A natural cubic spline undershoots a nonnegative table

Each displayed piece has endpoint values $S_0(0)=1$, $S_0(1)=0$, $S_1(0)=S_1(1)=0$, $S_2(0)=0$, and $S_2(1)=1$, so the pieces interpolate the given table. By [F4], their first derivatives are $S_0^{\mathrm{\prime }}=-6\mathrm{/}5+3t^2\mathrm{/}5$, $S_1^{\mathrm{\prime }}=-3\mathrm{/}5+6t\mathrm{/}5$, and $S_2^{\mathrm{\prime }}=3\mathrm{/}5+6t\mathrm{/}5-3t^2\mathrm{/}5$; the derivative values match at both shared knots. Their second derivatives are $S_0^{\mathrm{\prime }\mathrm{\prime }}=6t\mathrm{/}5$, $S_1^{\mathrm{\prime }\mathrm{\prime }}=6\mathrm{/}5$, and $S_2^{\mathrm{\prime }\mathrm{\prime }}=6\mathrm{/}5-6t\mathrm{/}5$; these match at shared knots, and $S_0^{\mathrm{\prime }\mathrm{\prime }}(0)=S_2^{\mathrm{\prime }\mathrm{\prime }}(1)=0$. By [F1], this is a natural cubic spline.

For $0<t<1$, $S_1(t)=-3t(1-t)\mathrm{/}5<0$, although every tabulated value is nonnegative. Thus the natural cubic spline undershoots the data range on the middle cell.

The secants are $d_0=-1$, $d_1=0$, and $d_2=1$. By [F2], the interior slopes at nodes $1$ and $2$ are both zero because each neighboring pair includes a zero secant; the Hermite piece joining the two zero values with zero slopes is constant zero. By [F3], the first and last PCHIP pieces remain between $1$ and $0$, and between $0$ and $1$, respectively. Hence PCHIP is nonnegative throughout the interval and does not reproduce the natural spline's undershoot.