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Order the interpolation nodes as $t_0<\cdots <t_n$, as required by their alternating nodal data. Write $w(x)={\prod }_{j=0}^n(x-t_j)$ and ${\ell }_i(x)=w(x)/[(x-t_i)w^{\prime }(t_i)]$ for the Lagrange cardinal polynomials. Differentiating Lagrange interpolation givesThe upper bound is the triangle inequality; equality follows by choosing each independent nodal value to be the sign of its weight. Thus this is the nodal derivative norm from Lagrange cardinal polynomials.

To identify the maximizing signs, put $P_i=w/(x-t_i)$, a monic polynomial of degree $n$. Since $\mathrm{sign}{w}^{\prime }(t_i)=(-1{)}^{n-i}$,For $0\le k<n$, let ${\xi }_{i,j}$ be the ordered roots of $P_i^{(k)}$, $1\le j\le n-k$. The roots of $P_i$ decrease componentwise as $i$ increases: deleting a later node replaces one retained node by an earlier one. The monotonicity of polynomial critical points in their roots, iterated $k$ times, therefore gives ${\xi }_{n,j}\le \cdots \le {\xi }_{0,j}$. Moreover, $P_n$ and $P_0$ weakly interlace, so the Markov interlacing lemma gives ${\xi }_{0,j}\le {\xi }_{n,j+1}$. Together,These root bands have disjoint interiors. At a fixed $x$, all but at most one band lie wholly on one side of $x$. In the possible remaining band, ${\xi }_{i,j}$ moves monotonically as $i$ increases. Since the leading coefficients of all $P_i^{(k)}$ are positive, the sequence $P_i^{(k)}(x)$ has at most one sign change after zero entries are ignored. The same holds for $(-1{)}^i{\ell }_i^{(k)}(x)$. If $k=n$, every $P_i^{(n)}=n!>0$, so the conclusion also holds.

The maximizing nodal signs can consequently be chosen as $\pm (-1{)}^i$ throughout, or as $\pm (-1{)}^i$ before one index $s$ and $\mp (-1{)}^i$ from that index onward. Zero weights can be assigned either sign to complete such a pattern. These are precisely the nodal values of $\pm p^{\ast }$ or $\pm q_s$, provingThe argument works for all real $x$, hence in particular for $[-1,1]$. For $k>n$ every derivative vanishes and the result is immediate.