Chebyshev cardinal interpolation reproduces polynomials and its norm is the cardinal sum

Put $z_i=x_{n-i}$, so $z_0<\cdots <z_n$ by [F1]. Under this reverse indexing, the supplier cardinal function at index $n-j$ is ${\mathrm{\ell }}_j$ by the specialization in [F1]. Applying [F3] gives its degree and cardinal values, including the one-node empty-product case $n=0$.

Put $L(x)={\sum }_j\mathrm{\mid }{\mathrm{\ell }}_{j,n}^{\sigma }(x)\mathrm{\mid }$. For every $g$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, the defining sum gives $\mathrm{\mid }{I}_n^{\sigma }g(x)\mathrm{\mid }\le L(x)$, hence ${\mathrm{\Lambda }}_n^{\sigma }\le {\sup }_{x}L(x)$.

Prescribe data $y_i=1$ at $i=n-j$ and $0$ otherwise on the increasingly ordered nodes $z_i$ from step 1.1. The polynomial ${\mathrm{\ell }}_j$ has these data by step 1.1. The supplier uniqueness theorem [F2] makes it the unique degree-at-most-$n$ polynomial with the required cardinal values.

Prescribe $y_i=p(z_i)$ on the ordered nodes in step 1.1. By [F1] and [F3], their cardinal sum is $I_n^{\sigma }p$ and has degree at most $n$. The polynomial $p$ itself interpolates those data, so the supplier uniqueness theorem [F2] gives $I_n^{\sigma }p=p$.

Fix $x$. Define $\mathrm{sgn}(t)=1$ for $t>0$, $-1$ for $t<0$, and $0$ for $t=0$, and prescribe $\mathrm{sgn}({\mathrm{\ell }}_{j,n}^{\sigma }(x))$ at each node. These values lie in $[-1,1]$ and satisfy $\mathrm{sgn}(t)t=\mathrm{\mid }t\mathrm{\mid }$ by cases. Sort nodes increasingly, interpolate these values linearly, and extend constantly to the interval endpoints; each interpolated value is a convex combination of endpoint values, so the resulting continuous $g_x$ has norm at most $1$. The cardinal sum gives $I_n^{\sigma }{g}_x(x)=L(x)$, so ${\mathrm{\Lambda }}_n^{\sigma }\ge \mathrm{\parallel }{I}_n^{\sigma }{g}_x{\mathrm{\parallel }}_{\mathrm{\infty }}\ge L(x)$. Taking the supremum in $x$ and using step 1.2 proves equality. For $n=0$ the same construction is constant on the interval.