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Properties of B-Splines 📂Fourier Analysis

Properties of B-Splines

Properties1

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

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

(b) ∫−∞∞Nm(x)dx=1\displaystyle \int _{-\infty} ^{\infty} N_{m}(x)dx=1

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

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

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

Explanation

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

Proof

(b)

  • Step 1. It holds when m=1m=1.

    By the definition of N1N_{1}, the following equation obviously holds.

    ∫−∞∞N1(x)dx=∫01dx=1 \int _{-\infty} ^{\infty} N_{1}(x)dx=\int _{0}^{1} dx = 1

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

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

    ∫−∞∞Nm(x)dx=1 \int _{-\infty} ^{\infty} N_{m}(x)dx=1

    holds. Then, by definition,

    ∫Nm+1(x)dx=∫Nm∗N1(x)dx=∫x∫tNm(x−t)N1(t)dtdx=∫tN1(t)∫xNm(x−t)dxdt=∫N1(t)dt=1 \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∈Nm\in N, the following holds.

    ∫−∞∞Nm(x)dx=1 \int _{-\infty} ^{\infty} N_{m}(x)dx=1

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(c)

  • Step 1. It holds when m=2m=2.

    As

    N2(x)={x0≤x≤1−x+21≤x≤20otherwise 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

    N2(x−k)= {x−kk≤x≤k+1−x+k+2k+1≤x≤k+20otherwiseN2(x−(k−1))= {x−k+1k−1≤x≤k−x+k+1k≤x≤k+10otherwise \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∈Zj\in \mathbb{Z}, it is j≤x0≤j+1j \le x_{0} \le j+1. Then,

    N2(x0−k)=0,k∈Z∖{j,j−1} N_{2}(x_{0}-k)=0,\quad k\in \mathbb{Z}\setminus \left\{ j,j-1 \right\}

    holds. Therefore,

    ∑k∈ZN2(x0−k)= N2(x0−j)+N2(x0−(j−1))= (x0−j)+(−x0+j+1)= 1 \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 x0x_{0}, thus

    ∑k∈ZN2(x−k)=1 \sum \limits _{k\in \mathbb{Z}} N_{2}(x-k) =1

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

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

    ∑k∈ZNm+1(x−k)= ∑k∈Z∫01Nm(x−k−t)dt= ∫01∑k∈ZNm(x−k−t)dt= ∫01dt= 1 \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 ∑k∈Z\sum _{k\in \mathbb{Z}}, as shown in Step 1., is a finite sum.

  • Step 3.

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

    ∑k∈ZNm(x−k)=1 \sum \limits _{k\in \mathbb{Z}} N_{m}(x-k) =1

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(c')

By the definition of N1N_{1},

∑k∈ZN1(x−k)={1x∈R∖Z2x∈Z \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}

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  1. Ole Christensen, Functions, Spaces, and Expansions: Mathematical Tools in Physics and Engineering (2010), p205-206 ↩︎