2.1. Notation

Let $\Delta$ be the open set of $\mathbb{R}^{D+1}$ given by $\Delta\,:\!=\, \big\{(m,x),\ m\in \mathbb{R},\ x \in\mathbb{R}^{d},\ m> x^1,\ x=\big(x^1,\ldots,x^d\big)\big\}.$ From now on, we use Einstein’s convention. The infinitesimal generator ${\mathcal L}$ of the diffusion X defined in (1) is the partial differential operator on the space $C^2(\mathbb{R}^d,\mathbb{R})$ given by

(2) \begin{align}{\mathcal{L}}= B^i\partial_{x_i}+{\frac{1}{2}} (A A^t)^{ij}\partial^2_{x_i,x_j},\end{align}

where $A^t$ denotes the transposed matrix.

Its adjoint operator is ${{\mathcal{L}}^*f= {\frac{1}{2}} \Sigma^{ij}\partial^2_{ij}f -\big[B^i-\partial_j\big(\Sigma^{ij}\big)\big]\partial_if -\big[\partial_iB^i-{\frac{1}{2}}\partial^2_{ij} \big(\Sigma^{ij}\big)\big]f}$ , where $\Sigma\,:\!=\,AA^t.$ In what follows, the operators ${\mathcal{L}}$ and ${\mathcal{L}}^* $ are extended to the space $C^2\big(\mathbb{R}^{d+1},\mathbb{R}\big)$ , for $\Phi \in C^2\big(\mathbb{R}^{d+1},\mathbb{R}\big)$ , as

\begin{equation*}{\mathcal{L}}(\Phi)(m,x)=B^i(x)\partial_{x_i}\Phi(m,x)+{\frac{1}{2}} \Sigma^{ij}(x)\partial^2_{x_i,x_j}\Phi(m,x)\end{equation*}

and

\begin{align*} {\mathcal{L}}^*&(\Phi)(m,x)= \\ &{\frac{1}{2}}\Sigma^{ij}(x) \partial^2_{ij} \Phi(m,x)-[B^i-\partial_j\big(\Sigma^{ij}\big)](x)\partial_{x_i}\Phi(m,x)+\left[{\frac{1}{2}} \partial^2_{x_i,x_j}\Sigma^{ij}-\partial_{x_i}B^i\right](x)\Phi(m,x).\end{align*}

We stress that these operators are degenerate, since no derivative with respect to the variable m appears.

Let $A^1(x)$ be the d-dimensional vector $A^1(x)=\big(A^1_j(x),\ j=1,\ldots,d\big)\in \mathbb{R}^d$ corresponding to the first column of A(x); similarly $A_j(x)$ denotes its jth row.

Recall that M denotes the running maximum of the first component of X, meaning $M_t=\sup_{0\leq s\leq t} \big\{X^1_s\big\}$ , and V is the $\mathbb{R}^{d+1}$ -valued process defined by $(V_t=(M_t, X_t), \forall t\geq 0).$ Finally, $\tilde x\in\mathbb{R}^{d-1}$ denotes the vector $(x_2,\ldots,x_d).$

In [Reference Coutin and Pontier9], under Assumptions (4) and (5) below, when the initial value is deterministic, $X_0=x_0 \in {\mathbb R}^d$ , the density of $V_t$ exists and is denoted by $p_V(.;\,t,x_0)$ . If $\mu_0$ is the distribution of $X_0$ , the density of the law of $V_t$ with respect to the Lebesgue measure on ${\mathbb R}^{d+1} $ is

(3) \begin{align}p_V(.;\,t,\mu_0)\,:\!=\,\int_{{\mathbb R}^d} p_V(.;\,t,x_0)d\mu_0(x_0).\end{align}

When there is no ambiguity, the dependency on $\mu_0$ is omitted.

Since $M_t \geq X^1_t,$ the support of $p_V(.;\,t,\mu_0)$ is contained in $\bar{\Delta}\,:\!=\,\left\{(m,x)\in {\mathbb R}^{d+1} | m\geq x^1\right\}.$

2.2. Main results

The aim of this article is to show that the density $p_V$ is a weak solution of a Fokker–Planck PDE. We assume that the coefficients B and A satisfy

(4) \begin{equation} B \in C^1_b\big(\mathbb{R}^d,\mathbb{R}^d\big) \quad \mbox{and} \quad {A\in C^2_b\big(\mathbb{R}^d,\mathbb{R}^{d\times d}\big)}\end{equation}

and that there exists a constant $c>0$ such that the Euclidean norm of any vector v satisfies

(5) \begin{align}c\|v\|^2\leq v^tA(x)A^t(x) v ,\,\,\forall v,x\in {\mathbb R}^d.\end{align}

Our first result will be established under the following hypothesis, which is a quite natural assumption on the regularity of $p_V$ in the neighborhood of the boundary of $\Delta$ , since the set of times where the process M increases is included in the set $\{t,\,\,M_t=X_t\}$ .

Theorem 2.2. Assume that A and B fulfil (4) and (5) and that (M, X) fulfils Hypothesis 2.1. Then, for any initial law $\mu_0$ and $F \in C^2_b\big({\mathbb R}^{d+1},{\mathbb R}\big)$ ,

(6) \begin{align}{\mathbb E}\left[F\big(M_t,X_t\big)\right)& ={\mathbb E} \left[ F\big(X^1_0,X_0\big)\right] + \int_0^t {\mathbb E} \left[ {\mathcal L}\left(F\right)(M_s,X_s)\right] ds\nonumber\\&+\frac{1}{2} {\int_0^t{\mathbb E}\left[\partial_mF\big(X^1_s,{X_s}\big)\|A^{1}(X_s)\|^2\frac{p_V\big(X^1_s,{X_s};\,s\big)}{p_{X}(X_s;\,s)}\right] ds.}\end{align}

Actually, $p_X$ is the solution of the PDE $ \partial_t p= {\mathcal L}^* p,\ p(.;\,0)=\mu_0,$ where

${{\mathcal{L}}^*f= {\frac{1}{2}} \Sigma^{ij}\partial^2_{ij}f -\big[B^i-\partial_j\big(\Sigma^{ij}\big)\big]\partial_if -\big[\partial_iB^i-{\frac{1}{2}}\partial^2_{ij} \big(\Sigma^{ij}\big)\big]f}$ . Let $a_{ij}\,:\!=\,\Sigma^{ij},$

$a_i\,:\!=\,\big[B^i-\partial_j\big(\Sigma^{ij}\big)\big]\partial_i,$ and $ {a_0\,:\!=\,\partial_iB^i-{\frac{1}{2}}\partial^2_{ij}\big(\Sigma^{ij}\big)}.$ Under Assumptions (4) and (5), the operator ${\mathcal{L}}^*$ satisfies all the assumptions of [Reference Garroni and Menaldi14, Theorem 3.5] (see (3.2), (3.3), and (3.4) on p. 177). As a consequence of [Reference Garroni and Menaldi14, Theorem 3.5], line 14, we have $p_X(x;\,s)>0$ .

The starting point of the proof of Theorem 2.2 is Itô’s formula: let F belong to $C^2_b\big(\mathbb{R}^{d+1},\mathbb{R}\big)$ . The process M is increasing; hence $V=(M, X)$ is a semi-martingale. Applying Itô’s formula to F(V) and taking the expectation of both sides, we have

\begin{align*}{\mathbb E} \left[F(V_t)\right]={\mathbb E} \left[F(V_0)\right] +\int_0^t {\mathbb E}\left[{\mathcal L}(F)(V_s)\right]ds+{\mathbb E}\left[ \int_0^t \partial_mF(V_s) dM_s\right].\end{align*}

The novelty comes from the third term in the right-hand side of the equation above. The following theorem, proved in Section 3, completes the proof of Theorem 2.2.

Theorem 2.3. Assume that A and B fulfil (4) and (5) and that (M, X) fulfils Hypothesis 2.1. For every $\Psi \in C^1_b\big({\mathbb R}^{d+1},{\mathbb R}\big)$ , let $F_\psi$ be the map

\begin{equation*}F_\psi\,:\,t\mapsto {\mathbb E} \left[\int_0^t \Psi(M_s,X_s)dM_s\right].\end{equation*}

Then $F_{\Psi}$ is absolutely continuous with respect to the Lebesgue measure, and its derivative is

\begin{align*}\dot{F_{\Psi}}(t)=\frac{1}{2}\int_{{\mathbb R}^d}{\Psi}(m,m,\tilde{x}) \|A^{1}(m,\tilde{x})\|^2p_V(m,m,\tilde{x};\,t) dm d\tilde{x}.\end{align*}

Observe that, as expressed in Theorem 2.2, this derivative can be written as

\begin{equation*}\frac{1}{2}{\mathbb E}\left[\Psi\big(X^1_t,{X_t}\big)\|A^{1}({X_t})\|^2\frac{p_V\big(X^1_t,{X_t};\,t\big)}{p_{X}( X_t;\,t)}\right].\end{equation*}

Theorem 2.4. Assume that $A=I_d$ and B satisfies Assumption (4). Then, for all $t>0$ , the distribution of the pair $\big(M_t,X_t\big)$ fulfils Hypothesis 2.1. As a consequence, for all $F\in C^2_b\big({\mathbb R}^{d+1},{\mathbb R}\big)$ ,

\begin{align*}&{\mathbb E}\left[F\big(M_t,X_t\big)\right] ={\mathbb E} \left[ F\big(X^1_0,X_0\big)\right] + \int_0^t {\mathbb E} \left[ {\mathcal L}\left(F\right)(M_s,X_s)\right] ds\\&+\frac{1}{2} \int_0^t{\mathbb E}\left[\partial_mF\big(X^1_s,{X_s}\big)\frac{p_V\big(X^1_s,{X_s};\,s\big)}{p_{X}( X_s;\,s)}\right] ds.\end{align*}

When $d=1$ , a Lamperti transformation leads to the following corollary.

Corollary 1. Assume that $d=1$ , and A and B satisfy (4) and (5). Then the density $p_V$ satisfies Hypothesis 2.1, so

\begin{align*}&{\mathbb E}\left[F\big(M_t,X_t\big)\right] ={\mathbb E} \left[ F(X_0,X_0)\right] + \int_0^t {\mathbb E} \left[ {\mathcal L}\left(F\right)(M_s,X_s)\right] ds\\&+\frac{1}{2} \int_0^t{\mathbb E}\left[A^2(X_s)\partial_mF(X_s,{X_s})\frac{p_V(X_s,{X_s};\,s)}{p_{X}( X_s;\,s)}\right] ds.\end{align*}

Remark 2.3. If $p_V$ is regular enough, and if the initial law of $X_0$ satisfies $\mu_0(dx)=f_0(x)dx$ , then Theorem 2.2 means that $p_V$ is a weak solution in the set $\Delta$ of $\partial_t p= {\mathcal{L}}^*p,$ where ${\mathcal{L}}^*f= {\frac{1}{2}} \Sigma^{ij}\partial^2_{ij}f -\big[B^i-\partial_j\big(\Sigma^{ij}\big)\big] \partial_if -\partial_iB^i-{\frac{1}{2}}\partial^2_{ij} \big(\Sigma^{ij}\big))f$ with boundary condition

(7) \begin{equation}B^1(m,\tilde x) p_V(m,m,\tilde x;\,s)=\partial_{x_k}\big(\Sigma^{1,k} p_V\big)(m,m,\tilde x;\,s)+{\frac{1}{2}} \partial_{m}\big(\|A^1\|^2 p_V\big)(m,m,\tilde x;\,s).\end{equation}

This result is proved in Appendix A.3.

This boundary condition also appears in [Reference Blanchet-Scalliet, Dorobantu and Gay4, Proposition 4, Equation (11)] (Ornstein–Uhlenbeck process). Finally, a similar PDE is studied in [Reference Garroni and Menaldi14, Chapter 1.2], where the authors establish the existence of a unique strong solution of the PDE, but in the case of a non-degenerate elliptic operator.

3.1. Tools for proving Proposition 3.1

Here we provide some estimates of the expectations of the increments of the processes X and M. Assumptions (4) and (5) allow us to introduce a constant K which denotes either $ \max\!(\|A\|_{\infty},\|B\|_{\infty})$ or $ \max\!(\|A\|_{\infty},\|B\|_{\infty},$ $\|\nabla A\|_\infty).$ Let $C_p$ be the constant in the Burkholder–Davis–Gundy inequality (cf. [Reference Bain and Crisan3, Theorem B.36]).

Lemma 3.1. Let A and B be bounded. Then, for all $ 0<h\leq 1,$ for all $ p\geq 1$ there exists a constant $C_{p,K}$ (depending only on p and K) such that

\begin{align*}{\sup_{t>0}}\,\mathbb{E} \left[ \sup_{0\leq s \leq h}\| X_{t+s} -X_t\|^p \right] \leq C_{p,K} h^{p/2}.\end{align*}

Proof. Using the fact that $(a+b)^p \leq 2^{p-1} \left[a^p +b^p \right],\,a,b\geq 0$ , one obtains

\begin{align*}0\leq \sup_{s \leq h} \| X_{t+s}-X_t \|^p \leq 2^{p-1}\left[\sup_{u \leq h} \left( \Big\|\int_t^{t+u} B(X_s) ds\Big\| \right)^p+ \sup_{u \leq h}\left( \Big\|\int_t^{t+u} A_j(X_s) dW^j_s \Big\|\right)^p \right].\end{align*}

If we take the expectation of both sides, the Burkholder–Davis–Gundy inequality implies

\begin{equation*}\mathbb{E}\Big[\sup_{s \leq h} \| X_{t+s}-X_t \|^p\Big] \leq 2^{p-1}(1+C_p)\mathbb{E}\left[ \left( \int_{t}^{t+h} \|B(X_s) \|ds \right)^p+\left( \int_t^{t+h} \|A(X_s)\|^2 ds\right)^{p/2} \right].\end{equation*}

Assumption (4) on B and A yields

\begin{equation*}\mathbb{E}\left[\sup_{s \leq h}{\| X_{t+s}-X_t \|^p}\right] \leq 2^{p-1}(1+C_p)\big(h^{p}K^p+ h^{p/2} K^p\big).\end{equation*}

Lemma 3.2. Let B and A satisfy Assumptions (4) and (5). Then, for all $ 0<h\leq 1,$ for all $ p\geq 1$ we get

(14) \begin{equation}{\sup_{t>0}}\,\mathbb{E}\big[| M_{t+h} -M_t|^p \big]\leq C_{p,K}h^{p/2}\,;\,\mathbb{E}\big[ | M_{t+h} -M_t|^p \big]=o\big(h^{p/2}\big).\end{equation}

Proof. Recall that

\begin{equation*}M_{t+h}-M_t= \left( \sup_{0 \leq u \leq h} \big(X_{t+u}^1 - X_t^1\big)+X_t^1 - M_t \right)_+,\end{equation*}

recalling $(x)_+=\max\!(x,0).$ For any $a \geq 0,$ one has ${( x-a)_+} \leq |x|{{\mathbf 1}_{\{x >a\}}}$ ; thus

\begin{align*}0 \leq M_{t+h}-M_t\leq \Big| \sup_{0 \leq u \leq h} \big(X_{t+u}^1- X_t^1\big)\Big| {\mathbf 1}_{\big\{\sup_{0 \leq u \leq h} \big(X_{t+u}^1- X_t^1\big)> M_t-X_t^1\big\}}.\end{align*}

The Cauchy–Schwarz inequality yields

\begin{align*}0 &\leq \mathbb{E} \left[ \left( M_{t+h}-M_t\right)^p \right]\\&\leq \sqrt{\mathbb{E}\left[\Big| \sup_{0 \leq u \leq h} \big(X_{t+u}^1- X_t^1\big)\Big|^{2p} \right] {\mathbb P} \bigg( \bigg\{\sup_{0 \leq u \leq h} \big(X_{t+u}^1- X_t^1\big)> M_t-X_t^1\bigg\}\bigg)}.\end{align*}

Replacing p by 2p in Lemma 3.1 leads to the inequality in (14), and the equality

\begin{equation*}\lim_{h\rightarrow 0}\sup_{0 \leq u \leq h} \big({X_{t+u}^1}- X_t^1\big)=0\end{equation*}

holds almost surely. According to [Reference Coutin and Pontier9, Theorem 1.1] extended to $X_0$ with law $\mu_0$ on $\mathbb{R}^d$ , the pair $\big(M_t,X_t\big)$ admits a density; thus ${\mathbb P} \{M_t-X_t^1=0\}=0$ holds almost surely. Therefore $E\! \left( \left[ M_{t+h}-M_t\right]^p \right)$ is bounded by the product of $h^{p/2}$ and a factor going to zero when h goes to 0, and this quantity is $o(h^{p/2})$ .

For any fixed t we recall the process $(X_{t,u},\ u \in [0,h])$ and the running maximum of its first component as follows:

(15) \begin{align} X_{t,u} \,:\!=\, {\sum_j A_j(X_t)} \hat W^j_u, \qquad M_{t,h}\,:\!=\,\sup _{0 \leq u \leq h} X_{t,u}^1.\end{align}

Lemma 3.3. Under Assumptions (4) and (5), for every $ p\geq 1$ there exists a constant $C_{p,K}$ such that for all $ t\leq T$ , for all $ h \in [0,1]$ ,

\begin{align*}\mathbb{E} \left[ \sup_{s \leq h} \big| X^1_{s+t}-X^1_t - X^1_{t,s}\big|^p \right]\leq C_{p,K}h^{p}.\end{align*}

Proof. By definition, recalling $\hat{W}_u\,:\!=\, W_{t+u} - W_t$ , $u \geq 0,$ we obtain

\begin{align*}X^1_{s+t} -X^1_t - X^1_{t,s} = {\int_0^s} {B^1(X_{u+t})}du +{\int_0^s} \left[ A^1(X_{u+t}) - A^1(X_t)\right] d{\hat{W}}_u.\end{align*}

Once again using the inequality $(a+b)^p \leq 2^{p-1}(a^p+b^p ),\ a,b\geq 0,$ we get

\begin{align*}&\sup_{0\leq s \leq h} \big| X^1_{s+t}-X^1_t - X^1_{t,s}\big|^p \leq\\& 2^{p-1}\left[\left( {\int_0^h} {\|B^1(X_{u+t})\|} du\right)^p +\sup_{0\leq s \leq h} \left\| {\int_0^s} \left[ A^1(X_{u+t}) - A^1(X_t)\right] d{\hat{W}}_u\right\|^p \right]. \end{align*}

Taking the expectation of both sides and applying the Burkholder–Davis–Gundy inequality yields, with $D_p=2^{p-1}(1+ C_p)$ ,

\begin{align*} & \mathbb{E} \left[\sup_{0\leq s \leq h} \big| X^1_{s+t}-X^1_t - X^1_{t,s}\big|^p \right]\\& \qquad \leq D_p\left(\mathbb{E}\left[{\int_0^h} {\|B^1(X_{u+t}) \|}du\right]^p +\mathbb{E} \left| {\int_0^h} \|A^1(X_{u+t}) - A^1(X_t)\|^2 du\right|^{p/2}\right). \end{align*}

The first term above is bounded by $K^ph^p$ , since B is bounded. The assumption that A belongs to $C^1_b\big(\mathbb{R}^d,\mathbb{R}^{d\times d}\big)$ and Jensen’s inequality imply that the second term is bounded by $K^ph^{p/2-1}{\int_0^h}E\|X_{u+t}-X_t\|^pdu$ ; thus

\begin{equation*}\mathbb{E}\left[\sup_{0\leq s \leq h} \big| X^1_{s+t}-X^1_t - X^1_{t,s}\big|^p \right] \leq D_p{K^ph^{p/2-1}\left(h^{p/2+1} + {\int_0^h}\mathbb{E} \|X_{u+t}- X_t\|^pdu\right).}\end{equation*}

From Lemma 3.1 we obtain the uniform upper bound $\mathbb{E} [\|X_{u+t}- X_t\|^p]\leq C_{p,K} u^{p/2}$ ; hence,

\begin{equation*}\mathbb{E} \left[\sup_{s \leq h} \big| X^1_{s+t}-X^1_t - X^1_{t,s}\big| ^p \right] \leq\frac{D_pK^pC_{p,K}}{\frac{p}{2}+1}h^p.\end{equation*}

Proof. First, observe that

(16) \begin{align}\forall a \in {\mathbb R},\,\,\left| \left(x-a\right)_+ -\left(y-a\right)_+ \right| \leq \left|x-y\right|\left[ {\mathbf 1}_{\{x> a\}} + {\mathbf 1}_{\{y> a\}} \right],\end{align}

and if f and g are functions on [0, T], then

\begin{align*}\forall s\in[0,T],\,\,f(s) -\sup_{0\leq u \leq T}g(u) \leq f(s) - g(s) \leq | f(s) - g(s)| \leq \sup_{v \leq T}| f(v) - g(v)|;\end{align*}

hence $\sup_{s \leq T} f(s) - \sup_{u \leq T}g(u) \leq \sup_{v \leq T}|f(v) - g(v)|.$ Here the roles of f and g are symmetrical, so $\sup_{s \leq T} g(s) - \sup_{u \leq T}f(u) \leq \sup_{v \leq T}| f(v) - g(v)|,$ and

(17) \begin{equation}\left| \sup_{s \leq T} g(s) - \sup_{u \leq T}f(u) \right| \leq \sup_{v \leq T}|f(v) - g(v)|.\end{equation}

We now consider $M_{t+h}-M_t =\left( \sup_{0 \leq u \leq h}\big(X_{u+t}^1 -X_t^1\big) -M_t + X_t^1\right)_{+};$ using (16),

\begin{align*}&\left| M_{t+h}- M_t - \left( M_{t,h} - M_t + X_t^1\right)_+ \right|\leq\\& \left| \sup_{0 \leq u \leq h}\big(X_{u+t}^1 -X_t^1\big) - M_{t,h} \right| \left[ {\mathbf 1}_{\big\{\sup_{0 \leq u \leq h}\big(X_{u+t}^1 -X_t^1\big) > M_t-{X^1_t}\big\}} + {\mathbf 1}_{\big\{M_{t,h}> M_t-X_t^1\big\}} \right].\end{align*}

Then, for any t fixed, we apply the inequality (17) to the maps $g\,:\,u\mapsto X_{{u+t}}^1 {-X_t}^1$ and

$f\,:\,u\mapsto X_{t,u}^1$ . Then

\begin{align*} &\left| M_{t+h}- M_t - \left( M_{t,h} - M_t + X_t^1\right)_+ \right| \leq \\& \quad \sup_{0\leq u \leq h} \left|X_{u+t}^1 -X_t^1 - X_{t,u}^1\right| \left[ {\mathbf 1}_{\big\{\sup_{0 \leq u \leq h}\big(X_{u+t}^1 -X_t^1\big) > M_t-X_t^1\big\}} + {\mathbf 1}_{\big\{M_{t,h}> M_t-X_t^1\big\}} \right].\end{align*}

From the Cauchy–Schwarz inequality and the fact that $(a+b)^2 \leq 2\big(a^2+b^2\big),$ we get

\begin{align*}&\mathbb{E} \left[ \left| M_{t+h}- M_t - \left( M_{t,h} - M_t + X_t^1\right)_+ \right|\right]\leq\\&\sqrt{2\mathbb{E}\left[\sup_{u \leq h} \left|X_{u+t}^1 -X_t^1 - X_{t,u}^1\right| ^2\right]\left( {\mathbb P} \Big\{\sup_{0 \leq u \leq h} \big(X_{u+t}^1 -X_t^1\big) > M_t-X_t^1\Big\}+{\mathbb P} \big\{M_{t,h}> M_t-X_t^1\big\} \right)}.\end{align*}

Lemma 3.3 with $p=2$ ensures that the map $h \mapsto h^{-1}\sqrt{2\mathbb{E} \left[ \sup_{u \leq h} \left|X_{u+t}^1-X_t^1 - X_{t,u}^1\right| ^2\right]}$ is uniformly bounded in t. Concerning the second factor, we have the following:

• • First, the almost sure continuity with respect to h ensures that the quantities $\lim_{h\rightarrow 0} \sup_{0 \leq u \leq h}\!\big(X_{u+t}^1 -X_t^1\big) $ and $\lim_{h\rightarrow 0} M_{t,h}$ are equal to 0.

• • Second, the law of the pair $\big(M_t,X_t\big)$ admits a density with respect to the Lebesgue measure on $\bar\Delta$ , according to [Reference Coutin and Pontier9, Theorem 1.1], so ${\mathbb P}\big( \big\{0= M_t-X_t^1\big\}\big)=0$ , and the limit of the second factor is equal to $0.$

This concludes the proof of the lemma.

Recall Definition (15): $X_{t,h}= A_j(X_t)\big[ W_{t+h}^j-W_t^j\big],\,{M_{t,h}=\mathbf{\sup_{0\leq u\leq h}} {X_{t,u}^1},}\,h \in [0,1].$

Lemma 3.5. Under Assumptions (4) and (5), with ${{\mathcal{H}}}$ defined in (13), we have

\begin{equation*}\mathbb{E}\left[ \big( M_{t,h} - M_t + X_t^1 \big)_+ | {\mathcal F}_t\right] =2\|A^1(X_t)\| \sqrt{h}{{\mathcal{H}}}\left(\frac{M_t -X_t^1}{\|A^{1}(X_t)\|\sqrt{h}}\right).\end{equation*}

Proof. For any t fixed, conditionally on ${\mathcal F}_t$ , the process $\big(X_{t,u}^1,\ u \in [0,h]\big)$ from (9) has the same law as $\big(\sqrt{h}\|A^{1}(X_t)\|\hat W_u,\,u\in [0,1]\big)$ , where $\hat W$ is a Brownian motion independent of ${\mathcal F}_t$ , and for any h, the random variable $M_{t,h}$ has the same law as $ \sqrt{h}\|A^{1}(X_t)\|\sup_{u \leq 1}\hat{W}_u.$

Following [Reference Jeanblanc, Yor and Chesney17, Section 3.1.3], the random variable $\sup_{u \leq 1}\hat W_u$ has the same law as $|G|$ , where G is a standard Gaussian variable (independent of ${\mathcal F}_t$ ) with density $\frac{2}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}{\mathbf 1}_{[0,+\infty [}(z)$ . Then, using the function ${{\mathcal{H}}}$ introduced in (13), we have

\begin{align*}\mathbb{E}\left[\big(M_{t,h} -\big(M_t -X_t^1\big)\big)_+ |{\mathcal F}_t \right]&= \int_0^{\infty} \left( \|A^{1}(X_t)\|\sqrt{h} z - \big(M_t -X_t^1\big)\right)_+ \frac{2}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz\\&= 2 \|A^{1}(X_t)\|\sqrt{h} {{\mathcal{H}}} \left( \frac{M_t -X_t^1}{\sqrt{h}\|A^{1}(X_t)\|} \right).\end{align*}

3.2. Proof of Proposition 3.1

Let $t>0$ . The key in this proof is to write the quantity

\begin{equation*}\mathbb{E}\left[\int_t^{t+h}\Psi(V_s)dM_s\right]-{2}\sqrt h\mathbb{E}\left[\Psi({V_t}){\|A^{1}({X_t})\|}{{\mathcal{H}}}\left(\frac{M_t-X_t^1}{\sqrt h \|A^{1}({X_t})\|}\right)\right]\end{equation*}

as the sum of three terms,

(18) \begin{align}&\mathbb{E}\Big[\int_t^{t+h}(\Psi(V_s)-\Psi(V_t))dM_s\Big]+{\mathbb{E}\Big[\Psi(V_t)\Big(\big(M_{t+h}-M_t\big)- \mathbb{E}\left[M_{t,h}-M_t+X_t^1\big)_+| {\mathcal F}_t \right]\Big)\Big]}\\&+\mathbb{E}\left[\Psi(V_t) {\mathbb{E}} \left[\big(M_{t,h}-M_t+X_t^1\big)_+\left| \right. {\mathcal F}_{t} \right]-{2}\sqrt h\Psi(V_t){\|A^{1}(X_t)\|}{{\mathcal{H}}}\left(\frac{M_t-X_t^1}{\sqrt h \|A^{1}(X_t)\|}\right)\right].\nonumber\end{align}

We now prove that each of the terms in the sum (18) are both o(h) and O(h) uniformly in time.

(a) Using Lemma 3.1, the third term is null.

(b) Concerning the second term, using the fact that $\Psi$ is bounded and Lemma 3.1(i), for all $ t\in [0,T]$

\begin{align*}&\left|\mathbb{E}\left[\Psi(V_t)[\big(M_{t+h}-M_t\big)- \mathbb{E}\big[\big(M_{t,h}-M_t+X_t^1\big)_+\left| \right. {\mathcal F}_{t} \big]\right]\right|\leq\\& \quad \|\Psi\|_\infty\left|\mathbb{E}\left[M_{t+h}-M_t- \mathbb{E}\big[\big(M_{t,h}-M_t+X_t^1\big)_+| {\mathcal F}_t\big]\right]\right|{\leq Ch\|\Psi\|_\infty},\end{align*}

as is required in (12). Moreover, using Lemma 3.4(ii),

\begin{align*}&\lim_{h\to 0}\frac{1}{h}\left|\mathbb{E}\left[\Psi(V_t)[\big(M_{t+h}-M_t\big)- \mathbb{E}\big[\big(M_{t,h}-M_t+X_t^1\big)_+\left| \right. {\mathcal F}_{t} \big]\right]\right|=0.\end{align*}

(c) Since $\nabla \Psi$ is bounded and the process M is increasing, the first term is bounded:

\begin{equation*}\mathbb{E}\left[\int_t^{t+h}[\Psi(V_s)-\Psi(V_t)]dM_s\right]\leq\|\nabla\Psi\|_\infty \mathbb{E}\bigg[\sup_{t\leq s\leq t+h}\|V_s-V_t\|\big(M_{t+h}-M_t\big)\bigg].\end{equation*}

Using the Cauchy–Schwarz inequality,

\begin{equation*}\mathbb{E}\left[\sup_{t\leq s\leq t+h}\|V_s-V_t\|\big(M_{t+h}-M_t\big)\right]\leq \sqrt{\mathbb{E}\bigg[\sup_{t\leq s\leq t+h}\|V_s-V_t\|^2\bigg]\mathbb{E}\big[\big(M_{t+h}-M_t\big)^2\big]}.\end{equation*}

Since $\|V_s-V_t\|^2=(M_s-M_t)^2+\|X_s-X_t\|^2,$ we obtain

\begin{equation*}\sup_{t\leq s\leq t+h}\|V_s-V_t\|^2\leq\big(M_{t+h}-M_t\big)^2+\sup_{t\leq s\leq t+h}\|X_s-X_t\|^2;\end{equation*}

hence

\begin{align*}&\mathbb{E}\left[\sup_{t\leq s\leq t+h}\|V_s-V_t\|\big(M_{t+h}-M_t\big)\right]\leq\\& \quad \sqrt{\mathbb{E}\big[\big(M_{t+h}-M_t\big)^2\big]+ \mathbb{E}\left[\sup_{t\leq s\leq t+h}\|X_s-X_t\|\right]^2}\sqrt{\mathbb{E}\big[\big(M_{t+h}-M_t\big)^2\big]}.\end{align*}

Lemmas 3.1 and 3.2 ( $p=2$ ) yield the fact that the first factor is $o\big(\sqrt h\big)$ and the second is $O\big(\sqrt h\big)$ uniformly with respect to $t\geq 0$ . Then $\mathbb{E}\big[\sup_{t\leq s\leq t+h}\|V_s-V_t\|\big(M_{t+h}-M_t\big)\big]$ is o(h) and O(h) uniformly with respect to $t\geq 0$ .

3.3. Proof of Proposition 3.2

(i) Recall that A and B fulfil (4), (5) and (M, X) fulfils Hypothesis 2.1. Then, using the density $p_V$ of the law of the pair $\big(M_t,X_t\big)$ , we have

\begin{align*}&{\mathbb E} \left[\Psi(V_t) \|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-{X^1_t}}{\sqrt{h}\|A^1(X_t)\|}\right) \right]\leq\\&\|\Psi\|_{\infty} \|A\|_{\infty}\int_{{\mathbb R}^{d+1}}{\mathcal H}\left( \frac{m-x^1}{\sqrt{h}\|A^1(x^1,\tilde{x})\|}\right) p_V\big(m,x^1,\tilde{x};\,t\big) dm\, dx^1\,d\tilde{x}.\end{align*}

The change of variable $x^1=m-u\sqrt{h}$ yields

(19) \begin{align}&\frac{\sqrt{h}}{h}{\mathbb E} \left[ \Psi(V_t)\|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-{X^1_t}}{\sqrt{h}\|A^1(X_t)\|}\right) \right]\leq \\&\|\Psi\|_{\infty} \|A^1\|_{\infty}\int_{{\mathbb R}^d\times[0,+\infty[} {\mathcal H}\left( \frac{u}{\|A^1\big(m-\sqrt{h}u,\tilde{x}\big)\|}\right) p_V\big(m,m-\sqrt{h}u,\tilde{x};\,t\big) dm\, d\tilde{x}\,du.\nonumber\end{align}

Since ${{\mathcal{H}}}$ is decreasing (Remark 3.1) and $0\leq h\leq 1,$ we have

\begin{equation*}{{\mathcal{H}}}\left( \frac{u}{\|A^1\big(m-\sqrt{h}u,\tilde{x}\big)\|}\right)\leq {{\mathcal{H}}}\left(\frac{u}{\|A^1\|_\infty}\right),\end{equation*}

so

\begin{align*}&\left|\frac{\sqrt{h}}{h}{\mathbb E} \left[\Psi(V_t) \|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-{X^1_t}}{\sqrt{h}\|A^1(X_t)\|}\right) \right]\right|\leq\\&\|\Psi\|_{\infty} \|A^1\|_{\infty}\int_{{\mathbb R}^d\times[0,+\infty[}{{\mathcal{H}}}\left(\frac{u}{\|A^1\|_\infty}\right)\sup_{r>0}p_V(m,m-r,\tilde{x};\,t) dm\, d\tilde{x}\,du.\end{align*}

Applying Tonelli’s theorem, computing the integral with respect to du in the right-hand side with $\int_0^{\infty}{{\mathcal{H}}}(v)dv=1/4$ (Remark 3.1), we obtain

\begin{align*}&\sup_{h>0}\left|\frac{\sqrt{h}}{h}{\mathbb E} \left[\Psi(V_t) \|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-{X^1_t}}{\sqrt{h}\|A^1(X_t)\|}\right) \right]\right|\\& \quad \leq\frac{1}{4} \|\Psi\|_{\infty} \|A^1\|^2_{\infty}\int_{{\mathbb R}^{d}}\sup_{r>0}p_V(m,m-r,\tilde{x};\,t) dm\, d\tilde{x}.\end{align*}

Using Hypothesis 2.1(i), we obtain that the map

\begin{equation*}t\mapsto \sup_{h>0}\left|\frac{\sqrt{h}}{h}{\mathbb E} \left[\Psi(V_t) \|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-X^1_t}{\sqrt{h}\|A^1(X_t)\|}\right) \right]\right|\end{equation*}

belongs to $ L^1([0,T], \mathbb{R})$ . Part (i) of Proposition 3.2 is proved.

(ii) For the proof of Part (ii), first note that

\begin{align*}&{\mathbb E} \left[ \Psi(V_t)\|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-X^1_t}{\sqrt{h}\|A^1(X_t)\|}\right)\right]\\&\quad =\int_{{\mathbb R}^{d+1}} \Psi(m,x) \|A^1(x)\| {\mathcal H} \left( \frac{m-x^1}{\sqrt{h}\|A^1(x)\|}\right) p_V(m,x;\,t) dm \,dx.\end{align*}

After the change of variable $x^1=m-u\sqrt{h},$ we obtain

(20) \begin{align}&\frac{\sqrt{h}}{h}{\mathbb E} \left[ \Psi(V_t)\|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-{X^1_t}}{\sqrt{h}\|A^1(X_t)\|}\right) \right] =\\&\int_{{\mathbb R}^d\times\mathbb R^+} \Psi\big(m,m-u\sqrt{h},\tilde{x}\big)\|A^1\big(m-u\sqrt{h},\tilde{x}\big)\|{\mathcal H}\left( \frac{u}{\|A^1\big(m-\sqrt{h}u,\tilde{x}\big)\|}\right)\nonumber\\& p_V\big(m,m-\sqrt{h}u,\tilde{x};\,t\big) dm\, d\tilde{x}\,du.\nonumber\end{align}

Using Lebesgue’s dominated convergence theorem, we let h go to 0 in (20) for $t>0$ , and using the fact that $\Psi,$ A, and ${\mathcal H}$ are continuous and Hypothesis 2.1(ii), we obtain

\begin{align*}&\lim_{h \rightarrow 0}\frac{\sqrt{h}}{h}{\mathbb E} \left[ \Psi(V_t)\|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-X_t}{\sqrt{h}\|A^1(X_t)\|}\right) \right]=\\&\int_{{\mathbb R}^{d}\times[0,+\infty[}\Psi(m,{m,\tilde{x}})\|A^1(m,\tilde{x})\| {{\mathcal{H}}}\left( \frac{u}{\| A^1({m,\tilde{x}})\|}\right) {p_V(m,m,\tilde{x};\,t)}dm\,d\tilde{x} \,du.\end{align*}

Using the change of variable $z=\frac{u}{\|A^1(m,\tilde{x})\|}$ and Remark 3.1 $\int_0^{\infty}{{\mathcal{H}}}(z)dz=1/4$ yields

\begin{align*}&\lim_{h \rightarrow 0}\frac{\sqrt{h}}{h}{\mathbb E} \left[ \Psi(V_t)\|A^1(X_t)\| {\mathcal H} \left( \frac{M_t-X^1_t}{\sqrt{h}\|A^1(X_t)\|}\right) \right]\\&\quad =\frac{1}{4}\int_{{\mathbb R}^{d}}\Psi(m,{m,\tilde x})\|A^1(m,\tilde{x})\|^2 p_V(m,{m,}\tilde{x};\,t)dm\,d\tilde{x}.\end{align*}

3.4. End of proof of Theorem 2.3

We recall Theorem 8.2 from Brezis [Reference Brezis5, p. 204]: let $f\in W^{1,1}(0,T)$ ; then f is almost surely equal to an absolutely continuous function. As a particular case, any $f\in W^{1,1}(0,T)\cap C(0,T)$ is absolutely continuous. Recall $F_{\psi}\,:\,t \mapsto {{\mathbb E}\left[\int_0 ^t \Psi(V_s)dM_s \right]}.$

Proof. Let $0 \leq s\leq t.$ Since $\Psi$ is bounded and M is non-decreasing,

\begin{align*}\left| F_{\Psi}(t)-F_{\Psi}(s)\right|=\left|{\mathbb E}\left[ \int_s^t \Psi(V_u) dM_u\right]\right|\leq \|\Psi\|_{\infty} {\mathbb E} [M_t-M_s].\end{align*}

The map $t\mapsto {\mathbb E}[M_t]$ being continuous, $F_{\Psi}$ is a continuous function on $\mathbb{R}^+.$

Lemma 3.7. Assume that A and B fulfil (4) and (5), (M, X) fulfils Hypothesis 2.1, and $\Psi \in C^1_b.$ Then for all $T>0$ , the map $F_{\psi}$ belongs to the Sobolev space $W^{1,1}(]0,T[)$ , and its weak derivative is

\begin{align*}\dot{F}_{\Psi}(t)\,:\!=\,\frac{1}{2}\int_{{\mathbb R}^{d}}\Psi(m,\tilde{x})\|A^1(m,\tilde{x})\|^2 p_V(m,m,\tilde{x};\,t)dmd\tilde{x}.\end{align*}

Proof. Let $g\,:\, [0,T]\rightarrow \mathbb{R}$ be $C^1$ with compact support $[\alpha,\beta]\subset(0,T).$ This means both functions g and $\dot g$ are continuous, hence bounded, and that moreover $g(\alpha)=g(\beta)=0.$ Note that

\begin{equation*}\dot{g}(t)= \lim_{h \rightarrow 0} \frac{g(t) -g(t-h)}{h},\,\,\forall t \in (0,T).\end{equation*}

Moreover,

\begin{equation*}\sup_{t \in [0,T]}\sup_{h \in [0,1]}|\frac{g(t) -g(t-h)}{h}| \leq \|\dot{g}\|_{\infty}.\end{equation*}

Observe that, since M is non-decreasing and the coefficients A and B are bounded,

\begin{equation*}\left| F_{\psi}(t)\right|\leq \|\Psi\|_{\infty} {\mathbb E} [M_T] <\infty.\end{equation*}

Then, using Lebesgue’s dominated convergence theorem, we have

\begin{align*}\int_0^T \dot{g}(s) F_{\psi}(s)ds=\int_0^T\lim_{h \rightarrow 0} \frac{g(s) -g(s-h)}{h}F_{\psi}(s)ds=\lim_{h \rightarrow 0}\int_0^T \frac{g(s) -g(s-h)}{h}F_{\psi}(s)ds.\end{align*}

Using the change of variable $u=s-h$ in the last integral, we have

\begin{align*}\int_0^T &\frac{g(s) -g(s-h)}{h}F_{\Psi}(s)ds=h^{-1}\int_0^T g(s) F_{\Psi}(s) ds -h^{-1}\int_{-h}^{T-h} g(u) F_{\Psi}(u+h)du\\&= \int_{0}^T g(s) \frac{F_{\Psi}(s) -F_{\Psi}(s+h)}{h}ds{-}h^{-1}\int_{-h}^0 g(s)F_{\Psi}(s+h) ds {+}h^{-1}\int_{T-h}^Tg(s) F_{\Psi}(s+h) ds.\end{align*}

Recalling that $supp(g)=[\alpha,\beta]\subset(0,T)$ , $gF_{\Psi}$ extended by 0 on $[\alpha,\beta]^c$ is bounded on [0, T], so $\lim_{s\rightarrow 0} g(s)=\lim_{s\rightarrow T} g(s)=0$ . Then

\begin{equation*}h^{-1}\int_{-h}^0 g(s)F_{\Psi}(s+h) ds= h^{-1}\int_{T-h}^Tg(s) F_{\psi}(s+h) ds=0\end{equation*}

as soon as $0<h\leq T-\beta$ ; thus

\begin{equation*}\lim_{h \rightarrow 0} \left[h^{-1}\int_{-h}^0 g(s)F_{\Psi}(s+h) ds\right] = \lim_{h \rightarrow 0}\left[ h^{-1}\int_{T-h}^Tg(s) F_{\psi}(s+h) ds\right]=0.\end{equation*}

Applying Lebesgue’s dominated convergence theorem yields that F admits a weak derivative:

\begin{align*}\int_0^T \dot{g}(s) F_{\psi}(s)ds=-\int_0^T g(s) \dot{F}_{\Psi}(s) ds.\end{align*}

Using Proposition 3.1(ii), we have

\begin{equation*}\lim_{h\to 0^+}\left({-}\frac{F_{\Psi}(t) -F_{\Psi}(t+h)}{h}- \frac{2}{\sqrt h}{\mathbb E}\left[ \Psi(V_t)\|A^1(X_t)\|{\mathcal H}\left( \frac{M_t-X^1_t}{\sqrt{h}\|A^1(X_t)\|}\right)\right]\right)=0.\end{equation*}

Using Proposition 3.2(ii),

\begin{align*}-\dot{F}_{\Psi}{(t{+})}\,:\!=\,\lim_{h \rightarrow 0,h>0}\frac{F_{\Psi}(t) -F_{\Psi}(t+h)}{h}=-\frac{1}{2}\!\int_{{\mathbb R}^{d}}\!\!\Psi(m,m,\tilde{x})\|A^1(m,\tilde{x})\|^2 p_V(m,m,\tilde{x};\,t)dmd\tilde{x},\end{align*}

and by Propositions 3.1(i) and 3.2(i),

\begin{align*}\sup_{h>0} \left| \frac{F_{\Psi}(t) -F_{\Psi}(t+h)}{h}\right| \in L^1([0,T],dt),\end{align*}

so $\dot{F}_{\Psi}\in L^1([0,T],\mathbb{R}).$ By [Reference Brezis5, Chapter 8, Section 2, p. 202], $F_{\Psi}$ belongs to $W^{1,1}(]0,T[,{\mathbb R}).$

We now finish the proof of Theorem 2.3. According to [Reference Brezis5, Theorem 8.2, p. 204], $F_{\psi}$ is equal almost surely to an absolutely continuous function. Since $F_{\Psi}$ is continuous (Lemma 3.6), the equality holds everywhere. Thus $F_{\Psi}$ is an absolutely continuous function, and its derivative is its right derivative.

4.1. Integral representation of the density: proof of Proposition 4.2

Let $t >0$ be fixed. First, we assume that $\mu_0=\delta_{x_0}$ , $x_0$ being fixed in ${\mathbb R}^d.$ By Lemma 4.2 below, and using the fact that B is bounded, for all $t\in [0,T]$ , the functions $ p^{k,\gamma}\in^\infty L\left([0,T], L^1\big(\mathbb{R}^{d+1}\big)\right)$ for $\gamma=\alpha,\beta$ .

Let $F\in C^1_b\big({\mathbb R}^{d+1},{\mathbb R}\big)$ with compact support. We will prove that

(22) \begin{equation}\mathbb{E}_\mathbb{P}[F\big(M_t,X_t\big)]=\int_{\mathbb{R}^{d+1}}F(m,x)\left(p^0-\sum_{k=m,x^1,\ldots,x^d}\big(p^{k,\alpha}+p^{k,\beta}\big)(m,x,t)\right)dmdx.\end{equation}

Using Malliavin calculus we obtain the following decomposition.

Lemma 4.1. We have

\begin{align*}{\mathbb E}_\mathbb{P} &\left[F\big(M_t,X_t\big)\right]= \int_{{\mathbb R}^{d+1}} F\big(x_0^1+b,x_0+a\big)p_{W^{*1},W}(b,a;\,t)dbda\nonumber\\&+ \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}\partial_mF \left(X^1_s + b,X_s+ a\right){\mathbf 1}_{\big\{M_s <X^1_s + b\big\}} B^1(X_s)p_{W^{*1},W}(b,a;\,t-s)dbda \right]ds\nonumber\\&+ \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{\partial_{k}} F\!\left(\max\! \left( M_s, X_s^1 + b\right),X_s+ a\right) B^k(X_s)p_{W^{*1},W}(b,a;\,t-s)dbda \right]ds.\end{align*}

Proof. Let Z be the exponential martingale solution of

(23) \begin{align}Z_t& =1+\int_0^t Z_s B^k\big(x_0+W_s\big)dW^k_s.\end{align}

As before, Einstein’s convention is used. Let ${\mathbb Q}=Z{\mathbb P}$ ; according to Girsanov’s theorem, using (23), $W_B\,:\!=\,\left(W_t^k-\int_0^t B^k\big(x_0+W_s\big) ds;\, k=1,\ldots,d\right)_{t\geq 0}$ is a ${\mathbb Q}$ -continuous martingale such that $\langle W_B\rangle_t=t$ for all $t \geq 0.$ That means that under ${\mathbb Q},$ $W_B$ is a d-dimensional Brownian motion. Then the distribution of X (resp. (M, X)) under ${\mathbb P}$ is the distribution of $W+x_0$ (resp. $(W^{1*}+x_0,W+x_0))$ under ${\mathbb Q}$ , and

(24) \begin{align}&{\mathbb E}_\mathbb{P}\left [ F\big(M_t,X_t\big)\right]={\mathbb E}_\mathbb{Q} \left [ F\big(x_0^1+W^{1*}_t,x_0+W_t\big)\right]= {\mathbb E}_\mathbb{P} \left [ F\big(x_0^1+W^{1*}_t,x_0+W_t\big)Z_t\right].\end{align}

Let $G\,:\!=\, F\big(x_0^1+W^{1*}_t,x_0+W_t\big)$ and $u\,:\!=\,ZB(x_0+W)$ ; using (23),

(25) \begin{align}\mathbb{E}_\mathbb{P}[F\big(M_t,X_t\big)]={\mathbb E}_\mathbb{P} \left [ F\big(x_0^1+W^{1*}_t,x_0+W_t\big)\right]+{\mathbb E}_\mathbb{P} \left [G\delta(u) \right].\end{align}

As a first step we will apply (50) (see the appendix) to the second term in (25). Thus we have to check that the pair $(G,u)\in {\mathbb D}^{1,2}\times {\mathbb L}^{1,2}$ . Since F is bounded and smooth, we have $G\in {\mathbb D}^{1,2}$ ; and according to Lemma A.1, the process u belongs to ${\mathbb L}^{1,2}$ .

Using (53) $\big({\tau\,:\!=\,\inf\big\{s, W_s^{1*} =W^{1*}_t\big\}}\big)$ , the pair $\big(W^{1*}_t,W_t\big)$ belongs to ${\mathbb D}^{1,2}$ with Malliavin gradient

\begin{align*}&D_sW^{1*}_t=\left({\mathbf 1}_{[0,\tau]}(s),0,\ldots,0\right),\,D_sW_t^k=\big(\delta_{j=k},\,\,j=1,\ldots,d\big){\mathbf 1}_{[0,t]}(s),\,\,k=1,\ldots,d.\end{align*}

Using the chain rule,

\begin{align*} \langle DG,u\rangle_{{\mathbb H}}&= \int_0^{t}\partial_m F\big({x^1_0}+W^{*1}_t,x_0+W_t\big) {\mathbf 1}_{\big\{W^{1*}_s<W^{1*}_t\big\}}B^1\big(x_0+W_s\big){Z_s} ds \\ &+ \int_0^{t}{\partial_{k}} F\big({x^1_0}+W^{*1}_t,x_0+W_t\big)B^k\big(x_0+W_s\big){Z_s}ds.\end{align*}

We are now in a position to apply (50), $E_\mathbb{P}[G\delta(u)]=E_\mathbb{P}[\langle DG,u\rangle_{\mathbb{H}}]$ :

(26) \begin{eqnarray}& & \mathbb{E}_\mathbb{P}[G\delta(u)]=\mathbb{E}_\mathbb{P}\left[\int_0^t \partial_m F \big(x_0^1+W^{1*}_t,x_0+W_t\big){\mathbf 1}_{\{W^{1*}_s <W^{1*}_t \}} B^1\big(x_0+W_s\big)Z_sds\right]\nonumber \\ & +\, &\mathbb{E}_\mathbb{P}\left[\int_0^t {\partial_{k}} F \big(x_0^1+ W^{1*}_t,x_0+W_t\big)B^k\big(x_0+W_s\big) Z_s ds\right].\end{eqnarray}

Plugging the identity (26) into the right-hand side of (25) and using Fubini’s theorem to commute the integrals in ds and $d\mathbb{P}$ , we obtain

(27) \begin{eqnarray}&&{\mathbb E}_\mathbb{P} \left [F\big(M_t,X_t\big)\right]={\mathbb E}_\mathbb{P} \left [ F\big(x_0^1+W^{1*}_t,x_0+W_t\big)\right]\\&+& \int_0^t {\mathbb E}_\mathbb{P}\left[\partial_m F \big(x_0^1+W^{1*}_t,x_0+W_t\big){\mathbf 1}_{\big\{W^{1*}_s <W^{1*}_t \big\}} Z_sB^1\big(x_0+W_s\big) \right]ds\nonumber \nonumber\\&+& \int_0^t {\mathbb E}_\mathbb{P}\left[ \partial_k F \big(x_0^1+W^{1*}_t,x_0+W_t\big)Z_sB^k\big(x_0+W_s\big) \right]ds.\nonumber\end{eqnarray}

As a second step, we use the independence of the increments of the Brownian motion in order to obtain the density of $\big(W^{1*}_{t-s},W_{t-s}\big)$ . Recall (9): $\hat W_{t-s}\,:\!=\,W_t-W_s$ and $\big(\hat W^1\big)^*_{t-s}= \max_{s \leq u \leq t} \left(W^1_u-W^1_s\right).$ Then

\begin{equation*}W^{1*}_t=\max\! \left( W^{1*}_s,W^1_s + \max_{s \leq u \leq t} \left(W^1_u-W^1_s\right)\right) =\max\! \left( W^{1*}_s, W^1_s +\big(\hat W^1\big)_{t-s}^*\right),\end{equation*}

so the expression (27) becomes

\begin{align*}&{\mathbb E}_\mathbb{P} \left [ F\big(M_t,X_t\big)\right]= {\mathbb E}_\mathbb{P} \left [F\big(x_0^1+W^{1*}_tx_0\,+,W_t\big)\right]+ \\& \int_0^t {\mathbb E}_\mathbb{P}\left[\partial_m F \left(x_0^1+ W^1_s +\big(\hat W^1\big)_{t-s}^*,x_0+W_s+ \hat W_{t-s}\right){\mathbf 1}_{\big\{W^{1*}_s <W^1_s + \big(\hat W^1\big)_{t-s}^*\big\}} Z_sB^1\big(x_0+W_s\big) \right]ds\nonumber\\&+ \int_0^t {\mathbb E}_\mathbb{P}\left[{\partial_{k}} F\!\left(\max\! \left( x_0^1+W^{1*}_s, x^1_0+W^1_s +\big(\hat W^1\big)_{t-s}^* \right),x_0+W_s+\hat W_{t-s}\right) Z_sB^k\big(x_0+W_s\big) \right]ds.\end{align*}

The random vector $\left(\big(\hat W^1\big)_{t-s}^*,\hat W_{t-s}\right)$ is independent of the $\sigma$ -field ${\mathcal{F}}_s$ and has the same distribution as the pair $\big(W^{1*}_{t-s},W_{t-s}\big).$ Let $p_{W^{*1},W}(.,.;\,t-s)$ be the density of its law, and express the expectation with this density:

\begin{align*}{\mathbb E}_\mathbb{P} \left [ F\big(M_t,X_t\big)\right] & = \int_{{\mathbb R}^{d+1}} F\big(x_0^1+b,x_0+a\big)p_{W^{*1},W}(b,a;\,t)dbda\\& \quad + \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}\partial_m F \big(x_0^1+ W^1_s + b,x_0\right.\\& \left. \qquad +\,W_s+ a\big){\mathbf 1}_{\big\{W^{1*}_s <W^1_s + b\big\}} Z_sB^1\big(x_0+W_s\big)p_{W^{*1},W}(b,a;\,t-s)dbda \right]ds\\& \quad + \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{\partial_{k}} F\big(x_0^1+\max\! \left( W^{1*}_s, W^1_s + b\right),x_0\right.\\& \left. \qquad +\,W_s+ a\big) Z_sB^k\big(x_0+W_s\big)p_{W^{*1},W}(b,a;\,t-s)dbda \right]ds. \end{align*}

Using Girsanov’s Theorem for $Z.\mathbb{P}=\mathbb{Q},$ since the law of (M, X) under $\mathbb{P}$ is the law of $\big(x^1_0+W^{1*},x_0+W\big)$ , under $\mathbb{Q}$ , using the equality (24), we have:

\begin{align*}{\mathbb E}_\mathbb{P} \left[F\big(M_t,X_t\big)\right]&= \int_{{\mathbb R}^{d+1}} F\big(x_0^1+b,x_0+a\big)p_{W^{*1},W}(b,a;\,t)dbda\nonumber\\&+ \int_0^t {\mathbb E}_\mathbb{P} \left[\int_{{\mathbb R}^{d+1}}\partial_mF \left(X^1_s + b,X_s+ a\right){\mathbf 1}_{\big\{M_s <X^1_s + b\big\}} B^1(X_s)p_{W^{*1},W}(b,a;\,t-s)dbda \right]ds\nonumber\\&+ \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{\partial_{k}} F\!\left(\max\! \left( M_s, X_s^1 + b\right),X_s+ a\right) B^k(X_s)p_{W^{*1},W}(b,a;\,t-s)dbda \right]ds.\end{align*}

We are now in a position to complete the proof of Proposition 4.2. Using some suitable translations of the variables (a, b), we have ${\mathbb E}_\mathbb{P} \left [F\big(M_t,X_t\big)\right]= {\sum_{k=0}^{d} I_k}+I_m,$ where

(28) \begin{align}&I_0=\int_{{\mathbb R}^{d+1}} F(b,a)p_{W^{*1},W}\big(b-x_0^1,a-x_0;\,t\big)dbda,\\&I_m=\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}\partial_m F(b,a){\mathbf 1}_{\{M_s < b\}} B^1(X_s)p_{W^{*1},W}\left(b-X_s^1,a-X_s;\,t-s\right)dbda \right]ds, \nonumber\end{align}

and for $k=1,\ldots,d,$

\begin{align*}I_k&= \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{{\partial_{k}}}F\!\left(\max\! \left( M_s, b\right),a\right) B^{k}(X_s)p_{W^{*1},W}\big(b-X_s^1,a-X_s;\,t-s\big)dbda \right]ds.\end{align*}

Since B, F, and its derivatives are bounded, all these integrals are finite. Using (54) in the appendix, the function $p_{W^{*1},W}(.,.;\,t)$ is $C^{\infty}$ on $\bar{\Delta}=\big\{(b,a),\,\,b{\geq} a^1_+, \,\,(a,b) \in {\mathbb R}^{d+1}\big\}$ . The aim is now to identify the terms $p^0$ , $p^{k,\alpha}$ , $p^{k,\beta}$ , ${\mathbf k=m,1,\ldots, d}$ , defined in Proposition 4.2.

Step 1. First we identify $p^0(b,a;\,t) $ as the factor of F(b, a) in the integrand of $I_0:$

\begin{equation*}p^0(b,a;\,t)= p_{W^{*1},W}\big(b-x_0^1,a-x_0;\,t\big).\end{equation*}

Step 2. We now deal with $I_k$ , $k=2,\ldots,d.$ Integrating by parts with respect to $a^k$ between $-\infty $ and $\infty$ in $I_k$ for $k=2,\ldots,d$ yields

\begin{align*}I_k&=-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}} F\!\left(\max\! \left( M_s,b\right),a\right) B^{k}(X_s)\partial_{k}p_{W^{*1},W}\big(b-{X^1_s},a-X_s;\,t-s\big)dbda \right]ds\nonumber \\ &=-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{{\mathbf 1}}_{\{b> M_s\}} F\!\left(b,a\right) B^{k}(X_s)\partial_{k}p_{W^{*1},W}\big(b-{X^1_s},a-X_s;\,t-s\big)dbda \right]ds \\ &-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}} {{\mathbf 1}}_{\{b< M_s\}}F\!\left( M_s,a\right) B^{k}(X_s)\partial_{k}p_{W^{*1},W}\big(b-{X^1_s},a-X_s;\,t-s\big)dbda \right]ds. \nonumber\end{align*}

We identify $-p^{k,\alpha}(b,a,t)$ inside the integral on the set $\{b>M_s\}.$ For the integral on the set $\{b<M_s\}$ , we introduce the density of $(M_s,X_s)$ and identify $-p^{k,\beta}(m,a;\,t)$ as the factor of F(m, a).

Step 3. Finally, we identify the terms $p^{m,\gamma}$ and $p^{1,\gamma}$ , $\gamma=\alpha, \beta$ , which come from the sum of $I_m$ and $I_1$ . Note that $p_{W^{*1},W}\!\left(b-X_s^1,a-X_s;\,t-s\right)=0$ on the set ${\big\{b<a^1\big\}}.$ Integrating by parts with respect to b between $\max\left({a^1},M_s\right) $ and $\infty$ in $I_m$ yields

(29) \begin{align}&I_m=-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^d}F\big( \max\!\big({a^1}, M_s\big),a\big)B^1(X_s)p_{W^{*1},W}\big(\max\!\big({a^1},M_s\big)-X_s^1,a-X_s;\,t-s\big)da\right]ds\nonumber\\[4pt]&-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{\mathbf 1}_{M_s<b}F\!\left( b,a\right)B^1(X_s)\partial_{m}p_{W^{*1},W}\left(b-X_s^1,a-X_s;\,t-s\right)dbda\right]ds.\end{align}

Integrating by parts with respect to $a^1$ between $-\infty $ and b in $I_1$ yields

(30) \begin{align}I_1&= \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d}}F\!\left( \max\left( M_s,b\right),b,\tilde a\right) B^1(X_s)p_{W^{*1},W}\big(b- X^1_s,b- X^1_s,\tilde a-\tilde X_s;\,t-s\big)dbd\tilde a \right]ds \nonumber \\&-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}} F\!\left(\max\! \left( M_s,b\right),a\right) B^1(X_s)\partial_{1}p_{W^{*1},W}\big(b- X^1_s,a-X_s;\,t-s\big)dbda \right]ds.\nonumber\\\end{align}

We then have the following:

1. (i) The term $p^{m,\beta}{(b,a,t)}$ comes from the second term in $I_m$ (Equation (29)) as the factor of F(b, a):

   \begin{equation*}-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}{\mathbf 1}_{M_s<b}F\!\left( b,a\right)B^1(X_s){\partial_m}p_{W^{*1},W}\big(b-X_s^1,a-X_s;\,t-s\big)dbda\right]ds\end{equation*}

2. (ii) The terms $-p^{1,\alpha}(b,a,t)$ and $-p^{1,\beta}(b,a;\,t)$ come from the second term in $I_1$ (Equation (30)):

   \begin{equation*}-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}} F\!\left(\max\! \left( b,M_s\right),a\right) B^1(X_s)\partial_{1}p_{W^{*1},W}\big(b-{X^1_s},a-X_s;\,t-s\big)dbda \right]ds.\end{equation*}
   Inside the integral on the set $\{M_s<b\}$ we identify $-p^{1,\alpha}(b,a,t)$ as the factor of $F(b,a)$ , and inside the integral on the set $\{M_s>b\}$ we identify $-p^{1,\beta}(b,a;t)$ as the factor of $F(M_s,a)$ .

3. (iii) The term $-p^{m,\alpha}(b,a,t)$ comes from the sum of first terms in $I_1$ (Equation (30)) and $I_m$ (Equation (29)).

Now we replace the variable b by $a_1$ and replace $dbd\tilde a$ by da in the first terms of $I_m$ and $I_1$ :

\begin{align*}I^1_m&=-\!\int_0^t {\mathbb E}_\mathbb{P}\!\left[\int_{{\mathbb R}^d}\!F\big( \max\!\big({a^1}, M_s\big),a\big)B^1(X_s)p_{W^{*1},W}\big(\max\!\big({a^1},M_s\big)-X_s^1,a-X_s;\,t-s\big)da\right]ds,\\I^1_1&= \int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d}}F\big( \max\!\big( M_s,a^1\big),a\big) B^1(X_s)p_{W^{*1},W}\big(a^1-{X^1_s},a-X_s;\,t-s\big)d a \right]ds.\end{align*}

Note that

\begin{align*}&-p_{W^{*1},W}\big(\max\!\big(a^1, M_s\big)-X_s^1,a-X_s;\,t-s\big)+p_{W^{*1},W}\big(a^1-X^1_s,a-X_s;\,t-s\big)\\&=\left[-p_{W^{*1},W}\big(M_s-X_s^1,a-X_s;\,t-s\big)+p_{W^{*1},W}\big(a^1-X_s^1,a-X_s;\,t-s\big)\right] {\mathbf 1}_{M_s >a^1}\\&=-\int_{a^1}^{M_s}{\partial_m}p_{W^{*1},W}\big(b-X^1_s, a-X_s,t-s\big){db} {\mathbf 1}_{M_s >a^1}.\end{align*}

Then the sum of $I_m^1$ and $I_1^1$ is

\begin{equation*}-\int_0^t {\mathbb E}_\mathbb{P}\left[\int_{{\mathbb R}^{d+1}}F(M_s,a)B^1(X_s){\partial_m}p_{W^{*1},W}\big(b-X_s^1,a-X_s;\,t-s\big) {\mathbf 1}_{M_s>b >a^1}dadb\right]ds.\end{equation*}

We introduce the density of the law of the pair $(M_s,X_s)$ and we identify $-p^{m,\alpha}(m,a;\,t)$ as the factor of $F(m,a).$

These three steps complete the proof of Proposition 4.2 when $\mu_0=\delta_{x_0}.$

Finally, when $\mu_0$ is the law of $X_0,$ we have $ p_V(m,w;\,t,\mu_0)=\int_{{\mathbb R}^d} p_V(m,x;\,t,\delta_{x_0}) \mu_0(dx_0).$ Then we can complete the proof of Proposition 4.2 for any initial law $\mu_0$ by integrating with respect to $\mu_0$ the expression obtained in (21) for $p_V\big(m,x;\,t,\delta_{x_0}\big)$ .

4.2. Proof of Proposition 4.1

Using some ideas used in [Reference Garroni and Menaldi14, Section V.3.2], let us introduce the following linear maps on $L^{\infty}\big([0, T],dt, L^1\big({\mathbb R}^{d+1},dmdx\big)\big),$ for $k=m,1,\ldots,d$ :

(31) \begin{align}&{{\mathcal{I}}^{k,\alpha}}[p](m,x;\,t)\,:\!=\,\int_0^t{\int_{\mathbb{R}^{d+1}} {\mathbf 1}_{b <m}B^{k}(a)\partial_{k}p_{W^{*1},W}\big(m-a^1,x-a;\,t-s\big)}p(b,a;\,s)dbdads,\\&{{\mathcal{I}}^{k,\beta}}[p](m,x;\,t)\,:\!=\,\int_0^t \int_{{\mathbb R}^{d+1}}{\mathbf 1}_{b<m} B^{k}(a)\partial_{k}p_{W^{*1},W}\big(b-a^1,x-a;\,t-s\big)p(m,a;\,s)dbda ds.\nonumber\end{align}

Let us introduce the following inductively defined functions:

(32) \begin{align}p_0(m,x;\,t,\mu_0)= \int_{\mathbb{R}^d}p_{W^{1*},W}\big(m-x_0^1,x-x_0;\,t\big)\mu_0(dx_0),\,\,p_n=-\sum_{k=m,1,\ldots,d}\left( {p^{k,\alpha}_n} +{p^{k,\beta}_n}\right),\end{align}

and for $k=m,1,\ldots,d$ , $j=\alpha,\beta$ , and $n \geq 1$ ,

\begin{equation*}{p_{n+1}^{k,j}}(m,x;\,t)\,:\!=\,{\mathcal I}^{k,j}[p_n](m,x;\,t).\end{equation*}

Let us define the operator

(33) \begin{equation}{\mathcal{I}}\,:\!=\,-\sum_{j=\alpha,\beta;\,k=m,1,\ldots,d}{\mathcal{I}}^{k,j}.\end{equation}

Moreover, one can observe that this means $p_{n+1}={\mathcal{I}}(p_n)$ , and Proposition 4.2 leads to $p_V=p_0+{\mathcal{I}}(p_V).$ Let

(34) \begin{align}{P_n} \,:\!=\, \sum_{k=0}^n p_k,\,\,n\geq 0.\end{align}

The proof is a consequence of the following two lemmas.

Lemma 4.2. Let $j=\alpha,\beta$ , $k=m,1,\ldots,d$ , and $T>0$ . The linear maps ${\mathcal I}^{k,j}$ are continuous on $L^{\infty}\big([0, T],dt, L^1\big({\mathbb R}^{d+1},dmdx\big)\big)$ : there exists a constant C such that for all $p \in L^{\infty}\big([0, T],dt, L^1\big({\mathbb R}^{d+1},dmdx\big)\big)$ ,

(35) \begin{align}\sup_{s \in [0,t]} \|{\mathcal I}^{k,j}[p](.,.;\,s)\|_{L^1\big({\mathbb R}^{d+1},dmdx\big)}\leq C \int_0^t \frac{1}{\sqrt{t-s}}\sup_{u \in [0,s]} \|p(.,.;\,u)\|_{L^1\big({\mathbb R}^{d+1},dmdx\big)}ds.\end{align}

As a consequence,

(36) \begin{equation}\sup_{s \in [0,t]} \|{\mathcal{I}}[p](.,.;\,s)\|_{L^1\big({\mathbb R}^{d+1},dmdx\big)}\leq 2(d+1)C \int_0^t \frac{1}{\sqrt{t-s}}\sup_{u \in [0,s]} \|p(.,.;\,u)\|_{L^1\big({\mathbb R}^{d+1},dmdx\big)}ds.\end{equation}

Proof. Let $T>0,$ $p \in L^{\infty}\big([0, T] \times L^1\big({\mathbb R}^{d+1},dmdx\big)\big)$ , and $t \in [0,T]$ , and let $\phi_{d+1}$ be the Gaussian law density restricted to the subset $\big\{b>a^1_+\big\}$ (up to a constant):

(37) \begin{align} \phi_{d+1}\big(b,b-a^1,\tilde a;\,2t\big) \,:\!=\,\frac{1}{{\sqrt{2 \pi t}^{d+1}}}{\mathbf 1}_{b>a^1_+}e^{- \frac{b^2 + \big(b-a^1\big)^2 + \|\tilde{a}\|^2}{4t}}. \end{align}

(i) Let $j=\alpha$ and $k=m,1,\ldots,d$ ; according to the definition of ${\mathcal I}^{k,\alpha}$ and the boundedness of B,

\begin{align*} \left| {\mathcal I}^{k,\alpha}[p](m,x;\,t)\right| \leq \|B\|_{\infty} \int_0^t \int_{{\mathbb R}^{d+1}}{\mathbf 1}_{b <m}| \partial_{k}p_{W^{*1},W}\big(m-a^1,x-a;\,t-s\big) p(b,a;\,s)|dbdads. \end{align*}

Using Lemma A.2, there exists a constant D such that for $k=m,1,\ldots,d$ ,

(38) \begin{align} |\partial_{k}p_{W^{*1},W}\left(b,a;\,t\right)| \leq \frac{D}{\sqrt t}\phi_{d+1}\big(b,b-a^1,\tilde a;\,2t\big). \end{align}

So

\begin{align*} &\left| {\mathcal I}^{k,\alpha}[p](m,x;\,t)\right| \\ & \quad \leq \|B\|_{\infty} \int_0^t \int_{{\mathbb R}^{d+1}} \frac{D}{\sqrt{t-s}}\phi_{d+1}\big(m-a^1,m-x^1,\tilde x-\tilde a;\,t-s\big)|p(b,a;\,s)|dbdads. \end{align*}

We perform an integration with respect to (m, x) using Tonelli’s theorem and omitting the indicator functions. Since $\phi_{d+1}$ is the density of a Gaussian law, we get the bound

\begin{align*} \left\| {\mathcal I}^{k,\alpha}[p](.,.;\,t)\right\|_{L^1\big({\mathbb R}^{d+1},dmdx\big)} \leq\, & D\|B\|_{\infty} \int_0^t \int_{{\mathbb R}^{d+1}}\frac{1}{\sqrt{t-s}}| p(b,a;\,s)|{db}dads \\ \leq\, & 2^{(d+1)/2}D\|B\|_{\infty} \int_0^t \frac{1}{\sqrt{t-s}}\sup_{u\leq s}\| p(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dbda\big)}ds, \end{align*}

which implies the inequality (35) when $j=\alpha.$

(ii) Let $j=\beta$ and $k=m,1,\ldots,d.$ According to the definition of ${\mathcal I}^{k,\beta}$ and the boundedness of B,

\begin{align*} \left| {\mathcal I}^{k,\beta}[p](m,x;\,t)\right| \leq \|B\|_{\infty} \int_0^t \int_{{\mathbb R}^{d+1}}{\mathbf 1}_{b <m}| \partial_{k}p_{W^{*1},W}\big(b-a^1,x-a;\,t-s\big)p(m,a;\,s)|dbdads. \end{align*}

Using (38) yields

\begin{equation*} \left| {\mathcal I}^{k,\beta}[p](m,x;\,t)\right| \leq \|B\|_{\infty} \int_0^t \int_{{\mathbb R}^{d+1}}\frac{D}{\sqrt{t-s}}\phi_{d+1}\big(b-a^1,b-x^1,\tilde x-\tilde a;\,2(t-s)\big)|p(m,a;\,s)|dbdads.\end{equation*}

We integrate with respect to x, then with respect to b, using Tonelli’s theorem and omitting the indicator functions, and using the fact that $\phi$ is the density of a Gaussian law. So the bound with respect to a multiplicative constant is

\begin{align*} \left\| {\mathcal I}^{k,\beta}[p](.,.;\,t)\right\|_{L^1\big({\mathbb R}^{d+1},dmdx\big)} \leq\, & D \|B\|_{\infty}2^{(d+1)/2} \int_0^t \int_{{\mathbb R}^{d+1}}\frac{1}{\sqrt{t-s}}| p(m,a;\,s)|dmdads\\ \leq\, & D \|B\|_{\infty}2^{(d+1)/2} \int_0^t \frac{1}{\sqrt{t-s}}\sup_{u\leq s}\| p(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmda\big)}ds, \end{align*}

which implies the inequality (35) for $j=\beta.$

Finally, the estimate (36) is obtained by adding the estimates (35) for $j=\alpha, \beta$ and $k=m, 1,\ldots,d.$

The following lemma is a consequence of (36) in Lemma 4.2.

Lemma 4.3. For all n,

(39) \begin{align} \sup_{u \leq t} \|p_n(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq (2(d+1)C)^{n} t^{n/2} \frac{\Gamma(1/2)^n}{\Gamma( 1+n/2)},\end{align}

(40) \begin{align}\sup_{u \leq t}\|(p_V -P_{n})(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq (2(d+1)C)^{n+1} t^{(n+1)/2} \frac{\Gamma(1/2)^{n+1}}{{\Gamma( (n+3)/2)}}.\end{align}

Proof. (i) For all $t>0,$ $p_0(.;\,t)$ is a probability density function, so (39) is satisfied for $n=0.$ We now assume that (39) is satisfied for n. Using $p_{n+1}={\mathcal{I}} [p_n]$ , (36), and the inductive assumption,

\begin{align*} \sup_{u \leq t} \|p_{n+1}(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq (2(d+1) C)^{n+1} \frac{\Gamma(1/2)^n}{\Gamma( 1+n/2)}\int_0^t \frac{\sqrt{s^n}}{\sqrt{t-s}}ds. \end{align*}

We perform the change of variable $s=tu$ and use

\begin{equation*}\int_0^1 u^{a-1}(1-u)^{b-1}du= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\end{equation*}

to obtain

\begin{align*} \sup_{u \leq t}\|p_{n+1}(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq (2(d+1) C)^{n+1} t^{(n+1)/2} \frac{\Gamma(1/2)^n}{\Gamma( 1+n/2)} \frac{\Gamma(1/2) \Gamma( 1+ n/2)}{\Gamma((n+3)/2)}, \end{align*}

which proves (39) for all n.

(ii) Noting that $P_0=p_0$ and $p_V-p_0= {\mathcal{I}}[p_V]$ and applying (36) to $p_V$ , we obtain

\begin{align*} \sup_{u \leq t}\|(p_V -P_{0})(.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq 2(d+1)C t^{1/2}. \end{align*}

But $\Gamma(1/2) / \Gamma(3/2)=2$ , so (40) is satisfied for $n=0.$

We now suppose that (40) is satisfied for n. Using

\begin{equation*}p_V-P_{n+1}=p_0+{\mathcal{I}}(p_V)- (p_0+{\mathcal{I}}(P_n))={\mathcal{I}}(p_V-P_n),\end{equation*}

the bound (36), and the induction assumption, we have

\begin{align*} \sup_{u \leq t}\|[p_{V}- P_{n+1}](.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq 2(d+1)C \int_0^t (2(d+1) C)^{n+1} \frac{\Gamma(1/2)^{n+1}}{\Gamma( (3+n)/2)} \frac{\sqrt{s^{n+1}}}{\sqrt{t-s}}ds. \end{align*}

We now perform the change of variable $s=tu$ and

\begin{equation*}\int_0^1 u^{a-1}(1-u)^{b-1}du= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\end{equation*}

with $a=(n+3)/2$ , $b={\frac{1}{2}}$ , to obtain

\begin{align*} \sup_{u \leq t}\|[p_V-P_{n+1}](.,.;\,u)\|_{L^1\big(\mathbb{R}^{d+1},dmdx\big)} \leq (2(d+1) C)^{n+2} t^{(n+2)/2} \frac{\Gamma(1/2)^{n+2}}{{\Gamma( (4+n)/2)}}, \end{align*}

which proves (40) for $n+1$ and thus for all n.

The series $\sum_n \frac{x^n}{\Gamma(n/2 +1)}$ is convergent for any x, so Proposition 4.3 is a consequence of Lemmas 4.2 and 4.3.

4.2.1. Upper bound of $p_V$ (i.e., Hypothesis 2.1(i)).

For all $T>0,$ $x_0 \in \mathbb{R}^d,$ and $p \in L^{\infty}([0,T], L^{1}\big(\mathbb{R}^{d+1},dmdx\big))$ whose support is contained in $\big\{(m,x),\ m>x_0^1, m>x^1\big\}$ , let us define

(41) \begin{align} N(p;\,t,x_0) \,:\!=\,\sup_{(m,x)\in \mathbb{R}^{d+1},\,\,m>x^1, m>x_0^1}\frac{|p(m,x;\,t)|}{\phi_{d+1}\big(m-x_0^1, m-x^1,\tilde{x}-\tilde{x}_0;\,2t\big)}. \end{align}

Proposition 4.4. For every $T>0$ there exists a constant $C_T$ , and for all n there exist constants $C_n= \frac{\left[\|B\|_\infty D{(2(d+1)) 2^{d/2}}{\Gamma(1/2)}\right]^n}{\Gamma(1+n/2)}$ , such that for all $x_0 \in \mathbb{R}^d$ , ${0<t\leq T},$ the following hold:

1. (i) $\left| p_n\big(m,x;\,t,x_0\big) \right| \leq C_nt^{n/2}\phi_{d+1}\big( m-x^1_0,m-x^1,\tilde{x}-\tilde{x}_0, 2t\big){\mathbf 1}_{m>\max\!\big(x^1,x^1_0\big)}$ ;

2. (ii) $\left|p_V\big(m,x;\,t,x_0\big) \right| \leq C_T\phi_{d+1}\big( m-x^1_0,m-x^1,\tilde{x}-\tilde{x}_0, 2t\big){\mathbf 1}_{m>\max\!\big(x^1,x^1_0\big)}$ ;

3. (iii) for every initial probability measure $\mu_0$ on ${\mathbb R}^d$ ,

   \begin{equation*}\sup_{u >0} p_V(m,m-u,\tilde{x},t;\, \mu_0) \in L^1([0,T] \times \mathbb{R}^d, dtdmd\tilde{x}).\end{equation*}

Note that Part (iii) is actually Hypothesis 2.1(i).

Proof. Part (ii) is a consequence of Part (i), since $p_V=\sum_{n=0}^{\infty}p_n,$ and the series $\sum_n \frac{1}{{\Gamma (1+n/2)}}x^n$ has an infinite radius of convergence (Proposition 4.3).

We prove Part (i) by induction on n, using Part (ii) of Lemma A.2:

\begin{align*} p_0\big(m,x;\,t,x_0\big) &\leq\frac{e^{-\frac{\big(m-x^1\big)^2}{4t}-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t} {- \frac{\big(m-x_0^1\big)^2}{4t}}}}{\sqrt{(2 \pi )^{d+1}{t^{d+1}}}}{\mathbf 1}_{m>\max\!\big(x^1,x^1_0\big)}\\&= \phi_{d+1}(m-x^1,m-x_0^1,\tilde x-\tilde x_0;\,2t){\mathbf 1}_{m>\max\!\big(x^1,x^1_0\big)},\end{align*}

so $N(p_0;\,t,x_0) \leq 1,$ which is Part (i) for $n=0,$ $C_0=1$ .

We assume Part (i) is true for $p_n,$ meaning $N(p_n;\,t,x_0) \leq C_nt^{n/2}.$ By definition, $p_{n+1}= {\mathcal I}[p_n]$ ; Lemma 4.4, proved below, yields

\begin{align*}N(p_{n+1};\,t,x_0)=N({\mathcal{I}}[p_n];\,t,x_0)&\leq 2(d+1)2^{d/2}\|B\|_{\infty}DC_n \int_0^t \frac{s^{n/2}}{\sqrt{2\pi (t-s)}}ds.\end{align*}

We perform the change of variable $s=tu$ to obtain

\begin{align*}N(p_{n+1};\,t,x_0)\leq \frac{2(d+1)2^{d/2}\|B\|_{\infty}D}{\sqrt{2\pi}}C_n {\big(\sqrt{t}\big)^{n+1}}\int_0^1 \frac{u^{n/2}}{\sqrt{1-u}} ds.\end{align*}

Using

\begin{equation*}\int_0^1 \frac{u^{n/2}}{\sqrt{1-u}} du= \frac{\Gamma((n+2)/2 )\Gamma(1/2)}{\Gamma((n+3)/2)}\end{equation*}

and the definition of $C_n$ , we have

\begin{align*}N(p_{n+1};\,t,x_0)\leq C_{n+1} \big(\sqrt{t}\big)^{n+1};\end{align*}

this completes the proof of Proposition 4.4(i).

Then, for all $x_0 \in \mathbb{R}^d$ and using $x^1=m-u,$

\begin{equation*}\sup_{u>0}p_V(m,m-u,\tilde x;\,t)\leq C_T\phi_{d+1}\big(m-x^1_0,0,\tilde{x}-\tilde{x}_0;\,2t\big) \in L^1\big([0,T]\times \mathbb{R}^d,dtdmd\tilde x\big).\end{equation*}

Since $p_V(m,x;\,t,\mu_0)=\int_{\mathbb{R}^d} p_V\big(m,x;\,t,x_0\big) \mu_0(dx_0)$ , Part (iii) is true.

Lemma 4.4. Let $T>0, $ $x_0 \in \mathbb{R}^d,$ $p \in L^{\infty}\big([0,T], dt, L^{1}\big(\mathbb{R}^{d+1},dmdx\big)\big)$ be such that the support of $p(.,.;\,t)$ is included in $\big\{(m,x),\ m>x_0^1, m>x^1\big\}$ and for all $s\in ]0,T]$ $N(p;\,s,x_0) <\infty. $ Then for $j=\alpha,$ $k=m,1,\ldots,d,$ the support of the function ${\mathcal I}^{j,k}[p](.;\,t)$ is included in $\big\{(m,x),\ m>x_0^1, m>x^1\big\}.$ Moreover, for all $t\in [0,T]$ , we have

\begin{align*}N({\mathcal I}[p];\,t,x_0)\leq 2(d+1)2^{d/2}\|B\|_{\infty}D \int_0^t \frac{1}{\sqrt{2\pi {(t-s)}}}N(p;\,s,x_0)ds,\,\,\forall t \in[0,T].\end{align*}

Proof. Let $T>0,$ $x_0\in\mathbb{R}^d,$ $p \in L^{\infty}\big([0,T], dt,L^{1}\big(\mathbb{R}^{d+1},dmdx\big)\big)$ be such that for all $t>0$ , the support of $p(.;\,t)$ is included in $\big\{(m,x),\ m>x_0^1, m>x^1\big\}.$

(i) For $j=\alpha,$ $k=m, 1,\ldots,d,$ using the definition of ${\mathcal I}^{\alpha,k}$ yields

\begin{align*} &{{\mathcal I}^{k,\alpha}}[p](m,x;\,t)\,:\!=\, \\ & \quad \int_0^t{\int_{\mathbb{R}^{d+1}} B^{k}(a)\partial_{k}p_{W^{*1},W}\big(m-a^1,x-a;\,t-s\big){\mathbf 1}_{x_0^1<b <m,m>x^1}p(b,a;\,s)dbda}ds. \end{align*}

So the support of ${\mathcal I}^{\alpha,k}[p](.;\,t)$ is included in $\big\{(m,x) \in \mathbb{R}^{d+1},\ m>\max\!\big(x^1_0,x^1\big)\big\}.$ From now on, we consider only (m, x) such that $m >\max\!\big(x^1,x^1_0\big).$

Let p be a function such that for all $s \in ]0,T]$ , $N(p;\,x_0,s)<\infty.$ The definition of ${\mathcal I}^{k,\alpha},$ the boundedness of B, the fact that $\partial_kp_{W^*,W}$ satisfies (38), and the definition (41) of $N(p;\,t,x_0)$ imply that

\begin{align*} \left| {\mathcal I}^{k,\alpha}[p](m,x;\,t) \right| \leq \|B\|_{\infty}\int_0^t\int_{{\mathbb R}^{d+1}}N(p;\,s,x_0) \frac{D}{\sqrt{(t-s)}} {\mathbf 1}_{m>x_1}{\mathbf 1}_{m>b>\max\!\big(a^1,x_0^1\big)} \\ \,\,\,\,\,\, \phi_{d+1}\big(m-a^1, m-x^1,\tilde{x}-\tilde{a};\,t-s\big) \phi_{d+1}\big(b-x_0^1,b-a^1,\tilde{a}-\tilde{x_0};\,s\big) db da ds. \end{align*}

We integrate in $\tilde{a}$ using Lemma A.3(ii) with $u=\tilde{x},$ $v=\tilde{a},$ $w=\tilde{x}_0$ and the fact that $\phi_{d+1}$ is a Gaussian probability density function:

(42) \begin{align} & \left| {\mathcal I}^{\alpha,k}[p](m,x;\,t) \right| \leq 2^{(d-1)/2}\|B\|_{\infty}D \\& \int_0^t\int_{{\mathbb R}^{2}}N(p;\,s,x_0) {\mathbf 1}_{m>b>\max\!\big(a^1,x_0^1\big)} \frac{e^{-\frac{\|\tilde{x}-\tilde{x_0}\|^2}{4t}}}{\sqrt{(2 \pi t)^{d-1}}} \frac{e^{-\frac{\big(m-a^1\big)^2}{4(t-s)}-\frac{\big(m-x^1\big)^2}{4(t-s)}}}{\sqrt{(2\pi)^2(t-s)^3}} \frac{e^{-\frac{\big(b-x_0^1\big)^2}{4s}-\frac{\big(b-a^1\big)^2}{4s}}}{\sqrt{(2\pi)^2 {s^2}}}db da^1 ds. \nonumber \end{align}

Using Part (i^′) of Lemma A.2 with $u=m,$ $v=a^1,$ $w=b,$ $k=1$ , we integrate in $da^1$ up to b:

\begin{align*}&\int_{-\infty}^b \frac{e^{-\frac{\big(m-a^1\big)^2}{4(t-s)}}}{\sqrt{2\pi(t-s)}}\frac{e^{-\frac{\big(b-a^1\big)^2}{4s}}}{\sqrt{(2\pi s)}}da^1=\frac{e^{\frac{-(m-b)^2}{4t}}}{\sqrt{2\pi t}}\Phi_G\left( \sqrt{\frac{s}{2t(t-s)}}(b-m)\right),\end{align*}

where

\begin{equation*} \Phi_G(u) =\int_{-\infty}^u e^{-z^2/2}dz\leq {\frac{1}{2}} e^{-u^2/2}\end{equation*}

for $u=b-m<0$ by Lemma A.3(iii). This yields the bound

\begin{equation*}\frac{e^{\frac{-(m-b)^2}{4t}}}{\sqrt{2\pi t}}e^{-\frac{s(b-m)^2}{4t(t-s)}}\end{equation*}

and

\begin{align*}2\int_{-\infty}^b \frac{e^{-\frac{\big(m-a^1\big)^2}{4(t-s)}}}{\sqrt{2\pi(t-s)}}\frac{e^{-\frac{\big(b-a^1\big)^2}{4s}}}{\sqrt{(2\pi s)}}da^1&\leq\frac{e^{-\frac{(m-b)^2}{4t}}}{\sqrt{2\pi t}}e^{-\frac{s(m-b)^2}{4t(t-s)}}= \frac{e^{-\frac{(m-b)^2}{4(t-s)}}}{\sqrt{2\pi t}}.\end{align*}

Plugging this inequality into (42) yields the following, with $C_{d,B}=2^{(d+1)/2}\|B\|_{\infty}D$ :

\begin{align*} \frac{\left| {\mathcal I}^{\alpha,k}[p](m,x;\,t)\right|}{C_{d,B}} \leq\int_0^t\int_{{\mathbb R}}N(p;\,s,x_0) {\mathbf 1}_{m>b>x_0^1} \frac{e^{-\frac{\|\tilde{x}-\tilde{x_0}\|^2}{4t}}}{\sqrt{(2 \pi t)^d}} \frac{e^{-\frac{(m-b)^2}{4(t-s)}-\frac{\big(m-x^1\big)^2}{4(t-s)}}}{\sqrt{2\pi(t-s)^2{s}}} e^{-\frac{\big(b-x_0^1\big)^2}{4s}}db ds. \end{align*}

Omitting the indicator functions, Lemma A.3(ii) with $u=m,\,v=b,\,w=x_0^1,\,k=1$ implies

\begin{equation*}\int_{b<m}\frac{e^{-\frac{(m-b)^2}{4(t-s)}}e^{-\frac{\big(b-x_0^1\big)^2}{4s}}}{\sqrt{2\pi(t-s)2\pi s}} db \leq\sqrt{\frac{2}{2\pi t}}e^{-\frac{\big(m-x_0^1\big)^2}{4t}}.\end{equation*}

Inserting this result, we obtain

\begin{align*} \left| {\mathcal I}^{\alpha,k}[p](m,x;\,t) \right| \leq \sqrt{2}C_{d,B}\int_0^tN(p;\,s,x_0) \frac{e^{-\frac{\|\tilde{x}-\tilde{x_0}\|^2}{4t}}}{\sqrt{(2 \pi t)^{d+1}}} \frac{e^{-\frac{\big(m-x_0^1\big)^2}{4t}-\frac{\big(m-x^1\big)^2}{4(t-s)}}}{\sqrt{2\pi{(t-s)}}} ds. \end{align*}

For $0<s<t,$ we have

\begin{equation*}e^{-\frac{\big(m-x^1\big)^2}{4(t-s)}}\leq e^{-\frac{\big(m-x^1\big)^2}{4t}},\end{equation*}

so

\begin{align*} \left| {\mathcal I}^{\alpha,k}[p](m,x;\,t) \right| \leq \sqrt{2} C_{d,B}\int_0^tN(p;\,s,x_0) \frac{e^{-\frac{\|\tilde{x}-\tilde{x_0}\|^2}{4t}}}{\sqrt{(2 \pi t)^{d+1}}} \frac{e^{-\frac{\big(m-x_0^1\big)^2}{4t}-\frac{\big(m-x^1\big)^2}{4t}}}{\sqrt{2\pi{(t-s)}}} ds. \end{align*}

Using the definition of $\phi_{d+1}$ we find

\begin{align*} \left| {\mathcal I}^{\alpha,k}[p](m,x;\,t) \right| \leq \sqrt{2} C_{d,B}\int_0^tN(p;\,s,x_0) \frac{\phi_{d+1}\big(m-x_0^1,m-x^1,\tilde{x}-\tilde{x_0};\,2t\big)}{\sqrt{2\pi {(t-s)}}} ds, \end{align*}

and with the definition of N, with respect to a multiplicative constant we have

\begin{align*}N( {\mathcal I}^{\alpha,k}[p],x_0,t) \leq \sqrt{2} C_{d,B} \int_0^t N(p;\,s,x_0) \frac{1}{\sqrt{2\pi{(t-s)}}} ds. \end{align*}

(ii) For $j=\beta,$ $k=m,1,\ldots,d$ , using the definition of ${\mathcal I}^{\beta,k}$ and the fact that the support of p is included in $\big\{(m,x),\ m>\max\!\big(x_0^1,x^1\big)\big\}$ yields

\begin{align*} &{\mathcal I}^{\beta,k}[p](m,x;\,t)= \\& \quad \int_0^t \int_{\mathbb{R}^{d+1}}{\mathbf 1}_{m>b>x^1, m>x_0^1,b>a^1} B^{k}(a) \partial_k p_{W^{1*},W}\big(b-a^1,x-a,t-s\big)p(m,a,s) da db ds. \end{align*}

Thus the support of $ {\mathcal I}^{\beta,k}[p](.;\,t)$ is included in $\big\{(m,x),\ m>\max\!\big(x^1,x^0\big)\big\}.$ From now on we consider only (m, x) satisfying $m>\max\!\big(x^1,x_0^1\big).$ From the definition of ${\mathcal I}^{\beta,k}$ and the boundedness of B, as well as the inequality (38) satisfied by $\partial_kp_{W^*,W}$ , namely

\begin{equation*}|\partial_k p_{W^{1*},W}\big(b-a^1,x-a,t-s\big)|\leq \frac{D}{\sqrt{t-s}}\phi_{d+1}\big(b-a^1,b-x^1,\tilde x-\tilde a,2(t-s)\big),\end{equation*}

and the definition of $N(p;\,t,x_0)$ , we obtain

\begin{align*} \left| {\mathcal I}^{\beta,k}[p](m,x;\,t)\right|\leq \|B\|_{\infty} D \int_0^t \int_{\mathbb{R}^{d+1}}{\mathbf 1}_{m>b>x^1, b>a^1} N(p;\,s,x_0)\\\frac{e^{- \frac{\big(b-a^1\big)^2}{4(t-s)}- \frac{\big(b-x^1\big)^2}{4(t-s)} - \frac{\|\tilde{x}-\tilde{a}\|^2}{4(t-s)}}}{\sqrt{(2\pi)^{d+1}(t-s)^{d+2}}}\frac{e^{- \frac{\big(m-x_0^1\big)^2}{4s}- \frac{\big(m-a^1\big)^2}{4s} -\frac{\|\tilde{x}_0 -\tilde{a}\|^2}{4s}}}{\sqrt{(2\pi)^{d+1}{s^{d+1}}}}da db ds. \end{align*}

We integrate in $\tilde{a} $ using Lemma A.3(ii) with $u=\tilde{x},$ $v=\tilde{a}$ , and $w=\tilde{x}_0$ :

(43) \begin{align} & \left| {\mathcal I}^{\beta,k}[p](m,x;\,t)\right|\leq C_{d,B}. \\ & \int_0^t \int_{\mathbb{R}^{2}}{\mathbf 1}_{m>b>x^1, b>a^1}\frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{(2 \pi t)^{d-1}}}N(p;\,s,x_0)\nonumber.\frac{e^{- \frac{\big(b-a^1\big)^2}{4(t-s)}- \frac{\big(b-x^1\big)^2}{4(t-s)}}}{\sqrt{(2\pi)^2(t-s)^{3}}}\frac{e^{- \frac{\big(m-x_0^1\big)^2}{4s}- \frac{\big(m-a^1\big)^2}{4s}}}{\sqrt{(2\pi)^{2}{s^{2}}}}da^1 db ds. \end{align}

Using Lemma A.3(i^′) for $u=b,$ $v=a^1,$ $w=m$ , $k=1$ , we have

\begin{align*} & \int_{-\infty}^b \frac{e^{- \frac{\big(b-a^1\big)^2}{4(t-s)}}}{\sqrt{2\pi(t-s)}}\frac{e^{- \frac{\big(m-a^1\big)^2}{4s}}}{\sqrt{2\pi s}}da^1= \sqrt{2}\frac{e^{- \frac{(b-m)^2}{4t}}}{\sqrt{2 \pi t}}\Phi_G\left( \sqrt{\frac{t}{4s(t-s)}}\left[ b - \left(\frac{s}{t} b + \frac{t-s}{t} m\right)\right] \right)\\ &= \frac{e^{- \frac{(b-m)^2}{4t}}}{\sqrt{2 \pi t}}\Phi_G\left( \sqrt{\frac{t-s}{4st}}[ b - m] \right)\leq \frac{e^{- \frac{(b-m)^2}{4t}}}{\sqrt{2 \pi t}}e^{- \frac{t-s}{4st}[ b - m]^2}= \frac{e^{- \frac{(b-m)^2}{4s}}}{\sqrt{2 \pi t}}, \end{align*}

the last bound coming from Lemma A.3(iii) since $b-m<0.$

We plug this estimate into (43):

\begin{align*}& \left| {\mathcal I}^{\beta,k}[p](m,x;\,t)\right|\leq \\& C_{d,B}\int_0^t \int_{\mathbb{R}}{\mathbf 1}_{m>b>x^1}\frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{(2 \pi t)^{d}}}N(p;\,s,x_0)\frac{e^{-\frac{(b-x^1)^2}{4(t-s)}}}{\sqrt{2\pi(t-s)^{2}}} \frac{e^{- \frac{\big(m-x_0^1\big)^2}{4s}-\frac{(b-m)^2}{4s}}}{\sqrt{2\pi s}} db ds. \end{align*}

We integrate with respect to b on $\mathbb{R}$ , and we use Lemma A.3(ii) with $u=x^1,$ $v=b$ , $w=m,$ $k=1$ :

\begin{align*}{\left| {\mathcal I}^{\beta,k}[p](m,x;\,t)\right|\leq \sqrt{2}C_{d,B} \int_0^t\frac{e^{-\frac{\big(m-x^1\big)^2}{4t}-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{(2 \pi t)^{d+1}}}N(p;\,s,x_0) \frac{e^{- \frac{\big(m-x_0^1\big)^2}{4s}}}{\sqrt{2\pi (t-s)}} ds}. \end{align*}

When $0<s<t,$ we have

\begin{equation*}e^{- \frac{\big(m-x_0^1\big)^2}{4s}}\leq e^{- \frac{\big(m-x_0^1\big)^2}{4t}},\end{equation*}

so

\begin{align*} \left| {\mathcal I}^{\beta,k}[p](m,x;\,t)\right|\leq \sqrt{2}C_{d,B}\int_0^t \frac{e^{-\frac{\big(m-x^1\big)^2}{4t}-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t} {- \frac{\big(m-x_0^1\big)^2}{4t}}}}{\sqrt{(2 \pi t)^{d+1}}}N(p;\,s,x_0) \frac{1}{\sqrt{2\pi (t-s)}} ds. \end{align*}

Under the integral we identify the factor $\phi_{d+1}\big(m-x^1_0,m-x^1, \tilde{x}-\tilde{x_0};\,2t\big)$ , so

\begin{align*} \left| {\mathcal I}^{\beta,k}[p](m,x;\,t)\right|\leq \sqrt{2}C_{d,B}\phi_{d+1}\big(m-x^1_0,m-x^1, \tilde{x}-\tilde{x_0};\,2t\big)\int_0^t N(p;\,s,x_0) \frac{1}{\sqrt{2\pi {(t-s)}}} db ds. \end{align*}

Finally, using the definition of N (Equation (41)), we have proved

\begin{align*}N\big( {\mathcal I}^{\beta,k}[p],x_0,t\big) \leq \sqrt{2} C_{d,B}\int_0^tN(p;\,s,x_0) \frac{1}{\sqrt{2\pi{(t-s)}}} ds, \end{align*}

which completes the proof of Lemma 4.4.

4.2.2. Proof of Hypothesis 2.1(ii), case $A=I_d$ .

Proposition 4.5. For any probability measure $\mu_0$ on ${\mathbb R}^d,$ for all $(m, \tilde{x},t)\in \mathbb{R}^d\times ]0,T],$ $u \mapsto p_V(m,m-u,\tilde{x},t)$ admits a limit when u goes to 0, $u>0.$

Proof. The proof is a consequence of the following three lemmas.

Lemma 4.5. Recall that $p^0(m,x;\,t)=\int_{{\mathbb R}^d}p_{W^{*1},W}\big(m-x_0^1,x-x_0;\,t\big)\mu_0(dx_0).$ We have

\begin{align*} \,\,\lim_{u\rightarrow 0,u>0} p^0(b,b-u,\tilde{a};\,t) =p^0(b,b,\tilde{a};\,t),\,\,\forall u>0, \,\,(b,\tilde{a}) \in {\mathbb R}^d, \quad {\forall t>0}.\end{align*}

Proof. We have

\begin{equation*}p^0(b,b-u,\tilde{x};\,t)={\int_{{\mathbb R}^d}} {2\frac{b+u-x_0^1}{\sqrt{(2\pi)^d{\mathbf{t^{d+1}}}}}e^{-\frac{\big(b+u-x_0^1\big)^2}{2t} -\frac{\|\tilde{x}-\tilde{x}_0\|^2}{2t}}}{{\mathbf 1}_{b \geq x_0^1,\,\,u \geq 0}\mu_0(dx_0)}.\end{equation*}

Then, since the integrand is dominated by

\begin{equation*}\frac{{D}}{\sqrt{(2 \pi)^d t^{d+1}}}\end{equation*}

and $\mu_0$ is a probability measure, using Lebesgue’s dominated convergence theorem yields

\begin{align*}\lim_{u \rightarrow 0,u>0} p^0(b,b-u,\tilde{x};\,t)=p^0(b,b,\tilde{x};\,t),\,\,\forall \,\,(b,\tilde{x}) \in {\mathbb R}^d, {\forall t>0}.\end{align*}

Lemma 4.6. For $k=m,1,\ldots,d$ , recall that

\begin{equation*}p^{k,\alpha}(m,x;\,t)=\int_0^t\int_{\mathbb{R}^{d+1}} {\mathbf 1}_{b<m} B^k(a) \partial_{k}p_{W^{*1}, W} (m-a^1,x-a, t-s) p_V(b,a;\,s) db da ds.\end{equation*}

The map $u \mapsto p^{k,\alpha}(m,m-u,\tilde{x};\,t)$ converges to $p^{k,\alpha}\big(m,m,\tilde{x};\,t\big)$ when u goes to $0^+.$

Proof. The proof will be a consequence of Lebesgue’s dominated convergence theorem. First, the map $u \mapsto {\mathbf 1}_{b<m} \partial_{k}p_{W^{*1}, W} \big(m-a^1,m-u-a^1,\tilde{x}-\tilde{a};\, t-s\big) p_V(b,a;\,s)$ converges to ${\mathbf 1}_{b<m} \partial_{k}p_{W^{*1}, W} \big(m-a^1,ma^1,\tilde{x}-\tilde{a};\, t-s\big) p_V(b,a;\,s)$ when u goes to $0^+.$

Second, it is dominated by

\begin{align*} &q^{k,\alpha}(m,\tilde{x},a,b;\,s,x_0)\,:\!=\, \\& \quad |B^k(a)|{\mathbf 1}_{b<m}\sup_{u>0} \left|\partial_{k}p_{W^{*1}, W} \big(m-a^1,m-u-a^1,\tilde{x}-\tilde{a};\, t-s\big) p_V(b,a;\,s,x_0)\right|.\end{align*}

We seek to prove that

(44) \begin{align}\int_0^T \int_{\mathbb{R}^{2d+1}}q^{k,\alpha}\big(m,\tilde{x},a,b;\,s,x_0\big) ds db da \mu_0(dx_0)<+\infty.\end{align}

According to the estimate (38) of $\partial_k p_{W^{*1},W}$ and the estimate (ii) of Proposition 4.4, we obtain

\begin{align*}q^{k,\alpha}\big(m,\tilde{x},a,b;\,s,x_0\big)&\leq\|B\|_{\infty}{\mathbf 1}_{m>b>a^1}\frac{D}{\sqrt{t-s} \sqrt{2\pi(t-s)}^{d+1}} \exp \bigg[ -\frac{\big(m-a^1\big)^2}{4(t-s)} -\frac{\|\tilde{x}-\tilde{a}\|^2}{4(t-s)} \bigg] \\ &\frac{C_T}{\sqrt{2\pi s}^{d+1}}\exp\bigg[- \frac{\big(b-x_0^1\big)^2}{4s} - \frac{\big(b-a^1\big)}{4s} - \frac{\|\tilde{x}_0-\tilde{a}\|^2}{4s}\bigg].\end{align*}

We integrate with respect to $\tilde{a}$ using Lemma A.3(ii) for $k=d+1,$ $u=\tilde{x},$ $v=\tilde{a}$ , and $w=\tilde{x}_0:$

\begin{align*}\int_{\mathbb{R}^{d-1}}q^{k,\alpha}\big(m,\tilde{x},a^1,b;\,s,x_0^1\big)d\tilde{a} &\leq {\mathbf 1}_{m>b>a^1}\frac{\|B\|_{\infty} C_T D2^{(d-1)/2}}{\sqrt{t-s} \sqrt{2\pi(t-s)}^{2} \sqrt{2\pi s}^{2}}\frac{e^{-\frac{|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d-1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& \exp \bigg[ -\frac{\big(m-a^1\big)^2}{4(t-s)} - \frac{\big(b-x_0^1\big)^2}{4s} - \frac{\big(b-a^1\big)^2}{4s} \bigg].\end{align*}

We integrate with respect to $a_1$ between $-\infty$ and b using Lemma A.3(i^′) for $u=m,$ $v=a^1$ , and $w=b$ :

\begin{align*}\int_{\mathbb{R}^{d}}{\mathbf 1}_{a^1<b}q^{k,\alpha}(m,\tilde{x},b,a;\,s;\,x_0)da &\leq {\mathbf 1}_{b<m}\frac{\|B\|_{\infty} C_T D2^{d/2}}{\sqrt{t-s} \sqrt{2\pi(t-s)} \sqrt{2\pi s}}\frac{e^{-\frac{|\tilde{x}-\tilde{x}_0\|^2}{4t}}-\frac{(b-m)^2}{4t}}{\sqrt{2 \pi t}^{d}}\\& \!\!\!\!\exp \bigg[ - \frac{\big(b-x_0^1\big)^2}{4s} \bigg]\Phi_G\left(\sqrt{\frac{t}{2s(t-s)}}\Big( b- \Big[\frac{s}{t}m + \frac{(t-s)}{t}b\Big]\Big)\right).\end{align*}

Note that

\begin{equation*} \sqrt{\frac{t}{2s(t-s)}}\Big( b- \Big[\frac{s}{t}m + \frac{(t-s)}{t}b\Big]\Big)= \sqrt{\frac{s}{2t(t-s)}}(b-m),\end{equation*}

and using Lemma A.3(iii),

\begin{align*}\int_{\mathbb{R}^{d}}q^{k,\alpha}(m,\tilde{x},b,a;\,s,x_0)da &\leq {\mathbf 1}_{b<m}\frac{\|B\|_{\infty} C_T D2^{d/2}}{\sqrt{t-s} \sqrt{2\pi(t-s)} \sqrt{2\pi s}}\frac{e^{-\frac{|\tilde{x}-\tilde{x}_0\|^2}{4t}-\frac{(b-m)^2}{4t}}}{\sqrt{2 \pi t}^{d}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& \exp \left[ - \frac{\big(b-x_0^1\big)^2}{4s}\right]\exp\left[ -\frac{s}{t(t-s)}\frac{(b-m)^2}{4}\right].\end{align*}

We observe that $\frac{1}{t} +\frac{s}{t(t-s)}=\frac{1}{t-s}$ , so that

\begin{equation*}\exp\left[-\frac{(b-m)^2}{4t}\right]\exp\left[ -\frac{s}{t(t-s)}\frac{(b-m)^2}{4}\right]=\exp\left[ -\frac{(b-m)^2}{4(t-s)}\right].\end{equation*}

We integrate with respect to b (neglecting the indicator function) using Lemma A.3(ii) for $u=m,$ $v=b$ , and $w=x_0^1$ , and $\exp\left[-\frac{\big(m-x^1_0\big)^2}{4t}\right]\leq 1$ :

\begin{align*}\int_{\mathbb{R}^{d+1}}q^{k,\alpha}(m,\tilde{x},b,a;\,s,x_0)dadb &\leq {\mathbf 1}_{m>x_0^1}\frac{\|B\|_{\infty} C_T D2^{(d+1)/2}}{\sqrt{t-s}}\frac{e^{-\frac{|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{t}^{d+1}}.\end{align*}

Since $\mu_0$ is a probability measure, we have

\begin{equation*}\int_0^t\int_{\mathbb{R}^{2d+1}}q^{k,\alpha}(m,\tilde{x},b,a;\,s,x_0)dadb\mu_0(dx_0)ds <+\infty.\end{equation*}

This is (44) and completes the proof of Lemma 4.6.

Lemma 4.7. For $k=m,1,\ldots,d$ , recall that

\begin{equation*}p^{k,\beta}(m,x;\,t)=\int_0^t\int_{\mathbb{R}^{d+1}} {\mathbf 1}_{b<m} B^{k}(a) \partial_{k}p_{W^{*1}, W} \big(b-a^1,x-a, t-s\big) p_V(m,a;\,s) db da ds.\end{equation*}

The map $u \mapsto p^{k,\beta}(m,m-u,\tilde{x};\,t)$ converges to 0 when u goes to $0^+.$

Proof. Using the estimate (38) of $\partial_kp_{W^{*1},W}$ and the estimate (ii) of Proposition 4.4 for $p_V,$ we dominate the integrand which defines $p^{k,\beta}(m,m-u,\tilde{x};\,t)$ by

\begin{align*} &q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s)\,:\!=\, \\& \quad {\mathbf 1}_{m-u<b<m,a^1<b}\frac{e^{-\frac{\big(b-a^1\big)^2}{4(t-s)} -\frac{(b-m+u)^2}{4(t-s)} -\frac{\|\tilde{x}-\tilde{a}\|^2}{4(t-s)} -\frac{\big(m-x_0^1\big)^2}{4s} - \frac{\big(m-a^1\big)}{4s} -\frac{\|\tilde{x}_0 -\tilde{a}\|^2}{4s}}}{\sqrt{t-s}\sqrt{2 \pi (t-s)}^{d+1}\sqrt{2 \pi s}^{d+1}}\end{align*}

up to a multiplicative constant, meaning that

(45) \begin{align}\left|p^{k,\beta}(m,m-u,\tilde{x};\,t)\right| \leq \|B\|_{\infty}\int_0^t \int_{{\mathbb R}^{2d+1}}q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) db da ds \mu_0(dx_0).\end{align}

We integrate with respect to $\tilde{a} $ using Lemma A.3(ii) with $u=\tilde{x},$ $v=\tilde{a}$ , and $w=\tilde{x}_0$ :

\begin{align*}\int_{R^{d-1}} q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) d\tilde{a} \leq \sqrt{2}^{d-1} \frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d-1}}\frac{e^{-\frac{\big(b-a^1\big)^2}{4(t-s)} -\frac{(b-m+u)^2}{4(t-s)} -\frac{\big(m-x_0^1\big)^2}{4s} - \frac{\big(m-a^1\big)}{4s}}}{\sqrt{t-s}\sqrt{2 \pi (t-s)}^{2}\sqrt{2 \pi s}^{2}}.\end{align*}

We integrate with respect to $a^1$ between $-\infty$ and b using Lemma A.3(i^′) for $u=b,$ $v=a^{1}$ , and $w=m$ :

\begin{align*}&\int_{R^{d}} {\mathbf 1}_{b>a^1}q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) da \leq\\&\frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d}}\frac{e^{-\frac{(b-m)^2}{4t} -\frac{(b-m+u)^2}{4(t-s)} -\frac{\big(m-x_0^1\big)^2}{4s}}}{\sqrt{t-s}\sqrt{2 \pi (t-s)}\sqrt{2 \pi s}}\Phi_G\left(\sqrt{\frac{t}{s(t-s)2}}( b-\frac{s}{t}b-\frac{t-s}{t}m)\right)\\&= \frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d}}\frac{e^{-\frac{(b-m)^2}{4t} -\frac{(b-m+u)^2}{4(t-s)} -\frac{\big(m-x_0^1\big)^2}{4s}}}{\sqrt{t-s}\sqrt{2 \pi (t-s)}\sqrt{2 \pi s}}\Phi_G\left(\sqrt{\frac{t-s}{2st}}( b-m)\right).\end{align*}

Since $b-m<0$ , using Lemma A.3(iii) we have

\begin{align*}\int_{R^{d}}{\mathbf 1}_{b>a^1} q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) da \leq\frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d}}\frac{e^{-\frac{(b-m)^2}{4t} -\frac{(b-m+u)^2}{4(t-s)} -\frac{\big(m-x_0^1\big)^2}{4s}}}{\sqrt{t-s}\sqrt{2 \pi (t-s)}\sqrt{2 \pi s}}e^{-\frac{t-s}{4st}( b-m)^2}.\end{align*}

Note that

\begin{equation*}e^{-\frac{(b-m)^2}{4t}}e^{-\frac{t-s}{4st}( b-m)^2\big)}=e^{-\frac{( b-m)^2\big)}{4s}}.\end{equation*}

We integrate this last bound with respect to b between $m-u$ and m using Lemma A.3(i^′) for the triplet $(m-u,b,m)$ and the fact that

\begin{equation*}\frac{s}{t}(m-u) +\frac{t-s}{t}m=\frac{s(m-u)+m(t-s)}{t},\end{equation*}

to obtain

\begin{eqnarray*}&\int_{R^{d+1}}{\mathbf 1}_{m-u<b<m,b<a^1} q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) da db\leq\frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d+1}}\frac{e^{-\frac{\big(m-x_0^1\big)^2}{4s}}}{\sqrt{t-s}}.\\&\left[\Phi_G\left(\sqrt{\frac{t}{2s(t-s)}}\left( m-\frac{s(m-u)+m(t-s)}{t}\right) \right)- \Phi_G\left( \sqrt{\frac{t}{2s(t-s)}}\left( m-u-\frac{s(m-u)+m(t-s)}{t}\right) \right)\right].\end{eqnarray*}

Then,

\begin{align*}&\int_{R^{d+1}}q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) da db\leq\\&\frac{e^{-\frac{\|\tilde{x}-\tilde{x}_0\|^2}{4t}}}{\sqrt{2 \pi t}^{d+1}}\frac{e^{-\frac{\big(m-x_0^1\big)^2}{4s}}}{\sqrt{t-s}}\left[\Phi_G\left( \sqrt{\frac{s}{2t(t-s)}}u \right)- \Phi_G\left({-} \sqrt{\frac{t-s}{2t(t-s)}} u \right)\right].\end{align*}

Note that

\begin{equation*}\lim_{u\rightarrow 0} \Phi_G\left( \sqrt{\frac{s}{2t(t-s)}}u \right)- \Phi_G\left({-} \sqrt{\frac{t-s}{2t(t-s)}} u \right)=0\end{equation*}

and

\begin{equation*}\left| \Phi_G\left( \sqrt{\frac{s}{2t(t-s)}}u \right)- \Phi_G\left({-} \sqrt{\frac{t-s}{2t(t-s)}} u \right)\right| \leq 1.\end{equation*}

Since $\mu$ is a probability measure, using Lebesgue’s dominated convergence theorem,

\begin{align*}\lim_{u\rightarrow 0^+} \int_0^t \int_{R^{2d+1}}{\mathbf 1}_{m-u<b<m,b<a^1} q^{k,\beta}(m,u,\tilde{x},a,b,x_0,s) da db \mu_0(dx_0) ds=0.\end{align*}

Finally, the estimate (45) yields $\lim_{u\rightarrow 0^+} p^{k,\beta}(m,m-u, \tilde x;\,t)=0.$