B-Spline basis function partition of unity

Question

I'm struggling to prove the partition of unity property for B-spline basis functions, which states that $$\sum_{i=r-p}^rN_{i,p}(\xi)=1$$, $\xi\in[\xi_r,\xi{r+1})$, where $p$ is the degree. I know that it shouldnt be difficult to prove using induction and i guess i just dont see the obvious. Can someone pls help me with this.

Answer 1

W.l.o.g. let $r=p$ throughout.

For $p=0$ the sum is $N_{0,0}(\xi)$ which equals $1$ for $\xi\in[\xi_0,\xi_1)$ by the definition:

$$N_{i,0}(\xi)=\begin{cases}1 & \xi_i \leq \xi < \xi_{i+1} \\ 0 & \text{else} \end{cases}$$

The recurrence relation for the remaining basis functions is

$$N_{i,p}(\xi)={\frac{\xi-\xi_{i}}{\xi_{i+p-1}-\xi_{i}}}N_{i, p-1}(\xi)+{\frac {\xi_{i+p}-\xi}{\xi_{i+p}-\xi_{i+1}}}N_{i+1,p-1}(\xi)$$

which reveals that $N_{i,p}$ is a linear combination of $N_{i,0},N_{i+1,0},\ \ldots\,,N_{i+p,0}$, which implies that $N_{i,p}(\xi)$ is nonzero only if $\xi\in[\xi_i,\xi_{i+p+1})$

Let $p\in\mathbb{N}$ and assume the sum is $1$ for all smaller values of $p$. Then the sum is

$$\begin{align} N_{0,p}(\xi)+N_{1,p}(\xi)+\ldots+N_{p,p}(\xi) =&{\frac{\xi-\xi_0}{\xi_{p-1}-\xi_0}}N_{0, p-1}(\xi)+{\frac {\xi_{p}-\xi}{\xi_{p}-\xi_{1}}}N_{1,p-1}(\xi)+{} \\&{\frac{\xi-\xi_1}{\xi_{p}-\xi_1}}N_{1, p-1}(\xi)+{\frac {\xi_{1+p}-\xi}{\xi_{1+p}-\xi_{2}}}N_{2,p-1}(\xi)+{} \\&\quad\ldots\quad\ldots\quad\ldots\quad\ldots\quad\ldots\quad\ldots\quad\ldots\ + \\&{\frac{\xi-\xi_p}{\xi_{2p-1}-\xi_p}}N_{p, p-1}(\xi)+{\frac {\xi_{2p}-\xi}{\xi_{2p}-\xi_{p+1}}}N_{p+1,p-1}(\xi) \end{align}$$

$\ \ \ N_{0,p-1}(\xi)$ is nonzero only if $\xi\in[\xi_0,\xi_p)\ \ \ \ \ \ \ \ \ $ so it's zero for $\xi\in[\xi_p,\xi_{p+1})$
$N_{p+1,p-1}(\xi)$ is nonzero only if $\xi\in[\xi_{p+1},\xi_{2p+1})$ so it's zero for $\xi\in[\xi_p,\xi_{p+1})$

Hence the sum is equal to

$$\begin{align} 0&+\left(\frac{\xi_{p}-\xi}{\xi_{p}-\xi_{1}}+\frac{\xi-\xi_1}{\xi_{p}-\xi_1}\right)N_{1,p-1}(\xi) \\&+\left(\frac {\xi_{1+p}-\xi}{\xi_{1+p}-\xi_{2}}+\frac{\xi-\xi_2}{\xi_{1+p}-\xi_2}\right)N_{2,p-1}(\xi)+{} \\&+\quad\ldots\quad\ldots\quad\ldots\quad\ldots\quad\ldots\quad\ldots \\&+\left(\frac {\xi_{2p-1}-\xi}{\xi_{2p-1}-\xi_p}+\frac{\xi-\xi_p}{\xi_{2p-1}-\xi_p}\right)N_{p, p-1}(\xi)+0\ \ =\ \ N_{1,p-1}(\xi)+N_{1,p-1}(\xi)+\ldots+N_{p,p-1}(\xi) \end{align}$$

which is equal to 1 by the induction assumption.