On the External Estimation of Reachable and Null-Controllable Limit Sets for Linear Discrete-Time Systems with a Summary Constraint on the Scalar Control

Abstract

The problem of constructing reachable and null-controllable sets for stationary linear discrete-time systems with a summary constraint on the scalar control is considered. For the case of quadratic constraints and a diagonalizable matrix of the system, these sets are built explicitly in the form of ellipsoids. In the general case, the limit reachable and null-controllable sets are represented as fixed points of a contraction mapping in the metric space of compact sets. On the basis of the method of simple iteration, a convergent procedure for constructing their external estimates with an indication of the a priori approximation error is proposed. Examples are given.

APPENDIX

Proof of Lemma 1. By the definition of the Minkowski functional, for any x ∈ ${{\mathbb{L}}_{2}}$ the following relations are true:

$$\begin{gathered} \mu (x,B\mathcal{U}) = \inf \{ t > 0{\kern 1pt} :x \in tB\mathcal{U}\} = \inf \{ t > 0{\kern 1pt} :\exists u \in t\mathcal{U},\;x = Bu\} \\ = \inf \{ t > 0{\kern 1pt} :\exists u \in {{B}^{{ - 1}}}(\{ x\} ),\;u \in t\mathcal{U}\} = \mathop {\inf }\limits_{u \in {{B}^{{ - 1}}}(\{ x\} )} \inf \{ t > 0{\kern 1pt} :u \in t\mathcal{U}\} = \mathop {\inf }\limits_{u \in {{B}^{{ - 1}}}(\{ x\} )} \mu (u,\mathcal{U}). \\ \end{gathered} $$

Lemma 1 is proven.

Proof of Lemma 2. Since, due to the Riesz theorem, the operator B is linear and bounded, then according to Lemma 1

$${{(\mu (x,B{{\mathcal{E}}_{p}}(\infty )))}^{p}} = {{\left( {\mathop {\inf }\limits_{u \in {{B}^{{ - 1}}}(\{ x\} )} \mu (u,{{\mathcal{E}}_{p}}(\infty ))} \right)}^{p}} = \mathop {\inf }\limits_{\substack{ Bu = x \\ u \in {{l}_{p}} }} \sum\limits_{k = 1}^\infty {{{{\left| {{{u}_{k}}} \right|}}^{p}}.} $$

(A.1)

To solve the optimization problem (A.1) we will use the Lagrange multiplier method for infinite-dimensional spaces [29]. The Lagrange function for λ ∈ ${{\mathbb{R}}^{n}}$ has the following form

$$L(u,\lambda ) = \sum\limits_{k = 1}^\infty {{{{\left| {{{u}_{k}}} \right|}}^{p}}} + {{\lambda }^{{\text{T}}}}(x - Bu).$$

Then the search for a minimum in the problem (A.1) is reduced to solving the system of equations

$$\left\{ \begin{gathered} \frac{{\partial L}}{{\partial {{u}_{k}}}} = 0,\;k \in \mathbb{N}, \hfill \\ Bu = x, \hfill \\ \end{gathered} \right.\quad \left\{ \begin{gathered} p{{I}_{p}}(u) - B{\kern 1pt} ^*{\kern 1pt} \lambda = 0, \hfill \\ Bu = x, \hfill \\ \end{gathered} \right.\quad \left\{ \begin{gathered} u = I_{p}^{{ - 1}}\left( {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right) = {{I}_{q}}\left( {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right), \hfill \\ Bu = x. \hfill \\ \end{gathered} \right.$$

Hence, taking into account (A.1) and the identity $\sum\limits_{k = 1}^\infty {{{{\left| {{{u}_{k}}} \right|}}^{p}}}$ = (u, I_p(u)) it follows that

$$(\mu {{(x,B{{\mathcal{E}}_{p}}(\infty ))}^{p}} = \left( {{{I}_{q}}\left( {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right),\,\,{{I}_{p}}\left( {{{I}_{q}}\left( {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right)} \right)} \right) = \left( {{{I}_{q}}\left( {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right),\,\,\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right) = \left\| {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right\|_{{{{l}_{q}}}}^{q},$$

$$B{{I}_{q}}\left( {\frac{1}{p}B{\kern 1pt} ^*{\kern 1pt} \lambda } \right) = x.$$

Redesignating $\frac{1}{p}\lambda$ by λ, we finally obtain

$$\mu (x,B{{\mathcal{E}}_{p}}(\infty )) = \left\| {B{\kern 1pt} ^*{\kern 1pt} \lambda } \right\|_{{{{l}_{q}}}}^{{\frac{q}{p}}} = \left\| {B{\kern 1pt} ^*{\kern 1pt} \lambda } \right\|_{{{{l}_{q}}}}^{{q - 1}},$$

where λ ∈ ${{\mathbb{R}}^{n}}$ is determined from (10).

Lemma 2 is proven.

Proof of Corollary 1. By definition, the operator I₂ is identical. Then the condition (10) take the form

$$BB{\kern 1pt} ^*{\kern 1pt} \lambda = x.$$

Since B is surjective, the operator BB* : ${{\mathbb{R}}^{n}} \to {{\mathbb{R}}^{n}}$ and the matrix generating it are invertible, which leads to the relation

$$\lambda = {{(BB^*)}^{{ - 1}}}x.$$

Taking into account Lemma 2 we obtain

$$\mu (x,B{{\mathcal{E}}_{2}}(\infty )) = {{\left\| {B{\kern 1pt} ^*{\kern 1pt} {{{(BB^*)}}^{{ - 1}}}x} \right\|}_{2}} = \sqrt {(B{\kern 1pt} ^*{\kern 1pt} {{{(BB^*)}}^{{ - 1}}}x,B{\kern 1pt} ^*{\kern 1pt} {{{(BB^*)}}^{{ - 1}}}x)} = \sqrt {{{x}^{{\text{T}}}}{{{(BB^*)}}^{{ - 1}}}x} .$$

Corollary 1 is proven.

Proof of Lemma 3. Let B = Y_∞. Then, taking into account the spectral decomposition of the matrix A = SΛS^–1, the following equalities are true for all k ∈ $\mathbb{N}$:

$$\begin{gathered} {{b}_{k}}b_{k}^{{\text{T}}} = {{A}^{{k - 1}}}b{{b}^{{\text{T}}}}{{({{A}^{{k - 1}}})}^{{\text{T}}}} = S{{\Lambda }^{{k - 1}}}{{S}^{{ - 1}}}b{{b}^{{\text{T}}}}{{({{S}^{{ - 1}}})}^{{\text{T}}}}{{\Lambda }^{{k - 1}}}{{S}^{{\text{T}}}} \\ = S\operatorname{diag} (\lambda _{1}^{{k - 1}}, \ldots ,\lambda _{n}^{{k - 1}})\left( {\begin{array}{*{20}{c}} {{{\alpha }_{{11}}}}& \ldots &{{{\alpha }_{{1n}}}} \\ \vdots & \ddots & \vdots \\ {{{\alpha }_{{n1}}}}& \ldots &{{{\alpha }_{{nn}}}} \end{array}} \right)\operatorname{diag} (\lambda _{1}^{{k - 1}}, \ldots ,\lambda _{n}^{{k - 1}}){{S}^{{\text{T}}}} \\ = S\left( {\begin{array}{*{20}{c}} {{{{({{\lambda }_{1}}{{\lambda }_{1}})}}^{{k - 1}}}{{\alpha }_{{11}}}}& \ldots &{{{{({{\lambda }_{1}}{{\lambda }_{n}})}}^{{k - 1}}}{{\alpha }_{{1n}}}} \\ \vdots & \ddots & \vdots \\ {{{{({{\lambda }_{n}}{{\lambda }_{1}})}}^{{k - 1}}}{{\alpha }_{{n1}}}}& \ldots &{{{{({{\lambda }_{n}}{{\lambda }_{n}})}}^{{k - 1}}}{{\alpha }_{{nn}}}} \end{array}} \right){{S}^{{\text{T}}}}. \\ \end{gathered} $$

Due to the inclusion b ∈ ${{\mathbb{L}}_{{ < 1}}}$, the coefficients α_ij are equal to zero for all i, j = $\overline {1,n}$ such that at least one of the two eigenvalues λ_i or λ_j turns out to be greater than or equal to 1:

$${{\alpha }_{{ij}}} = 0,\;\,{\text{if}}\,\;\left| {{{\lambda }_{i}}} \right|\; \geqslant \;1\;\,{\text{or}}\,\;\left| {{{\lambda }_{j}}} \right|\; \geqslant \;1.$$

(A.2)

Hence, taking into account the expression for the sum of an infinite decreasing geometric progression, we get the following equality:

$$\sum\limits_{k = 1}^\infty {{{{({{\lambda }_{i}}{{\lambda }_{j}})}}^{{k - 1}}}{{\alpha }_{{ij}}}} = \left\{ {\begin{array}{*{20}{l}} {\frac{{{{\alpha }_{{ij}}}}}{{1 - {{\lambda }_{i}}{{\lambda }_{j}}}},\;\,}&{\left| {{{\lambda }_{i}}} \right| < 1\;{\text{and}}\;\left| {{{\lambda }_{j}}} \right| < 1,} \\ {0,}&{\left| {{{\lambda }_{i}}} \right|\; \geqslant \;1\;{\text{or}}\;\left| {{{\lambda }_{j}}} \right|\; \geqslant \;1,} \end{array}} \right.$$

which, due to the (A.2), coincides with the definition of β_ij.

Then the following chain of equalities is true:

$$\begin{gathered} BB^* = \sum\limits_{k = 1}^\infty {{{b}_{k}}b_{k}^{{\text{T}}}} = S\left( {\begin{array}{*{20}{c}} {\sum\limits_{k = 1}^\infty {{{{({{\lambda }_{1}}{{\lambda }_{1}})}}^{{k - 1}}}{{\alpha }_{{11}}}} }& \ldots &{\sum\limits_{k = 1}^\infty {{{{({{\lambda }_{1}}{{\lambda }_{n}})}}^{{k - 1}}}{{\alpha }_{{1n}}}} } \\ \vdots & \ddots & \vdots \\ {\sum\limits_{k = 1}^\infty {{{{({{\lambda }_{n}}{{\lambda }_{1}})}}^{{k - 1}}}{{\alpha }_{{n1}}}} }& \ldots &{\sum\limits_{k = 1}^\infty {{{{({{\lambda }_{n}}{{\lambda }_{n}})}}^{{k - 1}}}{{\alpha }_{{nn}}}} } \end{array}} \right){{S}^{{\text{T}}}} \\ = S\left( {\begin{array}{*{20}{c}} {{{\beta }_{{11}}}}& \ldots &{{{\beta }_{{1n}}}} \\ \vdots & \ddots & \vdots \\ {{{\beta }_{{n1}}}}& \ldots &{{{\beta }_{{nn}}}} \end{array}} \right){{S}^{{\text{T}}}} = H. \\ \end{gathered} $$

This, taking into account Corollary 1, implies the equality ${{H}_{{\mathcal{Y},\infty }}}$ = H^–1.

The second item of Lemma 3 is proven in a similar way under the redesignation B = X_∞.

Lemma 3 is proven.

Proof of Lemma 4. The proof follows directly from (6) and (13).

Proof of Lemma 5. The proof follows directly from (7) and (13).

Proof of Lemma 6. Since all eigenvalues of the matrix A are strictly less than 1 in absolute value, according to [22, Theorem 5.6.12] the relation ||A^k|| $\xrightarrow{{k \to \infty }}$ 0 is true. Then, by definition of the limit there is M ∈ $\mathbb{N}$ for α_r ∈ [0; 1) such that ||A^M|| = ${{\sup }_{{{{{\left\| x \right\|}}_{r}}\,\leqslant \,1}}}{{\left\| {{{A}^{M}}x} \right\|}_{r}}$ < α_r. Since the inequality

$${{\left\| {{{A}^{M}}(x - y)} \right\|}_{r}}\;\leqslant \;\left\| {{{A}^{M}}} \right\|{{\left\| {x - y} \right\|}_{r}}\;\leqslant \;{{\alpha }_{r}}{{\left\| {x - y} \right\|}_{r}}$$

is true, A^M is a contraction with the contraction factor α_r ∈ [0; 1).

Let B ' = (b₁, b₂, …) ∈ $l_{q}^{n}$, C ' = (c₁, c₂, …) ∈ $l_{q}^{n}$. Then

$${{\left\| {{\mathbf{F}}_{{A,b}}^{{(M)}}(B{\kern 1pt} ') - {\mathbf{F}}_{{A,b}}^{{(M)}}(C{\kern 1pt} ')} \right\|}_{{l_{q}^{n}}}} = \left\| {(b,Ab, \ldots {{A}^{{M - 1}}}b,{{A}^{M}}{{b}_{1}},{{A}^{M}}{{b}_{2}}, \ldots )} \right.$$

$${{\left. { - \;(b,Ab, \ldots {{A}^{{M - 1}}}b,{{A}^{M}}{{c}_{1}},{{A}^{M}}{{c}_{2}}, \ldots )} \right\|}_{{l_{q}^{n}}}}$$

$$ = {{\left\| {(0, \ldots ,0,{{A}^{M}}({{b}_{1}} - {{c}_{1}}),{{A}^{M}}({{b}_{2}} - {{c}_{2}}), \ldots )} \right\|}_{{l_{q}^{n}}}}$$

$$ = {{\left\| {({{A}^{M}}({{b}_{1}} - {{c}_{1}}),{{A}^{M}}({{b}_{2}} - {{c}_{2}}), \ldots )} \right\|}_{{l_{q}^{n}}}}$$

$$ = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {({{A}^{M}}({{b}_{1}} - {{c}_{1}}),{{A}^{M}}({{b}_{2}} - {{c}_{2}}), \ldots )u} \right\|}_{r}}$$

$$ = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {\sum\limits_{k = 1}^\infty {{{A}^{M}}({{b}_{k}} - {{c}_{k}}){{u}_{k}}} } \right\|}_{r}} = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {{{A}^{M}}\sum\limits_{k = 1}^\infty {{{b}_{k}}{{u}_{k}}} - {{A}^{M}}\sum\limits_{k = 1}^\infty {{{c}_{k}}{{u}_{k}}} } \right\|}_{r}}$$

$$\leqslant \;\mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} {{\alpha }_{r}}{{\left\| {\sum\limits_{k = 1}^\infty {{{b}_{k}}{{u}_{k}}} - \sum\limits_{k = 1}^\infty {{{c}_{k}}{{u}_{k}}} } \right\|}_{r}} = {{\alpha }_{r}}{{\left\| {B{\kern 1pt} '\; - C{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}}.$$

Lemma 6 is proven.

Proof of Corollary 2. The mapping ${\mathbf{F}}_{{A,b}}^{{(M)}}$ is a contraction due to Theorem 6, and the space $l_{q}^{n}$ is complete. Then, by virtue of the Banach contraction mapping principle [25], there is a unique fixed-point of ${\mathbf{F}}_{{A,b}}^{{(M)}}$. Moreover, by construction, a fixed-point of the mapping ${\mathbf{F}}_{{A,b}}^{{}}$ must also be a fixed-point of the mapping ${\mathbf{F}}_{{A,b}}^{{(M)}}$. Taking into account Lemma 4 the unique fixed-point of ${\mathbf{F}}_{{A,b}}^{{(M)}}$ is Y_∞ ∈ $l_{q}^{n}$. Hence, Y_∞ is the unique fixed-point of ${\mathbf{F}}_{{A,b}}^{{}}$.

Note that the representation ${\mathbf{F}}_{{A,b}}^{{(NM)}}$(O) = Y_NM is valid, where by O  : l_p → $\mathbb{R}_{r}^{n}$ denotes the zero operator, which is identified with the zero sequence (0, 0, …) ∈ $l_{q}^{n}$. Then according to Theorem 6

$${{\left\| {{{Y}_{\infty }} - {{Y}_{{NM}}}} \right\|}_{{l_{q}^{n}}}} = {{\left\| {{\mathbf{F}}_{{A,b}}^{{(M)}}({{Y}_{\infty }}) - {\mathbf{F}}_{{A,b}}^{{(M)}}\left( {{\mathbf{F}}_{{A,b}}^{{(NM - M)}}(O)} \right)} \right\|}_{{l_{{^{q}}}^{n}}}}$$

$$\leqslant \;{{\alpha }_{r}}{{\left\| {{{Y}_{\infty }} - {\mathbf{F}}_{{A,b}}^{{(NM - M)}}(O)} \right\|}_{{l_{q}^{n}}}}$$

$$\leqslant \;{{\alpha }_{r}}{{\left\| {{{Y}_{\infty }} - {\mathbf{F}}_{{A,b}}^{{(NM)}}(O)} \right\|}_{{l_{q}^{n}}}} + {{\alpha }_{r}}{{\left\| {{\mathbf{F}}_{{A,b}}^{{(NM)}}(O) - {\mathbf{F}}_{{A,b}}^{{(NM - M)}}(O)} \right\|}_{{l_{q}^{n}}}}$$

$$\leqslant \;{{\alpha }_{r}}{{\left\| {{{Y}_{\infty }} - {\mathbf{F}}_{{A,b}}^{{(NM)}}(O)} \right\|}_{{l_{q}^{n}}}} + \alpha _{r}^{N}{{\left\| {{\mathbf{F}}_{{A,b}}^{{(NM)}}(O) - O} \right\|}_{{l_{q}^{n}}}}$$

$$ = {{\alpha }_{r}}{{\left\| {{{Y}_{\infty }} - {{Y}_{{NM}}}} \right\|}_{{l_{q}^{n}}}} + \alpha _{r}^{N}{{\left\| {{{Y}_{M}}} \right\|}_{{l_{q}^{n}}}},$$

$${{\left\| {{{Y}_{\infty }} - {{Y}_{{NM}}}} \right\|}_{{l_{q}^{n}}}}\;\leqslant \;\frac{{\alpha _{r}^{N}}}{{1 - {{\alpha }_{r}}}}{{\left\| {{{Y}_{M}}} \right\|}_{{l_{q}^{n}}}}.$$

Corollary 2 is proven.

Proof of Corollary 3. The proof follows from Corollary 2 by replacing A with A^–1, b with A^–1b in conjunction with Lemma 5 and the fact that the eigenvalues of A^–1 are inverse of the eigenvalues of A [28].

Proof of Lemma 7. Let’s consider the value

$${{\rho }_{H}}(B{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty ),C{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty ))$$

$$ = \max \left\{ {\mathop {\sup }\limits_{x \in B{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{y \in C{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty )} {{{\left\| {x - y} \right\|}}_{r}};\;\mathop {\sup }\limits_{x \in C{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{y \in B{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty )} {{{\left\| {x - y} \right\|}}_{r}}} \right\}$$

$$ = \max \left\{ {\mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} {{{\left\| {B{\kern 1pt} '{\kern 1pt} u - C{\kern 1pt} '{\kern 1pt} {v}} \right\|}}_{r}};\;\mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} {{{\left\| {C{\kern 1pt} '{\kern 1pt} u - B{\kern 1pt} '{\kern 1pt} {v}} \right\|}}_{r}}} \right\},$$

$$\mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {C{\kern 1pt} '{\kern 1pt} u - B{\kern 1pt} '{\kern 1pt} {v}} \right\|}_{r}} = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {B{\kern 1pt} '{\kern 1pt} u - C{\kern 1pt} '{\kern 1pt} {v}} \right\|}_{r}}$$

$$ = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {B{\kern 1pt} '{\kern 1pt} u - C{\kern 1pt} '{\kern 1pt} u + C{\kern 1pt} '{\kern 1pt} u - C{\kern 1pt} '{\kern 1pt} {v}} \right\|}_{r}}$$

$$\leqslant \;\mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} \left( {{{{\left\| {B{\kern 1pt} '{\kern 1pt} u - C{\kern 1pt} '{\kern 1pt} u} \right\|}}_{r}} + {{{\left\| {C{\kern 1pt} '{\kern 1pt} u - C{\kern 1pt} '{\kern 1pt} {v}} \right\|}}_{r}}} \right)$$

$$ = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \left( {{{{\left\| {(B{\kern 1pt} '\; - C{\kern 1pt} '{\kern 1pt} )u} \right\|}}_{r}} + \mathop {\inf }\limits_{{v} \in {{\mathcal{E}}_{p}}(\infty )} {{{\left\| {C{\kern 1pt} '{\kern 1pt} (u - {v})} \right\|}}_{r}}} \right)$$

$$ = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {(B{\kern 1pt} '\; - C{\kern 1pt} '{\kern 1pt} )u} \right\|}_{r}} = {{\left\| {B{\kern 1pt} '\; - C{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}}.$$

Finally we get that

$${{\rho }_{H}}(B{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty ),\;C{\kern 1pt} '{\kern 1pt} {{\mathcal{E}}_{p}}(\infty ))\;\leqslant \;{{\left\| {B{\kern 1pt} '\; - C{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}}.$$

Lemma 7 is proven.

Proof of Theorem 1. The proof follows directly from Corollary 2, the representations (11) and (8), and Lemma 7.

Proof of Theorem 2. The proof follows directly from Corollary 3, the representations (12) and (9), and Lemma 7.

Proof of Theorem 3. As is known [27], for any $\mathcal{X}$, $\mathcal{Y}$ ∈ ${{\mathbb{K}}_{n}}$ satisfying the condition ρ_H($\mathcal{X}$, $\mathcal{Y}$) $\leqslant$ R, the following inclusion is true:

$$\mathcal{X} \subset \mathcal{Y} + \mathcal{B}_{R}^{r}(0).$$

From here, taking into account Theorem 1, Theorem 3 follows.

Proof of Theorem 4. The proof is similar to the proof of Theorem 3, replacing Theorem 1 with Theorem 2.

Proof of Theorem 5. (1) Item 1 follows from the definition of the operator norm and the fact that the maximum of a convex function is achieved on the boundary of the convex set [22]:

$${{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}} = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} {{\left\| {B{\kern 1pt} '{\kern 1pt} u} \right\|}_{r}} = \mathop {\sup }\limits_{\sum\limits_{k = 1}^\infty {{{{\left| {{{u}_{k}}} \right|}}^{p}} = 1} } {{\left( {\sum\limits_{i = 1}^n {{{{\left| {\sum\limits_{k = 1}^\infty {{{b}_{{ik}}}{{u}_{k}}} } \right|}}^{r}}} } \right)}^{{\frac{1}{r}}}}$$

$$ = {{\left( {\mathop {\sup }\limits_{\sum\limits_{k = 1}^\infty {{{{\left| {{{u}_{k}}} \right|}}^{p}} = 1} } \sum\limits_{i = 1}^n {{{{\left| {\sum\limits_{k = 1}^M {{{b}_{{ik}}}{{u}_{k}}} } \right|}}^{r}}} } \right)}^{{\frac{1}{r}}}} = {{\left( {\mathop {\max }\limits_{\sum\limits_{k = 1}^M {{{{\left| {{{u}_{k}}} \right|}}^{p}} = 1} } \sum\limits_{i = 1}^n {{{{\left| {\sum\limits_{k = 1}^M {{{b}_{{ik}}}{{u}_{k}}} } \right|}}^{r}}} } \right)}^{{\frac{1}{r}}}}.$$

(2) By virtue of Hölder's inequality, item 2 follows from item 1:

$${{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}}\;\leqslant \;{{\left( {\mathop {\max }\limits_{\sum\limits_{k = 1}^M {{{{\left| {{{u}_{k}}} \right|}}^{p}} = 1} } \sum\limits_{i = 1}^n {{{{\left[ {{{{\left( {\sum\limits_{k = 1}^M {{{{\left| {{{b}_{{ik}}}} \right|}}^{q}}} } \right)}}^{{\frac{1}{q}}}}{{{\left( {\sum\limits_{k = 1}^M {{{{\left| {{{u}_{k}}} \right|}}^{p}}} } \right)}}^{{\frac{1}{p}}}}} \right]}}^{r}}} } \right)}^{{\frac{1}{r}}}} = {{\left( {\sum\limits_{i = 1}^n {{{{\left( {\sum\limits_{k = 1}^M {{{{\left| {{{b}_{{ik}}}} \right|}}^{q}}} } \right)}}^{{\frac{r}{q}}}}} } \right)}^{{\frac{1}{r}}}}.$$

(3) Also, for M = 1, item 3 follows from item 1:

$${{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}} = {{\left( {\mathop {\max }\limits_{{{{\left| {{{u}_{1}}} \right|}}^{p}} = 1} \sum\limits_{i = 1}^n {{{{\left| {{{b}_{{i1}}}{{u}_{1}}} \right|}}^{r}}} } \right)}^{{\frac{1}{r}}}} = {{\left( {\sum\limits_{i = 1}^n {{{{\left| {{{b}_{{i1}}}} \right|}}^{r}}} } \right)}^{{\frac{1}{r}}}}.$$

(4) For r = p = 2 the operator norm can be represented in terms of the scalar product in $\mathbb{R}_{2}^{n}$:

$$(x,y) = \sum\limits_{i = 1}^n {{{x}_{i}}{{y}_{i}}.} $$

Then, taking into account item 1, the following representation is true:

$${{\left( {{{{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}}_{{l_{q}^{n}}}}} \right)}^{2}} = \mathop {\max }\limits_{\substack{ u \in {{\mathbb{R}}^{M}}: \\ (u,u) = 1 } } (Bu,Bu) = \mathop {\max }\limits_{\substack{ u \in {{\mathbb{R}}^{M}}: \\ (u,u) = 1 } } (u,{{B}^{{\text{T}}}}Bu).$$

Due to the Lagrange multiplier method [29], the maximum point of the optimization problem under consideration u* ∈ ${{\mathbb{R}}^{M}}$ satisfies the following conditions:

$$\left\{ \begin{gathered} \nabla \left( {(u,{{B}^{{\text{T}}}}Bu) + \lambda (1 - (u,u))} \right) = 0, \hfill \\ (u,u) = 1, \hfill \\ \end{gathered} \right.\;\;\left\{ \begin{gathered} 2{{B}^{{\text{T}}}}B - 2\lambda u = 0, \hfill \\ (u,u) = 1, \hfill \\ \end{gathered} \right.\;\;\left\{ \begin{gathered} ({{B}^{{\text{T}}}}B - \lambda I)u = 0, \hfill \\ (u,u) = 1. \hfill \\ \end{gathered} \right.$$

Then, by definition, u* is a normed eigenvector of the matrix B^TB corresponding to the eigenvalue λ*, i.e.

$${{\left( {{{{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}}_{{l_{q}^{n}}}}} \right)}^{2}} = (u^*,{{B}^{{\text{T}}}}Bu^*) = (u^*,\lambda {\kern 1pt} ^*{\kern 1pt} u^*) = \lambda {\kern 1pt} ^*{\kern 1pt} .$$

Item 4 is completely proven.

(5) Item 5 follows from the representation of the operator norm B ' and the Riesz theorem on the norm of a linear and bounded functional in l_p [25]:

$${{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}} = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \mathop {\max }\limits_{i = \overline {1,n} } \left| {\sum\limits_{k = 1}^\infty {{{b}_{{ik}}}{{u}_{k}}} } \right| = \mathop {\max }\limits_{i = \overline {1,n} } {{\left( {\sum\limits_{k = 1}^\infty {{{{\left| {{{b}_{{ik}}}} \right|}}^{q}}} } \right)}^{{\frac{1}{q}}}} = \mathop {\max }\limits_{i = \overline {1,n} } {{\left( {\sum\limits_{k = 1}^M {{{{\left| {{{b}_{{ik}}}} \right|}}^{q}}} } \right)}^{{\frac{1}{q}}}}.$$

(6) To prove item 6, we take into account the representation |γ| = max{γ, –γ} for any γ ∈ $\mathbb{R}$ and consider the following chain of equalities:

$${{\left\| {B{\kern 1pt} '{\kern 1pt} } \right\|}_{{l_{q}^{n}}}} = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \sum\limits_{i = 1}^n {\left| {\sum\limits_{k = 1}^\infty {{{b}_{{ik}}}{{u}_{k}}} } \right|} = \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \sum\limits_{i = 1}^n {\mathop {\max }\limits_{{{\gamma }_{i}} \in \{ - 1;1\} } } \left( {{{\gamma }_{i}}\sum\limits_{k = 1}^\infty {{{b}_{{ik}}}{{u}_{k}}} } \right)$$

$$ = \mathop {\max }\limits_{\substack{ {{\gamma }_{i}} \in \{ - 1;1\} \\ i = \overline {1,n} }} \mathop {\sup }\limits_{u \in {{\mathcal{E}}_{p}}(\infty )} \left| {\sum\limits_{k = 1}^\infty {\left( {\sum\limits_{i = 1}^n {{{\gamma }_{i}}{{b}_{{ik}}}} } \right){{u}_{k}}} } \right| = \mathop {\max }\limits_{\substack{ {{\gamma }_{i}} \in \{ - 1;1\} \\ i = \overline {1,n} }} {{\left( {\sum\limits_{k = 1}^\infty {{{{\left| {\sum\limits_{i = 1}^n {{{\gamma }_{i}}{{b}_{{ik}}}} } \right|}}^{q}}} } \right)}^{{\frac{1}{q}}}}$$

$$ = \mathop {\max }\limits_{\substack{ {{\gamma }_{i}} \in \{ - 1;1\} \\ i = \overline {1,n} }} {{\left( {\sum\limits_{k = 1}^M {{{{\left| {\sum\limits_{i = 1}^n {{{\gamma }_{i}}{{b}_{{ik}}}} } \right|}}^{q}}} } \right)}^{{\frac{1}{q}}}}.$$

Theorem 5 is completely proven.