Let $X_t$, $Y_t$ be real-valued continuous stochastic processes with finite second moments such that$$E |X_t-Y_t|^2 \leq \int_0^t K(s) \left( E |X_s - Y_s |^2 + W_2^2( \mu_s, \nu_s) \right) ds, \quad t \in [0, T],$$where $K : [ 0, \infty ) \rightarrow [ 0, \infty )$ is locally integrable, $W_2$ denotes the $2$-Wasserstein distance, $s \mapsto \mu_s$, and $s \mapsto \nu_s$ are probability measures from the $2$-Wassertein space. I want to apply Gronwall's lemma to show that there exists a constant $C=C(T)$ such that$$E|X_t-Y_t|^2 \leq C(T) \int_0^t W_2^2( \mu_s, \nu_s) ds, \quad t \in [0, T]. $$By setting\begin{align}u(t) &= E|X_t-Y_t|^2,\\\beta(s) &= K(s),\\\alpha(t) &= \int_0^t K(s) W_2^2 ( \mu_s, \nu_s) ds,\end{align}and applying Gronwall's lemma we get\begin{align}u(t) &\leq \alpha(t) + \int_0^t \alpha ( s ) \beta( s ) e^{ \int_s^t \beta(r) dr } ds\\&\leq \alpha(t) \left( 1 + \int_0^t \beta( s ) e^{ \int_s^t \beta(r) dr } ds \right), \quad t \in [0, T],\end{align}where the second term in the product can be estimated from above by setting $t = T$. How can one further bound$$\alpha(t) = \int_0^t K(s) W_2^2 ( \mu_s, \nu_s) ds$$with$$C(T) \int_0^t W_2^2( \mu_s, \nu_s) ds$$to obtain the desired result. Note that we only know that $K$ is non-negative and locally integrable. Does one need an additional assumption to obtain the estimate?
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