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For distinct increasing knots, let ${\ell }_i(\cdot ,t)$ interpolate $(\cdot -t{)}_{+}^{k-1}$ on $t_i,\dots ,t_{i+k-1}$ and let ${\omega }_i(x)={\prod }_{r=1}^{k-1}(x-t_{i+r})$. Adjacent interpolants share $k-1$ values; their difference is a multiple of ${\omega }_i$. Its leading coefficient, by the divided difference recurrence, is the order-$k$ B-spline $N_i(t)$. Thus ${\ell }_{i+1}(x,t)-{\ell }_i(x,t)={\omega }_i(x)N_i(t)$. Telescoping proves the Marsden identity. Repeated knots require the well-defined confluent or limiting convention.

The Marsden dual functionals extract coefficients in a B-spline expansion. If ${\psi }_i$ is the knot polynomial in the Marsden identity, then on polynomials of degree at most $k-1$ one such functional isand the right-hand side is independent of $x$.

For order $k$ B-splines, comparison of coefficients in the Marsden identity giveson the basic knot interval. Here $e_m$ is the elementary symmetric polynomial, with $e_0=1$. The $m=0$ case is partition of unity; the $m=1$ coefficients are the Greville abscissae.

The ordinary interpolation formulation is unambiguous for strictly increasing knots; repeated knots require specified confluent interpolation or a limiting convention. First take distinct knots and $k\ge 2$, and for fixed $t$ write $F_t(u)=(u-t{)}_{+}^{k-1}$. The two degree-at-most-$k-1$ Lagrange interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at the $k-1$ shared knots $t_{i+1},\dots ,t_{i+k-1}$. Their difference is thereforeThis remains valid if their difference has lower degree: the coefficient is then zero. The leading coefficient of a degree-at-most-$k-1$ interpolant is its order-$(k-1)$ divided difference. Comparing leading coefficients and using the divided-difference recursion givesConsequently the Lee interpolation identity holds for every real $x,t$:No differentiability of the truncated power in its argument was needed for this distinct-node proof.

Sum the identity over $i=1,\dots ,n$. The right side telescopes, giving ${\sum }_i{\omega }_i(x)N_i(t)={\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. When $t_k<t<t_{n+1}$, all the nodes for ${\ell }_1$ lie below $t$, so its interpolation data are zero and ${\ell }_1\equiv 0$. All the nodes for ${\ell }_{n+1}$ lie above $t$, where $F_t(u)=(u-t{)}^{k-1}$ is already a polynomial of the admissible degree. Uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Therefore the Marsden identity isFor order one the empty product is one, the interpolants are constants and a consistent half-open interval convention for $(u-t{)}_{+}^0$ gives the same telescoping proof. Nondecreasing repeated-knot sequences can be treated by the appropriate confluent or limiting convention wherever that definition is valid; merely specifying repeated data values does not uniquely determine the interpolation polynomials. The basic open knot interval in the Marsden identity avoids endpoint convention ambiguity.

For fixed $t$, put $g_t(u)=(u-t{)}_{+}^{k-1}$, a truncated power function. Begin with distinct increasing knots. The two interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at their $k-1$ shared knots. Their difference has degree at most $k-1$, soFor $k=1$ the empty product is one and the same assertion holds for constant interpolants. The leading coefficient of a degree-at-most-$k-1$ interpolant on $k$ nodes is its order-$k-1$ divided difference, by the Newton interpolation polynomial. ThereforeThis proves the Lee interpolation identityThe identity holds for every real $x,t$, with the usual convention for the order-zero truncated power. Admissible repeated knots are handled by the corresponding confluent divided differences and consistent one-sided limits at breakpoints; taking these limits preserves the Lee interpolation identity.

Sum over $i$ to telescope:If $t_k<t<t_{n+1}$, the first $k$ knot values of $g_t$ are zero, so ${\ell }_1=0$. At the last $k$ knots $t_{n+1},\dots ,t_{n+k}$, the function $g_t$ agrees with the polynomial $(u-t{)}^{k-1}$, so uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the Marsden identity is

Let $d=k-1$ and let $e_m$ denote the elementary symmetric polynomial of degree $m$ in the $d$ interior knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expand the two sides of the Marsden identity as polynomials in $x$:andComparing the coefficient of $x^{d-m}$ and cancelling the common sign gives the monomial B-spline coefficientsEquivalently,In particular $a_i^{(0)}=1$, giving partition of unity on the interior interval. For $k\ge 2$, $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, and the highest-degree coefficient is the product of those interior knots. The formulas also cover $k=1,m=0$ through empty products.

The Korovkin theorem states that if positive linear operators on continuous functions $L_n:C[0,1]\to C[0,1]$ satisfythen $\Vert L_{n}f-f{\Vert }_{\infty }\to 0$ for every real continuous function $f$.

Let $r=k-1\ge 2$ and $h_n=|{\Delta }_n|$. For the B-splines in this question, $N_{i,n}\ge 0$ and their sum is one on the basic interval. Thus the Schoenberg spline operator $V_n$ is linear and positive. To verify the three test functions, suppress the index $n$ and write the Marsden identity with ${\omega }_i(x)={\prod }_{j=1}^r(x-t_{i+j})$. Comparing the coefficients of $x^r,x^{r-1},x^{r-2}$ gives the monomial B-spline coefficientswhereThe first coefficient is the partition of unity; the second is the Greville abscissa. Every internal knot and the sample point ${\tau }_i$ lie in $[t_i,t_{i+k}]$, of width at most $k{h}_n$. Consequently $|{\tau }_i-{\xi }_i|\le k{h}_n$. Since all these numbers lie in $[0,1]$, for any two internal knots $u,v$ we also haveAveraging gives $|{\tau }_i^2-q_i|\le 2k{h}_n$. Nonnegative weights of sum one now yieldHere the spline order $k$ is fixed, so all three errors tend to zero. The Korovkin theorem therefore provesNo special selection such as the Greville abscissae is required for the sample points. In fact the local-support error bound for a Schoenberg spline operator also gives $\Vert V_{n}f-f{\Vert }_{\infty }\le \omega (f,k{h}_n)$. Under the usual continuous-spline convention the operators have the codomain required by the stated theorem. If the permissive knot notation is taken to allow full interior multiplicities and hence discontinuous splines, the same conclusion in the uniform norm still follows from this direct bound, although each approximant need not then be continuous.

Set $r=k-1$, with $k\ge 2$ for the final linear-reproduction assertion, and use the normalized knot polynomial ${\psi }_i={\omega }_i/r!$. Differentiate the Marsden identity $r-j$ times with respect to the auxiliary variable $x$, for $0\le j\le r$. This givesThe Taylor polynomial expansion of an arbitrary polynomial $p$ of degree at most $r$ about $x$ is exact:Substitute the differentiated identity into each term and collect the B-spline coefficients. The result isThis is the normalized Marsden dual functional. There is no extra factorial outside the sum, because it is already included in ${\psi }_i$.

To prove independence of $x$ directly, differentiate the finite sum. The contribution from differentiating ${\psi }_i$ in term $j$ cancels the contribution from differentiating $p$ in term $j-1$. Explicitly,The two uncancelled endpoint terms vanish because $p^{(r+1)}={\psi }_i^{(r+1)}=0$. Thus the coefficient functional is constant in the auxiliary variable, as required.

The leading two terms of the knot polynomial giveFor a linear polynomial $p(t)=A+Bt$, only $j=0,1$ survive, whenceThe coefficient points $t_i^{\ast }$ are the Greville abscissae. Inserting these coefficients proves exact linear reproduction:All identities hold on the basic knot interval $[t_k,t_{n+1}]$, including the endpoints with the standard continuous endpoint convention. Repeated knots cause no difficulty in these polynomial differentiations.

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.

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:

Set $d=k-1$ and let $e_m(t_{i+1},\dots ,t_{i+d})$ be the elementary symmetric polynomial of degree $m$ in the interior knots, with $e_0=1$. Expand the Marsden identity in powers of $x$. The coefficient of $x^{d-m}$ on the left is $(-1{)}^m(\genfrac{}{}{0ex}{}{d}{m})t^m$; that coefficient in ${\omega }_i(x)$ is $(-1{)}^m{e}_m(t_{i+1},\dots ,t_{i+d})$. Comparing and cancelling the sign proves the monomial B-spline coefficients:For $m=0$ this gives partition of unity on the basic interval. For $m=1$ it gives the Greville abscissae, the arithmetic means of the $k-1$ interior knots. For the top degree it gives their product. When $k=1$, only $m=0$ occurs and the empty symmetric polynomial is one.

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 $C[0,1]$.

Put $m=k-1$, so the Marsden identity isDifferentiating $m-j$ times with respect to $x$ givesThe exact Taylor formula for a polynomial isSubstitution of the differentiated Marsden identities yieldswhere

Differentiating this expression for ${\lambda }_i$ produces two sums whose adjacent terms cancel. The uncancelled endpoints contain ${\psi }_i^{(k)}$ and $p^{(k)}$, both zero because the two functions have degree at most $k-1$. Thusso the Marsden dual functional ${\lambda }_i(p)$ is independent of the auxiliary point $x$.

Let $e_m$ denote the $m$th elementary symmetric polynomial of $t_{i+1},\dots ,t_{i+k-1}$. Expanding both sides of the Marsden identity in powers of $x$ and equating the coefficient of $x^{k-1-m}$ givesThus $a_{0,i}=1$, while $a_{1,i}$ and $a_{2,i}$ are respectively the means of the interior knots and of their pairwise products.