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\begin{center}
 \textbf{\Large
  OPTIMAL RECOVERY AND EXTREMUM THEORY
 }\\[5mm]
 \textit{\large
  GEORGII G. MAGARIL-IL'YAEV, KONSTANTIN YU. OSIPENKO,
  and
  VLADIMIR M. TIKHOMIROV\footnote{This research was carried out with the
financial support of the Russian Foundation for Basic Research (grants
99-01-01181, 02-01-00386), Federal Program on Support of Leading
Scientific Schools (grant 00--15--96109), INTAS-97-1050, and Research
Grant 12513 of the Royal Swedish Acad. of Sci.}
  %%%
 }\\[5mm]
\end{center}

\small
\noindent\textbf{Abstract.}
In this paper optimal recovery problems of linear functionals on classes of
smooth and analytic functions on the basis of linear information are
considered from the general viewpoint of extremum theory. A general result
about the connection of optimal recovery method with Lagrange multipliers
of some convex extremal problem is applied to the analysis of classical
recovery problems on the generalized Sobolev, Hardy, and Hardy--Sobolev
classes.

\vspace{2mm}
\noindent\textbf{Keywords:}
optimal recovery, Lagrange principle, Hardy spaces.

\vspace{2mm}
\noindent\textbf{2000 Mathematics Subject Classification:}
41A46, 30D55.

\normalsize
\noindent


\section{Setting of the problem and general theory}

Let $X$ and $Y$ be real or complex linear spaces and $x'$ a linear
functional on $X$. It is required to recover $x'$ (as exactly as possible)
on elements from some set (class) $A\subset X$ using the information $y=Fx$
where $F\colon A\to Y$ is a linear operator which is called an {\it
information operator}. Any function $\varphi\colon F(A)\to K$ where $K=
\mathbb R$ or $\mathbb C$ we call a {\it method of recovery of $x'$ on $A$
from the information $F$}. The error of recovery is given by
$$e(x',A,F,\varphi)=\sup_{x\in A}|\la x',x\ra-\varphi(Fx)|.$$
The value
\begin{equation}\label{1}
E(x',A,F)=\inf_{\varphi}e(x',A,F,\varphi)
\end{equation}
where the infimum is taken over all functions $\varphi\colon F(A)\to K$ is
called the {\it error of optimal recovery}. Any method $\widehat\varphi$ for
which the infimum in \eqref1 is attained we call an {\it optimal recovery
method}.

Examples of such recovery problems are the problem of best integration
methods (it is required to recover an integral of a function from some
class using information about values of the function and its derivatives at
a fixed system of points), the problem of recovery of a function value or a
value of its derivative at some given point using information about the
Fourier coefficients, Taylor coefficients, or values of the function at
some other points, etc.

The method of the solution of the optimal recovery problems which we
propose in this paper is based on the following concepts. In problem \eqref
1, for a convex and balanced set $A$, among optimal methods of recovery
there exists a linear method. Thus the infimum in \eqref1 may be taken over
linear functionals on $Y$. In other words, the value $E(x',A,F,)$ is the
value of the following convex problem
$$\sup_{x\in A}|\la x',x\ra-\la y',Fx\ra|\to\min,\quad y'\in Y'$$
(where $Y'$ is the algebraic dual of $Y$), which is dual to another convex
problem (see~\cite[p.~61]{MT}; here for definiteness $X$ and $Y$ are
complex linear spaces)
\begin{equation}\label{2}
\Re\la x',x\ra\to\max,\quad Fx=0,\quad x\in A,
\end{equation}
which we call an {\it associated problem\/} to \eqref1. Denote by
$$\mathcal L(x,\lambda,\lambda_0)=\lambda_0\Re\la x',x\ra+\Re\la\lambda,Fx
\ra$$
the Lagrange function of the problem \eqref2 where $\lambda_0\le0$ and $
\lambda\in Y'$ are the Lagrange multipliers. If there exists a solution to
\eqref2 then it follows from the general theory of extremum that the
Lagrange multipliers are connected with the solution of the dual problem,
i.e., with the optimal method of recovery. The explicit assertions are
contained in the following theorem.

\begin{theorem}\label{T1}
{\bf(the Lagrange principle for optimal recovery problems).}
Let $X$ and $Y$ be real or complex linear spaces, $A$ a convex balanced
subset of $X$, and $F\colon X\to Y$ a linear operator. Then the admissible
in $\eqref2$ point $\widehat x$ is a solution of this problem if and only
if there exists the Lagrange multiplier $\widehat\lambda\in Y'$ for which
\begin{equation}\label{3}
\min_{x\in A}\mathcal L(x,\widehat\lambda,-1)=\mathcal L(\widehat x,
\widehat\lambda,-1).
\end{equation}
In this case
$$\la x',x\ra\approx\la\widehat\lambda,Fx\ra$$
is an optimal method of recovery in $\eqref1$ and
$$E(x',A,F)=\Re\la x',\widehat x\ra.$$
\end{theorem}

\noindent
\textbf{Proof:}
We use the following algebraic version of the separation theorem: {\it Let
$C$ be a convex subset of a real linear space $X$, $\icr C\ne\emptyset
$\footnote{$\icr C$ is the set of algebraic relative interior points of $C
$. If $\aff C=\ov x+L_C$ (where $\ov x\in C$ and $L_C$ is a subspace of $X
$) is an affine hull of $C$, i.e., a minimal linear manifold containing $C
$, then $x_0\in\icr C$ if for any $x\in L_C$ there exists $\varepsilon=
\varepsilon(x)>0$ such that $[x_0,x_0+\varepsilon x]\subset C$.}, and $x_0
\notin\icr C$. Then there exists $x'\in X'$, $x'\ne 0$, such that
$$\inf_{x\in C}\la x',x\ra\ge\la x',x_0\ra$$
and $\la x',x\ra>\la x',x_0\ra$ for all $x\in\icr C$} (see~\cite[p.~38]
{MT}).

1. Necessity. Let $\widehat x$ be a solution of \eqref{2}. Suppose first
that $\Re\la x',\widehat x\ra=0$. We show that in this case there exists a
$\widehat\lambda\in Y'$ such that
\begin{equation}\label{i}
\Re\la x',x\ra=\Re\la\widehat\lambda,Fx\ra
\end{equation}
for all $x\in A$. From here evidently follows \eqref3. Note that since $A$
is balanced $\Re\la x',x\ra=0$ for all admissible $x$. Define the
functional $l$ on the subspace $F(\spa A)$ by the equality $l(y)=\Re\la x',
x\ra$ where $x\in F^{-1}(y)$. This definition is well-defined. Indeed, let
$x_1,x_2\in F^{-1}(y)$. Since $A$ is balanced it is absorbing in $\spa A$
and therefore there exists an $\alpha>0$ such that $\alpha(x_1-x_2)\in A$.
It is clear that $\alpha(x_1-x_2)\in F^{-1}(0)$ and consequently $\alpha(x_
1-x_2)$ is an admissible element in \eqref2. Thus $\Re\la x',\alpha(x_1-x_2
)\ra=0$, that is $\Re\la x',x_1\ra=\Re\la x',x_2\ra$. It is easy to verify
that $l$ is a linear functional. Denote by $\widehat\lambda$ any of its
extensions on the all $Y$. It is obvious that \eqref i is fulfilled with
this $\widehat\lambda$.

Assume that $\Re\la x',\widehat x\ra\ne0$. Denote by $Y_{\mathbb R}$ a real
linear space of elements from $Y$ with multiplication only by the real
numbers. Consider the set
$$C=\{\,(\alpha,y)\in\mathbb R\times Y_{\mathbb R}\mid\alpha=\Re\la x',x\ra
,\ y=Fx,\ x\in A\,\}.$$
It is easy to see that $C$ is a convex balanced set and, in particular, $(0
,0)\in\icr C$. It is also easy to verify that $(\Re\la x',\widehat x\ra,0)
\notin\icr C$. Then by the separation theorem there exist $(\widehat\lambda
_0,\widehat\lambda_{\mathbb R})\in\mathbb R\times Y^{\prime}_{\mathbb R}$
not all equal to zero such that
\begin{equation}\label{ii}
\widehat\lambda_0\alpha+\la\widehat\lambda_{\mathbb R},y\ra\ge\widehat
\lambda_0\Re\la x',\widehat x\ra,\quad\forall(\alpha,y)\in C,
\end{equation}
and
\begin{equation}\label{iii}
\widehat\lambda_0\alpha+\la\widehat\lambda_{\mathbb R},y\ra>\widehat\lambda
_0\Re\la x',\widehat x\ra,\quad\forall(\alpha,y)\in\icr C.
\end{equation}
Since $(0,0)\in\icr C$ it follows from \eqref{iii} that $\widehat\lambda_0
\ne0$. It is clear that $(\Re\la x',\widehat x\ra,0)\in C$ and consequently
$2^{-1}(\Re\la x',\widehat x\ra,0)\in C$. Substituting this in \eqref{ii}
we have that $2\widehat\lambda_0\le\widehat\lambda_0$, that is $\widehat
\lambda_0<0$. Let us assume that $\widehat\lambda_0=-1$.

Denote by $\widehat\lambda$ an element from $Y'$ for which
\begin{equation}\label{iv}
\la\widehat\lambda_{\mathbb R},y\ra=\Re\la\widehat\lambda,y\ra
\end{equation}
for all $y\in Y$. Let $x\in A$. Then
\begin{equation}\label{v}
(\Re\la x',x\ra,Fx)\in C
\end{equation}
and we have
\begin{multline*}
\mathcal L(x,\widehat\lambda,-1)=-\Re\la x',x\ra+\Re\la\widehat\lambda,Fx
\ra\buildrel{\eqref{iv}}\over=-\Re\la x',x\ra+\la\widehat\lambda_{\mathbb R
},Fx\ra\\
\buildrel{\eqref{ii},\eqref v}\over\ge-\Re\la x',\widehat x\ra\buildrel{F
\widehat x=0}\over=-\Re\la x',\widehat x\ra+\Re\la\widehat \lambda,F
\widehat x\ra=\mathcal L(\widehat x,\widehat\lambda,-1).
\end{multline*}

2. Sufficiency. Let \eqref3 be fulfilled and $x$ be an admissible point in
\eqref2. Then
$$-\Re\la x',x\ra=-\Re\la x',x\ra+\Re\la\widehat\lambda,Fx\ra\buildrel{
\eqref3}\over\ge-\Re\la x',\widehat x\ra+\Re\la\widehat\lambda,F\widehat x
\ra=-\Re\la x',\widehat x\ra,$$
i.e., $\widehat x$ is a solution of \eqref2.

Let us prove the second assertion of the theorem. Since $A$ is balanced
\eqref3 may be rewritten as follows
\begin{equation}\label{vi}
\max_{x\in A}|\la x',x\ra-\la\widehat\lambda,Fx\ra|=\Re\la x',\widehat x\ra
.
\end{equation}
Hence
\begin{equation}\label{vii}
E(x',A,F)\le\Re\la x',\widehat x\ra.
\end{equation}
Let us show that, in fact, we have here equality. Assume $x\in A$ and $Fx=0
$. Since $-x\in A$ for any method $\varphi$ we have
\begin{multline*}
2\Re\la x',x\ra\le 2|\la x',x\ra|=|\la x',x\ra-\varphi(0)+\varphi(0)+\la x'
,x\ra|\\
\le|\la x',x\ra-\varphi(0)|+|\la x',-x\ra-\varphi(0)|\le2\sup_{\substack{x
\in A\\Fx=0}}|\la x',x\ra-\varphi(0)|\le2\sup_{x\in A}|\la x',x\ra-\varphi(
Fx)|.
\end{multline*}
This means that the reverse inequality to \eqref{vii} holds and thus
$$E(x',A,F)=\Re\la x',\widehat x\ra.$$
It follows from this equality and \eqref{vi} that $\widehat\lambda$ is an
optimal method of recovery.
$\rule{2mm}{2mm}$

\begin{remark}
The existence of a linear optimal method of recovery in the problem \eqref1
was discovered for the first time by Smolyak \cite{Sm} for the real case
and convex centrally-symmetric set $A$ with $\dim\spa F(A)<\infty$. The
generalization of this result and corresponding literature may be found in
\cite{MOs}. Dual methods for the solution of the problem \eqref1 were used
by many authors (see \cite{Bo}, \cite{MiR}, \cite{TW}, \cite{MiR1}, \cite
{Ar}) but the exact connection between the problems \eqref1 and \eqref2,
which is that the optimal method of recovery is none other than the
Lagrange multiplier in \eqref2, was apparently used for the first time in
\cite{MT}.
\end{remark}

\begin{remark}\label{r2}
The constraints on $A$ in \eqref2 may be also described by a system of
equalities and/or inequalities. In this case some of them may be included
in the Lagrange function (with corresponding multipliers). But an optimal
method is always the Lagrange multiplier at the constraints related to the
information operator. The proof of this fact (which is more general only in
appearance) is just the same as that of Theorem~\ref{T1}.
\end{remark}

In Theorem~\ref{T1} the so-called Lagrange principle for convex extremal
problems with constraints defined by equalities and inclusions is
confirmed. This principle is in the fact that if a problem has a solution
then there exist Lagrange multipliers such that this solution is the
absolute minimum of the Lagrange function on the set of the remaining
constraints (not included in the Lagrange function). In the examples below
we use this principle as an heuristic method. Namely, using equality \eqref
3 we extract the information about what $\widehat x$ and $\widehat\lambda$
must be to satisfy the equality \eqref3. After that we use the sufficiency
of this condition and find the solution of \eqref2 and the optimal recovery
method. Sometimes, instead of using the sufficiency, it is easy to verify
the optimality of the obtained method directly.

In the next section using Theorem~\ref{T1} we prove a general result about
the optimal recovery of functions from classes defined by the convolution
with some kernels on the basis of information about Fourier coefficients.
In Section~3 we apply this result to classes defined by the convolution
with cyclic variation diminishing kernels. We list there several well-known
results which are particular cases of the considered problem.

In Section~4 using Theorem~\ref{T1} we obtain optimal recovery algorithms
for Hardy classes. These results are known, but we point them out in order
to demonstrate the general method from Theorem~\ref{T1}. In Sections~5 and
6 we obtain some new results related to optimal recovery methods from
Hardy--Sobolev classes.

\section{Optimal recovery of function values from Fourier coefficients}

Let $r\in\mathbb N$ and $1\le p\le\infty$. Denote by $W_p^r(\mathbb T)$ the
Sobolev class of functions $x(\cdot)$ defined on the unit circle $\mathbb T
$ (realized as the interval $[-\pi,\pi]$ with identified endpoints) whose
the $(r-1)$st derivative is absolutely continuous and $\|x^{(r)}(\cdot)\|_{
L_p(\mathbb T)}\le1$. In 1936 Favard proved that for all $n\in\mathbb N$
and for all functions $x(\cdot)\in W_{\infty}^r(\mathbb T)$ such that
$$\int_{\mathbb T}x(t)\cos kt\,dt=\int_{\mathbb T}x(t)\sin kt\,dt=0,\quad k
=0,1,\ldots,n-1,$$
the following exact inequality
\begin{equation}\label{4}
\|x(\cdot)\|_{C(\mathbb T)}\le\frac{K_r}{n^r}
\end{equation}
holds.
The numbers $K_r$ (known as the Favard constants) are defined by
$$K_r=\frac4\pi\sum_{j=0}^\infty\frac{(-1)^{j(r+1)}}{(2j+1)^{r+1}},\quad r
\in\mathbb Z_+.$$

Note that the case when $n=r=1$ was previously considered by H.~Bohr and
therefore \eqref4 is usually called the {\it Bohr--Favard inequality}.

It is obvious that the problem of the exact constant in \eqref4 is
equivalent (in view of shift-invariance of the norm) to the following:
$$x(0)\to\max,\quad a_0=\ldots=a_{n-1}=b_1=\ldots=b_{n-1}=0,\quad x(\cdot)
\in W_{\infty}^r(\mathbb T),$$
where $a_0,\ldots,b_{n-1}$ are the Fourier coefficients of $x(\cdot)$. This
problem has the form \eqref2 and hence it relates to the optimal recovery
problem of a function value at the point $0$ on the class $W_{\infty}^r(
\mathbb T)$ from Fourier coefficients. The same problem is closely related
to the problem of deviation of the class $W_{\infty}^r(\mathbb T)$ from the
space of trigonometric polynomials $\T_{n-1}$ of degree at most $n-1$.
Beginning from the Favard's result a lot of papers were devoted to these
subjects.

The recovery problem of a function value at some given point from the
Fourier coefficients on the class defined as the convolution of a real
kernel $K(\cdot)$ with functions from the unit ball of $L_p(\mathbb T)$
involves many particular cases. More precisely, let $K(\cdot)\in L_{p'}(
\mathbb T)$ ($1/p+1/p'=1$) and
$$\alpha_k=\frac1\pi\int_{\mathbb T}K(t)\cos kt\,dt,\quad k\in\mathbb Z_+,
\quad\beta_k=\frac1\pi\int_{\mathbb T}K(t)\sin kt\,dt,\quad k\in\mathbb N,
$$
be the Fourier coefficients of $K(\cdot)$. Assume that $\alpha_k^2+\beta_k^
2\ne0$ ($\beta_0=0$) with the exception of a finite (possibly empty) set $Q
\subset\mathbb Z_+$. Set $\T_Q=\spa\{\cos kt,\sin kt,\ k\in Q\}$ and
$$\mathcal W_p^K(\mathbb T,Q)=\Bigl\{\,x(\cdot)\mid x(\cdot)=y(\cdot)+\frac
1\pi\int_{\mathbb T}K(\cdot-t)u(t)\,dt,\ y(\cdot)\in\T_Q,\ u(\cdot)\in\T_Q^
{\bot},\ u(\cdot)\in L_p(\mathbb T)\,\Bigr\},$$
where $\T_Q^{\bot}$ is the annihilator of $\T_Q$. It is clear that $
\mathcal W_p^K(\mathbb T,Q)$ is a subspace of the space $C(\mathbb T)$ of
continuous functions on $\mathbb T$. The corresponding convolution class is
the set
$$W_p^K(\mathbb T,Q)=\{x(\cdot)\in\mathcal W_p^K(\mathbb T,Q)\mid\|u(\cdot)
\|_{L_p(\mathbb T)}\le1\}.$$

For instance, in the case of the Sobolev class $W_p^r(\mathbb T)$ we have $
Q=\{0\}$, $K(\cdot)=B_r(\cdot)$ where
$$B_r(t)=\sum_{k=1}^\infty\frac{\cos(kt-\pi r/2)}{k^r}$$
is the Bernoulli kernel.

Consider the problem of optimal recovery of a function $x(\cdot)$ at a
point $\theta\in\mathbb T$ on the class $W_p^K(\mathbb T,Q)$ from the
Fourier coefficients
$$a_k=\frac1\pi\int_{\mathbb T}x(t)\cos kt\,dt,\quad k=0,1,\ldots,n-1,\quad
b_k=\frac1\pi\int_{\mathbb T}x(t)\cos kt\,dt,\quad k=1,\ldots,n-1.$$
In accordance with the general notation we have $X=\mathcal W_p^K(\mathbb T
,Q)$, $A=W_p^K(\mathbb T,Q)$, $Y=\mathbb R^{2n-1}$, $Fx(\cdot)=\Four_nx(
\cdot)=(a_0,a_1,\ldots,a_{n-1},b_1,\ldots,b_{n-1})$, and $\la x',x(\cdot)
\ra=x(\theta)$.

Note that if $\{0,1,\ldots,n-1\}\setminus Q\ne\emptyset$, then it is easy
to check that the error of optimal recovery equals $+\infty$ and hence any
method is optimal. Therefore we assume that $Q\subset\{0,1,\ldots,n-1\}$.
Put $Q'=\{0,1,\ldots,n-1\}\setminus Q$.

For a normed linear space $X$, $x\in X$, and a nonempty subset $A$ of $X$
denote by $d(x,A,X)$ the deviation from $x$ to $A$ in the metric of $X$.

We say that a function $K(\cdot)\in L_1(\mathbb T)$ satisfies the {\it
Favard $\gamma$-property} (for a fixed $n\in\mathbb N$) if there exists a
polynomial $\widehat q(\cdot)\in\T_{n-1}$ and a number $\gamma\in[0,\pi/n)$
such that the function $(K(t)-\widehat q(t))\sin n(t+\gamma)$ is
nonnegative or nonpositive for almost all $t\in\mathbb T$. If $K(\cdot)$ is
a continuous function, then $\widehat q(\cdot)$ may be found as a
polynomial which interpolates $K(\cdot)$ at the zeros of $\sin n(\cdot+
\gamma)$.

The following theorem holds.

\begin{theorem}\label{T2}
{\bf(on optimal recovery from Fourier coefficients).}
Let $1<p\le\infty$ and
$$\widehat p(t)=\frac{A_0}2+\sum_{k=1}^{n-1}(A_k\cos kt+B_k\sin kt)$$
be a polynomial of the best approximation of $K(\cdot)$ by $\T_{n-1}$ in
the metric $L_{p'}(\mathbb T)$. Then
$$x(\theta)\approx\widehat\mu_0 a_0+\sum_{k=1}^{n-1}(\widehat\mu_k(\theta)a
_k+\widehat\nu_k(\theta)b_k),$$
where $\widehat\mu_0=1/2$ if $0\in Q$ and $\widehat\mu_0=A_0/(2\alpha_0)$
if $0\notin Q$; $\widehat\mu_k(\theta)=\cos k\theta$, $\widehat\nu_k(\theta
)=\sin k\theta$ if $k\in Q\setminus\{0\}$ and
\begin{align*}
\widehat\mu_k(\theta)&=\frac{(\alpha_k A_k+\beta_k B_k)\cos k\theta+(\alpha
_kB_k-\beta_kA_k)\sin k\theta}{\alpha_k^2+\beta_k^2},\\
\widehat\nu_k(\theta)&=\frac{(\beta_k A_k-\alpha_k B_k)\cos k\theta+(\alpha
_kA_k+\beta_kB_k)\sin k\theta}{\alpha_k^2+\beta_k^2}
\end{align*}
if $k\in Q'$, is an optimal method of recovery of $x(\theta)$ on the class
$W_p^K(\mathbb T,Q)$ from Fourier coefficients. Moreover,
$$E(x(\theta),W_p^K(\mathbb T,Q),\Four_n)=\frac1\pi d\left(K(\cdot),\T_{n-1
},L_{p'}(\mathbb T)\right).$$

If $p=\infty$ and $K(\cdot)$ satisfies the Favard $\gamma$-property, then
$$E(x(\theta),W_{\infty}^K(\mathbb T,Q),\Four_n)=\frac1\pi\left|\int_{
\mathbb T}K(t)\sign\sin n(t+\gamma)\,dt\right|.$$
\end{theorem}

Let us formulate a corollary from this theorem related to the deviation of
$W_p^K(\mathbb T,Q)$ from the subspace of trigonometric polynomials. Recall
that for a normed linear space $X$ and nonempty subsets $A$ and $C$ of $X$
the value
$$d(C,A,X)=\sup_{x\in C}d(x,A,X)$$
is called the {\it deviation of $C$ from $A$ in the metric $X$}.

The value
$$d^L(W_p^K(\mathbb T,Q),\T_{n-1},L_p(\mathbb T))=\infp_{\Lambda\vphantom{W
_p^K}}\sup_{x(\cdot)\in W_p^K(\mathbb T,Q)}\|x(\cdot)-\Lambda x(\cdot)\|_{L
_p(\mathbb T)},$$
where the infimum is taken over all linear operators $\Lambda\colon\mathcal
W_p^K(\mathbb T,Q)\to\T_{n-1}$, characterizes a best linear approximation
of $W_p^K(\mathbb T,Q)$ by trigonometric polynomials from $\T_{n-1}$. An
operator $\Lambda$ for which the infimum is attained is called an {\it
extremal method}.

Obviously,
$$d(W_p^K(\mathbb T,Q),\T_{n-1},L_p(\mathbb T))\le d^L(W_p^K(\mathbb T,Q),
\T_{n-1},L_p(\mathbb T)).$$

\begin{corollary*}\label{Cor}
If $K(\cdot)$ satisfies the Favard $\gamma$-property, then
$$d\left(W_\infty^K(\mathbb T,Q),\T_{n-1},C(\mathbb T)\right)=d^L\left(W_
\infty^K(\mathbb T,Q),\T_{n-1},C(\mathbb T)\right)=E(x(\theta),W_{\infty}^K
(\mathbb T,Q),\Four_n)$$
and the operator $\widehat\Lambda$ which associates $x(\cdot)\in\mathcal W_
\infty^K(\mathbb T,Q)$ with the polynomial
$$\widehat\mu_0 a_0+\sum_{k=1}^{n-1}(\widehat\mu_k(\theta)a_k+\widehat\nu_k
(\theta)b_k)$$
is extremal.
\end{corollary*}

\noindent
\textbf{Proof of Theorem~\ref{T2}:}
The problem associated with the considered problem of the optimal recovery
has the form
\begin{equation}\label{i1}
x(\theta)\to\max,\quad a_0=\ldots=a_{n-1}=b_1=\ldots=b_{n-1}=0,\quad x(
\cdot)\in W_p^K(\mathbb T).
\end{equation}
Its Lagrange function is
\begin{multline}\label{ii1}
\mathcal L(x(\cdot),\mu_0,\mu_1,\ldots,\mu_{n-1},\nu_1,\ldots,\nu_{n-1},
\lambda_0)=\lambda_0 x(\theta)+\frac{\mu_0}{\pi}\int_{\mathbb T}x(t)\,dt\\
+\frac1\pi\sum_{k=0}^{n-1}\int_{\mathbb T}(\mu_k\cos kt+\nu_k\sin kt)x(t)\,
dt,
\end{multline}
where $\lambda_0$, $\mu_k$, $k=0,1\ldots,n-1$, and $\nu_k$, $k=1,\ldots,n-1
$, are the Lagrange multipliers.

Further we argue heuristically. Namely, we set $\lambda_0=-1$ and use the
Lagrange principle formally. It allows us to understand how the solution of
\eqref{i1} is organized and what form the Lagrange multipliers have (which
determine an optimal method of recovery according to Theorem~\ref{T1}).
After that we use the sufficient conditions of Theorem~\ref{T1}.

Let $\widehat x(\cdot)$ be a solution of the problem \eqref{i1}. Then
(according to the Lagrange principle) there exist such numbers $\widehat\mu
_k$, $k=0,1\ldots,n-1$, and $\widehat\nu_k$, $k=1,\ldots,n-1$, that the
function $\mathcal L(x(\cdot),\widehat\mu_0,\widehat\mu_1,\ldots,\widehat
\mu_{n-1},\widehat\nu_1,\ldots,\widehat\nu_{n-1},-1)$ attains its absolute
minimum on $W_p^r(\mathbb T)$ at the point $\widehat x(\cdot)$ (for
simplicity we do not indicate that $\widehat\mu_k$ and $\widehat\nu_k$
depend on $\theta$).

For definiteness we assume that $0\in Q$. Substitute in $\mathcal L$
instead of $x(\cdot)$ its representation in terms of $u(\cdot)$ and
$$y(\cdot)=\gamma_0+\sum_{k\in Q\setminus\{0\}}(\gamma_k\cos kt+\delta_k
\sin kt)$$
(the coefficients $\gamma_k$ and $\delta_k$ are uniquely determined by $x(
\cdot)$ since they are the corresponding Fourier coefficients of $x(\cdot)
$). Denote by $\widehat\gamma_0$, $\widehat\gamma_k$, $\widehat\delta_k$, $
k\in Q\setminus\{0\}$, and $\widehat u(\cdot)$ the coefficients and the
function which correspond to $\widehat x(\cdot)$. By means of simple
calculations we obtain that the function
\begin{multline}\label{iii1}
-\gamma_0-\sum_{k\in Q\setminus\{0\}}(\gamma_k\cos k\theta+\delta_k\sin k
\theta)+2\widehat\mu_0\gamma_0+\sum_{k\in Q\setminus\{0\}}(\widehat\mu_k
\gamma_k+\widehat\nu_k\delta_k)\\
+\frac1\pi\int_{\mathbb T}\biggl(-K(\theta-t)+\sum_{k\in Q'}((\widehat\mu_k
\alpha_k+\widehat\nu_k\beta_k)\cos kt+(\widehat\nu_k\alpha_k-\widehat\mu_k
\beta_k)\sin kt)\biggr)u(t)\,dt
\end{multline}
attains its absolute minimum on the set
\begin{equation}\label{iv1}
\gamma_0,\gamma_k,\delta_k\in\mathbb R,\quad\frac1\pi\int_{\mathbb T}u(t)
\genfrac{}{}{0pt}{}{\cos kt}{\sin kt}dt=0,\ k\in Q,\quad\|u(\cdot)\|_{L_p(
\mathbb T)}\le1
\end{equation}
at the point $(\{\widehat\gamma_0,\widehat\gamma_k,\widehat\delta_k\}_{k\in
Q\setminus\{0\}},\widehat u(\cdot))$. Hence $\widehat\mu_0=1/2$ and $
\widehat\mu_k=\cos k\theta$, $\widehat\nu_k=\sin k\theta$, $k\in Q\setminus
\{0\}$.

The problem \eqref{iii1}--\eqref{iv1} is a problem of type \eqref{i1}
(the minimization of a linear functional on a convex balanced set). Its
Lagrange function may be written obviously (we set the multiplier at the
minimizing functional equals $1$ and do not include the last constraint in
\eqref{iv1}). Then according to the Lagrange principle there exist such $
\widehat c_0$, $\widehat c_k,\widehat d_k$, $k\in Q\setminus\{0\}$, that
the function
\begin{multline}\label{v1}
\frac1\pi\int_{\mathbb T}\biggl(-K(\theta-t)+\widehat c_0+\sum_{k\in Q
\setminus\{0\}}(\widehat c_k\cos kt+\widehat d_k\sin kt)\\
+\sum_{k\in Q'}((\widehat\mu_k\alpha_k+\widehat\nu_k\beta_k)\cos kt+(
\widehat\nu_k\alpha_k-\widehat\mu_k\beta_k)\sin kt)\biggr)u(t)\,dt
\end{multline}
attains the absolute minimum on the unite ball of $L_p(\mathbb T)$ at the
point $\widehat u(\cdot)$. If we denote by $L(\cdot)$ the multiplier
preceding $u(\cdot)$ under the integral sign, then it is clear that
$$\widehat u(\cdot)=-\|L(\cdot)\|^{1-p'}_{L_{p'}(\mathbb T)}|L(\cdot)|^{p'-
1}\sign L(\cdot).$$

Note that $\widehat u(\cdot)\in\T_{n-1}^\bot$. It follows from \eqref{iv1}
(when $k\in Q$) and the fact that for $k\in Q'$, \ $\alpha_k^2+\beta_k^2\ne
0$ and therefore the vanishing of Fourier coefficients of $\widehat x(\cdot
)$ implies the vanishing of corresponding Fourier coefficients of $\widehat
u(\cdot)$. Then in accordance with the criterion of the best approximation
in $L_{p'}(\mathbb T)$ we obtain that the polynomial
$$
\overline p(t)=\widehat c_0+\sum_{k\in Q\setminus\{0\}}(\widehat c_k\cos kt
+\widehat d_k\sin kt)
+\sum_{k\in Q'}((\widehat\mu_k\alpha_k+\widehat\nu_k\beta_k)\cos kt+(
\widehat\nu_k\alpha_k-\widehat\mu_k\beta_k)\sin kt)
$$
must be the best approximation polynomial for the function $t\to K(\theta-t
)$ by the subspace $\T_{n-1}$ in the metric $L_{p'}(\mathbb T)$.

Now we shall apply sufficient conditions. Let $\widehat p(\cdot)$ be the
polynomial mentioned in the statement of the theorem. Then $\widehat p(
\theta-\cdot)$ is the polynomial of the best approximation of $K(\theta-
\cdot)$ by the subspace $\T_{n-1}$ in the metric $L_{p'}(\mathbb T)$.
Choose multipliers $\widehat c_0$, $\widehat c_k$, $\widehat d_k$, $k\in Q
\setminus\{0\}$, and $\widehat\mu_k,\widehat\nu_k$, $k\in Q'$, so that $
\overline p(\cdot)=\widehat p(\theta-\cdot)$. We obtain just the same
formulae for these coefficients which are given in the theorem. With these
Lagrange multipliers the polynomial $\overline p(\cdot)$ is in fact the
polynomial of the best approximation of $K(\cdot)$ by the subspace $\T_{n-1
}$ in the metric $L_{p'}(\mathbb T)$. From the criterion of the best
approximation it follows that the function
$$\widehat u(\cdot)=-\|\widehat L(\cdot)\|^{1-p'}_{L_{p'}(\mathbb T)}|
\widehat L(\cdot)|^{p'-1}\sign\widehat L(\cdot),$$
where $\widehat L(\cdot)$ is $L(\cdot)$ with just defined Lagrange
multiplier, is orthogonal to $\T_{n-1}$ and evidently $\|\widehat u(\cdot)
\|_{L_p(\mathbb T)}=1$. Hence $\widehat u(\cdot)$ is admissible in \eqref
{iv1}. Put $\widehat\gamma_0=\widehat\gamma_k=\widehat\delta_k=0$, $
\widehat\mu_0=1/2$, $\widehat\mu_k=\cos k\theta$, and $\widehat\nu_k=\sin k
\theta$, $k\in Q\setminus\{0\}$. Then since $\widehat u(\cdot)$ is a
solution of \eqref{v1} (with corresponding multipliers) by Theorem~\ref{T1}
the point $(\{\widehat\gamma_0,\widehat\gamma_k,\widehat\delta_k\}_{k\in Q
\setminus\{0\}},\widehat u(\cdot))$ is a solution of the problem \eqref
{iii1}--\eqref{iv1}. This is equivalent to the fact that the function
$$\widehat x(\cdot)=\frac1\pi\int_{\mathbb T}K(\cdot-\tau)\widehat u(\tau)
\,d\tau$$
($\widehat y(\cdot)=0$ since $\widehat\gamma_0=\widehat\gamma_k=\widehat
\delta_k=0$, $k\in Q\setminus\{0\}$) gives the minimum of the Lagrange
function \eqref{ii1} with $\lambda_0=-1$ and Lagrange multipliers defined
above. Since $\widehat u(\cdot)\in\T_{n-1}^\bot$, by the same arguments as
above the Fourier coefficients of $\widehat x(\cdot)$ vanish for $k\in Q'$
and it means that $\widehat x(\cdot)$ is admissible in \eqref{i1}. Then by
Theorem~\ref{T1} it is a solution of this problem and the Lagrange
multipliers define an optimal method. The case when $0\in Q'$ is considered
analogously. The first part of the theorem is proved.

Further,
\begin{multline*}
\widehat x(\theta)=\frac1\pi\int_{\mathbb T}K(\theta-t)\widehat u(t)\,dt=
\frac1\pi\int_{\mathbb T}(K(\theta-t)-\widehat p(\theta-t))\widehat u(t)\,d
t\\
=\frac1\pi\|K(\cdot)-\widehat p(\cdot)\|_{L_{p'}(\mathbb T)}=\frac1\pi d(K(
\cdot),\T_{n-1},L_{p'}(\mathbb T)),
\end{multline*}
that is the quantity in the right-hand side is the value of the problem
\eqref{i1}. Hence and from Theorem~\ref{T1} the second assertion of the
theorem follows.

Let $p=\infty$ and $K(\cdot)$ satisfies the Favard $\gamma$-property. Put
$$\overline u(t)=\sign (K(t)-\widehat q(t)),$$
where $\widehat q(\cdot)$ is from the definition of the Favard $\gamma
$-property. Then
$$\overline u(t)=\varepsilon\sign\sin n(t+\gamma),\quad\varepsilon=1\text{
or }-1,$$
almost everywhere. Since it is clear that $\overline u(\cdot)\in\T_{n-1}^
\bot$, from the criterion of the best approximation it follows that $
\widehat q(\cdot)$ is the best approximation polynomial of $K(\cdot)$ by
the subspace $\T_{n-1}$ in the metric $L_1(\mathbb T)$. Consequently,
\begin{multline*}
d(K(\cdot),\T_{n-1},L_1(\mathbb T))=\int_{\mathbb T}|K(t)-\widehat q(t)|\,d
t=\varepsilon\int_{\mathbb T}(K(t)-\widehat q(t))\sign\sin n(t+\gamma)dt\\
=\left|\int_{\mathbb T}K(t)\sign\sin n(t+\gamma)\,dt\right|.
\end{multline*}
Together with the previous equality this proves the last assertion of the
theorem.
$\rule{2mm}{2mm}$

\noindent
\textbf{Proof of Corollary~\ref{Cor}:}
The upper bound. Set
$$\kappa=\left|\int_{\mathbb T}K(t)\sign\sin n(t+\gamma)\,dt\right|.$$
Let $x(\cdot)\in W_p^K(\mathbb T,Q)$. Then taking into account the last
assertion of Theorem~\ref{T2} we have
$$\|x(\cdot)-\widehat\Lambda x(\cdot)\|_{C(\mathbb T)}=\max_{\theta\in
\mathbb T}|x(\theta)-\widehat\Lambda x(\theta)|\le\max_{\theta\in\mathbb T}
E(x(\theta),W_p^K(\mathbb T,Q),\Four_n)=\frac\kappa\pi.$$
Hence
$$d^L(W_{\infty}^K(\mathbb T,Q),\T_{n-1},C(\mathbb T))\le E(x(\theta),W_{
\infty}^K(\mathbb T,Q),\Four_n)=\frac\kappa\pi.$$

The lower bound. Consider the function
$$\overline x(\cdot)=\frac1\pi\int_{\mathbb T}K(\cdot-\tau)\sign\sin n(\tau
-\gamma)\,d\tau.$$
Clearly, $\overline x(\cdot)\in W_{\infty}^K(\mathbb T,Q)$. This function
may be rewritten as follows
$$\overline x(t)=\frac1\pi\int_{\mathbb T}K(\tau)\sign\sin n(\tau+\gamma-t)
\,d\tau.$$
It is easily seen that
$$\overline x\left(\frac{k\pi}n\right)=\frac{(-1)^k}\pi\int_{\mathbb T}K(
\tau)\sign\sin n(\tau+\gamma)\,d\tau=\varepsilon(-1)^k\frac\kappa\pi,\quad
\varepsilon=1\text{ or }-1,$$
and in view of the fact that $|\overline x(t)|\le\kappa/\pi$ the function $
\overline x(\cdot)$ has $2n$-alternance on the period. By the Chebyshev
alternance theorem the trivial polynomial is its best approximation
polynomial by the subspace $\T_{n-1}$ in $C(\mathbb T)$. Therefore,
$$d(W_\infty^K(\mathbb T,Q),\T_{n-1},C(\mathbb T))\ge d(\overline x(\cdot),
\T_{n-1},C(\mathbb T))=\|\overline x(\cdot)\|_{C(\mathbb T)}=\frac\kappa\pi
.\quad\rule{2mm}{2mm}$$


\section{Cyclic variation diminishing kernels}

Denote by $\K(Q)$ the set of kernels $K(\cdot)\in L_1(\mathbb T)$ for which
for all $y(\cdot)\in\T_Q$ and all $u(\cdot)\in L_\infty(\mathbb T)$ such
that $u(\cdot)\perp\T_Q$ and $u(\cdot)\ne0$ the inequality
$$S(y(\cdot)+(K*x)(\cdot))\le S(u(\cdot))$$
holds, where $S(u(\cdot))$ is the number of sign changes of $u(\cdot)$ on
the period and
$$(K*u)(\cdot)=\frac1\pi\int_{\mathbb T}K(\cdot-t)u(t)\,dt.$$

For a function $u(\cdot)\in C(\mathbb T)$ denote by $\dist u(\cdot)$ the
length of the largest subinterval of $\mathbb T$ containing no zeros of $u(
\cdot)$. Denote by $\K(Q,\delta)$ the class of kernels $K(\cdot)\in L_1(
\mathbb T)$ for which for all $u(\cdot)\in L_\infty(\mathbb T)$ and $y(
\cdot)\in\T_Q$ such that $u(\cdot)\perp\T_Q$, $u(\cdot)\ne0$, and $\dist(y(
\cdot)+(K*u)(\cdot))<\delta$ the inequality
$$S(y(\cdot)+(K*u)(\cdot))\le S(u(\cdot))$$
holds, and moreover, if $(K*u)(\cdot)\in C^2(\mathbb T)$, then
$$Z_2(y(\cdot)+(K*u)(\cdot))\le S(u(\cdot)),$$
where $Z_2(u(\cdot))$ is the number of zeros of $u(\cdot)$ when multiple
zeros are counted twice and intervals on which the function vanishes
identically are discarded. Assume as before that $\alpha_k^2+\beta_k^2\ne0
$, $k\notin Q$, where $\alpha_k$ and $\beta_k$ are the Fourier coefficients
of $K(\cdot)$.

Suppose that
\begin{equation}\label{cond1}
K_j(\cdot)\in\K(Q_j,\delta_j),\quad j=1,\ldots,k,\qquad K_0(\cdot)\in\K(Q_0
).
\end{equation}
Set
\begin{equation}\label{cond2}
K(\cdot)=(K_k*\ldots*K_1*K_0)(\cdot),\quad Q=\bigcup_{j=0}^kQ_j,\quad\delta
=\min_{1\le j\le k}\delta_j.
\end{equation}
Consider some particular cases of the classes $W_\infty^K(\mathbb T,Q)$ for
such kernels.

1. Let $k=0$, $Q_0=\emptyset$. The kernels from the set $\K(\emptyset)$ are
called {\it cyclic variation diminishing kernels\/} or $CVD$-kernels.
The corresponding classes
$$W_\infty^K(\mathbb T,\emptyset)=\{\,x(\cdot)\mid x(\cdot)=(K*u)(\cdot),\
\|u(\cdot)\|_{L_\infty(\mathbb T)}\le1\,\}$$
were studied in \cite{Pi}. In particular, the kernel
$$K_\beta(t)=\frac12+\sum_{m=1}^\infty\frac{\cos mt}{\cosh m\beta}$$
is a $CVD$-kernel and the corresponding class $W_\infty^{K_\beta}(\mathbb T
,\emptyset)$ coincides with the class $h_\infty^\beta$ which is the set of
real, $2\pi$-periodic functions $f(\cdot)$ that can be analytically
continued to the strip $S_\beta=\{z\in\mathbb C\mid|\Im z|<\beta\}$ so that
$|\Re f(z)|\le1$ in this strip.

2. Let $P(D)$ be a differential polynomial of degree $r$ with constant real
coefficients
$$P(D)=D^r+a_{r-1}D^{r-1}+\ldots+a_0,\quad D=\frac d{dt}.$$
Set
$$K_P(t)=\frac12\sum_{\substack{m\in\mathbb Z\\P(im)\ne0}}\frac{e^{imt}}{P(
im)}.$$
For $Q=\{m\in\mathbb Z_+\mid P(im)=0\}$ the class $W_\infty^{K_P}(\mathbb T
,Q)$ coincides with the generalized Sobolev class which is the set of $2\pi
$-periodic functions $x(\cdot)$ with $(r-1)$st derivative absolutely
continuous and satisfying the condition
$$\|(P(D)x)(\cdot)\|_{L_\infty(\mathbb T)}\le1.$$
In particular, for $P(D)=D^r$ this class coincides with the standard
Sobolev class $W_\infty^r(\mathbb T)$.

In the general case a polynomial $P(D)$ can be represented in the following
form
\begin{equation}\label {qd}
P(D)=\prod_{j=1}^kP_j(D),
\end{equation}
where $P_j(D)$ are differential polynomials with real coefficients of
degrees at most $2$. It follows from \cite{Ng} that $K_{P_j}(\cdot)\in\K(Q_
j,\delta_j)$ where
\begin{equation}\label {mj}
Q_j=\{\,m\in\mathbb Z_+\mid P_j(im)=0\,\},\qquad\delta_j=\pi/h(P_j(\cdot)),
\end{equation}
and $h(P_j(\cdot))$ is the largest imaginary part of the zeros of the
polynomial $P_j(\cdot)$.

3. For a differential polynomial $P(D)$ with real coefficients let $h_{
\infty,\beta}^P$ be the class of $2\pi$-periodic real-valued functions $f(
\cdot)$ that can be analytically continued to the strip $S_\beta$
satisfying the condition $|\Re(P(D)f)(z)|\le1$ for all $z\in S_\beta$. Then
$h_{\infty,\beta}^P=W_\infty^{K_P}(\mathbb T,Q)$ where (using the notation
\eqref{qd}, \eqref{mj})
$$K_P(\cdot)=(K_{P_k}*\ldots*K_{P_1}*K_\beta)(\cdot),\quad Q=\bigcup_{j=1}^
kQ_j.$$

Set
$$h_n(t):=\sign\sin nt.$$

\begin{lemma*}
Assume that a kernel $K(\cdot)$ satisfies $\eqref{cond1}$ and $\eqref
{cond2}$. Then for all
\begin{equation}\label{nn}
n>\max\{\,\sup_{j\in Q}j,\,2\pi/\delta\,\}
\end{equation}
$K(\cdot)$ satisfies the Favard $\gamma$-property where $\gamma$ defined by
the condition
\begin{equation}\label{ga}
(K*h_n)(\gamma)=-\|(K*h_n)(\cdot)\|_{L_\infty(\mathbb T)}.
\end{equation}
\end{lemma*}

\noindent
\textbf{Proof:}
Consider the problem of optimal recovery of $x(\cdot)$ at the zero on the
class $W_p^K(\mathbb T,Q)$ from the Fourier coefficients of this function $
a_0,\ldots,a_{n-1},b_1,\ldots,b_{n-1}$. The associated problem has the form
$$x(0)\to\max,\quad a_0=\ldots=a_{n-1}=b_1=\ldots=b_{n-1}=0,\quad x(\cdot)
\in W_\infty^K(\mathbb T,Q).$$
It follows from \cite{Os} that under the conditions \eqref{nn}, \eqref{ga}
the function $(K*h_n)(\cdot-\gamma)$ is a solution of this problem. Assume
for definiteness that $0\in Q$. Similarly to the proof of Theorem~\ref{T2}
we obtain that there exist $\widehat c_k$, $\widehat d_k$, $k\in Q$, and $
\widehat\mu_k$, $\widehat\nu_k$, $k\in Q'$, such that the function
\begin{multline*}
\frac1\pi\int_{\mathbb T}\biggl(-K(-t)+\widehat c_0+\sum_{k\in Q\setminus\{
0\}}(\widehat c_k\cos kt+\widehat d_k\sin kt)\\
+\sum_{k\in Q'}((\widehat\mu_k\alpha_k+\widehat\nu_k\beta_k)\cos kt+(
\widehat\nu_k\alpha_k-\widehat\mu_k\beta_k)\sin kt)\biggr)u(t)\,dt
\end{multline*}
attains the absolute minimum on the unit ball of $L_\infty(\mathbb T)$ at
the point $\widehat u(\cdot)=h_n(\cdot+\gamma)$. Denoting by $L(\cdot)$ the
multiplier preceeding $u(\cdot)$ under the integral sign we obtain that $
\widehat u(\cdot)=-\sign L(\cdot)$. Changing the variable $t$ on $-t$ we
get the existence of polynomial $P(\cdot)\in\T_{n-1}$ such that $\sign(K(t)
-P(t))=-h_n(t+\gamma)$.
$\rule{2mm}{2mm}$

Thus for the classes $W_\infty^K(\mathbb T,Q)$ with kernels $K(\cdot)$
satisfying the conditions \eqref{cond1} and \eqref{cond2} the assertions of
Theorem~\ref{T2} (for $p=\infty$) and Corollary~\ref{Cor} hold.

We list several well-known results which are particular cases of the
assertions proved here. The inequality \eqref4 obtained by Favard \cite{Fa}
was used in \cite{Fa1} (and also independently by Akhiezer and M.~Krein
\cite{AK}) to prove the equality
\begin{equation}\label{FAK}
d(W_\infty^r(\mathbb T),\T_{n-1},C(\mathbb T))=\frac{K_r}{n^r},\quad r\in
\mathbb N.
\end{equation}
The class $W_\infty^r(\mathbb T)$ is defined by the convolution with the
Bernoulli kernel which satisfies the Favard $\gamma$-property for $\gamma=0
$ if $r$ is odd and $\gamma=\pi/(2n)$ if $r$ is even. For this class the
problem of optimal recovery from Fourier coefficients was solved by
Bojanov~\cite{Bo} who proved that
$$E(x(\theta),W_\infty^r(\mathbb T),\Four_n)=\frac{K_r}{n^r}.$$

The result of Favard--Akhiezer--Krein \eqref{FAK} was developed in several
directions. Partially this was elucidated in \cite{Akh}. Note the own
result of Akhiezer \cite{Akh1}
$$d(h_\infty^\beta,\T_{n-1},C(\mathbb T))=\frac4\pi\sum_{m=0}^\infty\frac{(
-1)^m}{(2m+1)\cosh(2m+1)n\beta}$$
and M. Krein \cite{Kr}
$$d(\Gamma_\infty^\rho,\T_{n-1},C(\mathbb T))=\frac4\pi\arctan\rho^n,$$
where $\Gamma_\infty^\rho$ is the class of functions $x(\cdot)$ represented
in the form $x(\cdot)=u(\rho,\cdot)$, $0<\rho<1$, with functions $u(r,t)$,
$0\le r<1$, $t\in\mathbb T$, harmonic in the unit ball and satisfying there
the condition $|u(r,t)|\le1$. The class $\Gamma_\infty^\rho$ coincides with
the class $W_\infty^{P_\rho}(\mathbb T,\emptyset)$ where
$$P_\rho(t)=\frac12\,\frac{1-\rho^2}{1-2\rho\cos t+\rho^2}$$
is the Poisson kernel which satisfies the Favard $\pi/(2n)$-property.

The problem of generalization of the Favard--Akhiezer--Krein result for
fractional $r$ was open for a long time. This problem was solved by Dzyadyk
\cite{Dz} and Sun Yongsheng \cite{Sun}, \cite{Sun1}. It turns out that the
Bernoulli kernel with fractional $r\ge1$ also satisfies the Favard $\gamma
$-property with $\gamma$ defined by the condition
$$\sum_{m=0}^\infty\frac{\cos((2m+1)\gamma-\pi r/2)}{(2m+1)^r}=0.$$
The following result holds:
$$d(W_\infty^r(\mathbb T),\T_{n-1},C(\mathbb T))=\frac4{\pi n^r}\left|\sum_
{m=0}^\infty\frac{\sin((2m+1)\gamma-\pi r/2)}{(2m+1)^r}\right|.$$


\section{The Hardy spaces}

Now we consider optimal recovery problems for classes of analytic
functions. First we give some definitions. Denote by $\mathcal H_{p}(D)$, $
1\le p\le\infty$, the Hardy space, i.e., the set of functions $f(\cdot)$
analytic in the unit disk $D=\{z\in\mathbb C\mid|z|<1\}$ and satisfying
\begin{gather*}
\sup_{0<r<1}\frac1{2\pi}\int_{\mathbb T}|f(re^{it})|^p\,dt=A_p^p<\infty,
\quad 1\le p<\infty,\\
\sup_{z\in D}|f(z)|=A_\infty<\infty,\quad p=\infty.
\end{gather*}
Every function $f(\cdot)\in\mathcal H_{p}(D)$ associates with the unique
function $\tilde f(\cdot)\in L_p(\partial D)$ ($\partial D$ is the boundary
of $D$) by the rule:
$$\tilde f\left(e^{it}\right)=\lim_{r\to1}f\left(re^{it}\right)$$
for almost all $t$. Moreover, $\|\tilde f(\cdot)\|_{L_p(\partial D)}=A_p$
and for all $z\in D$ the Cauchy formula
\begin{equation}\label{i2}
f^{(k)}(z)=\frac{k!}{2\pi i}\int_{\partial D}\frac{\tilde f(\zeta)}{(\zeta-
z)^{k+1}}\,d\zeta,\quad k\in\mathbb Z_+,
\end{equation}
holds.

The subset of $L_p(\partial D)$ that consists of all such functions $\tilde
f(\cdot)$ is the set of those functions from $L_p(\partial D)$ for which
\begin{equation}\label{ii2}
\frac1{2\pi i}\int_{\partial D}\tilde f(\zeta)\zeta^k\,d\zeta=0,\quad k\in
\mathbb N.
\end{equation}
For simplicity we shall use the same notation for $f(\cdot)\in\mathcal H_p(
D)$ and its boundary values.

The set
$$H_p(D)=\{f(\cdot)\in\mathcal H_p(D)\mid\|f(\cdot)\|_{L_p(\partial D)}\le1\}
$$
we call the {\it Hardy class}.

We consider the following recovery problem: recover a value of $f(\cdot)$
at a point $\tau\in D$ on the Hardy class $H_p(D)$ from the information
$$f(z_1),f'(z_1),\ldots,f^{(k_1-1)}(z_1),\ldots,f(z_n),f'(z_n),\ldots,f^{(k
_n-1)}(z_n),$$
where $z_1,\ldots,z_n$ are distinct points from the disk $D$. We are
interested in an optimal method of recovery.

In accordance with the general notation here $X=\mathcal H_p(D)$, $Y=
\mathbb C^N$, $N=k_1+\ldots+k_n$, $\Re\la x',f(\cdot)\ra=\Re f(\tau)$, $A=H
_p(D)$, and $F\colon A\to\mathbb C^N$,
$$Ff(\cdot)=\left(f(z_1),f'(z_1),\ldots,f^{(k_1-1)}(z_1),\ldots,f(z_n),f'(z
_n),\ldots,f^{(k_n-1)}(z_n)\right).$$
The associated problem has the form
\begin{equation}\label{as}
\Re f(\tau)\to\max,\quad f^{(k)}(z_j)=0,\ j=1,\ldots n,\ k=0,1,\ldots,k_j-1
,\quad f(\cdot)\in H_p(D).
\end{equation}
In contrast to the real case where a solution of associated problem and
optimal method were obtained simultaneously, here we first find a solution
of \eqref{as} directly and then use it to obtain an optimal recovery
method.

Since every function $f(\cdot)\in H_p(D)$ for which
$$f(z_1)=\ldots=f^{(k_1-1)}(z_1)=\ldots=f(z_n)=\ldots=f^{(k_n-1)}(z_n)=0$$
may be represented in the form $f(z)=B(z)g(z)$ where $g(\cdot)\in H_p(D)$
and
$$B(z)=\prod_{j=1}^n\left(\frac{z-z_j}{1-\ov z_jz}\right)^{k_j},$$
it suffices to find the extremum in the problem
$$\Re g(\tau)\to\max,\quad g(\cdot)\in H_p(D).$$
Evidently, for $p=\infty$ the function $\widehat g(z)\equiv1$ is extremal.
We prove that for $1\le p<\infty$  the function $\widehat g(z)=(1-|\tau|^2)
^{1/p}(1-\ov\tau z)^{-2/p}$ is extremal. By the residue theorem we have
\begin{equation}\label{res}
g(\tau)=\frac1{2\pi i}\int_{\partial D}\frac{g(\zeta)(1-|\tau|^2)^{(p-2)/p}
}{(\zeta-\tau)(1-\ov\tau\zeta)^{(p-2)/p}}\,d\zeta=\frac1{2\pi}\int_{\mathbb
T}\frac{g(e^{it})(1-|\tau|^2)^{(p-2)/p}}{(1-\tau e^{-it})(1-\ov\tau e^{it})
^{(p-2)/p}}\,dt.
\end{equation}

Applying the H\"older inequality to the last integral we obtain that
\begin{equation}\label{iv2}
|g(\tau)|\le(1-|\tau|^2)^{-1/p}
\end{equation}
for all $g(\cdot)\in H_p(D)$. Moreover, it follows from \eqref{res} that
$$(1-|\tau|^2)^{-1/p}=\widehat g(\tau)=\|\widehat g(\cdot)\|_{L_p(\partial
D)}^p(1-|\tau|^2)^{-1/p}.$$
Thus, $\widehat g(\cdot)\in H_p(\partial D)$ and for this function \eqref
{iv2} turns to equality. Consequently, the function
$$\widehat f(z)=e^{-i\arg B(\tau)}B(z)\widehat g(z)$$
is extremal in \eqref{as}.

In accordance with \eqref{i2} and \eqref{ii2} the problem \eqref{as} may be
rewritten as follows
\begin{multline}\label{iii2}
\Re\frac1{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-\tau}\,d\zeta\to
\max,\quad\frac{k!}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{(\zeta-z_j)^{k+
1}}\,d\zeta=0,\quad j=1,\ldots,n,\\
k=0,1,\ldots,k_j-1,\quad\frac1{2\pi i}\int_{\partial D}f(\zeta)\zeta^m\,d
\zeta=0,\quad m\in\mathbb Z_+,\quad\|f(\cdot)\|_{L_p(\partial D)}\le1.
\end{multline}

It is a convex problem. We apply the Lagrange principle to it noting that
the set $A$ (see \eqref2) is defined here by a countable number of
equalities and one inequality. We include the constraints of equality type
in the Lagrange function by ``natural" way without giving more precise
descriptions since, as it was said, we apply the Lagrange principle
heuristically. We note only that an optimal recovery method is defined by
multipliers at the constraints related to the information operator (see
Remark~\ref{r2}).

The Lagrange function of the problem \eqref{iii2} is
$$\mathcal L=\Re\frac1{2\pi}\int_{\mathbb T}\biggl(\frac{-1}{e^{it}-\tau}+
\sum_{j=1}^n\sum_{k=0}^{k_j-1}\frac{\mu_{jk}k!}{(e^{it}-z_j)^{k+1}}+\sum_{
m\ge0}\lambda_me^{imt}\biggr)f(e^{it})e^{it}\,dt,$$
where $\mu_{jk},\lambda_m\in\mathbb C$, $1\le j\le n$, $0\le k\le k_j-1$, $
m\ge0$. By the Lagrange principle there exist such $\widehat\mu_{jk},
\widehat\lambda_m\in\mathbb C$, $1\le j\le n$, $0\le k\le k_j-1$, $m\ge0$,
that $\mathcal L$ attains its minimum at the point $\widehat f(\cdot)$ on
the set $\{f(\cdot)\in L_p(\partial D)\mid\|f(\cdot)\|_{L_p(\partial D)}\le
1\}$. Hence it follows that for $z=e^{it}$ and $1<p<\infty$
\begin{equation}\label{LL}
L(z)=C\ov z\ov{\widehat f(z)}|\widehat f(z)|^{p-2}=\frac{C_1\ov{B(z)}}{(z-
\tau)(1-\ov\tau z)^{(p-2)/p}},
\end{equation}
where $L(z)$ is the expression in parentheses under the integral sign in
the Lagrange function. Since $\ov{B(z)}=B^{-1}(z)$ for $z=e^{it}$, we have
\begin{equation}\label{rr}
-\frac1{z-\tau}+\sum_{j=1}^n\sum_{k=0}^{k_j-1}\frac{\widehat\mu_{jk}k!}{(z-
z_j)^{k+1}}+\sum_{m\ge0}\widehat\lambda_mz^m=\frac{C_1}{B(z)(z-\tau)(1-\ov
\tau z)^{(p-2)/p}}.
\end{equation}
The function in the right-hand side of this equality is analytic in the
disk $D$ with the exception of points $\tau,z_1,\ldots,z_n$ where it has
poles. If we multiply the both sides of \eqref{rr} by $z-\tau$ and
substitute $z=\tau$, we get $C_1=-B(\tau)(1-|\tau|^2)^{(p-2)/p}$. In a
similar way we obtain
$$\widehat\mu_{jk}=\frac{B(\tau)(1-|\tau|^2)^{(p-2)/p}}{k!(k_j-k-1)!}\left(
\frac{(1-\ov z_jz)^{k_j}}{\omega_j(z)(\tau-z)(1-\ov\tau z)^{(p-2)/p}}\right
)^{(k_j-k-1)}_{\Big|z=z_j},$$
where
$$\omega_j(z)=\prod_{\substack{m=1\\m\ne j}}^n\left(\frac{z-z_m}{1-\ov z_mz
}\right)^{k_m}.$$

We prove now that the method defined by $\widehat\mu_{jk}$ (which are
well-defined for all $1\le p\le\infty$; for $p=\infty$ all expressions
involving $p$ are understood as the limit values as $p\to\infty$) is
optimal. Indeed, for all $1\le p\le\infty$ the equality \eqref{rr} holds
with some $\widehat\lambda_m$, $m\ge0$ (we do not need explicit expressions
for them). Let $1\le p<\infty$. Then for all $f\in H_p(D)$ taking into
account the last equality of \eqref{LL} and applying the H\"older
inequality we obtain
\begin{multline*}
\biggl|f(\tau)-\sum_{j=1}^n\sum_{k=0}^{k_j-1}\widehat\mu_{jk}f^{(k)}(z_j)
\biggr|\\
=\left|\frac1{2\pi i}\int_{\partial D}\left(\sum_{m\ge0}\widehat\lambda_mz^
m+\frac{B(\tau)}{(1-|\tau|^2)^{1/p}}\ov z\ov{\widehat f(z)}|\widehat f(z)|^
{p-2}\right)f(z)\,dz\right|\\
=\frac{|B(\tau)|}{(1-|\tau|^2)^{1/p}}\frac1{2\pi}\left|\int_{\mathbb T}\ov{
\widehat f(e^{it})}|\widehat f(e^{it})|^{p-2}f(e^{it})\,dt\right|\\
\le\frac{|B(\tau)|}{(1-|\tau|^2)^{1/p}}\|\widehat f(\cdot)\|_{L_p(\partial
D)}^{p-1}\|f(\cdot)\|_{L_p(\partial D)}\le|\widehat f(\tau)|.
\end{multline*}
For $p=\infty$ using the fact that the integral of the Poisson kernel
equals $1$ we have
\begin{multline*}
\biggl|f(\tau)-\sum_{j=1}^n\sum_{k=0}^{k_j-1}\widehat\mu_{jk}f^{(k)}(z_j)
\biggr|
=\left|\frac1{2\pi i}\int_{\partial D}\left(\sum_{m\ge0}\widehat\lambda_mz^
m+\frac{B(\tau)(1-|\tau|^2)}{B(z)(z-\tau)(1-\ov\tau z)}\right)f(z)\,dz
\right|\\
=|B(\tau)|\frac1{2\pi}\left|\int_{\mathbb T}\frac{1-|\tau|^2}{|1-\ov\tau e^
{it}|^2}f(e^{it})\,dt\right|\le|B(\tau)|=|\widehat f(\tau)|.
\end{multline*}

It follows from Theorem~\ref{T1} that the error of optimal recovery equals
$|\widehat f(\tau)|$. Thus it is proved that the method
$$f(\tau)\approx\sum_{j=1}^n\sum_{k=0}^{k_j-1}\widehat\mu_{jk}f^{(k)}(z_j)
$$
is optimal. In particular, for one point $z_1=0$ with the multiplicity $n$
(that is we consider the problem of optimal recovery from Taylor
coefficients at the zero) for $p=\infty$ we have
\begin{equation}\label{v2}
f(\tau)\approx\sum_{j=0}^{n-1}\tau^j(1-|\tau|^{2(n-j)})\frac{f^{(j)}(0)}{j!
}.
\end{equation}
The optimality of these methods was obtained, without using the Lagrange
principle, in \cite{Os1}, \cite{Os2} ($p=\infty$), and \cite{FM} ($1\le p<
\infty$).

\section{The Hardy--Sobolev spaces}

For $p=\infty$ we can obtain a more general result rather than formula
\eqref{v2}.

Denote by $H^r_\infty(D)$ the {\it Hardy--Sobolev\/} class which is the set
of all functions analytic in the unit disk $D$ for which $|f^{(r)}(z)|\le1
$, $z\in D$. Consider the problem of optimal recovery of $f(\cdot)\in H_
\infty^r(D)$ at a point $\tau$ (without loss of generality we may assume
that $\tau\in(0,1)$) from the information $f(0),f'(0),\ldots,f^{(n+r-1)}(0)
$.

From the equality
$$f(z)=S_{r-1}(z)+\int_0^z\frac{(z-\xi)^{r-1}}{(r-1)!}f^{(r)}(\xi)\,d\xi,
\quad S_{r-1}(z)=\sum_{k=0}^{r-1}\frac{f^{(k)}(0)}{k!}z^k,$$
and the Cauchy formula for $f^{(r)}(\xi)$ we have
\begin{equation}\label{vi2}
f(z)=S_{r-1}(z)+\frac1{2\pi}\int_{\mathbb T}\sum_{k=0}^\infty\frac{k!}{(k+r
)!}z^{k+r}e^{-ikt}f^{(r)}(e^{it})\,dt.
\end{equation}
Thus the associated problem can be written as follows
\begin{multline*}
\Re\left(S_{r-1}(\tau)+\frac1{2\pi}\int_{\mathbb T}\sum_{k=0}^\infty\frac{k
!}{(k+r)!}\tau^{k+r}e^{-ikt}f^{(r)}(e^{it})\,dt\right)\to\max,\\
f(0)=\ldots=f^{(r-1)}(0)=0,\quad\frac{k!}{2\pi}\int_{\mathbb T}f^{(r)}(e^{i
t})e^{-ikt}\,dt=0,\
k=0,\ldots,n-1,\\
\frac1{2\pi}\int_{\mathbb T}f^{(r)}(e^{it})e^{imt}\,dt=
0,\ m\in\mathbb N,\quad
\|f^{(r)}(\cdot)\|_{L_\infty(\partial D)}\le1.
\end{multline*}

We prove that the function
$$\widehat f(z)=\frac{n!}{(n+r)!}z^{n+r}$$
is extremal in this problem. Assume that there exists a function $f_0(\cdot
)\in H_\infty^r(D)$ for which $f_0(0)=\ldots=f^{(n+r-1)}(0)=0$ and $|f_0(
\tau)|>\widehat f(\tau)$. Without loss of generality we may assume that $f_
0(\tau)>0$. Since
$$g_0(z)=\frac{f_0(z)+\ov{f_0(\ov z)}}2$$
has the same properties as $f_0(\cdot)$ and is real on the real axis, we
assume from the very beginning that the function $\widehat f(\cdot)$ is
real on the real axis. Put
$$F(z):=\widehat f(z)-\rho f_0(z),\quad\rho=\frac{\widehat f_0(\tau)}{f_0(
\tau)}.$$
The function $F(\cdot)$ has at least $n+r+1$ zeros on the interval $(-1,1)$
taking into account multiplicities. Consequently, by Rolle's theorem $F^{(r
)}(\cdot)$ has at least $n+1$ zeros on this interval. For $z\in\partial D$
we have
$$|\widehat f^{(r)}(z)-F^{(r)}(z)|=\rho|f_0^{(r)}(z)|\le\rho<1\equiv|
\widehat f^{(r)}(z)|.$$
Since $\widehat f^{(r)}(\cdot)$ has exactly $n$ zeros in $D$ (counting
multiplicities) Rouche's theorem implies that the function $F^{(r)}(\cdot)$
must have the same number of zeros. This contradiction proves that the
function $\widehat f(\cdot)$ is extremal.

Now let us write out the Lagrange function of the considered problem
\begin{multline*}
\mathcal L=\Re\left(-S_{r-1}(\tau)+\sum_{k=0}^{r-1}\mu_kf^{(k)}(0)\right.\\
\left.+\frac1{2
\pi}\int_{\mathbb T}\left(-\sum_{k=0}^\infty\frac{k!}{(k+r)!}\tau^{k+r}e^{-
ikt}+\sum_{k=0}^{n-1}\mu_{k+r}k!e^{-ikt}+\sum_{m=1}^\infty\lambda_m
e^{imt}\right)f^{(r)}(e^{it})\,dt\right),
\end{multline*}
where $\mu_k,\lambda_m\in\mathbb C$, $0\le k\le n+r-1$, $m\in\mathbb N$.
According to the Lagrange principle there exist such $\widehat\mu_k,
\widehat\lambda_m\in\mathbb C$ that $\mathcal L$ attains its minimum at the
point $\widehat f(\cdot)$ on the set of functions for which $\|f^{(r)}(
\cdot)\|_{L_\infty(\partial D)}\le1$.

Clearly,
$$\widehat\mu_k=\frac1{k!}\tau^k,\quad k=0,\ldots,r-1.$$
Denoting by $L(\cdot)$ the expression in parentheses under the integral
sign we have
$$L(e^{it})=-\ov{\widehat f^{(r)}(e^{it})}|L(e^{it})|.$$
Consider the Fourier-series expansion of $|L(e^{it})|$
$$|L(e^{it})|=\sum_{k=-\infty}^\infty a_ke^{ikt}.$$
Taking into account the fact that it is a real function we have $a_{-k}=\ov
a_k$. Thus
\begin{multline*}
-\sum_{k=0}^\infty\frac{k!}{(k+r)!}\tau^{k+r}e^{-ikt}+\sum_{k=0}^{n-1}
\widehat\mu_{k+r}k!e^{-ikt}+\sum_{m=1}^\infty\widehat\lambda_me^{imt}\\
=-e^{-int}\left(a_0+\sum_{k=1}^\infty a_ke^{ikt}+\sum_{k=1}^\infty\ov a_ke^
{-ikt}\right).
\end{multline*}
Consequently,
\begin{align*}
\ov a_k&=\frac{(n+k)!}{(n+k+r)!}\tau^{n+k+r},\\
a_k&=\frac{(n-k)!}{(n-k+r)!}\tau^{n-k+r}-(n-k)!\widehat\mu_{n-k+r},\quad k=
1,\ldots,n.
\end{align*}
Hence we obtain
$$\widehat\mu_{n+r}=\frac{\tau^{m+r}}{(m+r)!}\left(1-\frac{(m+r)!}{m!}\frac
{(2n-m)!}{(2n-m+r)!}\tau^{2(n-m)}\right).$$
By the direct verification (similar to the case described above) it can be
proved that the method
\begin{equation}\label{opt}
f(\tau)\approx\sum_{k=0}^{r-1}\frac{f^{(k)}(0)}{k!}\tau^k+\sum_{k=r}^{n+r-1
}\left(1-\frac{k!}{(k-r)!}\frac{(2n+r-k)!}{(2n+2r-k)!}|\tau|^{2(n+r-k)}
\right)\frac{f^{(k)}(0)}{k!}\tau^k
\end{equation}
is optimal.

Thus we have proved the following theorem.

\begin{theorem}{\bf(on optimal recovery from Taylor coefficients).}
Let $r\in\mathbb Z_+$, $n\in\mathbb Z$, and $\tau\in D$. Then the method $
\eqref{opt}$ is optimal method of recovery on the class $W^rH_\infty(D)$
from the information
$$\Tay_{n+r}f(\cdot)=(f(0),f'(0),\ldots,f^{(n+r-1)}(0)).$$
Moreover,
$$E(f(\tau),W^rH_\infty,\Tay_{n+r})=\frac{n!}{(n+r)!}|\tau|^{n+r}.$$
\end{theorem}

For $r=1$ this result was obtained by Newman (see \cite[p.~42]{MiR}).

\section{Optimal recovery from the values at the equidistant system of
points on a circle}

Consider now the problem of optimal recovery of a value $f(\tau)$, $f(\cdot
)\in H^1_\infty(D)$, $\tau\in D$, from the values $f(\tau_j)$, $j=0,\ldots,
n-1$, where $\{\tau_j\}$ is the system of equidistant points on the circle
of the radius $0<\rho<1$: $\tau_j=\rho e^{ij2\pi/n}$. From \cite{HN} it
follows that the function
$$\widehat f(z):=\varepsilon\frac{z^n-\rho^n}n,\quad\varepsilon=e^{-i\arg(
\tau^n-\rho^n)},$$
is extremal in the associated problem
$$\Re f(\tau)\to\max,\quad f(\tau_j)=0,\quad j=0,\ldots,n-1,\quad f\in W^1H
_\infty(D).$$
In view of \eqref{vi2} this problem may be rewritten as follows
\begin{multline*}
\Re\left(f(0)+\frac1{2\pi}\int_{\mathbb T}\sum_{k=0}^\infty\frac{\tau^{k+1}
}{k+1}e^{-ikt}f'(e^{it})\,dt\right)\to\max,\\
f(0)+\frac1{2\pi}\int_{\mathbb T}\sum_{k=0}^\infty\frac{\tau_j^{k+1}}{k+1}e
^{-ikt}f'(e^{it})\,dt=0,\ j=0,\ldots,n-1,\\
\frac1{2\pi}\int_{\mathbb T}f'(e^{it})e^{imt}\,dt=0,\ m\in\mathbb N,\quad\|
f'(\cdot)\|_{L_\infty(\partial D)}\le1.
\end{multline*}

The Lagrange function of this problem is
\begin{multline*}
\mathcal L=\Re\left(-f(0)+\sum_{j=0}^{n-1}\mu_jf(0)\right.\\
\left.+\frac1{2\pi}\int_{\mathbb T}\left(-\sum_{k=0}^\infty\frac{\tau^{k+1}
}{k+1}e^{-ikt}+\sum_{j=0}^{n-1}\mu_j\sum_{k=0}^\infty\frac{\tau_j^{k+1}}{k+
1}e^{-ikt}+\sum_{m=1}^\infty\lambda_me^{imt}\right)f'(e^{it})\,dt\right),
\end{multline*}
where $\mu_j,\lambda_m\in\mathbb C$, $0\le j\le n-1$, $m\in\mathbb N$.
According to the Lagrange principle there exist such $\widehat\mu_j,
\widehat\lambda_m\in\mathbb C$ that $\mathcal L$ attains its minimum at the
point $\widehat f(\cdot)$ on the set of functions for which $\|f'(\cdot)\|_
{L_\infty(\partial D)}\le1$.

It is clear that
$$\sum_{j=0}^{n-1}\widehat\mu_j=1.$$
Denoting by $L(\cdot)$ the expression in parentheses under the integral
sign we have
$$L(e^{it})=-\ov{\widehat f'(e^{it})}|L(e^{it})|.$$
Consider the Fourier-series expansion of $|L(e^{it})|$
$$|L(e^{it})|=\sum_{k=-\infty}^\infty a_ke^{ikt}.$$
Taking into account the fact that it is a real function we have $a_{-k}=\ov
a_k$. Thus
$$\varepsilon e^{i(n-1)t}\left(\sum_{k=0}^\infty\nu_ke^{-ikt}-\sum_{m=1}^
\infty\widehat\lambda_me^{imt}\right)=a_0+\sum_{k=1}^\infty a_ke^{ikt}+\sum
_{k=1}^\infty\ov a_ke^{-ikt},$$
where
$$\nu_k=\frac1{k+1}\left(\tau^{k+1}-\sum_{j=0}^{n-1}\widehat\mu_j\tau_j^{k+
1}\right).$$
Hence $a_k=\varepsilon\nu_{n-k-1}$, $\ov a_k=\varepsilon\nu_{n+k-1}$, $k=1
,\ldots,n-1$.

Assume for simplicity that $\tau\in(-1,1)$. Put $\alpha=\tau/\rho$. Then
$$\frac1{n-k}\left(\alpha^{n-k}-\sum_{j=0}^{n-1}\widehat\mu_j\xi_j^{-k}
\right)=\frac{\rho^{2k}}{n+k}\left(\alpha^{n+k}-\sum_{j=0}^{n-1}\ov{
\widehat\mu}_j\ov\xi_j^k\right),$$
where $\xi_j=e^{ij2\pi/n}$, $j=0,\ldots,n-1$. Putting
$$b_k:=\sum_{j=0}^{n-1}\widehat\mu_j\xi_j^{-k}$$
we have
$$\sum_{j=0}^{n-1}\ov{\widehat\mu}_j\ov\xi_j^k=\sum_{j=0}^{n-1}\ov{\widehat
\mu}_j(\ov\xi_j)^{-(n-k)}=\ov b_{n-k}.$$
Thus we obtain the system
$$\frac1{n-k}(\alpha^{n-k}-b_k)=\frac{\rho^{2k}}{n+k}(\alpha^{n+k}-\ov b_{n
-k}),\quad k=1,\ldots,n-1.$$
Hence
$$\left\{
\begin{aligned}
\frac1{n-k}b_k-\frac{\rho^{2k}}{n+k}\ov b_{n-k}&=\frac{\alpha^{n-k}}{n-k}-
\frac{\alpha^{n+k}}{n+k}\rho^{2k},\\
\frac{\rho^{2(n-k)}}{2n-k}b_k-\frac1k\ov b_{n-k}&=\frac{\alpha^{2n-k}}{2n-k}
\rho^{2(n-k)}-\frac{\alpha^k}k.
\end{aligned}\right.$$
Consequently,
$$b_k=\frac{\alpha^{n-k}(1-q_k\alpha^n\rho^{2n})-p_k\alpha^k\rho^{2k}(1-
\alpha^n)}{1-q_k\rho^{2n}},\quad k=1,\ldots,n-1,$$
where
$$p_k=\frac{n-k}{n+k},\quad q_k=\frac k{2n-k}p_k.$$
Since $b_0=1$ we have
$$\widehat\mu_j=\frac1n\sum_{k=0}^{n-1}b_k\xi_k^j.$$
As above, the direct verification leads to the fact that the method
$$f(\tau)\approx\frac1n\sum_{j=0}^{n-1}\left(\sum_{k=0}^{n-1}b_k\xi_k^j
\right)f(\tau_j)$$
is optimal for the considered problem.


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\noindent
\textit{Georgii G. Magaril-Il'yaev}
\hfill
{\footnotesize
 \texttt{georg@magaril.mccme.ru}\\
Department of Higher Mathematics, Moscow State Institute of Radio
Engineering, Electronics and Automation (Technology University), pr.
Vernadskogo 78, 117454 Moscow, Russia
}

\noindent
\textit{Konstantin Yu. Osipenko}
\hfill
{\footnotesize
 \texttt{konst@osipenko.mccme.ru}\\
Department of Higher Mathematics, MATI --- Russian State Technological
University, Orshanskaya 3, 121552 Moscow, Russia
}

\noindent
\textit{Vladimir M. Tikhomirov}
\hfill
{\footnotesize
 \texttt{tikh@tikhomir.mccme.ru}\\
Department of General Control Problems, Moscow State University, Vorob'evy
Gory, 119899 Moscow, Russia
}
\end{document}