An integral estimate regarding tails of solutions in Di Castro–Kuusi–Palatucci Let $x_0 \in \mathbb{R}^n$ and suppose $u \colon B_r(x_0)\to(0,\infty)$ be an $L^p$ integrable function and let $k=\sup_{x \in B_r(x_0)} u(x)$.
Also let $\phi\in C_c^{\infty}(B_r(x_0))$ be such that $0 \le \phi\le 1$ in $B_r(x_0)$, $\phi \equiv 1$ on $B_{\frac{r}{2}}(x_0)$ and $|\nabla\phi| \le \frac{c}{r}$ in $B_r(x_0)$.
Then for $y\in \mathbb{R}^n\setminus B_r(x_0)$ and $x\in B_r(x_0)$, we have
\begin{multline}
\int_{\mathbb{R}^n\setminus B_r(x_0)}\int_{B_r(x_0)}|u(x)-u(y)|^{p-2}(u(x)-u(y))(u(x)-2k)\phi^p(x)\,dx dy \\
\ge \int_{\mathbb{R}^n\setminus B_r(x_0)}\int_{B_r(x_0)}k(u(y)-k)_{+}^{p-1}\phi^p(x)\,dx dy \\
-\int_{\mathbb{R}^n\setminus B_r(x_0)}\int_{B_r(x_0)}2k\chi_{\{u(y)<k\}}(y) \cdot (u(x)-u(y))_{+}^{p-1}\phi^p(x)\, dx dy.
\end{multline}
This has been written in this article on page 17. Can someone kindly help with how to get it?
 A: The indicator function $\chi_{\{u(y)<k\}}$ and the unchanged domain of integration suggests that the LHS integral ($I_2$) is being split into parts corresponding to the cases $u(y) \geq k$ and $u(y) < k$.
Indeed, if $u(y) \geq k$, then $u(y) \geq u(x)$ and so
$$
\lvert u(x) - u(y) \rvert^{p-2} (u(x) - u(y)) (u(x) - 2k)
= (u(y) - u(x))^{p-1} (2k - u(x))
\geq (u(y) - k)^{p-1} k,
$$
which matches the first RHS integral ($I_{2,1}$).
And if $u(y) < k$, we further split the cases $u(x) \geq u(y)$ and $u(x) < u(y)$, and additionally use the assumption from the lemma in question (Lemma 4.2) that $u \geq 0$, to obtain
\begin{multline*}
\lvert u(x) - u(y) \rvert^{p-2} (u(x) - u(y)) (u(x) - 2k)
\\
\begin{aligned}
&= (u(x) - u(y))_{+}^{p-1} (u(x) - 2k) + (u(y) - u(x))_{+}^{p-1} (2k - u(x))
\\ &\geq (u(x) - u(y))_{+}^{p-1} (u(x) - 2k)
\\ &\geq -2k (u(x) - u(y))_{+}^{p-1},
\end{aligned}
\end{multline*}
which is exactly what appears in the second RHS integral ($I_{2,2}$).
Taking the positive parts (the $+$ subscripts) ensures that in each case the other integral is $0$.
