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\begin{document}
\selectlanguage{english}
\title[On optimal extrapolation and interpolation]{On optimal extrapolation and interpolation
of fuzzy analytic functions}
\author{K. Yu.\ Osipenko}




\maketitle

\refstepcounter{section}
\section*{\S\arabic{section}. Introduction}

Let $\mathcal B$ denote the class of analytic functions on the unit disc $K=\{z:|z|<1\}$ with
modulus at most $1$. We consider the problem of finding the best approximation of a function $f\in \mathcal B$ at some point $z_0\in K$, knowing its values at the points $z_l,\ldots,z_n\in K$ with an error $\le\delta$, i.e. under the condition of knowing some values $f_1,\ldots,f_n$ which satisfy the inequalities
$$|f(z_j)-f_j|\le\delta,\quad j=1,\ldots,n.$$

We call
\begin{equation}\label{11}
r(z_0,z_l,\ldots,z_n,\delta)=\infp_S\sup_{f\in\mathcal B}\sup_{\substack{f_1,\ldots,f_n\\|f(z_j)-f_j|\le\delta,\ j=1,\ldots,n}}|f(z_0)-S(f_1,\ldots,f_n)|
\end{equation}
the error of the best approximation, where the infimum is taken on the set of all functions (called methods) $S\colon\mathbb C^n\to\mathbb C$. A method for which the infimum indicated in \eqref{11} is attained, is called optimal.

Let $E$ be a subset of $K$. We call
\begin{equation}\label{12}
r_n(\delta,E)=\infp_{z_l,\ldots,z_n\in E}\sup_{z_0\in E}r(z_0,z_l,\ldots,z_n,\delta)
\end{equation}
the error of the optimal interpolation on $E$ for $n$ points. Those knots for which the infimum in \eqref{12} is attained are called optimal knots of interpolation.

Problems \eqref{11}, \eqref{12} and similar ones were investigated for $\delta=0$ in \cite{1}, \cite{2}, \cite{5,6,7}, \cite{10,11,12}. The case $\delta\ge0$ of \eqref{11} was dealt with in \cite{10} (for $n=1$) and in \cite{8} (for $z_0,z_l,\ldots,z_n\in(-1,1)$). In these papers it was shown that
\begin{equation}\label{13}
r(z_0,z_l,\ldots,z_n,\delta)=\sup_{\substack{f\in\mathcal B\\|f(z_j)|\le\delta,\ j=1,\ldots,n}}|f(z_0)|
\end{equation}
(cf. also \cite{5}).

Set
\begin{gather}\label{14}
R_n(z_0,\delta,E)=\inf_{z_l,\ldots,z_n\in E}r(z_0,z_l,\ldots,z_n,\delta),\\
R(z_0,\delta,E)=\inf_nR_n(z_0,\delta,E).\label{15}
\end{gather}
It is worth while considering the problems \eqref{14} and \eqref{15} only for $z_0\notin E$ and $0<\delta<1$, since else their solution is obvious. We call $R_n(z_0,\delta,E)$ the error of the optimal extrapolation at $z_0$ for $n$ points from $E$, and the points where the infimum \eqref{14} is attained will be called optimal knots of extrapolation for the given $n$.

If there exist numbers $n$ for which the infimum \eqref{15} is attained, then the smallest of them is called the informativity order of $E$, and will be denoted by $\In(z_0,\delta,E)$. In the opposite case we say that $E$ has infinite informativity order. The optimal knots for $n=\In(z_0,\delta,E)$ will be called optimal knots of extrapolation on $E$ (for given $z_0$ and $\delta$). Thus, the informativity order of $E$ is the minimal number of points from $E$ giving the best possible extrapolation of a function in a point outside $E$ provided the function is known on $E$ with error $\delta$.

Problems \eqref{14} and \eqref{15} for $E=[-l,0]\subset(-1,1)$, $E=(-1,0)$ and $z_0\in(0,1)$ were investigated in \cite{9}. In the present paper we deal with the same problems for arbitrary closed sets $E\subset(-1,1)$ and $z_0\in(-1,1)\setminus E$, as well as with \eqref{12} for $E=[\alpha,\beta]\subset(-1,1)$ and for $\delta$ near to $1$.

\refstepcounter{section}
\section*{\S\arabic{section}. Optimal extrapolation}

Let $E$ be a closed subset of $K$. Put
$$\|f\|_E=\max_{z\in E}|f(z)|.$$
First we discuss the problem of Blaschke products
\begin{equation}\label{21}
B_n(z)=\lambda\prod_{j=1}^n\frac{z-\alpha_j}{1-\ov\alpha_jz},\quad|\alpha_j|<1,
\quad|\lambda|=1,
\end{equation}
of order $n$ with minimal norm
\begin{equation}\label{22}
\delta_n(E)=\inf_{B_n}\|B_n\|_E,
\end{equation}
on $E$.

The quantity $\delta_n(E)$ was dealt with in \cite{4}, \cite{6} for $E=[a,b]\subset(-1,1)$. In \cite{7} it was proved that for every closed set $E\subset K$ the inequalities
\begin{gather}\label{23}
\delta_n(E)\ge\exp\left[-\frac n{c(E)}\right],\quad n\ge1,\\
\delta_n(E)<\exp\left[-\frac n{c(E)+\varepsilon}\right]\label{24}
\end{gather}
hold for arbitrary $\varepsilon>0$ and for sufficiently large $n$, where $c(E)$ is the capacity of the condenser $(E,\mathbb C\setminus K)$. Let us also mention that problem \eqref{22} is closely connected with the problem of widths of classes of analytic functions (cf.\ \cite{2}).

We assume that$E\subset(-1,1)$. Let
$$\rho=\max_{z\in E}|z|,\quad\alpha^*=\min\{\rho,|\RE\alpha|\}\sign\RE\alpha.$$
It is easy to see that
\begin{equation}\label{25}
\left|\frac{z-\alpha}{1-\ov\alpha z}\right|\ge\left|\frac{z-\alpha^*}{1-\alpha^*z}\right|
\end{equation}
for all $z\in[-\rho,\rho]\supset E$ and all $\alpha\in K$. Hence, in determining the number \eqref{22}, we can confine ourselves to Blaschke products \eqref{21} with $\alpha_j\in[-\rho,\rho]$.

From the inequality
$$\left|\frac{z-\alpha}{1-\alpha z}-\frac{z-\beta}{1-\beta z}\right|\le\frac1{1-\rho^2}|\alpha-\beta|,$$
which holds for every $\alpha,\beta,z\in[-\rho,\rho]$, we obtain
\begin{multline*}
|B_n^1-B_n^2|=\left|B_{n-1}^1\left(\frac{z-\alpha_n}{1-\alpha_nz}-\frac{z-\beta_n}{1-\beta_nz}
\right)+(B_{n-1}^1-B_{n-1}^2)\frac{z-\beta_n}{1-\beta_nz}\right|\\
\le\frac1{1-\rho^2}|\alpha_n-\beta_n|+|B_{n-1}^1-B_{n-1}^2|\le\ldots\le\frac1{1-\rho^2}
\sum_{j=1}^n|\alpha_j-\beta_j|.
\end{multline*}
here
$$B_k^1=\prod_{j=1}^k\frac{z-\alpha_j}{1-\alpha_jz},\quad B_k^2=\prod_{j=1}^k\frac{z-\beta_j}{1-\beta_jz},\quad k=1,\ldots,n.$$
Hence it follows that the function
$$\varphi(\alpha_1,\ldots,\alpha_n):=\max_{z\in E}\biggl|\prod_{j=1}^k\frac{z-\alpha_j}{1-\alpha_jz}\biggr|$$
is continuous for $(\alpha_1,\ldots,\alpha_n)\in[-\rho,\rho]^n$ and, consequently, there exists a Blaschke product having real zeros, for which the infimum \eqref{22} is attained. As the inequality \eqref{25} is strict for $\alpha^*\ne\alpha$, every Blaschke product of order $n$ for which this infimum is attained has its real zeros in the interval $[-\rho,\rho]$.

\begin{lemma}\label{L21}
Suppose that the closed set $E\subset(-1,1)$ consists of more than $n$ elements and let $B_n^*(z)$ be a Blaschke product, normed by the condition $B_n^*(1)=1$, such that
$$\delta_n(E)=\|B^*\|_E.$$
If $f\in\mathcal B$ is a function real on the interval $(-1,1)$, such that there exist $n+1$ points $z_1<z_2<\ldots<z_{n+1}$ in $E$ with the property
$$f(z_j)f(z_{j+1})<0,\quad j=1,\ldots,n,$$
then
\begin{equation}\label{26}
\min_{1\le j\le n+1}|f(z_j)|\le\delta_n(E).
\end{equation}
Besides that, \eqref{26} turns into equality only if $f(z)\equiv\lambda B_n^*$ where $\lambda=\pm1$.
\end{lemma}

\begin{proof}
The results of \cite[\S10.2]{13} concerning the interpolation of functions from the class $\mathcal B$ imply the existence of a Blaschke product $B_m(z)$ of order $m\le n+1$ which is real on the real axis and satisfies
$$B_m(z_j)=f(z_j),\quad j=1,\ldots,n+1;$$
moreover, $f(z)\equiv B_m(z)$ if $m\le n$, and for $m=n+1$ one can choose $B_m(z)$ normed either by $B_m(1)=1$ or by $B_m(1)=-1$. Suppose $|f(z_j)|\ge\delta_n(E)$, $j=1,\ldots,n+1$. Consider the function
$$\varphi(z)=B_m(z)-\lambda B_n^*(z)\sign B_m(1),$$
where $\lambda=1$ if $m\le n$, and $\lambda=-1$, $B_m(1)=-\sign f(z_{n+1})$ if $m=n+1$. In virtue of its normalization, $B_n^*(z)$ is real in $(-1,1)$ whence
\begin{equation}\label{27}
(-1)^{j+n+1}\varphi(z_j)\sign f(z_{n+1})\ge0,\quad j=1,\ldots,n+1.
\end{equation}
Thus, $\varphi(z)$ has at least $n$ zeros (with multiplicity) in the interval $(-1,1)$. On the other hand, by \cite{8}, Lemma~1.1, we have for the number $k$ of roots of the function $\varphi(z)\not\equiv0$ in the disc $K$
\begin{equation}\label{28}
k\le(m+n-l)/2,
\end{equation}
where $l$ is the number of roots of $\varphi(z)$ in the unit disc. If $m<n$ or $m=n$ (in this
latter case $\varphi(-1)=\varphi(1)=0$ whence $l\ge2$), then \eqref{28} contradicts the fact that $k\ge n$. Hence in these cases the only possibility is $\varphi(z)\equiv0$ which means that $f(z)$ coincides with $B_n^*(z)$ up to signs. If $m=n+1$, then the normalization of $B_m(z)$ implies
$$\varphi(1)=-2\sign f(z_{n+1}).$$
Taking into account \eqref{27}, we obtain that the number of zeros of $\varphi(z)$ in the interval $(-1,1)$ is at least $n+1$. This contradicts \eqref{28}, which proves the lemma.
\end{proof}

\begin{lemma}\label{L22}
Suppose that the closed set $E\subset(-1,1)$ consists of more than $n$ points. Then the solution of problem \eqref{22} is unique up to a factor $\lambda$, $|\lambda|=1$. The Blaschke product $B_n^*$ normed by $B_n^*(1)=1$ is solution of this problem iff there exist $n+1$ points $z_1<\ldots<z_{n+1}$ in $E$ such that
\begin{equation}\label{29}
B_n^*(z_j)=(-1)^{n+j+1}\|B_n^*\|_E,\quad j=1,\ldots,n+1.
\end{equation}
\end{lemma}

\begin{proof}
Let $B_n^*$ be an extremal function for the problem \eqref{22}, normed by $B_n^*(1)=1$. It has been proved that the zeros of $B_n^*$ are real. Let $\alpha_1\le\ldots\le\alpha_n$ be these zeros. We are going to show that they are all different and that every interval $(\alpha_{j-1},\alpha_j)$, $j=1,\ldots,n+1$ ($\alpha_0=-1$, $\alpha_{n+1}=1$), contains some point of $E$ in which $|B_n^*|=\|B_n^*\|_E$. Suppose this is not the case, and let
$$\max_{z\in[\alpha_{k-1},\alpha_k]\cap E}|B_n^*(z)|=\delta<\|B_n^*\|_E$$
for some $k$, $1\le k\le n+1$ (putting $\delta=0$ if $[\alpha_{k-1},\alpha_k]\cap E=\emptyset$). Set
$$\varphi(z)=\prod_{j=k-1}^k\frac{z-\alpha_j}{1-\alpha_jz}.$$
Choose $\varepsilon>0$ subject to $\delta+\varepsilon<\|B_n^*\|_E$ and consider the function
$$B_n(z)=g(z)\prod_{\substack{j=1\\j\ne k-1,k}}^n\frac{z-\alpha_j}{1-\alpha_jz}\quad\mbox{where}\quad g(z)=\frac{\varphi(z)-\varepsilon}{1-\varepsilon\varphi(z)}.$$
It is not difficult to show that
$$g(z)=\prod_{j=k-1}^k\frac{z-\beta_j}{1-\beta_jz}$$
where $[\alpha_{k-1},\alpha_k]\subset[\beta_{k-1},\beta_k]$. From the inequalities
\begin{gather*}
|g(z)|\le|\varphi(z)|+\varepsilon,\quad z\in[\alpha_{k-1},\alpha_k],\\
|g(z)|<|\varphi(z)|,\quad z\in(-1,1)\setminus(\beta_{k-1},\beta_k),\\
|g(z)|<\varepsilon<\|B_n^*\|_E,\quad z\in(\beta_{k-1},\alpha_{k-1})\cup(\alpha_k,\beta_k),
\end{gather*}
it follows that $\|B_n\|_E<\|B_n^*\|_E$. The contradiction obtained proves the existence of
points $z_j\in(\alpha_{j-1},\alpha_j)\cap E$, $j=1,\ldots,n+1$, which satisfy \eqref{29}. The converse statement and unicity both follow from Lemma~\ref{L21}. This completes the proof of the lemma.
\end{proof}

\begin{theorem}\label{T21}
Let $E$ be a closed subset of $(-1,1)$. The following assertions hold:

$1)$ for $\delta_n(E)\le\delta<\delta_{n-1}(E)$ $($$\delta_0(E)=1$$)$ and for arbitrary $z_0\in(-1,1)\setminus E$ the inequality
\begin{equation}\label{210}
n\le\In(z_0,E,\delta)\le n+1
\end{equation}
holds. Besides that, if $z_0\in(-1,1)\setminus\co E$ --- where $\co E$ denotes the convex hull of $E$ ---, then $\In(z_0,E,\delta)=n$;

$2)$ for every $\delta\in(0,1)$ and $z_0\in(-1,1)\setminus E$ the inequality
$$\In(z_0,E,\delta)\ge c(E)\ln1/\delta$$
holds, and
$$\lim_{\delta\to0}\frac{\In(z_0,E,\delta)}{\ln1/\delta}=c(E);$$
here $c(E)$ is the capacity of the condenser $(E,\mathbb C\setminus K)$.
\end{theorem}

\begin{proof}
Consider the generalized Heins problem of finding
\begin{equation}\label{211}
\sup_{\substack{f\in\mathcal B\\|f(z)|\le\delta,\ z\in E}}|f(z_0)|
\end{equation}
(cf.\ \cite{3}, \cite{4}, \cite{8}, \cite{9}). From \cite{3} it follows that there exists a Blaschke product of finite order $m$ which solves this problem. In \cite{8} it was proved that for the extremality of the Blaschke product $B_m(z)$ normed by the condition $B_m(z_0)>0$ it is
necessary and sufficient that the inequality $|B_m(z)|\le\delta$, $z\in E$ is fulfilled, and that elements $z_1<\ldots<z_m$ of $E$ exist such that
\begin{equation}\label{212}
B_m(z_j)=\begin{cases}(-1)^{p+j}\delta,&j=1,\ldots,p,\\
(-1)^{p+j+1}\delta,&j=p+1,\ldots,m,\end{cases}
\end{equation}
where $p$ is defined by $z_0\in(z_p,z_{p+1})$ and $p=0$ if $z_0\in(-1,z_1)$, $p=m$ if $z_0\in(z_m,1)$. Thus, we have for arbitrary $y_1,\ldots,y_n\in E$
\begin{multline}\label{213}
\sup_{\substack{f\in\mathcal B\\|f(y_j)|\le\delta,\ j=1,\ldots,n}}|f(z_0)|\ge\sup_{\substack{f\in\mathcal B\\|f(z)|\le\delta,\ z\in E}}|f(z_0)|\\
=\sup_{\substack{f\in\mathcal B\\|f(z_j)|\le\delta,\ j=1,\ldots,m}}|f(z_0)|=B_m(z_0),
\end{multline}
Hence and from \eqref{13} follows
\begin{equation}\label{214}
R(z_0,\delta,E)=R_m(z_0,\delta,E).
\end{equation}
In \cite{9} the continuity of the left hand side of \eqref{213} as a function of $y_1,\ldots,y_n$ was proved. Since $E$ is closed, the existence of optimal knots of extrapolation for arbitrary $n$ follows.

Let $n<m$, and let $y_1,\ldots,y_n$ be optimal knots of extrapolation for $n$. From the uniqueness of the extremal function $f$ normed by the condition $f(z_0)>0$ in the problem of finding the supremum on the left side of \eqref{213}, and from the fact that the
extremal function is a Blaschke product of order not exceeding $n$, we obtain
$$R_n(z_0,\delta,E)>R_m(z_0,\delta,E).$$
Thus, $m$ is the smallest number which satisfies \eqref{214}, i.e. $\In(z_0,E,\delta)=m$.

Suppose $\delta_n(E)\le\delta<\delta_{n-1}(E)$. The obvious relationship
\begin{equation}\label{215}
\delta_m(E)\le\|B_m\|_E=\delta
\end{equation}
yields $m\ge n$. By \eqref{212}, for $B_m$ there is an $(m-1)$-point alternation on $E$. Using
Lemma~\ref{L21}, we obtain
\begin{equation}\label{216}
\delta<\delta_{m-2}(E).
\end{equation}
If $z_0\in(-1,1)\setminus\co E$, then the cardinality of the alternation is $m$ and we have
\begin{equation}\label{217}
\delta<\delta_{m-1}(E).
\end{equation}
From \eqref{216} it follows that $m\ge n+1$, and from \eqref{217} $m\ge n$. This proves the first assertion.

Now choose an arbitrary $\varepsilon>0$. From \eqref{215} and \eqref{216}, taking into account
\eqref{23} and \eqref{24}, we obtain for sufficiently small $\delta$
$$\exp\left[-\frac m{c(E)}\right]\le \delta<\exp\left[-\frac{m-2}{c(E)+\varepsilon}\right].$$
These inequalities yield assertion $2)$, which completes the proof.
\end{proof}

\refstepcounter{section}
\section*{\S\arabic{section}. Optimal interpolation}

Consider problem \eqref{12} with $E=[\alpha,\beta]\subset(-1,1)$. If $\delta\ge1$, then $r_n(\delta,E)=1$; therefore we shall take for granted $\delta<1$. Define the pseudohyperbolic distance of points of the unit disc $K$ by
$$\rho(u,v)=\left|\frac{u-v}{1-\ov vu}\right|.$$

If $W(z)$ is any conform transformation of the disc $K$, then we have $r_n(\delta,E)=r_n(\delta,W(E))$. There is a conform transformation of $K$ which maps the interval $[\alpha,\beta]$ onto the symmetric interval $[-k,k]$. As conform transformations of the disc do not change pseudohyperbolic distances, we have
$$\rho(\alpha,\beta)=\rho(-k,k)=\frac{2k}{1+k^2}.$$
Thus,
\begin{equation}\label{31}
r_n(\delta,[\alpha,\beta])=r_n(\delta,[-k,k]),
\end{equation}
where
$$k=\frac{\rho(\alpha,\beta)}{1+\sqrt{1-\rho^2(\alpha,\beta)}}.$$
Thereby we have proved that, for fixed $n$ and $\delta$, $r_n(\delta,[\alpha,\beta])$ depends only on the psendohyperbolic distance $\rho(\alpha,\beta)$.

From \cite{9} the continuity of the function
$$\varphi(z_1,\ldots,z_n)=\sup_{z_0\in[\alpha,\beta]}r(z_0,z_1,\ldots,z_n,\delta)$$
follows for $z_1,\ldots,z_n\in[\alpha,\beta]$. This implies the existence of optimal knots of interpolation.

Denote the quantity \eqref{12} for $n\ge2$ by $r_n^*(\delta,[\alpha,\beta])$ provided $z_1=\alpha$, $z_n=\beta$.

\begin{lemma}\label{L31}
Let $0\le\delta<1$ and $\rho(\alpha_1,\beta_1)>\rho(\alpha,\beta)$. Then
\begin{gather}\label{32}
r_n(\delta,[\alpha_1,\beta_1])>r_n(\delta,[\alpha,\beta]),\\
r_n^*(\delta,[\alpha_1,\beta_1])>r_n^*(\delta,[\alpha,\beta]).\label{33}
\end{gather}
\end{lemma}

\begin{proof}
First we prove \eqref{32}. In virtue of \eqref{31} it suffices to prove
$$r_n(\delta,[-k_1,k_1])>r_n(\delta,[-k,k])$$
for $k_1>k$. Let $u_1,\ldots,u_n$ be optimal knots of interpolation for the problem \eqref{12} with $E=[-k_1,k_1]$, and let $u_0\in E$ be such that
\begin{equation}\label{34}
r(u_0,u_1,\ldots,u_n,\delta)=r_n(\delta,[-k_1,k_1]).
\end{equation}
Consider the points $z_j=ku_j/k_1$, $j=0,\ldots,n$. Obviously, $z_j\in[-k,k]$. Let $f^*(z)$ be
an extremal function for the problem \eqref{13}. Then $|f^*(z)|\le\delta$, $j=1,\ldots,n$, and
$|f^*(z_0)|\ge r_n(\delta,[-k,k])$. Now consider the function $g(z)=f^*(kz/k_1)$. Note
that $g(u_j)=f(z_j)$, $j=0,\ldots,n$, furthermore, since $f^*(z)$ is a Blaschke product, $g(z)$
is not. Hence
\begin{equation}\label{35}
\sup_{\substack{f\in\mathcal B\\|f(u_j)|\le\delta,\ j=1,\ldots,n}}|f(z_0)|>|g(u_0)|\ge r_n(\delta,[-k,k]).
\end{equation}
Here the left hand side coincides with $r_n(\delta,[-k_1,k_1])$ in virtue of \eqref{13} and \eqref{34}. The inequality \eqref{33} can be proved analogously. The lemma is proved
\end{proof}

\begin{theorem}\label{T31}
Let $[\alpha,\beta]\subset(-1,1)$. Set $a=\arth\alpha$, $b=\arth\beta$, $d=b-a$. For
\begin{equation}\label{36}
\frac{\thh d\thh\dfrac{n-2}{n-1}d}{1+\sqrt{1-\thh^2d\thh^2\dfrac{n-2}{n-1}d}}\le\delta<1
\end{equation}
the identity
\begin{equation}\label{37}
r_n^*(\delta,[\alpha,\beta])=\frac{\thh^2\dfrac d{2(n-1)}+\delta}{1+\delta\thh^2\dfrac d{2(n-1)}}
\end{equation}
holds, and the only system of optimal knots is
\begin{equation}\label{38}
z_j^0=\thh\left[\frac{a+b}2-(n+1-2j)\frac d{2(n-1)}\right],\quad j=1,\ldots,n.
\end{equation}
\end{theorem}

\begin{proof}
The knots \eqref{38} are uniformly distributed on the segment $[\alpha,\beta]$ with respect to the pseudohyperbolic distance, i.e.
$$\rho(z_j^0,z_{j+1}^0)=\thh\dfrac d{n-1},\quad j=1,\ldots,n-1.$$
Consequently, in every system of knots $\alpha=z_1\le\ldots\le z_n=\beta$ one can find a pair $z_k$, $z_{k+1}$ such that
\begin{equation}\label{39}
\rho(z_k,z_{k+1})\ge\thh\dfrac d{n-1}.
\end{equation}
Consider the function
$$\varphi(z)=\frac{\delta-W(z)}{1-\delta W(z)},\quad\mbox{where}\quad W(z)=\prod_{j=k}^{k+1}
\frac{z-z_j}{1-z_jz}.$$
We want to prove that $r(z,z_1,\ldots,z_n,\delta)=\varphi(z)$ for $z\in[z_k,z_{k+1}]$. As $\varphi(z)$ is a Blaschke product of order $2$, it is sufficient to prove that $|\varphi(z_j)|\le\delta$, $j=1,\ldots,n$ (cf.\ \eqref{212}). By the monotonicity of $\varphi(z)$ on the segments $[\alpha,z_k]$, $[z_{k+1},\beta]$ we have to check only $\varphi(\alpha)\ge-\delta$, $\varphi(\beta)\ge-\delta$. We have
$$\varphi(\alpha)=\frac{\delta-W(\alpha)}{1-\delta W(\alpha)}=\frac{\delta-\rho(\alpha,z_k)
\rho(\alpha,z_{k+1})}{1-\delta\rho(\alpha,z_k)\rho(\alpha,z_{k+1})}.$$
As $\rho(\alpha,z_{k+1})\le\thh d$, and
\begin{multline*}
\rho(\alpha,z_k)=\frac{\rho(\alpha,z_{k+1})-\rho(z_k,z_{k+1})}{1-\rho(z_k,z_{k+1})
\rho(\alpha,z_{k+1})}\le\frac{\thh d-\thh d/(n-1)}{1-\thh d/(n-1)\thh d}\\
=\thh\frac{n-2}{n-1}d,
\end{multline*}
it follows
$$\varphi(\alpha)\ge\left(\delta-\thh d\thh\frac{n-2}{n-1}d\right)\bigg/\left(1-\delta\thh d\thh\frac{n-2}{n-1}d\right).$$
Taking into account \eqref{36}, we obtain $\varphi(\alpha)\ge-\delta$. Analogously, $\varphi(\beta)\ge-\delta$. It is not difficult to see that
$$\max_{z\in[z_k,z_{k+1}]}\varphi(z)=\frac{p^2+\delta}{1+\delta p^2},$$
where
$$p=\frac{\rho(z_k,z_{k+1})}{1+\sqrt{1-\rho^2(z_k,z_{k+1})}}=
\thh\frac{\arth\rho(z_k,z_{k+1})}2.$$
In virtue of \eqref{39}, the inequalities
\begin{equation}\label{310}
\sup_{z\in[\alpha,\beta]}r(z,z_1,\ldots,z_n,\delta)\ge\max_{z\in[z_k,z_{k+1}]}\varphi(z)\ge
\frac{\thh^2\dfrac d{2(n-1)}+\delta}{1+\delta\thh^2\dfrac d{2(n-1)}}
\end{equation}
hold. They turn into equalities (for every $k$) if $z_j=z_j^0$, $j=1,\ldots,n$. Thus, we have
proved \eqref{37} and the optimality of the knots \eqref{38}. Their uniqueness follows from the
observation that for any other system one can find a pair $z_k$, $z_{k+1}$ for which inequality \eqref{39} and, consequently, \eqref{310} are strict. This completes the proof.
\end{proof}

\begin{theorem}\label{T32}
With the notations of Theorem~$\ref{T31}$, let $\theta_n$ satisfy the equation
\begin{equation}\label{311}
\theta_n+2\arth\thh^2\frac{\theta_n}{2(n-1)}=d.
\end{equation}
Then
\begin{equation}\label{312}
r_n(\delta,[\alpha,\beta])=\left(\thh^2\frac{\theta_n}{2(n-1)}+\delta\right)\bigg/
\left(1+\delta\thh^2\frac {\theta_n}{2(n-1)}\right)
\end{equation}
holds for $\thh\theta_n/2\le\delta<1$, and the only system of optimal knots is
\begin{equation}\label{313}
z_j^0=\thh\left[\frac{a+b}2-(n+1-2j)\frac{\theta_n}{2(n-1)}\right],\quad j=1,\ldots,n.
\end{equation}
\end{theorem}

\begin{proof}
First of all note that for $\theta_n\in[0,+\infty)$ the left side of \eqref{311} increases monotonically from $0$ to $+\infty$. Therefore \eqref{311} has a single solution. Let $\alpha\le z_1\le\ldots\le z_n\le\beta$ be an arbitrary system of knots. Let us prove that for
\begin{equation}\label{314}
\rho(z_1,z_n)\le2\delta/(1+\delta^2)
\end{equation}
the identity
\begin{equation}\label{315}
\sup_{z\in[\alpha,z_1]}r(z,z_1,\ldots,z_n,\delta)=\frac{\rho(\alpha,z_1)+
\delta}{1+\delta\rho(\alpha,z_1)}
\end{equation}
holds. Put
$$\varphi(z)=\frac{\delta-W(z)}{1-\delta W(z)},\quad\mbox{where}\quad W(z)=\frac{z-z_1}{1-z_1z}.$$
By the monotonicity of $\varphi(z)$ we have for $z\in[z_1,z_n]$
$$\delta\ge\varphi(z)\ge\varphi(z_n)=\frac{\delta-\rho(z_1,z_n)}{1-\delta\rho(z_1,z_n)}
\ge-\delta.$$
Hence it follows that for $z\in[\alpha,z_1]$
$$r(z,z_1,\ldots,z_n,\delta)=\varphi(z).$$
This obviously implies \eqref{315}. Analogously, taking for granted \eqref{314}, one can prove
\begin{equation}\label{316}
\sup_{z\in[z_n,\beta]}r(z,z_1,\ldots,z_n,\delta)=\frac{\rho(z_n,\beta)+
\delta}{1+\delta\rho(z_n,\beta)}.
\end{equation}
From Theorem~\ref{T31} we obtain
\begin{equation}\label{317}
r_n^*(\delta,[z_1^0,z_n^0])=\sup_{z\in[z_1^0,z_n^0]}r(z,z_1^0,\ldots,z_n^0,\delta)=e_n(\delta),
\end{equation}
where $e_n(\delta)$ denotes the right side of \eqref{312}. In virtue of \eqref{313} and \eqref{311} we have
\begin{equation}\label{318}
\rho(\alpha,z_1^0)=\rho(z_n^0,\beta)=\thh\frac{d-\theta_n}2=\thh^2\frac{\theta_n}{2(n-1)}.
\end{equation}
As $\rho(z_1^0,z_n^0)=\thh\theta_n$, the inequality \eqref{314} holds by the condition $\thh\theta_n/2\le\delta$. Therefore \eqref{315}--\eqref{317} imply
\begin{equation}\label{319}
\sup_{z\in[\alpha,\beta]}r(z,z_1^0,\ldots,z_n^0,\delta)=e_n(\delta).
\end{equation}

Let $\alpha\le z_1\le\ldots\le z_n\le\beta$ be an arbitrary system of knots different from $z_1^0,\ldots,z_n^0$. If $\rho(z_1,z_n)>\rho(z_1^0,z_n^0)=\thh\theta_n$, then Lemma~\ref{L31} yields
$$r_n^*(\delta,[z_1,z_n])>r_n^*(\delta,[z_1^0,z_n^0]).$$
Hence and from \eqref{317},
\begin{equation}\label{320}
\sup_{z\in[\alpha,\beta]}r(z,z_1,\ldots,z_n,\delta)>e_n(\delta).
\end{equation}
If
\begin{equation}\label{321}
z_j=z_j^0,\quad j=1,n
\end{equation}
then, in consequence of the uniqueness of the system of optimal knots for $r_n^*(\delta,[z_1^0,z_n^0])$ proved in Theorem~\ref{T31}, the inequality \eqref{320} still holds.

Suppose $\rho(z_1,z_n)\le\thh\theta_n$ and at least one of the relations \eqref{321} does not hold. Then either $\rho(\alpha,z_1)>\rho(\alpha,z_1^0)$ or $\rho(z_n,\beta)>\rho(z_n^0,\beta)$. Let e.g. $\rho(\alpha,z_1)>\rho(\alpha,z_1^0)$. The condition $\rho(z_1,z_n)\le\thh\theta_n$ guarantees \eqref{314}. Therefore \eqref{315} and \eqref{318} imply
$$\sup_{z\in[\alpha,z_1]}r(z,z_1,\ldots,z_n,\delta)>e_n(\delta).$$
Hence \eqref{320} follows. The case $\rho(z_n,\beta)>\rho(z_n^0,\beta)$ is analogous. Thus, \eqref{320} holds for any system of knots other than $z_1^0,\ldots,z_n^0$ whence taking into consideration \eqref{319}, the identity \eqref{312} follows, as well as the uniqueness of the system \eqref{313}. The theorem is proved.
\end{proof}

If
$$\frac{\rho(\alpha,\beta)}{1+\sqrt{1-\rho^2(\alpha,\beta)}}\le\delta<1,$$
the conditions of Theorems~\ref{T31} and \ref{T32} are fulfilled for every $n$. From \eqref{37} and \eqref{312} it follows that in this case the asymptotic equations
\begin{gather*}
r_{n+1}^*(\delta,[\alpha,\beta])=\delta+\frac{(1-\delta^2)d^2}{4n^2}-
\frac{(1-\delta^2)(2+3\delta)d^4}{48n^4}+O\left(\frac1{n^6}\right),\\
r_{n+1}^*(\delta,[\alpha,\beta])-r_{n+1}(\delta,[\alpha,\beta])=\frac{(1-\delta^2)d^3}{4n^2}+
O\left(\frac1{n^6}\right),
\end{gather*}
hold, where $d=\arth\rho(\alpha,\beta)$.


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\bibitem{4} {\sc M.~Heins,} The problem of Milloux for functions analytic throughout the interior of the unit circle, {\it Amer. J. Math.}, {\bf67} (1945), 212--234.

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\selectlanguage{russian}
\small
\begin{center}
{\bf Об оптимальной экстраполяции и интерполяции неточно заданных аналитических функций}\\
\bigskip
\footnotesize К.~Ю.~ОСИПЕНКО
\end{center}

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\small
Для класса $\mathcal B$ аналитических в единичном круге функций, ограниченных по модулю единицей, погрешностью наилучшего приближения в точке $z_0$ по значениям в точках $z_l,\ldots,z_n$, заданным с погрешностью $\delta$, называется величина
$$r(z_0,z_l,\ldots,z_n,\delta)=\infp_S\sup_{f\in\mathcal B}\sup_{\substack{f_1,\ldots,f_n\\|f(z_j)-f_j|\le\delta\\ j=1,\ldots,n}}|f(z_0)-S(f_1,\ldots,f_n)|,$$
где нижняя грань берется по всевозможным функциям $S\colon\mathbb C^n\to\mathbb C$. Для $E\subset(-1,1)$ и $z_0(-1,1)\setminus E$ рассматривается задача о нахождении порядка информативности множества $E$, т.е. минимального $n$, на котором достигается нижняя грань в равенстве
$$R(z_0,\delta,E)=\inf_{n\vphantom{z_1E}}\inf_{z_l,\ldots,z_n\in E}r(z_0,z_l,\ldots,z_n,\delta).$$
Кроме того, при $\delta$, близких к $1$, решена задача о нахождении величины
$$r_n(\delta,E)=\infp_{z_l,\ldots,z_n\in E}\sup_{z_0\in E}r(z_0,z_l,\ldots,z_n,\delta)$$
и найдены узлы, на которых достигается нижняя грань.

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\noindent\footnotesize{К.~Ю.~ОСИПЕНКО\\
СССР, МОСКВА 103767\\
ПЕТРОВКА 27\\
МОСКОВСКИЙ АВИАЦИОННЫЙ ТЕХНОЛОГИЧЕСКИЙ\\ 
ИНСТИТУТ им. К.~Э.~ЦИОЛКОВСКОГО}
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