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\begin{document}

\title[Exact Values of $n$-widths and Optimal Quadratures]{Exact Values of $n$-widths and Optimal Quadratures on Classes of Bounded Analytic and Harmonic Functions}
\author{K. Yu.\ Osipenko}
\address{Department of Mathematics, Moscow Institute of Aviation Tech, Moscow 103767, Petrovka 27, Russia}

\begin{abstract}
In this paper we find some exact values of $n$-widths in the integral metric with the Chebyshev weight function for the classes of functions that are bounded and analytic or harmonic in
the interior of the ellipse with foci $\pm1$ and sum of semiaxes $c$. We also construct optimal quadrature formulas for these classes.
\end{abstract}


\maketitle


\section*{Introduction}

The Kolmogorov n-width of a subset A of a normed linear space is defined by
$$d_n(A,X):=\infp_{X_n}\sup_{x\in A}\infp_{y\in X_n}\|x-y\|,$$
where $X_n$ runs over all $n$-dimensional subspaces of $X_n$. If the infimum is attained by some $X_n$, then $X_n$ is called an optimal subspace for $d_n(A,X)$.

We will also study the linear $n$-width defined by
$$\lambda_n(A,X):=\infp_{P_n}\sup_{x\in A}\|x-P_ny\|,$$
where $P_n$ runs over all bounded linear operators mapping $X$ into $X$ whose range has dimension $n$ or less, and the Gel'fand $n$-width defined by
$$d^n(A,X):=\infp_{X^n}\sup_{x\in A\cap X^n}\|x\|,$$
where the infimum is taken over all subspaces $X^n$ of $X$ of codimension $n$. In the last definition we assume that $0\in A$. (The usually considered case is when $A$ is a convex and balanced set.) A detailed bibliography and history of the subject can be found in the book of A.~Pinkus \cite{1}.

Let $G$ be a domain in the complex plane. Let $H_\infty(G)$ be the space of bounded analytic functions on $G$ with the norm
$$\|f\|_{H_\infty(G)}:=\sup_{z\in G}|f(z)|.$$
The analogous space of harmonic functions we denote by $h_\infty(G)$. We shall write $H_\infty$ and $h_\infty$ if $G=D:=\{z\in\mathbb C:|z|<1\}$. Denote by $BX$ the closed unit ball of the normed space $X$.

Let $E\subset(-1,1)$ be a compact set and let $L_q(E,\mu)$ be the Lebesgue space with positive measure $\mu$ on $E$ and $1\le q\le\infty$. In Section 1 we obtain the values of the $n$-widths of $Bh_\infty$ in $L_q(E,\mu)$. We also find the exact value and two different optimal spaces for $d_n(Bh_\infty(\mathcal E_c),C[-1,1])$, where $\mathcal E_c$, is the interior of the ellipse with foci at the points $\pm1$ and sum $c$ of its semiaxes.

In section 2 we find the exact values of $n$-widths of $BH_\infty(\mathcal E_c)$ and $Bh_\infty(\mathcal E_c)$ in $L_q([-1,1],\mu)$ for $d\mu(x)=dx/\sqrt{1-x^2}$, $1\le q<\infty$. To obtain this result we solve some minimization problem with Blashke products. The solution of this problem allows us to construct optimal quadrature formulas for the classes $BH_\infty(\mathcal E_c)$ and $Bh_\infty(\mathcal E_c)$ and to improve the results of \cite{2,3}, where we proved that these formulas are optimal for sufficiently large $c$.

\section{$n$-widths of Harmonic Functions in $h_\infty$}

A Blashke product of degree $n$ is a function of the form

$$B(z)=\sigma\prod_{j=1}^n\frac{1-\alpha_j}{1-\overline\alpha_jz},\quad|\alpha_j|<1,\quad j=1,\ldots,n,\quad|\sigma|=1.$$

Denote by $\mathcal B_n$ the set of all Blashke products of degree $n$ or less and by $\mathcal B_n^0$ the ones with $\alpha_j\in(-1,1)$, $\sigma=\pm1$. Let $E$ be a compact subset of $D$ and $\mu$ a positive measure on $E$ such that $\mu(E)<\infty$. Denote by $L_q=L_q(E,\mu)$ the Lebesgue space of functions on $E$ with the usual norm $\|\cdot\|_q$.

It was proved by Fisher and Micchelli \cite{4} that
\begin{equation}\label{1}
d_n(BH_\infty,L_q)=\lambda_n(BH_\infty,L_q)=d^n(BH_\infty,L_q)=\inf_{B\in\mathcal B_n}\|B\|_q.
\end{equation}
We obtain a similar result for the class $Bh_\infty$.

\begin{theorem}\label{T1}
Let $E\subset(-1,1)$. Then for all $1\le q\le\infty$
$$d_n(Bh_\infty,L_q)=\lambda_n(Bh_\infty,L_q)=d^n(Bh_\infty,L_q)=\frac4\pi\inf_{B\in\mathcal B_n^0}\|\arctan B\|_q.$$
\end{theorem}

\begin{proof}
We use the scheme of proof from \cite{4}. Let $x_1,\ldots,x_n$ be any points in $(-1,1)$ and let $B(z)$ be the Blashke product with the zeros at these points. In \cite{3} an optimal recovery method was obtained for the functional $u(x)$, $u\in Bh_\infty$, $x\in(-1,1)$, based on the information $u(x_1),\ldots,u(x_n)$. It was also proved that the error of this method is equal to $(4/\pi)|\arctan B(x)|$. Thus there exist functions $g_1,\ldots,g_n\in C(E)$ such that for all $u\in Bh_\infty$ and all $x\in(-1,1)$
$$\biggl|u(x)-\sum_{j=1}^ng_j(x)u(x_j)\biggr|\le\frac4\pi|\arctan B(x)|,$$
where successive derivatives of $u$ at $x_j$ through order $r-1$ will appear if some $x_j$ coincide with order $r$. Hence
\begin{equation}\label{2}
d_n(Bh_\infty,L_q)\le\lambda_n(Bh_\infty,L_q)\le\frac4\pi\inf_{B\in\mathcal B_n^0}\|\arctan B\|_q.
\end{equation}

To obtain the reverse inequality we use the Borsuk Theorem (see, for example, \cite{1}). Fix points $x_0,\ldots,x_n\in(-1,1)$. Let $y=(y_0,\ldots,y_n\in\mathbb S^n:=\{y\in\mathbb R^{n+1}:\sum_{j=0}^ny_j^2=1\}$. Set
$$\rho(y):=\inf_{\substack{f\in H_\infty\\f(x_j)=y_j,\ j=0,\ldots,n}}\|f\|_{H_\infty}.$$
According to the classical Pick-Nevanlinna Theorem there is a unique Blashke product $B\in\mathcal B_n$ such that
\begin{equation}\label{3}
\rho(y)B(x_j)=y_j,\quad j=0,\ldots,n.
\end{equation}
Since $\overline{B(\overline z)}$ satisfies the Eqs. \eqref{3}, it follows that $B$ is real on the real axis. Denote by $\mathcal B_n^{\mathbb R}$ the set of all Blashke products $B\in\mathcal B_n$ which are real on the real axis and by $T$ the mapping
$$(Ty)(z):=\frac4\pi\arctan B(z),$$
where $B$ satisfies \eqref{3}. The function
$$w=\frac4\pi\arctan z$$
is a conformal mapping of $D$ on the strip $|\RE w|<1$. Therefore $\RE Ty\in Bh_\infty$ for every $y\in\mathbb S^n$. The continuity of $\rho(y)$ implies that T is a continuous and odd map of $\mathbb S^n$ into $L_q$.

Let $1<q<\infty$. Suppose that $X_n=\spa\{f_1,\ldots,f_n\}$ is an $n$-dimensional
subspace of $L_q$. For each $f\in L_q$ let $c_1(f),\ldots,c_n(f)$ be the coefficients of
$f_1,\ldots,f_n$, respectively, in the best approximation to $f$ from $X_n$. The
mapping $Sf:=(c_1(f),\ldots,c_n(f))$ is a continuous odd mapping of $L_q$ into $\mathbb R^n$. Thus $S\circ T$ is an odd, continuous map of $\mathbb S^n$ into $\mathbb R^n$. By the Borsuk Theorem there exists a $y^*\in\mathbb S^n$ for which $c_j(Ty^*)=0$, $j=1,\ldots,n$. Hence
$$\sup_{u\in Bh_\infty}\infp_{v\in X_n}\|u-v\|_q\ge\inf_{v\in X_n}\|Ty^*-v\|_q=\|Ty^*\|_q
\ge\frac4\pi\inf_{B\in\mathcal B_n^{\mathbb R}}\|\arctan B\|_q.$$
Since $X_n$, is arbitrary we have
$$d_n(Bh_\infty,L_q)\ge\frac4\pi\inf_{B\in\mathcal B_n^{\mathbb R}}\|\arctan B\|_q.$$
For every $\alpha+i\beta\in D$ and $x\in(-1,1)$
\begin{equation}\label{4}
\left|\frac{x-\alpha-i\beta}{1-(\alpha-i\beta)x}\right|\ge\left|\frac{x-\alpha}{1-\alpha x}\right|.
\end{equation}
Thus
$$d_n(Bh_\infty,L_q)\ge\frac4\pi\inf_{B\in\mathcal B_n^0}\|\arctan B\|_q.$$
The cases $q=1,\infty$ are established by passing to the limit as either $q\searrow1$
or $q\nearrow\infty$.

Now consider the case of $d^n$. Let $X^n$ be any subspace of $L_q$ of codimension $n$. Thus
$$X^n=\{\,u\in L_q:\langle f_j',u\rangle=0,\ j=1,\ldots,n\,\}$$
for some linearly independent and continuous functionals $f_j'$ on $L_q$. Denote by
$T'\colon\mathbb S^n\to\mathbb R^n$ the mapping
$$T'y:=(\langle f_1',Ty\rangle,\ldots,\langle f_n',Ty\rangle).$$
$T'$ is an odd and continuous map. By the Borsuk Theorem there exists a $y^*\in\mathbb S^n$ for which $T'y^*=0$. Since $Ty^*\in Bh_\infty$ we have
\begin{multline*}
\sup_{\substack{u\in Bh_\infty\\\langle f_j',u\rangle=0,\ j=1,\ldots,n}}\|u\|_q\ge\|Ty^*\|_q
\ge\frac4\pi\inf_{B\in\mathcal B_n^{\mathbb R}}\|\arctan B\|_q\\
=\frac4\pi\inf_{B\in\mathcal B_n^0}\|\arctan B\|_q.
\end{multline*}
As $X^n$ is arbitrary we find
$$d^n(Bh_\infty,L_q)\ge\frac4\pi\inf_{B\in\mathcal B_n^0}\|\arctan B\|_q.$$
The reverse inequality follows from the well-known inequality (see, for example, \cite{1})
$$\lambda_n(A,X)\ge d^n(A,X)$$
and \eqref{2}. The theorem is proved.
\end{proof}

We shall use the standard notation for the Jacobi elliptic function $w=\sn(z,k)$, which is defined from the equation
$$z=\int_0^w\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}.$$
Besides that we shall deal with the elliptic functions
$$\cn(z,k):=\sqrt{1-\sn^2(z,k)},\quad\dn(z,k):=\sqrt{1-k^2\sn^2(z,k)}$$
($\cn(0,k)=\dn(0,k)=1$) and complete elliptic integrals of the first kind with moduli $k$ and $k'=\sqrt{1-k^2}$,
$$K:=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}},\quad K':=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-k'^2t^2)}}.$$

From \cite{5} it follows that for every $k\in(0,1)$
$$\infp_{B\in\mathcal B_n\vphantom{\left[\sqrt k\right]}}\sup_{x\in\left[\sqrt k,\sqrt k\right]}|B(x)|=\sup_{x\in\left[\sqrt k,\sqrt k\right]}|Z_n(x)|
=2h^{n/4}\frac{\displaystyle\sum_{m=0}^\infty h^{nm(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty h^{nm}},$$
where $h=e^{-\pi K'/K}$,
\begin{gather*}
Z_n(z):=\prod_{j=1}^n\frac{z-z_j^0}{1-zz_j^0},\\
z_j^0:=\sqrt k\left[\left(\frac{2j-1}n-1\right)K,k\right],\quad j=1,\ldots,n.
\end{gather*}
As the function $\arctan z$ is monotone we have from Theorem~\ref{T1}
\begin{multline}\label{5}
d_n\left(Bh_\infty,C\left[-\sqrt k,\sqrt k\right]\right)=\lambda_n\left(Bh_\infty,C\left[-\sqrt k,\sqrt k\right]\right)\\
=d^n\left(Bh_\infty,C\left[-\sqrt k,\sqrt k\right]\right)
=\frac4\pi\arctan\left[2h^{n/4}\frac{\displaystyle\sum_{m=0}^\infty h^{nm(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty h^{nm^2}}\right].
\end{multline}
To rewrite the right hand side of \eqref{5} we need the following lemma.

\begin{lemma}\label{L1}
For all $h\in(0,1)$
$$\frac4\pi\arctan\left[2h\frac{\displaystyle\sum_{m=0}^\infty h^{4m(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty h^{4m^2}}\right]=\frac8\pi\sum_{m=0}^\infty
\frac{(-1)^m}{2m+1}\frac{h^{2m+1}}{1+h^{2(2m+1)}}.$$
\end{lemma}

\begin{proof}	
Determine $k$ by the equation
$$e^{-\pi K'/K}=h.$$
Then for real $x$ (see \cite{6})
$$\dn\left(\frac{Kx}\pi,k\right)=\frac\pi{2K}+\frac{2\pi}K\sum_{m=1}^\infty
\frac{h^m}{1+h^{2m}}\cos mx.$$
Hence
\begin{equation}\label{6}
\frac{4K}\pi\int_0^{\pi/2}\dn\left(\frac{Kx}\pi,k\right)\,dx=1+\frac8\pi\sum_{m=0}^\infty
\frac{(-1)^m}{2m+1}\frac{h^{2m+1}}{1+h^{2(2m+1)}}.
\end{equation}
It is easy to obtain that
$$\int\dn(t,k)\,dt=\arctan\frac{\sn(t,k)}{\cn(t,k)}+C.$$
Using the well-known equations from the theory of elliptic functions (see, for example, \cite{6})
\begin{gather*}
\sn(K/2,k)=\frac1{\sqrt{1+k'}},\quad\cn(K/2,k)=\frac{\sqrt{k'}}{\sqrt{1+k'}},\\
\sqrt{k'}=\frac{\displaystyle1+2\sum_{m=1}^\infty(-1)^mh^{m^2}}{\displaystyle1+2
\sum_{m=1}^\infty h^{m^2}},
\end{gather*}
we have by~\eqref{6}
\begin{multline*}
\frac8\pi\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\frac{h^{2m+1}}{1+h^{2(2m+1)}}=
\frac4\pi\int_0^{K/2}\dn(t,k)\,dt-1\\
=\frac4\pi\arctan\frac{\sn(t,k)}{\cn(t,k)}\bigg|_0^{K/2}-1=
\frac4\pi\left(\arctan\frac1{\sqrt{k'}}-\arctan1\right)\\
=\frac4\pi\arctan\frac{1-\sqrt{k'}}{1+\sqrt{k'}}=\frac4\pi\arctan\left[2h\frac{\displaystyle\sum_{m=0}^\infty h^{4m(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty h^{4m^2}}\right].
\end{multline*}
\end{proof}

The function
\begin{equation}\label{7}
\phi(w):=\sqrt k\sn\left(\frac{2K}\pi\arcsin w,k\right)
\end{equation}
maps $\mathcal E_c$ conformally on the unit disk $D$, and carries the interval $[-1,1]$ to the interval $[-\sqrt k,\sqrt k]$, where $k$ satisfies
\begin{equation}\label{8}
\frac{K'}K=\frac4\pi\log c.
\end{equation}
Note that the map $\phi$ carries the Chebyshev points
\begin{equation}\label{9}
x_j^0=\cos\frac{2j-1}{2n}\pi,\quad j=1,\ldots,n,
\end{equation}
to $z_j^0$. We obtain the following corollary by using this map, \eqref{5} and Lemma~\ref{L1}.

\begin{corollary}
For all $c>1$
\begin{multline*}
d_n\left(Bh_\infty(\mathcal E_c),C[-1,1]\right)=\lambda_n\left(Bh_\infty(\mathcal E_c),C[-1,1]\right)\\
=d^n\left(Bh_\infty(\mathcal E_c),C[-1,1]\right)=\frac8\pi\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\frac{c^{-(2m+1)n}}
{1+c^{-2(2m+1)n}}.
\end{multline*}
\end{corollary}

Denote by $A_0(\mathcal E_c)$ the class of functions $f$ which are analytic in $\mathcal E_c$, real on the real axis and satisfy
$$\RE f(z)|\le1,\quad z\in\mathcal E_c.$$
It was proved by N.~I.~Akhiezer \cite{7} that
\begin{multline}\label{10}
E_n(A_0(\mathcal E_c)):=\sup_{f\in A_0(\mathcal E_c)}\infp_{p\in\mathcal P_{n-1}}\|f-p\|_{C[-1,1]}\\
=\frac8\pi\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\frac{c^{-(2m+1)n}}
{1+c^{-2(2m+1)n}},
\end{multline}
where $\mathcal P_{n-1}$ is the set of all polynomials of degree $n-1$ or less.

It is easy to show that the restrictions on the real axis of $A_0(\mathcal E_c)$ and $Bh_\infty(\mathcal E_c)$ coincide. Thus Eq.~\eqref{10} is valid for $Bh_\infty(\mathcal E_c)$ and the space $\mathcal P_{n-1}$ is optimal subspace for $d_n\left(Bh_\infty(\mathcal E_c),C[-1,1]\right)$. By proving Theorem~\ref{T1} we see that the space
$$X_n:=\spa\{\,g_1(\phi(w)),\ldots,g_n(\phi(w))\,\},$$
where the functions $g_1,\ldots,g_n$ are determined by the points $z_j^0$, is also an optimal subspace. Akhiezer's result was brought to the author's attention by to V.~M.~Tikhomirov. He also conjectured that there is a sequence of optimal subspaces, like in the case of smooth functions (see \cite{8}).

\section{Exact Values of $n$-Widths in $L_q$ and Optimal Quadrature Formulas}

We first formulate a generalization of one result obtained by A.~Pinkus \cite{9} (see also \cite[p.~174]{1}). Let $h(t)$ be a piecewise continuous, $2\pi$-periodic function. Denote by $S_c(h)$ the number of sign changes of $h$. For a real, continuous, $2\pi$-periodic function $k$ set
$$(k*h)(x):=\frac1{2\pi}\int_0^{2\pi}k(x-t)h(t)\,dt.$$

The kernel $k$ is nondegenerate cyclic variation diminishing ($NCVD$) if $S_c(k*h)\le S_c(h)$ for all $h$, and
$$\dim\spa\{k(x_1-\cdot),\ldots,k(x_n-\cdot)\}=n$$
for every choice of $0\le x_1<\ldots<x_n<2\pi$ and all n. The kernel $k$ is said
to be strictly sign consistent of order $2l+1$ ($SSC_{2l+1}$) if
$$\sigma\det\left(k(x_j-y_m)\right)_{j,m=1}^{2l+1}>0$$
whenever $0\le x_1<\ldots<x_{2l+1}<2\pi$, $0\le y_1<\ldots<y_{2l+1}<2\pi$, and $\sigma=1$ or $-1$.

Set
$$\Lambda_{2m}:=\{\,\xi:\xi=(\xi_1,\ldots,\xi_{2m}),\ 0\le\xi_1\le\ldots\le\xi_{2m}<2\pi\,\}.$$
For each $\xi\in\Lambda_{2m}$ we define
$$h_\xi(t):=(-1)^j,\quad t\in[\xi_{j-1},\xi_j),\quad j=1,\ldots,2m+1,$$
where $\xi_0:=0$, $\xi_{2m+1}:=2\pi$. Denote by $h_m(t)$ the function $h_\xi$ for $\xi_j=(j-1)\pi/m$, $j=1,\ldots,2m$.

\begin{theorem}\label{T2}
Let $k$ be an $NCVD$ kernel and $\varphi$ a nonnegative function defined on $[0,C]$, where
$$C:=\frac1{2\pi}\int_0^{2\pi}|k(t)|\,dt.$$
Suppose that $\varphi'$ is an nonnegative, continuous, and strictly increasing function. Then
$$\inf_{\xi\in\Lambda_{2n}}\int_0^{2\pi}\varphi\left(\left|(k*h_\xi)(t)\right|\right)\,dt=
\int_0^{2\pi}\varphi\left(\left|(k*h_n)(t)\right|\right)\,dt.$$
Furthermore, if $k$ is $SSC_{2l+1}$, $l=0,1,\ldots,n$, and the infimum is attained by $\xi^*\in\Lambda_{2n}$, then $\xi_{j+1}^*-\xi_j^*=\pi/n$, $j=1,\ldots,2n-1$.
\end{theorem}

This theorem was proved by A.~Pinkus for $\varphi(x)=x^q$, $1\le q<\infty$. The general case is proven in a similar way. To count sign changes we only need to use the equation
$$\sign(a+b)=\sign(\varphi'(|a|)\sign a+\varphi'(|b|)\sign b)$$
instead of
$$\sign(a+b)=\sign(|a|^{q-1}\sign a+|b|^{q-1}\sign b),\quad1<q<\infty.$$

Set $D_H:=\{z\in\mathbb C:|\IM z|<H\}$. Denote by $A_H$ the class of all functions analytic in $D_H$, real and $\pi$-periodic on the real axis which satisfy
$$|\RE f(z)|\le1,\quad z\in D_H.$$
Each function $f\in A_H$ has the representation
$$f(z)=\frac1{2\pi}\int_0^{2\pi}K_H(z-t)\RE f(t+iH)\,dt,$$
where
$$K_H(z=1+2\sum_{j=1}^\infty\frac{\cos jz}{\cos jH}$$
and $K_H$ is $NCVD$ on $[0,2\pi)$ (see \cite{1}). Moreover, it was proved by W.~Forst \cite{10} that $K_H$ is $SSC_{2l+1}$ for all $l=0,1,\ldots$.

By Theorem~\ref{T2} with $k=K_H$ we solve some extremal problems which allow us to obtain exact values of $n$-widths and to construct optimal quadratures for $BH_\infty(\mathcal E_c)$ and $Bh_\infty(\mathcal E_c)$.

For $k\in(0,1)$ set
$$d\nu_k(z):=\frac\pi{2K}\frac{dz}{\sqrt{(k-z^2)(1-kz^2)}}.$$

\begin{theorem}\label{T3}
Let $\varphi$ be a function defined on $[0,1]$ which satisfies the assumptions of Theorem~$\ref{T2}$. Then for all $k\in(0,1)$
\begin{multline}\label{11}
\inf_{B\in\mathcal B_n^0}\int_{-\sqrt k}^{\sqrt k}\varphi\left(\frac4\pi\arctan|B(z)|\right)\,
d\nu_k(z)\\
=\frac\pi\Lambda\int_0^1\frac{\varphi\left(\dfrac4\pi\arctan(\sqrt\lambda t)\right)\,dt}{\sqrt{(1-t^2)(1-\lambda^2t^2)}},
\end{multline}
where
\begin{equation}\label{12}
\lambda=4h^{n/2}\left(\frac{\displaystyle\sum_{m=0}^\infty h^{nm(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty h^{nm^2}}\right)^2,\quad h=e^{-\pi K'/K},
\end{equation}
and $\Lambda$ is the complete elliptic integral of the first kind with modulus $\lambda$. Moreover, the functions $\pm Z_n$ are the only functions for which the infimum is attained.
\end{theorem}

\begin{proof}
Let $B\in\mathcal B_n^0$. Set
$$f(t):=\frac4\pi\arctan B\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right).$$
From the properties of the elliptic function $\sn(z,k)$ it follows that $f\in A_H$,
where $H=\pi K'/(4K)$. For $z=e^{i\theta}$ we have
\begin{multline*}
\RE B(z)=\sigma\RE\prod_{j=1}^n\frac{z-x_j}{1-x_jz}=\sigma\prod_{j=1}^n\frac1{|1-x_jz|^2}
\RE\prod_{j=1}^n\frac{(z-x_j)^2}z\\
=\sigma\prod_{j=1}^n\frac1{|1-x_jz|^2}\sum_{j=0}^nc_j\cos j\theta,
\end{multline*}
where $c_j\in\mathbb R$, $\sigma=1$ or $-1$. Thus the function $\RE B(e^{i\theta})$ has at most $2n$ zeros in $(0,2\pi)$. As $t$ runs from $0$ to $2\pi$ the point
$$z=\sqrt k\sn\left(\frac{2K}\pi(t+iH),k\right)$$
makes one rotation around the unit circle. Since for all $|z|=1$ and $z\ne\pm i$
$$\RE\left(\frac4\pi\arctan z\right)=\sign\RE z,$$
we have for almost all $t\in[0,2\pi]$
\begin{equation}\label{13}
\RE f(t+iH)=\sign\RE B\left(\sqrt k\sn\left(\frac{2K}\pi(t+iH),k\right)\right).
\end{equation}
Consequently there exists a $\xi\in\Lambda_{2n}$ for which
$$|f(t)|=\left|\left(K_H*h_\xi\right)(t)\right|.$$
By using the first fundamental transformation of degree $n$ (see \cite{6}) we can find
\begin{multline*}
Z_n\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)\\
=\begin{cases}
(-1)^m\sqrt\lambda\sn\left(\dfrac{2n\Lambda}\pi t+\Lambda,\lambda\right),&n=2m,\\
(-1)^m\sqrt\lambda\sn\left(\dfrac{2n\Lambda}\pi t,\lambda\right),&n=2m+1,\end{cases}
\end{multline*}
where $\lambda$ is determined by the equation
$$\frac{\Lambda'}\Lambda=n\frac{K'}K$$
($\Lambda'$	is the complete elliptic integral of the first kind with modulus $\lambda'=\sqrt{1-\lambda^2}$). From the standard equation
$$\sqrt\lambda=4h_1^{1/4}\frac{\displaystyle\sum_{m=0}^\infty h_1^{m(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty h_1^{m^2}},\quad h_1=e^{-\pi\Lambda'/\Lambda},$$
in the theory of elliptic functions, we obtain \eqref{12}. Set
$$f_n(t):=\frac4\pi\arctan\left[Z_n\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)\right].$$

Let $n=2m+1$. Then from the equation
$$\RE\sn(w+i\Lambda'/2,\lambda)=\frac{(1+\lambda)\sn(w,\lambda)}{1+\lambda^2\sn^2(w,\lambda)},$$
and \eqref{13} it follows
$$\RE f_n(t+iH)=(-1)^m\sign\sn\left(\dfrac{2n\Lambda}\pi t,\lambda\right).$$
Hence
$$f_n(t)=(-1)^m\left(K_H*h_n\right)(t).$$
It can be proved similarly that for $n=2m$
$$f_n(t)=(-1)^m\left(K_H*h_n\right)\left(t+\frac\pi{2n}\right).$$
Thus
\begin{equation}\label{14}
\left(K_H*h_n\right)(t)=\frac4\pi\arctan\left[\sqrt\lambda\sn\left(\frac{2n\Lambda}\pi t,\lambda\right)\right].
\end{equation}
Now from Theorem~\ref{T2} we have
\begin{multline*}
\inf_{B\in\mathcal B_n^0}\int_0^{2\pi}\varphi\left(\frac4\pi\arctan\left|B\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)\right|\right)\,dt\\
\ge\inf_{\xi\in\Lambda_{2n}}\int_0^{2\pi}\varphi\left(\left|
\left(K_H*h_\xi\right)(t)\right|\right)\,dt
=\int_0^{2\pi}\varphi\left(\left|
\left(K_H*h_n\right)(t)\right|\right)\,dt\\
=\int_0^{2\pi}\varphi\left(|f_n(t)|\right)\,dt.
\end{multline*}
In view of the equation $\sn(2K-w,k)=\sn(w,k)$ we will have
\begin{multline*}
\inf_{B\in\mathcal B_n^0}\int_{-\pi/2}^{\pi/2}\varphi\left(\frac4\pi\arctan\left|B\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)\right|\right)\,dt\\
=\int_{-\pi/2}^{\pi/2}\varphi\left(\left|
\left(K_H*h_n\right)(t)\right|\right)\,dt\\
=\frac\pi\Lambda\int_0^{\Lambda}\varphi\left(\frac4\pi\arctan\left(\sqrt\lambda\sn(z,\lambda)
\right)\right)\,dz.
\end{multline*}
Making the change of variables
\begin{equation}\label{15}
z=\sqrt k\sn\left(\frac{2K}\pi t,k\right),
\end{equation}
in the first integral and $t=\sn(z,\lambda)$ in the last one, we obtain \eqref{11}.

If the infimum in \eqref{11} is attained by any $B^*\in\mathcal B_n^0$ then from Theorem~\ref{T2}
there exists an $\alpha\in[0,\pi/n)$ and $\sigma=1$ or $-1$ for which
\begin{multline*}
\frac4\pi\arctan B^*\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)=\sigma(K_H*h_n)(t+\alpha)\\
=\sigma\frac4\pi\arctan\left[\sqrt\lambda \sn\left(\frac{2n\Lambda}\pi(t+\alpha),\lambda\right)\right].
\end{multline*}
Thus
$$B^*\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)=\sigma\sqrt\lambda \sn\left(\frac{2n\Lambda}\pi(t+\alpha),\lambda\right).$$
In view of the formula for $\sn(u+w,\lambda)$ we have
\begin{multline}\label{16}
B^*\left(\sqrt k\sn\left(\frac{2K}\pi t,k\right)\right)\\
=\frac{a\sn\left(\dfrac{2n\Lambda}\pi t,\lambda\right)+b\cn\left(\dfrac{2n\Lambda}\pi t,\lambda\right)\dn\left(\dfrac{2n\Lambda}\pi t,\lambda\right)}{1-c\sn^2\left(\dfrac{2n\Lambda}\pi t,\lambda\right)},
\end{multline}
where
$$b=\sigma\sqrt\lambda\sn\left(\dfrac{2n\Lambda}\pi\alpha,\lambda\right)$$
(the numbers $a$ and $c$ are irrelevant). Let $n=2m+1$. If we make the change of variable \eqref{15}, the left hand side of \eqref{16} and $\sn((2n\Lambda/\pi)t,\lambda)$ become rational functions. On the other hand, it is not difficult to show that
$$\cn\left(\dfrac{2n\Lambda}\pi t,\lambda\right)\dn\left(\dfrac{2n\Lambda}\pi t,\lambda\right)$$
is not a rational function (as a function of $z$). Therefore $b=0$ and consequently $\alpha=0$. This means that $B*=Z_n$ or $-Z_n$. The case $n=2m$ can be considered similarly. The theorem is proved.
\end{proof}

Set
\begin{align*}
I_{q0}(\lambda):=&\int_0^1\frac{t^q\,dt}{\sqrt{(1-t^2)(1-\lambda^2t^2)}},\\
I_{q1}(\lambda):=&\int_0^1\frac{\arctan^q(\sqrt\lambda t)\,dt}{\sqrt{(1-t^2)(1-\lambda^2t^2)}},\\
I_{q2}(\lambda):=&\int_0^1\frac{\arctan(\sqrt\lambda t)^q\,dt}{\sqrt{(1-t^2)(1-\lambda^2t^2)}}.
\end{align*}
Considering in Theorem~\ref{T3} the functions
$$\tan^q\frac\pi4x,\quad\left(\frac\pi4x\right)^q,\quad\arctan\left(\tan^q\frac\pi4x\right),\quad
1\le q<\infty,$$
as $\varphi$, we have

\begin{corollary}\label{C2}
For all $k\in(0,1)$ and $1\le q<\infty$
\begin{gather*}
\inf_{B\in\mathcal B_n^0}\int_{-\sqrt k}^{\sqrt k}|B(z)|^q\,d\nu_k(z)=\frac\pi\Lambda\lambda^
{q/2}I_{q0}(\lambda),\\
\inf_{B\in\mathcal B_n^0}\int_{-\sqrt k}^{\sqrt k}\arctan^q|B(z)|\,d\nu_k(z)=\frac\pi\Lambda I_{q1}(\lambda),\\
\inf_{B\in\mathcal B_n^0}\int_{-\sqrt k}^{\sqrt k}\arctan|B(z)|^q\,d\nu_k(z)=\frac\pi\Lambda I_{q2}(\lambda).
\end{gather*}
Furthermore, in every case the infimum is attained only by $\pm Z_n$.
\end{corollary}

Now we can prove the following result.

\begin{theorem}\label{T4}
Let $L_q=L_q([-1,1],\mu)$, $d\mu(x)=dx/\sqrt{1-x^2}$. For all $1\le q<\infty$ and $c>1$
\begin{multline*}
d_n\left(BH_\infty(\mathcal E_c),L_q\right)=\lambda_n\left(BH_\infty(\mathcal E_c),L_q\right)=
d^n\left(BH_\infty(\mathcal E_c),L_q\right)\\
=\sqrt\lambda\left(\frac\pi\Lambda I_{q0}(\lambda)\right)^{1/q}=2\left(\sqrt\pi\frac{\Gamma
\left(\dfrac{q+1}2\right)}{\Gamma\left(\dfrac q2+1\right)}\right)^{1/q}c^{-n}+O\left(c^{-5n}
\right),
\end{multline*}
\begin{multline*}
d_n\left(Bh_\infty(\mathcal E_c),L_q\right)=\lambda_n\left(Bh_\infty(\mathcal E_c),L_q\right)=
d^n\left(Bh_\infty(\mathcal E_c),L_q\right)\\
=\frac4\pi\left(\frac\pi\Lambda I_{q1}(\lambda)\right)^{1/q}=\frac8\pi\left(\sqrt\pi\frac{\Gamma
\left(\dfrac{q+1}2\right)}{\Gamma\left(\dfrac q2+1\right)}\right)^{1/q}c^{-n}+O\left(c^{-5n}
\right),
\end{multline*}
where $\Lambda$ is the complete elliptic integral of the first kind with modulus
\begin{equation}\label{17}
\lambda=4c^{-2n}\left[\frac{\displaystyle\sum_{m=0}^\infty c^{-4nm(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty c^{-4nm^2}}\right]^2.
\end{equation}
\end{theorem}

\begin{proof}
Let $z=\phi(w)$ be a conformal mapping of $\mathcal E_c$ on the unit disk $D$
determined by \eqref{7}. It is easy to show that
$$d\nu_k(z)=\frac{dw}{\sqrt{1-w^2}}.$$
Therefore by the mapping $\phi$, the original problem reduces to one of finding the exact values of the $n$-widths of $BH_\infty$ and $Bh_\infty$ in $L_q([-\sqrt k,\sqrt k],\nu_k)$. From \eqref{1} we have
$$d_n\left(BH_\infty,L_q([-\sqrt k,\sqrt k],\nu_k)\right)=\inf_{B\in\mathcal B_n}\left(\int_{-\sqrt k}^{\sqrt k}|B(z)|^q\,d\nu_k(z)\right)^{1/q}.$$
It follows from \eqref{4} that we can change $\mathcal B_n$ by $\mathcal B_n^0$ in the last equation. Thus by Corollary~\ref{C2}
$$d_n\left(BH_\infty(\mathcal E_c),L_q\right)=\sqrt\lambda\left(\frac\pi\Lambda I_{q0}(\lambda)\right)^{1/q}.$$
For the class $Bh_\infty(\mathcal E_c)$ from Theorem~\ref{T1} and Corollary~\ref{C2} we obtain
\begin{multline*}
d_n\left(Bh_\infty(\mathcal E_c),L_q\right)=\frac4\pi\inf_{B\in\mathcal B_n^0}\left(\int_{-\sqrt k}^{\sqrt k}\arctan^q|B(z)|\,d\nu_k(z)\right)^{1/q}\\
=\frac4\pi\left(\frac\pi\Lambda I_{q1}(\lambda)\right)^{1/q}.
\end{multline*}
The asymptotic equations follow from \eqref{17} and the well-known equations
\begin{gather*}
\int_0^1x^q(1-x^2)^{-1/2}\,dx=\frac{\sqrt\pi}2\frac{\Gamma
\left(\dfrac{q+1}2\right)}{\Gamma\left(\dfrac q2+1\right)},\\
\Lambda=\frac\pi2\biggl(1+2\sum_{m=1}^\infty h^{m^2}\biggr)^2,\quad h=e^{-\frac{\pi\Lambda'}
\Lambda}=c^{-4n}.
\end{gather*}
The theorem is proved.
\end{proof}

The $n$-widths of periodic functions, which are represented as a convolution with some kernel $k\in NCVD$, was studied by A.~Pinkus \cite{9} (see also \cite{1}). In particular, it follows from \cite{9}, that for all $1\le q\le\infty$
\begin{equation}\label{18}
d_{2n}(A_H,L_q)=\lambda_{2n}(A_H,L_q)=d^{2n}(A_H,L_q)=\|(K_H*h_n\|_q,
\end{equation}
where $\|\cdot\|_q$ is the usual norm in the space $L_q:=L_q[0,2\pi]$. Since the
function $K_H*h_n$ is found in direct form (see \eqref{14}) we can calculate the
exact values of these $n$-widths.

\begin{theorem}\label{T5}
For all $1\le q<\infty$
\begin{multline*}
d_{2n}(A_H,L_q)=\lambda_{2n}(A_H,L_q)=d^{2n}(A_H,L_q)=\frac4\pi\left(\frac{2\pi}\Lambda I_{q1}(\lambda)\right)^{1/q}\\
=\frac8\pi\left(2\sqrt\pi\frac{\Gamma
\left(\dfrac{q+1}2\right)}{\Gamma\left(\dfrac q2+1\right)}\right)^{1/q}e^{-Hn}+O\left(c^{-5Hn}
\right),
\end{multline*}
where
$$\lambda=4e^{-2Hn}\left[\frac{\displaystyle\sum_{m=0}^\infty e^{-4Hnm(m+1)}}{\displaystyle1+2\sum_{m=1}^\infty e^{-4Hnm^2}}\right]^2.$$
\end{theorem}

For $q=\infty$ we have from \eqref{14} and \eqref{18}
$$d_{2n}(A_H,L_\infty)=\lambda_{2n}(A_H,L_\infty)=d^{2n}(A_H,L_\infty)=
\frac4\pi\arctan\sqrt\lambda.$$
Now from Lemma~\ref{L1} and \eqref{17}
\begin{multline*}
d_{2n}(A_H,L_\infty)=\lambda_{2n}(A_H,L_\infty)=d^{2n}(A_H,L_\infty)\\
=\frac4\pi\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\frac1{\cosh[(2m+1)Hn]}.
\end{multline*}
These equations were previously calculated by V.~M.~Tikhomirov \cite{11} (the complete proof was given by W.~Forst \cite{10}).

Let us consider some applications of Theorem~\ref{T3} to optimal quadrature formulas. We are interested in the problem of approximate calculation of the integral
$$If:=\int_{-1}^1f(x)\frac{dx}{\sqrt{1-x^2}},$$
where $f\in W=BH_\infty(\mathcal E_c)$ or $Bh_\infty(\mathcal E_c)$, in terms of the values of $f$ and its derivatives at a system of knots. Denote by
$$\tau_\alpha:=\begin{pmatrix}
x_1,\ldots,x_n\\
\alpha_1,\ldots,\alpha_n\end{pmatrix},$$
a system of distinct knots	$x_1,\ldots,x_n\in[-1,1]$ with multiplicities $\alpha_1,\ldots,\alpha_n$.

The error of the best quadrature formula for a given system $\tau_\alpha$ is the
number
$$e(\tau_\alpha,W):=\infp_{a_{jm}}\sup_{f\in W}\biggl|If-\sum_{j=1}^n\sum_{m=0}^{\alpha_j}
a_{jm}f^{(m)}(x_j)\biggr|.$$
If $f\in Bh_\infty(\mathcal E_c)$ we mean by $f^{(m)}$ the partial derivative $\partial^mf/\partial x^m$. A quadrature formula is said to be best for a given system $\tau_\alpha$ if it realizes the infimum.

Set
$$e(\alpha,W):=\inf_{-1\le x_1<\ldots<x_n\le1}e(\tau_\alpha,W).$$
If the infimum is attained at the points $-1\le x_1^0<\ldots<x_n^0\le1$, then the best quadrature formula for this system of points, with multiplicities $\alpha=(\alpha_1,\ldots,\alpha_n)$, is said to be optimal for the given $\alpha$. The points $x_1^0<\ldots<x_n^0$ are also called optimal.

It was proved in \cite{2,3} that
\begin{gather*}
e(\alpha,BH_\infty(\mathcal E_c))=\inf_{-\sqrt k<z_1<\ldots<z_n<\sqrt k}\int_{-\sqrt k}^{\sqrt k}|B(z)|\,d\nu_k(z),\\
e(\alpha,Bh_\infty(\mathcal E_c))=\frac4\pi\inf_{-\sqrt k<z_1<\ldots<z_n<\sqrt k}\int_{-\sqrt k}^{\sqrt k}\arctan|B(z)|\,d\nu_k(z),
\end{gather*}
where
$$B(z):=\prod_{j=1}^n\left(\frac{z-z_j}{1-z_jz}\right)^{\mu_j},\quad\mu_j:=2[(\alpha_j+1)/2]$$
(here the brackets denote the integral part) and $k$ is determined by \eqref{8}. Using Corollary~\ref{C2} and the results of \cite{2,3} we obtain the following theorem.

\begin{theorem}\label{T6}
Let $q$ be an even positive integer. Then for all $c>1:$
\begin{itemize}
\item[(i)] for all $q-1\le\alpha_j\le q$
\begin{align*}
e(\alpha,BH_\infty(\mathcal E_c))&=\frac\pi\Lambda\lambda^{q/2}I_{q0}(\lambda)=2^{q/2}\pi
\frac{(q-1)!!}{(q/2)!}c^{-qn}+O\left(c^{-(q+4)n}\right),\\
e(\alpha,Bh_\infty(\mathcal E_c))&=\frac4\Lambda I_{q2}(\lambda)=2^{q/2+2}
\frac{(q-1)!!}{(q/2)!}c^{-qn}+O\left(c^{-(q+4)n}\right),
\end{align*}
\end{itemize}
where $\lambda$ is determined by \eqref{17}, and the unique system of optimal knots is the Chebyshev system \eqref{9}$;$
\begin{itemize}
\item[(ii)] for $\alpha_j\le2$ the quadrature formulas
\begin{align*}
\int_{-1}^1f(x)\frac{dx}{\sqrt{1-x^2}}&\approx\pi\frac{1-\Delta_n(c)}n\sum_{j=1}^nf\left(\cos
\frac{2j-1}{2n}\pi\right),\\
\int_{-1}^1f(x)\frac{dx}{\sqrt{1-x^2}}&\approx\pi\frac{1-\delta_n(c)}n\sum_{j=1}^nf\left(\cos
\frac{2j-1}{2n}\pi\right),
\end{align*}
\end{itemize}
where
\begin{gather*}
\Delta_n(c):=\frac{\lambda^2}\Lambda I_{40}(\lambda)=6c^{-4n}+O\left(c^{-8n}\right),\\
\delta_n(c):=2\frac{\lambda^2}\Lambda\int_0^1\frac{t^4\,dt}{(1+\lambda^2t^4)
\sqrt{(1-\lambda^2)(1-\lambda^2t^2)}}=12c^{-4n}+O\left(c^{-8n}\right),
\end{gather*}
are optimal for the classes $BH_\infty(\mathcal E_c)$ and $Bh_\infty(\mathcal E_c)$, respectively.
\end{theorem}


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\end{thebibliography}
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