对正定核函数Φ:X×X→R\Phi:X\times X\to\mathbb{R}Φ:X×X→R和采样点X:={
xi}i=1N⊂X\mathcal{X}\coloneqq\left\{x_i\right\}_{i=1}^N\subset XX:={
xi}i=1N⊂X,其对应的插值基函数应为
ui∗=∑j=1N[Φ−1]ijΦ(⋅,xj), u_i^\ast=\sum_{j=1}^N\left[\Phi^{-1}\right]_{ij}\Phi\left(\cdot,x_j\right), ui∗=j=1∑N[Φ−1]ijΦ(⋅,xj),
其中Φ:=[Φ(xi,xj)]ij\Phi\coloneqq\left[\Phi\left(x_i,x_j\right)\right]_{ij}Φ:=[Φ(xi,xj)]ij。由此,对任意f∈span‾⟨Φ(⋅,x)⟩x∈Xf\in \overline{\text{span}}\left\langle\Phi\left(\cdot,x\right)\right\rangle_{x\in X}f∈span⟨Φ(⋅,x)⟩x∈X,对应插值函数应为
Qf:=∑i=1Nf(xi)ui∗. \mathcal{Q}f\coloneqq\sum_{i=1}^Nf\left(x_i\right)u_i^\ast. Qf:=i=1∑Nf(xi)ui∗.
以下考察误差函数
Ef:=f−Qf. \mathcal{E}f\coloneqq f-\mathcal{Q}f. Ef:=f−Qf.
对任意x∈Xx\in Xx∈X,有
∣[Ef](x)∣=∣⟨f,Φ(⋅,x)−∑i=1Nui∗(x)Φ(⋅,xi)⟩∣≤∥f∥∥Φ(⋅,x)−∑i=1Nui∗(x)Φ(⋅,xi)∥. \begin{aligned} \left|\left[\mathcal{E}f\right]\left(x\right)\right|&=\left|\left\langle f,\Phi\left(\cdot,x\right)-\sum_{i=1}^Nu_i^\ast\left(x\right)\Phi\left(\cdot,x_i\right)\right\rangle\right|\\ &\leq\left\|f\right\|\left\|\Phi\left(\cdot,x\right)-\sum_{i=1}^Nu_i^\ast\left(x\right)\Phi\left(\cdot,x_i\right)\right\|. \end{aligned} ∣[Ef](x)∣=∣∣∣∣∣⟨f,Φ(⋅,x)−i=1∑Nui∗(x)Φ(⋅,xi)⟩∣∣∣∣∣≤∥f∥∥∥∥∥∥Φ(⋅,x)−i=1∑Nui∗(x)Φ(⋅,xi)∥∥∥∥∥.
遂以下研究函数
P(x)2:=∥Φ(⋅,x)−∑i=1Nui∗(x)Φ(⋅,xi)∥2=Φ(x,x)−2∑i=1NΦ(x,xi)ui∗(x)+∑i,j=1NΦ(xi,xj)ui∗(x)uj∗(x). \begin{aligned} P\left(x\right)^2\coloneqq{}& \left\|\Phi\left(\cdot,x\right)-\sum_{i=1}^Nu_i^\ast\left(x\right)\Phi\left(\cdot,x_i\right)\right\|^2\\ ={}&\Phi\left(x,x\right)-2\sum_{i=1}^N\Phi\left(x,x_i\r