Cubic spline

A cubic spline is a twice continuously differentiable piecewise polynomial of degree at most three whose polynomial pieces meet at specified knots.

A clamped cubic spline is a cubic spline whose endpoint first derivatives are prescribed. A clamped cubic spline interpolant additionally matches prescribed values at its spline knots. Distinct spline knots, arbitrary nodal values and arbitrary endpoint slopes determine one such interpolant. The minimum roughness property of a clamped cubic spline proves uniqueness. Unlike the natural cubic spline, this construction prescribes endpoint slopes rather than zero endpoint second derivatives.

Let $S$ be a clamped cubic spline interpolant and let $T$ have the same values at all spline knots and the same endpoint slopes. For $T\in C^2[x_0,x_n]$, or more generally $T^{\prime }$ an absolutely continuous function with square-integrable second derivative, the exact displayed identity holds. In particular $S$ uniquely minimizes the second derivative roughness penalty $J$ among these competitors.

To prove the identity, put $E=T-S$. On each knot interval, two applications of integration by parts give $\int S^{\prime \prime }{E}^{\prime \prime }=[S^{\prime \prime }{E}^{\prime }-S^{\prime \prime \prime }E]$, since the cubic piece has zero fourth derivative. Terms containing $E$ vanish at the spline knots; terms containing $E^{\prime }$ cancel at interior spline knots because $S^{\prime \prime }$ and $E^{\prime }$ are continuous. The endpoint terms vanish because $E^{\prime }$ is zero there. Thus $\int S^{\prime \prime }{E}^{\prime \prime }=0$, and expanding $(S^{\prime \prime }+E^{\prime \prime }{)}^2$ proves the identity. Equality makes $E^{\prime \prime }=0$, so $E$ is an affine function; a prescribed endpoint value and slope force $E=0$.

If $T$ also matches the first derivative of $S$ at every spline knot, both $E$ and $E^{\prime }$ vanish at the ends of each knot interval. For a sufficiently differentiable $T$, integration by parts separately on that interval gives the stronger local identity $\int E{T}^{(4)}=\int (E^{\prime \prime }{)}^2\ge 0$.

A natural cubic spline is linear beyond its two outer knots, equivalently its second derivative vanishes at those knots.

If $S$ is the natural cubic spline interpolating data $y_i$ at fixed distinct knots $x_i$, then $aS(x)+b+cx$ is the natural cubic spline interpolating $a{y}_i+b+c{x}_i$.

The transformed function is piecewise cubic, preserves continuity of the first two derivatives, and has endpoint second derivatives $a{S}^{\prime \prime }=0$. It takes exactly the transformed nodal values. Uniqueness of the natural spline therefore proves the claim. This permits affine changes in data or the addition of a linear trend without solving a new spline system.

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.

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.