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\begin{document}

\title{Optimal Recovery of the Derivative\\
of Periodic Analytic Functions\\
from Hardy Classes}
\author{Konstantin Yu.~Osipenko\thanks{This research was supported in part
by RFBR Grants \#96-01-00325 and \#96-15-96072.}\\[10pt]}

\date{}
\maketitle

\vspace{-40pt}
\begin{center}\small\it
Department of Mathematics, MATI --- Russian State University of Technology,
Moscow 103767, Petrovka 27, Russia
\end{center}

\begin{abstract}
Let $S_\beta:=\{z\in\bbbc:|\Im z|<\beta\}$. For $2\pi$-periodic functions
which are analytic in $S_\beta$ with $p$-integrable boundary values, we
construct an optimal method of recovery of $f'(\xi)$, $\xi\in S_\beta$,
using information about the values $f(z_1),\ldots,f(z_n)$, $z_j\in[0,2\pi)$
.
\end{abstract}

\begin{center}\sc
Introduction
\end{center}

Let $X$ and $Y$ be linear spaces, $L$ a linear functional on $X$, and $I
\colon X\to Y$ a linear operator (which is usually called an information
operator). Suppose that $W\subset X$. Consider the problem of the optimal
recovery of $Lx$, $x\in W$, on the basis of the information $Ix$. The value
\begin{equation}\label{pr}
e(L,W,I):=\infp_{F}\sup_{x\in W}|Lx-F(Ix)|,
\end{equation}
where $F\colon Y\to\bbbc$ are any functionals (not necessarily linear or
continuous) is called the intrinsic error. A functional $F_0$ for which
$$\sup_{x\in W}|Lx-F_0(Ix)|=e(L,W,I)$$
is said to be an optimal algorithm or optimal method.

General settings of recovery problems can be found in \cite{MR, MR2, TW,
MO}.

Denote by $\Hp$, $1\le p\le\infty$, the space of all $2\pi$-periodic
functions $f$, which are analytic in $S_\beta:=\{z\in\bbbc:|\Im z|<\beta\}$
and satisfy
$$\eqalign{\|f\|_{\Hp}&:=\sup_{0\le\eta<\beta}\left(\frac1{4\pi}\int_0^{2
\pi}(|f(t+i\eta)|^p+|f(t-i\eta)|^p)\,dt\right)^{1/p}<\infty,\cr
&\hspace{285pt}1\le p<\infty,\cr
\|f\|_{\Hi}&:=\sup_{z\in S_\beta}|f(z)|<\infty.}$$
Set
$$\hp:=\{\,f\in\Hp:\|f\|_{\Hp}\le1\,\}.$$
We will consider the problem (\ref{pr}) for $W=\hp$, $Lf=f'(\xi)$, $\xi\in
S_\beta$, and
$$If=(f(x_1),\ldots,f(x_n)),$$
where $x_j$ are distinct points from $\bbbt:=[0,2\pi)$. In this case we
denote the intrinsic error (\ref{pr}) by $e'(\xi,\hp,I)$. From the
well-known Smolyak's formula (the complex version of this result was proved
in Osipenko \cite{Os0}) it follows that
\begin{equation}\label{Sm}
e'(\xi,\hp,I)=\sup_\at{f\in\hp}{If=0}|f'(\xi)|.
\end{equation}

For the unit ball $H_p$ of the Hardy space of nonperiodic functions
analytic in the unit disk the analogous problem of optimal recovery was
solved in Micchelli, Rivlin \cite{MR,MR2} ($p=\infty$) and Osipenko,
Stessin \cite{OsS} ($1\le p<\infty$). The problem of recovery of $f^{(k)}(
\xi)$ in $H_p$ was considered in Osipenko \cite{Os}. An interesting
extremal problem concerning minimization of the intrinsic error by choosing
points $x_1,\ldots,x_n$ was studied by Rivlin, Ruscheweyh, Shaffer, Wirths
\cite{Ri}. Several results relating to optimal recovery of $f'(\xi)$, $f
\in H_p$, from inaccurate values of $f$ can be found in Osipenko, Stessin
\cite{OsS1,OsS2}. An optimal method of recovery of $f(\xi)$, $f\in\hp$, was
recently obtained by Osipenko, Wilderotter \cite{OsW}.

In Section 1 we construct an optimal method of recovery of $f'(\xi)$, $f\in
\hp$ and calculate the appropriate intrinsic error. In Section~2 we examine
the intrinsic error of optimal recovery for the classes $H_{\infty,\beta}$
and $H_{2,\beta}$ in the case where the values of functions are known at
equidistant nodes.

\begin{center}\sc
1. Optimal method of recovery
\end{center}

Extremal problems for periodic analytic functions are often solved in
terms of elliptic functions (see, for example, \cite{Os1,OsW}). We shall
recall some notions from this theory. The Jacobi elliptic function $w=\sn(z
,k)$ is defined by the equation
$$z=\int_0^w\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}.$$
We shall also 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 $k':=\sqrt{1-k^2}$:
$$K:=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}},\quad K':=\int_0^1\frac{dt
}{\sqrt{(1-t^2)(1-k'^2t^2)}}.$$
We always assume that the modulus $k$ is defined from the equation
$$\frac{\pi K'}{2K}=\beta.$$
It can be shown (see, for example, Akhiezer \cite{Ak}) that
$$k=4e^{-\beta}\left(\frac{\sum_{m=0}^\infty e^{-2\beta m(m+1)}}{1+2\sum_{m
=1}^\infty e^{-2\beta m^2}}\right)^2.$$
Henceforth we shall not note the dependence of the Jacobi elliptic
functions on the modulus $k$.

In what follows all expressions with $p$ for $p=\infty$ are considered in
their limits as $p\to\infty$.

Set
$$W(z):=k^{n/2}\prod_{j=1}^n\sn\kp(z-x_j),\quad\omega_j(z):=\prod_\at{s=1}{
s\ne j}^n\sn\kp(z-x_s).$$
Assume that $\xi\notin\{x_1,\ldots,x_n\}$ and consider the equation
\begin{equation}\label {gam}
\sn\gamma\left(\frac1{\sn(\gamma+K)}+k^2\frac{p-2}p\sn(\gamma+K)\right)=
\frac\pi K\frac{W'(\xi)}{W(\xi)}.
\end{equation}
Denote the function in the left hand side of (\ref{gam}) by $s(\gamma)$.
Since $s(\gamma)$ is a continuous function in $(-K,K)$ and $s(\gamma)\to\pm
\infty$ as $\gamma \to\pm K$ there exists a solution of (\ref{gam}) $\gamma
_0\in(-K,K)$. For $\xi\in\{x_1,\ldots,x_n\}$ put $\gamma_0=K$.

Set $x_0:=\xi-\pi\gamma_0/K$,
$$\displaylines{w(z):=k\sn\kp(z-\xi)\sn\kp(z-\xi+\pi),\cr
T_1:=\left\{\,\zeta\in\bbbt:\frac\pi{2kK}|W'(\zeta)|<\frac{p-1}p|W(\zeta)|
\,\right\},\qquad T_0:=\bbbt\setminus T_1,\cr
b:=\cases{\dfrac p{p-1}\,\dfrac\pi{2kK}\dfrac{W'(\xi)}{W(\xi)},&$\xi\in T_1
$, $n=2m$,\cr
\noalign{\vskip5pt}
\dfrac{W(\xi)\sign W'(\xi)}{\dfrac\pi{2kK}|W'(\xi)|+\sqrt{\left(\frac\pi{2kK
}W'(\xi)\right)^2-\dfrac{p-2}pW^2(\xi)}},&$\xi\in T_0$, $n=2m$,\cr
\noalign{\vskip5pt}
-w(x_0),&$n=2m-1$,}\cr
\noalign{\vskip5pt}
u_\xi(z):=\cases{1,&$\xi\in T_1$, $n=2m$,\cr
\noalign{\vskip2pt}
\dfrac{w(z)+b}{1+bw(z)},&$\xi\in T_0$, $n=2m$,\cr
\noalign{\vskip2pt}
\dfrac{w(z)+b}{1+bw(z)}\left(\sqrt k\sn\kp(z-x_0)\right)^{-1},&$n=2m-1$.}}
$$
\begin{theorem} For all $1\le p\le\infty$ the method
$$f'(\xi)\approx\sum_{j=1}^nc_j(\xi)f(x_j),$$
where for $\xi\ne x_j$
$$c_j(\xi)=-\frac{\pi\alpha(\xi)}{2k^{n/2+1}K}\,\frac{u_\xi(x_j)(1+bw(x_j))
^{\frac{2(p-1)}p}\dn^{\frac{2(p-1)}p}\kp(\xi-x_j)}{\omega_j(x_j)\sn^2\kp(
\xi-x_j)},$$
$$c_j(x_j)=\frac{\omega_j'(x_j)}{\omega_j(x_j)},$$
and
$$\alpha(\xi)=\cases{\dfrac{2kK^2}{\pi^2}\dfrac{W(\xi)}{u_\xi(\xi)},&$\xi
\notin\{x_1,\ldots,x_n\}$,\cr
\noalign{\vskip2pt}
\dfrac{2k^{\left[\frac{n+1}2\right]}K^2}{\pi^2}\omega_j(x_j),&$\xi=x_j$, $j
=1,\ldots,n$,}$$
is an optimal method of recovery on the class $\hp$.
Moreover, the following equality holds
$$e'(\xi,\hp,I)=\cases{\dfrac k2\left(\dfrac{2K}\pi\right)^{\frac{p+1}p}
\dfrac{|W(\xi)|}{|u_\xi(\xi)|}(1+b^2)^{\frac{p-1}p},&$\xi\notin\{x_1,\ldots
,x_n\}$,\cr
\noalign{\vskip2pt}
\dfrac{k^{\left[\frac{n+1}2\right]}}2\left(\dfrac{2K}\pi\right)^{\frac{p+1}p
}|\omega_j(x_j)|,&$\xi=x_j$.}$$
\end{theorem}
\begin{proof}
The function
$$v(z):=\sqrt k\sn\kp z$$
is analytic in $S_\beta$. Moreover, $v(z+2\pi)=-v(z)$ and $|v(x+i\beta)|
\equiv1$ for all $x\in\bbbr$. Thus $\overline{W(z)}=W^{-1}(z)$ for $z\in
\partial S_\beta$. Using the definition of $b$, it can be shown that $b\in[
-1,1]$. Consider the function
$$g(z):=\frac{w(z)+b}{1+bw(z)}\frac{W(z)}{u_\xi(z)}(1+bw(z))^{2/p}\dn^{2/p}
\kp(z-\xi).$$
Since $\dn\kp z$ and $w(z)$ are $2\pi$-periodic, $|w(z)|<1$, $z\in S_\beta
$, and $\dn\kp z$ does not vanish in $S_\beta$, $g\in\Hp$.

For $f\in\hp$ and $1\le p<\infty$ set
\eqlines{J}{Jf:=\frac{\alpha(\xi)}{4\pi}\int_0^{2\pi}\left(\overline{g(x+i
\beta)}|g(x+i\beta)|^{p-2}f(x+i\beta)\right.}
{\left.+\overline{g(x-i\beta)}|g(x-i\beta)|^{p-2}f(x-i\beta)\right)\,dx.}
Using the properties of elliptic functions, we have for all $x\in\bbbr$
\begin{equation}\label{dn}
\overline{\dn\kp(x\pm i\beta)}=\pm i\frac{\cn\kp(x\pm i\beta)}{\sn\kp(x
\pm i\beta)}.
\end{equation}
The element of integration in $Jf$ is a $2\pi$-periodic function.
Consequently, we can rewrite $Jf$ in the form
\eqlines{Jf}{Jf:=\frac{\alpha(\xi)}{4\pi i}\int_{\Gamma_\varepsilon}\frac{u
_\xi(z)(1+bw(z))^{\frac{2(p-1)}p}}{W(z)w(z)}\dn^{\frac{p-2}p}\kp(z-\xi)
\frac{\cn\kp(z-\xi)}{\sn\kp(z-\xi)}f(z)\,dz}
{=\frac{\alpha(\xi)}{4\pi i}\int_{\Gamma_\varepsilon}\frac{u_\xi(z)(1+bw(z)
)^{\frac{2(p-1)}p}\dn^{\frac{2(p-1)}p}\kp(z-\xi)}{kW(z)\sn^2\kp(z-\xi)}f(z)
\,dz,}
where $\Gamma_\varepsilon$ is the boundary of rectangle $-\varepsilon<\Re z
<2\pi-\varepsilon$, $|\Im z|<\beta$, and $\varepsilon$ is such that $\xi,x_
1,\ldots,x_{2n}$ lie inside this rectangle. Assume that $\xi\notin\{x_1,
\ldots,x_{2n}\}$. By the residue theorem
$$Jf=f'(\xi)+Cf(\xi)-\sum_{j=1}^{2n}c_j(\xi)f(x_j),$$
where
\lines{C=\frac{\alpha(\xi)}{2k}\lim_{z\to\xi}\left(\frac{(z-\xi)^2u_\xi(z)(
1+bw(z))^{\frac{2(p-1)}p}\dn^{\frac{2(p-1)}p}\kp(z-\xi)}{W(z)\sn^2\kp(z-\xi
)}\right)'}
{=\frac{W(\xi)}{u_\xi(\xi)}\left(\frac{u_\xi(z)(1+bw(z))^{\frac{2(p-1)}p}}{
W(z)}\right)'_{\Big|z=\xi}.}
It is not hard to check that $b$ is defined from the condition $C=0$. Thus
we have
\begin{equation}\label{pred}
Jf=f'(\xi)-\sum_{j=1}^nc_j(\xi)f(x_j).
\end{equation}

From (\ref J) and H\"older's inequality
$$|Jf|\le|\alpha(\xi)|\|g\|_{\Hp}^{p-1}.$$
Hence
$$e'(\xi,\hp,I)\le|\alpha(\xi)|\|g\|_{\Hp}^{p-1}.$$
On the other hand for $g_0:=g/\|g\|_{\Hp}$ we have $Jg_0=g_0'(\xi)$. Using
equality (\ref{Sm}), we obtain
$$e'(\xi,\hp,I)\ge|g_0'(\xi)|=|Jg_0|=|\alpha(\xi)|\|g\|_{\Hp}^{p-1}.$$
Thus
$$e'(\xi,\hp,I)=|\alpha(\xi)|\|g\|_{\Hp}^{p-1}.$$
To calculate $\|g\|_{\Hp}$ substitute $f(z)=g(z)$ in (\ref{Jf})
\eqlines{norm}
{\alpha(\xi)\|g\|_{\Hp}=\frac{\alpha(\xi)}{4\pi i}\int_{\Gamma_\varepsilon}
\frac{(w(z)+b)(1+bw(z))\dn^2\kp(z-\xi)}{k\sn^2\kp(z-\xi)}\,dz}{=\alpha(\xi)
\frac\pi{2K}(1+b^2)}
(we omit here some technical details concerned with the application of the
residue theorem). Consequently,
$$\|g\|_{\Hp}=\left(\frac\pi{2K}(1+b^2)\right)^{1/p}$$
and
$$e'(\xi,\hp,I)=\dfrac k2\left(\dfrac{2K}\pi\right)^{\frac{p+1}p}\dfrac{|W(
\xi)|}{|u_\xi(\xi)|}(1+b^2)^{\frac{p-1}p}.$$

If $\xi=x_j$, then $b=0$ and $g(z)=w(z)W(z)u_\xi^{-1}(z)\dn^{2/p}\kp(z-x_j)
$. In this case the assertion of the theorem can be obtained by the same
scheme.

For $p=\infty$ consider the integral
\lines{Jf:=\frac{\alpha(\xi)}{4\pi}\int_0^{2\pi}\left(\overline{g(x+i\beta)}
\varphi(x+i\beta)f(x+i\beta)\right.}
{\left.+\overline{g(x-i\beta)}\varphi(x-i\beta)f(x-i\beta)\right)\,dx,}
where
$$\varphi(z)=\left|(1+bw(z))\dn\kp(z-\xi)\right|^2.$$
The representation (\ref{pred}) follows from (\ref{dn}) and the residue
theorem. We have $$|Jf|\le|\alpha(\xi)|\|\varphi\|_{{\cal
H}_{1,\beta}}.$$ Taking in account that $\varphi(z)\ge0$, we obtain
$$|g'(\xi)|=|Jg|=|\alpha(\xi)|\|\varphi\|_{{\cal H}_{1,\beta}}.$$
Hence
$$e'(\xi,H_{\infty,\beta},I)=|\alpha(\xi)|\|\varphi\|_{{\cal H}_{1,\beta}}.
$$
Using the residue theorem, by analogy with (\ref{norm}) we obtain
$$\|\varphi\|_{{\cal H}_{1,\beta}}=\frac\pi{2K}(1+b^2).\quad\rule{5pt}
{12pt}$$
\end{proof}

Let us consider our problem in the case when $\xi=0$ and
$$If=I_hf:=(f(-h),f(h)),\quad h\in(0,\pi).$$
In other words we wish to construct an optimal formula of numerical
differentiation at the point $\xi=0$, using the information about values of
function at the points $\pm h$.

In this particular case we have
$$W(z)=k\sn\kp(z+h)\sn\kp(z-h),\quad W(0)=-k\sn^2\kp h,\quad W'(0)=0.$$
Moreover, $0\in T_1$ and $b=0$. Thus we obtain that an optimal method has
the form
$$f'(0)\approx\kp\frac{f(h)-f(-h)}{\sn\dfrac{2K}\pi h}\dn^{\frac{2(p-1)}p}
\kp h$$
and
$$e'(0,\hp,I_h)=\frac{k^2}2\left(\frac{2K}\pi\right)^{\frac{p+1}p}\sn^2\kp
h=k^22^{1/p}\left(\kp\right)^{\frac{3p+1}p}h^2+O(h^4).$$

\begin{center}\sc
2. Optimal recovery using an equidistant system of points
\end{center}

For optimal recovery of periodic functions the most natural system of
points is an equidistant system. We will estimate the error of optimal
recovery of the derivative from the information
$$If=I^{(2n)}f:=(f(t_1^0),\ldots,f(t_{2n}^0)),$$
where
$$t_j^0=(j-1)\frac\pi n,\quad j=1,\ldots,2n.$$
Set
$$e'_{2n}(\hp):=\sup_{\xi\in\bbbt}e'(\xi,\hp,I^{(2n)}).$$
\begin{theorem}
For all $\beta>0$
$$\displaylines{e'_{2n}(H_{\infty,\beta})=\sqrt\lambda\frac{2n\Lambda}\pi=2
ne^{-\beta n}+O(ne^{-5\beta n}),\cr
e'_{2n}(H_{2,\beta})=\sqrt{\frac{2K\lambda}\pi}\frac{2n\Lambda}\pi=\sqrt{
\frac{2K}\pi}2ne^{-\beta n}+O(ne^{-5\beta n}),}$$
where
$$\lambda=4e^{-2\beta n}\left(\frac{\sum_{m=0}^\infty e^{-4\beta nm(m+1)}}{
1+2\sum_{m =1}^\infty e^{-4\beta nm^2}}\right)^2$$
and $\Lambda$ is the complete elliptic integral of the first kind for
modulus $\lambda$.
\end{theorem}
\begin{proof}
Using the first principal transform of elliptic functions of degree $2n$
(see \cite{Ak}), we find
\multlines{W\left(z-\frac\pi{2n}\right)=k^n\prod_{j=1}^{2n}\sn\left(\kp z-
\frac{ 2j-1}{2n}K\right)}
{=(-1)^nk^n\prod_{j=1}^n\sn\left(\kp z-\frac{2j-1}{2n}K\right)\sn\left(\kp
z+\frac{2j-1}{2n}K\right)}
{=k^n\prod_{j=1}^n\frac{\sn^2\dfrac{2j-1}{2n}K-\sn^2\kp z}{1-k^2\sn^2\dfrac
{2j-1}{2n}K\sn^2\kp z}=\sqrt\lambda\sn\left(\frac{2n\Lambda}\pi z+\Lambda,
\lambda\right).}
Hence
$$W(z)=-\sqrt\lambda\sn\left(\frac{2n\Lambda}\pi z,\lambda\right).$$
In view of the equalities
$$\frac d{dt}\sn(t,\lambda)=\cn(t,\lambda)\dn(t,\lambda)=\sqrt{(1-\sn^2(t,
\lambda))(1-\lambda^2\sn^2(t,\lambda))},$$
from Theorem~1 we obtain
$$e'_{2n}(\hp)=\sup_{s\in[0,1]}\frac k2\left(\frac{2K}\pi\right)^{\frac{1+p
}p}\sqrt\lambda\Phi_p(s),$$
where
$$\displaylines{
\Phi_p(s)=\cases{s\left(1+\left(\dfrac{pa}{p-1}\right)^2\dfrac{(1-s^2)(1-
\lambda^2s^2)}{s^2}\right)^{\frac{p-1}p},&$s\in S_p$\cr
\gamma(s)\left(1+\dfrac{s^2}{\gamma^2(s)}\right)^{\frac{p-1}p},&$s\in[0,1]
\setminus S_p$}\cr
\noalign{\vskip5pt}
a=\frac{n\Lambda}{kK},\quad S_p=\left\{s\in[0,1]:a^2(1-s^2)(1-\lambda^2
s^2)<\left(\frac{p-1}p\right)^2s^2\right\},\cr
\noalign{\vskip5pt}
\gamma(s)=a\sqrt{(1-s^2)(1-\lambda^2s^2)}+\sqrt{a^2(1-s^2)(1-\lambda^2s^2)-
\frac{p-2}ps^2}.}$$

Let us begin with the case $p=2$. It is easy to check that
$$\Phi_2(s)=\sqrt{s^2+4a^2(1-s^2)(1-\lambda^2s^2)}.$$
From properties of the first principal transformations of elliptic
functions of degree $2n$ it follows that
\begin{equation}\label {a}
\frac{2n\Lambda}K=\prod_{j=1}^n\frac{\sn^2\left(\dfrac jnK\right)}{\sn^2
\left(\dfrac{2j-1}{2n}K\right)}>1.
\end{equation}
Hence $2a>1$ and
$$\Phi_2^2(s)\le s^2+4a^2(1-s^2)\le4a^2.$$
This estimate is attained for $s=0$. Thus
$$e'_{2n}(\hp)=\sqrt{\frac{2K}\pi}\sqrt\lambda\frac{2n\Lambda}\pi.$$
The asymptotic equality follows from the equations
$$\eqalign{\sqrt\lambda&=2e^{-\beta n}+O\left(e^{-5\beta n}\right),\cr
\Lambda&=\frac\pi2+O\left(e^{-4\beta n}\right).}$$

Let $p=\infty$. It can be easily shown that $S_\infty=(s^*,1]$ where $s^*$
is the unique solution of the equation
$$a^2(1-s^2)(1-\lambda^2s^2)=s^2.$$
We have
$$\Phi_\infty(s)=\cases{2a\sqrt{(1-s^2)(1-\lambda^2s^2)},&$s\in[0,s^*]$\cr
s+a^2\dfrac{(1-s^2)(1-\lambda^2s^2)}s,&$s\in(s^*,1]$}$$
Since the function
$$F(s):=s+a^2\dfrac{(1-s^2)(1-\lambda^2s^2)}s$$
is convex for $s\in(0,1)$ we obtain
$$\max_{s\in[s^*,1]}F(s)=\max\{F(s^*),F(1)\}=\max\{\Phi_\infty(s^*),1\}.$$
The function $\Phi_\infty(s)$ decreases while $s\in[0,s^*]$. Consequently
$$\max_{s\in[0,1]}\Phi_\infty(s)=\max\{\Phi_\infty(0),1\}=2a.\quad\rule{5pt}
{12pt}$$
\end{proof}


\begin{thebibliography}{99} \small

\bibitem{Ak}{\sc N. I. Akhiezer}, ``Elements of the Theory of Elliptic
Functions,'' Nauka, Moscow, 1970.

\bibitem{MO}{\sc G.~G.~Magaril-Il'yaev and K.~Yu.~Osipenko}, Optimal
recovery of functionals based on inaccurate data, {\it Mat. Zametki} {\bf
50} (1991), 85--93; English translation {\it Math. Notes} {\bf50} (1991),
1274--1279.

\bibitem{MR}{\sc C. A. Micchelli and T. J. Rivlin}, A survey of optimal
recovery, {\it in} ``Optimal Estimation in Approximation Theory''
(C.~A.~Micchelli and T.~J.~Rivlin, Eds.), pp.~1--54, Plenum Press, New
York, 1977.

\bibitem{MR2}{\sc C. A. Micchelli and T. J. Rivlin}, ``Lectures on Optimal
Recovery,'' Lecture Notes in Math., Vol.~1129, pp.~21--93, Springer--Verlag,
New York/Berlin, 1985.

\bibitem{Os0}{\sc K. Yu. Osipenko}, Best approximation of analytic functions
from information about their values at a finite number of points, {\it Mat.
Zametki} {\bf19}, (1976), 29--40; English translation {\it Math. Notes}
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\end{thebibliography}

\end{document}