Regularity of solutions to anisotropic nonlocal equations

Abstract

We study harmonic functions associated to systems of stochastic differential equations of the form $dX_t^i=A_{i1}(X_{t-})dZ_t^1+\cdots +A_{id}(X_{t-})dZ_t^d$, $i\in \{1,\dots ,d\}$, where $Z_t^j$ are independent one-dimensional symmetric stable processes of order $\alpha _j\in (0,2)$, $j\in \{1,\dots ,d\}$. In this article we prove Hölder regularity of bounded harmonic functions with respect to solutions to such systems.

1 Introduction

The consideration of stochastic processes with jumps and anisotropic behavior is natural and reasonable since such objects arise in several natural and financial models. In certain circumstances Lévy processes with jumps are more suitable to capture empirical facts that diffusion models do. See for instance [14] for examples of financial models with jumps.

In the nineteen fifties, De Giorgi [15] and Nash [29] independently prove an a-priori Hölder estimate for weak solutions u to second order equations of the form

$$\begin{aligned} {{\,\mathrm{div}\,}}(A(x)\nabla u(x))=0 \end{aligned}$$

for uniformly elliptic and measurable coefficients A. In [28], Moser proves Hölder continuity of weak solutions and gives a proof of an elliptic Harnack inequality for weak solutions to this equation. This article provides a new technique of how to derive an a-priori Hölder estimate from the Harnack inequality. For a large class of local operators, the Hölder continuity can be derived from the Harnack inequality, see for instance [19]. For a comprehensive introduction into Harnack inequalities, we refer the reader e.g. to [20].

The corresponding case of operators in non-divergence form is treated in by Krylov and Safonov in [23]. The authors develop a technique for proving Hölder regularity and the Harnack inequality for harmonic functions corresponding to non-divergence form elliptic operators. They take a probabilistic point of view and make use of the martingale problem to prove regularity estimates for harmonic functions. The main tool is a support theorem, which gives information about the topological support for solutions to the martingale problem associated to the corresponding operator. This technique is also used in [6] to prove similar results for nonlocal operators of the form

$$\begin{aligned} Lf(x)=\int _{\mathbb {R}^d\setminus \{0\}}[f(x+h)-f(x)-\mathbb {1}_{\{|h|\le 1\}}h\cdot \nabla f(x)]a(x,h)dh \end{aligned}$$

(1.1)

under suitable assumptions on the function a. In [4] Bass and Chen follow the same ideas to prove Hölder regularity for harmonic functions associated to solutions of systems of stochastic differential equations driven by Lévy processes with highly singular Lévy measures. In this work we extend the results obtained by Bass and Chen to a larger class of driving Lévy processes.

A one-dimensional Lévy process $(Y_t)_{t\ge 0}$ is called symmetric stable processes of order $\gamma \in$ (0,2) if its characteristic function is given by

$$\begin{aligned} {\mathbb {E}}e^{i\xi Y_t} = e^{-t|\xi |^\gamma }, \qquad \xi \in \mathbb {R}. \end{aligned}$$

The Lévy measure of such a process is given by $\nu (dh)=c_{\gamma }|h|^{-1-\gamma }\, dh,$ where $c_{\gamma }=2^{\gamma }\Gamma \left( \frac{1+\gamma }{2}\right) /\left| \Gamma \left( -\frac{\gamma }{2}\right) \right|$.

Let $d\in \mathbb {N}$ and $d\ge 2$. We assume that $Z_{t}^{i}, i=1,\dots ,d$, are independent one-dimensional symmetric stable processes of order $\alpha _{i} \in (0,2)$ and define $Z=(Z_t)_{t\ge 0}=(Z_t^{1},\dots ,Z_t^{d})_{t\ge 0}$.

The Lévy-measure of this process is supported on the coordinate axes and is given by

$$\begin{aligned} \nu (dw)= \sum _{k=1}^{d} \left( \frac{c_{\alpha _k}}{|w_k|^{1+\alpha _k}}dw_k \left( \prod _{j\ne k}\delta _{\{0\}}(dw_j) \right) \right) . \end{aligned}$$

Therefore $\nu (A)=0$ for every set $A\subset \mathbb {R}^d$, which has an empty intersection with the coordinate axes. The generator L of Z is given for $f\in C_b^2(\mathbb {R}^d)$ by the formula

$$\begin{aligned} Lf(x)=\sum _{k=1}^{d} \int _{\mathbb {R}\setminus \{0\}} (f(x+he_{k})-f(x)-\mathbb {1}_{\{|h|\le 1\}}\partial _kf(x)h)\frac{c_{{\alpha _k}}}{|h|^{1+\alpha _k}} dh. \end{aligned}$$

(1.2)

For a deeper discussion on Lévy processes and their generators we refer the reader to [30].

Let $x_0\in \mathbb {R}^d$ and $A:\mathbb {R}^d\rightarrow \mathbb {R}^{d\times d}$ a matrix-valued function. We consider the system of stochastic differential equations

$$\begin{aligned} \left\{ \begin{aligned}&dX^i_{t} = \sum _{j=1}^d A_{ij}(X_{t-})dZ_{t}^j, \\&X_0^i = x_0^i, \end{aligned} \right. \end{aligned}$$

(1.3)

where $X_{t-}=\lim \limits _{s\nearrow t} X_s$ is the left hand limit.

This system has been studied systematically in the case $\alpha _1=\alpha _2=\cdots =\alpha _d=\alpha \in (0,2)$ by Bass and Chen in the articles [3] and [4]. With the help of the martingale problem, Bass and Chen prove in [3] that for each $x_0\in \mathbb {R}^d$ there exists a unique weak solution $(X=(X_t^1,\dots ,X_t^d)_{t\ge 0}, {\mathbb {P}}^{x_0})$ to (1.3). Furthermore the authors prove that the family $\{X,{\mathbb {P}}^x,x\in \mathbb {R}^d\}$ forms a conservative strong Markov process on $\mathbb {R}^d$ whose semigroup maps bounded continuous functions to bounded continuous functions (see Theorem 1.1, [3]). Consequently it follows that

$$\begin{aligned} {\mathcal {L}}f(x)=\sum _{j=1}^{d} \int _{\mathbb {R}\setminus \{0\}} (f(x+a_{j}(x)h)-f(x)-h\mathbb {1}_{\{|h|\le 1\}}\nabla f(x) \cdot a_j(x))\frac{c_{\alpha }}{|h|^{1+\alpha }} dh \end{aligned}$$

coincides on $C_b^2(\mathbb {R}^d)$ with the generator for any weak solution to (1.3), where $a_j(x)$ denotes the j$^{th}$ column of the matrix A(x). In [4] the authors prove Hölder regularity of harmonic functions with respect to ${\mathcal {L}}$ and give a counter example which shows that the Harnack inequality for harmonic functions is not satisfied.

In this paper we do not study unique solvability of (1.3) but prove an a-priori regularity estimate for harmonic functions if unique solutions to the system exist. The following assumptions will be needed throughout the paper.

Assumption

• (i) For every $x\in \mathbb {R}^d$ the matrix A(x) is non-degenerate, that is $\det (A(x))\ne 0$.

• (ii) The functions $x\mapsto A_{ij}(x)$ and $x\mapsto A^{-1}_{ij}(x)$ are continuous and bounded for all $1\le i,j\le d$ and $x\in \mathbb {R}^d$.

• (iii) For any $x_0\in \mathbb {R}^d$, there exists a unique solution to the martingale problem for

  $$\begin{aligned} {\mathcal {L}}f(x)=\sum _{j=1}^{d} \int _{\mathbb {R}\setminus \{0\}} (f(x+a_{j}(x)h)-f(x)-h\mathbb {1}_{\{|h|\le 1\}}\nabla f(x) \cdot a_j(x))\frac{c_{\alpha _j}}{|h|^{1+\alpha _j}} dh \end{aligned}$$

  (1.4)

  started at $x_0$. The operator ${\mathcal {L}}$ coincides on $C_b^2(\mathbb {R}^d)$ with the generator for the weak solution to (1.3).

For a comprehensive introduction into the martingale problem we refer the reader to [16].

1.1 Notation

Let A be the matrix-valued function from (1.3). Let D be a Borel set. Throughout the paper $\varpi (D)$ denotes the modulus of continuity of A and we write $\Lambda (D)$ for the upper bound of A on D. We set $\alpha _{\min }:=\min \{\alpha _1,\dots ,\alpha _d\}$ and $\alpha _{\max }:=\max \{\alpha _1,\dots ,\alpha _d\}$. For $i\in \mathbb {N}$ we write $c_i$ for positive constants and additionally $c_i=c_i(\cdot )$ if we want to highlight all the quantities the constant depends on.

In order to deal with the anisotropy of the process we consider a corresponding scale of cubes.

Definition 1.1

Let $r\in (0,1]$ and $\alpha _1,\dots ,\alpha _d\in (0,2)$. For $k>0$, we define

For brevity we write $M_r(x)$ instead of $M^{1}_r(x)$.

Note that $M_r^k$ is increasing in k and r. For $z\in \mathbb {R}^d$ and $r\in (0,1]$, the set $M_r(z)$ is a ball with radius r and center z in the metric space $(\mathbb {R}^d,d)$, where

$$\begin{aligned} d(x,y)=\sup _{k\in \{1,\dots ,d\}} \{|x_k-y_k|^{\alpha _k/\alpha _{\max }}\mathbb {1}_{\{|x_k-y_k|\le 1\}}(x,y) + \mathbb {1}_{\{|x_k-y_k|> 1\}}(x,y)\}. \end{aligned}$$

This metric is useful for local considerations only, that is studies of balls with radii less or equal than one. The advantage of using these sets is the fact that they reflect the different jump intensities of the process Z and compensate them in an appropriate way, see for instance Proposition 2.4.

The purpose of this paper is to prove the following result.

Theorem 1.2

Let $r\in (0,1]$, $s>0$ and $x_0\in \mathbb {R}^d.$ Suppose h is bounded in $\mathbb {R}^d$ and harmonic in $M_r^{1+s}(x_0)$ with respect to X. Then there exist $c_1=c_1(\Lambda (M_r^{1+s}(x_0)),\varpi (M_r^{1+s}(x_0)))>0$ and $\beta =\beta (\Lambda (M_r^{1+s}(x_0)),\varpi (M_r^{1+s}(x_0)))>0$, independent of h and r, such that

$$\begin{aligned} |h(x)-h(y)|\le c_1 \left( \frac{|x-y|}{r^{\alpha _{\max }/\alpha _{\min }}} \right) ^{\beta } \sup \limits _{\mathbb {R}^d} |h(z)| \quad \text { for } x,y\in M_r(x_0). \end{aligned}$$

We want to emphasize, that in the case $\alpha _1=\cdots =\alpha _d$ the set $M_r(x_0)$ reduces to a cube with radius r and hence this result coincides with [4, Theorem 2.9], when one chooses cubes instead of balls.

Let us briefly discuss selected related results in the literature.

As previously mentioned, in [6] the authors study operators of the form (1.1) for coefficients $a:\mathbb {R}^d\times \mathbb {R}^d\rightarrow \mathbb {R}$ which are assumed to be symmetric in the second variable and satisfy $a(x,h)\asymp |h|^{-d-\alpha }$ for all $x,h\in \mathbb {R}^d$, where $\alpha \in (0,2)$. Using probabilistic techniques they prove a Harnack inequality and derive Hölder regularity estimates for bounded harmonic functions. The results of this work have been extended to more general kernels by several authors. For instance, in [5] the authors establish a Hölder estimate for harmonic functions to operators of the form (1.1), where they replace the jump measure $a(x,h)\, d h$ by a family of measures $n(x, d h)$, which is not required to have a density with respect to the Lebesgue meaure. Furthermore, [32] extends the method of [6] to prove the Harnack inequality for more general classes of Markov processes. In [7] the authors construct and study the heat kernel a class of highly anisotropic integro-differential operators, where the Lévy measure does not have to be absolutely continuous with respect to the Lebesgue measure.

This article studies regularity for operators in non-divergence form given by (1.4). Hölder regularity results have intensively been studied for linear and nonlinear nonlocal equations governed by operators in non-divergence form. [31] provides a purely analytic proof of Hölder continuity for harmonic functions with respect to a class of integro differential equations given by (1.1), where no symmetry on the kernel a is assumed. In [9], the authors study viscosity solutions to fully nonlinear integro-differential equations and prove a nonlocal version of the Aleksandrov-Bakelman-Pucci estimate, a Harnack inequality and a Hölder estimate. There are many more important results concerning Hölder estimates and Harnack inequalities for integro-differential equations in non-divergence form including [1, 8, 10, 22, 27] and [33]. Hölder regularity estimates have also been intensely studied for operators in divergence form. We would like to mention two works, where the corresponding jump intensities are similar to the ones we study in this article. In [12] and [13] the authors study nonlocal elliptic resp. parabolic equations for families of operators which can be of the form (1.2). They prove a weak Harnack inequality and Hölder regularity estimates for weak solutions to the corresponding equations.

Let us give a short survey to known results related to systems of stochastic differential equations given by (1.3). We first discuss some results in the case $\alpha _1=\dots = \alpha _d$. In [3] the authors prove unique weak solvability for (1.3). [4] shows Hölder regularity estimates for bounded harmonic functions. Furthermore, in [26] the authors prove the strong Feller property for the corresponding semigroup for (1.3). Sharp lower bounds for the transition densities for the process $Z_t=(Z_t^1,\dots ,Z_t^d)$ are studied in [17] and sharp upper bounds in [21].

The existence of a unique solution to the martingale problem for (1.3) in the case of different orders of differentiability, i.e. $\alpha _i\ne \alpha _j$ for $i\ne j$, is shown in [11] under the additional assumption that the matrix A is diagonal. [24] also studies the system (1.3) in the case of diagonal matrices A. The authors prove sharp two-sided estimates of the corresponding transition density $p^{A}(t,x,y)$ and prove Hölder and gradient estimates for the function $x\mapsto p^{A}(t,x,y)$. In [18] the authors study the existence of densities for solutions (1.3) with Hölder continuous coefficients. They allow for a wide class of Lévy processes including the anisotropic processes $Z_t$ with different orders of differentiability. In [25] the authors study systems of the form (1.3) where $Z_t^1,\dots ,Z_t^d$ are independent one-dimensional Lévy processes with characteristic exponents $\psi _1,\dots ,\psi _d$. Under scaling conditions and regularity properties on the characteristic function they prove semigroup properties for solutions.

1.2 Structure of the article

This article is organized as follows. In Sect. 2 we provide definitions and auxiliary results. We constitute sufficient preparation and study the behavior of the solution to the system. In Sect. 3 we study the topological support of the solution to the martingale problem associated to the system of stochastic differential equations. The aim of this section is to prove a support theorem. Sect. 4 contains the proof of Theorem 1.2.

2 Definitions and auxiliary results

In this section we provide important definitions and prove auxiliary results associated to the solution of the system (1.3).

Let $A^\tau (x)$ denote the transpose of the matrix A(x) and $(a_j^\tau (x))^{-1}$ the j$^{\text {th}}$ row of $(A^{\tau }(x))^{-1}$. For a Borel set D, we denote the first entrance time of the process X in D by $T_D:=\inf \{t\ge 0 :X_t\in D\}$ and the first exit time of X of D by $\tau _D:=\inf \{t\ge 0 :X_t\notin D\}$.

Let us first recall the definition of harmonicity with respect to a Markov process.

Definition 2.1

A bounded function $h:\mathbb {R}^d\rightarrow \mathbb {R}$ is called harmonic with respect to X in a domain $D\subset \mathbb {R}^d$ if for every bounded open set U with $U\Subset D$

$$\begin{aligned} h\left( X_{t\wedge \tau _U}\right) \text { is a } {\mathbb {P}}^x \text {-martingale for every } x\in U. \end{aligned}$$

For $R=M_s(y)$ we use the notation ${\widehat{R}}=M^3_s(y)$. The next Proposition is a pure geometrical statement and not related to the system of stochastic differential equations. We skip the proof and refer the reader to [2, Proposition V.7.2], which can be easily adjusted to our case.

Proposition 2.2

Let $r\in (0,1],q\in (0,1)$ and $x_0\in \mathbb {R}^d$. If $A\subset M_r(x_0)$ and $|A|<q$, then there exists a set $D\subset M_r(x_0)$ such that

• (1) D is the union of rectangles $\widehat{R_i}$ such that the interiors of the $R_i$ are pairwise disjoint,

• (2) $|A|\le |D\cap M_r(x_0)|$ and

• (3) for each i, $|A\cap R_i|>q|R_i|.$

Following the ideas of the proof of [6, Proposition 2.3], we next prove a Lévy system type formula.

Proposition 2.3

Suppose D and E are two Borel sets with ${{\,\mathrm{dist}\,}}(D,E)>0$. Then

$$\begin{aligned} \sum _{s\le t} \mathbb {1}_{\{X_{s-}\in D, X_s\in E\}} -\int _{0}^t \mathbb {1}_D (X_s) \int _E \sum _{k=1}^d \left( \frac{|(a_k^\tau (X_s))^{-1}|^{1+\alpha _k}c_{\alpha _k}}{|h_k - X_s^{k}|^{1+\alpha _k}}dh_k\left( \prod _{j\ne k}\delta _{\{X_s^{j}\}}(dh_j) \right) \right) ds \end{aligned}$$

is a ${\mathbb {P}}^x$-martingale for each x.

Proof

Let $f\in C_b^2(\mathbb {R}^d)$ with $f=0$ on D and $f=1$ on E. Moreover set

$$\begin{aligned} M_{t}^f:= f(X_t) -f(X_0)-\int _{0}^{t} {\mathcal {L}}f(X_s) ds. \end{aligned}$$

By Assumption (iii) for each $x\in \mathbb {R}^d$ the probability measure ${\mathbb {P}}^x$ is a solution to the martingale problem for ${\mathcal {L}}$. Since the stochastic integral with respect to a martingale is itself a martingale,

$$\begin{aligned} \int _{0}^t \mathbb {1}_D (X_{s-}) dM_s^f \end{aligned}$$

is a ${\mathbb {P}}^x$-martingale. Rewriting $f(X_t)-f(X_0) = \sum _{s\le t} (f(X_s)-f(X_{s-}))$ leads to

$$\begin{aligned} \sum _{s\le t} \left( \mathbb {1}_D (X_{s-}) (f(X_s)-f(X_{s-})) \right) - \int _{0}^t \mathbb {1}_D (X_{s-}) {\mathcal {L}} f(X_s) ds \end{aligned}$$

is a ${\mathbb {P}}^x$-martingale. Since $X_s\ne X_{s-}$ for only countably many values of s,

$$\begin{aligned} \sum _{s\le t} \left( \mathbb {1}_D(X_{s-}) (f(X_s)-f(X_{s-})) \right) - \int _{0}^{t} \mathbb {1}_D (X_s) {\mathcal {L}}f(X_s) ds \end{aligned}$$

(2.1)

is also a ${\mathbb {P}}^x$-martingale. Let $w=(w_1,\dots ,w_d)$ and $u=(u_1,\dots ,u_d)$. By definition of f, for $x\in D$ we have $f(x)=0$ and $\nabla f(x)=0$. Hence

$$\begin{aligned} {\mathcal {L}}f(x)&= \sum _{k=1}^{d} \int _{\mathbb {R}\setminus \{0\}} f(x+a_k(x)h) \frac{c_{\alpha _k}}{|h|^{1+\alpha _k}} dh \\&=\sum _{k=1}^d \int _{\mathbb {R}^d\setminus \{0\}} \left( f(x+A^{\tau }(x)w) \frac{c_{\alpha _k}}{|w|^{1+\alpha _k}}\left( \prod _{j\ne k} \delta _{\{0\}}(dw_j) \right) \right) dw_k \\&= \sum _{k=1}^d \int _{\mathbb {R}^d\setminus \{0\}} f(u) \frac{|(a_k^\tau (x))^{-1}|^{1+\alpha _k}c_{\alpha _k}}{|u-x|^{1+\alpha _k}}\left( \prod _{j\ne k} \delta _{\{x_j\}}(du_j) \right) du_k. \\ \end{aligned}$$

Note, that $c_{\alpha _k}/|h|^{1+\alpha _k}$ is integrable over h in the complement of any neighborhood of the origin for any $k\in \{1,\dots ,d\}$. Since D and E have a positive distance from each other, the sum in (2.1) is finite. Hence

$$\begin{aligned}&\sum _{s\le t}\left( \mathbb {1}_D (X_{s-}) (\mathbb {1}_E (X_s)-\mathbb {1}_E (X_s)) \right) \\&\quad - \int _{0}^t \mathbb {1}_D (X_s) \int _E \sum _{k=1}^d \left( \frac{|(a_j^{\tau }(X_s))^{-1}|^{1+\alpha _j}c_{\alpha _k}}{|h_k - X_s^{k}|^{1+\alpha _k}}dh_k\left( \prod _{j\ne k}\delta _{\{X_s^{j}\}}(dh_j) \right) \right) ds \end{aligned}$$

is a ${\mathbb {P}}^x$-martingale, which is equivalent to our assertion. $\square$

The next Proposition gives the behavior of the expected first exit time of the solution to (1.3) out of the set $M_r(\cdot )$. This Proposition highlights the advantage of $M_r(\cdot )$ and shows that the scaling of the cube in the different directions with respect to the jump intensity compensates the different jump intensities in the different directions.

Proposition 2.4

Let $x\in \mathbb {R}^d$ and $r\in (0,1].$ Then there exists a constant $c_1=c_1(\Lambda (M_r(x)),d)>0$ such that for all $z\in M_r(x)$

$$\begin{aligned} {\mathbb {E}}^{z}\left[ \tau _{M_r(x)}\right] \le c_1 r^{\alpha _{\max }}. \end{aligned}$$

Proof

First note

$$\begin{aligned} \begin{aligned} {\mathbb {E}}^z \left[ \tau _{M_r(y)}\right]&= {\mathbb {E}}^z \left[ \min \limits _{1\le i\le d} \inf \{ t\ge 0 : X_t^i \notin (y_i - r^{(\alpha _{\max }/\alpha _i)},y_i - r^{(\alpha _{\max }/\alpha _i)} ) \} \right] \\&\le \frac{1}{d} \sum _{i=1}^d {\mathbb {E}}^z \left[ \inf \{ t\ge 0 : X_t^i \notin (y_i - r^{(\alpha _{\max }/\alpha _i)},y_i - r^{(\alpha _{\max }/\alpha _i)} ) \} \right] \\&=: \frac{1}{d} \sum _{i=1}^d {\mathbb {E}}^z \left[ \Upsilon _i\right] . \end{aligned} \end{aligned}$$

(2.2)

Let $j\in \{1,\dots ,d\}$ be fixed but arbitrary. The aim is to show that there exists $c_2>0$ such that

$$\begin{aligned} {\mathbb {E}}^z (\Upsilon _j) \le c_2 r^{\alpha _{\max }}. \end{aligned}$$

(2.3)

Since we reduced the problem to a one-dimensional one, we may suppose by scaling $r=1$. Let

$$\begin{aligned} \kappa :=\inf \left\{ |A(x)e_j|: x\in \overline{M_1(x)} \right\} . \end{aligned}$$

By Assumption (i), we have $\kappa >0$. There exists a $c_3\in (0,1)$ with

$$\begin{aligned} {\mathbb {P}}^z(\exists s\in [0,1]:\Delta Z_s^j \in \mathbb {R}\setminus [-3/\kappa ,3/\kappa ])\ge c_3. \end{aligned}$$

The independence of the one-dimensional processes implies that with probability zero at least two of the $Z^i$’s make a jump at the same time. This leads to

$$\begin{aligned} {\mathbb {P}}^z(\exists s\in [0,1]:\Delta Z_s^j>\frac{3}{\kappa } \text { and } \Delta Z_s^i = 0 \text { for } i\in \{1,\dots ,d\}\setminus \{j\})\ge c_3. \end{aligned}$$

(2.4)

Our aim is to show that the probability of the process X for leaving $M_1(x)$ in the j$^{\text {th}}$ coordinate after time m is bounded in the following way

$$\begin{aligned} {\mathbb {P}}^{z}(\Upsilon _j>m) \le (1-k_j)^m \quad \text {for all } \ m\in \mathbb {N}. \end{aligned}$$

Suppose there exists $s\in [0,1]$ such that $\Delta Z_s^j > \frac{3}{\kappa }$, $\Delta Z_s^i=0$ for $i\in \{1,\dots ,d\}\setminus \{ j\}$, and $X_{s-}\in M_1(x).$ Then

$$\begin{aligned} |\Delta X_s^j|= |\Delta Z_s^j| \ |A(X_{s-})(e_j)| > 3. \end{aligned}$$

Note, that we leave $M_1(x)$ by this jump. By (2.4)

$$\begin{aligned} {\mathbb {P}}^{z}(\Upsilon _j\le 1) \ge c_3 \Leftrightarrow {\mathbb {P}}^{z}(\Upsilon _j>1)\le (1-c_3). \end{aligned}$$

Let $\{\theta _t : t\ge 0\}$ denote the shift operators for X. Now assume ${\mathbb {P}}^{z}(\Upsilon _j>m) \le (1-c_3)^m$. By the Markov property

$$\begin{aligned} {\mathbb {P}}^{z}(\Upsilon _j>m+1)&\le {\mathbb {P}}^{z}(\Upsilon _j>m ; \Upsilon _j \circ \theta _m>1 ) \\&= {\mathbb {E}}^{z}[{\mathbb {P}}^{X_m}(\Upsilon _j> 1); \Upsilon _j> m] \\&\le (1-c_3){\mathbb {P}}^{z}(\Upsilon _j>m) \\&\le (1-c_3)^{m+1}. \end{aligned}$$

Assertion (2.3) follows by

$$\begin{aligned} {\mathbb {E}}^{x}[\Upsilon _j] = \int _{0}^{\infty } {\mathbb {P}}^z(\Upsilon _j>t)dt \le \sum _{m=0}^\infty {\mathbb {P}}^z (\Upsilon _j>m)\le \sum _{m=0}^\infty (1-c_3)^m = c_2, \end{aligned}$$

where we used the fact that the sum on the right hand side is a geometric sum. Thus the assertion follows by (2.2) and (2.3). $\square$

We close this section by giving an estimate for leaving a rectangle with a comparatively big jump.

Proposition 2.5

Let $x\in \mathbb {R}^d$, $r\in (0,1]$ and $R\ge 2r$. There exists a constant $c_1=c_1(\Lambda (\mathbb {R}^d),d)>0$, such that for all $z\in M_r(x)$

$$\begin{aligned} {\mathbb {P}}^z (X_{\tau _{M_r(x)}}\notin M_R(x)) \le c_1 \left( \frac{r}{R}\right) ^{\alpha _{\max }}. \end{aligned}$$

Proof

Let

$$\begin{aligned} C_j:= \mathbb {R}\setminus [x_j-R^{\alpha _{\max }/\alpha _j},x_j+R^{\alpha _{\max }/\alpha _j}] \end{aligned}$$

and for $1\le j\le d$ let $k_j = \sup _{x\in \mathbb {R}} |(a_j^\tau (x))^{-1}|c_{\alpha _j}$. By Proposition 2.3 and optional stopping we get for $c_2=\sum _{j=1}^d ((2k_j2^\alpha _{\max })/c_{\alpha _j})\le 8d \sup _{x\in \mathbb {R}} |(a_j^\tau (x))^{-1}|$

$$\begin{aligned} {\mathbb {P}}^z \left( X_{t \wedge \tau _{M_r(x)}}\notin M_R(x)\right)&= {\mathbb {E}}^z \left[ \int _{0}^{t\wedge \tau _{M_r(x)}} \int _{M_R(x)^c} \sum _{j=1}^{d} \frac{|(a_j^\tau (X_s))^{-1}|c_{\alpha _j}}{|h_j-X_s^{j}|^{1+\alpha _j}} \left( \prod _{i\ne j}\delta _{\{X_s^{i}\}}(dh_i) \right) dh_j ds\right] \\&\le {\mathbb {E}}^z \left[ \int _{0}^{t\wedge \tau _{M_r(x)}} \sum _{j=1}^d \int _{C_j} \frac{k_j}{|h_j-X_s^{j}|^{1+\alpha _j}} dh_j ds \right] \\&\le {\mathbb {E}}^z \left[ \int _{0}^{t\wedge \tau _{M_r(x)}} \sum _{j=1}^d \int _{C_j} \frac{k_j}{|h_j-(x_j+r^{\alpha _{\max }/\alpha _j})|^{1+\alpha _j}} dh_j ds \right] \\&= {\mathbb {E}}^z [t\wedge \tau _{M_r(x))}] \sum _{j=1}^d \frac{2k_j}{\alpha _j (R^{\alpha _{\max }/\alpha _j}-r^{\alpha _{\max }/\alpha _j})^{\alpha _j}} \\&\le {\mathbb {E}}^z [t\wedge \tau _{M_r(x))}] \sum _{j=1}^d \frac{2k_j}{\alpha _j ((R/2)^{\alpha _{\max }/\alpha _j})^{\alpha _j}} = \frac{c_2}{R^{\alpha _{\max }}}{\mathbb {E}}^z [t\wedge \tau _{M_r(x))}]. \end{aligned}$$

Using the monotone convergence on the right and dominated convergence on the left, we have for $t\rightarrow \infty$

$$\begin{aligned} {\mathbb {P}}^z(X_{t \wedge \tau _{M_r(x)}}\notin M_R(x)) \le \frac{c_2}{R^{\alpha _{\max }}} {\mathbb {E}}^z (\tau _{M_r(x)}) \le c_2 c_3\left( \frac{r}{R}\right) ^{\alpha _{\max }}, \end{aligned}$$

where $c_3$ is the constant showing up in the estimate ${\mathbb {E}}^z (\tau _{M_r(x)})\le c_3 r^\alpha$ of Proposition 2.4. $\square$

3 The support theorem

In this section we prove the main ingredient for the proof of the Hölder regularity estimate for harmonic functions. The so-called support theorem states that sets of positive Lebesgue measure are hit with positive probability.

This theorem was first proved in [23] for the diffusion case. In the article [4], Bass and Chen prove the support theorem in the context of pure jump processes with singular and anisotropic kernels. They consider the system (1.3) in the case $\alpha _i=\alpha$ for all $i\in \{1,\dots ,d\}$ and use the technique by Krylov and Safonov to prove Hölder regularity with the help of the support theorem.

The idea we use to prove the support theorem is similar in spirit to the one in [4].

The following Lemma is a statement about the topological support of the law of the stopped process. It gives the existence of a bounded stopping time T such that with positive probability the stopped process stays in a small ball around its starting point up to time T, makes a jump along the k$^{\text {th}}$ coordinate axis and stays afterwards in a small ball.

Lemma 3.1

Let $r\in (0,1]$, $x_0\in \mathbb {R}^d, k\in \{1,\dots ,d\}, v_k=A(x_0)e_k,$ $\gamma \in (0,r^{\alpha _{\max }/\alpha _{\min }}), t_0>0$ and $\xi \in [-r^{\alpha _{\max }/\alpha _{\min }},r^{\alpha _{\max }/\alpha _{\min }}]$. There exists a constant $c_1>0= c_1(\gamma ,t_0,\xi ,r,\Lambda (M^{2}_r(x_0))), \varpi (M^{2}_r(x_0)))>0$ and a stopping time $T\le t_0$, such that

$$\begin{aligned} \begin{aligned} {\mathbb {P}}^{x_0} \left( \sup \limits _{s<T}|X_s-x_0|<\gamma \text { and } \sup \limits _{T\le s \le t_0} |X_s - (x_0+\xi v_k)|<\gamma \right) \ge c_1. \end{aligned} \end{aligned}$$

(3.1)

Proof

Let

$$\begin{aligned} \Vert A\Vert _\infty := 1 \vee \left( \sum _{i,j=1}^{d} \sup \limits _{x\in M^{2}_r(x_0)} |A_{ij}(x)|\right) . \end{aligned}$$

We assume $\xi \in [0,r^{\alpha _{\max }/\alpha _{\min }}]$. The case $\xi \in [-r^{\alpha _{\max }/\alpha _{\min }},0]$ can be proven similar. Let us first suppose $\xi \ge \gamma /(3\Vert A\Vert _\infty )$ and let $\beta \in (0,\xi )$, which will be chosen later. We decompose the process $Z_t^i$ in the following way:

$$\begin{aligned} {\widetilde{Z}}_t^i = \sum _{s\le t} \Delta Z_s^i \mathbb {1}_{\{|\Delta Z_s^i|>\beta \}}, \quad {\overline{Z}}_t^i = Z_t^i - {\widetilde{Z}}_t^i. \end{aligned}$$

Let $({\overline{X}}_t)_{t\ge 0}$ be the solution to

$$\begin{aligned} d{\overline{X}}^i_{t} = \sum _{j=1}^d A_{ij}({\overline{X}}_{t-})d{\overline{Z}}_{t}^j, X_0^i = x_0^i. \end{aligned}$$

The continuity of A allows us to find a $\ \delta < \gamma / (6\Vert A\Vert _\infty )$, such that

$$\begin{aligned} \begin{aligned} \sup \limits _{i,j}\sup \limits _{|x-x_0|<\delta } |A_{ij}(x)-A_{ij}(x_0)|<\frac{\gamma }{12d}. \end{aligned} \end{aligned}$$

(3.2)

Consider

$$\begin{aligned} C&= \left\{ \sup \limits _{s\le t_0}|{\overline{X}}_s - {\overline{X}}_0|\le \delta \right\} , \\ D&= \Big \{ {\widetilde{Z}}^k \text { has precisely one jump before time } t_0 \text { with jump size in } [\xi ,\xi +\delta ],\\&\qquad \Delta {\widetilde{Z}}_s^j = 0 \text { for all } s\le t_0 \text { and all } j\ne k \Big \}, \\ E&= \left\{ {\widetilde{Z}}_s^i=0 \text { for all } s\le t_0 \text { and } i=1,\dots ,d \right\} . \end{aligned}$$

Since A is bounded, we can find $c_2>0$, such that

$$\begin{aligned} \left[ {\overline{X}}^i, {\overline{X}}^i \right] _t\le c_2 \sum _{j=1}^d \left[ {\overline{Z}}^j, {\overline{Z}}^j\right] _t. \end{aligned}$$

Note, that $\beta \in (0,\xi )\subset (0,r^{\alpha _{\max }/\alpha _{\min }})\subset (0,1)$. Therefore, we get

$$\begin{aligned} {\mathbb {E}}^{x_0}\left[ {\overline{X}}^i,{\overline{X}}^i\right] _t&\le c_2 \sum _{j=1}^d {\mathbb {E}}^{x_0}\left[ {\overline{Z}}^j,{\overline{Z}}^j\right] _t = c_2 \sum _{j=1}^d \int _{0}^t\left( \int _{-\beta }^{\beta } \frac{c_{\alpha _j}h^2}{|h|^{1+\alpha _j}} dh\right) dt \le c_3 t d \beta ^{2-\alpha _{\max }}. \end{aligned}$$

By Tschebyscheff’s inequality and Doob’s inequality, we get

$$\begin{aligned} {\mathbb {P}}^{x_0}\left[ \sup \limits _{s\le t_0} |{\overline{X}}_s^i-{\overline{X}}_0^i|>\delta \right] \le \frac{1}{\delta ^2}{\mathbb {E}}^{x_0}\left[ \sup \limits _{s\le t_0} \left( {\overline{X}}_s^i - {\overline{X}}_0^i \right) ^2 \right]\le & {} \frac{1}{\delta ^2} 4 {\mathbb {E}}^{x_0}\left[ \left( {\overline{X}}_{t_0}^i - {\overline{X}}_0^i \right) ^2\right] \\\le & {} \frac{c_4 t_0 d \beta ^{2-\alpha _{\max }}}{\delta ^2}. \end{aligned}$$

Choose $\beta \in (0,\xi )$ such that

$$\begin{aligned} \begin{aligned} c_5 t_0 \beta ^{2-\alpha _{\max }} \le \frac{\delta ^2}{2d} \end{aligned} \end{aligned}$$

(3.3)

holds. Then by (3.3), we get

$$\begin{aligned} {\mathbb {P}}^{x_0}(C) = 1- {\mathbb {P}}^{x_0} \left( \sup \limits _{s\le t_0} |{\overline{X}}_s^i-{\overline{X}}_0^i|>\delta \right) \ge \frac{1}{2}. \end{aligned}$$

(3.4)

For ${\widetilde{Z}}^k$ to have a single jump before time $t_0$, and for that jump’s size to be in the interval $[\xi ,\xi +\delta ]$, then up to time $t_0$ ${\widetilde{Z}}^k_t$ must have

• (i) no negative jumps,

• (ii) no jumps whose size lies in $[\beta ,\xi ),$

• (iii) no jumps whose size lies in $(\xi +\delta ,\infty ),$

• (iv) precisely one jump whose size lies in the interval $[\xi ,\xi +\delta ]$.

We can use the fact, that ${\widetilde{Z}}^k$ is a compound Poisson process and use the knowledge about Poisson random measures. The events descriped in (i)-(iv) are the probabilities that Poisson random variables $P_1,P_2.P_3$ and $P_4$ of parameters $\lambda _1=c_6 t_0 \beta ^{-\alpha _k}$, $\lambda _2=c_6 t_0 (\beta ^{-\alpha _k}-\xi ^{-\alpha _k})$, $\lambda _3=c_6 t_0 (\xi +\delta )^{-\alpha _k}$, and $\lambda _4=c_6 t_0 (\xi ^{-\alpha _k} - (\xi +\delta )^{-\alpha _k})$, respectively, take the values 0, 0, 0,  and 1, respectively.

So there exists a constant $c_7=c_7(\alpha _k,t_0,\delta , \xi , \beta )>0$ such that

$$\begin{aligned} {\mathbb {P}}^{x_0}\big (&{\widetilde{Z}}^k \text { has a single jump before time}~ t_0, ~\text {and its size is in } [\xi ,\xi +\delta ]\big ) \ge c_7. \end{aligned}$$

For all $j\ne k$, the probability that ${\widetilde{Z}}^{j}$ does not have a jump before time $t_0$, is the probability that a Poisson random variable with parameter $2c_6 t_0\beta ^{-\alpha _j}$ is equal to 0. Using the indepence of ${\widetilde{Z}}^j$ for $j=1,\dots ,d$, we can find a $c_8>0$ such that

$$\begin{aligned} {\mathbb {P}}^{x_0}(\Delta {\widetilde{Z}}^j_s = 0 \text { for all } s\le t_0 \text { and all } j\ne k) \ge c_8. \end{aligned}$$

Thus we obtain

$$\begin{aligned} {\mathbb {P}}^{x_0}(D)\ge c_9 \end{aligned}$$

for a $c_9=c_9(\alpha _1,\dots ,\alpha _d,t_0,\delta , \xi , \beta )>0$. Furthermore the ${\overline{Z}}^i$’s are independent of the ${\widetilde{Z}}^j$’s for all $i,j\in \{1,\dots ,d\}$, so C and D are independent and we obtain

$$\begin{aligned} {\mathbb {P}}^{x_0}(C\cap D)\ge c_9/2. \end{aligned}$$

Similary we obtain

$$\begin{aligned} {\mathbb {P}}^{x_0}(E)\ge c_{10} \quad \text {and} \quad {\mathbb {P}}^{x_0}(C\cap E)\ge c_{11}. \end{aligned}$$

(3.5)

Let T be the time, when ${\widetilde{Z}}^k$ jumps the first time, i.e. $Z^k$ makes a jump greater then $\beta$. Then $Z_{s-}={\overline{Z}}_{s-}$ for all $s\le T$ and hence $X_{s-}={\overline{X}}_{s-}$ for all $s\le T.$ So up to time T, $X_s$ does not move away more than $\delta$ away from its starting point. Note $\Delta X_T = A(X_{T-})\Delta Z_T.$ By (3.2), we obtain on $C\cap D$

$$\begin{aligned} |X_T-(x_0+\xi v_k)|&\le |X_{T-}-x_0| + |\Delta X_T - \xi A(x_0)e_k))| \\&= |X_{T-}-x_0| + | A(X_{T-})\Delta Z_T - \xi A(x_0)e_k) | \\&\le |X_{T-}-x_0| + \xi |(A(X_{T-})-A(x_0))e_k| + |A(X_{T-})(\Delta Z_T -\xi e_k)| \\&\le \delta + \frac{\xi d \gamma }{12d} + \delta \Vert A\Vert _{\infty } \le \frac{\gamma }{6} \left( \frac{1}{\Vert A\Vert _\infty } + \frac{\gamma }{2} + 1\right) \le \frac{\gamma }{2}. \end{aligned}$$

Appling the strong Markov property at time T, we get by (3.5)

$$\begin{aligned} {\mathbb {P}}^{x_0}\left( \sup \limits _{T\le s\le T+t_0} |X_s-X_T|<\delta \right) \ge {\mathbb {P}}^{X_T} (C\cap E) \ge c_{11}. \end{aligned}$$

Note, that $|X_T-(x_0+\xi v_k)|<\gamma /2$ and $|X_s-X_T|<\delta$ for all $T\le s \le t_0$ imply $|X_s-(x_0+\xi v_k)|<\gamma .$

All in all we get by the strong Markov property

$$\begin{aligned} {\mathbb {P}}^{x_0} \left( \sup \limits _{s<T}|X_s-x_0|<\gamma \text { and } \sup \limits _{T\le s \le t_0} |X_s - (x_0+\xi v_k)|<\gamma \right) \ge \frac{c_{9}c_{11}}{2}, \end{aligned}$$

which proves the assertion.

Now suppose $\xi <\gamma /(3\Vert A\Vert _\infty$). Then $|x_0-(x_0+\xi v_k)|<\gamma /3$. We can choose $T\equiv 0$ and by (3.5) we get:

$$\begin{aligned} {\mathbb {P}}^{x_0}\left( \sup \limits _{s\ge t_0} |X_s-x_0|<\delta \right) \ge c_{11}, \end{aligned}$$

which finishes the proof. $\square$

We need two simple geometrical facts from the field of linear algebra, whose proofs can be found in [4] (Lemmas 2.4 and 2.5).

Lemma 3.2

Suppose u, v are two vectors in $\mathbb {R}^d$, $\eta \in (0,1)$, and p is the projection of v onto u. If $|p|\ge \eta |v|$, then

$$\begin{aligned} |v-p|\le \sqrt{1-\eta ^2}|v|. \end{aligned}$$

Lemma 3.3

Let v be a vector in $\mathbb {R}^d$, $u_k = Ae_k$, and $p_k$ the projection of v onto $u_k$ for $k=1,\dots ,d.$ Then there exists $\rho =\rho (\Lambda (\mathbb {R}^d))\in (0,1)$, such that for some k,

$$\begin{aligned} |v-p_k|\le \rho |v|. \end{aligned}$$

For a given time $t_1>0$ the following lemma shows that solutions stay with positive probability in an $\varepsilon$-tube around a given line segment on $[0,t_1]$. The case of $\alpha _1=\cdots =\alpha _d$ was considered in [4]. We follow their technique.

Lemma 3.4

Let $r\in (0,1]$, $x_0\in \mathbb {R}^d$, $t_1>0, \varepsilon \in (0,r^{\alpha _{\max }/\alpha _{\min }}), \xi \in (0,\varepsilon /4)$ and $\gamma >0.$ Moreover let $\psi :[0,t_1]\rightarrow \mathbb {R}^d$ be a line segment of length $\xi$ starting at $x_0$. Then there exists $c_1=c_1(\Lambda (M_r^2(x_0))), \varpi (M_r^2(x_0))),t_1, \varepsilon , \gamma )>0$, such that

$$\begin{aligned} {\mathbb {P}}^{x_0}\left( \sup \limits _{s\le t_1} |X_s - \psi (s)|<\varepsilon \text { and } |X_{t_1}-\psi (t_1)|<\gamma \right) \ge c_1. \end{aligned}$$

Proof

Note that $\varepsilon$ is chosen such that $B_{\varepsilon }(x_0)\subset M_r(x_0)$. Let $\rho \in (0,1)$ be such that the conclusion of Lemma 3.3 holds for all matrices $A=A(x)$ with $x\in M_r^{2}(x).$ Take $\gamma \in (0,\xi \wedge \rho )$ such that ${\widetilde{\rho }}:=\gamma + \rho <1$ and $n\ge 2$ sufficiently large, such that $({\widetilde{\rho }})^n < \gamma .$ Let $v_0:=\psi (t_1)-\psi (0)=\psi (t_1)-x_0,$ which has length $\xi .$ By Lemma 3.2, there exists a $k_0\in \{1,\dots ,d\}$ such that if $p_0$ is the projection of $v_0$ onto $A(x_0)e_{k_0}$, then $|v_0-p_0|\le \rho |v_0|.$ Note, that $|p_0|\le |v_0|=\xi .$ By Lemma 3.1 there exists $c_2>0$ and a stopping time $T_0\le t_1/n$ such that for

$$\begin{aligned} D_1:= \left\{ \sup \limits _{s<T_0}|X_s-x_0|<\gamma ^{n+1} \text { and } \sup \limits _{T_0\le s \le t_1/n}|X_s - (x_0+p_0)|<\gamma ^{n+1} \right\} . \end{aligned}$$

the estimate

$$\begin{aligned} {\mathbb {P}}^{x_0}(D_1)\ge c_2 \end{aligned}$$

holds. Since $\gamma <1$ and $\gamma ^n\le \gamma$ for all $n\in \mathbb {N}$, we have for $T_0 \le s \le t_1/n$

$$\begin{aligned} \begin{aligned} |\psi (t_1)-X_s|&\le |\psi (t_1)-(x_0+p_0)| + |(x_0+p_0)-X_s| \\&\le |v_0-p_0| + \gamma ^{n+1} = \rho \xi + \gamma ^{n+1} \le {\widetilde{\rho }}\xi \end{aligned} \end{aligned}$$

(3.6)

on $D_1$. Taking $s=t_1/n$, we have

$$\begin{aligned} |\psi (t_1)-X_{t_1/n}|\le {\widetilde{\rho }}\xi . \end{aligned}$$

Since ${\widetilde{\rho }}<1$ and $|\psi (t_1)-x_0|=|v_0|=\xi ,$ then (3.6) shows that on $D_1$

$$\begin{aligned} X_s\in B(x_0,2\xi )\subset B(x_0, \varepsilon /2) \quad \text {if } T_0\le s\le t_1/n. \end{aligned}$$

If $0\le s<T_0,$ then $|X_s-x_0|<\gamma ^{n+1}<\xi ,$ and so we have on $D_1$

$$\begin{aligned} \{ X_s, s\in [0,t_1/n] \}\subset B(x_0,2\xi )\subset B(x_0,\varepsilon /2). \end{aligned}$$

Now let $v_1:=\psi (t_1)-X_{t_1/n}.$ When $X_{t_1/n}\in B(x_0, \varepsilon /2),$ then by Lemma 3.3, there exists $k_1\in \{1,\dots ,d\}$ such that if $p_1$ is the projection of $v_1$ onto $A(X_{t_1/n})e_{k_1},$ then $|v_1-p_1|\le \rho |v_1|.$ Let $T_1\in [t_1/n, 2t_1/n]$ be a stopping time, determined by Lemma 3.1, and

$$\begin{aligned} D_2:= \left\{ \sup \limits _{t_1/n\le s<T_1}|X_s-X_{t_1/n}|<\gamma ^{n+1} \text { and } \sup \limits _{T_1\le s \le 2t_1/n}|X_s - (X_{t_1/n}+p_1)|<\gamma ^{n+1} \right\} . \end{aligned}$$

By the Markov property at the time $t_1/n$ and Lemma 3.1, there exists the same $c_2>0$ such that

$$\begin{aligned} {\mathbb {P}}^{x_0}(D_2|{\mathcal {F}}_{t_1/n})\ge c_2 \end{aligned}$$

on the event $\{ X_{t_1/n}\in B(x_0, \varepsilon /2) \}$ and hence especially on $D_1$. So

$$\begin{aligned} {\mathbb {P}}^{x_0}(D_1\cap D_2) \ge c_2 {\mathbb {P}}^{x_0}(D_1)\ge c_2^2. \end{aligned}$$

Let $s\in [T_1,2t_1/n]$. Then on $D_1\cap D_2$

$$\begin{aligned} |\psi (t_1)-X_s|&\le |\psi (t_1)-(X_{t_1/n}+p_1)|+|(X_{t_1/n}+p_1)-X_s| \\&\le \rho |v_1| + \gamma ^{n+1} \le \rho {\widetilde{\rho }}\xi +\gamma ^{n+1}\le ({\widetilde{\rho }})^2 \xi + \gamma ^{n+1}\\&\le {\widetilde{\rho }}^2\xi . \end{aligned}$$

In particular, by choosing $s=2t_1/n$, we get on $D_1\cap D_2$

$$\begin{aligned} |\psi (t_1)-X_{2t_1/n}| \le ({\widetilde{\rho }})^2\xi . \end{aligned}$$

On $D_1\cap D_2$, we have for $s\in [T_1,2t_1/n]$: $|\psi (t_1)-X_s|<\xi$ and $|\psi (t_1)-x_0|=\xi$, which implies

$$\begin{aligned} X_s\in B(x_0,2\xi )\subset B(x_0,\varepsilon /2)\quad \text { on } D_1\cap D_2. \end{aligned}$$

In particular,

$$\begin{aligned} |X_{2t_1/n}-x_0|<2\xi \quad \text { on } D_1\cap D_2. \end{aligned}$$

If $s\in [t_1/n,T_1]$, then $|X_s-X_{t_1/n}|<\xi$ and $|X_{t_1/n}-x_0|<2\xi$ on $D_1\cap D_2$, which yields to

$$\begin{aligned} X_s\in B(x_0,3\xi )\subset B(x_0,3\varepsilon /4)\quad \text { on } D_1\cap D_2. \end{aligned}$$

Let $v_2:=\psi (t_1)-X_{2t_1/n},$ and proceed as above to get events $D_3,\dots ,D_k$ for $k\le n$. At the k$^{\text {th}}$ stage

$$\begin{aligned} {\mathbb {P}}^{x_0}(D_k | F_{(k-1)t_1/n})\ge c_2 \quad \text { and so } \quad {\mathbb {P}}^{x_0}\left( \bigcap _{j=1}^k D_j\right) \ge c_2^k. \end{aligned}$$

For $kt_1/n \le T_k \le s \le (k+1)t_1/n$

$$\begin{aligned} |\psi (t_1)-X_s|\le ({\widetilde{\rho }})^{k+1}\xi <\xi ; \end{aligned}$$

on the event $\bigcap _{j=1}^k D_j.$ Since $|\psi (t_1)-x_0|=\xi .$

$$\begin{aligned} X_s\in B(x_0,2\xi )\subset B(x_0,\varepsilon /2)\quad \text {on } \bigcap _{j=1}^k D_j. \end{aligned}$$

If $kt_1/n \le s < T_k$, we obtain at the kth stage

$$\begin{aligned} |X_{kt_1/n}-x_0|<\varepsilon /2 \quad \text {on } \bigcap _{j=1}^k D_j. \end{aligned}$$

Thus

$$\begin{aligned} |X_s-x_0|\le |X_s-X_{kt_1/n}|+|X_{kt_1/n}-\psi (t_1)|+|\psi (t_1)-x_0|\le \gamma ^{n+1}+\xi +\xi < 3\xi , \end{aligned}$$

and therefore $X_s\in B(x_0,3\xi )\subset B(x_0,3\varepsilon /4).$ We continue this procedure n times to get events $D_1,\dots ,D_n$. On $\bigcap _{k=1}^n D_k$, we have

• (1) $X_s\in B(x_0,3\xi )$ for $s\le t_1$,

• (2) $|X_{t_1}-\psi (t_1)|<({\widetilde{\rho }})^n\xi<({\widetilde{\rho }})^n<\gamma ,$ and

• (3) $\displaystyle {\mathbb {P}}^{x_0}\left( \bigcap _{j=1}^n D_j\right) \ge c_2^n.$

For $s\in [0,t_1]$,

$$\begin{aligned} |X_s-\psi (s)|\le |X_s-x_0|+|x_0-\psi (s)|<3\xi +\xi < \frac{3}{4}\varepsilon + \frac{1}{4}\varepsilon =\varepsilon \quad \text { on } \bigcap _{j=1}^n D_j. \end{aligned}$$

Hence

$$\begin{aligned} {\mathbb {P}}^{x_0}\left( \sup \limits _{s\le t_1} |X_s - \psi (s)|<\varepsilon \text { and } |X_{t_1}-\psi (t_1)|<\gamma \right) \ge {\mathbb {P}}^{x_0}\left( \bigcap _{j=1}^n D_j\right) \ge c_2^n=: c_1. \end{aligned}$$

$\square$

We can now prove an important theorem, which will be the main ingredient in the proof of the Hölder regularity. It states that the solution to (1.3) stays with positive probability in a $\varepsilon$-tube around a given continuous function.

Theorem 3.5

Let $r\in (0,1], x_0\in \mathbb {R}^d, \varepsilon \in (0,r^{\alpha _{\max }/\alpha _{\min }}), t_0>0$ and $x_0\in \mathbb {R}^d$. Let $\varphi :[0,t_0]\rightarrow \mathbb {R}^d$ be continuous with $\varphi (0)=x_0$ and the image of $\varphi$ contained in $M_r(x_0).$ Then there exists $c_1=c_1(\Lambda (M^2_r(x_0)),\varpi (M^3_r(x_0)),\varphi ,\varepsilon , t_0)>0$ such that

$$\begin{aligned} {\mathbb {P}}^{x_0}\left( \sup \limits _{s\le t_0} |X_s - \varphi (s)|<\varepsilon \right) > c_1. \end{aligned}$$

Proof

Let $\varepsilon >0.$ We define

$$\begin{aligned} U:=\{x\in \mathbb {R}^d : \exists s\in [0,t_0] \text { such that } |x-\varphi (s)|<\varepsilon /2 \} \end{aligned}$$

and approximate $\varphi$ within U by a polygonal path. Hence we can assume that $\varphi$ is polygonal by changing $\varepsilon$ to $\varepsilon /2$ in the assertion. We subdivide $[0,t_0]$ into n subintervals of the same length for $n\ge 2$ such that

$$\begin{aligned} r:= L\left( \varphi \left( \left( \frac{kt_0}{n},\frac{(k+1)t_0}{n} \right) \right) \right) <\frac{\varepsilon }{4}, \end{aligned}$$

where L denotes the length of the line segment. Let

$$\begin{aligned} D_k := \left\{ \sup \limits _{(k-1)t_0/n \le s \le kt_0 / n} |X_s-\varphi (s)|<\frac{\varepsilon }{2} \text { and } |X_{kt_0/n}-\varphi (kt_0/n)|<\frac{\varepsilon }{4\sqrt{d}} \right\} . \end{aligned}$$

Using Lemma 3.4, there exists a constant $c_2>0$ such that

$$\begin{aligned} {\mathbb {P}}^{x_0} (D_1)\ge c_2. \end{aligned}$$

By the strong Markov property at time $t_0/n$ we get

$$\begin{aligned} {\mathbb {P}}^{x_0} (D_2|{\mathcal {F}}_{t_0/n})\ge c_2. \end{aligned}$$

Using the Iteration as in the proof of Lemma 3.4, we get for all $k\in \{1,\dots ,d\}$

$$\begin{aligned} {\mathbb {P}}^{x_0} (D_k|{\mathcal {F}}_{(k-1)t_0/n})\ge c_2 \quad \text {and} \quad {\mathbb {P}}^{x_0} \left( \bigcap _{k=1}^n D_k \right) \ge c_2^n. \end{aligned}$$

Hence the assertion follows by

$$\begin{aligned} {\mathbb {P}}^{x_0}\left( \sup \limits _{s\le t_0} |X_s - \varphi (s)|<\frac{\varepsilon }{2} \right) \ge {\mathbb {P}}^{x_0}\left( \bigcap _{k=1}^n D_k\right) \ge c_2^n=c_1. \end{aligned}$$

$\square$

We state two corollaries, which follow immediately from Theorem 3.5.

Corollary 3.6

Let $r\in (0,1]$, $\varepsilon \in (0,r^{\alpha _{\max }/\alpha _{\min }}/4)$, $k=1-(\varepsilon /r^{\alpha _{\max }/\alpha _{\min }})$, $\delta \in (\varepsilon ,r^{\alpha _{\max }/\alpha _{\min }}/2)$ and $x_0\in \mathbb {R}^d$. Moreover let $Q:=M_r(x_0), Q':=M_r^{k}(x_0)$ and $y\in \mathbb {R}^d$ such that $R:=M_r^{\delta }(y)\subset Q'$. There exists $c_1=c_1(\Lambda (Q),\varpi (Q),\varepsilon , \delta )>0$ such that

$$\begin{aligned} {\mathbb {P}}^{x}(T_R<\tau _Q)\ge c_1, \quad x\in Q'. \end{aligned}$$

Proof

Note, that

$$\begin{aligned} {{\,\mathrm{dist}\,}}(\partial Q, \partial Q') = \left| r^{\alpha _{\max }/\alpha _i} - kr^{\alpha _{\max }/\alpha _i}\right| = |\varepsilon \frac{r^{\alpha _{\max }/\alpha _i}}{r^{\alpha _{\max }/\alpha _{\min }}}|\ge \varepsilon . \end{aligned}$$

Let $x\in Q'$ be arbitrary and $\varphi :[0,t_0]\rightarrow \mathbb {R}^d$ be a polygonal path such that $\varphi (0)=x$ and $\varphi (t_0)=y$ and the image of $\varphi$ is contained in $Q'$. Then the assertion follows by Theorem 3.5 and

$$\begin{aligned} {\mathbb {P}}^x \left( \sup \limits _{s\le t_o} |X_s-\varphi (s)|<\varepsilon \right) \le {\mathbb {P}}^x (T_R<\tau _Q). \end{aligned}$$

$\square$

Corollary 3.7

Let $r\in (0,1]$, $x_0\in \mathbb {R}^d$ and $\varepsilon \in (0,r^{\alpha _{\max }/\alpha _{\min }}/4).$ For $x\in M_r(x_0)$, we define $R:=M_s(x)$ such that $R\subset M_r(x_0)=:M$ and ${{\,\mathrm{dist}\,}}(\partial R, \partial M)>\varepsilon$. Then there exists $\xi =\xi (\varepsilon , \Lambda (M), r, \varpi (M)\in (0,1)$ such that

$$\begin{aligned} {\mathbb {P}}^y(T_{R}<\tau _{M})\ge \xi =\xi (\epsilon ). \end{aligned}$$

for all $y\in M$ with ${{\,\mathrm{dist}\,}}(y,\partial M)>\varepsilon$.

Proof

Follows immediately by Corollary 3.6. $\square$

We now prove the main ingredient for the proof of the Hölder regularity. It states, that sets of positive Lebesgue measure are hit with positive probability.

Theorem 3.8

Let $r\in (0,1]$, $x_0\in \mathbb {R}^d$ and $M:=M_r(x_0).$ There exists a nondecreasing function $\varphi :(0,1)\rightarrow (0,1)$ such that

$$\begin{aligned} {\mathbb {P}}^x (T_A<\tau _{M})\ge \varphi (|A|) \end{aligned}$$

for all $x\in M_r^{1/2}(x_0)$ and all $A\subset M$ with $|A|>0$.

Proof

We will follow the proof of Theorem V.7.4 in [2]. Set

$$\begin{aligned} \varphi (\varepsilon )=\inf \left\{ {\mathbb {P}}^y (T_A<\tau _{M_R(z_0)}) : z_0\in \mathbb {R}^d, R>0, y\in M_R^{1/2}(z_0), |A|\ge \varepsilon |M_R(z_0)|, A\subset M_R(z_0). \right\} \end{aligned}$$

and

$$\begin{aligned} q_0:=\inf \{ \varepsilon : \varphi (\varepsilon )>0 \}. \end{aligned}$$

For $\epsilon$ sufficiently large, we know by Corollary 3.6$\varphi (\epsilon )>0$. We suppose $q_0>0$, and we will obtain our contradiction.

Since $q_0<1$, we can choose $1>q>q_0$ such that $(q+q^2)/2<q_0$. Moreover let

$$\begin{aligned} \eta := (q-q^2)/2, \quad \beta := \left( \frac{2^{1-d} q r^{-\sum _{i=1}^d (\alpha _{\max }/\alpha _i)}}{q+1}\right) ^{1/d} \quad \text {and} \quad \rho =\xi ((1-\beta )r^{\alpha _{\max }/\alpha _{\min }}/6), \end{aligned}$$

where $\xi$ is defined as in Corollary 3.7.

Let $z\in \mathbb {R}^d, R\in (0,1]$, $x\in M_R^{1/2}(z)$ and $A\subset M_R(z)$ such that

$$\begin{aligned} q-\eta< \frac{|A|}{|M_R(z)|}<q \ \text { and } \ {\mathbb {P}}^x(T_A<\tau _{M_R(z)})<\rho \varphi (q)^2. \end{aligned}$$

(3.7)

Without loss of generality, set $R=r$ and $z=x_0$. Hence

$$\begin{aligned} {\mathbb {P}}^x(T_A<\tau _{M_r(x_0)})<\rho \varphi (q)^2. \end{aligned}$$

By Proposition 2.2 there exists a set $D\subset M_r(x_0)$ such that

$$\begin{aligned} |A|\le q|D\cap M_r(x_0)|. \end{aligned}$$

Since $|A|>(q-\eta )|M_r(x_0)|$,

$$\begin{aligned} |D\cap M_r(x_0)| \ge \frac{|A|}{q}>\frac{(q-\eta )|M_r(x_0)|}{q} = \frac{(q+1)|M_r(x_0)|}{2}.\end{aligned}$$

Define $E=D\cap M_r^\beta (x_0)$. Since

$$\begin{aligned} \frac{(q+1)|M_{r}^\beta (x_0)|}{2}=\frac{(q+1)2^d \beta ^d r^{\sum _{i=1}^d \frac{\alpha _{\max }}{\alpha _i}}}{2} = q, \end{aligned}$$

we get $|E|>q.$ By the definition of $\varphi$, we have ${\mathbb {P}}^x(T_{E}<\tau _{M_r(x_0)})\ge \varphi (q).$

We will first show

$$\begin{aligned} {\mathbb {P}}^y (T_A<\tau _{M_r(x_0)})\ge \rho \varphi (q) \quad \text {for all } y\in E. \end{aligned}$$

(3.8)

Let $y\in \partial E$, then $y\in {\widehat{R}}_i$ for some $R_i\in {\mathcal {R}}$ and ${{\,\mathrm{dist}\,}}(y,\partial M_r(x_0))\ge (1-\beta )r^{\alpha _{\max }\alpha _{\min }}$. Define $R_i^*$ as the cube with the same center as $R_i$ but sidelength half as long. By Corollary 3.7

$$\begin{aligned} {\mathbb {P}}^y (T_{R_i^*}<\tau _{M_r(x_0)})\ge \rho . \end{aligned}$$

By Proposition 2.2 (3) for all $R_i\in {\mathcal {R}}$

$$\begin{aligned}|A\cap R_i|\ge q|R_i|\end{aligned}$$

and therefore

$$\begin{aligned}{\mathbb {P}}^{x_0}(T_{A\cap R_i}<\tau _{M_r(x_0)})\ge \varphi (q) \quad \text { for } x_0\in R_i^*.\end{aligned}$$

Using the strong Markov property, we have for all $y\in E$

$$\begin{aligned} {\mathbb {P}}^{y}(T_A<\tau _{M_r(x_0)})&\ge {\mathbb {E}}^{y} \left[ {\mathbb {P}}^{X_{T_{R_i^*}}}(T_A<\tau _{R_i}); T_{R_i^*}<\tau _{M_r(x_0)}\right] \\&\ge \rho \varphi (q). \end{aligned}$$

Now we get our contradiction by

$$\begin{aligned} {\mathbb {P}}^x(T_A<\tau _{M_r(x_0)})&\ge {\mathbb {P}}^x (T_{E}<T_A<\tau _{M_r(x_0)}) \\&\ge {\mathbb {E}}^x \left[ {\mathbb {P}}^{X_{T_{E}}}(T_A< \tau _{M_r(x_0)});T_{E}<\tau _{M_r(x_0)}\right] \\&\ge \rho \varphi (q) {\mathbb {P}}^x \left[ T_{E}<\tau _{M_r(x_0)}\right] \ge \rho \varphi (q)^2. \end{aligned}$$

$\square$

4 Proof of Theorem 1.2

In this section we prove our main result.

Proof

Let $S:=M_s(y)\subset M_r(x_0)$ and $A\subset S$ such that $3|A|\ge |S|.$ Those sets will be specified later. Set $k=1-(\varepsilon / r^{\alpha _{\max }/\alpha _{\min }})$ and $S':=M_s^k(y)$, where $\varepsilon$ is chosen such that $6|S\setminus S'|=|S|.$ Then

$$\begin{aligned} 6|A\cap S'| \ge |S|. \end{aligned}$$

Let ${\mathcal {R}}$ be a collection of N equal sized rectangles as in Definition 1.1 with disjoint interiors and ${\mathcal {R}}\subset S$. Moreover let ${\mathcal {R}}$ to be a covering of $S'$. For at least one rectangle $Q\in {\mathcal {R}}$

$$\begin{aligned} 6|A\cap S' \cap Q|\ge |Q|. \end{aligned}$$

Let $Q'$ be the rectangle with the same center as Q but each sidelength half as long. By Corollary 3.6 there exists a $c_2>0$ such that

$$\begin{aligned} {\mathbb {P}}^x(T_{Q'}<\tau _S) \ge c_2, \quad x\in M_s^{1/2}(y). \end{aligned}$$

(4.1)

Using Theorem 3.8 and the strong Markov property there exists a constant $c_3>0$ with

$$\begin{aligned} {\mathbb {P}}^x(T_A<\tau _S)\ge c_3, \quad x\in M_s^{1/2}(y). \end{aligned}$$

(4.2)

Let $R\ge 2r$. By Proposition 2.5 there exists a $c_4>0$ such that

$$\begin{aligned} {\mathbb {P}}^z \left( X_{\tau _{M_r(x_0)}}\notin M_R(x_0) \right) \le c_4 \left( \frac{r}{R} \right) ^\alpha _{\max } \quad \text {for all } z\in M_r(x_0). \end{aligned}$$

Let

$$\begin{aligned}\gamma := (1-c_3), \quad \text { } \quad \rho :=\left( \frac{c_3\gamma ^2}{4c_4} \right) ^{1/\alpha _{\max }}\wedge \left( \frac{\gamma }{2} \right) ^{1/\alpha _{\max }} \quad \text {and} \quad \beta := \frac{\log (\gamma )}{\log (\rho )}. \end{aligned}$$

By linearity it suffices to suppose $0\le h \le M$ on $\mathbb {R}^d.$ We first consider the case $r=1$.

Let $M_i = M_{\rho ^i}(x_0)$ and $\tau _i = \tau _{M_i}.$ We will show that for all $k\in \mathbb {N}_{0}$

$$\begin{aligned} \mathop {{{\,\mathrm{osc}\,}}}\limits _{M_k} h := \mathop {\sup }\limits _{M_k} h - \mathop {\inf }\limits _{M_k} h \le M\gamma ^k. \end{aligned}$$

(4.3)

To shorten notation, we set $a_i = \inf \nolimits _{M_i} h$ and $b_i = \sup \nolimits _{M_i} h.$ Assertion (4.3) will be will be proved by induction. Let $k\in \mathbb {N}$ be arbitrary but fixed. We suppose $b_i-a_i\le M\gamma ^i$ for all $i\le k$; then we need to show

$$\begin{aligned} b_{k+1}-a_{k+1}\le M\gamma ^{k+1}. \end{aligned}$$

(4.4)

By definition $M_{k+1}\subset M_k$ and therefore in particular $a_k\le h \le b_k$ on $M_{k+1}$. Set

$$\begin{aligned} A'=\{ z\in M_{k+1} : h(z)\le (a_k+b_k)/2 \}.\end{aligned}$$

Without loss of generality, assume $2|A'|\ge |M_{k+1}|.$ If this assumption does not hold, we consider $M-h$ instead of h. Let $A\subset A'$ be compact such that $3|A|\ge |M_{k+1}|.$ By (4.2) there exists a $c_3>0$ such that ${\mathbb {P}}^y (T_A<\tau _k)\ge c_3$ for all $y\in M_{k+1}$.

Let $\varepsilon >0$ and $y,z\in M_{k+1}$ such that $h(y)\ge b_{k+1}-\varepsilon$ and $h(z)\le a_{k+1}+\epsilon$.

Since h is harmonic, $h(X_t)$ is a martingale. We get by optimal stopping

$$\begin{aligned} h(y)-h(z) =\&{\mathbb {E}}^y \left( h(X_{T_A})-h(z) ; \tau _k>T_A \right) \\&+ {\mathbb {E}}^y \left( h(X_{\tau _k}) - h(z) ; \tau _k< T_A, X_{\tau _k}\in M_{k-1} \right) \\&+ \sum _{i=1}^\infty {\mathbb {E}}^y \left( h(X_{\tau _k}) - h(z) ; \tau _k < T_A, X_{\tau _k}\in M_{k-i-1}\setminus M_{k-i} \right) . \end{aligned}$$

We will now study these three components on the right hand side seperately. Note $h(z)\ge a_k \ge a_{k-i-1}$ for all $i\in \mathbb {N}$

• (1) In the first component, X enters $A\subset A'$ before leaving $M_k$. Hence

  $$\begin{aligned} {\mathbb {E}}^y ( h(X_{T_A})-h(z) ; T_A<\tau _k)\le {\mathbb {E}}^y \left( \frac{a_k+b_k}{2} - a_k; T_A<\tau _k \right) \\ = \frac{b_k-a_k}{2}{\mathbb {P}}^y (T_A<\tau _k) \le \frac{M\gamma ^k}{2} {\mathbb {P}}^y (T_A<\tau _k). \end{aligned}$$

• (2) In the component X leaves $M_k$ before entering A. While leaving $M_k$, X does not make a big jump in the following sense: X is at time $\tau _k$ in $M_{k-1}$. Hence in this case $h(X_{\tau _k})\le b_{k-1}.$ This yields to

  $$\begin{aligned} {\mathbb {E}}^y&\left( h(X_{\tau _k}) - h(z) ; \tau _k< T_A, X_{\tau _k}\in M_{k-1} \right) \le {\mathbb {E}}^y \left( b_{k-1}-a_{k-1} ; \tau _k< T_A, X_{\tau _k}\in M_{k-1} \right) \\&= (b_{k-1}-a_{k-1}) {\mathbb {P}}^y(\tau _k< T_A) \le M \gamma ^{k-1} (1-{\mathbb {P}}^y(T_A<\tau _k)). \end{aligned}$$

• (3) In the third component $X_{\tau _k}\in M_{k-i-1}$ for $i\in \mathbb {N}$. Therefore $h(\tau _k)\le b_{k-i-1}$.

  $$\begin{aligned} \sum _{i=1}^\infty&{\mathbb {E}}^y \left( h(X_{\tau _k}) - h(z) ; \tau _k< T_A, X_{\tau _k}\in M_{k-i-1}\setminus M_{k-i} \right) \\&\le \sum _{i=1}^\infty {\mathbb {E}}^y \left( (b_{k-i-1}-a_{k-i-1}) ; \tau _k < T_A, X_{\tau _k}\in M_{k-i-1}\setminus M_{k-i} \right) \\&= \sum _{i=1}^\infty (b_{k-i-1}-a_{k-i-1}){\mathbb {P}}^y(X_{\tau _k}\notin M_{k-i}) \le \sum _{i=1}^\infty M \gamma ^{k-i-1} c_4 \left( \frac{\rho ^k}{\rho ^{k-i}}\right) ^{\alpha _{\max }} \\&= c_4 M \gamma ^{k-1}\sum _{i=1}^\infty \left( \frac{\rho ^{\alpha _{\max }}}{\gamma }\right) ^i = c_4 M \gamma ^{k-1}\frac{\rho ^{\alpha _{\max }}}{\gamma -\rho ^{\alpha _{\max }}} \\&\le 2c_4 \gamma ^{k-2} M \rho ^{\alpha _{\max }} \le \frac{c_3M\gamma ^k}{2}, \end{aligned}$$

  where we used

  $$\begin{aligned} \frac{\rho ^{\alpha _{\max }}}{\gamma } \le \frac{1}{2} \quad \text {and} \quad \rho \le \left( \frac{\gamma }{2} \right) ^{1/\alpha _{\max }} \Leftrightarrow \frac{\rho ^{\alpha _{\max }}}{\gamma -\rho ^{\alpha _{\max }}} \le 2 \frac{\rho ^{\alpha _{\max }}}{\gamma }.\end{aligned}$$

Note, that the choice of $\gamma =1-c_3$ implies $\frac{\gamma -2}{2\gamma }c_3 + \frac{1}{\gamma }+\frac{c_3}{2}=\gamma .$ Hence

$$\begin{aligned} h(y)-h(z)&\le \frac{M\gamma ^k}{2} {\mathbb {P}}^y (T_A<\tau _k) + \frac{M\gamma ^{k-1}}{r^\beta } (1-{\mathbb {P}}^y (T_A<\tau _k)) + \frac{c_3M\gamma ^k}{2} \\&= M\gamma ^{k} \left( (\frac{\gamma -2}{2\gamma } {\mathbb {P}}^y (T_A<\tau _k)+\frac{1}{\gamma }+\frac{c_3}{2}\right) \\&\le M\gamma ^{k} \left( \frac{\gamma -2}{2\gamma } c_3 +\frac{1}{\gamma }+\frac{c_3}{2} \right) \\&= M\gamma ^{k+1}. \end{aligned}$$

We conclude that

$$\begin{aligned} b_{k+1}-a_{k+1}\le M\gamma ^{k+1}+2\varepsilon . \end{aligned}$$

Since $\varepsilon$ is arbitrary, this proves (4.4) and therefore (4.3). Let $x,y\in M_1(x_0)$ and choose $k\in \mathbb {N}_0$ such that $M^2_{\rho ^k}(x)$ is the smallest rectangle with $y\in M_{\rho ^k}(x)$.

Then $|x-y|\le 2\sqrt{d}\rho ^{k}$ and therefore

$$\begin{aligned}k\ge \frac{\log \left( \frac{|x-y|}{2\sqrt{d}}\right) }{\log (\rho )}\quad \text { for } y\in M_{\rho ^k}(x).\end{aligned}$$

Hence

$$\begin{aligned} |h(y)-h(x)|&\le M\gamma ^k = Me^{k \log (\gamma )} \\&\le M e^{(\log [|x-y|/(2\sqrt{d}))((\alpha _{\max }\log (\gamma ))/(\alpha \log (\rho )))} \\&= \frac{M |x-y|^{\log (\gamma )/\log (\rho )}}{2\sqrt{d}} = c_5 M |x-y|^\beta . \end{aligned}$$

Now let h be harmonic on $M_r^2(x_0)$. Then $h'(x):=h(r^{\alpha _{\max }/\alpha _{\min }}x)$ is harmonic on $M_1^2(x_0)$. Let $x,y\in M_1(x_0)$ and $x',y'\in M_r(x_0)$ such that

$$\begin{aligned}x=(x'_1/r^{\alpha _{\max }/\alpha _1},\dots ,x'_d/r^{\alpha _{\max }/\alpha _d}), \quad y=(y'_1/r^{\alpha _{\max }/\alpha _1},\dots ,y'_d/r^{\alpha _{\max }/\alpha _d})\in M_1(x_0).\end{aligned}$$

Then $|x-y| \le r^{\alpha _{\max }/\alpha _{\min }}|x'-y'|.$ Set ${\widetilde{x}}=x'/r^{\alpha _{\max }/\alpha _{\min }}$ and ${\widetilde{y}}=y'/r^{\alpha _{\max }/\alpha _{\min }}$. We conclude

$$\begin{aligned} |h(x')-h(y')|&= |h(r^{\alpha _{\max }/\alpha _{\min }}{\widetilde{x}})-h(r^{\alpha _{\max }/\alpha _{\min }}{\widetilde{y}})|= |h'({\widetilde{x}})-h'({\widetilde{y}})| \le c_1 |{\widetilde{x}}-{\widetilde{y}}|^\beta \sup \limits _{{\widetilde{z}}\in \mathbb {R}^d}|h({\widetilde{z}})|\\&= c_1 \left| \frac{x'}{r^{\alpha _{\max }/\alpha _{\min }}}-\frac{y'}{r^{\alpha _{\max }/\alpha _{\min }}}\right| ^\beta \sup \limits _{z\in \mathbb {R}^d}|h(z)| = c_1 \left( \frac{|x'-y'|}{r^{\alpha _{\max }/\alpha _{\min }}}\right) ^\beta \sup \limits _{z\in \mathbb {R}^d}|h(z)|. \end{aligned}$$

$\square$