Analytic ellipse data give exponential Chebyshev–Lobatto interpolation error

Put $L=\log \rho >0$. For $z=\theta -it$, $0\le t\le L$, let $r=e^t$, so $1\le r\le \rho$ by [F10]. The definitions in [F8] and Cartesian formula [F9] give $\cos (\theta -it)=A_r\cos \theta +i{B}_r\sin \theta$, where $A_r=(r+r^{-1})\mathrm{/}2$ and $B_r=(r-r^{-1})\mathrm{/}2$. For $1\le r\le \rho$, $A_r\le A_{\rho }$ and $B_r\le B_{\rho }$: the differences are $(\rho -r)(1-1\mathrm{/}(\rho r))\mathrm{/}2$ and $(\rho -r)(1+1\mathrm{/}(\rho r))\mathrm{/}2$. Therefore $x^2\mathrm{/}{A}_{\rho }^2+y^2\mathrm{/}{B}_{\rho }^2\le {\cos }^2\theta +{\sin }^2\theta =1$ for $r>1$, so the image lies in the filled ellipse $E_{\rho }$; at $r=1$ it lies on $[-1,1]\subset E_{\rho }$. The closed rectangle $R_0=\{-\pi \le \mathrm{\Re }z\le \pi ,-L\le \mathrm{\Im }z\le 0\}$ is compact and lies in the open set ${\cos }^{-1}(V)$, where $V$ is a neighborhood on which $f$ is holomorphic. Hence a slightly larger open rectangle $R$ still lies in ${\cos }^{-1}(V)$; it is convex and therefore star-shaped. By [F8], $g=f\circ \cos$ is holomorphic on $R$, and each $H_n(z)=g(z)e^{-inz}$ is holomorphic and continuous there by [F8]. On the lower edge $\mathrm{\mid }g(\theta -iL)\mathrm{\mid }\le M$.

For each $n\ge 1$ define $c_n=\frac{2}{\pi }{\int }_0^{\pi }f(\cos \theta )\cos (n\theta )d\theta$, and define $c_0=\frac{1}{\pi }{\int }_0^{\pi }f(\cos \theta )d\theta$. For real $\theta$, $g$ is even and $f(\cos \theta )$ is real; Euler's identity [F9] makes the sine part of ${\int }_{-\pi }^{\pi }g(\theta )e^{-in\theta }d\theta$ vanish, so this integral is $\pi c_n$. Also $H_n(z)=g(z)e^{-inz}$ is $2\pi$-periodic: $g$ is periodic by [F8]--[F9], and $e^{-2\pi in}=1$ by [F9]. Apply [F1] to the boundary of $R_0$; the vertical integrals cancel by periodicity, so the top and lower horizontal integrals, both oriented from left to right, are equal. On the lower edge, [F9] and [F11] give $\mathrm{\mid }{e}^{-in(\theta -iL)}\mathrm{\mid }=e^{-nL}={\rho }^{-n}$, hence $\mathrm{\mid }{c}_n\mathrm{\mid }\le 2M{\rho }^{-n}$. Also $\mathrm{\mid }{c}_0\mathrm{\mid }\le M$ by the defining integral and the bound $\mathrm{\mid }f\mathrm{\mid }\le M$.

The coefficient bounds in step 2.1 make ${\sum }_{n\ge 0}{c}_n{T}_n$ uniformly convergent, since [F2] and [F6] give $\mathrm{\mid }{T}_n(x)\mathrm{\mid }\le 1$. Uniform convergence permits termwise integration against each bounded $T_k$; orthogonality [F4] shows that this series and $f$ have the same coefficients as defined in step 2.1. Their continuous difference therefore pairs to zero with every Chebyshev mode, so [F4] gives $f={\sum }_{n\ge 0}{c}_n{T}_n$ on $[-1,1]$.

By [F3], every mode $T_n$ with $n>N$ interpolates to a Chebyshev mode. Each such mode has sup norm at most one by [F2] and [F6]. Thus the series tail and its interpolant each have sup norm at most ${\sum }_{n>N}\mathrm{\mid }{c}_n\mathrm{\mid }$, and $\mathrm{\parallel }f-I_N^{L}f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2{\sum }_{n>N}\mathrm{\mid }{c}_n\mathrm{\mid }\le 4M{\sum }_{n>N}{\rho }^{-n}=\frac{4M}{\rho -1}{\rho }^{-N},$ where [F7] evaluates the geometric tail.

Apply [F5] to the uniform error in step 4.1. This yields $\mathrm{\parallel }f-I_N^{L}f{\mathrm{\parallel }}_C\le 4\sqrt{\pi }M(\rho -1{)}^{-1}{\rho }^{-N}$, as claimed.