Well posedness and stochastic derivation of a diffusion-growth-fragmentation equation in a chemostat

Abstract

We study the existence and uniqueness of the solution of a non-linear coupled system constituted of a degenerate diffusion-growth-fragmentation equation and a differential equation, resulting from the modeling of bacterial growth in a chemostat. This system is derived, in a large population approximation, from a stochastic individual-based model where each individual is characterized by a non-negative trait whose dynamics is described by a diffusion process. Two uniqueness results are highlighted. They differ in their hypotheses related to the influence of the resource on individual trait dynamics, the main difficulty being the non-linearity due to this dependence and the degeneracy of the diffusion coefficient. Further we show by probabilistic arguments that the semi-group of the stochastic trait dynamics admits a density. We deduce that the diffusion-growth-fragmentation equation admits a function solution with a certain Besov regularity.

Appendices

Appendix

Proof of Theorem 2.1

Let us recall that Assumptions 2.1 and 2.2 hold.

1.1 Step 1: Control of tails

Since the individual trait is unbounded, we need to get a control on the measure valued process tails. We then have the following lemma that is adapted from [22].

Lemma A.1

Let $T>0$, if a subsequence of $\left\{ \left( \nu ^K_t\right) _{t\ge 0},K>1 \right\}$ converges in law in the Skorohod space $\mathbb {D}([0,T],(\mathcal {M}_F(\mathbb {R}_+),v))$ then its limit $(\nu _t)_{t\in [0,T]}$ satisfies

$$\begin{aligned} \mathbb {E}\left\{ \sup _{0\le t\le T}\left[ \left\langle \nu _t,1\right\rangle ^{1+\varrho } \right] \right\} < \infty . \end{aligned}$$

(A.1)

In addition, there exists a decreasing sequence of increasing functions $(f_n)_{n\ge 1}\subset \mathcal {C}_b^2(\mathbb {R}_+,\mathbb {R})$ such that $f_n=0$ on [0, n/2] and $f_n = 1$ on $[n,\infty )$, which satisfies

$$\begin{aligned} \lim _{n\rightarrow \infty }\limsup _{K\rightarrow \infty }\mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu ^K_t,f_n\right\rangle \right\} = 0. \end{aligned}$$

(A.2)

Proof of Lemma A.1

In order to show (A.1), we approximate the functions $x\mapsto 1$ by an increasing sequence of continuous and compact supported functions in order to use the convergence assumption for the vague topology. We then introduce the uniformly bounded sequence of continuous functions defined for $n\in \mathbb {N}^{\star }$ by

$$\begin{aligned} \phi _n(x) = \left\{ \begin{array}{ll} \displaystyle 1 &{}\text { if }0\le x< n-1, \\ \displaystyle n-x &{}\text { if }n-1\le x < n, \\ \displaystyle 0 &{}\text { if }x\ge n. \end{array}\right. \end{aligned}$$

that converges increasingly pointwise towards the constant function $x\mapsto 1$. It follows from the Fatou lemma that holds thanks to the convergence of the sequence of measure valued process for the vague topology, that we have for all $n\in \mathbb {N}$

$$\begin{aligned} \begin{array}{ll} \displaystyle \mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu _t,\phi _n\right\rangle ^{1+\varrho } \right\} &{}\le \liminf _{K\rightarrow \infty }\mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu ^K_t,\phi _n\right\rangle ^{1+\varrho } \right\} \\ &{} \le \liminf _{K\rightarrow \infty }\mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu ^K_t,1\right\rangle ^{1+\varrho } \right\} . \end{array} \end{aligned}$$

As $n\rightarrow \infty$ in this inequality, we obtain by monotony

$$\begin{aligned} \mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu _t,1\right\rangle ^{1+\varrho } \right\} \le \sup _K\mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu ^K_t,1\right\rangle ^{1+\varrho } \right\} <\infty . \end{aligned}$$

In order to show (A.2), we approximate the indicator of the outside of the compacts by a non-increasing and uniformly bounded sequence of regular functions. This allows us to control the tails of the distributions $\{\mathcal {Z}^K,K>1\}$. Let us then introduce an increasing function $f\in \mathcal {C}^2(\mathbb {R}_+,\mathbb {R})$ such that $f\equiv 0$ on [0, 1/2] and $f\equiv 1$ on $[1,\infty )$. We set

$$\begin{aligned} f_n(x) = f\left( \frac{x}{n} \right) ,~\forall n\ge 1 \end{aligned}$$

(A.3)

which belongs to $\mathcal {C}^2_b(\mathbb {R}_+,\mathbb {R})$ and whose derivatives have compact supports. It then follows from (2.12) that for all $s\le t\le T$

$$\begin{aligned}{} & {} \displaystyle \left\langle \nu ^K_s,f_n\right\rangle \le \left\langle \nu ^K_0,f_n\right\rangle + M^{K,f_n}_s \\{} & {} \qquad + \int _0^s\int _0^{\infty }\left\{ \zeta (x,R^K_u)f'_n(x) + D(x,R^K_s)f''_n(x) \right\} \nu ^K_u(dx)du \\{} & {} \qquad \displaystyle + 2\int _0^s\int _0^{\infty }\left\{ b(x,R^K_u)\int _0^1f_n(\alpha x)M(d\alpha ) \right\} \nu ^K_u(dx)du \\{} & {} \quad \displaystyle \le \left\langle \nu ^K_0,f_n\right\rangle + A^{K,f_n}_T + 2\Vert b\Vert _{\infty }\int _0^s\left\langle \nu ^K_u,f_n\right\rangle du + \sup _{0\le u\le T}\left| M^{K,f_n}_u\right| \end{aligned}$$

where

$$\begin{aligned} A^{K,f_n}_T:= C\int _0^T \left\langle \nu ^K_u,(1+x)(f'_n + \left| f''_n\right| )\right\rangle du. \end{aligned}$$

The sequence $(A^{K,f_n}_T)_{K>1}$ is uniformly integrable according to K and converges in law towards the random variable $C\int _0^T \left\langle \nu _u,(1+x)(f'_n + \left| f''_n\right| )\right\rangle du$ as $K\rightarrow \infty$ because the function $x\mapsto (1+x)(f'_n + \left| f''_n\right| )(x)$ is continuous with compact support. We have

$$\begin{aligned}{} & {} \mathbb {E}\left\{ \sup _{0\le s\le t}\left\langle \nu ^K_s,f_n\right\rangle \right\} \le \mathbb {E}\left\{ \left\langle \nu ^K_0,f_n\right\rangle + A^{K,f_n}_T + \sup _{0\le u\le T}\left| M^{K,f_n}_u\right| \right\} \\{} & {} \qquad + 2\Vert b\Vert _{\infty }\int _0^t\mathbb {E}\left\{ \sup _{0\le u\le s}\left\langle \nu ^K_u,f_n\right\rangle \right\} ds \end{aligned}$$

that induces thanks to Gronwall’s lemma that

$$\begin{aligned} \mathbb {E}\left\{ \sup _{0\le s\le t}\left\langle \nu ^K_s,f_n\right\rangle \right\} \le \mathbb {E}\left\{ \left\langle \nu ^K_0,f_n\right\rangle + A^{K,f_n}_T + \sup _{0\le u\le T}\left| M^{K,f_n}_u\right| \right\} e^{2\Vert b\Vert _{\infty }t}. \end{aligned}$$

As $K\rightarrow \infty$, it follows from (2.15) and the Doob inequality that the right term with the martingale converges towards 0. In addition the remaining terms in the expectation are uniformly integrable and converge in law, then

$$\begin{aligned}{} & {} \displaystyle \lim _{n\rightarrow \infty }\limsup _{K\rightarrow \infty }\mathbb {E}\left\{ \sup _{0\le s\le T}\left\langle \nu ^K_s,f_n\right\rangle \right\} \\{} & {} \quad \displaystyle \le e^{2\Vert b\Vert _{\infty }T}\lim _{n\rightarrow \infty }\mathbb {E}\left\{ \left\langle \nu _0,f_n\right\rangle \right. \\{} & {} \quad \left. + C\int _0^T\int _0^{\infty }\frac{(1+x)}{n}\bigg (f'\big (\frac{x}{n}\big ) + \frac{1}{n}\big |f''\big |\big (\frac{x}{n}\big ) \bigg )\nu _u(dx) du \right\} = 0. \end{aligned}$$

$\square$

1.2 Step 2: Tightness

Given the above Lemma A.1, the tightness of the sequence of laws $\Big \{\mathcal {Z}^K = \mathcal {L}(\nu ^K,R^K),K>1 \Big \}$ follows. It is adapted from [2], Theorem 7.4.

Proposition A.1

The sequence of laws $\left\{ \mathcal {Z}^K = \mathcal {L}(\nu ^K,R^K),K>1 \right\}$ is tight in the space of probability measures $\mathcal {P}(\mathbb {D}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},w\otimes |.|)))$ for all $T>0$.

Proof

Let us first endow the measure space $\mathcal {M}_F(\mathbb {R}_+)$ with its vague topology and prove the tightness property for the sequence $\left\{ \mathcal {Z}^K=\mathcal {L}\left( \nu ^K, R^K\right) ,K>1 \right\}$. By noticing that $\mathcal {C}^2_c(\mathbb {R}_+,\mathbb {R})$ is a dense subspace of $\mathcal {C}_0(\mathbb {R}_+,\mathbb {R})$ for the topology of uniform convergence on compact sets, it suffices according to [29] to show that for each test function $f\in \mathcal {C}_c^2(\mathbb {R}_+,\mathbb {R})$ the sequence of processes $\left\{ \left( \left\langle \nu ^K_t,f\right\rangle , R^K_t\right) _{t\ge 0},K>1 \right\}$ is tight in the Skorohod space $\mathbb {D}([0,T],\mathbb {R}^2)$. First, remember that we have the decomposition

$$\begin{aligned} \left\langle \nu ^K_t,f\right\rangle = \left\langle \nu ^K_0,f\right\rangle + V^{K,f}_t + M^{K,f}_t \end{aligned}$$

(see Lemma 2.1) and thanks to (2.15),

$$\begin{aligned} \lim _{K\rightarrow \infty }\mathbb {E}\left\langle M^{K,f}\right\rangle _t = 0 \,, \, \forall t\ge 0. \end{aligned}$$

It then directly follows from Doob’s inequality that the sequence of martingales $\left\{ M^{K,f},K>1 \right\}$ converges in $\mathbb {L}^2$ toward 0 locally uniformly in time. This sequence is then tight in the Skorohod space $\mathbb {D}([0,T],\mathbb {R})$, and all we need to prove now is the tightness of the other sequences $\{V^{K,f},K>1\}$ and $\{R^K,K>1\}$. For that purpose, we use the Aldous criterion (see [1]), that consists in showing the uniform controls

$$\begin{aligned} \begin{array}{l} \displaystyle \text {(a) } \sup _K\mathbb {E}\left[ \sup _{0\le t\le T}\left| V^{K,f}_t \right| \right]< \infty \,, ~~ \text {(b) } \sup _K\mathbb {E}\left[ \sup _{0\le t\le T}\left| R^K_t\right| \right] < \infty \end{array} \end{aligned}$$

and showing that the Aldous condition holds for $(R^K_t,V^{K,f}_t)_{t\ge 0}$. The assertions (a) and (b) are immediate. Indeed, $R^K_t \le r\vee R_0 = \bar{R}$ for any $t\ge 0$ and thanks to the boundedness of the test function,

$$\begin{aligned} \sup _{0\le t\le T} \left| V^{K,f}_t\right| \le C_f\sup _{0\le t\le T}\left[ \left\langle \nu ^K_t,1\right\rangle + \left| M^{K,f}_t\right| \right] \end{aligned}$$

that implies (a) thanks to (2.10) and (2.15). Let us now consider $\varepsilon >0$ and a stopping time $\tau$ that satisfies $\tau +\epsilon \le T$, then

$$\begin{aligned}{} & {} \displaystyle \mathbb {E}\left( \left| R^K_{\tau +\varepsilon }-R^K_{\tau } \right| \right) \le \mathbb {E}\left( \int _{\tau }^{\tau +\varepsilon }\left| r_{in}-R^K_s - \int _0^{\infty }\chi (x,R^K_s)\nu ^K_s(dx)\right| ds\right) \\{} & {} \quad \displaystyle \le \varepsilon \left[ r_{in}+\bar{R} + \Vert \chi \Vert _{\infty }\mathbb {E}\left( \sup _{0\le t\le T}\left\langle \nu ^K_t,1\right\rangle \right) \right] . \end{aligned}$$

By a similar computation, we obtain

$$\begin{aligned} \mathbb {E}\left( \left| V^{K,f}_{\tau +\varepsilon }-V^{K,f}_{\tau }\right| \right) \le C(f)\varepsilon \mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu ^K_t,1\right\rangle \right] , \end{aligned}$$

that implies the Aldous condition thanks to Lemma 2.1. The sequence of law $(\mathcal {Z}^K)_K$ is then tight and by the Prokhorov’s theorem one can extract from each sub-sequence, a sub-sub-sequence that converges in $\mathcal {P}(\mathbb {D}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},v\otimes |.|)))$. Let us denote by $\mathcal {Z}$ a limit value of this sequence, and for simplicity of the notation, $(\mathcal {Z}^K)_K$ a sub-sequence that converges towards $\mathcal {Z}$. For each $K>1$ we have

$$\begin{aligned} \sup _{0\le s\le T}\sup _{f\in L^{\infty }(\mathbb {R}_+),\Vert f\Vert _{\infty }\le 1}\left| \left\langle \nu ^K_{t},f\right\rangle - \left\langle \nu ^K_{t-},f\right\rangle \right| \le 3/K. \end{aligned}$$

(A.4)

We then deduce from the continuity of the mapping $\mu \mapsto \sup _{0\le t\le T}\left| \left\langle \mu _{t},f\right\rangle - \left\langle \mu _{t-},f\right\rangle \right|$ for $f\in \mathcal {C}_c(\mathcal {E}\times \mathbb {R}_+,\mathbb {R})$, that the limiting law $\mathcal {Z}$ only charges the set $\mathcal {C}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},v\otimes |.|))$. We are aiming to prove the same result for the weak topology on $\mathcal {M}_F(\mathbb {R}_+)$ and this is where the sequence of functions introduced in Lemma A.1 is usefull.

Let us denote by $(\nu ,R)$ a process of law $\mathcal {Z}$, then according to the above argument its trajectories are in the space $\mathcal {C}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},v\otimes |.|))$. As in [22], it follows from the Fatou lemma that for $n,l\in \mathbb {N}$,

$$\begin{aligned} \mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu _t,(1-f_l)f_n\right\rangle \right]\le & {} \liminf _{K\rightarrow \infty }\mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu ^K_t,(1-f_l)f_n\right\rangle \right] \\\le & {} \limsup _{K\rightarrow \infty }\mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu _t^K,f_n\right\rangle \right] . \end{aligned}$$

As $l\rightarrow \infty$ it follows from Beppo-Levi’s theorem applied to the left term, that

$$\begin{aligned} \mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu _t,f_n\right\rangle \right] \le \limsup _{K\rightarrow \infty }\mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu _t^K,f_n\right\rangle \right] \end{aligned}$$

and then by the Lemma A.1,

$$\begin{aligned} \lim _{n\rightarrow \infty }\mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu _t,f_n\right\rangle \right] \le \lim _{n\rightarrow \infty }\limsup _{K\rightarrow \infty }\mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu _t^K,f_n\right\rangle \right] = 0. \end{aligned}$$

One can therefore extract a sub-sequence $\left( \sup _{0\le t\le T}\langle \nu _t,f_{n_k}\rangle \right) _k$ that converges almost surely towards 0, and then the limit law $\mathcal {Z}$ only charges the set $\mathcal {C}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},w\otimes |.|))$. We deduce that the convergence of $\{\mathcal {Z}^K,K>1\}$ towards $\mathcal {Z}$ for the weak topology on $\mathcal {M}_F(\mathbb {R}_+)$ holds if the sequence of total masses $\left\{ \left\langle \nu ^K,1\right\rangle ,K>1\right\}$ converges in law towards $\left\langle \nu ,1\right\rangle$ in the Skorohod space $\mathbb {D}([0,T],\mathbb {R})$ (see [25]). So let us consider a Lipschitz continuous function $\Phi : \mathbb {D}([0,T],\mathbb {R})\rightarrow \mathbb {R}$, and $n\in \mathbb {N}$.

$$\begin{aligned} \displaystyle \left| \mathbb {E}\left[ \Phi \left( \left\langle \nu ^K,1\right\rangle \right) - \Phi \left( \left\langle \nu ,1\right\rangle \right) \right] \right|\le & {} \mathbb {E}\left| \Phi \left( \left\langle \nu ^K,1\right\rangle \right) - \Phi \left( \left\langle \nu ^K,1-f_n\right\rangle \right) \right| \\{} & {} \displaystyle +~\left| \mathbb {E}\left[ \Phi \left( \left\langle \nu ^K,1-f_n\right\rangle \right) - \Phi \left( \left\langle \nu ,1-f_n\right\rangle \right) \right] \right| \\{} & {} \displaystyle +\, \mathbb {E}\left| \Phi \left( \left\langle \nu ,1-f_n\right\rangle \right) - \Phi \left( \left\langle \nu ,1\right\rangle \right) \right| \\ \displaystyle\le & {} C_{\Phi }\mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu ^K_t,f_n\right\rangle + \sup _{0\le t\le T}\left\langle \nu _t,f_n\right\rangle \right\} \\{} & {} \displaystyle +\, \left| \mathbb {E}\left[ \Phi \left( \left\langle \nu ^K,1-f_n\right\rangle \right) - \Phi \left( \left\langle \nu ,1-f_n\right\rangle \right) \right] \right| \end{aligned}$$

The function $1-f_n$ being continuous with a compact support and the sequence $\{\nu ^K,K>1\}$ converging in law toward the limit process $\nu$ in $\mathbb {D}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},v\otimes |.|))$, we have

$$\begin{aligned} \lim _{K\rightarrow \infty }\left| \mathbb {E}\left[ \Phi \left( \left\langle \nu ^K,1-f_n\right\rangle \right) - \Phi \left( \left\langle \nu ,1-f_n\right\rangle \right) \right] \right| = 0. \end{aligned}$$

It follows that for all $n\in \mathbb {N}$,

$$\begin{aligned}{} & {} \displaystyle \limsup _{K\rightarrow \infty }\left| \mathbb {E}\left[ \Phi \left( \left\langle \nu ^K,1\right\rangle \right) - \Phi \left( \left\langle \nu ,1\right\rangle \right) \right] \right| \\{} & {} \quad \displaystyle \le C_{\Phi }\left[ \limsup _{K\rightarrow \infty }\mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu ^K_t,f_n\right\rangle \right\} + \mathbb {E}\left\{ \sup _{0\le t\le T}\left\langle \nu _t,f_n\right\rangle \right\} \right] . \end{aligned}$$

Thanks to Lemma A.1, that implies that as $n\rightarrow \infty$ in the right term,

$$\begin{aligned} \limsup _{K\rightarrow \infty }\left| \mathbb {E}\left[ \Phi \left( \left\langle \nu ^K,1\right\rangle \right) - \Phi \left( \left\langle \nu ,1\right\rangle \right) \right] \right| = 0. \end{aligned}$$

The sub-sequence $\{(\langle \nu ^K_t,1\rangle )_{t\in [0,T]},K>1 \}$ then converges in law towards the process $(\langle \nu _t,1\rangle )_{t\in [0,T]}$, and then the subsequence $(\mathcal {Z}^K)_K$ converges towards $\mathcal {Z}$ in the space of probability distributions $\mathcal {P}(\mathbb {D}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},w\otimes |.|)))$. That is enough to conclude that our initial sequence of laws is tight. $\square$

1.3 Step 3: Identification of a limiting process

We are here interested in the characterization of a process $(\nu _t,R_t)_{t\in [0,T]}$ whose law is a limiting value of the sequence $\{\mathcal {Z}^K,K>1\}$. Given the operator defined in (2.9), we have the following result

Proposition A.2

Let us denote by $\mathcal {Z}$ a limiting value of the sequence of distributions $\left\{ \mathcal {Z}^K,K>1\right\}$. Then a process $(\nu ,R)$ of law $\mathcal {Z}$ satisfies (2.16).

Proof

Let $f\in \mathcal {C}_c^2(\mathbb {R}_+,\mathbb {R})$ be a test function and $t\in [0,T]$ a fixed time. We introduce the following real-valued functions defined for $(\mu ,r)$ in the subset of $\mathcal {C}([0,T],(\mathcal {M}_F(\mathbb {R}_+)\times \mathbb {R},w\otimes |.|))$ constitued of processes that satisfy $|r| \le \bar{R}$

$$\begin{aligned} \displaystyle \Phi ^f_t(\mu ,r)= & {} \left\langle \mu _t,f\right\rangle - \left\langle \mu _0,f\right\rangle \\{} & {} - \int _0^t\int _0^{\infty }\left\{ \zeta (x,r_s)f'(x) + D(x,r_s)f''(x) \right\} \mu _s(dx)ds\\{} & {} \displaystyle -\int _0^t\int _0^{\infty }\big \{ b(x,r_s)\mathcal {G}[f](x) - d(x)f(x) \big \}\mu _s(dx)ds\\ \displaystyle \Phi _{t}(\mu ,r)= & {} r_t - r_0 - \int _0^t\left\{ r_{in}-r_s - \int _0^{\infty }\chi (x,r_s)\mu _s(dx)\right\} ds. \end{aligned}$$

We are aiming to show that

$$\begin{aligned} \mathbb {E}\left[ \left| \Phi ^f_t(\nu ,R)\right| ^2 + \left| \Phi _{t}(\nu ,R)\right| \right] = 0. \end{aligned}$$

(A.5)

For this purpose, we first note that the function $\Phi _{t}$ defined as above is continuous and

$$\begin{aligned} \Phi _{t}\left( \nu ^K,R^K\right) = 0. \end{aligned}$$

It then follows by the Fatou lemma that

$$\begin{aligned} \mathbb {E}\left| \Phi _{t}\left( \nu ,R\right) \right| \le \liminf _{K\rightarrow \infty }\mathbb {E}\left| \Phi _{t}\left( \nu ^K,R^K\right) \right| = 0. \end{aligned}$$

Futhermore, the terms in the integrals in the expression of the function $\Phi ^f_t$ are continuous and bounded, then this function is continuous. In addition, it follows from the decomposition (2.12) that

$$\begin{aligned} \Phi ^f_t\left( \nu ^K,R^K\right) = M^{K,f}_t \end{aligned}$$

and then thanks to the Fatou lemma and Doob’s inequality,

$$\begin{aligned} \displaystyle \mathbb {E}\left[ \left( \Phi ^f_t\left( \nu ,R\right) \right) ^2 \right]\le & {} \liminf _{K\rightarrow \infty }\mathbb {E}\left[ \left( \Phi ^f_t\left( \nu ^K,R^K\right) \right) ^2 \right] \\ \displaystyle\le & {} \liminf _{K\rightarrow \infty }\mathbb {E}\left[ \sup _{0\le s\le t}\left| M^{K,f}_s\right| ^2 \right] \le C \liminf _{K\rightarrow \infty }\mathbb {E}\left[ \left\langle M^{K,f}\right\rangle _t\right] . \end{aligned}$$

We now deduce from (2.15) that

$$\begin{aligned} \mathbb {E}\left[ \left( \Phi ^f_t\left( \nu ,R\right) \right) ^2 \right] \le \liminf _{K\rightarrow \infty } \frac{C}{K}\mathbb {E}\left[ \sup _{0\le t\le T}\left\langle \nu ^K_t,1\right\rangle \right] = 0. \end{aligned}$$

(A.6)

$\square$

Some properties of a specific SDE

Let us introduce two continuous functions $h(x,s),q(x,s)\in \mathcal {C}(\mathbb {R}_+\times [0,T],\mathbb {R})$ that satisfy the following assumption.

Assumption B.1

• (A.1) Boundary condition: for all $s\le [0, T]$,

  $$\begin{aligned} q(0,s) = 0 \text { and }h(0,s) \ge 0. \end{aligned}$$

• (A.2) Lipschitz condition: for all $x,y\ge 0$ and $s\in [0,T]$,

  $$\begin{aligned} |h(x,s) - h(y,s)| + |q(x,s) - q(y,s)| \le C|x-y|. \end{aligned}$$

• (A.3) The function q(x, s) is non negative.

Then we have the following results

Lemma B.1

There is weak existence and uniqueness, and strong existence for the stochastic differential equation

$$\begin{aligned} dZ_u = h(Z_u,u)du + \sqrt{q(Z_u,u)}dW_u, \forall u\le T. \end{aligned}$$

(B.1)

The solution $(Z^x_{s,u})_{u\in [s,T]}$ that starts at time $s<T$ at the position $x\ge 0$, has a zero local time at 0, is almost surely non negative at any time and satifies the moment estimate

$$\begin{aligned} \mathbb {E}\left\{ \sup _{s\le u\le T}\left( Z^x_{s,u}\right) ^p \right\} \le C_{p,T}(1+x^p),\, \forall p>0. \end{aligned}$$

(B.2)

In addition, for all $x_1,x_2\ge 0$ and $s_1,s_2\le T$, we have for all $u\le T$

$$\begin{aligned}{} & {} \mathbb {E}\left( \left| Z^{x_1}_{s_1,u} - Z^{x_2}_{s_2,u} \right| \right) \le C_{T}\left\{ |x_1-x_2| + (1+x_1+x_2)|s_1-s_2|\right. \nonumber \\{} & {} \quad \left. + |1+x_1+x_2|^{1/2}|s_1-s_2|^{1/2}\right\} , \end{aligned}$$

(B.3)

where $(Z^{x_1}_{s_1,u})_{u\in [s_1,T]}, (Z^{x_2}_{s_2,u})_{u\in [s_2,T]}$ are assumed to be built with the same Brownian motion, and satisfies $Z^{x_j}_{s_j,u} = x_j$ for all $u\le s_j$ with $j=1,2$.

Lemma B.2

Let $p(x)\in \mathcal {C}^1_b(\mathbb {R}_+,\mathbb {R})$ and $a(x,s)\in \mathcal {C}_b(\mathbb {R}_+\times [0,T],\mathbb {R})$ such that

$$\begin{aligned} |a(x,s)-a(y,s)|\le \bar{a}|x-y|, \, \forall x,y\ge 0, \forall s\le T, \end{aligned}$$

then for a given $t\le T$, the function defined by

$$\begin{aligned} w_s(x) = \mathbb {E}\left[ p(Z^x_{s,t})e^{\int _s^ta(Z^x_{s,u},u)du} \right] ,\, \forall (s,x)\in [0,t]\times \mathbb {R}_+ \end{aligned}$$

(B.4)

satisfy for all $s_1,s_2\le t$ and $x_1,x_2\ge 0$ the following inequality

$$\begin{aligned}{} & {} \displaystyle \left| w_{s_1}(x_1)-w_{s_2}(x_2) \right| \le C_T \left( \Vert p\Vert _{\infty } + \Vert p'\Vert _{\infty }\right) \nonumber \\{} & {} \quad \bigg \{ |x_1-x_2| + (1+x_1)^{1/2}|s_1-s_2|^{1/2}\displaystyle + (1+x_1)|s_1-s_2| \bigg \}. \end{aligned}$$

(B.5)

1.1 Proof of Lemma B.1

We split this proof into the several steps.

Existence and uniqueness According to [20], Theorem 1.1 it suffices to show that we have weak existence and pathwise uniqueness to conclude that there is weak existence and uniqueness, and strong existence. In addition it follows from the previous paragraph that this solution will be well defined at any time. The stochastic differential equation (B.1) is non homogeneous with continuous coefficients, then the coupled process $(Z_u,u)_{u\ge 0}$ satisfies an homogeneous stochastic differential equation with continuous coefficients. We then have weak existence for each initial condition $x\ge 0$ and initial time $s\in [0,T]$ thanks to [20], Theorem 2.3, p173. Furthermore, for $x,y\ge 0$ and $s\le u\le T$ we have by hypothesis

$$\begin{aligned} \big |h(x,u) - h(y,u)\big | + \big |\sqrt{q(x,u)} - \sqrt{q(y,u)}\big |^2 \le C |x-y|, \end{aligned}$$

thanks to the inequality $|\sqrt{a}-\sqrt{b}|\le \sqrt{|a-b|}\,,\, \forall a,b\ge 0$. We deduce from ( [35]) that there is pathwise uniqueness.

Moment estimate and positiveness Notice that the solution that we constructed in the previous paragraph is well defined until it becomes negative or explodes. In order to get well defined terms for all times, we introduce the new equation

$$\begin{aligned} d\tilde{Z}_u = h(|\tilde{Z}_u|,u)du + \sqrt{q(|\tilde{Z}_u|,u)}\,dW_u , \forall u\le T \end{aligned}$$

(B.6)

for which there is weak existence and pathwise uniqueness by an argument similar to the one in the previous paragraph, but only until explosion time. It follows from the continuity of the function h(x, s) and hypothesis (A.2) of Assumption B.1 that

$$\begin{aligned} |h(x,s)| \le C(1+x), \forall x\ge 0,\forall s\le T. \end{aligned}$$

For given $x\ge 0$ and $s\le t$, we consider the sequence of stopping times defined by

$$\begin{aligned} \lambda _N = \inf \{ v\ge s: |\tilde{Z}^x_{s,v}|> N \}\wedge T,\, N>1 \end{aligned}$$

that converges increasingly. Then for $p\ge 2$ and $v\in [s,T]$ we have

$$\begin{aligned} |\tilde{Z}^x_{s,v\wedge \lambda _N}|^p \le C_p\left\{ x^p + \int _s^{v\wedge \lambda _N}|h(|\tilde{Z}^x_{s,u}|,u)|^pdu + \left| \int _s^{v\wedge \lambda _N}\sqrt{q(|\tilde{Z}^x_{s,u}|,u)}\,dW_u\right| ^p \right\} \end{aligned}$$

that implies that

$$\begin{aligned} \displaystyle \sup _{s\le u\le v\wedge \lambda _N}|\tilde{Z}^x_{s,u}|^p\le & {} C_p\bigg \{x^p + \int _s^{v\wedge \lambda _N}|h(|\tilde{Z}^x_{s,u}|,u)|du \\{} & {} + \sup _{s\le u\le v}\bigg (\bigg |\int _s^{u\wedge \lambda _N}\sqrt{q(|\tilde{Z}^x_{s,u}|,u)}\,dW_u\bigg |^p\bigg ) \bigg \} \\ \displaystyle\le & {} C_p\bigg \{x^p + \int _s^{v\wedge \lambda _N}\left[ 1+|\tilde{Z}^x_{s,u}|^p\right] du \\{} & {} + \sup _{s\le u\le v}\bigg (\bigg |\int _s^{u\wedge \lambda _N}\sqrt{q(|\tilde{Z}^x_{s,u}|,u)}\,dW_u\bigg |^p\bigg ) \bigg \}. \end{aligned}$$

It follows from the Doob inequality that

$$\begin{aligned}{} & {} \displaystyle \mathbb {E}\left\{ \sup _{s\le u\le v}\left( \left| \int _s^{u\wedge \lambda _N}\sqrt{q(|\tilde{Z}^x_{s,u}|,u)}\,dW_u\right| ^p\right) \right\} \\{} & {} \quad \le C_p\mathbb {E}\left\{ \left( \int _s^{v\wedge \lambda _N}q(|\tilde{Z}^x_{s,u}|,u)du \right) ^{p/2} \right\} \\{} & {} \quad \displaystyle \le C_p\mathbb {E}\left\{ \left( \int _s^{v\wedge \lambda _N}|\tilde{Z}^x_{s,u}|du \right) ^{p/2} \right\} \\{} & {} \quad \displaystyle \le C_p\mathbb {E}\left\{ \left( \int _s^{v}|\tilde{Z}^x_{s,u\wedge \lambda _N}|du \right) ^{p/2} \right\} \\{} & {} \quad \displaystyle \le C_p\mathbb {E}\left\{ \int _s^{v}|\tilde{Z}^x_{s,u\wedge \lambda _N}|^{p/2}du \right\} \\{} & {} \quad \displaystyle \le C_p\mathbb {E}\left\{ \int _s^{v}\left[ 1+|\tilde{Z}^x_{s,u\wedge \lambda _N}|^{p}\right] du \right\} \end{aligned}$$

and then

$$\begin{aligned} \displaystyle \mathbb {E}\left[ \sup _{s\le u\le v\wedge \lambda _N}|\tilde{Z}^x_{s,u}|^p\right]\le & {} C_p\left\{ x^p + \mathbb {E}\left( \int _s^{v}\left[ 1+|\tilde{Z}^x_{s,u\wedge \lambda _N}|^p\right] du\right) \right\} \\ \displaystyle\le & {} C_p\left\{ x^p + \int _s^{v}\left[ 1+\mathbb {E}\left\{ \sup _{s\le z\le u\wedge \lambda _N}|\tilde{Z}^x_{s,z}|^p\right\} \right] du \right\} . \end{aligned}$$

Hence, we obtain for all $v\in [s,T]$

$$\begin{aligned}{} & {} 1+\mathbb {E}\left[ \sup _{s\le u\le v\wedge \lambda _N}|\tilde{Z}^x_{s,u}|^p\right] \le C_p\left\{ 1+x^p \right. \\{} & {} \quad \left. + \int _s^{v}\left[ 1+\mathbb {E}\left\{ \sup _{s\le z\le u\wedge \lambda _N}|\tilde{Z}^x_{s,z\wedge \lambda _N}|^p\right\} \right] du \right\} , \end{aligned}$$

and thanks to the Gronwall lemma,

$$\begin{aligned} \mathbb {E}\left[ \sup _{s\le u\le v\wedge \lambda _N}|\tilde{Z}^x_{s,u}|^p\right] \le C_p\left( 1+x^p\right) e^{C_p(v-s)}. \end{aligned}$$

If we assume that there exists $T_{0}<T$ such that $\mathbb {P}\left\{ \lim _N\lambda _N< T_0 \right\} > 0$, then for all $v\in [T_0,T]$, we will have

$$\begin{aligned} N^p\,\mathbb {P}\left\{ \lim _N\lambda _N< T_0 \right\} \le \mathbb {E}\left[ \sup _{s\le u\le v\wedge \lambda _N}|\tilde{Z}^x_{s,u}|^p\right] \le C_p\left( 1+x^p\right) e^{C_p(v-s)}. \end{aligned}$$

That is impossible because the left side term tends towards infinity as $N\rightarrow \infty$. It follows that $\lambda _N\xrightarrow []{a.s} T$ and by the monotone convergence theorem,

$$\begin{aligned} \mathbb {E}\left[ \sup _{s\le u\le v}|\tilde{Z}^x_{s,u}|^p\right] \le C_p\left( 1+x^p\right) e^{C_p(v-s)}\,,\, \forall v\in [s,T]. \end{aligned}$$

It then suffices to take $v = T$. The case $0< p< 2$ is obviously solved thanks to the Hölder inequality

$$\begin{aligned} \mathbb {E}\left[ \sup _{s\le u\le T}|\tilde{Z}^x_{s,u}|^p\right] \le \left( \mathbb {E}\left[ \sup _{s\le u\le T}|\tilde{Z}^x_{s,u}|^2\right] \right) ^{p/2} \le C_{p,T}(1+x^p). \end{aligned}$$

The process $(\tilde{Z}^x_{s,u})_{u\in [s,T]}$ is then well defined at any time and satisfies a moment estimate similar to (B.2). We will now show that this process is almost surely non negative, and then by uniqueness this will imply that it is a modification of the process $(Z^x_{s,u})_{u\in [s,T]}$. Indeed, the following decomposition holds

$$\begin{aligned} \tilde{Z}_{s,t}^x = (\tilde{Z}^x_{s,t})_+ - (\tilde{Z}^x_{s,t})_- , \forall t\in [s,T] \end{aligned}$$

(B.7)

and it follows from the Tanaka formula that

$$\begin{aligned} (\tilde{Z}^x_{s,t})_- = -\int _s^t1_{\{\tilde{Z}^x_{s,t}\le 0\}}d\tilde{Z}^x_{s,t} + \frac{1}{2}L^0_t(\tilde{Z}^x_{s,\cdot }). \end{aligned}$$

(B.8)

For $\varepsilon >0$,

$$\begin{aligned} \int _s^t 1_{\{0< \tilde{Z}^x_{s,u}\le \varepsilon \}}\frac{d\langle \tilde{Z}^x_{s,\cdot }\rangle _u}{\tilde{Z}^x_{s,u}} = \int _s^t1_{\{0< \tilde{Z}^x_{s,u}\le \varepsilon \}}\frac{q(\tilde{Z}^x_{s,u},u)}{\tilde{Z}^x_{s,u}}du < \infty \text { p.s,} \end{aligned}$$

since $(x,u)\mapsto q(x,u)/x$ is bounded. Then the local time $L^0_{t}(\tilde{Z}^x_{s,\cdot })$ is zero almost surely (see [28], Ch. IX, Lemma 3.3). Equation (B.8) becomes

$$\begin{aligned} (\tilde{Z}^x_{s,t})_- = -\int _s^t1_{\{\tilde{Z}^x_{s,u}\le 0\}}h(-\tilde{Z}^x_{s,u},u)du - \int _s^t1_{\{\tilde{Z}^x_{s,u}\le 0\}}\sqrt{q(-\tilde{Z}^x_{s,u},u)}\,dW_u \end{aligned}$$

where the second term in the right side of the equality is a true martingale thanks to the moment estimate that we previously stated. It then follows from the hypothesis (A.1) and (A.2) in Assumption B.1 that

$$\begin{aligned} \displaystyle \mathbb {E}\left[ (\tilde{Z}^x_{s,t})_- \right]= & {} -\mathbb {E}\left( \int _s^t1_{\{\tilde{Z}^x_{s,u}\le 0\}}h(-\tilde{Z}^x_{s,u},u)du\right) \\ \displaystyle\le & {} -\mathbb {E}\left( \int _s^t1_{\{\tilde{Z}^x_{s,u}\le 0\}}\left[ h(-\tilde{Z}^x_{s,u},u) - h(0,u)\right] du\right) \\ \displaystyle\le & {} C\mathbb {E}\left( \int _s^t1_{\{\tilde{Z}^x_{s,u}\le 0\}}\left| \tilde{Z}^x_{s,u}\right| du\right) \\ \displaystyle\le & {} C \int _s^t\mathbb {E}\left[ ( \tilde{Z}^x_{s,u})_-\right] du \end{aligned}$$

and hence by the Gronwall lemma, $\mathbb {E}\left[ (\tilde{Z}^x_{s,t})_- \right] = 0$ for any $t\in [s,T]$, thus $(\tilde{Z}^x_{s,t})_- = 0$ p.s. We deduce that the stochastic process $(\tilde{Z}^x_{s,t})_{t\in [s,T]}$ is almost surely non negative at any time, and it is possible to remove the absolute values in (B.6) to conclude by a uniqueness argument that

$$\begin{aligned} \forall t\in [s,T],\, \tilde{Z}^x_{s,t} = Z^x_{s,t} \text { p.s.} \end{aligned}$$

The stochastic process $(Z^x_{s,t})_{t\in [s,T]}$ is then almost surely non negative at any time, satisfies the moment estimate (B.2) and has a zero local time at 0.

Dependence on initial conditions Thanks to the strong existence, it is possible to construct two solutions $(Z^{x_1}_{s_1,v})_{v\in [s_1,T]}$ and $(Z^{x_2}_{s_2,v})_{v\in [s_2,T]}$ of (B.1) on the same filtered probability space and with the same Brownian motion, for all $x_1,x_2\ge 0$ and $0\le s_1\le s_2\le T$. Those solutions are well extended on [0, T] if we set

$$\begin{aligned} Z^{x_j}_{s_j,v} = x_j, \forall v\in [0,s_j], j=1,2. \end{aligned}$$

Hence, it follows from the Tanaka formula that for all $s_2\le v\le T$,

$$\begin{aligned} \left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right|= & {} \left| Z^{x_1}_{s_1,s_2} - x_2 \right| + \int _{s_2}^v\text {sign}(Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u})d(Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u})\nonumber \\{} & {} + l^0_v(Z^{x_1}_{s_1,\cdot }-Z^{x_2}_{s_2,\cdot }) \end{aligned}$$

(B.9)

where $l^0_v(Z^{x_1}_{s_1,\cdot }-Z^{x_2}_{s_2,\cdot })$ is the local time of the process $Z^{x_1}_{s_1,\cdot }-Z^{x_2}_{s_2,\cdot }$ at 0 up to time v. Furthermore, thanks to the inequality $|\sqrt{a}-\sqrt{b}|\le \sqrt{|a-b|}\,,\forall a,b\ge 0$ again, we have for $\varepsilon >0$,

$$\begin{aligned}{} & {} \displaystyle \int _{s_2}^v 1_{\{0< Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u}\le \varepsilon \}}\frac{d\left\langle Z^{x_1}_{s_1,\cdot }-Z^{x_2}_{s_2,\cdot }\right\rangle _u}{Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u}}\\{} & {} \quad = \int _{s_2}^v 1_{\{0< Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u}\le \varepsilon \}}\frac{\left| \sqrt{q(Z^{x_1}_{s_1,u},u)} - \sqrt{q(Z^{x_2}_{s_2,u},u)}\right| ^2}{Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u}}du\\{} & {} \quad \displaystyle \le C\int _{s_2}^v 1_{\{0< Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u}\le \varepsilon \}}du < \infty . \end{aligned}$$

We deduce thanks to [28], Ch. IX, lemma 3.3 that the local time $l^0_{v}(Z^{x_1}_{s_1,\cdot }-Z^{x_2}_{s_2,\cdot })$ is zero almost surely for all $v\in [s_2,T]$, and it follows from (B.9) that

$$\begin{aligned} \displaystyle \mathbb {E}\left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right|\le & {} \mathbb {E}\left( \left| Z^{x_1}_{s_1,s_2}-x_2\right| \right) \\{} & {} + \mathbb {E}\int _{s_2}^v\text {sign}(Z^{x_1}_{s_1,u}-Z^{x_2}_{s_2,u})\left[ h(Z^{x_1}_{s_1,u},u)-h(Z^{x_2}_{s_2,u},u) \right] du \\ \displaystyle\le & {} \mathbb {E}\left( \left| Z^{x_1}_{s_1,s_2}-x_2\right| \right) + C\int _{s_2}^v\mathbb {E}\left( \left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right| \right) du \end{aligned}$$

which implies by the Gronwall lemma that

$$\begin{aligned} \mathbb {E}\left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right| \le \mathbb {E}\left( \left| Z^{x_1}_{s_1,s_2}-x_2\right| \right) e^{C(v-s_2)}\, , \, \forall v\in [s_2,T]. \end{aligned}$$

(B.10)

We are now interested in what happens for $s_1\le v\le s_2$. Indeed,

$$\begin{aligned}{} & {} \displaystyle \mathbb {E}\left( \left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right| \right) = \mathbb {E}\left\{ \left| x_1 + \int _{s_1}^vh(Z^{x_1}_{s_1,u},u)du\right. \right. \\{} & {} \qquad \left. \left. + \int _{s_1}^v\sqrt{q(Z^{x_1}_{s_1,u},u)}\, dW_u - x_2 \right| \right\} \\{} & {} \quad \displaystyle \le \mathbb {E}\left\{ |x_1-x_2| + \int _{s_1}^v|h(Z^{x_1}_{s_1,u},u)|du + \left| \int _{s_1}^v\sqrt{q(Z^{x_1}_{s_1,u},u)}\, dW_u \right| \right\} \\{} & {} \quad \displaystyle \le |x_1-x_2| + C\int _{s_1}^{s_2}\left[ 1+\mathbb {E}(Z^{x_1}_{s_1,u})\right] du\\{} & {} \qquad + \mathbb {E}\left\{ \sup _{s_1\le v\le s_2}\left| \int _{s_1}^v\sqrt{q(Z^{x_1}_{s_1,u},u)}\, dW_u \right| \right\} \end{aligned}$$

and thanks to the Doob inequality,

$$\begin{aligned} \displaystyle \mathbb {E}\left\{ \sup _{s_1\le v\le s_2}\left| \int _{s_1}^v\sqrt{q(Z^{x_1}_{s_1,u},u)}\, dW_u \right| \right\}\le & {} \mathbb {E}\left\{ \left| \int _{s_1}^{s_2}q(Z^{x_1}_{s_1,u},u)\, du \right| ^{1/2} \right\} \\ \displaystyle\le & {} C\sqrt{\mathbb {E}\left( \int _{s_1}^{s_2}Z^{x_1}_{s_1,u}du\right) }\\ \displaystyle\le & {} C\sqrt{|s_1-s_2|\mathbb {E}\left( \sup _{s_1\le u\le T}Z^{x_1}_{s_1,u}\right) }. \end{aligned}$$

Hence, we have for all $v\in [s_1,s_2]$,

$$\begin{aligned}{} & {} \mathbb {E}\left( \left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right| \right) \le |x_1-x_2| + C\int _{s_1}^{s_2}\left[ 1+\mathbb {E}(Z^{x_1}_{s_1,u})\right] du\\{} & {} \quad + C\sqrt{|s_1-s_2|\mathbb {E}\left( \sup _{s_1\le u\le T}Z^{x_1}_{s_1,u}\right) } \end{aligned}$$

which implies thanks to (B.2) that

$$\begin{aligned} \mathbb {E}\left( \left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right| \right)\le & {} |x_1-x_2| + C_1(1+x_1)|s_1-s_2| \nonumber \\{} & {} + C_2|1+x_1|^{1/2}|s_1-s_2|^{1/2}, \forall v\in [s_1,s_2]. \end{aligned}$$

(B.11)

Notice that this inequality gives (B.3) for $v\in [s_1,s_2]$. Taking in particular $v=s_2$ and considering (B.10), we deduce that (B.3) also holds for $v\in [s_2,T]$. We finally obtain without any order on $s_1,s_2\in [0,T], x_1,x_2\ge 0$ that for all $v\in [s_1\wedge s_2, T]$

$$\begin{aligned} \mathbb {E}\left( \left| Z^{x_1}_{s_1,v}-Z^{x_2}_{s_2,v}\right| \right)\le & {} C_{T}\left\{ |x_1-x_2| + (1+x_1+x_2)|s_1-s_2| \right. \\{} & {} \left. + |1+x_1+x_2|^{1/2}|s_1-s_2|^{1/2}\right\} \end{aligned}$$

that is easily extended for $v\in [0,T]$.

1.2 Proof of Lemma B.2

Let us first introduce the decomposition

$$\begin{aligned} \left| w_{s_1}(x_1)-w_{s_2}(x_2) \right| \le |w_{s_1}(x_1)-w_{s_2}(x_1)| + |w_{s_2}(x_1)-w_{s_2}(x_2)| \end{aligned}$$

(B.12)

thanks to the triangular inequality. Furthermore, by considering that $s_1\le s_2$, we have

$$\begin{aligned}{} & {} \displaystyle \left| w_{s_1}(x_1) - w_{s_2}(x_1) \right| \le \mathbb {E}\left( \left| p(Z^{x_1}_{s_1,t})e^{\int _{s_1}^ta(Z^{x_1}_{s_1,u},u)du} - p(Z^{x_1}_{s_2,t})e^{\int _{s_2}^ta(Z^{x_1}_{s_2,u},u)du} \right| \right) \\{} & {} \quad \displaystyle \le \mathbb {E}\left( \left| p(Z^{x_1}_{s_1,t})-p(Z^{x_1}_{s_2,t})\right| e^{\int _{s_1}^ta(Z^{x_1}_{s_1,u},u)du}\right) \\{} & {} \qquad \displaystyle + \mathbb {E}\left( |p(Z^{x_1}_{s_2,t})|\left| e^{\int _{s_1}^ta(Z^{x_1}_{s_1,u},u)du} - e^{\int _{s_2}^ta(Z^{x_1}_{s_2,u},u)du}\right| \right) \\{} & {} \quad \displaystyle \le e^{\Vert a\Vert _{\infty }(t-s_1)}\Vert p'\Vert _{\infty }\mathbb {E}\left( \left| Z^{x_1}_{s_1,t}-Z^{x_1}_{s_2,t}\right| \right) \\{} & {} \qquad \displaystyle + \Vert p\Vert _{\infty }\left\{ \mathbb {E}\left( \left| e^{\int _{s_1}^ta(Z^{x_1}_{s_1,u},u)du} - e^{\int _{s_1}^ta(Z^{x_1}_{s_2,u},u)du}\right| \right) \right. \\{} & {} \qquad \displaystyle + \left. \mathbb {E}\left( \left| e^{\int _{s_1}^ta(Z^{x_1}_{s_2,u},u)du} - e^{\int _{s_2}^ta(Z^{x_1}_{s_2,u},u)du}\right| \right) \right\} \\{} & {} \quad \displaystyle \le e^{\Vert a\Vert _{\infty }(t-s_1)}\bigg [\Vert p'\Vert _{\infty }\mathbb {E}\left( \left| Z^{x_1}_{s_1,t}-Z^{x_1}_{s_2,t}\right| \right) \\{} & {} \qquad \displaystyle + \Vert p\Vert _{\infty }\bigg \{\mathbb {E}\left( \left| e^{\int _{s_1}^t\left[ a(Z^{x_1}_{s_1,u},u)-a(Z^{x_1}_{s_2,u},u) \right] du} - 1\right| \right) \\{} & {} \qquad + \mathbb {E}\left( \left| e^{-\int _{s_1}^{s_2}a(Z^{x_1}_{s_2,u},u)du} - 1\right| \right) \bigg \}\bigg ]. \end{aligned}$$

Let us set for all $s_1\le v\le t$

$$\begin{aligned} V^{x_1}_{s_1,s_2}(v) = e^{\int _{s_1}^v\left[ a(Z^{x_1}_{s_1,u},u) - a(Z^{x_1}_{s_2,u},u)\right] du} - 1, \end{aligned}$$

then

$$\begin{aligned} V^{x_1}_{s_1,s_2}(v)= & {} \int _{s_1}^v\left[ a(Z^{x_1}_{s_1,u},u) - a(Z^{x_1}_{s_2,u},u)\right] V^{x}_{s_1,s_2}(u)du \\{} & {} + \int _{s_1}^v\left[ a(Z^{x_1}_{s_1,u},u) - a(Z^{x_1}_{s_2,u},u)\right] du \end{aligned}$$

that implies that,

$$\begin{aligned} \left| V^{x_1}_{s_1,s_2}(v)\right| \le 2\Vert a\Vert _{\infty }\int _{s_1}^v\left| V^{x}_{s_1,s_2}(u)\right| du+ \bar{a}\int _{s_1}^t\left| Z^{x_1}_{s_1,u} - Z^{x_1}_{s_2,u} \right| du \end{aligned}$$

and thanks to the Gronwall lemma,

$$\begin{aligned} \left| V^{x_1}_{s_1,s_2}(v)\right| \le \bar{a} \, e^{2\Vert a\Vert _{\infty }(v-s_1)} \int _{s_1}^t\left| Z^{x_1}_{s_1,u} - Z^{x_1}_{s_2,u} \right| du \,,\, \forall v\in [s_1,t]. \end{aligned}$$

We show by a similar argument that

$$\begin{aligned} \left| e^{-\int _{s_1}^{s_2}a(Y^x_{s_2,u},u)du} - 1\right| \le e^{\Vert a\Vert _{\infty }|s_1-s_2|}\Vert a\Vert _{\infty }|s_1-s_2|. \end{aligned}$$

Hence, one can write for all $x_1\ge 0, s_1\le s_2 \le t$

$$\begin{aligned}{} & {} \displaystyle \left| w_{s_1}(x_1) - w_{s_2}(x_1) \right| \le e^{\Vert a\Vert _{\infty }(t-s_1)}\bigg \{ \Vert p'\Vert _{\infty }\mathbb {E}\left( \left| Z^{x_1}_{s_1,t}-Z^{x_1}_{s_2,t}\right| \right) \\{} & {} \quad \displaystyle +\,\Vert p\Vert _{\infty }\left[ \bar{a}\, e^{2\Vert a\Vert _{\infty }(t-s_1)} \int _{s_1}^t\mathbb {E}\left( \left| Z^{x_1}_{s_1,u} - Z^{x_1}_{s_2,u} \right| \right) du\right. \\{} & {} \quad \left. + e^{\Vert a\Vert _{\infty }|s_1-s_2|}\Vert a\Vert _{\infty }|s_1-s_2|\right] \bigg \} \end{aligned}$$

and thanks to Lemma B.1, we have for all $s_1,s_2\le t$

$$\begin{aligned}{} & {} \left| w_{s_1}(x_1) - w_{s_2}(x_1) \right| \le C_T \left( \Vert p\Vert _{\infty } + \Vert p'\Vert _{\infty }\right) \nonumber \\{} & {} \qquad \left\{ (1+x_1)^{1/2}|s_1-s_2|^{1/2} + (1+x_1)|s_1-s_2| \right\} . \end{aligned}$$

(B.13)

In addition, for $x_1,x_2\ge 0$ and $s_2\le t$,

$$\begin{aligned}{} & {} \displaystyle \left| w_{s_2}(x_1) - w_{s_2}(x_2) \right| \le \mathbb {E}\left( \left| p(Z^{x_1}_{s_2,t})e^{\int _{s_2}^ta(Z^{x_1}_{s_2,u},u)du} - p(Z^{x_2}_{s_2,t})e^{\int _{s_2}^ta(Z^{x_2}_{s_2,u},u)du} \right| \right) \\{} & {} \quad \displaystyle \le \mathbb {E}\left( \left| p(Z^{x_1}_{s_2,t})-p(Z^{x_2}_{s_2,t})\right| e^{\int _{s_2}^ta(Z^{x_1}_{s_2,u},u)du}\right) \\{} & {} \qquad \displaystyle + \mathbb {E}\left( |p(Z^{x_2}_{s_2,t})|e^{\int _{s_2}^ta(Z^{x_2}_{s_2,u},u)du}\left| e^{\int _{s_2}^t\left[ a(Z^{x_1}_{s_2,u},u) - a(Z^{x_2}_{s_2,u},u)\right] du} - 1\right| \right) \\{} & {} \quad \displaystyle \le e^{\Vert a\Vert _{\infty }(t-s_2)}\bigg \{\Vert p'\Vert _{\infty }\mathbb {E}\left( \left| Z^{x_1}_{s_2,t}-Z^{x_2}_{s_2,t}\right| \right) \\{} & {} \qquad \displaystyle + \Vert p\Vert _{\infty }\mathbb {E}\left( \left| e^{\int _{s_2}^t\left[ a(Z^{x_1}_{s_2,u},u) - a(Z^{x_2}_{s_2,u},u)\right] du} - 1\right| \right) \bigg \}. \end{aligned}$$

Let us now set for all $s_2\le v\le t$

$$\begin{aligned} U^{x_1,x_2}_{s_2}(v) = e^{\int _{s_2}^v\left[ a(Z^{x_1}_{s_2,u},u) - a(Z^{x_2}_{s_2,u},u)\right] du} - 1, \end{aligned}$$

then in the same way as for $V^{x_1}_{s_1,s_2}(v)$ we obtain

$$\begin{aligned} \left| U^{x_1,x_2}_{s_2}(v)\right| \le \bar{a}\, e^{2\Vert a\Vert _{\infty }(v-s_2)} \int _{s_2}^t\left| Z^{x_1}_{s_2,u} - Z^{x_2}_{s_2,u} \right| du \,,\, \forall v\in [s_2,t]. \end{aligned}$$

Hence for all $x_1,x_2\ge 0, s_2\le t$ we have

$$\begin{aligned} \left| w_{s_2}(x_1) - w_{s_2}(x_2) \right|\le & {} e^{\Vert a\Vert _{\infty }(t-s_2)}\bigg \{\Vert p'\Vert _{\infty }\mathbb {E}\left( \left| Z^{x_1}_{s_2,t}-Z^{x_2}_{s_2,t}\right| \right) \\{} & {} \displaystyle + \bar{a}\, e^{2\Vert a\Vert _{\infty }(t-s)} \Vert p\Vert _{\infty }\int _{s_2}^t\mathbb {E}\left( \left| Z^{x_1}_{s_2,u} - Z^{x_2}_{s_2,u} \right| \right) du \bigg \} \end{aligned}$$

and thanks to Lemma B.1,

$$\begin{aligned} \left| w_{s_2}(x_1) - w_{s_2}(x_2) \right| \le C_T\left( \Vert p\Vert _{\infty } + \Vert p'\Vert _{\infty }\right) |x_1-x_2|, \forall x_1,x_2\ge 0, \forall s_2\le t. \end{aligned}$$

(B.14)

Finally, the inequality (B.5) follows from (B.12), (B.13) and (B.14).

Difference operator and Besov spaces $B^s_{1,\infty }(\mathbb {R})$

For $\alpha \in \, ]0,1[$, we denote by $\mathscr {C}^{\alpha }_b(\mathbb {R})$ the Hölder-Zygmund space that is the set of real valued functions on $\mathbb {R}$ such that

$$\begin{aligned} \Vert f\Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})} = \Vert f\Vert _{\infty } + \sup _{x,y\in \mathbb {R},x\ne y}\frac{|f(x)-f(y)|}{|x-y|^{\alpha }} < \infty . \end{aligned}$$

We also introduce the difference operator defined for all $f:\mathbb {R}\rightarrow \mathbb {R}$, for given $m\ge 1$ integer and $h\in \mathbb {R}$ as

$$\begin{aligned} \forall x\in \mathbb {R}, \, \Delta ^1_hf(x) = f(x+h) - f(x)\,;\, \Delta ^m_hf(x) = \Delta ^1_h(\Delta ^{m-1}_hf)(x), \text { pour }m>1. \end{aligned}$$

that is, by a recurrence argument,

$$\begin{aligned} \Delta ^m_hf(x) = \sum _{j=0}^m(-1)^{m-j} \begin{pmatrix} m \\ j \end{pmatrix}f(x+jh) \end{aligned}$$

(C.1)

with the following properties.

Lemma C.1

Let $m\ge 1$ be an integer, $\alpha \in \, ]0,1[$, $h\in \mathbb {R}$ and $f: \mathbb {R}\rightarrow \mathbb {R}$, then:

• 1. If $f\in \mathscr {C}^{\alpha }_b(\mathbb {R})$,

  $$\begin{aligned} \Vert \Delta ^m_hf\Vert _{\infty } \le C_m |h|^{\alpha }\Vert f\Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})}. \end{aligned}$$

  (C.2)

• 2. If $f\in \mathscr {C}^{\alpha }_b(\mathbb {R})$,

  $$\begin{aligned} |\Delta ^m_hf(x) - \Delta ^m_hf(y)| \le C_m \Vert f\Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})} |x-y|^{\alpha } , \forall x,y\in \mathbb {R}. \end{aligned}$$

  (C.3)

• 3. If $f\in \mathcal {C}^m(\mathbb {R},\mathbb {R})$,

  $$\begin{aligned} \Vert \Delta ^m_hf\Vert _{\mathbb {L}^1} \le C_m |h|^m\Vert \partial _x^mf\Vert _{\mathbb {L}^1} \end{aligned}$$

  (C.4)

• 4. If $f\in \mathcal {C}_b(\mathbb {R},\mathbb {R})$ and $g: \mathbb {R}\rightarrow \mathbb {R}$ is an integrable function, then for all $a\in \mathbb {R}$,

  $$\begin{aligned} \int _{\mathbb {R}}\Delta ^m_hf(x + a) g(x)dx = \int _{\mathbb {R}}f(x + a)\Delta ^m_{-h}g(x)dx \end{aligned}$$

  (C.5)

Proof

For $0<\alpha <1$, $h\in \mathbb {R}$ and an integer $m\ge 1$,

• 1. If $f\in \mathscr {C}^{\alpha }_b(\mathbb {R})$, we proceed by a recurrence argument on $m\ge 1$. Indeed,

  $$\begin{aligned} \Vert \Delta ^1_hf\Vert _{\infty } = \sup _{x\in \mathbb {R}}\left| f(x+h)-f(x) \right| \le |h|^{\alpha }\Vert f\Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})} \end{aligned}$$

  and if we assume that the property holds at range m, we get

  $$\begin{aligned} \Vert \Delta ^{m+1}_hf\Vert _{\infty } = \sup _{x\in \mathbb {R}}\left| \Delta ^m_hf(x+h)-\Delta ^m_hf(x) \right| \le 2C_m|h|^{\alpha }\Vert f\Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})} \end{aligned}$$

  that corresponds to the property with $C_{m+1} = 2C_m$. We deduce that it holds for each integer $m\ge 1$.

• 2. If $f\in \mathscr {C}^{\alpha }_b(\mathbb {R})$, then thanks to (C.1) we get for all $x,y\in \mathbb {R}$,

  $$\begin{aligned} |\Delta ^m_hf(x) - \Delta ^m_hf(y)|\le & {} \sum _{j=0}^m \begin{pmatrix} m \\ j \end{pmatrix}|f(x+jh) - f(y+jh)| \\\le & {} 2^m\Vert f\Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})}|x-y|^{\alpha }. \end{aligned}$$

• 3. We proceed by a recurrence argument to show firstly that

  $$\begin{aligned} \mathbf {(P): }~~\forall m\ge 1, ~\Delta ^m_hf(x) = h^m\int _{0}^mH_m(t)\partial _x^mf(x+th)dt, \text { if }f\in \mathcal {C}^m(\mathbb {R},\mathbb {R}) \end{aligned}$$

  where $H_m$ is bounded and does not depend on f. Indeed for $f\in \mathcal {C}^1(\mathbb {R},\mathbb {R})$, then

  $$\begin{aligned} \Delta ^{1}_hf(x) = f(x+h)-f(x) = \int _x^{x+h}f'(t)dt = h\int _0^1f'(x+th)dt \end{aligned}$$

  and hence $H_1(t) = 1$. We now assume that the property holds at range m, then if $f\in \mathcal {C}^{m+1}(\mathbb {R},\mathbb {R})$,

  $$\begin{aligned} \displaystyle \Delta ^{m+1}_hf(x)= & {} \Delta _h^{m}f(x+h) - \Delta ^m_hf(x)\\ \displaystyle= & {} h^m\int _0^mH_m(t)\left[ \partial _x^mf(x+(1+t)h) - \partial _x^mf(x+th) \right] dt\\ \displaystyle= & {} h^m\int _0^mH_m(t)\int _{x+th}^{x+(1+t)h} \partial _x^{m+1}f(u)dudt\\ \displaystyle= & {} h^{m+1}\int _0^mH_m(t)\int _{t}^{1+t} \partial _x^{m+1}f(x+uh)dudt\\ \displaystyle= & {} h^{m+1}\int _0^mH_m(t)\int _{0}^{m+1} \partial _x^{m+1}f(x+uh)1_{\{u\in [t,1+t]\}}dudt\\ \displaystyle= & {} h^{m+1}\int _0^{m+1}\underbrace{\left( \int _0^mH_m(t)1_{\{u\in [t,1+t]\}}dt\right) }_{=\, H_{m+1}(u)} \partial _x^{m+1}f(x+uh)du \end{aligned}$$

  where the function $H_{m+1}$ is bounded and does not depend on f. We deduce that the property $\mathbf {(P)}$ holds, and then for all $f\in \mathcal {C}^m(\mathbb {R},\mathbb {R})$,

  $$\begin{aligned} \Vert \Delta ^m_hf\Vert _{\mathbb {L}^1} \le |h|^m\Vert H_m\Vert _{\infty }\int _{\mathbb {R}}\int _0^m|\partial _x^mf(x+th)|dtdx = m\Vert H_m\Vert _{\infty }|h|^m\Vert \partial _x^mf\Vert _{\mathbb {L}^1}. \end{aligned}$$

• 4. Let $f\in \mathcal {C}_b(\mathbb {R},\mathbb {R})$, and $g:\mathbb {R}\rightarrow \mathbb {R}$ an integrable function and $a\in \mathbb {R}$, then by changing variables in the integral

  $$\begin{aligned} \int _{\mathbb {R}}\Delta ^m_hf(x + a) g(x)dx= & {} \sum _{j=0}^m (-1)^{m-j} \begin{pmatrix} m \\ j \end{pmatrix}\int _{\mathbb {R}}f(x+a+jh)g(x)dx \\= & {} \sum _{j=0}^m (-1)^{m-j} \begin{pmatrix} m \\ j \end{pmatrix}\int _{\mathbb {R}}f(x+a)g(x - jh)dx \\= & {} \int _{\mathbb {R}}f(x+a)\Delta ^m_{-h}g(x)dx \end{aligned}$$

$\square$

The difference operator allows to define the Besov spaces $\mathcal {B}^s_{1,\infty }(\mathbb {R})$ as the set of function $f:\mathbb {R}\rightarrow \mathbb {R}$ that satisfy for $m>s$

$$\begin{aligned} \Vert f\Vert ^{(m)}_{\mathcal {B}^s_{1,\infty }(\mathbb {R})}:= \Vert f\Vert _{\mathbb {L}^1(\mathbb {R})} + \sup _{|h|\le 1}|h|^{-s}\Vert \Delta ^m_hf\Vert _{\mathbb {L}^1(\mathbb {R})} < \infty \end{aligned}$$

(see Triebel [33] Theorem 2.5.12, or [34] Theorem 2.6.1). The following result follows

Lemma C.2

Let $(E,\mathcal {F}_E,\mu )$ be a measured space where $\mu$ is a positive measure. Let $(x,y)\mapsto f(x,y)$ be an integrable function on $\mathbb {R}\times E$, and

$$\begin{aligned} \Phi _f(x) = \int _Ef(x,y)\mu (dy), \forall x\ge 0. \end{aligned}$$

Then we have for all $s>0$ and $m>s$ an integer,

$$\begin{aligned} \left\| \Phi _f \right\| ^{(m)}_{\mathcal {B}^s_{1,\infty }(\mathbb {R})} \le \int _E\Vert f(\cdot ,y)\Vert ^{(m)}_{\mathcal {B}^s_{1,\infty }(\mathbb {R})}\mu (dy). \end{aligned}$$

(C.6)

Proof

It follows from Fubini’s theorem that

$$\begin{aligned} \Vert \Phi _f\Vert _{\mathbb {L}^1(\mathbb {R})} = \int _{\mathbb {R}}\left| \int _E f(x,y) \mu (dy) \right| dx \le \int _E\Vert f(\cdot ,y)\Vert _{\mathbb {L}^1(\mathbb {R})}\mu (dy). \end{aligned}$$

Similarly, for any $s>0$, $m>s$ an integer and $h\in [-1,1]$,

$$\begin{aligned} \Vert \Delta ^m_h\Phi _f\Vert _{\mathbb {L}^1(\mathbb {R})} = \int _{\mathbb {R}}\left| \int _E \big (\Delta ^m_h f(\cdot ,y)\big )(x) \mu (dy) \right| dx \le \int _E\Vert \Delta ^m_hf(\cdot ,y)\Vert _{\mathbb {L}^1(\mathbb {R})}\mu (dy). \end{aligned}$$

The result directly follows by taking the supremum in h and summing the above inequalities. $\square$

Furthermore, the above definition of Besov spaces allows a sufficient condition for the existence of a density for random variables given by the following result

Lemma C.3

([30], Lemma A.1) Let X be a real valued random variable. If there are an integer $m\ge 1$, a real number $\theta >0$, a real $\alpha >0$, with $\alpha<\theta <m$, and a constant $K>0$ such that for every $\phi \in \mathscr {C}^{\alpha }_b(\mathbb {R})$ and $h\in \mathbb {R}$ with $|h|\le 1$,

$$\begin{aligned} \mathbb {E}\left[ \Delta ^m_h\phi (X) \right] \le K|h|^{\theta }\Vert \phi \Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})}, \end{aligned}$$

then X has a density $f_X$ with respect to the Lebesgue measure on $\mathbb {R}$. Moreover $f_X\in \mathcal {B}^{\theta -\alpha }_{1,\infty }(\mathbb {R})$ and

$$\begin{aligned} \Vert f_X\Vert ^{(m)}_{\mathcal {B}^{\theta -\alpha }_{1,\infty }(\mathbb {R})} \lesssim 1+K. \end{aligned}$$

The consequence that follows gives a sufficient condition that will is used in the paper

Lemma C.4

Let X be a real valued random variable. For a given non negative real valued function $\sigma (x)$ such that $\sigma (X)\in \mathbb {L}^1$ with $\mathbb {E}\big [ \sigma (X) \big ] \ne 0$, if there exists an integer $m\ge 1$, a real number $\theta >0$, a real $\alpha >0$ with $\alpha<\theta <m$, and a constant $K>0$ such that for every $\phi \in \mathscr {C}^{\alpha }_b(\mathbb {R})$ and $h\in \mathbb {R}$ with $|h|\le 1$,

$$\begin{aligned} \mathbb {E}\left[ \sigma (X)\Delta ^m_h\phi (X) \right] \le K |h|^{\theta }\Vert \phi \Vert _{\mathscr {C}^{\alpha }_b(\mathbb {R})}, \end{aligned}$$

(C.7)

then X admits a density $f_X$ on $\{ x: \sigma (x)\ne 0 \}$. In addition, if we denote by $f_X$ its density on this set, then the function $x\mapsto \sigma (x)f_X(x)$ is in the Besov space $\mathcal {B}^{\theta -\alpha }_{1,\infty }(\mathbb {R})$ and satisfies the bound

$$\begin{aligned} \Vert \sigma f_X\Vert ^{(m)}_{\mathcal {B}^{\theta -\alpha }_{1,\infty }(\mathbb {R})} \lesssim K + \mathbb {E}\left[ \sigma (X) \right] \end{aligned}$$

(C.8)

Proof

Let us denote by $\mu (dx)$ the law of X and set

$$\begin{aligned} \nu (dx) = \sigma (x)\mu (dx) \end{aligned}$$

that is a non negative finite measure by assumption. Then we have

$$\begin{aligned} \mathbb {E}\left[ \sigma (X)\Delta ^m_h\phi (X) \right] = \Vert \nu \Vert _{\mathbb {L}^1}\int _{\mathbb {R}}\Delta ^m_h\phi (x)\frac{\nu (dx)}{\Vert \nu \Vert _{\mathbb {L}^1}} = \Vert \nu \Vert _{\mathbb {L}^1} \mathbb {E}\left[ \Delta ^m_h\phi (Y) \right] \end{aligned}$$

where Y is a real valued random variable of law $\nu (dy)/\Vert \nu \Vert _{\mathbb {L}^1}$. This implies with (C.7) that

$$\begin{aligned} \mathbb {E}\left[ \Delta ^m_h\phi (Y) \right] \le \frac{K }{\Vert \nu \Vert _{\mathbb {L}^1}}|h|^{\theta }\Vert \phi \Vert _{\mathscr {C}_b^{\alpha }(\mathbb {R})} \end{aligned}$$

and it follows from Lemma C.3 that the probability measure $\nu (dy)/\Vert \nu \Vert _{\mathbb {L}^1}$ admits a density $g_X\in \mathcal {B}^{\theta -\alpha }_{1,\infty }(\mathbb {R})$ with respect to the Lebesgue measure on $\mathbb {R}$ that satisfies the bound

$$\begin{aligned} \Vert g_X\Vert ^{(m)}_{\mathcal {B}^{\theta -\alpha }_{1,\infty }(\mathbb {R})} \lesssim 1+\frac{K}{\Vert \nu \Vert _{\mathbb {L}^1}}. \end{aligned}$$

Then $\mu (dx)$ admits a density $f_X$ on the set $\{x: \sigma (x)\ne 0\}$ that is defined by

$$\begin{aligned} f_X(x) = \frac{\Vert \nu \Vert _{\mathbb {L}^1}}{\sigma (x)}g_X(x) \end{aligned}$$

and the bound (C.8) follows. $\square$

$\mathbb {L}^1$ estimates of derivatives of the Gaussian density

We are interested in the density of a centered Gaussian random variable with variance $\sigma ^2$, that is

$$\begin{aligned} g_{\sigma }(x) = \frac{1}{\sigma \sqrt{2\pi }}\exp \left( -\frac{x^2}{2\sigma ^2} \right) , \forall x\in \mathbb {R}. \end{aligned}$$

We have the following result

Lemma D.1

For each integer $m\ge 1$,

$$\begin{aligned} \Vert \partial ^m_xg_{\sigma }\Vert _{\mathbb {L}^1} \le \frac{C_m}{\sigma ^m} \end{aligned}$$

where the constant $C_m$ does not depend on $\sigma$.

Proof

Because of its exponential factor, let us set

$$\begin{aligned} \partial _x^mg_{\sigma }(x) = \frac{P^{\sigma }_m(x)}{\sigma ^m}\frac{1}{\sigma \sqrt{2\pi }}\exp \left( -\frac{x^2}{2\sigma ^2} \right) , \forall x\in \mathbb {R}, \forall m\ge 1. \end{aligned}$$

Then $P^{\sigma }_{1}(x) = - x/\sigma$ and

$$\begin{aligned} P^{\sigma }_{m+1} = \sigma \partial _xP_m^{\sigma } - \frac{x}{\sigma }P_m^{\sigma } \end{aligned}$$

that implies that each $P^{\sigma }_m$ is a polynomial, thus

$$\begin{aligned} P_m^{\sigma }(x) = \sum _{k=0}^mA^m_k(\sigma )x^k \end{aligned}$$

with

$$\begin{aligned} \displaystyle A^{m+1}_0(\sigma )= & {} \sigma A^{m}_1(\sigma ),\\ \displaystyle A^{m+1}_k(\sigma )= & {} \sigma (k+1)A^m_{k+1}(\sigma ) - \frac{1}{\sigma }A^m_{k-1}(\sigma ),\, k=1,\ldots ,m-1\\ \displaystyle A^{m+1}_m(\sigma )= & {} -\frac{1}{\sigma }A^{m}_{m-1}(\sigma )\\ \displaystyle A^{m+1}_{m+1}(\sigma )= & {} -\frac{1}{\sigma }A^{m}_m(\sigma ) \end{aligned}$$

We easily verify by a recurrence argument on $m\ge 1$ that

$$\begin{aligned} \forall m\ge 1, \, |A^m_k| \le \frac{C_m}{\sigma ^k} \text { for }k=1,\ldots ,m \end{aligned}$$

and then

$$\begin{aligned} \Vert \partial ^m_xg_{\sigma }\Vert _{\mathbb {L}^1} = \frac{1}{\sigma ^m}\int _{\mathbb {R}}\left| P_m^{\sigma }(x)\right| g_{\sigma }(x)dx \le \frac{1}{\sigma ^m}\sum _{k=0}^m\left| A^m_k(\sigma ) \right| \int _{\mathbb {R}}\left| x\right| ^kg_{\sigma }(x)dx \le \frac{C_m}{\sigma ^m} \end{aligned}$$

$\square$