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

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.

• Welcome to MO! Doesn’t Lipschitz continuity imply linear growth? Sep 9, 2022 at 13:18
• Hi @NawafBou-Rabee, thank you for your answer. Let me quote Karatzas Shreve: "Suppose that the coefficients $b(t,x), \sigma(t,x)$ satisfy the global Lipschitz and linear growth condition $\|b(t,x)-b(t,y)\|+\|\sigma(t,x)-\sigma(t,y)\| \leq K\|x-y\|$ and $\|b(t,x)\|^2+\|\sigma(t,x)\|^2 \leq K^2(1+\|x\|^2)$ for every $0 \leq t < \infty, x,y \in \mathbb{R}^d$". I think they would not state the latter, if it is implied by the former. I think the problem is, that you need one constant for all $t$. You just got me wondering, wether it is different in my finite-time setting....
– Blup
Sep 9, 2022 at 14:05
• Hi Blup, that was a comment, not an answer. Based on your feedback, I have provided a complete answer below. Basically the Theorem in the book is correct and indeed based on a classical result. Sep 12, 2022 at 14:36

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. $$\Box$$

• Hi @Nawaf Bou-Rabee! Thank you for your help. Today I was able to get my hands on the Evans-book and confirm your answer. You have a small typo in the last line of the last align, I think the last summand should be $4L^2|x-y|^2$. One might add that Evans also uses the Linear growth condition in his second step below align (13), but one can use the inequality in your second align there as well!
– Blup
Sep 12, 2022 at 14:45
• @Blup, thanks for the remark and accepting the answer! The typo has been corrected. If you found the answer helpful, you can upvote it using the arrow. I only mention this because you're a new user and might be unfamiliar with the point system in MO. Sep 12, 2022 at 14:55
• I find the answer helpful but I am a this new user that I do not have enough "reputation" to upvote any answer. I really don't like this system.
– Blup
Sep 12, 2022 at 18:21