For $p\ge 1$, let $\mathcal P_p(\mathbb R)$ be the set of probability measures on $\mathbb R$ of finite $p^{\rm th}$ moment. Denote by $W_p$ the Wasserstein metric of order $p$ and by $\preceq$ the increasing convex order on $\mathcal P_1(\mathbb R)$, i.e. $\mathcal P_1(\mathbb R)\ni \mu\preceq \nu\in \mathcal P_1(\mathbb R)$ if
$$\int_{\mathbb R} f d\mu \le \int_{\mathbb R} f d\nu \mbox{ holds for every convex function } f: \mathbb R\to\mathbb R \mbox{ of linear growth}.$$
My question is whether there exists a continuous function $g:\mathbb R_+\to\mathbb R_+$ with $g(0)=0$ such that :
for any $\mu, \nu\in \mathcal P_p(\mathbb R)$ of the same mean, there always exist $\lambda, \rho\in \mathcal P_p(\mathbb R)$ (depending on $\mu,\nu$) satisfying
$$\lambda\preceq \mu,\quad \lambda\preceq \nu,\quad \mu\preceq\rho,\quad \nu\preceq\rho\quad (\ast)$$
and
$$W_p(\lambda,\rho) \le g\big(W_p(\mu,\nu)\big)? \quad (\ast\ast)$$
PS : I am interested mostly for the cases $p=1$ and $p=2$. Further, it is known that $\preceq$ is a partial order and $\mu\preceq \nu$ iff
$$\int_{\mathbb R} x d\mu(x)=\int_{\mathbb R} x d\nu(x) \quad \mbox{and}\quad \int_{\mathbb R} |x-K| d\mu(x)=\int_{\mathbb R} |x-K| d\nu(x) \mbox{ for all } K\in\mathbb R.$$$$\int_{\mathbb R} x d\mu(x)=\int_{\mathbb R} x d\nu(x) \quad \mbox{and}\quad \int_{\mathbb R} |x-K| d\mu(x)\le\int_{\mathbb R} |x-K| d\nu(x) \mbox{ for all } K\in\mathbb R.$$
If finding such a function is not possible. A function satisfying locally the desired property is still welcome. Namely, for any $\mu\in \mathcal P_p(\mathbb R)$, find a function $g\equiv g_\mu$ such that $(\ast)$ and $(\ast\ast)$ are satisfied for all $\nu$ of the same mean of $\mu$ and with $W_p(\mu,\nu)\le 1$.
