Multidimensional spectral order for selfadjoint operators

TL;DR

This paper extends the spectral order concept to multidimensional settings for commuting selfadjoint operators, showing its preservation under spectral integral transformations and characterizing the order via operator inequalities.

Contribution

It introduces a multidimensional spectral order for families of commuting selfadjoint operators and explores its properties and characterizations.

Findings

Multidimensional spectral order is preserved by spectral integral transformations.

The spectral order for positive tuples is characterized by operator inequalities.

The multidimensional spectral order generalizes the product of one-dimensional spectral orders.

Abstract

The aim of this paper is to extend the notion of the spectral order for finite families of pairwise commuting bounded and unbounded selfadjoint operators in Hilbert space. It is shown that the multidimensional spectral order $\preccurlyeq$ is preserved by transformations represented by spectral integrals of separately increasing Borel functions on $\mathbb{R}^\kappa$. In particular, the $\kappa$-dimensional spectral order is the restriction of product of $\kappa$ spectral orders for selfadjoint operators. If $\mathbf{A}$ and $\mathbf{B}$ are positive $\kappa$-tuples of pairwise commuting selfadjoint operators, then relation $\mathbf{A}\preccurlyeq\mathbf{B}$ holds if and only if $\mathbf{A}^\alpha\leqslant \mathbf{B}^\alpha$ for every $\alpha\in\mathbb{Z}_+^\kappa$.

Equations

$$A={\int }_{\mathbb{R}}xE(\mathrm{d}x).$$

$$A_j={\int }_{{\mathbb{R}}^{\kappa }}{x}_{j}E(\mathrm{d}x),j=1,\dots ,\kappa ,$$

$$E_{A_k}({\Delta }_1)E_{A_j}({\Delta }_2)=E_{A_j}({\Delta }_2)E_{A_k}({\Delta }_1)$$

$$\mathscr{B}_{a}(A)=\bigcup_{c\in(0,\infty)}\;\bigcap_{n=0}^{\infty}\Big{\{}h\in{\mathscr{D}}^{\infty}(A)\colon\|A^{n}h\|\leqslant ca^{n}\Big{\}},\quad a\in\mathbb{R}_{+}.$$

$$\bigvee\big{\{}P_{M_{j}}\colon j\in J\big{\}}=P_{\,\bigvee_{j\in J}M_{j}}\quad\text{and}\quad\bigwedge\big{\{}P_{M_{j}}\colon j\in J\big{\}}=P_{\bigcap_{j\in J}M_{j}}.$$

$$E(\sigma )=\bigvee \{E(\tau ):\tau \subseteq \sigma ,\tau \text{ is compact in }X\}\text{ for every }\sigma \in \mathfrak{B}(X).$$

$$\bigvee \{E(\tau ):\tau \subseteq \sigma ,\tau \text{ is compact}\}⩽E(\sigma ).$$

$$|⟨h,E(\sigma )g⟩|$$

$$\displaystyle=\sup\big{\{}\langle E(\tau)h,h\rangle\colon\tau\subseteq\sigma,\tau\textmd{ is compact}\big{\}}\cdot\|g\|=0.$$

$$\phi (A)={\int }_{\mathbb{R}}\phi (x)E_A(\mathrm{d}x).$$

$$\mathcal{D}(M_{\phi }):=\{h\in L^2(X,\mu ):{\int }_X|\phi h{|}^2\text{d}\mu <\infty \},$$

$$M_{\phi }h:=\phi h,h\in \mathcal{D}(M_{\phi }).$$

$$E_{M_{\phi }}(\sigma )h={\chi }_{{\phi }^{-1}(\sigma )}h$$

$$e_{j}=(0,\ldots,0,\underbrace{1}\limits_{j},0,\ldots,0),\quad j=1,\ldots,\kappa.$$

$$(a,b]$$

$$[a,b]$$

$$(-\infty ,x]$$

$$\{y\in X:y⪯x\}\subseteq S,x\in S.$$

$$\downarrow\varOmega:=\bigcap\big{\{}S\in L(X,\preceq)\colon\varOmega\subseteq S\big{\}},$$

$$x=\underset{k\to \infty }{\lim }{x}_{n_k}{⩽}_{\iota }\underset{k\to \infty }{\lim }{y}_{n_k}=y.$$

$$x{⩽}_{\iota }y\Rightarrow \phi (x)⩽\phi (y)$$

$$\{\phi ⩽y\}\in L({\mathbb{R}}^{{\kappa }_1},{⩽}_{\iota })$$

$$d_{\Omega }(x)=\inf \{\Vert x-y\Vert :y\in \Omega \},x\in {\mathbb{R}}^{\kappa },$$

$$\mu (\Omega )=\sup \{\mu (D):D\subseteq \Omega ,D=\overline{D}\in L({\mathbb{R}}^{\kappa },{\le }_{\iota })\}.$$

$$\mu (\Omega )$$

$$⩽\sup \{\mu ({\downarrow }_{\iota }K):K\subseteq \Omega ,K\text{ is compact}\}$$

$$⩽\sup \{\mu (D):D\subseteq \Omega ,D=\overline{D}\in L({\mathbb{R}}^{\kappa },{\le }_{\iota })\}⩽\mu (\Omega ).$$

$${\nu }_1(\mathbb{R})={\mu }_1({\mathbb{R}}^{\kappa })={\mu }_2({\mathbb{R}}^{\kappa })={\nu }_2(\mathbb{R}).$$

$${\nu }_2((-\infty ,x])={\mu }_2(f^{-1}(-\infty ,x]))⩽{\mu }_1(f^{-1}(-\infty ,x]))={\nu }_1((-\infty ,x])$$

$${\int }_{\mathbb{R}}|x|\mathrm{d}{\nu }_2(x)={\int }_{{\mathbb{R}}^{\kappa }}|f|\mathrm{d}{\mu }_2<\infty .$$

$${\int }_{{\mathbb{R}}^{\kappa }}f\mathrm{d}{\mu }_1={\int }_{\mathbb{R}}x\mathrm{d}{\nu }_1(x)⩽{\int }_{\mathbb{R}}x\mathrm{d}{\nu }_2(x)={\int }_{{\mathbb{R}}^{\kappa }}f\mathrm{d}{\mu }_2.$$

$${\mu }_1({\mathbb{R}}^{\kappa }\backslash \Omega )={\int }_{{\mathbb{R}}^{\kappa }}{\chi }_{{\mathbb{R}}^{\kappa }\backslash \Omega }\mathrm{d}{\mu }_1\stackrel{\text{(ii)}}{⩽}{\int }_{{\mathbb{R}}^{\kappa }}{\chi }_{{\mathbb{R}}^{\kappa }\backslash \Omega }\mathrm{d}{\mu }_2={\mu }_2({\mathbb{R}}^{\kappa }\backslash \Omega ).$$

$${\mu }_2(D)⩽{\mu }_1(D),$$

Full text

The author was supported by the Ministry of Science and Higher Education of the Republic of Poland.

1. Introduction

Let ${\mathcal{H}}$ be a complex Hilbert space. Denote by $\mathbf{B}_{s}({\mathcal{H}})$ the set of all bounded selfadjoint operators on ${\mathcal{H}}$. If $A,B\in\mathbf{\mathbf{B}}_{s}({\mathcal{H}})$, then we write $A\leqslant B$ whenever $\langle Ah,h\rangle\leqslant\langle Bh,h\rangle$ for every $h\in{\mathcal{H}}$.

In [28] Sherman proved that $\mathcal{A}_{s}$ the set of all selfadjoint elements of a $C^{*}$-algebra $\mathcal{A}$ of bounded linear operators on ${\mathcal{H}}$ is a lattice with respect to $\leqslant$ if and only if $\mathcal{\mathcal{A}}$ is commutative. In fact, the more noncommutative $\mathcal{A}$ is the less lattice structure $\mathcal{A}_{s}$ has. The result of Kadison [18] shows that partially ordered set $(\mathbf{\mathbf{B}}_{s}({\mathcal{H}}),\leqslant)$ is an anti-lattice. This means that for any $A,B\in\mathbf{\mathbf{B}}_{s}({\mathcal{H}})$, the greatest lower bound of the set $\{A,B\}$ exists if and only if $A\leqslant B$ or $B\leqslant A$.

These results motivated Olson to introduce spectral order $\preccurlyeq$ for bounded selfadjoint operators [23]. One of the main result in [23] says that $(\mathbf{\mathbf{B}}_{s}({\mathcal{H}}),\preccurlyeq)$ is a conditionally complete lattice. It should also be mentioned that in contrast to the classical order $\leqslant$, the spectral order is not a vector order. Spectral order was considered in the context of matrix theory and von Neumann algebras (see [19, 2, 3, 1]). The spectral order also enabled to solve Dirichlet problem for operator-valued harmonic function in [14].

The spectral order has natural interpretation in the mathematical description of quantum mechanics (see [16]). This fact has led to increased interest in automorphisms of the spectral order. For instance, automorphisms of partially ordered set $(\mathcal{E}(H),\preccurlyeq)$, where $\mathcal{E}({\mathcal{H}})$ denote the set of all positive bounded operators in the closed unit ball of $\mathbf{B}({\mathcal{H}})$, were described in [22, 21]. In turn, automorphisms of subsets of positive selfadjoint operators were investigated in [30, 17]. Spectral order is also an important ingredient of the topos formulation of quantum theory (see for example [9, 10, 31, 12]).

In the previous paper [24] (see also [17, 30]) the spectral order was investigated in the context of unbounded selfadjoint operators. The aim of this paper is to extend the notion of the spectral order also for finite families of pairwise commuting bounded and unbounded selfadjoint operators. We start with the definition of the spectral order for selfadjoint operators. Let $A$ and $B$ be selfadjoint operators in ${\mathcal{H}}$. If $E_{A}$ and $E_{B}$ are spectral measures of $A$ and $B$ defined on Borel subsets of $\mathbb{R}$, then $A\preccurlyeq B$ if and only if $E_{B}((-\infty,x])\leqslant E_{A}((-\infty,x])$ for every $x\in\mathbb{R}$. As we see, the definition of spectral order depends on the fact that for every selfadjoint operator $A$, there exists a unique spectral measure $E$ on the $\sigma$-algebra of all Borel subsets of real line satisfying the condition

[TABLE]

If $\mathbf{A}=(A_{1},\ldots,A_{\kappa})$ is a $\kappa$-tuple of selfadjoint operators, $\kappa\in\mathbb{N}$, then similar representation

[TABLE]

holds for some Borel spectral measure $E$ on $\mathbb{R}^{\kappa}$ if and only if $A_{1},\ldots,A_{\kappa}$ pairwise commute. The set of all $\kappa$-tuples of pairwise commuting selfadjoint operators in ${\mathcal{H}}$ will be denoted by $\mathcal{S}_{c}({\mathcal{H}},\kappa)$. By definition, $\mathbf{A}=(A_{1},\ldots,A_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ if spectral measures $E_{A_{1}},\ldots,E_{A_{\kappa}}$ commute, i.e.

[TABLE]

for every Borel sets $\Delta_{1},\Delta_{2}\subset\mathbb{R}$ and $k,j\in\{1,\ldots,\kappa\}$. In fact, condition (1.1) determine the unique measure $E=E_{\mathbf{A}}$, which allow us to consider the multivariable spectral resolution $F_{\mathbf{A}}(x_{1},\ldots,x_{\kappa})=E_{\mathbf{A}}((-\infty,x_{1}]\times\ldots\times(-\infty,x_{\kappa}])$ for $x_{1},\ldots,x_{\kappa}\in\mathbb{R}$. In particular, if $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, then they can be ordered by $\mathbf{A}\preccurlyeq\mathbf{B}$ if $F_{\mathbf{B}}(x)\leqslant F_{\mathbf{A}}(x)$ for every $x\in\mathbb{R}^{\kappa}$. This is the first argument for the choice of $\mathcal{S}_{c}({\mathcal{H}},\kappa)$.

The second argument comes from quantum theory. Let $o_{1},\ldots,o_{\kappa}$ and $p_{1},\ldots,p_{\kappa}$ be two systems of commensurable observables represented by $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, respectively. Consider unit vector $h$ in ${\mathcal{H}}$, which is interpreted as the state of the system. Then the pairs $(h,\mathbf{A})$ and $(h,\mathbf{B})$ determine random vectors $X=(X_{1},\ldots,X_{\kappa})$ and $Y=(Y_{1},\ldots,Y_{\kappa})$, respectively. In particular, the distribution functions of $X$ and $Y$ are given by $F_{X}(x)=\langle F_{\mathbf{A}}(x)h,h\rangle$ and $F_{Y}(x)=\langle F_{\mathbf{B}}(x)h,h\rangle$, $x\in\mathbb{R}^{\kappa}$. Concluding, the relation $\mathbf{A}\preccurlyeq\mathbf{B}$ means that the corresponding distribution functions $F_{X}$ and $F_{Y}$ are pointwise ordered. For more details on quantum theory, we refer reader to [7].

In Section 5 we investigate fundamental properties of multidimensional spectral order. The main results state that the multidimensional spectral order is preserved by transformation represented by spectral integrals of separately increasing Borel functions on $\mathbb{R}^{\kappa}$ (see Theorem 5.6 and Theorem 5.7), which extends the known results for spectral order (confer [24, Proposition 6.2.],[22]). In particular, we obtain equivalent definition of multidimensional spectral order in Corollary 5.8, which is a counterpart of [13, Theorem 1.]. As the consequence of Theorem 5.6 we derive some information when the spectral order behaves like a vector order in Corollaries 5.12 and 5.13. According to Theorem 5.7 it is also possible to extend the multidimensional spectral order to $\mathcal{S}({\mathcal{H}},\kappa)$ the set of all $\kappa$-tuples of selfadjoint operators in ${\mathcal{H}}$ as a product order. It means that $\mathbf{A}\preccurlyeq\mathbf{B}$ if and only if $A_{j}\preccurlyeq B_{j}$ for every $j=1,\ldots,\kappa$ whenever $\mathbf{A},\mathbf{B}\in\mathcal{S}({\mathcal{H}},\kappa)$. Note that $(\mathcal{S}({\mathcal{H}},\kappa),\preccurlyeq)$ is a conditionally complete lattice since $(\mathcal{S}({\mathcal{H}},1),\preccurlyeq)$ is a conditionally complete lattice by [24, Corollary 5.4.]. In contrast to $(\mathcal{S}({\mathcal{H}},\kappa),\preccurlyeq)$, partially ordered set $(\mathcal{S}_{c}({\mathcal{H}},\kappa),\preccurlyeq)$ is not even a lattice (see Example 5.11). It should also be pointed out that in the case $\kappa=2$ the multidimensional spectral order on $\mathcal{S}_{c}({\mathcal{H}},2)$ can be identified with the spectral order for normal operators (see Example 5.14 and Proposition 5.15). This order was considered by Brenna and Flori in [8] in the context of the daseinisation of normal operators.

In Section 7 we discuss the relation between domains of monomials $\mathbf{A}^{\alpha}$ and $\mathbf{B}^{\alpha}$ for $\alpha\in\mathbb{Z}_{+}^{\kappa}$ provided that $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and $\mathbf{A}\preccurlyeq\mathbf{B}$. The role of operators $\mathbf{A}^{\alpha}$ and $\mathbf{B}^{\alpha}$ becomes more important if we consider positive $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. This question is studied in Section 8. One of the main result of this Section states that $\mathbf{A}\preccurlyeq\mathbf{B}$ if and only if $\mathbf{A}^{\alpha}\leqslant\mathbf{B}^{\alpha}$ for every $\alpha\in\mathbb{Z}_{+}^{\kappa}$. This generalize [23, Theorem 3.], which says that $A\preccurlyeq B$ if and only if $A^{n}\leqslant B^{n}$ for every $n=1,2,\ldots$, where $A$ and $B$ are bounded positive selfadjoint operators on ${\mathcal{H}}$.

2. Prerequisites

Denote by $\mathbb{Z}_{+}$, $\mathbb{N}$, $\mathbb{Q}$, $\mathbb{R}_{+}$, $\mathbb{R}$, and $\mathbb{C}$ the sets of nonnegative integers, positive integers, rational numbers, nonnegative real numbers, real numbers, and complex numbers, respectively. Set $\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty\}\cup\{\infty\}$. As usual, $\chi_{\sigma}$ stands for the characteristic function of a set $\sigma$ (it is clear from the context which set is the domain of definition of $\chi_{\sigma}$).

In what follows, ${\mathcal{H}}$ denotes a complex Hilbert space. By an operator in ${\mathcal{H}}$ we understand a linear mapping $A\colon{\mathcal{H}}\supseteq{\mathscr{D}}(A)\rightarrow{\mathcal{H}}$ defined on a linear subspace ${\mathscr{D}}(A)$ of ${\mathcal{H}}$, called the domain of $A$. Let $A$ be an operator in ${\mathcal{H}}$. We write ${\mathscr{N}}(A)$, ${\mathscr{R}}(A)$, $A^{*}$ and $\bar{A}$ for the kernel, the range, the adjoint and the closure of $A$, respectively (in case they exist). Set ${\mathscr{D}}^{\infty}(A)=\bigcap_{n=1}^{\infty}{\mathscr{D}}(A^{n})$ and

[TABLE]

Call a member of $\mathscr{B}(A):=\bigcup_{a\in\mathbb{R}_{+}}\mathscr{B}_{a}(A)$ a bounded vector of $A$ (cf. [11]). We say that an operator $A$ in ${\mathcal{H}}$ is positive if $\langle Ah,h\rangle\geqslant 0$ for every $h\in{\mathscr{D}}(A)$. A densely defined operator $A$ is said to be selfadjoint if $A=A^{*}$. If $A$ is a positive selfadjoint operator in ${\mathcal{H}}$, then its square root is denoted by $A^{1/2}$.

Denote by $\mathbf{B}({\mathcal{H}})$ the $C^{*}$-algebra of all bounded operators $A$ in ${\mathcal{H}}$ such that ${\mathscr{D}}(A)={\mathcal{H}}$. We write $I=I_{\mathcal{H}}$ for the identity operator on ${\mathcal{H}}$. It is well-known that the set $\mathbf{P}({\mathcal{H}})$ of all orthogonal projections on ${\mathcal{H}}$ equipped with the above-defined partial order “$\leqslant$” is a complete lattice. If $\mathcal{A}\subseteq\mathbf{P}({\mathcal{H}})$, then $\bigvee\mathcal{A}$ (respectively, $\bigwedge\mathcal{A}$) stands for the supremum (respectively, the infimum) of $\mathcal{A}$. Denote by $P_{M}$ the orthogonal projection of ${\mathcal{H}}$ onto a closed linear subspace $M$ of ${\mathcal{H}}$. If $\{M_{j}\}_{j\in J}$ is a family of subsets of ${\mathcal{H}}$, then $\bigvee_{j\in J}M_{j}$ stands for the smallest closed linear subspace of ${\mathcal{H}}$ containing $\bigcup_{j\in J}M_{j}$. Note that if each $M_{j}$ is a closed linear subspaces of ${\mathcal{H}}$, then

[TABLE]

We shall use the symbol “$\textsc{sot-}\lim$” to represent the limit in the strong operator topology on $\mathbf{B}({\mathcal{H}})$.

Let $E\colon{\mathfrak{B}}(X)\to\mathbf{B}({\mathcal{H}})$ be a spectral measure defined on the $\sigma$-algebra ${\mathfrak{B}}(X)$ of all Borel subsets of a topological space $X$. A set $Y\subseteq X$ is called a support of $E$, if $Y$ is the least (with respect to set inclusion) closed subset of $X$ such that $E(X\backslash Y)=0$. The support of $E$ is denoted by $\operatorname{supp}E$. It is known that every spectral measure $E$ on ${\mathfrak{B}}(X)$, where $X$ is a separable complete metric space, has closed support (cf. [5, page 129]). We say that $E$ is regular if

[TABLE]

For the sake of completeness, we prove the following Proposition (see also [20, Exercise 6.8.]).

Proposition 2.1**.**

If $X$ is a separable complete metric space and $E$ is a spectral measure on ${\mathfrak{B}}(X)$, then $E$ is regular.

Proof.

Let $\sigma\in{\mathfrak{B}}(X)$. It is evident that

[TABLE]

To establish the opposite inequality, take $h\in{\mathcal{H}}$ such that $h\perp E(\tau){\mathcal{H}}$ for every compact set $\tau\subseteq\sigma$. Applying [5, Subsection 1.3.22] we get that $\mu_{h}(\cdot):=\langle E(\cdot)h,h\rangle$ is an inner regular Borel measure (cf. [25]). Choose $g\in{\mathcal{H}}$. Then

[TABLE]

Thus $h\perp E(\sigma){\mathcal{H}}$, since $g$ was chosen arbitrarily. This completes the proof. ∎

Let us recall that $\operatorname{supp}E_{A}=\sigma(A)$ for every selfadjoint operator $A$ in ${\mathcal{H}}$, where $\sigma(A)$ denotes the spectrum of $A$ and $E_{A}$ is the spectral measure of $A$. In particular, a selfadjoint operator $A$ is positive if and only if $\operatorname{supp}E_{A}\subseteq[0,\infty)$ (cf. [5]). If $\varphi\colon\mathbb{R}\to\mathbb{R}$ is a Borel function, then we set

[TABLE]

The operator $\varphi(A)$ is selfadjoint. Moreover, if $\varphi\geqslant 0$ a.e. $[E]$, then $\varphi(A)$ is positive. For more information concerning Stone-von Neumann operator calculus $\varphi\mapsto\varphi(A)$, we refer the reader to [5, 27].

Given a positive selfadjoint operator $A$ in ${\mathcal{H}}$ and $s\in\mathbb{R}_{+}$, we define $A^{s}=\psi_{s}(A)$, where $\psi_{s}(x)=|x|^{s}\chi_{[0,\infty)}(x)$ for $x\in\mathbb{R}$ (with the convention that $0^{0}=1$). This definition agrees with the usual one for nonnegative integer exponents. If $A_{1}$ and $A_{2}$ are positive selfadjoint operators in ${\mathcal{H}}$ such that ${\mathscr{D}}(A_{2}^{1/2})\subseteq{\mathscr{D}}(A_{1}^{1/2})$ and $\|A_{1}^{1/2}h\|\leqslant\|A_{2}^{1/2}h\|$ for every $h\in{\mathscr{D}}(A_{2}^{1/2})$, then we write $A_{1}\leqslant A_{2}$ (cf. [27]). The last definition is easily seen to be consistent with that for bounded operators.

Finally, let $(X,\mathcal{A},\mu)$ be a measure space and $\varphi\colon X\to\mathbb{R}$ be an $\mathcal{A}$-measurable function. We denote by $M_{\varphi}$ the multiplication operator by $\varphi$ in $L^{2}(X,\mu)$, which is defined as follows:

[TABLE]

It is known that $M_{\varphi}$ is selfadjoint and

[TABLE]

for every $\sigma\in{\mathfrak{B}}(\mathbb{R})$ and $h\in L^{2}(X,\mu)$ (see [27, Example 5.3.]).

3. Separately increasing functions

Here we collect some facts about separately increasing functions we need in this paper.

Fix $\kappa\in\mathbb{N}$. In what follows, $\mathbb{R}^{\kappa}$ is regarded as the $\kappa$-dimensional Euclidean space. Denote by $\{e_{j}\}_{j=1}^{\kappa}$ the standard orthonormal basis of $\mathbb{R}^{\kappa}$, i.e.

[TABLE]

Throughout this paper, we adhere to the usual convention that if $a\in\overline{\mathbb{R}}^{\kappa}$, then the jth coordinate of $a$ is denoted by $a_{j}$; thus $a=(a_{1},\ldots,a_{\kappa})$. Let $a\in\overline{\mathbb{R}}^{\kappa}$ and $b\in\overline{\mathbb{R}}^{\kappa}$. We write $a\leqslant b$ (resp., $a<b$) if $a_{j}\leqslant b_{j}$ (resp., $a_{j}<b_{j}$) for all $j=1,\ldots,\kappa$. Set

[TABLE]

Similarly, we define $(-\infty,x)$, $[x,\infty)$ and $(x,\infty)$. Given $\iota\in\{1,\ldots,\kappa\}$, $a\in\mathbb{R}^{\kappa}$ and $b\in\mathbb{R}^{\kappa}$, we write $a\leqslant_{\iota}b$ if $a_{j}\leqslant b_{j}$ for $j=1,\ldots,\iota$ and $a_{j}=b_{j}$ for $j>\iota$. Note that $(\overline{\mathbb{R}}^{\kappa},\leqslant)$ and $(\mathbb{R}^{\kappa},\leqslant_{\iota})$ are partially ordered sets.

Let $(X,\preceq)$ be a partially ordered set. A set $S\subseteq X$ is called a lower set in $X$ if

[TABLE]

The collection of all lower sets in $X$ is denoted by $L(X,\preceq)$. Note that $\bigcap\Gamma\in L(X,\preceq)$ for any collection $\Gamma\subseteq L(X,\preceq)$. Hence, if $\varOmega\subseteq X$, then the following set

[TABLE]

is the smallest lower set in $X$ containing $\varOmega$. The following properties of lower sets are easily seen to be true.

Proposition 3.1**.**

If $(X,\preceq)$ is a partially ordered set and $\varOmega,\varOmega_{1},\varOmega_{2}\subseteq X$, then

1. (i)

$\downarrow\varOmega=\bigcup_{x\in\varOmega}\{y\in X\colon y\preceq x\}$, 2. (ii)

if $\varOmega_{1}\subseteq\varOmega_{2}$, then $\downarrow\varOmega_{1}\subseteq\;\downarrow\varOmega_{2}$.

In the case of the partially ordered set $(\mathbb{R}^{\kappa},\leqslant_{\iota})$, we write $\downarrow_{\iota}S$ in place of $\downarrow S$. The following fact is of some importance in this paper.

Proposition 3.2**.**

If $\varOmega\subseteq\mathbb{R}^{\kappa}$ is compact, then $\downarrow_{\iota}\varOmega$ is closed for every $\iota\in\{1,\ldots,\kappa\}$.

Proof.

Fix $x\in\overline{\downarrow_{\iota}\varOmega}$. Then there exists a sequence $\{x_{n}\}_{n=1}^{\infty}\subseteq\;\downarrow_{\iota}\varOmega$ such that $x_{n}\to x$ as $n\to\infty$. By Proposition 3.1(i), we can find a sequence $\{y_{n}\}_{n=1}^{\infty}\subseteq\varOmega$ such that $x_{n}\leqslant_{\iota}y_{n}$ for all $n\in\mathbb{N}$. By the compactness of $\varOmega$, there exists a subsequence $\{y_{n_{k}}\}_{k=1}^{\infty}$ of $\{y_{n}\}_{n=1}^{\infty}$ which converges to some $y\in\varOmega$. In particular, we have

[TABLE]

Applying Proposition 3.1(i) again, we see that $x\in\downarrow_{\iota}\varOmega$. ∎

Remark 3.3*.*

First, we note that if $\varOmega\neq\varnothing$, then the set $\downarrow_{\iota}\varOmega$ is always unbounded (see Proposition 3.1(i)). It is also worth pointing out that the assumption about compactness of $\varOmega$ in the Proposition 3.2 is essential. Indeed, if $\kappa=2$ and $\varOmega=\{(-\frac{1}{n},n)\colon n\in\mathbb{N}\}$, then by Proposition 3.1(i), $\downarrow_{2}\varOmega=(-\infty,0)\times\mathbb{R}$. This means that $\downarrow_{2}\varOmega$ is not closed, though $\varOmega$ is.

The following terminology will be frequently used in this paper.

Definition 3.4**.**

Let $\kappa_{1},\kappa_{2}\in\mathbb{N}$, $\iota\in\{1,\ldots,\kappa_{1}\}$ and $\varOmega\subseteq\mathbb{R}^{\kappa_{1}}$. We say that a function $\varphi\colon\varOmega\rightarrow\mathbb{R}^{\kappa_{2}}$ is $\iota$-increasing if

[TABLE]

The function $\varphi$ is called increasing if it is $\kappa_{1}$-increasing.

As shown below, lower sets can be used to characterize increasing functions.

Proposition 3.5**.**

Let $\kappa_{1},\kappa_{2}\in\mathbb{N}$ and $\iota\in\{1,\ldots,\kappa_{1}\}$. Then a function $\varphi\colon\mathbb{R}^{\kappa_{1}}\rightarrow\mathbb{R}^{\kappa_{2}}$ is $\iota$-increasing if and only if

[TABLE]

where $\{\varphi\leqslant y\}:=\{x\in\mathbb{R}^{\kappa_{1}}\colon\varphi(x)\leqslant y\}$.

Proof.

It is enough to prove the sufficiency. Take $x_{1},x_{2}\in\mathbb{R}^{\kappa_{1}}$ such that $x_{1}\leqslant_{\iota}x_{2}$. Set $\varOmega=\{\varphi\leqslant\varphi(x_{2})\}$. Since $\varOmega\in L(\mathbb{R}^{\kappa_{1}},\leqslant_{\iota})$ and $x_{2}\in\varOmega$, we see that $x_{1}\in\varOmega$. Thus $\varphi(x_{1})\leqslant\varphi(x_{2})$. ∎

Now we prove that the distance function to a $\iota$-lower set is $\iota$-increasing.

Proposition 3.6**.**

Let $\kappa\in\mathbb{N}$ and $\iota\in\{1,\ldots,\kappa\}$. Suppose $\|\cdot\|$ is a norm on $\mathbb{R}^{\kappa}$ and $\varOmega\in L(\mathbb{R}^{\kappa},\leqslant_{\iota})$. If $\varOmega\neq\varnothing$, then

1. (i)

the function $d_{\varOmega}\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ defined by

[TABLE]

is $\iota$-increasing, 2. (ii)

$\varOmega^{(\varepsilon)}:=\{d_{\varOmega}\leqslant\varepsilon\}\in L(\mathbb{R}^{\kappa},\leqslant_{\iota})$* for every $\varepsilon\in(0,\infty)$.*

Proof.

(i) Let $x,y\in\mathbb{R}^{\kappa}$ be such that $x\leqslant_{\iota}y$. Fix $\varepsilon\in(0,\infty)$ and take $y_{\varepsilon}\in\varOmega$ such that $\|y-y_{\varepsilon}\|<d_{\varOmega}(y)+\varepsilon$. Then $x_{\varepsilon}:=y_{\varepsilon}+x-y\leqslant_{\iota}y_{\varepsilon}$, which implies that $x_{\varepsilon}\in\varOmega$ (because $\varOmega\in L(\mathbb{R}^{\kappa},\leqslant_{\iota})$). Since $\|x-x_{\varepsilon}\|=\|y-y_{\varepsilon}\|<d_{\varOmega}(y)+\varepsilon$, we deduce that $d_{\varOmega}(x)<d_{\varOmega}(y)+\varepsilon$. This yields $d_{\varOmega}(x)\leqslant d_{\varOmega}(y)$.

(ii) This follows from (i) and Proposition 3.5. ∎

We conclude this section with an example showing that increasing function $f\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ may not be Borel (this is possible only for $\kappa\geqslant 2$, see [24, Lemma 6.1.]).

Example** 3.7****.**

Let $V$ be a subset of $\mathbb{R}$ which is not Borel measurable (cf. [25, Corollary, p. 53]). Set $S=\{(x,y)\in\mathbb{R}^{2}\colon x+y=0\}$, $S_{1}=\{(x,y)\in\mathbb{R}^{2}\colon x+y>0\}$ and $S_{2}=S\cap(V\times\mathbb{R})$. It is easily seen that the function $f:=\chi_{S_{1}\cup S_{2}}$ is increasing. If $f$ were a Borel function, then the set $f^{-1}(\{1\})=S_{1}\cup S_{2}$ would be a Borel subset of $\mathbb{R}^{2}$. This and the fact that $S_{1}$ is an open set in $\mathbb{R}^{2}$ which is disjoint from $S_{2}$ would imply that $S_{2}\in{\mathfrak{B}}(\mathbb{R}^{2})$. Since the function $\varphi\colon\mathbb{R}\ni x\rightarrow(x,-x)\in\mathbb{R}^{2}$ is continuous and $V=\varphi^{-1}(S_{2})$, this would give a contradiction.

4. Some integral inequalities on $\mathbb{R}^{\kappa}$

In this section we provide integral inequalities for separately incresing function. These inequalities are related to the multidimensional spectral order. The main result of this section is Theorem 4.2, which is a generalization of [24, Theorem 4.1.]. In the proof of this theorem, we will need the following lemma which characterizes regularity of lower sets.

Lemma 4.1**.**

Let $\mu$ be a finite Borel measure on $\mathbb{R}^{\kappa}$ and let $\iota\in\{1,\ldots,\kappa\}$. If $\varOmega\in L(\mathbb{R}^{\kappa},\leq_{\iota})\cap{\mathfrak{B}}(\mathbb{R}^{\kappa})$, then

[TABLE]

Proof.

Let $\varOmega\in L(\mathbb{R}^{\kappa},\leq_{\iota})\cap{\mathfrak{B}}(\mathbb{R}^{\kappa})$. Applying [25, Theorem 2.18.], Proposition 3.1, and Proposition 3.2, we get that

[TABLE]

This completes the proof. ∎

Theorem 4.2**.**

Let $\mu_{1}$ and $\mu_{2}$ be a finite Borel measure on $\mathbb{R}^{\kappa}$. We consider the following conditions:

1. (i)

$\mu_{2}(\varOmega)\leqslant\mu_{1}(\varOmega)$* for every $\varOmega\in L(\mathbb{R}^{\kappa},\leq_{\iota})\cap{\mathfrak{B}}(\mathbb{R}^{\kappa})$,* 2. (ii)

$\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{1}\leqslant\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{2}$* for every $\iota$-increasing Borel function $f\colon\mathbb{R}^{\kappa}\to[0,\infty)$,* 3. (iii)

$\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{1}\leqslant\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{2}$* for every $\iota$-increasing Borel function $f\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ such that $\int_{\mathbb{R}^{\kappa}}|f|\mathrm{d}\hskip 0.6458pt\mu_{2}<\infty$ *$($the integral $\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{1}$ may be -\infty$$), 4. (iv)

$\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{1}\leqslant\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{2}$* for every $\iota$-increasing bounded continuous function*

$f\colon\mathbb{R}^{\kappa}\to\mathbb{R}$, 5. (v)

$\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{1}\leqslant\int_{\mathbb{R}^{\kappa}}f\mathrm{d}\hskip 0.6458pt\mu_{2}$* for every nonnegative $\iota$-increasing bounded continuous function $f\colon\mathbb{R}^{\kappa}\to[0,\infty)$.*

Then

1. (a)

*condition (i) *$($*respectively (ii), (iii)$)$ implies that $\mu_{2}(\mathbb{R}^{\kappa})\leqslant\mu_{1}(\mathbb{R}^{\kappa})$ *$($respectively $\mu_{1}(\mathbb{R}^{\kappa})\leqslant\mu_{2}(\mathbb{R}^{\kappa})$, \mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$$). 2. (b)

Inequality $\mu_{1}(\mathbb{R}^{\kappa})\leqslant\mu_{2}(\mathbb{R}^{\kappa})$ and (i) imply that $\mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$. 3. (c)

If $\mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$, then all conditions (i)-(v) are equivalent.

Proof.

We only prove (c), since (a) and (b) are obvious. Assume that $\mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$.

(i)$\Rightarrow$(iii) Let $f$ be a real $\iota$-increasing Borel function such that $\int_{\mathbb{R}^{\kappa}}|f|\mathrm{d}\hskip 0.6458pt\mu_{2}<\infty$. Define $\nu_{j}(\sigma):=\mu_{j}(f^{-1}(\sigma))$ for $\sigma\in{\mathfrak{B}}(\mathbb{R})$ and $j=1,2$. It is evident that $\nu_{1}$ and $\nu_{2}$ are finite Borel measure on $\mathbb{R}$ and

[TABLE]

Moreover,

[TABLE]

for every $x\in\mathbb{R}$ since $f^{-1}((-\infty,x])$ is a lower set in $(\mathbb{R}^{\kappa},\leqslant_{\iota})$ by Proposition 3.5. Applying the measure transport theorem [15, p. 163, Theorem C.], we obtain

[TABLE]

Thus, by the measure transport theorem, (4.1), (4.2), and [24, Theorem 4.1.], we have

[TABLE]

(iii)$\Rightarrow$(ii), (iii)$\Rightarrow$(iv), and (iv)$\Rightarrow$(v) are obvious.

(ii)$\Rightarrow$(i) Let $\varOmega\in{\mathfrak{B}}(\mathbb{R}^{\kappa})\cap L(\mathbb{R}^{\kappa},\leq_{\iota})$. Then $f=\chi_{\mathbb{R}^{\kappa}\backslash\varOmega}$ is a $\iota$-increasing Borel function. By assumption we obtain

[TABLE]

Hence $\mu_{2}(\varOmega)\leqslant\mu_{1}(\varOmega)$, since $\mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$.

(v)$\Rightarrow$(i) According to Lemma 4.1 it is sufficient to show that

[TABLE]

for every closed set $D\in L(\mathbb{R}^{\kappa},\leq_{\iota})$.

Fix a closed set $D\in L(\mathbb{R}^{\kappa},\leq_{\iota})$. For every $n\in\mathbb{N}$, we define function $f_{n}\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ by $f_{n}(x):=\min\{1,nd_{D}(x)\}$, $x\in\mathbb{R}^{\kappa}$. It is easily seen that $0\leqslant f_{n}(x)\leqslant 1$, $f_{n}(x)\leqslant f_{n+1}(x)$ and $\lim_{n\to\infty}f_{n}(x)=\chi_{\mathbb{R}^{\kappa}\backslash D}(x)$ for every $x\in\mathbb{R}^{\kappa}$. Note that $f_{n}$’s are continuous and $\iota$-increasing, since $d_{D}$ is $\iota$-increasing by Proposition 3.6. It follows from Lebesgue’s monotone convergence theorem and assumption that

[TABLE]

This and equality $\mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$ imply that $\mu_{2}(D)\leqslant\mu_{1}(D)$. ∎

Remark 4.3*.*

Let $\mu_{1}$ and $\mu_{2}$ be finite nonnegative Borel measures on $\mathbb{R}^{\kappa}$ such that $\mu_{1}(\mathbb{R}^{\kappa})=\mu_{2}(\mathbb{R}^{\kappa})$. In this settings the condition

[TABLE]

does not have to imply conditions (i)-(v) in Theorem 4.2. Indeed, take two Borel measures $\mu_{1}$ and $\mu_{2}$ on $\mathbb{R}^{2}$ defined by the following formulas

[TABLE]

where $\delta_{(0,0)}$, $\delta_{(0,1)}$, $\delta_{(1,0)}$ and $\delta_{(1,1)}$ are Dirac measures on ${\mathfrak{B}}(\mathbb{R}^{2})$. It is evident, that $\mu_{1}(\mathbb{R}^{2})=\mu_{2}(\mathbb{R}^{2})=2$ and $\mu_{2}((-\infty,x])\leqslant\mu_{1}((-\infty,x])$ for every $x\in\mathbb{R}^{2}$. At the same time $\mu_{1}(S)<\mu_{2}(S)$, where $S=\{(x_{1},x_{2})\in\mathbb{R}^{2}\colon x_{1}+x_{2}\leqslant 1\}$. This means that measures $\mu_{1}$ and $\mu_{2}$ do not satisfy the condition (i). Hence all of conditions (i)-(v) in Theorem 4.2 do not hold.

5. Multidimensional spectral order

In this section we generalize the notion of spectral order to the case of finite families of pairwise commuting selfadjoint operators. First we give necessary background on joint spectral measures and multivariable resolution of the identity.

Let $\mathbf{B}_{cs}({\mathcal{H}},\kappa):=\mathcal{S}_{c}({\mathcal{H}},\kappa)\cap\mathbf{B}({\mathcal{H}})^{\kappa}$. If $\mathbf{A}=(A_{1},\ldots,A_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, then there exists the unique spectral measure $E_{\mathbf{A}}\colon{\mathfrak{B}}(\mathbb{R}^{\kappa})\to\mathbf{B}({\mathcal{H}})$, called the joint spectral measure of $\mathbf{A}$, such that

[TABLE]

In fact, $E_{\mathbf{A}}$ coincides with the product of the spectral measures of the operators $A_{1},\ldots,A_{\kappa}$ (cf. [5]). This relationship can be made more explicit by the following two equivalent equations

[TABLE]

where $\pi_{j}\colon\mathbb{R}^{\kappa}\ni(x_{1},\ldots,x_{\kappa})\rightarrow x_{j}\in\mathbb{R}$, and

[TABLE]

The joint spectral distribution $F_{\mathbf{A}}$ of $\mathbf{A}$ is defined by

[TABLE]

As in the one-variable case (cf. [5]), one can build the theory of multivariable resolutions of the identity independently of spectral measures. In Section A, for the reader’s convenience, we recall the definition of an abstract multivariable resolution of the identity and we outline an idea how to construct spectral measures on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$ by using multivariable resolution of the identity.

Let $\mathbf{A}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. Given a Borel function $\varphi\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ we define

[TABLE]

The operator $\varphi(\bf{A})$ is selfadjoint. Moreover, if $\varphi\geqslant 0$, then $\varphi(\bf{A})$ is positive.

After these preliminary remarks, we can state the definition of spectral order in multidimensional case.

Definition 5.1**.**

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. We write $\mathbf{A}\preccurlyeq\mathbf{B}$, if $F_{\mathbf{B}}(x)\leqslant F_{\mathbf{A}}(x)$ for every $x\in\mathbb{R}^{\kappa}$.

It is easily seen that the relation ”$\preccurlyeq$” is a partial order in $\mathcal{S}_{c}({\mathcal{H}},\kappa)$, since there is a one-to-one correspondence between the set of all resolutions of the identity on $\mathbb{R}^{\kappa}$ and the set $\mathcal{S}_{c}({\mathcal{H}},\kappa)$. This order will be called multidimensional spectral order. If $\kappa=1$, then we obtain the definition of spectral order for selfadjoint operators (cf. [23],[24]). Now we are going to investigate the fundamental properties of multidimensional spectral order ”$\preccurlyeq$”. First we are going to answer the question which Borel functions $\varphi\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ preserve the multidimensional spectral order, i.e. the map $\Phi\colon(\mathcal{S}_{c}({\mathcal{H}},\kappa),\preccurlyeq)\ni\mathbf{A}\to\varphi(\mathbf{A})\in(\mathcal{S}_{c}({\mathcal{H}},1),\preccurlyeq)$ is a morphism in the category of partially ordered set. We begin by showing that a Borel function $\varphi\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ preserving the multidimensional spectral order has to be increasing.

Proposition 5.2**.**

Let $\kappa\in\mathbb{N}$ and let ${\mathcal{H}}$ be a Hilbert space such that $\dim{\mathcal{H}}\geqslant 1$. If $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$ is a Borel function satisfying the following condition

[TABLE]

for every $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, then $\varphi$ is an increasing function.

Proof.

Fix $a=(a_{1},\ldots,a_{\kappa})\in\mathbb{R}^{\kappa}$ and $b=(b_{1},\ldots,b_{\kappa})\in\mathbb{R}^{\kappa}$ such that $a\leqslant b$. Define $A_{i}=a_{i}I$ and $B_{i}=b_{i}I$ for $i=1,\ldots,\kappa$. It is easily seen that $E_{\mathbf{A}}(\sigma)=\delta_{a}(\sigma)I$ and $E_{\mathbf{B}}(\sigma)=\delta_{b}(\sigma)I$ for every $\sigma\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$, where $\delta_{c}$ is a Dirac measure defined on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$ at the point $c\in\mathbb{R}^{\kappa}$. Hence $\mathbf{A}\preccurlyeq\mathbf{B}$ because $a\leqslant b$. Then, by the condition (5.5), $\varphi(a)I=\varphi(\mathbf{A})\preccurlyeq\varphi(\mathbf{B})=\varphi(b)I$. Thus $\varphi(a)\leqslant\varphi(b)$, since $\dim{\mathcal{H}}\geqslant 1$. This completes the proof. ∎

In fact, as we will see, every increasing Borel function $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$ preserves multidimensional spectral order. In order to prove this, we will need a few lemmata.

Lemma 5.3**.**

Let $\mathcal{M}$ be a $\sigma$-algebra on a set $X$ and let $E$ be a spectral measure on $\mathcal{M}$. If $\{V_{n}\}_{n=1}^{\infty}\subseteq\mathcal{M}$, then

[TABLE]

Proof.

Set $V_{\infty}:=\bigcup_{n=1}^{\infty}V_{n}$ and $M_{k}:=E(V_{k}){\mathcal{H}}$, for $k\in\mathbb{N}\cup\{\infty\}$. It suffices to show that $P_{M_{\infty}}=E(V_{\infty})\leqslant\bigvee_{n=1}^{\infty}E(V_{n})$, since $E(V_{n})\leqslant E(V_{\infty})$ for every $n\in\mathbb{N}$. By equation (2.1), we have $\bigvee_{n=1}^{\infty}E(V_{n})=\bigvee_{n=1}^{\infty}P_{M_{n}}=P_{\bigvee_{n=1}^{\infty}M_{n}}$. Thus we need to show that $M_{\infty}\subseteq\bigvee_{n=1}^{\infty}M_{n}$. Take $h\in M_{\infty}$ and let

[TABLE]

It is obvious that $V_{\infty}=\bigcup_{n=1}^{\infty}W_{n}$ and $W_{n}\cap W_{m}=\varnothing$ if $m\neq n$. Hence

[TABLE]

This completes the proof. ∎

Lemma 5.4**.**

Let $E$ be a spectral measure on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$, $\varOmega\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$ and $\iota\in\{1,\ldots,\kappa\}$. If $\varOmega\in L(\mathbb{R}^{\kappa},\leqslant_{\iota})$, then

[TABLE]

Proof.

Let $E_{0}(\varOmega):=\bigvee\{E(D)\colon D\subseteq\varOmega,D=\overline{D}\in L(\mathbb{R}^{\kappa},\leqslant_{\iota})\}$. It is obvious that $E_{0}(\varOmega)\leqslant E(\varOmega)$. For the opposite inequality, take any compact set $K\subseteq\varOmega$. Applying Proposition 3.2 and Proposition 3.1 (ii) we get that $\overline{\downarrow_{\iota}K}=\downarrow_{\iota}K\subseteq\downarrow_{\iota}\varOmega=\varOmega$ and

[TABLE]

This and corollary 2.1 imply that

[TABLE]

which completes the proof of equality (5.7). ∎

We need some additional notation. Let $\iota\in\mathbb{N}$ such that $\iota<\kappa$. If $\mathbf{C}=(C_{1},\ldots,C_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, then we define $\mathbf{C}^{\prime}:=(C_{1},\ldots,C_{\iota})\in\mathcal{S}_{c}({\mathcal{H}},\iota)$ and $\mathbf{C}^{\prime\prime}:=(C_{\iota+1},\ldots,C_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa-\iota)$. Applying spectral theorem for finitely many commuting selfadjoint operators and measure transport theorem [5, Theorem 5.4.10.], we can deduce that $E_{\mathbf{C}^{\prime}}$ and $E_{\mathbf{C}^{\prime\prime}}$ commute and

[TABLE]

Lemma 5.5**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa}),\mathbf{B}=(B_{1},\ldots,B_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and let $\iota\in\{1,\ldots,\kappa\}$. Assume that $A_{j}=B_{j}$ for every $j=\iota+1,\ldots,\kappa$, whenever $\iota<\kappa$. If $\mathbf{A}\preccurlyeq\mathbf{B}$, then

[TABLE]

for every $\varOmega\in{\mathfrak{B}}(\mathbb{R}^{\kappa})\cap L(\mathbb{R}^{\kappa},\leqslant_{\iota})$.

Proof.

The proof of inequality (5.9) will be divided into several steps. We are going to deal only with the case $\iota<\kappa$, since the proof in the case $\iota=\kappa$ is similar.

Step 1. Inequality (5.9) holds for $\varOmega=(-\infty,x^{\prime}]\times\varOmega^{\prime\prime}$, where $x^{\prime}\in\mathbb{R}^{\iota}$ and $\varOmega^{\prime\prime}\in{\mathfrak{B}}(\mathbb{R}^{\kappa-\iota})$.

Equation (5.8), Lemma 5.3, and inequality $\mathbf{A}\preccurlyeq\mathbf{B}$ imply that

[TABLE]

By assumption, $E_{\mathbf{B}^{\prime\prime}}=E_{\mathbf{A}^{\prime\prime}}$. Therefore,

[TABLE]

Step 2. Inequality (5.9) holds, if $\varOmega=\bigcup_{n=1}^{\infty}(-\infty,x_{n}]\times\varOmega^{\prime\prime}_{n}$, where $x_{n}\in\mathbb{R}^{\iota}$ and $\varOmega^{\prime\prime}_{n}\in{\mathfrak{B}}(\mathbb{R}^{\kappa-\iota})$ for all $n\in\mathbb{N}$.

Indeed, by Lemma 5.3 and Step 1., we get that

[TABLE]

Step 3. Assume that $\varOmega$ is a lower set in $(\mathbb{R}^{\kappa},\leqslant_{\iota})$. Let $\|\cdot\|=\|\cdot\|_{m}$, $m\in\mathbb{N}$, be a norm on $\mathbb{R}^{m}$ given by $\|x\|_{m}=\sum_{j=1}^{m}|x_{j}|$ for every $x\in\mathbb{R}^{m}$. We are going to show that

[TABLE]

for every $\varepsilon>0$, where $\mathbb{B}_{m}\left(x,r\right)$ denote open ball $\{y\in\mathbb{R}^{m}\colon\|y-x\|_{m}<r\}$ for $x\in\mathbb{R}^{m}$ and $r>0$.

Let $x\in\varOmega$. By the density of $\mathbb{Q}^{\kappa}$, we find $q\in\mathbb{B}_{\kappa}\left(x,\varepsilon\right)\cap\mathbb{Q}^{\kappa}$ such that $x<q$. Hence $q\in\varOmega^{(\varepsilon)}$ and $x\in(-\infty,q^{\prime}]\times\mathbb{B}_{\kappa-\iota}\left(q^{\prime\prime},\varepsilon\right)$. This proves the first inclusion.

To prove the second inclusion fix $q=(q^{\prime},q^{\prime\prime})\in(\mathbb{Q}^{\iota}\times\mathbb{Q}^{\kappa-\iota})\cap\varOmega^{(\varepsilon)}$. Then

[TABLE]

since $d_{\varOmega}(q)\leqslant\varepsilon$. Hence, by Proposition 3.1 and Proposition 3.6,

[TABLE]

This gives the second inclusion.

Step 4. Inequality (5.9) holds whenever $\varOmega$ is a closed lower set in $(\mathbb{R}^{\kappa},\leqslant_{\iota})$.

Let $Q_{q,n}:=(-\infty,q^{\prime}]\times\mathbb{B}_{\kappa-\iota}\left(q^{\prime\prime},\frac{1}{n}\right)$, where $q=(q^{\prime},q^{\prime\prime})\in(\mathbb{Q}^{\iota}\times\mathbb{Q}^{\kappa-\iota})\cap\varOmega^{(\frac{1}{n})}$ and $n\in\mathbb{N}$. By (5.10) and Step 2., we obtain that

[TABLE]

Note that $\{\varOmega^{(\frac{1}{n})}\}_{n=1}^{\infty}$ is a decreasing family of sets (with respect to set inclusion). What is more, $\varOmega=\bigcap_{\epsilon\in(0,\infty)}\varOmega^{(\epsilon)}$, since $\varOmega$ is closed. Hence

[TABLE]

Step 5. Application of Step 4. and Lemma 5.4 completes the proof of the inequality (5.9) for an arbitrary Borel lower set $\varOmega$ in $(\mathbb{R}^{\kappa},\leqslant_{\iota})$. ∎

Theorem 5.6**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa}),\mathbf{B}=(B_{1},\ldots,B_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, and $\varphi\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ be a Borel function. Assume that $\bf{A}\preccurlyeq\bf{B}$. If

1. (a)

the function $\varphi$ is increasing,

or there exist $\iota\in\mathbb{N}$ such that $\iota<\kappa$, and $\varOmega\in{\mathfrak{B}}(\mathbb{R}^{\kappa-\iota})$ satisfying the following conditions

1. (a’)

$A_{j}=B_{j}$* for every $j=\iota+1,\ldots,\kappa$,* 2. (b’)

the function $\varphi$ is $\iota$-increasing on $\mathbb{R}^{\iota}\times\varOmega$, 3. (c’)

$E_{\mathbf{A}^{\prime\prime}}(\mathbb{R}^{\kappa-\iota}\setminus\varOmega)=E_{\mathbf{B}^{\prime\prime}}(\mathbb{R}^{\kappa-\iota}\setminus\varOmega)=0$,

then $\varphi(\bf{A})\preccurlyeq\varphi(\bf{B})$.

Proof.

We are going to prove only the second case (i.e. $\iota<\kappa$), since similar line of reasoning applies in the first case. First let us observe, that the set

[TABLE]

is a lower set in $(\mathbb{R}^{\kappa},\leqslant_{\iota})$ for every $x\in\mathbb{R}$ by the assumption (b’). Using measure transport theorem, equality (5.8), and (c’), we get that

[TABLE]

where $\mathbf{C}=\mathbf{A},\mathbf{B}$. Thus, by Lemma 5.5,

[TABLE]

for every $x\in\mathbb{R}$. This completes the proof. ∎

Theorem 5.7**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa}),\mathbf{B}=(B_{1},\ldots,B_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. Then the following conditions are equivalent:

1. (i)

$\bf{A}\preccurlyeq\bf{B}$, 2. (ii)

$\varphi(\bf{A})\preccurlyeq\varphi(\bf{B})$* for every increasing Borel function $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$,* 3. (iii)

$A_{j}\preccurlyeq B_{j}$* for every $j=1,\ldots,\kappa$.*

Proof.

(i)$\Rightarrow$(ii) It follows directly from Theorem 5.6.

(ii)$\Rightarrow$(iii) Apply equation (5.1) for the function $\pi_{j}$, which is increasing and Borel for every $j=1,\ldots,\kappa$.

(iii)$\Rightarrow$(i) Fix an arbitrary $x=(x_{1},\ldots,x_{\kappa})\in\mathbb{R}^{\kappa}$. By the definition of the multidimensional spectral order for selfadjoint operators and equality (5.3), we get that

[TABLE]

for every $x_{1},\ldots,x_{\kappa}\in\mathbb{R}$. This completes the proof. ∎

The following corollary give us an alternative definition for the multidimensional spectral order (see [13, Theorem 1.] for bounded selfadjoint operators and [24, Theorem 6.5.] for unbounded selfadjoint operators) in terms of standard order ”$\leqslant$” for bounded selfadjoint operators and the spectral integrals of increasing Borel functions.

Corollary 5.8**.**

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. Then the following conditions are equivalent:

1. (i)

$\bf{A}\preccurlyeq\bf{B}$, 2. (ii)

$\varphi(\bf{A})\leqslant\varphi(\bf{B})$* for every bounded increasing continuous function $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$,* 3. (iii)

$\varphi(\bf{A})\leqslant\varphi(\bf{B})$* for every bounded increasing Borel function $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$.*

Proof.

(i)$\Rightarrow$(iii) Notice that $\varphi(\bf{A}),\varphi(\bf{B})\in\mathbf{B}({\mathcal{H}})$ for every bounded Borel function $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$ and apply Theorem 5.7 and [24, Proposition 6.3.].

(iii)$\Rightarrow$(ii) It is obvious.

(ii)$\Rightarrow$(i) In view of Theorem 5.7, it is sufficient to show that $A_{j}\preccurlyeq B_{j}$ for every $j=1,\ldots,\kappa$. Fix $j\in\{1,\ldots\kappa\}$. Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be an arbitrary continuous and bounded increasing function. In this case $\varphi:=f\circ\pi_{j}\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$ is a continuous and bounded increasing function. By [5, Lemma 6.5.2.] and condition (ii) we get that

[TABLE]

Since this inequality holds for every bounded continuous increasing function $f\colon\mathbb{R}\rightarrow\mathbb{R}$, we deduce that $A_{j}\preccurlyeq B_{j}$ by [24, Theorem 6.5.]. ∎

Remark 5.9*.*

Note that there is another proof of the implication (ii)$\Rightarrow$(i). Indeed, similarly as in the proof of the implication (ii)$\Rightarrow$(i) in [24, Theorem 6.5.], we obtain that

[TABLE]

for every bounded increasing continuous function $\varphi\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}$. Thus, by Theorem 4.2, $\langle F_{\mathbf{B}}(x)h,h\rangle\leqslant\langle F_{\mathbf{A}}(x)h,h\rangle$ for every $x\in\mathbb{R}^{\kappa}$ and $h\in{\mathcal{H}}$. This means that $\mathbf{A}\preccurlyeq\mathbf{B}$.

Let $\mathbf{A}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and $\varphi=(\varphi_{1},\ldots,\varphi_{\iota})\colon\mathbb{R}^{\kappa}\to\mathbb{R}^{\iota}$ be a Borel function. Set

[TABLE]

It is evident that $\varphi(\mathbf{A})\in\mathcal{S}_{c}({\mathcal{H}},\iota)$.

Corollary 5.10**.**

Let $\kappa,\iota\in\mathbb{N}$. Suppose that $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and $\mathbf{A}\preccurlyeq\mathbf{B}$. If $\varphi\colon\mathbb{R}^{\kappa}\to\mathbb{R}^{\iota}$ is an increasing Borel function, then $\varphi(\mathbf{A})\preccurlyeq\varphi(\mathbf{B})$.

Proof.

First let us observe, that the functions $\varphi_{j}:=\pi_{j}\circ\varphi$, where $j=1,\ldots,\iota$, are increasing, since $\varphi$ is increasing. Applying Theorem 5.7 to $\varphi_{j}$ we get that $\varphi_{j}(\mathbf{A})\preccurlyeq\varphi_{j}(\mathbf{B})$ for $j=1,\ldots,\iota$. By Theorem 5.7, this is equivalent to the inequality $\varphi(\mathbf{A})\preccurlyeq\varphi(\mathbf{B})$. ∎

Now let us turn our attention to two questions concerning the multidimensional spectral order. The first problem is whether the partially ordered sets $(\mathcal{S}_{c}({\mathcal{H}},\kappa),\preccurlyeq)$ and $(\mathbf{B}_{cs}({\mathcal{H}},\kappa),\preccurlyeq)$ are a conditionally complete lattice if $\kappa>1$. The second question that we study is whether spectral order satisfies some kind of vector order properties.

The first question is answered in the negative by the following example. In fact, the sets $(\mathcal{S}_{c}({\mathcal{H}},\kappa),\preccurlyeq)$ and $(\mathbf{B}_{cs}({\mathcal{H}},\kappa),\preccurlyeq)$ are not even a lattice for $\kappa>1$.

Example** 5.11****.**

Let ${\mathcal{H}}=\mathbb{C}^{3}$ be a Hilbert space with the standard orthonormal basis $\{(1,0,0),(0,1,0),(0,0,1)\}$. Then $(\mathcal{S}({\mathcal{H}},2),\preccurlyeq)$ and $(\mathbf{B}_{cs}({\mathcal{H}},2),\preccurlyeq)$ are not a lattice. Indeed, let $M_{1}$, $M_{2}$, $N_{1}$, and $N_{2}$ are defined to be the linear subspaces of $\mathbb{C}^{3}$ of the form

[TABLE]

Define $\mathbf{A}=(A_{1},A_{2})$ and $\mathbf{B}=(B_{1},B_{2})$ by

[TABLE]

It is obvious that $\mathbf{A},\mathbf{B}\in\mathbf{B}_{cs}({\mathcal{H}},2)$. Suppose that $\mathbf{C}=(C_{1},C_{2})$ is the infimum of $\{\mathbf{A},\mathbf{B}\}$ in $(\mathbf{B}_{cs}({\mathcal{H}},2),\preccurlyeq)$. Then, by Theorem 5.7 and the definition of infimum, we obtain that

[TABLE]

On the other hand, $(P_{M_{1}\cap N_{1}},0),\,(0,P_{M_{2}\cap N_{2}})\in\mathbf{B}_{cs}({\mathcal{H}},2)$. Moreover, the set $\{\mathbf{A},\mathbf{B}\}$ is bounded below by $(P_{M_{1}\cap N_{1}},0)$ and $(0,P_{M_{2}\cap N_{2}})$ with respect to the multidimensional spectral order. Hence, by the definition of infimum and Theorem 5.7, we deduce that

[TABLE]

Inequalities (5.12) and (5.13) imply that $\mathbf{C}=(P_{M_{1}\cap N_{1}},P_{M_{2}\cap N_{2}})\notin B_{cs}({\mathcal{H}},2)$ since

[TABLE]

do not commute. Thus $(\mathbf{B}_{cs}({\mathcal{H}},2),\preccurlyeq)$ is not a lattice. The same argument shows that $(\mathcal{S}_{c}({\mathcal{H}},2),\preccurlyeq)$ is not a lattice too.

Now we are going to consider the second question. As shown by Olson (cf. [23]) the relation $0\preccurlyeq A-B$ may not imply $A\preccurlyeq B$ for arbitrary bounded selfadjoint operators $A,B$ unless $A$ and $B$ commute. Thus the spectral order is not a vector order. It is known also that $A\preccurlyeq B$ imply $\lambda A\preccurlyeq\lambda B$ whenever $\lambda\in[0,\infty)$. It appears that this properties can be strengthened, which is the consequence of Theorem 5.7 and Theorem 5.6.

Corollary 5.12**.**

Let $\mathbf{A}=(A_{1},A_{2}),\,\mathbf{B}=(B_{1},B_{2})\in\mathcal{S}_{c}({\mathcal{H}},2)$. Assume that $A_{1}\preccurlyeq B_{1}$ and $A_{2}\preccurlyeq B_{2}$. Then

[TABLE]

Proof.

Note that $\mathbf{A}\preccurlyeq\mathbf{B}$ by assumptions and Theorem 5.7. Let $\varphi\colon\mathbb{R}^{2}\rightarrow\mathbb{R}$ be a function defined by $\varphi(x_{1},x_{2}):=x_{1}+x_{2}$ for $x_{1},x_{2}\in\mathbb{R}$. By [5, Theorem 5.4.7] and Theorem 5.7 applied to the function $\varphi$, which is increasing, we get that

[TABLE]

This completes the proof of (5.14). ∎

Corollary 5.13**.**

Suppose that $A,B,C$ are selfadjoint operators in ${\mathcal{H}}$. Assume also that $C$ is positive and commutes with $A$ and $B$. If $A\preccurlyeq B$, then

[TABLE]

Proof.

Let $\mathbf{A}=(A,C)$ and $\mathbf{B}=(B,C)$. By Theorem 5.7 (iii), $\mathbf{A}\preccurlyeq\mathbf{B}$. Take a function $\psi\colon\mathbb{R}^{2}\rightarrow\mathbb{R}$ given by $\psi(x_{1},x_{2}):=x_{1}\cdot x_{2}$ for $x_{1},x_{2}\in\mathbb{R}$. It is easily seen that the function $\psi$ is 1-increasing on $\mathbb{R}\times[0,\infty)$ and $E_{C}((-\infty,0))=0$. Thus, by Theorem 5.6 and [5, Theorem 5.4.7], we obtain that

[TABLE]

∎

We close this section with the example of an order for normal operators. In [8] Brenna and Flori defined spectral order for bounded normal operators. The definition of this order can be easily adopted also in the case of unbounded normal operators.

Example** 5.14****.**

Let $T$ be a normal operator in ${\mathcal{H}}$, i.e. densely defined operator such that ${\mathscr{D}}(T)={\mathscr{D}}(T^{*})$ and $\|Th\|=\|T^{*}h\|$ for every $h\in{\mathscr{D}}(T)$. The notation $\mathcal{N}({\mathcal{H}})$ will mean the set of all normal operators in ${\mathcal{H}}$ and $\mathbf{B}_{n}({\mathcal{H}}):=\mathcal{N}({\mathcal{H}})\cap\mathbf{B}({\mathcal{H}})$. Denote by $E_{T}$ the spectral measure of $T$ on ${\mathfrak{B}}(\mathbb{C})$, where $\mathbb{C}\simeq\mathbb{R}^{2}$ via $\rho(x,y)=x+iy$, $x,y\in\mathbb{R}$. Consider the real and imaginary part of $T$ defined by

[TABLE]

Then

[TABLE]

Note that $\mathrm{Re}\,T$ and $\mathrm{Im}\,T$ are selfadjoint. Moreover, from the [5, Theorem 5.4.10.] we derive

[TABLE]

Therefore $(\mathrm{Re}\,T,\mathrm{Im}\,T)\in\mathcal{S}_{c}({\mathcal{H}},2)$. Now, let $T_{1},T_{2}\in\mathcal{N}({\mathcal{H}})$. Then we write

[TABLE]

Applying equations (5.3) and (5.4) we can rewrite this as

[TABLE]

The following propositions shows that the spectral order for pairs of commuting selfadjoint operators and the spectral order for normal operators are isomorphic in the category of partially ordered sets.

Proposition 5.15**.**

The mappings $\Psi$ and $\Psi_{0}$ are order isomorphisms, where $\Psi\colon(\mathcal{N}({\mathcal{H}}),\preccurlyeq)\ni T\to(\mathrm{Re}\,T,\mathrm{Im}\,T)\in(\mathcal{S}_{c}({\mathcal{H}},2),\preccurlyeq)$ and $\Psi_{0}\colon(\mathbf{B}_{n}({\mathcal{H}}),\preccurlyeq)\ni T\to(\mathrm{Re}\,T,\mathrm{Im}\,T)\in(\mathbf{B}_{cs}({\mathcal{H}},2),\preccurlyeq)$.

Proof.

Let $\mathbf{A}=(A_{1},A_{2})\in\mathcal{S}_{c}({\mathcal{H}},2)$. Define the measure $E_{\mathbf{A}}\circ\rho^{-1}$ on ${\mathfrak{B}}(\mathbb{C})$ by $E_{\mathbf{A}}\circ\rho^{-1}(\Delta)=E_{\mathbf{A}}(\rho^{-1}(\Delta))$, $\Delta\in{\mathfrak{B}}(\mathbb{C})$. Then $T_{\mathbf{A}}=\int_{\mathbb{C}}zE_{\mathbf{A}}\circ\rho^{-1}(\mathrm{d}\hskip 0.6458ptz)$ is a normal operator in ${\mathcal{H}}$ such that $\mathrm{Re}\,T_{\mathbf{A}}=A_{1}$ and $\mathrm{Im}\,T_{\mathbf{A}}=A_{2}$. Indeed, we have

[TABLE]

and

[TABLE]

In partcular, $\Psi$ is bijective. According to the definition of the multidimensional spectral order and (5.15), $\Psi$ is order preserving. At the end, note that $\Psi(\mathbf{B}_{n}({\mathcal{H}}))=\mathbf{B}_{cs}({\mathcal{H}},2)$, which completes the proof. ∎

6. Joint bounded vectors

The aim of this section is to describe the joint resolution of the identity of $\mathbf{A}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ in terms of joint bounded vectors. The main result of this section Proposition 6.3 will be used in Section 8 to characterize multidimensional spectral order in the case of positive operators.

First we are going to recall the definition of joint bounded vectors. Let $x=(x_{1},\ldots,x_{\kappa})\in\mathbb{R}^{\kappa}$ and $\alpha\in[0,\infty)^{\kappa}$. Set $x^{\alpha}:=x_{1}^{\alpha_{1}}\cdot\ldots\cdot x_{\kappa}^{\alpha_{\kappa}}$ provided that $x_{j}^{\alpha_{j}}$ are well-defined for every $j=1,\ldots,\kappa$. Define functions $\varphi_{\alpha}\colon\mathbb{R}^{\kappa}\to\mathbb{R}$, $\alpha\in\mathbb{Z}^{\kappa}_{+}$, and $\psi_{\beta}\colon\mathbb{R}^{\kappa}\to\mathbb{R}$, $\beta\in[0,\infty)^{\kappa}$, by the following formulas

[TABLE]

The monomial $\mathbf{A}^{\alpha}$ is defined by

[TABLE]

Similarly, the fractional power $\mathbf{A}^{\alpha}$ is given by

[TABLE]

Note that $\psi_{\alpha}(\mathbf{A})=\varphi_{\alpha}(\mathbf{A})$, provided that $\alpha\in\mathbb{Z}_{+}^{\kappa}$ and $\operatorname{supp}E_{\mathbf{A}}\subseteq[0,\infty)^{\kappa}$.

Let $\mathbf{A}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. We define the following sets

[TABLE]

We say that $h\in{\mathcal{H}}$ is a joint bounded vector of $\mathbf{A}$ if $h\in\mathscr{B}(\mathbf{A})$. More information on joint bounded vectors can be found in [26]. For the sake of completeness, Proposition 6.3 will be preceded by the following two Lemmata (see also [26, Theorem 1.13.]).

Lemma 6.1**.**

If $\mathbf{A}=(A_{1},\ldots,A_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, then

[TABLE]

Proof.

Applying equalities (5.2) and (5.6) we obtain that

[TABLE]

what proves the claim. ∎

If $\operatorname{supp}E_{\mathbf{A}}\subseteq[0,\infty)^{\kappa}$, then we say that $\mathbf{A}$ is positive.

Lemma 6.2**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. Then

[TABLE]

Moreover, if $\mathbf{A}$ is positive, then

[TABLE]

Proof.

Let $D:=\displaystyle\bigcap_{j=1}^{\kappa}{\mathscr{D}}^{\infty}(A_{j})$. We have to show that $D\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$ since inclusion ${\mathscr{D}}^{\infty}(\mathbf{A})\subseteq D$ is obvious.

Let $h\in D$ and $\alpha\in\mathbb{Z}_{+}^{\kappa}$. Consider a measure $\mu_{h}$ on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$ defined by $\mu_{h}(\sigma):=\langle E_{\mathbf{A}}(\sigma)h,h\rangle$ for $\sigma\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$. Using the definition of $D$ we derive that $\int_{\mathbb{R}^{\kappa}}x_{j}^{2\kappa\alpha_{j}}\mathrm{d}\hskip 0.6458pt\mu_{h}(x)<\infty$ for $j=1,\ldots,\kappa$. This and [6, Corollary 2.11.5] imply that $\int_{\mathbb{R}^{\kappa}}x^{2\alpha}\mathrm{d}\hskip 0.6458pt\mu_{h}(x)<\infty$. Thus $h\in{\mathscr{D}}(\mathbf{A}^{\alpha})$ for every $\alpha\in\mathbb{Z}_{+}^{\kappa}$. Consequently, $h\in{\mathscr{D}}^{\infty}(\mathbf{A})$.

For the ”moreover” part, consider only the inclusion $\subseteq$ since the inclusion $\supseteq$ is evident. Let $h\in{\mathscr{D}}^{\infty}(\mathbf{A})$ and $\alpha\in[0,\infty)^{\kappa}$. Without loss of generality we may assume that $\|h\|=1$. Take $\beta\in\mathbb{N}^{\kappa}$ such that $\kappa\alpha\leqslant\beta$. By assumption, $\int_{[0,\infty)^{\kappa}}x_{j}^{2\beta_{j}}\mathrm{d}\hskip 0.6458pt\mu_{h}(x)<\infty$. Applying [6, Corollary 2.11.5] and Jensen inequality, we obtain that

[TABLE]

Hence $h\in\bigcap_{\alpha\in[0,\infty)^{\kappa}}{\mathscr{D}}(\mathbf{A}^{\alpha})$, which completes proof. ∎

In the sequel we will use the following notation $\mathscr{D}^{\Lambda}(\mathbf{A}):=\bigcap_{\alpha\in\Lambda}{\mathscr{D}}(\mathbf{A}^{\alpha})$ for $\Lambda\subseteq[0,\infty)^{\kappa}$ and $|\alpha|:=\alpha_{1}+\ldots+\alpha_{\kappa}$ for $\alpha=(\alpha_{1},\ldots,\alpha_{\kappa})\in[0,\infty)^{\kappa}$.

Proposition 6.3**.**

Let $\mathbf{A}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive, $h\in{\mathcal{H}}$, and $a\in[0,\infty)^{\kappa}$. Assume that $\Lambda\subseteq[0,\infty)^{\kappa}$ satisfies the condition

[TABLE]

Then the following conditions are equivalent:

1. (i)

$h\in\mathscr{B}_{a}(\mathbf{A})$, 2. (ii)

$h\in{\mathscr{R}}(F_{\mathbf{A}}(a))$, 3. (iii)

$h\in{\mathscr{D}}^{\infty}(\mathbf{A})$* and there exists real number $c>0$ such that*

[TABLE]

Moreover, if $h\in\mathscr{D}^{\Lambda}(\mathbf{A})\cap\left(\bigcup_{j=1}^{\kappa}{\mathscr{N}}(A_{j})\right)^{\perp}$ and there exists real number $c>0$ such that

[TABLE]

then $h\in{\mathscr{R}}(F_{\mathbf{A}}(a))$.

Proof.

Let $\mu_{h}(\sigma):=\langle E_{\mathbf{A}}(\sigma)h,h\rangle$ for every $\sigma\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$.

(i)$\Rightarrow$(ii) By Lemma 6.2, $h\in{\mathscr{D}}^{\infty}(A_{j})$ for every $j=1,\ldots,\kappa$. At the same time $\|A_{j}^{n}h\|=\|\mathbf{A}^{ne_{j}}h\|\leqslant ca_{j}^{n}$ for every $n\in\mathbb{Z}_{+}$. This implies that $h\in\mathscr{B}_{a_{j}}(A_{j})$. Hence, by [24, Proposition 5.1.], $h\in{\mathscr{R}}(F_{A_{j}}(a_{j}))$ for every $j=1,\ldots,\kappa$. Thus $h\in{\mathscr{R}}(F_{\mathbf{A}}(a))$, since $F_{\mathbf{A}}(a)=F_{A_{1}}(a_{1})\ldots F_{A_{\kappa}}(a_{\kappa})$.

(ii)$\Rightarrow$(iii) Let $\alpha\in[0,\infty)^{\kappa}$. Then, by (ii) and positivity of $\mathbf{A}$, we obtain that $\operatorname{supp}\mu_{h}\subseteq[0,a]$ and

[TABLE]

Thus $h\in{\mathscr{D}}(\mathbf{A}^{\alpha})$ and

[TABLE]

(iii)$\Rightarrow$(i) It is obvious.

For the ”moreover” part, assume that $h\in\mathscr{D}^{\Lambda}(\mathbf{A})\cap\left(\bigcup_{j=1}^{\kappa}{\mathscr{N}}(A_{j})\right)^{\perp}$ satisfies equation (6.4). Suppose that $h\notin{\mathscr{R}}(F_{\mathbf{A}}(a))$. In particular, $\mu_{h}(\mathbb{R}^{\kappa}\backslash(-\infty,a])>0$. Applying [5, Theorem 6.1.3] and (5.2) we obtain that

[TABLE]

What is more $E_{\mathbf{A}}(\mathbb{R}^{\kappa}\backslash[0,\infty)^{\kappa})=0$, since $\mathbf{A}$ is positive. Thus

[TABLE]

Put

[TABLE]

for $j\in\{1,\ldots,\kappa\}$ and $n\in\mathbb{N}$. Then

[TABLE]

This together with (6.5) allow us to choose $j\in\{1,\ldots,\kappa\}$ and $n\in\mathbb{N}$ such that

[TABLE]

Fix $\alpha\in\Lambda$. Applying inequality (6.4) we obtain that

[TABLE]

Hence

[TABLE]

By condition (6.3), we can find a sequence $\{\alpha(m)\}_{m=1}^{\infty}\subseteq\Lambda$ such that

[TABLE]

where $\alpha(m)=(\alpha_{1}(m),\ldots,\alpha_{\kappa}(m))$. Then inequality (6.6) implies that

[TABLE]

which is a contradiction. Thus $h\in{\mathscr{R}}(F_{\mathbf{A}}(a))$. ∎

Assume that $\mathbf{A}$ and $\Lambda$ satisfy the assumptions of Proposition 6.3. The following example shows that without the condition (6.4) it may happen that $h\in\mathscr{D}^{\Lambda}(\mathbf{A})\cap\left(\bigcup_{j=1}^{\kappa}{\mathscr{N}}(A_{j})\right)^{\perp}$ but $h\notin{\mathscr{D}}^{\infty}(\mathbf{A})$.

Example** 6.4****.**

Let $\Lambda:=[0,\infty)\times(0,\infty)$ and ${\mathcal{H}}=L^{2}([1,\infty),m)$, where $m$ is Lebesgue measure on Borel subsets of $[1,\infty)$. Define $\mathbf{A}=\left(M_{\varphi_{1}},M_{\varphi_{2}}\right)$, where the functions $\varphi_{1},\varphi_{2}\colon[1,\infty)\rightarrow\mathbb{R}$ are given by $\varphi_{1}(x)=x$ and $\varphi_{2}(x)=e^{-x}$ for $x\in\mathbb{R}$. Then, by Lemma 7.6, $\mathbf{A}\in\mathcal{S}_{c}({\mathcal{H}},2)$. $\mathbf{A}$ is also positive since $\varphi_{j}(x)\geqslant 0$ for every $x\in\mathbb{R}$ and $j=1,2$. Let $h(x):=\frac{1}{x}$ for $x\in[1,\infty)$. It is evident that $h\in{\mathcal{H}}$ and the set $\Lambda$ satisfies the condition (6.3).

We are going to show that $h\in\mathscr{D}^{\Lambda}(\mathbf{A})\backslash{\mathscr{D}}^{\infty}(\mathbf{A})$. If $\alpha\in\Lambda$, then $\alpha_{2}>0$. Hence

[TABLE]

Thus, by Lemma 7.6 (iii), $h\in{\mathscr{D}}(M_{\varphi_{1}^{\alpha_{1}}\varphi_{2}^{\alpha_{2}}})={\mathscr{D}}(\mathbf{A}^{\alpha})$. On the other hand,

[TABLE]

which means that $h\notin{\mathscr{D}}(A_{1})\supseteq{\mathscr{D}}^{\infty}(\mathbf{A})$, which completes the proof.

7. Monomials and multidimensional spectral order

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$. In this section we discuss the relations between inequality $\mathbf{A}\preccurlyeq\mathbf{B}$ and the domains of $\mathbf{A}^{\alpha}$ and $\mathbf{B}^{\alpha}$ for $\alpha\in\mathbb{Z}^{\kappa}$. In Theorem 7.3, the appropriate condition guaranteeing the inclusion ${\mathscr{D}}(\mathbf{B}^{\alpha})\subseteq{\mathscr{D}}(\mathbf{A}^{\alpha})$ will be formulated.

We begin our consideration by some facts about positive and negative parts of selfadjoint operators. Let $A$ be a selfadjoint operator in ${\mathcal{H}}$. We denote the positive part and the negative part of $A$ by $A_{+}$ and $A_{-}$, respectively. By the definition,

[TABLE]

where the functions $f_{\pm}\colon\mathbb{R}\to\mathbb{R}$ are given by

[TABLE]

It is known that $A_{+}$ and $A_{-}$ are selfadjoint. Moreover, $A_{+}$ is positive. Since the functions $f_{+}$ and $f_{-}$ are increasing, the following property holds.

Lemma 7.1**.**

If $A$ and $B$ are selfadjoint operators in ${\mathcal{H}}$ such that $A\preccurlyeq B$, then $A_{+}\preccurlyeq B_{+}$ and $A_{-}\preccurlyeq B_{-}$.

Let $\varepsilon=(\varepsilon_{1},\ldots,\varepsilon_{\kappa})\in\{-,+\}^{\kappa}$. The function $f_{\varepsilon_{1}}\times\ldots\times f_{\varepsilon_{\kappa}}\colon\mathbb{R}^{\kappa}\rightarrow\mathbb{R}^{\kappa}$, where $f_{\pm}$ are given by formulas (7.1), will be dentoted by $f_{\varepsilon}$. In particular,

[TABLE]

In the sequel we will use the following multi-sign $\dagger=(\dagger_{1},\ldots,\dagger_{\kappa})\in\{-,+\}^{\kappa}$, where $\dagger_{j}=+$ for $j=1,\ldots,\kappa$. For $\mathbf{C}=(C_{1},\ldots,C_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, we define $\mathbf{C}_{\varepsilon}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ by

[TABLE]

Then, by (5.11), (7.2), and [5, Lemma 6.5.2.], we derive that

[TABLE]

Moreover, by equality $f_{\varepsilon_{j}}(C_{j})=(\pi_{j}\circ f_{\varepsilon})(\mathbf{C})$ and [5, Theorem 5.4.10], we get that

[TABLE]

Thus, by [5, Theorem 6.5.1],

[TABLE]

Lemma 7.2**.**

If $\mathbf{C}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$, then

[TABLE]

Proof.

Set $\mathbb{R}_{-}:=(-\infty,0]$. If $\varepsilon=(\varepsilon_{1},\ldots,\varepsilon_{\kappa})\in\{-,+\}^{\kappa}$, then we define $\mathbb{R}_{\varepsilon}^{\kappa}:=\mathbb{R}_{\varepsilon_{1}}\times\ldots\times\mathbb{R}_{\varepsilon_{\kappa}}$.

Let $h\in{\mathcal{H}}$. Applying equality (7.4) we get that

[TABLE]

Thus

[TABLE]

which gives (7.5), since $h\in{\mathcal{H}}$ was chosen arbitrarily. ∎

The following theorem answers question raised at the beginning of this section.

Theorem 7.3**.**

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and let $\alpha\in\mathbb{Z}_{+}^{\kappa}$. If $\mathbf{A}\preccurlyeq\mathbf{B}$ and

[TABLE]

then

[TABLE]

Proof.

First observe that the function $f_{\varepsilon}$ is increasing. Hence, by Corollary 5.10 and assumptions, we get that $\mathbf{A}_{\varepsilon}=f_{\varepsilon}(\mathbf{A})\preccurlyeq f_{\varepsilon}(\mathbf{B})=\mathbf{B}_{\varepsilon}$.

Now we will prove the inclusion (7.7). It follows from the condition (7.6) that ${\mathscr{D}}((\mathbf{A}_{\varepsilon})^{\alpha})={\mathcal{H}}$ for every $\varepsilon\neq\dagger$. This and the equality (7.5) imply that

[TABLE]

Note that $\operatorname{supp}E_{\mathbf{A}_{\dagger}}\cup\operatorname{supp}E_{\mathbf{B}_{\dagger}}\subseteq[0,\infty)^{\kappa}$ by Corollary 6.1 and the definition of $\mathbf{A}_{\dagger}$ and $\mathbf{B}_{\dagger}$. Hence, by Theorem 5.7, we have $(\mathbf{A}_{\dagger})^{\alpha}=\psi_{\alpha}(\mathbf{A}_{\dagger})\preccurlyeq\psi_{\alpha}(\mathbf{B}_{\dagger})=(\mathbf{B}_{\dagger})^{\alpha}$, since the function $\psi_{\alpha}$ is increasing. Moreover, the operators $(\mathbf{A}_{\dagger})^{\alpha}$ and $(\mathbf{B}_{\dagger})^{\alpha}$ are positive. Applying [24, Proposition 7.1.] we infer that ${\mathscr{D}}((\mathbf{B}_{\dagger})^{\alpha})\subseteq{\mathscr{D}}((\mathbf{A}_{\dagger})^{\alpha})$. Eventually, using equalities (7.5) and (7.8), we deduce that

[TABLE]

which completes the proof. ∎

Remark 7.4*.*

If $\kappa=1$ and $\alpha=\alpha_{1}=1$, then Theorem 7.3 implies the second part of [24, Proposition 6.3.]. Indeed, the condition $A_{-}\in\mathbf{B}({\mathcal{H}})$ holds if and only if $A$ is bounded from below.

We are now going to discuss in more detail the condition (7.6). It may happen that the inclusion (7.7) may not hold, even if the condition (7.6) does not hold for only one $\varepsilon\in\{-,+\}^{\kappa}\backslash\{\dagger\}$. This will be illustrated by Example 7.8. The example will be preceded by two propositions describing spectral order for multiplication operators.

Proposition 7.5**.**

Let $(X,\mathcal{A},\mu)$ be a measure space. Assume that $\mu$ is $\sigma$-finite. If $\varphi,\psi\colon X\to\mathbb{R}$ are $\mathcal{A}$-measurable, then the following conditions are equivalent:

1. (i)

$M_{\varphi}\preccurlyeq M_{\psi}$, 2. (ii)

$\langle M_{\varphi}h,h\rangle\leqslant\langle M_{\psi}h,h\rangle$* for every $h\in{\mathscr{D}}(M_{\varphi})\cap{\mathscr{D}}(M_{\psi})$,* 3. (iii)

$\varphi\leqslant\psi$* a.e. $[\mu]$.*

Proof.

(i)$\Rightarrow$(ii) It follows from [24, Proposition 6.3.].

(ii)$\Rightarrow$(iii) Suppose that for some $N\in\mathbb{N}$ there exists an $\mathcal{A}$-measurable set $\Delta\subset\{x\in X\colon-N<\psi(x)<\varphi(x)<N\}$ such that $0<\mu(\Delta)<\infty$. Then

[TABLE]

On the other hand, $h:=\chi_{\Delta}\in{\mathscr{D}}(M_{\varphi})\cap{\mathscr{D}}(M_{\psi})$. Thus, by assumption,

[TABLE]

which is a contradiction. Hence $\varphi\leqslant\psi$ a.e. $[\mu]$, since $\mu$ is $\sigma$-finite.

(iii)$\Rightarrow$(i) Let $t\in\mathbb{R}$. Note that

[TABLE]

By assumption, $\mu(\{x\in X\colon\psi(x)<\varphi(x)\})=0$. Thus, by [27, Example 4.3.] and (7.9), we obtain that

[TABLE]

for every $h\in L^{2}(X,\mu)$. Hence $F_{M_{\psi}}(t)\leqslant F_{M_{\varphi}}(t)$ for every $t\in\mathbb{R}$, which completes the proof. ∎

For the sake of completeness, we include the following lemma.

Lemma 7.6**.**

Let $(X,\mathcal{A},\mu)$ be a measure space. Assume that $\varphi_{j}\colon X\to\mathbb{R}$ is an $\mathcal{A}$-measurable function for every $j=1,\ldots,\kappa$. If $\varphi=(\varphi_{1},\ldots,\varphi_{\kappa})\colon X\to\mathbb{R}^{\kappa}$ and $\mathbf{M}_{\varphi}:=(M_{\varphi_{1}},\ldots,M_{\varphi_{\kappa}})$, then

1. (i)

$\mathbf{M}_{\varphi}\in\mathcal{S}_{c}(L^{2}(X,\mu),\kappa)$, 2. (ii)

for every $\sigma\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$ and $h\in L^{2}(X,\mu)$

[TABLE] 3. (iii)

if $f\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ is a Borel function, then

[TABLE]

Proof.

(i) Let $\sigma,\tau\in{\mathfrak{B}}(\mathbb{R})$ and $h\in L^{2}(X,\mu)$. Applying equation (2.2) we obtain that

[TABLE]

Hence $\mathbf{M}_{\varphi}\in\mathcal{S}_{c}(L^{2}(X,\mu),\kappa)$.

(ii) Let us define $E(\sigma)h:=\chi_{\varphi^{-1}(\sigma)}h$ for $\sigma\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$ and $h\in L^{2}(X,\mu)$. Take $\sigma=\sigma_{1}\times\ldots\times\sigma_{\kappa}$, where $\sigma_{1},\ldots,\sigma_{\kappa}\in{\mathfrak{B}}(\mathbb{R})$. Then, by (2.2) and (5.3),

[TABLE]

Hence $E_{\mathbf{M}_{\varphi}}(\sigma)=E(\sigma)$ for every $\sigma\in{\mathfrak{B}}(\mathbb{R}^{\kappa})$, since $E$ is a spectral measure on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$ and $({\mathfrak{B}}(\mathbb{R}))^{\kappa}$ generates $\sigma$-algebra ${\mathfrak{B}}(\mathbb{R}^{\kappa})$.

(iii) Using [5, Theorem 5.4.10.]transport theorem, (7.10), and (2.2), we obtain

[TABLE]

This implies (7.11), since there is a one to one correspondence between spectral measures and selfadjoint operators. ∎

Combining Proposition 7.5, Lemma 7.6, and Theorem 5.7, we obtain the following proposition.

Proposition 7.7**.**

Let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite measure space. Assume that $\varphi_{j},\psi_{j}\colon X\to\mathbb{R}$ are $\mathcal{A}$-measurable functions for every $j=1,\ldots,\kappa$. Consider $\varphi=(\varphi_{1},\ldots,\varphi_{\kappa})\colon X\to\mathbb{R}^{\kappa}$ and $\psi=(\psi_{1},\ldots,\psi_{\kappa})\colon X\to\mathbb{R}^{\kappa}$. Then

[TABLE]

Example** 7.8****.**

Let ${\mathcal{H}}=L^{2}(\mathbb{R}^{\kappa},m_{\kappa})$, where $m_{\kappa}$ denotes Lebesgue measure on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$, and let $\varepsilon=(\varepsilon_{1},\ldots,\varepsilon_{\kappa})\in\{-,+\}^{\kappa}\backslash\{\dagger\}$. Fix $j_{0}\in\{1,\ldots,\kappa\}$ such that $\varepsilon_{j_{0}}=-$. Define the functions $\varphi_{j},\psi_{j}\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ for $j\in\{1,\ldots,\kappa\}$ by

[TABLE]

[TABLE]

and

[TABLE]

It is evident that $\varphi_{j}$ and $\psi_{j}$ are Borel function for every $j=1,\ldots,\kappa$. Let $\mathbf{A}:=M_{\varphi}$ and $\mathbf{B}:=M_{\psi}$, where $\varphi=(\varphi_{1},\ldots,\varphi_{\kappa})$ and $\psi=(\psi_{1},\ldots,\psi_{\kappa})$. According to Lemma 7.6, $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$.

We are going to show that $\mathbf{A}$ and $\mathbf{B}$ satisfy the following conditions:

1. (i)

$\mathbf{A}\preccurlyeq\mathbf{B}$, 2. (ii)

$(\mathbf{A}_{\delta})^{\alpha}\in\mathbf{B}({\mathcal{H}})$ for every $\delta\in\{-,+\}^{\kappa}\backslash\{\varepsilon\}$ and for all $\alpha\in\mathbb{N}^{\kappa}$, 3. (iii)

$(\mathbf{A}_{\varepsilon})^{\alpha}\notin\mathbf{B}({\mathcal{H}})$ for every $\alpha\in\mathbb{N}^{\kappa}$, 4. (iv)

${\mathscr{D}}(\mathbf{B}^{\alpha})\not\subseteq{\mathscr{D}}(\mathbf{A}^{\alpha})$ for every $\alpha\in\mathbb{N}^{\kappa}$.

(i) This is the consequence of Proposition 7.7, since $\varphi_{j}(x)\leqslant\psi_{j}(x)$ for every $x\in\mathbb{R}^{\kappa}$ and $j=1,\ldots,\kappa$.

(ii) We can find $i\in\{1,\ldots,\kappa\}$ such that $\delta_{i}\neq\varepsilon_{i}$, since $\delta=(\delta_{1},\ldots,\delta_{\kappa})\neq\varepsilon$. Then

[TABLE]

By definition (7.3), Lemma 7.6 (iii), and equality (7.14), we obtain

[TABLE]

(iii) Similarly to (ii), we show that

[TABLE]

since the function $\prod_{j=1}^{\kappa}(f_{\varepsilon_{j}}\circ\varphi_{j})^{\alpha_{j}}(x)=(-1)^{\alpha_{j_{0}}}x^{\alpha-\alpha_{j_{0}}e_{j_{0}}}(x_{j_{0}}^{2}+1)^{\alpha_{j_{0}}}\chi_{\mathbb{R}^{\kappa}_{\varepsilon}}(x)$ is not essentially bounded with respect to Lebesgue measure.

(iv) Using Lemma 7.6 (iii), we get that

[TABLE]

Define a function $h\colon\mathbb{R}^{\kappa}\to\mathbb{R}$ by the following formula

[TABLE]

where $\Delta=[-1,1]\times\ldots\times[-1,1]\times\underbrace{(-\infty,0]}_{j_{0}}\times[-1,1]\times\ldots\times[-1,1]$. Note that $h\in{\mathscr{D}}(\mathbf{B}^{\alpha})\backslash{\mathscr{D}}(\mathbf{A}^{\alpha})$. Indeed, by the definition of $h$ and Fubini theorem, we get

[TABLE]

Thus $h\in{\mathscr{D}}(M_{\Pi_{j=1}^{\kappa}(\psi_{j})^{\alpha_{j}}})$. At the same time

[TABLE]

since $\alpha_{j_{0}}\neq 0$. This means that $h\notin{\mathscr{D}}(M_{\Pi_{j=1}^{\kappa}(\varphi_{j})^{\alpha_{j}}})$.

8. Positive $\kappa$-tuples

In this section we study the multidimensional spectral order in the case of positive $\kappa$-tuples of pairwise commuting selfadjoint operators. We begin by the following proposition, which generalizes [24, Proposition 7.1.].

Proposition 8.1**.**

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and let $\alpha\in[0,\infty)^{\kappa}$. Suppose that $\mathbf{A}$ and $\mathbf{B}$ are positive. If $\mathbf{A}\preccurlyeq\mathbf{B}$, then the following conditions hold:

1. (i)

$\mathbf{A}^{\alpha}\preccurlyeq\mathbf{B}^{\alpha}$, 2. (ii)

${\mathscr{D}}(\mathbf{B}^{\alpha})\subseteq{\mathscr{D}}(\mathbf{A}^{\alpha})$, 3. (iii)

$\|\mathbf{A}^{\alpha}h\|\leqslant\|\mathbf{B}^{\alpha}h\|$* for every $h\in{\mathscr{D}}(\mathbf{B}^{\alpha})$,* 4. (iv)

$\langle\mathbf{A}^{\alpha}h,h\rangle\leqslant\langle\mathbf{B}^{\alpha}h,h\rangle$* for every $h\in{\mathscr{D}}(\mathbf{B}^{\alpha})$,* 5. (v)

$\mathbf{A}^{\alpha}\leqslant\mathbf{B}^{\alpha}$.

Moreover ${\mathscr{D}}^{\infty}(\mathbf{B})\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$ and $\mathscr{B}(\mathbf{B})\subseteq\mathscr{B}(\mathbf{A})$.

Proof.

(i) Apply Theorem 5.7 for function $\psi_{\alpha}$ defined by equality (6.1), which is increasing.

(ii), (iii), (iv) and (v) follows from (i) and [24, Proposition 7.1.] applied for $s=1$ and positive operators $\mathbf{A}^{\alpha}$ and $\mathbf{B}^{\alpha}$.

To prove the ”moreover” part let us note that inclusion ${\mathscr{D}}^{\infty}(\mathbf{B})\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$ follows from (ii) and Lemma 6.2. If $h\in\mathscr{B}_{a}(\mathbf{B})$ for some $a\in(0,\infty)^{\kappa}$, then using (iii) and Proposition 6.3 (b) we can find real number $c>0$ such that

[TABLE]

Applying Proposition 6.3 once more we get that $h\in\mathscr{B}_{a}(\mathbf{A})$, which gives us the second inclusion. ∎

The following two theorems Theorem 8.2 and Theorem 8.3, which are counterparts of [24, Theorem 7.4.], characterize multidimensional spectral order for positive $\kappa$-tuples in terms of appropriate inequalities for $\mathcal{C}^{\infty}$-vectors. In Theorem 8.3 we assume additionally that one of the compared $\kappa$-tuples consists of injective operators.

Theorem 8.2**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa}),\mathbf{B}=(B_{1},\ldots,B_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive. Then the following conditions are equivalent:

1. (i)

$\mathbf{A}\preccurlyeq\mathbf{B}$, 2. (ii)

${\mathscr{D}}^{\infty}(\mathbf{B})\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$* and $L_{\mathbf{A}/\mathbf{B}}(h)\leqslant 1$ for every $h\in{\mathscr{D}}^{\infty}(\mathbf{B})$,* 3. (iii)

$\mathscr{B}(\mathbf{B})\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$* and $L_{\mathbf{A}/\mathbf{B}}(h)\leqslant 1$ for every $h\in\mathscr{B}(\mathbf{B})$,* 4. (iv)

$\mathscr{B}(\mathbf{B})\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$* and $L_{A_{j}/B_{j}}(h)\leqslant 1$ for every $h\in\mathscr{B}(\mathbf{B})$ and $j=1,\ldots,\kappa$,* 5. (v)

$\mathscr{B}(\mathbf{B})\subseteq\mathscr{B}(\mathbf{A})$* and $L_{\mathbf{A}/\mathbf{B}}(h)\leqslant 1$ for every $h\in\mathscr{B}(\mathbf{B})$,*

where111We use the convention that $0/0=0$ and $a/0=\infty$ if $a>0$.

[TABLE]

Proof.

(i)$\Rightarrow$(ii) and (i)$\Rightarrow$(v) follow from Proposition 8.1.

Implications (ii)$\Rightarrow$(iii), (iii)$\Rightarrow$(iv), and (v)$\Rightarrow$(iii) are obvious.

(iv)$\Rightarrow$(i) According to Theorem 5.7, it suffices to show that

[TABLE]

since $\mathbf{A},\mathbf{B}$ are positive. Fix $j\in\{1,\ldots,\kappa\}$ and $x\geqslant 0$. Let $h\in{\mathscr{R}}(F_{B_{j}}(x))$. Define $h_{k}:=E_{\mathbf{B}}(\Delta_{k})h$ for $k\in\mathbb{N}$, where

[TABLE]

By Proposition 6.3, $h_{k}\in\mathscr{B}(\mathbf{B})$ for every $k\in\mathbb{N}$. In particular, by the assumption, Lemma 6.2, and [24, Proposition A.1.], $h_{k}\in\mathscr{B}(B_{j})\cap{\mathscr{D}}^{\infty}(A_{j})$ and $\lim_{n\to\infty}\sqrt[n]{\|A_{j}^{n}h_{k}\|/\|B_{j}^{n}h_{k}\|}=L_{A_{j}/B_{j}}(h_{k})\leqslant 1$ for every $k\in\mathbb{N}$. This implies that there is $N\in\mathbb{N}$ such that $\|A_{j}^{n}h_{k}\|=\|B_{j}^{n}h_{k}\|=0$ or $\|B_{j}^{n}h_{k}\|\neq 0$ for $n>N$. Then, by [29, Lemma 8.], we have

[TABLE]

Hence, by [29, Lemma 8.] and [24, Proposition 5.1.], $h_{k}\in\mathscr{B}_{x}(A_{j})={\mathscr{R}}(F_{A_{j}}(x))$ for every $k\in\mathbb{N}$. Note that $h_{k}\to h$ as $k\to\infty$, since $h\in{\mathscr{R}}(F_{B_{j}}(x))$ and \bigcup_{k\in\mathbb{N}}\Delta_{k}=[0,\infty)\times\ldots\times\underbrace{[0,x]}\limits_{j}\times\ldots\times[0,\infty). Thus $h\in{\mathscr{R}}(F_{A_{j}}(x))$ as ${\mathscr{R}}(F_{A_{j}}(x))$ is closed, which proves (8.1). ∎

Theorem 8.3**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa}),\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive. Assume that ${\mathscr{N}}(A_{j})=\{0\}$ for $j=1,\ldots,\kappa$. If $\Lambda\subseteq[0,\infty)^{\kappa}$ satisfies the condition (6.3), then the following conditions are equivalent:

1. (i)

$\mathbf{A}\preccurlyeq\mathbf{B}$, 2. (ii)

${\mathscr{D}}^{\infty}(\mathbf{B})\subseteq{\mathscr{D}}^{\infty}(\mathbf{A})$* and $L_{\mathbf{A}/\mathbf{B}}(h)\leqslant 1$ for every $h\in{\mathscr{D}}^{\infty}(\mathbf{B})$,* 3. (iii)

$\mathscr{B}(\mathbf{B})\subseteq\mathscr{D}^{\Lambda}(\mathbf{A})$* and $L_{\mathbf{A}/\mathbf{B}}^{\Lambda}(h)\leqslant 1$ for every $h\in\mathscr{B}(\mathbf{B})$,* 4. (iv)

$\mathscr{B}(\mathbf{B})\subseteq\mathscr{B}(\mathbf{A})$* and $L_{\mathbf{A}/\mathbf{B}}^{\Lambda}(h)\leqslant 1$ for every $h\in\mathscr{B}(\mathbf{B})$,*

where

[TABLE]

Proof.

(i)$\Rightarrow$(ii) This follows from Proposition 8.1 and equality (6.2).

(ii)$\Rightarrow$(iii) It is obvious.

(iii)$\Rightarrow$(i) It suffices to verify, that $F_{\mathbf{B}}(a)\leqslant F_{\mathbf{A}}(a)$ for every $a\in[0,\infty)^{\kappa}$, since $\mathbf{A}$ and $\mathbf{B}$ are positive. Fix $a\in[0,\infty)^{\kappa}$ and $h\in{\mathscr{R}}(F_{\mathbf{B}}(a))$. Applying Proposition 6.3 we get that

[TABLE]

for some real number $c>0$. By the assumptions

[TABLE]

Fix $\varepsilon>0$. Since $L_{\mathbf{A}/\mathbf{B}}^{\Lambda}(h)\leqslant 1$, there exists a positive number $M\in\mathbb{N}$ such that

[TABLE]

In particular,

[TABLE]

for every $\alpha\in\Lambda_{M}$. Note that the set $\Lambda_{M}$ satifies the condition (6.3). Hence, by (8.2) and Proposition 6.3, we obtain that $h\in{\mathscr{R}}(F_{\mathbf{A}}((1+\varepsilon)a))$. Since $\varepsilon>0$ was chosen arbitrarily, we infer that $h\in{\mathscr{R}}(F_{\mathbf{A}}(a))$.

(i)$\Rightarrow$(iv) It follows from Proposition 8.1.

(iv)$\Rightarrow$(iii) Note that $\mathscr{B}(\mathbf{A})\subseteq\mathscr{D}^{\Lambda}(\mathbf{A})$ which is the consequence of equality (6.2). ∎

The next proposition describes the multidimensional spectral order for positive elements in $\mathcal{S}_{c}({\mathcal{H}},\kappa)$ in terms of the classical order ”$\leqslant$”.

Proposition 8.4**.**

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive. Suppose that there exists a family $\{r_{\alpha}\}_{\alpha\in[0,\infty)^{\kappa}}\subseteq[1,\infty)$ such that

[TABLE]

Then the following conditions are equivalent:

1. (i)

$\mathbf{A}\preccurlyeq\mathbf{B}$, 2. (ii)

$\mathbf{A}^{\alpha}\preccurlyeq\mathbf{B}^{\alpha}$* for every $\alpha\in[0,\infty)^{\kappa}$,* 3. (iii)

$\mathbf{A}^{\alpha}\leqslant\mathbf{B}^{\alpha}$* for every $\alpha\in[0,\infty)^{\kappa}$,* 4. (iv)

$\mathbf{A}^{\alpha}\leqslant r_{\alpha}\mathbf{B}^{\alpha}$* for every $\alpha\in[0,\infty)^{\kappa}$.*

Proof.

(i)$\Rightarrow$(ii) and (ii)$\Rightarrow$(iii) follow from Proposition 8.1.

(iii)$\Rightarrow$(iv) It is obvious.

(iv)$\Rightarrow$(i) Let $h\in\mathscr{B}(\mathbf{B})$. Then, by assumption and equality (6.2), $h\in{\mathscr{D}}(\mathbf{B}^{\frac{\alpha}{2}})\subseteq{\mathscr{D}}(\mathbf{A}^{\frac{\alpha}{2}})$ and

[TABLE]

In particular, $h\in{\mathscr{D}}^{\infty}(\mathbf{A})$. Applying (8.4) and (8.3) we obtain

[TABLE]

Hence $\mathbf{A}\preccurlyeq\mathbf{B}$ by Theorem 8.2. ∎

Repeating the proof of Proposition 8.4 and applying Theorem 8.3 instead of Theorem 8.2, we have the following proposition.

Proposition 8.5**.**

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive and let $\Lambda\subseteq[0,\infty)^{\kappa}$ such that ${\mathscr{N}}(A_{j})=\{0\}$ for $j=1,\ldots,\kappa$, Suppose that the set $\Lambda\subseteq[0,\infty)^{\kappa}$ satisfies condition (6.3), and there exists a family $\{r_{\alpha}\}_{\alpha\in\Lambda}\subseteq[1,\infty)$ satisfying the following condition

[TABLE]

Then $\mathbf{A}\preccurlyeq\mathbf{B}$ if and only if $\mathbf{A}^{\alpha}\leqslant r_{\alpha}\mathbf{B}^{\alpha}$ for every $\alpha\in\Lambda$.

Let $\mathbf{A},\mathbf{B}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive. Let us consider the set

[TABLE]

Proposition 8.1 tells us, that $\Lambda(\mathbf{A},\mathbf{B})=[0,\infty)^{\kappa}$ if $\mathbf{A}\preccurlyeq\mathbf{B}$. Taking into account [24, Corollary 7.6.], we are going to determine what are sufficient and necessary condition on the set $\Lambda(\mathbf{A},\mathbf{B})$ for the relation $\mathbf{A}\preccurlyeq\mathbf{B}$ to hold.

Proposition 8.6**.**

Let $\mathbf{A}=(A_{1},\ldots,A_{\kappa}),\mathbf{B}=(B_{1},\ldots,B_{\kappa})\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ be positive. Cosider the following conditions:

1. (i)

$\mathbf{A}\preccurlyeq\mathbf{B}$, 2. (ii)

the set $\Lambda(\mathbf{A},\mathbf{B})\cap\{se_{j}\colon s\in[0,\infty)\}$ is unbounded for every $j=1,\ldots,\kappa$. 3. (iii)

[TABLE]

Then conditions (i) and (ii) are equivalent. Moreover, if ${\mathscr{N}}(A_{j})=\{0\}$ for $j=1,\ldots,\kappa$, then all the conditions (i)-(iii) are equivalent.

Proof.

(i)$\Rightarrow$(ii) It is a consequence of the Proposition 8.1.

(ii)$\Rightarrow$(i) Fix $j\in\{1,\ldots,\kappa\}$. By [5, Lemma 6.5.2] we get that $C_{j}^{s}=\mathbf{C}^{se_{j}}$ for every positive $\mathbf{C}\in\mathcal{S}_{c}({\mathcal{H}},\kappa)$ and for every $s\in[0,\infty)$, since $\psi_{se_{j}}(x)=\psi_{s}\circ\pi_{j}(x)$ for every $x\in[0,\infty)^{\kappa}$. Thus the set $\{s\in[0,\infty)\colon A_{j}^{s}\leqslant B_{j}^{s}\}$ is infinite for every $j=1,\ldots,\kappa$. From [24, Corollary 7.6] we infer that $A_{j}\preccurlyeq B_{j}$ for every $j=1,\ldots,\kappa$. Hence $\mathbf{A}\preccurlyeq\mathbf{B}$ by Theorem 5.7.

Now assume that ${\mathscr{N}}(A_{j})=\{0\}$ for $j=1,\ldots,\kappa$. The equivalence (i)$\Leftrightarrow$(iii) follows from Proposition 8.5 applied for $\Lambda=\Lambda(\mathbf{A},\mathbf{B})$ and $\{r_{\alpha}\}_{\alpha\in\Lambda}$, where $r_{\alpha}=1$ for every $\alpha\in\Lambda$. ∎

Note that in the case $\kappa=1$ conditions (ii) and (iii) coincide. In particular, injectivity of $A_{1}$ can be omitted by [24, Corollary 7.6]. On the other hand, the condition (8.5) may be to weak to ensure inequality $\mathbf{A}\preccurlyeq\mathbf{B}$ if $\kappa>1$. This can happen if some of the operators $A_{1},\ldots,A_{\kappa}$ are not injective.

Example** 8.7****.**

Let ${\mathcal{H}}=\mathbb{C}^{2}$ be a Hilbert space with standard orthonormal basis $\{(1,0),(0,1)\}$ and let $\theta\in[1,\infty)$. Consider the $2\times 2$ matrices $A_{1}$, $A_{2}$, $B_{\theta,1}$, $B_{\theta,2}$ given by

[TABLE]

Let us define $\mathbf{A}:=(A_{1},A_{2})$ and $\mathbf{B}_{\theta}:=(B_{\theta,1},B_{\theta,2})$. It is evident that $\mathbf{A},\mathbf{B}_{\theta}\in\mathbf{B}_{cs}({\mathcal{H}},2)$ are positive for $\theta\in[1,\infty)$, ${\mathscr{N}}(A_{1})={\mathcal{H}}$, and ${\mathscr{N}}(A_{2})=\{0\}$. We are going to show, that $\mathbf{A}$ and $\mathbf{B}_{\theta}$ satisfy the following conditions:

1. (i)

for every $k\in\mathbb{N}$ there exixts $\theta_{k}\in(2,\infty)$ such that

[TABLE] 2. (ii)

$\mathbf{A}\preccurlyeq\mathbf{B}_{\theta}$ if and only if $\theta=2$.

In particular, if $\theta\neq 2$, then $\mathbf{A}\not\preccurlyeq\mathbf{B}_{\theta}$ although $\mathbf{A}$ and $\mathbf{B}_{\theta}$ satisfy condition (8.5).

Proof.

(i) First, if $(s,t)\in(0,\infty)\times[0,\infty)$, then $A_{1}^{s}A_{2}^{t}=0$ and $B_{\theta,1}^{s}B_{\theta,2}^{t}=B_{\theta,2}^{t}\geqslant 0$. In particular $(s,t)\in\Lambda(\mathbf{A},\mathbf{B}_{\theta})$. On the other hand, by [24, Proposition 8.3.] there exists $\theta_{k}\in(2,\infty)$ such that

[TABLE]

This implies that

[TABLE]

By Löwner-Heinz inequality we get that $A_{2}^{s}=(A_{2}^{k})^{\frac{s}{k}}\leqslant(B_{\theta,2}^{k})^{\frac{s}{k}}=B_{\theta,2}^{s}$ for every $s\in[0,k]$. Thus

[TABLE]

(ii) We deduce from Theorem 5.7 and Corollary 5.12 that the inequality $\mathbf{A}\preccurlyeq\mathbf{B}_{\theta}$ holds if and only if $A=A_{2}-I_{\mathcal{H}}\preccurlyeq B_{\theta,2}-I_{\mathcal{H}}=B_{\theta}$, since $A_{1}\preccurlyeq B_{\theta,1}$. The application of [24, Lemma 8.2.] completes the proof.∎

Appendix A Resolutions of the identity on $\mathbb{R}^{\kappa}$

In this section, we recall the definition of an abstract multivariable resolution of the identity and we outline an idea how to construct spectral measures on $\mathbb{R}^{\kappa}$ by using multivariable resolutions of the identity (see [4] for the case of scalar measures).

For $c,d\in\mathbb{R}$ and $j\in\{1,\ldots,\kappa\}$, we define the difference operator $\bigtriangleup_{d,c}^{(j)}$ acting on operator-valued functions $F\colon\mathbb{R}^{\kappa}\to\mathbf{B}({\mathcal{H}})$ by

[TABLE]

for all $x_{1},\ldots,x_{\kappa}\in\mathbb{R}$. If $F\colon\mathbb{R}^{\kappa}\to\mathbf{B}({\mathcal{H}})$, then we set

[TABLE]

We say that an operator-valued function $F\colon\mathbb{R}^{\kappa}\to\mathbf{P}({\mathcal{H}})$ is a resolution of the identity on $\mathbb{R}^{\kappa}$ if $F$ satisfies the following three conditions222Given a sequence $\{a_{n}\}_{n=1}^{\infty}\subseteq\mathbb{R}^{\kappa}$ and $a\in\overline{\mathbb{R}}^{\kappa}$, we write $a_{n}\searrow a$ (resp., $a_{n}\nearrow a$) if $\{a_{n}\}_{n=1}^{\infty}$ is monotonically decreasing (resp., increasing) with respect to the partial order “$\leqslant$” in $\mathbb{R}^{\kappa}$, and convergent to $a$. If $a=(-\infty,\ldots,-\infty)$ (resp., $a=(\infty,\ldots,\infty)$), we also write $a_{n}\searrow-\infty$ (resp., $a_{n}\nearrow\infty$).:

1. (A)

\big{(}F(a,b]\big{)}(x)\in\mathbf{P}({\mathcal{H}}) for all $a,b,x\in\mathbb{R}^{\kappa}$ such that $a\leqslant b$, 2. (B)

$\textsc{sot-}\lim\limits_{\begin{subarray}{c}x\to x_{0}\\ x_{0}\leqslant x\end{subarray}}F(x)=F(x_{0})$ for every $x_{0}\in\mathbb{R}^{\kappa}$, 3. (C)

if $\{a_{n}\}_{n=1}^{\infty},\{b_{n}\}_{n=1}^{\infty}\subseteq\mathbb{R}^{\kappa}$ are such that $a_{n}\searrow-\infty$ and $b_{n}\nearrow\infty$, then

[TABLE]

If $E$ is a spectral measure on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$, then the function $F\colon\mathbb{R}^{\kappa}\to\mathbf{P}({\mathcal{H}})$ given by

[TABLE]

is a resolution of the identity. In fact, the equation (A.1) defines a one-to-one correspondence between the set of all resolutions of the identity on $\mathbb{R}^{\kappa}$ and the set of all spectral measures on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$, which can be deduced from the following theorem.

Theorem A.1**.**

Let $F$ be a resolution of the identity on $\mathbb{R}^{\kappa}$. Then there exists the unique spectral measure on ${\mathfrak{B}}(\mathbb{R}^{\kappa})$ such that

[TABLE]

The sketch of the proof.

The uniqueness. The uniqueness is a consequence of the fact that the family $\{(a,b]\colon a,b\in\mathbb{R}^{\kappa},\>a\leqslant b\}$ generates the $\sigma$-algebra ${\mathfrak{B}}(\mathbb{R}^{\kappa})$.

The existence. Denote by $\mathcal{F}_{0}(\mathbb{R}^{\kappa})$ the collection of all finite disjoint sums of sets of the form $(a,b]\cap\mathbb{R}^{\kappa}$, where $a,b\in\overline{\mathbb{R}}^{\kappa}$. Clearly, $\mathcal{F}_{0}(\mathbb{R}^{\kappa})$ is the algebra generated by the family $\mathcal{I}=\{(a,b]\cap\mathbb{R}^{\kappa}\colon a,b\in\overline{\mathbb{R}}^{\kappa}\}$. Now we define the function $E^{0}\colon\mathcal{F}_{0}(\mathbb{R}^{\kappa})\to\mathbf{P}({\mathcal{H}})$ in few steps:

Step 1. If $a,b\in\mathbb{R}^{\kappa}$ and $a\leqslant b$, then $E^{0}((a,b]):=F(a,b]$.

Step 2. If $a,b\in\overline{\mathbb{R}}^{\kappa}$ and $a\leqslant b$, then $E^{0}((a,b]):=\textsc{sot-}\lim F(a_{n},b_{n}]$, where $a_{n},b_{n}\in\mathbb{R}^{\kappa}$ for $\mathbb{N}$, $a_{n}\searrow a$ and $b_{n}\nearrow b$. The limit exists since $F(a,b]\leqslant F(c,d]$ whenever $c\leqslant a\leqslant b\leqslant d$.

Step 3. Now if $J=\bigcup_{j=1}^{n}J_{j}$, where $J_{1},\ldots,J_{n}\in\mathcal{I}$ are pairwise disjoint, then we set $E^{0}(J):=\sum_{j=1}^{n}E^{0}(J_{j})$. $E^{0}$ is well-defined and additive. Moreover, $E^{0}(\mathbb{R}^{\kappa})=I$ by (C).

By [5, Theorem 5.2.3.] it is sufficient to show that $E^{0}$ is $\sigma$-additive. Fix $h\in{\mathcal{H}}$. Let $\mu_{h}^{0}(J):=\langle E^{0}(J)h,h\rangle$ for $J\in\mathcal{F}_{0}(\mathbb{R}^{\kappa})$ and $F_{h}(x):=\langle F(x)h,h\rangle$ for $x\in\mathbb{R}^{\kappa}$. By the conditions (A) and (B), $F_{h}$ is a distribution function on $\mathbb{R}^{\kappa}$. Moreover, $\mu_{h}^{0}((a,b])=F_{h}(a,b]$ for $a,b\in\mathbb{R}^{\kappa},\>a\leqslant b$. By [4, Theorem 1.4.9.] there exists a Borel measure $\mu_{h}$ such that $\mu_{h}((a,b])=F_{h}(a,b]$ for $a,b\in\mathbb{R}^{\kappa},\>a\leqslant b$. In particular, $\mu_{h}|_{\mathcal{F}_{0}(\mathbb{R}^{\kappa})}=\mu_{h}^{0}$. Thus $\mu_{h}^{0}$ is $\sigma$-additive. Therefore $E^{0}$ is also $\sigma$-additive. ∎

Acknowledgements

I would like to thank Professor Jan Stochel for his helpful advice, which improved the paper.