Uniform meshes refute the C^(d+1)-seminorm-only Floater–Hormann rate

Fix $d\ge 0$. For every integer $n\ge 2d+2$, take $h=1\mathrm{/}n$, $x_j=jh$ on $[0,1]$, and query $x=h\mathrm{/}2$. This family has $h\mathrm{/}{h}_{\min }=1$. From [F1], after multiplying every weight by the same positive factor $h^{d}d!$, its weights are ${\tilde{w}}_j=(-1{)}^{j-d}{\delta }_j,{\delta }_j={\sum }_{i=\max (0,j-d)}^{\min (j,n-d)}\left(\genfrac{}{}{0px}{}{d}{j-i}\right)\mathrm{.}$ Indeed, on a uniform mesh each reciprocal-distance product in [F1] equals $h^{-d}\mathrm{/}((j-i)!(i+d-j)!)$. Thus $1\le {\delta }_j\le 2^d$, and ${\delta }_j=2^d$ for $d\le j\le n-d$.

Define $\varphi (t)=\{\begin{array}{ll}(1-16t^2{)}^{d+2},& \mathrm{\mid }t\mathrm{\mid }\le 1\mathrm{/}4,\\ 0,& \mathrm{\mid }t\mathrm{\mid }>1\mathrm{/}4.\end{array}$ At both support endpoints the polynomial has a zero of multiplicity $d+2$, so its derivatives through order $d+1$ vanish there and agree with the zero extension; hence $\varphi \in C^{d+1}(\mathbb{R})$ and $\varphi (0)=1$. Expanding the polynomial and differentiating each term gives $\mathrm{\parallel }{\varphi }^{(d+1)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le M_d:=4^{d+1}{\sum }_{k=\lceil (d+1)\mathrm{/}2\rceil }^{d+2}\left(\genfrac{}{}{0px}{}{d+2}{k}\right)\frac{(2k)!}{(2k-d-1)!}<\mathrm{\infty }\mathrm{.}$ This is a finite bound depending only on $d$, and its $k=d+2$ term shows $M_d>0$.

Let $D(x)={\sum }_{j=0}^n{\tilde{w}}_j\mathrm{/}(x-x_j)$ be the denominator at the query. If $d=0$, its $j=0$ term has magnitude $2\mathrm{/}h$, and the remaining terms form an alternating decreasing sequence beginning with magnitude $2\mathrm{/}h$, so $\mathrm{\mid }D(x)\mathrm{\mid }\le 4\mathrm{/}h$. If $d\ge 1$, the endpoint blocks $j=0,\dots ,d-1$ and $j=n-d+1,\dots ,n$ each contain $d$ terms of magnitude at most $2^{d+1}\mathrm{/}h$. The middle block $j=d,\dots ,n-d$ has constant ${\delta }_j=2^d$ and alternating decreasing magnitudes, so its sum has magnitude at most $2^{d+1}\mathrm{/}h$. Hence in both cases $0<\mathrm{\mid }D(x)\mathrm{\mid }\le \frac{2^d(4d+4)}{h},$ The strict lower inequality follows from [F2].

The real number ${\tilde{w}}_j\mathrm{/}(x-x_j)$ is nonzero. Define ${\sigma }_j=1$ if it is positive and ${\sigma }_j=-1$ if it is negative, and set $f_n(t)=h^{d+1}{\sum }_{j=0}^n{\sigma }_j\varphi \text{\hspace{0.17em}⁣}\left(\frac{t-x_j}{h}\right),0\le t\le 1.$ The supports have radius $h\mathrm{/}4$ about distinct nodes, so they are disjoint; the query $x=h\mathrm{/}2$ is outside every support. Therefore $f_n(x)=0$, $f_n(x_j)=h^{d+1}{\sigma }_j$, $f_n\in C^{d+1}([0,1])$, and $\mathrm{\parallel }{f}_n^{(d+1)}{\mathrm{\parallel }}_{\mathrm{\infty }}\le M_d$ by [F3] and the derivative estimate in step 1.2.

By the barycentric quotient formula [F4] and the signs ${\sigma }_j$, the numerator at $x$ is $h^{d+1}{\sum }_j\mathrm{\mid }{\tilde{w}}_j\mathrm{/}(x-x_j)\mathrm{\mid }$. Restricting to $j=d+1,\dots ,n-d$ and using ${\delta }_j=2^d$ and $\mathrm{\mid }x-x_j\mathrm{\mid }=h(j-\frac{1}{2})$, step 2.1 gives $\mathrm{\mid }{f}_n(x)-r_d(x)\mathrm{\mid }\ge \frac{h^{d+1}}{4d+4}{\sum }_{j=d+1}^{n-d}\frac{1}{j-\frac{1}{2}}\mathrm{.}$ Since $t\mapsto 1\mathrm{/}(t-\frac{1}{2})$ is decreasing, ${\sum }_{j=d+1}^{n-d}\frac{1}{j-\frac{1}{2}}\ge {\int }_{d+1}^{n-d+1}\frac{dt}{t-\frac{1}{2}}=\log \frac{n-d+\frac{1}{2}}{d+\frac{1}{2}}\mathrm{.}$ This lower bound grows without limit as $n$ increases.

By [F2], $f_n-r_d$ is continuous on $[0,1]$, so [F5] makes its uniform norm finite. It follows that $\frac{\mathrm{\parallel }{f}_n-r_d{\mathrm{\parallel }}_{\mathrm{\infty }}}{h^{d+1}{M}_d}\ge \frac{\log ((n-d+\frac{1}{2})\mathrm{/}(d+\frac{1}{2}))}{(4d+4)M_d}⟶\mathrm{\infty }(n\to \mathrm{\infty })\mathrm{.}$ Every mesh in the family has ratio one, but no fixed $C(d,1)$ can satisfy the refuted inequality. Thus it fails even at $\rho =1$ for each fixed $d$.