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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 degree-two Cardinal B-spline basis on knots $1,\dots ,n+3$, sampling at $x_i=i+2$ gives the upper-bidiagonal B-spline collocation matrix $A=(I+S)/2$, where $S$ is the one-step upper shift. Since $S^n=0$, $A^{-1}=2{\sum }_{r=0}^{n-1}(-S{)}^r$ and its maximum absolute row sum is $2n$. The B-spline interpolation operator norm estimate with the order-three stability constant $d_3=3$ gives the displayed linear upper and lower bounds. The lower bound proves failure of uniform boundedness.

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.

Uniform knots give translation invariance $N_{i,k}(x)=N_{0,k}(x-i)$. Put $v_{k,j}=N_{0,k}(j)$ and set missing values to zero. The recurrence in part (a) gives the exact numerical recursionThe order-two starting row is $(0,1,0)$. For example, the order-four middle value is $(2\cdot \frac{1}{2}+2\cdot \frac{1}{2})/3=2/3$, and the order-six central value is $(3\cdot \frac{11}{24}+3\cdot \frac{11}{24})/5=66/120$. Iteration gives the requested array, listing $N_{0,k}(0),\dots ,N_{0,k}(k)$:The symmetry follows from reflection of the uniform Cardinal B-spline about the midpoint of its support. The row sums equal one, consistent with sampling the uniform partition of unity. These are partition-normalized values; normalizing each spline to have peak one would produce a different array.

There are no numerical continuity bounds determined by $w$ and $a$ alone: the subdivision mask coefficients matter. First check constant reproduction, ${\sum }_j{M}_{ϵ-aj}=1$ for every residue $ϵ$, and convergence. For the usual stationary scalar scheme, two useful calculations are a uniform contraction test and a spectral obstruction test. Higher smoothness is tested with derived difference subdivision masks.

Write $M(z)={\sum }_j{M}_j{z}^j$ and $s_a(z)=1+z+\cdots +z^{a-1}$. Using backward differences $\Delta P_i=P_i-P_{i-1}$, the generating-function identityproves the subdivision difference scheme relation $\Delta S_M=S_{M/s_a}\Delta$. Iterating gives ${\Delta }^r{S}_M=S_{M/s_a^r}{\Delta }^r$. To test a candidate $C^r$ limit, suppose the requisite sum rules hold, so $M/s_a^{r+1}$ is a finite Laurent polynomial. The scaled derivative controls $q_i^{\ell }=a^{r\ell }{\Delta }^r{P}_i^{\ell }$ are refined by the subdivision maskwhose residue sums are one. Their first differences are refined byThe powers of $a$ are essential because each refinement shrinks parameter intervals by $a$.

A lower continuity bound comes from proving uniform contraction. The infinity norm of one difference-refinement step is bounded byIf $q<1$, adjacent derivative controls contract geometrically at every location. More generally, form the $p$-step subdivision maskand test ${\max }_{ϵ\mathrm{mod}{a}^p}{\sum }_j|(B_r^{[p]}{)}_{ϵ-a^{p}j}|<1$. This proves convergence of the constant-preserving derivative scheme to a continuous limit. One justification is to compare consecutive piecewise-linear interpolants: their difference is bounded by a constant times the maximum adjacent-control difference, and the geometrically decreasing bound is summable. To identify the limit as $P^{(r)}$, use fixed-degree Cardinal B-spline interpolants of the original refined controls. Their $r$th derivatives are sums of the scaled finite differences; convergence of both functions and derivatives gives $P\in C^r$. Testing successive $r$ supplies a guaranteed lower bound.

An upper continuity bound comes from an observable noncontracting shape mode. Build finite local subdivision matrices for $B_r$ on a refinement-invariant control neighborhood. For a digit $ϵ$ their entries are $(A_{ϵ}{)}_{ij}=(B_r{)}_{ϵ+i-aj}$, since the new index is $ak+ϵ+i$. Restrict to the realizable difference space and discard modes that do not affect the limit. A word ${ϵ}_1,\dots ,{ϵ}_p$ corresponds to a nested parameter location. If its product has an observable eigenvalue $\lambda$ with $|\lambda |>1$, repetition of this word makes some scaled derivative differences grow rather than tend to zero. For a stable limit representation, $C^r$ regularity would force these differences to vanish; hence general-position data cannot have $C^r$ continuity. A modulus-one mode requires examination of whether it actually persists, rather than an automatic rounding rule.

The two calculations can be organized through the joint spectral radius $\hat{\rho }$ of these restricted matrices:These norm and spectral bounds for subdivision regularity make the test computable from the subdivision mask. The right bound certifies all-word contraction when it is less than one; the left bound detects a repeated-word obstruction when it exceeds one. Longer products improve the tests. A single matrix's spectrum is not a substitute for controlling arbitrary products. The product estimates also appear in Charina's analysis of subdivision regularity; the strict contraction and observable-mode arguments above explain their respective roles.

Let $C_k(u)$ be the order-$k$ Cardinal B-spline supported on $[0,k]$. The Cox-de Boor recurrence on unit-spaced knots readsStart with $C_1=1_{[0,1)}$. The order-two hat satisfies $C_2(1)=1$ and vanishes at its support endpoints. Substituting these values into the recurrence gives $C_3(1)=C_3(2)=1/2$ and thenThe continuous cubic spline is zero at $0,4$ and outside its support. Since $N_j(t)=C_4(t-j)$ and $x_i=i+2$, this yields the midpoint cubic spline collocation valuesOnly indices $1\le j\le n$ are retained. In particular no nonexistent spline is added to an endpoint row.

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 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.

The unit-direction univariate box spline is the density of the sum of $m$ independent uniform unit-interval coordinates. It is the order-$m$ Cardinal B-spline: its closed support is $[0,m]$, it is strictly positive on the interior, and its polynomial degree is $m-1$. For $m\ge 2$ it is $C^{m-2}$, generally not $C^{m-1}$. The recurrence $B_m(x)={\int }_0^1{B}_{m-1}(x-u)du$ proves its support and positivity. Starting with a half-open unit interval, integrating the translate partition recursively proves ${\sum }_{j\in \mathbb{Z}}{B}_m(x-j)=1$. Orthogonal projection onto the unit diagonal rescales support width from $m$ to $\sqrt{m}$.