is a constant depending only on the inputs $u$, $b$, and $p$. The function $\varphi_{\epsilon}$ is independent of $v$. Furthermore, it holds $$\begin{aligned}
\label{eq:W1infbounds}\|\varphi_{\epsilon}\|_{L^{\infty}_{t,x}} &\leq M, \\
\|\{\nabla_{t,x},\nabla^{2}_{x}\}\varphi_{\epsilon}\|_{L^{1}_{t,x}} & \leq M \\
 \label{eq:uniflowerBounds}\varphi_{\epsilon} &\geq \min \left \{1,1/(2p) \right \}.\end{aligned}$$

For the proof, it will be useful to introduce the following notation. For two radii, $0 \leq r<R$, we use $A_{r,R}$ to denote the closed annulus $$A_{r,R}=\left \{x \in \R^{d} \mid \quad r \leq |x| \leq R \right \}.$$

Let $p>0$ be given. We will select $\epsilon_{0}(p)<1$ in the course of the proof.\
*Step 1: Reduction to a bounded domain*\
By Hypothesis \[hyp:drift:subCritical\], we may choose a radius $R>0$ such that $u_{\epsilon}$ is supported in $[0,T] \times B_{R}$ for all $\epsilon<1$. Given $R$, we may build a smooth function $\widehat{\varphi}$ independent of $\epsilon$ such that $\widehat{\varphi}=0$ on $B_{R}$, $\widehat{\varphi}=1$ outside of $B_{2R}$, and $$\label{eq:boundOnPhiHat}
\|\nabla^{2}_{x}\widehat{\varphi}\|_{L^{\infty}(\R^{d})} + \|\nabla_{x} \widehat{\varphi}\|_{L^{\infty}(\R^{d})} \leq C_{1} \max \left \{R^{-1},R^{-2}\right \},$$ where $C_{1}$ is a universal constant. Since $\widehat{\varphi}$ vanishes on the support of $u_{\epsilon}$, it follows that for all $\epsilon<1$ $$\begin{aligned}
\label{eq:trivSubSol1}u_{\epsilon} \cdot \nabla_{x}\widehat{\varphi} + p |\Div_{x} u_{\epsilon}| \widehat{\varphi} + \frac{1}{2}b^{2}\Delta_{x} \widehat{\varphi} &\leq \frac{1}{2}\Lambda C_{1}\max \left \{R^{-1},R^{-2}\right \}.\\
\label{eq:trivLowBd}1_{\R^{d} \setminus B_{2R}}(\widehat{\varphi}-1)&\geq 0.\end{aligned}$$\
*Step 2: Decomposition of the 