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The centered subdivision mask $[1,2,1]/2$ retains old control points and inserts edge midpoints. Its control polygon is unchanged as a geometric curve by refinement. Its basic limit is the hat $\varphi (t)=\max (1-|t|,0)$, with support $[-1,1]$. This linear spline representation reproduces constants and linear functions, and is $C^0$ but generally not $C^1$ at its spline knots.

An order-$k$ B-spline $N_{i,k}$ is a local, nonnegative spline determined by $k+1$ consecutive entries $t_i,\dots ,t_{i+k}$ of a nondecreasing spline knot sequence. Its polynomial pieces have degree at most $k-1$, and its support is contained in $[t_i,t_{i+k}]$. The Cox-de Boor recurrence gives a constructive definition from interval indicators.^[ref]

For example, the order-two B-spline with spline knots $0,1,2$ is the triangular hatOnly nearby B-splines are nonzero at a given point. This local support makes spline interpolation matrices and spline Gram matrices sparse. On the basic knot interval, a completed B-spline basis has partition of unity; spline knot multiplicities determine the smoothness across joins.

A clamped cubic spline is a cubic spline whose endpoint first derivatives are prescribed. A clamped cubic spline interpolant additionally matches prescribed values at its spline knots. Distinct spline knots, arbitrary nodal values and arbitrary endpoint slopes determine one such interpolant. The minimum roughness property of a clamped cubic spline proves uniqueness. Unlike the natural cubic spline, this construction prescribes endpoint slopes rather than zero endpoint second derivatives.

For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.

For a nondecreasing spline knot sequence $(t_i)$, initialize the order-one B-splines by $N_{i,1}(x)=1_{[t_i,t_{i+1})}(x)$. For $k\ge 2$, the Cox-de Boor recurrence isA term with zero denominator is assigned zero. For clamped endpoint evaluation, take the appropriate one-sided continuous value rather than applying the half-open order-one convention beyond the final knot.^[ref]

Each order-$k$ B-spline is assembled from two neighboring order-$(k-1)$ B-splines. On their active spans the weights are nonnegative, which proves nonnegativity inductively and keeps the support local. For simple equally spaced spline knots and $k=2$, this produces the triangular hat. Evaluating a whole spline at one point can combine these same local affine steps through De Boor's algorithm.

A de Boor point is a control point in a B-spline representation. For a degree-two uniform segment with three active de Boor points $D_0,D_1,D_2$, the equivalent Bézier curve handles are $(D_0+D_1)/2,D_1,(D_1+D_2)/2$. The next segment shares $D_1,D_2$ and introduces one new handle. This formula assumes simple equally spaced spline knots; clamped or repeated spline knots have different endpoint conversions.

The fourth derivative is the derivative of the third derivative. It vanishes for a polynomial of polynomial degree at most three. For a cubic spline, this statement applies inside each knot interval; a jump of the third derivative can prevent a classical fourth derivative from existing at a spline knot.

A linear spline is a continuous piecewise linear function on a specified spline knot sequence. For uniform integer spline knots, the hat basis in the displayed representation interpolates the data $s(i)=c_i$. The derivative is constant on each open span and can jump at a knot. Coinciding neighboring slopes remove a particular jump, but generic data give only $C^0$ regularity.

For the distinct-knot linear B-spline hats, let $h_i=t_{i+1}-t_i>0$. The mixed-normalization spline Gram matrix has entriesOnly indices in the finite basis are retained. This matrix has a row strict diagonal dominance margin at least $1/3$, so the inverse infinity-norm bound from diagonal dominance gives $\Vert G^{-1}{\Vert }_{{\ell }^{\infty }}\le 3$. Equidistant spline knots give off-diagonal entries $1/6$. Both pieces of every hat are retained; repeated endpoint spline knots require different boundary formulas.

Let $S$ be a clamped cubic spline interpolant and let $T$ have the same values at all spline knots and the same endpoint slopes. For $T\in C^2[x_0,x_n]$, or more generally $T^{\prime }$ an absolutely continuous function with square-integrable second derivative, the exact displayed identity holds. In particular $S$ uniquely minimizes the second derivative roughness penalty $J$ among these competitors.

To prove the identity, put $E=T-S$. On each knot interval, two applications of integration by parts give $\int S^{\prime \prime }{E}^{\prime \prime }=[S^{\prime \prime }{E}^{\prime }-S^{\prime \prime \prime }E]$, since the cubic piece has zero fourth derivative. Terms containing $E$ vanish at the spline knots; terms containing $E^{\prime }$ cancel at interior spline knots because $S^{\prime \prime }$ and $E^{\prime }$ are continuous. The endpoint terms vanish because $E^{\prime }$ is zero there. Thus $\int S^{\prime \prime }{E}^{\prime \prime }=0$, and expanding $(S^{\prime \prime }+E^{\prime \prime }{)}^2$ proves the identity. Equality makes $E^{\prime \prime }=0$, so $E$ is an affine function; a prescribed endpoint value and slope force $E=0$.

If $T$ also matches the first derivative of $S$ at every spline knot, both $E$ and $E^{\prime }$ vanish at the ends of each knot interval. For a sufficiently differentiable $T$, integration by parts separately on that interval gives the stronger local identity $\int E{T}^{(4)}=\int (E^{\prime \prime }{)}^2\ge 0$.

Solutions to all three questions of the paper dated 10 June 2015. The constructive solid geometry calculations use closed half-spaces with nonpositive defining functions inside; the B-spline calculations use unit-spaced simple spline knots and explicitly oriented parameters.^[ref]

Choose a quadratic uniform B-spline segment on a unit spline knot interval. This choice is useful because adjacent pieces can later share de Boor points at simple, equally spaced spline knots. Its three active basis pieces on local $t\in [0,1]$ areThey follow from the Cox-de Boor recurrence for degree two and consecutive integer spline knots. If the three de Boor points are $D_0,D_1,D_2$, then the equivalent Bézier curve handles areOne can verify this conversion simply by substituting these expressions into the Bernstein basis formula and collecting each $D_i$. Solving givesThe desired matrix representation is thereforeThe left row is again the monomial row, the middle matrix contains the monomial coefficients of $b_0,b_1,b_2$, and the right matrix lists the de Boor points. Unlike the Bézier curve basis matrix, the constant row averages two handles. A clamped B-spline with repeated end spline knots could instead reuse $A,B,C$ directly; the simple uniform convention above specifies which representation is being used and supplies the shared handles required for the extension.

Assign one unit parameter interval to each piece. ThenThere is no ambiguity at $u=1$, since both pieces give the origin. For a single B-spline representation take degree two, the four de Boor pointsand the equally spaced spline knot vector $\tau =(-2,-1,0,1,2,3,4)$. With $N_{i,2}$ denoting the normalized degree-two B-spline on this vector, the complete valid parameter interval is $[{\tau }_2,{\tau }_4]=[0,2]$ andOn $[0,1]$ the active handles are $D_0,D_1,D_2$, and on $[1,2]$ they are $D_1,D_2,D_3$. Shared handles and a simple interior spline knot are what merge the two pieces into one uniform B-spline. The exterior spline knots support the basis; they do not extend the interval being drawn.

In particular, the requested horizontal pieces and their vertical counterparts are

Fix an explicit stationary-grid convention. For subdivision arity $m>1$, write the univariate rule asA column of the subdivision matrix is one translated subdivision mask; the next column is translated by $m$ fine-grid rows. Thus the column displacement is the arity. If $a_k$ is active from $k=\alpha$ to $k=\beta$, the subdivision mask width is $w=\beta -\alpha +1$ fine-grid positions, after removing zero padding. Equivalently the mask span is $\beta -\alpha$ fine intervals. This is different from counting old control points in a row stencil: that count measures how many controls influence one newly created point. Stating the convention avoids confusing these two widths.

For a convergent subdivision curve scheme, its support is the influence region of one original control point, equivalently the support of a function of the basic limit $\varphi$ obtained from data $p_j={\delta }_{j0}$. Translation gives the representation $P(t)={\sum }_j{p}_j\varphi (t-j)$. Descendant indices of index zero have parameter positionsThus the limiting support enclosure is $[\alpha /(m-1),\beta /(m-1)]$, as in the support of a stationary subdivision scheme. Endpoint activity gives equality for the ordinary positive convergent masks under discussion.

The functional precision set of a subdivision scheme consists of functions exactly recovered from their uniformly sampled data using the scheme's consistent parameter placement. Polynomial precision records the exactly reproduced polynomial space. Sampling-grid spacing and origin must be specified; the usual arbitrary-grid precision requirement uses every spacing and shift, rather than merely membership in the fixed-grid reconstruction space.

Center the given mask at indices $-1,0,1$. Its binary linear interpolatory subdivision rule isEach new control point is either an old point or an edge midpoint. Hence every refined control polygon is the same curve with extra vertices. Starting from scalar impulse data, the limit is explicitlyThe mask width is three positions, its mask span is two, its largest row stencil uses two old control points, and its subdivision arity is two. Translating the mask indexing to $0,1,2$ translates the basic support to $[0,2]$; the support width remains two original grid intervals.

On $[j,j+1]$, reconstruction is $(j+1-t)p_j+(t-j)p_{j+1}$. Constant and affine samples are therefore recovered exactly. A non-affine polynomial cannot agree with a linear expression on a whole open interval. Thus its polynomial functional precision set of a subdivision scheme isFor continuous functions reproduced at every translated sampling grid, the same conclusion holds: agreement with all chord interpolants forces the affine interpolation identity, hence affinity. At one fixed grid, the larger exactly representable class is the space of linear splines with those spline knots; this is not a higher polynomial precision claim.

For arbitrary data, the one-sided derivatives at $j$ are $p_j-p_{j-1}$ and $p_{j+1}-p_j$. They need not coincide. Therefore the generic limit is $C^0$, not $C^1$. Its first derivative is piecewise constant; exceptional affine data, or individual matching neighboring slopes, can be smoother. This conclusion follows directly from the limit formula, not just from a heuristic mask-width test.

A tensor-product surface basis is formed from two independent univariate basis families ${\varphi }_i(u)$ and ${\psi }_j(v)$ on separate parameter intervals. Given a rectangular control net $p_{ij}$, its parametric surface isThe parameters and indices vary independently over a product domain. The representation can be evaluated first in either parameter, giving a convenient computational advantage. Assume a finite basis, or locally finite spline bases, so differentiation and summation below are justified locally.

For positivity, if each factor basis is nonnegative, then ${\varphi }_i(u){\psi }_j(v)\ge 0$. If both factors are strictly positive at a point, their product is strictly positive there. Nonnegative weights are the usual geometric meaning of positivity; zero boundary weights are allowed. With partition of unity, this gives the local convex hull enclosure of the active control points.

For continuity, suppose the first family is $C^r$ and the second $C^s$, including matching derivatives across their respective spline knots. On every polynomial piece, and by matching at the knot lines,Every factor on the right is continuous in its own variable, so its product is jointly continuous. Thus all these mixed derivatives are continuous, in particular the surface is jointly $C^{\min (r,s)}$. The separate coordinate smoothness can be better: $r$ orders in $u$ and $s$ in $v$. For piecewise polynomials of degrees $d_u,d_v$, an interior knot of multiplicity $k$ in the first factor normally gives $C^{d_u-k}$ matching in $u$, inherited along that knot line; similarly for $v$. The statement is an inherited lower bound. Special control points can cancel derivative jumps, so a particular surface may be smoother than a generic member of its basis.

For summation to unity, independent summation factors exactly:This also explains why translations of all control points translate the surface by the same vector, as in affine equivariance of a geometric basis. These three arguments establish tensor-product inheritance of geometric basis properties directly from the factor identities.

A triangular Bézier patch is a non-tensor-product example. For barycentric parameters $u,v,w\ge 0$ with $u+v+w=1$, a quadratic patch isThe six basis functions have total degree two and sum to $(u+v+w{)}^2=1$. Their index constraint and triangular parameter domain do not arise from two independently indexed univariate factors on a rectangle. An individual triangular polynomial could be embedded in another larger representation, but this triangular basis and control net are not a tensor-product surface definition.

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 $[i,i+k]$.

A cubic spline belongs to $C^2[a,b]$ and is a polynomial of degree at most three on each interval separated by the spline knots. A natural cubic spline is additionally an affine function on the exterior intervals $[a,x_1]$ and $[x_n,b]$. On $[x_1,x_n]$ this is equivalent to the endpoint conditions $s^{\prime \prime }(x_1)=s^{\prime \prime }(x_n)=0$; these conditions remain part of the definition when an outer knot coincides with $a$ or $b$.

Let $s$ be the natural cubic spline interpolant of the prescribed values, and let $v$ be any other interpolating $C^2$ function. Set $u=v-s$, so $u(x_i)=0$. On the interval $[x_i,x_{i+1}]$, integration by parts twice givesbecause the fourth derivative of each cubic piece is zero. The second boundary term is zero because $u$ vanishes at the spline knots. After summing the first terms, the interior contributions cancel by continuity of $s^{\prime \prime }$ and $u^{\prime }$, and the two remaining terms vanish by the natural endpoint conditions. On the exterior intervals $s^{\prime \prime }=0$. Thus ${\int }_a^b{s}^{\prime \prime }{u}^{\prime \prime }=0$, and the second derivative roughness penalty satisfiesIf equality holds, the continuous nonnegative function $(u^{\prime \prime }{)}^2$ has zero integral, so $u^{\prime \prime }=0$ everywhere. Hence $u$ is an affine function, and its zeros at two distinct spline knots force $u=0$. This proves the minimum roughness property of the natural cubic spline interpolant and its equality case: the unique minimum-roughness interpolant is $s$.

For the fixed-design nonparametric regression fit, use the penalized least squares criterionLet $s_z$ be the natural cubic spline interpolant with value vector $z=(z_1,\dots ,z_n{)}^T$. Replacing $v$ by $s_z$ for $z_i=v(x_i)$ leaves the residual sum unchanged and reduces the second derivative roughness penalty, strictly unless $v=s_z$. It therefore remains to minimize the finite-dimensional quadratic formThe supplied spline roughness penalty matrix $K$ is positive semidefinite. For every nonzero vector $w$,Consequently $I+\lambda K$ is a positive-definite matrix, and $Q_{\lambda }$ is strictly convex. Its normal equations for linear least squares have the unique solutionThe preceding strict reduction also proves uniqueness over the whole class $C^2[a,b]$, not just over the natural cubic splines. This is the cubic smoothing spline. If the residual criterion is divided by $n$ instead, the corresponding formula is $(I+n\lambda K{)}^{-1}Y$, reflecting a change in the numerical convention for the smoothing parameter.

The independence and specified error mean and variance are not needed for existence or uniqueness of this minimizer. They do imply, with smoothing matrix $A_{\lambda }=(I+\lambda K{)}^{-1}$ and $g_X=(g(x_i){)}_i$, that $\mathbb{E}\hat{z}=A_{\lambda }{g}_X$ and $\mathrm{Cov}(\hat{z})={\sigma }^2{A}_{\lambda }^2$. The null space of $K$ consists of sampled affine functions: these have zero roughness, and conversely zero roughness forces their natural cubic spline interpolant to be an affine function. Thus these functions are reproduced without bias of an estimator, while other components are smoothed.

Write a degree-$p$ B-spline as $C(t)={\sum }_i{P}_i{N}_{i,p}(t)$ and locate its spline knot span. There are several equivalent exact evaluation methods.

Evaluate the at most $p+1$ nonzero basis functions by the Cox-de Boor recurrence, then form their weighted sum of control points. Alternatively apply De Boor's algorithm directly to the $p+1$ active control points. In a span $[t_k,t_{k+1})$, initialize $d_j^{(0)}=P_{k-p+j}$ and repeatedly interpolate:for $r=1,\dots ,p$ and $j=p,p-1,\dots ,r$. The last point $d_p^{(p)}$ is $C(t)$; zero knot intervals are handled by the standard limiting convention or an appropriate neighboring nonempty span.

A third method precomputes the local polynomial coefficients on each span and evaluates them by Horner's method. For a uniform cubic B-spline, a fixed four-by-four basis matrix gives these coefficients. A fourth method uses knot insertion to express the relevant span as a Bézier curve, followed by De Casteljau's algorithm. Both interpolation algorithms also split the curve at the requested parameter. Repeated subdivision supplies arbitrarily accurate polygonal evaluation, although a finite subdivision mesh is generally an approximation rather than the exact requested point. Basis summation, de Boor interpolation, local polynomial evaluation and Bézier extraction are equivalent exact representations.

Fix $t$ and let $F_t(u)=(u-t{)}_{+}^{k-1}$. Initially assume distinct spline knots, so the ordinary polynomial interpolation definitions are unambiguous. The two interpolants ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree with $F_t$ at their $k-1$ shared nodes $t_{i+1},\dots ,t_{i+k-1}$. Their difference has degree at most $k-1$ and all these roots, henceFor $k=1$ the product is empty and the same statement concerns constants. The leading coefficient of an interpolating polynomial is its highest-order divided difference, so the divided difference recurrence givesThis proves Lee's formula:Summing over $i$ telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. If $t_k<t<t_{n+1}$, the first interpolation nodes $t_1,\dots ,t_k$ are all smaller than $t$, so their truncated-power data vanish and ${\ell }_1=0$. The last nodes $t_{n+1},\dots ,t_{n+k}$ are larger than $t$, so their data are values of the polynomial $(u-t{)}^{k-1}$ and uniqueness gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the telescoped identity isthe Marsden identity, valid for every real $x$.

If repeated spline knots are allowed, ordinary value interpolation at repeated nodes does not specify ${\ell }_i$ uniquely. The standard extension uses confluent divided differences and Hermite interpolation, or limits of distinct knots: the same polynomial identity then persists wherever these quantities are defined. At a maximally repeated knot where the truncated power lacks the required derivative, the appropriate one-sided spline convention must be fixed; an unspecified classical derivative is not part of the proof. For distinct knots, the printed identity requires no such qualification.

Put $d=k-1$ and let $e_m$ be the elementary symmetric polynomial of degree $m$ in the $d$ interior spline knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expanding the knot polynomial givesCompare the coefficient of $x^{d-m}$ in the Marsden identity. The left coefficient is $(-1{)}^m(\genfrac{}{}{0ex}{}{d}{m})t^m$; the right coefficient is $(-1{)}^m{\sum }_i{e}_m(t_{i+1},\dots ,t_{i+d})N_i(t)$. Cancelling the common sign therefore gives the monomial B-spline coefficients:Equivalently, for $m\ge 1$ the numerator is ${\sum }_{1\le r_1<\cdots <r_m\le k-1}{t}_{i+r_1}\cdots t_{i+r_m}$. In particular $a_i^{(0)}=1$, so the order-zero case reproduces constants, and for $k\ge 2$ the linear coefficients are $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, the Greville abscissae. At $m=k-1$, the coefficient is the product of all interior knots. Each formula holds on the basic knot interval used in the Marsden identity.