About the Method for Constructing External Estimates of the Limit Controllability Set for the Linear Discrete-Time System with Bounded Control

Abstract

This article discusses the problem of constructing an external estimate of the limit set of controllability for a linear discrete system with convex control constraints. We have proposed a decomposition method that allows us to reduce the problem for the initial system to subsystems of smaller dimension by switching to the normal Jordan basis of the matrix of the system. The statement about the structure of the reference hyperplane to the limit set of controllability is formulated and proved. A method for constructing an external estimate of the limit set of controllability with an arbitrary order of accuracy in the sense of the Hausdorff distance is proposed based on the principle of contraction mappings. The paper provides examples.

APPENDIX

Proof of Lemma 1. Denote the initial states of the system (A₁, ${{\mathcal{U}}_{1}}$) and (A₂, ${{\mathcal{U}}_{2}}$) by x_0,е1 ∈ ${{\mathbb{R}}^{{{{n}_{1}}}}}$ and x_0, 2 ∈ ${{\mathbb{R}}^{{{{n}_{2}}}}}$, respectively. Then x₀ = $\left( \begin{gathered} {{x}_{{0,1}}} \\ {{x}_{{0,2}}} \\ \end{gathered} \right)$ is the initial state of the system (A, $\mathcal{U}$).

By (1) it is true that for all N ∈ $\mathbb{N}$

$$x(N) = {{A}^{N}}{{x}_{0}} + {{A}^{{N - 1}}}u(0) + {{A}^{{N - 2}}}u(1) + \ldots + u(N - 1)$$

$$ = \left( {\begin{array}{*{20}{c}} {A_{1}^{N}}&O \\ O&{A_{2}^{N}} \end{array}} \right)\left( \begin{gathered} {{x}_{{0,1}}} \\ {{x}_{{0,2}}} \\ \end{gathered} \right) + \left( {\begin{array}{*{20}{c}} {A_{1}^{{N - 1}}}&O \\ O&{A_{2}^{{N - 1}}} \end{array}} \right)\left( \begin{gathered} {{u}_{1}}(0) \\ {{u}_{2}}(0) \\ \end{gathered} \right) + \ldots + \left( \begin{gathered} {{u}_{1}}(N - 1) \\ {{u}_{2}}(N - 1) \\ \end{gathered} \right)$$

$$ = \left( \begin{gathered} A_{1}^{N}{{x}_{{0,1}}} + A_{1}^{{N - 1}}{{u}_{1}}(0) + \ldots + {{u}_{1}}(N - 1) \\ A_{2}^{N}{{x}_{{0,2}}} + A_{2}^{{N - 1}}{{u}_{2}}(0) + \ldots + {{u}_{2}}(N - 1) \\ \end{gathered} \right) = \left( \begin{gathered} {{x}_{1}}(N) \\ {{x}_{2}}(N) \\ \end{gathered} \right).$$

Then x(N) = 0 if and only if there exist u₁(0), …, u₁(N – 1) ∈ ${{\mathcal{U}}_{1}}$ and u₂(0), …, u₂(N – 1) ∈ ${{\mathcal{U}}_{2}}$ such that x₁(N) = 0, x₂(N) = 0. The equality data is, by virtue of (2), equivalent to including x_0, 1 ∈ ${{\mathcal{X}}_{1}}$(N), x_0, 2 ∈ ${{\mathcal{X}}_{2}}$(N). Hence,

$$\mathcal{X}(N) = {{\mathcal{X}}_{1}}(N)\,\, \times \,\,{{\mathcal{X}}_{2}}(N).$$

Let x₀ ∈ ${{\mathcal{X}}_{\infty }}$. Then according to (3) there exists $\tilde {N} \in \mathbb{N}$ such that

$${{x}_{0}} = \mathcal{X}(\tilde {N}) = {{\mathcal{X}}_{1}}(\tilde {N})\,\, \times \,\,{{\mathcal{X}}_{2}}(\tilde {N}) \subset \bigcup\limits_{N = 0}^\infty {{{\mathcal{X}}_{1}}(N)} \,\, \times \,\,\bigcup\limits_{N = 0}^\infty {{{\mathcal{X}}_{2}}(N)} = {{\mathcal{X}}_{{1,\infty }}}\,\, \times \,\,{{\mathcal{X}}_{{2,\infty }}}.$$

Then ${{\mathcal{X}}_{\infty }}$ ⊂ ${{\mathcal{X}}_{{1,\infty }}}$ × ${{\mathcal{X}}_{{2,\infty }}}$.

Let x₀ ∈ ${{\mathcal{X}}_{{1,\infty }}}$ × ${{\mathcal{X}}_{{2,\infty }}}$. Therefore, there are ${{\tilde {N}}_{1}}$, ${{\tilde {N}}_{2}} \in \mathbb{N}$ such that x₀ ∈ ${{\mathcal{X}}_{1}}({{\tilde {N}}_{1}})$ × ${{\mathcal{X}}_{2}}({{\tilde {N}}_{2}})$ ⊂ ${{\mathcal{X}}_{1}}(\tilde {N})$ × ${{\mathcal{X}}_{2}}(\tilde {N})$, where $\tilde {N}$ = max{${{\tilde {N}}_{1}}$, ${{\tilde {N}}_{2}}$}. Then, by point 1 of Lemma 1

$${{x}_{0}} \in \mathcal{X}(\tilde {N}) \subset \bigcup\limits_{N = 0}^\infty {\mathcal{X}(N)} = {{\mathcal{X}}_{\infty }}.$$

It follows that ${{\mathcal{X}}_{{1,\infty }}}$ × ${{\mathcal{X}}_{{2,\infty }}} \subset {{\mathcal{X}}_{\infty }}$.

Finally, we get that ${{\mathcal{X}}_{\infty }}$ = ${{\mathcal{X}}_{{1,\infty }}}$ × ${{\mathcal{X}}_{{2,\infty }}}$. Lemma 1 is proved.

Proof of Lemma 2. Let $\{ y(k)\} _{{k = 0}}^{N}$ be the trajectory of the system (${{S}^{{ - 1}}}AS$, ${{S}^{{ - 1}}}\mathcal{U}$), i.e., y(N) according to (1) for the initial state y₀ ∈ ${{\mathbb{R}}^{n}}$ admits the following representation:

$$y(N) = {{S}^{{ - 1}}}ASy(N - 1) + {v}(N - 1) = \ldots = {{S}^{{ - 1}}}{{A}^{N}}S{{y}_{0}} + {{S}^{{ - 1}}}{{A}^{{N - 1}}}S{v}(0) + \ldots + {v}(N - 1),$$

where ${v}$(0), …, ${v}$(N – 1) ∈ ${{S}^{{ - 1}}}\mathcal{U}$.

By virtue of (2) y₀ ∈ $\mathcal{Y}(N)$ if and only if y(N) = 0, i.e.,

$${{S}^{{ - 1}}}{{A}^{N}}S{{y}_{0}} + {{S}^{{ - 1}}}{{A}^{{N - 1}}}S{v}(0) + \ldots + {v}(N - 1) = 0,$$

$${{A}^{N}}S{{y}_{0}} + {{A}^{{N - 1}}}S{v}(0) + \ldots + S{v}(N - 1) = 0,$$

which, by virtue of (2), is equivalent to including Sy₀ ∈ $\mathcal{X}(N)$, since by construction $S{v}$(0), …, $S{v}$(N – 1) ∈ $\mathcal{U}$. Whence follows the equality $\mathcal{X}(N)$ = $S\mathcal{Y}(N)$.

Let x₀ ∈ ${{\mathcal{X}}_{\infty }}$. By (3), there exists $\tilde {N} \in \mathbb{N} \cup \{ 0\}$ such that x₀ ∈ $\tilde {\mathcal{X}}(\tilde {N})$, which is equivalent to including x₀ ∈ $S\mathcal{Y}(\tilde {N})$ according to point 1 of Lemma 2. Hence S^–1x₀ ∈ $\mathcal{Y}(\tilde {N})$. Then S^–1x₀ ∈ $\bigcup\nolimits_{N = 0}^\infty {\mathcal{Y}(N)}$ = ${{\mathcal{Y}}_{\infty }}$, i.e., x₀ ∈ $S{{\mathcal{Y}}_{\infty }}$. Then ${{\mathcal{X}}_{\infty }}$ ⊂ $S{{\mathcal{Y}}_{\infty }}$.

Let x₀ ∈ $S{{\mathcal{Y}}_{\infty }}$, then S^–1x₀ ∈ ${{\mathcal{Y}}_{\infty }}$. By (3) there exists $\tilde {N} \in \mathbb{N} \cup \{ 0\}$ such that S^–1x₀ ∈ $\mathcal{Y}(\tilde {N})$. Then x₀ ∈ $S\mathcal{Y}(\tilde {N})$, which is equivalent to including x₀ ⊂ $\mathcal{X}(\tilde {N})$ by point 1 of Lemma 2. According to relations (3), the inclusion x₀ ∈ ${{\mathcal{X}}_{\infty }}$ is also true. Then $S{{\mathcal{Y}}_{\infty }}$ ⊂ ${{\mathcal{X}}_{\infty }}$.

Finally, we get that $S{{\mathcal{Y}}_{\infty }}$ = ${{\mathcal{X}}_{\infty }}$. Lemma 2 is proved.

Proof of Lemma 3. Let x₀ ∈ ${{\mathbb{R}}^{{{{n}_{1}}}}} \times {{\mathcal{X}}_{{2,\infty }}}$. Then x₀ = $\left( \begin{gathered} {{x}_{{0,1}}} \\ {{x}_{{0,2}}} \\ \end{gathered} \right)$, where x_0, 1 ∈ ${{\mathbb{R}}^{{{{n}_{1}}}}}$, x_0, 2 ∈ ${{\mathcal{X}}_{{2,\infty }}}$, whence according to (3) there exists $\tilde {N} \in \mathbb{N} \cup \{ 0\}$ such that x_0, 2 ∈ ${{\mathcal{X}}_{2}}(\tilde {N})$, which according to (2) is equivalent to the existence of $u_{2}^{*}$(0), …, $u_{2}^{*}$(N – 1) ∈ ${{\mathcal{U}}_{2}}$ such that x₂($\tilde {N}$) = 0. Then for the system (A, $\mathcal{U}$) there are u(0), …, u($\tilde {N}$ – 1) ∈ $\mathcal{U}$ such that u(k) = $\left( \begin{gathered} {{u}_{1}}(k) \\ u_{2}^{*}(k) \\ \end{gathered} \right)$, k = $\overline {0,\tilde {N} - 1}$. By (1), x($\tilde {N}$) has the representation

$$x(\tilde {N}) = \left( \begin{gathered} A_{1}^{{\tilde {N}}}{{x}_{{0,1}}} + A_{1}^{{\tilde {N} - 1}}{{u}_{1}}(0) + \ldots + {{u}_{1}}(\tilde {N} - 1) \\ 0 \\ \end{gathered} \right) = \left( \begin{gathered} {{{\tilde {x}}}_{1}} \\ 0 \\ \end{gathered} \right).$$

According to Lemma 1, it suffices to show that there exists ${{\mathcal{U}}_{1}} \subset {{\mathbb{R}}^{{{{n}_{1}}}}}$ such that ${{\mathcal{U}}_{1}}$ × {0} ⊂ $\mathcal{U}$ and ${{\mathcal{X}}_{{1,\infty }}}$ = ${{\mathbb{R}}^{{{{n}_{1}}}}}$, where ${{\mathcal{X}}_{{1,\infty }}}$ is the limit set of 0-controllability of the system (A₁, ${{\mathcal{U}}_{1}}$).

Denote by S ∈ ${{\mathbb{R}}^{{{{n}_{1}} \times {{n}_{1}}}}}$ the transition matrix to the normal Jordan basis of the matrix A₁. Since 0 ∈ int$\mathcal{U}$, there exists u_max > 0 such that S[–u_max; ${{u}_{{\max }}}{{]}^{{{{n}_{1}}}}}$ × {0} ⊂ $\mathcal{U}$. Moreover, due to the non-degeneracy of the matrix S and Lemma 2, the equality ${{\mathcal{X}}_{{1,\infty }}}$ = ${{\mathbb{R}}^{{{{n}_{1}}}}}$ is true for the case ${{\mathcal{U}}_{1}}$ = S[–u_max; ${{u}_{{\max }}}{{]}^{{{{n}_{1}}}}}$ if and only if ${{S}^{{ - 1}}}{{\mathcal{X}}_{{1,\infty }}}$ = ${{\mathbb{R}}^{{{{n}_{1}}}}}$, where ${{S}^{{ - 1}}}{{\mathcal{X}}_{{1,\infty }}}$ is the limit set of 0-controllability of the system (S^–1A₁S, [–u_max; ${{u}_{{\max }}}{{]}^{{{{n}_{1}}}}}$). Moreover, according to the normal Jordan form theorem [25], the equality

$${{S}^{{ - 1}}}{{A}_{1}}S = \left( {\begin{array}{*{20}{c}} {{{J}_{1}}}&0&0& \ldots \\ 0&{{{J}_{2}}}&0& \ldots \\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &{{{J}_{{{{{\tilde {n}}}_{1}}}}}} \end{array}} \right),$$

where the Jordan cells J_i corresponding to the real eigenvalues λ_i ∈ $\mathbb{R}$ of the matrix A₁ have the form

$${{J}_{i}} = \left( {\begin{array}{*{20}{c}} {{{\lambda }_{i}}}&1&0& \ldots &0 \\ 0&{{{\lambda }_{i}}}&1& \ldots &0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0& \ldots &{{{\lambda }_{i}}} \end{array}} \right) \in {{\mathbb{R}}^{{{{m}_{i}} \times {{m}_{i}}}}},$$

(A.1)

and the Jordan cells J_i corresponding to the complex eigenvalues λ_i ∈ $\mathbb{C}$ of the matrix A₁ have the form

$${{J}_{i}} = \left( {\begin{array}{*{20}{c}} {{{r}_{i}}{{A}_{{{{\varphi }_{i}}}}}}&I&0& \ldots &0 \\ 0&{{{r}_{i}}{{A}_{{{{\varphi }_{i}}}}}}&I& \ldots &0 \\ \vdots & \vdots & \ddots &{}& \vdots \\ 0&0& \ldots &{{{r}_{i}}{{A}_{{{{\varphi }_{i}}}}}}&I \\ 0&0&0& \ldots &{{{r}_{i}}{{A}_{{{{\varphi }_{i}}}}}} \end{array}} \right) \in {{\mathbb{R}}^{{2{{m}_{i}} \times 2{{m}_{i}}}}},\quad {{A}_{{{{\varphi }_{i}}}}} = \left( {\begin{array}{*{20}{c}} {\cos {{\varphi }_{i}}}&{\sin {{\varphi }_{i}}} \\ { - \sin {{\varphi }_{i}}}&{\cos {{\varphi }_{i}}} \end{array}} \right),$$

(A.2)

where r_i = |λ_i|, φ_i = arg(λ_i).

Hence, by virtue of Lemma 1, it suffices to show that for |λ_i| $\leqslant$ 1, the limit sets of null-controllability of the system are (J_i, [–u_max; ${{u}_{{\max }}}{{]}^{{{{m}_{i}}}}}$) for the case of (A.1) and the systems (J_i, [–u_max; ${{u}_{{\max }}}{{]}^{{2{{m}_{i}}}}}$) for the case of (A.2) coincide with ${{\mathbb{R}}^{{{{m}_{i}}}}}$ and ${{\mathbb{R}}^{{2{{m}_{i}}}}}$ correspondingly, for all i = $\overline {1,{{{\tilde {n}}}_{1}}}$.

Let J ∈ ${{\mathbb{R}}^{{m \times m}}}$ satisfy (A.1). Then for all N $\geqslant$ m the following relations hold

$${{J}^{N}} = \left( {\begin{array}{*{20}{c}} {{{\lambda }^{N}}}&{N{{\lambda }^{{N - 1}}}}&{C_{N}^{2}{{\lambda }^{{N - 2}}}}& \ldots &{C_{N}^{{m - 1}}{{\lambda }^{{N - m + 1}}}} \\ 0&{{{\lambda }^{N}}}&{N{{\lambda }^{{N - 1}}}}& \ldots &{C_{N}^{{m - 2}}{{\lambda }^{{N - m + 2}}}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0& \ldots &{{{\lambda }^{N}}} \end{array}} \right),$$

(A.3)

where we denote the number of combinations of N choose m by $C_{N}^{m}$:

$$C_{N}^{m} = \frac{{N!}}{{(N - m)!m!}}.$$

Denote by {y(k), ${v}$(k – 1), ${{y}_{0}}\} _{{k = 1}}^{N}$ the process of controlling the system (J, [–u_max; u_max]^m). Hence

$$y(N) = {{J}^{N}}{{y}_{0}} + \sum\limits_{k = 0}^{N - 1} {{{J}^{k}}{v}(N - k - 1).} $$

If we denote z₀ = J^Ny₀, then by (A.3) it is right for each ith coordinate of z₀, that

$${{z}_{{0,i}}} = \sum\limits_{j = 0}^{m - i} {{{\lambda }^{{N - j}}}C_{N}^{j}{{y}_{{0,j + i}}}} ,\quad i = \overline {1,m} .$$

Let us assume, that |λ| < 1. Then for all N $\geqslant$ 2m the following relations hold

$$\left| {{{z}_{{0,i}}}} \right|\;\leqslant \;\sum\limits_{j = 0}^{m - i} {\left| {{{\lambda }^{{N - j}}}{{y}_{{0,j + i}}}C_{N}^{j}} \right|} \;\leqslant \;\sum\limits_{j = 0}^{m - 1} {\left| {{{\lambda }^{{N - j}}}} \right|\left| {\mathop {\max }\limits_{i = \overline {1,m} } {{y}_{{0,i}}}} \right|\left| {C_{N}^{j}} \right|} $$

$$\leqslant \;m\left| {{{\lambda }^{{N - m + 1}}}} \right|\mathop {\max }\limits_{i = \overline {1,m} } \left| {{{y}_{{0,i}}}} \right|C_{N}^{{m - 1}}\;\leqslant \;m\mathop {\max }\limits_{i = \overline {1,m} } \left| {{{y}_{{0,i}}}} \right|{{\left| \lambda \right|}^{{N - m + 1}}}\frac{{N \cdot (N - 1) \cdot \ldots \cdot (N - m + 2)}}{{(m - 1)!}}$$

$$\leqslant \;m\mathop {\max }\limits_{i = \overline {1,m} } \left| {{{y}_{{0,i}}}} \right|{{\left| \lambda \right|}^{{N - m + 1}}}\frac{{{{N}^{{m - 1}}}}}{{(m - 1)!}}\;\xrightarrow{{N \to \infty }}\;0.$$

Then there exists $\tilde {N} \in \mathbb{N}$ such, that for all i = $\overline {1,m}$

$$ - {{u}_{{\max }}} < {{z}_{{0,i}}} < {{u}_{{\max }}}.$$

Let us take ${v}$(0) = … = ${v}(\tilde {N} - 2)$ = 0 and ${v}(\tilde {N} - 1)$ = –z₀ ∈ [–u_max; u_max]^m. Then we obtain y($\tilde {N}$) = 0, i.e., y₀ ∈ $\mathcal{Y}(\tilde {N})$. Therefore, by choosing arbitrary y₀ ∈ ${{\mathbb{R}}^{m}}$ and Eq. (3) we obtain the result ${{\mathcal{Y}}_{\infty }}$ = ${{\mathbb{R}}^{m}}$.

Let us assume, that |λ| = 1. Then by (A.3) for some N_m ∈ $\mathbb{N}$ and mth coordinate of y(N_m) it is right that

$${{y}_{m}}({{N}_{m}}) = {{\lambda }^{{{{N}_{m}}}}}{{y}_{m}}(0) + \sum\limits_{k = 0}^{{{N}_{m}} - 1} {{{\lambda }^{k}}{{{v}}_{m}}({{N}_{m}} - k - 1).} $$

Here we choose N_m ∈ $\mathbb{N}$, requiring |y_m(0)| $\leqslant$ N_mu_max. Then we take ${v}$(0), …, ${v}$(N_m – 1) ∈ [–u_max; u_max]^m in accordance with the following condition:

$${{{v}}_{m}}({{N}_{m}} - k - 1) = - \frac{{{{\lambda }^{{{{N}_{m}}}}}{{y}_{m}}(0)}}{{{{\lambda }^{k}}{{N}_{m}}}} \in [ - {{u}_{{\max }}};{{u}_{{\max }}}],\quad k = \overline {0,{{N}_{m}} - 1} .$$

We obtain

$${{y}_{m}}({{N}_{m}}) = {{\lambda }^{{{{N}_{m}}}}}{{y}_{m}}(0) + \sum\limits_{k = 0}^{{{N}_{m}} - 1} {{{\lambda }^{k}}\frac{{ - {{\lambda }^{{{{N}_{m}}}}}{{y}_{m}}(0)}}{{{{\lambda }^{k}}{{N}_{m}}}}} = 0.$$

Let us assume that for some N ∈ $\mathbb{N}$ and i = $\overline {1,m - 1}$ it is right that y_m(N) = … = ${{y}_{{m - i + 1}}}(N)$ = 0. Then if ${{{v}}_{m}}$(N + N_m – i – k – 1) = … = ${{{v}}_{{m - i + 1}}}$(N + N_m – i – k – 1) = 0, k = $\overline {0,{{N}_{{m - i}}} - 1}$ is right, then by (A.3) the following result is straightforward

$${{y}_{{m - i}}}(N + {{N}_{{m - i}}}) = {{\lambda }^{{{{N}_{{m - i}}}}}}{{y}_{{m - i}}}(N) + \sum\limits_{k = 0}^{{{N}_{{m - i}}} - 1} {{{\lambda }^{k}}{{{v}}_{{m - i}}}(N + {{N}_{{m - i}}} - k - 1),} $$

where ${{N}_{{m - i}}} \in \mathbb{N}$ is chosen from the condition |y_m – i(N)| $\leqslant$ ${{N}_{{m - i}}}{{u}_{{\max }}}$. Then we determine ${v}$(N), …, ${v}$(N + ${{N}_{{m - i}}}$ – 1) ∈ [–u_max; u_max]^m in order that

$${{{v}}_{{m - i}}}(N + {{N}_{{m - i}}} - k - 1) = \frac{{ - {{\lambda }^{{{{N}_{{m - i}}}}}}{{y}_{{m - i}}}(N)}}{{{{N}_{{m - i}}}{{\lambda }^{k}}}},\quad k = \overline {0,{{N}_{{m - i}}} - 1} .$$

We obtain

$${{y}_{{m - i}}}(N + {{N}_{{m - i}}}) = {{\lambda }^{{{{N}_{{m - i}}}}}}{{y}_{{m - i}}}(N) + \sum\limits_{k = 0}^{{{N}_{{m - i}}} - 1} {{{\lambda }^{k}}\frac{{ - {{\lambda }^{{{{N}_{{m - i}}}}}}{{y}_{{m - i}}}(N)}}{{{{N}_{{m - i}}}{{\lambda }^{k}}}}} = 0,$$

$${{y}_{m}}(N + {{N}_{{m - i}}}) = \ldots = {{y}_{{m - i + 1}}}(N + {{N}_{{m - i}}}) = 0.$$

Then there exists N ∈ $\mathbb{N}$ such, that y(N) = 0, i.e., y₀ ∈ $\mathcal{Y}(N)$ by the method of mathematical induction. Therefore, we obtain the result ${{\mathcal{Y}}_{\infty }}$ = ${{\mathbb{R}}^{m}}$ by choosing arbitrary y₀ ∈ ${{\mathbb{R}}^{m}}$ and Eq. (3).

Let J ∈ ${{\mathbb{R}}^{{2m \times 2m}}}$ satisfy the case (A.2). Then for all N $\geqslant$ m the following relations hold

$${{J}^{N}} = \left( {\begin{array}{*{20}{c}} {{{r}^{N}}{{A}_{{N\varphi }}}}&{N{{r}^{{N - 1}}}{{A}_{{(N - 1)\varphi }}}}& \ldots &{C_{N}^{{m - 1}}{{r}^{{N - m + 1}}}{{A}_{{(N - m + 1)\varphi }}}} \\ 0&{{{r}^{N}}{{A}_{{N\varphi }}}}& \ldots &{C_{N}^{{m - 2}}{{r}^{{N - m + 2}}}{{A}_{{(N - m + 2)\varphi }}}} \\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &{{{r}^{N}}{{A}_{{N\varphi }}}} \end{array}} \right).$$

(A.4)

Denote the process of controlling the system (J, [–u_max; u_max]^2m) by {y(k), ${v}$(k – 1), ${{y}_{0}}\} _{{k = 1}}^{N}$. Hence

$$y(N) = {{J}^{N}}{{y}_{0}} + \sum\limits_{k = 0}^{N - 1} {{{J}^{k}}{v}(N - k - 1).} $$

If we denote z₀ = J^Ny₀, then by (A.4) for each ith two-dimensional subvector z₀ it is right, that

$${{z}_{{0,i}}} = \sum\limits_{j = 0}^{m - i} {C_{N}^{j}{{r}^{{N - j}}}{{A}_{{(N - j)\varphi }}}{{y}_{{0,j + i}}}} \in {{\mathbb{R}}^{2}},\quad i = \overline {1,m} ,$$

where z₀ = $(z_{{0,1}}^{{\text{T}}}$, …, $z_{{0,m}}^{{\text{T}}}{{)}^{{\text{T}}}}$, y₀ = $(y_{{0,1}}^{{\text{T}}}$, …, $y_{{0,m}}^{{\text{T}}}{{)}^{{\text{T}}}}$.

Let us assume, that r < 1. Then for all N $\geqslant$ 2m the following relations hold

$$\left\| {{{z}_{{0,i}}}} \right\|\;\leqslant \;\sum\limits_{j = 0}^{m - i} {\left\| {{{r}^{{N - j}}}{{A}_{{(N - j)\varphi }}}{{y}_{{0,j + i}}}C_{N}^{j}} \right\|} \;\leqslant \;\sum\limits_{j = 0}^{m - 1} {{{r}^{{N - j}}}\left\| {{{A}_{{(N - j)\varphi }}}{{y}_{{0,i}}}} \right\|C_{N}^{j}} $$

$$\leqslant \;\sum\limits_{j = 0}^{m - 1} {{{r}^{{N - j}}}\mathop {\max }\limits_{i = \overline {1,m} } \left\| {{{y}_{{0,i}}}} \right\|C_{N}^{j}} \;\leqslant \;m{{r}^{{N - m + 1}}}\mathop {\max }\limits_{i = \overline {1,m} } \left\| {{{y}_{{0,i}}}} \right\|\frac{{N(N - 1) \cdot \ldots \cdot (N - m + 2)}}{{(m - 1)!}}$$

$$\leqslant \;m{{r}^{{N - m + 1}}}\mathop {\max }\limits_{i = \overline {1,m} } \left\| {{{y}_{{0,i}}}} \right\|\frac{{{{N}^{{m - 1}}}}}{{(m - 1)!}}\;\xrightarrow{{N \to \infty }}\;0.$$

Then there exists $\tilde {N} \in \mathbb{N}$ such, that for all i = $\overline {1,m}$ it is right that

$$\left\| {{{z}_{{0,i}}}} \right\| < {{u}_{{\max }}}.$$

Let us determine ${v}$(0) = … = ${v}$($\tilde {N}$ – 2) = 0 and ${v}$($\tilde {N}$ – 1) = –z₀ ∈ [–u_max; u_max]^2m. We obtain the result y($\tilde {N}$) = 0, i.e., y₀ ∈ $\mathcal{Y}(\tilde {N})$. Therefore, we obtain the result ${{\mathcal{Y}}_{\infty }}$ = ${{\mathbb{R}}^{{2m}}}$ by choosing arbitrary y₀ ∈ ${{\mathbb{R}}^{{2m}}}$ and Eq. (3).

Let us assume, that r = 1. Then by (A.4) for some N_m ∈ $\mathbb{N}$ and mth two-dimensional subvector y(N_m) it is true, that

$${{y}_{m}}({{N}_{m}}) = {{A}_{{{{N}_{m}}\varphi }}}{{y}_{m}}(0) + \sum\limits_{k = 0}^{{{N}_{m}} - 1} {{{A}_{{k\varphi }}}{{{v}}_{m}}({{N}_{m}} - k - 1).} $$

Let us take N_m ∈ $\mathbb{N}$ which hold the inequality ||y_m(0)|| $\leqslant$ N_mu_max. Then we choose ${v}$(0), …, ${v}$(N_m – 1) ∈ [‒u_max; u_max]^2m in accordance with the equality

$${{{v}}_{m}}({{N}_{m}} - k - 1) = - \frac{{{{A}_{{({{N}_{m}} - k)\varphi }}}{{y}_{m}}(0)}}{{{{N}_{m}}}} \in {{[ - {{u}_{{\max }}};{{u}_{{\max }}}]}^{{2m}}},\quad k = \overline {0,{{N}_{m}} - 1} .$$

We obtain

$${{y}_{m}}({{N}_{m}}) = {{A}_{{{{N}_{m}}\varphi }}}{{y}_{m}}(0) + \sum\limits_{k = 0}^{{{N}_{m}} - 1} {\frac{{ - {{A}_{{{{N}_{m}}\varphi }}}{{y}_{m}}(0)}}{{{{N}_{m}}}}} = 0.$$

Let us assume that for some N ∈ $\mathbb{N}$ and i = $\overline {1,m - 1}$ the relation y_m(N) = … = ${{y}_{{m - i + 1}}}(N)$ = 0 is correct. Then if ${{{v}}_{m}}$(N + ${{N}_{{m - i}}}$ – k – 1) = … = ${{{v}}_{{m - i + 1}}}$(N + ${{N}_{{m - i}}}$ – k – 1) = 0, k = $\overline {0,{{N}_{{m - i}}} - 1}$, then according to the (A.4)

$${{y}_{{m - i}}}(N + {{N}_{{m - i}}}) = {{A}_{{{{N}_{{m - i}}}\varphi }}}{{y}_{{m - i}}}(N) + \sum\limits_{k = 0}^{{{N}_{{m - i}}} - 1} {{{A}_{{k\varphi }}}{{{v}}_{{m - i}}}(N + {{N}_{{m - i}}} - k - 1),} $$

where ${{N}_{{m - i}}}$ ∈ $\mathbb{N}$ is selected from the condition ||${{y}_{{m - i}}}$(N)|| $\leqslant$ ${{N}_{{m - i}}}{{u}_{{\max }}}$. Then we define ${v}$(N), …, ${v}$(N + ${{N}_{{m - i}}}$ – 1) ∈ [–u_max; u_max]^2m in order that

$${{{v}}_{{m - i}}}(N + {{N}_{{m - i}}} - k - 1) = \frac{{ - {{A}_{{({{N}_{{m - i}}} - k)\varphi }}}{{y}_{{m - i}}}(N)}}{{{{N}_{{m - i}}}}},\quad k = \overline {0,{{N}_{{m - i}}} - 1} .$$

We obtain

$${{y}_{{m - i}}}(N + {{N}_{{m - i}}}) = {{A}_{{{{N}_{{m - i}}}\varphi }}}{{y}_{{m - i}}}(N) + \sum\limits_{k = 0}^{{{N}_{{m - i}}} - 1} {\frac{{ - {{A}_{{{{N}_{{m - i}}}\varphi }}}{{y}_{{m - i}}}(N)}}{{{{N}_{{m - i}}}}}} = 0,$$

$${{y}_{m}}(N + {{N}_{{m - i}}}) = \ldots = {{y}_{{m - i + 1}}}(N + {{N}_{{m - i}}}) = 0.$$

Then there exists N ∈ $\mathbb{N}$ such, that y(N) = 0, i.e., y₀ ∈ $\mathcal{Y}(N)$ by the method of mathematical induction. Therefore, we obtain the result ${{\mathcal{Y}}_{\infty }}$ = ${{\mathbb{R}}^{{2m}}}$ by choosing arbitrary y₀ ∈ ${{\mathbb{R}}^{{2m}}}$ and Eq. (3).

Hence Lemma 3 is proved.

Proof of Theorem 1. Let x₀ ∈ ${{\mathcal{X}}_{\infty }}$, which by (3) is equivalent to the existence of N ∈ $\mathbb{N}$ ∪ {0} such, that x₀ ∈ $\mathcal{X}$(N). By (2) there exist u(0), …, u(N – 1) ∈ $\mathcal{U}$ such, that x(N) = 0. Based on (1), for all h ∈ ∂B₁(0), ε > 0 the following relations hold

$$\tilde {x}(N) = {{A}^{N}}({{x}_{0}} + \varepsilon h) + {{A}^{{N - 1}}}u(0) + \ldots + u(N - 1)$$

$$ = {{A}^{N}}{{x}_{0}} + {{A}^{{N - 1}}}u(0) + \ldots + u(N - 1) + {{A}^{N}}h\varepsilon = x(N) + {{A}^{N}}h\varepsilon = {{A}^{N}}h\varepsilon ,$$

$$\tilde {x}(N + 1) = A\tilde {x}(N) + u(N) = {{A}^{{N + 1}}}h\varepsilon + u(N),$$

where u(N) ∈ $\mathcal{U}$. Considering 0 ∈ int $\mathcal{U}$, there exists δ > 0 such, that O_δ(0) ⊂ $\mathcal{U}$, and ε > 0 such, that $\varepsilon {{A}^{{N + 1}}}{{B}_{1}}(0)$ ⊂ O_δ(0). Let us take

$$u(N) = - {{A}^{{N + 1}}}h\varepsilon \in \varepsilon {{A}^{{N + 1}}}{{B}_{1}}(0) \subset {{O}_{\delta }}(0) \subset \mathcal{U}.$$

Then $\tilde {x}$(N + 1) = 0, i.e., for all h ∈ B₁(0) it is true that x₀ + εh ∈ $\mathcal{X}$(N + 1). As a result, B_ε(x₀) ⊂ $\mathcal{X}$(N + 1) ⊂ ${{\mathcal{X}}_{\infty }}$, i.e., x₀ ∈ int ${{\mathcal{X}}_{\infty }}$. Hence, ${{\mathcal{X}}_{\infty }}$, is open.

Let x_0, 1, x_0, 2 ∈ ${{\mathcal{X}}_{\infty }}$, α ∈ [0; 1]. Then there exists N ∈ $\mathbb{N}$ ∪ {0} such, that x_0, 1, x_0, 2 ∈ $\mathcal{X}$(N), i.e., there exist u¹(0), u¹(1), …, u¹(N – 1), u²(0), u²(1), …, u²(N – 1) ∈ $\mathcal{U}$ such, that x¹(N) = 0, x²(N) = 0. According to (1) it is true that

$$0 = {{x}^{1}}(N) = {{A}^{N}}{{x}_{{0,1}}} + {{A}^{{N - 1}}}{{u}^{1}}(0) + {{A}^{{N - 2}}}{{u}^{1}}(1) + \ldots + {{u}^{1}}(N - 1),$$

$$0 = {{x}^{2}}(N) = {{A}^{N}}{{x}_{{0,2}}} + {{A}^{{N - 1}}}{{u}^{2}}(0) + {{A}^{{N - 2}}}{{u}^{2}}(1) + \ldots + {{u}^{2}}(N - 1),$$

$$0 = \alpha {{x}^{1}}(N) = \alpha {{A}^{N}}{{x}_{{0,1}}} + \sum\limits_{k = 0}^{N - 1} {\alpha {{A}^{k}}{{u}^{1}}(N - k - 1),} $$

$$0 = (1 - \alpha ){{x}^{2}}(N) = (1 - \alpha ){{A}^{N}}{{x}_{{0,2}}} + \sum\limits_{k = 0}^{N - 1} {(1 - \alpha ){{A}^{k}}{{u}^{2}}(N - k - 1),} $$

$$0 = {{A}^{N}}(\alpha {{x}_{{0,1}}} + (1 - \alpha ){{x}_{{0,2}}}) + \sum\limits_{k = 0}^{N - 1} {{{A}^{k}}(\alpha {{u}^{1}}(N - k - 1) + (1 - \alpha ){{u}^{2}}(N - k - 1)).} $$

According to the convexity of $\mathcal{U}$ the relations ${v}$(N – k – 1) = αu¹(N – k – 1) + (1 – α)u²(N – k – 1) ∈ $\mathcal{U}$, k = $\overline {0,N - 1}$ are correct. Then αx_0, 1 + (1 – α)x_0, 2 ∈ $\mathcal{X}$(N) ⊂ ${{\mathcal{X}}_{\infty }}$, from which it follows that ${{\mathcal{X}}_{\infty }}$ is convex.

Theorem 1 is proved.

Proof of Lemma 4. Let x₀ ∈ ${{\mathcal{X}}_{\infty }}$. Then by (3) there exists N ∈ $\mathbb{N}$ ∪ {0} such, that x₀ ∈ $\mathcal{X}$(N). According to (2) there exist u(0), u(1), …, u(N – 1) ∈ $\mathcal{U}$ such, that x(N) = 0. Based on (1), it is right that

$$0 = x(N) = {{A}^{N}}{{x}_{0}} + \sum\limits_{k = 0}^{N - 1} {{{A}^{k}}u(N - k - 1),} $$

$$0 = {{x}_{0}} + {{A}^{{ - N}}}\sum\limits_{k = 0}^{N - 1} {{{A}^{k}}u(N - k - 1),} $$

$${{x}_{0}} = - \sum\limits_{k = 0}^{N - 1} {{{A}^{{ - N + k}}}u(N - k - 1)} = - \sum\limits_{k = 1}^N {{{A}^{{ - k}}}u(k - 1).} $$

Let p belong to ${{\mathbb{R}}^{n}}$\{0}. Then

$$(p,{{x}_{0}}) = \left( {p,\,\, - \sum\limits_{k = 1}^N {{{A}^{{ - k}}}u(k - 1)} } \right) = \sum\limits_{k = 1}^N {\left( {p,\,\, - {{A}^{{ - k}}}u(k - 1)} \right)} $$

$$ = \sum\limits_{k = 1}^N {\left( {{{{\left( { - {{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u(k - 1)} \right)} \;\leqslant \;\sum\limits_{k = 1}^N {\mathop {\max }\limits_{{{u}_{k}} \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,{{u}_{k}}} \right).} $$

Since 0 ∈ $\mathcal{U}$, then for all k ∈ $\mathbb{N}$

$$\mathop {\max }\limits_{{{u}_{k}} \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,{{u}_{k}}} \right)\; \geqslant \;0.$$

Then

$$(p,{{x}_{0}})\;\leqslant \;\sum\limits_{}^{} {\mathop {\max }\limits_{{{u}_{k}} \in \mathcal{U}} \left. {\left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}} \right)p,\,\,{{u}_{k}}} \right),} $$

i.e., x₀ ∈ ${{\mathcal{H}}_{p}}$. It follows that ${{\mathcal{X}}_{\infty }} \subset {{\mathcal{H}}_{p}}$.

Let us consider the following quantity for some p ∈ ${{\mathbb{R}}^{n}}$\{0}:

$$(p,x\text{*}) = \sum\limits_{k = 1}^\infty {\left( {p, - {{A}^{{ - k}}}u_{k}^{*}} \right)} = \sum\limits_{k = 1}^\infty {\left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u_{k}^{*}} \right)} = \sum\limits_{k = 1}^\infty {\mathop {\max }\limits_{{{u}_{k}} \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,{{u}_{k}}} \right).} $$

Then x* ∈ $\partial {{\mathcal{H}}_{p}}$.

Since all eigenvalues of the matrix A are strictly greater than 1 in absolute value, then the series $\sum\nolimits_{k = 1}^\infty {{{A}^{{ - k}}}u_{k}^{*}}$ converges. Then x_N = $- \sum\limits_{k = 1}^N {{{A}^{{ - k}}}u_{k}^{*}} \xrightarrow{{N \to \infty }}x\text{*}$. Demonstrate that x_N ∈ $\mathcal{X}$(N) ⊂ ${{\mathcal{X}}_{\infty }}$. Let us take u(k) = $u_{{k + 1}}^{*}$ ∈ $\mathcal{U}$. Then

$$x(N) = {{A}^{N}}{{x}_{N}} + \sum\limits_{k = 0}^{N - 1} {{{A}^{k}}u(N - k - 1)} = - \sum\limits_{k = 1}^N {{{A}^{{N - k}}}u_{k}^{*}} + \sum\limits_{k = 0}^{N - 1} {{{A}^{k}}u_{{N - k - 1}}^{*}} = 0.$$

Then x_N ∈ $\mathcal{X}$(N) ⊂ ${{\mathcal{X}}_{\infty }}$ for all N ∈ $\mathbb{N}$. It follows that x* = $\mathop {\lim }\limits_{N \to \infty } {{x}_{N}} \in \overline {{{\mathcal{X}}_{\infty }}}$.

Lemma 4 is proved.

Corollary 2. Let all eigenvalues of the matrix A ∈ ${{\mathbb{R}}^{{n \times n}}}$ be strictly greater than 1 in absolute value, ${{\mathcal{X}}_{\infty }}$ is defined by (3).

Then for all p ∈ ${{\mathbb{R}}^{n}}$\{0} it is right that:

(1) ${{\mathcal{X}}_{\infty }} \subset {{\mathcal{H}}_{{ - p}}} = \left\{ {x \in {{\mathbb{R}}^{n}}{\kern 1pt} {\kern 1pt} :\;(p,x)\; \geqslant \;\sum\limits_{k = 1}^\infty {\mathop {\min }\limits_{{{u}_{k}} \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,{{u}_{k}}} \right)} } \right\}$;

(2) $x\text{*} = - \sum\limits_{k = 1}^\infty {{{A}^{{ - k}}}u_{k}^{*}} \in \overline {{{\mathcal{X}}_{\infty }}} \cap \partial {{\mathcal{H}}_{{ - p}}}$, where

$$u_{k}^{*} = \arg \mathop {\min }\limits_{{{u}_{k}} \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,{{u}_{k}}} \right).$$

Proof of Corollary 2. For proving that it is sufficient to consider the conditions of Lemma 4 for the vector –p.

Proof of Lemma 5. Let p = (0 … 0 1 0 … 0)^T ∈ ${{\mathbb{R}}^{n}}$, where 1 corresponds to the ith coordinate of the vector p. Then for arbitrary k ∈ $\mathbb{N}$

$${{A}^{{ - k}}} = \left( {\begin{array}{*{20}{c}} {{{\lambda }^{{ - k}}}}&{( - 1)k{{\lambda }^{{ - k - 1}}}}&{{{{( - 1)}}^{2}}k(k + 1){{\lambda }^{{ - k - 2}}}\frac{1}{2}}& \ldots &{{{{( - 1)}}^{{n - 1}}}\frac{{(k + n - 2)!}}{{(n - 1)!(k - 1)!}}{{\lambda }^{{ - k - n + 1}}}} \\ 0&{{{\lambda }^{{ - k}}}}&{( - 1)k{{\lambda }^{{ - k - 1}}}}& \ldots &{{{{( - 1)}}^{{n - 2}}}\frac{{(k + n - 3)!}}{{(n - 2)!(k - 1)!}}{{\lambda }^{{ - k - n + 2}}}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0& \ldots &{{{\lambda }^{{ - k}}}} \end{array}} \right),$$

$$ - {{({{A}^{{ - k}}})}^{{\text{T}}}}p = - \left( \begin{gathered} 0 \\ \vdots \\ 0 \\ {{\lambda }^{{ - k}}} \\ ( - 1)k{{\lambda }^{{ - k - 1}}} \\ \vdots \\ {{( - 1)}^{{n - i}}}\frac{{(k + n - i - 1)!}}{{(k - 1)!(n - i)!}}{{\lambda }^{{ - k - n + i}}} \\ \end{gathered} \right),$$

$$\left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = - \left( {{{\lambda }^{{ - k}}}{{u}_{i}} - k{{\lambda }^{{ - k - 1}}}{{u}_{{i + 1}}} + \ldots + {{{( - 1)}}^{{n - i}}}{{\lambda }^{{ - k - n + i}}}C_{{k + n - i - 1}}^{{n - i}}{{u}_{n}}} \right)$$

$$ = - \sum\limits_{j = 0}^{n - i} {{{\lambda }^{{ - k - j}}}{{{( - 1)}}^{j}}{{u}_{{j + i}}}C_{{k + j - 1}}^{j}} = \sum\limits_{j = 0}^{n - i} {{{\lambda }^{{ - k - j}}}{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i}}}C_{{k + j - 1}}^{j}.} $$

Let us consider the case λ > 1. Since u_i ∈ [u_i, min; u_i, max], then for all k ∈ $\mathbb{N}$ the following inequalities hold

$$\left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right)\;\leqslant \;\sum\limits_{j = 0}^{n - i} {{{\lambda }^{{ - k - j}}}C_{{k + j - 1}}^{j}\max \left\{ {{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\min }}};\,\,{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\max }}}} \right\},} $$

$$ - \left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right)\;\leqslant \; - \sum\limits_{j = 0}^{n - i} {{{\lambda }^{{ - k - j}}}C_{{k + j - 1}}^{j}\min \left\{ {{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\min }}};\,\,{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\max }}}} \right\}.} $$

Additionally, the following equalities hold

$$\sum\limits_{k = 1}^\infty {\sum\limits_{j = 0}^{n - i} {{{\lambda }^{{ - k - j}}}C_{{k + j - 1}}^{j}} } = \sum\limits_{j = 0}^{n - i} {\sum\limits_{k = 1}^\infty {{{\lambda }^{{ - k - j}}}C_{{k + j - 1}}^{j}} } = \sum\limits_{j = 0}^{n - i} {\frac{1}{{{{{(\lambda - 1)}}^{{j + 1}}}}}.} $$

By Lemma 4 and Corollary 2, these equalities imply, that for all x ∈ ${{\mathcal{X}}_{\infty }}$ inequalities hold

$$\sum\limits_{j = 0}^{n - i} {\frac{{\min \left\{ {{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\min }}};\,\,{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\max }}}} \right\}}}{{{{{(\lambda - 1)}}^{{j + 1}}}}}} \;\leqslant \;(p,\,\,x)$$

$$\leqslant \;\sum\limits_{j = 0}^{n - i} {\frac{{\max \left\{ {{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\min }}};\,\,{{{( - 1)}}^{{j + 1}}}{{u}_{{j + i,\max }}}} \right\}}}{{{{{(\lambda - 1)}}^{{j + 1}}}}}.} $$

Since ${{\mathcal{X}}_{\infty }}$ is open according to Theorem 1, these inequalities strictly hold, i.e.,

$${{\mathcal{X}}_{\infty }} \subset \bigcap\limits_{i = 1}^n {\left\{ {x \in {{\mathbb{R}}^{n}}{\kern 1pt} :\;{{x}_{i}} \in ({{x}_{{i,\min }}};{{x}_{{i,\max }}})} \right\}.} $$

Let us consider the case λ < –1. For all k ∈ $\mathbb{N}$ it is true that

$$\left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = \sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - k - j}}}{{{( - 1)}}^{{ - k - j + j + 1}}}{{u}_{{j + i}}}C_{{k + j - 1}}^{j}} = \sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - k - j}}}{{{( - 1)}}^{{ - k + 1}}}{{u}_{{j + i}}}C_{{k + j - 1}}^{j}.} $$

Then

$$\mathop {\max }\limits_{u \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - (2k - 1)}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = \mathop {\max }\limits_{u \in \mathcal{U}} \left( {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - (2k - 1) - j}}}{{{( - 1)}}^{{ - (2k - 1) + 1}}}{{u}_{{j + i}}}C_{{(2k - 1) + j - 1}}^{j}} } \right)$$

$$ = \sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - (2k - 1) - j}}}{{u}_{{i + j,\max }}}C_{{(2k - 1) + j - 1}}^{j},} $$

$$\mathop {\max }\limits_{u \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - 2k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = \mathop {\max }\limits_{u \in \mathcal{U}} \left( {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - 2k - j}}}{{{( - 1)}}^{{ - 2k + 1}}}{{u}_{{j + i}}}C_{{2k + j - 1}}^{j}} } \right)$$

$$ = \sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - 2k - j}}}( - {{u}_{{i + j,\min }}})C_{{2k + j - 1}}^{j}.} $$

Then by Lemma 4 for all x ∈ ${{\mathcal{X}}_{\infty }}$ it is right, that

$$(p,x)\;\leqslant \;\sum\limits_{k = 1}^\infty {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - (2k - 1) - j}}}{{u}_{{i + j,\max }}}C_{{(2k - 1) + j - 1}}^{j}} } - \sum\limits_{k = 1}^\infty {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - 2k - j}}}{{u}_{{i + j,\min }}}C_{{2k + j - 1}}^{j}} } $$

$$ = \sum\limits_{j = 0}^{n - i} {{{u}_{{i + j,\max }}}\left( {\frac{1}{{2{{{\left( {\left| \lambda \right| + 1} \right)}}^{{j + 1}}}}} + \frac{1}{{2{{{\left( {\left| \lambda \right| - 1} \right)}}^{{j + 1}}}}}} \right)} $$

$$ - \;\sum\limits_{j = 0}^{n - i} {{{u}_{{i + j,\min }}}\left( {\frac{1}{{2{{{\left( {\left| \lambda \right| - 1} \right)}}^{{j + 1}}}}} + \frac{1}{{2{{{\left( {\left| \lambda \right| + 1} \right)}}^{{j + 1}}}}}} \right)} = {{x}_{{i,\max }}}.$$

Similarly

$$\mathop {\min }\limits_{u \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - (2k - 1)}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = \mathop {\min }\limits_{u \in \mathcal{U}} \left( {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - (2k - 1) - j}}}{{{( - 1)}}^{{ - (2k - 1) + 1}}}{{u}_{{j + i}}}C_{{(2k - 1) + j - 1}}^{j}} } \right)$$

$$ = \sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - (2k - 1) - j}}}{{u}_{{j + i,\min }}}C_{{(2k - 1) + j - 1}}^{j},} $$

$$\mathop {\min }\limits_{u \in \mathcal{U}} \left( { - {{{\left( {{{A}^{{ - 2k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = \mathop {\min }\limits_{u \in \mathcal{U}} \left( {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - 2k - j}}}{{{( - 1)}}^{{ - 2k + 1}}}{{u}_{{j + i}}}C_{{2k + j - 1}}^{j}} } \right)$$

$$ = \sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - 2k - j}}}( - {{u}_{{j + i,\max }}})C_{{2k + j - 1}}^{j}.} $$

Then by Corollary 2 for all x ∈ ${{\mathcal{X}}_{\infty }}$ it is right that

$$(p,x)\; \geqslant \;\sum\limits_{k = 1}^\infty {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - (2k - 1) - j}}}{{u}_{{i + j,\min }}}C_{{(2k - 1) + j - 1}}^{j}} } - \sum\limits_{k = 1}^\infty {\sum\limits_{j = 0}^{n - i} {{{{\left| \lambda \right|}}^{{ - 2k - j}}}{{u}_{{i + j,\max }}}C_{{2k + j - 1}}^{j}} } $$

$$ = \sum\limits_{j = 0}^{n - i} {{{u}_{{i + j,\min }}}\left( {\frac{1}{{2{{{\left( {\left| \lambda \right| + 1} \right)}}^{{j + 1}}}}} + \frac{1}{{2{{{\left( {\left| \lambda \right| - 1} \right)}}^{{j + 1}}}}}} \right)} $$

$$ - \;\sum\limits_{j = 0}^{n - i} {{{u}_{{i + j,\max }}}\left( {\frac{1}{{2{{{\left( {\left| \lambda \right| - 1} \right)}}^{{j + 1}}}}} + \frac{1}{{2{{{\left( {\left| \lambda \right| + 1} \right)}}^{{j + 1}}}}}} \right)} = {{x}_{{i,\min }}}.$$

Since ${{\mathcal{X}}_{\infty }}$ is open according to Theorem 1

$${{x}_{{i,\min }}} < (p,x) < {{x}_{{i,\max }}},$$

$${{\mathcal{X}}_{\infty }} \subset \bigcap\limits_{i = 1}^n {\left\{ {x \in {{\mathbb{R}}^{n}}{\kern 1pt} :\;{{x}_{i}} \in ({{x}_{{i,\min }}};{{x}_{{i,\max }}})} \right\}.} $$

Lemma 5 is proved.

Proof of Lemma 6. Let p = (0 0 … ${{\tilde {p}}^{{\text{T}}}}$ … 0)^T ∈ ${{\mathbb{R}}^{{2n}}}$, $\tilde {p}$ = (p₁ p₂)^T ∈ ${{\mathbb{R}}^{2}}$, $p_{1}^{2}$ + $p_{2}^{2}$ = 1, where $\tilde {p}$ corresponds to the (2i – 1)th and 2ith coordinates of the vector p. Then for arbitrary k ∈ $\mathbb{N}$ it is true that

$${{A}^{{ - k}}} = \left( {\begin{array}{*{20}{c}} {{{r}^{{ - k}}}{{A}_{{ - k\varphi }}}}&{ - k{{r}^{{ - k - 1}}}{{A}_{{( - k - 1)\varphi }}}}& \ldots &{{{{( - 1)}}^{{n - 1}}}C_{{n + k - 2}}^{{n - 1}}{{r}^{{ - k - n + 1}}}{{A}_{{( - k - n + 1)\varphi }}}} \\ 0&{{{r}^{{ - k}}}{{A}_{{ - k\varphi }}}}& \ldots &{{{{( - 1)}}^{{n - 2}}}C_{{n + k - 3}}^{{n - 2}}{{r}^{{ - k - n + 2}}}{{A}_{{( - k - n + 2)\varphi }}}} \\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \ldots &{{{r}^{{ - k}}}{{A}_{{ - k\varphi }}}} \end{array}} \right),$$

$$ - {{\left( {{{A}^{{ - k}}}} \right)}^{{\text{T}}}}p = \left( \begin{gathered} 0 \\ \vdots \\ 0 \\ {{r}^{{ - k}}}{{A}_{{ - k\varphi }}}\tilde {p} \\ - k{{r}^{{ - k - 1}}}{{A}_{{( - k - 1)\varphi }}}\tilde {p} \\ \vdots \\ {{( - 1)}^{{n - i}}}C_{{k + n - i - 1}}^{{n - i}}{{r}^{{ - k - n + i}}}{{A}_{{( - k - n + i)\varphi }}}\tilde {p} \\ \end{gathered} \right).$$

Let uⁱ ∈ ${{\mathbb{R}}^{2}}$, i = $\overline {1,n}$, u = (u^1T, ..., u^nT)^T ∈ ${{\mathbb{R}}^{{2n}}}$. Then

$$\left( { - {{{\left( {{{A}^{{ - k}}}} \right)}}^{{\text{T}}}}p,\,\,u} \right) = - \left( {{{r}^{{ - k}}}\left( {{{A}_{{ - k\varphi }}}\tilde {p},{{u}^{i}}} \right) + ( - 1)k{{r}^{{ - k - 1}}}\left( {{{A}_{{( - k - 1)\varphi }}}\tilde {p},{{u}^{{i + 1}}}} \right)} \right. + \ldots $$

$$ + \;{{( - 1)}^{{n - i}}}C_{{k + n - i - 1}}^{{n - i}}{{r}^{{ - k - n + i}}}\left. {\left( {{{A}_{{( - k - n + i)\varphi }}}\tilde {p},{{u}^{n}}} \right)} \right)$$

$$ = - \sum\limits_{j = 0}^{n - i} {{{r}^{{ - k - j}}}\left( {{{A}_{{( - k - j)\varphi }}}\tilde {p},{{u}^{{i + j}}}} \right)C_{{k + j - 1}}^{j}} \;\leqslant \;\sum\limits_{j = 0}^{n - i} {{{r}^{{ - k - j}}}\left\| {({{A}_{{( - k - j)\varphi }}}\tilde {p}} \right\|\left\| {{{u}^{{i + j}}}} \right\|C_{{k + j - 1}}^{j}} $$

$$ = \sum\limits_{j = 0}^{n - i} {{{r}^{{ - k - j}}}\left\| {{{u}^{{i + j}}}} \right\|C_{{k + j - 1}}^{j}} \;\leqslant \;\sum\limits_{j = 0}^{n - i} {{{r}^{{ - k - j}}}{{r}_{{i + j,\max }}}C_{{k + j - 1}}^{j}.} $$

Then by Lemma 4 for arbitrary x ∈ ${{\mathcal{X}}_{\infty }}$ it is right that

$$(p,x)\;\leqslant \;\sum\limits_{k = 1}^\infty {\sum\limits_{j = 0}^{n - i} {{{r}^{{ - k - j}}}{{r}_{{i + j,\max }}}C_{{k + j - 1}}^{j}} } = \sum\limits_{j = 0}^{n - i} {\sum\limits_{k = 1}^\infty {{{r}^{{ - k - j}}}{{r}_{{i + j,\max }}}C_{{k + j - 1}}^{j}} } = \sum\limits_{j = 0}^{n - i} {\frac{{{{r}_{{i + j,\max }}}}}{{{{{(r - 1)}}^{{j + 1}}}}}.} $$

Since ${{\mathcal{X}}_{\infty }}$ according to Theorem 1 is open, then

$${{\mathcal{X}}_{\infty }} \subset \bigcap\limits_{i = 1}^n {\left\{ {x \in {{\mathbb{R}}^{{2n}}}{\kern 1pt} :\;{{{\left\| {{{{({{x}_{{2i - 1}}}\;{{x}_{{2i}}})}}^{{\text{T}}}}} \right\|}}_{{{{\mathbb{R}}^{2}}}}} < {{R}_{{i,\max }}}} \right\}.} $$

Lemma 6 is proved.

Proof of Theorem 2. Let us consider for some B ∈ ${{\mathbb{R}}^{{n \times n}}}$ and $\mathcal{C} \in {{\mathbb{K}}_{n}}$ the mapping

$$\tilde {T}(\mathcal{X}) = B\mathcal{X} + \mathcal{C}.$$

Let us demonstrate, that if B  : ${{\mathbb{R}}^{n}} \to {{\mathbb{R}}^{n}}$ is a contraction mapping with the compression ratio β ∈ [0; 1), then $\tilde {T}$ : ${{\mathbb{K}}_{n}} \to {{\mathbb{K}}_{n}}$ is also a contraction mapping.

$${{\rho }_{H}}(\tilde {T}(\mathcal{X}),\tilde {T}(\mathcal{Y})) = \max \left\{ {\mathop {\sup }\limits_{x \in \tilde {T}(\mathcal{X})} \mathop {\inf }\limits_{y \in \tilde {T}(\mathcal{Y})} \rho (x,y);\,\,\,\mathop {\sup }\limits_{y \in \tilde {T}(\mathcal{Y})} \mathop {\inf }\limits_{x \in \tilde {T}(\mathcal{X})} \rho (x,y)} \right\}$$

$$ = \max \left\{ {\mathop {\sup }\limits_{\substack{ x \in \mathcal{X} \\ {{c}_{1}} \in \mathcal{C} }} \mathop {\inf }\limits_{\substack{ y \in \mathcal{Y} \\ {{c}_{2}} \in \mathcal{C} }} \left\| {Bx + {{c}_{1}} - By - {{c}_{2}}} \right\|;\,\,\,\mathop {\sup }\limits_{\substack{ y \in \mathcal{Y} \\ {{c}_{2}} \in \mathcal{C} }} \mathop {\inf }\limits_{\substack{ x \in \mathcal{X} \\ {{c}_{1}} \in \mathcal{C} }} \left\| {Bx + {{c}_{1}} - By - {{c}_{2}}} \right\|} \right\}$$

$$\leqslant \;\max \left\{ {\mathop {\sup }\limits_{\substack{ x \in \mathcal{X} \\ {{c}_{1}} \in \mathcal{C} }} \mathop {\inf }\limits_{\substack{ y \in \mathcal{Y} \\ {{c}_{2}} \in \mathcal{C} }} \left( {\left\| {B(x - y)} \right\| + \left\| {{{c}_{1}} - {{c}_{2}}} \right\|} \right);\,\,\,\mathop {\sup }\limits_{\substack{ y \in \mathcal{Y} \\ {{c}_{2}} \in \mathcal{C} }} \mathop {\inf }\limits_{\substack{ x \in \mathcal{X} \\ {{c}_{1}} \in \mathcal{C} }} \left( {\left\| {B(x - y)} \right\| + \left\| {{{c}_{1}} - {{c}_{2}}} \right\|} \right)} \right\}$$

$$ = \max \left\{ {\mathop {\sup }\limits_{x \in \mathcal{X}} \mathop {\inf }\limits_{y \in \mathcal{Y}} \left\| {B(x - y)} \right\|;\,\,\,\mathop {\sup }\limits_{y \in \mathcal{Y}} \mathop {\inf }\limits_{x \in \mathcal{X}} \left\| {B(x - y)} \right\|} \right\}$$

$$\leqslant \;\max \left\{ {\mathop {\sup }\limits_{x \in \mathcal{X}} \mathop {\inf }\limits_{y \in \mathcal{Y}} \beta \left\| {x - y} \right\|;\,\,\,\mathop {\sup }\limits_{y \in \mathcal{Y}} \mathop {\inf }\limits_{x \in \mathcal{X}} \beta \left\| {x - y} \right\|} \right\} = \beta {{\rho }_{H}}(\mathcal{X},\mathcal{Y}).$$

Then $\tilde {T}$ is a contraction mapping with the compression ratio β.

According to (4) the following equality holds

$${{T}^{M}}(\mathcal{X}) = {{A}^{{ - M}}}\mathcal{X} + \sum\limits_{k = 1}^M {\left( { - {{A}^{{ - k}}}\mathcal{U}} \right)} = \tilde {T}(\mathcal{X}),$$

where B = A^–M, $\mathcal{C}$ = $\sum\nolimits_{k = 1}^M {( - {{A}^{{ - k}}}\mathcal{U})}$.

Since all eigenvalues of the matrix A are strictly greater than 1 in absolute value, then all eigenvalues of  the matrix A^–1 are strictly less than 1 in absolute value. Then according to the [24, Theorem 5.6.12] ||A^–k|| $\xrightarrow{{k \to \infty }}$ 0. Then by definition of the limit for α ∈ [0; 1) there exists M ∈ $\mathbb{N}$ such, that ||A^–M|| < α. As the following inequality

$$\left\| {{{A}^{{ - M}}}(x - y)} \right\|\;\leqslant \;\left\| {{{A}^{{ - M}}}} \right\| \cdot \left\| {x - y} \right\| < \alpha \left\| {x - y} \right\|,$$

is true A^–M is a contraction mapping with the compression ratio α ∈ [0; 1). Then T^M: ${{\mathbb{K}}_{n}} \to {{\mathbb{K}}_{n}}$ is also a contraction mapping with the compression ratio α.

By virtue of Lemma 7 for all N ∈ $\mathbb{N}$ it is right, that $\mathcal{X}$(N) ⊂ $\mathcal{X}$(N + 1), in addition, $\mathcal{X}$(N) is a compact set. Then according to the ([27], Corollary A.3.4)

$${{\rho }_{H}}({{\mathcal{X}}_{\infty }},\mathcal{X}(N))\;\xrightarrow{{N \to \infty }}\;0.$$

(A.5)

In contrast, by virtue of ([27], Theorem A.3.9) the metric space (${{\mathbb{K}}_{n}}$, ρ_H) is complete. Then the contraction mapping $\tilde {T}$ has a unique fixed point $\mathcal{X}\text{*} \in {{\mathbb{K}}_{n}}$, which can be computed by fixed point iteration method:

$$\mathcal{X}\text{*} = \mathop {\lim }\limits_{N \to \infty } \underbrace {(\tilde {T} \circ \ldots \circ \tilde {T})}_N(\mathcal{X}),$$

(A.6)

where $\mathcal{X} \in {{\mathbb{K}}_{n}}$ is arbitrary. Let us take $\mathcal{X}$ = {0}. Then by virtue of Lemma 7

$$\underbrace {(\tilde {T} \circ \ldots \circ \tilde {T})}_N(\{ 0\} ) = - \sum\limits_{k = 1}^{MN} {{{A}^{{ - k}}}\mathcal{U}} = \mathcal{X}(NM).$$

According to the uniqueness of the limit point and formulae (A.5) and (A.6)

$$\overline {{{\mathcal{X}}_{\infty }}} = \overline {\bigcup\limits_{N = 0}^\infty {\mathcal{X}(N)} } = \mathcal{X}\text{*}{\kern 1pt} .$$

The error in the fixed point iteration method can be estimated by the following formula [28]:

$${{\rho }_{H}}\left( {\overline {{{\mathcal{X}}_{\infty }}} ,\,\,\mathcal{X}(NM)} \right)\;\leqslant \;\frac{{{{\alpha }^{N}}}}{{1 - \alpha }}{{\rho }_{H}}(\mathcal{X}(M),\,\,\{ 0\} ).$$

Theorem 2 is proved.

Proof of Theorem 3. By virtue of the point 3 of Theorem 2

$${{\rho }_{H}}\left( {\overline {{{\mathcal{X}}_{\infty }}} ,\,\,\mathcal{X}(NM)} \right)\;\leqslant \;\frac{{\alpha _{p}^{N}}}{{1 - \alpha _{p}^{N}}}{{\rho }_{H}}(\mathcal{X}(M),\,\,\{ 0\} ) = {{R}_{p}},\quad p \in \{ 1,2,\infty \} .$$

Then according to the definition of the Hausdorff distance

$${{\mathcal{X}}_{\infty }} \subset \overline {{{\mathcal{X}}_{\infty }}} \subset \mathcal{X}(NM) + {{B}_{{{{R}_{p}}}}}(0),$$

where

$${{B}_{{{{R}_{1}}}}}(0) = {\text{conv}}\left\{ {{{{(\underbrace {0, \ldots ,0}_i,r,0, \ldots ,0)}}^{{\text{T}}}}{\kern 1pt} :\;r \in \{ - {{R}_{1}},{{R}_{1}}\} ,i = \overline {0,n - 1} } \right\},$$

$${{B}_{{{{R}_{2}}}}}(0) = \left\{ {x \in {{\mathbb{R}}^{n}}{\kern 1pt} :\;\sqrt {\sum\limits_{i = 1}^n {{{{\left| {{{x}_{i}}} \right|}}^{2}}} } \;\leqslant \;{{R}_{2}}} \right\},$$

$${{B}_{{{{R}_{\infty }}}}}(0) = \left\{ {x \in {{\mathbb{R}}^{n}}{\kern 1pt} :\;\mathop {\max }\limits_{i = \overline {1,n} } \left| {{{x}_{i}}} \right|\;\leqslant \;{{R}_{\infty }}} \right\}.$$

Theorem 3 is proved.