 Solution

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.