On Some Properties of Sets of Bounded Controllability for Stationary Linear Discrete Systems with Total Control Constraints

Abstract

We consider the problem of constructing reachable sets, i.e., sets of terminal states into which a system can be transferred from the origin in a fixed time, and 0-controllability sets, i.e., sets of initial states, from which a system can be transferred to the origin in a fixed time, for stationary linear discrete systems with a total control constraint. The representation of reachable and 0-controllable sets as linear transformations of superellipsoidal sets of finite and infinite dimensions is proved. A constructive method for describing the desired sets based on the apparatus of supporting half-planes is proposed, including for the limit sets of reachability and controllability. In the case of Euclidean spaces, the description is obtained explicitly. Examples are given. For a three-dimensional satellite motion control system in a near-circular orbit, reachability sets are modeled.

APPENDIX

For arbitrary $x \in {{\mathbb{R}}^{n}}$ and a convex body $\mathcal{U} \in {{\mathbb{R}}^{n}}$ containing 0 as an interior point, we denote the Minkowski functional by $\mu (x,\mathcal{U})$ ([15], Section 3, §2, Ch. III):

$$\mu (x,\mathcal{U}) = \inf \{ t > 0:x \in t\mathcal{U}\} .$$

An arbitrary closed convex body $\mathcal{U} \in {{\mathbb{R}}^{n}}$, containing 0 as an interior point admits the following representation due to the Minkowski theorem [15, Theorem 3, Section 3, §2, Ch. III]:

$$\mathcal{U} = \{ x \in {{\mathbb{R}}^{n}}:\mu (x,\mathcal{U}) \leqslant 1\} .$$

(A.1)

Lemma 12. Assume ${{\mathcal{U}}_{1}},{{\mathcal{U}}_{2}} \subset {{\mathbb{R}}^{n}}$ are convex and compact bodies containing 0 as an interior point. Then the inclusion ${{\mathcal{U}}_{1}} \subset {{\mathcal{U}}_{2}}$ is true if and only if for an arbitrary $x \in {{\mathbb{R}}^{n}}$ the following inequality is fulfilled:

$$\mu (x,{{\mathcal{U}}_{1}}) \geqslant \mu (x,{{\mathcal{U}}_{2}}).$$

Proof of Lemma 12. Assume ${{\mathcal{U}}_{1}} \subset {{\mathcal{U}}_{2}}$, $x \in {{\mathbb{R}}^{n}}$. Then by the definition of the Minkowski functional

$$x \in \mu (x,{{\mathcal{U}}_{1}}){{\mathcal{U}}_{1}} \subset \mu (x,{{\mathcal{U}}_{1}}){{\mathcal{U}}_{2}},$$

$$\mu (x,{{\mathcal{U}}_{1}}) \geqslant \inf \{ t > 0:x \in t{{\mathcal{U}}_{2}}\} = \mu (x,{{\mathcal{U}}_{2}}).$$

We assume that for all $x \in {{\mathbb{R}}^{n}}$ the following inequality is valid:

$$\mu (x,{{\mathcal{U}}_{1}}) \geqslant \mu (x,{{\mathcal{U}}_{2}}).$$

Then due to (A.1)

$${{\mathcal{U}}_{1}} = \{ x \in {{\mathbb{R}}^{n}}:\mu (x,{{\mathcal{U}}_{1}}) \leqslant 1\} \subset \{ x \in {{\mathbb{R}}^{n}}:\mu (x,{{\mathcal{U}}_{2}}) \leqslant 1\} = {{\mathcal{U}}_{2}}.$$

Lemma 12 is proved.

Lemma 13. Assume ${{p}_{2}} > {{p}_{1}} > 0$. Then for any $x \in {{\mathbb{R}}^{n}}{{\backslash }}\{ 0\}$, the following inequality is valid:

$${{\left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{{{{p}_{1}}}}}} \right)}^{{1/{{p}_{1}}}}} \geqslant {{\left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{{{{p}_{2}}}}}} \right)}^{{1/{{p}_{2}}}}}.$$

Proof of Lemma 13. For an arbitrary $x \in {{\mathbb{R}}^{n}}{{\backslash }}\{ 0\}$, we consider the function $f:(0; + \infty ) \to \mathbb{R}$:

$$f(p) = {{\left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)}^{{1/p}}}.$$

Function f on the domain of definition is continuously differentiable. Then, to prove Lemma 13, it suffices to show that $f{\kern 1pt} '(p) \leqslant 0$ for all $p > 0$. We denote by $\overline {\ln } {\text{|}}t{\text{|}}$ a function of the following form:

$$\overline {\ln } {\text{|}}t{\text{|}} = \left\{ \begin{gathered} \ln {\text{|}}t{\text{|}}, \quad t \ne 0, \hfill \\ 0, \quad t = 0. \hfill \\ \end{gathered} \right.$$

$$f{\kern 1pt} '(p) = \frac{\partial }{{\partial p}}\left( {\exp \left\{ {\frac{1}{p}\ln \left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)} \right\}} \right)$$

$$\, = \exp \left\{ {\frac{1}{p}\ln \left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)} \right\}\left( { - \frac{1}{{{{p}^{2}}}}\ln \left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right) + \frac{{\sum\limits_{i = 1}^n ({\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}\overline {\ln } {\text{|}}{{x}_{i}}{\text{|}})}}{{p\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}}}} \right)$$

$$\, = {{\left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)}^{{1/p - 1}}}\frac{1}{{{{p}^{2}}}}\left( { - \left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)\ln \left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right) + \sum\limits_{i = 1}^n \left( {{\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}\overline {\ln } {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)} \right)$$

$$\, = {{\left( {\sum\limits_{i = 1}^n {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}} \right)}^{{1/p - 1}}}\frac{1}{{{{p}^{2}}}}\sum\limits_{i = 1}^n \left[ {{\text{|}}{{x}_{i}}{{{\text{|}}}^{p}}\left( {\overline {\ln } {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}} - \ln \left( {\sum\limits_{j = 1}^n {\text{|}}{{x}_{j}}{{{\text{|}}}^{p}}} \right)} \right)} \right] \leqslant 0,$$

since for all $i = \overline {1,n}$

$$\overline {\ln } {\text{|}}{{x}_{i}}{{{\text{|}}}^{p}} \leqslant \ln \left( {\sum\limits_{j = 1}^n {\text{|}}{{x}_{j}}{{{\text{|}}}^{p}}} \right).$$

Lemma 13 is proved.

Proof of Lemma 1. By Lemma 13, for each $u \in {{\mathbb{R}}^{N}}{{\backslash }}\{ 0\}$, the following inequality is valid:

$${{\left( {\sum\limits_{j = 1}^N {\text{|}}{{u}_{j}}{{{\text{|}}}^{{{{p}_{1}}}}}} \right)}^{{1/{{p}_{1}}}}} \geqslant {{\left( {\sum\limits_{j = 1}^N {\text{|}}{{u}_{j}}{{{\text{|}}}^{{{{p}_{2}}}}}} \right)}^{{1/{{p}_{2}}}}}.$$

Since for an arbitrary $p \in (1; + \infty )$,

$$\mu (u,{{\mathcal{E}}_{p}}(N)) = {{\left( {\sum\limits_{j = 1}^N {\text{|}}{{u}_{j}}{{{\text{|}}}^{p}}} \right)}^{{1/p}}},$$

according to Lemma 12, the inclusion ${{\mathcal{E}}_{{{{p}_{1}}}}}(N) \subset {{\mathcal{E}}_{{{{p}_{2}}}}}(N)$ is correct. Due to Representation (2.1), the inclusion ${{\mathcal{U}}_{{{{p}_{1}}}}}({{B}_{N}}) \subset {{\mathcal{U}}_{{{{p}_{2}}}}}({{B}_{N}})$ is also valid.

Lemma 1 is proved.

Proof of Lemma 2. Assume ${{x}^{1}},{{x}^{2}} \in {{\mathcal{U}}_{p}}({{B}_{N}})$ and $\lambda \in (0;1)$. By (2.1), there are ${{u}^{1}},{{u}^{2}} \in {{\mathcal{E}}_{p}}(N)$ such that ${{x}^{1}} = {{B}_{N}}{{u}^{1}}$ and ${{x}^{2}} = {{B}_{N}}{{u}^{2}}$. Since ${{\mathcal{E}}_{p}}(N)$ is convex, $\lambda {{u}^{1}} + (1 - \lambda ){{u}^{2}} \in {{\mathcal{E}}_{p}}(N)$. Taking into account (2.1),

$$\lambda {{x}^{1}} + (1 - \lambda ){{x}^{2}} = {{B}_{N}}(\lambda {{u}^{1}} + (1 - \lambda ){{u}^{2}}) \in {{B}_{n}}{{\mathcal{E}}_{p}}(N) = {{\mathcal{U}}_{p}}({{B}_{N}}),$$

which means ${{\mathcal{U}}_{p}}({{B}_{N}})$ is convex.

We assume that ${\text{rank}}{{B}_{N}} = n$. Since at $p \in (1; + \infty )$ the set ${{\mathcal{E}}_{p}}(N)$ is strictly convex, λu¹ + $(1 - \lambda ){{u}^{2}} \in {\text{int}} \mathcal{U}$; i.e., there is an open sphere ${{O}_{\varepsilon }}(\lambda {{u}^{1}} + (1 - \lambda ){{u}^{2}}) \subset {{\mathcal{E}}_{p}}(N)$. Then by (2.1)

$${{B}_{N}}{{O}_{\varepsilon }}(\lambda {{u}^{1}} + (1 - \lambda ){{u}^{2}}) \subset \mathcal{U}_{p}({{B}_{N}}).$$

Since ${\text{rank}}{{B}_{N}} = n$, there is an open sphere

$${{O}_{\delta }}({{B}_{N}}(\lambda {{u}^{1}} + (1 - \lambda ){{u}^{2}})) \subset {{B}_{N}}{{O}_{\varepsilon }}(\lambda {{u}^{1}} + (1 - \lambda ){{u}^{2}}) \subset {{\mathcal{U}}_{p}}({{B}_{N}}),$$

i.e., $\lambda {{x}^{1}} + (1 - \lambda ){{x}^{2}} \in {\text{int}}{{\mathcal{U}}_{p}}({{B}_{N}})$. Hence, ${{\mathcal{U}}_{p}}({{B}_{N}})$ is strictly convex by definition.

Closure and limitation follow from the fact that, by definition, ${{\mathcal{U}}_{p}}({{B}_{N}})$ is the image of the closed and bounded set ${{\mathcal{E}}_{p}}(N)$ under the action of a continuous linear mapping ${{B}_{N}}:{{\mathbb{R}}^{N}} \to {{\mathbb{R}}^{n}}$ ([15], Section 2, §5, Ch. IV).

Lemma 2 is proved.

Proof of Lemma 3. As shown in [4], Lemma 5 for an arbitrary convex and compact set $\mathcal{U} \subset {{\mathbb{R}}^{N}}$, a linear bounded operator $A:{{\mathbb{R}}^{N}} \to {{\mathbb{R}}^{n}}$, and point $u \in \mathcal{U}$, the following equality is valid:

$$\mathcal{N}(Au,A\mathcal{U}) = ({{A}^{{\text{T}}}}{{)}^{{ - 1}}}(\mathcal{N}(u,\mathcal{U})) \cup (\ker {{A}^{{\text{T}}}}{{\backslash }}\{ 0\} ).$$

If in this equation we put $A = {{B}_{N}}$, $\mathcal{U} = {{\mathcal{E}}_{p}}(N)$, we get the relation

$$\mathcal{N}({{B}_{N}}u,{{\mathcal{U}}_{p}}({{B}_{N}})) = (B_{N}^{{\text{T}}}{{)}^{{ - 1}}}(\mathcal{N}(u,{{\mathcal{E}}_{p}}(N))) \cup (\ker B_{N}^{{\text{T}}}{{\backslash }}\{ 0\} ).$$

(A.2)

Since at $p \in (1; + \infty )$ the functional of the Minkowski set ${{\mathcal{E}}_{p}}(N)$ is a smooth function with respect to the first argument

$$\mu (u,{{\mathcal{E}}_{p}}(N)) = {{\left( {\sum\limits_{j = 1}^N {\text{|}}{{u}_{j}}{{{\text{|}}}^{p}}} \right)}^{{1/p}}},$$

the normal cone of the set ${{\mathcal{E}}_{p}}(N)$ at any boundary point $u \in \partial {{\mathcal{E}}_{p}}(N)$ is the conical hull of the gradient of the Minkowski functional at point u ([12], Theorem 25.6):

$$\mathcal{N}(u,{{\mathcal{E}}_{p}}(N)) = {\text{cone}}\left\{ {{{\nabla }_{u}}\mu (u,{{\mathcal{E}}_{p}}(N))} \right\}{{\backslash }}\{ 0\} = \left\{ {\alpha {{{\left( {\begin{array}{*{20}{c}} {{\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}}}& \ldots &{{\text{|}}{{u}_{N}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{N}}} \end{array}} \right)}}^{{\text{T}}}}:\alpha > 0} \right\}.$$

Taking into account the fact that ${\text{rank}}{{B}_{N}} = n$ and $\ker B_{N}^{{\text{T}}} = \{ 0\}$, due to (A.2), the following equalities are valid:

$$\mathcal{N}({{B}_{N}}u,{{\mathcal{U}}_{p}}({{B}_{N}})) = (B_{N}^{{\text{T}}}{{)}^{{ - 1}}}(\{ \alpha {{(\begin{array}{*{20}{c}} {{\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}}}& \ldots &{{\text{|}}{{u}_{N}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{N}}} \end{array})}^{{\text{T}}}}:\alpha > 0\} )$$

$$\, = \bigcup\limits_{\alpha > 0} {{(B_{N}^{{\text{T}}})}^{{ - 1}}}(\{ \alpha {{(\begin{array}{*{20}{c}} {{\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}}}& \ldots &{{\text{|}}{{u}_{N}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{N}}} \end{array})}^{{\text{T}}}}\} )$$

$$\, = {\text{cone}}\{ f \in {{\mathbb{R}}^{n}}:B_{N}^{{\text{T}}}f = {{(\begin{array}{*{20}{c}} {{\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}}}& \ldots &{{\text{|}}{{u}_{N}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{N}}} \end{array})}^{{\text{T}}}}\} {{\backslash }}\{ 0\} .$$

Since the ${\text{rank}}{{B}_{N}} = {\text{rank}}B_{N}^{{\text{T}}} = n$, the equation

$$B_{N}^{{\text{T}}}f = {{\left( {\begin{array}{*{20}{c}} {{\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}}}& \ldots &{{\text{|}}{{u}_{N}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{N}}} \end{array}} \right)}^{{\text{T}}}}$$

has either one or zero solutions. In other words,

$$\dim \mathcal{N}({{B}_{N}}u,{{\mathcal{U}}_{p}}({{B}_{N}})) \leqslant 1.$$

Meanwhile, in the case ${{B}_{N}}u \in \partial {{\mathcal{U}}_{p}}({{B}_{N}})$, taking into account the definition of a normal cone, the following lower bound holds:

$$\mathcal{N}({{B}_{N}}u,{{\mathcal{U}}_{p}}({{B}_{N}})) \ne \varnothing , \quad \dim \mathcal{N}({{B}_{N}}u,{{\mathcal{U}}_{p}}({{B}_{N}})) \geqslant 1.$$

This implies the first assertion of Lemma 3.

To prove the second assertion of Lemma 3, according to the definition of the support vector, it suffices to show that $x{\kern 1pt} {\text{*}}(f)$ is the maximum point (f, x) given that $x \in {{\mathcal{U}}_{p}}({{B}_{N}})$.

Given that

$$B_{N}^{{\text{T}}}f = {{\left( {\begin{array}{*{20}{c}} {(f,{{b}_{1}})}& \ldots &{(f,{{b}_{N}})} \end{array}} \right)}^{{\text{T}}}},$$

we get the following chain of equalities:

$$\arg \mathop {\max }\limits_{x \in {{\mathcal{U}}_{p}}({{B}_{N}})} (f,x) = {{B}_{N}}\arg \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(N)} (f,{{B}_{N}}u) = {{B}_{N}}\arg \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(N)} (B_{N}^{{\text{T}}}f,u)$$

$$\, = \frac{{\left( {\begin{array}{*{20}{c}} {{{b}_{1}}}& \ldots &{{{b}_{N}}} \end{array}} \right)}}{{{{{\left( {\sum\limits_{i = 1}^N {\text{|}}(f,{{b}_{i}}){{{\text{|}}}^{q}}} \right)}}^{{1/p}}}}}\left( {\begin{array}{*{20}{c}} {{\text{|}}(f,{{b}_{1}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{1}})} \\ \vdots \\ {{\text{|}}(f,{{b}_{N}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{N}})} \end{array}} \right) = x{\kern 1pt} {\text{*}}(f).$$

The second assertion of Lemma 3 is proved.

Proof of Lemma 4. We visualize $B{\kern 1pt} '$ in the following way:

$$B{\kern 1pt} 'u = {{\left( {({{y}^{1}},u), \ldots ,({{y}^{n}},u)} \right)}^{{\text{T}}}},\quad u \in {{l}_{p}},$$

where ${{y}^{1}}, \ldots ,{{y}^{n}}$ are some linear functionals.

Assume the inclusion $B{\kern 1pt} ' \in l_{q}^{n}$ is true. By ([15], Section 3, §3, Ch. IV) for all $i = \overline {1,n}$, functionals ${{y}^{i}}:{{l}_{p}} \to \mathbb{R}$ are bounded and defined on all l_p, i.e.,

$$\mathop {\sup }\limits_{\parallel u{{\parallel }_{{{{l}_{p}}}}} \leqslant 1} {\text{|}}({{y}^{i}},u){\text{|}} = {{C}_{i}} < \infty ,\quad {{\mathcal{D}}_{{B{\kern 1pt} '}}} = {{l}_{p}}.$$

Hence, according to the definition of a bounded operator ([15], Section 2, §5, Ch. IV), the boundedness of the set ${{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ in ${{\mathbb{R}}^{n}}$ in the sense of the following norm follows:

$${\text{||}}x{\text{|}}{{{\text{|}}}_{\infty }} = \mathop {\max }\limits_{i = \overline {1,n} } {\text{|}}{{x}_{i}}{\text{|}}.$$

Since ${{\mathbb{R}}^{n}}$ is a finite-dimensional space, all the norms in it are equivalent ([15], Par. 3, Ch. III) and, consequently, ${{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ is bounded in ${{\mathbb{R}}^{n}}$ in the sense of any norm.

Assume $B{\kern 1pt} ' \notin l_{q}^{n}$, which is equivalent to the existence of $j = \overline {1,n}$ such that ${{y}^{j}}\notin{{l}_{q}}$. According to ([15], Section 3, §2, Ch. IV), y^j is not a bounded functional in l_p, which, taking into account ([15], Section 2, §1, Ch. IV), implies its unboundedness on ${{\mathcal{E}}_{p}}(\infty )$. Then there is a sequence ${{\{ {{u}^{{(N)}}}\} }_{{N \in \mathbb{N}}}} \subset {{\mathcal{E}}_{p}}(\infty )$ on which the functional y^j is defined, but at the same time unbounded; i.e., for all C > 0, we can pick a number ${{N}_{C}} \in \mathbb{N}$ such that

$${\text{|}}({{y}^{j}},{{u}^{{({{N}_{C}})}}}){\text{|}} = \left| {\sum\limits_{k = 1}^\infty y_{k}^{j}u_{k}^{{({{N}_{C}})}}} \right| > 2C.$$

However, due to the convergence of the series, there exists ${{K}_{C}} \in \mathbb{N}$ such that

$$\left| {\sum\limits_{k = {{K}_{C}} + 1}^\infty y_{k}^{j}u_{k}^{{({{N}_{C}})}}} \right| < C.$$

Then

$$\left| {\sum\limits_{k = 1}^{{{K}_{C}}} y_{k}^{j}u_{k}^{{({{N}_{C}})}}} \right| \geqslant \left| {\sum\limits_{k = 1}^\infty y_{k}^{j}u_{k}^{{({{N}_{C}})}}} \right| - \left| {\sum\limits_{k = {{K}_{C}} + 1}^\infty y_{k}^{j}u_{k}^{{({{N}_{C}})}}} \right| > C.$$

(A.3)

Assume

$${{\tilde {u}}^{{(M)}}} = (u_{1}^{{({{N}_{M}})}}, \ldots ,u_{{{{K}_{M}}}}^{{({{N}_{M}})}},0, \ldots ).$$

Then the inclusion is true ${{\{ {{\tilde {u}}^{{(M)}}}\} }_{{M \in \mathbb{N}}}} \in {{\mathcal{E}}_{p}}(\infty ) \cap {{\mathcal{D}}_{{B{\kern 1pt} '}}}$, because for each $M \in \mathbb{N}$, the subsequence ${{\tilde {u}}^{{(M)}}}$ is finite. However, taking into account (A.3), the following relations are valid:

$${{\left\| {B{\kern 1pt} '{{{\tilde {u}}}^{{(M)}}}} \right\|}_{\infty }} = \mathop {\max }\limits_{i = \overline {1,n} } \left| {\left( {{{y}^{i}},{{{\tilde {u}}}^{{(M)}}}} \right)} \right| \geqslant \left| {({{y}^{j}},{{{\tilde {u}}}^{{(M)}}})} \right| = \left| {\sum\limits_{k = 1}^{{{K}_{M}}} y_{k}^{j}u_{k}^{{({{N}_{M}})}}} \right| > M.$$

Since ${{\left\{ {B{\kern 1pt} '{{{\tilde {u}}}^{{(M)}}}} \right\}}_{{M \in \mathbb{N}}}} \subset {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$, according to (2.4), the set ${{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ is unbounded in the sense of the norm ${{\left\| {\, \cdot \,} \right\|}_{\infty }}$, and therefore unbounded in ${{\mathbb{R}}^{n}}$ in the sense of any norm by their equivalence.

Lemma 4 is proved.

Proof of Lemma 5. Assume $x \in \bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}})$; i.e., there exists $\tilde {N} \in \mathbb{N}$ such that $x \in {{\mathcal{U}}_{p}}({{B}_{{\tilde {N}}}})$. From this, by (2.1), there exists ${{\left( {{{u}_{1}}, \ldots ,{{u}_{{\tilde {N}}}}} \right)}^{{\text{T}}}} \in {{\mathcal{E}}_{p}}(\tilde {N})$, satisfying the equality

$$x = \sum\limits_{j = 1}^{\tilde {N}} {{u}_{j}}{{b}_{j}}.$$

However, then $({{u}_{1}}, \ldots ,{{u}_{{\tilde {N}}}},0, \ldots ) \in {{\mathcal{E}}_{p}}(\infty )$, which, by (2.4), is equivalent to the inclusion $x \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$. Hence it follows that

$$\bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}) \subset {{\mathcal{U}}_{p}}(B{\kern 1pt} ').$$

Since by the condition $B{\kern 1pt} ':{{l}_{p}} \to {{\mathbb{R}}^{n}}$ is a bounded operator, taking into account (2.4), the set ${{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ is closed, since it is the image of the closed set ${{\mathcal{E}}_{p}}(\infty )$. Then

$$\overline {\bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}})} \subset \overline {{{\mathcal{U}}_{p}}(B{\kern 1pt} ')} = {{\mathcal{U}}_{p}}(B{\kern 1pt} ').$$

An arbitrary $x \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ is the limit point of the set $\bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}})$. According to (2.4), there exists $u = ({{u}_{1}},{{u}_{2}}, \ldots ) \in {{\mathcal{E}}_{p}}(\infty )$ such that $x = B{\kern 1pt} 'u$. For each $N \in \mathbb{N}$, we define ${{u}^{{(N)}}} \in {{\mathbb{R}}^{N}}$:

$${{u}^{{(N)}}} = ({{u}_{1}}, \ldots ,{{u}_{N}}{{)}^{{\text{T}}}}.$$

By construction ${{u}^{{(N)}}} \in {{\mathcal{E}}_{p}}(N)$. Then, by (2.1), the following inclusion is true:

$${{\{ {{B}_{N}}{{u}^{{(N)}}}\} }_{{N \in \mathbb{N}}}} \subset \bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}) \subset {{\mathbb{R}}^{n}}.$$

At the same time, in the sense of any norm in ${{\mathbb{R}}^{n}}$, taking into account the Hölder inequality, the following relations are valid:

$${\text{||}}B{\kern 1pt} 'u - {{B}_{N}}{{u}^{{(N)}}}{\text{||}} = \left\| {\sum\limits_{k = N + 1}^\infty {{b}_{k}}{{u}_{k}}} \right\| \leqslant \sum\limits_{k = N + 1}^\infty {\text{||}}{{b}_{k}}{\text{||}}\,{\text{|}}{{u}_{k}}{\text{|}} \leqslant {{\left( {\sum\limits_{k = N + 1}^\infty {\text{||}}{{b}_{k}}{\text{|}}{{{\text{|}}}^{q}}} \right)}^{{1/q}}}{{\left( {\sum\limits_{k = N + 1}^\infty {\text{|}}{{u}_{k}}{{{\text{|}}}^{p}}} \right)}^{{1/p}}}\xrightarrow{{N \to \infty }}0.$$

Convergence holds because both factors are residues of convergent series ([23], Section 4, §1, Ch. III), where the series $\sum\limits_{k = 1}^\infty {\text{||}}{{b}_{k}}{\text{|}}{{{\text{|}}}^{q}}$ converges because $B{\kern 1pt} ' \in l_{q}^{n}$ and the row $\sum\limits_{k = 1}^\infty {\text{|}}{{u}_{k}}{{{\text{|}}}^{p}}$ converges because $u \in {{\mathcal{E}}_{p}}(\infty )$. We obtain

$${{\mathcal{U}}_{p}}(B{\kern 1pt} ') \subset \overline {\bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}})} .$$

Lemma 5 is proved.

Proof of Lemma 6. Assume ${{y}^{i}} = ({{b}_{{i1}}},{{b}_{{i2}}}, \ldots ) \in {{l}_{q}}$ is not finite for some $i = \overline {1,n}$. We calculate the norm $\tilde {u} \in {{l}_{p}}$:

$${\text{||}}\tilde {u}{\text{|}}{{{\text{|}}}_{{{{l}_{p}}}}} = {{\left( {\sum\limits_{j = 1}^\infty {{{\left| {\frac{{{\text{|}}{{b}_{{ij}}}{{{\text{|}}}^{{q - 1}}}{\text{sign}}{{b}_{{ij}}}}}{{{\text{||}}{{y}^{i}}{\text{||}}_{{{{l}_{q}}}}^{{q/p}}}}} \right|}}^{p}}} \right)}^{{1/p}}} = \frac{{{{{\left( {\sum\limits_{j = 1}^\infty {\text{|}}{{b}_{{ij}}}{{{\text{|}}}^{{qp\left( {1 - 1/q} \right)}}}} \right)}}^{{1/p}}}}}{{{\text{||}}{{y}^{i}}{\text{||}}_{{{{l}_{q}}}}^{{q/p}}}} = \frac{{{{{\left( {\sum\limits_{j = 1}^\infty {\text{|}}{{b}_{{ij}}}{{{\text{|}}}^{q}}} \right)}}^{{1/q \cdot q/p}}}}}{{{\text{||}}{{y}^{i}}{\text{||}}_{{{{l}_{q}}}}^{{q/p}}}} = 1.$$

Then by definition $\tilde {u} \in {{\mathcal{E}}_{p}}(\infty )$ and due to (2.4)

$$\tilde {x} \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ').$$

We consider the following hyperplane:

$${{\mathcal{H}}_{i}} = \left\{ {u \in {{l}_{p}}:\sum\limits_{j = 1}^\infty {{b}_{{ij}}}{{u}_{j}} = {\text{||}}{{y}^{i}}{\text{|}}{{{\text{|}}}_{{{{l}_{q}}}}}} \right\} \subset {{l}_{p}}.$$

By construction

$${{\tilde {x}}_{i}} = ({{y}^{i}},\tilde {u}) = \sum\limits_{j = 1}^\infty \frac{{{\text{|}}{{b}_{{ij}}}{{{\text{|}}}^{{q - 1}}}{{b}_{{ij}}}{\text{sign}}{{b}_{{ij}}}}}{{{\text{||}}{{y}^{i}}{\text{||}}_{{{{l}_{q}}}}^{{q/p}}}} = \sum\limits_{j = 1}^\infty \frac{{{\text{|}}{{b}_{{ij}}}{{{\text{|}}}^{q}}}}{{{\text{||}}{{y}^{i}}{\text{||}}_{{{{l}_{q}}}}^{{q/p}}}} = {\text{||}}{{y}^{i}}{\text{||}}_{{{{l}_{q}}}}^{{q - q/p}} = {\text{||}}{{y}^{i}}{\text{|}}{{{\text{|}}}_{{{{l}_{q}}}}}.$$

Due to the Hölder inequality ([15], Section 2, §1, Ch. IV),

$$\mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} ({{y}^{i}},u) \leqslant \mathop {\max }\limits_{\parallel u{{\parallel }_{{{{l}_{p}}}}} \leqslant 1} {\text{||}}{{y}^{i}}{\text{|}}{{{\text{|}}}_{{{{l}_{q}}}}}{\text{||}}u{\text{|}}{{{\text{|}}}_{{{{l}_{p}}}}} \leqslant {\text{||}}{{y}^{i}}{\text{|}}{{{\text{|}}}_{{{{l}_{q}}}}} = ({{y}^{i}},\tilde {u}).$$

Then the hyperplane ${{\mathcal{H}}_{i}}$ is the support to the set ${{\mathcal{E}}_{p}}(\infty )$ at point $\tilde {u} \in \partial {{\mathcal{E}}_{p}}(\infty )$. Since ${{\mathcal{E}}_{p}}(\infty )$ is a strictly convex set with $p \in (1; + \infty )$, $\tilde {u}$ is the only point of contact ([4], Lemma 3):

$$\{ \tilde {u}\} = {{\mathcal{H}}_{i}} \cap {{\mathcal{E}}_{p}}(\infty ).$$

In other words, for any $u \in {{\mathcal{E}}_{p}}(\infty ){{\backslash }}\{ \tilde {u}\}$, the inequality ${{(B{\kern 1pt} 'u)}_{i}} = ({{y}_{i}},u) < ({{y}_{i}},\tilde {u}) = {{\tilde {x}}_{i}}$ holds. Hence, $\tilde {x} \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ has a unique preimage under the mapping $B{\kern 1pt} '$; however, there is no $N \in \mathbb{N}$ such that $x \in {{\mathcal{U}}_{p}}({{B}_{N}})$, because the sequences yⁱ and $\tilde {u}$ are not finite by assumption. We finally obtain that

$$\tilde {x}\notin\bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}).$$

Lemma 6 is proved.

Proof of Corollary 1. The proof of Corollary 1 follows directly from Lemma 6 and the fact that if all ${{y}^{1}}, \ldots ,{{y}^{n}} \in {{l}_{q}}$, there is an ${{N}_{0}} \in \mathbb{N}$ such that

$${{\mathcal{U}}_{p}}({{B}_{{{{N}_{0}}}}}) = {{\mathcal{U}}_{p}}({{B}_{N}}) = {{\mathcal{U}}_{p}}(B{\kern 1pt} '),\quad N \geqslant {{N}_{0}}.$$

Lemma 14 ([3]). For any $f \in {{l}_{q}}{{\backslash }}\{ 0\}$, vector

$$u\text{*} = \frac{1}{{{\text{||}}f{\text{||}}_{{{{l}_{q}}}}^{{q/p}}}}\left( {{\text{|}}{{f}_{1}}{{{\text{|}}}^{{q - 1}}}{\text{sign}}{{f}_{1}},{\text{|}}{{f}_{2}}{{{\text{|}}}^{{q - 1}}}{\text{sign}}{{f}_{2}}, \ldots } \right)$$

is the only solution to the optimization problem

$$(f,u) \to \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} .$$

Proof of Lemma 7. We consider an arbitrary

$$\tilde {x} \in \bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}).$$

We assume that the condition $\dim {\text{Lin}}\{ {{b}_{N}},{{b}_{{N + 1}}}, \ldots \} = n$ is fulfilled for all $N \in \mathbb{N}$. Then there exists $\tilde {N} \in \mathbb{N}$ such that $\tilde {x} \in {{\mathcal{U}}_{p}}({{B}_{{\tilde {N}}}})$, which by (2.1) and (2.4), is equivalent to the existence of $\tilde {u} = ({{\tilde {u}}_{1}}, \ldots ,{{\tilde {u}}_{{\tilde {N}}}},0, \ldots ) \in {{\mathcal{E}}_{p}}(\infty )$ such that the following equality is fulfilled:

$$\tilde {x} = {{B}_{{\tilde {N}}}}{{\left( {\begin{array}{*{20}{c}} {{{{\tilde {u}}}_{1}}}& \ldots &{{{{\tilde {u}}}_{{\tilde {N}}}}} \end{array}} \right)}^{{\text{T}}}} = B{\kern 1pt} '\tilde {u}.$$

We consider an arbitrary $f \in {{\mathbb{R}}^{n}}{{\backslash }}\{ 0\}$. Then, taking into account Lemma 14,

$$\mathop {\max }\limits_{x \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')} (f,x) = \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} (f,B{\kern 1pt} 'u) = \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \sum\limits_{j = 1}^\infty (f,{{b}_{j}}){{u}_{j}} = {{\left( {\sum\limits_{j = 1}^\infty {\text{|}}(f,{{b}_{j}}){{{\text{|}}}^{q}}} \right)}^{{1/q}}},$$

$$u\text{*} = {\text{arg}} \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \sum\limits_{j = 1}^\infty (f,{{b}_{j}}){{u}_{j}} = \frac{1}{{{{{\left( {\sum\limits_{j = 1}^\infty {\text{|}}(f,{{b}_{j}}){{{\text{|}}}^{q}}} \right)}}^{{1/p}}}}}\left( {{\text{|}}(f,{{b}_{1}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{1}}),{\text{|}}(f,{{b}_{2}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{2}}), \ldots } \right).$$

In this case, u* is the only maximum point; i.e., for all $u \in {{\mathcal{E}}_{p}}(\infty ){{\backslash }}\{ u\text{*}\}$, the following relations are valid:

$$(f,B{\kern 1pt} 'u\text{*}) = \sum\limits_{j = 1}^\infty (f,{{b}_{j}})u_{j}^{*} > \sum\limits_{j = 1}^\infty (f,{{b}_{j}}){{u}_{j}} = (f,B{\kern 1pt} 'u).$$

(A.4)

Therefore, there is no $u \in {{\mathcal{E}}_{p}}(\infty ){{\backslash }}\{ u\text{*}\}$ such that $B{\kern 1pt} 'u\text{*} = B{\kern 1pt} 'u$. The equality $B{\kern 1pt} 'u\text{*} = B{\kern 1pt} '\tilde {u}$ is true if and only if $u\text{*} = \tilde {u}$. This is equivalent to the following conditions:

$$(f,{{b}_{j}}) = 0,\quad j > \tilde {N},$$

$$f \in {\text{Li}}{{{\text{n}}}^{ \bot }}\{ {{b}_{{\tilde {N} + 1}}},{{b}_{{\tilde {N} + 2}}}, \ldots \} ,$$

which, due to the condition $f \ne 0$ is equivalent to the inequalities

$$\dim {\text{Li}}{{{\text{n}}}^{ \bot }}\{ {{b}_{{\tilde {N} + 1}}},{{b}_{{\tilde {N} + 2}}}, \ldots \} \geqslant 1,$$

$$\dim {\text{Lin}}\{ {{b}_{{\tilde {N} + 1}}},{{b}_{{\tilde {N} + 2}}}, \ldots \} < n.$$

We get a contradiction. In other words, equality $u{\text{*}} = \tilde {u}$ is impossible, which, taking into account (A.4) and the definition of a normal cone, means that for all $f \in {{\mathbb{R}}^{n}}{{\backslash }}\{ 0\}$

$$\mathop {\max }\limits_{x \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')} (f,x) = (f,B{\kern 1pt} 'u\text{*}) > (f,B{\kern 1pt} '\tilde {u}) = (f,\tilde {x}),$$

$$\mathcal{N}(\tilde {x},{{\mathcal{U}}_{p}}(B{\kern 1pt} ')) = \varnothing .$$

Then from the definition of a normal cone, $\tilde {x} \in {\text{int}} {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$, i.e.

$$\bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}) \subset {\text{int}} {{\mathcal{U}}_{p}}(B{\kern 1pt} '),$$

which, taking Lemma 5 into consideration, leads to the equality

$${\text{int}} {{\mathcal{U}}_{p}}(B{\kern 1pt} ') = \bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}).$$

Assume there exists $N \in \mathbb{N}$ such that

$$\dim {\text{Lin}}\{ {{b}_{{N + 1}}},{{b}_{{N + 2}}}, \ldots \} < n.$$

Then

$$\dim {\text{Li}}{{{\text{n}}}^{ \bot }}\{ {{b}_{{N + 1}}},{{b}_{{N + 2}}}, \ldots \} \geqslant 1,$$

$${\text{Li}}{{{\text{n}}}^{ \bot }}\{ {{b}_{{N + 1}}},{{b}_{{N + 2}}}, \ldots \} {{\backslash }}\{ 0\} \ne \varnothing .$$

We choose an arbitrary $f \in {\text{Li}}{{{\text{n}}}^{ \bot }}\{ {{b}_{{N + 1}}},{{b}_{{N + 2}}}, \ldots \} {{\backslash }}\{ 0\}$ and build $u\text{*} \in {{\mathcal{E}}_{p}}(\infty )$ of the following form:

$$u\text{*} = \frac{1}{{{{{\left( {\sum\limits_{j = 1}^\infty {\text{|}}(f,{{b}_{j}}){{{\text{|}}}^{q}}} \right)}}^{{1/p}}}}}\left( {{\text{|}}(f,{{b}_{1}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{1}}),{\text{|}}(f,{{b}_{2}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{2}}), \ldots } \right)$$

$$\, = \frac{1}{{{{{\left( {\sum\limits_{j = 1}^N {\text{|}}(f,{{b}_{j}}){{{\text{|}}}^{q}}} \right)}}^{{1/p}}}}}\left( {{\text{|}}(f,{{b}_{1}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{1}}), \ldots ,{\text{|}}(f,{{b}_{N}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{N}}),0, \ldots } \right).$$

By Lemma 14

$$(f,B{\kern 1pt} 'u\text{*}) = \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \sum\limits_{j = 1}^\infty (f,{{b}_{j}}){{u}_{j}} = \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} (f,B{\kern 1pt} 'u) = \mathop {\max }\limits_{x \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')} (f,x).$$

Since, taking into account (2.4), $B{\kern 1pt} 'u\text{*} \in {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$, according to the definition of a normal cone, $f \in \mathcal{N}(B{\kern 1pt} 'u\text{*},{{\mathcal{U}}_{p}}(B{\kern 1pt} '))$, from which it follows that

$$B{\kern 1pt} 'u\text{*} \in \partial {{\mathcal{U}}_{p}}(B{\kern 1pt} ').$$

However, by construction u* is finite and $B{\kern 1pt} 'u\text{*} \in {{\mathcal{U}}_{p}}({{B}_{N}})$. Then,

$$B{\kern 1pt} 'u\text{*} \in \bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}){{\backslash \text{int}}} {{\mathcal{U}}_{p}}(B{\kern 1pt} '),$$

$${\text{int}} {{\mathcal{U}}_{p}}(B{\kern 1pt} ') \ne \bigcup\limits_{N = 1}^\infty {{\mathcal{U}}_{p}}({{B}_{N}}).$$

Lemma 7 is proved.

Proof of Lemma 8. By ([15], Section 3, §2, Ch. IV), $B{\kern 1pt} ':{{l}_{p}} \to {{\mathbb{R}}^{n}}$ is a linear and bounded operator. According to [4], Lemma 5, for the strictly convex set ${{\mathcal{E}}_{p}}(\infty )$ and arbitrary $u \in {{\mathcal{E}}_{p}}(\infty )$, the equality

$$\mathcal{N}(B{\kern 1pt} 'u,B{\kern 1pt} '{{\mathcal{E}}_{p}}(\infty )) = (B{\kern 1pt} '\text{*}{{)}^{{ - 1}}}(\mathcal{N}(u,{{\mathcal{E}}_{p}}(\infty ))) \cup (\ker B{\kern 1pt} '\text{*}{{\backslash }}\{ 0\} ),$$

where $B{\kern 1pt} '\text{*}:{{\mathbb{R}}^{n}} \to {{l}_{q}}$ denotes the operator adjoint to $B{\kern 1pt} '$ ([15], Section 5, §5, Ch. IV), is valid. By (2.2), the condition $\ker B{\kern 1pt} '\text{*} = \{ 0\}$ is fulfilled. Just as demonstrated in [4] and Example 1,

$$\mathcal{N}(u,{{\mathcal{E}}_{p}}(\infty )) = {\text{cone}}\{ ({\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}},{\text{|}}{{u}_{2}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{2}}, \ldots )\} {{\backslash }}\{ 0\} .$$

Then, taking into account Representation (2.4),

$$\mathcal{N}(B{\kern 1pt} 'u,{{\mathcal{U}}_{p}}(B{\kern 1pt} ')) = (B{\kern 1pt} '\text{*}{{)}^{{ - 1}}}\left( {\left\{ {\alpha ({\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}},{\text{|}}{{u}_{2}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{2}}, \ldots ):\alpha > 0} \right\}} \right)$$

$$\, = \bigcup\limits_{\alpha > 0} {{(B{\kern 1pt} '\text{*})}^{{ - 1}}}\left( {\left\{ {\alpha ({\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}},{\text{|}}{{u}_{2}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{2}}, \ldots )} \right\}} \right)$$

$$\, = {\text{cone}}\left\{ {f \in {{\mathbb{R}}^{n}}:B{\kern 1pt} '\text{*}f = ({\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}},{\text{|}}{{u}_{2}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{2}}, \ldots )} \right\}{{\backslash }}\{ 0\} .$$

Since condition (2.2) is valid, the equation

$$B{\kern 1pt} '\text{*}f = ({\text{|}}{{u}_{1}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{1}},{\text{|}}{{u}_{2}}{{{\text{|}}}^{{p - 1}}}{\text{sign}}{{u}_{2}}, \ldots )$$

has either one or zero solutions, i.e.,

$$\dim \mathcal{N}(B{\kern 1pt} 'u,{{\mathcal{U}}_{p}}(B{\kern 1pt} ')) \leqslant 1.$$

Meanwhile, in the case $B{\kern 1pt} 'u \in \partial {{\mathcal{U}}_{p}}(B{\kern 1pt} ')$, taking into account the definition of a normal cone, the following lower bound holds:

$$\mathcal{N}(B{\kern 1pt} 'u,{{\mathcal{U}}_{p}}(B{\kern 1pt} ')) \ne \varnothing ,$$

$$\dim \mathcal{N}(B{\kern 1pt} 'u,{{\mathcal{U}}_{p}}(B{\kern 1pt} ')) \geqslant 1.$$

This implies the first assertion of Lemma 8.

However,

$$\arg \mathop {\max }\limits_{x \in {{\mathcal{U}}_{p}}(B')} (f,x) = B{\kern 1pt} '\arg \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} (f,B{\kern 1pt} 'u) = B{\kern 1pt} '\arg \mathop {\max }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} (B{\kern 1pt} '{\kern 1pt} {\text{*}}f,u)$$

$$\, = \frac{1}{{{{{\left( {\sum\limits_{i = 1}^\infty {\text{|}}(f,{{b}_{i}}){{{\text{|}}}^{q}}} \right)}}^{{1/p}}}}}B{\kern 1pt} '\left( {{\text{|}}(f,{{b}_{1}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{1}}),{\text{|}}(f,{{b}_{2}}){{{\text{|}}}^{{q - 1}}}{\text{sign}}(f,{{b}_{2}}), \ldots } \right).$$

Here the last equality is due to Lemma 14.

Given definition $B{\kern 1pt} '$, the second assertion of Lemma 8 is proved.

Proof of Lemma 9. The proof of the lemma follows directly from definitions (1.3) and (1.4), as well as the definition of sets ${{\mathcal{E}}_{p}}(N)$ and ${{\mathcal{U}}_{p}}({{B}_{N}})$.

Proof of Theorem 1. Point 1 follows from the definition of the reachability (1.3) and controllability (1.4) sets; point 2, from Lemmas 1 and 9; point 3, from Lemma 2; and point 4, from Lemmas 2 and 3, as well as point 1 of Lemma 9. Similarly, point 5 follows from the same lemmas when using point 2 of Lemma 9.

Theorem 1 is proved.

Proof of Theorem 2. We consider the boundedness of ${{\mathcal{Y}}_{{p,\infty }}}$. Assume ${{h}_{1}}, \ldots ,{{h}_{n}} \subset {{\mathbb{R}}^{n}}$ is the Jordan basis of matrix A, and vector b admits the decomposition $b = {{\alpha }_{1}}{{h}_{1}} + \ldots + {{\alpha }_{n}}{{h}_{n}}$. We denote by $S \in {{\mathbb{R}}^{{n \times n}}}$ the nonsingular matrix defining the similarity transformation of A to its real Jordan form ([20], Section 3.4, Ch. 3):

$${{S}^{{ - 1}}}AS = J = \left( {\begin{array}{*{20}{c}} {{{J}_{1}}}&0& \cdots &0 \\ 0&{{{J}_{2}}}& \cdots &0 \\ \vdots & \vdots & \ddots & \vdots \\ 0&0& \cdots &{{{J}_{m}}} \end{array}} \right),$$

(A.5)

where each block J_i corresponds to either a real eigenvalue ${{\lambda }_{i}} \in \mathbb{R}$ of matrix A and takes the form

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

(A.6)

or corresponds to a pair of complex conjugate eigenvalues $r{{{\text{e}}}^{{ \pm i\varphi }}} \in \mathbb{C}$ of matrix A and takes the form

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

(A.7)

Note that in the case (A.7), n_i is even. In addition, due to Representation (A.5), the inclusions $(b,Ab,{{A}^{2}}b, \ldots ) \in l_{q}^{n}$ and $({{S}^{{ - 1}}}b,J{{S}^{{ - 1}}}b,{{J}^{2}}{{S}^{{ - 1}}}b, \ldots ) \in l_{q}^{n}$ are equivalent.

According to the definition of the Jordan basis, ${{S}^{{ - 1}}}b = ({{\alpha }_{1}}, \ldots ,{{\alpha }_{n}}{{)}^{{\text{T}}}}$. We group the coordinates of the vector ${{S}^{{ - 1}}}b$ in accordance with the dimensions and arrangement of the Jordan cells in expansion (A.5) and introduce the following notation:

$${{S}^{{ - 1}}}b = ({{\alpha }_{{1,1}}}, \ldots ,{{\alpha }_{{1,{{n}_{1}}}}}, \ldots ,{{\alpha }_{{m,1}}}, \ldots ,{{\alpha }_{{m,{{n}_{m}}}}}{{)}^{{\text{T}}}},$$

$${{J}^{k}}{{S}^{{ - 1}}}b = ({{\alpha }_{{1,1}}}(k), \ldots ,{{\alpha }_{{1,{{n}_{1}}}}}(k), \ldots ,{{\alpha }_{{m,1}}}(k), \ldots ,{{\alpha }_{{m,{{n}_{m}}}}}(k{{))}^{{\text{T}}}},\quad k \in \mathbb{N}.$$

Taking into account expansion (A.5), the following relation is valid for any $i = \overline {1,m}$:

$$\left( {\begin{array}{*{20}{c}} {{{\alpha }_{{i,1}}}(k)} \\ \vdots \\ {{{\alpha }_{{i,{{n}_{i}}}}}(k)} \end{array}} \right) = J_{i}^{k}\left( {\begin{array}{*{20}{c}} {{{\alpha }_{{i,1}}}} \\ \vdots \\ {{{\alpha }_{{i,{{n}_{i}}}}}} \end{array}} \right).$$

1. In the case (A.6), for all $k \geqslant {{n}_{i}} - 1$, the following representation is valid:

$$J_{i}^{k} = \left( {\begin{array}{*{20}{c}} {\lambda _{i}^{k}}&{C_{k}^{1}\lambda _{i}^{{k - 1}}}&{C_{k}^{2}\lambda _{i}^{{k - 2}}}& \cdots &{C_{k}^{{{{n}_{i}} - 1}}\lambda _{i}^{{k - {{n}_{i}} + 1}}} \\ 0&{\lambda _{i}^{k}}&{C_{k}^{1}\lambda _{i}^{{k - 1}}}& \cdots &{C_{k}^{{{{n}_{i}} - 2}}\lambda _{i}^{{k - {{n}_{i}} + 2}}} \\ 0&0&{\lambda _{i}^{k}}& \cdots &{C_{k}^{{{{n}_{i}} - 3}}\lambda _{i}^{{k - {{n}_{i}} + 3}}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0& \cdots &{\lambda _{i}^{k}} \end{array}} \right),$$

where, hereinafter, the number of combinations of k by j is denoted through $C_{k}^{j}$:

$$C_{k}^{j} = \frac{{k!}}{{(k - j)!j!}}.$$

We get

$${{\alpha }_{{i,j}}}(k) = \sum\limits_{l = j}^{{{n}_{i}}} {{\alpha }_{{i,l}}}C_{k}^{{l - j}}\lambda _{i}^{{k - l + j}},\quad j = \overline {1,{{n}_{i}}} .$$

If ${\text{|}}{{\lambda }_{i}}{\text{|}} < 1$ and $k > 2({{n}_{i}} - 1)$, then the following estimates are valid:

$${\text{|}}{{\alpha }_{{i,j}}}(k){\text{|}} \leqslant \sum\limits_{l = j}^{{{n}_{i}}} {\text{|}}{{\alpha }_{{i,l}}}{\text{|}}C_{k}^{{l - j}}{\text{|}}{{\lambda }_{i}}{{{\text{|}}}^{{k - l + j}}} \leqslant \sum\limits_{l = 1}^{{{n}_{i}}} {\text{|}}{{\alpha }_{{i,l}}}{\text{|}}C_{k}^{{{{n}_{i}} - 1}}{\text{|}}{{\lambda }_{i}}{{{\text{|}}}^{{k - {{n}_{i}} + 1}}}$$

$$ \leqslant \frac{{\sum\limits_{l = 1}^{{{n}_{i}}} {\text{|}}{{\alpha }_{{i,l}}}{\text{|}}}}{{({{n}_{i}} - 1)!}}\frac{{k!}}{{(k - {{n}_{i}} + 1)!}}{\text{|}}{{\lambda }_{i}}{{{\text{|}}}^{{k - {{n}_{i}} + 1}}} \leqslant \frac{{\sum\limits_{l = 1}^{{{n}_{i}}} {\text{|}}{{\alpha }_{{i,l}}}{\text{|}}}}{{({{n}_{i}} - 1)!}}{{k}^{{{{n}_{i}} - 1}}}{\text{|}}{{\lambda }_{i}}{{{\text{|}}}^{{k - {{n}_{i}} + 1}}}.$$

From where, given the definition of space l_q and the convergence criteria for the numerical series ([23], Section 4, §1, Ch. III), it follows that $({{\alpha }_{{i,j}}},{{\alpha }_{{i,j}}}(1),{{\alpha }_{{i,j}}}(2), \ldots ) \in {{l}_{q}}$.

Assume ${\text{|}}{{\lambda }_{i}}{\text{|}} \geqslant 1$ and there exists $j = \overline {1,{{n}_{i}}}$ such that ${{\alpha }_{{i,j}}} \ne 0$. Without loss of generality, we will assume that $j = {{n}_{i}}$ or ${{\alpha }_{{i,j + 1}}} = \ldots = {{\alpha }_{{i,{{n}_{i}}}}} = 0$. Then

$${{\alpha }_{{i,j}}}(k) = {{\alpha }_{{i,j}}}\lambda _{i}^{k},$$

i.e., $({{\alpha }_{{i,j}}},{{\alpha }_{{i,j}}}(1),{{\alpha }_{{i,j}}}(2), \ldots )\notin{{l}_{q}}$.

If ${{\alpha }_{{i,1}}} = \ldots = {{\alpha }_{{i,{{n}_{i}}}}} = 0$, then for any values ${{\lambda }_{i}} \in \mathbb{R}$ and $k \in \mathbb{N}$

$${{\alpha }_{{i,j}}}(k) = 0,\quad j = \overline {1,{{n}_{i}}} ,$$

i.e., $({{\alpha }_{{i,j}}},{{\alpha }_{{i,j}}}(1),{{\alpha }_{{i,j}}}(2), \ldots ) = 0 \in {{l}_{q}}$.

We obtain that in the case (A.6), the inclusion

$$\left( {\begin{array}{*{20}{c}} {{{\alpha }_{{i,1}}}}&{{{\alpha }_{{i,1}}}(1)}&{{{\alpha }_{{i,1}}}(2)\,\, \cdots } \\ \vdots & \vdots & \vdots \\ {{{\alpha }_{{i,{{n}_{i}}}}}}&{{{\alpha }_{{i,{{n}_{i}}}}}(1)}&{{{\alpha }_{{i,{{n}_{i}}}}}(2)\,\, \cdots } \end{array}} \right) \in l_{q}^{{{{n}_{i}}}}$$

is true if and only if ${\text{|}}{{\lambda }_{i}}{\text{|}} < 1$ or ${{\alpha }_{{i,1}}} = \ldots = {{\alpha }_{{i,{{n}_{i}}}}} = 0$.

2. In the case (A.7), we take into account that n_i is even, and introduce the notation

$${{\tilde {n}}_{i}} = \frac{{{{n}_{i}}}}{2},\quad {{\tilde {\alpha }}_{{i,j}}} = \left( {\begin{array}{*{20}{c}} {{{\alpha }_{{i,2j - 1}}}} \\ {{{\alpha }_{{i,2j}}}} \end{array}} \right),\quad {{\tilde {\alpha }}_{{i,j}}}(k) = \left( {\begin{array}{*{20}{c}} {{{\alpha }_{{i,2j - 1}}}(k)} \\ {{{\alpha }_{{i,2j}}}(k)} \end{array}} \right),\quad j = \overline {1,{{{\tilde {n}}}_{i}}} .$$

For all $k \geqslant {{\tilde {n}}_{i}} - 1$, the following representation is valid:

$$J_{i}^{k} = \left( {\begin{array}{*{20}{c}} {r_{i}^{k}{{A}_{{k\varphi }}}}&{C_{k}^{1}r_{i}^{{k - 1}}{{A}_{{(k - 1)\varphi }}}}&{C_{k}^{2}{{A}_{{(k - 2)\varphi }}}r_{i}^{{k - 2}}}& \cdots &{C_{k}^{{{{{\tilde {n}}}_{i}} - 1}}r_{i}^{{k - {{{\tilde {n}}}_{i}} + 1}}{{A}_{{(k - {{{\tilde {n}}}_{i}} + 1)\varphi }}}} \\ 0&{r_{i}^{k}{{A}_{{k\varphi }}}}&{C_{k}^{i}r_{i}^{{k - 1}}{{A}_{{(k - 1)\varphi }}}}& \cdots &{C_{k}^{{{{{\tilde {n}}}_{i}} - 2}}r_{i}^{{k - {{{\tilde {n}}}_{i}} + 2}}{{A}_{{(k - {{{\tilde {n}}}_{i}} + 2)\varphi }}}} \\ 0&0&{r_{i}^{k}{{A}_{{k\varphi }}}}& \cdots &{C_{k}^{{{{{\tilde {n}}}_{i}} - 3}}r_{i}^{{k - {{{\tilde {n}}}_{i}} + 3}}{{A}_{{(k - {{{\tilde {n}}}_{i}} + 3)\varphi }}}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0& \cdots &{r_{i}^{k}{{A}_{{k\varphi }}}} \end{array}} \right).$$

Therefore, we get the representation

$${{\tilde {\alpha }}_{{i,j}}}(k) = \sum\limits_{l = j}^{{{{\tilde {n}}}_{i}}} C_{k}^{{l - j}}r_{i}^{{k - l + j}}{{A}_{{(k - l + j)\varphi }}}{{\tilde {\alpha }}_{{i,l}}},\quad j = \overline {1,{{{\tilde {n}}}_{i}}} .$$

If ${{r}_{i}} < 1$ and $k > 2({{\tilde {n}}_{i}} - 1)$, then the following estimates are valid:

$$\max \{ {\text{|}}{{\alpha }_{{i,2j - 1}}}(k){\text{|}},{\text{|}}{{\alpha }_{{i,2j}}}(k){\text{|}}\} \leqslant {\text{||}}{{\tilde {\alpha }}_{{i,j}}}(k){\text{||}} \leqslant \sum\limits_{l = j}^{{{{\tilde {n}}}_{i}}} {\text{||}}{{A}_{{(k - l + j)\varphi }}}{{\tilde {\alpha }}_{{i,l}}}{\text{||}}C_{k}^{{l - j}}r_{i}^{{k - l + j}}$$

$$ \leqslant \sum\limits_{l = 1}^{{{{\tilde {n}}}_{i}}} {\text{||}}{{\tilde {\alpha }}_{{i,l}}}{\text{||}}C_{k}^{{{{{\tilde {n}}}_{i}} - 1}}r_{i}^{{k - {{{\tilde {n}}}_{i}} + 1}} \leqslant \frac{{\sum\limits_{l = 1}^{{{{\tilde {n}}}_{i}}} {\text{||}}{{{\tilde {\alpha }}}_{{i,l}}}{\text{||}}}}{{({{{\tilde {n}}}_{i}} - 1)!}}\frac{{k!}}{{(k - {{{\tilde {n}}}_{i}} + 1)!}}r_{i}^{{k - {{{\tilde {n}}}_{i}} + 1}} \leqslant \frac{{\sum\limits_{l = 1}^{{{{\tilde {n}}}_{i}}} {\text{||}}{{{\tilde {\alpha }}}_{{i,l}}}{\text{||}}}}{{({{{\tilde {n}}}_{i}} - 1)!}}{{k}^{{{{{\tilde {n}}}_{i}} - 1}}}r_{i}^{{k - {{{\tilde {n}}}_{i}} + 1}}.$$

From this, given the definition of space l_q and the convergence criteria for the numerical series ([23], Section 4, §1. Ch. III), it follows that $({{\alpha }_{{i,2j - 1}}},{{\alpha }_{{i,2j - 1}}}(1),{{\alpha }_{{i,2j - 1}}}(2), \ldots ) \in {{l}_{q}}$ and $({{\alpha }_{{i,2j}}},{{\alpha }_{{i,2j}}}(1),{{\alpha }_{{i,2j}}}(2), \ldots ) \in {{l}_{q}}$.

Assume ${{r}_{i}} \geqslant 1$ and there exists $j = \overline {1,{{{\tilde {n}}}_{i}}}$ such that ${\text{||}}{{\tilde {\alpha }}_{{i,j}}}{\text{||}} \ne 0$. Without loss of generality, we will assume that $j = {{\tilde {n}}_{i}}$ or ${\text{||}}{{\tilde {\alpha }}_{{i,j + 1}}}{\text{||}} = \ldots = {\text{||}}{{\tilde {\alpha }}_{{i,{{{\tilde {n}}}_{i}}}}}{\text{||}} = 0$. Then

$${\text{||}}{{\tilde {\alpha }}_{{i,j}}}(k){\text{||}} = {\text{||}}{{A}_{{k\varphi }}}{{\tilde {\alpha }}_{{i,j}}}{\text{||}}r_{i}^{k} = {\text{||}}{{\tilde {\alpha }}_{{i,j}}}{\text{||}}r_{i}^{k},$$

i.e., $({\text{||}}{{\tilde {\alpha }}_{{i,j}}}{\text{||}},{\text{||}}{{\tilde {\alpha }}_{{i,j}}}(1){\text{||}},{\text{||}}{{\tilde {\alpha }}_{{i,j}}}(2){\text{||}}, \ldots )\notin{{l}_{q}}$. Hence, taking into account the Minkowski inequality, it follows that at least one sequence $({{\alpha }_{{i,2j - 1}}},{{\alpha }_{{i,2j - 1}}}(1),{{\alpha }_{{i,2j - 1}}}(2), \ldots )$ or $({{\alpha }_{{i,2j}}},{{\alpha }_{{i,2j}}}(1),{{\alpha }_{{i,2j}}}(2), \ldots )$ does not belong to l_q.

If ${{\alpha }_{{i,1}}} = \ldots = {{\alpha }_{{i,{{n}_{i}}}}} = 0$, then for any values ${{r}_{i}} \geqslant 0$, $\varphi \in [0;2\pi )$, and $k \in \mathbb{N}$,

$${{\alpha }_{{i,j}}}(k) = 0,\quad j = \overline {1,{{n}_{i}}} ,$$

i.e., $({{\alpha }_{{i,j}}},{{\alpha }_{{i,j}}}(1),{{\alpha }_{{i,j}}}(2), \ldots ) = 0 \in {{l}_{q}}$.

We obtain that in the case (A.7) the inclusion

$$\left( {\begin{array}{*{20}{c}} {{{\alpha }_{{i,1}}}}&{{{\alpha }_{{i,1}}}(1)}&{{{\alpha }_{{i,1}}}(2)}& \cdots \\ \vdots & \vdots & \vdots &{} \\ {{{\alpha }_{{i,{{n}_{i}}}}}}&{{{\alpha }_{{i,{{n}_{i}}}}}(1)}&{{{\alpha }_{{i,{{n}_{i}}}}}(2)}& \cdots \end{array}} \right) \in l_{q}^{{{{n}_{i}}}}$$

is true if and only if ${{r}_{i}} < 1$ or ${{\alpha }_{{i,1}}} = \ldots = {{\alpha }_{{i,{{n}_{i}}}}} = 0$.

Finally, the inclusions $(b,Ab,{{A}^{2}}b, \ldots ) \in l_{q}^{n}$ and $({{S}^{{ - 1}}}b,J{{S}^{{ - 1}}}b,{{J}^{2}}{{S}^{{ - 1}}}b, \ldots ) \in l_{q}^{n}$, equivalent, according to (A.5), are valid if and only if ${{\alpha }_{{i,1}}} = \ldots = {{\alpha }_{{i,{{n}_{i}}}}} = 0$ for all $i = \overline {1,m}$, which correspond to the eigenvalues of matrix A modulo not less than 1, i.e., $b \in {{\mathbb{L}}_{{ < 1}}}$. By Lemmas 4 and 5, and point 1 of Lemma 9, this fact is equivalent to the boundedness of ${{\mathcal{Y}}_{{p,\infty }}}$.

Taking into account definition (1.5) and Lemma 5, the set ${{\mathcal{Y}}_{{p,\infty }}}$ coincides up to closure with ${{\mathcal{U}}_{p}}(B{\kern 1pt} ')$ if $B{\kern 1pt} ' = (b,Ab,{{A}^{2}}b, \ldots )$. Then the rest of the properties of ${{\mathcal{Y}}_{{p,\infty }}}$ follow from Lemma 8. Moreover, the expression for the support function follows directly from its definition:

$$s(f,{{\mathcal{Y}}_{{p,\infty }}}) = \mathop {\sup }\limits_{x \in {{\mathcal{Y}}_{{p,\infty }}}} (f,x) = (f,y\text{*}(f)) = {{\left( {\sum\limits_{i = 0}^\infty {\text{|}}(f,{{A}^{i}}b){{{\text{|}}}^{q}}} \right)}^{{1/q}}}.$$

Theorem 2 is proved.

Proof of Theorem 3. We consider the issues of the boundedness of ${{\mathcal{X}}_{{p,\infty }}}$. If detA = 0, then matrix A has an eigenvector $h \in {{\mathbb{R}}^{n}}$ corresponding to the eigenvalue λ = 0. Then, according to (1.4), the inclusion ${\text{Lin}}\{ h\} \subset {{\mathcal{X}}_{p}}(1)$, which, taking into account (1.6), leads to the inclusion ${\text{Lin}}\{ h\} \subset {{\mathcal{X}}_{{p,\infty }}}$, from which it follows that ${{\mathcal{X}}_{{p,\infty }}}$ is not bounded. If $\det A \ne 0$, then there is a matrix A^–1 whose eigenvalues are mutually inverse to the eigenvalues of A ([20], Ch. I). Hence it is clear that the proof of Theorem 3 is similar to the proof of Theorem 2 if we replace A by A^–1.

Proof of Corollary 2. The proof of Corollary 2 follows from Theorems 2 and 3 and the fact that $\det A \ne 0$ if and only if all the eigenvalues of matrix A are nonzero ([20], Assertion 1.1.7).

We denote by $H\text{*}:{{l}_{2}} \to {{l}_{2}}$ the operator adjoint to the linear and bounded operator $H:{{l}_{2}} \to {{l}_{2}}$ ([15], Section 6, §5, Ch. IV).

Lemma 15. Assume linear and bounded operators ${{H}_{{11}}}:{{\mathbb{R}}^{n}} \to {{\mathbb{R}}^{n}}$, ${{H}_{{12}}}:{{\mathbb{R}}^{n}} \to {{l}_{2}}$, and ${{H}_{{22}}}:{{l}_{2}} \to {{l}_{2}}$, such that $H:{{l}_{2}} \to {{l}_{2}}$ is a linear, bounded, positive definite, and self-adjoint operator, where

$$Hu = ({{H}_{{11}}}({{u}_{1}}, \ldots ,{{u}_{n}}) + H_{{12}}^{*}({{u}_{{n + 1}}},{{u}_{{n + 2}}}, \ldots ),{{H}_{{12}}}({{u}_{1}}, \ldots ,{{u}_{n}}) + H_{{22}}^{*}({{u}_{{n + 1}}},{{u}_{{n + 2}}}, \ldots )),$$

$$Pu = ({{u}_{1}}, \ldots ,{{u}_{n}}{{)}^{{\text{T}}}},$$

$$\mathcal{U} = \{ u \in {{l}_{2}}:(u,Hu) \leqslant 1\} .$$

Then

$$P\mathcal{U} = \{ x \in {{\mathbb{R}}^{n}}:(x,({{H}_{{11}}} - H_{{12}}^{*}H_{{22}}^{{ - 1}}{{H}_{{12}}})x) \leqslant 1\} .$$

Proof of Lemma 15. By the positive definiteness and self-adjointness of operator H, operator H₂₂ is also positive definite and self-adjoint. Then $(u,{{H}_{{22}}}u) > 0$ for all $u \in {{l}_{2}}{{\backslash }}\{ 0\}$, and we invert H₂₂ ([16], Section 4, Ch. X). We consider the chain of equalities

$$P\mathcal{U} = \left\{ {x \in {{\mathbb{R}}^{n}}:\exists u \in {{l}_{2}}:\left( {({{x}_{1}}, \ldots ,{{x}_{n}},{{u}_{1}},{{u}_{2}}, \ldots ),H({{x}_{1}}, \ldots ,{{x}_{n}},{{u}_{1}},{{u}_{2}}, \ldots )} \right) \leqslant 1} \right\}$$

$$\, = \left\{ {x \in {{\mathbb{R}}^{n}}:\exists u \in {{l}_{2}}:(x,{{H}_{{11}}}x) + 2({{H}_{{12}}}x,u) + (u,{{H}_{{22}}}u) \leqslant 1} \right\}$$

$$\, = \left\{ {x \in {{\mathbb{R}}^{n}}:\mathop {\min }\limits_{u \in {{l}_{2}}} \left( {(x,{{H}_{{11}}}x) + 2({{H}_{{12}}}x,u) + (u,{{H}_{{22}}}u)} \right) \leqslant 1} \right\}$$

$$\, = \{ x \in {{\mathbb{R}}^{n}}:\mathop {\min }\limits_{u \in {{l}_{2}}} \left( {(x,{{H}_{{11}}}x) - (H_{{22}}^{{ - 1}}{{H}_{{12}}}x,{{H}_{{12}}}x)} \right. + \left. {(u + H_{{22}}^{{ - 1}}{{H}_{{12}}}x,{{H}_{{22}}}(u + H_{{22}}^{{ - 1}}{{H}_{{12}}}x))} \right) \leqslant 1\} $$

$$\, = \{ x \in {{\mathbb{R}}^{n}}:(x,{{H}_{{11}}}x) - (x,H_{{12}}^{*}H_{{22}}^{{ - 1}}{{H}_{{12}}}x) \leqslant 1\} = \{ x \in {{\mathbb{R}}^{n}}:(x,({{H}_{{11}}} - H_{{12}}^{*}H_{{22}}^{{ - 1}}{{H}_{{12}}})x) \leqslant 1\} .$$

Lemma 15 is proved.

Proof of Lemma 10. For the case N > n, we choose $B{\kern 1pt} ' = ({{B}_{N}},0, \ldots ):{{l}_{p}} \to {{\mathbb{R}}^{n}}$, and we define the matrix ${{\hat {B}}_{N}} = ({{b}_{{n + 1}}}, \ldots ,{{b}_{N}}) \in {{\mathbb{R}}^{{n \times (N - n)}}}$ and operator $\tilde {B}:{{l}_{p}} \to {{l}_{p}}$:

$$\tilde {B}u = (B{\kern 1pt} 'u,{{u}_{{n + 1}}},{{u}_{{n + 2}}}, \ldots ).$$

Due to (2.2), operator $\tilde {B}$ is inversible. Then

$$\tilde {B}{{\mathcal{E}}_{2}}(N) = \left\{ {u \in {{l}_{2}}:({{{\tilde {B}}}^{{ - 1}}}u,{{{\tilde {B}}}^{{ - 1}}}u) \leqslant 1} \right\} = \left\{ {u \in {{l}_{2}}:(u,({{{\tilde {B}}}^{{ - 1}}}){\kern 1pt} {\text{*}}{{{\tilde {B}}}^{{ - 1}}}u) \leqslant 1} \right\}.$$

Since

$${{\tilde {B}}^{{ - 1}}}u = (B_{n}^{{ - 1}}({{u}_{1}}, \ldots ,{{u}_{n}}) - B_{n}^{{ - 1}}B({{u}_{{n + 1}}},{{u}_{{n + 2}}}, \ldots ),{{u}_{{n + 1}}},{{u}_{{n + 2}}}, \ldots ),$$

where $B = (0,0, \ldots )$ for $N = n$ and $B = ({{\hat {B}}_{N}},0, \ldots )$ for N > n, the positive definite and self-adjoint operator $H = ({{\tilde {B}}^{{ - 1}}}){\kern 1pt} {\text{*}}{{\tilde {B}}^{{ - 1}}}$ admits the representation described in Lemma 15:

$${{H}_{{11}}} = {{(B_{n}^{{ - 1}})}^{{\text{T}}}}B_{n}^{{ - 1}} = {{H}_{n}}, $$

$${{H}_{{12}}} = - B{\kern 1pt} {\text{*}}{{(B_{n}^{{ - 1}})}^{{\text{T}}}}B_{n}^{{ - 1}} = - B\text{*}{{H}_{n}},$$

$${{H}_{{22}}} = B{\kern 1pt} {\text{*}}{{(B_{n}^{{ - 1}})}^{{\text{T}}}}B_{n}^{{ - 1}}B + I = B{\kern 1pt} {\text{*}}{{H}_{n}}B + I.$$

Hence, according to Lemma 15, the representation of the set ${{\mathcal{U}}_{2}}({{B}_{N}})$ for the case N = n follows directly. For the case N > n, taking into account the fact that, due to the finiteness of sequence B the following relation is valid

$$Bu = {{\hat {B}}_{N}}{{\left( {\begin{array}{*{20}{c}} {{{u}_{{n + 1}}}}& \ldots &{{{u}_{N}}} \end{array}} \right)}^{{\text{T}}}},$$

and the following representation follows:

$${{\mathcal{U}}_{2}}({{B}_{N}}) = \{ x \in {{\mathbb{R}}^{n}}:(x,({{H}_{n}} - {{H}_{n}}{{\hat {B}}_{N}}{{(\hat {B}_{N}^{{\text{T}}}{{H}_{n}}{{\hat {B}}_{N}} + I)}^{{ - 1}}}\hat {B}_{N}^{{\text{T}}}{{H}_{n}})x) \leqslant 1\} .$$

Moreover, according to the Sherman–Morrison–Woodbury identity ([24], §5.1, Ch. 5), the following equality is valid:

$${{H}_{n}} - {{H}_{n}}{{\hat {B}}_{N}}{{(\hat {B}_{N}^{{\text{T}}}{{H}_{n}}{{\hat {B}}_{N}} + I)}^{{ - 1}}}\hat {B}_{N}^{{\text{T}}}{{H}_{n}} = {{({{B}_{N}}B_{N}^{{\text{T}}})}^{{ - 1}}} = {{H}_{N}}.$$

Lemma 10 is proved.

Lemma 16. Assume $H \in {{\mathbb{R}}^{{n \times n}}}$ is a symmetric and positive definite matrix, $b \in {{\mathbb{R}}^{n}}$ is an arbitrary vector, and $\mathcal{X} \subset {{\mathbb{R}}^{n}}$ is an ellipsoid:

$$\mathcal{X} = \{ x \in {{\mathbb{R}}^{n}}:{{x}^{{\text{T}}}}Hx \leqslant 1\} .$$

Then

$$\mathcal{Y} = \left\{ {{{\alpha }_{1}}b + {{\alpha }_{2}}x:x \in \mathcal{X}\alpha _{1}^{2} + \alpha _{2}^{2} \leqslant 1} \right\} = \left\{ {x \in {{\mathbb{R}}^{n}}:{{x}^{{\text{T}}}}\left( {H - \frac{{Hb{{b}^{{\text{T}}}}H}}{{1 + {{b}^{{\text{T}}}}Hb}}} \right)x \leqslant 1} \right\}.$$

Proof of Lemma 16. By construction $\mathcal{Y}$ is a convex set and $0 \in {\text{int}} \mathcal{Y}$. Let us construct the Minkowski functional $\mu (y,\mathcal{Y})$. According to the definition, the calculation of the Minkowski functional is reduced to solving the following equivalent optimization problems:

$$\mu (y,\mathcal{Y}) = \min \{ \gamma > 0:y = {{\alpha }_{1}}b + {{\alpha }_{2}}x, {{x}^{{\text{T}}}}Hx = 1, \alpha _{1}^{2} + \alpha _{2}^{2} = {{\gamma }^{2}}\} $$

$$\, = \min \{ \sqrt {\alpha _{1}^{2} + \alpha _{2}^{2}} :{{\alpha }_{2}}x = {{\alpha }_{1}}b - y, \,\,{{({{\alpha }_{2}}x)}^{{\text{T}}}}H({{\alpha }_{2}}x) = \alpha _{2}^{2}\} $$

$$\, = \min \{ \sqrt {\alpha _{1}^{2} + \alpha _{2}^{2}} :{{({{\alpha }_{1}}b - y)}^{{\text{T}}}}H({{\alpha }_{1}}b - y) = \alpha _{2}^{2}\} = \mathop {\min }\limits_{{{\alpha }_{1}} \in \mathbb{R}} \sqrt {\alpha _{1}^{2} + {{{({{\alpha }_{1}}b - y)}}^{{\text{T}}}}H({{\alpha }_{1}}b - y)} $$

$$\, = \mathop {\min }\limits_{{{\alpha }_{1}} \in \mathbb{R}} \sqrt {\alpha _{1}^{2}(1 + {{b}^{{\text{T}}}}Hb) - 2{{\alpha }_{1}}{{b}^{{\text{T}}}}Hy + {{y}^{{\text{T}}}}Hy} $$

$$\, = \sqrt {{{{\left( {\frac{{{{b}^{{\text{T}}}}Hy}}{{1 + {{b}^{{\text{T}}}}Hb}}} \right)}}^{2}}(1 + {{b}^{{\text{T}}}}Hb) - 2\frac{{{{b}^{{\text{T}}}}Hy}}{{1 + {{b}^{{\text{T}}}}Hb}}{{b}^{{\text{T}}}}Hy + {{y}^{{\text{T}}}}Hy} $$

$$\, = \sqrt {{{y}^{{\text{T}}}}Hy - \frac{{{{y}^{{\text{T}}}}Hb{{b}^{{\text{T}}}}Hy}}{{1 + {{b}^{{\text{T}}}}Hb}}} = \sqrt {{{y}^{{\text{T}}}}\left( {H - \frac{{{{y}^{{\text{T}}}}Hb{{b}^{{\text{T}}}}Hy}}{{1 + {{b}^{{\text{T}}}}Hb}}} \right)y} .$$

Taking into account (A.1),

$$\mathcal{Y} = \{ y \in {{\mathbb{R}}^{n}}:\mu (y,\mathcal{Y}) \leqslant 1\} = \{ y \in {{\mathbb{R}}^{n}}:{{(\mu (y,\mathcal{Y}))}^{2}} \leqslant 1\} .$$

Lemma 16 is proved.

Proof of Lemma 11. Representations (4.1) are well-posed due to Lemmas 9 and 10. Taking (2.1) into consideration, the following chain of equalities is valid:

$${{\mathcal{Y}}_{2}}(N + 1) = \left\{ {\sum\limits_{k = 0}^N {{A}^{k}}b{{u}_{k}}:\sum\limits_{k = 0}^N u_{k}^{2} \leqslant 1} \right\} = \left\{ {{{u}_{0}}b + A\sum\limits_{k = 0}^{N - 1} {{A}^{k}}b{{u}_{{k + 1}}}:u_{0}^{2} + \sum\limits_{k = 1}^N u_{k}^{2} \leqslant 1} \right\}$$

$$\, = \left\{ {{{u}_{0}}b + \alpha x:x \in A{{\mathcal{Y}}_{2}}(N),u_{0}^{2} + {{\alpha }^{2}} \leqslant 1} \right\} = \left\{ {y \in {{\mathbb{R}}^{n}}:\left( {y,{{H}_{{\mathcal{Y},N + 1}}}y} \right) \leqslant 1} \right\}.$$

Here the last equality is due to Lemma 16 and the equality

$$A{{\mathcal{Y}}_{2}}(N) = \{ x \in {{\mathbb{R}}^{n}}:(x,({{A}^{{ - 1}}}{{)}^{{\text{T}}}}{{H}_{{\mathcal{Y},N}}}{{A}^{{ - 1}}}x) \leqslant 1\} .$$

By Lemma 9, the description of ${{\mathcal{X}}_{2}}(N + 1)$ is constructed similarly with the replacement of A by A^–1 and b by ${{A}^{{ - 1}}}b$.

Lemma 11 is proved.

Proof of Theorem 4. The representation of ${{\overline {\mathcal{Y}} }_{{2,\infty }}}$ and ${{\overline {\mathcal{X}} }_{{2,\infty }}}$ in the form of ellipsoidal sets if they are bounded follows from Lemma 15, taking into account Lemma 9 and Assumption (1.2).

According to Lemma 9 and representation (2.4), we have the equalities

$${{\overline {\mathcal{Y}} }_{{2,\infty }}} = \left\{ {\sum\limits_{k = 0}^\infty {{A}^{k}}b{{u}_{k}}:\sum\limits_{k = 0}^\infty u_{k}^{2} \leqslant 1} \right\} = \left\{ {{{u}_{0}}b + A\sum\limits_{k = 0}^\infty {{A}^{k}}b{{u}_{{k + 1}}}:u_{0}^{2} + \sum\limits_{k = 1}^\infty u_{k}^{2} \leqslant 1} \right\}$$

$$\, = \left\{ {{{u}_{0}}b + \alpha x:x \in A{{{\overline {\mathcal{Y}} }}_{{2,\infty }}},u_{0}^{2} + {{\alpha }^{2}} \leqslant 1} \right\},$$

$${{\overline {\mathcal{X}} }_{{2,\infty }}} = \left\{ {\sum\limits_{k = 1}^\infty {{A}^{{ - k}}}b{{u}_{k}}:\sum\limits_{k = 1}^\infty u_{k}^{2} \leqslant 1} \right\} = \left\{ {{{u}_{1}}{{A}^{{ - 1}}}b + {{A}^{{ - 1}}}\sum\limits_{k = 1}^\infty {{A}^{{ - k}}}b{{u}_{{k + 1}}}:u_{1}^{2} + \sum\limits_{k = 2}^\infty u_{k}^{2} \leqslant 1} \right\}$$

$$\, = \left\{ {{{u}_{1}}{{A}^{{ - 1}}}b + \alpha x:x \in {{A}^{{ - 1}}}{{{\overline {\mathcal{X}} }}_{{2,\infty }}},u_{1}^{2} + {{\alpha }^{2}} \leqslant 1} \right\}.$$

From this, by Lemma 16, the representations for ${{H}_{{\mathcal{Y},\infty }}}$ and ${{H}_{{\mathcal{X},\infty }}}$ follow through the corresponding algebraic discrete Riccati equations.

Theorem 4 is proved.