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\begin{document}
\begin{frontmatter}
\title{Recovery of Derivatives for Functions Defined on the Semiaxis}
\author{K.~Yu.~Osipenko\fnref{myfootnote1}}
%\cortext[cor]{Corresponding author}
\fntext[myfootnote1]{The research was carried out with the financial support of the Russian Foundation for Basic Research (grant no.\ 17-01-00649)}
\address{Moscow State University,\\
Institute for Information Transmission Problems,
Russian Academy of Sciences, Moscow}
\ead{kosipenko@yahoo.com}
\begin{abstract}
The paper is concerned with the recovery problem of derivatives at the origin from noisy information about functions defined on the semiaxis for the Sobolev class. The problem of S.~B.~Stechkin about approximation of derivatives by bounded
linear functionals is also studied.
\end{abstract}

\begin{keyword}
optimal recovery \sep inequalities for derivatives \sep approximation by bounded
linear functionals
\MSC[2010] 41A65 \sep 41A44 \sep 41A17 \sep 49N30
\end{keyword}

\end{frontmatter}


\section{Setting of Problems}


Let $\WW$, $n\in\mathbb N$, be the Sobolev space of functions $x\cd\in\Lt$ such that the $(n-1)$-st derivative is locally absolutely continuous and $x^{(n)}\cd\in\Lt$. Set
$$\ww=\{\,x\cd\in\WW:\|x^{(n)}\cd\|_{\Lt}\le1\,\}.$$

We consider the problem of recovery of $x^{(k)}(0)$, $0\le k<n$, on the class $\ww$ by inaccurate information about $x\cd$. We assume that for every function $x\cd\in\ww$ we know $y\cd\in\Lt$ such that $$\|x\cd-y\cd\|_{\Lt}\le\delta,\quad\delta>0.$$
For a given $y\cd$ we have to construct an approximate value of $x^{(k)}(0)$.

As recovery methods we consider all possible mappings $m\colon\Lt\to\mathbb R$. The error of a method $m$ is defined as follows
$$e_k(\ww,\delta,m)=\sup_{\substack{x\cd\in\ww,\ y\cd\in\Lt\\\|x\cd-y\cd\|_{\Lt}\le\delta}}|x^{(k)}(0)-m(y\cd)|.$$
The quantity
$$E_k(\ww,\delta)=\inf_{m\colon\Lt\to\mathbb R}e_k(\ww,\delta,m)$$
is known as the optimal recovery error and a method for which this infimum is attained is called optimal.

Note that optimal recovery problems are closely connected with statistical estimation problems. Details may be found in \cite{Donoho}.

We also study the problem of best approximation of $x^{(k)}(0)$ on the class $\ww$ by linear continuous functionals on $\Lt$ with the norm not greater than some fixed positive number $N$ (this problem is known as Stechkin's problem). It relies on finding the value
$$S_k(\ww,N)=\infp_{\substack{y^*\in\Lt\\||y^*\|_{\Lt}\le N}}\sup_{x\cd\in\ww}|x^{(k)}(0)-\la y^*,x\cd\ra|,$$
and also a functional delivering the lower bound which is called extremal.

Solutions of the formulated problems are closely connected with the problems of exact constants in Kolmogorov-type inequalities for derivatives. For the semiaxis the corresponding results which we essentially use here where obtained in \cite{Ka} and \cite{Lu}. For $\mathbb R$ the analogous problems of recovery and approximation by bounded linear functionals were considered in \cite{MO}. The range of problems connected with Stechkin's problem was elucidated in the survey paper \cite
{Ar1}.


\section{Main Results}


It may be obtained from \cite{Lu} that there exists a function $\wx\cd\in\mathcal W_2^{2n}(\mathbb R_+)$ such that for all $x\cd\in\WW$ the following equality
\begin{equation}\label{eq}
x^{(k)}(0)=\ii x(t)\ov{\wx(t)}\,dt+\ii x^{(n)}(t)\ov{\wx^{(n)}(t)}\,dt
\end{equation}
holds. We give the explicit form of $\wx\cd$ and prove the corresponding result for completeness.

Put
$$\lambda_j=e^{(n+2j-1)\frac{i\pi}{2n}},\quad j=1,\ldots,n,\quad
A=\begin{pmatrix}
\lambda_1^n&\ldots&\lambda_n^n\\
\hdotsfor{3}\\
\lambda_1^{2n-1}&\ldots&\lambda_n^{2n-1}\end{pmatrix},$$
and $A_{s,j}$ is the cofactor of $\lambda_j^{n+s-1}$. Denote by $|A|$ the determinant of matrix $A$.

\begin{lemma}
Set
$$\wx(t)=\frac{(-1)^{n-k}}{|A|}\sum_{j=1}^nA_{n-k,j}e^{\lambda_jt}.$$
Then for all $x\cd\in\WW$ \eqref{eq} holds, moreover,
\begin{equation}\label{xkk}
\wx^{(k)}(0)=\wA^2,
\end{equation}
where
$$\wA=\frac1{\sin^{1/2}((2k+1)\alpha)}\prod_{j=1}^k\cot j\alpha,\quad\alpha=\frac\pi{2n}.$$
\end{lemma}

\begin{proof}
Since $\lambda_j^{2n}=(-1)^{n-1}$ for all $j=1,\ldots,n$ we have
$$\wx(t)+(-1)^n\wx^{(2n)}(t)=0.$$
In view of the fact that $\RE\lambda_j<0$ for all $j=1,\ldots,n$, $\wx\cd\in\mathcal W_2^{2n}(\mathbb R_+)$.
Thus, for every $x\cd\in\WW$ we have
$$\ii x(t)\ov{\wx(t)}\,dt+(-1)^n\ii x(t)\ov{\wx^{(2n)}(t)}\,dt=0.$$
Using integration by parts we obtain
\begin{multline}\label{aa}
\ii x(t)\ov{\wx(t)}\,dt+(-1)^n\sum_{p=1}^n(-1)^px^{(p-1)}(0)\ov{\wx^{(2n-p)}(0)}\\
+\ii x^{(n)}(t)\ov{\wx^{(n)}(t)}\,dt=0.
\end{multline}
For all $s,p=1,\ldots,n$ we have
$$\sum_{j=1}^nA_{s,j}\lambda_j^{n+p-1}=\delta_{p,s}|A|,$$
where $\delta_{p,s}$ is the Kronecker delta. Consequently,
\begin{equation}\label{ab}
\wx^{(2n-p)}(0)=\frac{(-1)^{n-k}}{|A|}\sum_{j=1}^nA_{n-k,j}\lambda_j^{2n-p}=(-1)^{n-k}
\delta_{n-p+1,n-k}.
\end{equation}
Substituting \eqref{ab} in \eqref{aa} we obtain \eqref{eq}.

Now let us calculate $\wx^{(k)}(0)$. We have
$$\wx^{(k)}(0)=\frac{(-1)^{n-k}}{|A|}\sum_{j=1}^nA_{n-k,j}\lambda_j^k=(-1)^{n-k}\frac{|A_k|}{|A|},$$
where
$$A_k=\begin{pmatrix}
\lambda_1^n&\ldots&\lambda_n^n\\
\hdotsfor{3}\\
\lambda_1^{2n-k-2}&\ldots&\lambda_n^{2n-k-2}\\
\lambda_1^{k}&\ldots&\lambda_n^{k}\\
\lambda_1^{2n-k}&\ldots&\lambda_n^{2n-k}\\
\hdotsfor{3}\\
\lambda_1^{2n-1}&\ldots&\lambda_n^{2n-1}\end{pmatrix}.$$
Since
$$\lambda_j^m=\sigma_m\mu_m^{j-1},$$
where
$$|\sigma_m|=1,\quad\mu_m=e^{i\frac{\pi m}n},$$
matrices $A$ and $A_k$ may be rewritten in the forms
\begin{align*}
A&=\begin{pmatrix}
\sigma_n&\ldots&\sigma_n\mu_n^{n-1}\\
\hdotsfor{3}\\
\sigma_{2n-1}&\ldots&\sigma_{2n-1}\mu_{2n-1}^{n-1}\end{pmatrix},\\[8pt]
A_k&=\begin{pmatrix}
\sigma_n&\ldots&\sigma_n\mu_n^{n-1}\\
\hdotsfor{3}\\
\sigma_{2n-k-2}&\ldots&\sigma_{2n-k-2}\mu_{2n-k-2}^{n-1}\\
\sigma_k&\ldots&\sigma_k\mu_k^{n-1}\\
\sigma_{2n-k}&\ldots&\sigma_{2n-k}\mu_{2n-k}^{n-1}\\
\hdotsfor{3}\\
\sigma_{2n-1}&\ldots&\sigma_{2n-1}\mu_{2n-1}^{n-1}\end{pmatrix}.
\end{align*}
If we put $x\cd=\wx\cd$ in \eqref{eq}, then we obtain
\begin{equation}\label{xk}
\wx^{(k)}(0)=\|\wx\cd\|_{\Lt}^2+\|\wx^{(n)}\cd\|_{\Lt}^2.
\end{equation}
Thus, $\wx^{(k)}(0)>0$. Using this fact and the formula for the Vandermonde determinant we have
$$\wx^{(k)}(0)=\prod_{\substack{j=1\\j\ne n-k}}^n\frac{|\mu_k-\mu_{n+j-1}|}{|\mu_{2n-k-1}-\mu_{n+j-1}|}.$$
Since
$$|\mu_p-\mu_s|=2|\sin(p-s)\alpha|,$$
we obtain
\begin{multline*}
\wx^{(k)}(0)=\prod_{\substack{j=1\\j\ne n-k}}^n\frac{|\sin(n+j-k-1)\alpha|}{|\sin(n+j+k)\alpha|}\\
=\frac1{\sin((2k+1)\alpha)}
\frac{\prod_{j=k+1}^{k+n}\sin j\alpha}{\prod_{j=1}^k\sin j\alpha\prod_{j=1}^{n-k-1}\sin j\alpha}\\
=\frac1{\sin((2k+1)\alpha)}
\frac{\prod_{j=k+1}^{n-1}\sin j\alpha\prod_{j=n+1}^{n+k}\sin j\alpha}{\prod_{j=1}^k\sin j\alpha\prod_{j=1}^{n-k-1}\sin j\alpha}\\=
\frac1{\sin((2k+1)\alpha)}\prod_{j=1}^k\cot j\alpha
\frac{\prod_{j=k+1}^{n-1}\sin j\alpha}{\prod_{j=1}^{n-k-1}\sin j\alpha}\\=
\frac1{\sin((2k+1)\alpha)}\prod_{j=1}^k\cot j\alpha\prod_{j=1}^{n-k-1}\cot j\alpha=\frac1{\sin((2k+1)\alpha)}\prod_{j=1}^k\cot^2j\alpha.
\end{multline*}

\end{proof}

\begin{theorem}\label{T1}
The following equality
$$E_k(\ww,\delta)=\wA\left(\frac{2n}{2n-2k-1}\right)^{\frac{2n-2k-1}{4n}}
\left(\frac{2n}{2k+1}\right)^{\frac{2k+1}{4n}}\delta^{\frac{2n-2k-1}{2n}}$$
holds. Moreover,
\begin{equation}\label{met}
\wm(y)=\beta^{k+1}\ii y(t)\ov{\wx(\beta t)}\,dt,
\end{equation}
where
$$\beta=\left(\frac{2n-2k-1}{2k+1}\right)^{\frac1{2n}}\delta^{-\frac1n},$$
is the optimal method of recovery.
\end{theorem}

\begin{proof}
From \eqref{eq} by the Cauchy-Schwarz inequality we obtain
\begin{multline*}
|x^{(k)}(0)|\le\left(\|\wx\cd\|_{\Lt}^2+\|\wx^{(n)}\cd\|_{\Lt}^2\right)^{1/2}\\
\times\left(\|x\cd\|_{\Lt}^2+\|x^{(n)}\cd\|_{\Lt}^2\right)^{1/2}.
\end{multline*}
Taking into account \eqref{xk} we have
\begin{equation}\label{ixk}
|x^{(k)}(0)|\le \wA\left(\|x\cd\|_{\Lt}^2+\|x^{(n)}\cd\|_{\Lt}^2\right)^{1/2}.
\end{equation}
Put $y(t)=\wx(bt)$, $b>0$. Then
$$\|y\cd\|_{\Lt}^2=\frac1b\|\wx\cd\|_{\Lt}^2,\quad\|y^{(n)}\cd\|_{\Lt}^2=b^{2n-1}\|\wx^{(n)}\cd\|
_{\Lt}^2.$$
Substituting $y\cd$ into \eqref{ixk} we obtain that for all $b>0$ the inequality
$$b^k\wA^2\le \wA\left(\frac1b\|\wx\cd\|_{\Lt}^2+b^{2n-1}\|\wx^{(n)}\cd\|_{\Lt}^2\right)^{1/2}$$
holds. In view of \eqref{xk} and \eqref{xkk} we get that for all $b>0$ the inequality $f(b)\ge0$ is fulfilled, where
$$f(b)=b^{2n}\|\wx^{(n)}\cd\|_{\Lt}^2-\wA^2b^{2k+1}+\wA^2-\|\wx^{(n)}\cd\|_{\Lt}^2.$$
It is easily seen that $f\cd$ has the unique minimum on $\mathbb R_+$
$$b_0=\left(\frac{(2k+1)\wA^2}{2n\|\wx^{(n)}\cd\|_{\Lt}^2}\right)^{\frac1{2n-2k-1}}.$$
On the other hand, $f(1)=0$. Consequently, $b_0=1$. Thus,
$$\|\wx^{(n)}\cd\|_{\Lt}=\wA\sqrt{\frac{2k+1}{2n}}.$$
It follows from \eqref{xk} and \eqref{xkk} that
$$\|\wx\cd\|_{\Lt}=\wA\sqrt{\frac{2n-2k-1}{2n}}.$$

Put $\wx_1(t)=\alpha\wx(\beta t)$, $\alpha,\beta>0$. Choose $\alpha$ and $\beta$ such that
$\|\wx_1\cd\|_{\Lt}=\delta$ and  $\|\wx_1^{(n)}\cd\|_{\Lt}=1$. We have
$$\alpha^2\beta^{-1}\|\wx\cd\|_{\Lt}^2=\delta^2,\quad\alpha^2\beta^{2n-1}\|\wx^{(n)}\cd\|
_{\Lt}^2=1.$$
Hence,
\begin{align*}
\alpha&=\wA^{-1}\sqrt{\frac{2n}{2n-2k-1}}\left(\frac{2n-2k-1}{2k+1}\right)^{\frac1{4n}}
\delta^{1-\frac1{2n}},\\
\beta&=\left(\frac{2n-2k-1}{2k+1}\right)^{\frac1{2n}}\delta^{-\frac1n}.
\end{align*}

Substituting $\wx(t)=\alpha^{-1}\wx_1(t/\beta)$ in \eqref{eq} we obtain
$$x^{(k)}(0)=\alpha^{-1}\ii x(t)\ov{\wx_1(t/\beta)}\,dt+\alpha^{-1}\beta^{-n}\ii x^{(n)}(t)\ov{\wx_1^{(n)}(t/\beta)}\,dt.$$
We change variables $t=\beta s$ and put $z(s)=x(\beta s)$. Then we have that for all
$z\cd\in\WW$ the equality
\begin{equation}\label{eq1}
z^{(k)}(0)=\lambda_1\ii z(s)\ov{\wx_1(s)}\,ds+\lambda_2\ii z^{(n)}(s)\ov{\wx_1^{(n)}(s)}\,ds
\end{equation}
holds with
$$\lambda_1=\frac{\beta^{k+1}}\alpha,\quad\lambda_2=\frac1{\alpha\beta^{2n-k-1}}.$$

It follows from general results about optimal recovery of linear functionals (see, for example, \cite{MO1}) that
\begin{equation}\label{sm}
E_k(\ww,\delta)=\sup_{\substack{z\cd\in\ww\\\|z\cd\|_{\Lt}\le\delta}}|z^{(k)}(0)|.
\end{equation}
From \eqref{eq1} by the Cauchy-Schwarz inequality we obtain
\begin{equation}\label{lb}
E_k(\ww,\delta)\le\lambda_1\delta^2+\lambda_2.
\end{equation}


Let us estimate the error of method \eqref{met}, which may be written in the following way
$$\wm(y)=\lambda_1\ii y(t)\ov{\wx_1(t)}\,dt.$$
Suppose that $z\cd\in\ww$ and $\|z\cd-y\cd\|_{\Lt}\le\delta$. Taking into account \eqref{eq1} we have
\begin{multline*}
|z^{(k)}(0)-\wm(y)|\\
=\biggl|z^{(k)}(0)-\lambda_1\ii z(t)\ov{\wx_1(t)}\,dt+\lambda_1\ii (z(t)-y(t))\ov{\wx_1(t)}\,dt\biggr|\\
=\biggl|\lambda_1\ii (z(t)-y(t))\ov{\wx_1(t)}\,dt+\lambda_2\ii z^{(n)}(t)\ov{\wx_1^{(n)}(t)}\,dt\biggr|\le\lambda_1\delta^2+\lambda_2.
\end{multline*}
Consequently,
\begin{equation}\label{ub}
E_k(\ww,\delta)\le e_k(\ww,\delta,\wm)\le\lambda_1\delta^2+\lambda_2.
\end{equation}
The last inequality together with \eqref{lb} gives
\begin{multline*}
E_k(\ww,\delta)=\lambda_1\delta^2+\lambda_2\\
=\wA\left(\frac{2n}{2n-2k-1}\right)^{\frac{2n-2k-1}{4n}}
\left(\frac{2n}{2k+1}\right)^{\frac{2k+1}{4n}}\delta^{\frac{2n-2k-1}{2n}}.
\end{multline*}
Inequality \eqref{ub} implies also that $\wm$ is the optimal method of recovery.
\end{proof}

Note that the exact solution of extremal problem \eqref{sm} gives us the exact inequality
$$|x^{(k)}(0)|\le K_{nk}\|x\cd\|_{\Lt}^{\frac{2n-2k-1}{2n}}\|x^{(n)}\cd\|_{\Lt}^{\frac{2k+1}{2n}},$$

where
$$K_{nk}=\wA\left(\frac{2n}{2n-2k-1}\right)^{\frac{2n-2k-1}{4n}}
\left(\frac{2n}{2k+1}\right)^{\frac{2k+1}{4n}}.$$
It may be also obtained from exact inequality \eqref{ixk} by Proposition 4 from \cite[p.~119]{Th}.

We now proceed to the Stechkin problem.

\begin{theorem}
The following equality
$$S_k(\ww,N)=\wA^{\frac{2n}{2k+1}}\sqrt{\frac{2k+1}{2n-2k-1}}\left(\frac{2n-2k-1}{2n}\right)^{\frac n{2k+1}}N^{-\frac{2n-2k-1}{2k+1}}$$
holds. The functional
$$\la\wy^*,x\cd\ra=\beta_N^{k+1}\ii x(t)\ov{\wx(\beta_N t)}\,dt$$
where
$$\beta_N=\left(\frac{2n}{2n-2k-1}\right)^{\frac1{2k+1}}\left(\frac N{\wA}\right)^{\frac2{2k+1}}$$
is extremal.
\end{theorem}

\begin{proof}
As was proved in the optimal
recovery problem among all optimal methods there exists a method defined by
a linear continuous functional, therefore
\begin{multline*}
E_k(\ww,\delta)=\infp_{N>0}\infp_{\|y^*\|_{\Lt}\le N}\sup_{\substack{x\cd
\in\ww,\ y\cd\in\Lt\\\|x\cd-y\cd\|_{\Lt}\le\delta}}|x^{(k)}(0)-\la y^*,y
\cd\ra|\\
\le\infp_{\|y^*\|_{\Lt}\le N}\sup_{x\cd\in\ww}|x^{(k)}(0)-\la y^*,x\cd\ra|+
\delta N=S_k(\ww,N)+\delta N.
\end{multline*}
Consequently, for all $N>0$
\begin{equation}\label{SAr}
S_k(\ww,N)\ge E_k(\ww,\delta)-\delta N.
\end{equation}
We define the linear functional $\wy^*$ as follows
$$\la\wy^*,x\cd\ra=\lambda_1\ii x(t)\ov{\wx_1(t)}\,dt.$$
Then $\|\wy^*\|_{\Lt}=\lambda_1\delta$. If we choose $\delta$ such that $N=\lambda_1\delta$, then it follows from \eqref{SAr} that
$$S_k(\ww,N)\ge\lambda_2.$$
On the other hand, in view of \eqref{eq1} we have
$$S_k(\ww,N)\le\sup_{x\cd\in\ww}|x^{(k)}(0)-\la\wy^*,x\cd\ra|=
\lambda_2.$$
Consequently, $S_k(\ww,N)=\lambda_2$. If $N=\lambda_1\delta$, then
$$\delta=\delta_N=\sqrt{\frac{2n-2k-1}{2k+1}}\left(\frac{2n-2k-1}{2n}\right)^{\frac n{2k+1}}
\left(\frac{\wA}N\right)^{\frac{2n}{2k+1}}.$$
For $\delta=\delta_N$ we have
$$\lambda_2=\wA^{\frac{2n}{2k+1}}\sqrt{\frac{2k+1}{2n-2k-1}}\left(\frac{2n-2k-1}{2n}\right)^{\frac n{2k+1}}N^{-\frac{2n-2k-1}{2k+1}}.$$

The functional $\wy^*$ may be written in the following way
$$\la\wy^*,x\cd\ra=\beta^{k+1}\ii x(t)\ov{\wx(\beta t)}\,dt$$
where $\beta$ is defined in Theorem~\ref{T1}. For $\delta=\delta_N$
$$\beta=\beta_N=\left(\frac{2n}{2n-2k-1}\right)^{\frac1{2k+1}}\left(\frac N{\wA}\right)^{\frac2{2k+1}}.$$
\end{proof}

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\end{thebibliography}





\end{document}
