**3. Chebyshev Type Inequalities Involving Tracy-Singh Products of Operators**

From now on, let Ω be an LCH space equipped with a finite Radon measure *μ*. Let A = (*At*)*<sup>t</sup>*∈Ω, B = (*Bt*)*<sup>t</sup>*∈Ω, C = (*Ct*)*<sup>t</sup>*∈<sup>Ω</sup> and D = (*Dt*)*<sup>t</sup>*∈<sup>Ω</sup> be continuous fields of Hilbert space operators.

**Definition 2.** *The fields* A *and* B *are said to have the synchronous Tracy-Singh property if, for all s*, *t* ∈ Ω*,*

$$(A\_l - A\_s) \boxtimes (B\_l - B\_s) \quad \text{\textquotedblleft} \text{ } 0.\tag{6}$$

*They are said to have the opposite-synchronous Tracy-Singh property if the reverse of* (6) *holds for all s*, *t* ∈ Ω*.*

**Theorem 1.** *Let* A *and* B *be bounded continuous fields of Hermitian operators in* B(H) *and* <sup>B</sup>(K)*, respectively, and let α* : Ω → [0, ∞) *be a bounded measurable function.*

*1. If* A *and* B *have the synchronous Tracy-Singh property, then*

$$\int\_{\Omega} a(\mathbf{s}) d\mu(\mathbf{s}) \int\_{\Omega} a(\mathbf{t}) (A\_{\mathbf{t}} \boxtimes \mathcal{B}\_{\mathbf{t}}) d\mu(\mathbf{t}) \, \geqslant \int\_{\Omega} a(\mathbf{t}) A\_{\mathbf{t}} d\mu(\mathbf{t}) \boxtimes \int\_{\Omega} a(\mathbf{s}) \mathcal{B}\_{\mathbf{t}} d\mu(\mathbf{s}).\tag{7}$$

*2. If* A *and* B *have the opposite-synchronous Tracy-Singh property, then the reverse of* (7) *holds.*

**Proof.** By using Lemma 1, Proposition 1 and Fubini's Theorem [13], we have

Ω *α*(*s*)*dμ*(*s*) Ω *α*(*t*)(*At Bt*)*dμ*(*t*) − Ω *α*(*t*)*Atdμ*(*t*) Ω *α*(*s*)*Bsdμ*(*s*) = Ω2 *α*(*s*)*α*(*t*)(*At Bt*)*dμ*(*t*)*dμ*(*s*) − Ω2 *α*(*t*)*α*(*s*)(*At Bs*)*dμ*(*t*)*dμ*(*s*) = 1 2 Ω2 [*α*(*s*)*α*(*t*)(*At Bt*) − *α*(*t*)*α*(*s*)(*At Bs*)] *dμ*(*t*)*dμ*(*s*) + 1 2 Ω2 [*α*(*t*)*α*(*s*)(*As Bs*) − *α*(*s*)*α*(*t*)(*As Bt*)] *dμ*(*s*)*dμ*(*t*) = 1 2 Ω2 *α*(*s*)*α*(*t*)[(*At* − *As*) (*Bt* − *Bs*)] *dμ*(*t*)*dμ*(*s*).

For the case 1, we have

$$\int\_{\Omega^2} a(\mathbf{s}) a(\mathbf{t}) \left[ (A\_\mathbf{t} - A\_\mathbf{s}) \boxtimes (B\_\mathbf{t} - B\_\mathbf{s}) \right] d\mu(\mathbf{t}) d\mu(\mathbf{s}) \quad \gtrapprox \ 0 \tag{8}$$

and thus (7) holds. For another case, we ge<sup>t</sup> the reverse of (8) and, thus, the reverse of (7) holds.

**Remark 1.** *In Theorem 1 and other results in this paper, we may assume that* Ω *is a compact Hausdorff space. In this case, every continuous field on* Ω *is automatically bounded.*

The next corollary is a discrete version of Theorem 1.

**Corollary 1.** *Let Ai*, *Bi be Hermitian operators and let ωi be nonnegative numbers for each i* = 1, ... , *n. Let* A = (*<sup>A</sup>*1,..., *An*) *and* B = (*<sup>B</sup>*1,..., *Bn*)*.*

*1. If* A *and* B *have the synchronous Tracy-Singh property, then*

$$\sum\_{i=1}^{n} \omega\_i \sum\_{i=1}^{n} \omega\_i (A\_i \boxtimes B\_i) \quad \geqslant \left(\sum\_{i=1}^{n} \omega\_i A\_i\right) \boxtimes \left(\sum\_{i=1}^{n} \omega\_i B\_i\right). \tag{9}$$

*2. If* A *and* B *have the opposite-synchronous Tracy-Singh property, then the reverse of* (9) *holds.*

**Proof.** From the previous theorem, set Ω = {1, ... , *n*} equipped with the counting measure and *α*(*i*) = *ωi* for all *i* = 1, . . . , *n*.

### **4. Chebyshev Integral Inequalities Concerning Weighted Pythagorean Means of Operators**

Throughout this section, the space Ω is equipped with a total ordering .

**Definition 3.** *We say that a field* A *is increasing (decreasing, resp.) whenever s t implies As At (As* - *At, resp.).*

**Definition 4.** *Two ordered pairs* (*<sup>X</sup>*1, *<sup>X</sup>*2) *and* (*<sup>Y</sup>*1,*Y*2) *of Hermitian operators are said to have the synchronous property if either*

$$X\_i \ll\_i Y\_i \text{ for } i = 1, 2, \text{ or } \ X\_i \gg\_i Y\_i \text{ for } i = 1, 2.$$

*The pairs* (*<sup>X</sup>*1, *<sup>X</sup>*2) *and* (*<sup>Y</sup>*1,*Y*2) *are said to have the opposite-synchronous property if either*

> *X*1 *Y*1 *and X*2 - *Y*2, *or X*1 - *Y*1 *and X*2 *Y*2.

**Definition 5.** *Let* A, B, C, D *be continuous fields of Hermitian operators. Two ordered pairs* (A, B) *and* (C, D) *are said to have the synchronous monotone property if* (*At*, *Bt*) *and* (*Ct*, *Dt*) *have the synchronous property for all t* ∈ Ω*. They are said to have the opposite-synchronous monotone property if* (*At*, *Bt*) *and* (*Ct*, *Dt*) *have the opposite-synchronous property for all t* ∈ Ω*.*

Let us recall the notions of weighted classical Pythagorean means for operators. Indeed, they are generalizations of three famous symmetric operator means as follows. For any *w* ∈ [0, 1], the *w*-weighted arithmetic mean of *A*, *B* ∈ B(H) is defined by

$$A \,\mathrm{\,\,\mathrm{w}}\,B\,=\,(1-w)A + wB.$$

The *w*-weighted geometric mean and *w*-weighted harmonic mean of *A*, *B* ∈ B(H)++ are defined respectively by

$$\begin{aligned} A\sharp\_{\overline{w}}B &= \, \_A\overset{\mathbb{I}}{\,}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\overline{w}}A^{\frac{1}{2}}\, \_\prime \\ A \, \_wB &= \, \_0\left[(1-w)A^{-1} + wB^{-1}\right]^{-1} .\end{aligned}$$

.

For any *A*, *B* ∈ <sup>B</sup>(H)+, we define the *w*-weighted geometric mean and *w*-weighted harmonic mean of *A* and *B* to be

$$A\sharp\_{\mathbb{B}}B = \lim\_{\varepsilon \to 0^{+}} (A + \varepsilon I)\sharp\_{\mathbb{B}}(B + \varepsilon I).$$

$$A \mathop{\!\!\!\!\_{\mathbb{B}}B} = \lim\_{\varepsilon \to 0^{+}} (A + \varepsilon I) \mathop{\!\!\!\_{\mathbb{B}}(B + \varepsilon I)}\nolimits$$

respectively. Here, the limits are taken in the strong-operator topology.

**Lemma 4** (see e.g., [14])**.** *The weighted geometric means, weighted arithmetic means and weighted harmonic means for operators are monotone in the sense that if A*1 *A*2 *and B*1 *B*2*, then A*1*σB*1 *A*2*σB*2 *where σ is any of <sup>w</sup>*, !*<sup>w</sup>*,*w.*

**Lemma 5** ([15])**.** *Let A*, *B*, *C*, *D* ∈ B(H)+ *and w* ∈ [0, 1]*. Then*

$$((A \boxtimes B) \sharp\_w (C \boxtimes D) \, = \, (A \sharp\_w C) \boxtimes (B \sharp\_w D) \,.$$

**Theorem 2.** *Let* A, B, C, D *be bounded continuous fields in* B(H)+ *and let α* : Ω → [0, ∞) *be a bounded measurable function.*

*1. If* A, B, C, D *are either all increasing, or all decreasing then*

$$\begin{split} \int\_{\Omega} \mathfrak{a}(\mathbf{s}) d\mu(\mathbf{s}) \int\_{\Omega} \mathfrak{a}(t) [(A\_t \boxtimes B\_t)\sharp\_w(\mathbb{C}\_t \boxtimes D\_t)] d\mu(t) \\ \quad \quad \quad \quad \quad \int\_{\Omega} \mathfrak{a}(t) (A\_t \sharp\_w \mathbb{C}\_t) d\mu(t) \boxtimes \int\_{\Omega} \mathfrak{a}(\mathbf{s}) (B\_t \sharp\_w D\_s) d\mu(\mathbf{s}). \end{split} \tag{10}$$

	- *2.1* A, C *are increasing and* B, D *are decreasing, or*
	- *2.2* A, C *are decreasing and* B, D *are increasing.*

**Proof.** Let *s*, *t* ∈ Ω and assume without loss of generally that *s t*. By applying Lemmas 1 and 5, Proposition 1, and Fubini's Theorem [13], we have

Ω*<sup>α</sup>*(*s*)*dμ*(*s*) Ω *α*(*t*)[(*At Bt*)*w*(*Ct Dt*)]*dμ*(*t*) − Ω *α*(*t*)(*AtwCt*)*dμ*(*t*) Ω *α*(*s*)(*BswDs*)*dμ*(*s*) = Ω2 *α*(*s*)*α*(*t*)[(*At Bt*)*w*(*Ct Dt*)]*dμ*(*t*)*dμ*(*s*) − Ω2 *α*(*t*)*α*(*s*)[(*AtwCt*) (*BswDs*)]*dμ*(*t*)*dμ*(*s*) = Ω2 *α*(*s*)*α*(*t*)[(*AtwCt*) (*BtwDt*)]*dμ*(*t*)*dμ*(*s*) − Ω2 *α*(*t*)*α*(*s*)[(*AtwCt*) (*BswDs*)]*dμ*(*t*)*dμ*(*s*) = 1 2 Ω2 *α*(*s*)*α*(*t*)[(*AtwCt*) (*BtwDt*) − (*AtwCt*) (*BswDs*)]*dμ*(*t*)*dμ*(*s*) + 1 2 Ω2 *α*(*t*)*α*(*s*)[(*AswCs*) (*BswDs*) − (*AswCs*) (*BtwDt*)]*dμ*(*s*)*dμ*(*t*) = 1 2 Ω2 *α*(*s*)*α*(*t*)[*AtwCt* − *AswCs*] [*BtwDt* − *Bs<sup>w</sup>Ds*]*dμ*(*t*)*dμ*(*s*).

If A, B, C, D are all increasing, we have by Lemma 4 that *AtwCt* - *AswCs* and *BtwDt* - *Bs<sup>w</sup>Ds*. If A, B, C, D are all decreasing, we have *AtwCt AswCs* and *BtwDt Bs<sup>w</sup>Ds*. Both cases lead to the same conclusion that

$$(A\_t \sharp\_w C\_t - A\_s \sharp\_w C\_s) \boxtimes (B\_t \sharp\_w D\_t - B\_s \sharp\_w D\_s) \quad \gtrless \ 0,$$

and hence (10) holds. The cases 2.1 and 2.2 yield the same conclusion that

$$(A\_t \sharp\_{\mathbb{H}} C\_t - A\_s \sharp\_{\mathbb{H}} C\_s) \boxtimes (B\_t \sharp\_{\mathbb{H}} D\_t - B\_s \sharp\_{\mathbb{H}} D\_s) \prec\_\circ 0.$$

and hence the reverse of (10) holds.

**Lemma 6.** *Let A*, *B*, *C*, *D be Hermitian operators in* B(H) *and w* ∈ [0, 1]*.*

*1. If* (*<sup>A</sup>*, *B*) *and* (*<sup>C</sup>*, *D*) *have the synchronous property, then*

$$(A \boxtimes B) \sqcap\_{\mathfrak{w}} (C \boxtimes D) \quad \not\simeq (A \sqcap\_{\mathfrak{w}} C) \boxtimes (B \sqcap\_{\mathfrak{w}} D). \tag{11}$$

*2. If* (*<sup>A</sup>*, *B*) *and* (*<sup>C</sup>*, *D*) *have the opposite-synchronous property, then the reverse of* (11) *holds.*

**Proof.** For the synchronous case, we have by using positivity of the Tracy-Singh product (Lemma 1) that (*A* − *C*) (*B* − *D*) - 0. Applying Lemma 1, we obtain

$$\begin{split} 0 &\leqslant w(1-w)\left[\left(A\_{1}-B\_{1}\right)\boxtimes \left(A\_{2}-B\_{2}\right)\right] \\ &= w(1-w)\left[A\_{1}\boxtimes A\_{2}-A\_{1}\boxtimes B\_{2}-B\_{1}\boxtimes A\_{2}+B\_{1}\boxtimes B\_{2}\right] \\ &= \left[(1-w)\left(A\_{1}\boxtimes A\_{2}\right)+w\left(B\_{1}\boxtimes B\_{2}\right)\right]-\left[(1-w)A\_{1}+wB\_{1}\right]\boxtimes \left[(1-w)A\_{2}+wB\_{2}\right] \\ &= \left[\left(A\_{1}\boxtimes A\_{2}\right)\bigtri\_{w}\left(B\_{1}\boxtimes B\_{2}\right)\right]-\left[\left(A\_{1}\boxtimes w\right)\boxtimes \left(A\_{2}\right)\bigtri\_{w}B\_{2}\right]. \end{split}$$

Thus (*<sup>A</sup>*1 *w <sup>B</sup>*1) (*<sup>A</sup>*2 *w <sup>B</sup>*2) (*<sup>A</sup>*1 *<sup>A</sup>*2) *w* (*<sup>B</sup>*1 *<sup>B</sup>*2).

For the opposite-synchronous case, we have (*<sup>A</sup>*1 − *<sup>B</sup>*1) (*<sup>A</sup>*2 − *<sup>B</sup>*2) 0 and hence the reverse of inequality (11) holds.

**Theorem 3.** *Let* A, B, C, D *be bounded continuous fields of operators in* <sup>B</sup>(H)+*, let α* : Ω → [0, ∞) *be a bounded measurable function.*

*1. If* (A, B) *and* (C, D) *have the synchronous monotone property and all of* A, B, C, D *are either increasing or decreasing, then*

$$\begin{split} \int\_{\Omega} a(s) d\mu(s) \int\_{\Omega} a(t) [(A\_t \boxtimes B\_t) \nabla\_w (\mathbb{C}\_t \boxtimes D\_t)] d\mu(t) \\ \quad \geqslant \int\_{\Omega} a(t) (A\_t \nabla\_w \mathbb{C}\_t) d\mu(t) \boxtimes \int\_{\Omega} a(s) (B\_s \nabla\_w D\_s) d\mu(s). \end{split} \tag{12}$$

*2. If* (A, B) *and* (C, D) *have the opposite-synchronous monotone property and if either*


*then the reverse of* (12) *holds.*

**Proof.** Let *s*, *t* ∈ Ω and assume without loss of generally that *s t*. First, we consider the case 1. We have by using Lemmas 1 and 6, proposition 1, and Fubini's Theorem [13] that

Ω *α*(*s*)*dμ*(*s*) Ω *α*(*t*)[(*At Bt*)*w*(*Ct Dt*)]*dμ*(*t*) − Ω *<sup>α</sup>*(*t*)(*AtwCt*)*dμ*(*t*) Ω *<sup>α</sup>*(*s*)(*BswDs*)*dμ*(*s*) = Ω2 *α*(*s*)*α*(*t*)[(*At Bt*)*w*(*Ct Dt*)]*dμ*(*t*)*dμ*(*s*) − Ω2 *<sup>α</sup>*(*t*)*α*(*s*)[(*AtwCt*) (*BswDs*)]*dμ*(*t*)*dμ*(*s*) - Ω2 *<sup>α</sup>*(*s*)*α*(*t*)[(*AtwCt*) (*BtwDt*)]*dμ*(*t*)*dμ*(*s*) − Ω2 *<sup>α</sup>*(*t*)*α*(*s*)[(*AtwCt*) (*BswDs*)]*dμ*(*t*)*dμ*(*s*) = Ω2 *<sup>α</sup>*(*s*)*α*(*t*)[(*AtwCt*) (*BtwDt*) − (*AtwCt*) (*BswDs*)]*dμ*(*t*)*dμ*(*s*) = 1 2 Ω2 *<sup>α</sup>*(*s*)*α*(*t*)[(*AtwCt*) − (*AswCs*)] [(*BtwDt*) − (*BswDs*)]*dμ*(*t*)*dμ*(*s*).

Now, by Lemmas 1 and 4, we have

$$(A\_l \nabla\_{\mathcal{W}} \mathcal{C}\_l - A\_s \nabla\_{\mathcal{W}} \mathcal{C}\_s) \boxtimes (B\_l \nabla\_{\mathcal{W}} D\_l - B\_s \nabla\_{\mathcal{W}} D\_s) \not\supset \mathcal{O}$$

and hence (12) holds. The case 2 can be similarly proven.

**Lemma 7.** *Let A*, *B*, *C*, *D be positive operators in* B(H) *and w* ∈ [0, 1]*.*

*1. If* (*<sup>A</sup>*, *B*) *and* (*<sup>C</sup>*, *D*) *are synchronous, then*

$$(A \boxplus B) \upharpoonright\_{\mathcal{W}} (\mathcal{C} \boxplus D) \quad \lessgtr (A \mathrel{\mathop{\cdot}}\_{\mathcal{W}} \mathcal{C}) \boxtimes (B \mathrel{\mathop{\cdot}}\_{\mathcal{W}} D).\tag{13}$$

*2. If* (*<sup>A</sup>*, *B*) *and* (*<sup>C</sup>*, *D*) *are opposite-synchronous, then the reverse of* (13) *holds.*

**Proof.** Assume that (*<sup>A</sup>*, *B*) and (*<sup>C</sup>*, *D*) are synchronous. By continuity, we may assume that *A*, *B*, *C*, *D* > 0. We have

$$(A^{-1} - \mathbb{C}^{-1}) \boxtimes (B^{-1} - D^{-1}) \quad \text{>} \quad \text{(14)}$$

Using Lemma 1 and (14), we ge<sup>t</sup>

$$\begin{split} 0 &\leqslant \ w(1-w)A^{-1}\boxtimes B^{-1} + w(1-w)\mathbb{C}^{-1}\boxtimes D^{-1} - w(1-w)A^{-1}\boxtimes D^{-1} - w(1-w)\mathbb{C}^{-1}\boxtimes B^{-1} \\ &= \left[(1-w) - (1-w)^{2}\right]A^{-1}\boxtimes B^{-1} + (w-w^{2})\mathbb{C}^{-1}\boxtimes D^{-1} - w(1-w)A^{-1}\boxtimes D^{-1} \\ &- w(1-w)\mathbb{C}^{-1}\boxtimes B^{-1} \\ &= \left(A^{-1}\boxtimes B^{-1}\right)\operatorname{\vee}\_{w}\left(\mathbb{C}^{-1}\boxtimes D^{-1}\right) - \left(A^{-1}\operatorname{\vee}\_{w}\mathbb{C}^{-1}\right)\boxtimes \left(B^{-1}\operatorname{\vee}\_{w}D^{-1}\right). \end{split}$$

This implies that

$$\left(\left(A^{-1}\boxtimes B^{-1}\right)\operatorname{\boldsymbol{\nabla}\_{\operatorname{\boldsymbol{w}}}}\left(\operatorname{\boldsymbol{C}}^{-1}\boxtimes D^{-1}\right)\right)\geqslant\left(A^{-1}\operatorname{\boldsymbol{\nabla}\_{\operatorname{\boldsymbol{w}}}}\operatorname{\boldsymbol{C}}^{-1}\right)\boxtimes\left(B^{-1}\operatorname{\boldsymbol{\nabla}\_{\operatorname{\boldsymbol{w}}}}\operatorname{\boldsymbol{D}}^{-1}\right)\geqslant$$

Hence,

$$\begin{split} \left( (A \boxtimes B) \, !\_{\text{w}} \left( \mathbb{C} \boxtimes D \right) \right) &= \left\{ \left( A \boxtimes B \right)^{-1} \, \middle\vert \, \mathbb{C} \boxtimes D \right\vert^{-1} \right\}^{-1} \\ &= \left\{ \left( A^{-1} \boxtimes B^{-1} \right) \, \middle\vert \, \mathbb{C}^{-1} \boxtimes D^{-1} \right\rfloor \right\}^{-1} \\ &\leqslant \left\{ \left( A^{-1} \, \middle\vert \, \mathbb{C}\_{\text{w}} \mathbb{C}^{-1} \right) \boxtimes \left( B^{-1} \, \middle\vert \, \mathbb{C}\_{\text{w}} D^{-1} \right) \right\}^{-1} \\ &= \left( A^{-1} \, \middle\vert \, \mathbb{C}\_{\text{w}} \mathbb{C}^{-1} \right)^{-1} \boxtimes \left( B^{-1} \, \middle\vert \, \mathbb{C}\_{\text{w}} D^{-1} \right)^{-1} \\ &= \left( A \, !\_{\text{w}} \mathbb{C} \right) \boxtimes \left( B \, !\_{\text{w}} D \right) . \end{split}$$

For the opposite-synchronous case, we have

$$(A^{-1} - \mathbb{C}^{-1}) \boxtimes (B^{-1} - D^{-1}) \preccurlyeq 0$$

and hence the reverse of (13) holds.

**Theorem 4.** *Let* A, B, C, D *be bounded continuous fields of operators in* B(H)+ *and α* : Ω → [0, ∞) *be a bounded measurable function.*

*1. If* (A, B) *and* (C, D) *have the opposite-synchronous monotone property and if all of* A, B, C, D *are either increasing or decreasing, then*

$$\begin{split} \int\_{\Omega} a(\mathbf{s}) d\mu(\mathbf{s}) \int\_{\Omega} a(t) [(A\_t \boxtimes B\_t) \mathop{\mathsf{T}}\_{\mathbf{w}} (\mathbb{C}\_t \boxtimes D\_t)] d\mu(t) \\ \quad \quad \quad \quad \quad \int\_{\Omega} a(t) (A\_t \mathop{\mathsf{T}}\_{\mathbf{w}} \mathop{\mathsf{C}}\_{\mathbf{t}}) d\mu(\mathbf{t}) \mathop{\boxtimes} \int\_{\Omega} a(\mathbf{s}) (B\_s \mathop{\mathsf{T}}\_{\mathbf{w}} D\_s) d\mu(\mathbf{s}). \end{split} \tag{15}$$

	- *2.1* A, C *are both increasing, and* B, D *are both decreasing, or*
	- *2.2* A, C *are both decreasing and* B, D *are both increasing,*

*then the reverse of* (15) *holds.*

**Proof.** Let *s*, *t* ∈ Ω with *s t*. If the pairs (A, B) and (C, D) are opposite-synchronous, then we have by applying Lemmas 1 and 7, Proposition 1, and Fubini's Theorem [13] that

$$\begin{split} &\int\_{\Omega} a(s) d\mu(s) \int\_{\Omega} a(t) [(A\_{t} \boxtimes B\_{t}) \mathop{!}\_{w} (\mathbb{C}\_{t} \boxtimes D\_{t})] d\mu(t) - \int\_{\Omega} a(t) (A\_{t} \mathop{!}\_{w} \mathop{!}\_{C}) d\mu(t) \boxtimes \int\_{\Omega} a(s) (B\_{s} \mathop{!}\_{w} D\_{s}) d\mu(s) \\ &= \iint\_{\Omega^{2}} a(s) a(t) [(A\_{t} \boxtimes B\_{t}) \mathop{!}\_{w} (\mathbb{C}\_{t} \boxtimes D\_{t})] d\mu(t) d\mu(s) \\ &\quad - \iint\_{\Omega^{2}} a(t) a(s) [(A\_{t} \mathop{!}\_{w} \mathop{!}\_{C}) \boxtimes (B\_{s} \mathop{!}\_{w} D\_{s})] d\mu(t) d\mu(s) \\ &\geqslant \iint\_{\Omega^{2}} a(s) a(t) [(A\_{t} \mathop{!}\_{w} \mathop{!}\_{C}) \boxtimes (B\_{t} \mathop{!}\_{w} D\_{t})] d\mu(t) d\mu(s) \\ &\quad - \iint\_{\Omega^{2}} a(t) a(s) [(A\_{t} \mathop{!}\_{w} \mathop{!}\_{C}) \boxtimes (B\_{s} \mathop{!}\_{w} D\_{s})] d\mu(t) d\mu(s) \\ &= \frac{1}{2} \int\_{\Omega^{2}} a(s) a(t) [(A\_{t} \mathop{!}\_{w} \mathop{!}\_{C} - A\_{s} \mathop{!}\_{w} \mathop{!}\_{C}]] \boxtimes [B\_{t} \mathop{!}\_{w} D\_{t} - B\_{s} \mathop{!}\_{w} D\_{s}] d\mu(t) d\mu(s). \end{split}$$

For the case 1, we have, by Lemmas 1 and 4,

$$(A\_{\mathcal{I}} \, !\_{\mathrm{w}} \, \mathbb{C}\_{\mathcal{I}} - A\_{\mathbb{s}} \, \, !\_{\mathrm{w}} \, \mathbb{C}\_{\mathcal{S}}) \boxtimes (B\_{\mathcal{I}} \, \, !\_{\mathrm{w}} \, D\_{\mathcal{I}} - B\_{\mathbb{s}} \, \, !\_{\mathrm{w}} \, D\_{\mathcal{S}}) \; \gtrless 001$$

and hence (15) holds. Another assertion can be proved in a similar manner to that of the second assertion in Theorem 3.
