B-spline basis properties and polynomial reproduction

For $p=0$, support, piecewise degree, and nonnegativity follow from the indicator definition in [F1]. Inductively, each degree-$p$ term is a lower-degree spline multiplied by an affine factor; on the lower-degree spline's support that factor is nonnegative, and the two supports lie inside $[{\xi }_j,{\xi }_{j+p+1}]$. Thus support, degree at most $p$, and nonnegativity hold for every $p$.

Let the distinct interior knots be ${\eta }_1<\cdots <{\eta }_s$ with multiplicities ${\mu }_1,\dots ,{\mu }_s$. For $k\ge 1$ set $u_{r,k}(x)=(x-{\eta }_r{)}^k{\mathbf{1}}_{[{\eta }_r,\mathrm{\infty })}(x)$ and set $u_{r,0}(x)={\mathbf{1}}_{[{\eta }_r,\mathrm{\infty })}(x)$; the value at ${\eta }_r$ follows the right-cell convention of [F1]. The functions $1,x,\dots ,x^p$ together with $u_{r,k}$ for $p-{\mu }_r+1\le k\le p$ span $S_{p,\xi }$: start with the polynomial on the first interval, then at each knot the difference of adjacent pieces has its derivatives through $p-{\mu }_r$ equal to zero, so it is a linear combination of the listed powers, including the right-continuous step when ${\mu }_r=p+1$. They are independent because restriction to successive knot intervals first forces the initial polynomial coefficients, then each new knot's coefficients, to vanish. Thus $\dim S_{p,\xi }=p+1+{\sum }_r{\mu }_r=n$.

Define ${\psi }_{j,0}(y)=1$ and ${\psi }_{j,d}(y)={\prod }_{r=1}^d(y-{\xi }_{j+r})$. When ${\xi }_{j+d}>{\xi }_j$, factoring ${\psi }_{j,d-1}$ gives $(y-x){\psi }_{j,d-1}(y)=\frac{x-{\xi }_j}{{\xi }_{j+d}-{\xi }_j}{\psi }_{j,d}(y)+\frac{{\xi }_{j+d}-x}{{\xi }_{j+d}-{\xi }_j}{\psi }_{j-1,d}(y)$. Fix a nonempty cell $[{\xi }_m,{\xi }_{m+1})$. At degree zero its sole nonzero indicator is $B_{m,0,\xi }$, so the local identity holds. Inductively, the degree-$(d-1)$ local sum has indices $m-d+1\le j\le m$. For every such index, ${\xi }_j\le {\xi }_m<{\xi }_{m+1}\le {\xi }_{j+d}$, so the displayed identity has a strictly positive denominator. Multiply that local sum by $y-x$, apply the identity, and regroup with [F1]. The two extra lower-degree functions $B_{m-d,d-1,\xi }$ and $B_{m+1,d-1,\xi }$ vanish on the cell by support, even when their supports are empty; their recurrence terms are zero under [F1]. Thus $(y-x{)}^d={\sum }_{i=m-d}^m{\psi }_{i,d}(y)B_{i,d,\xi }(x)$ for $0\le d\le p$, and in particular for $d=p$.

Differentiate the identity in step 2.1 exactly $p-k$ times in $y$ and set $y=0$, for each $0\le k\le p$. Since $\frac{d^{p-k}}{d{y}^{p-k}}(y-x{)}^p{\mathrm{\mid }}_{y=0}=\frac{p!}{k!}(-x{)}^k$, this yields $x^k={\sum }_{i=m-p}^m[(-1{)}^k\frac{k!}{p!}{\psi }_{i,p}^{(p-k)}(0)]B_{i,p,\xi }(x)\mathrm{.}$ The coefficients do not depend on $m$, so this is a single expansion valid on every knot interval. Linear combination proves reproduction of every member of ${\mathcal{P}}_p$; taking left limits extends the expansion to $b$.

Fix an open nonempty knot interval $({\xi }_m,{\xi }_{m+1})$ and $p\ge 1$. The $p+1$ dual polynomials ${\psi }_{i,p}$, $m-p\le i\le m$, form a basis of ${\mathcal{P}}_p$ in $y$: differentiating the identity of step 2.1 in $x$ at an interior point $x_0$ shows that their span contains $(y-x_0{)}^k$ for every $0\le k\le p$. For $m-p+1\le i\le m$, the span ${\xi }_{i+p}-{\xi }_i$ is strictly positive and factoring gives ${\psi }_{i,p}-{\psi }_{i-1,p}=-({\xi }_{i+p}-{\xi }_i){\psi }_{i,p-1}$. Differentiate the degree-$p$ local identity in $x$ and use the degree-$(p-1)$ identity and this difference formula to obtain ${\sum }_{i=m-p}^m{B}_{i,p,\xi }^{\mathrm{\prime }}{\psi }_{i,p}=p{\sum }_{i=m-p+1}^m\frac{B_{i,p-1,\xi }}{{\xi }_{i+p}-{\xi }_i}({\psi }_{i,p}-{\psi }_{i-1,p})$. Comparison in the dual basis yields $B_{j,p,\xi }^{\mathrm{\prime }}=p(\frac{B_{j,p-1,\xi }}{{\xi }_{j+p}-{\xi }_j}-\frac{B_{j+1,p-1,\xi }}{{\xi }_{j+p+1}-{\xi }_{j+1}})$ on this interval. The boundary lower-degree terms vanish by support; any zero denominator corresponds to an identically zero lower-degree spline and its term is zero. For indices outside the active range both sides vanish by support.

Taking $k=0$ in step 3.1 gives ${\sum }_{i=m-p}^m{B}_{i,p,\xi }(x)=1$. Step 3.1 also shows that these $p+1$ active functions span ${\mathcal{P}}_p$ on the interval. Because $\dim {\mathcal{P}}_p=p+1$, they are linearly independent there and are a local basis.

Induct on $p$. If the knot multiplicity $\mu \le p-1$, the lower-degree splines in the recurrence are continuous there by induction, so the recurrence makes $B_{j,p,\xi }$ continuous; the derivative formula then makes its derivative $C^{p-1-\mu }$, hence $B_{j,p,\xi }\in C^{p-\mu }$. If $\mu =p$, a B-spline whose support crosses the repeated knot has one nonzero piece on either side, namely $(x-{\xi }_j{)}^p\mathrm{/}({\xi }_{j+p}-{\xi }_j{)}^p$ and $({\xi }_{j+p+1}-x{)}^p\mathrm{/}({\xi }_{j+p+1}-{\xi }_{j+1}{)}^p$; both one-sided limits equal $1$, while any B-spline not crossing it has zero endpoint value there. If $\mu =p+1$, the required order is $C^{-1}$ and no matching is asserted.

The smoothness and piecewise-degree properties from steps 1.1 and 4.2 put every B-spline in $S_{p,\xi }$. If a linear combination of all $n$ B-splines is zero, then on every nonempty knot interval its active coefficients vanish by the local basis property in step 4.1. Each B-spline has a nonempty support interval by the clamped-knot hypothesis, so every coefficient occurs on at least one such interval and is zero. The B-splines are therefore $n$ independent members of the $n$-dimensional space in step 1.2, hence form its basis. The unique global expansion includes every polynomial from step 3.1.