I)^*{\geqslant}0,$$ since the adjoint operator preserves positivity.

\[T105\] For $y\in Y$ and $\Psi\in{\mathcal P}_1(Y)$, define $b_\Psi(y)$ as in . Assume that $v_\Psi\colon Y\to{\mathbb{R}^{d}}$ satisfies (v1)-(v3) and ${\mathcal T}(x,\Psi)\colon {\operatorname{Lip}}(U)\to{\operatorname{Lip}}(U)$ satisfies (T0)-(T3). Then

-   for every $R>0$, for every $\Psi\in {\mathcal P}(B_R^Y)$, and for every $y^1,y^2\in B_R^Y$, there exists $L_R>0$ such that $$\label{T106}
    \lVert b_\Psi(y^1)-b_\Psi(y^2)\rVert_{\overline Y}{\leqslant}L_R \lVert y^1-y^2\rVert_{\overline Y}\,;$$

-   for every $R>0$, for every $\Psi^1,\Psi^2\in {\mathcal P}(B_R^Y)$, and for every $y\in B_R^Y$, there exists $L_R>0$ such that $$\label{T117}
    \lVert b_{\Psi^1}(y)-b_{\Psi^2}(y)\rVert_{\overline Y}{\leqslant}L_R {\mathcal W}_1(\Psi^1,\Psi^2);$$

-   for every $R>0$, there exists $\theta>0$ such that for every $y\in B_R^Y$ and for every $\Psi\in {\mathcal P}(B_R^Y)$ $$\label{T109}
    y+\theta b_\Psi(y)\in Y;$$

-   there exists $M>0$ such that for every $y\in Y$ and for every $\Psi\in {\mathcal P}_1(Y)$ there holds $$\label{T110}
    \lVert b_{\Psi}(y)\rVert_{\overline Y}{\leqslant}M \big(1+ \lVert y\rVert_{\overline Y} +m_1(\Psi)\big)\,.$$

Property (i) is a consequence of and ; property (ii) follows from and . Observing that for all $y=(x,\lambda)\in Y$ we have that $\lVert \lambda\rVert_{{\mathrm{BL}}}{\leqslant}1$, we obtain from and . Finally, to prove (iii), since ${\mathbb{R}^{d}}$ is convex, we simply have to show that for every $R>0$, there exists $\theta>0$ such that for every $y=(x,\lambda)\in B_R^Y$ and for every $\Psi\in {\mathcal P}(B_R^Y)$ $$\label{T121}
\lambda+\theta {\mathcal T}^*(x,\Psi)\lambda\in {\mathcal 