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For an order-$k$ B-spline with $k\ge 2$, its Greville abscissa is the arithmetic mean of its $k-1$ interior knots. These are precisely the coefficients reproducing the linear function in the monomial B-spline coefficients formula. The irregular plural is Greville abscissae.

The Korovkin theorem says that a sequence of positive linear operators on continuous functions on $[0,1]$ converges uniformly to the identity on every continuous function if it does so on $1$, $x$ and $x^2$. Necessity is immediate; the quadratic barrier argument establishing sufficiency is given in Solution 6.

There is an endpoint misprint in the original PDF: the repeated knots at the right end must equal $1$, not $0$. Otherwise the advertised nondecreasing spline knot sequence on $[0,1]$ does not exist. Use the corrected clamped spline knot sequences, with fixed order $k$ and maximum gap $h_n\to 0$. Let $m_n$ denote the number of B-splines for the $n$th sequence. Nonnegativity and partition of unity show that the Schoenberg spline operator $V_n$ is positive and $V_{n}1=1$.

The monomial B-spline coefficients reproduce the first two nonconstant test polynomials. PutThe pair sum uses $r<s$, as in the PDF; the converted TeX incorrectly includes $r=s$. Every sampling point ${\tau }_j$ and every knot entering these coefficients lie in $[t_j,t_{j+k}]$, whose length is at most $k{h}_n$. Thus $|{\tau }_j-a_{1,j}|\le k{h}_n$. Also all these numbers belong to $[0,1]$, soThe positive B-spline weights giveTogether with $V_{n}1=1$, the three limits proveUnder the usual interior knot multiplicities at most $k-1$, the splines are continuous and the quoted form of the Korovkin theorem applies directly. If full interior multiplicity is allowed, the output can be discontinuous. The same positive-operator quadratic barrier proof still works with bounded output functions and the supremum norm, and proves exactly the stated uniform error limit without asserting continuity of each output. Indeed the stronger local-support error bound for a Schoenberg spline operator, $\Vert V_{n}g-g{\Vert }_{\infty }\le \omega (g,k{h}_n)$, follows by comparing $g(x)$ with each active sample value. The restriction $k\ge 3$ is needed for the displayed quadratic reproduction argument, not for this direct support estimate.

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.

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.

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.