Chebyshev–Lobatto interpolation has an algebraic weighted spectral-Sobolev rate

Let $w_n=(1+n^2{)}^s$. By [F1] the weighted coefficient sum ${\sum }_{n\ge 0}{w}_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2$ is finite. For $n\ge 1$, monotonicity of $x^{-s}$ from [F1] gives $w_n^{-1}\le n^{-2s}$, so [F6] implies ${\sum }_{n\ge 0}{w}_n^{-1}<\mathrm{\infty }$. For every finite $Q$, apply [F7] to $\sqrt{w_n}\mathrm{\mid }{c}_n\mathrm{\mid }$ and $w_n^{-1\mathrm{/}2}$, $0\le n\le Q$; this bounds ${\sum }_{n=0}^Q\mathrm{\mid }{c}_n\mathrm{\mid }$ uniformly in $Q$. Nonnegative partial sums therefore give ${\sum }_{n\ge 0}\mathrm{\mid }{c}_n\mathrm{\mid }<\mathrm{\infty }$ by A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum. By [F5], $\mathrm{\mid }{T}_n(x)\mathrm{\mid }\le 1$ on $[-1,1]$, so the series converges uniformly and its nodal values are defined.

Write $u=S_N+R_N$, where $S_N={\sum }_{n=0}^N{c}_n{T}_n$ and $R_N={\sum }_{n>N}{c}_n{T}_n$. The Lagrange operator [F3] reproduces $S_N$ and is continuous in the uniform norm; therefore $u-I_N^{L}u=R_N-I_N^L{R}_N$. By [F2], uniform convergence and the finite cardinal sum give $I_N^L{R}_N={\sum }_{r=0}^N{b}_r{T}_r$, where $b_r={\sum }_{n>N:{\rho }_N(n)=r}{c}_n$.

Fix $r\in \{0,\dots ,N\}$. Since $n\equiv r$ or $-r(\mathrm{m}\mathrm{o}\mathrm{d}2N)$, every index in its alias class has form $2qN+r$ or $2qN-r$. When $n>N$ the corresponding integer $q$ is positive. For each $q$ there are at most two candidates; the plus candidate is at least $qN$, and the minus candidate is at least $(2q-1)N\ge qN$. Thus $1+n^2\ge (qN{)}^2$; since $x^{-s}$ is nonincreasing for $s>0$, and by [F8], $(1+n^2{)}^{-s}\le (qN{)}^{-2s}=N^{-2s}{q}^{-2s}$. For every integer $Q\ge 2$, monotonicity gives $q^{-2s}\le {\int }_{q-1}^q{x}^{-2s}dx$ for $2\le q\le Q$, so [F9] yields ${\sum }_{q=1}^Q{q}^{-2s}\le 1+{\int }_1^Q{x}^{-2s}dx=1+\frac{1-Q^{1-2s}}{2s-1}\le 1+\frac{1}{2s-1}$. By nonnegativity and [F9], the infinite sum has the same bound. Therefore ${\sum }_{n>N:{\rho }_N(n)=r}(1+n^2{)}^{-s}\le 2(1+\frac{1}{2s-1})N^{-2s}$.

Let $A_s=1+1\mathrm{/}(2s-1)$. Apply [F7] to finite partial sums over an alias class, use step 3.1, and pass to the limit; the coefficient sums converge absolutely by step 1.1. This gives $\mathrm{\mid }{b}_r{\mathrm{\mid }}^2\le 2A_s{N}^{-2s}{\sum }_{n>N:{\rho }_N(n)=r}(1+n^2{)}^s\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{.}$ The alias classes partition $n>N$. By [F4], $\mathrm{\parallel }{I}_N^L{R}_N{\mathrm{\parallel }}_C^2\le \pi {\sum }_{r=0}^N\mathrm{\mid }{b}_r{\mathrm{\mid }}^2\le 2\pi A_s{N}^{-2s}{\sum }_{n>N}(1+n^2{)}^s\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\le 4A_s{N}^{-2s}\mathrm{\parallel }u{\mathrm{\parallel }}_{H_C^s}^2,$ where the last inequality uses [F1]. Also [F1] and [F4] give $\mathrm{\parallel }{R}_N{\mathrm{\parallel }}_C^2=\frac{\pi }{2}{\sum }_{n>N}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\le (N+1{)}^{-2s}\mathrm{\parallel }u{\mathrm{\parallel }}_{H_C^s}^2\mathrm{.}$ Thus $\mathrm{\parallel }{R}_N{\mathrm{\parallel }}_C\le (N+1{)}^{-s}\mathrm{\parallel }u{\mathrm{\parallel }}_{H_C^s}$.

By step 2.1 and the triangle inequality, $\mathrm{\parallel }u-I_N^{L}u{\mathrm{\parallel }}_C\le \mathrm{\parallel }{R}_N{\mathrm{\parallel }}_C+\mathrm{\parallel }{I}_N^L{R}_N{\mathrm{\parallel }}_C\le (1+2\sqrt{A_s})N^{-s}\mathrm{\parallel }u{\mathrm{\parallel }}_{H_C^s}$ because $(N+1{)}^{-s}\le N^{-s}$. This is the claimed estimate.