Cardinal B-spline

A cardinal B-spline is a B-spline whose knots are consecutive integers.

Sampling consecutive order-six Cardinal B-splines at their support midpoints gives the displayed finite B-spline collocation matrix. No exterior columns are retained. Its row strict diagonal dominance margin is at least $(66-2\cdot 26-2)/120=1/10$, so the inverse infinity-norm bound from diagonal dominance gives $\Vert A^{-1}{\Vert }_{{\ell }^{\infty }}\le 10$. The uniform knot samples are $N_{0,6}(0),\dots ,N_{0,6}(6)=(0,1,26,66,26,1,0)/120$, obtained recursively from the Cox-de Boor recurrence.

This degree-two Cardinal B-spline has three consecutive unit knot intervals as its support. Its first and last polynomial pieces are $(t-i{)}^2/2$ and $(i+3-t{)}^2/2$; on the middle interval it equals $3/4-(t-i-3/2{)}^2$. Its two interior integer values are $1/2$, and its endpoint values are zero. The maximum is $3/4$, emphasizing that partition normalization does not mean unit maximum height.

Sampling order-three Cardinal B-splines $N_j$ at $x_i=i+3/2$ gives $N_i(x_i)=3/4$, $N_{i-1}(x_i)=N_{i+1}(x_i)=1/8$, and all other entries zero. The B-spline collocation matrix therefore has row strict diagonal dominance margin at least $1/2$, giving inverse operator norm at most two. More precisely, with $q=3-2\sqrt{2}$, its absolute inverse row sums are $v_i=2[1-(q^i+q^{n+1-i})/(1+q^{n+1})]$. To see this, change signs by $D_{ii}=(-1{)}^i$: $DAD$ has negative off-diagonal entries and a nonnegative inverse given by a convergent Neumann series. Its inverse row sums solve $6v_i-v_{i-1}-v_{i+1}=8$, $v_0=v_{n+1}=0$. Thus $\Vert A^{-1}{\Vert }_{\infty }=v_{\lfloor (n+1)/2\rfloor }<2$, tending to two as the dimension grows.

A cardinal cubic B-spline is a degree-three Cardinal B-spline. At the three interior integer knots, its nonzero normalized values are $1/6$, $4/6$, and $1/6$.

For the unit-spaced cubic cardinal B-spline on $[0,4]$, the four scalar cubic Bernstein basis coefficient rows are $(0,0,0,1/6)$, $(1/6,1/3,2/3,2/3)$, $(2/3,2/3,1/3,1/6)$ and $(1/6,0,0,0)$. Each row applies on its interval with local parameter in $[0,1]$. They result from the compactly supported $C^2$ truncated-cube combination $\frac{1}{6}{\sum }_{j=0}^4(-1{)}^j(\genfrac{}{}{0ex}{}{4}{j})(x-j{)}_{+}^3$. Each graph piece is a cubic Bézier curve with abscissa controls $j,j+1/3,j+2/3,j+1$.

Sampling consecutive Cardinal cubic B-splines at the centers of their supports gives diagonal values $2/3$ and adjacent values $1/6$, with all other values zero. Finite truncation gives the displayed B-spline collocation matrix, including its endpoint rows without wrapping or adding boundary conditions. Its inverse has alternating signs. The exact inverse norm of midpoint cubic spline collocation is less than three for each finite dimension and approaches three as the dimension grows.

Conjugate the collocation matrix by the diagonal matrix with entries $(-1{)}^i$. The resulting negative-off-diagonal matrix has a nonnegative inverse, as a convergent Neumann series shows. The absolute inverse row sums are therefore the entries of its inverse applied to the all-ones vector. They solve $4u_i-u_{i-1}-u_{i+1}=6$ with $u_0=u_{n+1}=0$, whose solution is displayed. The maximum occurs at the central row or central two rows and equals the inverse operator norm on ${\ell }^{\infty }$. Consequently the associated spline interpolation operator has norm at most three independently of dimension.