Finding an existence and uniqueness result of a strong solution of Lipschitz SDEs

Question

I try to understand the Proof of Theorem 4.21 in Carmona Delarue (2018). In the following, what I don't understand:

Processes are assumed to be defined on a complete filtered probability space $(\Omega, \mathcal F, \mathbb F=(\mathcal F_t)_{t \in [0,T]},\mathbb P)$ supporting a $d$-dimensional Wiener process $W=(W_t)_{t\in [0,T]}$ wrt. $\mathbb F$, the filtration $\mathbb{F}$ satisfying the usual conditions. We denote by $\mathbb{H}^{2,n}$ the Hilbert space $$\mathbb{H}^{2,n} = \{ Z \in \mathbb{H}^{0,n}: \mathbb{E} \int_0^T |Z_s|^2 dx < \infty \},$$ where $\mathbb{H}^{0,n}$ stands for the collection of all $\mathbb{R}^n$-valued progressively measurable processes on $[0,T]$.

Assume that, for each $(x,\mu) \in \mathbb{R}^d \times \mathcal{P}_2(\mathbb{R}^d)$, the processes $B(\cdot,\cdot,x,\mu):[0,T]\times \Omega \to \mathbb{R}^d, (t,\omega) \mapsto B(t,\omega,x,\mu)$ and $\Sigma (\cdot,\cdot,x,\mu):[0,T]\times \Omega \to \mathbb{R}^{d \times d}, (t,\omega) \mapsto \Sigma(t,\omega,x,\mu)$ are $\mathbb{F}$-progressively measurable and belong to $\mathbb{H}^{2,d}$ and $\mathbb{H}^{2,d\times d}$ respectively. Furthermore, assume that for any $t \in [0,T], \omega \in \Omega, x, x' \in \mathbb{R}^d$ and $\mu, \mu'\in \mathcal{P}_2(\mathbb{R}^d)$, $$|B(t,x,\mu)-B(t,x',\mu')|+|\Sigma(t,x,\mu)-\Sigma(t,x',\mu')| \leq L (|x-x'|+W_2(\mu,\mu')).$$

Temporarily fix some $\mu = (\mu_t)_{t \in [0,T]} \in \mathcal{C}([0,T],\mathcal{P}_2(\mathbb{R}^d))$ and let $X_0 \in L^2(\Omega,\mathcal{F}_0,\mathbb{P},\mathbb{R}^d)$. Then "the classical existence result for Lipschitz SDE guarantees existence and uniqueness of a strong solution of the classical stochastic differential equation with random coefficients": $$dX_t = B(t,X_t,\mu_t)dt + \Sigma(t,X_t,\mu_t)dW_t$$

My problem: I don't find such a classical existence result. I think the $\mu$ part is mostly irrelevant here. It comes from McKean-Vlasov setting that is treated originally. I didn't want to omit it here in case it is relevant in some way. Crossing it out simplifies things a little. Still, when I look for the standard existence result (e.g. Karatzas Shreve, Chapter 5, Theorem 2.9) there is always some linear growth condition like $$ \|B(t,x)\|^2 + \|\Sigma(t,x)\|^2 \leq K^2(1+\|x\|^2).$$ I do not see, why this is implied by the conditions.

Answer 1

Following the proof of Theorem 4.21, fix the environment $\mu$, and additionally suppress the dependence of the SDE coefficients on $\mu$, so that the nonlinear SDE reduces to a classical one.

Claim: In the classical existence/uniqueness theorem for SDEs, the standard linear growth condition can be relaxed to: there exists $y\in \mathbb{R}^d$ such that for all $T>0$ $$ E \int_0^T ( \|B(t,y)\|^2 + \|\Sigma(t,y)\|^2 ) dt < \infty \;.\tag{$\star$} \label{A1} $$

Note that Asssumption (A1) in Theorem 4.21 is actually stronger than \eqref{A1}, since (A1) holds for all $y\in \mathbb{R}^d$.

Proof. Recall that the linear growth condition is used to prove that the SDE solution is a real-valued, progressively measurable process such that $E \int_0^T \| X_s \|^2 dt < \infty$, i.e., $X \in \mathbb{L}^2_d(0,T)$; see, e.g., Chapter 5. Here we show that \eqref{A1} can play the same role. Indeed, let $X^{k}$ denote the $k$th Picard iterate in the standard existence/uniqueness proof. Then \begin{align*} &E\|X_t^{k+1}\|^2 \le 3 \left(E\|X_0\|^2 + E \| \int_0^T B(s,X_s^k) ds \|^2 + E \| \int_0^T \Sigma(s, X_s^k) dW_s \|^2 \right) \\ &~~\le 3 \left(E\|X_0\|^2 + T E \int_0^T \| B(s,X_s^k) \|^2 ds + E \int_0^T \| \Sigma(s, X_s^k)\|^2 ds \right) \;. \tag{$1$} \label{1} \end{align*} Since the SDE coefficients are uniformly Lipschitz continuous in space, for any $x \in \mathbb{R}^d$ and $s \in [0,T]$, \begin{align*} & \| B(s,x) \|^2 + \| \Sigma(s, x)\|^2 \le 2 ( \| B(s,x) - B(s,y)\|^2 + \|B(s,y)\|^2) + 2 ( \| \Sigma(s,x)- \Sigma(s,y) \|^2 + \| \Sigma(s,y) \|^2 ) \\ &~~ \le 2 ( \|B(s,y)\|^2 + \| \Sigma(s,y) \|^2 ) + 4 L^2 |x-y|^2 \end{align*} Inserting this bound into \eqref{1} and invoking \eqref{A1} proves the claim. $\Box$