Let $\Lambda:\mathbb{R}^d\to \mathbb{R}^d$, be Lipschitz. Is the convolution of $\Lambda$ with a measure Lipshchitz? That is do we have
$$ \int_{\mathbb{R}^d} \| \Lambda*\rho(x) - \Lambda*\mu(x)\|^2 ~ d\rho(x) \leq C W_2^2(\mu,\rho) $$
for some $C>0$, and $W_2$ the Wasserstein metric, and $\mu,\rho$ probability measures. My attempt : let $Y\sim \rho, Z\sim \mu$
\begin{align} \int_{\mathbb{R}^d} \| \Lambda*\rho(x) - \Lambda*\mu(x)\|^2 ~ d\rho(x) =& \int_{\mathbb{R}^d} \| \int_{\mathbb{R}^d}\Lambda(x-y)\big(\rho(y)-\mu(y)\big) dy\|^2 ~ d\rho(x) \\ =& \int_{\mathbb{R}^d} \| \mathbb{E}[\Lambda(x-Y)-\Lambda(x-Z)] \|^2 ~ d\rho(x) \\ \leq&\int_{\mathbb{R}^d}\mathbb{E}[\|\Lambda(x-Y)-\Lambda(x-Z)\|^2] ~ d\rho(x) \\ \leq&C \int_{\mathbb{R}^d}\mathbb{E}[\|Y-Z\|^2] ~ d\rho(x)=C \mathbb{E}[\|Y-Z\|^2]. \end{align}
Is this ok?