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On an infinite regular triangular grid, height data constant along one edge direction remain constant along that direction under Loop subdivision. Across rows the induced rule is $g_{2j}^{\prime }=(g_{j-1}+6g_j+g_{j+1})/8$, $g_{2j+1}^{\prime }=(g_j+g_{j+1})/2$, the centered Cardinal cubic B-spline rule.

// Target: analysis.bigb

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.

For the centered Cardinal cubic B-spline $\beta$ with knot values $1/6,4/6,1/6$, write $s(x)={\sum }_{j=-1}^{m+1}{c}_j\beta (x/h-j)$ on $[0,mh]$. interpolation gives $c_{i-1}+4c_i+c_{i+1}=6u(ih)$ for $i=0,\dots ,m$. Since ${\beta }^{\prime \prime }(0)=-2$ and ${\beta }^{\prime \prime }(\pm 1)=1$, the natural endpoint conditions $s^{\prime \prime }(0)=s^{\prime \prime }(mh)=0$ give the displayed extra two rows. These supply the two equations missing from interpolation alone.

A $C^2$ piecewise cubic supported on $[0,ph]$ has the truncated-power representation with indices $0,\dots ,p$. Vanishing for $x\ge ph$ imposesafter expanding its cubic exterior polynomial; powers of $h$ cancel. For $p\le 3$, the Vandermonde matrix columns are independent, forcing every coefficient to be zero. For $p=4$, coefficients proportional to $(1,-4,6,-4,1)$ satisfy these conditions because a fourth finite difference annihilates cubics.

Normalize to get the Cardinal cubic B-splineAt its five support knots its values are $0,1/6,2/3,1/6,0$, so their sum is one. Thus four intervals are necessary and sufficient.

For $h=1/2$, the five translates active in the interior of $[0,1]=[0,2h]$ are indexed $-3,-2,-1,0,1$. Write $s(x)={\sum }_{j=-3}^1{c}_j{N}_j^3(x)$. Nodal interpolation givesAt the left endpoint the first derivatives of the three active functions are $-1/(2h),0,1/(2h)$, and their second derivatives are $1/h^2,-2/h^2,1/h^2$. The two remaining equations areTogether these five equations form the requested system; insert $h=1/2$ if numerical coefficients are desired. The first, fourth and fifth determine the first three coefficients, and the other two then determine $c_0,c_1$, so the system is nonsingular.

The source only assumes $u\in C[0,1]$, which need not have either endpoint derivative. The Hermite conditions require those derivatives to exist, or to be supplied as separate interpolation data. The nodal conditions alone remain meaningful for every continuous $u$.

Use the standard meaning of a cubic spline: a $C^2$ function that is cubic on each unit interval. If it vanishes on both exterior intervals, continuity of the function and its first two derivatives gives zero endpoint jets at zero and at $k$.

Such a spline can be represented through its third-derivative jumps asFor $x\ge k$, this must be the zero polynomial. Expanding in powers of $x$ yields the four moment conditionsIf $k\le 3$, the columns of this Vandermonde system are linearly independent, so every $c_j=0$. No nonzero spline has such short support.

For $k=4$, a nonzero null vector is $(1,-4,6,-4,1)$, since the fourth finite difference annihilates every cubic. Thus the Cardinal cubic B-spline isIts knot values at $0,1,2,3,4$ are $0,1/6,2/3,1/6,0$, whose sum is one. It is nonzero and supported on $[0,4]$, proving the minimal value $k=4$. Equivalently, on $[0,1]$ it is $x^3/6$, on $[1,2]$ it is $[-3x^3+12x^2-12x+4]/6$, and the other half follows by symmetry $\varphi (4-x)=\varphi (x)$.

Scale the equal intervals to unit length. A nonzero compactly supported cubic B-spline with maximum continuity is $C^2$, so its fourth derivative is a sum of impulses at the knots. Equivalently it is a combination of truncated cubes,To vanish to the right of its support, the polynomial tail must be identically zero. Expanding that tail requires ${\sum }_j{a}_j{j}^r=0$ for $r=0,1,2,3$. Four or fewer distinct knots give a full-rank Vandermonde matrix and only the zero solution. Hence a nonzero function requires at least five knots, or four intervals. At knots $0,1,2,3,4$, the one-dimensional nullspace has coefficients proportional to $(1,-4,6,-4,1)$. The unscaled combination integrates to six on $[0,4]$, so unit-integral cardinal normalization gives:This is the Cardinal cubic B-spline, nonnegative with support $[0,4]$. To obtain its cubic B-spline Bézier extraction, use local parameter $t=x-j\in [0,1]$ on interval $[j,j+1]$. The four power-form numerators, with a common divisor six, areFor $a+bt+c{t}^2+d{t}^3$, the cubic Bernstein basis coefficients are $a$, $a+b/3$, $a+2b/3+c/3$, $a+b+c+d$. Applying this conversion givesEach scalar row describes a cubic Bézier curve graph piece with abscissa controls $j,j+1/3,j+2/3,j+1$. The truncated-cube construction proves value, first- and second-derivative agreement at all joins, including the zero exterior. Minimum support and maximum continuity determine the basis up to a scalar; the displayed normalization fixes that remaining freedom.

A norm estimate controls the worst possible refined differences for every choice of the initial controls. Submultiplicativity turns a length-$p$ contraction into a bound for arbitrary long products, so it proves a certain number of continuous derivatives. Its absolute values can lose cancellations; failure of the estimate is inconclusive. It therefore gives a lower bound on smoothness.

Conversely, an observable eigenmode of a repeated local product exhibits a particular sequence of shrinking neighborhoods with inadequate decay. General-position controls have a nonzero coefficient in that mode, so greater derivative continuity is impossible. Special symmetric or polynomial data may suppress the bad mode; this is why the general-position qualification matters. The obstruction gives an upper bound on smoothness, while failure to find an obstruction proves nothing. A lower bound on spectral growth becomes an upper bound on smoothness because large surviving differences correspond to rough behavior.

For example, the binary Cardinal cubic B-spline subdivision mask has symbol $M(z)=(1+z{)}^4/8$, up to an irrelevant index shift. At derivative order $r=2$, the first-difference subdivision mask is $B_2=(1+z)/2$, whose maximum parity sum is $1/2$: the limit is at least $C^2$. At $r=3$, $B_3=1$, and the surviving difference mode does not decay. For generic controls the third derivative is piecewise constant with nonzero jumps, so the subdivision curve is not $C^3$. Here the lower and upper tests determine the exact integer continuity, but they need not coincide for an arbitrary subdivision mask. Stability, sum rules and borderline modes must be handled explicitly before interpreting any numerical estimate as a continuity level.

For integer knots, the recurrence simplifies toOrder-two splines are triangular: $N_{i,2}(i)=N_{i,2}(i+2)=0$ and $N_{i,2}(i+1)=1$. Applying the recurrence once more givesFor example $N_{0,3}(1)=\frac{1}{2}{N}_{0,2}(1)+N_{1,2}(1)=1/2$ and $N_{0,3}(2)=N_{0,2}(2)+\frac{1}{2}{N}_{1,2}(2)=1/2$. Thus the Cardinal cubic B-spline values areThereforeIn particular its actual maximum is $2/3$ in this standard normalization; dividing by that maximum would change the requested recurrence normalization.

Use unit knot spacing and the centered Cardinal cubic B-spline convention $C(t)={\sum }_i{P}_i{M}_3(t-i)$, where $M_3$ is supported on $[-2,2]$. On the span $[i,i+1]$, with $s=t-i$, the active control points are $P_{i-1},P_i,P_{i+1},P_{i+2}$, and their B-spline weights areFor $t=11/4$, take $i=2$, $s=3/4$. The weights and their derivatives areThe PDF has $P_3=(0,0,0)$ and $P_4=(0,1,0)$, so the point and tangent vector areThus, under this centered convention,The original PDF specifies neither the knot origin nor its indexing relative to $P_i$. If instead the convention associates span $[i,i+1]$ with $P_i,P_{i+1},P_{i+2},P_{i+3}$, the identical local weights multiply $P_2,P_3,P_4,P_5$ at the requested parameter, givingThese are parameter translations of the same uniform construction, rather than contradictory evaluations with the same knots. A specified knot convention is necessary to select the numerical pair uniquely; changing the knot spacing would also rescale the derivative.

The transverse row sequence in the extrusion invariance of Loop subdivision obeysIn the convention $g_k^{\prime }={\sum }_j{a}_{k-2j}{g}_j$, the centered univariate subdivision mask isThe even coefficients sum to one and the odd coefficients sum to one; the whole binary mask sums to two. Its generating expression is $a(z)=z^{-2}(1+z{)}^4/8$, identifying the Cardinal cubic B-spline refinement rule. Its limit is $G(y)={\sum }_j{g}_j{M}_3(y-j)$ for the centered unit-grid cubic basis $M_3$. The curve across rows is this cubic B-spline; extrusion generates the whole limit surface. The along-extrusion direction itself has constant height, so its profile is a straight line rather than an additional nonconstant cubic curve.

Let the three grid-edge vectors be $v_1=a$, $v_2=b$, $v_3=b-a$, and define the constant-coefficient differential operator $D={\sum }_{r=1}^3(v_r\cdot \nabla {)}^2$. For a stencil at grid spacing $h$, its displacement vectors $\delta$ about the new vertex have vanishing first and third moments by central symmetry. All four stencil types have the same second momentFor the old-vertex stencil, each pair $\pm h{v}_r$ contributes $h^2{v}_r{v}_r^T/8$. For an edge parallel to $a$, the endpoint displacements are $\pm ha/2$ with weights $3/8$, and the opposite displacements are $\pm h(b-a/2)$ with weights $1/8$. Their moment is $h^2[a{a}^T/4+b{b}^T/4-(a{b}^T+b{a}^T)/8]$, precisely the same expression; the other edge orientations follow by symmetry.

If $f$ is a multivariate polynomial of total degree at most three, its Taylor expansion is exact through the third-order term. Thus every refined vertex samples the single polynomialThe polynomial $Df$ has degree at most one, so subsequent refinement reproduces it exactly. Starting at unit spacing and successively halving $h$, the limit therefore isThe correction lowers degree by two, leaving the leading homogeneous term unchanged. Hence every sampled polynomial of degree at most three generates a polynomial of that same degree, even though quadratics and cubics need not be reproduced identically. For the extruded cases this gives $y^2\mapsto y^2+1/3$ and $y^3\mapsto y^3+y$.

Degree four is not guaranteed. Extruded data $g_j=j^4$ reduce to the Cardinal cubic B-spline $G(y)={\sum }_j{j}^4{M}_3(y-j)$, which is piecewise cubic and hence cannot be a single polynomial of degree four. It is not a lower-degree polynomial either: at integer rows, $G(k)=[(k-1{)}^4+4k^4+(k+1{)}^4]/6=k^4+2k^2+1/3$, which cannot agree with a cubic at all integers. This counterexample excludes every guaranteed degree $d\ge 4$. ConsequentlyThis distinguishes polynomial generation from exact reproduction and again refers to all polynomials through the stated degree.

For Cardinal cubic B-splines, the values at integer knots areand all other knot values vanish. Since $x_i=t_{i+2}$,This tridiagonal matrix is strictly diagonally dominant. The standard inverse bound for such a matrix givesThe minimum denominator is $4/6-2/6=1/3$, soPart (b) then yields

For the three unit-grid edge vectors $v_1,v_2,v_3$ and a multivariate polynomial $f$ of total degree at most three, sampled regular-grid Loop subdivision generates the graph of $f+\frac{1}{12}{\sum }_{r=1}^3(v_r\cdot \nabla {)}^{2}f$. Thus affine polynomials are reproduced exactly and all polynomials through degree three retain their degree. Fourth-degree data constant along one edge direction reduce to a Cardinal cubic B-spline, whose limit need not be a polynomial.