Minimum roughness property of the natural cubic spline interpolant ($J(\tilde{g})=J(g)+J(\tilde{g}-g)$)

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.

The minimum roughness property of the natural cubic spline interpolant extends to interpolants with an absolutely continuous first derivative and square-integrable second derivative. For $h=f-g$ vanishing at the knots, integration by parts on each cubic interval gives $\int g^{\prime \prime }{h}^{\prime \prime }=0$: the knot boundary terms cancel by continuity, the exterior terms vanish by natural boundary conditions, and the remaining constant-third-derivative term vanishes because $h$ is zero at the interval endpoints. Expanding the square proves minimality. Equality makes $h$ affine; its zeros at two distinct knots force $h=0$, proving uniqueness. Competitors with infinite roughness cannot improve the finite spline penalty.