(f-r)=0$ everywhere; since $w>0$, $r=f$ pointwise, so the best rational function is unique.

Since $q>0$ on compact $I$, [F10] gives $q_0={\min }_{I}q>0$. For all sufficiently small $\epsilon >0$, $q+\epsilon v>0$ on $I$. Reduce $(p+\epsilon u,q+\epsilon v)$ by [F7]. Its reduced denominator has no zero on $I$, since it divides $q+\epsilon v$; it is continuous by [F8] and has constant sign by [F9], so a common sign change makes it positive and a further positive scaling gives coefficient norm $1$. Thus $r_{\epsilon }=(p+\epsilon u)\mathrm{/}(q+\epsilon v)$, as a rational function, is represented by an admissible member of ${\mathcal{R}}_{m,n}^{+}(I)$. Its weighted residual is $e_{\epsilon }=e-\epsilon h_{\epsilon }$, where $h_{\epsilon }:=wg\mathrm{/}[q(q+\epsilon v)]$. On the active-extrema set, $h_0=wg\mathrm{/}{q}^2$ has the same sign as $e$ and is bounded away from zero. By [F8] and [F10], $h_0$ and $e$ are continuous and bounded; the denominator bound implies $h_{\epsilon }\to h_0$ uniformly as $\epsilon \downarrow 0$. Hence there are a neighborhood $U$ of that set and a constant $c>0$ such that $\mathrm{\mid }e\mathrm{\mid }>E\mathrm{/}2$, $e$ and $h_{\epsilon }$ have the same sign, and $\mathrm{\mid }{h}_{\epsilon }\mathrm{\mid }\ge c$ on $U$ for all sufficiently small $\epsilon$. Shrink $\epsilon$ further so $\epsilon \mathrm{\parallel }{h}_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty },I}<E\mathrm{/}2$; then $\epsilon \mathrm{\mid }{h}_{\epsilon }\mathrm{\mid }<\mathrm{\mid }e\mathrm{\mid }$ on $U$ and $\mathrm{\mid }{e}_{\epsilon }\mathrm{\mid }=\mathrm{\mid }e\mathrm{\mid }-\epsilon \mathrm{\mid }{h}_{\epsilon }\mathrm{\mid }\le E-\epsilon c<E$. The compact complement $I\setminus U$, if nonempty, contains no active extremum, so [F10] gives $\mathrm{\mid }e\mathrm{\mid }\le E-\eta$ there for some $\eta >0$; a further small choice of $\epsilon$ makes $\epsilon \mathrm{\parallel }{h}_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty },I}<\eta \mathrm{/}2$. Hence $\mathrm{\parallel }{e}_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty },I}<E$, contradicting best approximation. This proves necessity.

Suppose the best error $E$ is positive and $r,s=p^{\mathrm{\prime }}\mathrm{/}{q}^{\mathrm{\prime }}$ are both best. By step 2.1, $r$ has $N+2$ alternating nodes. At each such node, $w(s-r)=e_r-e_s$ has weak sign $\sigma (-1{)}^i$, because $\mathrm{\mid }{e}_s\mathrm{\mid }\le E$. The numerator $g=p^{\mathrm{\prime }}q-p{q}^{\mathrm{\prime }}$ lies in ${\mathcal{P}}_N$ by [F3], and its values have the same weak alternating signs since $wq{q}^{\mathrm{\prime }}>0$. Thus $\epsilon (-1{)}^{i}g(x_i)\ge 0$ with $\epsilon =\sigma$, so [F5] gives $g=0$ and hence $r=s$.

Steps 1.1 and 2.1 prove the two directions of the criterion for positive error; steps 1.3 and 3.1 prove uniqueness in the zero- and positive-error cases. Existence is [F2].