Existence, uniqueness, and spline reduction for cubic smoothing

Let $g\in W^{2,2}(a,b)$ and use the continuous representative in [F2] for $g$ and $g^{\mathrm{\prime }}$. The Countable Choice convention in [F1] and the choice hypotheses for [F2] are available by [F5] and [A1]. Interpolate the values $g(t_j)$ by the natural cubic $s$ from [F3], and put $h=g-s$. Then $h(t_j)=0$ for every $j$. On each interval between distinct knots, $s^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ is constant; on the two linear tails, $s^{\mathrm{\prime }\mathrm{\prime }}=s^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=0$.

The natural spline with knots $t_1,\dots ,t_r$ is uniquely determined by its nodal vector $z=(s(t_1),\dots ,s(t_r))\in {\mathbb{R}}^r$ by [F3]. The tridiagonal system in the interpolation proof is linear in its data, and the interval formula is linear in the nodal values and second derivatives; hence $z\mapsto s_z^{\mathrm{\prime }\mathrm{\prime }}$ is linear into the finite-dimensional space of piecewise-linear functions. Therefore $R(z)=\int \mathrm{\mid }{s}_z^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}^2$ is a finite nonnegative quadratic form. In these coordinates $J(s)={\sum }_j{\sum }_{i:x_i=t_j}(y_i-z_j{)}^2+\lambda R(z)$. The data term has positive-definite diagonal quadratic part because every $t_j$ occurs at least once. Adding the nonnegative quadratic form $\lambda R$ leaves a positive-definite quadratic objective, which has a unique minimizer by completing the square.

On each knot interval, the product $s^{\mathrm{\prime }\mathrm{\prime }}{h}^{\mathrm{\prime }}$ is absolutely continuous, so the fundamental theorem of calculus gives $\int s^{\mathrm{\prime }\mathrm{\prime }}{h}^{\mathrm{\prime }\mathrm{\prime }}=[s^{\mathrm{\prime }\mathrm{\prime }}{h}^{\mathrm{\prime }}]-\int s^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{h}^{\mathrm{\prime }}$. Summing over the intervals, the interior boundary terms cancel because $s^{\mathrm{\prime }\mathrm{\prime }}$ and $h^{\mathrm{\prime }}$ are continuous. At each outer endpoint, $s^{\mathrm{\prime }\mathrm{\prime }}=0$ either on a linear tail or by the natural condition when that endpoint is a knot. The remaining terms are $-s^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}_{(t_j,t_{j+1})}(h(t_{j+1})-h(t_j))=0$. Thus ${\int }_a^b{s}^{\mathrm{\prime }\mathrm{\prime }}{h}^{\mathrm{\prime }\mathrm{\prime }}=0$.

Since $g=s+h$, expanding the square and applying step 2.1 yields ${\int }_a^b\mathrm{\mid }{g}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}^2={\int }_a^b\mathrm{\mid }{s}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}^2+{\int }_a^b\mathrm{\mid }{h}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{.}$ Also $s(x_i)=g(x_i)$ for every observed input, so their data residual terms agree. Therefore $J(g)=J(s)+\lambda {\int }_a^b\mathrm{\mid }{h}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}^2\ge J(s)$, with equality only if $h^{\mathrm{\prime }\mathrm{\prime }}=0$ almost everywhere.

Let $s_{\ast }$ be this finite-dimensional minimizer. Step 3.1 shows $J(g)\ge J(s_{\ast })$ for every $g\in W^{2,2}$. If equality holds, then $h^{\mathrm{\prime }\mathrm{\prime }}=0$ almost everywhere, so by [F2] $h$ is affine; its zero values at the two distinct knots $t_1,t_r$ force $h=0$. Thus $s_{\ast }$ is the unique minimizer over the full Sobolev class, and its natural cubic form and linear tails are those of [F3].