PCHIP loses third-order value accuracy next to a smooth extremum

The secants on $[-2h,-h]$, $[-h,0]$, $[0,h]$, and $[h,2h]$ are $-3h$, $-h$, $h$, and $3h$. At $x=-h$, the equal-width weights are $3h,3h$, so [F1] gives $m(-h)=6h\mathrm{/}(3h\mathrm{/}(-3h)+3h\mathrm{/}(-h))=-3h\mathrm{/}2$. At $x=0$ the adjacent secants have opposite signs, so [F1] gives $m(0)=0$.

With $t=(x+h)\mathrm{/}h$ on $[-h,0]$, [F2] gives $P_h(x)\mathrm{/}{h}^2=H_{00}(t)-\frac{3}{2}{H}_{10}(t)=1-\frac{3}{2}t+\frac{1}{2}{t}^3$, whereas $f(x)\mathrm{/}{h}^2=(1-t{)}^2$. Therefore $(P_h(x)-f(x))\mathrm{/}{h}^2=\frac{1}{2}t(1-t{)}^2$.

On this cell, step 2.1 gives the error $E(x)=\mathrm{\mid }{P}_h(x)-f(x)\mathrm{\mid }=\frac{h^2}{2}t(1-t{)}^2$, since $0\le t\le 1$. This expression is a polynomial in $x$ through $t=(x+h)\mathrm{/}h$, so [F4] makes it differentiable, hence continuous, on the cell. As $h>0$, the interval $[-h,0]$ is nonempty and compact by [F5]; [F6] therefore gives an attained maximum of $E$ there. Since $t=1\mathrm{/}3$ corresponds to a point of the cell and step 2.1 gives $E=2h^2\mathrm{/}27$ at that point, the maximum is at least $2h^2\mathrm{/}27$. Any uniform bound $\mathrm{\mid }{P}_h-f\mathrm{\mid }\le C{h}^3$ on these grids with $C$ independent of $n$ would imply $2\mathrm{/}27\le Ch=C\mathrm{/}n$ for all sufficiently large $n$, which is impossible as $n\to \mathrm{\infty }$. The hypothesis $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\ge c>0$ in [F3] also fails because [F4] gives $f^{\mathrm{\prime }}(0)=0$.