**5. Chebyshev-Grüss Inequaities via Oscillations**

Throughout this section, let Ω be an LCH space equipped with a probability Radon measure *μ*. For any continuous field A = (*At*)*<sup>t</sup>*∈<sup>Ω</sup> in B(H) and B = (*Bt*)*<sup>t</sup>*∈<sup>Ω</sup> in <sup>B</sup>(K), we define

$$\mathcal{A} \boxtimes \mathcal{B} = (A\_l \boxtimes B\_l)\_{t \in \Omega \prime} \quad \mathcal{T}(\mathcal{A}) = \int\_{\Omega} A\_l d\mu(t),$$

$$\text{osc}(\mathcal{A}) = \max \{ \|A\_l - A\_s\| : (t, s) \in \text{supp}(\mu \times \mu) \}.$$

Here, we recall that the support of the product measure *μ* × *μ* is defined by

$$\text{supp}(\mu \times \mu) \, = \{ (t, s) \in \Omega^2 \, : \, (\mu \times \mu)(G) > 0 \text{ for all open sets } G \subseteq \Omega^2 \text{ containing } (t, s) \}.$$

We call osc(A) the oscillation of the field A.

**Theorem 5.** *Let* A = (*At*)*<sup>t</sup>*∈<sup>Ω</sup> *and* B = (*Bt*)*<sup>t</sup>*∈<sup>Ω</sup> *be continuous fields of Hermitian operators in* B(H) *and* <sup>B</sup>(K)*, respectively. Then*

$$\mathcal{Z}(\mathcal{A}\boxtimes\mathcal{B}) - \mathcal{Z}(\mathcal{A})\boxtimes\mathcal{Z}(\mathcal{B}) \leqslant \frac{1}{2}\operatorname{osc}(\mathcal{A})\cdot\operatorname{osc}(\mathcal{B})(\mu\times\mu)(\Omega^{2}\backslash\Delta)(l\_{\overline{\mathbb{K}}}\boxtimes l\_{\overline{\mathbb{K}}}),\tag{16}$$

*where* Δ = {(*<sup>t</sup>*, *t*) : *t* ∈ <sup>Ω</sup>}*.*

**Proof.** We have by using Lemma 1, Proposition 1 and Fubini's Theorem [13] that

$$\begin{split} \mathcal{Z}(\mathcal{A}\boxtimes\mathcal{B}) - \mathcal{Z}(\mathcal{A})\boxtimes\mathcal{Z}\mathcal{Q} &= \int\_{\Omega} d\mu(s) \int\_{\Omega} A\_{t} \boxtimes B\_{t} d\mu(t) - \int\_{\Omega} A\_{t} d\mu(t) \boxtimes \int\_{\Omega} B\_{t} d\mu(s) \\ &= \iint\_{\Omega^{2}} A\_{t} \boxtimes B\_{t} d\mu(t) d\mu(s) - \iint\_{\Omega^{2}} A\_{t} \boxtimes B\_{s} d\mu(t) d\mu(s) \\ &= \frac{1}{2} \iint\_{\Omega^{2}} (A\_{t} \boxtimes B\_{t} - A\_{t} \boxtimes B\_{s} + A\_{s} \boxtimes B\_{s} - A\_{s} \boxtimes B\_{t}) d\mu(t) d\mu(s) \\ &= \frac{1}{2} \iint (A\_{t} - A\_{s}) \boxtimes (B\_{t} - B\_{s}) d\mu(t) d\mu(s) \\ &\leqslant \frac{1}{2} \operatorname{osc}(\mathcal{A}) \cdot \operatorname{osc}(\mathcal{B}) (\mu \times \mu)(\Omega^{2} \backslash \Delta) (I\_{\mathbb{E}} \boxtimes I\_{\mathbb{E}}). \qed \end{split}$$

**Corollary 2.** *Let Ai* ∈ B(H) *and Bi* ∈ B(K) *be Hermitian operators for all i* = 1, . . . , *n. Then*

$$\sum\_{i=1}^{n} (A\_i \boxtimes B\_i) - \left(\sum\_{i=1}^{n} A\_i\right) \boxtimes \left(\sum\_{i=1}^{n} B\_i\right) \leqslant \frac{n(n-1)}{2} \max\_{1 \leqslant i,j \leqslant n} \|A\_i - A\_j\| \cdot \max\_{1 \leqslant i,j \leqslant n} \|B\_i - B\_j\| (I\_{\boxplus} \boxtimes I\_{\mathbb{K}}) \cdot A\_i$$

**Proof.** Set Ω = {1, . . . , *n*} equipped with the counting measure. We have

$$(\mu \times \mu)(\Omega^2 \backslash \Delta) = \frac{n(n-1)}{2}, \quad \text{supp}(\mu \times \mu) = \Omega \times \Omega$$

and thus

$$\text{osc}(A\_1, \dots, A\_n) = \max\_{1 \le i, j \le n} \|A\_i - A\_j\|, \quad \text{osc}(B\_1, \dots, B\_n) = \max\_{1 \le i, j \le n} \|B\_i - B\_j\|. \quad \square$$

**Example 1.** *Let* Ω = [0, 1]*, w* ∈ Ω *and* 0 < *α* 1*. Consider the probability Radon measure μ* = *αλ* + (1 − *<sup>α</sup>*)*<sup>δ</sup>w, where λ is Lebesgue measure on* Ω *and δw is the Dirac measure at w. Set*

$$\mathcal{Z}(A) := \int\_0^1 A\_t d\mu(t) \, = \,\, a \int\_0^1 A\_t d\lambda(t) + (1 - \alpha) A\_{\text{av}}$$

*We have*

$$\mu \times \mu = \mathfrak{a}^2(\lambda \times \lambda) + \mathfrak{a}(1-\mathfrak{a})\left(\lambda \times \delta\_{\mathfrak{w}}\right) + (1-\mathfrak{a})\mathfrak{a}\left(\delta\_{\mathfrak{w}} \times \lambda\right) + (1-\mathfrak{a})^2\left(\delta\_{\mathfrak{w}} \times \delta\_{\mathfrak{w}}\right) \dots$$

*Then* supp(*μ* × *μ*)=[0, 1] × [0, 1] *and* (*μ* × *μ*)[0, 1]<sup>2</sup>\Δ = *α*(<sup>2</sup> − *<sup>α</sup>*)*. For any continuous fields* A = (*At*)*<sup>t</sup>*∈<sup>Ω</sup> *and* B = (*Bt*)*<sup>t</sup>*∈<sup>Ω</sup> *of Hermitian operators, the inequality* (16) *becomes*

$$\mathbb{E}(\mathcal{X}(\mathcal{A}\boxtimes\mathcal{B}) - \mathcal{Z}(\mathcal{A})\boxtimes\mathcal{Z}(\mathcal{B}) \leqslant \frac{1}{2}a(2-a)\max\_{0\leqslant s,t\leqslant 1} \|A\_{\mathcal{I}} - A\_{\mathcal{S}}\| \cdot \max\_{0\leqslant t,s\leqslant 1} \|B\_{\mathcal{I}} - B\_{\mathcal{S}}\| (I\_{\mathcal{H}}\boxtimes I\_{\mathcal{K}}).$$
