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\begin{document}
\begin{frontmatter}
\title{Optimal Recovery of Operators and Multidimensional Carlson Type Inequalities}
\author{K.~Yu.~Osipenko\corref{cor}\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 nos.\  14-01-00456, 14-01-00744)}
\address{MATI --- Russian State Technological University\\
Institute for Information Transmission Problems,
Russian Academy of Sciences, Moscow}
\ead{kosipenko@yahoo.com}
\begin{abstract}
The paper is concerned with recovery problems of linear multiplier-type operators from noisy information on weighted classes of functions. Optimal methods of recovery are constructed. The dual extremal problem is closely connected with Carlson type inequalities. \end{abstract}

\begin{keyword}
optimal recovery \sep linear operator \sep Fourier transform \sep inequalities for derivatives
\MSC[2010] 41A65 \sep 41A46 \sep 49N30
\end{keyword}

\end{frontmatter}

\section{General Setting}

Let $T$ be a nonempty set, $\Sigma$ be the $\sigma$-algebra of subsets of $T$, and $\mu$ be a nonnegative $\sigma$-additive measure on $\Sigma$. We denote by $L_p(T,\Sigma,\mu)$ (or simply $\lp$) the set of all $\Sigma$-measurable functions with values in $\mathbb R$ or in $\mathbb C$ for which
\begin{align*}
\|x\cd\|_{\lp}&=\biggl(\int_T|x(t)|^p\,d\mu(t)\biggr)^{1/p}<\infty,\quad1\le p<\infty,\\
\|x\cd\|_{L_\infty(T,\mu)}&=\vraisup_{t\in T}|x(t)|<\infty,\quad p=\infty.
\end{align*}

Put
\begin{gather*}
\mathcal W=\{\,x\cd\in\lp:\|\varphi\cd x\cd\|_{\lr}<\infty\,\},\\
W=\{\,x\cd\in\mathcal W:\|\varphi\cd x\cd\|_{\lr}\le1\,\},
\end{gather*}
where $1\le p,r\le\infty$, and $\varphi\cd$ is a measurable function on $T$.
Consider the problem of recovery of operator $\Lambda\colon\mathcal W\to L_q(T,\mu)$, $1\le q\le\infty$, defined by equality $\Lambda x\cd=\psi\cd x\cd$, where $\psi\cd$ is a measurable function on $T$, on the class $W$ by the information about functions $x\cd\in W$ given inaccurately. More precisely, we assume that for any function $x\cd\in W$ we know $y\cd\in\lpn$, where $T_0$ is not empty $\mu$-measurable subset of $T$, such that $\|x\cd-y\cd\|_{\lpn}\le\delta$, $\delta\ge0$. We want to approximate the value $\Lambda x\cd$ knowing $y\cd$.

As recovery methods we consider all possible mappings $$m\colon\lpn\to\lqq.$$
The error of a~method~$m$ is defined as
$$e(p,q,r,m)=\sup_{\substack{x\cd\in W,\ y\cd\in\lpn\\\|x\cd-y\cd\|_{\lpn}\le\delta}}\|\Lambda x\cd-m(y)\cd\|_{\lqq}.$$

The quantity
\begin{equation}\label{p1}
E(p,q,r)=\inf_{m\colon\lpn\to\lqq}e(p,q,r,m)
\end{equation}
is known as the optimal recovery error, and a method on which this infimum is attained is called optimal.

Various settings of optimal recovery theory and examples of such problems may be found in \cite{MO}, \cite{MR}, \cite{TW}, \cite{TWW},  \cite{Pl}, \cite{Os}. Much of them are devoted to optimal recovery of linear functionals. There are not so many results about optimal recovery of linear operators when non-Euclidean metrics is used (\cite[Theorem 12 on p. 45]{MR}, \cite{MO1}, \cite{Os1}). In \cite{Os1} we considered problem \eqref{p1} when any two of $p$, $q$, and $r$ coincide. Here we analyze the case when all metrics can be different and $1\le q<p,r<\infty$. We construct optimal method of recovery, find its error, and apply this result to obtain exact constants in Carlson type inequalities. The case $p=\infty$ and/or $r=\infty$ requires a slightly different approach. Some particular results of such kind may be found in \cite{MOs3} ($T=\mathbb Z$) and \cite{MOs4} ($T=\mathbb R$).

\section{Main results}

Let $\chi_0\cd$ be the characteristic function of the
set $T_0$:
$$\chi_0(t)=\begin{cases}1,&t\in T_0,\\
0,&t\notin T_0.\end{cases}$$

\begin{theorem}\label{T1}
Let $1\le q<p,r<\infty$, $\lambda_1,\lambda_2\ge0$, $\lambda_1+\lambda_2>0$, $\varphi(t)\ne0$ for almost all $t\in T\setminus T_0$, $\wx(t)=\wx(t,\lambda_1,\lambda_2)\ge0$ be a solution of equation
\begin{equation}\label{ko1}
-q|\psi(t)|^q+p\lambda_1x^{p-q}(t)\chi_0(t)+r\lambda_2|\varphi(t)|^
rx^{r-q}(t)=0,
\end{equation}
and $\lambda_1$, $\lambda_2$ such that
\begin{equation}\label{nez}
\begin{aligned}
\iTn\wx^p(t)\,d\mu(t)\le\delta^p,&\quad\iT|\varphi(t)|^r
\wx^r(t)\,d\mu(t)\le1,\\
\lambda_1\biggl(\iTn\wx^p(t)\,d\mu(t)-\delta^p\biggr)=0,&\quad\lambda_2\biggr(\iT|\varphi(t)|^r
\wx^r(t)\,d\mu(t)-1\biggr)=0,
\end{aligned}
\end{equation}
and $\lambda_2>0$, if $T\setminus T_0\ne\emptyset$.
Then
$$E(p,q,r)=\left(\frac{p\lambda_1\delta^p+r\lambda_2}q\right)^{1/q},$$
and the method
\begin{equation}\label{met}
\wm(y)(t)=\begin{cases}q^{-1}p\lambda_1\wx^{p-q}(t)|\psi(t)|^{-q}
\psi(t)y(t),&t\in T_0,\ \psi(t)\ne0,\\
0,&\mbox{otherwise},\end{cases}
\end{equation}
is optimal recovery method.
\end{theorem}

To prove this theorem we need some preliminary results.

\begin{lemma}\label{L1}
\begin{equation}\label{lb}
E(p,q,r)\ge\sup_{\substack{x\cd\in W\\\|x\cd\|_{\lpn}\le\delta}}\|\Lambda x\cd\|_{\lqq}.
\end{equation}
\end{lemma}

The lower bound of type \eqref{lb} is the well-known result which is usually applied to obtain the error of optimal recovery. In more or less general forms it was proved in many papers (see, for example, \cite{Os1}).

The extremal problem which arises on the right-hand side of \eqref{lb}, known as the dual problem, may be written as
\begin{multline}\label{du}
\|\psi\cd x\cd\|_{\lqq}\to\max,\quad\|x\cd\|_{\lpn}\le\delta,\\
\|\varphi\cd x\cd\|_{\lr}\le1.
\end{multline}
For $T_0=T\subset\mathbb R^n$ and $q=1$ problem \eqref{du}
was examined in \cite{Ar} in connection with Stechkin's problem.

We give a straightforward result (resembling the sufficient conditions
in the Kuhn-Tucker theorem), which we will require in solving dual problems similar to \eqref{du}.

Let $f_j\colon A\to\mathbb R$, $j=0,1,\ldots,n$, be functions defined on some set $A$. Consider the extremal problem
\begin{equation}\label{ex0}
f_0(x)\to\max,\quad f_j(x)\le0,\quad j=1,\ldots,n,\quad x\in A,
\end{equation}
and write down its Lagrange function
$$\mL(x,\lambda)=-f_0(x)+\sum_{j=1}^n\lambda_jf_j(x),\quad
\lambda=(\lambda_1,\ldots,\lambda_n).$$

\begin{lemma}[\cite{Os1}]\label{L2}
Assume that there exist $\wl_j\ge0$, $j=1,\ldots,n$, and an element $\wx\in A$,
admissible for problem \eqref{ex0}, such that
\begin{align*}
(a)&\quad\min_{x\in A}\mL(x,\wl)=\mL(\wx,\wl),\quad\wl=(\wl_1,\ldots,\wl_n),\\
(b)&\quad\sum_{j=1}^n\wl_jf_j(\wx)=0.
\end{align*}
Then $\wx$ is an extremal element for problem \eqref{ex0}.
\end{lemma}

Put
$$F(u,v,\alpha)=-((1-\alpha)u+\alpha v)^q+av^p+bu^r,\quad u,v\ge0,\quad\alpha\in[0,1],$$
where $a,b\ge0$, and $1\le p,q,r<\infty$.

\begin{lemma}\label{L3}
For all $a,b\ge0$, $a+b>0$, and all $1\le q<p,r<\infty$, there exists the unique solution $\wu>0$ of the equation
\begin{equation}\label{lem}
-q+pau^{p-q}+rbu^{r-q}=0.
\end{equation}
Moreover, for all $u,v\ge0$ and $\alpha=q^{-1}pa\wu^{p-q}=1-q^{-1}rb\wu^{r-q}$
\begin{equation}\label{FF}
F(\wu,\wu,\alpha)\le F(u,v,\alpha).
\end{equation}
In particular, for all $u\ge0$
$$-\wu^q+a\wu^p+b\wu^r\le-u^q+au^p+bu^r.$$
\end{lemma}

\begin{proof}
The existence of the unique solution of \eqref{lem} follows from the fact that the continuous function $f(u)=pau^{p-q}+rbu^{r-q}$ increases monotonically from $0$ to $+\infty$.

Let us prove \eqref{FF}. The cases $a=0$ or $b=0$ are easily obtained by finding the minimum of $F(u,v,0)=-u^q+bu^r$ if $a=0$ or $F(u,v,1)=-v^q+av^p$ if $b=0$. Assume that $a,b>0$. Then $\alpha\in(0,1)$. Let
$$C>\max\{a^{-\frac1{p-q}},b^{-\frac1{r-q}}\}.$$
Then for $u\ge C$ and $v\le u$ we have
\begin{equation}\label{F1}
F(u,v,\alpha)\ge-u^q+bu^r=u^q(-1+bu^{r-q})>0.
\end{equation}
If $v\ge C$ and $v\ge u$, then
\begin{equation}\label{F2}
F(u,v,\alpha)\ge-v^q+av^p=v^q(-1+av^{p-q})>0.
\end{equation}
Since $F(0,0,\alpha)=0$ we obtain that
$$\inf_{(u,v)\in\mathbb R^2_+}F(u,v,\alpha)=\inf_{\substack{0\le u\le C\\0\le v\le C}}F(u,v,\alpha).$$
It follows from the Weierstrass extreme value theorem that there exist $0\le u_0\le C$ and $0\le v_0\le C$ such that
$$\inf_{(u,v)\in\mathbb R^2_+}F(u,v,\alpha)=F(u_0,v_0,\alpha).$$
In view of \eqref{F1} and \eqref{F2} $u_0<C$ and $v_0<C$.
We have
\begin{multline*}
F_u(u,v,\alpha)=-q((1-\alpha)u+\alpha v)^{q-1}(1-\alpha)+rbu^{r-1}\\
=rb(-((1-\alpha)u+\alpha v)^{q-1}\wu^{r-q}+u^{r-1}).
\end{multline*}
Thus, for any $v_0\ge0$ and sufficiently small $u>0$ \ $F_u(u,v_0,\alpha)<0$. Consequently,
$$F(u,v_0,\alpha)<F(0,v_0,\alpha)$$
for sufficiently small $u$. It means that $0<u_0<C$. The similar arguments show that $0<v_0<C$. Hence
$$F_u(u_0,v_0,\alpha)=F_v(u_0,v_0,\alpha)=0.$$
Since
\begin{multline*}
F_v(u,v,\alpha)=-q((1-\alpha)u+\alpha v)^{q-1}\alpha+pav^{p-1}\\
=pa(-((1-\alpha)u+\alpha v)^{q-1}\wu^{p-q}+v^{p-1})
\end{multline*}
we have
\begin{align}\label{lem1}
-((1-\alpha)u_0+\alpha v_0)^{q-1}\wu^{r-q}+u_0^{r-1}&=0,\\
-((1-\alpha)u_0+\alpha v_0)^{q-1}\wu^{p-q}+v_0^{p-1}&=0.\label{lem11}
\end{align}
Consequently,
$$\frac{u_0^{r-1}}{v_0^{p-1}}=\wu^{r-p}.$$
Suppose that $p\le r$. Substituting
\begin{equation}\label{u0v0}
u_0=\wu^{\frac{r-p}{r-1}}{v_0}^{\frac{p-1}{r-1}}
\end{equation}
into \eqref{lem11}, we obtain the equality
$$(\alpha v_0+(1-\alpha)\wu^{\frac{r-p}{r-1}}{v_0}^{\frac{p-1}{r-1}})
^{q-1}\wu^{p-q}=v_0^{p-1}.$$
This equality may be rewritten in the form
\begin{equation}\label{tt}
(\alpha+(1-\alpha)t^{\frac{p-r}{r-1}})^{q-1}=t^{p-q},
\end{equation}
where $t=v_0\wu^{-1}$. It is easily seen that \eqref{tt} has the unique solution $t=1$. Consequently, $v_0=\wu$ and it follows by \eqref{u0v0} that $u_0=\wu$.

If $p>r$, then we substitute
$$v_0=\wu^{\frac{p-r}{p-1}}{u_0}^{\frac{r-1}{p-1}}$$
into \eqref{lem1}. Similar to the previous case we obtain the equality which may be written in the form
\begin{equation}\label{ss}
(\alpha s^{\frac{r-p}{p-1}}+1-\alpha)^{q-1}=s^{r-q},
\end{equation}
where $s=u_0\wu^{-1}$. The unique solution of \eqref{ss} is $s=1$. Thus, for the case when $p>r$ we have the same solution of \eqref{lem1}, \eqref{lem11} $u_0=v_0=\wu$. Hence, for all $u,v\ge0$
$$F(u,v,\alpha)\ge\inf_{(u,v)\in\mathbb R^2_+}F(u,v,\alpha)=F(\wu,\wu,\alpha).$$
\end{proof}

\begin{proof}[Proof of Theorem~$\ref{T1}$]
\

1. Lower estimate. The extremal problem \eqref{du} (for convenience, we raise the quantity to be maximized to the $q$-th power) is as follows:
\begin{multline}\label{ex}
\iT|\psi(t)x(t)|^q\,d\mu(t)\to\max,\quad\iTn|x(t)|^p\,d\mu(t)\le\delta^p,\\
\iT|\varphi(t)x(t)|^r\,d\mu(t)\le1.
\end{multline}
The Lagrange function for this problem reads as
$$\mathcal L(x\cd,\lambda_1,\lambda_2)=\int_T L\bigl(t,x(t),\lambda_1,\lambda_2\bigr)\,d\mu(t),$$
where
$$L(t,x,\lambda_1,\lambda_2)=-|\psi(t)x|^q+\lambda_1|x|^p\chi_0(t)+
\lambda_2|\varphi(t)x|^r.$$
If $t\in T$ such that $\psi(t)=0$, then evidently $\wx(t)=0$ and for those $t$ for all $x\cd\in\mathcal W$
$$L(t,0,\lambda_1,\lambda_2)\le L(t,x(t),\lambda_1,\lambda_2).$$
Using this fact and Lemma~\ref{L3}, we obtain that there is the unique solution $\wx\cd$ of \eqref{ko1} and, moreover, for almost all $t\in T$ and all $x\cd\in\mathcal W$
$$L(t,\wx(t),\lambda_1,\lambda_2)\le L(t,x(t),\lambda_1,\lambda_2).$$
Consequently,
$$\mL(\wx\cd,\lambda_1,\lambda_2)\le\mL(x\cd,\lambda_1,\lambda_2).$$
Taking into account \eqref{nez} we obtain by Lemma~\ref{L2} that $\wx\cd$ is the extremal function in \eqref{ex}. It follows by \eqref{lb} that
$$E(p,q,r)\ge\biggl(\iT|\psi(t)|^q\wx^q(t)\,d\mu(t)\biggr)^{1/q}.$$
From \eqref{ko1} we have
$$|\psi(t)|^q\wx^q(t)=q^{-1}p\lambda_1\wx^p(t)\chi_{T_0}(t)+q^{-1}r
\lambda_2|\varphi(t)|^r\wx^r(t).$$
Integrating this equality over the set $T$, we obtain
\begin{equation}\label{iTT}
\iT|\psi(t)|^q\wx^q(t)\,d\mu(t)=\frac{p\lambda_1\delta^p+r\lambda_2}q.
\end{equation}
Thus,
$$E(p,q,r)\ge\left(\frac{p\lambda_1\delta^p+r\lambda_2}q\right)^{1/q}.$$

2. Upper estimate. To estimate the error of method \eqref{met} we need to find the value of the extremal problem:
\begin{multline}\label{La}
\iTn|\psi(t)x(t)-\psi(t)\alpha(t)y(t)|^q\,d\mu(t)+\int_{T\setminus T_0}|\psi(t)x(t)|^q\,d\mu(t)\to\max,\\\iTn|x(t)-y(t)|^p\,d\mu(t)\le
\delta^p,\quad\iT|\varphi(t)x(t)|^r\,d\mu(t)\le1,
\end{multline}
where
\begin{equation}\label{alp}
\alpha(t)=\begin{cases}q^{-1}p\lambda_1\wx^{p-q}(t)|\psi(t)|^{-q},&t\in T_0,\ \psi(t)\ne0,\\
0,&\mbox{otherwise}.\end{cases}
\end{equation}
Taking
$$z(t)=\begin{cases}x(t)-y(t),&t\in T_0,\\
0,&t\in T\setminus T_0,\end{cases}$$
we rewrite \eqref{La} as follows:
\begin{multline*}
\iT|\psi(t)|^q|(1-\alpha(t))x(t)+\alpha(t)z(t)|^q\,d\mu(t)\to\max,\\
\iTn|z(t)|^p\,d\mu(t)\le\delta^p,\quad\iT|\varphi(t)x(t)|^r\,d\mu(t)\le1.
\end{multline*}
The value of this problem does not exceed the value of the problem
\begin{multline}\label{uv}
\iT|\psi(t)|^q((1-\alpha(t))u(t)+\alpha(t)v(t))^q\,d\mu(t)\to\max,\\\iTn v^p(t)\,d\mu(t)\le\delta^p,\quad
\iT|\varphi(t)|^ru^r(t)\,d\mu(t)\le1,\\ u(t)\ge0,\ v(t)\ge0\quad\mbox{for almost all }\ t\in T.
\end{multline}
The Lagrange function for this problem is
$$\mL_1(u\cd,v\cd,\mu_1,\mu_2)=\iT L_1(t,u(t),v(t),\mu_1,\mu_2)\,d\mu(t),$$
where
\begin{multline*}
L_1(t,u,v,\mu_1,\mu_2)=-|\psi(t)|^q((1-\alpha(t))u+\alpha(t)v)^q\\
+\mu_1v^p\chi_0(t)+\mu_2
|\varphi(t)|^ru^r.
\end{multline*}
By Lemma~\ref{L3} we have
$$L_1(t,\wx(t),\wx(t),\lambda_1,\lambda_2)\le L_1(t,u(t),v(t),\lambda_1,\lambda_2).$$
Thus,
$$\mL_1(\wx\cd,\wx\cd,\lambda_1,\lambda_2)\le\mL_1(u\cd,v\cd,
\lambda_1,\lambda_2).$$
It follows by Lemma~\ref{L2} that functions $u(t)=v(t)=\wx(t)$ are extremal in \eqref{uv}. Consequently,
%\begin{multline*}
$$e(p,q,r,\wm)\le\biggl(\iT|\psi(t)|^q\wx^q(t)\,d\mu(t)\biggr)^{1/q}
=\left(\frac{p\lambda_1\delta^p+r\lambda_2}q\right)^{1/q}\le E(p,q,r).$$
%\end{multline*}
It means that the method \eqref{met} is optimal and the optimal recovery error is as stated.
\end{proof}

Note that if conditions of Theorem~\ref{T1} hold we proved the equality
\begin{equation}\label{extr}
E(p,q,r)=\sup_{\substack{\|x\cd\|_{\lpn}\le\delta\\\|\varphi\cd x\cd\|_{\lr}\le1}}\|\psi\cd x\cd\|_{\lqq}.
\end{equation}

\begin{corollary}\label{Co1}
Let $1\le q<p,r<\infty$, $\varphi(t)\ne0$ for almost all $t\in T$, and
$$0<\iT\left|\frac{\psi(t)}{\varphi(t)}\right|^{\frac{qr}{r-q}}\,d\mu(t)<\infty,\quad\iTn
\left(\frac{|\psi(t)|^q}{|\varphi(t)|^r}\right)^{\frac p{r-q}}\,d\mu(t)<\infty.$$
Then for all
\begin{gather*}
\delta\ge\frac{\biggl(\displaystyle\iTn
\left(\frac{|\psi(t)|^q}{|\varphi(t)|^r}\right)^{\frac p{r-q}}\,d\mu(t)\biggr)^{1/p}}
{\biggl(\displaystyle\iT\left|\frac{\psi(t)}{\varphi(t)}\right|^{\frac{qr}{r-q}}\,d\mu(t)
\biggr)^{1/r}}\\
E(p,q,r)=\biggl(\iT\left|\frac{\psi(t)}{\varphi(t)}\right|^{\frac{qr}{r-q}}\,d\mu(t)
\biggr)^{\frac{r-q}{qr}},
\end{gather*}
and the method $\wm(y)(t)=0$ is optimal recovery method.
\end{corollary}

\begin{proof}
It suffices to check that $\lambda_1=0$ and
$$\lambda_2=\frac qr\biggl(\iT\left|\frac{\psi(t)}{\varphi(t)}\right|^{\frac{qr}{r-q}}\,d\mu(t)
\biggr)^{\frac{r-q}r}$$
satisfy the conditions of Theorem~\ref{T1}.
\end{proof}

\begin{corollary}\label{Co2}
Let $1\le q<p,r<\infty$, $T_0=T$, and
$$0<\iT|\varphi(t)|^r||\psi(t)|^{\frac{qr}{p-q}}\,d\mu(t)<\infty,\quad\iT
|\psi(t)|^{\frac{qp}{p-q}}\,d\mu(t)<\infty.$$
Then for all
\begin{gather*}
\delta\le\frac{\biggl(\displaystyle\iT
|\psi(t)|^{\frac{qp}{p-q}}\,d\mu(t)\biggr)^{1/p}}
{\biggl(\displaystyle\iT|\varphi(t)|^r||\psi(t)|^{\frac{qr}{p-q}}\,d\mu(t)
\biggr)^{1/r}}\\
E(p,q,r)=\delta\biggl(\iT|\psi(t)|^{\frac{qp}{p-q}}\,d\mu(t)\biggr)^{\frac{p-q}{qp}},
\end{gather*}
and the method $\wm(y)(t)=\psi(t)y(t)$ is optimal recovery method.
\end{corollary}

\begin{proof}
It suffices to check that
$$\lambda_1=\frac q{p\delta^{p-q}}\biggl(\iT
|\psi(t)|^{\frac{qp}{p-q}}\,d\mu(t)
\biggr)^{\frac{p-q}p}$$
and $\lambda_2=0$ satisfy the conditions of Theorem~\ref{T1}.
\end{proof}

Note that assumption \eqref{nez} need not be satisfied in all cases. For example, in the trivial case $\delta=0$, $T_0=T$, and $\psi(t)=1$ there are no such $\lambda_1$ and $\lambda_2$ which satisfy \eqref{nez}.

Let us consider the problem of optimal recovery of the linear functional
$$Lx=\int_T\psi(t)x(t)\,d\mu(t)$$
on the class $W$, knowing $y\cd\in\lpn$, $T_0\subset T$, such that $\|x\cd-y\cd\|_{\lpn}\le\delta$, $\delta\ge0$. In this case as recovery methods we consider all possible mappings $m\colon\lpn\to\mathbb C\ \mbox{or}\ \mathbb R$. The error of a~method~$m$ is defined as
$$e_1(p,r,m)=\sup_{\substack{x\cd\in W,\ y\cd\in\lpn\\\|x\cd-y\cd\|_{\lpn}\le\delta}}|Lx-m(y)|.$$
The quantity
\begin{equation}\label{E1}
E_1(p,r)=\inf_{m\colon\lpn\to\mathbb C(\mathbb R)}e_1(q,r,m)
\end{equation}
is optimal recovery error, and a method on which this infimum is attained is called optimal.

\begin{theorem1}\label{T1'}
Let $1<p,r<\infty$, $\lambda_1,\lambda_2\ge0$, $\lambda_1+\lambda_2>0$, $\varphi(t)\ne0$ for almost all $t\in T\setminus T_0$, $\wx(t)=\wx(t,\lambda_1,\lambda_2)\ge0$ be a solution of equation
%\begin{equation}\label{ko1}
$$-|\psi(t)|+p\lambda_1x^{p-1}(t)\chi_0(t)+r\lambda_2|\varphi(t)|^
rx^{r-1}(t)=0,$$
%\end{equation}
and $\lambda_1$, $\lambda_2$ such that conditions \eqref{nez} are fulfilled, and $\lambda_2>0$, if $T\setminus T_0\ne\emptyset$. Then
$$E_1(p,r)=p\lambda_1\delta^p+r\lambda_2,$$
and the method
\begin{equation}\label{m'}
\wm(y)=p\lambda_1\int_{T_0}\wx^{p-1}(t)\varepsilon(t)y(t)\,d\mu(t),
\end{equation}
where
$$\varepsilon(t)=\begin{cases}\dfrac{\psi(t)}{|\psi(t)|},&\psi(t)\ne0,\\
1,&\psi(t)=0,\end{cases}$$
is optimal recovery method.
\end{theorem1}

\begin{proof}
For the functional case it is known (see, for example, \cite{MOs2}) that
$$E_1(p,r)=\sup_{\substack{x\cd\in W\\\|x\cd\|_{\lpn}\le\delta}}\biggl|\int_T\psi(t)x(t)\,d\mu(t)\biggr|.$$
Put $\whx\cd=\ov{\varepsilon\cd}\wx\cd$. It follows by \eqref{nez} that $\whx\cd\in W$ and $\|\whx\cd\|_{\lpn}\le\delta$. Taking into account \eqref{iTT}, we obtain
$$E_1(p,r)\ge\biggl|\iT\psi(t)\whx(t)\,d\mu(t)\biggr|=
\iT|\psi(t)|\wx(t)\,d\mu(t)=p\lambda_1\delta^p+r\lambda_2.$$

Now we estimate the error of method \eqref{m'}. We have
\begin{multline*}
e_1(p,r,\wm)=\sup_{\substack{x\cd\in W,\ y\cd\in\lpn\\\|x\cd-y\cd\|_{\lpn}\le\delta}}\biggl|\iT\psi(t)x(t)\,d\mu(t)-
\wm(y)\biggr|\\
\le\sup_{\substack{x\cd\in W,\ z\cd\in\lpn\\\|z\cd\|_{\lpn}\le\delta}}\iT|\psi(t)||
(1-\alpha(t))x(t)+\alpha(t)z(t)|\,d\mu(t),
\end{multline*}
where $\alpha\cd$ is defined by \eqref{alp} for $q=1$. It follows from the proof of Theorem~\ref{T1} that
$$E_1(p,r)\le e_1(p,r,\wm)\le\iT|\psi(t)|\wx(t)\,d\mu(t)=p\lambda_1\delta^p+r\lambda_2.$$
\end{proof}

One can easily obtain analogs of Corollaries~\ref{Co1} and \ref{Co2} for problem \eqref{E1}.

\section{The case of homogenous weight functions}

Let $T$ be a cone in a linear space, $T_0=T$, $|\psi\cd|$ and $|\varphi\cd|$ be homogenous functions of degrees $\eta$, $\nu$, respectively, $\varphi(t)\ne0$ and $\psi(t)\ne0$ for almost all $t\in T$, and $\mu\cd$ be a homogenous measure of degree $d$.  We assume, again, that $1\le p<q,r<\infty$. For $k\in[0,1)$ the function $k^{\frac1{p-q}}(1-k)^{-\frac1{r-q}}$ increases monotonically from $0$ to $+\infty$. Consequently, for all $z\in T$ such that $\varphi(z)\ne0$ and $\psi(z)\ne0$ (if $p<r$), there exists $k(z)$ for which
\begin{equation}\label{kz}
\frac{k^{\frac1{p-q}}(z)}{(1-k(z))^{\frac1{r-q}}}=\frac{|\psi(z)|
^{\frac{q(p-r)}{(p-q)(r-q)}}}{|\varphi(z)|^{\frac r{r-q}}}.
\end{equation}
Thus, the function $k(z)$ is well defined for almost all $z\in T$.

\begin{theorem}\label{T2}
Let $1\le q<p,r<\infty$, $\varphi(t),\psi(t)\ne0$ for almost all $t\in T$, and $\nu+d(1/r-1/p)\ne0$. Assume that
\begin{align*}
I_1&=\iT|\psi(z)|^{\frac{qp}{p-q}}k^{\frac p{p-q}}(z)\,d\mu(z)<\infty,\\
I_2&=\iT|\psi(z)|^{\frac{qr}{p-q}}|\varphi(z)|^rk^{\frac r{p-q}}(z)\,d\mu(z)<\infty.
\end{align*}
Then
$$E(p,q,r)=\delta^\gamma
I_1^{-\gamma/p}I_2^{-(1-\gamma)/r}(I_1+I_2)^{1/q},$$
where
\begin{equation}\label{ga}
\gamma=\frac{\nu-\eta-d(1/q-1/r)}{\nu+d(1/r-1/p)},
\end{equation}
and the method
$$\wm(y)(t)=k(\xi t)\psi(t)y(t),$$
where
\begin{equation}\label{yy}
\xi=\left(\delta I_1^{-1/p}I_2^{1/r}\right)^{\frac1{\nu+d(1/r-1/p)}},
\end{equation}
is optimal recovery method.
\end{theorem}

\begin{proof}
Put
$$\wx(t)=\left(\frac{q|\psi(t)|^q}{p\lambda_1}\right)^{\frac1{p-q}}k^{\frac1{p-q}}(\xi t),$$
where $\lambda_1>0$ will be specified later. We show that $\wx\cd$ satisfies \eqref{ko1}, where
\begin{equation}\label{l2}
\lambda_2=r^{-1}q^{\frac{p-r}{p-q}}\left(p\lambda_1\right)^
{\frac{r-q}{p-q}}\xi^{\nu r-\eta\frac{q(p-r)}{p-q}}.
\end{equation}
We have
$$p\lambda_1\wx^{p-q}(t)=q|\psi(t)|^qk(\xi t),$$
and further,
$$r\lambda_2|\varphi(t)|^r\wx^{r-q}(t)=r\lambda_2|\varphi(t)|^r
\left(\frac{q|\psi(t)|^q}{p\lambda_1}\right)^{\frac{r-q}{p-q}}k^
{\frac{r-q}{p-q}}(\xi t).$$
Since $|\varphi\cd|$ and $|\psi\cd|$ are homogenous it follows by \eqref{kz} that
$$k^{\frac{r-q}{p-q}}(\xi t)=\frac{|\psi(\xi t)|^{\frac{q(p-r)}{p-q}}}{|\varphi(\xi t)|^r}
(1-k(\xi t))=\xi^{\eta\frac{q(p-r)}{p-q}-\nu r}\frac{|\psi(t)|^{\frac{q(p-r)}{p-q}}}{|\varphi(t)|^r}(1-k(\xi t)).$$
Thus,
\begin{multline*}
r\lambda_2|\varphi(t)|^r\wx^{r-q}(t)=
r\lambda_2\left(\frac q{p\lambda_1}\right)^{\frac{r-q}{p-q}}\xi^{\eta\frac{q(p-r)}{p-q}-\nu r}|\psi(t)|^q(1-k(\xi t))\\
=q|\psi(t)|^q(1-k(\xi t))=q|\psi(t)|^q-p\lambda_1\wx^{p-q}(t).
\end{multline*}
Now we show that for
\begin{equation}\label{l1}
\lambda_1=\frac qpI_1^{\frac{p-q}p}\xi^{-\eta q-d\frac{p-q}p}\delta^{q-p}
\end{equation}
the equalities
$$\iT\wx^p(t)\,d\mu(t)=\delta^p,\quad\iT|\varphi(t)|^r\wx^r(t)\,d\mu(t)
=1$$
hold. In view of the definition of $\wx\cd$ we need to check that
\begin{align*}
\iT\left(\frac{q|\psi(t)|^q}{p\lambda_1}\right)^{\frac p{p-q}}k^{\frac p{p-q}}(\xi t)\,d\mu(t)&
=\delta^p,\\
\iT|\varphi(t)|^r\left(\frac{q|\psi(t)|^q}{p\lambda_1}\right)^{\frac r{p-q}}k^{\frac r{p-q}}(\xi t)\,d\mu(t)&
=1.
\end{align*}
Changing $z=\xi t$ and taking into account that functions $|\psi\cd|$, $|\varphi\cd|$ with the measure $\mu\cd$ are homogenous, we obtain
\begin{align*}
\left(\frac q{p\lambda_1}\right)^{\frac p{p-q}}I_1&
=\delta^p\xi^{\frac{\eta qp}{p-q}+d},\\
\left(\frac q{p\lambda_1}\right)^{\frac r{p-q}}I_2&
=\xi^{\frac{\eta qr}{p-q}+\nu r+d}.
\end{align*}
The validity of these equalities immediately follows from the definitions of $\lambda_1$ and $\xi$.

It follows by Theorem~\ref{T1}, \eqref{l1}, \eqref{l2}, and \eqref{yy} that
\begin{multline*}
E^q(p,q,r)=\frac{p\lambda_1\delta^p+r\lambda_2}q=I_1^{\frac{p-q}p}
\xi^{-\eta q-d\frac{p-q}p}\delta^q+\left(\frac{p\lambda_1}q\right)^{\frac{r-q}{p-q}}
\xi^{\nu r-\eta\frac{q(p-r)}{p-q}}\\
=\delta^{q\gamma}I_1^{-q\gamma/p}I_2^{-q(1-\gamma)/r}(I_1+I_2).
\end{multline*}
Moreover, the same theorem states that the method
$$\wm(y)(t)=q^{-1}p\lambda_1\wx^{p-q}(t)|\psi(t)|^{-q}\psi(t)y(t)=
k(\xi t)\psi(t)y(t)$$
is optimal.
\end{proof}

It follows by Theorem~\ref{T2} and \eqref{extr} that for all $x\cd\in\mathcal W$ such that $\|\varphi\cd x\cd\|_{\lr}\le1$ the exact inequality
\begin{equation}\label{eqeq}
\|\psi\cd x\cd\|_{\lqq}\le C\|x\cd\|_{\lp}^\gamma
\end{equation}
holds, where
%\begin{equation}\label{CC}
$$C=I_1^{-\gamma/p}I_2^{-(1-\gamma)/r}(I_1+I_2)^{1/q}.$$
%\end{equation}
(Here and later the exactness means that $C$ cannot be replaced by any other constant smaller than $C$).

From \eqref{eqeq} the following exact inequality can be easily obtained
\begin{equation}\label{CC}
\|\psi\cd x\cd\|_{\lqq}\le C\|x\cd\|_{\lp}^\gamma
\|\varphi\cd x\cd\|_{\lr}^{1-\gamma},
\end{equation}
which holds for all $x\cd\in\mathcal W$, $x\cd\ne0$.

Let $|w\cd|$, $|w_0\cd|$, and $|w_1\cd|$ be homogenous functions of degrees $\theta$, $\theta_0$, and $\theta_1$, respectively. We assume that $w(t),w_0(t),w_1(t)\ne0$ for almost all $t\in T$ and $1\le q<p,r<\infty$. Then for almost all $z\in T$ such that $w(z),w_0(z),w_1(z)\ne0$ there exists $\wk(z)$ satisfying
$$\frac{\wk^{\frac1{p-q}}(z)}{(1-\wk(z))^{\frac1{r-q}}}=\left|\frac{w(z)}{w_1(z)}\right|
^{\frac r{r-q}}\left|\frac{w_0(z)}{w(z)}\right|^{\frac p{p-q}}.$$
Put
\begin{equation}\label{theta}
\wt=\theta+d/q,\quad\wt_0=\theta_0+d/p,\quad\wt_1=\theta_1+d/r.
\end{equation}

\begin{corollary}\label{c1}
Let $1\le q<p,r<\infty$, $w(t),w_0(t),w_1(t)\ne0$ for almost all $t\in T$, and $\wt_0\ne\wt_1$. Assume that
\begin{align*}
\wII_1&=\iT\left|\frac{w(z)}{w_0(z)}\right|^{\frac{qp}{p-q}}\wk^{\frac p{p-q}}(z)\,d\mu(z)<\infty,\\
\wII_2&=\iT\frac{|w(z)|^{\frac{qr}{p-q}}}{|w_0(z)|^{\frac{pr}{p-q}}}|w_1(z)|^r\wk^{\frac r{p-q}}(z)\,d\mu(z)<\infty.
\end{align*}
Then for all $x\cd\ne0$ such that $w_0\cd x\cd\in L_p(T,\mu)$ and $w_1\cd x\cd\in L_r(T,\mu)$ the exact inequality
\begin{equation}\label{CCC}
\|w\cd x\cd\|_{\lqq}\le\wC\|w_0\cd x\cd\|_{\lp)}^{\wg}\|w_1\cd x\cd\|_{\lr}^{1-\wg}
\end{equation}
holds; here
$$\wC=\wII_1^{-\wg/p}\wII_2^{-(1-\wg)/r}(\wII_1+\wII_2)^{1/q},\quad
\wg=\frac{\wt_1-\wt}{\wt_1-\wt_0}.$$
\end{corollary}

\begin{proof}
Put
$$\psi(x)=\frac{w(x)}{w_0(x)},\quad\varphi(x)=\frac{w_1(x)}{w_0(x)}.$$
Then $|\psi\cd|$ and $|\varphi\cd|$ are homogeneous functions of degrees $\eta=\theta-\theta_0$ and $\nu=\theta_1-\theta_0$, respectively. It follows by \eqref{CC} that for all $x\cd\in\mathcal W$, $x\cd\ne0$, the exact inequality
$$\|\psi\cd x\cd\|_{\lqq}\le\wC\|x\cd\|_{\lp}^{\wg}\|\varphi\cd x\cd\|_{\lr}^{1-\wg}$$
holds. Substituting $x\cd=w_0\cd y\cd$, we obtain \eqref{CCC}.
\end{proof}

The well-known Carlson inequality \cite{Ca}
\begin{equation}\label{Car}
\|x(t)\|_{L_1(\mathbb R_+)}\le\sqrt\pi\|x(t)\|_{L_2(\mathbb R_+)}^{1/2}
\|tx(t)\|_{L_2(\mathbb R_+)}^{1/2}
\end{equation}
was generalized in many directions (see \cite{Le}, \cite{An}, \cite{B}). Inequality \eqref{CCC} is also a generalization of the Carlson inequality.


Let $1\le p<q,r<\infty$, $T$ be a cone in $\mathbb R^d$, $d\mu(t)=dt$, $|\psi\cd|$ and $|\varphi\cd|$ be homogenous functions of degrees $\eta$, $\nu$, respectively, $\varphi(t)\ne0$ and $\psi(t)\ne0$ for almost all $t\in T$. Thus $\mu\cd$ is a homogeneous measure of degree $d$. Consider the polar transformation
$$\arraycolsep=0.08em
\begin{array}{rcl}
x_1&=&\rho\cos\omega_1,\\
x_2&=&\rho\sin\omega_1\cos\omega_2,\\
\hdotsfor{3}\\
x_{d-1}&=&\rho\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}\cos\omega_{d-1},\\
x_d&=&\rho\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}\sin\omega_{d-1}.
\end{array}$$
Set $\omega=(\omega_1,\ldots,\omega_{d-1})$,
\begin{equation}\label{wtw}
\begin{aligned}
\wps(\omega)&=\rho^{-\eta}|\psi(\rho\cos\omega_1,\ldots,
\rho\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}\sin\omega_{d-1})|,\\
\wva(\omega)&=\rho^{-\nu}|\varphi(\rho\cos\omega_1,\ldots,
\rho\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}\sin\omega_{d-1})|.
\end{aligned}
\end{equation}
Denote by $\Omega$ the range of $\omega$. Since $T$ is a cone, $\Omega$ does not depend on $\rho$. Put
$$J(\omega)=\sin^{d-2}\omega_1\sin^{d-3}\omega_2\ldots\sin\omega_{d-2}.$$
By \eqref{kz} we obtain the following equality for $k\cd$:
\begin{equation}\label{kkr}
\frac{k^{\frac1{p-q}}(\rho,\omega)}
{(1-k(\rho,\omega))^{\frac1{r-q}}}=\rho^{\frac{\eta q(p-r)-\nu r(p-q)}{(p-q)(r-q)}}\frac{\wps
^{\frac{q(p-r)}{(p-q)(r-q)}}(\omega)}{\wva^{\frac r{r-q}}(\omega)}.
\end{equation}
Assume that $\gamma\in(0,1)$, where $\gamma$ is defined by \eqref{ga}. Put
\begin{equation}\label{qu}
\frac1{q^*}=\frac1q-\frac\gamma p-\frac{1-\gamma}r.
\end{equation}
It is easy to verify that $q^*>q\ge1$. Moreover,
$$q^*=\frac{pqr(\nu+d(1/r-1/p))}{\nu r(p-q)-\eta q(p-r)}.$$

\begin{theorem}\label{T3}
Let $1\le q<p,r<\infty$, $\gamma\in(0,1)$, and $\wva(\omega),\wps(\omega)\ne0$ for almost all $\omega\in\Omega$.
Assume that
$$I=\int_\Omega\frac{\wps^{q^*}(\omega)}{\wva^{q^*(1-\gamma)}(\omega)}
J(\omega)\,d\omega<\infty.$$
Then
$$E(p,q,r)=C_1\delta^\gamma,$$
where
$$C_1=\gamma^{-\frac\gamma p}
(1-\gamma)^{-\frac{1-\gamma}r}\Biggl(\frac{B\left(q^*\gamma /p,q^*(1-\gamma)/r\right)I}{|\nu+d(1/r-1/p)|(\gamma r+(1-\gamma)p)}\Biggr)^{1/q^*},$$
where $B(\cdot,\cdot)$ is the beta-function. Moreover, the method
$$\wm(y)(t)=k\left(\xi_1^{\frac1{\nu+d(1/r-1/p)}}t\right)\psi(t)y(t),$$
where
$$\xi_1=\delta\left(\gamma^{q-r}(1-\gamma)^{p-q}C_1^{p-r}\right)^{\frac
{q^*}{pqr}},$$
is optimal recovery method.
\end{theorem}

\begin{proof}
Using Theorem~\ref{T2}, we obtain
\begin{multline*}
I_1=\iT|\psi(z)|^{\frac{qp}{p-q}}k^{\frac p{p-q}}(z)\,dz\\
=\int_\Omega\wps^{\frac{qp}{p-q}}(\omega)J(\omega)\,d\omega\int_0^{+\infty}
\rho^{\frac{\eta qp}{p-q}
+d-1}k^{\frac p{p-q}}(\rho,\omega)\,d\rho.
\end{multline*}
By \eqref{kkr} we have
\begin{equation}\label{rho}
\rho^{\nu r(p-q)-\eta q(p-r)}=\frac
{(1-k(\rho,\omega))^{p-q}}{k^{r-q}(\rho,\omega)}\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}.
\end{equation}

Fixing $\omega$, we pass to $k$
\begin{multline*}
d\rho^{\frac{\eta qp}{p-q}
+d}=\left(\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}\right)^\zeta\,d\frac
{(1-k)^{(p-q)\zeta}}{k^{(r-q)\zeta}}\\
=-\zeta\left(\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}\right)^\zeta\frac
{(1-k)^{(p-q)\zeta-1}}{k^{(r-q)\zeta+1}}(r-q+(p-r)k)\,dk,
\end{multline*}
where
$$\zeta=\frac{\eta qp+d(p-q)}{(p-q)(\nu r(p-q)-\eta q(p-r))}=\frac{q^*(1-\gamma)}{r(p-q)}.$$
Consequently,
\begin{multline*}
\int_0^{+\infty}
\rho^{\frac{\eta qp}{p-q}
+d-1}k^{\frac p{p-q}}(\rho,\omega)\,d\rho\\
=\frac{p-q}{\eta qp+d(p-q)}\int_0^{+\infty}
k^{\frac p{p-q}}(\rho,\omega)\,d\rho^{\frac{\eta qp}{p-q}
+d}\\
=\frac1{|\nu r(p-q)-\eta q(p-r)|}\left(\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}\right)^\zeta(K_1+K_2),
\end{multline*}
where
\begin{align*}
K_1&=(r-q)\int_0^1k^{\wpp}(1-k)^{\wqq-1}\,dk=(r-q)B(\wpp+1,\wqq),\\
K_2&=(p-r)\int_0^1k^{\wpp+1}(1-k)^{\wqq-1}\,dk=(p-r)B(\wpp+2,\wqq)\\
&\hspace{176pt}=(p-r)\frac{\wpp+1}{\wpp+\wqq+1}B(\wpp+1,\wqq),\\
\wpp&=\frac{qr(\nu-\eta)-d(r-q)}{\nu r(p-q)-\eta q(p-r)}=q^*\frac{\gamma}p,\quad\wqq=\frac{\eta qp+d(p-q)}{\nu r(p-q)-\eta q(p-r)}=q^*\frac{1-\gamma}r.
\end{align*}
Thus,
\begin{multline*}
K_1+K_2=p\frac{\nu r(p-q)-\eta q(p-r)}{\nu pr+d(p-r)}B(\wpp+1,\wqq)=\frac{pq}{q^*}B(\wpp+1,\wqq)\\
=\frac{q\gamma}{q^*}\left(\frac\gamma p+\frac{1-\gamma}r\right)^{-1}B(\wpp,\wqq).
\end{multline*}

The analogous calculations give
\begin{multline*}
I_2=\iT|\psi(z)|^{\frac{qr}{p-q}}|\varphi(z)|^rk^{\frac r{p-q}}(z)\,d\mu(z)\\=
\int_\Omega\wps^{\frac{qr}{p-q}}(\omega)\wva^r(\omega)J(\omega)\,
d\omega\int_0^{+\infty}\rho^{\frac{\eta qr}{p-q}+\nu r
+d-1}k^{\frac r{p-q}}(\rho,\omega)\,d\rho.
\end{multline*}
Fixing $\omega$, we pass to $k$
\begin{multline*}
d\rho^{\frac{\eta qr}{p-q}+\nu r
+d}=\left(\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}\right)^{\zeta_1}\,d\frac
{(1-k)^{(p-q)\zeta_1}}{k^{(r-q)\zeta_1}}\\
=-\zeta_1\left(\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}\right)^{\zeta_1}\frac
{(1-k)^{(p-q)\zeta_1-1}}{k^{(r-q)\zeta_1+1}}(r-q+(p-r)k)\,dk,
\end{multline*}
where
$$\zeta_1=\frac{\eta qr+(\nu r+d)(p-q)}{(p-q)(\nu r(p-q)-\eta q(p-r))}=\frac{q^*(1-\gamma)}{r(p-q)}+\frac1{p-q}.$$

We have
\begin{multline*}
\int_0^{+\infty}\rho^{\frac{\eta qr}{p-q}+\nu r
+d-1}k^{\frac r{p-q}}
(\rho,\omega)\,d\rho\\=\frac{p-q}{\eta qr+(\nu r+d)(p-q)}\int_0^{+\infty}
k^{\frac r{p-q}}(\rho,\omega)\,d\rho^{\frac{\eta qr}{p-q}
+\nu r+d}\\
=\frac1{|\nu r(p-q)-\eta q(p-r)|}\left(\frac{\wps
^{q(p-r)}(\omega)}{\wva^{r(p-q)}(\omega)}\right)^{\zeta_1}(L_1+L_2),
\end{multline*}
where
\begin{align*}
L_1&=(r-q)\int_0^1k^{\wpp-1}(1-k)^{\wqq}\,dk=(r-q)B(\wpp,\wqq+1),\\
L_2&=(p-r)\int_0^1k^{\wpp}(1-k)^{\wqq}\,dk=(p-r)B(\wpp+1,\wqq+1)\\
&\hspace{176pt}=(p-r)\frac{\wpp}{\wpp+\wqq+1}B(\wpp,\wqq+1).
\end{align*}
Thus,
\begin{multline*}
L_1+L_2=r\frac{\nu r(p-q)-\eta q(p-r)}{\nu pr+d(p-r)}B(\wpp,\wqq+1)=\frac{qr}{q^*}B(\wpp,\wqq+1)\\
=\frac{q(1-\gamma)}{q^*}\left(\frac\gamma p+\frac{1-\gamma}r\right)^{-1}B(\wpp,\wqq).
\end{multline*}
We obtain
\begin{align*}
I_1&=\frac\gamma{pr|\nu+d(1/r-1/p)|}\left(\frac\gamma p+\frac{1-\gamma}r\right)^{-1}B(\wpp,\wqq)I,\\ I_2&=\frac{1-\gamma}{pr|\nu+d(1/r-1/p)|}\left(\frac\gamma p+\frac{1-\gamma}r\right)^{-1}B(\wpp,\wqq)I.
\end{align*}
It remains to apply Theorem~\ref{T2}.
\end{proof}

Note that for $d=1$ we have $I=1$ when $T=\mathbb R_+$ and $I=2$ when $T=\mathbb R$.

Assume that $|w\cd|$, $|w_0\cd|$, and $|w_1\cd|$ are homogenous functions of degrees $\theta$, $\theta_0$, and $\theta_1$, respectively. Define $\wtw\cd$, $\wtw_0\cd$, $\wtw_1\cd$ by the analogy with \eqref{wtw}.

From Theorem~\ref{T2} (analogously to Corollary~\ref{c1}) we immediately obtain

\begin{corollary}[\cite{B}\footnote{The exact constant in \cite{B} (formula (10)) was given with a misprint.}]\label{c2}
Suppose that $w(t),w_0(t),w_1(t)\ne0$ for almost all $t\in T$, $1\le q<p,r<\infty$, $\wg\in(0,1)$, where
$$\wg=\frac{\wt_1-\wt}{\wt_1-\wt_0},$$
and $\wt$, $\wt_0$, and $\wt_1$ are defined by \eqref{theta}. Moreover, assume that
$$\wII=\int_\Omega\frac{\wtw^{\vq}(\omega)}{\wtw_0^{\vq\wg}(\omega)
\wtw_1^{\vq(1-\wg)}(\omega)}J(\omega)\,d\omega<\infty,$$
where
$$\frac1{\vq}=\frac1q-\frac{\wg}p-\frac{1-\wg}r.$$
Then the exact inequality
\begin{equation}\label{CCC1}
\|w\cd x\cd\|_{\lqq}\le\wC_1\|w_0\cd x\cd\|_{\lp)}^{\wg}\|w_1\cd x\cd\|_{\lr}^{1-\wg}
\end{equation}
holds; here
$$\wC_1=\wg^{-\frac{\wg}p}(1-\wg)^{-\frac{1-\wg}r}\Biggl(
\frac{B\left(\vq\wg/p,\vq(1-\wg)/r\right)\wII}{|\theta_1-\theta_0|(\wg r+(1-\wg)p)}\Biggr)^{1/\vq}.$$
\end{corollary}

Put
$$w_0(t)=1,\quad w_1(t)=t^{1-(\lambda+1)/p},\quad w_2(t)=t^{1+(\mu-1)/q}.$$
From Corollary~\ref{c2} we obtain

\begin{corollary}[\cite{Le}]
Let $1<p,q<\infty$ and $\lambda,\mu>0$. Put
$$\alpha=\frac\mu{p\mu+q\lambda},\quad\beta=\frac\lambda{p\mu+q\lambda}.$$
Then the exact inequality
$$\|x(t)\|_{L_1(\mathbb R_+)}\le C\|t^{1-(\lambda+1)/p}x(t)\|_{L_p(\mathbb R_+)}^{p\alpha}
\|t^{1+(\mu-1)/q}x(t)\|_{L_q(\mathbb R_+)}^{q\beta}$$
holds; here
$$C=\frac1{(p\alpha)^\alpha(q\beta)^\beta}\left(\frac1{\lambda+\mu}
B\left(\frac\alpha{1-\alpha-\beta},\frac\beta{1-\alpha-\beta}\right)
\right)^{1-\alpha-\beta}.$$
\end{corollary}

Using Theorem~$1'$ and calculations from the proofs of Theorems~\ref{T2} and \ref{T3} we obtain

\begin{theorem3}\label{T3'}
Let $1<p,r<\infty$, $\wva(\omega),\wps(\omega)\ne0$ for almost all $\omega\in\Omega$ and $\gamma$, $q^*$, $I$, $k\cd$, $C_1$, $\xi_1$ as above but for $q=1$. Assume that $\gamma\in(0,1)$ and $I<\infty$.
Then
$$E_1(p,r)=C_1\delta^\gamma.$$
Moreover, the method
$$\wm(y)=\int_T k\left(\xi_1^{\frac1{\nu+d(1/r-1/p)}}t\right)\psi(t)y(t)\,d\mu(t)$$
is optimal recovery method.
\end{theorem3}

\section{Optimal recovery of functions from a noisy Fourier transform}

Let $S$ be the Schwartz space of rapidly decreasing $C^\infty$-functions on $\mathbb R$, $S'$ the corresponding space of distributions, and let $F\colon S'\to S'$ be the Fourier transform. We let $\mathcal F_p$ denote the space of distribution $x\cd$ in $S'$ for which
$$\|x\cd\|_p=\left(\iR|Fx(t)|^p\,dt\right)^{1/p}
<\infty,\quad1\le p<\infty.$$
We set
\begin{gather*}
\mathcal F_p^n=\{\,x\cd\in S':\|x^{(n)}\cd\|_p<\infty\,\},\\
F_p^n=\{\,x\cd\in\mathcal F^n_p:\|x^{(n)}\cd\|_p\le1\,\}.
\end{gather*}

Assume that the Fourier transform of a function $x\cd\in F_r^n\cap\mathcal F_p$ is known on $\mathbb R$ to within $\delta>0$ in the metric of $\Lp$. In other words, we know a function $y\cd\in\Lp$ such that $\|Fx\cd-y\cd\|_{\Lp}\le\delta$. How should we best use this information to recover the $l$th derivative of the function
in the metric $\mathcal F_q$, $0\le l<n$? By recovery methods here we mean all possible mappings $m\colon\Lp\to\mathcal F_q$. The error of a method is, by definition, the quantity
$$e_{p,q,r}(m)=\sup_{\substack{x\cd\in F_r^n\cap\mathcal F_p,\ y\cd\in\Lp\\\|Fx\cd-y\cd\|_{\Lps}\le\delta}}\|x^{(l)}\cd-m(y)\cd\|_q.$$
The optimal recovery error is defined as follows:
$$E_{p,q,r}=\inf_{m\colon\Lp\to\mathcal F_q}e_{p,q,r}(m).$$
A method on which this lower bound is attained is called optimal.

It is readily checked that this problem is a special case of the general problem \eqref{p1} with $T=T_0=\mathbb R$, $\psi(t)=(it)^l$, $\varphi(t)=(it)^n$.

The cases 1) $1\le q=r<p<\infty$, 2) $1\le q=p<r<\infty$, 3) $1\le q=p=r<\infty$, and 4) $1\le q<p=r<\infty$ were studied in \cite{Os1}.

For the case $1\le q<p,r<\infty$ we can apply Theorem~\ref{T3}. In this case
$$\frac{k^{\frac1{p-q}}(t)}
{(1-k(t))^{\frac1{r-q}}}=|t|^{\frac{lq(p-r)-nr(p-q)}{(p-q)(r-q)}},\quad
\gamma=\frac{n-l-1/q+1/r}{n+1/r-1/p},$$
and $I=2$. It is easy to verify that if $n>l+1/q-1/r$, then $\gamma\in(0,1)$. Thus, it follows by Theorem~\ref{T3}

\begin{theorem}\label{T4}
Let $1\le q<p,r<\infty$ and $n>l+1/q-1/r$. Then
\begin{equation}\label{Epqr}
E_{p,q,r}=C_1\delta^\gamma,
\end{equation}
where
$$C_1=\gamma^{-\frac\gamma p}
(1-\gamma)^{-\frac{1-\gamma}r}\Biggl(\frac{2B\left(q^*\gamma /p,q^*(1-\gamma)/r\right)}{(n+1/r-1/p)(\gamma r+(1-\gamma)p)}\Biggr)^{1/q^*}$$
and $q^*$ is defined by \eqref{qu}. Moreover, the method $\wm(y)\cd=F^{-1}Y_y\cd$ is optimal, where
$$Y_y(t)=(it)^lk\left(\xi_1^{\frac1{n+1/r-1/p}}t\right)y(t),\quad
\xi_1=\delta\left(\gamma^{q-r}(1-\gamma)^{p-q}C_1^{p-r}\right)^{\frac
{q^*}{pqr}}.$$
\end{theorem}

Note that case 4) immediately follows from Theorem~\ref{T4} for $p=r$. In cases 1)--3) the optimal recovery error coincides with the limits $\lim_{r\to q}E_{p,q,r}$, $\lim_{p\to q}E_{p,q,r}$, $\lim_{p\to q}E_{p,q,p}$, respectively, where $E_{p,q,r}$ is given by \eqref{Epqr}.

\section{Optimal recovery of derivatives and generalized Carlson-Levin-Taikov inequalities}

For functions $x\cd\in\Lt$ whose $(n-1)$st derivative is locally absolutely continuous and $0\le k\le n-1$, L. V. Taikov \cite{Ta} obtained exact inequality
$$|x^{(k)}(0)|\le K\|x\cd\|_{\Lt}^{\frac{2n-2k-1}{2n}}\|x^{(n)}\cd\|_{\Lt}^
{\frac{2k+1}{2n}},$$
where
$$K=\left(\frac{2k+1}{2n-2k-1}\right)^{\frac{2n-2k-1}{4n}}\left((2k+1)
\sin\frac{2k+1}{2n}\pi\right)^{-1/2}.$$

Passing to the Fourier transform we have the following equivalent inequality
\begin{multline*}
\biggl|\frac1{2\pi}\int_{\mathbb R}t^kFx(t)\,dt\biggr|\le K\biggl(\frac1{2\pi}\int_{\mathbb R}|Fx(t)|^2\,dt\biggr)^{\frac{2n-2k-1}{4n}}\\
\times\biggl(\frac1{2\pi}\int_{\mathbb R}t^{2n}|Fx(t)|^2\,dt\biggr)^{\frac{2k+1}{4n}}.
\end{multline*}
Set $g(t)=t^kFx(t)$. Then we obtain the following inequality
\begin{multline*}
\biggl|\int_{\mathbb R}g(t)\,dt\biggr|\le K\sqrt{2\pi}\biggl(\int_{\mathbb R}t^{-2k}|g(t)|^2\,dt\biggr)^{\frac{2n-2k-1}{4n}}\\
\times\biggl(\int_{\mathbb R}t^{2(n-k)}|g(t)|^2\,dt\biggr)^{\frac{2k+1}{4n}}.
\end{multline*}

Put $p=q=2$, $\lambda=2k+1$, and $\mu=2n-2k-1$. Then by Corollary~\ref{c2} we have
\begin{multline*}
\int_0^\infty|g(t)|\,dt\le C\biggl(\int_0^\infty t^{-2k}|g(t)|^2\,dt\biggr)^{\frac{2n-2k-1}{4n}}\\
\times\biggl(\int_0^\infty t^{2(n-k)}|g(t)|^2\,dt\biggr)^{\frac{2k+1}{4n}},
\end{multline*}
where
$$C=\left(\frac{2k+1}{2n-2k-1}\right)^{\frac{2n-2k-1}{4n}}(2k+1)^{-1/2}
B^{1/2}\left(\frac{2n-2k-1}{2n},\frac{2k+1}{2n}\right).$$
Since
$$B\left(1-\frac{2k+1}{2n},\frac{2k+1}{2n}\right)=
\frac\pi{\sin\dfrac{2k+1}{2n}\pi}$$
we have
$$C=\sqrt\pi\left(\frac{2k+1}{2n-2k-1}\right)^{\frac{2n-2k-1}{4n}}\left((2k+1)
\sin\frac{2k+1}{2n}\pi\right)^{-1/2}.$$
From the inequality
$$a_1b_1+a_2b_2\le2^{1-s-t}(a_1^{1/r}+a_2^{1/r})^r
(b_1^{1/s}+b_2^{1/s})^s$$
it follows that
\begin{multline*}
\int_{\mathbb R}|g(t)|\,dt=\int_{-\infty}^0|g(t)|\,dt+\int_0^\infty
|g(t)|\,dt\\
\le C\biggl(\int_{-\infty}^0 t^{-2k}|g(t)|^2\,dt\biggr)^{\frac{2n-2k-1}{4n}}\biggl(\int_{-\infty}^0 t^{2(n-k)}|g(t)|^2\,dt\biggr)^{\frac{2k+1}{4n}}\\
+C\biggl(\int_0^\infty t^{-2k}|g(t)|^2\,dt\biggr)^{\frac{2n-2k-1}{4n}}\biggl(\int_0^\infty t^{2(n-k)}|g(t)|^2\,dt\biggr)^{\frac{2k+1}{4n}}\\
\le\sqrt2C\biggl(\int_{\mathbb R} t^{-2k}|g(t)|^2\,dt\biggr)^{\frac{2n-2k-1}{4n}}\biggl(\int_{\mathbb R} t^{2(n-k)}|g(t)|^2\,dt\biggr)^{\frac{2k+1}{4n}}.
\end{multline*}
Thus Taikov's inequality follows from Levin's inequality.

This inequality is closely connected with the problem of optimal recovery of derivatives from inaccurate information about the Fourier transform (see \cite{MOs}). We consider such problem in multidimensional case.

Consider linear operators
$D_1\colon\ld\to\ld\cap C(\mathbb R^d)$ and $D_2\colon\ld\to\ld$ ($D_1$ and $D_2$ are not necessary differentiation operators). Put
$$W=\{\,x\cd\in\ld:\|D_2x\cd\|_{\ld}\le1\,\}.$$
We consider the problem of optimal recovery of $D_1x(\tau)$, $\tau\in\mathbb R^d$, on the class $W$ from the information about $x\cd$, given inaccurately in $\ld$-metric.

As recovery methods we consider all possible mappings $m\colon\ld\to\mathbb C$ or $\mathbb R$. The error of a~method~$m$ is defined as
$$e(m)=\sup_{\substack{x\cd\in W,\ y\cd\in\ld\\\|x\cd-y\cd\|_{\ld}\le\delta}}|D_1x(\tau)-m(y)|.$$

The quantity
\begin{equation}\label{ETa}
E=\inf_{m\colon\ld\to\mathbb C(\mathbb R)}e(m)
\end{equation}
is known as the optimal recovery error, and a method on which this infimum is attained is called optimal.

For the case when $d=1$, $D_1x\cd=x^{(k)}\cd$, and $D_2x\cd=x^{(n)}\cd$, $0\le k<n$, similar problems were considered in \cite{MOs}.

Let $d_1(t)$ and $d_2\cd$ be measurable functions on $R^d$. Put
$$X=\{\,x\cd\in\ld:d_2\cd Fx\cd\in\ld\,\}.$$
We define the operator $D_2$ as follows
$$D_2x\cd=F^{-1}(d_2\cd Fx\cd)\cd.$$
Assume that $d_1\cd Fx\cd\in\ld$ for all $x\cd\in X$ and the operator $D_1$ which is defined by the equality
$$D_1x\cd=F^{-1}(d_1\cd Fx\cd)\cd$$
maps $X$ to $\ld\cap C(\mathbb R^d)$.

Let $|d_1\cd|$ and $|d_2\cd|$ be homogenous functions of degrees $k$, $n$, respectively ($k$ and $n$ are not necessarily integer), $d_j(t)\ne0$, $j=1,2$, for almost all $t\in\mathbb R^d$. Put
\begin{align*}
\wdd_1(\omega)&=\rho^{-k}|d_1(\rho\cos\omega_1,\ldots,
\rho\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}\sin\omega_{d-1})|,
\\
\wdd_2(\omega)&=\rho^{-n}|d_2(\rho\cos\omega_1,\ldots,
\rho\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}\sin\omega_{d-1})|.
\end{align*}

By Plancherel's theorem we have
\begin{gather*}
W=\biggl\{\,x\cd\in\ld:\frac1{(2\pi)^d}\int_{\mathbb R^d}|d_2(t)Fx(t)|^2\,dt\le1\,\biggr\},\\
\|x\cd-y\cd\|_{\ld}=\frac1{(2\pi)^{d/2}}\|Fx\cd-Fy\cd\|_{\ld}.
\end{gather*}
Moreover,
$$D_1x(\tau)=\frac1{(2\pi)^d}\int_{\mathbb R^d}d_1(t)Fx(t)e^{i\la\tau,t\ra}\,dt,$$
where $\la\tau,t\ra=\tau_1t_1+\ldots+\tau_dt_d$.
Thus we obtain problem \eqref{E1} with $p=r=2$, $\delta_1=\delta(2\pi)^{d/2}$,
$$\psi(t)=\frac1{(2\pi)^d}d_1(t)e^{i\la\tau,t\ra},\quad\varphi(t)=
\frac1{(2\pi)^{d/2}}d_2(t).$$

By Theorem~$3'$ we have

\begin{theorem}\label{TT}
Let $k\ge0$ and $n>k+d/2$.
Assume that
$$I=\int_{\Pi_{d-1}}\frac{{\wdd_1\,\!\!}^2(\omega)}{{\wdd_2\,\!\!}
^{\frac{2k+d}n}(\omega)}
J(\omega)\,d\omega<\infty,\quad\Pi_{d-1}=[0,\pi]^{d-2}\times[0,2\pi].$$
Then
$$E=\frac{(\pi I)^{1/2}}{(2\pi)^{d/2}}K_d(k,n)\delta^{\frac{2n-2k-d}{2n}},$$ where
$$K_d(k,n)=\left(\frac{2k+d}{2n-2k-d}
\right)^{\frac{2n-2k-d}{4n}}\left((2k+d)
\sin\frac{2k+d}{2n}\pi\right)^{-1/2}.$$
Moreover, the method
$$\wm(y)=\frac1{(2\pi)^d}\int_{\mathbb R^d}d_1(t)\left(1+\frac{\delta^2(2k+d)}{(2\pi)^d(2n-2k-d)}
\right)^{-1}y(t)e^{i\la\tau,t\ra}\,dt$$
is optimal recovery method.
\end{theorem}

By this theorem analogously to \eqref{CC}  we obtain the exact inequality
$$|D_1x(\tau)|\le\frac{(\pi I)^{1/2}}{(2\pi)^{d/2}}K_d(k,n)\|x\cd\|_{\ld}^{\frac{2n-2k-d}{2n}}
\|D_2x\cd\|_{\ld}^{\frac{2k+d}{2n}}$$
or
\begin{equation}\label{Exa}
\|D_1x\cd\|_{L_\infty(\mathbb R^d)}\le\frac{(\pi I)^{1/2}}{(2\pi)^{d/2}}K_d(k,n)\|x\cd\|_{\ld}^{\frac{2n-2k-d}{2n}}
\|D_2x\cd\|_{\ld}^{\frac{2k+d}{2n}}.
\end{equation}

Now we consider some examples. Define the operator $(-\Delta)^{n/2}$, $n\ge0$, as follows
$$(-\Delta)^{n/2}x\cd=F^{-1}(|t|^n Fx(t))\cd.$$
Put $d_1(t)=|t|^k$ and $d_2(t)=|t|^n$. Then problem \eqref{ETa} is the problem of optimal recovery of $(-\Delta)^{k/2}x(\tau)$ on the class
$$W=\{\,x\cd\in\ld:\|(-\Delta)^{n/2}x\cd\|_{\ld}\le1\,\}$$
by the inaccurate information about $x\cd$.

By Theorem~\ref{TT} we obtain

\begin{corollary}
Let $n>k+d/2$. Then
$$E=C_d(k,n)\delta^{\frac{2n-2k-d}{2n}},\quad
C_d(k,n)=\frac{K_d(k,n)}{(2^{d-1}\pi^{d/2-1}\Gamma(d/2))^{1/2}},$$
and the method
$$\wm(y)=\frac1{(2\pi)^d}\int_{\mathbb R^d}|t|^k\left(1+\frac{\delta^2(2k+d)}{(2\pi)^d(2n-2k-d)}
\right)^{-1}y(t)e^{i\la\tau,t\ra}\,dt$$
is optimal.
\end{corollary}

By \eqref{Exa} we get the exact inequality
$$\|(-\Delta)^{k/2}x\cd\|_{L_\infty(\mathbb R^d)}\le C_d(k,n)\|x\cd\|_{\ld}^{\frac{2n-2k-d}{2n}}
\|(-\Delta)^{n/2}x\cd\|_{\ld}^{\frac{2k+d}{2n}}.$$

Consider one more example. Let $\alpha=(\alpha_1,\ldots,\alpha_d)\in\mathbb Z_+^d$. We define $D^\alpha$ (the derivative of order $\alpha$) as follows:
$$D^\alpha x\cd=F^{-1}((it)^\alpha Fx(t))\cd,$$
where $(it)^\alpha=(it_1)^{\alpha_1}\cdots(it_d)^{\alpha_d}$. Let $D_1=D^\alpha$ and $D_2=(-\Delta)^{n/2}$. Then \eqref{ETa} is the problem of optimal recovery of $D^\alpha x(\tau)$ on the class $W$
by the inaccurate information about $x\cd$.

From the well-known Dirichlet formula we have
$$\int_{\substack{x_1\ge0,\ldots,x_d\ge0\\
x_1^2+\ldots+x_d^2\le1}}x_1^{p_1-1}\ldots x_d^{p_d-1}\,dx_1\ldots dx_d
=\frac{\Gamma(p_1/2)\ldots\Gamma(p_d/2)}
{2^d\Gamma(p_1/2+\ldots+p_d/2+1)},$$
$p_1,\ldots,p_d>0$. Using this formula and passing to the polar transformation we obtain
$$I(p_1,\ldots,p_d)=\int_{\Pi_{d-1}}\Phi(\omega,p_1,\ldots,p_d)
J(\omega)\,d\omega=2\frac{\Gamma(p_1/2)\ldots\Gamma(p_d/2)}
{\Gamma(p_1/2+\ldots+p_d/2)},$$
where
\begin{multline*}
\Phi(\omega,p_1,\ldots,p_d)=|\cos\omega_1|^{p_1-1}|\sin\omega_1
\cos\omega_2|^{p_2-1}\times\ldots\\
\times|\sin\omega_1\sin\omega_2\ldots
\sin\omega_{d-2}\cos\omega_{d-1}|^{p_{d-1}-1}\\
\times|\sin\omega_1\sin\omega_2\ldots\sin\omega_{d-2}
\sin\omega_{d-1}|^{p_d-1}.
\end{multline*}
Thus for $d_1(t)=(it)^\alpha$ and $d_2(t)=|t|^n$ we have
$$I=I(2\alpha_1+1,\ldots,2\alpha_d+1)=2\frac{\Gamma(\alpha_1+1/2)
\ldots\Gamma(\alpha_d+1/2)}
{\Gamma(|\alpha|+d/2)},$$
where $|\alpha|=\alpha_1+\ldots\alpha_d$.

\begin{corollary}
Let $n>|\alpha|+d/2$. Then
$$E=C_{d,\alpha}(n)\delta^{\frac{2n-2|\alpha|-d}{2n}},$$
where
$$C_{d,\alpha}(n)=\frac{K_d(|\alpha|,n)}{(2\pi)^{(d-1)/2}}\left(
\frac{\Gamma(\alpha_1+1/2)\ldots\Gamma(\alpha_d+1/2)}
{\Gamma(|\alpha|+d/2)}\right)^{1/2},$$
and the method
$$\wm(y)=\frac1{(2\pi)^d}\int_{\mathbb R^d}(it)^\alpha\left(1+\frac{\delta^2(2|\alpha|+d)}
{(2\pi)^d(2n-2|\alpha|-d)}\right)^{-1}y(t)e^{i\la\tau,t\ra}\,dt$$
is optimal.
\end{corollary}

The exact inequality in this case has the form:
$$\|D^\alpha x\cd\|_{L_\infty(\mathbb R^d)}\le C_{d,\alpha}(n)\|x\cd\|_{\ld}^{\frac{2n-2|\alpha|-d}{2n}}
\|(-\Delta)^{n/2}x\cd\|_{\ld}^{\frac{2|\alpha|+d}{2n}}.$$

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\end{document}

