Infinite-dimensional polynomial processes

Abstract

We introduce polynomial processes taking values in an arbitrary Banach space ${B}$ via their infinitesimal generator $L$ and the associated martingale problem. We obtain two representations of the (conditional) moments in terms of solutions of a system of ODEs on the truncated tensor algebra of dual respectively bidual spaces. We illustrate how the well-known moment formulas for finite-dimensional or probability-measure-valued polynomial processes can be deduced in this general framework. As an application, we consider polynomial forward variance curve models which appear in particular as Markovian lifts of (rough) Bergomi-type volatility models. Moreover, we show that the signature process of a $d$-dimensional Brownian motion is polynomial and derive its expected value via the polynomial approach.

Appendices

Appendix A: Auxiliary results

We collect here auxiliary results which are needed in Sect. 2.

Lemma A.1

Fix $p({y})=\sum _{j=0}^{k}\langle {a}_{j},{y}^{\otimes j}\rangle$ for ${a}_{j}\in ({B}^{\otimes j})^{*}$. Then $p({y})=0$ for all ${y}\in {B}$ if and only if ${a}_{j}=0$ for all $j$.

Proof

Note that $\langle {a}_{j},{y}^{\otimes j}\rangle =0$ for all ${y}\in {B}$ if and only if $\langle {a}_{j},\tilde{{y}}\rangle =0$ for all $\tilde{{y}}\in {B}^{\otimes j}$ and thus if and only if ${a}_{j}=0$. If $p({y})=0$ for all ${y}\in {B}$, then ${a}_{0}=p(0)=0$. Proceeding inductively, we can prove that $\langle {a}_{j},{y}^{ \otimes j}\rangle =\lim _{{\varepsilon }\to 0}{\varepsilon }^{-j}p({ \varepsilon }{y})=0$ and the “only if” part follows. The “if” part is clear. □

Lemma A.2

Let $L\colon P^{D}\to P$ be an ${\mathcal{S}}$-polynomial operator. Then there exists a $k$th dual operator corresponding to $L$ for each $k\in {\mathbb{N}}$.

Proof

We claim that there exists a ${B}$-polynomial operator $\widetilde{L}:P^{D}\to P$ such that $(\widetilde{L}p)|_{\mathcal{S}}=(Lp)|_{\mathcal{S}}$ for each $p\in P^{D}$. In other words, we claim that for each $p\in P^{D}$, there exists a $q_{p}\in P$ such that $q_{p}=Lp$ on ${\mathcal{S}}$, $\deg (q_{p}) \le \deg (p)$ and $p\mapsto q_{p}$ is linear. If this is the case, the claim follows by Lemma A.1.

We proceed by induction. Set $P^{D}_{k}:=\{p\in P^{D}:\deg (p)\leq k\}$ and for all $p\in P^{D}_{0}$, set $\widetilde{L}p\equiv q$, where $q\in {\mathbb{R}}$ satisfies $(Lp)|_{\mathcal{S}}\equiv q$. Clearly, $\widetilde{L}|_{P^{D}_{0}}$ is linear, satisfies the ${B}$-polynomial property and $(\widetilde{L}p)|_{\mathcal{S}}=(Lp)|_{\mathcal{S}}$ for all $p\in P^{D}_{0}$. Fix $k\in {\mathbb{N}}$ and suppose that $\widetilde{L}|_{P^{D}_{k}}$ is linear, satisfies the ${B}$-polynomial property and $(\widetilde{L}p)|_{\mathcal{S}}=(Lp)|_{\mathcal{S}}$ for all $p\in P^{D}_{k}$. Consider the set of all pairs $(V,{\mathcal{L}})$ of a vector space $V$ such that $P^{D}_{k}\subseteq V \subseteq P^{D}_{k+1}$ and ${\mathcal{L}}: V\to P$ is a linear extension of $\widetilde{L}|_{P^{D}_{k}}$ satisfying the ${B}$-polynomial property and such that $({\mathcal{L}}p)|_{\mathcal{S}}= (Lp)|_{\mathcal{S}}$ for all $p\in V$. By a standard application of Zorn’s lemma, this set has a maximal element $(V,{\mathcal{L}})$ with respect to the order relation given by

$$ (V_{1},{\mathcal{L}}_{1})\text{ ``} {{\leq }} \text{'' }(V_{2},{ \mathcal{L}}_{2})\qquad :\Longleftrightarrow \qquad V_{1}\subseteq V_{2} \text{ and ${\mathcal{L}}_{2}|_{V_{1}}={\mathcal{L}}_{1}$.} $$

Assume by way of contradiction that $V\neq P^{D}_{k+1}$ and pick $p\in P^{D}_{k+1}\setminus V$. Since $L$ is ${\mathcal{S}}$-polynomial, there exists a $q\in P$ such that $(Lp)|_{\mathcal{S}}=q$ and $\deg (q)\leq k+1$. Set ${\mathcal{L}}p:=q$ and extend ℒ to $V+{\mathbb{R}}p$ linearly. Linearity of $L$ and ℒ yields that $({\mathcal{L}}\tilde{p})|_{\mathcal{S}}=(L\tilde{p})|_{\mathcal{S}}$ for all $\tilde{p}\in V+{\mathbb{R}}p$. Finally, noting that $\deg (\tilde{p}+\alpha p)=k+1$ for all $\tilde{p}\in V$ and $\alpha \neq 0$, we can conclude that

$$ \deg \big({\mathcal{L}}(\tilde{p}+\alpha p)\big)=\deg ({\mathcal{L}} \tilde{p} +\alpha q)\leq \max \big(\deg ({\mathcal{L}}\tilde{p}), \deg ( \alpha q)\big)\leq k+1 $$

contradicts the maximality of $(V,{\mathcal{L}})$. Setting $\widetilde{L} p:= {\mathcal{L}}p$ for all $p\in P^{D}_{k+1}$ concludes the proof. □

Appendix B: Proofs of Propositions 4.7 and 4.10

We collect here the proofs of the VIX moment formulas stated in Sect. 4.4.

Proof of Proposition 4.7

Recall that by Remark 4.5 (iii), the process $(\lambda _{t})_{t \geq 0}$ solves the martingale problem for the ${B}$-polynomial operator $L:P^{D}\to P$. We follow now the scheme outlined in Sect. 4.4.1. Throughout the proof, we use the following inequalities. Set $C_{t}:=e^{\frac{k(k-1)}{2}\sum _{\ell =1}^{m}\int _{0}^{t+\Delta }K_{\ell }(\tau )^{2}d\tau }$ and note that $e^{\int _{0}^{t}V_{k}(x+ \tau 1)d\tau }\leq C_{t}<\infty$ for each $t>0$. Observe also that by Benth and Krühner [12, Lemma 3.2], we have

$$ |{y}(x)|\leq {\overline{c}}\| {y}\|_{\alpha }\qquad \text{and}\qquad \frac{1}{\Delta }\int _{0}^{\Delta }{y}'(x+t)dx\leq \|{y}\|_{\alpha }$$

(B.1)

for each ${y}\in {B}$, for some constant ${\overline{c}}>0$.

(i) We observe that $M_{n}\vec{y}=({\mathcal{M}}_{0}{y}_{0},\ldots ,{\mathcal{M}}_{n}{y}_{n})$ with

$$\begin{aligned} {\mathcal{M}}_{k}{y}( x)&=1^{\top }\nabla {y}(x)+V_{k}(x){y}(x),\qquad {y} \in \! \big(\!\operatorname{dom} (\mathcal{A})\big)^{\otimes k}, \end{aligned}$$

where $\mathcal{A}$ is given by (4.4).

(ii) Observe that for each ${y}\in (\operatorname{dom}(\mathcal{A}))^{\otimes k}$, the corresponding ODE (4.10) can be seen as a PDE corresponding to the Cauchy problem with potential $V_{k}$ of a pure drift process $dX_{t}= 1 dt$ on $\mathbb{R}^{k}$. The Feynman–Kac formula thus yields that its classical strong solution on $({\mathbb{R}}_{+}\setminus \{0\})^{k}\times [0,T]$ is given by $Z_{t}y( x):={y}(x+t1)e^{\int _{0}^{t}V_{k}(x+\tau 1)d\tau }$. Moreover, by (4.9), the definition of $C_{t}$ and (B.1), we get

$$ |\langle \widehat{{a}}^{\otimes k},Z_{t}{y}^{\otimes k}\rangle _{\alpha }| =\bigg|\frac{1}{\Delta ^{k}}\int _{[0,\Delta ]^{k}}Z_{t}{y}^{ \otimes k}(\vec{x})d\vec{x}\bigg| \leq {\overline{c}}^{k}C_{t} \|{y}\|_{\alpha }^{k} $$

for each ${y}\in {B}$. This implies that $\langle \widehat{{a}}^{\otimes k},Z_{t}{y}\rangle _{\alpha }\leq { \overline{c}}^{k}C_{t} \|y\|_{\times }$ for each ${y}\in {B}^{\otimes k}$ and thus (4.11).

(iii) Due to the above estimate, we can define a candidate solution ${a}_{t} \in ({B}^{\otimes k})^{*}$ (in the sense of Definition 3.3) to $\partial _{t}{a}_{t}={\mathcal{L}}_{k}{a}_{t}$ for ${a}_{0}=\widehat{{a}}^{\otimes k}$ as

$$ \langle {a}_{t},{y}\rangle _{\alpha }:=\langle \widehat{{a}}^{\otimes k},Z_{t}{y} \rangle _{\alpha }=\frac{1}{\Delta ^{k}}\int _{[0,\Delta ]^{k}}{y}( x+t1)e^{ \int _{0}^{t}V_{k}( x+\tau 1)d\tau }d x,\qquad {y}\in {B}^{\otimes k}. $$

(iv) To conclude the proof, we need to check that $({a}_{t})_{t\geq 0}$ satisfies the conditions of the dual moment formula in Theorem 3.4 (i)–(iii). Observe that $({a}_{t})_{t\geq 0}$ satisfies condition (i) if and only if $({a}_{t})_{t\geq 0}\subseteq {\mathcal{D}}({\mathcal{L}}_{k})$ and

$$\begin{aligned} \langle {a}_{t}, y^{\otimes k }\rangle _{\alpha }=\langle \widehat{{a}}, y^{\otimes k }\rangle _{\alpha } +\int _{0}^{t} \langle \mathcal{L}_{k} a_{s}, y^{\otimes k }\rangle _{\alpha }ds \end{aligned}$$

(B.2)

for all $y \in B$. We prove now the following claims.

Claim 1: ${a}_{t}\in {\mathcal{D}}({\mathcal{L}}_{k})$. To show this, we fix $t\geq 0$ and construct a sequence ${a}_{n}=\sum _{i=1}^{n}\alpha _{n,i}{a}_{n,i}^{\otimes k}$ such that ${a}_{n,i}\in D$, $\|{a}_{n}-{a}_{t}\|_{*k}\to 0$ and $\|{\mathcal{L}}_{k} {a}_{n}-{\mathcal{L}}_{k} {a}_{t}\|_{*k}\to 0$, where ${\mathcal{L}}_{k} {a}_{t}\in ({B}^{\otimes k})^{*}$ is uniquely determined by

$$ \langle {\mathcal{L}}_{k} {a}_{t},{y}^{\otimes k}\rangle _{\alpha }= \frac{1}{\Delta ^{k}}\int _{[0,\Delta ]^{k}}{\mathcal{M}}_{k}{y}^{ \otimes k}(x+t1)F(x)dx ,\qquad {y}\in \operatorname{dom}(\mathcal{A}), $$

(B.3)

for $F( x):=e^{\int _{0}^{t}V_{k}( x+\tau 1)d\tau }$. Observe that the assumptions on $K$ guarantee that $F$ is symmetric and continuous on $[0,\Delta ]^{k}$. By the Stone–Weierstrass theorem, we can therefore find a sequence $F_{n}=\sum _{i=1}^{n}\alpha _{n,i}F_{n,i}^{\otimes k}$, $n \in {\mathbb{N}}$, such that $F_{n,i}\in C([0,\Delta ])$ and ${\varepsilon }_{n}:=\|F_{n}-F\|_{\infty ,k}\to 0$ for $n\to \infty$, where $\|{\,\cdot \,}\|_{\infty ,k}$ denotes the supremum norm on $[0,\Delta ]^{k}$. Define ${a}_{n,i}\in {B}$ such that

$$ \langle {a}_{n,i},{y}\rangle _{\alpha }:=\frac{1}{\Delta }\int _{0}^{\Delta }{y}(x+t)F_{n,i}(x)dx $$

for each ${y}\in {B}$. Using (B.1), we obtain the estimates

$$\begin{aligned} |\langle {a}_{n,i},{\mathcal{A}}{y}\rangle _{\alpha }|&\leq \frac{1}{\Delta }\int _{0}^{\Delta }|{\mathcal{A}}{y}(x+t)|dx\|F_{n,i}\|_{ \infty ,1}\leq \|{y}\|_{\alpha }\|F_{n,i}\|_{\infty ,1}, \\ |\langle \langle {a}_{n,i},K_{\ell }{y}\rangle \rangle _{\alpha }| &= \lim _{z\to 0}|\langle {a}_{n,i},K_{\ell }({\,\cdot \,}+z){y}({\,\cdot \,}+z)\rangle _{\alpha }| \\ &\leq \frac{1}{\Delta }\int _{0}^{\Delta }|K_{\ell }(x+t){y}(x+t)||F_{n,i}(x)|dx \\ &\leq {\overline{c}}\|{y}\|_{\alpha }\|F_{n,i}\|_{\infty ,1} \frac{1}{\Delta }\int _{t}^{\Delta +t} K_{\ell }(x)^{2}dx, \end{aligned}$$

from which we conclude that ${a}_{n,i}\in D$. Next, using again (B.1), we have for each ${y}\in {B}$ that

$$ |\langle {a}_{n}-{a}_{t},{y}^{\otimes k}\rangle _{\alpha }|\leq \bigg( \frac{1}{\Delta }\int _{0}^{\Delta }|y(x+t)|dx\bigg)^{k}{\varepsilon }_{n} \leq {\overline{c}}^{k}\|y\|_{\alpha }^{k}{\varepsilon }_{n} $$

which by (4.3) proves that $\| {a}_{n}-{a}_{t}\|_{*k}\leq {\overline{c}}^{k}{\varepsilon }_{n}$ and thus that $\|{a}_{n}-{a}_{t}\|_{*k}\to 0$ for $n \to \infty$. Since for each ${y}\in \operatorname{dom}(\mathcal{A})$, we have again by (B.1) that

$$\begin{aligned} |\langle {\mathcal{L}}_{k} {a}_{n}-{\mathcal{L}}_{k} {a}_{t},{y}^{ \otimes k}\rangle _{\alpha }| &\leq \bigg(\frac{1}{\Delta ^{k}}\int _{[0, \Delta ]^{k}}|{\mathcal{M}}_{k}{y}^{\otimes k}( x+t1)|d x\bigg) { \varepsilon }_{n} \\ &\leq \bigg(k{\overline{c}}^{k-1} +\frac{k(k-1)}{2}{\overline{c}}^{k} \frac{1}{\Delta }\sum _{\ell =1}^{m}\int _{t} ^{\Delta +t} K_{\ell }(x)^{2}d x\bigg)\|y\|_{\alpha }^{k}{\varepsilon }_{n} \end{aligned}$$

and $\operatorname{dom}(\mathcal{A})$ is dense in ${B}$, we can also conclude that $\|{\mathcal{L}}_{k} {a}_{n}-{\mathcal{L}}_{k} {a}_{t}\|_{*k}\to 0$. As a side product of this computation, we also get

$$ |\langle {\mathcal{L}}_{k} {a}_{t},{y}\rangle _{\alpha }|\leq c_{t} \|{y} \|_{\times },\qquad {y}\in {B}^{\otimes k}, $$

(B.4)

where $c_{t}:=(k{\overline{c}}^{k-1} +\frac{k(k-1)}{2}{\overline{c}}^{k} \frac{1}{\Delta }\sum _{\ell =1}^{m}\int _{t} ^{\Delta +t} K_{\ell }(x)^{2}d x)\|F\|_{\infty ,k}$.

Claim 2: $\partial _{t}{a}_{t}={\mathcal{L}}_{k}{a}_{t}$. Fix ${y}\in \operatorname{dom}(\mathcal{A})$ and define $m_{t}:=Z_{t}{y}^{\otimes k}$. Observe that for all $x\in (0,\Delta ]^{k}$ and $t\in [0,T]$, we have

$$\begin{aligned} |\partial _{t}m_{t}( x)| &=\big|\partial _{t}\big({y}^{\otimes k}(x+t1)e^{ \int _{0}^{t}V_{k}(x+\tau 1)d\tau }\big)\big| \\ &=\bigg|\bigg(\sum _{i=1}^{k} y'(x_{i}+t)\prod _{j\neq i}{y}(x_{j}+t)+ \prod _{i=1}^{k}{y}(x_{i}+t)V_{k}(x+t1)\bigg) \\ &\qquad \times e^{\int _{0}^{t}V_{k}(x+\tau 1)d\tau }\bigg| \\ &\leq \big(k \|{y}'\|_{\alpha }\|{y}\|_{\alpha }^{k-1}{\overline{c}}^{k}+{ \overline{c}}^{k}\overline{V}_{k}(x)\|{y}\|_{\alpha }^{k}\big)C_{T}, \end{aligned}$$

where $\overline{V}_{k}( x):=\sup _{t\in [0,T]}V_{k}(x+1t)$. Condition (4.12) then implies that the map $x\mapsto \sup _{t\in [0,T]}|\partial _{t}m_{t}( x)|$ is dominated by an integrable function. Hence the Leibniz integral rule and (B.3) yield

$$\begin{aligned} \partial _{t}\langle {a}_{t},{y}^{\otimes k}\rangle _{\alpha }&= \frac{1}{\Delta ^{k}}\int _{[0,\Delta ]^{k}}\partial _{t}m_{t}( x) d x \\ &=\frac{1}{\Delta ^{k}}\int _{[0,\Delta ]^{k}}\partial _{t}\big({y}^{ \otimes k}(x+1t)e^{\int _{0}^{t}V_{k}(x+\tau 1)d\tau }\big) d x \\ &=\frac{1}{\Delta ^{k}}\int _{[0,\Delta ]^{k}} {\mathcal{M}}_{k}{y}^{ \otimes k}( x +1t)e^{\int _{0}^{t}V_{k}(x+\tau 1)d\tau } d x \\ &=\langle {\mathcal{L}}_{k}{a}_{t},{y}^{\otimes k}\rangle _{\alpha } \end{aligned}$$

for each ${y}\in \operatorname{dom}(\mathcal{A})$. Since $(\langle {\mathcal{L}}_{k}{a}_{t},{y}^{\otimes k}\rangle _{\alpha })_{t \geq 0}$ is continuous, (B.2) follows by the fundamental theorem of calculus. By (B.4), this result can be extended to all ${y}\in {B}$ and the claim follows.

Claim 3: Conditions (ii) and (iii) of Theorem 3.4hold. We apply Lemma 3.17. The moment condition is satisfied by assumption and (3.8) holds since $\sup _{t\in [0,T]}c_{t}<\infty$ with $c_{t}$ given in (B.4).

The result now follows from Theorem 3.4. □

Proof of Proposition 4.10

Recall that by Remark 4.5 (iii), the process $(\lambda _{t})_{t \geq 0}$ solves the martingale problem for the ${B}$-polynomial operator $L:P^{D}\to P$. We follow again the scheme of Sect. 4.4.1. Throughout the proof, we need the following inequalities. Set $C_{t}:=e^{t{\overline{c}}^{k}\|V_{k}\|_{\times }}$ and note that for each $f:[0,t]\to [0,\Delta +t]^{k}$ and each $t>0$, it holds that $e^{\int _{0}^{t}V_{k}(f(\tau ))d\tau }\leq C_{t}<\infty$.

(i) We observe that $M_{n}\vec{y}=({\mathcal{M}}_{0}{y}_{0},\ldots ,{\mathcal{M}}_{n}{y}_{n})$ for

$$ {\mathcal{M}}_{k}{y}(x)=1^{\top }\nabla {y}(x)+\int {y}( x+\xi )\nu (x,d \xi ),\qquad {y}\in \big(\operatorname{dom}(\mathcal{A})\big)^{ \otimes k}, $$

where $\mathcal{A}$ is given by (4.4).

(ii) We now prove that $m_{t}=Z_{t}y(x):={\mathbb{E}}[e^{\int _{0}^{t}V_{k}(X^{(k)}_{\tau })d \tau }{y}(X^{(k)}_{t})|X^{(k)}_{0}=x]$ solves (4.10) (seen as a PDE) on ${\mathbb{R}}_{+}^{k}\times [0,T]$ for each ${y}\in {\mathbb{R}}+C^{1}_{0}({\mathbb{R}}_{+}^{k})$ and thus, since $(\operatorname{dom}({\mathcal{A}}))^{\otimes k}\subseteq {\mathbb{R}}+C^{1}_{0}({ \mathbb{R}}_{+}^{k})$, for each $y\in (\operatorname{dom}({\mathcal{A}}))^{\otimes k}$. Indeed, this follows from the Feynman–Kac formula. To see this, note that ${\mathcal{M}}_{k}{y}( x)={\mathcal{G}}^{(k)} {y}( x)+V_{k}( x){y}(x)$. Moreover, by the assumptions on $K$, we know that the map $x \mapsto \int (\| \xi \|^{2} \wedge 1) \nu (x, d\xi )$ is continuous and bounded. This yields the existence of a unique solution to the martingale problem for ${\mathcal{G}}^{(k)}:{\mathbb{R}}+C^{1}_{0}({\mathbb{R}}_{+}^{k})\to { \mathbb{R}}+C_{0}({\mathbb{R}}_{+}^{k})$ (see e.g. Jacod and Shiryaev [45, Theorem III.2.34]). Moreover, by Itô’s formula, $Z_{t}y$ is differentiable with respect to $t$ whence the claim follows.

In particular, $(Z_{t})_{t \geq 0}$ is a strongly continuous semigroup on ${\mathbb{R}}+C_{0}({\mathbb{R}}_{+}^{k})$ with generator $\mathcal{M}_{k}$, implying that

$$\begin{aligned} \lim _{{\varepsilon }\to 0}\frac{1}{{\varepsilon }}\|Z_{t+{ \varepsilon }}y-Z_{t}y-{\varepsilon }{\mathcal{M}}_{k} Z_{t}y\|_{\times }=0 \end{aligned}$$

(B.5)

for each $y \in {\mathbb{R}}+C^{1}_{0}({\mathbb{R}}_{+}^{k})$. Finally, since

$$ {\mathbb{P}}\big[X^{(k)}_{t}\in [0,\Delta +t]^{k} \big| X^{(k)}_{0} \sim {\mathcal{U}}([0,\Delta ]^{k})\big]=1, $$

we get from (4.9) and (B.1) that

$$ |\langle \widehat{{a}}^{\otimes k},Z_{t}{y}^{\otimes k}\rangle _{\alpha }| =\big|{\mathbb{E}}\big[e^{\int _{0}^{t}V_{k}(X^{(k)}_{\tau })d \tau }{y}^{\otimes k}(X^{(k)}_{t})\big| X^{(k)}_{0}\sim {\mathcal{U}}([0, \Delta ]^{k})\big]\big| $$

which is bounded by ${\overline{c}}^{k}C_{t} \|{y}\|_{\alpha }^{k}$ for ${y}\in {\mathbb{R}}+C_{0}({\mathbb{R}}_{+}^{k})$, and thus for ${y} \in (\operatorname{dom}({\mathcal{A}}))^{\otimes k}$. This proves (4.11).

(iii) Due to (4.11) and using the notation introduced in (4.9), we can now define a candidate solution ${a}_{t}\in ({B}^{\otimes k})^{*}$ (in the sense of Definition 3.3) to $\partial _{t}{a}_{t}={\mathcal{L}}_{k}{a}_{t}$ for ${a}_{0}=\widehat{{a}}^{\otimes k}$ as

$$ \langle {a}_{t},{y}\rangle _{\alpha }:=\langle \widehat{{a}}^{\otimes k},Z_{t} {y}\rangle _{\alpha },\qquad {y}\in {B}^{\otimes k}. $$

(iv) In order to conclude the proof we just need to check that $({a}_{t})_{t\geq 0}$ satisfies the conditions of the dual moment formula as given in Theorem 3.4 (i)–(iii). Similarly as in the proof of Proposition 4.7, they consist in the following properties.

Claim 1: $\partial _{t}\langle {a}_{t},{y}^{\otimes k}\rangle _{\alpha }= \langle {a}_{t},{\mathcal{M}}_{k} {y}^{\otimes k}\rangle _{\alpha }$. Observe that fixing ${y}\in {\mathbb{R}}+C_{0}^{1}({\mathbb{R}}_{+}^{k})$ and using that

$$ \langle {a}_{t+{\varepsilon }},{y}\rangle _{\alpha }=\langle \widehat{{a}}^{\otimes k},Z_{t+{\varepsilon }}{y}\rangle _{\alpha }= \langle \widehat{{a}}^{\otimes k},Z_{t}Z_{\varepsilon }{y}\rangle _{\alpha }=\langle {a}_{t}, Z_{\varepsilon }{y}\rangle _{\alpha }, $$

we can compute by (B.5) that

$$\begin{aligned} \frac{1}{{\varepsilon }}|\langle {a}_{t+{\varepsilon }},{y}\rangle _{\alpha }-\langle {a}_{t},{y}\rangle _{\alpha }-{\varepsilon }\langle {a}_{t},{ \mathcal{M}}_{k} {y}\rangle _{\alpha }| &=\frac{1}{{\varepsilon }}| \langle {a}_{t},Z_{{\varepsilon }}{y}\rangle _{\alpha }\!-\langle {a}_{t},{y} \rangle _{\alpha }\!-{\varepsilon }\langle {a}_{t},{\mathcal{M}}_{k} {y} \rangle _{\alpha }| \\ & \leq \|{a}_{t}\|_{*k}\frac{1}{{\varepsilon }}\|Z_{{\varepsilon }}{y}-{y}-{ \varepsilon }{\mathcal{M}}_{k} {y}\|_{\times } \end{aligned}$$

which goes to 0. Hence $\partial _{t}\langle {a}_{t},{y}\rangle _{\alpha }=\langle {a}_{t},{ \mathcal{M}}_{k} {y}\rangle _{\alpha }$ for all $y \in {\mathbb{R}}+C_{0}^{1}({\mathbb{R}}_{+}^{k})$.

Claim 2: ${a}_{t}\in {\mathcal{D}}({\mathcal{L}}_{k})$. Observe that since ${y}\mapsto \int {y}(x+\xi )\nu (x,d\xi )$ is a bounded linear operator, we have that ${\mathcal{D}}({\mathcal{L}}_{k})$ coincides with

$$ {\mathcal{D}}({{\mathcal{A}}^{*}\otimes {\mathrm{Id}}^{\otimes (k-1)}}):= \{{a}\in ({{B}^{\otimes k}})^{*}\colon \langle {a},\nabla {y}^{ \otimes k}\rangle _{\alpha }\leq C_{a}\|{y}\|_{\alpha }^{k},\ {y}\in \operatorname{dom}({\mathcal{A}})\}. $$

Fix ${y}\in {\mathbb{R}}+C_{0}^{1}({\mathbb{R}}_{+})$ and note that

$$ \langle {a}_{t},{\mathcal{M}}_{k}{y}^{\otimes k}\rangle _{\alpha }= \partial _{t}\langle {a}_{t},{{y}^{\otimes k}}\rangle _{\alpha }= \partial _{t}\langle \widehat{{a}}^{\otimes k},Z_{t}{{y}^{\otimes k}} \rangle _{\alpha }=\langle \widehat{{a}}^{\otimes k},{\mathcal{M}}_{k}Z_{t}{{y}^{ \otimes k}}\rangle _{\alpha }. $$

Since ${\mathcal{M}}_{k}Z_{t}{{y}^{\otimes k}}\in {\mathbb{R}}+C_{0}({ \mathbb{R}}_{+}^{k})$ we know that $1^{\top }\nabla Z_{t}{{y}^{\otimes k}}\in {\mathbb{R}}+C_{0}({\mathbb{R}}_{+}^{k})$. By the fundamental theorem of calculus, we can thus compute

$$\begin{aligned} &\langle \widehat{{a}}^{\otimes k},{\mathcal{M}}_{k}Z_{t}{{y}^{ \otimes k}}\rangle _{\alpha }\\ &=\bigg\langle \widehat{{a}}^{\otimes k},1^{\top }\nabla Z_{t}{{y}^{ \otimes k}}+\int Z_{t}{y}^{\otimes k}({\,\cdot \,}+\xi )\nu ({\, \cdot \,},d\xi )\bigg\rangle _{\alpha }\\ &= \frac{k}{\Delta }\langle \widehat{{a}}^{\otimes (k-1)}, Z_{t}{{y}^{ \otimes k}}(\Delta ,{\,\cdot \,})\rangle _{\alpha }-\frac{k}{\Delta }\langle \widehat{{a}}^{\otimes (k-1)}, Z_{t}{{y}^{\otimes k}}(0,{\, \cdot \,})\rangle _{\alpha }\\ &\phantom{=:}+\bigg\langle \widehat{{a}}^{\otimes k},\int Z_{t}{y}^{\otimes k}({ \,\cdot \,}+\xi )\nu ({\,\cdot \,},d\xi )\bigg\rangle _{\alpha }\\ &= \frac{k}{\Delta }{\mathbb{E}}\big[e^{\int _{0}^{t}V_{k}(X^{(k)}_{\tau })d \tau }{y}^{\otimes k}(X^{(k)}_{t}) \big| X^{(k)}_{0}=\delta _{\Delta }\otimes {\mathcal{U}}([0,\Delta ]^{k-1})\big] \\ &\phantom{=:}-\frac{k}{\Delta }{\mathbb{E}}\big[e^{\int _{0}^{t}V_{k}(X^{(k)}_{\tau })d \tau }{y}^{\otimes k}(X^{(k)}_{t}) \big| X^{(k)}_{0}=\delta _{0} \otimes {\mathcal{U}}([0,\Delta ]^{k-1})\big] \\ &\phantom{=:}+\frac{k(k-1)}{2\Delta ^{2}}\bigg(\int _{0}^{\Delta }K(x)dx\bigg)^{2} \\ &\phantom{=:+}\times {\mathbb{E}}\big[e^{\int _{0}^{t}V_{k}(X^{(k)}_{\tau })d\tau }{y}^{ \otimes k}(X^{(k)}_{t}) \big| X^{(k)}_{0}=\delta _{0}\otimes \delta _{0} \otimes {\mathcal{U}}([0,\Delta ]^{k-2})\big]. \end{aligned}$$

An inspection of this expression yields

$$ | \langle {a}_{t},{\mathcal{M}}_{k}{y}^{\otimes k}\rangle _{\alpha }| \leq \bigg(\frac{2k}{\Delta }+\frac{k(k-1)}{2\Delta ^{2}}\int _{0}^{\Delta }K(x)^{2}dx\bigg) C_{t} {\overline{c}}^{k}\|{y}\|^{k}_{\alpha }=:c_{t} \|{y}\|^{k}_{\alpha }$$

for all $t\in [0,T]$. Since $\operatorname{dom}(\mathcal{A})\subseteq {\mathbb{R}}+C^{1}_{0}({ \mathbb{R}}_{+})$, the claim follows by noting that

$$ \langle {a}_{t},\nabla {y}^{\otimes k}\rangle _{\alpha }\leq \bigg(c_{t}+ \|{a}_{t}\|_{*k}\frac{k(k-1)}{2}\|K\|_{\alpha }^{2}\bigg)\|{y}\|_{\alpha }^{k}. $$

Claim 3: Conditions (ii) and (iii) of Theorem 3.4hold. We again apply Lemma 3.17. The moment condition is satisfied by assumption, and (3.8) holds because we have $\sup _{t\in [0,T]}c_{t}<\infty$ with $c_{t}$ given just above. □