Natural cubic spline interpolant ($N\mathbf{v},(N\mathbf{v})(x_i)=v_i$)

For at least two distinct knots $x_1<\cdots <x_n$ and values $v_i$, the unique natural cubic spline matching those values and linear beyond $x_1,x_n$. Its interpolation map is a linear map by uniqueness. Over a larger domain $[a,b]$, the two exterior pieces remain linear. With only one knot, arbitrary slopes of affine functions make uniqueness false.

Let $b_i=N{e}_i$ be the interpolating cardinal basis and set ${\Gamma }_{ij}={\int }_a^b{b}_i^{\prime \prime }{b}_j^{\prime \prime }dx$. Linearity gives $J(N\mathbf{v})={\mathbf{v}}^T\Gamma \mathbf{v}$, and this Gram matrix is a positive semidefinite matrix. Its null space consists precisely of vectors $(\alpha +\beta x_i{)}_i$, by the null-space statement for the second derivative roughness penalty. Thus its rank is $n-2$, including the zero matrix when $n=2$.

For $g=N\mathbf{v}$ and any $\tilde{g}\in C^2[a,b]$ taking the same values at at least two distinct knots, $J(\tilde{g})=J(g)+J(\tilde{g}-g)$. To prove it, set $r=\tilde{g}-g$, integrate $g^{\prime \prime }{r}^{\prime \prime }$ by parts on each polynomial interval and use $g^{(4)}=0$, $r(x_i)=0$, continuous $g^{\prime \prime }$ and linear exterior pieces. Boundary terms cancel. Expanding the square proves the identity. Equality forces $r^{\prime \prime }=0$, so $r$ is an affine function and its two prescribed zeros make it identically zero.