📂Fourier Analysis

Properties of B-Splines

Properties¹

The B-spline of order $m\in \mathbb{N}$, denoted $N_{m}$, satisfies the following properties.

(a) $\mathrm{supp}N_{m}=[0,m] \quad \text{and} \quad N_{m}(x)>0 \text{ for } x\in(0,m)$

(b) $\displaystyle \int _{-\infty} ^{\infty} N_{m}(x)dx=1$

(c) For $m\ge 2$, the following equation holds.

$$ \begin{equation} \sum \limits_{k \in \mathbb{Z}} N_{m}(x-k)=1,\quad \forall x\in \mathbb{R} \end{equation} $$

(c’) When $m=1$, the above equation holds for $x\in \mathbb{R}\setminus \mathbb{Z}$.

Explanation

(c) In other words, $\left\{ N_{m}(x-k) \right\}_{k}$ means that it is a partition of unity.

Proof

(b)

• Step 1. It holds when $m=1$.

  By the definition of $N_{1}$, the following equation obviously holds.

  $$ \int _{-\infty} ^{\infty} N_{1}(x)dx=\int _{0}^{1} dx = 1 $$

• Step 2. If it holds when $m$, then it also holds when $m+1$.

  Assuming that for some $m\in \mathbb{N}$,

  $$ \int _{-\infty} ^{\infty} N_{m}(x)dx=1 $$

  holds. Then, by definition,

  $$ \begin{align*} \int N_{m+1}(x)dx &= \int N_{m} \ast N_{1}(x)dx \\ &= \int_{x} \int_{t}N_{m}(x-t)N_{1}(t)dtdx \\ &= \int_{t}N_{1}(t) \int_{x}N_{m}(x-t)dxdt \\ &= \int N_{1}(t)dt \\ &= 1 \end{align*} $$

• Step 3.

  By mathematical induction, for any $m\in N$, the following holds.

  $$ \int _{-\infty} ^{\infty} N_{m}(x)dx=1 $$

■

(c)

• Step 1. It holds when $m=2$.

  As

  $$ N_{2}(x) = \begin{cases} x & 0\le x \le 1 \\ -x+2 & 1 \le x \le 2 \\ 0 & \text{otherwise} \end{cases} $$

  it follows that

  $$ \begin{align*} N_{2}(x-k) =&\ \begin{cases} x-k & k\le x \le k+ 1 \\ -x+k+2 & k+1 \le x \le k+2 \\ 0 & \text{otherwise} \end{cases} \\ N_{2}(x-(k-1)) =&\ \begin{cases} x-k+1 & k-1\le x \le k \\ -x+k+1 & k \le x \le k+1 \\ 0 & \text{otherwise} \end{cases} \end{align*} $$

  Suppose for some $j\in \mathbb{Z}$, it is $j \le x_{0} \le j+1$. Then,

  $$ N_{2}(x_{0}-k)=0,\quad k\in \mathbb{Z}\setminus \left\{ j,j-1 \right\} $$

  holds. Therefore,

  $$ \begin{align*} \sum \limits _{k\in \mathbb{Z}} N_{2}(x_{0}-k) =&\ N_{2}(x_{0}-j)+N_{2}(x_{0}-(j-1)) \\ =&\ (x_{0}-j)+(-x_{0}+j+1) \\ =&\ 1 \end{align*} $$

  This holds for any $x_{0}$, thus

  $$ \sum \limits _{k\in \mathbb{Z}} N_{2}(x-k) =1 $$

• Step 2. If it holds when $m$, then it also holds when $m+1$.

  Assuming that for some $m\in \mathbb{N}$, $(1)$ holds. Then, by definition, the following holds.

  $$ \begin{align*} \sum \limits _{k\in \mathbb{Z}} N_{m+1}(x-k) =&\ \sum \limits _{k\in \mathbb{Z}} \int_{0} ^{1} N_{m}(x-k-t)dt \\ =&\ \int_{0} ^{1} \sum \limits _{k\in \mathbb{Z}} N_{m}(x-k-t)dt \\ =&\ \int_{0}^{1} dt \\ =&\ 1 \end{align*} $$

  The second equality follows because $\sum _{k\in \mathbb{Z}}$, as shown in Step 1., is a finite sum.

• Step 3.

  By mathematical induction, for any $m\in \mathbb{N}$, the following holds.

  $$ \sum \limits _{k\in \mathbb{Z}} N_{m}(x-k) =1 $$

■

(c')

By the definition of $N_{1}$,

$$ \sum \limits _{k\in \mathbb{Z}} N_{1}(x-k) =\begin{cases} 1 & x\in \mathbb{R}\setminus\mathbb{Z} \\ 2 & x \in \mathbb{Z} \end{cases} $$

■