G(t) \alpha \right) \; : \; \gamma \in [-1,1] \right \} \subseteq D(A^*)\\
D^{1,-} W(t,x)= \emptyset
\end{array}
\right .$$

So we have to verify that $W$ is a subsolution on $\left\langle
\alpha,x\right\rangle=0$. If $W - \varphi - g$ attains a maximum at $(t,x)$ with $\left\langle
\alpha,x\right\rangle=0$ we have that $p{\stackrel{def}{=}}(p_1 + p_2){\stackrel{def}{=}}D(\varphi
+ g)(t,x)\in \left \{ \gamma G(t) \alpha \; : \; \gamma \in [-1,1]
\right \} \subseteq D(A^*)$. From the definition of test1 function $p_1=D\varphi(t,x)\in D(A^*)$ so $\eta(t)g_0'(|x|)\frac{x}{|x|}=p_2=Dg(t,x)\in D(A^*)$. $W(\cdot,x)$ is a $C^1$ function and then, recalling that ${\left\langle}\bar \alpha (t)
,x {\right\rangle}_t={\left\langle}G'(t) \alpha ,x {\right\rangle}=0$, we have $$\label{eq:exampleestimate0}
\partial_t (\varphi +g)(t,x)=\partial_t W(t,x)=\frac{1}{2}\left( R^*\bar\alpha(t)\right)^2,$$ and for $p=\gamma \bar\alpha(t)$ we have $$\label{eq:exampleestimate1} \inf_{u\in U} \left ( {\left\langle}Ru,p{\right\rangle}+
\frac{1}{2} u^{2} \right) = - \frac{1}{2} \gamma^2 \left(
R^*\bar\alpha(t)\right)^2$$ Moreover, recalling that $g_0'(|x|)\geq 0$ and $-A^*$ is monotone, we have $$\begin{gathered}
\label{eq:exampleestimate2} {\left\langle}A^* p_1, x {\right\rangle}= {\left\langle}A^*(p-p_2),
x {\right\rangle}= {\left\langle}A^* \gamma G(t) \alpha, x {\right\rangle}- \frac{g_0'(|x|)}{|x|}
{\left\langle}A^* x, x {\right\rangle}\geq\\ \geq \gamma G(t) {\left\langle}A^* \alpha, x {\right\rangle}=
0\end{gathered}$$ So, by (\[eq:exampleestimate0\]), (\[eq:exampleestimate1\]) and (\[eq:exampleestimate2\]), $$\begin{gathered}
\partial_t (\varphi +g)(t,x) + {\left\langle}A^* p_1 ,x{\right\rangle}- \left| \left\langle \alpha,x\right\rangle\right| + \\
+\inf_{u\in U} \left ( {\left\langle}Ru,D(\varphi +g)(t,x){\right\rangle}+ \frac{1}{2}
u^{2} \right) \geq \frac{1}{2} (1-\gamma^2) \left(
R^*\bar\alpha(s)\right)^2 \geq 0\end{gathered}$$ 