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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

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: