*2.1. Tracy-Singh Product for Operators*

Let *A* ∈ B(H) and *B* ∈ <sup>B</sup>(K). Recall that the tensor product of *A* and *B*, denoted by *A* ⊗ *B*, is a unique bounded linear operator on the tensor product space H ⊗ K such that

$$(A \otimes B)(\mathfrak{x} \otimes \mathfrak{y}) = A\mathfrak{x} \otimes B\mathfrak{y}, \quad \forall \mathfrak{x} \in \mathbb{H}, \forall \mathfrak{y} \in \mathbb{K}.$$

When H = K = C, the tensor product of operators becomes the Kronecker product of matrices.

**Definition 1.** *Let A* = [*Aij*]*<sup>m</sup>*,*<sup>m</sup> i*,*j*=1 ∈ B(H) *and B* = [*Bkl*]*<sup>n</sup>*,*<sup>n</sup> k*,*l*=1 ∈ <sup>B</sup>(K)*. The Tracy-Singh product of A and B is defined to be in the form*

$$A \boxtimes B \quad = \left[ \left[ A\_{ij} \otimes B\_{kl} \right]\_{kl} \right]\_{ij'} \tag{5}$$

*which is a bounded linear operator from m i*=1 *n k*=1 H*i* ⊗ K*k into itself.*

When *m* = *n* = 1, the Tracy-Singh product *A B* is the tensor product *A* ⊗ *B*. If H*i* = K*j* = C for all *i*, *j*, the above definition becomes the usual Tracy-Singh product for complex matrices.

**Lemma 1** ([10,11])**.** *Let A*, *B*, *C*, *D be compatible operators. Then*



Let Ω be a locally compact Hausdorff (LCH) space equipped with a finite Radon measure *μ*. A family A = (*At*)*<sup>t</sup>*∈<sup>Ω</sup> of operators in B(H) is said to be bounded if there is a constant *M* > 0 for which *At M* for all *t* ∈ Ω. The family A is said to be a continuous field if parametrization *t* → *At* is norm-continuous

on Ω. Every continuous field A = (*At*)*<sup>t</sup>*∈<sup>Ω</sup> can have the Bochner integral Ω *Atdμ*(*t*) if the norm function *t* → *At* possess the Lebesgue integrability. In this case, the resulting integral is a unique element in B(H) such that

$$\left(\phi\left(\int\_{\Omega} A\_t d\mu(t)\right)\right) = \int\_{\Omega} \phi\left(A\_t\right) d\mu(t)$$

for every bounded linear functional *φ* on <sup>B</sup>(H).

**Lemma 2** (e.g., [12])**.** *Let* (<sup>X</sup>, ·X) *be a Banach space and* (<sup>Γ</sup>, *υ*) *a finite measure space. Then a measurable function f* : Γ → X *is Bochner integrable if and only if its norm function f is Lebesgue integrable.*

**Lemma 3** (e.g., [12])**.** *Let f* : Γ → X *be a Bochner integrable function. If ϕ* : X → Y *is a bounded linear operator, then the composition ϕ* ◦ *f is Bochner integrable and*

Γ(*ϕ* ◦ *f*)*dυ* = *ϕ* Γ *f d<sup>υ</sup>*.

**Proposition 1.** *Let* (*At*)*<sup>t</sup>*∈<sup>Ω</sup> *be a bounded continuous field of operators in* <sup>B</sup>(H)*. Then for any X* ∈ <sup>B</sup>(K)*,*

$$\int\_{\Omega} A\_t d\mu(t) \boxtimes X = \int\_{\Omega} (A\_t \boxtimes X) d\mu(t).$$

**Proof.** Since the map *t* → *At* is continuous and bounded, it is Bochner integrable on Ω. Note that the map *T* → *T X* is linear and bounded by Lemma 1. Now, Lemma 3 implies that the map *t* → *At X* is Bochner integrable on Ω and

$$\int\_{\Omega} A\_t d\mu(t) \boxtimes X = \int\_{\Omega} (A\_t \boxtimes X) d\mu(t).$$

for all *X* ∈ <sup>B</sup>(K).
