Uniform cone domains admit stable polynomial norming stencils

The evaluations at points of the open unit cone $C=\{z:0<\mathrm{\mid }z\mathrm{\mid }<1,z_1>\mathrm{\mid }z\mathrm{\mid }\cos \theta \}$ span the dual of ${\mathrm{\Pi }}_{m-1}({\mathbb{R}}^d)$. Indeed, if fewer than $J$ independent evaluations have been chosen, the map from ${\mathrm{\Pi }}_{m-1}$ to their finite value list has nonzero kernel by [F3]; choose a nonzero polynomial $p$ in that kernel. Suppose, for contradiction, that $p$ vanishes throughout $C$. Since $C$ is nonempty and open (it contains $e_1\mathrm{/}2$), it contains a box $I_1\times \cdots \times I_d$ of nonempty open intervals. Write $p$ as a polynomial in its first coordinate, of degree at most $m-1$. For every fixed tuple of the other coordinates in their intervals, this univariate polynomial vanishes at the $m$ distinct points $a_1+\frac{k}{m+1}(b_1-a_1)$, $1\le k\le m$, where $I_1=(a_1,b_1)$. These points lie in $I_1$ by [F4], so the root bound [F2] forces every coefficient to vanish at that tuple. Applying the same argument inductively to each coefficient polynomial on $I_2\times \cdots \times I_d$ shows every coefficient is identically zero, contrary to the choice of $p$. Thus $p$ is nonzero at some point of $C$, and evaluation there increases the independent list. After the finite number $J$ of steps, choose $z_1,\dots ,z_J\in C$ with invertible evaluation matrix.

For the chosen nodes, the weights representing evaluation at zero are the row $\mathbf{p}(0{)}^{\mathsf{T}}{E}^{-1}$, where $E$ is their evaluation matrix and $\mathbf{p}$ is the monomial vector. Choose $0<\epsilon <1$; if $J>1$, also require it to be smaller than one third of the minimum distance between distinct reference nodes. On the bounded perturbation neighborhood, every monomial entry changes by at most a constant times $\epsilon$ by telescoping its finite product; the determinant, being a finite signed sum of products of entries, therefore changes by at most $C\epsilon$. Choose $\epsilon$ smaller if necessary so $C\epsilon <\mathrm{\mid }\det E\mathrm{\mid }\mathrm{/}2$. Then $\mathrm{\mid }\det E^{\mathrm{\prime }}\mathrm{\mid }\ge \mathrm{\mid }\det E\mathrm{\mid }\mathrm{/}2$ for every perturbed matrix $E^{\mathrm{\prime }}$. Its cofactors are bounded there because the entries are bounded, so the adjugate formula $E^{\mathrm{\prime }-1}=\mathrm{adj}(E^{\mathrm{\prime }})\mathrm{/}\det E^{\mathrm{\prime }}$ gives a uniform bound $B$ on the sum of absolute evaluation-at-zero weights.

Choose $L>1\mathrm{/}\epsilon$ and set $h_0\le r_0\mathrm{/}L$. For the fixed $x$, use [F1] to choose $Q_x$. Each target $x+Lh{Q}_x{z}_j$ lies in $\mathrm{\Omega }$ when $h\le h_0$, so the fill-distance definition supplies a site $y_j\in X$ within distance $h$. The normalized points $z_j^{\mathrm{\prime }}=(Lh{)}^{-1}{Q}_x^{-1}(y_j-x)$ then satisfy $\mathrm{\mid }{z}_j^{\mathrm{\prime }}-z_j\mathrm{\mid }\le 1\mathrm{/}L<\epsilon$; hence they are distinct and their evaluation weights have sum at most $B$.

For any $p\in {\mathrm{\Pi }}_{m-1}({\mathbb{R}}^d)$, the affine pullback $q(z)=p(x+Lh{Q}_{x}z)$ has degree at most $m-1$. The weights from step 3.1 represent evaluation at zero for every such polynomial, so $p(x)=q(0)={\sum }_j{\lambda }_{j}q(z_j^{\mathrm{\prime }})={\sum }_j{\lambda }_{j}p(y_j)$. Also $\mathrm{\mid }{y}_j-x\mathrm{\mid }\le Lh\mathrm{\mid }{z}_j\mathrm{\mid }+h\le (L+1)h$. Taking $R=L+1$ proves the bounds with constants independent of $X,h,x$; no separation condition entered the construction.