 Discrete minimax interpolation on ordered nodes

For $n+2$ increasingly ordered nodes $y_j$, let $r_j$ be their Lagrange cardinal polynomials, $t=\sum f(y_j)r_j$, and $r=\sum (-1{)}^j{r}_j$. The leading coefficient of $r_j$ has sign $(-1{)}^{n+2-j}$, so all summands in the leading coefficient of $r$ have the same nonzero sign. There is a unique $\lambda$ canceling degree $n+1$ in $t-\lambda r$. Its errors at the nodes are $\lambda (-1{)}^j$, and a better degree-at-most-$n$ approximation would have a difference polynomial with at least $n+1$ roots. Thus it minimizes the maximum node error. Increasing order matters: for the permutation $(0,1/3,1,2/3)$ the alternating interpolant is already the quadratic $-1+9x(1-x)$, so the leading-coefficient assertion for $n=2$ fails.