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Use $s_j=t_{i+j}$ again and take $k\ge 2$. Set $g(s)=(s-t{)}_{+}^{k-2}$, so $(s-t)g(s)=(s-t{)}_{+}^{k-1}$. LetThe Leibniz rule for divided differences, applied to the linear factor $s-t$, readsMultiply by $s_k-s_0$ and substitute $G$ to get $(t-s_0)H+(s_k-t)K$. Converting $H$ and $K$ to the lower-order normalized B-splines proves the Cox-de Boor recurrence:It starts with $N_{i,1}=1_{[t_i,t_{i+1})}$. The coefficients are nonnegative wherever the corresponding lower-order B-spline is nonzero, so induction proves nonnegativity.

For later use, extend the finite knot sequence to a strictly increasing sequence in both directions. The full order-one sum is one. When the Cox-de Boor recurrence is summed, the two coefficients of each fixed lower-order B-spline add to one. Induction therefore gives a full partition of unity. Removing terms proves the subpartition of unity for B-splines, ${\sum }_{i=1}^n{N}_{i,k}(t)\le 1$. For repeated knots, the same statement follows by a limit of strictly increasing knot sequences, with the usual convention that a recurrence term with zero denominator is zero; the bound holds on each open knot interval and for the standard one-sided endpoint values. The terminology “$L_{\infty }$-normalized” refers to this standard basis normalization; it does not claim that each individual B-spline has supremum norm exactly one.

For the usual spline interpolation problem the sites are distinct and ordered; the Schoenberg–Whitney theorem then makes $A_x$ invertible under the stated support positivity. More generally, the following argument applies whenever the interpolation map in the question is well defined and $A_x$ is invertible.

Factor the spline interpolation operator as $P_x=B{A}_x^{-1}R$, whereSampling has operator norm at most one from the supremum norm to ${\ell }_{\infty }$. Nonnegativity and the subpartition of unity for B-splines giveConsequently $\Vert P_x\Vert \le \Vert A_x^{-1}{\Vert }_{{\ell }_{\infty }}$.

For the reverse bound, select a row $r$ of $A_x^{-1}$ with maximal absolute row sum, and choose $y_j=\mathrm{sgn}((A_x^{-1}{)}_{rj})$, assigning any value of modulus at most one when the entry is zero. Then $\Vert y{\Vert }_{{\ell }_{\infty }}=1$ andAt distinct sites there is a continuous piecewise linear function $f$ with $f(x_j)=y_j$ and $\Vert f{\Vert }_{\infty }=1$: interpolate linearly between the ordered sites and extend constantly to the endpoints. The given uniform-norm stability of a B-spline basis therefore impliesThus the B-spline interpolation operator norm satisfiesPositivity $N_i(x_i)>0$ alone, without the usual distinct-site convention, is insufficient to define an interpolation operator. For example, take order-two B-splines with knots $0,1,2,3$ and $x_1=x_2=3/2$. Both diagonal values are $1/2$, but the B-spline collocation matrix has identical rows $(1/2,1/2)$ and is singular. The displayed proof uses the well-defined interpolation map posited in the question; ordered distinct sites supply its standard existence guarantee.

Use the usual strictly increasing spline knot sequence, and initially assume $k\ge 2$. The explicit formula for the order-$k$ divided difference givesEach summand is a truncated power function in $t$, with its junction at $t_j$. It is a piecewise polynomial function of degree at most $k-1$ and has $k-2$ continuous derivatives. Finite summation preserves these properties.

For $t\ge t_{i+k}$ all summands vanish. For $t\le t_i$, all spline knot data come from the polynomial $u\mapsto (u-t{)}^{k-1}$. An order-$k$ divided difference annihilates a polynomial of smaller degree, so $M_i(t)=0$ there too. Thus the support is contained in $[t_i,t_{i+k}]$.

On the last open spline knot interval only the $j=i+k$ term survives; it is a nonzero multiple of $(t_{i+k}-t{)}^{k-1}$. On the first open spline knot interval, compare the truncated spline knot data with the untruncated polynomial data: only the $j=i$ value differs. Since the untruncated divided difference is zero, the result is a nonzero multiple of $(t_i-t{)}^{k-1}$. Consequently both outer pieces are nonzero, giving exact closed support. The Cox-de Boor recurrence used below also proves positivity throughout the open support. ThereforeThis is simple-knot B-spline regularity. For $k=1$ the B-splines are interval indicators; they are piecewise constant and need not be continuous. The notation $C^{k-2}$ should then be read as no continuity requirement, not as a classical negative-order differentiability space.

The divided difference is a finite linear combination of spline knot values, so it commutes with the finite integral. For every spline knot $t_j$ in the defining interval,It follows thatbecause the order-$k$ divided difference of a monic degree-$k$ polynomial is its leading coefficient. This proves the unit-integral normalization of a B-spline.

For the partition of unity, let $N_{i,r}$ denote the partition-normalized order-$r$ B-spline. The Cox-de Boor recurrence isIt follows by substituting the explicit divided difference formula into both sides and collecting each truncated power function. The starting case is $N_{i,1}=1_{[t_i,t_{i+1})}$, with endpoint conventions irrelevant when $r\ge 2$. The two weights are nonnegative wherever their respective lower-order B-splines are nonzero. Induction therefore gives $N_{i,r}\ge 0$ and strict positivity in its open support.

Extend the finite spline knot sequence strictly in both directions, with no finite accumulation. At each point only finitely many B-splines are nonzero. For the full extended family the sum is one for order one. Summing the Cox-de Boor recurrence, the combined coefficient of $N_{j,r-1}$ isInduction gives a full partition of unity at every order. On $t_k<t<t_{n+1}$, all extended order-$k$ functions with index $i\le 0$ or $i\ge n+1$ vanish by their supports. HenceThe same argument gives the useful global subpartition of unity for B-splines, $0\le {\sum }_{i=1}^n{N}_i(t)\le 1$, including outside the basic spline knot interval.

The orthogonal projection has range $\mathcal{S}$, not all of $C[0,1]$: the latter is an infinite-dimensional vector space whereas $\mathcal{S}$ is a finite-dimensional vector space. For a concrete counterexample, $e^x$ cannot belong to any finite-degree piecewise polynomial space, since all its derivatives are nonzero on every open interval. For $k\ge 2$, the B-splines are continuous, so the orthogonal projection is a map $C[0,1]\to \mathcal{S}\subset C[0,1]$. For order one it instead has a generally discontinuous piecewise constant range. For example, a single interval indicator on $[1/4,3/4)$ is the projection of the constant function one onto its span, and is not continuous.

Write $s^{\ast }={\sum }_j{a}_j{N}_j$. Since $M_i$ is a positive scalar multiple of $N_i$, the operator normal equations $⟨f-s^{\ast },N_i⟩=0$ are equivalent toThe mixed-normalization spline Gram matrix is invertible: $G=DH$ where $D_{ii}=k/(t_{i+k}-t_i)>0$ and $H_{ij}=⟨N_i,N_j⟩$ is the ordinary positive-definite matrix of inner products of a linearly independent basis. The positivity and unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines givesHere the operator norm of a matrix on ${\ell }^{\infty }$ is its maximum absolute row sum. Consequently,This proves the maximum-norm bound for spline projection.

We derive the Cox-de Boor recurrence from the Leibniz rule for divided differences. Fix $t$, let $u_j=t_{i+j}$, and setAway from the spline knots, $h(s)=(s-t{)}_{+}^{k-1}$. IntroduceThe defining divided-difference recurrence says $(u_k-u_0)D=R-L$. Since $s-t=(s-u_0)+(u_0-t)$ and the first factor is linear, the Leibniz rule for divided differences givesMultiplication by $u_k-u_0$ consequently yieldsReplace $L$ and $R$ by the corresponding lower-order B-splines divided by their support widths. This provesThe denominators are positive for the distinct spline knots of this question. The recursion starts with $N_{i,1}$, the interval indicator on $[t_i,t_{i+1})$. For $k\ge 2$, values at the spline knots follow by the continuous extension of the left side; the usual half-open convention for the order-one factors gives the same result. This proves the recurrence rather than assuming it as a definition.

Write $Ba={\sum }_{j=1}^n{a}_j{N}_j$, and let $Rf=(f(x_1),\dots ,f(x_n){)}^T$. We use the standard interpolation convention of distinct ordered sites $x_1<\cdots <x_n$. Together with $N_i(x_i)>0$, the Schoenberg–Whitney theorem makes the B-spline collocation matrix $A$ invertible. Equivalently, its invertibility is implicit in the existence of the interpolation operator for every data vector.

If distinctness is not understood, the printed positivity condition alone is insufficient. For the order-two basis on spline knots $1,2,3,4$, take $x_1=x_2=5/2$. Both diagonal B-spline values are $1/2>0$, but the two rows of $A$ coincide. Arbitrary data cannot then be interpolated uniquely. The norm statement below concerns the intended well-defined interpolation operator.

Spline interpolation determines the coefficient vector by $Aa=Rf$, and henceNonnegativity and the subpartition of unity for B-splines giveFor completeness, extend the finite spline knot sequence beyond both ends. For the full sequence the order-one interval indicators sum to one. Summing the Cox-de Boor recurrence and shifting the index in its second term combines the two coefficients of each lower-order B-spline to one, so induction gives partition of unity at every order. Our finite collection is a subset of that nonnegative collection, and thus has sum at most one. Repeated spline knots are handled by the standard zero-term convention or a knot limit. This argument controls the entire interval, not only the basic knot interval.

Since $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, the operator norm upper bound isHere the matrix operator norm is the maximum absolute row sum. For the lower bound, choose a row of $A^{-1}$ with maximum absolute row sum and a data vector $y$ whose components are the signs of that row's entries. Then $\Vert y{\Vert }_{{\ell }^{\infty }}=1$ andA continuous piecewise-linear function taking these values at the distinct ordered sites, and constant outside their range, has supremum norm one. Apply the given uniform-norm stability of a B-spline basis to its coefficient vector:ThereforeThe construction of the bounded continuous data extension is what permits the matrix norm to give a lower bound for a function-space operator norm.

Write $g_t(y)=(y-t{)}_{+}^{k-1}$. Begin with distinct increasing knots so ordinary polynomial interpolation is unambiguous. The two interpolants agree at the $k-1$ common knots $t_{i+1},\dots ,t_{i+k-1}$. Their difference is consequently a scalar multiple of the monic polynomial ${\omega }_i$ of degree $k-1$.

The leading coefficient of an interpolant is its highest divided difference. Thus this scalar isThe equality is precisely the divided-difference recurrence. This proves the Lee interpolation identity:For $k=1$ the common-knot product is empty; the same argument is a difference of constants. Interpret $(y-t{)}_{+}^0$ as $1_{y>t}$ to obtain the usual half-open order-one B-splines. Repeated knots use the standard confluent or limiting interpretation, where the interpolation data are defined.

Sum over $i=1,\dots ,n$. The right side telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. When $t_k<t<t_{n+1}$, the first interpolation nodes all lie below $t$, so ${\ell }_1=0$. The last interpolation nodes all lie above $t$, so ${\ell }_{n+1}$ interpolates the polynomial $(x-t{)}^{k-1}$ of degree $k-1$ and equals it identically. Hence the Marsden identity follows:

Use the standard partition normalization of B-splines, not rescaling each spline to have supremum exactly one. Let $h_i=t_{i+k}-t_i>0$. Then $M_i=k{N}_i/h_i$. In the ordinary real inner product, the mixed-normalization spline Gram matrix is $G=D^{-1}H$, where $D=\mathrm{diag}(h_i/k)$ and $H_{ij}=(N_i,N_j)$. Since the $N_i$ form a basis, $H$ is positive definite; $D$ is invertible, so $G$ is invertible. The undeclared upper index $h$ in the printed matrix should be $n$.

We need the standard positivity and normalization properties with their justification. The Cox-de Boor recurrence iswith zero-width terms interpreted as zero. It starts from nonnegative interval indicators. On each lower-order spline's support, its recurrence coefficients are nonnegative, so induction gives $N_i\ge 0$. Extend the finite knots outside their endpoints to a full knot sequence. In the full sum, coefficients of each lower-order spline add to one, so the recurrence preserves partition of unity. Our finite collection is a subset of nonnegative B-splines; therefore ${\sum }_{i=1}^n{N}_i(t)\le 1$ everywhere. The Marsden identity gives equality on the basic knot interval.

The divided-difference definition also gives support in $[t_i,t_{i+k}]$: below the left endpoint the knot-data function is a polynomial of degree $k-1$, whose order-$k$ divided difference vanishes; above the right endpoint the knot data vanish. Choose $a<t_i$ and $b>t_{i+k}$. Integrating in $t$ and then taking the divided difference in its knot variable givesbecause an order-$k$ divided difference of a monic degree-$k$ polynomial is one. The support lies inside $[0,1]$, so ${\int }_0^1{M}_i=1$.

Orthogonality of $f-s^{\ast }$ to the spline space gives $Ga=r$, with $r_i=(M_i,f)$. Since $M_i\ge 0$ and has unit integral, $|r_i|\le \Vert f{\Vert }_{\infty }$. The induced matrix maximum norm is the absolute row-sum norm, soTaking the supremum over nonzero $f$ proves the maximum-norm bound for spline projection:The range is $\mathcal{S}$, so the printed phrase “onto $C[0,1]$” must mean into $C[0,1]$, and onto $\mathcal{S}$. Continuity of the range requires continuous B-splines, for example $k\ge 2$ and internal multiplicities at most $k-1$. For order-one or discontinuous B-splines, the same bound is valid with codomain $L^{\infty }[0,1]$ rather than $C[0,1]$.

Fix $i$. If $t<t_i$, all knot values of the function $x\mapsto (x-t{)}_{+}^{k-1}$ agree with a polynomial of degree $k-1$. Its order-$k$ divided difference vanishes. If $t>t_{i+k}$, all the knot values are zero, so the same divided difference again vanishes. Thus the support is finite:The support and Gram bandwidth of B-splines also holds at repeated knots: the same conclusion follows with the standard limiting convention. If $j\ge i+k$, then $t_j\ge t_{i+k}$, so the two supports have no positive-length overlap. An endpoint intersection contributes zero to the integral defining $G_{ij}$. The case $i\ge j+k$ is identical. HenceFor strictly increasing knots, B-splines whose indices differ by $k-1$ overlap on a positive-length interval and their product is positive there, so this universal bound is sharp. In standard matrix terminology the half-bandwidth is $k-1$ and the full bandwidth is $2k-1$; the paper's integer $d$ is $k$.

Write the interpolating spline asThe interpolation equations are $A_{\mathbf{x}}c=f_{\mathbf{x}}$, where $(f_{\mathbf{x}}{)}_i=f(x_i)$, soThe B-splines are nonnegative and form a partition of unity on the spline interval. ConsequentlyTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the operator norm bound