= Solution
There are no numerical <continuity> bounds determined by $w$ and $a$ alone: the <subdivision mask> coefficients matter. First check constant reproduction, $\sum_jM_{\epsilon-aj}=1$ for every residue $\epsilon$, 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 $M(z)=\sum_jM_jz^j$ and $s_a(z)=1+z+\cdots+z^{a-1}$. Using backward differences $\Delta P_i=P_i-P_{i-1}$, the generating-function identity
$$
(1-z)M(z)=\frac{M(z)}{s_a(z)}(1-z^a)
$$
proves the <subdivision difference scheme> relation $\Delta S_M=S_{M/s_a}\Delta$. Iterating gives $\Delta^rS_M=S_{M/s_a^r}\Delta^r$. 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 $q_i^\ell=a^{r\ell}\Delta^rP_i^\ell$ are refined by the <subdivision mask>
$$
D_r(z)=a^r\frac{M(z)}{s_a(z)^r},
$$
whose residue sums are one. Their first differences are refined by
$$
B_r(z)=a^r\frac{M(z)}{s_a(z)^{r+1}}.
$$
The powers of $a$ are essential because each refinement shrinks parameter intervals by $a$.
\b[A lower <continuity> bound comes from proving uniform contraction.] The infinity norm of one difference-refinement step is bounded by
$$
q=\max_{0\le\epsilon<a}\sum_j|(B_r)_{\epsilon-aj}|.
$$
If $q<1$, adjacent <derivative> controls contract geometrically at every location. More generally, form the $p$-step <subdivision mask>
$$
B_r^{[p]}(z)=\prod_{\nu=0}^{p-1}B_r(z^{a^\nu})
$$
and test $\max_{\epsilon\bmod a^p}\sum_j|(B_r^{[p]})_{\epsilon-a^pj}|<1$. 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>.
\b[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 $\epsilon$ their entries are $(A_\epsilon)_{ij}=(B_r)_{\epsilon+i-aj}$, since the new index is $ak+\epsilon+i$. Restrict to the realizable difference space and discard modes that do not affect the limit. A word $\epsilon_1,\ldots,\epsilon_p$ 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> $\widehat\rho$ of these restricted <matrices>:
$$
\boxed{\max_{|\epsilon|=p}\rho(A_{\epsilon_p}\cdots A_{\epsilon_1})^{1/p}\le\widehat\rho\le\max_{|\epsilon|=p}\|A_{\epsilon_p}\cdots A_{\epsilon_1}\|^{1/p}.}
$$
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 https://arxiv.org/pdf/1202.2765[Charina's analysis of subdivision regularity]; the strict contraction and observable-mode arguments above explain their respective roles.
Back to article page
