Proof We denote by $H_1$ the set of vectors h of H such that the function $X \to \mathbb {C}$ , $x \mapsto \langle \alpha _x ,h \rangle _H$ is measurable. It is obvious that $H_1$ is a closed subspace of H. It is clear that any vector of H which is orthogonal to the range of $\alpha $ belongs to $H_1$ , i.e., $\alpha (X)^\perp \subset H_1$ . We infer that $\overline {\mathop {\mathrm {span}\,}\nolimits \alpha (X)}^\perp \subset H_1$ . Since the partial functions of $\varphi $ are measurable, for any $y \in X$ , the map $X \to \mathbb {C}$ , $x \mapsto \langle \alpha _x,\alpha _{y} \rangle _H$ is measurable. So we also have the inclusion $\alpha (X) \subset H_1$ , hence $\overline {\mathop {\mathrm {span}\,}\nolimits \alpha (X)} \subset H_1$ . We conclude that $H \subset H_1$ .

Proof Fix $x_0 \in X$ . Recall that by [Reference Bekka, de la Pierre and ValetteBCR84, Lemma 2.1, p. 74], the function ${\varphi \colon X \times X \to \mathbb {R}}$ defined by

(3.1) $$ \begin{align} \varphi(x,y) =\frac{1}{2}\big[\psi(x,x_0)+\psi(x_0,y)-\psi(x,y)\big], \quad x,y \in X, \end{align} $$

is a positive-definite kernel. Since $\psi $ is measurable, this kernel is measurable. So there exist a real Hilbert space H and a function $\alpha \colon X \to H$ satisfying (2.19), which is weak measurable by Lemma 3.1. Now, for any $x,y \in X$ , we have

$$ \begin{align*} \| \alpha_x - \alpha_y \|_H^2 = \| \alpha_x \|_H^2+ \| \alpha_y \|_H^2-2 \langle \alpha_x,\alpha_y \rangle_H \overset{(2.19)}{=} \varphi(x,x)+\varphi(y,y)-2\varphi(x,y) \\ &\overset{(3.1)}{=}\frac{1}{2}\big(\psi(x,x_0)+\psi(x_0,x)-\psi(x,x)\big)+\frac{1}{2}\big(\psi(y,x_0)+\psi(x_0,y)-\psi(y,y)\big)\\ &\qquad-\big(\psi(x,x_0)+\psi(x_0,y)-\psi(x,y)\big) =\psi(x,y).\\[-37pt] \end{align*} $$

Proof The self-adjointness property is equivalent to

$$ \begin{align*}\operatorname{\mathrm{Tr}}\big(M_\varphi(K_f)K_g^*\big) =\operatorname{\mathrm{Tr}}\big(K_f(M_\varphi(K_g))^*\big), \quad \text{i.e.,} \quad \operatorname{\mathrm{Tr}}\big(K_{\varphi f}K_g^*\big) =\operatorname{\mathrm{Tr}}\big(K_f(K_{\varphi g})^*\big) \end{align*} $$

for any $f,g \in {\mathrm {L}}^2(X \times X)$ . Since ${\mathrm {L}}^2(X \times X) \to S^2_X$ , $f \mapsto K_f$ is an isometry from the Hilbert space ${\mathrm {L}}^2(X \times X)$ onto the Hilbert space $S^2_X$ of Hilbert–Schmidt operators on ${\mathrm {L}}^2(X)$ , that means that

$$ \begin{align*}\iint_{X \times X} \varphi(x,y)f(x,y)\overline{g(x,y)} \mathop{}\mathopen{}\mathrm{d} x\mathop{}\mathopen{}\mathrm{d} y =\iint_{X \times X} f(x,y) \overline{\varphi(x,y)g(x,y)} \mathop{}\mathopen{}\mathrm{d} x\mathop{}\mathopen{}\mathrm{d} y. \end{align*} $$

Note that $f\overline {g}$ belongs to ${\mathrm {L}}^1(X \times X)$ and that each function of ${\mathrm {L}}^1(X \times X)$ has this form. By duality, we deduce that $\varphi =\overline {\varphi }$ almost everywhere.

Proof For any $f \in {\mathrm {L}}^2(X \times X)$ , we let $R(K_f) \overset {\mathrm {def}}{=} K_{\check {f}}$ . On the one hand, for any $f,g \in {\mathrm {L}}^2(X \times X)$ , the Hilbert–Schmidt operator $K_{f}K_g \colon {\mathrm {L}}^2(X) \to {\mathrm {L}}^2(X)$ is associated with the function h defined by (2.17), and for almost all $(x,y) \in X \times X$ , we have

$$ \begin{align*}\check{h}(x,y)= \int_X f(y,z)g(z,x) \mathop{}\mathopen{}\mathrm{d} z. \end{align*} $$

On the other hand, by (2.17), the Hilbert–Schmidt operator $K_{\check {g}}K_{\check {f}} \colon {\mathrm {L}}^2(X) \to {\mathrm {L}}^2(X)$ is associated with the function

$$ \begin{align*}(x,y) \mapsto \int_X \check{g}(x,z)\check{f}(z,y) \mathop{}\mathopen{}\mathrm{d} z=\int_X f(y,z) g(z,x) \mathop{}\mathopen{}\mathrm{d} z. \end{align*} $$

We obtain that $R(K_fK_g)=R(K_g)R(K_f)$ . Moreover, we have

$$ \begin{align*}(R(K_{f}))^* =\big(K_{\check{f}}\big)^* \overset{(2.16)}{=} K_{\overline{f}} =R(K_{\check{\overline{f}}}) \overset{(2.16)}{=} R((K_f)^*). \end{align*} $$

We conclude that the map $S^2_X \to S^2_X$ , $K_f \mapsto K_{\check {f}}$ is an involutive $*$ -antiautomorphism. From the formula $(K_f)^*=K_{\check {\overline {f}}}$ , it is not difficult to see that the Banach adjoint of $K_f$ is $K_{\check {f}}$ . Hence, $ \| K_f \|_{\mathrm {B}({\mathrm {L}}^2(X))}= \| K_{\check {f}} \|_{\mathrm {B}({\mathrm {L}}^2(X))}$ . We deduce an involutive $*$ -antiautomorphism $R \colon S^\infty _X \to S^\infty _X$ . By [Reference Blecher and Le MerdyBLM04, Lemma A.2.2, p. 360], we deduce a unique weak* continuous extension $R \colon (S^\infty _X)^{**} \to \mathrm {B}({\mathrm {L}}^2(X))$ . Moreover, by [Reference Blecher and Le MerdyBLM04, Section 2.5.5, p. 79], this map is a $*$ -antiautomorphism where the double dual $(S^\infty _X)^{**}$ is equipped with the Arens product. Finally, note that by [Reference RicardPal01, Theorem 9.1.39, p. 838], the algebra $(S^\infty _X)^{**}$ identifies with the algebra $\mathrm {B}({\mathrm {L}}^2(X))$ .

Proof Recall that the map ${\mathrm {L}}^2(X \times X) \to S^2_X$ , $f \mapsto K_f$ is an isometry from the Hilbert space ${\mathrm {L}}^2(X \times X)$ onto the Hilbert space $S^2_X$ of Hilbert–Schmidt operators on ${\mathrm {L}}^2(X)$ . For any $K_f,K_g \in S^1_X$ , we deduce that

$$ \begin{align*} \big\langle M_{\varphi}(K_f),K_g \big\rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} =\big\langle K_{\varphi f},K_g \big\rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} \overset{(4.2)}{=} \operatorname{\mathrm{Tr}}(R(K_{\varphi f})K_g) \\ &=\operatorname{\mathrm{Tr}}(K_{\check{\varphi} \check{f}}K_g) =\iint_{X \times X} \varphi(y,x)f(y,x) g(y,x) \mathop{}\mathopen{}\mathrm{d} x\mathop{}\mathopen{}\mathrm{d} y. \end{align*} $$

Moreover, we have

$$ \begin{align*} \big\langle K_f,K_{\varphi g} \big\rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} \overset{(4.2)}{=} \operatorname{\mathrm{Tr}}(R(K_f) K_{\varphi g})=\operatorname{\mathrm{Tr}}(K_{\check{f}} K_{\varphi g}) \\ &=\iint_{X \times X} f(y,x)\varphi(y,x) g(y,x) \mathop{}\mathopen{}\mathrm{d} x\mathop{}\mathopen{}\mathrm{d} y. \end{align*} $$

We conclude by density.

Proof It is elementary to check that a $*$ -antiautomorphism is necessarily unital. If the measurable Schur multiplier $M_\varphi \colon \mathrm {B}({\mathrm {L}}^2(X)) \to \mathrm {B}({\mathrm {L}}^2(X))$ is unital, then for any $y \in S^1_X$ , we have $\big \langle M_{\varphi }(1),y \big \rangle _{\mathrm {B}({\mathrm {L}}^2(X)), S^1_X} \overset {(4.3)}{=} \big \langle 1,M_{\varphi }(y) \big \rangle _{\mathrm {B}({\mathrm {L}}^2(X)), S^1_X}$ , that is, $\operatorname {\mathrm {Tr}} y=\operatorname {\mathrm {Tr}} M_{\varphi }(y)$ . We conclude that $M_\varphi $ is trace-preserving.

If the measurable Schur multiplier $M_\varphi \colon \mathrm {B}({\mathrm {L}}^2(X)) \to \mathrm {B}({\mathrm {L}}^2(X))$ is trace-preserving, then for any $x \in S^1_X$ , we have

$$ \begin{align*} \operatorname{\mathrm{Tr}} (R(M_\varphi(1))x) \overset{(4.2)}{=} \big\langle M_{\varphi}(1),x \big\rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} \overset{(4.3)}{=} \big\langle 1,M_{\varphi}(x) \big\rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} \\ &\overset{(4.2)}{=} \operatorname{\mathrm{Tr}}(R(1)M_\varphi(x)) =\operatorname{\mathrm{Tr}}(R(1)x). \end{align*} $$

We conclude by duality that $M_\varphi (1)=1$ .

Proof 1. $\Rightarrow $ 2. It is obvious.

2. $\Rightarrow $ 1. That is essentially a standard application of [Reference TodorovSTT14, Lemma 4.3]. See also [Reference van Neerven, Veraar and WeisTod14, Exercise 4.15].

2. $\Rightarrow $ 3. Let $\xi \in {\mathrm {L}}^1(X)$ . We can write $\xi =\xi _1\xi _2$ for some functions $\xi _1,\xi _2 \in {\mathrm {L}}^2(X)$ . Note that the operator $K_{\overline {\xi _1} \otimes \xi _1} \colon {\mathrm {L}}^2(X) \to {\mathrm {L}}^2(X)$ is positive. Indeed, for any $\eta \in {\mathrm {L}}^2(X)$ , we have

$$ \begin{align*} \big\langle K_{\overline{\xi_1} \otimes \xi_1}(\eta),\eta \big\rangle_{{\mathrm{L}}^2(X)} \overset{(2.15)}{=} \bigg(\int_X \xi_1 \eta \bigg)\big\langle \overline{\xi_1} ,\eta \big\rangle_{{\mathrm{L}}^2(X)} \\ &=\overline{\big\langle \overline{\xi_1} ,\eta \big\rangle_{{\mathrm{L}}^2(X)}} \big\langle \overline{\xi_1} ,\eta \big\rangle_{{\mathrm{L}}^2(X)} =\big|\big\langle \overline{\xi_1} ,\eta \big\rangle_{{\mathrm{L}}^2(X)} \big|^2 \geqslant 0. \end{align*} $$

We deduce that $\langle M_\varphi (K_{\overline {\xi _1} \otimes \xi _1})\xi _2,\xi _2 \rangle _{{\mathrm {L}}^2(X)} \geqslant 0$ . Finally, it suffices to observe that

$$ \begin{align*} &\big\langle M_\varphi(K_{\overline{\xi_1} \otimes \xi_1})\xi_2,\xi_2 \big\rangle_{{\mathrm{L}}^2(X)} =\Big\langle K_{\varphi(\overline{\xi_1} \otimes \xi_1)} \xi_2,\xi_2 \Big\rangle_{{\mathrm{L}}^2(X)} =\int_X \big(K_{\varphi(\overline{\xi_1} \otimes \xi_1)} \xi_2\big)(x)\overline{\xi_2(x)} \mathop{}\mathopen{}\mathrm{d} x \\ &\!\!\!\quad\overset{(2.15)}{=} \int_X \bigg(\int_X \varphi(x,y)\overline{\xi_1(x)} \xi_1(y)\xi_2(y)\mathop{}\mathopen{}\mathrm{d} y\bigg)\overline{\xi_2(x)} \mathop{}\mathopen{}\mathrm{d} x \\ &\,\quad=\iint_{X \times X} \varphi(x,y) \overline{\xi_1(x)} \xi_1(y)\xi_2(y) \overline{\xi_2(x)} \mathop{}\mathopen{}\mathrm{d} x \mathop{}\mathopen{}\mathrm{d} y =\iint_{X \times X} \varphi(x,y) \overline{\xi(x)} \xi(y) \mathop{}\mathopen{}\mathrm{d} x \mathop{}\mathopen{}\mathrm{d} y. \end{align*} $$

So (3.2) is satisfied.

Proof 1. For any $t \geqslant 0$ , let $\phi _t \colon X \times X \to \mathbb {C}$ be the continuous symbol of the Schur multiplier $T_t$ . Since the map $T_t$ is self-adjoint, the function $\phi _t$ is real-valued by Proposition 4.2. By Corollary 4.6, we have $\phi _t(x,x)=1$ for any $x \in X$ and any $t \geqslant 0$ . By Proposition 4.7, each $\phi _t$ is an integrally positive-definite kernel. We deduce by Proposition 3.6 that each $\phi _t$ is a positive-definite kernel. Finally, we have $\phi _0=1$ .

For any $t,t' \geqslant 0$ and any $f \in {\mathrm {L}}^2(X \times X)$ , the relation $T_tT_{t'}(K_{f})=T_{t+t'}(K_f)$ gives the equality $K_{\phi _{t}\phi _{t'} f}=K_{\phi _{t+t'}f}$ . We infer that $\phi _t\phi _{t'}f=\phi _{t+t'}f$ in the space ${\mathrm {L}}^2(X \times X)$ . It is apparent that $(\phi _t)_{t \geqslant 0}$ defines a semigroup $(\mathrm {Mult}_t)_{t \geqslant 0}$ of multiplication operators acting on ${\mathrm {L}}^2(X \times X)$ defined by

(4.5) $$ \begin{align} \mathrm{Mult}_t(f) \overset{\mathrm{def}}{=} \phi_tf, \quad t \geqslant 0, f \in {\mathrm{L}}^2(X \times X). \end{align} $$

Note that (4.1) implies that $(\mathrm {Mult}_t)_{t \geqslant 0}$ is a contraction semigroup. We will show the weak continuity of this semigroup. For any functions $f,g \in {\mathrm {L}}^2(X \times X)$ such that $K_g \in S^1_X$ , we have using the weak* continuity of the semigroup $(T_t)_{t \geqslant 0}$

$$ \begin{align*} \big\langle \mathrm{Mult}_t(f),g \big\rangle_{{\mathrm{L}}^2(X \times X)} \overset{(4.5)}{=} \langle \phi_tf,g \rangle_{{\mathrm{L}}^2(X \times X)} =\big\langle K_{\phi_t f}, K_g \big\rangle_{S^2_X} \\ &=\big\langle T_{t}(K_f), K_g \big\rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} \xrightarrow[t \to 0]{} \langle K_f, K_g \rangle_{\mathrm{B}({\mathrm{L}}^2(X)), S^1_X} =\langle f,g \rangle_{{\mathrm{L}}^2(X \times X)} \end{align*} $$

(here, we make a slight abuse of notation for the definition of the brackets). Note that the set of functions $g \in {\mathrm {L}}^2(X \times X)$ such that $K_g \in S^1_X$ is dense in ${\mathrm {L}}^2(X \times X)$ since $S^1_X$ is dense in the space $S^2_X$ . Now, with [Reference MegginsonMeg98, Exercise 2.71(a), p. 234], it is clear (using the contractivity of the semigroup) that this semigroup is weak continuous. By [Reference Engel and NagelEnN00, Theorem 5.8, p. 40], this semigroup is even strongly continuous. By [Reference Engel and NagelEnN00, Proposition 4.12, p. 32], there exists a measurable function $\psi \colon X \times X \to \mathbb {R}$ such that for almost all $x,y \in X$ ,

$$ \begin{align*}\phi_t(x,y) =\mathrm{e}^{-t\psi(x,y)}, \quad t \geqslant 0. \end{align*} $$

We deduce that for almost all $x,y \in X$ , we have if $t>0$ ,

$$ \begin{align*}\psi(x,y) =-\frac{1}{t}\log \phi_t(x,y). \end{align*} $$

Since $\phi _t$ is continuous, we can suppose that $\psi $ is continuous. Moreover, for any $x \in X$ , we have

$$ \begin{align*}\psi(x,x) =-\frac{1}{t}\log \phi_t(x,x) =0. \end{align*} $$

By Schoenberg’s theorem (Theorem 2.5), the function $\psi $ defines a real-valued negative definite kernel which vanishes on the diagonal of $X \times X$ . Finally, we use the characterization of continuous real-valued negative-definite kernels of [Reference Berg, Christensen and ResselBHV08, Theorem C.2.3].

2. For any $t \geqslant 0$ , let $\phi _t \colon X \times X \to \mathbb {C}$ be the symbol of $T_t$ . Since the map $T_t$ is self-adjoint, the function $\phi _t$ is real-valued almost everywhere by Proposition 4.2. Consequently, we can suppose that each function $\phi _t$ is real-valued. By Proposition 4.7, each $\phi _t$ is an integrally positive-definite kernel. Consequently, we can suppose by Proposition 3.3 that the function $\phi _t$ is a real-valued positive-negative kernel and bounded (by 1).

By Corollary 4.6, we have $\phi _t(x,x)=1$ for almost all $x \in X$ and any $t \geqslant 0$ . For any $t>0$ , let $A_t$ be the subset of X such that $\phi _t(x,x) \not =1$ . The subset $\{ (x,x) : x \in A_t \}$ is a subset of $A_t \times A_t$ , hence negligible. By modifying $\phi _t$ on $A_t$ , we can suppose that $\phi _t(x,x) =1$ for any $x \in X$ . Indeed, it is clear by observing the definition (2.18) that the property positive-definite is not lost by this change.

The end of the proof is similar to the proof of the first point replacing [Reference Berg, Christensen and ResselBHV08, Theorem C.2.3] by Lemma 3.2.

Proof For almost all $x \in X$ , we have

$$ \begin{align*} &\big((\mathrm{Mult}_{g} K_f \mathrm{Mult}_{h})(\xi)\big)(x) =\big((\mathrm{Mult}_{g} K_f)(h\xi)\big)(x) \\ &\quad\!\!\!\overset{(2.15)}{=} \bigg(\mathrm{Mult}_{g}\bigg(\int_{X} f(\cdot,y)h(y)\xi(y) \mathop{}\mathopen{}\mathrm{d} y\bigg)\bigg)(x)\\ &\,\quad=\bigg(g(\cdot)\int_{X} f(\cdot,y)h(y)\xi(y) \mathop{}\mathopen{}\mathrm{d} y\bigg)(x) =\int_{X} g(x)h(y)f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x.\\[-40pt] \end{align*} $$

Proof Let $\mathrm {W} \colon {\mathrm {L}}^2_{\mathbb {R}}(\mathbb {R},H) \to {\mathrm {L}}^0(\Omega )$ be an ${\mathrm {L}}^2_{\mathbb {R}}(\mathbb {R},H)$ -isonormal process on a probability space $(\Omega ,\mu )$ as in (2.4). We define the von Neumann algebra $M \overset {\mathrm {def}}{=} {\mathrm {L}}^\infty (\Omega ) \overline \otimes \mathrm {B}({\mathrm {L}}^2(X))$ . Note that M is an injective von Neumann algebra. We equip this von Neumann algebra with the normal semifinite faithful trace $\tau _M \overset {\mathrm {def}}{=} \int _{\Omega } \cdot \otimes \operatorname {\mathrm {Tr}}$ . By [Reference Steen, Todorov and TurowskaSak71, Theorem 1.22.13], we have a $*$ -isomorphism $M={\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ . We define the canonical unital normal injective $*$ -homomorphism

(5.3) $$ \begin{align} \begin{array}{cccc} J \colon & \mathrm{B}({\mathrm{L}}^2(X)) & \longrightarrow & {\mathrm{L}}^\infty(\Omega) \overline\otimes \mathrm{B}({\mathrm{L}}^2(X)) \\ & z & \longmapsto & 1 \otimes z \\ \end{array}. \end{align} $$

It is clear that the map J preserves the traces. We denote by $\mathbb {E} \colon {\mathrm {L}}^\infty (\Omega ) \overline \otimes \mathrm {B}({\mathrm {L}}^2(X)) \to \mathrm {B}({\mathrm {L}}^2(X))$ the canonical trace-preserving normal faithful conditional expectation of M onto $\mathrm {B}({\mathrm {L}}^2(X))$ . Note that $\mathbb {E}=\int _\Omega \cdot \otimes \mathrm {Id}_{\mathrm {B}({\mathrm {L}}^2(X))}$ . For almost all $\omega \in \Omega $ and any $t \geqslant 0$ , we introduce with (2.7)

(5.4) $$ \begin{align} k_{t,\omega}(x) \overset{\mathrm{def}}{=} \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x))(\omega)}, \quad x \in X. \end{align} $$

Let $(f_i)$ be an orthonormal basis of the separable Hilbert space ${\mathrm {L}}^2_{\mathbb {R}}(\mathbb {R}^+)$ . By the Pettis measurability theorem [Reference Hytönen, van Neerven, Veraar and WeisHvNVW16, Theorem 1.1.20, p. 10], there exists a negligible subset N of X such that $\alpha (X-N)$ is subset of a separable closed subspace $H_0$ of H. Let $(e_j)$ be an orthonormal basis of $H_0$ . For any $t>0$ fixed, we can write almost everywhere (with [Reference Hytönen, van Neerven, Veraar and WeisHvNVW18, Corollary 6.4.4, pp. 35–36])

$$ \begin{align*} (\mathrm{W}_t(\alpha_x))(\omega) \overset{(2.4)(2.7)}{=} \sum_{i,j \in I} \gamma_{i,j}(\omega) \big\langle 1_{[0,t]} \otimes \alpha_x,f_i \otimes e_j \big\rangle_{{\mathrm{L}}^2_{\mathbb{R}}(\mathbb{R}^+,H)} \\ &=\sum_{i,j \in I} \gamma_{i,j}(\omega) \big\langle 1_{[0,t]},f_i \big\rangle_{} \langle \alpha_x, e_j \rangle_H. \end{align*} $$

We conclude that the function $X \times \Omega \to \mathbb {C}$ , $(x,\omega ) \mapsto k_{t,\omega }(x)$ is measurable.

If $t<0$ , we define $k_{t,\omega }(x)$ by the same formula but with $\mathrm {W}_t(\alpha _x) \overset {\mathrm {def}}{=} \mathrm {W}(-1_{[t,0]} \otimes \alpha _x)$ . For any $t \in \mathbb {R}$ , we can define an element $V_t$ of ${\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ by

(5.5) $$ \begin{align} V_t(\omega) \overset{\mathrm{def}}{=} \mathrm{Mult}_{k_{t,\omega}}. \end{align} $$

Indeed, the function $V_t$ is weak* measurable since if we consider a normal linear form $f \colon \mathrm {B}({\mathrm {L}}^2(X)) \to \mathbb {C}$ , $z \mapsto \sum _{n \geqslant 1} \langle z(\xi _n), \eta _n \rangle _{{\mathrm {L}}^2(X)}$ with $\sum _{n \geqslant 1} ( \| \xi _n \|_{{\mathrm {L}}^2(X)}+ \| \eta _n \|_{{\mathrm {L}}^2(X)}) < \infty $ [Reference LiLi92, Proposition 1.2.2, p. 9], we see that

$$ \begin{align*}f\big(\mathrm{Mult}_{k_{t,\omega}}\big) =\sum_{n \geqslant 1} \big\langle \mathrm{Mult}_{k_{t,\omega}}\xi_n, \eta_n \big\rangle_{{\mathrm{L}}^2(X)} =\sum_{n \geqslant 1} \int_X k_{t,\omega} \xi_n \eta_n \end{align*} $$

is measurable since each integral defines a measurable function $\omega \to \int _X k_{t,\omega } \xi _n \eta _n$ by Fubini’s theorem. Note that $V_t$ is an unitary element of ${\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ . Now, for any $t \in \mathbb {R}$ , we define the linear map

(5.6) $$ \begin{align} \begin{array}{cccc} \mathcal{U}_t \colon & {\mathrm{L}}^\infty\big(\Omega,\mathrm{B}({\mathrm{L}}^2(X))\big) & \longrightarrow & {\mathrm{L}}^\infty\big(\Omega,\mathrm{B}({\mathrm{L}}^2(X))\big) \\ & g & \longmapsto & V_t gV_t^* \\ \end{array}. \end{align} $$

It is obvious that each map $\mathcal {U}_t$ is a trace-preserving $*$ -automorphism of the von Neumann algebra M. Moreover, for any elements $\xi ,\eta $ of the Hilbert space ${\mathrm {L}}^2(\Omega ,{\mathrm {L}}^2(X))={\mathrm {L}}^2(\Omega \times X)$ , we obtain usingFootnote ⁷ the first part of Lemma 2.1 since $\overline {\xi } \eta \in {\mathrm {L}}^1(\Omega \times X)$

$$ \begin{align*} \big\langle V_t(\xi), \eta \big\rangle_{{\mathrm{L}}^2(\Omega,{\mathrm{L}}^2(X))} =\int_{\Omega \times X} \overline{V_t(\xi)(\omega,x)} \eta(\omega,x) \mathop{}\mathopen{}\mathrm{d}\mu(\omega) \mathop{}\mathopen{}\mathrm{d} x \\ &\!\!\!\overset{(5.5)}{=} \int_{\Omega \times X} \overline{k_{t,\omega}(x)\xi(\omega,x)} \eta(\omega,x) \mathop{}\mathopen{}\mathrm{d}\mu(\omega) \mathop{}\mathopen{}\mathrm{d} x \\ &\!\!\!\overset{(5.4)}{=} \int_{\Omega \times X} \mathrm{e}^{-\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x))(\omega)} \overline{\xi(\omega,x)} \eta(\omega,x) \mathop{}\mathopen{}\mathrm{d}\mu(\omega) \mathop{}\mathopen{}\mathrm{d} x \\ &\xrightarrow[t \to 0]{} \int_{\Omega \times X} \mathrm{e}^{-\sqrt{2}{\mathrm{i}} (\mathrm{W}_{t_0}(\alpha_x))(\omega)}\overline{\xi(\omega,x)} \eta(\omega,x)\mathop{}\mathopen{}\mathrm{d}\mu(\omega)\mathop{}\mathopen{}\mathrm{d} x =\langle V_{t_0}(\xi), \eta \rangle_{{\mathrm{L}}^2(\Omega,{\mathrm{L}}^2(X))}. \end{align*} $$

So $(V_t)_{t \in \mathbb {R}}$ is a weak continuous family of unitaries hence a strongly continuous family since by [Reference TodorovStr81, p. 41] the weak operator topology and the strong operator topology coincide on the unitary group. By composition on bounded subsets, the map $\mathbb {R} \to {\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ , $t \mapsto \mathcal {U}_t(g)$ is strong operator continuous for any $g \in {\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ and clearly even strong* operator continuous, hence weak* continuous. We conclude that $(\mathcal {U}_t)_{t \in \mathbb {R}}$ is a point-weak* continuous family of $*$ -automorphisms. For any $f \in {\mathrm {L}}^2(X \times X)$ and any $t \in \mathbb {R}$ , we have

(5.7) $$ \begin{align} \mathcal{U}_t(1 \otimes K_f) \overset{(5.6)}{=} \omega \mapsto (V_t (1 \otimes K_f) V_t^*)(\omega) =\omega \mapsto V_t(\omega) K_f V_t(\omega)^* \\ &\overset{(5.5)}{=} \omega \mapsto \mathrm{Mult}_{k_{t,\omega}} K_f (\mathrm{Mult}_{k_{t,\omega}})^* \nonumber \overset{(5.1)}{=} \omega \mapsto K\big((x,y) \mapsto k_{t,\omega}(x)\overline{k_{t,\omega}(y)}f(x,y) \big) \nonumber \\ &\overset{(5.4)}{=} \omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x-\alpha_y))(\omega)} f(x,y) \big). \nonumber \end{align} $$

In particular, for almost all $\omega \in \Omega $ , any $\xi \in {\mathrm {L}}^2(X)$ , and almost all $y \in X$ , we have

(5.8) $$ \begin{align} \big(\big[(\mathcal{U}_t(1 \otimes K_f))(\omega)\big](\xi)\big)(y) \overset{(2.15)}{=} \int_X \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x-\alpha_y))(\omega)} f(x,y) \xi(x) \mathop{}\mathopen{}\mathrm{d} x. \end{align} $$

For any $t \in \mathbb {R}$ , we introduce the right shift $\mathcal {S}_t \colon {\mathrm {L}}^2_{\mathbb {R}}(\mathbb {R}) \to {\mathrm {L}}^2_{\mathbb {R}}(\mathbb {R})$ . We consider the operator $S_t \overset {\mathrm {def}}{=} \Gamma _\infty (\mathcal {S}_t) \otimes \mathrm {Id}_{\mathrm {B}({\mathrm {L}}^2(X))} \colon {\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X))) \to {\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ . Essentially, by the second part of Lemma 2.1, $(S_t)_{t \in \mathbb {R}}$ is a weak* continuous semigroup of $*$ -automorphisms. We have

(5.9) $$ \begin{align} S_t\big(\mathrm{e}^{{\mathrm{i}} \mathrm{W}(1_{[s,s'[} \otimes h)} \otimes z\big) \overset{(2.6)}{=} \mathrm{e}^{{\mathrm{i}} \mathrm{W}(1_{[s+t,s'+t[} \otimes h)} \otimes z, \quad t \in \mathbb{R}, h \in H, z \in \mathrm{B}({\mathrm{L}}^2(X)), s<s'. \end{align} $$

For any $t \in \mathbb {R}$ , we define the $*$ -automorphism $U_t \overset {\mathrm {def}}{=} \mathcal {U}_t S_t$ . We will prove that $(U_t)_{t \in \mathbb {R}}$ is a group of operators. On the one hand, for any $t,t' \in \mathbb {R}$ , we have

$$ \begin{align*} U_{t'}U_t\big(\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s,s']} \otimes h)} \otimes 1\big) =\mathcal{U}_{t'} S_{t'} \mathcal{U}_t S_t\big(\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s,s']} \otimes h)} \otimes 1\big) \\ &\!\!\!\!\overset{(5.9)}{=} \mathcal{U}_{t'} S_{t'} \mathcal{U}_t\big(\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s+t,s'+t]} \otimes h)} \otimes 1\big) \\ &=\mathcal{U}_{t'} S_{t'} \big(\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s+t,s'+t]} \otimes h)} \otimes 1\big) \overset{(5.9)}{=} \mathcal{U}_{t'}\big(\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s+t+t',s'+t+t']} \otimes h)} \otimes 1\big) \\ &=\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s+t+t',s'+t+t']} \otimes h)} \otimes 1 =U_{t+t'}\big(\mathrm{e}^{{\mathrm{i}}\mathrm{W}(1_{[s,s']} \otimes h)} \otimes 1\big). \end{align*} $$

On the other hand, for any $t,t' \geqslant 0$ and any $\xi \in {\mathrm {L}}^2(X)$ , we have

$$ \begin{align*} U_{t'}U_t(1 \otimes K_f) =\mathcal{U}_{t'} S_{t'}\mathcal{U}_tS_t(1 \otimes K_f) \\ &\!\!\!\overset{(5.9)}{=} \mathcal{U}_{t'} S_{t'}\mathcal{U}_t (1 \otimes K_f) \\ &\!\!\!\overset{(5.7)}{=} \mathcal{U}_{t'} S_{t'} \big(\omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x-\alpha_y))(\omega)} f(x,y)\big)\big)\\ &=\mathcal{U}_{t'} \big(\omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(1_{[t',t+t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y)\big)\big)\\ &\!\!\!\!\!\!\!\overset{(5.6)(5.5)}{=} \omega \mapsto \mathrm{Mult}_{k_{t',\omega}} K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(1_{[t',t+t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y)\big) \\ &(\mathrm{Mult}_{k_{t',\omega}})^*\\ &\!\!\!\overset{(5.1)}{=} \omega \mapsto K\big((x,y) \mapsto k_{t',\omega}(x)\overline{k_{t',\omega}}(y)\mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(1_{[t',t+t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y) \big) \\ &\!\!\!\overset{(5.4)}{=} \omega \mapsto K\big((x,y) \mapsto\mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_{t'}(\alpha_x-\alpha_y))(\omega)}\mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(1_{[t',t+t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y) \big)\\ &= \omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_{t+t'}(\alpha_x-\alpha_y))(\omega)} f(x,y) \big) \\ &\!\!\!\overset{(5.8)}{=} \mathcal{U}_{t+t'}(1 \otimes K_f) =U_{t+t'}(1 \otimes K_f). \end{align*} $$

Similarly for any $t \leqslant 0$ and any $t' \geqslant 0$ such that $t' \leqslant -t$ and any $\xi \in {\mathrm {L}}^2(X)$ , we have

$$ \begin{align*} U_{t'}U_t(1 \otimes K_f) =\mathcal{U}_{t'} S_{t'}\mathcal{U}_tS_t(1 \otimes K_f) \overset{(5.9)}{=} \mathcal{U}_{t'} S_{t'}\mathcal{U}_t (1 \otimes K_f) \\ &\!\!\!\overset{(5.7)}{=} \mathcal{U}_{t'} S_{t'} \big(\omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(-1_{[t,0]} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y)\big)\big)\\ &=\mathcal{U}_{t'} \big(\omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(-1_{[t'+t,t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y)\big)\big)\\ &\!\!\!\!\!\!\!\!\!\overset{(5.6)(5.5)}{=} \omega \mapsto \mathrm{Mult}_{k_{t',\omega}} K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(-1_{[t'+t,t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y)\big) \\ &\!\!\!(\mathrm{Mult}_{k_{t',\omega}})^*\\ &\!\!\!\overset{(5.1)}{=} \omega \mapsto K\big((x,y) \mapsto k_{t',\omega}(x)\overline{k_{t',\omega}}(y)\mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(-1_{[t'+t,t']} \otimes (\alpha_x-\alpha_y)))(\omega)} f(x,y) \big) \\ &\!\!\!\overset{(5.4)}{=} \omega \mapsto K\big((x,y) \mapsto\mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(1_{[0,t']} \otimes (\alpha_x-\alpha_y)(\omega)}\mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(-1_{[t'+t,t']} \otimes (\alpha_x-\alpha_y)))(\omega)} \\ &\!\!\!f(x,y) \big)\\ &=\omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}(-1_{[t'+t,0]} \otimes (\alpha_x-\alpha_y)(\omega)} f(x,y) \big) \quad \text{since } t+t' \leqslant 0 \leqslant t' \\ &\!\!\!\overset{(5.8)}{=} \mathcal{U}_{t+t'}(1 \otimes K_f) =U_{t+t'}(1 \otimes K_f). \end{align*} $$

The remaining cases are left to the reader. With these computations, it is easy to check that $(U_t)_{t \in \mathbb {R}}$ is a weak* continuous group of $*$ -automorphisms (consider the preadjoint maps of the $U_t$ ’s and use [Reference Engel and NagelEnN00, Lemme, B.15] to prove the weak* continuity of the group).

For any $f \in {\mathrm {L}}^2(X \times X)$ , any $\xi \in {\mathrm {L}}^2(X)$ , all $t \geqslant 0$ , and almost all y of X, we finally haveFootnote ⁸

$$ \begin{align*} \big(\big[\mathbb{E} U_tJ(K_f)\big](\xi)\big)(y) \overset{(5.3)}{=} \big(\big[\mathbb{E} U_t(1 \otimes K_f)\big](\xi)\big)(y) =\big(\big[\mathbb{E} \mathcal{U}_t S_t(1 \otimes K_f)\big](\xi)\big)(y) \\ &= \big(\big[\mathbb{E} \mathcal{U}_t(1 \otimes K_f)\big](\xi)\big)(y) =\bigg(\bigg[\int_{\Omega} \big(\mathcal{U}_t(1 \otimes K_f)\big)(\omega) \mathop{}\mathopen{}\mathrm{d} \mu(\omega)\bigg](\xi)\bigg)(y) \\ &=\int_{\Omega} \big(\big[ (\mathcal{U}_t(1 \otimes K_f))(\omega)\big](\xi)\big)(y) \mathop{}\mathopen{}\mathrm{d} \mu(\omega)\\ &\!\!\kern-2pt\overset{(5.8)}{=} \int_{\Omega}\bigg(\int_{X} \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x-\alpha_y))(\omega)} f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x\bigg) \mathop{}\mathopen{}\mathrm{d} \mu(\omega)\\ &=\int_{X} \bigg(\int_{\Omega} \mathrm{e}^{\sqrt{2}{\mathrm{i}} (\mathrm{W}_t(\alpha_x-\alpha_y))(\omega)} \mathop{}\mathopen{}\mathrm{d} \mu(\omega) \bigg) f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x \\ &\!\!\!\!\kern-1pt\overset{(2.13)}{=} \int_{X} \mathrm{e}^{-t \| \alpha_x-\alpha_y \|_{H}^2} f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x \\ &\!\!\kern-2pt\overset{(4.4)}{=} \int_{X} \phi_t(x,y)f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x \overset{(2.15)}{=}\big(K_{\phi_t f}(\xi)\big)(y). \end{align*} $$

Hence, we have $\big (\mathbb {E} U_tJ(K_f)\big )(\xi )=\big (K_{\phi _t f}(\xi )$ and finally $\mathbb {E} U_tJ(K_f)=K_{\phi _t f}$ . By weak* density, we deduce the existence of $T_t$ and (5.2). For any $t,t' \geqslant 0$ and all $x,y \in X$ , we have

$$ \begin{align*}\mathrm{e}^{-t \| \alpha_x-\alpha_y \|_{H}^2}\mathrm{e}^{-t' \| \alpha_x-\alpha_y \|_{H}^2} =\mathrm{e}^{-(t+t') \| \alpha_x-\alpha_y \|_{H}^2}. \end{align*} $$

We deduce that $(T_t)_{t \geqslant 0}$ is a semigroup. Furthermore, the function (4.4) is real-valued. By Proposition 4.2, we conclude that each $T_t$ is self-adjoint. The factorization (5.2) shows that each $T_t$ is completely positive and unital. The weak* continuity is a consequence of the one of the group $(U_t)_{t \in \mathbb {R}}$ .

Proof Let $\mathrm {W} \colon {\mathrm {L}}^2_{\mathbb {R}}(\mathbb {R}^+,H) \to {\mathrm {L}}^0(\Omega )$ be an H-cylindrical Brownian motion on a probability space $(\Omega ,\mu )$ (see Section 2). As in the proof of Theorem 5.2, for almost all $\omega \in \Omega $ and any $t \geqslant 0$ , we introduce the element (5.4) and for any $t \geqslant 0$ the element $V_t$ of ${\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ defined by (5.5). Recall that $V_t$ is a unitary element of the von Neumann algebra $N \overset {\mathrm {def}}{=} {\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)))$ , which is clearly injective. Again with the $*$ -homomorphism J defined in (5.3) and the $*$ -automorphism $\mathcal {U}_t$ defined in (5.6), we define for any $t \geqslant 0$ the unital normal injective $*$ -homomorphism

(6.2) $$ \begin{align} \begin{array}{cccc} \pi_t\overset{\mathrm{def}}{=} \mathcal{U}_tJ \colon & \mathrm{B}({\mathrm{L}}^2(X)) & \longrightarrow & {\mathrm{L}}^\infty\big(\Omega,\mathrm{B}({\mathrm{L}}^2(X))\big) \\ & z & \longmapsto & V_t (1 \otimes z)V_t^* \\ \end{array}. \end{align} $$

It is obvious that $\pi _t$ is trace-preserving. For any $t \geqslant 0$ , we also define the canonical normal conditional expectations $\mathbb {E}_{\mathscr {F}_t} \colon {\mathrm {L}}^\infty (\Omega ) \to {\mathrm {L}}^\infty (\Omega ,\mathscr {F}_t)$ and $\mathbb {E}_t \overset {\mathrm {def}}{=}\mathbb {E}_{\mathscr {F}_t} \otimes \mathrm {Id}_{\mathrm {B}({\mathrm {L}}^2(X))} \colon {\mathrm {L}}^\infty (\Omega ,\mathrm {B}({\mathrm {L}}^2(X)) \to {\mathrm {L}}^\infty (\Omega ,\mathscr {F}_t,\mathrm {B}({\mathrm {L}}^2(X))$ where the $\sigma $ -algebra $\mathscr {F}_t$ is defined in (2.8). We introduce the von Neumann algebra $N_t \overset {\mathrm {def}}{=} {\mathrm {L}}^\infty (\Omega ,\mathscr {F}_t,\mathrm {B}({\mathrm {L}}^2(X))$ . Note that the range of $\pi _t$ is included in $N_t$ .

For almost all $\omega \in \Omega $ , the Hilbert–Schmidt operator

$$ \begin{align*}\big(\pi_t(K_f)\big)(\omega) \overset{(6.2)(5.3)}{=} \big(\mathcal{U}_t(1 \otimes K_f)\big)(\omega) \end{align*} $$

is associated by the computation (5.7) with the function

(6.3) $$ \begin{align} (x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}(\mathrm{W}_t(\alpha_x-\alpha_y))(\omega)} f(x,y). \end{align} $$

Similarly, for any $0 \leqslant s \leqslant t$ and almost all $\omega \in \Omega $ , the operator

$$ \begin{align*} \big(\pi_s(T_{t-s}(K_f))\big)(\omega) =\big(\pi_s(K_{\phi_{t-s}f}) \big)(\omega) \end{align*} $$

is Hilbert–Schmidt and associated with the function

(6.4) $$ \begin{align} (x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}(\mathrm{W}_s(\alpha_x-\alpha_y))(\omega)} \mathrm{e}^{-(t-s) \| \alpha_x-\alpha_y \|_{H}^2}f(x,y). \end{align} $$

Now, recall that $(\mathrm {W}_t(h))_{t \geqslant 0}$ is a Brownian motion for any fixed $h \in H$ . Hence, for any $0 \leqslant s \leqslant t$ and any $x,y \in X$ , the random variable

(6.5) $$ \begin{align} \mathrm{W}\big(1_{]s,t]} \otimes (\alpha_x-\alpha_y)\big) =\mathrm{W}\big(1_{[0,t]} \otimes (\alpha_x-\alpha_y)\big)-\mathrm{W}\big(1_{[0,s]} \otimes (\alpha_x-\alpha_y)\big) \\ &\overset{(2.7)}{=} \mathrm{W}_t(\alpha_x-\alpha_y)-\mathrm{W}_s(\alpha_x-\alpha_y) \nonumber \end{align} $$

is independent by (2.12) from the $\sigma $ -algebra $\mathscr {F}_s \overset {(2.8)}{=} \sigma \big (\mathrm {W}_r(h) : r \in [0,s], h \in H\big )$ . Consequently, the random variable $\mathrm {e}^{\sqrt {2}{\mathrm {i}}\mathrm {W}(1_{]s,t]}\otimes (\alpha _x-\alpha _y))}$ is also independent from the $\sigma $ -algebra $\mathscr {F}_s$ . Using [Reference Hytönen, van Neerven, Veraar and WeisHvNVW16, Proposition 2.6.31] in the fourth equality since each random variable $\mathrm {W}_s(\alpha _x)$ is $\mathscr {F}_s$ -measurable, we finally obtain, for any $0 \leqslant s \leqslant t$ and any $f \in {\mathrm {L}}^2(X \times X)$ ,

$$ \begin{align*} \mathbb{E}_s\pi_t(K_f) \overset{(6.3)}{=} \mathbb{E}_s\Big(\omega \mapsto \big[K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}} \mathrm{W}_t(\alpha_x-\alpha_y)(\omega)} f(x,y)\big)\big]\Big) \\ &\overset{(6.5)}{=} \mathbb{E}_s\Big(\omega \mapsto \big[K\big((x,y) \mapsto\mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}_s(\alpha_x-\alpha_y)(\omega)}\mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big]\Big)\\ &\overset{(5.1)}{=} \mathbb{E}_s\Big(\omega \mapsto \mathrm{Mult}_{k_{s,\omega}}\big[K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big]\mathrm{Mult}_{\overline{k_{s,\omega}}}\Big)\\ &=\big[\omega \mapsto \mathrm{Mult}_{k_{s,\omega}}\big] \mathbb{E}_s\big(\omega \mapsto \big[K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big) \\ &\big[\omega \mapsto \mathrm{Mult}_{\overline{k_{s,\omega}}} \big]\\ &\overset{(2.14)}{=}\big[\omega \mapsto \mathrm{Mult}_{k_{s,\omega}}\big] \mathbb{E}\big(\omega \mapsto \big[ K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big]\big) \\ &\big[\omega \mapsto \mathrm{Mult}_{\overline{k_{s,\omega}}} \big]. \end{align*} $$

Now, for any $\xi \in {\mathrm {L}}^2(X)$ and almost all $y \in Y$ , we see that

$$ \begin{align*} \mathbb{E}\big(\omega \mapsto \big[K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big]\big) (\xi)(y) \\ &=\bigg(\int_\Omega K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big) \mathop{}\mathopen{}\mathrm{d} \mu(\omega)\bigg) (\xi)(y) \\ &=\int_X \bigg(\int_\Omega \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}\mathop{}\mathopen{}\mathrm{d} \mu(\omega)\bigg)f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x \\ &\!\!\!\!\kern-1pt\overset{(2.13)}{=} \int_X \mathrm{e}^{-(t-s) \| \alpha_x-\alpha_y \|_{H}^2}f(x,y)\xi(x) \mathop{}\mathopen{}\mathrm{d} x. \end{align*} $$

Consequently, the previous expectation is

$$ \begin{align*} \mathbb{E}\big(\omega \mapsto \big[K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}(1_{]s,t]}\otimes (\alpha_x-\alpha_y))(\omega)}f(x,y)\big)\big]\big) \\ &=K\big((x,y) \mapsto \mathrm{e}^{-(t-s) \| \alpha_x-\alpha_y \|_{H}^2}f(x,y)\big). \end{align*} $$

Finally, we obtain

$$ \begin{align*} \mathbb{E}_s\pi_t(K_f) =\big[\omega \mapsto \mathrm{Mult}_{k_{s,\omega}}\big] K\big((x,y) \mapsto \mathrm{e}^{-(t-s) \| \alpha_x-\alpha_y \|_{H}^2}f(x,y)\big) \big[\omega \mapsto \mathrm{Mult}_{\overline{k_{s,\omega}}} \big]\\ &\overset{(5.1)}{=} \omega \mapsto K\big((x,y) \mapsto \mathrm{e}^{\sqrt{2}{\mathrm{i}}\mathrm{W}_s(\alpha_x-\alpha_y)(\omega)}\mathrm{e}^{-(t-s) \| \alpha_x-\alpha_y \|_{H}^2}f(x,y)\big) \\ &\overset{(6.4)}{=} \pi_s(T_{t-s}(K_f)). \end{align*} $$

By weak* density, the proof is complete.