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Fix a clamped nondecreasing spline knot sequence $t_1,\dots ,t_{n+k}$ on $[a,b]$, with $t_1=\cdots =t_k=a$ and $t_{n+1}=\cdots =t_{n+k}=b$. For continuous splines assume interior multiplicities at most $k-1$, and assume $t_i<t_{i+k}$. An order-$k$ spline has degree at most $k-1$ on each nonempty knot interval; at a knot of multiplicity $r$ it has $k-1-r$ continuous derivatives. Its dimension is $n$, and a normalized B-spline basis $N_1,\dots ,N_n$ has local support, nonnegative values and sum one on $[a,b]$.

These properties are visible in the Cox-de Boor recurrence. Starting with interval indicators at order one, setwith a term of zero denominator defined to be zero. On each contributing support the weights are nonnegative, and summing the recurrence makes the coefficients of each lower-order B-spline add to one. This gives partition of unity and $\mathrm{supp}{N}_{i,k}=[t_i,t_{i+k}]$, with endpoint values defined by limits.

At distinct ordered sites $x_1<\cdots <x_n$, interpolation is the linear systemIt has a unique solution for every data vector exactly when the B-spline collocation matrix is invertible. The Schoenberg–Whitney theorem characterizes this by $N_i(x_i)>0$ for every $i$; away from clamped endpoints this is $t_i<x_i<t_{i+k}$. At the clamped endpoints the first and last B-splines have value one, so the endpoint version uses this positive-value formulation rather than the strict inequalities. Both ordering and the support conditions matter. A spline space with interior knots is generally not a Chebyshev system on the entire interval: a locally supported B-spline vanishes on an interval. Thus arbitrary distinct nodes need not permit interpolation. Merely specifying all data sites as knots also need not fix the remaining freedom; for example, cubic interpolation with simple interior knots at $m-2$ data sites has dimension $m+2$, and two extra boundary conditions are needed to interpolate $m$ values uniquely.

The positivity governing this existence theorem is total nonnegativity of B-spline collocation matrices: every square minor with increasing row and column indices is nonnegative. In spline terminology this is often called total positivity, but it permits zero minors caused by disjoint supports. It is not the stronger assertion that every minor is strictly positive. A useful proof uses knot insertion. Each refinement step changes coefficients bywith the unchanged endpoint pieces. The refinement matrix is nonnegative bidiagonal, and its minors are nonnegative: any nonzero determinant has its allowed diagonal matching and is a product of nonnegative entries. Products preserve this property by the Cauchy–Binet formula. Insert each sampling site to full multiplicity, so evaluation at that site becomes selection of an ordered refined coefficient. The collocation matrix is then an ordered row submatrix of the refinement product, proving total nonnegativity. The strict support criterion is the strict part of this argument: an ordered path through the consecutive support intervals exists exactly when every diagonal B-spline value is positive, giving the Schoenberg–Whitney theorem determinant criterion.

If $A$ is invertible, its determinant is positive and the adjugate identity yieldsThis checkerboard inverse of a totally nonnegative matrix is the central stability fact. With alternating data $e_i=(-1{)}^i$, the absolute row sums satisfyFor general data, the cardinal splines are ${\ell }_i(x)={\sum }_j{N}_j(x)(A^{-1}{)}_{ji}$. The spline interpolation operator and its exact Lebesgue constant of interpolation areThe equality for the operator norm follows by extending the signs at distinct sites to a continuous unit-norm data function, as in Solution 2. That solution also shows the last inequality can be very strict. Stability concerns the range function as well as its coefficients.

An optimal interpolation set must be specified relative to an objective. For the normalized B-spline coefficients, every admissible sampling set satisfiessince $c(s)=A_{\mathbf{x}}^{-1}(s(x_i))$ and sampling cannot increase the supremum norm. If a unit-norm spline $s_{\ast }$ has $n$ ordered alternating extrema $s_{\ast }(x_i)=(-1{)}^i$ at admissible sites, the checkerboard identity givesTaking the largest row shows $\Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }\le \kappa (\mathcal{S})$, so equality holds. This proves the optimal spline coefficient interpolation sites property and the Chebyshev spline coefficient and dual norm equality. In particular, for the cubic space of Solution 1 the alternating extrema $-1,-2/3,0,2/3,1$ satisfy the support conditions. Its coefficients have maximum magnitude $11/2$, so these sites are optimal for coefficient recovery andThe argument proves optimality whenever this admissible equioscillating spline is available; it does not identify extrema of an arbitrary spline as optimal sites, or transfer coefficient optimality automatically to the Lebesgue constant of interpolation.

For the intrinsic interpolation objective, minimize ${\Lambda }_{\mathbf{x}}$ itself. There is a minimizer for a fixed finite-dimensional continuous spline space. Indeed the sites lie in a compact ordered simplex. On admissible sets, inverse entries and hence ${\Lambda }_{\mathbf{x}}$ vary continuously. As a collocation matrix tends to a singular one, its inverse norm diverges. The uniform-norm stability of a B-spline basis gives the complementary boundchoose sample signs attaining a largest inverse row, extend them to a continuous unit-norm function, and use $\Vert Tc{\Vert }_{\infty }\ge \Vert c{\Vert }_{\infty }/\kappa (\mathcal{S})$. Thus a minimizing sequence with bounded Lebesgue constants of interpolation cannot approach a singular site set. A convergent subsequence has an admissible limit attaining the minimum.

Fekete interpolation sites for a continuous function space offer another useful, constructive criterion: maximize the absolute evaluation determinant. A nonzero maximum exists by compactness and linear independence. Replacing one row by evaluation at $x$ multiplies the determinant by the corresponding cardinal function, so maximality gives $|{\ell }_i(x)|\le 1$, and therefore ${\Lambda }_{\mathbf{x}}\le n$. These sites are determinant-optimal, with a proven interpolation bound; one should not claim without further argument that they minimize ${\Lambda }_{\mathbf{x}}$. Greville abscissae provide inexpensive admissible sites for the usual continuous clamped spline spaces of order at least two. Admissibility alone does not give the best stability bound.

Finally, the coefficient condition number of a normalized B-spline basis is bounded by a constant depending only on order. To see the mechanism, for each coefficient choose a longest nonempty knot cell in that B-spline's support. If its length is $h$ and the support width is $W$, then $W/h\le k$. On that cell the spline is a degree-at-most-$k-1$ polynomial; repeated Markov polynomial inequalities bound its $r$th derivative by $C_k{h}^{-r}\Vert s{\Vert }_{\infty }$. In the De Boor–Fix spline coefficient functional, the accompanying derivative of the knot polynomial is bounded by $C_k{W}^r$. Each term is consequently at most $C_k(W/h{)}^r\Vert s{\Vert }_{\infty }$, yielding $\Vert c(s){\Vert }_{\infty }\le C_k\Vert s{\Vert }_{\infty }$ independently of the number and spacing of knots. This is basis stability; it does not bound $\Vert A_{\mathbf{x}}^{-1}\Vert$ at arbitrarily poor sampling sites.

For approximation, reproduction gives the Lebesgue inequality for approximation projectorsAs the maximum knot gap tends to zero with fixed order, spline quasi-interpolation gives an approximant with error at most $\omega (f,kh)$, so the infimum tends to zero. A sequence of uniformly bounded interpolation linear operators therefore converges to every continuous target. Local support and a stable choice of sites make spline interpolation effective; mesh refinement without control of the interpolation operator norm is not by itself a convergence proof.

A solid body transform is a rigid transformation: it preserves distances within the object. In three-dimensional coordinates it has the form $T(x)=Rx+b$, with translation $b\in {\mathbb{R}}^3$ and an orthogonal matrix $R$. For a proper physical rotation, the coefficient constraints areEquivalently the columns of $R$ are an orthonormal basis with positive orientation. A reflection also preserves distance but has $\det R=-1$ and is excluded when “solid-body motion” means a rotation and translation. There are three rotational and three translational degrees of freedom, not twelve independent matrix coefficients.

Computationally use the homogeneous coordinates representationThe last row is $(0,0,0,1)$; matrix multiplication composes transforms. Alternatively store a unit quaternion together with $b$, maintaining its unit norm. A transpose gives the rotational inverse: $T^{-1}(y)=R^T(y-b)$.

Write the cubic Bézier curve as $P(t)={\sum }_{i=0}^3{B}_i^3(t)p_i$, where $B_i^3(t)=\left(\genfrac{}{}{0ex}{}{3}{i}\right)(1-t{)}^{3-i}{t}^i$. The binomial theorem gives ${\sum }_i{B}_i^3(t)=1$. Transforming the controls producesThis proves pointwise equivariance, including the original parameter values; it is stronger than a statement that the two curves merely have the same shape.

Orthogonality was not used in the calculation. The identical proof applies to every affine map $T(x)=Ax+b$, including scales, shears and even singular affine maps. It applies to every representation $P(u)={\sum }_i{\varphi }_i(u)p_i$ with scalar, control-independent partition of unity basis functions: Bézier curves of every degree, B-splines, surfaces built on a tensor-product surface basis, triangular Bézier patches, and more general linear control-point representations. Positivity is not needed for affine equivariance of a geometric basis. For linear stationary subdivision matrices with row sums one, each refinement step commutes with the affine transformation; taking a convergent limit proves the corresponding assertion for subdivision curves and subdivision surfaces. Rules with geometry-dependent nonlinear weights require separate equivariance tests and are not covered automatically.

The affine class is the largest class that preserves every such Euclidean-control construction with unchanged basis weights, under the usual continuity assumption. In fact the degree-one case would forcefor all $x,y$ and $0\le t\le 1$. This segment identity, followed by continuity, forces $T$ to be affine. Nonlinear transforms such as perspective division generally fail it.

There is a further extension when the representation itself is enlarged: apply a projective transformation linearly to homogeneous controls $(w_i{p}_i,w_i)$, then dehomogenize the weighted sum. The transformed objects are rational Bézier curves or non-uniform rational B-splines, with transformed weights as well as transformed Euclidean controls. This works wherever the new denominator is nonzero. It does not assert projective invariance of ordinary polynomial Bézier curves with their old weights.

Use the standard partition normalization of B-splines, with order $k$ meaning degree $k-1$. The Cox-de Boor recurrence on arbitrary increasing spline knots isstarting with $N_{i,1}(x)=1_{[t_i,t_{i+1})}(x)$. Terms with zero denominator are interpreted as zero in the usual repeated-knot convention. For the uniform knots here, $t_i=i$, this reduces toThe index of the second lower-order B-spline is $i+1$, not $i-1$. Both lower-order supports combine to give $[i,i+k]$.

Use the standard $L_{\infty }$-oriented partition normalization of B-splines, in contrast to the unit-integral normalization. Thus $N_i\ge 0$ and ${\sum }_i{N}_i(t)\le 1$ on $[0,1]$, with equality for a complete basis on its basic knot interval. This terminology does not mean that each basis function's maximum is exactly one; for example an order-three cardinal spline has maximum $3/4$. Assume the basis has positive support on the integration interval and is continuous, as is needed for a projection taking values in $C[0,1]$.

LetThe ordinary Gram matrix $H$ is positive definite by linear independence. The appropriate row-normalized, or mixed-normalization spline Gram matrix, isHere $M_i$ is a unit-integral nonnegative spline. Unlike $H$, $G$ need not be symmetric. If the full support lies inside the interval, $d_i=(t_{i+k}-t_i)/k$; otherwise use the actual interval integral rather than that full-support formula.

The normal equations for the $L_2$ orthogonal projection $P_{\mathcal{S}}f={\sum }_i{a}_i{N}_i$ say $Ha=c$, with $c_i=\int f{N}_i$. Divide each row by $d_i$ to obtainNonnegativity and unit integral give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines then givesTaking the supremum over $t$ and then over nonzero $f$ proves the uniform norm bound for B-spline orthogonal projection:The row normalization is essential; replacing $G$ with the ordinary Gram matrix would not be the same asserted estimate. The operator is defined on $C[0,1]$ and maps into its finite-dimensional spline subspace, onto $\mathcal{S}$, not onto all of $C[0,1]$ as one phrase in the PDF says. Its range follows both from the formula and from reproducing every $s\in \mathcal{S}$.

If “$L_{\infty }$-normalized” were instead read literally as individually peak-one splines, their sum can exceed one. The basis-dependent factor $K=\Vert {\sum }_i{N}_i{\Vert }_{\infty }$ must then be retained: $\Vert P_{\mathcal{S}}{\Vert }_{\infty }\le K\Vert (D^{-1}H{)}^{-1}{\Vert }_{{\ell }^{\infty }}$. Equivalently the displayed bound without an outside factor uses $G=K^{-1}{D}^{-1}H$. The standard partition-normalized convention has $K\le 1$ and yields the sharper bound proved above.

In a convenient cardinal parameter normalization, take the unit $m$-cube and project by the sum of its coordinates. Its pushforward density is the unit-direction univariate box splineEquivalently, with $B_1=1_{[0,1)}$ at its discontinuity points,Here there are $m$ convolution factors; some conventions instead index the spline by its degree $m-1$. The half-open convention for $m=1$ assigns knot values consistently and does not change the pushforward measure.

Support and positivity. A sum of $m$ numbers in $[0,1]$ lies in $[0,m]$, so the closed support is finite. For $0<x<m$, the interval of $u\in (0,1)$ with $0<x-u<m-1$ has positive length. Induction in the convolution recurrence therefore gives $B_m(x)>0$ in the interior. For $m\ge 2$, $B_m(0)=B_m(m)=0$: strict positivity includes the open support interval, not its endpoints.

Partition of unity. The translates of $B_1$ cover the line exactly once under the half-open convention. Nonnegativity permits interchange of sum and integral, and induction givesThus $\boxed{\sum _j{B}_m(x-j)=1}$, a locally finite sum. Each $B_m$ also has integral one, although partition of unity does not mean its maximum value is one.

For $m\ge 2$, repeated integration gives the explicit truncated-power formOn each unit knot interval this is a polynomial of degree at most $m-1$, and on $(0,1)$ it is $x^{m-1}/(m-1)!$, so the degree is attained. Each truncated power has $m-2$ continuous derivatives; the derivative of order $m-1$ jumps at the first knot. Therefore the dependence on the projected box dimension isFor $m=1$ the pieces are constants and the function is discontinuous at the knots. If geometric width is measured under orthogonal projection onto the unit diagonal instead, the projected coordinate is $x/\sqrt{m}$, so the physical support width is $\sqrt{m}$; its unit-mass density is $\sqrt{m}{B}_m(\sqrt{m}x)$. These are different scalings of the same cardinal spline, not different continuity or degree claims.

For a bivariate box spline, choose a spanning direction matrix $\Xi =({\xi }_1,\dots ,{\xi }_m)$ in the plane and project the unit cube by $u\mapsto \Xi u$. Define its normalization byThe support is the zonotope ${\sum }_r[0,1]{\xi }_r$, and the polynomial pieces generally have total degree $m-2$. The grid comes from the chosen direction families and their lattice translates. Two independent directions, with repetitions, give tensor-product B-splines on a parallelogram grid. Three families such as $e_1,e_2,e_1+e_2$ give a three-direction triangular grid; an affine change of basis can make it the usual equilateral triangular grid. The four families $e_1,e_2,e_1+e_2,e_1-e_2$ give the criss-cross square grid and the quadratic four-direction box spline. Changing the direction matrix and repeating directions therefore changes the grid, support and smoothness while retaining the box-projection construction.

Write $N_{i,k}$ for the order-$k$ B-spline. At order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, and the finite sum is one on its basic knot interval. For larger order, the Cox-de Boor recurrence iswith a zero-denominator term defined as zero. Sum over $i=1,\dots ,n$. For each interior lower-order index $j=2,\dots ,n$, the combined coefficient isThe two unmatched lower-order terms have supports $[t_1,t_k]$ and $[t_{n+1},t_{n+k}]$, respectively, and vanish on the open basic interval $(t_k,t_{n+1})$. Induction therefore givesFor simple knots and $k\ge 2$, continuity extends the identity to both endpoints. In the order-one case and at admissible repeated endpoint knots, use the one-sided value from the basic interval. Extending the knot sequence in both directions also proves the subpartition of unity for B-splines, $0\le {\sum }_i{N}_i\le 1$ everywhere: the finite sum is a subset of a nonnegative full partition. That weaker global inequality will be needed in Question 7(a).

For distinct ordered knots, put $h_i=t_{i+1}-t_i>0$. The linear B-spline $N_i$ is the triangular hat supported on $[t_i,t_{i+2}]$, equal to $(t-t_i)/h_i$ on its rising segment and $(t_{i+2}-t)/h_{i+1}$ on its falling segment. Its integral-normalized partner is $M_i=2N_i/(h_i+h_{i+1})$. ThusFor example, the left overlap is ${\int }_0^{h_i}(u/h_i)(1-u/h_i)du=h_i/6$. Nonadjacent hats have disjoint interiors of their supports, so their inner product is zero. The linear-spline mixed Gram matrix therefore has rowRetain only column indices in $1,\dots ,n$: the first or last row has no neighbor outside this range. Admissible repeated endpoint knots are obtained by a limit with zero spacings, provided $h_i+h_{i+1}>0$; a degenerate hat with zero support width is not a basis element.

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 degree-$p$ B-spline basis keeps the positive partition of unity and convex hull properties while introducing local support. At a generic parameter value, only $p+1$ consecutive basis functions are nonzero. Moving one control point therefore changes only its support interval, unlike a generic interior Bézier curve control, which affects the whole polynomial segment.

Many low-degree pieces can represent a long curve with automatic continuity. At a knot of multiplicity $m$, the generic continuity is $C^{p-m}$; simple cubic knots give $C^2$. Knot positions and multiplicities control the placement of detail and joins. Knot insertion can refine the representation without changing the curve, and local refinement avoids raising the global polynomial degree. Thus B-splines combine geometric bounds, local editing and controlled smoothness with a fixed low degree. Clamped end knots can also give endpoint interpolation.

Let $N_{ij}(u,v)=B_i(u)C_j(v)$. Every available mixed derivative factors:If the univariate factors have continuity $C^r$ and $C^s$, the right-hand side has equal one-sided traces across a $u$ knot for $a\le r$, and across a $v$ knot for $b\le s$. Their products therefore remain continuous, including where knot lines cross. The basis and its surface inherit coordinate-wise continuity orders $r,s$, and joint $C^{\min (r,s)}$ continuity. In particular, degree-$p$ and degree-$q$ B-splines at simple knots give $C^{p-1}$ and $C^{q-1}$ continuity in the respective directions. Higher continuity can occur through special coefficient cancellation, but is not generally guaranteed.

Nonnegativity passes to products. Their partition of unity follows fromThus the surface remains in the convex hull of its active control points and commutes with application of affine functions. The support is the Cartesian product of the two univariate supports, giving local rectangular influence regions and efficient local evaluation. Linear precision also passes independently in each coordinate. Clamped endpoint interpolation gives boundary curves determined by the outer rows/columns and corner interpolation; parameter-line curves inherit the corresponding univariate properties. These are the relevant inherited properties, rather than a claim of arbitrary global variation-diminishing behavior for surfaces.

Let $y=(f(x_1),\dots ,f(x_n){)}^T$ and write the interpolating spline as $P_{x}f={\sum }_{j=1}^n{c}_j{N}_j$. The interpolation equations are $A_{x}c=y$, so $c=A_x^{-1}y$. Since $P_x$ is defined as a unique interpolant for arbitrary data, its B-spline collocation matrix is invertible. Ordered sites satisfying diagonal positivity ensure this by question 7. Positivity alone without admissible ordering does not: two equal sites in overlapping basis supports could give two identical rows. The norm argument below uses the existence of the operator stipulated here.

Standard partition-normalized B-splines satisfy $N_j\ge 0$ and ${\sum }_{j=1}^n{N}_j\le 1$. The latter is a subpartition of unity for B-splines: extending the knot sequence gives a full partition of unity, and discarding basis functions cannot increase its sum. HenceTaking the supremum over $t$ and then over unit-norm $f$ gives the B-spline interpolation operator norm estimateHere the matrix norm is the induced maximum-row-sum norm, not the maximum of its individual entries. The proof only needs nonnegative basis functions, their sum at most one, and invertible interpolation.

First take $k\ge 2$, so the simple-knot B-splines are continuous, vanish at both support endpoints, and are strictly positive inside their supports. Thus the diagonal condition is $t_i<x_i<t_{i+k}$. We prove both directions of the Schoenberg–Whitney theorem, deriving the weighted zero-count fact needed for sufficiency rather than relying on an ambiguous coefficient-free hint.

For necessity, suppose $N_i(x_i)=0$. If $x_i\le t_i$, then for every row $r\le i$ and column $j\ge i$ we have $x_r\le x_i\le t_i\le t_j$, hence $N_j(x_r)=0$. The first $i$ rows have nonzero entries in at most $i-1$ columns and are linearly dependent. If instead $x_i\ge t_{i+k}$, then for $r\ge i$ and $j\le i$ we have $x_r\ge x_i\ge t_{i+k}\ge t_{j+k}$, so those entries vanish. The last $n-i+1$ rows have nonzero entries in at most $n-i$ columns and are again dependent. In either case the B-spline collocation matrix is singular. Thus invertibility implies the diagonal positivity condition.

We establish two elementary spline facts for the converse. For local linear independence of B-splines, extend the strictly increasing knots in both directions. On a cell $(t_j,t_{j+1})$ the $k$ active basis functions have indices $j-k+1,\dots ,j$. Their knot polynomials are ${\psi }_i(y)={\prod }_{\ell =1}^{k-1}(y-t_{i+\ell })$. The Cox-de Boor recurrence proves Marsden's identityHere is the induction behind that identity. For order one the indicator functions sum to one. On passing from order $k-1$ to $k$, collect the coefficient of each lower-order basis function. With $u=t_i$, $v=t_{i+k-1}$, its extra factor isThus the identity at one order multiplies by $y-x$ at the next.

The $k$ polynomials ${\psi }_i$ are independent: evaluate successively at $y=t_j,t_{j-1},\dots ,t_{j-k+1}$. The first evaluation isolates ${\psi }_j$; after its coefficient has been eliminated the next isolates ${\psi }_{j-1}$, and so on, with nonzero diagonal factors because the knots are distinct. Comparing powers of $y$ in the identity then shows that the active $N_i$ span all polynomials of degree at most $k-1$ on the cell. There are $k$ of them, so they are independent. In particular, a spline vanishes on a whole open cell exactly when all its active coefficients vanish. Padding the finite coefficient vector with zeros makes this fact apply also near the outer knots.

Next let $s={\sum }_{i=p}^q{c}_i{N}_i$ be a real spline on $I=(t_p,t_{q+k})$, with no identically zero knot cell there. The simple-knot B-spline regularity gives $s\in C^{k-2}$, and its derivatives through order $k-2$ vanish at both endpoints because it is supported on the closed interval. Suppose it has $Z$ distinct interior zeros. Its zero set initially has $Z+2$ connected components, counting the endpoints; the zeros are isolated because no cell vanishes identically.

A useful form of Rolle's theorem counts zero components, including any zero intervals that arise after differentiation. Between two successive zero components a real differentiable function is nonzero and has an interior extremum, supplying a zero of its derivative. These derivative-zero components are distinct: a zero interval connecting two such extrema would make the original function constant across an intervening zero and nonzero gap, which is impossible. If its derivative also vanishes at both endpoints, those supply two further distinct components. Thus differentiation increases the zero-component count by at least one. Apply this successively through derivative order $k-2$; every required endpoint derivative vanishes. No derivative in this process can vanish identically: that would make $s$ a global polynomial of degree below $k-1$, with all its endpoint derivatives through that degree zero, forcing $s=0$. The final derivative has at least $Z+k$ zero components.

That final derivative is continuous and piecewise affine on $q-p+k$ knot cells. Each closed affine cell can meet at most one zero component: two different zeros would force the whole cell to vanish, joining them into one component. Every zero component meets a cell, so there are at most $q-p+k$ of them. For $k=2$ this argument applies directly without differentiation. We have proved the compact-support spline zero countThis proof also accommodates zero intervals of the derivatives; counting them as infinitely many distinct zeros would invalidate a naive repeated-Rolle argument.

Now assume $N_i(x_i)>0$ for all $i$ and suppose $A_{x}c=0$ for a nonzero coefficient vector. Since $A_x$ is real, a nonzero real null vector can be chosen. Set $s={\sum }_i{c}_i{N}_i$, so $s(x_i)=0$ for all $i$. Consider one connected component of the union of the open supports of the terms with $c_i\ne 0$. If its first and last indices are $p,q$, the component is exactly $I=(t_p,t_{q+k})$. On this interval $s$ equals $s_0={\sum }_{i=p}^q{c}_i{N}_i$. Each knot cell in $I$ has at least one active nonzero coefficient, so local independence ensures $s_0$ does not vanish on any whole cell.

For every $p\le i\le q$, diagonal positivity gives $t_i<x_i<t_{i+k}$, and this places $x_i$ inside $I$. Thus $s_0$ has the $q-p+1$ distinct zeros $x_p,\dots ,x_q$, contradicting the upper bound $q-p$. There is no nonzero null vector, proving sufficiency. Together the two directions giveFor completeness, order $k=1$ consists of disjoint interval indicators, with the usual left-closed, right-open convention. An invertible ordered collocation requires exactly one site in each interval; ordering makes this equivalent to $N_i(x_i)>0$, and the matrix is then the identity. This handles possible sites at left knots without incorrectly replacing positivity by a strict interior inequality for order one.

For order $k\ge 2$, set $g(u)=(u-t{)}_{+}^{k-2}$. Then $(u-t)g(u)=(u-t{)}_{+}^{k-1}$. Apply the given divided difference product identity with nodes $t_i,\dots ,t_{i+k}$ and multiply by the length $t_{i+k}-t_i$ of their knot interval:Using the definition of each lower-order normalized B-spline, the two divided differences are $N_{i,k-1}(t)/(t_{i+k-1}-t_i)$ and $N_{i+1,k-1}(t)/(t_{i+k}-t_{i+1})$. Hence the Cox-de Boor recurrence isFor completeness, the product identity itself follows fromand the ordinary divided difference recurrence. Substitution yields the stated convex combination of the two order-$k-1$ divided differences.

The initial order-one splines are the indicators $N_{i,1}(t)=1_{[t_i,t_{i+1})}(t)$. For repeated knots, a term with zero knot-span denominator is defined as zero; this agrees with the standard limiting basis convention. The recurrence uses the partition of unity normalization with $0\le N_{i,k}\le 1$, not a separate rescaling of each spline to make its actual supremum exactly one. The original PDF recurrence has the denominators above; several of these indices and signs were corrupted in the converted TeX.

The factorization of the binary subdivision mask identifies eight box spline directions. Apart from the monomial that centers the support, it isThe four normalized factors correspond respectively to averaging in the directions $(1,0),(0,1),(1,1),(1,-1)$, each with multiplicity two. The factor four is the determinant of the binary dilation in two dimensions. The centering monomial translates the resulting box spline and changes neither its polynomial degree nor its smoothness. Thus this is the twice-smoothed four-direction box spline, rather than a tensor-product bicubic B-spline.

Choose the knot origin so the span most influenced by $P_i$ is $[i,i+1]$, and use its local parameter $t\in [0,1]$. The three active uniform quadratic B-spline basis functions giveThe central coefficient reaches its maximum at $t=1/2$, so this is indeed the span centered on the strongest influence of $P_i$. Express it in the quadratic Bernstein basis:Matching constant, linear and quadratic coefficients gives the quadratic Bézier curve controlsFor example, $C_i^{\prime }(0)=2(B_1-B_0)=P_i-P_{i-1}$ and $C_i^{\prime }(1)=2(B_2-B_1)=P_{i+1}-P_i$, checking the endpoint derivatives and the $C^1$ joins of neighboring spans.

Use the quartic treble-knot representation of a quadratic B-spline. Specify the indexing as well as the controls: for the interior of the long curve, take the quartic spline knot sequenceWith the standard B-spline convention that control $Q_j$ multiplies $N_{j,4}$ supported on $[{\tau }_j,{\tau }_{j+5}]$, the answer isIn other words, keep the three interior quartic Bézier curve controls $E_{1,i},E_{2,i},E_{3,i}$ for each span in that order. The offset $3i-1$ matters for the stated knot indexing; shifting all indices together gives an equivalent convention.

To prove the representation, insert each integer interior knot once, changing multiplicity three to four. A degree-four knot of multiplicity four separates Bézier curve spans while leaving their shared endpoint common. The three interior controls $E_{1,i},E_{2,i},E_{3,i}$ remain, and knot insertion constructs the shared endpoint from its neighboring controls with weights $1/2,1/2$, since adjacent knot intervals have equal length. In fact,The right endpoint is obtained in the same way. Hence every extracted quartic span has exactly the five controls proved in part (b). Knot insertion leaves the curve unchanged, so the assembled quartic B-spline is identical to the original quadratic one on every span.

There is also a derivative check at each join:This is why treble knots are appropriate: degree four with interior knot multiplicity three permits $C^{4-3}=C^1$ continuity, exactly the generic continuity of the uniform quadratic B-spline. For a finite clamped curve, include its endpoint controls and endpoint knot multiplicities using the same degree elevation of Bernstein coefficients and knot insertion; the displayed formulas describe the long interior requested here, without imposing an unstated end condition.