Let $ h\in C^1(\mathbb R)$ such that $ h’$ is Lipschitz continuous and $ $ L\varphi:=-h’\varphi’+\varphi”\;\;\;\text{for }\varphi\in C^2(\mathbb R).$ $ Now, let $ (X_t)_{t\ge0}$ be the unique strong solution of $ $ {\rm d}X_t=-h'(X_t){\rm d}t+\sqrt2{\rm d}W_t\tag1,$ $ where $ (W_t)_{t\ge0}$ is a Brownian motion.

I’ve read (in this paper, below Assumption 2.4) that if $ V:\mathbb R\to[0,\infty)$ with $ $ V(x)\xrightarrow{x\to\infty}\infty\tag2$ $ and $ a,d>0$ with $ $ LV\le-aV+d\tag3,$ $ then $ $ \operatorname E\left[V(X_t)\mid\mathcal F_s^X\right]\le e^{-a(t-s)}V(X_s)+\frac da(1-e^{-a(t-s)})\tag4.$ $ Why does $ (4)$ hold?

Obviously, $ (4)$ is an application of (the Itō formula and) a Gronwall-type lemma. Actually, it’s precisely Theorem 6 here. However, in order to apply that theorem, we should need that the process $ \left(\operatorname E\left[V(X_t)\mid\mathcal F_s^X\right]\right)_{t\ge0}$ is continuous. This shouldn’t hold, unless $ V$ is (continuous and) bounded (which would allow an application of Lebesgue’s dominated convergence theorem). But $ V$ is clearly assumed to be unbounded by $ (2)$ . So, what am I missing?

(Clearly, the authors of the paper are missing assumptions on $ V$ anyway. In order for $ (3)$ to make sense, $ V$ needs to be twice differentiable (at least in some weak sense).