Proof. The significant parts of this proof follows the steps of [7]. Set, \(A:=\{x\in B_\rho(\xi):u\leq k\}\). Consider \(|A|>0\), otherwise the result will be obvious. Define \[\begin{align} u:=\begin{cases} &\min\{u,l\}-k,\,\,\,\,\,\,\,\mathrm{if }\,u>k,\\ & 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{ if }\,u\leq k. \end{cases} \end{align}\] Then, we have \[\begin{align} \int_{B_\rho(\xi)}|u-u_{B_\rho(\xi)}|dx=\int_{B_\rho(\xi)\setminus A}|u-u_{B_\rho(\xi)}|dx+\int_{A}|u_{B_\rho(\xi)}|dx, \end{align}\] where \(u_{B_\rho(\xi)}:=\fint_{B_\rho(\xi)}|u|dx\). Consequently, \[\begin{align} \label{eq2462} |u_{B_\rho(\xi)}|\leq \frac{1}{|A|}\int_{B_\rho(\xi)}|u-u_{B_\rho(\xi)}|dx. \end{align}\tag{4}\] Also, it is obvious that \[\begin{align} \label{eq2463} \int_{B_\rho(\xi)}|u|dx=\int_{\{u>l\}\cap B_\rho(\xi)}|u|dx+\int_{\{k<u\leq l\}\cap B_\rho(\xi)}|u|dx. \end{align}\tag{5}\] Now, with the help of (5 ) \[\begin{align} \label{eq2464} (l-k)|\{u>l\}\cap B_\rho(\xi)|\leq \int_{B_\rho(\xi)}|u|dx&\leq\int_{B_\rho(\xi)}|u-u_{B_\rho(\xi)}|dx+\int_{B_\rho(\xi)}|u_{B_\rho(\xi)}|dx\nonumber\\ &\leq 2\frac{|B_\rho(\xi)|}{|A|}\int_{B_\rho(\xi)}|u-u_{B_\rho(\xi)}|dx. \end{align}\tag{6}\] At this point, we recall the Poincaré type inequality in Heisenberg group (see 6) i.e. \[\begin{align} \int_{B_\rho(\xi)}|u-u_{B_\rho(\xi)}|^pdx\leq C(Q,p)\rho^p\int_{B_\rho(\xi)}|\nabla_{\mathbb{H}^N}u|^pdx, \end{align}\] for all \(u\in W^{1,p}(B_\rho(\xi))\) with \(1\leq p<\infty\). Finally, considering this Poincaré inequality with \(p=1\) on the right hand side of (6 ), we obtain the result. ◻

Proof. This result can be proved similarly to Theorem C.2 and Theorem C.3 of [23] replacing \(s\) with \(1\) as we are considering \(u\in W^{1,2}(B)\). For this purpose, we need to consider Sobolev embedding result (Theorem 7). Moreover, note that, in that context we need an extension result for Sobolev spaces in Heisenberg group which can be found in the thesis of Nhieu [37]. ◻

Proof. We omit this proof as it is similar to the proof of [38]. ◻

Proof. This result follows the outline of the proof of [23], which is valid for fractional Sobolev spaces. Since \(Q>2\), then for \((r,m)\in \left(\frac{2Q}{Q-2},2\right)\), (?? ) follows from Theorem 7. Otherwise, if we take \(\theta=\frac{2}{m}\), then (?? ) holds due to (?? ). Then proceeding in a similar manner to [23], we obtain the required inequality. ◻

Proof. Most of the proof of this result is done in the respective part of [23] by changing \(p=2\). The only part which we need to estimate rigorously is the third term of the left hand side of (2 ). For this reason we only present all the estimations of the concerned part only. Moreover, we prove this result only for the subsolution case since all the processes for a supersolution are same. Before proceeding, we need to borrow some notations and constructions from the proof of [23]. These are as follows

• Let us take \(t_1=t_0-\theta_2\) and take an arbitrary \(t_2\) in \((t_0-\theta_2,t_0)\).

• For every \(t\in (0,T)\), choose \(\phi=\phi(x,t)\) whose spacial support contained in \(B_\frac{r+l}{2}(\xi_0)\).

Using \(\phi^\varepsilon(\xi,t)\) as a test function in the weak subsolution formula (2 ) \[\begin{align} \label{eq3461} &\int_{B_r(\xi_0)}u(\xi,t_2)\phi^\varepsilon(\xi,t_2)d\xi-\int_{B_r(\xi_0)}u(\xi,t_1)\phi^\varepsilon(\xi,t_1)d\xi\nonumber\\ &\quad\int_{t_1}^{t_2}\int_{B_r(\xi_0)}u(\xi,t)\phi_t^\varepsilon(\xi,t)d\xi\,dt+\underbrace{\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)\nabla_{\mathbb{H}^N}\phi^\varepsilon(\xi,t)\,d\xi\,dt}_{I^\varepsilon} \nonumber\\ &\qquad+\int_{t_1}^{t_2}\iint_{\mathbb{H}^N\times\mathbb{H}^N}\frac{1}{2}(u(\xi,t)-u(\eta,t))(\phi^\varepsilon(\xi,t)-\phi^\varepsilon(\eta,t))K(\xi,\eta,t)\;d\eta\,d\xi \,dt\leq 0. \end{align}\tag{7}\] As we discussed, we will estimate only \(I^\varepsilon\). Consider the smooth cut-off function \(\psi\) on \(B_r(\xi_0)\) such that \(0\leq \psi\leq 1\) in \(\mathbb{H}^N\) and \[\begin{align} \begin{cases} & supp \psi= B_\frac{r+l}{2}(\xi_0),\\ &\psi\equiv 1\;\;\; inB_l(\xi_0),\\ & \nabla_{\mathbb{H}^N}\psi\leq \frac{C}{r-l}. \end{cases} \end{align}\] Next, we construct text function \(\phi(\xi,t)\) as \[\phi(\xi,t)=w^\varepsilon_+(\xi,t)\psi^2(\xi)\nu^2(t)=w^\varepsilon_+\psi^2\nu^2(\xi,t),\] where \(w^\varepsilon_+(\xi,t):=(u^\varepsilon(\xi,t)-g^\varepsilon(t)-k)_+,\;\;w_+(\xi,t):=(u(\xi,t)-g(t)-k)_+\) and \(g\) is denoted by \(g(t):=\mathfrak{C}R^{-2}\int_{t_0-R^{2}}^t\mathrm{Tail}(u_+(\varsigma);R,\xi_0)d\varsigma\). Further, for simplicity we denote \[\begin{align} \psi^i(\xi)\nu^j(t)\; as \,\psi^i\nu^j(\xi,t)\, for some non-negative real numbers i,j. \end{align}\] Using this test function, \(I^\varepsilon\) becomes \[\begin{align} \label{eq3462} I&=\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)\nabla_{\mathbb{H}^N}(w^\varepsilon_+\psi^2\nu^2)^\varepsilon(\xi,t)\,d\xi\,dt\nonumber\\ &=\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)((\nabla_{\mathbb{H}^N}w^\varepsilon_+)\psi^2\nu^2)^\varepsilon(\xi,t)+2\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)(w^\varepsilon_+\psi(\nabla_{\mathbb{H}^N}\psi)\nu^2)^\varepsilon(\xi,t)\nonumber\\ &\geq \underbrace{\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)((\nabla_{\mathbb{H}^N}w^\varepsilon_+)\psi^2\nu^2)^\varepsilon(\xi,t)\;d\xi\,dt}_{D_1^\varepsilon}-2\underbrace{\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)(w^\varepsilon_+\psi(\nabla_{\mathbb{H}^N}\psi)\nu^2)^\varepsilon(\xi,t)\;d\xi\,dt.}_{D_2^\varepsilon} \end{align}\tag{8}\] Passage to limit as \(\varepsilon\rightarrow 0\) : Next, we estimate \(D_i^\varepsilon\) for \(i=1,2\) successively. Begin with \(D_1^\varepsilon\), using Hölder inequality, \[\begin{align} \label{eq3463} \int_{t_1}^{t_2}&\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)((\nabla_{\mathbb{H}^N}w^\varepsilon_+)\psi^2\nu^2)^\varepsilon(\xi,t)\;d\xi\,dt- \int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)((\nabla_{\mathbb{H}^N}w_+)\psi^2\nu^2)(\xi,t)\;d\xi\,dt\nonumber\\ &\leq \Bigg(\int_{t_1}^{t_2}\int_{B_r(\xi_0)}|\nabla_{\mathbb{H}^N}u(\xi,t)|^2\,d\xi\,dt\Bigg)^{1/2}\Bigg(\int_{t_1}^{t_2}\int_{B_r(\xi_0)}|((\nabla_{\mathbb{H}^N}w^\varepsilon_+)\psi^2\nu^2)^\varepsilon-((\nabla_{\mathbb{H}^N}w_+)\psi^2\nu^2)|^2(\xi,t)\,d\xi\,dt\Bigg)^{1/2}\nonumber\\ &\stackrel{Lemma \,\ref{L2461}}{\rightarrow} 0,\;\;\;\; as\varepsilon\rightarrow 0 \end{align}\tag{9}\]

The second step makes sense since \(\nabla_{\mathbb{H}^N}u\in L^2(t_1,t_2;L^2(B_r(\xi_0))\). Next, by a similar argument, we have the following \[\begin{align} \label{eq3464} D_2^\varepsilon\rightarrow D_2:= \int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)(w_+\psi(\nabla_{\mathbb{H}^N}\psi)\nu^2)(\xi,t)\;d\xi\,dt. \end{align}\tag{10}\]
Combining results of (9 ) and (10 ) into (8 ), we have for \(\varepsilon\rightarrow 0\), \[\begin{align} &\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)((\nabla_{\mathbb{H}^N}w_+)\psi^2\nu^2)(\xi,t)\;d\xi\,dt\\ &\qquad -2\int_{t_1}^{t_2}\int_{B_r(\xi_0)}\nabla_{\mathbb{H}^N}u(\xi,t)(w_+\psi(\nabla_{\mathbb{H}^N}\psi)\nu^2)(\xi,t)\;d\xi\,dt\leq I\longleftarrow I^\varepsilon \end{align}\] Next, using the Young’s inequality, we can rewrite last inequality as follows \[\begin{align} \label{eq3465} I&\geq \int_{t_1}^{t_2}\int_{B_r(\xi_0)}(|\nabla_{\mathbb{H}^N}w_+|^2\psi^2\nu^2)(\xi,t)\;d\xi\,dt-\epsilon \int_{t_1}^{t_2}\int_{B_r(\xi_0)}(|\nabla_{\mathbb{H}^N}w_+|^2\psi^2\nu^2)(\xi,t)\;d\xi\;dt\nonumber\\ &\quad - C(\epsilon,2)\int_{t_1}^{t_2}\int_{B_r(\xi_0)}(w_+^2(\nabla_{\mathbb{H}^N}\psi)^2\nu^2)(\xi,t)\;d\xi\;dt\nonumber\\ & \geq \frac{1}{2} \int_{t_1}^{t_2}\int_{B_r(\xi_0)}(|\nabla_{\mathbb{H}^N}w_+|^2\psi^2\nu^2)(\xi,t)\;d\xi\,dt-c\int_{t_1}^{t_2}\int_{B_r(\xi_0)}(w_+^2(\nabla_{\mathbb{H}^N}\psi)^2\nu^2)(\xi,t)\;d\xi\;dt \end{align}\tag{11}\] with \(\epsilon=\frac{1}{2}\). Hence we have the desired result. ◻

Proof. This lemma is essentially an extension of the Lemma 4.1 of [23] to the mixed local-nonlocal case. As most of the calculations are similar, we shall only mention the changes. Assume \[\begin{align} k\geq \Big[ \fint_{t_0-R^{2}}^{t_0} \left( \fint_{B_R(\xi_0)} u_+(\eta,t) \,d\eta\right)^2\,dt \Big]^\frac{1}{2}, \end{align}\] and insert \(\theta_2=r^{2}\) into the energy inequality (?? ) to get \[\begin{align} &\mathop{\mathrm{ess\,sup}}_{t_0-r^{2s}<t<t_0}\int_{B_r(\xi_0)}(w_+^2\psi^p\nu^2)(\xi,t)d\xi+\int_{t_0-\theta_2}^{t_0}\int_{B_r(\xi_0)}(|\nabla_{\mathbb{H}^N}w_\pm|^2\psi^2\nu^2)(\xi,t)\;d\xi\,dt\\ &\qquad \leq C\Bigg(\frac{r^{1-s}}{r-l}\Bigg)^2\|w_+\|^2_{L^2(t_0-r^{2},t_0;L^2(B_r(\xi_0)))}+ \underbrace{\int_{t_0-r^{2}}^{t_0}\int_{B_r(\xi_0)}(w_+^2\psi^2)(\xi,t)\nu(t)\partial_t\nu(t)\;d\xi\,dt}_{I}\\ &\quad\quad\quad+\underbrace{CR^{-2s}\left(\frac{R}{r-l}\right)^{Q+2s}\int_{t_0-r^{2}}^{t_0}\int_{B_r(\xi_0)}w_+(\xi,t)\psi^2(\xi)\nu^2(t)\Bigg(\fint_{B_R(\xi_0)} u_+(\eta,t)\,d\eta\Bigg)d\xi\,dt}_{II}\\ & \quad\quad\quad\quad+C\underbrace{\int_{t_0-r^{2}}^{t_0}\int_{B_r(\xi_0)}(w_\pm^2(\nabla_{\mathbb{H}^N}\psi)^2\nu^2)(\xi,t)\;d\xi\;dt.}_{III} \end{align}\] The estimation of \(I\) follows from the fact that \(\partial_t\nu<\frac{C}{r^2}\) that is \[\begin{align} I\leq \frac{C}{r^2}\int_{t_0-r^{2}}^{t_0}\int_{B_r(\xi_0)}w^2_+(\xi,t)\,d\xi \,dt. \end{align}\] Next, using the fact \(|\nabla_{\mathbb{H}^N}\psi|\leq \frac{C}{r-l}\), we have in \(III\), \[\begin{align} III\leq \frac{C}{(r-l)^2}\int_{t_0-r^{2}}^{t_0}\int_{B_r(\xi_0)}w^2_+(\xi,t)\,d\xi \,dt. \end{align}\] Now, we try to estimate \(II\). This estimation is inspired from the corresponding steps of lemma 4.1 in [23]. Estimation of II: We start this estimation by noting that \[\begin{align} \mathcal{G}(t):=\fint_{B_R(\xi_0)}u_+(\eta,t)\,d\eta\in L^{\infty,2}_{loc}(\Omega_T). \end{align}\] Now, using the Jensen, Hölder and Youngs’ inequality, \(II\) implies, \[\begin{align} II&=CR^{-2s}\left(\frac{R}{r-l}\right)^{Q+2s}\int_{t_0-r^2}^{t_0}\int_{B_r(\xi_0)}\mathcal{G}(t)(w_+\psi^2\nu^2)(\xi,t)\,d\xi\\ &\leq CR^{-2s}\left(\frac{R}{r-l}\right)^{Q+2s}\|\mathcal{G}\|_{L^{\infty,2}(z_0+\mathcal{Q}^\ominus_r)}\|w_+\psi\nu\|_{L^{1,2}(z_0+\mathcal{Q}^\ominus_r)}\\ &\leq C \left(\frac{R}{r-l}\right)^{Q+2s}\|kR^{-s}w_+\psi\nu\|_{L^{1,2}(z_0+\mathcal{Q}^\ominus_r)}\\ &\leq \frac{Ck^2}{\epsilon}\left(\frac{R}{r-l}\right)^{2(Q+2s)}\frac{1}{R^s}\|\chi_{\{u>g(.)+k\}}\|_{L^{1,2}(z_0+\mathcal{Q}^\ominus_r)}+\frac{\epsilon}{R^s}\|\left(w_+\psi\nu\right)^2\|_{L^{1,2}(z_0+\mathcal{Q}^\ominus_r)}\\ &\leq \frac{Ck^2}{\epsilon}\left(\frac{R}{r-l}\right)^{2(Q+2s)}r^{-1}\left(\int_{t_0-r^2}^{t_0}|\mathcal{A}^g_+(t;\xi_0,r,k)|^2\right)^\frac{1}{2}+\epsilon\underbrace{\frac{1}{R^s}\|\left(w_+\psi\nu\right)^2\|_{L^{1,2}(z_0+\mathcal{Q}^\ominus_r)}.}_{IV} \end{align}\] In the sequence of calculations, we use the fact that \(\frac{r}{R}<1\) and \(r<1\) at the last inequality. Once again using these facts and employing the Hölder inequality in \(IV\) to obtain the following \[\begin{align} IV&\leq Cr^Q\left(\frac{r}{R^s}\right)\left(\fint_{t_0-r^2}^{t_0}\left(\fint_{B_r}\left(w_+\psi\nu\right)^2\right)^2\right)^\frac{1}{2}\\ &\leq Cr^Q\left(\frac{r}{R}\right)^s\left(\fint_{t_0-r^2}^{t_0}\left(\fint_{B_r}\left(w_+\psi\nu\right)^{\hat{q}}\right)^\frac{2}{1+\kappa}\right)^\frac{1}{2}\\ &\leq Cr^Q\left(\frac{r}{R}\right)^s\left(\fint_{t_0-r^2}^{t_0}\left(\fint_{B_r}\left(w_+\psi\nu\right)^{\hat{q}}\right)^2\right)^\frac{2}{2\hat{q}}\\ &\leq C\|w_+\psi\nu\|^2_{L^{\hat{q},2\hat{q}}(z_0+\mathcal{Q}^\ominus_r)}. \end{align}\] Next, we use Theorem 8 such that \[\begin{align} IV\leq C\Bigg(&\underbrace{\iint_{z_0+\mathcal{Q}^\ominus_r}|\nabla_{\mathbb{H}^N}(w_+\psi\nu)|^2\,d\xi dt}_{V}+r^{-2}\iint_{z_0+\mathcal{Q}^\ominus_r}\left(w_+\psi\nu\right)^2(\xi,t)\,d\xi dt\\ & +\mathop{\mathrm{ess\,sup}}_{t_0-r^2<t<t_0}\int_{B_r(\xi_0)}\left(w_+\psi\nu\right)^2(\xi,t)\,d\xi\Bigg). \end{align}\] Next, using the fact that \(|\nabla_{\mathbb{H}^N}\psi|\leq \frac{C}{r-l}\) in \(V\), we have the following, \[\begin{align} V\leq \iint_{z_0+\mathcal{Q}^\ominus_r}\left(|\nabla_{\mathbb{H}^N}w_+|\psi\nu\right)^2\,d\xi dt+\frac{C}{(r-l)^2}\iint_{z_0+\mathcal{Q}^\ominus_r}w_+^2(\xi,t)\,d\xi dt. \end{align}\] Then, \(II\) will be as follows, \[\begin{align} II&\leq Ck^2\left(\frac{R}{r-l}\right)^{2(Q+2s)}R^{2(1-s)}r^{-1}\left(\int_{t_0-r^2}^{t_0}|\mathcal{A}^g_+(t;\xi_0,r,k)|^2\right)^\frac{1}{2}\\ &\quad\quad +\underbrace{\iint_{z_0+\mathcal{Q}^\ominus_r}\left(|\nabla_{\mathbb{H}^N}w_+|\psi\nu\right)^2\,d\xi dt}_{VI}+\frac{C}{(r-l)^2}\iint_{z_0+\mathcal{Q}^\ominus_r}w_+^2(\xi,t)\,d\xi dt\\ &\quad\quad\quad\quad+\underbrace{r^{-2}\iint_{z_0+\mathcal{Q}^\ominus_r}\left(w_+\psi\nu\right)^2(\xi,t)\,d\xi dt}_{VII}+\underbrace{\mathop{\mathrm{ess\,sup}}_{t_0-r^2<t<t_0}\int_{B_r(\xi_0)}\left(w_+\psi\nu\right)^2(\xi,t)\,d\xi}_{VIII}. \end{align}\] Merging \(VII\) with the estimate \(I\), sending \(VI,VIII\) to the left, using the facts \((r-l)^2\leq (r^2-l^2)\) and combining all the estimates, we have our result. ◻

Proof. For the sake of simplicity, we assume \((\xi_0,t_0)=(0,0)\). We proceed this with the supersolution. The sub counterpart follows the similar steps. For \(n=0,1,2,...\), we define \(\rho_n,\tilde{\rho}_n,B_n,\tilde{B}_n,\mathcal{Q}_n^\ominus(\delta),\tilde{\mathcal{Q}}_n^\ominus(\delta)\) as follows \[\begin{align} \begin{cases} & \rho_n=\frac{\rho}{2}+\frac{\rho}{2^{n+1}},\;\;\tilde{\rho}_n=\frac{\rho_n+\rho_{n+1}}{2},\\ & B_n=B_{\rho_n}(0),\,\,\tilde{B}_n=B_{\tilde{\rho}_n}(0),\\ &\mathcal{Q}_n^\ominus(\delta)=B_n\times(-\delta\rho_n^2,0],\,\,\tilde{\mathcal{Q}}_n^\ominus(\delta)=\tilde{B}_n\times(-\delta\tilde{\rho}_n^2,0]. \end{cases} \end{align}\] Here we consider the Caccioppoli estimate in Remark 10 with domain \(B_n\) and parabolic cylinder \(\mathcal{Q}^\ominus_n(\delta)\) to \(w_+(\xi,t)=(u-g-k_n)_+\) with the time dependent level \(g(t)+k_n\) where for \(n=0,1,2,...\) \[\begin{align} k_n=\mu^+-\lambda_n\omega,\,\,\,where\lambda_n=\frac{\lambda}{2}+\frac{\lambda}{2^{n+1}}, \end{align}\] and \[g(t)=\mathfrak{C}\int_{-R^2}^t\int_{\mathbb{H}^N\setminus B_R}\frac{|u_+(\eta,\tau)|}{|\eta^{-1}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta d\tau,\] where positive constant \(\mathfrak{C}\) to be chosen later. Moreover two cut-off functions \(\psi\) and \(\nu\) in \(B_n\) and \((-\delta\rho_n^2,0]\) respectively \[\begin{align} \psi=\begin{cases} & 1\,\,\,\mathrm{in } B_{n+1}\\ & 0\,\,\,\mathrm{in } \tilde{B}^c_n \end{cases} \,\,\,\,\,\,\,\,\,\,\mathrm{such that } |\nabla_{\mathbb{H}^N}\psi|\leq \gamma\frac{2^{(n+2)}}{\rho}. \end{align}\] and \[\begin{align} \nu =\begin{cases} & 1\,\,\,\mathrm{in } (-\delta\rho_{n+1}^2,0]\\ & 0\,\,\,\mathrm{in } (-\delta\tilde{\rho}_n^2,0]^c \end{cases} \,\,\,\,\,\,\,\,\,\,\mathrm{such that } |\partial_t\nu|\leq \gamma\frac{2^{2(n+2)}}{\delta\rho^2}. \end{align}\] Considering all these entities, (?? ) implies \[\begin{align} \label{eq5461} &\mathop{\mathrm{ess\,sup}}_{-\delta\rho^2_{n+1}<t<0}\int_{B_{n+1}}w^2_+(\xi,t)\,d\xi+\iint_{\mathcal{Q}_{n+1}^\ominus(\delta)}|\nabla_{\mathbb{H}^N}w_+|^2\,d\xi dt\nonumber\\ &\quad +\int_{-\delta\rho^2_{n+1}}\iint_{B_{n+1}\times B_{n+1}}\frac{|w_+(\xi,t)-w_+(\eta,t)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\nonumber\\ &\quad\quad\leq\gamma \int_{-\delta\rho_n^2}^0\iint_{B_n\times B_n}\max\left\{w^2_+(\xi,t),w^2_+(\eta,t)\right\}\frac{|\psi(\xi)-\psi(\eta)|^2}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\nu^2(t)\,d\xi d\eta dt\nonumber\\ &\quad\quad\quad +\gamma\iint_{\mathcal{Q}^\ominus_n(\delta)}\left(w_+\nabla_{\mathbb{H}^N}\psi\,\nu\right)^2(\xi,t)\,d\xi dt+ \gamma\iint_{\mathcal{Q}^\ominus_n(\delta)}\left(w_+^2\psi^2\nu\,\partial_t\nu\right)(\xi,t)\,d\xi dt\\ &\quad\quad\quad\quad+\iint_{\mathcal{Q}^\ominus_n(\delta)}(w_+\psi^2\nu^2)(\xi,t)\left\{\int_{\mathbb{H}^N\setminus B_n}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}d\xi dt\nonumber\\ &\quad\quad\quad\quad\quad-\mathfrak{C}\iint_{\mathcal{Q}^\ominus_n(\delta)}g'(t)(w_+\psi^2\nu^2)(\xi,t)\,d\xi dt\nonumber\\ &\quad\quad\quad\quad\quad\quad =: I_1+I_2+I_3+I_4+I_5.\nonumber \end{align}\tag{25}\] For the terms \(I_2\) and \(I_3\), we have the following by using the definition \(\mathcal{A}_n:=\{(\xi,t)\in\mathcal{Q}_n^\ominus(\delta):u(\xi,t)-g(t)>k_n\}\) \[\begin{align} I_2\leq \gamma\frac{2^{2n}}{\rho^2}(\lambda\omega)^2|\mathcal{A}_n|, \end{align}\] and \[\begin{align} I_3\leq \gamma\frac{2^{2n}}{\delta\rho^2}(\lambda\omega)^2|\mathcal{A}_n|. \end{align}\] In the case of \(I_1\), using assumption of our lemma \[\begin{align} I_1&\leq \gamma \frac{2^{2n}}{\rho^2}\,2(\lambda\omega)^2 \int_{\delta\rho_n^2}^0\iint_{B_n\times B_n}\frac{\chi_{\{u(\xi,t)-g(t)>k_n\}}}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2(s-1)}}\,d\xi d\eta dt\\ &=\gamma \frac{2^{2n}}{\rho^2}\,2(\lambda\omega)^2\iint_{\mathcal{Q}_n^\ominus(\delta)}\chi_{\{u(\xi,t)-g(t)>k_n\}}\,d\xi\left[\int_{B_n}\frac{1}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2(s-1)}}\right]\\ &\leq \gamma\frac{2^{2n}}{\rho^2}(\lambda\omega)^2|\mathcal{A}_n|. \end{align}\] In the last line we use the fact that \(\frac{1}{\rho^{2s}}\leq \frac{1}{\rho^{2}}\) due to \(\rho<1\). Next, we want to estimate \(I_4\). Prior to this estimate, we note the following observations

• If \(|\xi|<\rho_{n+1}\) and \(|\eta|>\rho_n\), then \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{\rho_{n+1}}{\rho_n}\geq \frac{1}{2^{n+2}}. \end{align}\]

• If \(|\xi|\leq \rho\) and \(|\eta|>R\), then \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq \frac{1}{2},\,\,\,\,provided \rho\leq\frac{R}{2} . \end{align}\]

Considering all the entities with the fact that \(u\geq 0\) a.e. in \(\mathcal{Q}\), \(I_4\) implies \[\begin{align} \label{eq5462} I_4&=\iint_{\mathcal{Q}^\ominus_n(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\left\{\int_{\mathbb{H}^N\setminus B_R}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta+\int_{\mathbb{H}^N\setminus B_n}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}d\xi dt\nonumber\\ &\leq \gamma 2^{(Q+2s)n}\iint_{\mathcal{Q}^\ominus_n(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\left\{\int_{B_R\setminus B_n}\frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}dt\nonumber\\ &\quad\quad+\gamma\iint_{\mathcal{Q}^\ominus_n(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\left\{\int_{\mathbb{H}^N\setminus B_R}\frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}dt\nonumber\\ &\leq \gamma 2^{(Q+2s)n}\frac{\lambda\omega}{\rho^{2s}}\iint_{\mathcal{Q}_n^\ominus(\delta)}w_+(\xi,t)\,d\xi+\gamma \iint_{\mathcal{Q}_n^\ominus(\delta)}g'(t)\left(w_+\psi^2\nu^2\right)\,d\xi dt\nonumber\\ &\leq \gamma 2^{(Q+2s)n}\frac{(\lambda\omega)^2}{\rho^{2}}|\mathcal{A}_n|+\gamma \int_{\mathcal{Q}_n^\ominus(\delta)}g'(t)\left(w_+\psi^2\nu^2\right)\, d\xi dt. \end{align}\tag{26}\] In the second to last step, we employ Lemma 4 and in the last step, we use definition of \(\mathcal{A}_n\). Next, changing the arbitrary constant from \(\gamma\) to \(\mathfrak{C}\) in the last line of (26 ), we can cancel out \(I_5\). Hence, combining all the estimate into (25 ), we have \[\begin{align} \label{eq5463} \mathop{\mathrm{ess\,sup}}_{-\delta\rho^2_{n+1}<t<0}\int_{B_{n+1}}&w^2_+\,d\xi+\iint_{\mathcal{Q}_{n+1}^\ominus(\delta)}|\nabla_{\mathbb{H}^N}w_+|^2\,d\xi dt\nonumber\\ &\leq \gamma \frac{2^{n(Q+2s+2)}}{\delta\rho^2}(\lambda\omega)^2|\mathcal{A}_n|. \end{align}\tag{27}\] Consequently, using Hölder inequality, (27 ) and parabolic Sobolev inequality \[\begin{align} \frac{\lambda\omega}{2^{n+2}}|\mathcal{A}_{n+1}|&\leq\iint_{\mathcal{Q}_n^\ominus(\delta)}w_+(\xi,t)\,d\xi dt\\ & \leq \left(\iint_{\mathcal{Q}^\ominus_n(\delta)}w_+^{2\kappa}\,d\xi dt\right)^\frac{1}{2\kappa}|\mathcal{A}_n|^{1-\frac{1}{2\kappa}}\\ &\leq \gamma \left(\rho^2\iint_{\mathcal{Q}^\ominus_n(\delta)}|\nabla_{\mathbb{H}^N}u(\xi,t)|^2\,d\xi dt+\iint_{\mathcal{Q}^\ominus_n(\delta)}|u(\xi,t)|^2\,d\xi dt\right)^\frac{1}{2\kappa}\\ &\quad \quad\times\left(\mathop{\mathrm{ess\,sup}}_{-\delta\rho_n^2<t<0}\fint_{B_n}w_+^2\,d\xi\right)^\frac{1}{Q\kappa}|\mathcal{A}_n|^{1-\frac{1}{2\kappa}}\\ &\leq \gamma b^n\delta^{-\left[\frac{1}{\kappa}+\frac{2}{Q\kappa}\right]}\rho^{-(Q+2)\frac{1}{Q\kappa}}(\lambda\omega)^{2\left[\frac{1}{Q\kappa}+\frac{1}{2\kappa}\right]}|\mathcal{A}_n|^{1+\frac{1}{\delta\kappa}}, \end{align}\] where \(b=b(Q)\). In the course of these calculations, we have denoted \(\frac{Q}{Q-2}\) as \(\kappa\). Now, the previous estimate leads to a recursive inequality by \(Y_n=\frac{|\mathcal{A}_n|}{|\mathcal{Q}^\ominus_n(\delta)|}\), viz, there is some constant (depending only on data) such that \[\begin{align} Y_{n+1}\leq \gamma(2b)^n\delta^{-\left[\frac{1}{\kappa}+\frac{2}{Q\kappa}\right]}Y_n^{1+\frac{1}{\delta\kappa}}. \end{align}\] Hence by using the convergence Lemma 5, taking \(g(T_2)\leq \frac{1}{4}\lambda\omega\) and redefining \(4\mathfrak{C}\) as \(\mathfrak{C}\) (as done in [15]), we have the result. ◻

Proof. Without loss of any generality, we assume \((\xi_0,t_0)=(0,0)\). Unlike the previous lemma, this relies on point wise initial data. We want to prove this lemma using the energy estimate (?? ) but on the forward-in-time domain. Moreover, as far as cut-off function is concerned, we only consider time-independent cut-off function i.e. \(\partial_t\nu\equiv 0\). Consider \(w_+\) and \(g(t)\) ass in the previous lemma. Let us introduce \(\rho_n,\tilde{\rho}_n,B_n,\tilde{B}_n,\mathcal{Q}^\oplus_n(\delta),\tilde{\mathcal{Q}}^\oplus_n(\delta),k_n,\lambda_n\) as follows \[\begin{align} \begin{cases} \rho_n=\frac{\rho}{2}+\frac{\rho}{2^{n+1}},\,\,\tilde{\rho}_n=\frac{\rho_n+\rho_{n+1}}{2},\\ B_n=B_{\rho_n}(0),\,\,\tilde{B}_n=B_{\tilde{\rho}_n}(0),\\ \mathcal{Q}^\oplus_n(\delta)=B_n\times (0,\delta\rho_n^2],\,\,\tilde{\mathcal{Q}}^\oplus_n(\delta)=\tilde{B}_n\times (0,\delta\tilde{\rho}_n^2],\\ \lambda_n=\frac{\lambda}{2}+\frac{\lambda}{2^{n+1}},\,\,k_n=\mu^+-\lambda_n\omega. \end{cases} \end{align}\] Moreover, cut-off function \(\psi(x)\) in \(B_n\) is defined as \[\begin{align} \psi\equiv \begin{cases} 1\,\,\,\,onB_{n+1}\\ 0\,\,\,\,in\tilde{B}^c_n \end{cases} \,\,\,\,\,\,such that |\nabla_{\mathbb{H}^N}\psi|\leq \frac{\gamma 2^{n+4}}{\rho}. \end{align}\] As a result, we have \[\begin{align} \mathop{\mathrm{ess\,sup}}_{0<t<\delta\rho_{n+1}^2}&\int_{B_{n+1}}w^2_+\,d\xi+\iint_{\mathcal{Q}_n^\oplus(\delta)}|\nabla_{\mathbb{H}^N}w_+|^2\,d\xi dt\\ &\leq \gamma\int_0^{\delta\rho_n^2}\iint_{B_n\times B_n}\max\left\{w^2_+(\xi,t),w_+^2(\eta,t)\right\}\frac{|\psi(\xi)-\psi(\eta)|^2}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\\ &\quad\quad+\gamma\iint_{\mathcal{Q}^\oplus_n(\delta)}\left(w_+\nabla_{\mathbb{H}^N}\psi\,\nu\right)^2\,d\xi dt+\iint_{\mathcal{Q}^\oplus_n(\delta)}\left(w_+\psi^2\nu^2\right)\,d\xi\left\{\int_{\mathbb{H}^N\setminus B_n}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}dt\\ &\quad\quad\quad\quad -\mathfrak{C}\iint_{\mathcal{Q}^\oplus_n(\delta)}g'(t)(w_+\psi^2\nu^2)(\xi,t)\,d\xi dt. \end{align}\] Proceeding as in the previous lemma with suitable choice of \(\mathfrak{C}\) and \(\gamma\), we conclude that the left hand side of the previous inequality is dominated by \[\begin{align} \gamma\frac{2^{(Q+2s+2)n}}{\delta\rho^2}(\lambda\omega)^2|\mathcal{A}_n|. \end{align}\] where we use the definition \(\mathcal{A}_n:=\{(\xi,t)\in\mathcal{Q}_n^\oplus(\delta):u(\xi,t)-g(t)>k_n\}\). After running the De Giorgi iteration (as done in Lemma 12) on the last estimation, there exists a constant \(\iota_0\) (depending only on data), such that with the choice of \(\delta=\iota_0\), we have \[\begin{align} u(\xi,t)-g(t)\leq \mu^+-\frac{1}{2}\xi\omega\,\,\,\mathrm{ a.e. in } B_{\frac{\rho}{2}}\times (0,\iota_0\rho^2). \end{align}\] Consequently, if we impose \(g(T_2)\leq \frac{\xi\omega}{4}\) after redefining \(\mathfrak{C}=4\mathfrak{C}\), then \[\begin{align} u\leq -\frac{\xi\omega}{4}+\mu^+\,\,\,\mathrm{ a.e. in } B_{\frac{\rho}{2}}\times (0,\iota_0\rho^2). \end{align}\] ◻

Proof. For the sake of simplicity, we assume \((\xi_0,t_0)=(0,0)\). We prove this lemma for the subsolution case. For convenience, let us denote \(aM:=-a\lambda\omega+\mu^+\) for any real number \(a\), and for any \(t>0\) \[\begin{align} \mathcal{A}_{aM,\rho}(t):=\{u(\cdot,t)>aM\}\cap B_\rho, \end{align}\] then it is evident that \[\begin{align} \label{eq5465} |\mathcal{A}_{aM,\rho}(0)|\leq (1-\alpha)|B_\rho|. \end{align}\tag{28}\] Moreover, we set a time-independent cut-off function \(\psi\) on \(B_\rho\) as follows \[\begin{align} \psi\equiv \begin{cases} 1\,\,\,\, onB_{(1-\sigma)\rho},\\ 0\,\,\,\, onB^c_{(1-\sigma/2)\rho}, \end{cases} \,\,\,\, such that |\nabla_{\mathbb{H}^N}\psi|\leq \frac{\gamma}{\sigma\rho}. \end{align}\] Next, consider the truncated function \(w_+=(u-M)_+\) and examine the energy estimate (?? ) over the domain \(B_\rho\times (0,\delta\rho^2]\). Then, for all \(t\in (0,\delta\rho^2]\), we have \[\begin{align} \label{eq5466} \int_{B(1-\sigma)\rho}w^2_+(\xi,t)\,d\xi&\leq \int_{B_\rho}w^2_+(\xi,0)\,d\xi+\gamma\int_0^{\delta\rho^2}\int_{B_\rho}\left(w^2_+\,\nabla_{\mathbb{H}^N}\psi\right)^2(\xi,t)\,d\xi\nonumber\\ &\quad+\gamma\int_0^{\delta\rho^2}\max\{w^2_+(\xi,t),w^2_+(\eta,t)\}\frac{|\psi(\xi)-\psi(\eta)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\\ &\quad\quad+\gamma\int_0^{\delta\rho^2}\int_{B_\rho}w_+(\xi,t)\left\{\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,d\xi dt\nonumber\\ &\quad\quad\quad =:I_1+I_2+I_3+I_4,\nonumber \end{align}\tag{29}\] where \(\gamma=\gamma(\texttt{data})\). Now, we estimate the terms on the right hand side of previous inequality one by one. Consider \(I_3\) and \(I_2\) successively. \[\begin{align} I_3&\leq \gamma\delta\rho^2\left(\frac{\lambda\omega}{\sigma\rho}\right)^2\iint_{B_\rho\times B_\rho}\frac{1}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s-2}}\,d\xi dt\\ &=\gamma\delta \left(\frac{\lambda\omega}{\sigma}\right)^2|B_\rho|, \end{align}\] and \[\begin{align} I_2&\leq \gamma\left(\frac{\lambda\omega}{\sigma\rho}\right)^2\delta\rho^2|B_\rho|\\ &=\gamma\delta\left(\frac{\lambda\omega}{\sigma}\right)^2|B_\rho|. \end{align}\] Before proceeding towards \(I_4\), we observe that for \(\xi\in supp \psi\) i.e. \(|\xi|\leq (1-\frac{\sigma}{2})\rho\) and \(|\eta|\geq \rho\), we have \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{|\xi|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq \frac{\sigma}{2}. \end{align}\] Also, when \(|\eta|_{\mathbb{H}^N}\geq R\) and \(|\xi|_{\mathbb{H}^N}\leq \rho\), we have \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq\,\,\,\mathrm{ provided }\,\rho\leq \frac{R}{2}. \end{align}\] So, \(I_4\) implies with the enforcement of \[\begin{align} \frac{1}{\delta}R^{-2}\int_{T_1}^{T_2}\mathrm{Tail }[(u-\mu^\pm)_\pm(t);R,\xi_0]\leq\lambda\omega, \end{align}\] we have \[\begin{align} I_4&\leq 2^{Q+2s}\frac{\lambda\omega}{\sigma^{Q+2s}}|B_\rho|\left\{\int_0^{\delta\rho^2}\int_{B_R\setminus B_\rho} \frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta dt+\int_0^{\delta\rho^2}\int_{\mathbb{H}^N\setminus B_R} \frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta dt\right\}\\ &\leq \gamma \frac{\lambda\omega}{\sigma^{Q+2}}|B_\rho|\left\{\gamma\delta\lambda\omega+\int_0^{\delta\rho^2}\int_{\mathbb{H}^N\setminus B_R} \frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta dt\right\}\\ &\leq \gamma \frac{\delta(\lambda\omega)^2}{\sigma^{Q+2}}|B_\rho|. \end{align}\] Combining all the estimates into (29 ), we have \[\begin{align} \int_{B(1-\sigma)\rho}w^2_+(\xi,t)\,d\xi&\leq \int_{B_\rho}w^2_+(\xi,0)\,d\xi+\gamma\frac{\delta(\lambda\omega)^2}{\sigma^{Q+2}}|B_\rho|\\ &\leq \left[(1-\alpha)+\frac{\gamma\delta}{\sigma^{Q+2}}\right](\lambda\omega)^2|B_\rho|. \end{align}\] for all \(t\in(0,\delta\rho^2]\). In the last step of the above estimates, we used (28 ). The left-hand side of the above inequality can be estimated for \(\varepsilon\in(0,1)\), \[\begin{align} \int_{B(1-\sigma)\rho}w^2_+(\xi,t)\,d\xi&\geq \int_{B_{(1-\sigma)\rho}\cap\{u(\cdot,t)>\varepsilon M\}}w^2_+(\xi,t)\,d\xi\\ &\geq (1-\varepsilon)^2(\lambda\omega)^2|\mathcal{A}_{\varepsilon M,(1-\sigma)\rho}(t)|. \end{align}\] Also \[\begin{align} |\mathcal{A}_{\varepsilon M,\rho}(t)|&=\left|\mathcal{A}_{\varepsilon M,(1-\sigma)\rho}(t)\cup\left(\mathcal{A}_{\varepsilon M,\rho}(t)-\mathcal{A}_{\varepsilon M,(1-\sigma)\rho}(t)\right)\right|\\ &\leq |\mathcal{A}_{\varepsilon M,(1-\sigma)\rho}(t)|+|B_\rho-B_{(1-\sigma)\rho}|\\ & \leq |\mathcal{A}_{\varepsilon M,(1-\sigma)\rho}(t)|+Q\sigma|B_\rho|. \end{align}\] Now, proceeding as in [15], with a suitable choice of parameters \(\sigma=\frac{\alpha}{8Q},\,\delta=\frac{1}{8\gamma}\sigma^{Q+2}\alpha,\,(1-\varepsilon)\geq \sqrt{\frac{1-\frac{3}{4}\alpha}{1-\frac{1}{2}\alpha}}\), we have our result. ◻

Proof. For convenience, we will assume \((\xi_0,t_0)=(0,0)\). Consider the larger cylinder \[\begin{align} \mathcal{Q}_{8\rho}(\delta)=\mathcal{Q}^\oplus_{8\rho}(\delta)\cup \mathcal{Q}^\ominus_{8\rho}(\delta)=B_{8\rho}(0)\times(-\delta\rho^2,\delta\rho^2] \end{align}\] and the truncated function \[\begin{align} w_+=(u-g-M_n)_+,\,\,\,\,\,\mathrm{ where } M_n:=-\frac{\varepsilon\lambda\omega}{2^n}+\mu^+,\,g(t)=\mathfrak{C}\int_{-R^2}^t\int_{\mathbb{H}^N\setminus B_R}\frac{|u_+(\eta,\tau)|}{|\eta^{-1}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta d\tau, \end{align}\] for \(n=0,1,2,3,...\) and \(\varepsilon\) is as in previous Lemma 14. The nonnegative, piecewise smooth cut-off functions \(\psi\) and \(\nu\) are defined as follows \[\begin{align} \psi\equiv\begin{cases} 1\,\,\, onB_{4\rho},\\ 0\,\,\, onB^c_{8\rho}, \end{cases} \,\,\,\,\,\, such that |\nabla_{\mathbb{H}^N}\psi|\leq \frac{\gamma}{\rho}, \end{align}\] and \[\begin{align} \nu\equiv\begin{cases} 1\,\,\, on \, t\geq 0,\\ 0\,\,\, on \,t\leq -\delta\rho^2, \end{cases} \,\,\,\,\,\, such that |\partial_t\nu|\leq \frac{\gamma}{\delta\rho^2}. \end{align}\] Using all this entities in (?? ), we have \[\begin{align} \label{eq5467} \iint_{\mathcal{Q}_{4\rho}^\oplus(\delta)}|\nabla_{\mathbb{H}^N}w_+|^2\,d\xi dt&\leq \gamma\int_{-\delta\rho^2}^{\delta\rho^2}\iint_{B_{8\rho}\times B_{8\rho}}\max\left\{w^2_+(\xi,t),w^2_+(\eta,t)\right\}\frac{|\psi(\xi)-\psi(\eta)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\nonumber\\ &\quad+ \gamma\iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\,\nabla_{\mathbb{H}^N}\psi\,\nu\right)^2(\xi,t)\,d\xi dt+\iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w^2_+\psi^2\nu\,\partial_t\nu\right)(\xi,t)\,d\xi dt\nonumber\\ &\quad\quad+\iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\left\{\int_{\mathbb{H}^N\setminus B_{8\rho}}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,dt\\ &\quad\quad\quad-\mathfrak{C}\iint_{\mathcal{Q}^\ominus_n}g'(t)(w_+\psi^2\nu^2)(\xi,t)\,d\xi dt.\nonumber \end{align}\tag{30}\] Note that we remove the first, second and third terms on the left hand side of (?? ) due to choice of cut-off function and non-negativeness of the terms . Next, we estimate each term on the right hand side of (30 ) one by one. The first term on the right hand side of (30 ) is estimated as \[\begin{align} \int_{-\delta\rho^2}^{\delta\rho^2}\iint_{B_{8\rho}\times B_{8\rho}}&\max\left\{w^2_+(\xi,t),w^2_+(\eta,t)\right\}\frac{|\psi(\xi)-\psi(\eta)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt \\&\leq \frac{\gamma}{\rho^2}\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2\,(2\delta\rho^2)\iint_{B_{8\rho}\times B_{8\rho}}\frac{\chi_{\{u(\xi,t)-g(t)>M_n\}}}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2(s-1)}}\,d\xi d\eta\\ &\leq \gamma \delta\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|B_{4\rho}|. \end{align}\] In these course of calculations, we have used the measure theoretical assumptions of our lemma and \(|B_{8\rho}|=\mathcal{C}|B_{4\rho}|\). The second and third terms of the right hand side of (30 ) can be dealt with as follows with the help of the measure theoretical information, \(|B_{8\rho}|=\mathcal{C}|B_{4\rho}|\) and properties of cut-off functions \[\begin{align} \iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\,\nabla_{\mathbb{H}^N}\psi\,\nu\right)^2(\xi,t)\,d\xi dt&\leq \frac{\gamma}{\rho^2}\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|\mathcal{Q}_{8\rho}(\delta)|\\ &\leq\gamma\delta\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|B_{4\rho}|, \end{align}\] and \[\begin{align} \iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w^2_+\psi^2\,\partial_t\nu\right)(\xi,t)\,d\xi dt&\leq \frac{\gamma}{\delta\rho^2}\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|\mathcal{Q}_{8\rho}(\delta)|\\ &\leq \gamma \left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|B_{4\rho}|. \end{align}\] The second to last term on the right hand side of (30 ) is estimated by employing tail estimation. Prior to the estimation we want to mention the following observations

• If \(|\xi|_{\mathbb{H}^N}\leq 4\rho\), \(|\eta|_{\mathbb{H}^N}\geq 8\rho\), then \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{|\xi|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{4\rho}{8\rho}=\frac{1}{2}. \end{align}\]

• If \(|\xi|\leq 4\rho\), \(|\eta|\geq R\) and suppose \(\rho\leq \frac{R}{8}\), then \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{|\xi|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{4\rho}{R}\geq\frac{1}{2}. \end{align}\]

We have the following estimate. \[\begin{align} \iint_{\mathcal{Q}_{8\rho}(\delta)}&\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\left\{\int_{\mathbb{H}^N\setminus B_{8\rho}}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,dt\\ &=\iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\left\{\int_{\mathbb{H}^N\setminus B_R}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta+\int_{B_R\setminus B_{8\rho}}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,dt\\ &\leq \gamma \iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\left\{\int_{\mathbb{H}^N\setminus B_R}\frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta+\int_{B_R\setminus B_{8\rho}}\frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,dt\\ &\leq \gamma \iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\int_{\mathbb{H}^N\setminus B_R}\frac{u_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta dt\\ &\quad \quad +\gamma \iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\int_{B_R\setminus B_{8\rho}}\frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta dt\\ &\leq \gamma\delta\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|B_{4\rho}|+\gamma \iint_{\mathcal{Q}_{8\rho}(\delta)}\left(w_+\psi^2\nu^2\right)(\xi,t)\,d\xi\int_{\mathbb{H}^N\setminus B_R}\frac{u_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta dt. \end{align}\] In this series of calculations, we have used Lemma 4 and the property \(|B_{8\rho}|=\mathcal{C}|B_{4\rho}|\). Moreover changing \(\gamma\) to \(\mathfrak{C}\) in the last integral of the last step to cancel out the last term on the right hand side of (30 ). Now, combining all the estimates into (30 ) and considering \(\delta<1\), we have \[\begin{align} \iint_{\mathcal{Q}_{4\rho}^\oplus(\delta)}|\nabla_{\mathbb{H}^N}w_+|^2\,d\xi dt\leq \gamma\left(\frac{\varepsilon\lambda\omega}{2^n}\right)^2|B_{4\rho}|, \end{align}\] where \(\gamma\) depends upon only data. Next, we recall De Giorgi isoperimetric inequality for the levels \[\begin{align} l= M_{n+1}=-\frac{\varepsilon\lambda\omega}{2^{n+1}}+\mu^+,\,\,\mathrm{ and } \,\,k=M_n=-\frac{\varepsilon\lambda\omega}{2^{n}}+\mu^+\,\,\mathrm{ for } n=0,1,2,3,... \end{align}\] that is \[\begin{align} \label{eq5469} \frac{\varepsilon\lambda\omega}{2^{n+1}}|\mathcal{A}_{M_{n+1},4\rho}(t)|\leq \gamma\frac{\mathcal{C}\rho}{\alpha}\int_{B_{4\rho}\cap\{M_n<u(\cdot,t)-g(t)<M_{n+1}\}}|\nabla_{\mathbb{H}^N}u|\,d\xi. \end{align}\tag{31}\] In this calculation, we have made use of the inequality \[\begin{align} \left|\{u(\cdot,t)-g(t)\leq -\varepsilon\lambda\omega+\mu^+\}\cap B_{4\rho}\right|\geq \left|\{u(\cdot,t)\leq -\varepsilon\lambda\omega+\mu^+\}\cap B_{4\rho}\right|\geq \frac{\alpha}{\mathcal{C}}|B_{4\rho}|. \end{align}\] This can be deduced from the measure theoretical information in the hypothesis, Lemma 14 and the property \(|B_{4\rho}|=\mathcal{C}|B_\rho|\). Now, integrate (31 ) with respect to time over \((0,\delta\rho^2)\) and set \[\begin{align} |\mathcal{A}_n|=|\{u-g>M_n\}\cap\mathcal{Q}_{4\rho}^\oplus(\delta)|=\int_0^{\delta\rho^2}|\mathcal{A}_{M_n,4\rho}(\tau)|\,d\tau. \end{align}\] Then, (31 ) yields with help of Hölder inequality, \[\begin{align} \frac{\varepsilon\lambda\omega}{2^{n+1}}|\mathcal{A}_{n+1}|&\leq \tilde{\gamma}\rho\iint_{\mathcal{Q}_{4\rho}^\oplus(\delta)\cap\{M_n<u(\cdot,t)-g(t)<M_{n+1}\}}|\nabla_{\mathbb{H}^N}u|\,d\xi dt\\ &\leq \tilde{\gamma}\rho\left(\iint_{\mathcal{Q}_{4\rho}^\oplus(\delta)}|\nabla_{\mathbb{H}^N}w_+|^2\,d\xi dt\right)^\frac{1}{2}|\mathcal{A}_n-\mathcal{A}_{n+1}|^\frac{1}{2}\\ &\leq 2\tilde{\gamma}\,\frac{\varepsilon\lambda\omega}{2^{n+1}}\sqrt{|\mathcal{Q}_{4\rho}^\oplus(\delta)|}\left(|\mathcal{A}_n|-|\mathcal{A}_{n+1}|\right)^\frac{1}{2},\,\,\,\,\,\mathrm{ for } n=0,1,2,3,..., \end{align}\] where \(\tilde{\gamma}:=\frac{\mathcal{C}\gamma}{\alpha}\) is depending upon data and \(\alpha\). Squaring both side of the above inequality, we have \[\begin{align} |\mathcal{A}_{n+1}|^2\leq (2\tilde{\gamma})^2|\mathcal{Q}_{4\rho}^\oplus(\delta)|^2\left(|\mathcal{A}_n|-|\mathcal{A}_{n+1}|\right). \end{align}\]Next, adding all these inequalities for \(n=0,1,2,3,...,n^*-1\), where \(n^*\) is a positive integer (to be determined later). Since \(\mathcal{A}_{n^*}\subset \mathcal{A}_i\) for any positive integer \(i<n^*\). So the estimate yields \[\begin{align} n^*|\mathcal{A}_{n^*}|^2\leq \sum_{n=0}^{n^*-1}|\mathcal{A}_{n+1}|^2&\leq (2\tilde{\gamma)}^2|\mathcal{Q}_{4\rho}^\oplus(\delta)|\sum_{n=0}^{\infty}\left(|\mathcal{A}_n|-|\mathcal{A}_{n+1}|\right)\\ &\leq (2\tilde{\gamma})^2|\mathcal{Q}_{4\rho}^\oplus(\delta)|^2. \end{align}\] Hence \[\begin{align} \label{eq54610} |\mathcal{A}_{n^*}|^2\leq \frac{(2\tilde{\gamma})^2}{n^*}|\mathcal{Q}_{4\rho}^\oplus(\delta)|^2. \end{align}\tag{32}\] Thus, considering fixed \(\iota\in (0,1)\), one can choose \(n^*\) so large that \[\begin{align} \label{eq54611} \frac{|\{u-g>-\varepsilon_\iota\lambda\omega+\mu^+\}\cap\mathcal{Q}_{4\rho}^\oplus(\delta)|}{|\mathcal{Q}_{4\rho}^\oplus(\delta)|}<\iota,\,\,\,\,\mathrm{ for }\,\,\, \frac{2\tilde{\gamma}}{\sqrt{n^*}}\leq\iota,\,\,\,\mathrm{ and }\,\,\varepsilon_\iota=\frac{\varepsilon}{2^{n^*}}. \end{align}\tag{33}\] Next, consider the De Giorgi Lemma 12 over forward parabolic cylinder \(\mathcal{Q}_{4\rho}^\oplus(\delta)\) with replacement of \(\lambda\omega\) with \(\varepsilon_\iota\lambda\omega\). In comparison with Lemma 12, both \(\iota\) depends only on data and \(\delta\) (that is fixed term \(\alpha\)). When such \(\iota\) is determined and fixed in terms of data, we calculate \(n^*\) by the indicated procedure. The conditions for (33 ) are satisfied in that case as well. Also the conditions of Lemma 12 hold. Then, from the conclusion of Lemma 12, we have \[\begin{align} u(\xi,t)-g(t)\leq -\frac{1}{2}\varepsilon_\iota\lambda\omega+\mu^+\,\,\,\,\,\mathrm{ a.e. in }\,\, B_{4\rho}\times (\frac{1}{2}\delta\rho^2,\delta\rho^2]. \end{align}\] Now considering \(\hat{\gamma}:=\max\{4\mathfrak{C},\frac{1}{\delta},\frac{4}{\varepsilon_\iota}\}\) and considering \(g(T_2)\leq \frac{1}{4}\varepsilon_\iota\lambda\omega\), we have \[\begin{align} u\leq -\varkappa\lambda\omega+\mu^+\,\,\,\,\,\mathrm{ a.e. in }\,\, B_{4\rho}\times (\frac{1}{2}\delta\rho^2,\delta\rho^2], \end{align}\] where \(\varkappa=\frac{1}{4}\varepsilon_\iota\). ◻

Proof. This proof will follow the steps of [12]. Considering lemmas 14 and 15, note that it is sufficient to show that there exists some \(\bar{\varepsilon}>0\) depending on \(\texttt{data}, \alpha\) and \(K\), such that \[\begin{align} \label{eq54612} \left|\{\pm(\mu^\pm-u+g(t))\geq \bar{\varepsilon}M\}\cap B_\rho(\xi_0)\right|\geq \alpha|B_\rho(\xi_0)|, \end{align}\tag{34}\] where \(g(t)\) as in Lemma 12. We prove this lemma for the subsolution case and choose \[\begin{align} aM:=-aL+\mu^+,\,\,\,\,\mathrm{ for any } a. \end{align}\] For the supersolution case, similar steps will be followed with some required sign changes.
Step I: Consider \(\mathcal{A}_{aM,\rho}(t)=\left\{u(\cdot,t)-g(t)>aM\right\}\cap B_\rho\). Choose truncated function \(w_+:=(u-g(t)-M)_+\), where \(g(t)\) as in Lemma 12. Proceeding as in (29 ) in Lemma 14, for \(t\in (0,\delta\rho^2]\), we have

\[\begin{align} \label{eq54613} \int_{B(1-\sigma)\rho}w^2_+(\xi,t)\,d\xi&\leq \gamma\int_{B_\rho}w^2_+(\xi,0)\,d\xi+\gamma\int_0^{\delta\rho^2}\int_{B_\rho}\left(w^2_+\,\nabla_{\mathbb{H}^N}\psi\right)^2(\xi,t)\,d\xi\nonumber\\ &\quad+\gamma\int_0^{\delta\rho^2}\max\{w^2_+(\xi,t),w^2_+(\eta,t)\}\frac{|\psi(\xi)-\psi(\eta)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\nonumber\\ &\quad\quad+\gamma\int_0^{\delta\rho^2}\int_{B_\rho}w_+(\xi,t)\left\{\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,d\xi dt\\ &\quad\quad\quad -\mathfrak{C}\iint_{\mathcal{Q}_\rho^\oplus(\delta)}g'(t)(w_+\psi^2)(\xi,t)\,d\xi dt\nonumber\\ &\quad\quad\quad\quad =:I_1+I_2+I_3+I_4+I_5,\nonumber \end{align}\tag{35}\] where we set a time-independent cut-off function \(\psi\) on \(B_\rho\) as follows \[\begin{align} \psi\equiv \begin{cases} 1\,\,\,\, onB_{(1-\sigma)\rho},\\ 0\,\,\,\, onB^c_{(1-\sigma/2)\rho}, \end{cases} \,\,\,\, such that |\nabla_{\mathbb{H}^N}\psi|\leq \frac{\gamma}{\sigma\rho}. \end{align}\] Now, we estimate the terms on the right-hand side of previous inequality one by one. Consider \(I_3\) and \(I_2\) successively, \[\begin{align} I_3&\leq \gamma\left(\frac{L}{\sigma\rho}\right)^2\int_0^{\delta\rho^2}\iint_{B_\rho\times B_\rho}\frac{\chi_{\mathcal{A}_{M,\rho}(t)}}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s-2}}\,d\xi dt\\ &=\gamma\delta \left(\frac{L}{\sigma}\right)^2\frac{\left|\left\{u-g>M\right\}\cap \mathcal{Q}_\rho^\oplus(\delta)\right|}{|\mathcal{Q}_\rho^\oplus(\delta)|}|B_\rho|, \end{align}\] and similarly \[\begin{align} I_2&\leq \gamma\delta \left(\frac{L}{\sigma}\right)^2\frac{\left|\left\{u-g>M\right\}\cap \mathcal{Q}_\rho^\oplus(\delta)\right|}{|\mathcal{Q}_\rho^\oplus(\delta)|}|B_\rho|. \end{align}\] Before proceeding towards \(I_4\), we observe that for \(\xi\in supp \psi\) i.e. \(|\xi|\leq (1-\frac{\sigma}{2})\rho\) and \(|\eta|\geq \rho\), we have \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq 1-\frac{|\xi|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq \frac{\sigma}{2}. \end{align}\] Also, when \(|\eta|_{\mathbb{H}^N}\geq R\) and \(|\xi|_{\mathbb{H}^N}\leq \rho\), we have \[\begin{align} \frac{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}}{|\eta|_{\mathbb{H}^N}}\geq\frac{1}{2}\,\,\,\mathrm{ provided }\,\rho\leq \frac{R}{2}. \end{align}\] So, \(I_4\) implies \[\begin{align} I_4&\leq 2^{Q+2s}\frac{1}{\sigma^{Q+2s}}\int_0^{\delta\rho^2}\int_{B_\rho}w_+\,d\xi\left\{\int_{B_R\setminus B_\rho} \frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta +\int_{\mathbb{H}^N\setminus B_R} \frac{w_+(\eta,t)}{|\eta|_{\mathbb{H}^N}^{Q+2s}}\,d\eta \right\}dt. \end{align}\] Similar to the respective part of Lemma 12, we can cancel out the second integral on the right hand of previous inequality with \(I_5\) by suitable choice of \(\mathfrak{C}\) and \(\gamma\). Then rest part on the right hand side of the above integral can be dominated by \[\begin{align} \gamma\delta\frac{L^2}{\sigma^{Q+2}}\frac{\left|\left\{u-g>M\right\}\cap \mathcal{Q}_\rho^\oplus(\delta)\right|}{|\mathcal{Q}_\rho^\oplus(\delta)|}|B_\rho|. \end{align}\] The left hand side of (35 ) has the following estimation for \(\varepsilon\in (0,1)\) \[\begin{align} \int_{B_{(1-\sigma)\rho}}w^2_+(\xi,t)\,d\xi \geq C(1-\varepsilon)^2L^2|\mathcal{A}_{\varepsilon M,(1-\sigma)\rho}(t)|, \end{align}\] where \(C\) is some positive constant. Combining all the estimates and proceeding as in Lemma 14, we have \[\begin{align} \label{eq54614} \left|\mathcal{A}_{\varepsilon M,\rho}(t)\right|\leq \frac{1}{(1-\varepsilon)^2}\left[(1-\alpha)+\frac{\gamma\delta}{\sigma^{Q+2}}\frac{\left|\left\{u-g>M\right\}\cap \mathcal{Q}_\rho^\oplus(\delta)\right|}{|\mathcal{Q}_\rho^\oplus(\delta)|}+Q\sigma\right]|B_\rho| \end{align}\tag{36}\] Choose \(\sigma\in (0,1)\), so that \[\begin{align} \sigma=\delta^\frac{1}{Q+3}\left(\frac{\left|\left\{u-g>M\right\}\cap \mathcal{Q}_\rho^\oplus(\delta)\right|}{|\mathcal{Q}_\rho^\oplus(\delta)|}\right)^\frac{1}{Q+3}. \end{align}\] Then, (36 ) implies, \[\begin{align} \left|\mathcal{A}_{\varepsilon M,\rho}(t)\right|\leq \frac{1}{(1-\varepsilon)^2}\left[(1-\alpha)+\gamma\delta^\frac{1}{Q+3}\left(\frac{\left|\left\{u-g>M\right\}\cap \mathcal{Q}_\rho^\oplus(\delta)\right|}{|\mathcal{Q}_\rho^\oplus(\delta)|}\right)^\frac{1}{Q+3}\right]|B_\rho|. \end{align}\] Further, choose \[\begin{align} \label{eq54615} \begin{cases} \delta=\left(\frac{\alpha}{8\gamma}\right)^{Q+3},\\ \frac{1}{(1-\epsilon)^2}\leq \frac{1-\frac{3}{4}\alpha}{1-\frac{7}{8}\alpha}, \end{cases} \end{align}\tag{37}\] then \[\begin{align} \label{eq54616} \left|\mathcal{A}_{\varepsilon M,\rho}(t)\right|\leq\left(1-\frac{3}{4}\alpha\right)|B_\rho|, \end{align}\tag{38}\] for any \(0<t\leq \delta\rho^2=:t_1\).
Step II: Set \[\begin{align} \begin{cases} L_1:=\varepsilon L,\,\,\varepsilon_1:=\frac{1}{2^{n_1+i_1+1}},\,\,M_1:=\varepsilon M=-L_1+\mu^+,\,\,\mathcal{M}_1:=-\frac{L_1}{2^{i_1+1}}+\mu^+,\\ \mathcal{Q}_1:=B_\rho\times [t_1,t_2]\,\,\mathrm{ with }\,t_2:=t_1+\delta\rho^2, \end{cases} \end{align}\] where both natural numbers \(i_1,n_1\) will be calculated later. Next, write (35 ) with these parameters and truncated function \(w_+:=(u-g(t)-\mathcal{M}_1)_+\) over the cylinder \(\mathcal{Q}_1\) to have for any \(t\in (t_1,t_2)\), \[\begin{align} \label{eq54617} \int_{B(1-\sigma_1)\rho}w^2_+(\xi,t)\,d\xi&\leq \int_{B_\rho}w^2_+(\xi,t_1)\,d\xi+\gamma\int_{t_1}^{t_2}\int_{B_\rho}\left(w^2_+\,\nabla_{\mathbb{H}^N}\psi\right)^2(\xi,t)\,d\xi\nonumber\\ &\quad+\gamma\int_{t_1}^{t_2}\max\{w^2_+(\xi,t),w^2_+(\eta,t)\}\frac{|\psi(\xi)-\psi(\eta)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\nonumber\\ &\quad\quad+\gamma\int_{t_1}^{t_2}\int_{B_\rho}w_+(\xi,t)\left\{\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}\,d\xi dt\\ &\quad\quad\quad -\mathfrak{C}\iint_{\mathcal{Q}_1}g'(t)(w_+\psi^2)(\xi,t)\,d\xi dt\nonumber\\ &\quad\quad\quad\quad =:I_1+I_2+I_3+I_4+I_5\nonumber, \end{align}\tag{39}\] for some \(\sigma_1\in (0,1)\) to be determined later. Recalling the measure theoretical information (38 ) at time level \(t=t_1\), we have \[\begin{align} I_1\leq \left(\frac{L_1}{2^{i_1+1}}\right)^2\left|\{u(\cdot,t_1)-g(t_1)>\mathcal{M}_1\}\cap B_\rho \right|&\leq \left(\frac{L_1}{2^{i_1+1}}\right)^2\left|\{u(\cdot,t_1)-g(t_1)> M_1\}\cap B_\rho \right|\\ &\leq \left(1-\frac{3}{4}\alpha\right) \left(\frac{L_1}{2^{i_1+1}}\right)^2|B_\rho|. \end{align}\] Rest of the right hand side of (39 ) can be estimated as in Step I to have \[\begin{align} I_2+I_3+I_4+I_5\leq \frac{\gamma(\texttt{data})\delta}{\sigma_1^{Q+2}}\left(\frac{L_1}{2^{i_1+1}}\right)^2\frac{\left|\left\{u-g>\mathcal{M}_1\right\}\cap \mathcal{Q}_1\right|}{|\mathcal{Q}_1|}|B_\rho|. \end{align}\] Combining all the estimates into (39 ), we have \[\begin{align} \label{eq54618} \int_{B(1-\sigma_1)\rho}w^2_+(\xi,t)\,d\xi\leq \left[\left(1-\frac{3}{4}\alpha\right)+\frac{\gamma(\texttt{data})\delta}{\sigma_1^{Q+2}}\frac{\left|\left\{u-g>\mathcal{M}_1\right\}\cap \mathcal{Q}_1\right|}{|\mathcal{Q}_1|}\right] \left(\frac{L_1}{2^{i_1+1}}\right)^2|B_\rho|. \end{align}\tag{40}\] Similar to the Step I, the left hand side of (40 ) can be estimated , \[\begin{align} \int_{B(1-\sigma_1)\rho}w^2_+(\xi,t)\,d\xi &\geq \int_{\mathcal{A}_{\varepsilon_1M_1,(1-\sigma_1)\rho}(t)}\left[(\varepsilon_1 M_1-\mathcal{M}_1)\right]\,d\xi\\ & \geq (1-2^{-n_1})^2\left(\frac{L_1}{2^{i_1+1}}\right)^2\left|\mathcal{A}_{\varepsilon_1M_1,(1-\sigma_1)\rho}(t)\right|, \end{align}\] for some constant \(C\). Therefore inserting this into (40 ), we get , for every \(t_1\leq t\leq t_2\) \[\begin{align} \label{eq54619} \left|\mathcal{A}_{\varepsilon_1M_1,\rho}(t)\right|\leq \frac{1}{(1-2^{-n_1})^2}\left[\left(1-\frac{3}{4}\alpha\right)+\frac{\gamma(\texttt{data})\delta}{\sigma_1^{Q+2}}\frac{\left|\left\{u-g>M_1\right\}\cap \mathcal{Q}_1\right|}{|\mathcal{Q}_1|}+Q\sigma_1\right]|B_\rho|. \end{align}\tag{41}\] Choose \(\sigma_1\) as, \[\begin{align} \sigma_1=\delta^{\frac{1}{Q+3}}\left(\frac{\left|\left\{u-g>\mathcal{M}_1\right\}\cap \mathcal{Q}_1\right|}{|\mathcal{Q}_1|}\right)^{\frac{1}{Q+3}}. \end{align}\] Then (41 ) implies, \[\begin{align} \left|\mathcal{A}_{\varepsilon_1M_1,\rho}(t)\right|\leq \frac{1}{(1-2^{-n_1})^2}\left[\left(1-\frac{3}{4}\alpha\right)+\gamma\delta^{\frac{1}{Q+3}}\left(\frac{\left|\left\{u-g>\mathcal{M}_1\right\}\cap \mathcal{Q}_1\right|}{|\mathcal{Q}_1|}\right)^{\frac{1}{Q+3}}\right]|B_\rho|. \end{align}\] At this point, keeping the parameters \(L_1,\varepsilon_1\) in hand, we recall the arguments and steps that lead to (32 ) in Lemma 15, we get \[\begin{align} \frac{\left|\left\{u-g>M_1\right\}\cap \mathcal{Q}_1\right|}{|\mathcal{Q}_1|}\leq \frac{\gamma}{\alpha\sqrt{i_1}}\leq \frac{\gamma}{\alpha\delta\sqrt{i_1}}, \end{align}\] for some constant \(\gamma=\gamma(\texttt{data})\). Therefore, the previous estimate looks like \[\begin{align} \left|\mathcal{A}_{\varepsilon_1M_1,\rho}(t)\right|\leq \frac{1}{(1-2^{-n_1})^2}\left[\left(1-\frac{3}{4}\alpha\right)+\frac{\gamma(\texttt{data)}}{\alpha^{\frac{1}{Q+3}}i_1^{\frac{1}{2(Q+3)}}}\right]|B_\rho|. \end{align}\] For given \(K\), we choose \(i_1,n_1\in \mathbb{N}\) in such a way that \[\begin{align} \frac{\gamma(\texttt{data)}}{\alpha^{\frac{1}{Q+3}}i_1^{\frac{1}{2(Q+3)}}(1-2^{-n_1})^2}\leq \frac{\alpha\delta}{8K},\,\,\,\,\,\,\frac{\left(1-\frac{3}{4}\alpha\right)}{(1-2^{-n_1})^2}\leq 1-\frac{3}{4}\alpha+ \frac{\alpha\delta}{8K}, \end{align}\] to obtain, for all \(t_1\leq t\leq t_2\) \[\begin{align} \left|\mathcal{A}_{\varepsilon_1M_1,\rho}(t)\right|\leq\left[1-\frac{3}{4}\alpha+ \frac{\alpha\delta}{4K}\right]|B_\rho|. \end{align}\]
Step III: In this step, we begin the induction process. Fix a natural number \(j(>2)\in\mathbb{N}\). In Step II, \(i=1\) case has been done and we assume that procedure has been done inductively upto \((j-1)-\)th steps; in turn, three sequences \(\{\varepsilon_i\}_{i=1}^{j-1}\), \(\{L_i\}_{i=1}^{j-1}\) and \(\{t_i\}_{i=1}^{j-1}\) have been selected up to the \((j-1)-\)th and assume that \[\begin{align} \label{eq54620} \left|\mathcal{A}_{\varepsilon_{j-1}M_{j-1},\rho}(t)\right|\leq\left[1-\frac{3}{4}\alpha+ (j-1)\frac{\alpha\delta}{4K}\right]|B_\rho|,\,\,\,\,\,\mathrm{ for all } t\in [t_{j-1},t_j], \end{align}\tag{42}\] where \(t_j:=t_{j-1}+\delta\rho^2\) and \(\varepsilon_{j-1}:=\frac{1}{2^{n_{j-1}+i_{j-1}+1}}\) with \(i_{j-1},n_{j-1}\in\mathbb{N}\) being large enough. Also note that the parameters \(\delta\) and \(\alpha\) have been determined in Step I. Now set \[\begin{align} \begin{cases} L_j:=\varepsilon L_{j-1},\,\varepsilon_j:=\frac{1}{2^{n_j+i_j+1}},\,\,M_j:=\varepsilon^j M=-L_j+\mu^+,\,\,\mathcal{M}_j:=-\frac{L_j}{2^{i_j+1}}+\mu^+,\\ \mathcal{Q}_j:=B_\rho\times [t_j,t_{j+1}]\,\,\mathrm{ with } t_{j+1}:=t_j+\delta\rho^2, \end{cases} \end{align}\] where both natural numbers \(n_j,i_j\) are large enough so that \[\begin{align} n_j\geq n_{j-1}\,\,\mathrm{ and }\,\,i_j\geq n_{j-1}+i_{j-1}. \end{align}\] Write (35 ) with these parameters and truncated function \(w_+:=(u-g(t)-\mathcal{M}_j)_+\) over the cylinder \(\mathcal{Q}_j\) and estimate similar to Step II, we get \[\begin{align} \label{eq54621} \int_{B_{(1-\sigma_j)\rho}}w^2_+(\xi,t)\,d\xi\leq \int_{B_\rho}w^2_+(\xi,t_j)\,d\xi+ \frac{\gamma(\texttt{data})\delta}{\sigma_j^{Q+2}}\left(\frac{L_j}{2^{i_j+1}}\right)^2\frac{\left|\left\{u-g>\mathcal{M}_j\right\}\cap \mathcal{Q}_j\right|}{|\mathcal{Q}_j|}|B_\rho|, \end{align}\tag{43}\] where \(\sigma_j\in (0,1)\) to be determined later. Using the measure theoretical information (42 ) at time level \(t=t_j\), the first integral on the right hand side of (43 ) can be estimated as \[\begin{align} \int_{B_\rho}w^2_+(\xi,t_j)\,d\xi&\leq \left(\frac{L_j}{2^{i_j+1}}\right)^2\left|\{u(\cdot,t_j)-g(t_j)>\mathcal{M}_j\}\cap B_\rho \right|\\ & \leq \left(\frac{L_j}{2^{i_j+1}}\right)^2\left|\{u(\cdot,t_j)-g(t_j)>-\frac{L_{j-1}}{2^{n_{j-1}+i_{j-1}+1}}+\mu^+\}\cap B_\rho \right|\\ &\leq \left(\frac{L_j}{2^{i_j+1}}\right)^2\left|\mathcal{A}_{\varepsilon_{j-1}M_{j-1},\rho}(t)\right|\\ & \leq \left(\frac{L_j}{2^{i_j+1}}\right)^2\left[1-\frac{3}{4}\alpha+ (j-1)\frac{\alpha\delta}{4K}\right]|B_\rho|. \end{align}\] Similar to what was done before, the left hand side of (43 ) can be estimated as \[\begin{align} \int_{B_{(1-\sigma_j)\rho}}w^2_+(\xi,t)\,d\xi&\geq \int_{\mathcal{A}_{\varepsilon_j M_j,(1-\sigma_j)\rho}(t)}w^2_+(\xi,t)\,d\xi\\ &\geq (\varepsilon_jM_j-\mathcal{M}_j)^2|\mathcal{A}_{\varepsilon_j M_j,(1-\sigma_j)\rho}(t)|\\ &\geq (1-2^{-n_j})^2\left(\frac{L_j}{2^{i_j+1}}\right)^2\left[|\mathcal{A}_{\varepsilon_j M_j,\rho}(t)|-Q\sigma_j|B_\rho|\right]. \end{align}\] Combining all these estimates into (43 ), we have \[\begin{align} |\mathcal{A}_{\varepsilon_j M_j,\rho}(t)|\leq \frac{1}{(1-2^{-n_j})^2}\left[1-\frac{3}{4}\alpha+ (j-1)\frac{\alpha\delta}{4K}+\frac{\gamma\delta}{\sigma_j^{Q+2}}\frac{\left|\left\{u-g>\mathcal{M}_j\right\}\cap \mathcal{Q}_j\right|}{|\mathcal{Q}_j|}+Q\sigma_j\right]|B_\rho|. \end{align}\] Choose \(\sigma_j\in (0,1)\) as \[\begin{align} \sigma_j=\delta^{\frac{1}{Q+3}}\left(\frac{\left|\left\{u-g>\mathcal{M}_j\right\}\cap \mathcal{Q}_j\right|}{|\mathcal{Q}_j|}\right)^{\frac{1}{Q+3}}. \end{align}\] Then it implies that \[\begin{align} |\mathcal{A}_{\varepsilon_j M_j,\rho}(t)|\leq \frac{1}{(1-2^{-n_j})^2}\left[1-\frac{3}{4}\alpha+ (j-1)\frac{\alpha\delta}{4K}+\gamma\delta^{\frac{1}{Q+3}}\left(\frac{\left|\left\{u-g>\mathcal{M}_j\right\}\cap \mathcal{Q}_j\right|}{|\mathcal{Q}_j|}\right)^{\frac{1}{Q+3}}\right]|B_\rho|. \end{align}\] Again similar to Step II, considering Lemma 15, we have \[\begin{align} \frac{\left|\left\{u-g>\mathcal{M}_j\right\}\cap \mathcal{Q}_j\right|}{|\mathcal{Q}_j|}\leq \frac{\gamma}{\alpha\delta\sqrt{i_j}}, \end{align}\] for some constant \(\gamma=\gamma(\texttt{data})\). Then, we get \[\begin{align} \left|\mathcal{A}_{\varepsilon_jM_j,\rho}(t)\right|\leq \frac{1}{(1-2^{-n_j})^2}\left[1-\frac{3}{4}\alpha+(j-1)\frac{\alpha\delta}{4K}+\frac{\gamma}{\alpha^{\frac{1}{Q+3}}i_j^{\frac{1}{2(Q+3)}}}\right]|B_\rho|. \end{align}\] Finally for given \(K\), we choose \(i_j,n_j\in\mathbb{N}\) in such a way that \[\begin{align} \frac{\gamma}{\alpha^{\frac{1}{Q+3}}i_j^{\frac{1}{2(Q+3)}}(1-2^{-n_j})^2}\leq \frac{\alpha\delta}{8K}\,\,\mathrm{ and }\,\,\frac{1-\frac{3}{4}\alpha+(j-1)\frac{\alpha\delta}{4K}}{(1-2^{-n_j})^2}\leq 1-\frac{3}{4}\alpha+(j-1)\frac{\alpha\delta}{4K}+\frac{\alpha\delta}{8K}. \end{align}\] Consequently, these all yield, for every \(t\in[t_j,t_{j+1}]\), \[\begin{align} \label{eq54622} \left|\mathcal{A}_{\varepsilon_jM_j,\rho}(t)\right|\leq\left[1-\frac{3}{4}\alpha+j\frac{\alpha\delta}{4K}\right]|B_\rho|. \end{align}\tag{44}\] Hence, induction procedure is complete.
Step IV: Repeating the procedure which leads to (44 ) up to \(j\in\mathbb{N}\) satisfying \[\begin{align} \frac{K}{\delta}-1<j\leq\frac{K}{\delta} \end{align}\] and choose \(\bar{\varepsilon}\in (0,1)\) in such a way that \[\begin{align} \bar{\varepsilon}L=\varepsilon_jL_j=\frac{L_j}{2^{n_j+i_j+1}}. \end{align}\] Therefore, we get \[\begin{align} |\mathcal{A}_{\bar{\varepsilon}M,\rho}(t)|\leq \left(1-\frac{\alpha}{2}\right)|B_\rho|, \end{align}\] for all \(0<t\leq \sum_{i=1}^{j+1}(t_i-t_{i-1})=(j+1)\delta\rho^2\) which eventually implies that \[\begin{align} &|\{u(\cdot,t)-g(t)\leq -\bar{\varepsilon}L+\mu^+\}\cap B_\rho|\geq \frac{\alpha}{2}|B_\rho|, \end{align}\] for all \(0<t\leq K\rho^2\). ◻

Proof. Let, \(0<k<\frac{\rho_0-\rho}{2m\rho}\), for some constant \(m>1\). Let \(\left\{ B_{k\rho}(\xi)\right\}_{x\in B_\rho(\xi_0)}\) be a covering of \(B_\rho(\xi_0)\), then there exists disjoints sequence \(\left\{ B_{k\rho}(\xi_i)\right\}_{\xi_i\in B_\rho(\xi_0)}\) (see [45]), for all \(i\in I\), such that \[\begin{align} \label{eq6461} B_\rho(\xi_0)\subset\cup_{i\in I}B_{2k\rho}(\xi_0)\subset B_{\rho_0}(x_0). \end{align}\tag{45}\] It is obvious that \(B_{2mk\rho}(\xi_i)\subset B_R(\xi_0)\). Moreover, the ball \(B_{2mk\rho}(\xi_i)\) have bounded overlap with bound independent of \(k\), due to the measure doubling property. Next, let us define \[\begin{align} \begin{cases} D^+:=\left\{i\in I: \left|\{u>c\}\cap B_{2k\rho}(\xi_0)\right|>\frac{\alpha}{2\mathcal{C}}|B_{2k\rho}(\xi_0)|\right\},\\ D^-:=\left\{i\in I: \left|\{u>c\}\cap B_{2k\rho}(\xi_0)\right|\leq \frac{\alpha}{2\mathcal{C}}|B_{2k\rho}(\xi_0)|\right\}. \end{cases} \end{align}\] By assumption, we get \[\begin{align} \alpha|B_\rho(\xi_0)|&\leq \left|\left\{u(.)> c\right\}\cap B_\rho(\xi_0)\right|\\ &\leq \sum_{i\in D^+}\left|\left\{u(.)> c\right\}\cap B_{2k\rho}(\xi_0)\right|+\sum_{i\in D^-}\left|\left\{u(.)> c\right\}\cap B_{2k\rho}(\xi_0)\right|\\ & \leq\underbrace{ \sum_{i\in D^+}\left|\left\{u(.)> c\right\}\cap B_{2k\rho}(\xi_0)\right|}_{A}+\frac{\alpha}{2\mathcal{C}}\sum_{i\in D^-}\left|B_{2k\rho}(\xi_i)\right|\\ &\leq A+\frac{\alpha}{2}\sum_{i\in D^-}\left|B_{k\rho}(\xi_i)\right|\\ & \leq A+\frac{\alpha}{2} \left|B_{(1+k)\rho}(\xi_0)\right|. \end{align}\] The last inequality follows by the fact that \(\cup_{i\in I}B_{k\rho}(\xi_i)\subset B_{(1+k)\rho}(\xi_0)\). Consequently, we have \[\begin{align} \label{eq6462} \frac{\alpha}{2}\left\{|B_\rho(\xi_0)|-|B_{(1+k)\rho}(\xi_0)\setminus B_{k\rho}(\xi_0)|\right\}\leq A. \end{align}\tag{46}\] Whenever [eqCI] and [eqCII] are true, there might be two outcomes namely either \[\begin{align} \label{I} \left|\left\{u>\lambda c\right\}\cap B_{k\rho}(\xi_i)\right|>(1-\delta)|B_{k\rho}(\xi_i)|, \end{align}\tag{47}\] or \[\begin{align} \label{II} \left|\left\{u>\lambda c\right\}\cap B_{k\rho}(\xi_i)\right|\leq(1-\delta)|B_{k\rho}(\xi_i)|. \end{align}\tag{48}\] Now, if (I) holds, then let \(\xi_*\in B_\rho(\xi_0)\) be the center of \(B_{k\rho}\) and the result follows. If (II) is true for \(i\in D^+\), then \[\begin{align} \label{eq6463} \left|\left\{u\leq\lambda c\right\}\cap B_{k\rho}(\xi_i)\right|>\delta|B_{k\rho}(\xi_i)|. \end{align}\tag{49}\] For every \(\xi\in \left\{u(.)>c\right\}\cap B_{2k\rho}(\xi_i)\) and \(\eta\in \left\{u(.)\leq\lambda c\right\}\cap B_{k\rho}(\xi_i)\), we have \[\begin{align} u(\xi)-u(\eta)>c(1-\lambda)>0, \end{align}\] and \[\begin{align} |{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}&\leq |{\xi_i}^{-1}\circ{\xi}|_{\mathbb{H}^N}+|{\eta}^{-1}\circ{\xi_i}|_{\mathbb{H}^N}\\ &\leq 3k\rho. \end{align}\] Therefore, summing up over \(D^+\) and using the bounded overlapping property, assumptions [eqCII], (46 ) and (49 ), we have by Lemma 7 \[\begin{align} \frac{\alpha\delta}{2}(1-\lambda)^2&\left[|B_\rho(\xi_0)|-|B_{(1+k)\rho}(\xi_0)\setminus B_\rho(\xi_0)\right]\\ &\leq C \sum_{i\in D^+}\int_{\{u(.)<\lambda c\}\cap B_{k\rho}(\xi_i)}\int_{\{u(.)> c\}\cap B_{2k\rho}(\xi_i)}\frac{(3k\rho)^{Q+2s}}{|B_{k\rho}(\xi_i)|}\frac{|u(\xi-u(\eta)|^2}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\, d\xi d\eta\\ &\leq 3^{Q+2s}C(k\rho)^{2s}\iint_{B_\rho(\xi_0)\times B_\rho(\xi_0)} \frac{|u(\xi-u(\eta)|^2}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\, d\xi d\eta\\ & \leq 3^{Q+2s}CB(k\rho_0)^{2s}, \end{align}\] where \(C=C(Q,s)\) is a constant. Consequently, sending \(k\rightarrow0\) in the last inequality renders that \[\begin{align} \frac{\alpha\delta}{2}(1-\lambda)^2|B_\rho(\xi_0)|=0, \end{align}\] which contradicts that \(|B_\rho(\xi_0)|>0\). ◻

Proof. We prove this result through several steps, inspired from the proof of the Proposition 5.3 in [12].
Step I Due to the assumptions, Lemma 14, there exists \(\varepsilon,\delta\in (0,1)\) (depending upon data) such that \[\begin{align} \left|\{u(t)\geq \varepsilon M\}\cap B_\rho(\xi_0)\right|\geq \frac{\alpha}{2}|B_\rho(\xi_0)|, \end{align}\] whenever \[\begin{align} \frac{1}{\delta}R^{-2}\int_{T_1}^{T_2}\mathrm{Tail }[u_-(t);R,\xi_0]\,dt\leq M, \end{align}\] for all \(t\in [t_0,t_0+\delta\rho^2]\). Set two parabolic cylinders as follows \[\begin{align} & \mathcal{Q}_1:= B_{2\rho}(\xi_0)\times (t_0+\frac{1}{2}\delta\rho^2,t_0+\delta\rho^2],\\ & \mathcal{Q}_2:= B_{4\rho}(\xi_0)\times (t_0,t_0+\delta\rho^2]. \end{align}\] Taking into account the truncated function \((u(\xi,t)-M)_-\) in the energy estimate (?? ) and proceeding as in Lemma 12, we achieve the following \[\begin{align} \label{eq6465} \int_{\frac{1}{2}\delta\rho^2}^{\delta\rho^2}\int_{B_{2\rho}(\xi_0)}|\nabla_{\mathbb{H}^N}(u-M)_-|^2\,d\xi dt\leq \gamma M^2|B_{2\rho}(\xi_0)|, \end{align}\tag{50}\] where \(\gamma\) is a constant depending only on \(\texttt{data}\).
Step II We start this step with rescaled function \[\begin{align} w(\xi,t):=\frac{1}{\varepsilon}\left(1-\frac{\left(u(\xi,t)-M\right)_-}{M}\right). \end{align}\] Using this rescaled version in the previous inequality (50 ), it follows that \[\begin{align} \int_{\frac{1}{2}\delta\rho^2}^{\delta\rho^2}\fint_{B_{2\rho}(\xi_0)}|\nabla_{\mathbb{H}^N}w(\xi,t)|^2\,d\xi dt\leq \frac{\gamma}{\varepsilon^2}, \end{align}\] and \[\begin{align} \left|\{w(t)\geq 1\}\cap B_\rho(\xi_0)\right|\geq \frac{\alpha}{2}\left|B_\rho(\xi_0)\right|, \end{align}\] for all \(t\in \left(t_0+\frac{1}{2}\delta\rho^2,t_0+\delta\rho^2\right]\). Then there exists a time level \(t_1\in \left(t_0+\frac{1}{2}\delta\rho^2,t_0+\delta\rho^2\right]\) such that the previous inequality implies \[\begin{align} \fint_{B_{2\rho}(\xi_0)}|\nabla_{\mathbb{H}^N}w(\xi,t_1)|^2\,d\xi dt\leq \frac{\gamma}{\delta\rho^2\varepsilon^2}, \end{align}\] and \[\begin{align} \left|\{w(t_1)\geq 1\}\cap B_\rho(\xi_0)\right|\geq \frac{\alpha}{2}\left|B_\rho(\xi_0)\right|. \end{align}\] Next, we are in stage to apply Clustering Lemma [L6.1] with \(\delta=\lambda=1\) and \(c=1\). This implies that there exists a point \(\xi_*\in B_\rho(\xi_0)\) and a scale parameter \(k=k(Q,s,\alpha,\varepsilon)\in (0,1)\) such that \[\begin{align} &\left|\left\{w(t_1)>\frac{1}{2}\right\}\cap B_{k\rho}(\xi_*)\right|>\frac{|B_{k\rho}(\xi_*)}{2}>\frac{\alpha}{2}|B_{k\rho}(\xi_*)|\\ \iff &\left|\left\{w(t_1)>\frac{1}{2}\varepsilon M\right\}\cap B_{k\rho}(\xi_*)\right|>\frac{\alpha}{2}|B_{k\rho}(\xi_*)| . \end{align}\] It is useful to note that without loss of any generality, the parameter \(k>0\) can be stipulated to satisfy \(k<\frac{1}{4}\).
Step III Applying Lemma 15 with \(\alpha=\frac{1}{2},\,(\xi_0,t_0)=(\xi_*,t_1)\) and \(\lambda\omega\) replaced with \(\frac{\varepsilon M}{2}\), then there exists \(\varkappa_1,\delta\in (0,1)\), both depending only on data, such that \[\begin{align} \label{eq6466} u(t_1)\geq \varkappa_1\frac{\varepsilon M}{2}\,\,\,\,\mathrm{ a.e. in } B_{2k\rho}(\xi_*)\times \left(t_1+\frac{\delta_1}{2}(k\rho)^2,t_1+\delta_1(k\rho)^2\right], \end{align}\tag{51}\] whenever \[\begin{align} \label{eq6467} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail }[u_-(t);R,\xi_0]\,dt\leq \varkappa_1\frac{\varepsilon M}{2}. \end{align}\tag{52}\] Then, like Step II, by (51 ) we have, at the time level \(t_2=t_1+\delta_1(k\rho)^2\), \[\begin{align} \left|\left\{u(t_2)\geq \varkappa_1\frac{\varepsilon M}{2}\right\}\cap B_{2k\rho}(\xi_*)\right|=|B_{2k\rho}(\xi_*)|. \end{align}\] Recalling, Lemma 15 with \(\varkappa_2, \delta_2\in (0,1)\), both depending only on data, such that \[\begin{align} \label{eq6468} u(t)\geq \varkappa_2\varkappa_1\frac{\varepsilon M}{2}\,\,\,\,\mathrm{ a.e. in } B_{4k\rho}(\xi_*)\times \left(t_2+\frac{\delta_2}{2}(2k\rho)^2, t_2+\delta_2(2k\rho)^2\right], \end{align}\tag{53}\] whenever \[\begin{align} \label{eq6469} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail }[u_-(t);R,\xi_0]\,dt\leq \varkappa_1\varkappa_2\frac{\varepsilon M}{2}. \end{align}\tag{54}\] Let, \(\hat{n}\in \mathbb{N}\setminus \{1,2\}\) so that \[\begin{align} 2^{2-\hat{n}}<k\leq 2^{3-\hat{n}}\,\,\,\mathrm{and}\,\,\,B_{2\rho}(\xi_0)\subset B_{2^{\hat{n}}k\rho}(\xi_*). \end{align}\] Continuing the same procedures which led to the inequality (53 ) and (54 ), we have the sequences of parameters viz. \(\{\varkappa_i\}_{i=1}^{\hat{n}}\), \(\{\delta_i\}_{i=1}^{\hat{n}}\) in \((0,1)\), depending on data, such that \[\begin{align} \label{eq64610} u(t)\geq \left(\Pi_{i=1}^{\hat{n}}\varkappa_i\right)\frac{\varepsilon}{2}M\,\,\,\,\mathrm{ a.e. in } B_{2^{\hat{n}}k\rho}(\xi_*)\times\left(t_{\hat{n}}+\frac{\delta_{\hat{n}}}{2}\left(2^{\hat{n}-1}k\rho\right)^2, t_{\hat{n}}+\delta_{\hat{n}}\left(2^{\hat{n}-1}k\rho\right)^2\right], \end{align}\tag{55}\] whenever \[\begin{align} \label{eq64611} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail }[u_-(t);R,\xi_0]\,dt\leq \left(\Pi_{i=1}^{\hat{n}}\varkappa_i\right)\frac{\varepsilon M}{2}, \end{align}\tag{56}\] where \(t_{\hat{n}}:=t_{\hat{n}-1}+\delta_{\hat{n}-1}\left(2^{\hat{n}-2}k\rho\right)^2\). Unlike, the both \(\varkappa_1\) and \(\delta_1\), the others are independent of \(\alpha\). Moreover, conditions (52 ) and (54 ) can be verified from (56 ). From the proofs of Lemma 14 and 15, we can stipulate \(\delta_1\) and \(\varkappa_1\) to fulfill \[\begin{align} \delta_1\leq \frac{\alpha^{Q+3}}{8\gamma\,(8Q)^{Q+2}},\,\,\,\,\,\,\,\,\,\,\,\varkappa_1\leq \frac{1}{2^{n_1}}\left[1-\left(\frac{1-\frac{3}{4}\alpha}{1-\frac{1}{2}\alpha}\right)^\frac{1}{2}\right], \end{align}\] where \(n_1\in \mathbb{N}\) is chosen in such a way that \(\frac{\gamma(\texttt{data},\alpha,\delta_1)}{\sqrt{n_1}}\leq \iota\) where \(\iota=\iota(Q,\alpha)\in (0,1)\) comes from Lemma 12. Additionally, note that \(\delta_i\) and \(\varkappa_i\) would be allowed to choose commonly as \[\begin{align} \delta_i=\frac{1}{8\gamma.\,(8Q)^{Q+2}},\,\,\,\,\,\,\,\,\,\,\,\varkappa_i\leq\frac{1}{2^{n_2}}\left[1-\left(\frac{1}{8}\right)^\frac{1}{2}\right], \end{align}\] for all \(i\in \{2,3,4,...,\hat{n}\}\) and \(n_2\) satisfies \(\frac{\gamma(\texttt{data},\alpha,\delta_1)}{\sqrt{n_2}}\leq \iota\). It is easy to see that, for all \(i\in\{2,3,4,...,\hat{n}\}\), \(\delta_1\leq \delta_i\) and \(\varkappa_1\leq\varkappa_i\). Therefore redefining these parameters as, \[\begin{align} \delta_i\equiv \delta_*:=\frac{\alpha^{Q+3}}{10^6\,(8Q)^{Q+2}},\,\,\,\,\,\,\,\,\,\,\,\varkappa_i\equiv\varkappa_*\leq \frac{1}{2^{n_0}}\left[1-\left(\frac{1-\frac{3}{4}\alpha}{1-\frac{1}{2}\alpha}\right)^\frac{1}{2}\right]\leq \frac{1}{8}, \end{align}\] for some large \(n_0\in\mathbb{N}\). Next, recalling all the observations and defining recursively \(\{t_i\}_{i=2}^{\hat{n}+1}\) by \[\begin{align} t_i:=t_{i-1}+\delta_{\hat{n}-1}\left(2^{i-2}k\rho\right)^2, \end{align}\] we finally conclude that if \[\begin{align} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,\xi_0]\,dt\leq \frac{1}{2}\varkappa_*^{\hat{n}}\varepsilon M, \end{align}\] then, \[\begin{align} \label{eq64612} u(t)\geq \frac{1}{2}\varkappa_*^{\hat{n}}\varepsilon M\,\,\,\,\mathrm{a.e. in } B_{2^{\hat{n}}k\rho}(\xi_*)\times \left(t_{\hat{n}+1}(\xi_*)(2^{\hat{n}-1}k\rho)^2, t_{\hat{n}+1}\right]. \end{align}\tag{57}\] By prior assumption \(4\leq 2^{\hat{n}}k\leq 8\), \[\begin{align} \varkappa_*^{\hat{n}}=\left(2^{\hat{n}}\right)^{\log_2{\varkappa_*}}\begin{cases} >\left(\frac{k}{4}\right)^{\log_2(1/\varkappa_*)},\\ \leq \left(\frac{k}{8}\right)^{\log_2(1/\varkappa_*)}. \end{cases} \end{align}\] Define \(d=d(\texttt{data},\alpha)\) and \(\varkappa_0=\varkappa_0(\texttt{data},\alpha)\in (0,1)\) by \[\begin{align} d:=\log_\alpha\left[\left(\frac{k}{8}\right)^{\log_2(1/\varkappa_*)}\right],\,\,\,\,\,\,\varkappa_0:=\frac{\varepsilon}{2}. \end{align}\] Then the result in (57 ) along with the tail are estimated as \[\begin{align} \frac{1}{2}\varkappa_*^{\hat{n}}\varepsilon M\geq \varkappa_0\alpha^dM, \end{align}\] and \[\begin{align} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,\xi_0]\,dt\leq \frac{1}{2}\varkappa_*^{\hat{n}}\varepsilon M&\leq \frac{1}{2}\left(\frac{k}{8}\right)^{\log_2(1/\varkappa_*)}\varepsilon M\\ &=\varkappa_0\alpha^dM, \end{align}\] respectively. As a result, we recall the fact \(B_{2\rho}(\xi_0)\subset B_{2^{\hat{n}}k\rho}(\xi_*)\) and merge the previous results with display (57 ) to have for every \(t_{\hat{n}+1}-\frac{\delta_*}{2}\left(2^{\hat{n}-1}k\rho\right)^2<t\leq t_{\hat{n}+1}\), \[\begin{align} \label{eq64613} u(t)\geq \varkappa_0\alpha^dM-R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,\xi_0]\,dt\,\,\,\,\,\,\mathrm{ a.e. in } B_{2\rho}(\xi_0). \end{align}\tag{58}\] Here, note that time interval still depends on \(\hat{n}\). So we need to find some subinterval that is independent of \(\hat{n}\). For this purpose a direct computation can be done as follows \[\begin{align} t_{\hat{n}+1}&=t_1+\delta_*\sum_{i=2}^{\hat{n}+1}\left(2^{i-2}k\rho\right)^2\\ &=t_1+\delta_*\rho^2\frac{(2^{\hat{n}-1}k)^2-k^2}{2^2-1}. \end{align}\] Without any loss of generality, by considering \(k\) sufficiently small as required, we may assume that \(\log_2(1/k)\) is an integer and choose \(\hat{n}=3+\log_2(1/k)\). Then we have \[\begin{align} t_{\hat{n}+1}=t_1+\left(\delta_*\frac{4^2-k^2}{3}\right)\rho^2, \end{align}\] and \[\begin{align} t_{\hat{n}+1}-\frac{\delta_*}{2}\left(2^{\hat{n}-1}k\rho\right)^2&=t_1+\delta_*\rho^2\left[\frac{4^2-k^2}{3}-\frac{1}{2}\left(2^{\hat{n}-1}k\right)^2\right]\\ &=t_1++\delta_*\rho^2\left[\frac{4^2-k^2}{3}-\frac{1}{2}\left(4\right)^2\right]\\ &<t_1+\delta_*\rho^2\frac{4^2-k^2}{3}. \end{align}\] Take \(\hat{\delta}_*:=\delta_*\frac{4^2-k^2}{3}\), which depends only on data and \(\alpha\). Next, (58 ) yields that for every \(t_1+\frac{1}{2}\hat{\delta}_*\rho^2<t\leq t_1+\hat{\delta}_*\rho^2\) \[\begin{align} \label{eq64614} u(t)\geq \varkappa_0\alpha^dM-R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,\xi_0]\,dt\,\,\,\,\,\mathrm{ a.e. in } B_{2\rho}(\xi_0). \end{align}\tag{59}\] Moreover, note that the selection of \(\delta_*\) assure that \[\begin{align} \hat{\delta}_*=\delta_*\frac{4^2-k^2}{3}=\frac{\alpha^{Q+3}}{10^6(8Q)^{Q+2}}\frac{4^2-k^2}{3}<1. \end{align}\]
Step IV The result in (59 ) is similar to our required result but in a different time interval. We need to shift the qualitative location of \(t_1\) to the original time layer \(t_0\) by the use of the result in Proposition 11. From (59 ), we can write \[\begin{align} \left|\left\{u\left(t_1+\frac{1}{2}\hat{\delta}_*\rho^2\right)\geq\varkappa_0\alpha^dM\right\}\cap B_{2\rho}(\xi_0)\right|=|B_{2\rho}(\xi_0)|, \end{align}\] therefore, from the result in Proposition 11 with \(\alpha=1\) and for any \(K>0\), there exists \(\hat{\varkappa}_0\equiv\hat{\varkappa}_0(\texttt{data},K)\in (0,1)\) such that \[\begin{align} u(t)\geq \hat{\varkappa}_0(\varkappa_0\alpha^dM)-R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,\xi_0]\,dt\,\,\,\,\mathrm{ a.e. in } B_{2\rho}(\xi_0), \end{align}\] for every time \(t_1+\frac{1}{2}\hat{\delta}_*\rho^2\leq t\leq t_1+\hat{\delta}_*\rho^2+K\rho^2\). Now, recalling the range of \(t_1\) such that for all \(\theta\in (1/2,1]\) such that \(t_1=t_0+\theta\delta\rho^2\) and taking \[\begin{align} K:=\theta\delta+\hat{\delta}_*=:\frac{\delta_0}{2},\,\,\,\,\tilde{\varkappa}:=\hat{\varkappa}_0\varkappa_0, \end{align}\] we obtain our result. Also note that \(\delta_0,\tilde{\varkappa}\) depend upon \(\texttt{data}\) and \(\alpha\). ◻

Proof. For simplicity, we assume \((\xi_0,t_0)=(0,0)\). Consider time independent cut-off function \(\psi\) in \(B_\rho\) such that \[\begin{align} \psi\equiv\begin{cases} 1\,\,\,\,\mathrm{ on } B_\rho,\\ 0\,\,\,\,\mathrm{ on } B_{\frac{3}{2}\rho}, \end{cases} \,\,\,\,\mathrm{ such that }|\nabla_{\mathbb{H}^N}\psi|\leq \frac{\gamma}{\rho}, \end{align}\] where \(\gamma=\gamma(\texttt{data)}\). For any \(M>0\), consider the truncated function \(w_\pm:=(u-M)_\pm\) and use this in the energy estimate (?? ) over the cylinder \(\mathcal{Q}_\rho^\ominus\) where \(\mathcal{Q}_\rho^\ominus=B_\rho\times(-\rho^2,0]\), we have \[\begin{align} \label{eq64616} \int_{\mathcal{Q}_\rho^\ominus}&w_-(\xi,t)d\xi\left(\int_{B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}d\eta\right)dt+\int_{\mathcal{Q}_\rho^\ominus}w_-(\xi,t)d\xi\left(\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}d\eta\right)dt\nonumber\\ &\leq \gamma\iint_{\mathcal{Q}_\rho^\ominus}\left(w^2_-\nabla_{\mathbb{H}^N}\psi\right)^2(\xi,t)\,d\xi dt+\gamma\int_{-\rho^2}^0\iint_{B_\rho\times B_\rho}\max\{w^2_-(\xi,t),w^2_-(\eta,t)\}\frac{|\psi(\xi)-\psi(\eta)|^2}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\nonumber\\ &\quad+\gamma\int_{\mathcal{Q}_\rho^\ominus}w_-(\xi,t)d\xi\left(\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_-(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right)dt+\int_{B_\rho}w^2_-(\xi,-\rho^2)\, d\xi. \end{align}\tag{61}\] Estimating like its counterpart in Lemma 14, we obtain that the right hand side of the previous inequality is dominated by \[\begin{align} \gamma M^2|B_\rho|, \end{align}\] for some constant \(\gamma=\gamma(\texttt{data})\). Recalling the non-negative property of \(u\) in \(\mathcal{Q}_\rho^\ominus\), then we have from the inequality (61 ), \[\begin{align} \gamma M^2|B_\rho|&\geq \int_{\mathcal{Q}_\rho^\ominus}w_-(\xi,t)d\xi\left(\int_{B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}d\eta\right)dt\\ &\geq \gamma\inf_{t\in (-\rho^2,0]}\int_{B_\rho}w_-(\xi,t)\,d\xi\left(\int_{\mathcal{Q}\rho^\ominus}\frac{u_+}{(2\rho)^{Q+2s}}\,d\xi dt-M\right)\\ &=\gamma \left(\int_{\mathcal{Q}\rho^\ominus}\frac{u_+}{(2\rho)^{Q+2s}}\,d\xi dt\right)\inf_{t\in (-\rho^2,0]}\int_{\{u(\cdot,t)\leq M\}\cap B_\rho}w_-(\xi,t)\,d\xi-\gamma M^2|B_\rho|\\ &\geq \gamma \left(\int_{\mathcal{Q}\rho^\ominus}\frac{u_+}{(2\rho)^{Q+2s}}\,d\xi dt\right) \inf_{t\in (-\rho^2,0]}\left|\left\{u(\cdot,t)\leq M\right\}\cap B_\rho\right|\,M-\gamma M^2|B_\rho|. \end{align}\] In this sequence of calculations, we use the non-negativity of \(\int_{\mathcal{Q}_\rho^\ominus}w_-(\xi,t)d\xi\left(\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}d\eta\right)dt\) in the first line and the second line follows by considering \(\rho<1\) and the simple observations \[\begin{align} u_+\leq \gamma\left((u-M)_++M\right)\,\,\,\mathrm{ and }\,\,\,|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}\leq 2\rho\,\,\, \mathrm{ for all }x,y\in B_\rho. \end{align}\] Finally, adjusting all the constants, we have our result from previous inequality. ◻

Proof. Without loss of any generality, consider \((\xi_0,t_0)=(0,0)\). We mainly use the inequality (61 ) with the property that \(\int_{\mathcal{Q}_\rho^\ominus}w_-(\xi,t)d\xi\left(\int_{B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}d\eta\right)dt\geq 0\). Since \(|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}\leq 2|\eta|_{\mathbb{H}^N}\) for all \(\xi\in B_\rho\) and \(\eta\in \mathbb{H}^N\setminus B_\rho\),upon using the estimate \(u_+\leq \gamma((u-M)_++M)\), we deduce that \[\begin{align} \gamma M^2|B_\rho|&\geq \int_{\mathcal{Q}_\rho^\ominus}w_-(\xi,t)d\xi\left(\int_{\mathbb{H}^N\setminus B_\rho}\frac{w_+(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}d\eta\right)dt\\ &\geq \frac{\gamma(\texttt{data})}{2^{Q+2s}} \inf_{t\in(-\rho^2,0]}\int_{B_\rho}w_-(\xi,t)\,d\xi\left(\int_{-\rho^2}^0\int_{\mathbb{H}^N\setminus B_\rho}\frac{u_+(\eta,t)}{|\eta|^{Q+2s}}d\eta dt-M\right)\\ & = \frac{\gamma(\texttt{data})}{2^{Q+2s}}\left[\inf_{t\in (-\rho^2,0]}\left|\left\{u(\cdot,t)\leq M\right\}\cap B_\rho\right|\, M\left(\fint_{-\rho^2}^0\mathrm{Tail}[u_+(t);\rho,0]\, dt\right)-M^2|B_\rho|\right]. \end{align}\] Consequently, adjusting all the relevant constants, we have \[\begin{align} \inf_{t\in (-\rho^2,0]}\left|\left\{u(\cdot,t)\leq \frac{M}{2}\right\}\cap B_\rho\right|\leq \frac{\gamma M}{\fint_{-\rho^2}^0\mathrm{Tail}[u_+(t);\rho,0]\, dt}|B_\rho|. \end{align}\] ◻

Proof. This lemma mainly follow the steps of the De Giorgi Lemma 12. For simplicity \((\xi_0,t_0)=(0,0)\). Let us introduce \(\rho_n,\tilde{\rho}_n,B_n,\tilde{B}_n,\mathcal{Q}_n^\oplus,\tilde{\mathcal{Q}}_n^\oplus,\lambda_n\) as follows \[\begin{align} \begin{cases} \rho_n=\rho+\frac{\rho}{2^{n+1}},\,\tilde{\rho}_n=\frac{\rho_n+\rho_{n+1}}{2},\\ B_n=B_{\rho_n}(0),\,\tilde{B}_n=B_{\tilde{\rho}_n}(0),\\ \mathcal{Q}_n^\oplus=B_n\times(0,\delta\rho_n^2],\,\tilde{\mathcal{Q}}_n^\oplus=\tilde{B}_n\times(0,\delta\tilde{\rho}_n^2],\\ \lambda_n=\lambda+\frac{\lambda}{2^{n+1}}. \end{cases} \end{align}\] Unlike Lemma 12, we consider time independent cut-off function \(\psi\) as follows \[\begin{align} \psi\equiv\begin{cases} 1\,\,\,\,\mathrm{ on } B_{n+1},\\ 0\,\,\,\,\mathrm{ on } B^c_n, \end{cases} \,\,\,\,\mathrm{ such that }|\nabla_{\mathbb{H}^N}\psi|\leq \gamma\frac{2^{n+4}}{\rho}. \end{align}\] Set truncated function \[\begin{align} w_-:=\left(u+g(t)-\lambda_n\right)_-\,\,\,\,\mathrm{ where } \,g(t)=\mathfrak{C}\int_{R^2}^t\int_{\mathbb{H}^N\setminus B_R}\frac{|u(\eta,t)|}{|\eta|^{Q+2s}_{\mathbb{H}^N}}\,d\eta dt, \end{align}\] where positive constant \(\mathfrak{C}\) to be chosen later. Moreover, we define \[\begin{align} &\mathcal{A}_n:=\left\{u(\xi,t)+g(t)<\lambda_n\right\}\cap\mathcal{Q}_n^\oplus,\\ &\mathcal{A}_n(t):=\left\{u(\xi,t)+g(t)<\lambda_n\right\}\cap B_n\,\,\,\,\mathrm{ for }\, t\in (0,\delta\rho^2]. \end{align}\] Considering all these entities in (?? ), we have \[\begin{align} \label{eq64617} \mathop{\mathrm{ess\,sup}}_{0<t\leq \delta\rho_{n+1}^2}&\int_{B_{n+1}}w^2_-(\xi,t)\,d\xi+\iint_{\mathcal{Q}_{n+1}}|\nabla_{\mathbb{H}^N}w_-|^2\,d\xi dt\nonumber\\ &\leq \gamma\int_0^{\delta\rho^2}\iint_{B_n\times B_n}\max\{w^2_-(\xi,t),w^2_-(\eta,t)\}\frac{|\psi(\xi)-\psi(\eta)|}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\xi d\eta dt\nonumber\\ &\quad+\gamma \iint_{\mathcal{Q}_n^\oplus}(w_-\nabla_{\mathbb{H}^N}\psi\nu)^2\,d\xi dt+\iint_{\mathcal{Q}_n^\oplus}(w_-\psi^2\nu^2)\,d\xi\left\{\frac{w_-(\eta,t)}{|{\eta}^{-1}\circ{\xi}|_{\mathbb{H}^N}^{Q+2s}}\,d\eta\right\}dt\nonumber\\ &\quad\quad-\mathfrak{C}\iint_{\mathcal{Q}_n^\oplus}g'(t)(w_-\psi^2\nu^2)(\xi,t)\,d\xi dt. \end{align}\tag{62}\]

Next, considering the respective calculations in Lemma 13 with suitable choice of \(\mathfrak{C}\) and \(\gamma\), we conclude that the left hand side of the previous inequality is dominated by \[\begin{align} \gamma\frac{2^{(Q+2s+2)n}}{\delta\rho^2}\lambda^2|\mathcal{A}_n|. \end{align}\] After that by similar calculations as in Lemma 12 with the forward in time interval, we have the recursive inequality \[\begin{align} Y_{n+1}\leq \gamma (2b)^n\delta^{-\left[\frac{1}{\kappa}+\frac{2}{q\kappa}\right]}Y_n, \end{align}\] where \(b=b(Q),\,\gamma=\gamma(\texttt{data}),\,\kappa=\frac{Q}{Q-2}\) and \(Y_n=\frac{|\mathcal{A}_n|}{|\mathcal{Q}_n^\oplus|}.\)
At this stage, we consider two possibilities. Case I: For some \(n_0\in\mathbb{N}\), let \[\begin{align} \frac{|\mathcal{A}_{n_0}|}{|\mathcal{Q}_{n_0}^\oplus|}\leq \iota_*. \end{align}\] This yields the following \[\begin{align} &\left|\{u+g(t)\leq \lambda_{n_0}\}\cap \mathcal{Q}_{n_0}^\oplus\right|\leq \iota_*|\mathcal{Q}_{n_0}^\oplus|\\ \implies&\left|\{u+g(t)\leq \lambda\}\cap (0,\delta\rho^2)+\mathcal{Q}_\rho^\ominus(\delta)\right|\leq \iota_*2^Q|(0,\delta\rho^2)+\mathcal{Q}_\rho^\ominus(\delta)|. \end{align}\] We get the last line because \(\rho< \rho_{n_0}\leq 2\) and \(\lambda\leq \lambda_{n_0}\). We choose \(\iota_*\) sufficiently small such that \[\begin{align} \label{eq64619} \iota_*\leq 2^{-Q}\iota, \end{align}\tag{63}\] where \(\iota\) is chosen as in Lemma 12. Then we have \[\begin{align} \label{eq64620} \left|\{u+g(t)\leq \lambda\}\cap (0,\delta\rho^2)+\mathcal{Q}_\rho^\ominus(\delta)\right|\leq \iota|(0,\delta\rho^2)+\mathcal{Q}_\rho^\ominus(\delta)|. \end{align}\tag{64}\] At this point, by tail condition and (64 ), we are allowed to apply Lemma 12 with the center \((0,\delta\rho^2)\) such that \[\begin{align} u+g(t)\geq \frac{\lambda}{4}\,\,\,\,\mathrm{a.e. in } B_\frac{\rho}{2}\times \left(\frac{3}{4}\delta\rho^2,\delta\rho^2\right]. \end{align}\] Considering the tail condition with \(g(T_2)\leq \frac{1}{8}\lambda\), we have \[\begin{align} \label{eq64621} u\geq \frac{\lambda}{8}\,\,\,\,\mathrm{a.e. in } B_\frac{\rho}{2}\times \left(\frac{3}{4}\delta\rho^2,\delta\rho^2\right]. \end{align}\tag{65}\] Case II: If for any \(n\in\mathbb{N}\), \[\begin{align} \frac{|\mathcal{A}_n|}{|\mathcal{Q}_n^\oplus|}> \iota_*. \end{align}\] Then applying convergence result (Lemma 5), we have \(\iota\in (0,1)\) as in Lemma 12 such that \[\begin{align} Y_0\leq \iota, \end{align}\] and \(Y_n\rightarrow 0\) as \(n\rightarrow 0\). Therefore, \[\begin{align} u+g(t)\geq \lambda\,\,\,\,\mathrm{a.e. in } B_\frac{\rho}{2}\times \left(0,\delta\rho^2\right]. \end{align}\] Considering the tail condition with \(g(T_2)\leq \frac{1}{8}\lambda\), we have \[\begin{align} \label{eq64622} u\geq \frac{7\lambda}{8}\,\,\,\,\mathrm{a.e. in } B_\frac{\rho}{2}\times \left(0,\delta\rho^2\right] \end{align}\tag{66}\] Finally, combining both (65 ) and (66 ), we have our result. ◻

Proof. Without loss of generality, let us consider \((\xi_0,t_0)=(0,0)\). Recalling the Lemma 17 with redefining \(M\), such that \[\begin{align} \inf_{(-(2\rho)^2,0]}\left|\{u(\cdot,t)\leq M\}\cap B_{2\rho}\right|\leq \frac{\gamma(\mathrm{\texttt{data}}) M}{\fint_{-{(2\rho)}^2}^0\mathrm{Tail}[u_+(t);2\rho,0]\, dt}|B_{2\rho}|, \end{align}\] provided that \[\begin{align} \label{eq64623} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,0]\, dt\leq M. \end{align}\tag{67}\] Take \(\iota\in (0,1)\) as in Lemma 12. Then choose \(M>0\) in such a way that \[\begin{align} &\frac{\gamma M}{\fint_{-{(2\rho)}^2}^0\mathrm{Tail}[u_+(t);2\rho,0]\, dt}\leq \iota\\ \implies&\frac{\gamma M}{\left[\left(\mathchoice {{\setbox 0=\displaystyle{\textstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\textstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\textstyle{\scriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\scriptstyle{\scriptscriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptscriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\scriptscriptstyle{\scriptscriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptscriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} \!\iint_{\mathcal{Q}_{2\rho}^\ominus}u_+^2\,d\xi dt\right)^\frac{1}{2}+\left(\fint_{-{(2\rho)}^2}^{0}\left(\fint_{B_{2\rho}}u_+(\xi,t)\,d\xi\right)^2\,dt\right)^\frac{1}{2}+\fint_{-{(2\rho)}^2}^0\mathrm{Tail}[u_+(t);2\rho,0]\, dt+1\right]}\leq \iota. \end{align}\] Set \(M\) as \[\begin{align} M=\frac{\iota}{4\gamma}\left[\left(\mathchoice {{\setbox 0=\displaystyle{\textstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\textstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\textstyle{\scriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\scriptstyle{\scriptscriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptscriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\scriptscriptstyle{\scriptscriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptscriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} \!\iint_{\mathcal{Q}_{2\rho}^\ominus}u_+^2\,d\xi dt\right)^\frac{1}{2}+\left(\fint_{-{(2\rho)}^2}^{0}\left(\fint_{B_{2\rho}}u_+(\xi,t)\,d\xi\right)^2\,dt\right)^\frac{1}{2}+\fint_{-{(2\rho)}^2}^0\mathrm{Tail}[u_+(t);2\rho,0]\, dt+1\right]. \end{align}\] Due to choice of this \(M\), there exists \(t^*\in (-(2\rho)^2,0]\) such that \[\begin{align} \inf_{(-(2\rho)^2,0]}\left|\{u(\cdot,t)\leq M\}\cap B_{2\rho}\right|\leq \iota|B_{2\rho}|. \end{align}\] Therefore, applying the Lemma 18, we have \[\begin{align} \label{eq64624} u\geq \frac{1}{8}M\,\,\,\,\mathrm{ a.e in }\, B_{\frac{\rho}{2}}\times (t^*+\frac{3}{4}\delta\rho^2,t^*+\delta\rho^2], \end{align}\tag{68}\] where \(\delta\in (0,1)\) depends upon data and \(\alpha\) with \(B_{2\rho}\times\left(-(2\rho)^2,\delta\rho^2\right]\subseteq \mathcal{Q}\Subset \Omega_T\) and \[\begin{align} \label{eq64625} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,0]\, dt\leq \frac{M}{8}. \end{align}\tag{69}\] Let, \(\bar{t}\in (t^*+\frac{3}{4}\delta\rho^2,t^*+\delta\rho^2]\), then from (68 ), we have \[\begin{align} \left|\{u(\cdot,\bar{t})\geq \frac{1}{4} M\}\cap B_{2\rho}\right|=|B_{2\rho}|. \end{align}\] Since (69 ) automatically implies (67 ), so we can proceed with the tail condition in (69 ). Next, we can use Proposition 12, i.e., the power-like expansion of positivity with \(\alpha=1\). Thus, we have some \(\tilde{\varkappa}\in (0,1)\) and \(\delta_0\) depending only on data, such that \[\begin{align} u\geq \tilde{\varkappa} M\,\,\,\,\mathrm{ a.e. in } B_\rho\times (t^*+\frac{3}{4}\delta\rho^2+\frac{1}{2}\delta_0\rho^2,t^*+\delta\rho^2+\delta_0\rho^2+K\rho^2], \end{align}\] where \(K>0\) (see in Step IV of Proposition 12) will be chosen later and provided \[\begin{align} R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,0]\, dt\leq \tilde{\varkappa}M. \end{align}\] Take minimum between \(\frac{1}{8}\) and \(\tilde{\varkappa}\) and proceed with this tail condition. Take \(K=32\), then for any \(t^*\in(-(2\rho)^2,0]\), it is obvious to have \[\begin{align} \left(\frac{5}{4}\rho^2,(4\rho)^2\right]\subseteq (t^*+\frac{3}{4}\delta\rho^2+\frac{1}{2}\delta_0\rho^2,t^*+\delta\rho^2+\delta_0\rho^2+K\rho^2], \end{align}\] provided \(B_{2\rho}\times\left(-(2\rho)^2, 2(4\rho)^2\right]\subseteq\mathcal{Q}\Subset \Omega_T\). Therefore, due to the choice of \(K\), we can conclude that \[\begin{align} u\geq \frac{\iota\tilde{\varkappa}}{2\gamma}&\Bigg[\left(\mathchoice {{\setbox 0=\displaystyle{\textstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\textstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\textstyle{\scriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\scriptstyle{\scriptscriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptscriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} {{\setbox 0=\scriptscriptstyle{\scriptscriptstyle{-}\mkern-3.5mu{-}}{\iint} \vcenter{\scriptscriptstyle{-}\mkern-3.5mu{-}}\kern-.51\wd 0}} \!\iint_{\mathcal{Q}_{2\rho}^\ominus}u_+^2\,d\xi dt\right)^\frac{1}{2}+\left(\fint_{-{(2\rho)}^2}^{0}\left(\fint_{B_{2\rho}}u_+(\xi,t)\,d\xi\right)^2\,dt\right)^\frac{1}{2}\\&\quad+\fint_{-{(2\rho)}^2}^0\mathrm{Tail}[u_+(t);2\rho,0]\, dt+1\Bigg] - R^{-2}\int_{T_1}^{T_2}\mathrm{Tail}[u_-(t);R,0]\, dt, \end{align}\] for a.e. in \(B_\rho\times \left(\frac{5}{4}\rho^2,(4\rho)^2\right]\), provided that \[\begin{align} B_{2\rho}\times\left(-(2\rho)^2, 2(4\rho)^2\right]\subseteq\mathcal{Q}\Subset \Omega_T. \end{align}\] Hence, we can conclude the result by taking \(\varkappa=\frac{\iota\tilde{\varkappa}}{2\gamma}\). ◻