April 09, 2026
We establish Harnack inequalities for viscosity solutions of a class of degenerate fully nonlinear anisotropic elliptic equations exhibiting non-standard growth conditions. A primary example of such operators is the degenerate anisotropic \((p_i)\)-Laplacian. Our approach relies on the sliding paraboloid method, adapted with suitably chosen anisotropic functions to derive the basic measure estimates. A central contribution of this work is the development of a doubling property, achieved through the explicit construction of a novel barrier function. By combining these tools with the intrinsic geometry techniques introduced in [1], [2], we prove the intrinsic Harnack inequality for this class of operators under appropriate conditions on the exponents \((p_i)\).
In this paper, we establish Harnack inequalities for viscosity solutions of the following class of inequalities in nondivergence form: \[\begin{align} \label{PDE} \begin{cases} \mathcal{M}^-_{\lambda,\Lambda} \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu\sum_i|D_iu|^{p_i-1} \leq c_0, \\ \mathcal{M}^+_{\lambda,\Lambda} \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) +\mu\sum_i|D_iu|^{p_i-1} \geq -c_0, \end{cases} \quad \text{in } \Omega, \end{align}\tag{1}\] where \(2 \leq p_1 \leq \cdots \leq p_n\), \(0<\lambda \leq \Lambda\), \(c_0 \geq0\), \(\mu\geq0\), and \(\left[|z_i|^{\frac{p_i-2}{2}}\right] =\operatorname{diag}(|z_i|^{\frac{p_i-2}{2}})\) is the diagonal matrix with entries \(|z_i|^{\frac{p_i-2}{2}}\) on the diagonal.
This structure encompasses several examples of equations including:
the degenerate anisotropic-\((p_i)\)-Laplacian equation. \[\begin{align} \label{piLap} \tilde{\Delta}_{(p_i)}u(x)= \sum_i \frac{1}{p_i-1}\partial_i(|\partial_iu|^{p_i-2}\partial_{i}u)=\sum_i |\partial_iu|^{p_i-2}\partial_{ii}u = f(x). \end{align}\tag{2}\]
For some symmetric matrix function \((a_{ij}(x))\) with eigenvalues in \([\lambda,\Lambda]\) and some \(|b_i(x)|\leq\mu\), \[\begin{align} \sum_{i,j} a_{ij}(x)|\partial_iu|^{\frac{p_i-2}{2}}|\partial_ju|^{\frac{p_j-2}{2}} \partial_{ij}u + \sum_i b_i(x)|\partial_iu|^{p_i-2}\partial_iu= f(x). \end{align}\]
The degenerate anisotropic-\((p_i)\)-Laplacian 2 has attracted considerable attention and has been the subject of extensive research in recent decades. This interest stems from two key features that render the equation particularly challenging to analyze and make the derivation of regularity results difficult.
The first is its anisotropic degeneracy, which is more degenerate than that of the standard \(p\)-Laplacian \(\Delta_{p} u = \sum_i \partial_i(|\nabla u|^{p-2} \partial_iu)\). While the degeneracy of the \(p\)-laplacian occurs when the gradient is identically zero, the anisotropic \((p_i)\)-Laplacian degenerates if just one component of the gradient, \(\partial_iu\), is zero for some coordinate direction \(i\). Hence, even when the exponents \(p_i\) are all equal, several important regularity properties remained open until recently; the Lipschitz regularity for weak solutions is established in [3] and the authors of [4], [5] proved Lipschitz regularity for viscosity solutions. Moreover, while the basic regularity such as Harnack inequality for weak solutions can be obtained by the classical De Giorgi-Nash-Moser theory for \(p\)-growth, the corresponding result for viscosity solutions is proved recently in [6]. For additional related results, see [7], [8].
The second feature is that the equation exhibits non-standard growth, which means that the equation arises from the Euler-Lagrange equation of a following type of functional: \[\begin{align} w \mapsto \int F(Dw) dx, \quad \text{ where } \quad |z|^{p_1} \lesssim F(z) \lesssim |z|^{p_1}+|z|^{p_n}. \end{align}\] Functionals of this type have been intensively studied since the foundational works [9], [10] and they now constitute a central topic in the regularity theory of nonlinear elliptic equations. See [11], [12] for broad surveys of the field and [13] for viscosity results.
A crucial aspect in establishing regularity for problems with non-standard growth is the smallness of gap between \(p_1\) and \(p_n\). Without this condition, counterexamples to regularity are known, see the comment (2) below Corollary 1. The optimal gap for the boundedness of weak solutions of 2 shown in [14], [15] is \[\begin{align} \label{optpcond} p_n \leq \frac{n\overline{p}}{n-\overline{p}}, \quad \text{ where } \overline{p} = \left( \frac{1}{n}\sum_i\frac{1}{p_i} \right)^{-1}. \end{align}\tag{3}\] See also [16]–[20] for more results about boundedness. Moreover, Lipschitz regularity is recently established in [21] for bounded weak solutions of 2 without any assumption on the closeness of \(p_1\) and \(p_n\), and the analogous result for viscosity solutions is proved in [22]. See also [23], [24].
Despite significant progress, establishing Hölder regularity and the Harnack inequality for 2 under optimal conditions on the exponents \((p_i)\) remains an open problem. For weak solutions involving rough coefficients, Hölder regularity has been demonstrated for specific cases of \((p_i)\), satisfying 3 and \(p_1<p_2 =\cdots =p_n\) (or \(p_1=\cdots =p_{n-1}<p_n\)), as shown in [25], [26] and [27]. Moreover, when the exponents \(p_i\) are all distinct, the authors of [28] established Hölder regularity, but under the condition on \((p_i)\) as \[\begin{align} \label{pwcond} p_n - p_1 \leq \frac{1}{q} \quad \text{ with for some large } q=q(n,(p_i),\lambda,\Lambda). \end{align}\tag{4}\] See also [29]–[33]. The primary techniques rely on De Giorgi-type techniques and the intrinsic geometry framework introduced in [1], [34], to address the inhomogeneous scaling of the equations.
In case of weak solutions of 2 without variable coefficients, the Harnack inequality is established in [35] under the condition on \((p_i)\) as \[\begin{align} \label{ppcond} 2<p_1, \quad p_n \leq \overline{p}\left( 1+\frac{1}{n}\right), \end{align}\tag{5}\] which matches with a parabolic range. The novel approach used in [35] is constructing a self-similar Barenblatt solution by abstract functional-analytic method and using it as a barrier function.
However, to the best of our knowledge, there are no known results about Harnack inequality for viscosity solutions of 1 . The methods typically employed for weak solutions are not applicable in this context. In particular, the absence of a Sobolev inequality precludes the use of De Giorgi-type approaches, and the lack of an abstract functional-analytic framework prevents the construction of barrier functions. Despite these technical challenges, we are able to establish an intrinsic Harnack inequality for viscosity solutions of 1 under appropriate conditions on the exponents \((p_i)\).
We now state the main theorem.
Theorem 1. Let \(u \in C(\Omega)\) be nonnegative and satisfy 1 in the viscosity sense in \(\Omega\). Assume that \(2\leq p_1\leq \cdots \leq p_n\) and that \[\begin{align} \label{pcond} \frac{p_n-1}{p_1-1} \leq \frac{p_n}{\overline{p}} \frac{n}{n-\frac{\lambda}{\Lambda}} \quad \text{ where } \quad \overline{p} = \left( \frac{1}{n}\sum_i \frac{1}{p_i}\right)^{-1}. \end{align}\qquad{(1)}\] Then there exist constants \(C_0,R_0>1\) and \(\epsilon_1,\epsilon_2 >0\) depending on \(n,(p_i),\lambda\) and \(\Lambda\) such that if \(u(0)>0\), the followings hold:
\(K_{R_0r}(u(0)) \subset \Omega\), \(c_0 \in [0,\epsilon_1 u(0)^{p_n-1}]\) and \(\mu \in [0,\mu_1]\) where \(\mu_1=\mu_1(u(0),n,(p_i),\lambda,\Lambda)>0\), then for any \(r \in(0,1]\), \[\begin{align} \sup_{K_r(u(0))}u\leq C_0u(0). \end{align}\]
\(K_{R_0r}(C_0u(0)) \subset \Omega\), \(c_0 \in [0,\epsilon_2u(0)^{p_n-1}]\) and \(\mu \in [0,\mu_2]\) where \(\mu_2=\mu_2(u(0),n,(p_i),\lambda,\Lambda)>0\), then for any \(r \in(0,1]\), \[\begin{align} \inf_{K_r(C_0u(0))}u \geq \frac{1}{C_0}u(0) . \end{align}\]
Here, \[\begin{align} K_r(M) :=\{|x_i| \leq r^{\alpha_i}M^{\beta_i}\} \quad \text{ with } \quad \alpha_i = \frac{1}{p_i}, \quad \beta_i = -\left( \frac{p_n-p_i}{p_i} \right). \end{align}\]
A direct consequence of the main theorem is the following Hölder regularity.
Corollary 1. Let \(u \in C(\Omega)\) be a viscosity solutions of 1 in \(\Omega\) under the same assumptions on \((p_i)\) as in Theorem 1. Then \(u\) is locally Hölder continuous in \(\Omega\) with a universal exponent \(\alpha=\alpha(n,(p_i),\lambda,\Lambda) \in (0,1)\).
We gives some comments of the theorem.
Through scaling, we can always assume that \(c_0\) and \(\mu\) are sufficiently small, provided the radius \(r\) is chosen small enough. Furthermore, in the case where \(c_0=\mu=0\), the results hold for any \(r>0\). See Remark 1.
When the gap between \(p_1\) and \(p_n\) is sufficiently large, then there exist counterexamples that fail to satisfy the Harnack inequality provided in [9], [36]. Specifically, for \(n\geq 6\), \(p_1=\cdots = p_{n-1} =2\) and \(p_n=4\), then an unbounded function \[\begin{align} u(x) = \sqrt{\frac{n-4}{8}}\frac{x_n^2}{\left|\sum_{i=1}^{n-1}x_i^2\right|^{1/2}} \geq0 \end{align}\] is a classical solution (and thus a viscosity solution) to the equation \[\begin{align} \sum_{i=1}^{n-1}\partial_{ii}u + |\partial_nu|^2\partial_{nn}u = 0 \quad \text{ in } \mathbb{R}^n \setminus \bigcap_{i=1}^{n-1}\{x_i = 0\}. \end{align}\] In particular, \(u=0\) in \(\{x_n=0\},\) and therefore the Harnack inequality fails.
We provide several remarks concerning the main structural condition ?? on the exponents \((p_i)\). The proof presented in this paper appears to extend naturally to the singular/degenerate case, when \(1<p_i<2\) for some \(i\). The reason we restrict ourselves to the degenerate case \(p_1 \geq 2\) is that the theory of viscosity solutions has not yet been developed for the singular case.
It is currently unclear whether the ratio \(\Lambda/\lambda\) and the smallest exponent \(p_1\) are truly essential to the Harnack inequality. Our assumption ?? is more restrictive than the optimal condition for boundedness given in 3 . Moreover, our result may be interpreted as providing a more explicit characterization of the parameter \(q\) in 4 as studied in [28].
In the special case where \(\Lambda/\lambda = 1\) and \(c_0 = \mu = 0\), the equation 1 reduces to the anisotropic \((p_i)\)-Laplacian 2 . Comparing our condition ?? (with \(\Lambda/\lambda = 1\)) to the condition 5 found in [35], we observe that neither condition is uniformly stronger than the other. For instance, if \(p_1 < p_2 = \cdots = p_n\), then ?? is more restrictive than 5 . On the other hand, if \(p_1 = \cdots = p_{n-1} < p_n\), then 5 is more restrictive than ?? .
We introduce main ideas and novelties of our proof. Due to the non-homogeneous scaling of the equation 1 , it is essential to employ the intrinsic scaling technique, originally introduced in [1], [34]. In that work, this approach was used to establish the Harnack inequality for the parabolic \(p\)-Lapalcian. A recent extension to viscosity solutions of the corresponding parabolic problem was achieved in [2]. The core idea is that, when controlling the superlevel set \(\{u>M\}\), we use the intrinsic cube \(K_r(M)\) instead of the standard cube \(Q_r\), which matches with the natural scaling of the equation at height level \(M\). A major challenge arises because when \(M\) is large, \(K_r(M)\) becomes very flat in most directions, which severely complicates any covering or iteration argument used in the proof. Moreover, while the intrinsic geometry of parabolic \(p\)-Laplacian involves distinct scalings in only two directions (time and space), the intrinsic cube \(K_r(M)\) for equation 1 can exhibit different growth rates in every coordinate direction, making the situation far more intricate. Despite these difficulties, by suitably adapting a technique developed in [2], we are able to derive a Krylov-Safanov type regularity result.
First step of the proof is the basic measure estimates, Lemma 3. We employ the sliding paraboloid method, originally introduced by Cabre [37] and developed by Savin [38]. The procedure involves sliding a paraboloid upward from below until it first touches the graph of the solution \(u\). Then the touching points exhibit certain favorable properties, and the measure of these touching points can be estimated in terms of the measure of the corresponding vertex points by applying the area formula. However, the standard isotropic paraboloid is incompatible with the strong anisotropic degeneracy inherent in equation 1 . Instead, we adapt an anisotropic paraboloid of the form: \[\begin{align} \varphi(x) = K_1 |x_1|^{\frac{p_1}{p_1-1}} + \cdots +K_n |x_n|^{\frac{p_n}{p_n-1}}, \end{align}\] for some \(K_i>0\), chosen to align with the coordinate-wise degeneracy of the equation, following the approach introduced in [6], [39].
The second part of the proof is the doubling property, stated in Lemma 7. This requires constructing a suitable barrier function, which constitutes the most technically demanding part of the argument. We explicitly build an anisotropic barrier of the form: \[\begin{align} \Phi(x) =(|b_1x_1|^{a_1} + \cdots+|b_nx_n|^{a_n})^{-\gamma}. \end{align}\] By appropriately selecting parameters \(\alpha_i>1\), \(\beta_i>0\) and \(\gamma>0\), we verify in Lemma 5 that the barrier function is a subsolution to ?? provided the exponents \((p_i)\) satisfy our assumption ?? . Exploiting the scaling property of the barrier function and applying the comparison principle to the solution, we then deduce the desired doubling property. This explicit anisotropic barrier construction appears to be novel in the setting of the degenerate/singular anisotropic \((p_i)\)-Laplacian. We anticipate that this construction may also turn out to be useful in establishing the Harnack inequality for weak solutions. Combining these two fundamental results with a covering argument based on the intrinsic cubes \(K_r(M)\), we obtain \(L^\epsilon\) estimates, Theorem 2, which states that the algebraic decay of the level sets \(\{u>M\}\). These estimates are the final critical component required to complete the proof of the Harnack inequality, Theorem 1.
The paper is organized as follows. In Section 2 we introduce the notations and some preliminaries. In Section 3 we prove the basic measure estimate, Lemma 3. In Section 4 we construct an anisotropic barrier function and in Section 5, we prove the doubling property using the barrier function. In Section 6, we establish the \(L^\epsilon\) estimates for supersolutions, and we prove the main theorem, Theorem 1, in Section 7.
Throughout this paper, we write a point \(x \in \mathbb{R}^n\) as \(x=(x_1,\cdots ,x_n)\). We denote by \(S(n)\) the space of symmetric \(n \times n\) real matrices and \(I_n \in S(n)\) denotes the identity matrix. For \(r>0\) and \(x_0 \in \mathbb{R}^n\), the standard cube is defined as \(Q_r(x_0) = \{ x \in \mathbb{R}^n : |x_i-(x_0)_i| <r \text{ for any } i\in\{ 1, \cdots,n\}\}\) and we write \(Q_r = Q_r(0)\).
Moreover, we define the intrinsic cube adapted to the anisotropic scaling of the equation as \[\begin{align} aK_r(M,x_0) := \{x \in \mathbb{R}^n : |x_i-(x_0)_i| \leq ar^{\alpha_i}M^{\beta_i}\} \quad \text{ where } \quad \alpha_i = \frac{1}{p_i}, \;\beta_i = -\left( \frac{p_n-p_i}{p_i} \right), \end{align}\] for \(a,r,M>0\) and \(x_0 \in \mathbb{R}^n\). We abbreviated as \(K_r(M,x_0)=1 \cdot K_r(M,x_0)\) and \(K_r(M)=K_r(M,0)\). Observe that \(\alpha_1 \geq \cdots \geq \alpha_n >0\) and \(\beta_1 \leq \cdots \leq \beta_n=0\). Thus, for \(0<N\leq M\) and \(0<r \leq s\), we have \(K_r(M) \subset K_s(N)\).
For \(a=(a_i) \in \mathbb{R}^n\), we always write the diagonal matrix with entries \(a_i\) as \[\begin{align} \left[a_i\right] := \operatorname{diag}(a_1,\cdots,a_n) \in S(n). \end{align}\] We provide the definitions and properties of the Pucci extremal operators (see [40]).
Definition 1. For given \(0<\lambda \leq \Lambda\), the Pucci operators \(\mathcal{M}_{\lambda,\Lambda}^{\pm} : S(n) \rightarrow \mathbb{R}\) are defined as follows: \[\begin{align} \mathcal{M}_{\lambda,\Lambda}^{+}(M) := \lambda \sum_{e_i(M)<0}e_i(M) + \Lambda \sum_{e_i(M)>0}e_i(M), \quad \mathcal{M}_{\lambda,\Lambda}^{-}(M) := \Lambda \sum_{e_i(M)<0}e_i(M) + \lambda \sum_{e_i(M)>0}e_i(M), \end{align}\] where \(e_i(M)\)’s are the eigenvalues of \(M\). We also abbreviate \(\mathcal{M}^{\pm}_{\lambda,\Lambda}\) as \(\mathcal{M}^{\pm}\).
Proposition 1. For any \(M,N \in S(n)\), we have
*\(\mathcal{M}^{-}(-M) = -\mathcal{M}^{+}(M)\).*
\(\mathcal{M}^{-}(M) +\mathcal{M}^{-}(N)\leq \mathcal{M}^{-}(M+N) \leq \mathcal{M}^{-}(M) +\mathcal{M}^{+}(N).\)
We also introduce the definition of the inequalities 1 in the viscosity sense, see [40], [41].
Definition 2. We say that \(u \in C(\Omega)\) satisfies \[\mathcal{M}^\pm_{\lambda,\Lambda} \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) \pm\mu\sum_i|D_iu|^{p_i-1} \leq c_0, \quad\text{in}\;\Omega \quad (\text{resp.} \geq)\] in the viscosity sense, if for any \(x_0 \in \Omega\) and any test function \(\psi \in C^2(\Omega)\) such that \(u-\psi\) has a local minimum (resp. maximum) at \(x_0\), then \[\mathcal{M}^\pm_{\lambda,\Lambda} \left( \left[|D_i\psi(x_0)|^{\frac{p_i-2}{2}}\right] D^2\psi(x_0) \left[|D_i\psi(x_0)|^{\frac{p_i-2}{2}}\right] \right) \pm\mu\sum_i|D_i\psi(x_0)|^{p_i-1} \leq c_0 \quad (\text{resp.} \geq).\]
Since the equation 1 is nonhomogeneous, we need to employ an intrinsic scaling framework to properly account for this inhomogeneity of the equation.
Remark 1 (Scaling). For \(r,M>0\), if \(u \in C(K_r(M))\) satisfies \[\begin{align} \label{scalrmkpde} \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu\sum_i|D_iu|^{p_i-1} \leq c_0 \quad \text{ in }K_r(M), \end{align}\tag{6}\] then the function \(v \in C(K_1(1))\) defined as \[\begin{align} v(x) = \frac{u(r^{\alpha_i}M^{\beta_i} x_i)}{M} \quad \text{ with } \quad \alpha_i = \frac{1}{p_i}, \quad \beta_i = -\left( \frac{p_n-p_i}{p_i} \right), \end{align}\] satisfies the following inequality \[\begin{align} \mathcal{M}^- \left( \left[|D_iv|^{\frac{p_i-2}{2}}\right] D^2v \left[|D_iv|^{\frac{p_i-2}{2}}\right] \right) -\mu\sum_i \frac{r^{1/p_i}}{M^{(p_n-p_i)/p_i}}|D_iv|^{p_i-1} \leq \frac{r}{M^{p_n-1}}c_0 \quad \text{ in }K_1(1). \end{align}\] In particular, when \(r\leq1\) and \(M\geq1\), then \(v\) satisfies the same inequality 6 in \(K_1(1)\). Moreover, if \(c_0=\mu=0\), then then \(v\) satisfies the same inequality for any \(r,M>0\). Choosing small enough \(r \leq 1\), we can assume that \(c_0\) and \(\mu\) are arbitrarily small.
We introduce comparison principle for viscosity solutions, see [41].
Lemma 1 (Comparison principle). Let \(F=F(r,p,X) \in C(\mathbb{R}\times \mathbb{R}^n \times S(n))\) be degenerate elliptic and strictly decreasing in \(r\). More precisely, assume:
\(F(r,p,X) \leq F(r,p,Y)\) for any \(X \leq Y\).
There exists \(\gamma>0\) such that \(\gamma(r-s) \leq F(s,p,X)- F(r,p,Y)\) for any \(s \leq r\).
Then \(F\) satisfies the comparison principle: if \(u \in USC(\overline{\Omega})\) and \(v \in LSC(\overline{\Omega})\) are viscosity subsolutions and supersolutions, respectively, such that \(F(v,Dv,D^2v) \leq 0 \leq F(u,Du,D^2u)\) in \(\Omega\), and \(u \leq v\) in \(\partial \Omega\), then \(u \leq v\) in \(\Omega\).
We conclude this section by stating the Vitali covering lemma for the intrinsic cubes.
Lemma 2. Fix \(M>0\) and let \(\mathcal{F} = \{K_i=K_{r_i}(M,x_i)\}\) be a collection of intrinsic cubes with \(x_i \in \mathbb{R}^n\) and \(r_i \in (0,1]\). Then there exists a countable subcollection \(\mathcal{G} = \{K_j\} \subset \mathcal{F}\) of pairwise disjoint cubes such that \[\begin{align} \bigcup_{K_j \in \mathcal{G}} 5K_j \supset \bigcup_{K_j \in \mathcal{F}} K_i. \end{align}\]
Proof. We may assume that \(\mathcal{F}\) is countable since \(\mathbb{R}^n\) is Lindelöf space. For \(k \geq 0\), we define the partition of \(\mathcal{F}\) as \[\begin{align} \mathcal{F}_k = \{ K_i \in \mathcal{F}: 2^{-k-1} < r_i \leq 2^{-k} \}. \end{align}\] We now construct the desired subfamily \(\mathcal{G}\) inductively. We define \(\tilde{\mathcal{F}}_1 \subset\mathcal{F}_1\) as a maximal family of disjoint cubes in \(\mathcal{F}_1\). We also define \(\tilde{\mathcal{F}}_k\subset\mathcal{F}_k\) for \(k>1\) as inductive way. Let \(\tilde{\mathcal{F}}_j\) is defined for \(1 \leq j \leq k-1\). Then for a subfamily \[\begin{align} \mathcal{H}_k := \{ K_i \in F_k : K_i \cap K = \emptyset \quad \text{ for any } K \in \bigcup_{j=1}^{k-1} \tilde{\mathcal{F}}_j \}, \end{align}\] we define \(\tilde{\mathcal{F}}_k\) as a maximal family of disjoint cubes in \(\mathcal{H}_k\). Then we set \(\mathcal{G} = \bigcup_{k=1}^{\infty}\mathcal{F}_k\). By construction, the cubes in \(\mathcal{G}\) are pairwise disjoint. To verify the covering property, for \(K_l \in \mathcal{F}_k\) with some \(k \geq 0\), we claim that \(K_l \subset \bigcup_{K_j \in \tilde{\mathcal{F}}_k} 5K_j\). Note that if \(K_l \notin \tilde{\mathcal{F}}_k\), then there exists \(K_m \in \bigcup_{j=1}^{k} \tilde{\mathcal{F}}_j\) satisfying \(K_l \cap K_m \neq \emptyset\) by maximality of \(\tilde{\mathcal{F}}_k\). Since \(K_l \in \mathcal{F}_k\) implies \(2^{-k-1} < r_l \leq 2^{-k}\) and \(K_m \in \bigcup_{j=1}^{k} \tilde{\mathcal{F}}_j\) implies \(2^{-k-1}\leq r_m \leq 1\), we obtain \(r_l \leq 2 r_m\). We prove the claim by showing that \(K_l \subset 5K_m\). For \(x \in K_l\) and \(y \in K_l \cap K_m\), we have \[\begin{align} |(x-x_m)_i| &\leq |(x-y)_i| +|(y-x_m)_i| \leq 2r_l^{\alpha_i}M^{\beta_i} +r_m^{\alpha_i}M^{\beta_i} \leq 5r_m^{\alpha_i}M^{\beta_i}. \end{align}\] Thus, \(x \in 5K_m\), which implies \(K_l \subset 5K_m\). This completes the proof. ◻
The goal of this section is to prove a basic measure estimate lemma using the sliding paraboloid method. We adapt this approach by replacing standard quadratic paraboloids with a specialized class of anisotropic test functions 7 , which is designed to match the anisotropic degeneracy of the operator. However, the function defined in 7 fails to be \(C^2\) on the coordinate hyperplanes \(\bigcup\{x_i=0\}\). To overcome this technical issue, we rely on the “restriction to slices” argument introduced in [6], which reduces the analysis of the singular set to lower-dimensional coordinate slices where the test function remains smooth. Since this procedure carries over to our setting without any essential modification, we omit the detailed repetition of the argument here and, for the sake of clarity and brevity, simply assume throughout the proof that the solution \(u\) is \(C^2\).
Note that the assumption ?? on \((p_i)\) is not required for the validity of the following lemma.
Lemma 3. For any \(\epsilon>0\), there exist \(r_0 \in (0,1)\), \(M_0>1\), and \(\delta_0 \in (0,1)\) depending only on \(n,(p_i),\lambda, \Lambda\) and \(\epsilon\) such that if \(u \in C(K_1(M_0))\) is nonnegative in \(K_1(M_0)\) and satisfies \[\begin{align} \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\Lambda\sum|D_iu|^{p_i-1} \leq \epsilon \quad \text{ in } K_1(M_0), \end{align}\] with \[\begin{align} \inf_{K_{r_0}(1)}u \leq 1, \end{align}\] then we have \[|\{u <M_0\} \cap K_1(M_0)\}| \geq \delta_0|K_1(M_0)|.\]
Proof. For the sake of clarity, we present the argument under the assumption that \(u \in C^2\); the full argument is given in [6]. Given that \(\inf_{K_{r_0}(1)}u \leq 1\), there exists a point \(x_0 \in \overline{K_{r_0}(1)}\) such that \(u(x_0) \leq 1\). We now apply the sliding paraboloid method, by using the following anisotropic ‘paraboloid’ defined as \[\begin{align} \label{anipara} \varphi(x) = -\sum_i K^{\frac{1}{p_i-1}} (p_i-1)^{\frac{1}{p_i-1}} \frac{p_i-1}{p_i} |x_i|^{\frac{p_i}{p_i-1}}, \end{align}\tag{7}\] where \(K \geq 1\) is a constant to be determined. Observe that \(\frac{1}{2}\leq(p_i-1)^{\frac{1}{p_i-1}} \frac{p_i-1}{p_i}\leq 2\). We define \(V=K_{r_0}(1)\) as the vertex set, with \(r_0<1\) to be determined. We slide the paraboloid with vertex in \(V\) from below until it touches the graph of \(u\). The touching point set is then defined as \[\begin{align} T=\{x\in\overline{K_{1}(M_0)}: \exists y \in V \text{ such that } u(x) - \varphi(x-y) = \min_{z\in \overline{K_{1}(M_0)}} u(z)-\varphi(z-y) \}. \end{align}\] For \(y \in V\) and \(z \in \partial K_1(M_0)\), we have \[\begin{align} u(z)-\varphi(z-y) &\geq \frac{1}{2}\min_i \{K^{\frac{1}{p_i-1}} |M_0^{-\frac{p_n-p_i}{p_i}}-r_0^{\frac{1}{p_i}}|^{\frac{p_i}{p_i-1}}\} \\ &\geq \frac{1}{4}\min_i \{K^{\frac{1}{p_i-1}} M_0^{-\frac{p_n-p_i}{p_i-1}}\} \\ &= \frac{1}{4} M_0, \end{align}\] by choosing \(K=M_0^{p_n-1}\) and \(r_0 \leq M_0^{-(p_n-p_1)}A^{-p_n}\) with a sufficiently large constant \(A \geq 2\), which implies \(|M_0^{-\frac{p_n-p_i}{p_i}}-r_0^{\frac{1}{p_i}}| \geq \frac{1}{2}M_0^{-\frac{p_n-p_i}{p_i}}\). Conversely, for \(y \in V\) and \(x_0 \in \overline{K_{r_0}(1)}\), we have \[\begin{align} u(x_0)-\varphi(x_0-y) &\leq 1+2\sum_i K^{\frac{1}{p_i-1}} |2r_0^{\frac{1}{p_i}}|^{\frac{p_i}{p_i-1}} \\ &\leq 1+8 \sum_i(Kr_0)^{\frac{1}{p_i-1}} \\ &\leq 1+ 8 \sum_i (M_0^{p_1-1}A^{-p_n})^{\frac{1}{p_i-1}} \\ &\leq 1+8n MA^{-\frac{p_n}{p_n-1}} \leq \frac{1}{4}M_0, \end{align}\] by choosing large \(A > 64n\) and \(M_0 >8\). These estimates imply that the minimum must be attained in the interior. Therefore, we have \(T \subset \{ u <M_0\} \cap K_1(M_0)\).
For each \(x\in T\) there is a unique vertex point \(y\in V\) such that \(\varphi(z-y)\) touches \(u\) from below at \(x\), so we define the mapping \(m :T\rightarrow V\) as \(m(x)=y\). We also have \[\begin{align} \tag{8} D_iu(x)&= D_i\varphi(x-y) = -(K(p_i-1))^{\frac{1}{p_i-1}}|x_i-y_i|^{\frac{1}{p_i-1}} \frac{x_i-y_i}{|x_i-y_i|}, \\ D^2u(x) &\geq D^2\varphi(x-y) = -\left[K^{\frac{1}{p_i-1}}(p_i-1)^{-\frac{p_i-2}{p_i-1}}|x_i-y_i|^{-\frac{p_i-2}{p_i-1}}\right]. \tag{9} \end{align}\] Note that \(D^2\varphi(x-y)\) is not well-defined whenever \(x_i=y_i\) for some coordinate \(i\). Thus, almost every \(x\in T\), we assume that \(x_i \neq y_i\) for all \(i \in \{1,\cdots,n\}\). For the more delicate case in which \(x_i = y_i\) for some \(i\), we refer to [6]. Using 8 , the map \(m\) can be written explicitly as \[\begin{align} y_i = x_i + \frac{1}{K(p_i-1)}|D_iu|^{p_i-2}D_iu(x). \end{align}\] Differentiating this equation yields \[\begin{align} D_xy= I_n + \frac{1}{K}\left[|D_iu|^{p_i-2}\right]D^2u. \end{align}\] By the identity \(\det (I_n+MN)=\det(I_n+NM)\) for \(n\times n\) matrices \(M\) and \(N\), we have \[\begin{align} \det D_xy&= \det \left(I_n + \frac{1}{K}\left[|D_iu|^{p_i-2}\right]D^2u \right) \\ &= \det \left(I_n + \frac{1}{K}\left[|D_iu|^{\frac{p_i-2}{2}}\right]D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right)=: \det B. \end{align}\] A direct calculation using 8 and 9 shows that \[\begin{align} -\frac{1}{K}\left[|D_i\varphi|^{\frac{p_i-2}{2}}\right]D^2\varphi \left[|D_i\varphi|^{\frac{p_i-2}{2}}\right] =I_n , \end{align}\] which implies \[\begin{align} B &=I_n + \frac{1}{K}\left[|D_iu|^{\frac{p_i-2}{2}}\right]D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \\ & =\frac{1}{K}\left[|D_iu|^{\frac{p_i-2}{2}}\right](D^2u-D^2\varphi) \left[|D_iu|^{\frac{p_i-2}{2}}\right] \geq0. \end{align}\] From 8 , we also have \(|D_iu|^{p_i-1} \leq 2K(p_i-1)\). Therefore, \[\begin{align} \lambda \operatorname{tr}\left(B\right) &= \mathcal{M}^-(B) \\ & \leq \mathcal{M}^{+}(I_n) + \frac{1}{K}\mathcal{M}^{-}\left(\left[|D_iu|^{\frac{p_i-2}{2}}\right]D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right]\right) \\ &\leq n\Lambda + \frac{1}{K}(\epsilon + \Lambda\sum|D_iu|^{p_i-1}) \\ &\leq C(\epsilon+1), \end{align}\] where \(C\) depends only on \(n,(p_i),\lambda\) and \(\Lambda\). Consequently, \[\begin{align} \det D_xy = \det B \leq \left( \frac{\operatorname{tr}\left(B\right)}{n} \right)^n \leq C(\epsilon+1)^n. \end{align}\] By the area formula, we obtain \[\begin{align} |K_{r_0}(1)|=|V| = \int_{T} \det(D_xy) \,dx \leq \int_T C(1+\epsilon)^n \, dx \leq C(1+\epsilon)^n |T|. \end{align}\] Since \(T \subset \{u<M_0\} \cap K_1(M_0)\), it follows that \[\begin{align} |\{u<M_0\} \cap K_1(M_0)| \geq |T| \geq \frac{|K_{r_0}(1)|}{C(1+\epsilon)^n} =: \delta_0|K_1(M_0)|, \end{align}\] which concludes the proof. ◻
In this section, we construct an explicit barrier function under the assumption ?? on exponents \((p_i)\). We emphasize that this is the only part of the paper where the condition ?? is actually required. Consequently, if one were able to construct a suitable barrier function under a more general assumption on the exponents \((p_i)\), the same proof strategy would immediately yield the intrinsic Harnack inequality under these broader conditions. As a preliminary step, we first prove a simple auxiliary lemma that will play a key role in the construction of the barrier function.
Lemma 4. For any \(k_i >0\), \(h_i>0\) and \(\tau_i \geq0\) with \(i\in \{1,\cdots,n\}\) satisfying \(\sum_{i=1}^n\tau_i=1\), the following inequality holds: \[\begin{align} \sum_{i=1}^n \frac{1}{h_i^{k_i}} \tau_i^{k_i}(\tau_i-h_i) \geq 1-\sum_{i=1}^nh_i. \end{align}\]
Proof. Observe that \((\tau_i^{k_i} - h_i^{k_i})(\tau_i-h_i) \geq0\) for any \(k_i>0\), \(h_i>0\), and \(\tau_i \geq0\). Therefore, we obtain \[\begin{align} \sum_i \frac{1}{h_i^{k_i}} \tau_i^{k_i}(\tau_i-h_i) &= \sum_i \frac{1}{h_i^{k_i}} (\tau_i^{k_i}-h_i^{k_i})(\tau_i-h_i) +\sum_i (\tau_i-h_i) \\ &\geq \sum_i\tau_i - \sum_ih_i = 1-\sum_i h_i. \end{align}\] ◻
For \(a_i >1\) and \(b_i>0\) with \(i\in \{1,\cdots,n\}\), we define an anisotropic function \(|bx|_a : \mathbb{R}^n \rightarrow \mathbb{R}^n\) as \[\begin{align} |bx|_a = |b_1x_1|^{a_1} + \cdots+|b_nx_n|^{a_n}. \end{align}\] Direct computation yields the gradient and Hessian: \[\begin{align} D_i |bx|_a &= a_ib_i|b_ix_i|^{a_i-1} \frac{x_i}{|x_i|}, \\ D^2|bx|_a &= \left[a_i(a_i-1)b_i^2 |b_ix_i|^{a_i-2}\right]. \end{align}\] We define our candidate barrier function by \[\begin{align} \Phi(x) = \frac{1}{\gamma}|bx|_a^{-\gamma} = \frac{1}{\gamma} (|b_1x_1|^{a_1} + \cdots+|b_nx_n|^{a_n})^{-\gamma}, \end{align}\] where \(a_i>1\), \(b_i >0\) and \(\gamma>0\) are parameters to be determined. Note that \(\Phi\) is \(C^1(\mathbb{R}^n\setminus\{0\})\), but may fail to be \(C^2\) on the coordinate hyperplanes \(\bigcup_i\{x_i=0\}\). The main result of this section is the following lemma, which shows that \(\Phi\) serves as a viscosity subsolution of the following equation.
Lemma 5. If \((p_i)\) satisfy ?? , then for any \(R>1\), there exists \(\mu<\Lambda\) depending on \(n,(p_i),\lambda,\Lambda\) such that \[\begin{align} \label{Phiequ} \mathcal{M}^-_{\lambda,\Lambda} \left( \left[|D_i\Phi|^{\frac{p_i-2}{2}}\right] D^2\Phi \left[|D_i\Phi|^{\frac{p_i-2}{2}}\right] \right) -\mu\sum_i|D_i\Phi|^{p_i-1} > \Phi^{d_0} \quad \text{ in }Q_{R} \setminus\{0\} \end{align}\qquad{(2)}\] in viscosity sense with \(d_0 > p_n-1\).
Proof. Observe that \[\begin{align} D\Phi&= -|bx|_a^{-(\gamma+1)} D|bx|_a ,\\ D^2 \Phi &= (r+1)|bx|_a^{-(\gamma+2)} (D|bx|_a \otimes D|bx|_a) -|bx|_a^{-(\gamma+1)}D^2|bx|_a. \end{align}\] This implies \[\begin{align} \left[|D_i\Phi|^{\frac{p_i-2}{2}}\right] D^2\Phi \left[|D_i\Phi|^{\frac{p_i-2}{2}}\right] =&(\gamma+1)|bx|_a^{-(\gamma+2)}(A \otimes A) \\ & -|bx|_a^{-(\gamma+1)} \left[|bx|_a^{-(\gamma+1)(p_i-2)}|D_i|bx|_a|^{p_i-2} D_{ii}|bx|_a\right], \end{align}\] where \[\begin{align} A=(A_i)=\left(|bx|_a^{-(\gamma+1)\frac{p_i-2}{2}}|D|bx|_a|^{\frac{p_i-2}{2}}D|bx|_a\right) \in \mathbb{R}^n. \end{align}\] Using Proposition 1 and the fact that the only nonzero eigenvalue of \(A\otimes A\) is \(|A|^2 = \sum_i|A_i|^2\), we obtain \[\begin{align} \mathcal{M}^-& \left( \left[|D_i\Phi|^{\frac{p_i-2}{2}}\right] D^2\Phi \left[|D_i\Phi|^{\frac{p_i-2}{2}}\right] \right) -\mu\sum_i|D_i\Phi|^{p_i-1} \\ \geq& \mathcal{M}^- \left( (\gamma+1)|bx|_a^{-(\gamma+2)}(A \otimes A)\right) - \mathcal{M}^+\left(|bx|_a^{-(\gamma+1)} \left[|bx|_a^{-(\gamma+1)(p_i-2)}|D_i|bx|_a|^{p_i-2} D_{ii}|bx|_a\right] \right) -\mu\sum_i|D_i\Phi|^{p_i-1}\\ \geq &\lambda (\gamma+1) |bx|_a^{-(\gamma+2)} \sum_i |bx|_a^{-(\gamma+1)(p_i-2)}|D_i|bx|_a|^{p_i} \\ &-\Lambda|bx|_a^{-(\gamma+1)}\sum_i|bx|_a^{-(\gamma+1)(p_i-2)} |D_i|bx|_a|^{p_i-2} D_{ii}|bx|_a -\mu \sum_i |bx|_a^{-(\gamma+1)(p_i-1)}|D_i|bx|_a|^{p_i-1}\\ =&\lambda(\gamma+1)|bx|_a^{-(\gamma+2)} \sum_i (a_ib_i)^{p_i} |bx|_a^{-(\gamma+1)(p_i-2)}|b_ix_i|^{(a_i-1)p_i} \\ &-\Lambda|bx|_a^{-(\gamma+1)}\sum_i(a_ib_i)^{p_i}|bx|_a^{-(\gamma+1)(p_i-2)} \left(1-\frac{1}{a_i}\right) |b_ix_i|^{(a_i-1)(p_i-2) +(a_i-2)}\\ &-\mu \sum_i |bx|_a^{-(\gamma+1)(p_i-1)}(a_ib_i)^{p_i-1}|b_ix_i|^{(a_i-1)(p_i-1)}\\ =&|bx|_a^{-(\gamma+1)} \sum_i (a_ib_i)^{p_i} |bx|_a^{-(\gamma+1)(p_i-2)}|b_ix_i|^{(a_i-1)p_i-a_i} \left(\lambda(\gamma+1) \frac{|b_ix_i|^{\alpha_i}}{|bx|_a} - \Lambda\left( 1- \frac{1-(\mu/\Lambda)|x_i|}{a_i} \right)\right) =: (I). \end{align}\] Now we choose the exponents \(a_i\) in such a way that the terms \(|bx|_a^{-(\gamma+1)(p_i-2)}|b_ix_i|^{(a_i-1)p_i-a_i}\) uniformly comparable across all coordinates \(i\) by arranging that each of them behaves like \(|bx|_a^{-d}\) for some fixed \(d \in \mathbb{R}\). We choose \(a_i>0\) such that \[\begin{align} -(\gamma+1)(p_i-2) + \left( 1 -\frac{1}{a_i} \right)p_i -1 = -d. \end{align}\] Then it follows that \[\begin{align} a_i := \frac{p_i}{d+1-\gamma(p_i-2)}>0 \;\Longleftrightarrow \;d>\gamma(p_i-2)-1. \end{align}\] Moreover, we also require \((a_i-1)p_i-a_i =:a_ik_i\) to be positive, which gives \[\begin{align} k_i := (\gamma+1)(p_i-2)-d>0 \;\Longleftrightarrow \;d<(\gamma+1)(p_i-2). \end{align}\] Note that the condition \(d<(\gamma+1)(p_i-2)\) implies \(a_i >\frac{p_i}{p_i-1} >1\). Since \(\max_i\gamma(p_i-2)-1 = \gamma(p_n-2)-1\), and \(\min_i(\gamma+1)(p_i-2)=(\gamma+1)(p_1-2)\), we need to choose \(d \in \mathbb{R}\) satisfying \[\begin{align} \label{dcond} \gamma(p_n-2)-1 < d<(\gamma+1)(p_1-2). \end{align}\tag{10}\] For such a constant \(d\) to exist, the following condition on \(\gamma>0\) is necessary: \[\begin{align} \label{gamcond1} \gamma(p_n-2)-1 < (\gamma+1)(p_1-2) \Longleftrightarrow \gamma< \frac{p_1-1}{p_n-p_1}. \end{align}\tag{11}\] Then for \(x \in Q_{R} \setminus\{0\}\) and \(\mu \leq \frac{\delta\Lambda}{R}\) for some \(\delta<1\), it follows that \[\begin{align} (I) & \geq |bx|_a^{-(\gamma+1)} \sum_i (a_ib_i)^{p_i} |bx|_a^{-d} \left(\frac{|b_ix_i|^{\alpha_i}}{|bx|_a}\right)^{k_i} \left(\lambda(\gamma+1) \frac{|b_ix_i|^{\alpha_i}}{|bx|_a} - \Lambda\left( 1- \frac{(1-\delta)}{a_i} \right)\right) \\ & \geq|bx|_a^{-(d+\gamma+1)} \lambda(\gamma+1)\sum_i (a_ib_i)^{p_i} \tau_i^{k_i} \left( \tau_i - h_i \right), \end{align}\] where \[\begin{align} \tau_i := \frac{|b_ix_i|^{\alpha_i}}{|bx|_a} \geq0 \quad \text{ and } \quad h_i := \frac{\Lambda}{(1+\gamma)\lambda}\left( 1- \frac{(1-\delta)}{a_i}\right)>0. \end{align}\] Observe that \(\sum\tau_i = 1\) by the definition of \(|bx|_a\). We choose \(b_i>0\) by \(b_i := \frac{K^{1/p_i}}{a_ih_i^{k_i/p_i}}>0\) for some \(K>0\), then \((a_ib_i)^{p_i} =\frac{K}{h_i^{k_i}}\). Applying Lemma 4, we deduce \[\begin{align} (I) &\geq|bx|_a^{-(d+\gamma+1)} K\lambda (\gamma+1)\sum_i \frac{1}{h_i^{k_i}} \tau_i^{k_i} \left( \tau_i - h_i \right) \\ &\geq\Phi^{d_0} K\lambda \gamma^{-d_0}(\gamma+1) \left(1-\sum_i h_i\right), \end{align}\] where \(d_0 := 1+\frac{d+1}{\gamma}>p_n-1\). Using 10 , we obtain \[\begin{align} \sum_i h_i &= \sum_i \frac{\Lambda}{(1+\gamma)\lambda}\left( 1- (1-\delta) \frac{d+1-\gamma(p_i-2)}{p_i} \right) \\ &= \frac{n\Lambda}{(1+\gamma)\lambda} \left( 1+(1-\delta)\left(\gamma -\frac{d+1+2\gamma}{\overline{p}} \right) \right) \\ &\leq \frac{n\Lambda}{(1+\gamma)\lambda} \left( 1- \gamma(1-\delta)\left(\frac{p_n}{\overline{p}}-1 \right) \right), \end{align}\] where \(\overline{p} = \left( \frac{1}{n}\sum_i \frac{1}{p_i}\right)^{-1}\). Thus, we get \[\begin{align} 1-\sum_ih_i >0 \;\Longleftrightarrow \;\gamma>\frac{n-\frac{\lambda}{\Lambda}}{\frac{\lambda}{\Lambda} + n(1-\delta)(\frac{p_n}{\overline{p}}-1)}. \end{align}\] Combining this with 11 , we need to choose \(\gamma\) and \(\delta>0\) small enough such that \[\begin{align} \frac{n-\frac{\lambda}{\Lambda}}{\frac{\lambda}{\Lambda} + n(1-\delta)(\frac{p_n}{\overline{p}}-1)}<\gamma< \frac{p_1-1}{p_n-p_1}. \end{align}\] For such \(\gamma\) and sufficiently small \(\delta>0\) to exist, we need the following condition on \((p_i)\): \[\begin{align} \frac{n-\frac{\lambda}{\Lambda}}{\frac{\lambda}{\Lambda} + n(\frac{p_n}{\overline{p}}-1)}< \frac{p_1-1}{p_n-p_1}, \end{align}\] which is equivalent to the condition ?? . Finally, we choose \(K > \frac{\gamma^{d_0}}{\lambda(\gamma+1)(1-\sum h_i)}>0\), then we conclude that \((I) > \Phi^{d_0}\), which implies ?? . However, since \(\Phi\) may not be \(C^2\) in \(\bigcup\{x_i=0\}\), the inequality ?? does not necessarily hold in the classical sense. Nevertheless, we can prove that \(\Phi\) satisfies ?? in the viscosity sense. If there is a smooth test function \(\phi\) which touches \(\Phi\) from below at a point \(y\) with \(y_i=0\) for some \(i\), then we have \(D_i\phi(y)=0\). As a result, \(D_{ii}\phi(y)\) plays no role in the evaluation of the operator allowing us to perform essentially the same direct computation as in the non-degenerate case. Consequently, we conclude that \(\Phi\) serves as a viscosity subsolution to ?? in \(Q_{R} \setminus\{0\}\). ◻
We also require the following scaling property of the barrier function \(\Phi\).
Lemma 6. For any \(r>0\), we have \[\begin{align} \lim_{L\rightarrow\infty} \frac{1}{L} \sup_{\mathbb{R}^n \setminus K_r(L)} \Phi =0, \quad \lim_{m\rightarrow0} \frac{1}{m} \inf_{ K_r(m)} \Phi =\infty. \end{align}\]
Proof. By direct calculation, \[\begin{align} \frac{1}{L} \sup_{\mathbb{R}^n \setminus K_r(L)} \Phi &= \frac{c}{L} (\inf_{\mathbb{R}^n \setminus K_r(L)} |bx|_a)^{-\gamma} = \frac{c}{L} (\min_{i} |b_ir^{\alpha_i} L^{\beta_i}|^{a_i})^{-\gamma}\\ &= \frac{c}{L} (\min_{i} c_i L^{-\frac{p_n-p_i}{d+1-\gamma(p_i-2)}})^{-\gamma} = \frac{c}{L} \max_i c_i L^{\frac{\gamma(p_n-p_i)}{d+1-\gamma(p_i-2)}} \\ &= \max_i c_i L^{-\frac{d+1-\gamma(p_n-2)}{d+1-\gamma(p_i-2)}} \overset{L\rightarrow \infty}{\longrightarrow} 0, \end{align}\] since \(d+1-\gamma(p_n-2)>0\) by 10 . Similarly, \[\begin{align} \frac{1}{m} \inf_{ K_r(m)} \Phi &= \frac{c}{m} (\sup_{K_r(m)} |bx|_a)^{-\gamma} \leq \frac{c}{m} (\max_{i} n|b_ir^{\alpha_i} m^{\beta_i}|^{a_i})^{-\gamma}\\ &= \frac{c}{m} (\max_{i} c_i m^{-\frac{p_n-p_i}{d+1-\gamma(p_i-2)}})^{-\gamma} = \frac{c}{m} \min_i c_i m^{\frac{\gamma(p_n-p_i)}{d+1-\gamma(p_i-2)}} \\ &= \min_i c_i m^{-\frac{d+1-\gamma(p_n-2)}{d+1-\gamma(p_i-2)}} \overset{m\rightarrow 0}{\longrightarrow} \infty. \end{align}\] ◻
In this section, we establish the doubling property for viscosity supersolutions by utilizing the barrier function constructed in Section 4. For the remainder of this paper, we assume that the exponents \((p_i)\) satisfy the assumption ?? throughout.
Lemma 7. There exist \(m_0 \in(0,1)\), \(L_0>1\), \(R_1>1\), \(\mu_0 \in(0,1)\) and \(\epsilon_0 \in (0,1)\) such that if \(u \in C(K_{R_1}(m_0))\) is nonnegative in \(K_{R_1}(m_0)\), and satisfies \[\begin{align} \label{doubpde} \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum|D_iu|^{p_i-1} \leq \epsilon_0 \quad \text{ in } K_{R_1}(m_0), \end{align}\qquad{(3)}\] then \[\begin{align} u>L_0m_0 \quad \text{ in }K_{r_0}(L_0m_0), \quad \Longrightarrow \quad u>m_0 \quad \text{ in } K_1(m_0), \end{align}\] where \(r_0 =r_0(\epsilon_0) \in(0,1)\) is as in the Lemma 3 with \(\epsilon=\epsilon_0\).
Proof. We find a barrier function \(\Psi \in C(K_{R_1}(m_0) \setminus K_{r_0}(L_0m_0))\) that satisfies the following properties:
\(\mathcal{M}^- \left( \left[|D_i\Psi|^{\frac{p_i-2}{2}}\right] D^2\Psi \left[|D_i\Psi|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum|D_i\Psi|^{p_i-1} -\Psi^{d_0}> \epsilon_0\) in \(K_{R_1}(m_0) \setminus K_{r_0}(L_0m_0)\).
\(\Psi < 0 \leq u\) in \(\partial K_{R_1}(m_0)\).
\(\Psi < L_0m_0 < u\) in \(\partial K_{r_0}(L_0m_0)\).
\(\Psi > m_0\) in \(K_{1}(m_0)\).
By Lemma 6, there exists \(m_0 \in(0,1)\) such that \(\frac{1}{m_0} \inf_{ K_1(m_0)} \Phi >2\). We define the barrier as a translation of our previous function \[\Psi(x) = \Phi(x) - m_0.\] Then since \(\Phi >2m_0\) in \(K_1(m_0)\), we immediately obtain condition \((4)\). Moreover, we choose large \(R_1>1\) such that \(\{ \Phi \geq m_0\} \subset K_{R_1}(m_0)\), which implies condition \((2)\). Using Lemma 6 again, we choose \(L_0>\frac{1}{m_0}\) satisfying \(\frac{1}{L_0m_0} \sup_{\mathbb{R}^n \setminus K_{r_0}(L_0m_0)} \Phi <1\), which guarantees condition \((3)\). Finally, by Lemma 5, there exists a constant \(\mu_0>0\) such that \[\begin{align} \mathcal{M}^-\left( \left[|D_i\Psi|^{\frac{p_i-2}{2}}\right] D^2\Psi \left[|D_i\Psi|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum_i|D_i\Psi|^{p_i-1} > (\Psi+m_0)^{d_0} \geq \epsilon_0\Psi+\epsilon_0 \quad \text{ in }K_{R_1}(m_0) \setminus \{0\}, \end{align}\] where \(\epsilon_0 = m_0^{d_0}\). This yields condition \((1)\). Note that since \(u \geq0\), we have \[\begin{align} F(u,Du,D^2u) := \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum|D_iu|^{p_i-1} -\epsilon_0u\leq \epsilon_0 \quad \text{ in }K_{R_1}(m_0). \end{align}\] Since \(F(r,p,X)\) is strictly decreasing in \(r\) and degenerate elliptic, \(F\) satisfies the comparison principle (Lemma 1). Thus, since \(F(u,Du,D^2u)\leq\epsilon_0 \leq F(\Psi,D\Psi,D^2\Psi)\) in \(K_{R_1}(m_0) \setminus K_{r_0}(L_0m_0)\) and \(\Psi \leq u\) in \(\partial(K_{R_1}(m_0) \setminus K_{r_0}(L_0m_0))\), we have \(\Psi \leq u\) in \(K_{R_1}(m_0) \setminus K_{r_0}(L_0m_0)\). Therefore, we obtain \(m_0 < \Psi <u\) in \(K_1(m_0)\), which completes the proof. ◻
By iteratively applying the previous result, we obtain the following result.
Lemma 8. Let \(m_0 \in(0,1)\), \(L_0>1\), \(R_1>1\), \(\mu_0\in (0,1)\) and \(\epsilon_0 \in(0,1)\) be as in the previous lemma. If \(u \in C(K_{R_1}(m_0))\) is nonnegative and satisfies ?? in \(K_{R_1}(m_0)\), then \[\begin{align} \label{doubkres} u>L_0^km_0 \quad \text{ in }K_{r_0^k}(L_0^km_0), \quad \Longrightarrow \quad u>m_0 \quad \text{ in } K_1(m_0). \end{align}\qquad{(4)}\] for any \(k \in \mathbb{N}\).
Proof. We proceed by induction on \(k\). The case of \(k=1\) follows directly from Lemma 7. We assume that ?? is true for some \(k \in\mathbb{N}\) and prove it for \(k+1\). Suppose that \(u>L_0^{k+1}m_0\) in \(K_{r_0^{k+1}}(L_0^{k+1}m_0)\). We define a rescaled function \[\begin{align} v(x) = \frac{u((r_0^k)^{\alpha_i}(L_0^k)^{\beta_i} x_i)}{L_0^k}. \end{align}\] This scaling maps the domain so that \(v \in C( K_{R_1/r_0^k}(m_0/L_0^k))\). Observe that \(K_{R_1}(m_0)\subset K_{R_1/r_0^k}(m_0/L_0^k)\) and by Remark 1, the function \(v\) is again nonnegative and satisfies ?? in \(K_{R_1}(m_0)\). Moreover, the inductive assumption \(u>L_0^{k+1}m_0\) in \(K_{r_0^{k+1}}(L_0^{k+1}m_0)\) translates directly to \(v > L_0m_0\) in \(K_{r_0}(L_0m_0)\). Applying Lemma 7, we obtain \(v > m_0\) in \(K_{1}(m_0)\). Rescaling back, this implies \(u>L_0^km_0\) in \(K_{r_0^k}(L_0^km_0)\). Therefore, by the induction assumption that ?? is true for \(k\), we conclude that \(u>m_0\) in \(K_1(m_0)\), which completes the proof. ◻
We conclude this section by proving the following type of doubling property.
Lemma 9. Let \(m_0 \in(0,1)\), \(R_1>1\), \(\mu_0\in (0,1)\) and \(\epsilon_0 \in(0,1)\) be as in the previous lemma. For any fixed \(\nu >1\), there exist \(r_1 \in (0,1)\) and \(k_0 \in \mathbb{N}\) such that if \(u \in C(K_{R_1}(m_0))\) is nonnegative and satisfies ?? in \(K_{R_1}(m_0)\), then \[\begin{align} \label{doubkres2} u>\nu^km_0 \quad \text{ in }K_{r_1^k}(\nu^km_0), \quad \Longrightarrow \quad u>m_0 \quad \text{ in } K_1(m_0), \end{align}\qquad{(5)}\] for any \(k \geq k_0\).
Proof. If \(\nu\geq L_0\), then the desired implication follows immediately from Lemma 8. Thus, we assume that \(\nu < L_0\). We first fix \(\tilde{k}_1 \in \mathbb{N}\) such that \(r_0^{\tilde{k}_1} L_0^{p_n-p_1} < 1\). Next, we define \(r_1 = \nu^{-\theta} \in (0,1)\) with a sufficiently small \(\theta \in (0,1)\) such that \[\begin{align} \label{thetacond} r_0L_0^{\theta} <1, \quad \text{ and } \quad r_0^{\tilde{k}_1}L_0^{\theta(\tilde{k}_1+1)} L_0^{p_n-p_1}\leq 1. \end{align}\tag{12}\] We select \(k_0 \in \mathbb{N}\) such that \(L_0^{\tilde{k}_1} \leq \nu^{k_0} < L_0^{\tilde{k}_1+1}\). Then for any \(k \geq k_0\), there exists \(\tilde{k} \geq \tilde{k}_1\) such that \(L_0^{\tilde{k}} \leq \nu^{k} < L_0^{\tilde{k}+1}\). Therefore, recalling the fact that \(K_r(N) \supset K_r(M)\) for any \(0<N\leq M\) and \(r>0\), we have \[\begin{align} u>\nu^km_0 \geq L_0^{\tilde{k}}m_0 \quad \text{ in }K_{r_1^k}(\nu^km_0) \supset K_{r_1^k}(L_0^{\tilde{k}+1}m_0). \end{align}\] We claim that \(K_{r_1^k}(L_0^{\tilde{k}+1}m_0) \supset K_{r_0^{\tilde{k}}}(L_0^{\tilde{k}}m_0)\). Observe that since 12 and \(\tilde{k} \geq \tilde{k}_1\), we have \[\begin{align} \frac{r_0^{\tilde{k}}}{r_1^k}L_0^{p_n-p_i}=r_0^{\tilde{k}}\nu^{\theta k}L_0^{p_n-p_i} \leq r_0^{\tilde{k}}L_0^{\theta(\tilde{k}+1)} L_0^{p_n-p_i}\leq 1, \end{align}\] for any coordinate index \(i\), and so \[\begin{align} (r_1^k)^{\alpha_i} (L_0^{\tilde{k}+1}m_0)^{\beta_i} \geq (r_0^{\tilde{k}})^{\alpha_i} (L_0^{\tilde{k}}m_0)^{\beta_i}, \end{align}\] which implies the claim. Therefore, we have \(u> L_0^{\tilde{k}}m_0\) in \(K_{r_0^{\tilde{k}}}(L_0^{\tilde{k}}m_0)\). Applying Lemma 8, we conclude that \(u >m_0\) in \(K_1(m_0)\). ◻
In this section, we prove the \(L^\epsilon\) estimate. We first prove the following lemma by combining the basic measure estimate (Lemma 3) and the doubling property (Lemma 7).
Lemma 10. Let \(m_0 \in(0,1)\), \(R_1>1\), \(\mu_0\in (0,1)\) and \(\epsilon_0 \in(0,1)\) be as in Lemma 7. There exist \(L_1\geq L_0\) and \(\delta_1 \in(0,1)\) such that if \(r \in(0,1)\) and \(u \in C(K_{R_1}(m_0))\) is nonnegative and satisfies ?? in \(K_{R_1}(m_0)\) with \[\begin{align} \inf_{K_r(m_0)} u \leq m_0, \end{align}\] then \[\begin{align} |\{u <L_1m_0\} \cap K_r(m_0)\}| \geq \delta_1|K_r(m_0)|. \end{align}\]
Proof. We define a rescaled function \(v \in C(K_{R_1/r}(m_0))\) as \[\begin{align} v(x) = u(r^{\alpha_i} x_i). \end{align}\] Then \(v\) is nonnegative and a supersolution of ?? in \(K_{R_1}(m_0) \subset K_{R_1/r}(m_0)\) by Remark 1. Moreover, since \(\inf_{K_1(m_0)} v \leq m_0\), we have \(\inf_{K_{r_0}(L_0m_0)} v \leq L_0m_0\) by Lemma 7. We also define a second scaled function \(w \in C(K_{R_1}(1/L_0))\) as \[\begin{align} w(x) = \frac{v( (L_0m_0)^{\beta_i}x)}{L_0m_0}. \end{align}\] Then \(w\) is also nonnegative and satisfies ?? in \(K_1(M_0) \subset K_{R_1}(1/L_0)\). Furthermore, since \(\inf_{K_{r_0}(L_0m_0)} v \leq L_0m_0\), we have \(\inf_{K_{r_0}(1)} w \leq 1\). Applying Lemma 3 yields \[\begin{align} |\{w <M_0\} \cap K_1(M_0)\}| \geq \delta_0|K_1(M_0)|. \end{align}\] Scaling back to the original function \(u\), we obtain \[\begin{align} |\{u <M_0L_0m_0\} \cap K_r(M_0L_0m_0)\}| \geq \delta_0|K_r(M_0L_0m_0)|. \end{align}\] Letting \(L_1 = M_0L_0\) and using \(K_r(L_1m_0) \subset K_r(m_0)\), we have \[\begin{align} |\{u <L_1m_0\} \cap K_r(m_0)\}| &\geq |\{u <L_1m_0\} \cap K_r(L_1m_0)\}| \\ &\geq \frac{\delta_0|K_r(L_1m_0)|}{|K_r(m_0)|}|K_r(m_0)| = \delta_1|K_r(m_0)|, \end{align}\] with \(\delta_1 =\delta_0 \prod_{i}L_1^{\beta_i}\). ◻
Using the previous lemma and the Vitali covering lemma (Lemma 2), we now establish the \(L^\epsilon\) estimate. This result quantifies the decay of the level set for a supersolution \(u\).
Theorem 2 (\(L^\epsilon\) estimate). Let \(m_0 \in(0,1)\), \(L_1>1\), \(\mu_0\in (0,1)\) and \(\epsilon_0 \in(0,1)\) be as in the previous lemma. There exist \(\eta \in(0,1)\), \(R_2>1\) and \(k_1 \in \mathbb{N}\) such that if \(u \in C(K_{R_2}(m_0))\) is nonnegative in \(K_{R_2}(m_0)\), and satisfies \[\begin{align} \label{Lepspde} \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum|D_iu|^{p_i-1} \leq \epsilon_0 \quad \text{ in } K_{R_2}(m_0), \end{align}\qquad{(6)}\] with \(u(0) \leq m_0\), then \[\begin{align} \left|\{u \geq L_1^km_0\} \cap \frac{2}{3}K_1(m_0)\right| \leq \eta^k\left|\frac{2}{3}K_1(m_0)\right| \quad \text{ for any } k\geq k_1. \end{align}\]
Proof. We choose \(R_2>R_1\) such that \(R_2^{1/p_n} \geq R_1^{1/p_n}+1\). This choice makes all intrinsic cubes in the proof remain within the domain \(K_{R_2}(m_0)\). We then fix \(\rho >1\) satisfying \(\rho(1-L_0^{\beta_m}) \geq 1\), where \(\beta_m<0\) is the largest exponent among the set \((\beta_i)\) such that \(\beta_i\neq0\). We also select sufficiently large \(k_1 \in \mathbb{N}\) such that \(\rho L_0^{k_1\beta_m} \leq \frac{1}{3}\) and \(\sum_{l=k_1}^\infty r_0^{l/p_n} \leq \frac{1}{3}\). We define shrinking cubes \(Q_k\) as \[\begin{align} Q_k := \left\{|x_i| \leq m_0^{\beta_i}\left(\frac{2}{3}+\rho L_0^{k\beta_i} \right) \text{ for }\beta_i\neq 0, \quad |x_j| \leq \frac{2}{3} + \sum_{l=k}^{\infty} r_0^{l/p_n}\text{ for }\beta_j = 0 \right\}. \end{align}\] Then we have \(Q_{k+1} \subset Q_k\), \(Q_{k_1} \subset K_1(m_0)\) and \(\lim_{k\rightarrow \infty} Q_k = \frac{2}{3}K_1(m_0)\). For \(k \geq k_1\), we define a level set \[\begin{align} A_k := \{ u \geq L_1^k m_0\} \cap Q_k. \end{align}\] For any \(x_0 \in Q_{k+1} \cap A_k\), let \(r_{x_0}>0\) be the largest radius such that \(K_{r_{x_0}}(L_1^km_0,x_0) \subset A_k\). We claim that \[\begin{align} K_{r_0^k}(L_0^km_0,x_0) \subset Q_k, \quad \text{ and } \quad\inf_{K_{r_0^k}(L_0^km_0,x_0)}u < L_1^km_0. \end{align}\] This immediately implies \(K_{r_0^k}(L_0^km_0,x_0) \not\subset A_k\), and therefore \(r_{x_0} < r_0^k <1\).
To prove the claim, first note that for \(\beta_i \neq 0\), \[\begin{align} m_0^{\beta_i}\left(\frac{2}{3} + \rho L_0^{(k+1)\beta_i} \right) + (r_0^k)^{\alpha_i} (L_0^km_0)^{\beta_i} \leq m_0^{\beta_i}\left(\frac{2}{3} + L_0^{k\beta_i} (\rho L_0^{\beta_i} +1)\right) \leq m_0^{\beta_i}\left(\frac{2}{3} + \rho L_0^{k\beta_i}\right), \end{align}\] and for \(\beta_j = 0\), \[\begin{align} \left(\frac{2}{3} + \sum_{l=k+1}^{\infty} r_0^{l/p_n} \right) + (r_0^{k})^{a_j} \leq \left(\frac{2}{3} + \sum_{l=k}^{\infty} r_0^{l/p_n} \right). \end{align}\] Hence we get \(K_{r_0^k}(L_0^km_0,x_0) \subset Q_k\). For the second part, since \(u(0) \leq m_0\) and \(x_0 \in K_1(m_0)\), we have \(\inf_{K_1(m_0,x_0)}u \leq m_0\). By the choice of \(R_2\), we get \(K_{R_1}(m_0,x_0) \subset K_{R_2}(m_0)\). Applying Lemma 8 yields \(\inf_{K_{r_0^k}(L_0^km_0,x_0)}u \leq L_0^km_0 < L_1^km_0\), which completes the claim.
Consequently, we obtain the collection of cube \(\mathcal{F} = \{K_{r_{x_0}}(L_1^km_0,x_0) \}_{x \in Q_{k+1}\cap A_k}\) which is a covering of \(Q_{k+1}\cap A_k\). Using the Vitali covering lemma (Lemma 2) we find a countable disjoint subcollection \(\mathcal{G} = \{K_l=K_{r_{l}}(L_1^km_0,x_l) \}_{l\in\mathbb{N}}\) such that \(\bigcup_l5K_l \supset Q_{k+1}\cap A_k\). By the construction, we have \[\begin{align} K_l \subset A_k, \quad\text{ and }\quad \inf_{K_l} u \leq L_1^km_0. \end{align}\] We now claim that for any \(l \in\mathbb{N}\), \[\begin{align} |K_l \cap A_k\setminus \tilde{A}_{k+1}| \geq \delta_1|K_l|, \end{align}\] where \[\begin{align} \tilde{A}_{k+1} = \{ u\geq L_1^{k+1}m_0 \} \cap Q_k. \end{align}\] To prove this, we consider the rescaled function \(v\) as \[\begin{align} v(x) = \frac{u((L_1^k)^{\beta_i}(x-x_l))}{L_1^k}. \end{align}\] Since \(K_{R_1}(m_0,x_l) \subset K_{R_2}(m_0)\), \(v\) is nonnegative and satisfies ?? in \(K_{R_1}(m_0)\subset K_{R_1}(m_0/L_1^k)\). Moreover, since \(\inf_{K_{r_{l}}(L_1^km_0,x_l)} u \leq L_1^km_0\), we have \(\inf_{K_{r_l}(m_0)}v \leq m_0\). Applying Lemma 10, we have \[\begin{align} |\{v <L_1m_0\} \cap K_{r_l}(m_0)\}| \geq \delta_1|K_{r_l}(m_0)|. \end{align}\] Scaling back to \(u\), this becomes \[\begin{align} |\{u <L_1^{k+1}m_0\} \cap K_l\}| \geq \delta_1|K_l|, \end{align}\] which proves the claim.
Since the cubes \(K_l\) are pairwise disjoint and \(\bigcup_l5K_l \supset Q_{k+1}\cap A_k\), we find that \[\begin{align} |Q_{k+1} \cap A_k| &\leq \sum_l |5K_l| \leq C \sum_i |K_l| \\ &\leq C \sum_l |K_l \cap A_k \setminus\tilde{A}_{k+1}| \leq C|A_k\setminus\tilde{A}_{k+1}|\\ &\leq C(|Q_{k+1} \cap A_k \setminus A_{k+1}| + |Q_k| - |Q_{k+1}|). \end{align}\] Next, using the elementary inequality \[\prod_{i}(s_i+t_i) - \prod_{i} t_i \leq \sum_{i} s_i \prod_{j\neq i} (s_j+t_j),\] for any \(s_i,t_i \geq 0\), together with \(\frac{2}{3}K_1(m_0) \subset Q_k \subset K_1(m_0)\), we obtain \[\begin{align} |Q_k|-|Q_{k+1}| &\leq C \left(\sum_{\{i: \beta_i\neq0\}} m_0^{\beta_i}(\rho L_0^{k\beta_i} - \rho L_0^{(k+1)\beta_i}) +\sum_{\{i: \beta_j=0\}} (\sum_{l=k}^{\infty} r_0^{l/p_n}-\sum_{l=k+1}^{\infty} r_0^{l/p_n})\right) \\ &\leq C(L_0^{k\beta_m} + r_0^{k/p_n}). \end{align}\] Combining the above estimates yields \[\begin{align} |Q_{k+1} \cap A_{k+1}| &= |Q_{k+1} \cap A_k| - |Q_{k+1} \cap A_k \setminus A_{k+1}| \\ &\leq \left(1-\frac{1}{C}\right)|Q_{k+1} \cap A_k| + |Q_k| - |Q_{k+1}| \\ &\leq \left(1-\frac{1}{C}\right)|Q_{k} \cap A_k| + C(L_0^{k\beta_m} + r_0^{k/p_n}). \end{align}\] Applying a standard iteration argument to this recursive inequality completes the proof of the theorem. ◻
The following corollary is a direct consequence of Theorem 2 and will be used later in the proof of the Harnack inequality.
Corollary 2. Under the same assumptions as in Theorem 2, there exists \(L_2>1\) such that \[\begin{align} \left|\{u \geq L_2m_0\} \cap \frac{2}{3}K_1(m_0)\right| \leq \frac{1}{4^{n+1}}\left|\frac{2}{3}K_1(m_0)\right|. \end{align}\]
In this section, we conclude the proof of the main result, Theorem 1, by adapting the argument from [40].
Lemma 11. Let \(u \in C(K_{R_2}(m_0))\) be nonnegative in \(K_{R_2}(m_0)\) and satisfy \[\begin{align} \label{harnpde} \begin{cases} \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum|D_iu|^{p_i-1} \leq \epsilon_0 \\ \mathcal{M}^+ \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) +\mu_0\sum|D_iu|^{p_i-1} \geq -\epsilon_0 \end{cases} \quad \text{ in } K_{R_2}(m_0), \end{align}\qquad{(7)}\] with \(u(0) \leq m_0\). Then there exists \(k_2 \in \mathbb{N}\) such that if there exists \(x_0 \in K_1(m_0)\) with \[\begin{align} u(x_0) \geq 2\nu^{k-1} L_2m_0, \end{align}\] where \(\nu = \frac{L_2}{L_2-1/2}\) for some \(k \geq k_2\), then \[\begin{align} \sup_{K_{R_{2}r_1^k}(\nu^km_0,x_0)} u \geq 2\nu^kL_2m_0, \end{align}\] where \(r_1 =r_1(\nu)\) is as in Lemma 9.
Proof. Let \(k_2 \geq k_1\) with \(1+r_1^{k_2}(R_2^{\alpha_n} +1) \leq R_2^{\alpha_n}\) ensuring all subsequent intrinsic cubes remain within the domain \(K_{R_2}(m_0)\). We argue by contradiction and assume that for some \(k \geq k_2\), \[\begin{align} \sup_{K_{R_{2}r_1^k}(\nu^km_0,x_0)} u < 2\nu^kL_2m_0. \end{align}\] Observe that the choice of \(k_2\) implies \(K_{R_{2}r_1^k}(\nu^km_0,x_0) \subset K_{R_2}(m_0)\). To derive a contradiction, we define a scaled function \(v\) as follows: \[\begin{align} v(x) = \frac{2\nu^kL_2m_0 - u((r_1^k)^{\alpha_i}(\nu^k)^{\beta_i}(x_i-(x_0)_i)}{\nu^k}. \end{align}\] The function \(v\) is nonnegative and satisfies ?? in \(K_{R_2}(m_0)\). Since \(v(0) \leq m_0\), we apply Corollary 2 to find that \[\begin{align} \left|\{v \geq L_2m_0\} \cap \frac{2}{3}K_1(m_0)\right| \leq \frac{1}{4^{n+1}}\left|\frac{2}{3}K_1(m_0)\right|. \end{align}\] Scaling back to \(u\) implies \[\begin{align} \label{harnmeas1} \left|\{u \leq \nu^k L_2m_0\} \cap \frac{2}{3}K_{r^k_1}(\nu^km_0,x_0)\right| \leq \frac{1}{4^{n+1}}\left|\frac{2}{3}K_{r^k_1}(\nu^km_0)\right|. \end{align}\tag{13}\] On the other hand, since \(u(0) \leq m_0\), we have \(\inf_{K_1(m_0,x_0)}u \leq m_0\). Using \(K_{R_1}(m_0,x_0) \subset K_{R_2}(m_0)\) and applying Lemma 9, we obtain \[\begin{align} \inf_{K_{r_1^k}(v^km_0,x_0)}u \leq \nu^k m_0. \end{align}\] Consequently, there exists a point \(\overline{x} \in K_{r_1^k}(v^km_0,x_0)\) satisfying \(u(\overline{x}) \leq \nu^k m_0\). Note that \(K_{R_2r_1^k}(v^km_0,\overline{x}) \subset K_{R_2}(m_0)\) by the choice of \(k_2\). We then define a second scaled function \[\begin{align} w(x) = \frac{u((r_1^k)^{\alpha_i}(\nu^k)^{\beta_i}(x_i-\overline{x}_i))}{ \nu^k}. \end{align}\] Then, \(w\) is nonnegative and satisfies ?? in \(K_{R_2}(m_0)\). Since \(w(0) \leq m_0\), Corollary 2 yields \[\begin{align} \left|\{w \geq L_2m_0\} \cap \frac{2}{3}K_1(m_0)\right| \leq \frac{1}{4^{n+1}}\left|\frac{2}{3}K_1(m_0)\right|. \end{align}\] Scaling back to \(u\), we get \[\begin{align} \label{harnmeas2} \left|\{u \geq \nu^k L_2m_0\} \cap \frac{2}{3}K_{r^k_1}(\nu^km_0,\overline{x})\right| \leq \frac{1}{4^{n+1}}\left|\frac{2}{3}K_{r^k_1}(\nu^km_0)\right|. \end{align}\tag{14}\] Note that since \(\overline{x} \in K_{r_1^k}(v^km_0,x_0)\), we have \[\begin{align} \frac{1}{6}K_{r_1^k}(v^km_0,\tilde{x}) \subset \frac{2}{3}K_{r_1^k}(v^km_0,x_0) \cap \frac{2}{3}K_{r_1^k}(v^km_0,\overline{x}), \end{align}\] where \(\tilde{x} = \frac{x_0+\overline{x}}{2}\). Therefore, the estimates 13 and 14 imply \[\begin{align} \left|\{u \leq \nu^k L_2m_0\} \cap \frac{1}{6}K_{r_1^k}(v^km_0,\tilde{x})\right| \leq \frac{1}{4}\left|\frac{1}{6}K_{r^k_1}(\nu^km_0)\right|, \\ \left|\{u \geq\nu^k L_2m_0\} \cap \frac{1}{6}K_{r_1^k}(v^km_0,\tilde{x})\right| \leq \frac{1}{4}\left|\frac{1}{6}K_{r^k_1}(\nu^km_0)\right|. \end{align}\] Summing the above two inequalities, we obtain \(\left|\frac{1}{6}K_{r^k_1}(\nu^km_0)\right| \leq \frac{1}{2}\left|\frac{1}{6}K_{r^k_1}(\nu^km_0)\right|\). This constitutes a contradiction, thereby completing the proof. ◻
Lemma 12. Let \(u \in C(K_{R_3}(m_0))\) be nonnegative in \(K_{R_3}(m_0)\) and satisfy \[\begin{align} \label{harnpde2} \begin{cases} \mathcal{M}^- \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) -\mu_0\sum|D_iu|^{p_i-1} \leq \epsilon_0 \\ \mathcal{M}^+ \left( \left[|D_iu|^{\frac{p_i-2}{2}}\right] D^2u \left[|D_iu|^{\frac{p_i-2}{2}}\right] \right) +\mu_0\sum|D_iu|^{p_i-1} \geq -\epsilon_0 \end{cases} \quad \text{ in } K_{R_3}(m_0), \end{align}\qquad{(8)}\] for some universal \(R_3 > R_2\). Then there exists \(C_0>1\) such that \[\begin{align} u(0) \leq m_0 \quad &\Longrightarrow \quad \sup_{K_{1/2}(m_0)} u \leq C_0m_0, \\ u(0) \geq C_0m_0 \quad &\Longrightarrow \quad \inf_{K_{1/2}(m_0)} u \geq m_0. \end{align}\]
Proof. We begin by choosing \(k_3 \geq k_2\) such that \(\sum_{k=k_3}^\infty R_2^{\alpha_1}r_0^{k\alpha_n} < 1-\left(\frac{1}{2}\right)^{\alpha_n}\). Our purpose is to prove that if \(u \geq 0\) in \(K_{R_2}(m_0)\) and \(u(0) \leq m_0\), then there holds \[\begin{align} \label{harnfirst} \sup_{K_{1/2}(m_0)} u < 2 \nu^{k_3}L_2m_0 = C_0m_0. \end{align}\tag{15}\] We proceed by contradiction. Suppose there exists a point \(x_0 \in K_{1/2}(m_0)\) such that \[\begin{align} u(x_0) \geq 2 \nu^{k_3} L_2m_0. \end{align}\] Then by Lemma 11, there exists \(x_1 \in K_{R_{1}r_1^{k_3}}(\nu^{k_3}m_0,x_0)\) such that \[\begin{align} u(x_1) \geq 2 \nu^{k_3+1} L_2m_0. \end{align}\] Note that \(x_1 \in K_1(m_0)\) since \(\nu^{\beta_i} \leq 1\) and \[\begin{align} |(x_1)_i|\leq m_0^{\beta_i}\left( \frac{1}{2^{\alpha_i}} + R_2^{\alpha_i} r_1^{k_3\alpha_i} \right) \leq m_0^{\beta_i}. \end{align}\]
By induction, we claim that there exists a sequence of points \(x_{l} \in K_{R_{1}r_1^{k_3+l-1}}(\nu^{k_3+l-1}m_0,x_{l-1})\) for \(l \geq 1\) satisfying \[\begin{align} \label{xlcond} x_{l} \in K_1(m_0) \quad \text{ and } \quad u(x_l) \geq 2 \nu^{k_3 + l} L_2 m_0. \end{align}\tag{16}\] With the base case \(l =1\) completed, we prove 16 for \(l+1\) assuming that 16 holds for every \(k \leq l\). Using Lemma 11, there exists \(x_{l+1} \in K_{R_{1}r_1^{k_3+l}}(\nu^{k_3+l}m_0,x_{l})\) satisfying \(u(x_{l+1}) \geq 2 \nu^{k_3 + l+1} L_2 m_0\). Moreover, using \(x_{k} \in K_{R_{1}r_1^{k_3+k-1}}(\nu^{k_3+k-1}m_0,x_{k-1})\) for any \(k \leq l\) and \(\nu^{\beta_i} \leq 1\), we obtain \[\begin{align} |(x_{l+1})_i| &\leq |(x_0)_i| + \sum_{k=1}^{l+1}|(x_k)_i-(x_{k-1})_i| \\ &\leq m_0^{\beta_i}\left( \frac{1}{2^{\alpha_i}} + \sum_{k=k_3}^{k_3+l} R_2^{\alpha_i} r_1^{k\alpha_i} \right) \leq m_0^{\beta_i}. \end{align}\] This confirms that \(x_{l+1} \in K_1(m_0)\) and the claim is proved. Therefore, as \(l \rightarrow \infty\), \(x_l\) converges to some point \(x_\infty \in \overline{K_1(m_0)}\). However, the continuity of \(u\) implies that \(u(x_\infty)\) must be finite, which contradicts the fact that \(u(x_l) \rightarrow \infty\) for \(l \rightarrow \infty\) as required by 16 . This establishes the first part of the lemma with \(C_0 = 2 \nu^{k_3}L_2>1\).
The second part is proved by contradiction. Let \(R_3>R_2\) be chosen such that \(R_3^{1/p_n} \geq R_2^{1/p_n}+1\). Suppose that \(u(0) \geq C_0m_0\), but there exists \(x_0 \in K_{1/2}(m_0)\) such that \(u(x_0) < m_0\). Since \(K_{R_2}(m_0,x_0) \subset K_{R_3}(m_0)\) by the choice of \(R_3\), the previous argument 15 yields \(\sup_{K_{1/2}(m_0,x_0)} u < C_0m_0\). However, this contradicts with our initial assumption \(u(0) > C_0m_0\), which completes the proof. ◻
With the preceding lemma at hand, we are ready to prove the main theorem, Theorem 1.
Proof of Theorem 1. We set \(C_0 = 2 \nu^{k_3}L_2\) as in Lemma 12 and \(R_0 = 2R_3\). Assume that \(u\) is defined in \(K_{R_0r}(u(0)) \subset \Omega\). We define a rescaled function \(v \in C(K_{R_3}(m_0))\) as \[\begin{align} v(x) = \frac{u((2r)^{\alpha_i}M_1^{\beta_i}x)}{M_1}, \end{align}\] where \(M_1 = \frac{u(0)}{m_0}>0\). Then \(v\) is nonnegative in \(K_{R_3}(m_0)\), and satisfies \[\begin{align} \label{harnpdescal} \begin{cases} \mathcal{M}^- \left( \left[|D_iv|^{\frac{p_i-2}{2}}\right] D^2v \left[|D_iv|^{\frac{p_i-2}{2}}\right] \right) -\mu\sum_i \frac{(2r)^{1/p_i}}{M^{(p_n-p_i)/p_i}}|D_iv|^{p_i-1} \leq \frac{2r}{M^{p_n-1}}c_0 \\ \mathcal{M}^+ \left( \left[|D_iv|^{\frac{p_i-2}{2}}\right] D^2v \left[|D_iv|^{\frac{p_i-2}{2}}\right] \right) +\mu\sum_i \frac{(2r)^{1/p_i}}{M^{(p_n-p_i)/p_i}}|D_iv|^{p_i-1} \geq -\frac{2r}{M^{p_n-1}}c_0 \end{cases} \text{ in }K_{R_3}(m_0) \end{align}\tag{17}\] with \(M=M_1\). Also, we have \(v(0) \leq m_0\). By selecting \(\epsilon_1 = \frac{2}{m_0^{p_n-1}}\epsilon_0\), we have \(\frac{2}{M_1^{p_n-1}}c_0 \leq \epsilon_0\). Moreover, we choose small enough \(\mu_1>0\) such that \(\mu_1 \frac{2^{1/p_i}}{M_1^{(p_n-p_i)/p_i}}\leq \mu_0\) for any \(i\). Then \(v\) satisfies ?? , so we obtain \[\begin{align} \sup_{K_{1/2}(m_0)} v \leq C_0m_0, \end{align}\] by Lemma 12. Scaling back to \(u\), we get \[\begin{align} \sup_{K_{r}(u(0))} u \leq C_0u(0). \end{align}\] which completes the first part of the theorem.
For the second part, assume that \(u\) is defined in \(K_{R_0r}(C_0u(0)) \subset \Omega\). We define a second rescaled function \(w \in C(K_{R_3}(m_0))\) as \[\begin{align} w(x) = \frac{u((2r)^{\alpha_i}M_2^{\beta_i}x)}{M_2}, \end{align}\] where \(M_2 = \frac{u(0)}{C_0m_0}>0\). Then \(w\) is nonnegative in \(K_{R_3}(m_0)\), and satisfies 17 with \(M=M_2\). Moreover, we have \(w(0) \geq C_0m_0\). We choose \(\epsilon_2 = \frac{2}{(C_0m_0)^{p_n-1}}\epsilon_0\) and sufficiently small \(\mu_2 >0\) satisfying \(\mu_1 \frac{2^{1/p_i}}{M_1^{(p_n-p_i)/p_i}}\leq \mu_0\) for any \(i\). Then \(w\) satisfies ?? , so we obtain \[\begin{align} \inf_{K_{1/2}(m_0)} w \geq m_0, \end{align}\] by Lemma 12. Scaling back to \(u\), we get \[\begin{align} \inf_{K_{r}(C_0u(0))} u \geq \frac{1}{C_0}u(0). \end{align}\] This completes the proof of Theorem 1. ◻
Data Availability Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.
Conflict of interest The authors declared that they have no conflict of interest to this work.
S.-S. Byun was supported by Mid-Career Bridging Program through Seoul National University. H. Kim was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government [Grant No. 2022R1A2C1009312].↩︎