On Recovery of the Singular Differential Laplace—Bessel Operator from the Fourier–Bessel Transform

Abstract

This paper is devoted to the problem of the best recovery of a fractional power of the B-elliptic operator of a function on ${\mathbb{R}}_{+}^N$ by its Fourier–Bessel transform known approximately on a convex set with the estimate of the difference between Fourier–Bessel transform of the function and its approximation in the metric $L_{\infty }$. The optimal recovery method has been found. This method does not use the Fourier–Bessel transform values beyond a ball centered at the origin.

1. Introduction

The problem of recovering fractional powers of operators was developed in the works [1,2,3] by G.G. Magaril-Ilyaev, K.Y. Osipenko and E.O. Sivkova. In these papers a Laplace operator was considered and the main instrument was a classical Fourier transform. We further developed these results for a more general Laplace—Bessel singular differential operator based on Fourier–Bessel transform. According to these authors, the formulation of the question of finding the error of optimal recovery and the optimal method on a class of elements ideologically goes back to Kolmogorov’s work on the cross-sections of functional classes [4]. The formulation of the optimal recovery problem (but in a much simpler case than the one given here) belongs to Smolyak [5].

There are well-known situations when special attributes of the Laplace operator and Fourier transform can be carried onto elliptic singular differential operators comprising the Bessel operator. In such cases, the Fourier–Bessel transform is used for research. The theory of singular differential operators accommodating the Bessel operator, as well as functional spaces generated with such operators and the Fourier–Bessel transformation, has taken complete shape in the works of I.A. Kipriyanov and their disciples (see [6,7,8,9,10,11,12]). In this paper, we transfer the technique and results of [3] to the case of the Fourier–Bessel transform and the singular B-elliptic operator. We base our results on $L_{\infty }$–estimates for a difference between Fourier–Bessel transform and its approximation on a convex subset with an error. The method, based on $L_2^{\gamma }$–estimates, was constructed in [13].

2. Necessary Definitions

We consider a part of the Euclidean space ${\mathbb{R}}^N$

$${\mathbb{R}}_{+}^N=\{\chi =({\chi }^{\prime },{\chi }^{″}),$$

$${\chi }^{\prime }=({\chi }_1,\dots ,{\chi }_n),{\chi }^{″}=({\chi }_{n+1},\dots ,{\chi }_N),{\chi }_1⩾0,\dots ,{\chi }_n⩾0\},$$

where $1⩽n⩽N$.

Let ${\Xi }^{+}\subset {\mathbb{R}}_{+}^N$ be a domain abutting to the hyperplanes $x_1=0,\dots ,x_n=0$. Let the boundary of ${\Xi }^{+}$ be a union of two parts: ${\Gamma }^{+}$ in ${\mathbb{R}}_{+}^N$ and ${\Gamma }_0$ in the hyperplanes $x_1=0,\dots ,x_n=0$. We assume that ${\Gamma }_0\subset {\Xi }^{+}$; however, we treat ${\Xi }^{+}$ as a domain.

Let ${\Xi }_{\delta }^{+}$ be a subset of ${\Xi }^{+}$. Let us assume that all points of ${\Xi }_{\delta }^{+}$ are located at a distance not more than $\delta$ from ${\Gamma }^{+}$. In this case we call ${\Xi }_{\delta }^{+}$ a symmetrically interior (s-interior) sub-domain of the set ${\Xi }^{+}$.

Let ${\Xi }^{-}$ be a set constructed from ${\Xi }^{+}$ with symmetry in relation to $x^{\prime }=0$, $\Xi ={\Xi }^{+}\cup {\Xi }^{-}\subset {\mathbb{R}}^N$.

Let us denote by $C_{ev}^{\ell }\left({\Xi }^{+}\right)$ the set of all functions obeying the below listed properties. They are necessary for correct definitions of the Laplace–Bessel differential operator and Fourier–Bessel transform on proper functions.

1. Any function $\psi \in C_{ev}^{\ell }\left({\Xi }^{+}\right)$, and all of its partial derivatives of every order not more than ℓ, are continuous in ${\Xi }^{+}$.

2. Even continuations of any function $\psi \in C_{ev}^{\ell }\left({\Xi }^{+}\right)$ in relation to $x^{\prime }=0$ still belongs to the class $C^{\ell }(\Xi )$.

In addition, $C_{ev}^{\infty }\left({\Xi }^{+}\right)={\displaystyle \bigcap _{l=0}^{\infty }}{C}_{ev}^{\ell }\left({\Xi }^{+}\right)$.

We say that functions admitting symmetrical even continuation in relation to the corresponding variables while keeping smoothness are even with respect to these variables (see [6]).

Let us denote by $C_{ev,0}^{\ell }\left({\Xi }^{+}\right)$ the set of all functions $\psi \in C_{ev}^{\ell }\left({\Xi }^{+}\right)$ that equal zero beyond some s-interior sub-domain of ${\Xi }^{+}$. Let $\gamma =({\gamma }_1,\cdots ,{\gamma }_n),{\left({\chi }^{\prime }\right)}^{\gamma }={\displaystyle \prod _{j=1}^n}{\chi }_j^{{\gamma }_j},$ where ${\gamma }_j>0$, $j=1,\cdots ,n$. We will write sometimes ${\chi }^{\gamma }$ instead of ${\left({\chi }^{\prime }\right)}^{\gamma }$, if this does not cause misunderstandings.

Let us denote by $L_p^{\gamma }\left({\Xi }^{+}\right)$ a closure of $C_{ev}\left({\Xi }^{+}\right)$ by the norm

$${\parallel g(•)\parallel }_{L_p^{\gamma }\left({\Xi }^{+}\right)}={\left[\underset{{\Xi }^{+}}{\int }{\left|g\left(\chi \right)\right|}^p{\left({\chi }^{\prime }\right)}^{\gamma }d\chi \right]}^{1/p}.$$

If ${\Xi }^{+}$ = $R_{+}^N$, we can write $L_p^{\gamma }$ without the symbol $R_{+}^N$. Let $L_{\infty }^{\gamma }\left({\Xi }^{+}\right)$ be a closure of $C_{ev}\left({\Xi }^{+}\right)$ by the norm

$${\parallel g\parallel }_{L_{\infty }^{\gamma }\left({\Xi }^{+}\right)}=\parallel g\left(x\right){\left(x^{\prime }\right)}^{\gamma }{\parallel }_{L_{\infty }\left({\Xi }^{+}\right)}=\mathrm{ess}\sup \left|g\left(x\right)x^{\gamma }\right|.$$

Let $L_{p,loc}^{\gamma }\left({\Xi }^{+}\right)$ be the set of all such functions f, that

$$\underset{{\Xi }_{\delta }^{+}}{\int }{\left|f\left(\chi \right)\right|}^p{\left({\chi }^{\prime }\right)}^{\gamma }d\chi <+\infty$$

for all s-interior sub-domains ${\Xi }_{\delta }^{+}$ of the domain ${\Xi }^{+}$.

Let us introduce the spaces below.

Let ${\mathcal{D}}_{ev}\left({\Xi }^{+}\right)\left({\mathcal{E}}_{ev}\left({\Xi }^{+}\right)\right)$ be the set of all constrictions of even functions with respect to $x^{\prime }=0$ in the space $\mathcal{D}(\Xi )\left(\mathcal{E}\right(\Xi \left)\right)$ to the set ${\Xi }^{+}$ (test functions) with this topology, induced by the topology in $\mathcal{D}(\Xi )\left(\mathcal{E}\right(\Xi \left)\right)$, ${\mathcal{D}}_{ev}={\mathcal{D}}_{ev}\left({\mathbb{R}}_{+}^N\right)$.

Let ${\mathcal{S}}_{ev}$ be the linear space of all functions (also test functions) $\psi \left(x\right)\in C_{ev}^{\infty }\left({\mathbb{R}}_{+}^N\right)$ that tend to zero, as well as all their derivatives of all order, faster than any power of ${\left|x\right|}^{-1}$ as $\left|x\right|\to \infty$. The topology in ${\mathcal{S}}_{ev}$ is introduced as being the same as in the space $\mathcal{S}$ (see [14,15,16,17,18,19]).

We define the space of distributions ${\mathcal{D}}_{ev}^{\prime }\left({\Xi }^{+}\right)({\mathcal{E}}_{ev}^{\prime }\left({\Xi }^{+}\right),{\mathcal{S}}_{ev}^{\prime })$ as the dual space of ${\mathcal{D}}_{ev}\left({\Xi }^{+}\right)({\mathcal{E}}_{ev}\left({\Xi }^{+}\right),{\mathcal{S}}_{ev})$ with the weak topology. The designation

$${⟨g\left(x\right),\psi \left(x\right)⟩}_{\gamma }=⟨g\left(x\right),\psi \left(x\right)⟩$$

(1)

means the action of a distribution g on a test function $\psi$.

We do not make any difference between a function $g\left(x\right)\in L_{1,loc}^{\gamma }\left({\Xi }^{+}\right)$ and the functional $g\in {\mathcal{D}}_{ev}^{\prime }\left({\Xi }^{+}\right)$ called regular, acting by the formula

$$⟨g\left(x\right),\psi \left(x\right)⟩=\underset{{\Xi }^{+}}{\int }g\left(\chi \right)\psi \left(\chi \right){\left({\chi }^{\prime }\right)}^{\gamma }d\chi .$$

(2)

If a functional in ${\mathcal{D}}_{ev}^{\prime }\left({\Xi }^{+}\right)$ is not regular, we call it singular.

Let us introduce the direct and inverse mixed Fourier–Bessel transforms in ${\mathcal{S}}_{ev}$ by the formulas

$$F_{\gamma }\psi =F_{\gamma }\left[\psi ({\chi }^{\prime },{\chi }^{″})\right]\left(\xi \right)=$$

$$=\underset{R_{+}^N}{\int }\psi \left(\chi \right)\prod _{\kappa =1}^n{j}_{{\nu }_{\kappa }}\left({\xi }_{\kappa }{\chi }_{\kappa }\right)e^{-i{\chi }^{″}•{\xi }^{″}}{\left({\chi }^{\prime }\right)}^{\gamma }d\chi =$$

$$={\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }\prod _{\kappa =1}^n{\Gamma }^2({\nu }_{\kappa }+1)F_{\gamma }^{-1}\left[\psi ({\chi }^{\prime },-{\chi }^{″})\right]\left(\xi \right),$$

$$F_{\gamma }^{-1}\left[\psi \right]\left(\chi \right)=\frac{1}{{\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }{\displaystyle \prod _{\kappa =1}^n}{\Gamma }^2({\nu }_{\kappa }+1)}\times$$

$$\times \underset{R_{+}^N}{\int }\psi \left(\xi \right)\prod _{\kappa =1}^n{j}_{{\nu }_{\kappa }}\left({\xi }_{\kappa }{\chi }_{\kappa }\right)e^{i{\chi }^{″}•{\xi }^{″}}{\left({\xi }^{\prime }\right)}^{\gamma }d\xi =$$

$$=\frac{1}{{\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }{\displaystyle \prod _{\kappa =1}^n}{\Gamma }^2({\nu }_{\kappa }+1)}{F}_{\gamma }\left[\psi ({\xi }^{\prime },-{\xi }^{″})\right]\left(\chi \right),$$

where

$${\chi }^{\prime }•{\xi }^{\prime }={\chi }_1{\xi }_1+\dots +{\chi }_n{\xi }_n,{\chi }^{″}•{\xi }^{″}={\chi }_{n+1}{\xi }_{n+1}+\dots +{\chi }_N{\xi }_N,$$

$$\left|\nu \right|={\nu }_1+\dots +{\nu }_n,$$

$$j_{{\nu }_{\kappa }}\left(z_{\kappa }\right)=\frac{2^{{\nu }_{\kappa }}\Gamma ({\nu }_{\kappa }+1)}{z_{\kappa }^{{\nu }_{\kappa }}}{J}_{{\nu }_{\kappa }}\left(z_{\kappa }\right)=$$

$$=\Gamma ({\nu }_{\kappa }+1)\sum _{m=1}^{\infty }\frac{{(-1)}^m{z}_{\kappa }^{2m}}{2^{2m}m!\Gamma (m+{\nu }_{\kappa }+1)},$$

$\Gamma (•)$ is the Euler gamma function, $J_{{\nu }_k}(•)$ is the Bessel function of the first kind, ${\nu }_k=({\gamma }_k-1)/2,k=1,\cdots ,n$.

Theorem 1

(The Parseval–Plancherel theorem for the Fourier–Bessel transform [7]). The formula

$${\parallel \psi \parallel }_{L_2^{\gamma }}={\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }\prod _{\kappa =1}^n{\Gamma }^2({\nu }_{\kappa }+1){\parallel \widehat{\psi }\parallel }_{L_2^{\gamma }},\widehat{\psi }=F_{\gamma }\left[\psi \right]$$

(the Parseval–Plancherel formula) holds.

We define the Fourier–Bessel transform of a distribution g by the formula

$${⟨F_{\gamma }\left[g\right],\psi ⟩}_{\gamma }={⟨g,F_{\gamma }\left[\psi \right]⟩}_{\gamma },$$

where $\psi \in S$.

The B-elliptic Laplace–Bessel operator ${\Delta }_B$ is defined by the formula (see [6])

$${\Delta }_{B}u=\sum _{k=1}^n\left(\frac{{\partial }^{2}u}{\partial x_k^2}+\frac{{\gamma }_k}{x_k}\frac{\partial u}{\partial x_k}\right)+\sum _{k=n+1}^N\frac{{\partial }^{2}u}{\partial x_k^2}.$$

We will use the denotations below:

$$\Pi ={\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }\prod _{\kappa =1}^n{\Gamma }^2({\nu }_{\kappa }+1)=2^{N+\left|\gamma \right|}{\pi }^{N-n}\prod _{\kappa =1}^n{\Gamma }^2(({\gamma }_{\kappa }+1)/2),$$

(3)

$$B(\zeta ,\rho )=\{\chi \in {\mathbb{R}}^N:|\chi -\zeta |⩽\rho \},$$

(4)

$$B_{+}(\zeta ,\rho )=\{\chi \in {\mathbb{R}}_{+}^N:|\chi -\zeta |⩽\rho \},$$

(5)

$$r_{\Xi }=sup\{\rho >0:B(0,\rho )\subset \Xi \},$$

(6)

$$\widehat{r}=\widehat{r}(\alpha ,ϵ,\delta )=$$

$$={\left(\frac{\left(\right|\gamma |+N+2\alpha )\Gamma \left(\left(\right|\gamma |+N)/2\right){\displaystyle \prod _{\kappa =1}^n}\Gamma \left(({\gamma }_{\kappa }+1)/2\right)}{2^{-\mid \gamma \mid -N+n+1}{\pi }^{(n-N)/2}{\delta }^2}\right)}^{\frac{1}{\left|\gamma \right|+N+2\alpha }},$$

(7)

$$r_0=min\{\widehat{r},r_{\Xi }\}.$$

(8)

3. Problem Statement

We define the fractional degree of the operator ${\Delta }_B$ using the equality (see [20])

$${(-{\Delta }_B)}^{\alpha /2}f\left(x\right)=F_{\gamma }^{-1}{\left(\right|\xi |}^{\alpha }{F}_{\gamma }f\left(\xi \right))\left(x\right),$$

where $\alpha >0$. Let us introduce the functional spaces

$${\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right)=$$

$$=\{f(•)\in {\mathcal{S}}_{ev}^{\prime }:{(-{\Delta }_B)}^{\alpha /2}f(•)\in L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right),F_{\gamma }\left[f(•)\right]\in L_{\infty }\left({\mathbb{R}}_{+}^N\right)\},$$

$$W_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right)=\{f(•)\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right):\parallel {(-{\Delta }_B)}^{\alpha /2}f(•){\parallel }_{L_2\left({\mathbb{R}}_{+}^N\right)}⩽1\}.$$

Let us consider a measurable non-empty bounded subset G in ${\mathbb{R}}_{+}^N$. Let $0<ϵ<\alpha ,\delta >0$. Suppose that a function $F_{\gamma }f\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right)$ is known approximately, namely, we know only such a function $g(•)\in L_{\infty }\left(G\right)$ that

$$\Vert F_{\gamma }{f(•)-g(•)\Vert }_{L_{\infty }\left(G\right)}\le \delta .$$

Based on this information, we aim to recover the functions f and ${(-{\Delta }_B)}^{ϵ/2}f$ in the best possible way. We consider the function $g(•)\in L_{\infty }\left(G\right)$ as an approximation of the constriction $F_{\gamma }f(•){|}_G$ with the error $\delta$ in the $L_{\infty }\left(G\right)$–metric.

As in the papers [2,3], we call any measurable mapping

$$\mu :L_{\infty }\left(G\right)⟶L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$$

recovering method. Its error we define as

$$e\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right),G,\delta ,\mu \right)=$$

$$=\underset{\mathbf{U}(\alpha ,G,\delta )}{sup}{\Vert {(-{\Delta }_B)}^{ϵ/2}f(•)-\mu \left(g(•)\right)(•)\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)},$$

where

$$\mathbf{U}(\alpha ,G,\delta )=$$

$$=\left\{\left(f(•)\in W_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right),g(•)\in L_{\infty }^{\gamma }\left(G\right)\right):{\Vert F_{\gamma }f(•)-g(•)\Vert }_{L_{\infty }\left(G\right)}\le \delta \right\}.$$

The value

$$E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),G,\delta \right)=\underset{\mu }{inf}e\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right),G,\delta ,\mu \right)$$

$$=\underset{\mu }{inf}\underset{\mathbf{U}(\alpha ,G,\delta )}{sup}{\Vert {(-{\Delta }_B)}^{ϵ/2}f(•)-\mu \left(g(•)\right)(•)\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}$$

is called the optimal recovery error (supremum is over all methods $\mu :L_{\infty }\left(G\right)⟶L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$).

The mappings $\mu$, on which the lower bound is reached, we call the optimal recovery methods.

4. Lower Estimate of the Optimal Recovery Error Value

The proofs of both lemmas in this section are carried out according to the scheme suggested in the papers [13,21], with the difference that the initial error estimate is given here in the $L_{\infty }$–metric and in those papers it is given in the $L_2^{\gamma }$–metric.

Let us explore an auxiliary task.

Extremal task ${\mathrm{I}}_{\infty }$.

$$\begin{array}{c}\Vert {(-{\Delta }_B)}^{ϵ/2}{f\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}⟶max,{\Vert F_{\gamma }f\Vert }_{L_{\infty }\left(G\right)}\le \delta ,\\ {\Vert {(-{\Delta }_B)}^{\alpha /2}f\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}\le 1,f\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right).\end{array}$$

(9)

Lemma 1.

The optimal recovery error $E\left({(-\Delta )}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),G,\delta \right)$ is not lower than the value of Extremal task ${\mathrm{I}}_{\infty }$.

Proof. 

Let $u_0(•)$ be an acceptable function of extremal task $I_{\infty }$; in other words, $u_0(•)$ satisfies the constraints of this problem. Then the function $-u_0(•)$ is also acceptable. Let $\mu :L_{\infty }\left(G\right)⟶L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$ be any fixed method and $\mu \left(0\right)$ be an image of the zero element of the space $L_{\infty }\left(G\right)$ with the mapping $\mu$. Then

$$2\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}={\Vert 2{(-{\Delta }_B)}^{ϵ/2}{u}_0(•)\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}=$$

$$=\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•)+{(-{\Delta }_B)}^{ϵ/2}{u}_0(•){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}=$$

$$=\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•)-{(-{\Delta }_B)}^{ϵ/2}(-u_0(•)){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}=$$

$$=\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•)-\mu \left(0\right)(•)+\mu \left(0\right)(•)-{(-{\Delta }_B)}^{ϵ/2}(-u_0(•)){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}⩽$$

$$⩽\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•)-\mu \left(0\right)(•){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}+$$

$$+\Vert {(-{\Delta }_B)}^{ϵ/2}(-u_0)(•)-\mu \left(0\right)(•){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}⩽$$

$$⩽2\underset{\begin{array}{c}\Vert F_{\gamma }{f(•)\Vert }_{L_{\infty }\left(G\right)}⩽\delta ,\\ \Vert {(-{\Delta }_B)}^{\alpha /2}f(•){\Vert }_{L_{2,\gamma }\left({\mathbb{R}}_{+}^N\right)}⩽1\end{array}}{sup}{\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•)-\mu \left(0\right)(•)\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}⩽$$

$$⩽2\underset{\begin{array}{c}\Vert F_{\gamma }{f(•)-g(•)\Vert }_{L_{\infty }\left(G\right)}\le \delta ,\\ \Vert {(-{\Delta }_B)}^{\alpha /2}f(•){\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}⩽1\end{array}}{sup}{\Vert {(-{\Delta }_B)}^{ϵ/2}{u}_0(•)-\mu \left(g\right)(•)\Vert }_{L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)}.$$

Now we can pass to a supremum over all acceptable functions of extremal task $I_{\infty }$ on the left side of this inequality and for all mappings (methods) $\mu$ on the right side. Thus, we complete the proof. □

Let us explore another extremal task.

Extremal task${\mathrm{I}}_{\infty }^2$.

$$\begin{array}{c}{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}|F_{\gamma }{f\left(\xi \right)|}^{2}d\xi ⟶max,{\parallel F_{\gamma }f\parallel }_{L_{\infty }\left(G\right)}⩽\delta ,\\ {\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi \le 1,f(•)\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right).\end{array}$$

Lemma 2.

The squared value of extremal task $I_{\infty }$ is equal to the value of extremal task ${\mathrm{I}}_{\infty }^2$.

Proof. 

The affirmation of the lemma follows from the Parseval–Plancherel theorem for the Fourier–Bessel transform.

Furthermore, we assume that $\Xi ={\Xi }^{+}\cup {\Xi }^{-}$ is convex. □

Theorem 2.

If $0\notin {\Xi }^{+}$, then

$$E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right),{\Xi }^{+},\delta \right)=+\infty .$$

Proof. 

Assume that $0\notin {\Xi }^{+}$. In this case we can apply the finite dimensional separability theorem (for instance [22]) and separate the origin from a convex set $\Xi$ and, as a consequence, from ${\Xi }^{+}$. It means that there exists such a vector $\eta =({\eta }_1,\dots ,{\eta }_N)\in {\mathbb{R}}_{+}^N,$ $\left|\eta \right|=\sqrt{{\eta }_1^2+\cdots +{\eta }_N^2}=1,$ that

$$\underset{\xi \in {\Xi }^{+}}{sup}(\eta ,\xi )\le 0.$$

For arbitrarily small $\epsilon >0$, we introduce a ball $B_{\epsilon }=B(\epsilon \eta ,\epsilon /2)$. If $\xi \in B_{\epsilon }$, then

$$(\xi -\epsilon \eta ,\xi -\epsilon \eta )={\left|\xi \right|}^2+{\epsilon }^2{\left|\eta \right|}^2-2\epsilon (\xi ,\eta )\le {\epsilon }^2/4$$

and, taking into account that $\left|\eta \right|=1$, we get

$$(\xi ,\eta )\ge \frac{{\left|\xi \right|}^2}{2\epsilon }+\frac{3\epsilon }{8}>0.$$

Thus, $B_{\epsilon }\cap {\Xi }^{+}=\oslash$.

Let us introduce a function $f_{\epsilon }(•)$ such that

$$F_{\gamma }{f}_{\epsilon }\left(\xi \right)=\left\{\begin{array}{c}{\Pi }^{1/2}{\left(\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi \right)}^{-1/2},\mathrm{if}\xi \in B_{\epsilon },\hfill \\ 0,\mathrm{if}\xi \notin B_{\epsilon }.\hfill \end{array}\right.$$

Obviously, the function $F_{\gamma }{f}_{\epsilon }$ has a bounded support and, therefore, $F_{\gamma }{f}_{\epsilon }\in L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$ and $F_{\gamma }{f}_{\epsilon }\in L_{\infty }\left({\mathbb{R}}_{+}^N\right)$. Thus, $f_{\epsilon }\in L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$. Moreover, the function $\xi \to -{\left|\xi \right|}^{\alpha }{F}_{\gamma }{f}_{\epsilon }\left(\xi \right)$ belongs to $L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$. Therefore, $f_{\epsilon }\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right)$, because ${(-\Delta )}^{\alpha /2}{f}_{\epsilon }(•)\in L_2^{\gamma }\left({\mathbb{R}}_{+}^N\right)$. It is also easy to show that the function $f_{\epsilon }(•)$ satisfies other conditions of Problem ${\mathrm{I}}^2$. Let $\xi \in B_{\epsilon }$. Then $\left|\xi \right|=|\xi -\epsilon \eta +\epsilon \eta |\le |\xi -\epsilon \eta |+\left|\epsilon \eta \right|\le 3\epsilon /2$. Taking this fact into account, we get

$$\frac{1}{{\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }{\displaystyle \prod _{k=1}^n}{\Gamma }^2({\nu }_k+1)}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi =$$

$$=\frac{\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}d\xi }{\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi }=\frac{\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ+2\alpha -2\alpha }d\xi }{\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi }⩾$$

$$⩾{\left(\frac{3}{2}\epsilon \right)}^{-2(\alpha -ϵ)}\frac{\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi }{\underset{B_{\epsilon }}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi }={\left(\frac{3}{2}\epsilon \right)}^{-2(\alpha -ϵ)}.$$

Since $\epsilon$ is arbitrary small, we get that the value of the objective functional in extremal task ${\mathrm{I}}^2$ and, therefore, in the original recovery problem, can be made arbitrarily large. The proof is completed. □

Lemma 3.

Let $r_{\Xi }<\widehat{r}$. The lower estimate

$$E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),{\Xi }^{+},\delta \right)⩾$$

$$⩾\left(\frac{2^{2-N-2\left|\nu \right|}{\pi }^{(n-N)/2}{\delta }^2(\alpha -ϵ)r_{\Xi }^{\left|\gamma \right|+N+2ϵ}}{\left(\right|\gamma |+N+2ϵ)\left(\right|\gamma |+N+2\alpha )\Gamma \left(\left(\right|\gamma |+N)/2\right){\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right)}\right.+$$

$$+{\left.\frac{1}{r_{\Xi }^{2(\alpha -ϵ)}}\right)}^{1/2}$$

(10)

holds.

Proof. 

The intersection of the boundary of semi-ball $B_{+}(0,r_{\Xi })\subset {\Xi }^{+}$ and the boundary of ${\Xi }^{+}$ is non-empty. Assume that ${\xi }_0$ belongs to this intersection. Then ${\xi }_0\notin \mathrm{int}{\Xi }^{+}$. Hence, there is the ability to separate the point ${\xi }_0$ from the convex set $\mathrm{int}{\Xi }^{+}$, i.e., there is a vector $\eta \in {\mathbb{R}}_{+}^N$, such that $\left|\eta \right|=1$ and $\underset{\xi \in {\Xi }^{+}}{sup}(\eta ,\xi )\le (\eta ,{\xi }_0)$. Explore a sequence of points ${\xi }_{\kappa }={\xi }_0+(1/\kappa )\eta$ $\kappa =1,2,3.\cdots$, and a sequence of balls $B({\xi }_{\kappa },1/\left(2\kappa \right))$. Every one of these balls does not intersect with ${\Xi }^{+}$. Indeed, if $\xi \in B({\xi }_{\kappa },1/\left(2\kappa \right)$, then $(\eta ,\xi )>(\eta ,{\xi }_0)$, that is $\xi \notin {\Xi }^{+}$. Let

$$V_{\kappa }=\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }d\xi .$$

Let us explore a sequence of functions $f_{\kappa }(•)\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right)$ having the Fourier–Bessel transform of the form

$$F_{\gamma }{f}_{\kappa }\left(\xi \right)=\left\{\begin{array}{c}{\sigma }^{1/2}{\Pi }^{1/2}{V}_{\kappa }^{-1/2}{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-\alpha },\mathrm{if}\xi \in B({\xi }_{\kappa },1/\left(2\kappa \right)),\hfill \\ \delta ,\mathrm{if}\xi \in B(0,r_{\Xi }),\hfill \\ 0,\mathrm{else}.\hfill \end{array}\right.$$

where

$$\sigma =1-{\Pi }^{-1}\delta \underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi .$$

From the last two conditions we get $\parallel F_{\gamma }{f}_{\kappa }{\parallel }_{L_{\infty }(\Xi )}⩽\delta$. Let us note that, if $\xi \in B({\xi }_{\kappa },1/\left(2\kappa \right))$, then

$$\left|\xi \right|=|\xi -{\xi }_0-\frac{1}{\kappa }\eta +{\xi }_0+\frac{1}{\kappa }\le \frac{1}{2\kappa }+\left|{\xi }_0\right|+\frac{1}{\kappa }=r_{\Xi }+\frac{3}{2\kappa }.$$

Hence,

$${\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi =$$

$$={\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi +$$

$$+{\Pi }^{-1}\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi =$$

$$={\delta }^2{\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi +$$

$$+\sigma {\Pi }^{-1}\Pi V_{\kappa }^{-1}{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi \le$$

$$⩽\delta {\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi +$$

$$+\sigma V_{\kappa }^{-1}{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{2\alpha }\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }d\xi =1.$$

It means that functions $f_{\kappa }$ are acceptable in problem ${\mathrm{I}}_{\infty }^2$.

Let us match “the Lagrange function” $\mathcal{L}\left(f\right(•\left)\right)$ to the problem ${\mathrm{I}}_{\infty }^2$

$$\mathcal{L}\left(f\right)={\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{(-|\xi |}^{2ϵ}+p\left(\xi \right)+{\lambda \left|\xi \right|}^{2\alpha }\left)\right|F_{\gamma }{f\left(\xi \right)|}^{2}d\xi ,$$

(11)

where $\lambda =r_{\Xi }^{-2(\alpha -ϵ)}$,

$$0⩽p\left(\xi \right)=\left\{\begin{array}{c}{\left|\xi \right|}^{2ϵ}-\lambda {\left|\xi \right|}^{2\alpha },\mathrm{if}\xi \in B_{+}(0,r_{\Xi }),\hfill \\ 0,\mathrm{if}\xi \in {\mathbb{R}}_{+}^N\setminus B_{+}(0,r_{\Xi }).\hfill \end{array}\right.$$

For every ${\mathrm{I}}_{\infty }^2$-acceptable function, we have

$${\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}|F_{\gamma }{f\left(\xi \right)|}^{2}d\xi ={\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi -$$

$$-{\Pi }^{-1}\underset{{\Xi }^{+}}{\int }{\xi }^{\gamma }p\left(\xi \right)\left(|F_{\gamma }{f\left(\xi \right)|}^2-{\delta }^2\right)d\xi -$$

$$-\lambda \left({\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi -1\right)=$$

$$=-\mathcal{L}\left(f\right)+{\delta }^2{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi +\lambda ⩽{\delta }^2{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi +\lambda .$$

In particular, the value of extremal task ${\mathrm{I}}_{\infty }^2$ is not more than

$${\delta }^2{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi +\lambda .$$

For every $\kappa$, we have

$$\mathcal{L}\left(f_{\kappa }\right)={\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{(-|\xi |}^{2ϵ}+p\left(\xi \right)+{\lambda \left|\xi \right|}^{2\alpha }\left)\right|F_{\gamma }{f}_{\kappa }\left(\xi \right){|}^{2}d\xi =$$

$$=\sigma V_{\kappa }^{-1}{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{(-|\xi |}^{2ϵ}+{\lambda \left|\xi \right|}^{2\alpha })d\xi .$$

For $\xi \in B({\xi }_{\kappa },1/\left(2\kappa \right))$, we have

$$r_{\Xi }=|{\xi }_0|=|{\xi }_0-\xi +\xi -\frac{1}{\kappa }\eta +\frac{1}{\kappa }\eta |\le |\xi |+|{\xi }_0+\frac{1}{\kappa }\eta -\xi |+|\frac{1}{\kappa }\eta |\le |\xi |+\frac{1}{2\kappa }+\frac{1}{\kappa },$$

whence we get

$$\left|\xi \right|\ge r_{\Xi }-\frac{3}{2\kappa }.$$

Therefore,

$$\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{(-|\xi |}^{2ϵ})d\xi ⩽-{\left(r_{\Xi }-\frac{3}{2\kappa }\right)}^{2ϵ}{V}_{\kappa }.$$

On the other hand, we showed earlier that for $\xi \in B({\xi }_{\kappa },1/\left(2\kappa \right))$

$$\left|\xi \right|⩽r_{\Xi }+\frac{3}{2\kappa }.$$

Therefore,

$$\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{\left(\right|\xi |}^{2\alpha })d\xi ⩽{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{2\alpha }{V}_{\kappa }.$$

> From the last two inequalities, we have

$$0⩽\mathcal{L}\left(f_{\kappa }\right)⩽\sigma \left(-{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{2ϵ}+\lambda \right).$$

Since $\lambda =r_{\Xi }^{2(ϵ-\alpha )}$, we obtain

$$\underset{\kappa \to \infty }{lim}\mathcal{L}\left(f_{\kappa }\right)=0.$$

(12)

Let us return to the integral

$${\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }|F_{\gamma }{f}_{\kappa }{\left(\xi \right)|}^{2}d\xi ={\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi +$$

$$+{\Pi }^{-1}\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi =$$

$$={\delta }^2{\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi +$$

$$+\sigma {\Pi }^{-1}\Pi V_{\kappa }^{-1}{\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }\underset{B({\xi }_{\kappa },1/\left(2\kappa \right))}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi ⩾$$

$$⩾{\delta }^2{\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi +$$

$$+\sigma {\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }{\left(r_{\Xi }-\frac{3}{2\kappa }\right)}^{2\alpha }.$$

This sequence tends to 1 when $\kappa \to \infty$. Hence, since

$${\delta }^2{\Pi }^{-1}\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi +\sigma {\left(r_{\Xi }+\frac{3}{2\kappa }\right)}^{-2\alpha }{\left(r_{\Xi }-\frac{3}{2\kappa }\right)}^{2\alpha }⩽$$

$$⩽{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi ⩽1,$$

we have

$$\underset{\kappa \to \infty }{lim}{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi =1.$$

(13)

Based on Formulas (12) and (13), we get

$$\underset{\kappa \to \infty }{lim}{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi =\lambda +\underset{\kappa \to \infty }{lim}{\delta }^2{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi +$$

$$+\lambda \underset{\kappa \to \infty }{lim}\left({\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{f}_{\kappa }\left(\xi \right)\right|}^{2}d\xi -1\right)-\underset{\kappa \to \infty }{lim}\mathcal{L}\left(f_{\kappa }\right)+$$

$$+\underset{\kappa \to \infty }{lim}{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)\left(\right|F_{\gamma }{f}_{\kappa }\left(\xi \right){|}^2-{\delta }^2)d\xi =$$

$$=\lambda +{\delta }^2{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi .$$

This fact means that the value

$$\lambda +r_{\Xi }^{-2\gamma }{\delta }^2{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi$$

(14)

is the value of the problem ${\mathrm{I}}_{\infty }^2$, and the square root from (14) is the value of the problem ${\mathrm{I}}_{\infty }$.

The integral

$$\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}d\xi$$

is a special case of the integral calculated in the book [23] (p. 613, Formula (4.635.1)). In our case, we get

$$\underset{B_{+}(0,r_{\Xi })}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}d\xi =\frac{r_{\Xi }^{\left|\gamma \right|+N+2ϵ}{\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right){\pi }^{(N-n)/2}}{2^{n-1}\left(\right|\gamma |+N+2ϵ)\Gamma \left(\left(\right|\gamma |+N)/2\right)}.$$

(15)

Now we can calculate the value of the problem ${\mathrm{I}}_{\infty }^2$, substituting (15) into (14):

$$\lambda +{\delta }^2{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi =$$

$$\begin{gathered} =\frac{2^{2-N-2\left|\nu \right|}{\pi }^{(n-N)/2}{\delta }^2(\alpha -ϵ)r_{\Xi }^{\left|\gamma \right|+N+2ϵ}}{\left(\right|\gamma |+N+2ϵ)\left(\right|\gamma |+N+2\alpha )\Gamma \left(\left(\right|\gamma |+N)/2\right){\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right)}+ \\ +\frac{1}{r_{\Xi }^{2(\alpha -ϵ)}}. \end{gathered}$$

(16)

> From here, using Lemmas 1 and 2, we get a lower estimate of the error of optimal recovery (10). The lemma is proved. □

Lemma 4.

Let $r_{\Xi }>\widehat{r}$. The lower estimate

$$E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),{\Xi }^{+},\delta \right)⩾$$

$$⩾\sqrt{\frac{\left|\gamma \right|+N+2\alpha }{\left|\gamma \right|+N+2ϵ}}{\left(\frac{2^{-\left|\gamma \right|-N+n+1}{\pi }^{(n-N)/2}{\delta }^2}{\Gamma (N+|\gamma \left|\right)/2)\prod _{j=1}^n\Gamma ({\nu }_j+1)}\right)}^{\frac{\alpha -ϵ}{\left|\gamma \right|+N+2\alpha }}$$

(17)

holds.

Proof. 

Explore a functions $u_0(•)\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right)$ whose Fourier–Bessel transform has the form

$$F_{\gamma }{u}_0\left(\xi \right)=\left\{\begin{array}{c}\delta ,\mathrm{if}\xi \in B_{+}(0,\widehat{r}),\hfill \\ 0,\mathrm{else}.\hfill \end{array}\right.$$

Obviously, $\parallel F_{\gamma }{u}_0{\parallel }_{L_{\infty }(\Xi )}⩽\delta$.

Let us again match “the Lagrange function” $\mathcal{L}\left(f\right)\left(f\right(•\left)\right)$ to the problem ${\mathrm{I}}_{\infty }^2$

$$\mathcal{L}\left(f\right)={\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{(-|\xi |}^{2ϵ}+\widehat{p}\left(\xi \right)+\widehat{\lambda }{\left|\xi \right|}^{2\alpha }\left)\right|F_{\gamma }{f\left(\xi \right)|}^{2}d\xi ,$$

(18)

where $\widehat{\lambda }={\widehat{r}}^{-2(\alpha -ϵ)}$,

$$0⩽\widehat{p}\left(\xi \right)=\left\{\begin{array}{c}{\left|\xi \right|}^{2ϵ}-\widehat{\lambda }{\left|\xi \right|}^{2\alpha },\mathrm{if}\xi \in B_{+}(0,\widehat{r}),\hfill \\ 0,\mathrm{if}\xi \in {\mathbb{R}}_{+}^N\setminus B_{+}(0,\widehat{r}).\hfill \end{array}\right.$$

Obviously, $\mathcal{L}\left(u_0\right)=0$. Moreover, taking into account (15), we get

$${\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }|F_{\gamma }{f\left(\xi \right)|}^{2}d\xi ={\delta }^2{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }d\xi =$$

$$=\frac{{\delta }^2{\widehat{r}}^{\left|\gamma \right|+N+2\alpha }{\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right){\pi }^{(N-n)/2}}{2^{n-1}\left(\right|\gamma |+N+2\alpha )\Gamma \left(\left(\right|\gamma |+N)/2\right){\left(2\pi \right)}^{N-n}{2}^{2\mid \nu \mid }{\displaystyle \prod _{k=1}^n}{\Gamma }^2({\nu }_k+1)}=1.$$

(19)

It means that function $u_0$ is acceptable in problem ${\mathrm{I}}_{\infty }^2$. Given these facts, for any ${\mathrm{I}}_{\infty }^2$-acceptable function, we have

$${\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}|F_{\gamma }{f\left(\xi \right)|}^{2}d\xi ⩽{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi -$$

$$-{\Pi }^{-1}\underset{{\Xi }^{+}}{\int }{\xi }^{\gamma }\widehat{p}\left(\xi \right)\left(|F_{\gamma }{f\left(\xi \right)|}^2-{\delta }^2\right)d\xi -$$

$$-\widehat{\lambda }\left({\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi -1\right)=$$

$$=-\mathcal{L}\left(f\right)+{\delta }^2{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }p\left(\xi \right)d\xi +\lambda ⩽$$

$$⩽-\mathcal{L}\left(u_0\right)+{\delta }^2{\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }\widehat{p}\left(\xi \right)d\xi +\widehat{\lambda }=$$

$$={\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }{u}_0\left(\xi \right)\right|}^{2}d\xi -$$

$$-{\Pi }^{-1}\underset{B_{+}(0,\widehat{r})}{\int }{\xi }^{\gamma }\widehat{p}\left(\xi \right)\left(\right|F_{\gamma }{u}_0\left(\xi \right){|}^2-{\delta }^2)d\xi +$$

$$+\widehat{\lambda }\left({\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }{u}_0\left(\xi \right)\right|}^{2}d\xi -1\right)=$$

$$={\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }{u}_0\left(\xi \right)\right|}^{2}d\xi .$$

This means that $u_0$ is the solution to problem ${\mathrm{I}}_{\infty }^2$. The value of this problem is equal to

$${\Pi }^{-1}\underset{R_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}|F_{\gamma }{u}_0{\left(\xi \right)|}^{2}d\xi ={\delta }^2{\Pi }^{-1}\underset{B_{+}(0,\widehat{r})}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}d\xi =$$

$$={\delta }^2{\Pi }^{-1}\underset{B_{+}(0,\widehat{r})}{\int }{\xi }^{\gamma }{\left(\right|\xi |}^{2ϵ}-\widehat{\lambda }{\left|\xi \right|}^{2\alpha })d\xi +\widehat{\lambda }=$$

$$={\delta }^2{\Pi }^{-1}\left(\frac{{\widehat{r}}^{\left|\gamma \right|+N+2ϵ}{\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right){\pi }^{(N-n)/2}}{2^{n-1}\left(\right|\gamma |+N+2ϵ)\Gamma \left(\left(\right|\gamma |+N)/2\right)}\right.-$$

$$-\left.\widehat{\lambda }\frac{{\widehat{r}}^{\left|\gamma \right|+N+2\alpha }{\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right){\pi }^{(N-n)/2}}{2^{n-1}\left(\right|\gamma |+N+2\alpha )\Gamma \left(\left(\right|\gamma |+N)/2\right)}\right)+\widehat{\lambda }=$$

$$=\frac{\left|\gamma \right|+N+2\alpha }{\left|\gamma \right|+N+2ϵ}{\left(\frac{\Gamma (N+|\gamma \left|\right)/2){\displaystyle \prod _{j=1}^n}\Gamma ({\nu }_j+1)}{2^{-\left|\gamma \right|-N+n+1}{\pi }^{(n-N)/2}{\delta }^2}\right)}^{\frac{2(ϵ-\alpha )}{\left|\gamma \right|+N+2\alpha }}$$

> From here, using Lemmas 1 and 2, we get a lower estimate of the error of optimal recovery (17). The lemma is proved. □

5. Upper Error Estimation and Optimal Recovery Method

We explore one more auxiliary extremal task.

Extremal task E.

$$\begin{gathered} \Vert {(-{\Delta }_B)}^{ϵ/2}f(•)-{\widehat{\mu }}_r\left(g(•)\right)(•){\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}⟶max, \\ {\Vert F_{\gamma }f(•)-g(•)\Vert }_{L_{\infty }\left({\Xi }_{+}\right)}\le \delta , \\ \Vert {(-{\Delta }_B)}^{\alpha /2}f(•){\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}\le 1,g(•)\in L_{\infty }\left({\Xi }_{+}\right),f(•)\in {\mathcal{W}}_{\infty ,2}^{\gamma ,\alpha }\left({\mathbb{R}}_{+}^N\right). \end{gathered}$$

The optimality of the method ${\widehat{\mu }}_r$ from the statement of the theorem means that the value of extremal task E is equal to $E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),{\Xi }_{+},\delta \right)$.

We will look for the optimal method among linear mappings, which in Fourier images act according to the rule $F_{\gamma }\mu \left(g\right)=ag$ with a vanishing outside the ball $B(0,r_0)$ function a. Let m be a mapping of this kind. Then

$$\Vert {(-{\Delta }_B)}^{ϵ/2}f(•)-{\widehat{\mu }}_r\left(g(•)\right)(•){\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}^2=$$

$$\begin{gathered} ={\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }{\left|\right|\xi |}^{ϵ}{F}_{\gamma }{f\left(\xi \right)-a\left(\xi \right)g\left(\xi \right)|}^{2}d\xi + \\ +{\Pi }^{-1}\underset{\left|\xi \right|⩾r_0}{\int }{\xi }^{\gamma }{\left|\right|\xi |}^{ϵ}{F}_{\gamma }{f\left(\xi \right)|}^{2}d\xi . \end{gathered}$$

(20)

Let $\tilde{\lambda }=\lambda$, $\tilde{p}=p$, when $r_{\Xi }<\widehat{r}$, $\tilde{\lambda }=\widehat{\lambda }$ and $\tilde{p}=\widehat{p}$, when $r_{\Xi }⩾\widehat{r}$. According to the Cauchy–Bunyakovsky inequality, we obtain for the integrand function in the second integral:

$${\left|{\left|\xi \right|}^{2ϵ}{F}_{\gamma }f\left(\xi \right)-a\left(\xi \right)g\left(\xi \right)\right|}^2=$$

$${\left|\frac{a\left(\xi \right)\sqrt{\widehat{p}\left(\xi \right)}}{\sqrt{\widehat{p}\left(\xi \right)}}(F_{\gamma }f\left(\xi \right)-g\left(\xi \right))+\frac{{\left|\xi \right|}^{ϵ}-a\left(\xi \right)}{\sqrt{\tilde{\lambda }}{\left|\xi \right|}^{\alpha }}\sqrt{\tilde{\lambda }}{\left|\xi \right|}^{\alpha }{F}_{\gamma }f\left(\xi \right)\right|}^2⩽$$

$$⩽\left(\frac{{\left|a\left(\xi \right)\right|}^2}{\tilde{p}\left(\xi \right)}+\frac{{\left|\right|\xi |}^{ϵ}{-a\left(\xi \right)|}^2}{\tilde{\lambda }{\left|\xi \right|}^{2\alpha }}\right)\times$$

$$\times \left(\tilde{p}\left(\xi \right)\right|F_{\gamma }f\left(\xi \right)-g\left(\xi \right){|}^2+\tilde{\lambda }\left|\xi {|}^{2\alpha }\right|F_{\gamma }f\left(\xi \right){|}^2).$$

For any $\xi$ the minimum value of the expression

$$J\left(a\right)=\frac{{\left|a\left(\xi \right)\right|}^2}{\tilde{p}\left(\xi \right)}+\frac{{\left|\right|\xi |}^{ϵ}{-a\left(\xi \right)|}^2}{\tilde{\lambda }{\left|\xi \right|}^{2\alpha }}$$

is reached at the point

$$\widehat{a}={\left|\xi \right|}^{ϵ}\left(1-{\left(\frac{\left|\xi \right|}{r_0}\right)}^{2(\alpha -ϵ)}\right).$$

(21)

This minimum value is equal to 1. Substituting (21) into the first term of (20) and taking into account the first constraint of the problem E, we get

$${\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }{\left|\right|\xi |}^{ϵ}{F}_{\gamma }{f\left(\xi \right)-a\left(\xi \right)g\left(\xi \right)|}^{2}d\xi ⩽$$

$$⩽{\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }(\tilde{p}\left(\xi \right)|F_{\gamma }{f\left(\xi \right)-g\left(\xi \right)|}^2+\tilde{\lambda }{\left|\xi \right|}^{2\alpha }|F_{\gamma }f\left(\xi \right){|}^2)d\xi ⩽$$

$$⩽{\delta }^2{\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }\tilde{p}\left(\xi \right)d\xi +\tilde{\lambda }{\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi .$$

(22)

For the second term of (20), we have

$${\Pi }^{-1}\underset{\left|\xi \right|⩾r_0}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ}{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi ⩽$$

$$⩽{\Pi }^{-1}\underset{\left|\xi \right|⩾r_0}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2ϵ-2\alpha }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi ⩽$$

$$⩽\tilde{\lambda }{\Pi }^{-1}\underset{\left|\xi \right|⩾r_0}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi .$$

(23)

Adding up (22) and (23), we obtain the next upper estimation

$$\Vert {(-{\Delta }_B)}^{ϵ/2}f(•)-{\widehat{m}}_r\left(g(•)\right)(•){\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}^2⩽$$

$$⩽{\delta }^2{\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }\tilde{p}\left(\xi \right)d\xi +\tilde{\lambda }{\Pi }^{-1}\underset{{\mathbb{R}}_{+}^N}{\int }{\xi }^{\gamma }{\left|\xi \right|}^{2\alpha }{\left|F_{\gamma }f\left(\xi \right)\right|}^{2}d\xi =$$

$$={\delta }^2{\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }\tilde{p}\left(\xi \right)d\xi +\tilde{\lambda }{\Vert {(-{\Delta }_B)}^{\alpha /2}f(•)\Vert }_{L_2^{\gamma }\left(R_{+}^N\right)}^2⩽$$

$$⩽{\delta }^2{\Pi }^{-1}\underset{\left|\xi \right|⩽r_0}{\int }{\xi }^{\gamma }\tilde{p}\left(\xi \right)d\xi +\tilde{\lambda }.$$

The obtained upper estimate of the squared error for the constructed method does not exceed the squared error of the optimal recovery method. This means that the constructed method is optimal. Thus, we proved the next result.

Theorem 3.

If $0\in \Xi$ and $r_{\Xi }<\widehat{r}$, then

$$E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),{\Xi }^{+},\delta \right)=$$

$$=\left(\frac{2^{2-N-2\left|\nu \right|}{\pi }^{(n-N)/2}{\delta }^2(\alpha -ϵ)r_{\Xi }^{\left|\gamma \right|+N+2ϵ}}{\left(\right|\gamma |+N+2ϵ)\left(\right|\gamma |+N+2\alpha )\Gamma \left(\left(\right|\gamma |+N)/2\right){\displaystyle \prod _{j=1}^n}\Gamma \left(({\gamma }_j+1)/2\right)}\right.+$$

$$+{\left.\frac{1}{r_{\Xi }^{2(\alpha -ϵ)}}\right)}^{1/2}.$$

If $r_{\Xi }>\widehat{r}$, then

$$E\left({(-{\Delta }_B)}^{ϵ/2},W_{\infty ,2}^{\gamma ,\alpha }\left(R_{+}^N\right),{\Xi }^{+},\delta \right)=$$

$$=\sqrt{\frac{\left|\gamma \right|+N+2\alpha }{\left|\gamma \right|+N+2ϵ}}{\left(\frac{2^{-\left|\gamma \right|-N+n+1}{\pi }^{(n-N)/2}{\delta }^2}{\Gamma (N+|\gamma \left|\right)/2){\displaystyle \prod _{j=1}^n}\Gamma ({\nu }_j+1)}\right)}^{\frac{\alpha -ϵ}{\left|\gamma \right|+N+2\alpha }}.$$

The method

$$\mu \left(g\right)={\Pi }^{-1}\underset{B{(0,r_0)}_{+}}{\int }{\left|\xi \right|}^{ϵ}\left(1-{\left(\frac{\left|\xi \right|}{r_0}\right)}^{2(\alpha -ϵ)}\right)\times$$

$$\times g\left(\xi \right){\displaystyle \prod _{k=1}^n}{j}_{{\nu }_k}\left({\xi }_k{x}_k\right)e^{i{x}^{\prime \prime }•{\xi }^{\prime \prime }}{\left({\xi }^{\prime }\right)}^{\gamma }d\xi$$

is optimal.