Splines and Series Methods — Examples

The only nonzero degree-zero spline on $(0,1)$ is $B_{3,0}=1$. Applying the recurrence [F1] first at degree one gives $B_{2,1}=1-x$ and $B_{3,1}=x$. Applying it again gives $B_{1,2}=(1-x{)}^2$, $B_{2,2}=x(1-x)+x(1-x)=2x(1-x)$, and $B_{3,2}=x^2$. Each is nonnegative on $[0,1]$ and their sum is $(1-x{)}^2+2x(1-x)+x^2=1$.

For $q(x)=a+bx+c{x}^2$, choose coefficients $a$, $a+b\mathrm{/}2$, and $a+b+c$. Their B-spline combination is $a(1-x{)}^2+2(a+b\mathrm{/}2)x(1-x)+(a+b+c)x^2=a+bx+c{x}^2\mathrm{.}$ Thus every quadratic is reproduced; uniqueness of the coefficient vector follows from the basis assertion [F2].

For two knots, there are no interior second derivatives in the system of [F2]. The endpoint second derivatives are both zero, so the cubic segment has affine second derivative vanishing at both ends, hence second derivative identically zero. The segment is the unique line through the two values, and its linear extension on any data-free tail is the same line.

For the three-knot data, $h_1=h_2=1$ and the secant slopes are $d_1=1$, $d_2=-1$. The one interior system equation is $4M_2=6(d_2-d_1)=-12$, so $M_2=-3$ while $M_1=M_3=0$.

Substitution in the interval formula gives the displayed pieces. They interpolate $0,1,0$; their first derivatives at $1$ are both $0$, their second derivatives at $1$ are both $-3$, and the second derivatives at $0,2$ are $0$. Thus they join to a natural cubic spline, which is unique by [F2].

For every $c$, both terms in [F1] vanish at $g_c$, so $J(g_c)=0$. The criterion is a sum of nonnegative terms, hence zero is its minimum.

If $c\mathrm{\ne }d$, then $g_c(1)=1+c\mathrm{/}2\mathrm{\ne }1+d\mathrm{/}2=g_d(1)$, so $g_c$ and $g_d$ are distinct minimizers. The one-observation extension is not unique, proving the lower bound of two distinct inputs is necessary.

Use the disjoint cells $I_j=[t_j,t_{j+1})$, $0\le j<J$, with the final cell also containing $1$, and put $J_j=I_j$. For any continuous input $g$ and $x\in [t_j,t_{j+1}]$, $I_{h}g(x)=\frac{t_{j+1}-x}{h}g(t_j)+\frac{x-t_j}{h}g(t_{j+1})$. The nonnegative weights sum to one. Continuity makes the endpoint values limits from within $I_j$ and gives $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I_j)}={\max }_{[t_j,t_{j+1}]}\mathrm{\mid }g\mathrm{\mid }$, so $\mathrm{\parallel }{I}_{h}g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I_j)}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(J_j)}$. Continuous inputs agreeing almost everywhere on $J_j$ agree on its closure and hence have the same interpolant on $I_j$. The map is real-linear, reproduces every affine function, and is a continuous piecewise-linear spline by [F1]. Each neighborhood is one cell and each point belongs to exactly one neighborhood. Thus, as $J$ varies, the maps $I_h$ form a mesh-uniform family as in [F2] with $p=l=1$, $q=\mathrm{\infty }$, $M=1$, and $C_Q=1$.

Let $M_2=\mathrm{\parallel }{D}^{2}f{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,1)}$. On a cell $[u,v]$ of length $h$, put $a_x=x-u$ and $b_x=v-x$. By [F3] applied to $f$ and $Df$, let $f^{\ast }$ and $g^{\ast }$ be their absolutely continuous representatives; $(g^{\ast }{)}^{\mathrm{\prime }}=D^{2}f$ almost everywhere and $f^{\ast }(x)-f^{\ast }(u)={\int }_u^x{g}^{\ast }(s)ds$. The absolutely continuous function $t\mapsto (x-t)g^{\ast }(t)$ has derivative $-g^{\ast }(t)+(x-t)D^{2}f(t)$ almost everywhere, so the fundamental theorem for absolutely continuous functions gives $f^{\ast }(x)=f^{\ast }(u)+a_x{g}^{\ast }(u)+{\int }_u^x(x-t)D^{2}f(t)dt$, and the same formula at $v$. Since $I_{h}f(x)=f^{\ast }(u)+(a_x\mathrm{/}h)(f^{\ast }(v)-f^{\ast }(u))$, subtraction and simplification yield $f^{\ast }(x)-I_{h}f(x)=-{\int }_u^x\frac{b_x(t-u)}{h}{D}^{2}f(t)dt-{\int }_x^v\frac{a_x(v-t)}{h}{D}^{2}f(t)dt$. Hence $\mathrm{\mid }{f}^{\ast }(x)-I_{h}f(x)\mathrm{\mid }\le M_2\{b_x{a}_x^2\mathrm{/}(2h)+a_x{b}_x^2\mathrm{/}(2h)\}=M_2{a}_x{b}_x\mathrm{/}2\le M_2{h}^2\mathrm{/}8$, since $a_x+b_x=h$ and $a_x{b}_x\le h^2\mathrm{/}4$. This holds on every cell and at every point, so taking the supremum gives $\mathrm{\parallel }f-I_{h}f{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,1)}\le h^2\mathrm{\parallel }{D}^{2}f{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,1)}\mathrm{/}8$.

On the clamped knot vector $(0,0,1,1)$, the only nonzero degree-zero spline on $(0,1)$ is $B_{2,0}=1$; the recurrence in [F3] gives $B_{1,1}=1-x$ and $B_{2,1}=x$, so $p(1\mathrm{/}2)=(1\mathrm{/}2,1\mathrm{/}2)$. Therefore $p(1\mathrm{/}2{)}^T\widehat{\beta }=(1\mathrm{/}2,1\mathrm{/}2)(0,2{)}^T=1$, the training residual and empirical training loss are zero, and the fitted vector equals the projection in [F1]. The chosen coefficient vector is one admissible representation of it.

The independent test squared error has expectation $E\{\widehat{g}(X^{\mathrm{\prime }})-g_0(X^{\mathrm{\prime }}){\}}^2={\int }_0^1(2x-1{)}^{2}dx=\frac{4}{3}-2+1=\frac{1}{3}>0.$ Thus zero training loss does not imply zero independent test loss, consistent with the scope of [F2].