Proof. It suffices to verify that the weight w is a locally integrable function in $\mathbb{R}^{n}$ under these above cases. Specifically, it only needs to prove that for any r > 0, there holds $\mu(B_{r}) \lt \infty$. Recall the following elemental inequalities: for $a,b\geq0$,

(2.1)\begin{align} \begin{cases} \frac{a^{q}+b^{q}}{2}\leq(a+b)^{q}\leq a^{q}+b^{q},&0\leq q\leq1,\\ a^{q}+b^{q}\leq(a+b)^{q}\leq2^{q-1}(a^{q}+b^{q}),&q \gt 1. \end{cases} \end{align}

Step 1. Consider the case when $\theta_{2}\geq0$. Then we have from Equation (2.1) that

\begin{align*} \mu(B_{r})=&2\int_{B_{r}\cap\{x_{n} \gt 0\}}\left|x^{\prime}\right|^{\theta_{1}}\left|x\right|^{\theta_{2}}\,{\rm d}x\sim\int_{B_{r}^{\prime}}\left|x^{\prime}\right|^{\theta_{1}}\,{\rm d}x^{\prime}\int_{0}^{\sqrt{r^{2}-|x^{\prime}|^{2}}}\left(\left|x^{\prime}\right|^{\theta_{2}}+x_{n}^{\theta_{2}}\right)\,{\rm d}x_{n}\notag\\ \sim&\int^{r}_{0}\left(s^{n+\theta_{1}+\theta_{2}-2}\left(r^{2}-s^{2}\right)^{\frac{1}{2}}+s^{n+\theta_{1}-2}\left(r^{2}-s^{2}\right)^{\frac{\theta_{2}+1}{2}}\right)\,{\rm d}s\notag\\ \sim&\,r^{n+\theta_{1}+\theta_{2}}\int^{1}_{0}\left(s^{n+\theta_{1}+\theta_{2}-2}\left(1-s\right)^{\frac{1}{2}}+s^{n+\theta_{1}-2}\left(1-s\right)^{\frac{\theta_{2}+1}{2}}\right)\,{\rm d}s. \end{align*}

Observe that this type of integration is called Beta function. It makes sense if and only if $n+\theta_{1}+\theta_{2}-1 \gt 0$ and $n+\theta_{1}-1 \gt 0$. Then the conclusion is proved in the case when $(\theta_{1},\theta_{2})\in\mathcal{A}$.

Step 2. Consider $\theta_{2} \lt 0$. Then it follows from Equation (2.1) that

(2.2)\begin{align} \mu(B_{r})=&2\int_{B_{r}\cap\{x_{n} \gt 0\}}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\notag\\ \sim&\int_{B_{r}^{\prime}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{\sqrt{r^{2}-|x^{\prime}|^{2}}}_{0}\frac{1}{|x^{\prime}|^{-\theta_{2}}+x_{n}^{-\theta_{2}}}\,{\rm d}x_{n}. \end{align}

For the last integration term in Equation (2.2), we further split it as follows:

\begin{align*} I_{1}=&\int_{B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{|x^{\prime}|}_{0}\frac{1}{|x^{\prime}|^{-\theta_{2}}+x_{n}^{-\theta_{2}}}\,{\rm d}x_{n},\notag\\ I_{2}=&\int_{B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{\sqrt{r^{2}-|x^{\prime}|^{2}}}_{|x^{\prime}|}\frac{1}{|x^{\prime}|^{-\theta_{2}}+x_{n}^{-\theta_{2}}}\,{\rm d}x_{n},\notag\\ I_{3}=&\int_{B_{r}^{\prime}\setminus B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{\sqrt{r^{2}-|x^{\prime}|^{2}}}_{0}\frac{1}{|x^{\prime}|^{-\theta_{2}}+x_{n}^{-\theta_{2}}}\,{\rm d}x_{n}. \end{align*}

Observe that $|x^{\prime}|^{-\theta_{2}}\leq|x^{\prime}|^{-\theta_{2}}+x_{n}^{-\theta_{2}}\leq2|x^{\prime}|^{-\theta_{2}}$ if $0\leq x_{n}\leq |x^{\prime}|$. Then for the first term I ₁, we have

\begin{align*} I_{1}\sim&\int_{B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{|x^{\prime}|}_{0}|x^{\prime}|^{\theta_{2}}\,{\rm d}x_{n}\sim\int^{\frac{r}{\sqrt{2}}}_{0}s^{n+\theta_{1}+\theta_{2}-1}\,{\rm d}s\notag\\ \sim&r^{n+\theta_{1}+\theta_{2}}\int^{1}_{0}s^{n+\theta_{1}+\theta_{2}-1}\,{\rm d}s. \end{align*}

This integration makes sense iff $n+\theta_{1}+\theta_{2} \gt 0$.

With regard to the second term I ₂, we divide it into two cases to discuss as follow.

Case 1. If $\theta_{2}=-1$, then

\begin{align*} I_{2}\sim&\int_{B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{\sqrt{r^{2}-|x^{\prime}|^{2}}}_{|x^{\prime}|}x_{n}^{-1}\,{\rm d}x_{n}\sim\int^{\frac{r}{\sqrt{2}}}_{0}s^{n+\theta_{1}-2}\,\ln\frac{\sqrt{r^{2}-s^{2}}}{s}\,{\rm d}s\notag\\ \sim&\,r^{n+\theta_{1}-1}\int^{\frac{1}{\sqrt{2}}}_{0}s^{n+\theta_{1}-2}\,\ln\frac{\sqrt{1-s^{2}}}{s}\,{\rm d}s\sim-r^{n+\theta_{1}-1}\int^{\frac{1}{\sqrt{2}}}_{0}s^{n+\theta_{1}-2}\,\ln s\,{\rm d}s\notag\\ \sim&-r^{n+\theta_{1}+\theta_{2}}\bigg(s^{n+\theta_{1}+\theta_{2}}\,\ln s\Big|^{\frac{1}{\sqrt{2}}}_{0}-\int^{\frac{1}{\sqrt{2}}}_{0}s^{n+\theta_{1}+\theta_{2}-1}\,{\rm d}s\bigg). \end{align*}

It makes sense iff $n+\theta_{1}+\theta_{2} \gt 0$.

Case 2. If $\theta_{2}\neq-1$, then

\begin{align*} I_{2}\sim&\int_{B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{\sqrt{r^{2}-|x^{\prime}|^{2}}}_{|x^{\prime}|}x_{n}^{\theta_{2}}\,{\rm d}x_{n}\notag\\ \sim&\int^{\frac{1}{\sqrt{2}}r}_{0}s^{n+\theta_{1}-2}\left(\left(r^{2}-s^{2}\right)^{\frac{\theta_{2}+1}{2}}-s^{\theta_{2}+1}\right)\,{\rm d}s\notag\\ \sim&\,r^{n+\theta_{1}+\theta_{2}}\int^{\frac{1}{\sqrt{2}}}_{0}s^{n+\theta_{1}-2}\left(\left(1-s^{2}\right)^{\frac{\theta_{2}+1}{2}}-s^{\theta_{2}+1}\right)\,{\rm d}s. \end{align*}

Note that $\min\{1,2^{-\frac{\theta_{2}+1}{2}}\}\leq|(1-s^{2})^{\frac{\theta_{2}+1}{2}}|\leq\max\{1,2^{-\frac{\theta_{2}+1}{2}}\}$ in $[0,\frac{1}{\sqrt{2}}].$ Then it makes sense iff $\theta_{1} \gt -(n-1)$ and $n+\theta_{1}+\theta_{2} \gt 0$.

The last term I ₃ remains to be analyzed. Note that $|x^{\prime}|\geq\sqrt{r^{2}-|x^{\prime}|^{2}}\geq x_{n}$ if $\frac{1}{\sqrt{2}}r\leq|x^{\prime}|\leq r$ and $0\leq x_{n}\leq\sqrt{r^{2}-|x^{\prime}|^{2}}$. Then we deduce

\begin{align*} I_{3}\sim&\int_{B_{r}^{\prime}\setminus B^{\prime}_{\frac{1}{\sqrt{2}}r}}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\int^{\sqrt{r^{2}-|x^{\prime}|^{2}}}_{0}|x^{\prime}|^{\theta_{2}}\,{\rm d}x_{n}\notag\\ \sim&\int_{\frac{1}{\sqrt{2}}r}^{r}s^{n+\theta_{1}+\theta_{2}-2}\sqrt{r^{2}-s^{2}}\,{\rm d}s\sim r^{n+\theta_{1}+\theta_{2}}\int_{\frac{1}{\sqrt{2}}}^{1}s^{n+\theta_{1}+\theta_{2}-2}\sqrt{1-s^{2}}\,{\rm d}s\notag\\ \sim&r^{n+\theta_{1}+\theta_{2}}, \end{align*}

where we used the fact that the integrand $s^{n+\theta_{1}+\theta_{2}}\sqrt{1-s^{2}}$ has no singular point in $[\frac{1}{\sqrt{2}},1].$ Consequently, combining these above facts, we obtain that if $(\theta_{1},\theta_{2})\in\mathcal{B}$, then ${\rm d}\mu$ is a Radon measure. The proof is complete.

Proof. For any $\bar{x}\in\mathbb{R}^{n}$ and R > 0, we divide all balls $B_{R}(\bar{x})$ into two types as follows: the first type satisfies $|\bar{x}|\geq3R$ and the second type satisfies $|\bar{x}| \lt 3R$.

Step 1. Consider the case when $|\bar{x}|\geq 3R$. Observe that

(2.3)\begin{align} \int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\geq \begin{cases} (|\bar{x}|-R)^{\theta_{2}}\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x,&\text{if }\theta_{2}\geq0,\\ (|\bar{x}|+R)^{\theta_{2}}\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x,&\text{if }\theta_{2} \lt 0, \end{cases} \end{align}

and

(2.4)\begin{align} \int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq \begin{cases} (|\bar{x}|+2R)^{\theta_{2}}\int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x,&\text{if }\theta_{2}\geq0,\\ (|\bar{x}|-2R)^{\theta_{2}}\int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x,&\text{if }\theta_{2} \lt 0. \end{cases} \end{align}

On one hand, we have

\begin{align*} \int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x=&2\int_{B_{R}^{\prime}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\sqrt{R^{2}-|x^{\prime}-\bar{x}^{\prime}|^{2}}\,{\rm d}x^{\prime}\geq\sqrt{3}R\int_{B^{\prime}_{R/2}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}. \end{align*}

Observe that $|x^{\prime}|^{\theta_{1}}$ increases radially if $\theta_{1}\geq0$, while it decreases radially for $\theta_{1} \lt 0$. Then we obtain that

1. (i) for $|\bar{x}^{\prime}|\geq\frac{3}{2}R$, then

   \begin{align*} \int_{B^{\prime}_{R/2}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\geq&\omega_{n-1}\left(\frac{R}{2}\right)^{n-1} \begin{cases} \left(|\bar{x}^{\prime}|-R/2\right)^{\theta_{1}},&\text{if }\theta_{1}\geq0,\\ \left(|\bar{x}^{\prime}|+R/2\right)^{\theta_{1}},&\text{if }\theta_{1} \lt 0; \end{cases} \end{align*}

2. (ii) for $|\bar{x}^{\prime}| \lt \frac{3}{2}R,$ then

   \begin{align*} \int_{B^{\prime}_{R/2}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\geq& \begin{cases} \int_{B^{\prime}_{R/2}(0^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime},&\text{if }\theta_{1}\geq0,\\ \int_{B^{\prime}_{R/2}\left(\frac{3}{2}R\frac{\bar{x}^{\prime}}{|\bar{x}^{\prime}|}\right)}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime},&\text{if }\theta_{1} \lt 0 \end{cases}\notag\\ \geq&\omega_{n-1}R^{n-1+\theta_{1}} \begin{cases} \frac{n-1}{2^{n-1+\theta_{1}}(n-1+\theta_{1})},&\text{if }\theta_{1}\geq0,\\ \frac{1}{2^{n-1}},&\text{if }\theta_{1} \lt 0. \end{cases} \end{align*}

On the other hand, we have

(2.5)\begin{align} \int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x=&2\int_{B_{2R}^{\prime}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\sqrt{4R^{2}-|x^{\prime}-\bar{x}^{\prime}|^{2}}\,{\rm d}x^{\prime}\notag\\ \leq&4R\int_{B^{\prime}_{2R}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}. \end{align}

By the same argument as above, we deduce that

1. (1) for $|\bar{x}^{\prime}|\geq3R$, then

   (2.6)\begin{align} \int_{B^{\prime}_{2R}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\leq&\omega_{n-1}(2R)^{n-1} \begin{cases} \left(|\bar{x}^{\prime}|+2R\right)^{\theta_{1}},&\text{if }\theta_{1}\geq0,\\ \left(|\bar{x}^{\prime}|-2R\right)^{\theta_{1}},&\text{if }\theta_{1} \lt 0; \end{cases} \end{align}

2. (2) for $|\bar{x}^{\prime}| \lt 3R$, then

   (2.7)\begin{align} \int_{B^{\prime}_{2R}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\leq&\int_{B^{\prime}_{5R}(0^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}=\frac{(n-1)\omega_{n-1}}{n-1+\theta_{1}}(5R)^{n-1+\theta_{1}}. \end{align}

   Note that if $\theta_{1}\geq0$, then

   \begin{align*} \begin{cases} |\bar{x}^{\prime}|+2R\leq2(|\bar{x}^{\prime}|-R/2),&\text{for }|\bar{x}^{\prime}|\geq3R,\\ R \lt |\bar{x}^{\prime}|-R/2 \lt 5R/2,&\text{for }3R/2 \lt |\bar{x}^{\prime}| \lt 3R, \end{cases} \end{align*}
   while, if $\theta_{1} \lt 0$,
   \begin{align*} \begin{cases} |\bar{x}^{\prime}|-2R\geq\frac{2}{7}(|\bar{x}^{\prime}|+R/2),&\text{for }|\bar{x}^{\prime}|\geq3R,\\ 2R \lt |\bar{x}^{\prime}|+R/2 \lt 7R/2,&\text{for }3R/2 \lt |\bar{x}^{\prime}| \lt 3R. \end{cases} \end{align*}

   Then combining these above facts, we obtain

   (2.8)\begin{align} \int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x\leq C(n,\theta_{1},\theta_{2})\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x. \end{align}

   Since $|\bar{x}|\geq3R$, then $|\bar{x}|+2R\leq4(|\bar{x}-R|)$ and $|\bar{x}|+R\leq4(|\bar{x}|-2R)$. This, in combination with Equations (2.3)–(2.8), reads that for $|\bar{x}|\geq3R,$

   (2.9)\begin{align} \int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq C(n,\theta_{1},\theta_{2})\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x. \end{align}

Step 2. Let $|\bar{x}| \lt 3R$. Then we have

\begin{align*} \int_{B_{2R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq \int_{B_{5R}(0)}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq C(n,\theta_{1},\theta_{2})R^{n+\theta_{1}+\theta_{2}}. \end{align*}

First, if $\theta_{1} \lt 0$, then

\begin{align*} \int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\geq&\int_{B_{R}(\bar{x})}|x|^{\theta_{1}+\theta_{2}}\,{\rm d}x\notag\\ \geq& \begin{cases} \int_{B_{R}(0)}|x|^{\theta_{1}+\theta_{2}}\,{\rm d}x,&\text{if }\theta_{2}\geq-\theta_{1},\\ \int_{B_{R}\left(3R\frac{\bar{x}}{|\bar{x}|}\right)}|x|^{\theta_{1}+\theta_{2}}\,{\rm d}x,&\text{if } \theta_{2} \lt -\theta_{1} \end{cases}\notag\\ \geq&R^{n+\theta_{1}+\theta_{2}} \begin{cases} \frac{n\omega_{n}}{n+\theta_{1}+\theta_{2}},&\text{if }\theta_{2}\geq-\theta_{1},\\ 2^{\theta_{1}+\theta_{2}}\omega_{n},&\text{if } \theta_{2} \lt -\theta_{1}, \end{cases} \end{align*}

where we utilized the fact that $|x|^{\theta_{1}+\theta_{2}}$ is radially increasing if $\theta_{1}+\theta_{2}\geq0$ and radially decreasing if $\theta_{1}+\theta_{2} \lt 0$.

Second, if $\theta_{1}\geq0$, we discuss as follows:

1. (i) for $\theta_{2}\geq0$, similarly as before, we have

   \begin{align*} \int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\geq&\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}+\theta_{2}}\,{\rm d}x\geq\sqrt{3}R\int_{B^{\prime}_{R}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}+\theta_{2}}\,{\rm d}x^{\prime}\notag\\ \geq&\int_{B_{R}^{\prime}(0^{\prime})}|x^{\prime}|^{\theta_{1}+\theta_{2}}\,{\rm d}x^{\prime}=\frac{(n-1)\omega_{n-1}}{n+\theta_{1}+\theta_{2}-1}R^{n+\theta_{1}+\theta_{2}}; \end{align*}

2. (ii) for $\theta_{2} \lt 0$, then

   \begin{align*} \int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x=&\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}+\theta_{2}}\left(\frac{|x^{\prime}|}{|x|}\right)^{-\theta_{2}}\,{\rm d}x\notag\\ \geq&\frac{8^{\theta_{2}}}{2^{\theta_{1}+\theta_{2}}}R^{\theta_{1}+\theta_{2}}\int_{B_{R}(\bar{x})\cap\{|x^{\prime}| \gt R/2\}}\,{\rm d}x\notag\\ \geq&\frac{8^{\theta_{2}}}{2^{\theta_{1}+\theta_{2}}}R^{\theta_{1}+\theta_{2}}\int_{B_{R}(0)\cap\{|x^{\prime}| \gt R/2\}}\,{\rm d}x\notag\\ \geq&\frac{8^{\theta_{2}}(\omega_{n}-2^{2-n}\omega_{n-1})}{2^{\theta_{1}+\theta_{2}}}R^{n+\theta_{1}+\theta_{2}}. \end{align*}

   Then combining these aforementioned facts, we obtain that Equation (2.9) also holds if $|\bar{x}| \lt 3R.$ The proof is complete.

Proof. For $1 \lt q \lt \infty,$ according to the definition of A _q-weight, it needs to verify the following inequality:

(2.10)\begin{align} \rlap{\mbox{--}}\kern-3.5pt\int_{B}w\,{\rm d}x\left(\rlap{\mbox{--}}\kern-3.5pt\int_{B}w^{-\frac{1}{q-1}}\,{\rm d}x\right)^{q-1}\leq C(n,q,\theta_{1},\theta_{2}) \end{align}

for any ball $B\subset\mathbb{R}^{n}$. For any R > 0 and $\bar{x}\in\mathbb{R}^{n}$, the ball $B_{R}(\bar{x})$ must belong to one of the following two types: $|\bar{x}|\geq3R$ and $|\bar{x}| \lt 3R$. On the one hand, if $|\bar{x}|\geq3R$, then we have

(2.11)\begin{align} \frac{2}{3}|\bar{x}|\leq|\bar{x}|-R\leq|x|\leq|\bar{x}|+R\leq\frac{4}{3}|\bar{x}|, \quad\text{for }x\in B_{R}(\bar{x}). \end{align}

Applying Equations (2.4)–(2.7) with $B_{2R}(\bar{x})$ and $B^{\prime}_{2R}(\bar{x}^{\prime})$ replaced by $B_{R}(\bar{x})$ and $B^{\prime}_{R}(\bar{x}^{\prime})$, it follows from Equation (2.11) that

\begin{align*} \int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq&C(\theta_{2})|\bar{x}|^{\theta_{2}}\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x\leq C(\theta_{2})|\bar{x}|^{\theta_{2}+1}\int_{B_{R}^{\prime}(\bar{x}^{\prime})}|x^{\prime}|^{\theta_{1}}\,{\rm d}x^{\prime}\notag\\ \leq&C(n,\theta_{1},\theta_{2})|\bar{x}|^{n+\theta_{1}+\theta_{2}} \end{align*}

and

\begin{align*} \int_{B_{R}(\bar{x})}|x^{\prime}|^{-\frac{\theta_{1}}{q-1}}|x|^{-\frac{\theta_{2}}{q-1}}\,{\rm d}x\leq&C(q,\theta_{2})|\bar{x}|^{-\frac{\theta_{2}}{q-1}}\int_{B_{R}(\bar{x})}|x^{\prime}|^{-\frac{\theta_{1}}{q-1}}\,{\rm d}x\notag\\ \leq& C(q,\theta_{2})|\bar{x}|^{-\frac{\theta_{2}}{q-1}+1}\int_{B_{R}^{\prime}(\bar{x}^{\prime})}|x^{\prime}|^{-\frac{\theta_{1}}{q-1}}\,{\rm d}x^{\prime}\notag\\ \leq& C(n,q,\theta_{1},\theta_{2})|\bar{x}|^{n-\frac{\theta_{1}+\theta_{2}}{p-1}}, \end{align*}

where we require that $-(n-1) \lt \theta_{1} \lt (n-1)(q-1)$ and $\theta_{2}\in\mathbb{R}$. Combining these two relations, we obtain that Equation (2.10) holds in the case of $|\bar{x}|\geq3R$.

On the other hand, if $|\bar{x}| \lt 3R$, we have $|x|\leq4R$ for $x\in B_{R}(\bar{x})$. Therefore, it follows from Lemma 2.1 that

\begin{align*} \rlap{\mbox{--}}\kern-3.5pt\int_{B_{R}(\bar{x})}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq 4^{n}\rlap{\mbox{--}}\kern-3.5pt\int_{B_{4R}(0)}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\leq C(n,\theta_{1},\theta_{2})R^{\theta_{1}+\theta_{2}} \end{align*}

and

\begin{align*} \left(\rlap{\mbox{--}}\kern-3.5pt\int_{B_{R}(\bar{x})}|x^{\prime}|^{-\frac{\theta_{1}}{q-1}}|x|^{-\frac{\theta_{2}}{q-1}}\,{\rm d}x\right)^{q-1}\leq& 4^{n(q-1)}\left(\rlap{\mbox{--}}\kern-3.5pt\int_{B_{4R}(0)}|x^{\prime}|^{-\frac{\theta_{1}}{q-1}}|x|^{-\frac{\theta_{2}}{q-1}}\,{\rm d}x\right)^{q-1}\notag\\ \leq& C(n,q,\theta_{1},\theta_{2})R^{-\theta_{1}-\theta_{2}}, \end{align*}

where these two relations hold if $(\theta_{1},\theta_{2}),(-\frac{\theta_{1}}{q-1},-\frac{\theta_{2}}{q-1})\in\mathcal{A}\cup\mathcal{B}$, that is, $(\theta_{1},\theta_{2})\in(\mathcal{A}\cup\mathcal{B})\cap(\mathcal{C}_{q}\cup\mathcal{D}_{q})$. Therefore, Equation (2.11) holds for any $B\subset\mathbb{R}^{n}$. The proof is complete.

Proof. Step 1. Set

\begin{align*} u_{1}=\inf\{u,l\}-\inf\{u,k\},\quad\bar{u}_{1}=\frac{1}{\mu(B_{R})}\int_{B_{R}}u_{1}\,{\rm d}\mu. \end{align*}

First, we have

\begin{align*} \int_{\{u_{1}=0\}\cap B_{R}}\bar{u}_{1}^{q}\,{\rm d}\mu=&\frac{1}{(\mu(B_{R}))^{q}}\left(\int_{B_{R}}u_{1}\,{\rm d}\mu\right)^{q}\int_{\{u\leq k\}\cap B_{R}}{\rm d}\mu\notag\\ \geq&\frac{C(l-k)^{q}}{R^{q(n+\theta_{1}+\theta_{2})}}\left(\int_{\{u\geq l\}\cap B_{R}}{\rm d}\mu\right)^{q}\int_{\{u\leq k\}\cap B_{R}}{\rm d}\mu. \end{align*}

Second, it follows from Corollary 2.8 that

\begin{align*} \int_{\{u_{1}=0\}\cap B_{R}}|\bar{u}_{1}|^{q}\,{\rm d}\mu\leq&\int_{B_{R}}|u_{1}-\bar{u}_{1}|^{q}\,{\rm d}\mu\leq CR^{q}\int_{B_{R}}|\nabla u_{1}|^{q}\,{\rm d}\mu\notag\\ =&CR^{q}\int_{\{k \lt u \lt l\}\cap B_{R}}|\nabla u|^{q}\,{\rm d}\mu. \end{align*}

The proof of Equation (2.13) is finished.

Step 2. Denote

\begin{align*} u_{2}=\sup\{u,l\}-\sup\{u,k\},\quad\bar{u}_{2}=\frac{\int_{B_{R}}u_{2}\,{\rm d}\mu}{|B_{R}|_{\mu}}. \end{align*}

By the same argument as before, we have

\begin{align*} \int_{\{u_{2}=0\}\cap B_{R}}\bar{u}_{2}^{q}\,{\rm d}\mu=&\frac{1}{(\mu(B_{R}))^{q}}\left(\int_{B_{R}}u_{2}\,{\rm d}\mu\right)^{q}\int_{\{u\geq l\}\cap B_{R}}{\rm d}\mu\notag\\ \geq&\frac{C(l-k)^{q}}{R^{q(n+\theta_{1}+\theta_{2})}}\left(\int_{\{u\leq k\}\cap B_{R}}{\rm d}\mu\right)^{q}\int_{\{u\geq l\}\cap B_{R}}{\rm d}\mu \end{align*}

and

\begin{align*} \int_{\{u_{2}=0\}\cap B_{R}}|\bar{u}_{2}|^{q}\,{\rm d}\mu\leq&\int_{B_{R}}|u_{2}-\bar{u}_{2}|^{q}\,{\rm d}\mu\leq CR^{q}\int_{B_{R}}|\nabla u_{2}|^{q}\,{\rm d}\mu\notag\\ =&CR^{q}\int_{\{k \lt u \lt l\}\cap B_{R}}|\nabla u|^{q}\,{\rm d}\mu. \end{align*}

The proof is complete.

Proof. First, pick test function $\varphi=v\eta^{p}$ if $v=(u-k)^{+}$. Since

\begin{align*} 0=\int_{B_{R}(x_{0})}Aw|\nabla u|^{p-2}\nabla u\cdot\nabla\varphi=\int_{B_{R}(x_{0})}Aw|\nabla v|^{p-2}\nabla v\cdot\nabla\varphi, \end{align*}

then it follows from Young’s inequality that

(3.1)\begin{align} &\lambda\int_{B_{R}(x_{0})}|\nabla v|^{p}\eta^{p}w\,{\rm d}x\notag\\ &\leq\int_{B_{R}(x_{0})}Aw|\nabla v|^{p}\eta^{p}=-p\int_{B_{R}(x_{0})}Aw|\nabla v|^{p-2}\nabla v\cdot\nabla\eta v\eta^{p-1}\notag\\ &\leq p\lambda\int_{B_{R}(x_{0})}|\eta\nabla v|^{p-1}|v\nabla\eta|w\,{\rm d}x\notag\\ &\leq\frac{\lambda}{2}\int_{B_{R}(x_{0})}|\nabla v|^{p}\eta^{p}w\,{\rm d}x+C\int_{B_{R}(x_{0})}|v|^{p}|\nabla\eta|^{p}w\,{\rm d}x, \end{align}

which yields that

\begin{align*} \int_{B_{R}(x_{0})}|\nabla(v\eta)|^{p}w\,{\rm d}x\leq&2^{p-1}\int_{B_{R}(x_{0})}\big(|\eta\nabla v|^{p}+|v\nabla\eta|^{p}\big)w\,{\rm d}x\notag\\ \leq& C\int_{B_{R}(x_{0})}|\nabla\eta|^{p}|v|^{p}w\,{\rm d}x. \end{align*}

Second, choose test function $\varphi=-v\eta^{2}$ if $v=(u-k)^{-}$. Then we have

\begin{align*} 0=\int_{B_{R}(x_{0})}Aw|\nabla u|^{p-2}\nabla u\cdot\nabla\varphi=-\int_{B_{R}(x_{0})}Aw|\nabla v|^{p-2}\nabla v\cdot\nabla\varphi. \end{align*}

Therefore, in exactly the same way to Equation (3.1), we obtain that Lemma 3.1 holds.

Proof. Step 1. For ɛ > 0 and $i=0,1,2,\ldots,$ let

\begin{align*} r_{i}=\frac{R}{2}+\frac{R}{2^{i+1}},\quad k_{i}=M-\varepsilon+\frac{\varepsilon}{2}(1-2^{-i}). \end{align*}

Take a cutoff function $\eta_{i}\in C_{0}^{\infty}(B_{r_{i}})$, satisfying that $\eta_{i}=1$ in $B_{r_{i+1}}$, $0\leq\eta_{i}\leq1$, $|\nabla\eta_{i}|\leq C(r_{i}-r_{i+1})^{-1}$ in $B_{r_{i}}$. For $k\in[m,M]$ and $\rho\in(0,R]$, write $v_{i}=(u-k_{i})^{+}$ and $A(k,\rho)=\{x\in B_{\rho}: u \gt k\}$. By Theorem 1.1 in [Reference Li and Yan30], we have the following anisotropic Caffarelli–Kohn–Nirenberg type inequality:

\begin{align*} \|u\|_{L^{\frac{(n+\theta_{1}+\theta_{2})p}{n+\theta_{1}+\theta_{2}-p}}(B_{R},w)}\leq C(n,p,\theta_{1},\theta_{2})\|\nabla u\|_{L^{p}(B_{R},w)},\quad \forall u\in W_{0}^{1,p}(B_{R},w), \end{align*}

which, together with Lemma 3.1, reads that

\begin{align*} \left(\int_{B_{R}}|\eta_{i}v_{i}|^{p\chi}w\,{\rm d}x\right)^{\frac{1}{\chi}}\leq C\int_{B_{R}}|v_{i}|^{p}|\nabla\eta_{i}|^{p}w\,{\rm d}x,\quad\chi=\frac{n+\theta_{1}+\theta_{2}}{n+\theta_{1}+\theta_{2}-p}. \end{align*}

Since

\begin{align*} \int_{B_{R}}|v_{i}|^{p}|\nabla\eta_{i}|^{p}w\,{\rm d}x\leq\frac{C(M-k_{i})^{p}}{(r_{i}-r_{i+1})^{p}}|A(k_{i},r_{i})|_{\mu} \end{align*}

and

\begin{align*} \left(\int_{B_{R}}|\eta_{i}v_{i}|^{p\chi}w\,{\rm d}x\right)^{\frac{1}{\chi}}\geq(k_{i+1}-k_{i})^{p}|A(k_{i+1},r_{i+1})|_{\mu}^{\frac{1}{\chi}}, \end{align*}

it then follows that there exists a positive constant $i_{0}=i_{0}(n,p,\theta_{1},\theta_{2}) \gt 0$ such that for $i\geq i_{0}$,

\begin{align*} \frac{|A(k_{i+1},r_{i+1})|_{\mu}}{|B_{R}|_{\mu}}\leq&(C2^{2p(i+2)})^{\chi}R^{(\theta_{1}+\theta_{2}+n)(\chi-1)-p\chi}\left(\frac{|A(k_{i},r_{i})|_{\mu}}{|B_{R}|_{\mu}}\right)^{\chi}\notag\\ =&(C2^{2p(i+2)})^{\chi}\left(\frac{|A(k_{i},r_{i})|_{\mu}}{|B_{R}|_{\mu}}\right)^{\chi}\notag\\ \leq&\prod\limits^{i}_{s=0}(C2^{2p(i+2-s)})^{\chi^{s+1}}\left(\frac{|A(k_{0},r_{0})|_{\mu}}{|B_{R}|_{\mu}}\right)^{\chi^{i+1}}\notag\\ \leq&(C^{\ast})^{i+1}\left(\frac{|A(k_{0},r_{0})|_{\mu}}{|Q_{R}|_{\mu}}\right)^{\chi(i+1)}. \end{align*}

Fix $\tau_{0}=(C^{\ast})^{-\chi}$. Then we deduce that for any ɛ > 0 and $0 \lt \tau \lt \tau_{0}$, if Equation (3.2) holds, then

\begin{align*} \frac{|A(k_{i+1},r_{i+1})|_{\mu}}{|B_{R}|_{\mu}}\leq (C^{\ast}\tau^{\chi})^{i+1}=\left(\frac{\tau}{\tau_{0}}\right)^{\chi(i+1)}\rightarrow0,\quad\mathrm{as}\;i\rightarrow\infty. \end{align*}

Hence, Equation (3.3) is proved.

Step 2. Similarly as above, set

\begin{align*} r_{i}=\frac{R}{2}+\frac{R}{2^{i+1}},\quad \tilde{k}_{i}=m+\varepsilon-\frac{\varepsilon}{2}(1-2^{-i}),\quad i\geq0. \end{align*}

For $k\in[m,M]$ and $\rho\in(0,R]$, let $\tilde{v}_{i}=(u-\tilde{k}_{i})^{-}$ and $\tilde{A}(k,\rho)=\{x\in B_{\rho}: u \lt k\}$. Then applying the proof of Equation (3.3) with minor modification, we obtain that Equation (3.4) also holds. The proof is finished.

Proof. Step 1. For $i\geq0$, let $k_{i}=M-\frac{\varepsilon}{2^{i}}$ and $A(k_{i},R)=B_{R}\cap\{u \gt k_{i}\}.$ From Equation (2.13), we know that for $q \gt 1,$

(3.8)\begin{align} &(k_{i+1}-k_{i})^{q}|A(k_{i+1},R)|_{\mu}^{q}|B_{R}\setminus A(k_{i},R)|_{\mu}\notag\\ &\leq CR^{q(n+\theta_{1}+\theta_{2}+1)}\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{q}w\,{\rm d}x. \end{align}

Using Equation (3.5), we have

\begin{align*} |B_{R}\setminus A(k_{i},R)|_{\mu}\geq\gamma|B_{R}|_{\mu}=C(n,\theta_{1},\theta_{2})\gamma R^{n+\theta_{1}+\theta_{2}}. \end{align*}

This, together with Equation (3.8), shows that

\begin{align*} &|A(k_{i+1},R)|_{\mu}\leq\frac{C2^{i+1}}{\varepsilon\sqrt[q]{\gamma}}R^{\frac{(n+\theta_{1}+\theta_{2})(q-1)}{q}+1}\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{q}w\,{\rm d}x\right)^{\frac{1}{q}}. \end{align*}

Since $1 \lt q \lt p \lt n+\theta_{1}+\theta_{2}$, we then have from Hölder’s inequality that

\begin{align*} &\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{q}w\,{\rm d}x\notag\\ &\leq\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{p}w\,{\rm d}x\right)^{\frac{q}{p}}\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}w\,{\rm d}x\right)^{\frac{p-q}{p}}\notag\\ &\leq\left(\int_{B_{R}}|\nabla(u-k_{i})^{+}|^{p}w\,{\rm d}x\right)^{\frac{q}{p}}\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}w\,{\rm d}x\right)^{\frac{p-q}{p}}. \end{align*}

Choose a cutoff function $\eta\in C_{0}^{\infty}(B_{2R})$ satisfying that

(3.9)\begin{align} \eta=1\;\mathrm{in}\;B_{R},\quad0\leq\eta\leq1,\;|\nabla\eta|\leq\frac{C(n)}{R}\;\,\mathrm{in}\;B_{2R}. \end{align}

It then follows from Lemma 3.1 that

\begin{align*} &\left(\int_{B_{R}}|\nabla (u-k_{i})^{+}|^{p}w\,{\rm d}x\right)^{\frac{1}{p}}\leq C\left(\int_{B_{2R}}|(u-k_{i})^{+}|^{p}|\nabla\eta|^{p}w\,{\rm d}x\right)^{\frac{1}{p}}\leq\frac{C\varepsilon}{2^{i}}R^{\frac{n+\theta_{1}+\theta_{2}-p}{p}}. \end{align*}

A combination of these above facts shows that

\begin{align*} |A(k_{i+1},R)|_{\mu}\leq&\frac{C}{\sqrt[q]{\gamma}}R^{(n+\theta_{1}+\theta_{2})(1-\frac{p-q}{pq})}|A(k_{i},R)\setminus A(k_{i+1},R)|_{\mu}^{\frac{p-q}{pq}}. \end{align*}

This leads to that for $j\geq1$,

\begin{align*} j|A(k_{j},R)|_{\nu}^{\frac{pq}{p-q}}\leq&\sum^{j-1}_{i=0}|A(k_{i+1},R)|_{\mu}^{\frac{pq}{p-q}}\leq\frac{C}{\gamma^{\frac{p}{p-q}}}R^{(n+\theta_{1}+\theta_{2})\left(\frac{pq}{p-q}-1\right)}|B_{R}|_{\mu}\notag\\ \leq&\frac{C}{\gamma^{\frac{p}{p-q}}}|B_{R}|_{\mu}^{\frac{pq}{p-q}}. \end{align*}

Then Equation (3.6) is proved.

Step 2. For $i\geq0$, denote $\tilde{k}_{i}=m+\frac{\varepsilon}{2^{i}}$ and $\tilde{A}(k_{i},R)=B_{R}\cap\{u \lt k_{i}\}.$ In light of Equation (2.14), we see that for $q \gt 1,$

\begin{align*} &(\tilde{k}_{i}-\tilde{k}_{i+1})^{q}|\tilde{A}(\tilde{k}_{i+1},R)|_{\mu}^{q}|B_{R}\setminus \tilde{A}(\tilde{k}_{i},R)|_{\mu}\notag\\ &\leq CR^{q(n+\theta_{1}+\theta_{2}+1)}\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|\nabla u|^{q}w\,{\rm d}x. \end{align*}

From Equation (3.7), we have

\begin{align*} |B_{R}\setminus \tilde{A}(\tilde{k}_{i},R)|_{\mu}\geq\gamma|B_{R}|_{\mu}=C(n,\theta_{1},\theta_{2})\gamma R^{n+\theta_{1}+\theta_{2}}. \end{align*}

Hence, we obtain

\begin{align*} &|\tilde{A}(\tilde{k}_{i+1},R)|_{\mu}\leq\frac{C2^{i+1}}{\varepsilon\sqrt[q]{\gamma}}R^{\frac{(n+\theta_{1}+\theta_{2})(q-1)}{q}+1}\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|\nabla u|^{q}w\,{\rm d}x\right)^{\frac{1}{q}}. \end{align*}

Analogously as before, it follows from Hölder’s inequality and Lemma 3.1 that for $1 \lt q \lt p \lt n+\theta_{1}+\theta_{2}$

\begin{align*} &\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|\nabla u|^{q}w\,{\rm d}x\right)^{\frac{1}{q}}\notag\\ &\leq\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|\nabla u|^{p}w\,{\rm d}x\right)^{\frac{1}{p}}\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}w\,{\rm d}x\right)^{\frac{p-q}{pq}}\notag\\ &\leq\left(\int_{B_{R}}|\nabla(u-\tilde{k}_{i})^{-}|^{p}w\,{\rm d}x\right)^{\frac{1}{p}}\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}w\,{\rm d}x\right)^{\frac{p-q}{pq}}\notag\\ &\leq C\left(\int_{B_{2R}}|(u-\tilde{k}_{i})^{-}|^{p}|\nabla\eta|^{p}w\,{\rm d}x\right)^{\frac{1}{p}}|\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)|_{\mu}^{\frac{p-q}{pq}}\notag\\ &\leq\frac{C\varepsilon}{2^{i}}R^{\frac{n+\theta_{1}+\theta_{2}-p}{p}}|\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)|_{\mu}^{\frac{p-q}{pq}}, \end{align*}

where η is given by Equation (3.9). Then we obtain

\begin{align*} |\tilde{A}(\tilde{k}_{i+1},R)|_{\mu}\leq&\frac{C}{\sqrt[q]{\gamma}}R^{(n+\theta_{1}+\theta_{2})(1-\frac{p-q}{pq})}|\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)|_{\mu}^{\frac{p-q}{pq}}, \end{align*}

and thus,

\begin{align*} j|\tilde{A}(\tilde{k}_{j},R)|_{\nu}^{\frac{pq}{p-q}}\leq&\sum^{j-1}_{i=0}|\tilde{A}(\tilde{k}_{i+1},R)|_{\mu}^{\frac{pq}{p-q}}\leq\frac{C}{\gamma^{\frac{p}{p-q}}}R^{(n+\theta_{1}+\theta_{2})\left(\frac{pq}{p-q}-1\right)}|B_{R}|_{\mu}\notag\\ \leq&\frac{C}{\gamma^{\frac{p}{p-q}}}|B_{R}|_{\mu}^{\frac{pq}{p-q}}\quad \text{for }j\geq1. \end{align*}

The proof is complete.

Proof. Applying Lemmas 3.2 and 3.4, we obtain that Corollary 3.5 holds. In particular, in the case of $\theta_{1}=0$, $\theta_{2} \gt -(n-1)$, $1 \lt p \lt n+\theta_{2}$, we fix $q=\frac{p+1}{2}$ in Lemma 3.4.

Proof of Theorem 1.1

For $0 \lt R\leq\frac{1}{2}$, denote

\begin{align*} \overline{\mu}(R)=\sup_{B_{R}}u,\quad\underline{\mu}(R)=\inf_{B_{R}}u,\quad \omega(R)=\overline{\mu}(R)-\underline{\mu}(R). \end{align*}

Note that one of the following two statements must hold: either

(3.10)\begin{align} |\{x\in B_{R}:u \gt \overline{\mu}(R)-2^{-1}\omega(R)\}|_{\mu}\leq\frac{1}{2}|B_{R}|_{\mu} \end{align}

or

(3.11)\begin{align} |\{x\in B_{R}:u \lt \underline{\mu}(R)+2^{-1}\omega(R)\}|_{\mu}\leq\frac{1}{2}|B_{R}|_{\mu}. \end{align}

Using Corollary 3.5 with $\gamma=\frac{1}{2}$, we derive that there is a large constant $k_{0} \gt 1$ such that

\begin{align*} \overline{\mu}(R/2)\leq\overline{\mu}(R)-\frac{\omega(R)}{2^{k_{0}+1}},\quad\text{when Equation~(3.10) holds} \end{align*}

and

\begin{align*} \underline{\mu}(R/2)\geq\underline{\mu}(R)+\frac{\omega(R)}{2^{k_{0}+1}},\quad\text{when Equation~(3.11) holds.} \end{align*}

In either case, we have

\begin{align*} \omega(R/2)\leq\left(1-\frac{1}{2^{k_{0}+1}}\right)\omega(R)=\frac{1}{2^{\alpha}}\omega(R),\quad\mathrm{with}\;\alpha=-\frac{\ln\big(1-\frac{1}{2^{k_{0}+1}}\big)}{\ln2}. \end{align*}

Observe that for each $0 \lt R\leq\frac{1}{2}$, there exists an integer $l\geq1$ such that $2^{-(l+1)} \lt R\leq2^{-l}$. Since $\omega(R)$ is non-decreasing with respect to R, we then have

\begin{align*} \omega(R)\leq\omega(2^{-l})\leq 2^{-(l-1)\alpha}\omega(2^{-1})=4^{\alpha}2^{-(l+1)\alpha}\omega(2^{-1})\leq CR^{\alpha}, \end{align*}

where $C=C(n,p,q,\theta_{1},\theta_{2},\lambda,\overline{M}).$ The proof is complete.

Proof of Theorem 1.4

First, by applying the proof of Theorem 1.1 with a slight modification, we also obtain that there exist a small constant $0 \lt \alpha=\alpha(n,p,\theta_{2},\lambda) \lt 1$ and a large constant $C=C(n,p,\theta_{2},\lambda,\overline{M}) \gt 0$ such that

(3.12)\begin{align} |u(x)-u(0)|\leq C|x|^{\alpha}\quad\text{for all}\;x\in B_{1/2}. \end{align}

For $R\in(0,1/2)$, $y\in Q_{1/R}$, denote

\begin{align*} u_{R}(y)=u(Ry),\qquad A_{R}(y)=A(Ry). \end{align*}

Hence, u _R is the solution of

\begin{align*} \mathrm{div}(A_{R}|y|^{\theta_{2}}\nabla u_{R})=0\quad\mathrm{in}\;Q_{1/R}. \end{align*}

After the change of variables, we see that this equation becomes degenerate elliptic equation in $B_{1/2}(\bar{y})$ for any $\bar{y}\in\partial B_{1}$. For any two given points $x,\tilde{x}\in B_{1/2},$ let $|\tilde{x}|\leq|x|$ without loss of generality. Denote $R=|x|$. By the interior Hölder estimate for degenerate elliptic equation, we derive that there exist two constants $0 \lt \beta=\beta(n,p,\theta_{2},\lambda) \lt 1$ and $0 \lt C=C(n,p,\theta_{2},\lambda,\overline{M})$ such that for any $\bar{y}\in\partial B_{1}$,

(3.13)\begin{align} |u_{R}(y)-u_{R}(\bar{y})|\leq C|y-\bar{y}|^{\beta},\quad\forall y\in B_{1/2}(\bar{y}). \end{align}

Consequently, for $|x-\tilde{x}|\leq R^{2}$, we have from Equation (3.13) that

\begin{align*} |u(x)-u(\tilde{x})|=&\left|u_{R}(x/R)-u_{R}(\tilde{x}/R)\right|\leq C|(x-\tilde{x})/R|^{\beta}\leq C|x-\tilde{x}|^{\beta/2}, \end{align*}

while, for $|x-\tilde{x}| \gt R^{2}$, we deduce from Equation (3.12) that

\begin{align*} |u(x)-u(\tilde{x})|\leq&|u(x)-u(0)|+|u(0)-u(\tilde{x})|\leq C\big(R^{\alpha}+|\tilde{x}|^{\alpha}\big)\leq CR^{\alpha}\leq C|x-\tilde{x}|^{\frac{\alpha}{2}}. \end{align*}

Therefore, the proof of Theorem 1.4 is complete.

Proof. Applying the anisotropic version of the Caffarelli–Kohn–Nirenberg inequality in [Reference Li and Yan30], we obtain that for any $u\in W_{0}^{1,2}(B_{R},w_{2})$,

\begin{align*} &\left(\int_{B_{R}}|u|^{\frac{2(n+\theta_{1}+\theta_{2})}{n+\theta_{1}+\theta_{2}-2}}|x^{\prime}|^{\frac{\theta_{3}(n+\theta_{1}+\theta_{2})-2\theta_{1}}{n+\theta_{1}+\theta_{2}-2}}|x|^{\frac{\theta_{4}(n+\theta_{1}+\theta_{2})-2\theta_{2}}{n+\theta_{1}+\theta_{2}-2}}\,{\rm d}x\right)^{\frac{n+\theta_{1}+\theta_{2}-2}{n+\theta_{1}+\theta_{2}}}\notag\\ &\leq C\int_{B_{R}}|\nabla u|^{2}|x^{\prime}|^{\theta_{3}}|x|^{\theta_{4}}\,{\rm d}x. \end{align*}

This, in combination with the Hölder’s inequality, leads to

(4.1)\begin{align} &\int_{B_{R}}|u|^{2\chi}|x^{\prime}|^{\theta_{3}}|x|^{\theta_{4}}\,{\rm d}x\notag\\ &=\int_{B_{R}}|u|^{2}|x^{\prime}|^{\theta_{3}-\theta_{1}(\chi-1)}|x^{\prime}|^{\theta_{4}-\theta_{2}(\chi-1)}|u|^{2(\chi-1)}|x^{\prime}|^{\theta_{1}(\chi-1)}|x|^{\theta_{2}(\chi-1)}\,{\rm d}x\notag\\ &\leq\left(\int_{B_{R}}|u|^{\frac{2}{2-\chi}}|x^{\prime}|^{\frac{\theta_{3}-\theta_{1}(\chi-1)}{2-\chi}}|x|^{\frac{\theta_{4}-\theta_{2}(\chi-1)}{2-\chi}}\,{\rm d}x\right)^{2-\chi}\left(\int_{B_{R}}|u|^{2}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\right)^{\chi-1}\notag\\ &\leq C\int_{B_{R}}|\nabla u|^{2}|x^{\prime}|^{\theta_{3}}|x|^{\theta_{4}}\,{\rm d}x\left(\int_{B_{R}}|u|^{2}|x^{\prime}|^{\theta_{1}}|x|^{\theta_{2}}\,{\rm d}x\right)^{\chi-1}. \end{align}

Then integrating Equation (4.1) from t ₁ to t ₂, it follows from Young’s inequality that

\begin{align*} \left(\int^{t_{2}}_{t_{1}}\int_{B_{R}}|u|^{2\chi}w_{2}\right)^{\frac{1}{\chi}}\leq& C\left(\sup\limits_{t\in(t_{1},t_{2})}\int_{B_{R}}|u|^{2}w_{1}\,{\rm d}x\right)^{\frac{\chi-1}{\chi}}\left(\int_{B_{R}}|\nabla u|^{2}w_{2}\,{\rm d}x\right)^{\frac{1}{\chi}}\notag\\ \leq&C\left(\sup\limits_{t\in(t_{1},t_{2})}\int_{B_{R}}|u|^{2}w_{1}\,{\rm d}x+\int^{t_{2}}_{t_{1}}\int_{B_{R}}|\nabla u|^{2}w_{2}\,{\rm d}x\,{\rm d}t\right). \end{align*}

The proof is complete.

Proof. Choose test function $\varphi=v\eta^{2}$. By denseness, we obtain that for $t_{1}\leq s\leq t_{2}$,

\begin{align*} \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\partial_{t}u^{p}v\eta^{2}\,{\rm d}x\,{\rm d}t+\int^{s}_{t_{1}}\int_{B_{R}(x_{0})}Aw_{2}\nabla u\nabla(v\eta^{2})\,{\rm d}x\,{\rm d}t=0. \end{align*}

On the one hand,

\begin{align*} \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\partial_{t}u^{p}v\eta^{2}dxdt=&p \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}v\eta^{2}u^{p-1}\partial_{t}v\,dxdt\notag\\ =&p \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\eta^{2}[(v+k)^{p}-k(v+k)^{p-1}]\partial_{t}v\,dxdt\notag\\ =&p \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\eta^{2}\partial_{t}\mathcal{H}, \end{align*}

where

\begin{align*} \mathcal{H}:=\frac{(v+k)^{p+1}}{p+1}-\frac{k(v+k)^{p}}{p}+\frac{k^{p+1}}{p(p+1)}. \end{align*}

Remark that the last term $\frac{k^{p+1}}{p(p+1)}$ in $\mathcal{H}$ is added to keep it non-negative. In fact, by Taylor expansion, we obtain

\begin{align*} \frac{(v+k)^{p+1}}{p+1}=&\frac{k^{p+1}}{p+1}\left(1+\frac{v}{k}\right)^{p+1}\notag\\ =&\frac{k^{p+1}}{p+1}\left(1+(p+1)\frac{v}{k}+\frac{p(p+1)}{2}\frac{v^{2}}{k^{2}}+\frac{(p-1)p(p+1)}{6}\frac{v^{3}}{k^{3}}+O\left(\frac{v^{4}}{k^{4}}\right)\right), \end{align*}

and

\begin{align*} \frac{k(v+k)^{p}}{p}=&\frac{k^{p+1}}{p}\left(1+\frac{v}{k}\right)^{p}\notag\\ =&\frac{k^{p+1}}{p}\left(1+p\frac{v}{k}+\frac{p(p-1)}{2}\frac{v^{2}}{k^{2}}+\frac{p(p-1)(p-2)}{6}\frac{v^{3}}{k^{3}}+O\left(\frac{v^{4}}{k^{4}}\right)\right). \end{align*}

A consequence of these two relations shows that

\begin{align*} \frac{(v+k)^{p+1}}{p+1}-\frac{k(v+k)^{p}}{p}=-\frac{k^{p+1}}{p(p+1)}+\frac{1}{2}k^{p-1}v^{2}+\frac{p-1}{3}k^{p-2}v^{3}+O\left(\frac{v^{4}}{k^{4}}\right), \end{align*}

which yields that

(4.2)\begin{align} 0\leq\mathcal{H}-\frac{1}{2}k^{p-1}v^{2}\leq C(p,\overline{m},\overline{M})v^{3}. \end{align}

In light of Equation (4.2), it follows from integration by parts that

\begin{align*} &\int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\partial_{t}u^{p}v\eta^{2}\,{\rm d}x\,{\rm d}t\notag\\ &\geq\frac{p}{2}k^{p-1}\int_{B_{R}(x_{0})}w_{1}\eta^{2}(x,s)v^{2}(x,s)\,{\rm d}x-\frac{p}{2}k^{p-1}\int_{B_{R}(x_{0})}w_{1}\eta^{2}(x,t_{1})v^{2}(x,t_{1})\,{\rm d}x\notag\\ &\quad-C\int_{B_{R}(x_{0})}w_{1}\eta^{2}(x,t_{1})v^{3}(x,t_{1})\,{\rm d}x-C\int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\eta|\partial_{t}\eta|v^{2}\,{\rm d}x\,{\rm d}t. \end{align*}

On the other hand, utilizing Young’s inequality, we have

\begin{align*} &\int_{B_{R}(x_{0})\times(t_{1},s)}Aw_{2}\nabla u\nabla(v\eta^{2})\,{\rm d}x\,{\rm d}t\notag\\ &=\int_{B_{R}(x_{0})\times(t_{1},s)}Aw_{2}\nabla v(\eta^{2}\nabla v+2v\eta\nabla\eta)\,{\rm d}x\,{\rm d}t\notag\\ &\geq\frac{1}{\lambda}\int_{B_{R}(x_{0})\times(t_{1},s)}|\eta\nabla v|^{2}w_{2}\,{\rm d}x\,{\rm d}t-C\int_{B_{R}(x_{0})\times(t_{1},s)}v\eta|\nabla\eta||\nabla v|w_{2}\,{\rm d}x\,{\rm d}t\notag\\ &\geq\frac{1}{2\lambda}\int_{B_{R}(x_{0})\times(t_{1},s)}|\eta\nabla v|^{2}w_{2}\,{\rm d}x\,{\rm d}t-C\int_{B_{R}(x_{0})\times(t_{1},s)}v^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t\notag\\ &\geq\frac{1}{4\lambda}\int_{B_{R}(x_{0})\times(t_{1},s)}|\nabla(\eta v)|^{2}w_{2}\,{\rm d}x\,{\rm d}t-C\int_{B_{R}(x_{0})\times(t_{1},s)}v^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t, \end{align*}

where in the last inequality, we used the following elementary inequality:

\begin{align*} |a-b|^{2}\geq\frac{1}{2}|a|^{2}-|b|^{2}\quad\text{for any } a,b\in\mathbb{R}^{n}. \end{align*}

Therefore, the first inequality in Lemma 4.2 holds.

The proof of the second inequality in Lemma 4.2 is analogous by picking test function $\varphi=-\tilde{v}\eta^{2}$. Then we obtain

\begin{align*} -\int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\partial_{t}u^{p}\tilde{v}\eta^{2}=&p \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\tilde{v}\eta^{2}u^{p-1}\,\partial_{t}\tilde{v}\notag\\ =&p \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\eta^{2}[-(k-\tilde{v})^{p}+k(k-\tilde{v})^{p-1}]\,\partial_{t}\tilde{v}\notag\\ =&p \int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\eta^{2}\,\partial_{t}\widetilde{\mathcal{H}}, \end{align*}

where

\begin{align*} \widetilde{\mathcal{H}}=\frac{(k-\tilde{v})^{p+1}}{p+1}-\frac{k(k-\tilde{v})^{p}}{p}+\frac{k^{p+1}}{p(p+1)}. \end{align*}

Similarly as before, it follows from Taylor expansion that

\begin{align*} -C(p,\overline{m},\overline{M})\tilde{v}^{3}\leq\widetilde{\mathcal{H}}-\frac{1}{2}k^{p-1}\tilde{v}^{2}\leq0, \end{align*}

which reads that

\begin{align*} &\int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\partial_{t}u^{p}v\eta^{2}\,{\rm d}x\,{\rm d}t\notag\\ &\geq\frac{p}{2}k^{p-1}\int_{B_{R}(x_{0})}w_{1}\eta^{2}(x,s)\tilde{v}^{2}(x,s)\,{\rm d}x-C\int_{B_{R}(x_{0})}w_{1}\eta^{2}(x,s)\tilde{v}^{3}(x,s)\,{\rm d}x\notag\\ &\quad-\frac{p}{2}k^{q-1}\int_{B_{R}(x_{0})}w_{1}\eta^{2}(x,t_{1})\tilde{v}^{2}(x,t_{1})\,{\rm d}x-C\int^{s}_{t_{1}}\int_{B_{R}(x_{0})}w_{1}\eta|\partial_{t}\eta|\tilde{v}^{2}\,{\rm d}x\,{\rm d}t. \end{align*}

By the same argument as before, we have

\begin{align*} &-\int_{B_{R}(x_{0})\times(t_{1},s)}Aw_{2}\nabla u\nabla(\tilde{v}\eta^{2})\,{\rm d}x\,{\rm d}t=\int_{B_{R}(x_{0})\times(t_{1},s)}Aw_{2}\nabla\tilde{v}\nabla(\tilde{v}\eta^{2})\,{\rm d}x\,{\rm d}t\notag\\ &\geq\frac{1}{4\lambda}\int_{B_{R}(x_{0})\times(t_{1},s)}|\nabla(\eta\tilde{v})|^{2}w_{2}\,{\rm d}x\,{\rm d}t-C\int_{B_{R}(x_{0})\times(t_{1},s)}\tilde{v}^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t. \end{align*}

The proof is complete.

Proof. Without loss of generality, let $t_{0}=0$.

Step 1. For ɛ > 0 and $i=0,1,2,\ldots,$ set

\begin{align*} r_{i}=\frac{R}{2}+\frac{R}{2^{i+1}},\qquad k_{i}=M-\varepsilon+\frac{\varepsilon}{2}(1-2^{-i}). \end{align*}

Choose a cutoff function $\eta_{i}\in C_{0}^{\infty}(Q_{r_{i}})$ such that

\begin{align*} \eta_{i}=1\;\mathrm{in}\;Q_{r_{i+1}},\quad 0\leq\eta_{i}\leq1,\;|\nabla\eta_{i}|\leq\frac{C}{r_{i}-r_{i+1}},\;|\partial_{t}\eta_{i}|\leq\frac{C}{r_{i}^{\theta_{1}+\theta_{2}}-r_{i+1}^{\theta_{1}+\theta_{2}}}\;\,\mathrm{in}\;Q_{r_{i}}. \end{align*}

Denote $v_{i}=(u-k_{i})^{+}$ and $A(k,\rho)=\{(x,t)\in Q_{\rho}: u \gt k\}$ for $k\in[m,M]$ and $\rho\in(0,R]$. Then combining Proposition 4.1 and Lemma 4.2, we deduce

(4.7)\begin{align} \left(\int_{Q_{R}}|\eta_{i}v_{i}|^{2\chi}w_{2}\right)^{\frac{1}{\chi}}\leq C\int_{Q_{R}}\left(|\nabla\eta_{i}|^{2}+|\partial_{t}\eta_{i}||x^{\prime}|^{\theta_{1}-\theta_{3}}|x|^{\theta_{2}-\theta_{4}}\right)v_{i}^{2}w_{2}, \end{align}

where $\chi=(n+\theta_{1}+\theta_{2}+2)(n+\theta_{1}+\theta_{2})^{-1}$. Note that

(4.8)\begin{align} |\partial_{t}\eta_{i}|R^{\theta_{1}+\theta_{2}-\theta_{3}-\theta_{4}}\leq\frac{CR^{2-\theta_{3}-\theta_{4}}}{(r_{i}-r_{i+1})^{2}}=\frac{C}{(r_{i}-r_{i+1})^{2}}. \end{align}

Therefore, we have

\begin{align*} \int_{Q_{R}}\left(|\nabla\eta_{i}|^{2}+|\partial_{t}\eta_{i}||x^{\prime}|^{\theta_{1}-\theta_{3}}|x|^{\theta_{2}-\theta_{4}}\right)v_{i}^{2}w_{2}\leq\frac{C(M-k_{i})^{2}}{(r_{i}-r_{i+1})^{2}}|A(k_{i},r_{i})|_{\nu_{w_{2}}} \end{align*}

and

\begin{align*} \left(\int_{Q_{R}}|\eta_{i}v_{i}|^{2\chi}w_{2}\,{\rm d}x\,{\rm d}t\right)^{\frac{1}{\chi}}\geq(k_{i+1}-k_{i})^{2}|A(k_{i+1},r_{i+1})|_{\nu_{w_{2}}}^{\frac{1}{\chi}}. \end{align*}

Define

\begin{align*} F_{i}:=\frac{|A(k_{i},r_{i})|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}. \end{align*}

Then we have

(4.9)\begin{align} F_{i+1}\leq&(C2^{4(i+2)})^{\chi}R^{(n+\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4})(\chi-1)-2\chi}F_{i}^{\chi}\notag\\ =&(C2^{4(i+2)})^{\chi}R^{\frac{2(\theta_{3}+\theta_{4}-2)}{n+\theta_{1}+\theta_{2}}}F_{i}^{\chi}=(C2^{4(i+2)})^{\chi}F_{i}^{\chi}\notag\\ \leq&\prod\limits^{i}_{s=0}(C2^{4(i+2-s)})^{\chi^{s+1}}F_{0}^{\chi^{i+1}}. \end{align}

Observe from Equations (4.8)–(4.9) that the value of $\theta_{3}+\theta_{4}$ must be chosen to be 2. A consequence of Equations (4.3) and (4.9) shows that there exists a constant $i_{0}=i_{0}(n,\theta_{1},\theta_{2},\theta_{3}) \gt 0$ such that if $i\geq i_{0}$,

\begin{align*} F_{i+1}\leq (C^{\ast})^{i+1}\left(\frac{|A(k_{0},r_{0})|_{\nu_{w_{1}}}}{|Q_{R}|_{\nu_{w_{1}}}}\right)^{\frac{\theta_{3}}{\theta_{1}}\chi^{i+1}}\leq (C^{\ast})^{i+1}\left(\frac{|A(k_{0},r_{0})|_{\nu_{w_{1}}}}{|Q_{R}|_{\nu_{w_{1}}}}\right)^{\frac{\theta_{3}}{\theta_{1}}\chi(i+1)}. \end{align*}

By taking $\tau_{0}=(C^{\ast})^{-\frac{\theta_{1}}{\theta_{3}}\chi}$, we obtain that for any ɛ > 0 and $0 \lt \tau \lt \tau_{0}$, if Equation (4.4) holds, then

\begin{align*} F_{i+1}\leq \left(C^{\ast}\tau^{\frac{\theta_{3}}{\theta_{1}}\chi}\right)^{i+1}=\left(\frac{\tau}{\tau_{0}}\right)^{\frac{\theta_{3}}{\theta_{1}}\chi(i+1)}\rightarrow0\quad\mathrm{as}\;i\rightarrow\infty. \end{align*}

That is, Equation (4.5) holds.

Step 2. Analogously as before, pick

\begin{align*} r_{i}=\frac{R}{2}+\frac{R}{2^{i+1}},\quad \tilde{k}_{i}=m+\varepsilon-\frac{\varepsilon}{2}(1-2^{-i}),\quad i\geq0. \end{align*}

Let $\varepsilon_{0}=\frac{1}{C_{0}}$, where C ₀ is given in Lemma 4.2. Denote $\tilde{v}_{i}=(u-\tilde{k}_{i})^{-}$. Then we obtain that for any $0 \lt \varepsilon\leq\varepsilon_{0}$,

\begin{align*} \tilde{v}_{i}^{2}-C_{0}\tilde{v}_{i}^{3}\geq(1-C_{0}\varepsilon)\tilde{v}_{i}^{2}\geq0, \end{align*}

which implies that Equation (4.7) holds with v _i replaced by $\tilde{v}_{i}$. Then following the left proof of Equation (4.5) above, we deduce that Equation (4.6) holds. The proof is complete.

Proof. Step 1. For $i\geq0$, denote $k_{i}=M_{a}-\frac{\varepsilon}{2^{i}}$ and

\begin{align*} A(k_{i},R;t)=B_{R}\cap\{u(\cdot,t) \gt k_{i}\},\qquad A(k_{i},R)=(B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}])\cap\{u \gt k_{i}\}. \end{align*}

It then follows from Equation (2.13) that for $1 \lt q \lt 2,$

(4.11)\begin{align} &(k_{i+1}-k_{i})^{q}|A(k_{i+1},R;t)|_{\mu_{w_{1}}}^{q}|B_{R}\setminus A(k_{i},R;t)|_{\mu_{w_{1}}}\notag\\ &\leq CR^{q(n+\theta_{1}+\theta_{2}+1)}\int_{A(k_{i},R;t)\setminus A(k_{i+1},R;t)}|\nabla u|^{q}w_{1}\,{\rm d}x. \end{align}

From the assumed condition, we have

\begin{align*} |B_{R}\setminus A(k_{i},R;t)|_{\mu_{w_{1}}}\geq\gamma|B_{R}|_{\mu_{w_{1}}}=C(n,\theta_{1})\gamma R^{n+\theta_{1}+\theta_{2}}. \end{align*}

Substituting this into Equation (4.11) and integrating from t ₀ to $t_{0}+a R^{\theta_{1}+\theta_{2}}$, we deduce from Hölder’s inequality that

\begin{align*} &\int^{t_{0}+aR^{\theta_{1}+\theta_{2}}}_{t_{0}}|A(k_{i+1},R;t)|_{\mu_{w_{1}}}\,{\rm d}t\notag\\ &\leq\frac{C2^{i+1}}{\varepsilon\sqrt[q]{\gamma}}R^{\frac{(n+\theta_{1}+\theta_{2})(q-1)}{q}+1}\int_{t_{0}}^{t_{0}+a R^{\theta_{1}+\theta_{2}}}\left(\int_{A(k_{i},R;t)\setminus A(k_{i+1},R;t)}|\nabla u|^{q}w_{1}\,{\rm d}x\right)^{\frac{1}{q}}\,{\rm d}t\notag\\ &\leq\frac{C2^{i+1}a^{\frac{q-1}{q}}}{\varepsilon\sqrt[q]{\gamma}}R^{\frac{(n+2\theta_{1}+2\theta_{2})(q-1)}{q}+1}\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{q}w_{1}\,{\rm d}x\,{\rm d}t\right)^{\frac{1}{q}}. \end{align*}

In light of $1 \lt q \lt 2$, it follows from Hölder’s inequality again that

\begin{align*} &\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{q}w_{1}\right)^{\frac{1}{q}}\notag\\ &\leq\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|\nabla u|^{2}w_{2}\right)^{\frac{1}{2}}\left(\int_{A(k_{i},R)\setminus A(k_{i+1},R)}|x'|^{\frac{2\theta_{1}-q\theta_{3}}{2-q}}|x|^{\frac{2\theta_{2}-q\theta_{4}}{2-q}}\right)^{\frac{2-q}{2q}}\notag\\ &\leq R^{\frac{\theta_{1}+\theta_{2}-\theta_{3}-\theta_{4}}{2}}|A(k_{i},R)\setminus A(k_{i+1},R)|_{\nu_{w_{1}}}^{\frac{2-q}{2q}}\left(\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}|\nabla (u-k_{i})^{+}|^{2}w_{2}\right)^{\frac{1}{2}}. \end{align*}

Pick a cutoff function $\eta\in C_{0}^{\infty}(B_{2R})$ such that

(4.12)\begin{align} \eta=1\;\mathrm{in}\;B_{R},\quad0\leq\eta\leq1,\;|\nabla\eta|\leq\frac{C(n)}{R}\;\,\mathrm{in}\;B_{2R}. \end{align}

Then from Lemma 4.2, we deduce

\begin{align*} &\left(\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}|\nabla (u-k_{i})^{+}|^{2}w_{2}\right)^{\frac{1}{2}}\notag\\ &\leq C\left(\int_{B_{2R}}|(u-k_{i})^{+}(x,t_{0})|^{2}w_{1}+\frac{1}{R^{2}}\int_{B_{2R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}|(u-k_{i})^{+}|^{2}w_{2}\right)^{\frac{1}{2}}\notag\\ &\leq\frac{C\varepsilon}{2^{i}}R^{\frac{n+\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}-2}{2}}, \end{align*}

where in the last inequality, we used the assumed condition that $0\leq\theta_{3}+\theta_{4}\leq2$. Therefore, combining these above facts, we obtain

\begin{align*} |A(k_{i+1},R)|_{\nu_{w_{1}}}\leq&\frac{Ca^{\frac{q-1}{q}}}{\sqrt[q]{\gamma}}R^{\frac{(n+2\theta_{1}+2\theta_{2})(3q-2)}{2q}}\left|A(k_{i},R)\setminus A(k_{i+1},R)\right|_{\nu_{w_{1}}}^{\frac{2-q}{2q}}, \end{align*}

which yields that for $j\geq1$,

\begin{align*} j|A(k_{j},R)|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}\leq&\sum^{j-1}_{i=0}|A(k_{i+1},R)|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}\notag\\ \leq&\frac{Ca^{\frac{2(q-1)}{2-q}}}{\gamma^{\frac{2}{2-q}}}R^{\frac{(n+2\theta_{1}+2\theta_{2})(3q-2)}{2-q}}|B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]|_{\nu_{w_{1}}}\notag\\ \leq&\frac{C}{\gamma^{\frac{2}{2-q}}a^{\frac{q}{2-q}}}|B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}. \end{align*}

Then Equation (4.10) holds.

Step 2. For $i\geq0$, write

\begin{align*} \tilde{k}_{i}=m_{a}+\frac{\varepsilon}{2^{i}} \end{align*}

and

\begin{align*} \tilde{A}(k_{i},R;t)=B_{R}\cap\{u(\cdot,t) \lt k_{i}\},\qquad \tilde{A}(k_{i},R)=(B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}])\cap\{u \lt k_{i}\}. \end{align*}

Using Equation (2.14), we obtain that for $1 \lt q \lt 2,$

\begin{align*} &(\tilde{k}_{i}-\tilde{k}_{i+1})^{q}|\tilde{A}(\tilde{k}_{i+1},R;t)|_{\mu_{w_{1}}}^{q}|B_{R}\setminus \tilde{A}(\tilde{k}_{i},R;t)|_{\mu_{w_{1}}}\notag\\ &\leq CR^{q(n+\theta_{1}+\theta_{2}+1)}\int_{\tilde{A}(\tilde{k}_{i},R;t)\setminus \tilde{A}(\tilde{k}_{i+1},R;t)}|\nabla u|^{q}w_{1}\,{\rm d}x. \end{align*}

Observe by the assumed condition that

\begin{align*} |B_{R}\setminus \tilde{A}(\tilde{k}_{i},R;t)|_{\mu_{w_{1}}}\geq\gamma|B_{R}|_{\mu_{w_{1}}}=C(n,\theta_{1},\theta_{2})\gamma R^{n+\theta_{1}+\theta_{2}}. \end{align*}

Analogously as above, integrating from t ₀ to $t_{0}+a R^{\theta_{1}+\theta_{2}}$ and using Hölder’s inequality, we have

\begin{align*} &\int^{t_{0}+aR^{\theta_{1}+\theta_{2}}}_{t_{0}}|\tilde{A}(\tilde{k}_{i+1},R;t)|_{\mu_{w_{1}}}\,{\rm d}t\notag\\ &\leq\frac{C2^{i+1}a^{\frac{q-1}{q}}}{\varepsilon\sqrt[q]{\gamma}}R^{\frac{(n+2\theta_{1}+2\theta_{2})(q-1)}{q}+1}\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|\nabla u|^{q}w_{1}\,{\rm d}x\,{\rm d}t\right)^{\frac{1}{q}} \end{align*}

and

\begin{align*} &\int_{\tilde{A}(\tilde{k}_{i},R)\setminus A(\tilde{k}_{i+1},R)}|\nabla u|^{q}w_{1}\notag\\ &\leq\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|\nabla u|^{2}w_{2}\right)^{\frac{q}{2}}\left(\int_{\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)}|x'|^{\frac{2\theta_{1}-q\theta_{3}}{2-q}}|x|^{\frac{2\theta_{2}-q\theta_{4}}{2-q}}\right)^{\frac{2-q}{2}}\notag\\ &\leq R^{\frac{q(\theta_{1}+\theta_{2}-\theta_{3}-\theta_{4})}{2}}|\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)|_{\nu_{w_{1}}}^{\frac{2-q}{2q}}\left(\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}|\nabla (u-\tilde{k}_{i})^{-}|^{2}w_{2}\right)^{\frac{q}{2}}. \end{align*}

For any $0 \lt \varepsilon\leq\varepsilon_{0}=C_{0}^{-1}$ with C ₀ given in Lemma 4.2, we know that

\begin{align*} [(u-\tilde{k}_{i})^{-}]^{2}-C_{0}[(u-\tilde{k}_{i})^{-}]^{3}\geq(1-C_{0}\varepsilon)[(u-\tilde{k}_{i})^{-}]^{2}\geq0. \end{align*}

Therefore, in view of $0\leq\theta_{3}+\theta_{4}\leq2$ and applying Lemma 4.2 with η defined by Equation (4.12), we derive

\begin{align*} &\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}|\nabla (u-\tilde{k}_{i})^{-}|^{2}w_{2}\notag\\ &\leq C\left(\int_{B_{2R}}|(u-\tilde{k}_{i})^{-}(x,t_{0})|^{2}w_{1}+\frac{1}{R^{2}}\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}|(u-\tilde{k}_{i})^{-}|^{2}w_{2}\right)\notag\\ &\leq\frac{C\varepsilon^{2}}{4^{i}}R^{n+\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}-2}. \end{align*}

Then we deduce

\begin{align*} |\tilde{A}(\tilde{k}_{i+1},R)|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}\leq&\frac{Ca^{\frac{2(q-1)}{2-q}}}{\gamma^{\frac{2}{2-q}}}R^{\frac{(n+2\theta_{1}+2\theta_{2})(3q-2)}{2-q}}|\tilde{A}(\tilde{k}_{i},R)\setminus \tilde{A}(\tilde{k}_{i+1},R)|_{\nu_{w_{1}}}. \end{align*}

This leads to that for $j\geq1$,

\begin{align*} j|\tilde{A}(\tilde{k}_{j},R)|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}\leq&\sum^{j-1}_{i=0}|\tilde{A}(\tilde{k}_{i+1},R)|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}\notag\\ \leq&\frac{Ca^{\frac{2(q-1)}{2-q}}}{\gamma^{\frac{2}{2-q}}}R^{\frac{(n+2\theta_{1}+2\theta_{2})(3q-2)}{2-q}}|B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]|_{\nu_{w_{1}}}\notag\\ \leq&\frac{C}{\gamma^{\frac{2}{2-q}}a^{\frac{q}{2-q}}}|B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]|_{\nu_{w_{1}}}^{\frac{2q}{2-q}}. \end{align*}

The proof is complete.

Proof. Step 1. For $a\in(0,1]$ and $k\in[\overline{m},\overline{M}]$, define

\begin{align*} A^{a}(k,R)=(B_{R}\times[t_{0},t_{0}+a R^{\theta_{1}+\theta_{2}}])\cap\{u \gt k\}. \end{align*}

Take a smooth cutoff function $\eta\in C^{\infty}_{0}(B_{R})$ satisfying that η = 1 in $B_{\sigma R}$, where $\sigma\in(0,1)$ to be determined later. Set $k_{1} \gt 1$. Denote $v=(u-(M_{1}-\varepsilon))^{+}$. From Lemma 4.2, we obtain

(4.17)\begin{align} &\sup\limits_{t\in(t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}})}\int_{B_{R}}v^{2}\eta^{2}w_{1}\,{\rm d}x\notag\\ &\leq\int_{B_{R}}(v^{2}+Cv^{3})\eta^{2}w_{1}\,{\rm d}x\big|_{t_{0}}+C\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}v^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t. \end{align}

Observe that for $t\in[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]$,

\begin{align*} \int_{B_{R}}v^{2}\eta^{2}w_{1}\,{\rm d}x\big|_{t}\geq\varepsilon^{2}(1-2^{-k_{1}})^{2}|B_{\sigma R}\cap\{u(x,t) \gt M_{1}-2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}, \end{align*}

and by Equation (4.13),

\begin{align*} \int_{B_{R}}(v^{2}+Cv^{3})\eta^{2}w_{1}\,{\rm d}x\big|_{t_{0}}\leq&\varepsilon^{2}(1+C\varepsilon)|\{x\in B_{R}: u(x,t_{0}) \gt M_{1}-\varepsilon\}|_{\mu_{w_{1}}}\notag\\ \leq&\varepsilon^{2}(1+C\varepsilon)(1-\gamma)|B_{R}|_{\mu_{w_{1}}}, \end{align*}

and

\begin{align*} \int^{t_{0}+aR^{\theta_{1}+\theta_{2}}}_{t_{0}}\int_{B_{R}}v^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t\leq&\frac{C\varepsilon^{2}}{(1-\sigma)^{2}R^{2}}|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}\notag\\ \leq&\frac{C\varepsilon^{2}R^{\theta_{3}+\theta_{4}-2}}{(1-\sigma)^{2}}|B_{R}|_{\mu_{w_{1}}}\frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\notag\\ =&\frac{C\varepsilon^{2}}{(1-\sigma)^{2}}|B_{R}|_{\mu_{w_{1}}}\frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}, \end{align*}

where we utilized the assumed condition of $\theta_{3}+\theta_{4}=2$. A consequence of these facts gives that for $t\in[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]$,

\begin{align*} &|B_{\sigma R}\cap\{u(x,t) \gt M_{1}-2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}\notag\\ &\leq|B_{R}|_{\mu_{w_{1}}}\left(\frac{(1+C\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}+\frac{C}{(1-\sigma)^{2}}\frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\right), \end{align*}

which, together with the fact that $|B_{R}\setminus B_{\sigma R}|_{\mu_{w_{1}}}\leq C(1-\sigma)|B_{R}|_{\mu_{w_{1}}}$, reads that

\begin{align*} &\frac{|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}}{|B_{R}|_{\mu_{w_{1}}}}\notag\\ &\leq \frac{(1+C\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}+\frac{C}{(1-\sigma)^{2}}\bigg((1-\sigma)^{3}+\frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\bigg). \end{align*}

Pick σ such that

\begin{align*} (1-\sigma)^{3}=\frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}, \end{align*}

which yields that for $t\in[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}],$

(4.18)\begin{align} &\frac{|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}}{|B_{R}|_{\mu_{w_{1}}}}\notag\\ &\leq \frac{(1+\overline{C}\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}+\overline{C}\bigg(\frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\bigg)^{\frac{1}{3}}, \end{align}

where $\overline{C}=\overline{C}(n,p,\theta_{1},\theta_{2},\lambda,\overline{m},\overline{M}).$ Note that

\begin{align*} \frac{|A^{a}(M_{1}-\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\leq a. \end{align*}

Take a small positive constant a such that a ⁻¹ is an integer and

\begin{align*} \overline{C}a^{\frac{1}{3}}\leq\frac{\gamma}{8}. \end{align*}

By fixing the value of a, we now divide the time interval $[t_{0},t_{0}+R^{\theta_{1}+\theta_{2}}]$ into finite small intervals. Denote $N=a^{-1}$ and $t_{i}=t_{0}+iaR^{\theta_{1}+\theta_{2}}$, $i=1,2,\ldots,N.$

Claim that there exist a small positive constant $\bar{\varepsilon}_{0}$ and a large positive constant $k_{0} \gt 1$ depending only on $n,p,\theta_{1},\theta_{2},\lambda,\gamma,\overline{m},\overline{M}$ such that for any $0 \lt \varepsilon\leq\bar{\varepsilon}_{0}$ and $k_{1}\geq k_{0}$,

(4.19)\begin{align} \frac{(1+\overline{C}\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}\leq1-\gamma+\frac{\gamma}{8N}. \end{align}

In fact, since $(1-t)^{-2}\leq(1+6t)$ for $t\in(0,\frac{1}{2})$, then

\begin{align*} \frac{1+\overline{C}\varepsilon}{(1-2^{-k_{1}})^{2}}\leq(1+\overline{C}\varepsilon)(1+6\cdot2^{-k_{1}}). \end{align*}

Let $\overline{C}\varepsilon=6\cdot2^{-k_{1}}$. Then we have

\begin{align*} \frac{1+\overline{C}\varepsilon}{(1-2^{-k_{1}})^{2}}\leq1+2\overline{C}\varepsilon+\overline{C}^{2}\varepsilon^{2}. \end{align*}

Pick

(4.20)\begin{align} \bar{\varepsilon}_{0}=\frac{-2\overline{C}+\sqrt{4\overline{C}^{2}+\frac{\gamma}{2N(1-\gamma)}}}{2\overline{C}^{2}},\qquad k_{0}=-\frac{\ln2}{\ln(\overline{C}\bar{\varepsilon}_{0})-\ln6}. \end{align}

Then we obtain that for any $0 \lt \varepsilon\leq\bar{\varepsilon}_{0}$ and $k_{1}\geq k_{0}$,

\begin{align*} \frac{1+\overline{C}\varepsilon}{(1-2^{-k_{1}})^{2}}\leq&1+2\overline{C}\bar{\varepsilon}_{0}+\overline{C}^{2}\bar{\varepsilon}^{2}_{0}=1+\frac{\gamma}{8N(1-\gamma)}. \end{align*}

That is, Equation (4.19) holds.

Consequently, it follows from Equation (4.18) that for $0 \lt \varepsilon\leq\bar{\varepsilon}_{0}$, $k_{1}\geq k_{0}$ and $t\in[t_{0},t_{1}]$,

\begin{align*} &\frac{|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}}{|B_{R}|_{\mu_{w_{1}}}}\leq1-\left(\frac{7}{8}-\frac{1}{8N}\right)\gamma. \end{align*}

Then applying Lemma 4.5, we deduce from Equation (4.3) that for any $k_{2} \gt k_{1}\geq k_{0}$,

\begin{align*} \frac{|A^{a}(M_{1}-2^{-k_{2}}\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\leq& C\left(\frac{|A^{a}(M_{1}-2^{-k_{2}}\varepsilon,R)|_{\nu_{w_{1}}}}{|Q_{R}|_{\nu_{w_{1}}}}\right)^{\frac{\theta_{3}}{\theta_{1}}}\notag\\ \leq& \widehat{C}\left(\frac{\sqrt{a}}{\sqrt[q]{\gamma}(k_{2}-k_{1})^{\frac{2-q}{2q}}}\right)^{\frac{\theta_{3}}{\theta_{1}}}, \end{align*}

where $\widehat{C}=\widehat{C}(n,p,q,\theta_{1},\theta_{2},\theta_{3},\lambda,\overline{m},\overline{M}).$ Pick

\begin{align*} k_{2}=k_{1}+a^{\frac{q}{2-q}}\gamma^{-\frac{2}{2-q}}\left(\frac{\gamma}{8N\overline{C}\sqrt[3]{\widehat{C}}}\right)^{-\frac{6\theta_{1}q}{\theta_{3}(2-q)}}. \end{align*}

Then we have

(4.21)\begin{align} \overline{C}\left(\frac{|A^{a}(M_{1}-2^{-k_{2}}\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\right)^{\frac{1}{3}}\leq\frac{\gamma}{8N}. \end{align}

Choose $k_{1}=k_{0}$ and $l_{1}=k_{1}+k_{2}$. By letting $2^{-k_{2}}\varepsilon$ substitute for ɛ in Equation (4.18), we have

\begin{align*} \sup\limits_{t\in[t_{0},t_{1}]}|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-l_{1}}\varepsilon\}|_{\mu_{w_{1}}}\leq\Big(1-\gamma+\frac{\gamma}{4N}\Big)|B_{R}|_{\mu_{w_{1}}}. \end{align*}

Then it can be inductively proved that there exist a strictly increasing integer set $\{l_{i}\}_{i=1}^{N}$ such that for $i=1,2,\ldots,N$,

\begin{align*} \sup\limits_{t\in[t_{i-1},t_{i}]}|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-l_{i}}\varepsilon\}|_{\mu_{w_{1}}}\leq\Big(1-\gamma+\frac{i\gamma}{4N}\Big)|B_{R}|_{\mu_{w_{1}}}. \end{align*}

In fact, let the above relation hold in interval $[t_{i-1},t_{i}]$ and then prove that it also holds in the next interval $[t_{i},t_{i+1}]$. For simplicity, denote $\varepsilon_{i}=2^{-l_{i}}\varepsilon$ and $\gamma_{i}=\gamma(1-\frac{i}{4N}).$ Then the assumption implies that

\begin{align*} |B_{R}\cap\{u(x,t_{i}) \gt M_{1}-\varepsilon_{i}\}|_{\mu_{w_{1}}}\leq(1-\gamma_{i})|B_{R}|_{\mu_{w_{1}}}. \end{align*}

By the same argument as in Equation (4.18), it follows from Equations (4.19)–(4.21) that for $\bar{k}_{1}\geq k_{0}$ and $t\in[t_{i},t_{i+1}],$

\begin{align*} &\frac{|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-\bar{k}_{1}}\varepsilon_{i}\}|_{\mu_{w_{1}}}}{|B_{R}|_{\mu_{w_{1}}}}\notag\\ &\leq \frac{(1+\overline{C}\varepsilon_{i})(1-\gamma_{i})}{(1-2^{-\bar{k}_{1}})^{2}}+\overline{C}\bigg(\frac{|A^{a}(M_{1}-\varepsilon_{i},R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\bigg)^{\frac{1}{3}}\notag\\ &\leq1-\gamma_{i}+\frac{\gamma_{i}}{8N}+\overline{C}\bigg(\frac{|A^{a}(M_{1}-2^{-l_{i}}\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\bigg)^{\frac{1}{3}}\notag\\ &\leq1-\gamma_{i}+\frac{\gamma_{i}}{8N}+\frac{\gamma}{8N}\notag\\ & \lt 1-\gamma+\frac{i+1}{4N}\gamma, \end{align*}

where $\overline{C}=\overline{C}(n,p,\theta_{1},\theta_{2},\lambda,\overline{m},\overline{M})$ is defined above and in the third inequality, we used the fact that $l_{i}\geq l_{1} \gt k_{2}$. By taking $\bar{k}_{1}=k_{0}$ and $l_{i+1}=l_{i}+\bar{k}_{1}$, we obtain

\begin{align*} \sup\limits_{t\in[t_{i},t_{i+1}]}|B_{R}\cap\{u(x,t) \gt M_{1}-2^{-l_{i+1}}\varepsilon\}|_{\mu_{w_{1}}}\leq\Big(1-\gamma+\frac{(i+1)\gamma}{4N}\Big)|B_{R}|_{\mu_{w_{1}}}. \end{align*}

Then picking

(4.22)\begin{align} \bar{l}_{0}:=&l_{N}=l_{1}+(N-1)k_{0}\notag\\ =&(N+1)k_{0}++a^{\frac{q}{2-q}}\gamma^{-\frac{2}{2-q}}\left(\frac{\gamma}{8N\overline{C}\sqrt[3]{\widehat{C}}}\right)^{-\frac{6\theta_{1}q}{\theta_{3}(2-q)}}, \end{align}

we obtain that Equation (4.14) holds.

Step 2. For $0 \lt a\leq1$ and $\overline{m}\leq k\leq\overline{M}$, let

\begin{align*} \tilde{A}^{a}(k,R)=(B_{R}\times[t_{0},t_{0}+a R^{\theta_{1}+\theta_{2}}])\cap\{u \lt k\}. \end{align*}

Define $\tilde{v}=(u-(m_{1}+\varepsilon))^{-}$. A direct application of Lemma 4.2 gives that

\begin{align*} &\sup\limits_{t\in(t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}})}\int_{B_{R}}(\tilde{v}^{2}-C_{0}\tilde{v}^{3})\eta^{2}w_{1}\,{\rm d}x\notag\\ &\leq\int_{B_{R}}\tilde{v}^{2}\eta^{2}w_{1}\,{\rm d}x\big|_{t_{0}}+C_{0}\int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}\tilde{v}^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t, \end{align*}

where $C_{0}=C_{0}(n,p,\lambda,\overline{m},\overline{M})$ and η is defined in Equation (4.17). Pick a small constant $0 \lt \bar{\varepsilon}_{1}\leq(2C_{0})^{-1}$, which implies that $1-C_{0}\bar{\varepsilon}_{1}\geq\frac{1}{2}$. Then we obtain that for $t_{0} \lt t \lt t_{0}+aR^{\theta_{1}+\theta_{2}}$, $0 \lt \varepsilon\leq\bar{\varepsilon}_{0}$ and $k_{1} \gt 1$,

\begin{align*} \int_{B_{R}}(\tilde{v}^{2}-C_{0}\tilde{v}^{3})\eta^{2}w_{1}\,{\rm d}x\big|_{t}\geq(1-C_{0}\varepsilon)\varepsilon^{2}(1-2^{-k_{1}})^{2}|B_{\sigma R}\cap\{u(x,t) \lt m_{1}+2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}, \end{align*}

and in view of Equation (4.15),

\begin{align*} \int_{B_{R}}\tilde{v}^{2}\eta^{2}w_{1}\,{\rm d}x\big|_{t_{0}}\leq&\varepsilon^{2}|\{x\in B_{R}: u(x,t_{0}) \lt m_{1}+\varepsilon\}|_{\mu_{w_{1}}}\leq\varepsilon^{2}(1-\gamma)|B_{R}|_{\mu_{w_{1}}}, \end{align*}

and

\begin{align*} \int_{B_{R}\times[t_{0},t_{0}+aR^{\theta_{1}+\theta_{2}}]}\tilde{v}^{2}|\nabla\eta|^{2}w_{2}\,{\rm d}x\,{\rm d}t\leq&\frac{C\varepsilon^{2}}{(1-\sigma)^{2}R^{2}}|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}\notag\\ \leq&\frac{C\varepsilon^{2}R^{\theta_{3}+\theta_{4}-2}}{(1-\sigma)^{2}}|B_{R}|_{\mu_{w_{1}}}\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\notag\\ =&\frac{C\varepsilon^{2}}{(1-\sigma)^{2}}|B_{R}|_{\mu_{w_{1}}}\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}. \end{align*}

Therefore, we deduce that for $t_{0}\leq t\leq t_{0}+aR^{\theta_{1}+\theta_{2}}$,

\begin{align*} &|B_{\sigma R}\cap\{u(x,t) \lt m_{1}+2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}\notag\\ &\leq|B_{R}|_{\mu_{w_{1}}}\left(\frac{1-\gamma}{(1-C\varepsilon)(1-2^{-k_{1}})^{2}}+\frac{C}{(1-\sigma)^{2}}\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\right)\notag\\ &\leq|B_{R}|_{\mu_{w_{1}}}\left(\frac{(1+C\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}+\frac{C}{(1-\sigma)^{2}}\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\right), \end{align*}

and thus,

\begin{align*} &\frac{|B_{R}\cap\{u(x,t) \lt m_{1}+2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}}{|B_{R}|_{\mu_{w_{1}}}}\notag\\ &\leq \frac{(1+C\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}+\frac{C}{(1-\sigma)^{2}}\bigg((1-\sigma)^{3}+\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\bigg). \end{align*}

Take σ such that

\begin{align*} (1-\sigma)^{3}=\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}. \end{align*}

Then we obtain that for $t_{0}\leq t\leq t_{0}+aR^{\theta_{1}+\theta_{2}},$

\begin{align*} &\frac{|B_{R}\cap\{u(x,t) \lt m_{1}+2^{-k_{1}}\varepsilon\}|_{\mu_{w_{1}}}}{|B_{R}|_{\mu_{w_{1}}}}\notag\\ &\leq \frac{(1+\overline{C}\varepsilon)(1-\gamma)}{(1-2^{-k_{1}})^{2}}+\overline{C}\bigg(\frac{|\tilde{A}^{a}(m_{1}+\varepsilon,R)|_{\nu_{w_{2}}}}{|Q_{R}|_{\nu_{w_{2}}}}\bigg)^{\frac{1}{3}}, \end{align*}

where $\overline{C}=\overline{C}(n,p,\theta_{1},\theta_{2},\lambda,\overline{m},\overline{M}).$ Consequently, by the same argument as in the left proof of Equation (4.14) above, we deduce that Equation (4.16) holds. The proof is complete.

Proof. Applying Lemma 4.3, Lemma 4.5 with a = 1 and Lemma 4.7, we obtain that Corollary 4.9 holds. In particular, we fix $q=\frac{3}{2}$ under the assumed conditions in Theorem 1.9.

Proof of Theorem 1.6

Pick a sufficiently large constant $\kappa_{0}\geq2$ such that

\begin{align*} \frac{\overline{M}-\overline{m}}{\kappa_{0}} \lt \tilde{\varepsilon}_{0}, \end{align*}

where $\tilde{\varepsilon}_{0}$ is given by Corollary 4.9 with $\gamma=\frac{1}{2}$. For $0 \lt R\leq\frac{1}{2}$ and $-\frac{1}{4} \lt t_{0} \lt 0$, define

\begin{align*} \overline{\mu}(R)=\sup\limits_{(x,t)\in Q_{R}(0,t_{0})}u(x,t),\quad\underline{\mu}(R)=\inf\limits_{(x,t)\in Q_{R}(0,t_{0})}u(x,t),\quad \omega(R)=\overline{\mu}(R)-\underline{\mu}(R). \end{align*}

Observe that there is at least one inequality holding in terms of the following two inequalities:

(4.23)\begin{align} |\{x\in B_{R/2}:u(x,t_{0}-(R/2)^{\theta_{1}+\theta_{2}}) \gt \overline{\mu}(R)-\kappa_{0}^{-1}\omega(R)\}|_{\mu_{w_{1}}}\leq\frac{1}{2}|B_{R/2}|_{\mu_{w_{1}}}, \end{align}

and

(4.24)\begin{align} |\{x\in B_{R/2}:u(x,t_{0}-(R/2)^{\theta_{1}+\theta_{2}}) \lt \underline{\mu}(R)+\kappa_{0}^{-1}\omega(R)\}|_{\mu_{w_{1}}}\leq\frac{1}{2}|B_{R/2}|_{\mu_{w_{1}}}. \end{align}

From Corollary 4.9, it follows that there exists a large constant $l_{0} \gt 1$ such that

\begin{align*} \overline{\mu}(R/4)\leq\overline{\mu}(R)-\frac{\omega(R)}{\kappa_{0}2^{l_{0}}}\quad\text{if Equation~(4.23) holds,} \end{align*}

and

\begin{align*} \underline{\mu}(R/4)\geq\underline{\mu}(R)+\frac{\omega(R)}{\kappa_{0}2^{l_{0}}}\quad\text{if Equation~(4.24) holds.} \end{align*}

In both cases, we have

\begin{align*} \omega(R/4)\leq\left(1-\frac{1}{\kappa_{0}2^{l_{0}}}\right)\omega(R)=\frac{1}{4^{\alpha}}\omega(R),\quad\mathrm{with}\;\alpha=-\frac{\ln\big(1-\frac{1}{\kappa_{0}2^{l_{0}}}\big)}{\ln4}. \end{align*}

Note that for any $0 \lt R\leq\frac{1}{2}$, there is an integer k such that $4^{-(k+1)}\cdot2^{-1} \lt R\leq4^{-k}\cdot2^{-1}$. In light of the fact that $\omega(R)$ is non-decreasing in R, it follows that

\begin{align*} \omega(R)\leq\omega(4^{-k}\cdot2^{-1})\leq 4^{-k\alpha}\omega(2^{-1})=8^{\alpha}(4^{-(k+1)}\cdot2^{-1})^{\alpha}\omega(2^{-1})\leq CR^{\alpha}, \end{align*}

where $C=C(n,p,q,\theta_{1},\theta_{2},\theta_{3},\lambda,\overline{m},\overline{M})$. Therefore, for any $(x,t)\in B_{1/2}\times(-1/4,t_{0})$, we obtain that

1. (i) if $|t-t_{0}|\leq2^{-(\theta_{1}+\theta_{2})}$, then

   \begin{align*} |u(x,t)-u(0,t_{0})|\leq&|u(x,t)-u(x,t_{0})|+|u(x,t_{0})-u(0,t_{0})|\\ &\leq C\left(|t-t_{0}|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}+|x|^{\alpha}\right)\notag\\ \leq& C\left(|x|+|t-t_{0}|^{\frac{1}{\theta_{1}+\theta_{2}}}\right)^{\alpha}; \end{align*}

2. (ii) if $|t-t_{0}| \gt 2^{-(\theta_{1}+\theta_{2})}$, there exists a set $\{t_{i}\}_{i=1}^{N}$ such that $t \lt t_{1}\leq\cdots\leq t_{N} \lt t_{0}$,

   \begin{align*} |u(x,t)-u(0,t_{0})|\leq&|u(x,t)-u(x,t_{1})|+|u(x,t_{1})-u(x,t_{0})|+|u(x,t_{0})-u(0,t_{0})|\notag\\ \leq&C\left(|t-t_{1}|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}+|t_{1}-t_{0}|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}+|x|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}\right)\notag\\ \leq&C\left(|x|+|t-t_{0}|^{\frac{1}{\theta_{1}+\theta_{2}}}\right)^{\alpha}\quad\text{if}\;N=1, \end{align*}
   and
   \begin{align*} &|u(x,t)-u(0,t_{0})|\notag\\ &\leq|u(x,t)-u(x,t_{1})|+\sum^{N-1}_{i=1}|u(x,t_{i})-u(x,t_{i+1})|\notag\\ &\quad+|u(x,t_{N})-u(x,t_{0})|+|u(x,t_{0})-u(0,t_{0})|\notag\\ &\leq C\left(|t-t_{1}|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}+\sum^{N-1}_{i=1}|t_{i}-t_{i+1}|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}+|t_{N}-t_{0}|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}+|x|^{\frac{\alpha}{\theta_{1}+\theta_{2}}}\right)\notag\\ &\leq C\left(|x|+|t-t_{0}|^{\frac{1}{\theta_{1}+\theta_{2}}}\right)^{\alpha}\quad \text{if}\;N\geq2. \end{align*}

   The proof is complete.

Proof of Theorem 1.9

To begin with, applying the aforementioned proof of Theorem 1.6 with minor modification, we also obtain that there exists a small constant $0 \lt \alpha \lt 1$ and a large constant C > 0, both depending only on $n,p,\theta_{2},\lambda,\overline{m},\overline{M},$ such that for any $t_{0}\in (-1/4,0)$,

(4.25)\begin{align} |u(x,t)-u(0,t_{0})|\leq C\left(|x|+\sqrt[\theta_{2}]{|t-t_{0}|}\right)^{\alpha},\quad\forall (x,t)\in B_{1/2}\times(-1/4,t_{0}]. \end{align}

For $R\in(0,1/2)$, $(y,s)\in Q_{1/R}$, define

\begin{align*} u_{R}(y,s)=u(Ry,R^{\theta_{2}}s),\qquad A_{R}(y)=A(Ry). \end{align*}

Therefore, u _R verifies

\begin{align*} |y|^{\theta_{2}}\partial_{s}u^{q}_{R}-\mathrm{div}\left(A_{R}|y|^{2}\nabla u_{R}\right)=0\quad\mathrm{in}\;Q_{1/R}. \end{align*}

By the change of variables, we obtain that this equation keeps uniformly parabolic in $B_{1/2}(\bar{y})\times(-R^{-\theta_{2}},0)$ for any $\bar{y}\in\partial B_{1}$.

For any $(x,t),(\tilde{x},\tilde{t})\in B_{1/2}\times(-1/4,0),$ let $|\tilde{x}|\leq|x|$ without loss of generality. Write $R=|x|$. It then follows from the interior Hölder estimates for uniformly parabolic equations that there exist two constants $0 \lt \beta=\beta(n,p,\theta_{2},\lambda,\overline{m},\overline{M}) \lt 1$ and $0 \lt C=C(n,p,\theta_{2},\lambda,\overline{m},\overline{M})$ such that for any $\bar{y}\in\partial B_{1}$ and $\bar{s}\in(-4^{-1} R^{-\theta_{2}},0)$,

(4.26)\begin{align} |u_{R}(y,s)-u_{R}(\bar{y},\bar{s})|\leq C(|y-\bar{y}|+\sqrt{|s-\bar{s}|})^{\beta}, \end{align}

for any (y, s) satisfying that $|y-\bar{y}|+\sqrt{|s-\bar{s}|} \lt 1/2$.

Observe that for any $(x,t),(\tilde{x},\tilde{t})\in B_{1/2}\times(-1/4,0),$

\begin{align*} |u(x,t)-u(\tilde{x},\tilde{t})|\leq|u(x,t)-u(x,\tilde{t})|+|u(x,\tilde{t})-u(\tilde{x},\tilde{t})|. \end{align*}

On the one hand, if $|t-\tilde{t}|\leq R^{2\theta_{2}}$, then we deduce from Equation (4.26) that

\begin{align*} |u(x,t)-u(x,\tilde{t})|\leq&\left|u_{R}(x/R,t/R^{\theta_{2}})-u_{R}(x/R,\tilde{t}/R^{\theta_{2}})\right|\notag\\ \leq&C|(t-\tilde{t})/R^{\theta_{2}}|^{\beta/2}\leq C|t-\tilde{t}|^{\beta/4}, \end{align*}

while, if $|t-\tilde{t}| \gt R^{2\theta_{2}}$, then we have from Equation (4.25) that

\begin{align*} &|u(x,t)-u(x,\tilde{t})|\notag\\ &\leq|u(x,t)-u(0,t)|+|u(0,t)-u(0,\tilde{t})|+|u(0,\tilde{t})-u(x,\tilde{t})|\notag\\ &\leq C\big(R^{\alpha}+|t-\tilde{t}|^{\frac{\alpha}{\theta_{2}}}\big)\leq C|t-\tilde{t}|^{\frac{\alpha}{2\theta_{2}}}. \end{align*}

On the other hand, if $|x-\tilde{x}|\leq R^{2}$, then it follows from Equation (4.26) that

\begin{align*} |u(x,\tilde{t})-u(\tilde{x},\tilde{t})|=&\left|u_{R}(x/R,\tilde{t}/R^{\theta_{2}})-u_{R}(\tilde{x}/R,\tilde{t}/R^{\theta_{2}}\big)\right|\notag\\ \leq&C|(x-\tilde{x})/R|^{\beta}\leq C|x-\tilde{x}|^{\beta/2}, \end{align*}

while, if $|x-\tilde{x}| \gt R^{2}$, then we see from Equation (4.25) that

\begin{align*} |u(x,\tilde{t})-u(\tilde{x},\tilde{t})|\leq&|u(x,\tilde{t})-u(0,\tilde{t})|+|u(0,\tilde{t})-u(\tilde{x},\tilde{t})|\notag\\ \leq& C\big(R^{\alpha}+|\tilde{x}|^{\alpha}\big)\leq CR^{\alpha}\leq C|x-\tilde{x}|^{\frac{\alpha}{2}}. \end{align*}

Consequently, we complete the proof of Theorem 1.9.