You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'In this paper we consider Harnack inequalities with respect to a symmetric $\alpha$-stable Lévy process $X$ in $\mathbb{R}^d$, $\alpha \in (0,2)$, $d\geq 2$. We study the example from the article [@bg-sz-1]. There, the authors have associated the Harnack inequality with the relative Kato condition, which is a condition on the Lévy measure. By checking the condition, in the case $\alpha \in (0,1)$, they have established that the Harnack inequality does not hold. We give an alternative proof of this fact, using the setting of [@bg-sz-1]. We define the harmonic functions explicitly. For a given starting point of the process, we examine the probability of hitting a certain set at the first exit time of a unit ball. Moreover, we also examine the weak Harnack inequality for a certain class of symmetric $\alpha$-stable Lévy processes. We consider a symmetric $\alpha$-stable Lévy process, $\alpha \in (0,2)$, for which a spherical part $\mu$ of the Lévy measure is a spectral measure. In addition, we assume that $\mu$ is absolutely continuous with respect to the uniform measure $\sigma$ on the sphere and impose certain bounds on the corresponding density. Eventually, we show that the weak Harnack inequality holds.'
address: 'Fakultät für Mathematik, Universität Bielefeld, Universitätsstraße 25, 33615 Bielefeld'
author:
- MARINA SERTIC
title: HARNACK INEQUALITIES FOR SYMMETRIC STABLE LÉVY PROCESSES
---

=1

[^1]

Acknowledgment {#acknowledgment .unnumbered}
--------------

The author is thankful to Moritz Kassmann and Mateusz Kwaśnicki for valuable discussions.

Introduction
============

In the paper, we use the notation $$\begin{aligned}
 B(x_0,r)=\{x : |x-x_0| \leq r\}, r\geq 0,\end{aligned}$$ and $$\begin{aligned}
 B_{r}=B(0,r), r\geq 0.\end{aligned}$$

We consider *a symmetric $\alpha$-stable Lévy process*, which has the characteristic function of the form $$\begin{aligned}
\label{k_fun}
 \mathbb{E}^{0}\big[e^{i u\cdot X_t}\big]=e^{-t\Phi(u)}, u \in \mathbb{R}^d,\ t\geq 0,\end{aligned}$$ where the characteristic exponent $\Phi$ is given by $$\begin{aligned}
\label{k_e}
\Phi(u)=\int_{\mathbb{S}^{d-1}} |u\cdot \xi|^\alpha \mu (\mathrm{d}\xi).\end{aligned}$$ The measure $\mu$ is symmetric, finite and non-zero on $\mathbb{S}^{d-1}$ (see [@sato], Theorem 14.13). Le the measure $\mu$ be absolutely continuous with respect to the uniform surface measure on $\mathbb{S}^{d-1}$ and denote its density by $f_\mu$.

*The potential density (or the heat kernel) $p(t,x,y)=p(t,y-x)$* is determined by the Fourier transform $$\begin{aligned}
 \int_{\mathbb{R}^d} e^{i\xi\cdot x}\ p(t,x)\ \mathrm{d}x=e^{-t\Phi(\xi)},\ \xi \in \mathbb{R}^d,\ t \geq 0.\end{aligned}$$

*The Green function (or the potential kernel)* is defined by $$\begin{aligned}
\label{Gr_f}
 G(x,y)=\int_{0}^\infty p(t,x,y)\ \mathrm{d}t,\ x,y \in \mathbb{R}^d.\end{aligned}$$

Let $D$ be an open set, $D\subset \mathbb{R}^d$. *The Green function of $X^D$* is defined by $$\begin{aligned}
\label{Gr_f_D}
G_D(x,y)=G(x,y)-\mathbb{E}^x[G(X_{\tau_D}, y)],\ x,y\in D.\end{aligned}$$

Harnack Inequality
==================

A non-negative Borel measurable function $u:\mathbb{R}^d \to \mathbb{R}$ is *harmonic in an open set $D\subset \mathbb{R}^d$* with respect to the process $X$ if $$\begin{aligned}
 u(x)=\mathbb{E}^{x}[u(X_{\tau_U})],\ x\in U,\end{aligned}$$ for every bounded open set $U$ such that $\bar{U}\subset D$. Here, $\tau_U=\inf\{t>0 : X_t \notin U\}$ is the first exit time from the set $U$.

*The Harnack inequality* for a symmetric $\alpha$-stable Lévy processes $X$ holds true if there is a constant $K\geq 1$ such that for every non-negative function $u\colon \mathbb{R}^d\rightarrow \mathbb{R}$ which is harmonic in $B_1$ with respect to $X$, the inequality $$\begin{aligned}
u(x)\leq K\cdot u(y),\ x,y \in B_{1/2}.\end{aligned}$$ holds true.

Construction of the Sequence of Harmonic Functions
--------------------------------------------------

[\[lema\]]{} Let $c>1$. There are sequences $(\alpha_n)$ and $(\beta_n)$ of positive numbers such that: $$\begin{aligned}
&A=\sum\limits_{n=1}^\infty \alpha_n<\frac{\pi}{2},\ \ B=\sum\limits_{n=1}^\infty \beta_n<\frac{\pi}{2},\\
&\lim_{n \to \infty} (A_n+B_n)=A+B=\frac{\pi}{2},\\
&\lim_{n \to \infty}\frac{\tan(A_n + B_{n+1})}{\tan(A_n + B_n)} = +\infty \label{uvjet},\\
&(c+2)\cdot \tan(A_n + B_{n+1}) \leq (c-1)\cdot \tan(A_{n+1} + B_{n+1}),\ n \in \mathbb{N}\label{uvjet_d},\end{aligned}$$ where $A_n$ and $B_n$ denote the partial sum of the sequences $(\alpha_n)$ and $(\beta_n)$, respectively, i.e. $A_n= \sum\limits_{k=1}^n \alpha_k$, $B_n= \sum\limits_{k=1}^n \beta_k$.

We construct inductively sequences $(\alpha_n)$ and $(\beta_n)$ of positive numbers which fulfill the conditions of the lemma. Set $K=(c+2)/(c-1)$.

Choose $\alpha_1$ and $\beta_1$ so that $$\begin{aligned}
 \alpha_1 + \beta_1 <  \pi/2.\end{aligned}$$ Next, choose $\beta_2$ and $\alpha_2$ so that: $$\begin{aligned}
&\beta_2 < \pi/2 - (A_1 + B_1), \\
&\arctan\big(K\cdot \tan(A_1+B_2)\big) - (A_1 + B_2) <\alpha_2 < \pi/2 - (A_1 + B_2).\end{aligned}$$ Assume $\beta_1, \ldots, \beta_n$, $\alpha_1, \ldots, \alpha_n$ are chosen. Choose $\beta_{n+1}$ and $\alpha_{n+1}$ so that: $$\begin{aligned}
&\beta_{n+1} < \pi/2 - (A_n + B_n), \\
&n+1 \leq \frac{\tan(A_n + B_{n+1})}{\tan(A_n + B_n)},\\
&\arctan\big(K\cdot \tan(A_n+B_{n+1})\big) - (A_n + B_{n+1}) < \alpha_{n+1}\\
&\alpha_{n+1}< \pi/2 - (A_n + B_{n+1}).\end{aligned}$$

Notice that the choice of the sequence $(\beta_n)$ is possible, since $$\begin{aligned}
 \lim_{h \nearrow \pi/2 - c} \frac{\tan(h+c)}{\tan(c)} = +\infty,\end{aligned}$$ for fixed $c \in (0,\frac{\pi}{2})$.

Furthermore, the choice of of the sequence $(\alpha_n)$ is possible due to the choice of $(\beta_{n})$.

Now, from the choice of $(\alpha_n)$, we have $$\begin{aligned}
 \arctan\big(K\cdot \tan(A_n+B_{n+1})\big) <A_{n+1} + B_{n+1} < \pi/2,\end{aligned}$$ which, together with the choice of $(\beta_n)$, implies $$\begin{aligned}
 \lim_{n \to \infty} (A_n+B_n)=\frac{\pi}{2}.\end{aligned}$$

At this point, we would like to mention that none of the following sequences satisfies the property from the Lemma \[lema\].

1.  $\alpha_k = 2^{-k}, \ \beta_k = \frac{1}{k(k+1)}$

2.  $\alpha_k = a^{-k}, \ \beta_k = b^{-k}$,   $1 < b \leq a$,

3.  $\alpha_k =  \frac{1}{k(k+1)} , \ \beta_k = k^{-1-\delta},\ \delta > 0$,

4.  $\alpha_k = 2^{-(k^3)}, \ \beta_k = 2^{-(k^2)}$,\
    $\alpha_k = a^{-(k^3)}, \ \beta_k = b^{-(k^2)}$,$a>1,b>1$

The details can be found in Lemma \[lema\_pr\] (Appendix \[Ap\]).

Let $\delta, x \in (0,1)$ and $(\alpha_n)$ and $(\beta_n)$ be sequences as in Lemma \[lema\]. We define the sets $$\begin{aligned}
S_n^{\delta}(x)&=\{(\delta,y):A_{n-1}+B_n<\arctan\Big(\frac{y}{x+\delta}\Big)<A_n+B_n\},\\
L_n^{\delta}(x)&=\bigcup_{z\in [x,1]} S_n^{\delta}(z).\end{aligned}$$

Notice that, by the definitions of $S_n^{\delta}(x)$ and $L_n^{\delta}(x)$, we have $$\begin{aligned}
\label{L_n}
L_n^{\delta}(x)=\{(\delta,y):(x+\delta)&\cdot\tan(A_{n-1}+B_n)<y<(1+\delta)\tan(A_n+B_n)\}.\end{aligned}$$

The sets $S_n^{\delta}(x)$ and $L_n^{\delta}(x)$ are illustrated at the pictures that follow.

![image](S_n_delta)

![image](L_n_delta)

$S_n^{\delta}(x)$ describes the set on the line $\{(x,y): x=\delta, y>0\}$, which is seen from the point $(-x,0)$ by the cone $K_n$, and $L_n^{\delta}(x)$ describes the set on the line $\{(x,y):
x=\delta, y>0\}$ which is seen from the set $[-1,-x]\times \{0\}$ by the cone $K_n$.

Let $(\alpha_n)$ and $(\beta_n)$ be sequences as in Lemma \[lema\]. For every $\delta\in (0,1)$ and every $x \in (0,1)$ there is $n_0=n_0(\delta,x)$ such that for every $n\geq n_0$ $$L_n^{\delta}(x)\cap L_{n+1}^{\delta}(x)=\emptyset.$$

Let $\delta, x\in (0,1)$. Recall, by , we have $$\begin{aligned}
L_n^{\delta}(x)=\{(\delta,y):(x+\delta)&\cdot\tan(A_{n-1}+B_n)<y<(1+\delta)\tan(A_n+B_n)\}.\end{aligned}$$ Now, from Lemma \[lema\], condition , it follows that there is $n_0=n_0(\delta,x)$ such that for every $n\geq n_0$, $$\begin{aligned}
L_n^{\delta}(x)\cap
L_{n+1}^{\delta}(x)=\emptyset,\end{aligned}$$ which proves the proposition.

Let $\delta \in(0,1)$ and $(\alpha_n)$ and $(\beta_n)$ be sequences as in Lemma \[lema\]. For $z=(z_1, z_2), y=(y_1, y_2)\in \mathbb{R}^2$, we define the sets $$\begin{aligned}
\tilde{S}_n^{\delta}(z)&=\{(\delta,u):A_{n-1}+B_n<\arctan\Big(\frac{u-z_2}{\delta+z_1}\Big)<A_n+B_n\},\\
\tilde{L}_n^{\delta}(y)&=\bigcup_{\{z:z_1\in [y_1,1], z_2=y_2\}}\tilde{S}_n^{\delta}(z).\end{aligned}$$

$\tilde{S}_n^{\delta}(z_1,z_2)$ describes the set on the line $\{(x,u): x=\delta, u>0\}$, which is seen from the point $(-z_1,z_2)$ by the cone $K_n$, and $\tilde{L}_n^{\delta}(y_1,y_2)$ describes the set on the line $\{(x,u):
x=\delta, u>0\}$, which is seen from the set $[-1,-y_1]\times \{y_2\}$ by the cone $K_n$.

Let $w=(w_1,w_2)\in\mathbb{R}^2$. Notice that by the definitions of $\tilde{S}_n^{\delta}(z)$ and $\tilde{L}_n^{\delta}(y)$, we have $$\label{til_L_n}
\begin{aligned}
\tilde{L}_n^{\delta}(w)=\{(\delta,u): &(w_1+\delta)\cdot\tan(A_{n-1}+B_n)+w_2 <\\
                      &<u<(1+\delta)\cdot\tan(A_n+B_n)+w_2\}.
\end{aligned}$$

By means of the two propositions below, we will define the sequence of harmonic functions so that the Harnack inequality does not hold. To be more precise, we will construct the sequence of sets $(B_n)$ and consequently the sequence of functions $(u_n)$, so that $$\begin{aligned}
 \lim_{n\rightarrow \infty}\frac{u_n(1/2,0)}{u_n(0,0)}=0.\end{aligned}$$

![image](prop4)

[\[prop4\]]{} Let $(\alpha_n)$, $(\beta_n)$ be sequences as in Lemma \[lema\]. There are sequences $(\delta_n)$, $(x_n)$, $(y_n)$, $(c_n)$ and $(d_n)$ of positive numbers such that $$\begin{aligned}
&V_n\subset \tilde{S}_{n+1}^{\delta_n}(-x_1,x_2)\cap \Delta_n^{\delta_n},\end{aligned}$$ for every $x=(x_1, x_2)\in U_n$, where $$\begin{aligned}
U_n&\coloneqq (-y_n, y_n)^2,\\
V_n&\coloneqq \{\delta_n\}\times(c_n,d_n).\end{aligned}$$

Notice that the picture above is in the connection with the statement.

For $n\in\mathbb{N}$, define $$\begin{aligned}
 I_n&\coloneqq 2+\tan(A_n+B_{n+1})+\tan(A_{n+1}+B_{n+1}),\label{def_I_n}\\
 P_n&\coloneqq \tan(A_{n+1}+B_{n+1})-\tan(A_n+B_{n+1},)\label{def_P_n}\\
 P_n'&\coloneqq \tan(A_n+B_{n+1})-\tan(A_n+B_n).\label{def_P_n'}\end{aligned}$$ Furthermore, set $$\begin{aligned}
\delta_n &\coloneqq \frac{I_n\cdot (1+\tan(A_n+B_n))\cdot (1+\tan(A_{n+1}+B_{n+1}))}{P_n^2\cdot(1+\tan(A_n+B_{n+1}))+P_n' I_n \cdot(1+\tan(A_{n+1}+B_{n+1}))},\label{def_del_n}\\
x_n&\coloneqq 2\delta_n\cdot\frac{1+\tan(A_n+B_{n+1})}{\tan(A_n+B_{n+1})}\cdot \frac{P_n}{I_n}+\frac{1}{\tan(A_n+B_{n+1})},\label{def_x_n}\\
y_n&\coloneqq \frac{1+\tan(A_n+B_n)-\delta_n\cdot P_n'}{1+\tan(A_n+B_{n+1})},\label{def_y_n}\\
c_n&\coloneqq (\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n,\label{def_c_n}\\
d_n&\coloneqq (\delta_n -y_n)\cdot \tan(A_{n+1}+B_{n+1})-y_n\label{def_d_n}.\end{aligned}$$

For the convenience, let us denote the denominator of $\delta_n$ by $H_n$, $$\begin{aligned}
 H_n\coloneqq P_n^2\cdot(1+\tan(A_n+B_{n+1}))+P_n' I_n \cdot(1+\tan(A_{n+1}+B_{n+1}))\label{def_H_n}.\end{aligned}$$

First, from Lemma \[lema4\] (Appendix \[Ap\]), we have that $I_n$, $P_n$, $P_n'$, $\delta_n$, $x_n$, $y_n$, $c_n$ and $d_n$ are positive and $c_n<d_n$.

Let us prove $V_n\subset\Delta_n^{\delta_n}$. What is more, we show $V_n=\Delta_n^{\delta_n}$. Recall, we defined the set $\Delta_{n}^{\delta_n}$ in the way $$\begin{aligned}
\Delta_{n}^{\delta_n}=\{(y_1,y_2):\ &y_1=\delta_n,
(\delta_n+1)\cdot\tan(A_n+B_n)+1 <\\
& <y_2<(\delta_n+x_n)\cdot\tan(A_n+B_{n+1})-1\}.\end{aligned}$$

The definition yields $$\begin{aligned}
 (\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n=(\delta_n+1)\cdot\tan(A_n+B_n)+1.\end{aligned}$$

Therefore, using the equality $$\begin{aligned}
c_n=(\delta_n+1)\cdot\tan(A_n+B_n)+1\end{aligned}$$ holds.

By Lemma \[lema4\] (Appendix \[Ap\]) we have $d_n=(\delta_n+x_n)\cdot\tan(A_n+B_{n+1})-1$, hence $V_n=\Delta_{n}^{\delta_n}$.

Now, for $x=(x_1, x_2)\in U_n$, we prove $V_n\subset \tilde{S}_{n+1}^{\delta_n}(-x_1,x_2)$. Because of geometrical reasoning, it is enough to show $$\begin{aligned}
V_n\subset \tilde{S}_{n+1}^{\delta_n}(-y_n,-y_n)\cap 
\tilde{S}_{n+1}^{\delta_n}(y_n,y_n).\end{aligned}$$

Let $p_1^n$ and $p_2^n$ be the lines given by $$\begin{aligned}
 &y-y_n=\tan(A_n+B_{n+1})\cdot(x+y_n),\\
 &y+y_n=\tan(A_{n+1}+B_{n+1})\cdot(x-y_n),\end{aligned}$$ respectively. As in the picture, $p_1^n$ and $p_2^n$ are the lines through the points $(-y_n,y_n)$ and $(y_n,-y_n)$, respectively.

Denote by $P_1^{n}$ and $P_2^{n}$ the points which determine the intersection of lines $p_1^n$ and $p_2^n$ with the set $\{(y_1,y_2): y_1=\delta_n, y_2>0\}$: $$\begin{aligned}
 P_1^{n}&=(x_{P_1}, y_{P_1})=(\delta_n, (\delta_n + y_n)\cdot \tan(A_n+B_{n+1})+y_n),\\
 P_2^{n}&=(x_{P_2}, y_{P_2})=(\delta_n, (\delta_n-y_n)\cdot\tan(A_{n+1}+B_{n+1})-y_n).\end{aligned}$$

Notice that $P_1^{n}$ and $P_2^{n}$ determine the lower boundary point and the upper boundary point of the sets $\tilde{S}_{n+1}^{\delta_n}(y_n,y_n)$ and $\tilde{S}_{n+1}^{\delta_n}(-y_n,-y_n)$, respectively. Notice that showing $V_n= \{\delta_n\}\times (y_{P_1},y_{P_2})$ finishes the proof. Clearly, the definitions and yield $$\begin{aligned}
 c_n&=y_{P_1},\\
 d_n&=y_{P_2},\end{aligned}$$ and hence the proposition.

![image](prop5)

[\[prop5\]]{} Let $(\alpha_n)$, $(\beta_n)$ be sequences as in the Lemma \[lema\] and let $(I_n)$, $(P_n)$, $(P_n')$, $(\delta_n)$, $(x_n)$, $(y_n)$, $(c_n)$, $(d_n)$ and $(U_n)$ be as in the Proposition \[prop4\]. There exist sequences $(\delta_n')$ and $(B(S_n, r_n))$, where:

1.  $(\delta_n')$ is such that $\delta_n'<\delta_n$, for every $n\in\mathbb{N}$,

2.  $B(S_n, r_n)$ is a ball with the center $S_n=(x_{S_n}, y_{S_n})$ and radius $r_n$, such that $$B(S_n, r_n)\subset \bigcup_{\{\varepsilon : \delta_n'<\varepsilon <\delta_n\}}
    \Big(\tilde{S}_{n+1}^{\varepsilon}(-x_1,x_2)\cap \Delta_n^{\varepsilon}\Big),$$ for every $x=(x_1, x_2)\in U_n$.

Notions from the Proposition are illustrated at the picture above.

For $n\in\mathbb{N}$, define $$\begin{aligned}
\delta_n'&\coloneqq \frac{\delta_n\cdot(1+\tan(A_n+B_{n+1}))+y_n I_n}{1+\tan(A_{n+1}+B_{n+1})},\label{def_del_n_c}\\
J_n&\coloneqq 2\tan(A_n+B_{n+1})\cdot\tan(A_{n+1}+B_{n+1})\notag\\
    &\phantom{=\ } +\tan(A_n+B_{n+1})+\tan(A_{n+1}+B_{n+1}).\label{def_J_n}\end{aligned}$$ Set $$\begin{aligned}
&x_{S_n}\coloneqq \frac{\delta_n+y_n}{2}\cdot\frac{I_n}{1+\tan(A_{n+1}+B_{n+1})},\label{def_x_S}\\
&y_{S_n}\coloneqq \frac{\delta_n+y_n}{2}\cdot\frac{J_n}{1+\tan(A_{n+1}+B_{n+1})},\label{def_y_S}\\
&r_n\coloneqq \frac{1}{2}\cdot\frac{\delta_n  P_n-y_n 
I_n}{1+\tan(A_{n+1}+B_{n+1})}\label{def_r_n}.\end{aligned}$$ By the first part of Lemma \[lema5\] (Appendix \[Ap\]), $$\begin{aligned}
B(S_n, r_n)\subset Q_n:=(\delta_n', \delta_n)\times (c_n,
c_n+(\delta_n-\delta_n')).\end{aligned}$$ Now, it is enough to show that for every $x=(x_1, x_2)\in U_n$ $$\begin{aligned}
Q_n \subset \bigcup_{\{\varepsilon : \delta_n'<\varepsilon <\delta_n\}}\Big( \tilde{S}_{n+1}^{\varepsilon}(-x_1,x_2)\cap \Delta_n^{\varepsilon}\Big).\end{aligned}$$

Let $z=(z_1,z_2)\in Q_n$. We show $$\begin{aligned}
z \in \tilde{S}_{n+1}^{\varepsilon}(-x_1,x_2)\cap \Delta_n^{\varepsilon},\end{aligned}$$ for some $\delta_n'<\varepsilon<\delta_n$.

Recall, $$\begin{aligned}
\Delta_{n}^{\varepsilon}=\{(y_1,y_2):\ &y_1=\varepsilon,
(\varepsilon+1)\cdot\tan(A_n+B_n)+1<\\
&<y_2<(\varepsilon+x_n)\cdot\tan(A_n+B_{n+1})-1\},\end{aligned}$$ so in order to show $z\in \Delta_{n}^{\varepsilon}$, we prove that $$\begin{aligned}
c_n&\geq(\varepsilon+1)\cdot\tan(A_n+B_n)+1,\\
c_n+(\delta_n-\delta_n')&\leq(\varepsilon+x_n)\cdot\tan(A_n+B_{n+1})-1,\end{aligned}$$ for some $\delta_n'<\varepsilon<\delta_n$. Since the proof is quite technical, we spell out the details in the second part of Lemma \[lema5\] (Appendix \[Ap\]).

Now, for $x=(x_1, x_2)\in U_n$, let us prove $z \in \tilde{S}_{n+1}^{\varepsilon}(-x_1,x_2)$. Due to geometrical reasoning, it is enought to show that $$\begin{aligned}
z\in \tilde{S}_{n+1}^{\varepsilon}(-y_n,-y_n)\cap \tilde{S}_{n+1}^{\varepsilon}(y_n,y_n).\end{aligned}$$

As before, let $p_1^n$ and $p_2^n$ denote the lines $$\begin{aligned}
 &y-y_n=\tan(A_n+B_{n+1})\cdot(x+y_n),\\
 &y+y_n=\tan(A_{n+1}+B_{n+1})\cdot(x-y_n),\end{aligned}$$ respectively. As before, the lines $p_1^n$ and $p_2^n$ are the lines through the points $(-y_n,y_n)$ and $(y_n,-y_n)$, respectively.

Denote by $P_1^{n}$ and $P_2^{n}$ the points which determine the intersection of $p_1^n$ and $p_2^n$ with the set $\{(y_1,y_2): y_1=\varepsilon, y_2>0\}$: $$\begin{aligned}
 P_1^{n}&=(x_{P_1}, y_{P_1})=(\varepsilon, (\varepsilon + y_n)\cdot \tan(A_n+B_{n+1})+y_n),\label{y_P_1}\\
 P_2^{n}&=(x_{P_2}, y_{P_2})=(\varepsilon,(\varepsilon-y_n)\cdot\tan(A_{n+1}+B_{n+1})-y_n)\label{y_P_2}.\end{aligned}$$

Notice that $P_1^{n}$ and $P_2^{n}$ determine the lower boundary point and the upper boundary point of sets $\tilde{S}_{n+1}^{\varepsilon}(y_n,y_n)$ and $\tilde{S}_{n+1}^{\varepsilon}(-y_n,-y_n)$, respectively. If we show $z_2\in  (y_{P_1},y_{P_2})$, the proposition is proved.

To this end, we prove $$\begin{aligned}
c_n &\geq y_{P_1},\\
c_n + (\delta_n-\delta_n')&\leq y_{P_2},\end{aligned}$$ and the details can be found in the Lemma \[har1\] in Appendix \[Ap\].

Harnack Inequality
------------------

In the subsection that follows, we give a proof of the fact that Harnack inequality does not hold. ![image](hf)

Let $X$ be a symmetric $\alpha$-stable Lévy process in $\mathbb{R}^2$, $0<\alpha<1$, with the characteristic function of the form $$\begin{aligned}
\mathbb{E}^0 e^ {iu\cdot X_t} =  e^{-t \Phi(u)}, \ u \in\mathbb{R}^2,\end{aligned}$$ where $$\begin{aligned}
\Phi(u)=\int_{\mathbb{S}^{1}} |u\cdot \xi|^\alpha \mu(\mathrm{d}\xi).\end{aligned}$$ We define the spectral measure $\mu$ appropriately, below in Theorem \[har\_thm\_mat\]. Furthermore, for a given starting point $x\in B_1$ of the process $X$, we examine the probability $$\begin{aligned}
\mathbb{P}^{x}(X_{\tau_{B_1}} \in B_n),\ n\in \mathbb{N}\end{aligned}$$ of hitting the set $B_n=B(S_n,r_n)$ at the first exit time of a unit ball $B_1$, where for every $n\in \mathbb{N}$, $B_n$ are as in Proposition \[prop5\]. We choose the points $w_0=(1/2,0)$, $0=(0,0)$ and define the harmonic functions $$\begin{aligned}
 u_n(x)= \mathbb{P}^{x}(X_{\tau_{B_1}} \in B_n),\ n\in\mathbb{N},\ x\in B_1.\end{aligned}$$ We show $$\begin{aligned}
 \frac{u_n(w_0)}{u_n(0)}\leq c(\alpha)\cdot
a_n b_n ,\end{aligned}$$ for sequences $(a_n)$ and $(b_n)$ with certain properties and eventually obtain the desired conclusion.

Constants are positive real numbers, which exact value may vary from one line to the other. For the convenience they will not be explicitly stated throughout the proof.

Before we prove the Theorem \[har\_thm\_mat\], we state the proposition for the Green function estimate of the unit ball $B_1$ for $X$ (see [@bg-sz-1], Theorem 2).

\[Gr\_f\_est\] Let $X$ be a symmetric $\alpha$-stable Lévy process in $\mathbb{R}^d$, $0<\alpha<2$, with the characteristic function of the form $$\begin{aligned}
\mathbb{E}^0 e^ {iu\cdot X_t} =  e^{-t \Phi(u)}, \ u \in\mathbb{R}^d,\end{aligned}$$ where $$\begin{aligned}
\Phi(u)=\int_{\mathbb{S}^{d-1}} |u\cdot \xi|^\alpha \mu(\mathrm{d}\xi),\end{aligned}$$ and $\mu$ is a finite, symmetric measure on $\mathbb{S}^{d-1}$ such that $\mu(S^{d-1})>0$. Let $\mu$ be absolutely continuous with respect to the surface measure on $\mathbb{S}^{d-1}$, and denote by $f_\mu$ its density, for which the inequality $$\begin{aligned}
 0\leq f_\mu(\xi)\leq m,\ \xi\in\mathbb{S}^{d-1},\end{aligned}$$ holds, for some $m>0$. Then the estimate for the Green function $G_{B_1}$ $$\begin{aligned}
c^{-1}\cdot s(y)\cdot|x-y|^{\alpha-d} \leq G_{B_1}(x,y)\leq c\cdot s(y)\cdot|x-y|^{\alpha-d},\end{aligned}$$ holds, for all $|x|<1/2, |y|<1$. Here, $s(y)=\mathbb{E}^y [\tau_{B_1}]$ is the expected time spent in $B_1$, if the process starts from $y\in B_1$.

\[har\_thm\_mat\] Let $X$ be a symmetric $\alpha$-stable Lévy process in $\mathbb{R}^2$,\
$0<\alpha<1$, with the characteristic function of the form $$\begin{aligned}
\mathbb{E}^0 e^ {iu\cdot X_t} =  e^{-t \Phi(u)}, \ u \in\mathbb{R}^2,\end{aligned}$$ where $$\begin{aligned}
\Phi(u)=\int_{\mathbb{S}^{1}} |u\cdot \xi|^\alpha \mu(\mathrm{d}\xi).\end{aligned}$$ Let the measure $\mu$ on $\mathbb{S}^{1}$ be absolutely continuous with respect to the surface measure on $\mathbb{S}^{1}$, and denote by $f_\mu$ its density. Set $$\begin{aligned}
 f_\mu&=\mathbf{1}_{\bigcup_{n\geq 1} (B_{\xi_n,r_n} \cup \ B_{-\xi_n,r_n})},\end{aligned}$$ where $$\begin{aligned}
  &B_{\xi_n,r_n}=B(\xi_n,r_n)\cap \mathbb{S}^1,\\
  &r_n=2\cdot\sin(\alpha_n/4),\\
  &\xi_n=\big(\cos(A_{n-1}+B_n+\alpha_n/2\big), \sin\big(A_{n-1}+B_n+\alpha_n/2)\big) \in \mathbb{S}^1, n\in \mathbb{N},\end{aligned}$$ where $(\alpha_n)$ and $(\beta_n)$ are as in Lemma \[lema\]. Then the Harnack inequality for $X$ does not hold.

Let $$\begin{aligned}
u_n(x)&= \mathbb{P}^{x}(X_{\tau_B} \in B_n),\ n\in \mathbb{N},\ x\in B(0,1),\\
O_n&=\big((-x_n,\delta_n)\times(-1,1)\big)\cap B_1,\end{aligned}$$ where $B_n=B(S_n,r_n)$, $x_n$, $\delta_n$ are introduced in the Propositions \[prop4\] and \[prop5\], and set $w_0=(1/2,0)$ and $0=(0,0)$.

By the strong Markov property, every function $u_n$ is harmonic in $B_1$ with respect to $X$.

Namely, for a set $U\subset \bar{U} \subset B_1$, $$\begin{aligned}
\mathbb{E}^x[u_n(X_{\tau_U})]&=\mathbb{E}^x[\mathbb{P}^{X_{\tau_U}}[X_{\tau_B}\in B_n]]\\
&= \mathbb{P}^x[X_{\tau_B}\circ \theta_{\tau_U}\in B_n]\\
&=\mathbb{P}^x[X_{\tau_B}\in B_n]\\
&=u_n(x).\end{aligned}$$

Using the Lévy system formula ([@chen]) it follows $$\begin{aligned}
&\mathbb{E}^{w_0}\Bigg[\sum_{s\leq t\wedge \tau_B}\mathbf{1}_{\{X_{s-}\in O_n, X_{s} \in B_n\}}\Bigg]\\
&=\mathbb{E}^{w_0}\Bigg[\int_{0}^{t\wedge \tau_B}\mathbf{1}_{O_n}(X_s)\Big(\int_{B_n} f_{\nu}(y-X_s)\ \mathrm{d}y\Big)\ \mathrm{d}s\Bigg] \\
&\leq\mathbb{E}^{w_0}\Bigg[\int_{0}^{t\wedge \tau_B}\mathbf{1}_{O_n}(X_s)\Big(\int_{B_n} |y-X_s|^{-\alpha-2}\ \mathrm{d}y\Big)\ \mathrm{d}s\Bigg]\\
&\leq |B_n|\cdot[(\delta_n+1)\cdot\tan(A_n+B_n)]^{-\alpha-2}\cdot\mathbb{E}^{w_0}\Bigg[\int_0^{t\wedge\tau_B} \mathbf{1}_{O_n}(X_s)\ \mathrm{d}s\Bigg]\\
&\leq |B_n|\cdot [(\delta_n+1)\cdot\tan(A_n+B_n)]^{-\alpha-2}\cdot \mathbb{E}^{w_0}\Bigg[\int_0^{\infty} \mathbf{1}_{O_n}(X_s)\ \mathrm{d}s\Bigg].\end{aligned}$$

The estimates of the potential kernel ([@sz-1]) imply $$\begin{aligned}
\mathbb{E}^{w_0}\Bigg[\int_0^{\infty} \mathbf{1}_{O_n}(X_s)\ \mathrm{d}s\Bigg]&\leq c(\alpha)\cdot \int_ {O_n} |y-w_0|^{\alpha-2}\ \mathrm{d}y\\
&\leq c(\alpha)\cdot(\delta_n+ x_n).\end{aligned}$$ From here we have $$\begin{aligned}
&\mathbb{E}^{w_0}\Bigg[\sum_{s\leq t\wedge \tau_B}\mathbf{1}_{\{X_{s-}\in O_n, X_{s} \in B_n\}}\Bigg]\\
&\leq c(\alpha)\cdot|B_n|\cdot [(\delta_n+1)\cdot\tan(A_n+B_n)]^{-\alpha-2}\cdot (\delta_n+x_n).\end{aligned}$$

Letting $t\rightarrow\infty$, by the dominated convergence theorem, we obtain $$\begin{aligned}
\label{gornja_og_alt}
u_n(w_0)
\leq
c_1(\alpha,n)\cdot(\delta_n+ x_n),\end{aligned}$$ where $$\begin{aligned}
c_1(\alpha,n)=c(\alpha)\cdot|B_n|\cdot [(\delta_n+1)\cdot\tan(A_n+B_n)]^{-\alpha-2}.\end{aligned}$$

Let us compute the lower bound on $u_n(0)$.

By the Lévy system formula (see [@chen]) and the construction, it follows that $$\begin{aligned}
u_n(0)&=\mathbb{P}^{0}(X_{\tau_B} \in B_n) \\
&\geq \mathbb{E}^{0}\Bigg[\sum_{s\leq t\wedge \tau_B}\mathbf{1}_{\{X_{s-}\in U_n, X_{s} \in B_n\}}\Bigg]\\
&=\mathbb{E}^{0}\Bigg[\int_{0}^{t\wedge\tau_B}\mathbf{1}_{U_n}(X_s)\Big(\int_{B_n} f_{\nu}(y-X_s)\ \mathrm{d}y\Big)\ \mathrm{d}s\Bigg]\\
&=\mathbb{E}^{0}\Bigg[\int_{0}^{t\wedge\tau_B}\mathbf{1}_{U_n}(X_s)\Big(\int_{B_n} |y-X_s|^{-\alpha-2}\ \mathrm{d}y\Big)\ \mathrm{d}s\Bigg]\\
&\geq \bar{c}_2(\alpha,n)\cdot\mathbb{E}^{0}\Bigg[\int_{0}^{t\wedge\tau_B}\mathbf{1}_{U_n}(X_s)\ \mathrm{d}s\Bigg],\end{aligned}$$ where $$\begin{aligned}
 \bar{c}_2(\alpha,n)=|B_n|\cdot[2\delta_n-\delta_n'+2y_n+ 1 +(\delta_n+1)\cdot\tan(A_n+B_n)]^{-\alpha-2}.\end{aligned}$$

Letting $t\rightarrow\infty$, by the dominated convergence theorem, it follows $$\begin{aligned}
u_n(0)\geq \bar{c}_2(\alpha,n)\cdot\mathbb{E}^{0}\Bigg[\int_{0}^{\tau_B}\mathbf{1}_{U_n}(X_s)\ \mathrm{d}s\Bigg].\end{aligned}$$

Using the the estimate of the Green function from [@bg-sz-1] there is $c(\alpha)$ so that $$\begin{aligned}
\mathbb{E}^{0}\Bigg[\int_{0}^{\tau_B}\mathbf{1}_{U_n}(X_s)\ \mathrm{d}s\Bigg]&=\mathbb{E}^{0}\Bigg[\int_{0}^{\infty}\mathbf{1}_{\{X_s\in U_n,\tau_B>s\}}\ \mathrm{d}s\Bigg]\\
&=\int_{0}^{\infty}\mathbb{P}^0(X_s\in U_n,\tau_B>s)\ \mathrm{d}s\\
&=\int_{U_n}G_B(0,y)\ \mathrm{d}y \\
&\geq \int_{U_n}s(y)|y|^{\alpha-2}\ \mathrm{d}y\\
&\geq c(\alpha)\cdot (1-2y_n)^\alpha \int_{U_n}|y|^{\alpha-2}\ \mathrm{d}y \\
&\geq c(\alpha)\cdot (1-2y_n)^\alpha \int_{B(0,y_n/2)}|y|^{\alpha-2}\ \mathrm{d}y \\
&\geq c(\alpha)\cdot \big(1-2 y_n\big)^{\alpha}\big(y_n\big)^{\alpha}.\end{aligned}$$

From here we obtain the lower bound $$\begin{aligned}
\label{donja_alt}
u_n(0)\geq c_2(\alpha,n)\cdot \big(1-2 y_n\big)^{\alpha}\big(y_n\big)^{\alpha},\end{aligned}$$ where $$\begin{aligned}
&c_2(\alpha,n)=c(\alpha)|B_n|\cdot
[2\delta_n-\delta_n'+2y_n+ 1 +(\delta_n+1)\cdot\tan(A_n+B_n)]^{-\alpha-2}.\end{aligned}$$

Combining inequalities and , we obtain:

$$\begin{aligned}
\label{nejedn_alt}
\frac{u_n(w_0)}{u_n(0)}\leq c(\alpha)\cdot a_n b_n,\end{aligned}$$

where $$\begin{aligned}
a_n&=\Bigg(\frac{2\delta_n-\delta_n'+2y_n+1}{(\delta_n+1)\cdot\tan(A_n+B_n)}+1\Bigg)^{\alpha+2},\\
b_n&=\frac{\delta_n+x_n}{y_n^\alpha\cdot (1-2 y_n)^\alpha}.\end{aligned}$$ Due to the construction, $$\begin{aligned}
\label{limes_1_alt}
\lim_{n\rightarrow \infty} a_n=1.\end{aligned}$$

Using $\eqref{uvjet}$ and $\eqref{uvjet_d}$, we obtain $$\begin{aligned}
\label{limsup_alt}
\limsup_{n\rightarrow \infty} \frac{\delta_n+x_n}{y_n}<\infty.\end{aligned}$$ Since the proof of is similar to the proof of Lemma \[har\] (Appendix), we skip it. Since $0<\alpha<1$, $$\begin{aligned}
b_n&=\frac{\delta_n+x_n}{y_n^\alpha\cdot (1-2 y_n)^\alpha}\\
&=\frac{1}{(1-2 y_n)^\alpha}\cdot \frac{\delta_n+x_n}{y_n}\cdot y_n^{1-\alpha}.\end{aligned}$$

Therefore, $$\begin{aligned}
\label{limes_2_alt}
\lim_{n\rightarrow \infty} b_n=0.\end{aligned}$$

To conclude, $\eqref{nejedn_alt}$, $\eqref{limes_1_alt}$ and $\eqref{limes_2_alt}$ imply $$\begin{aligned}
\frac{u_n(0)}{u_n(w_0)}\end{aligned}$$ can be made as large as we like by taking $n$ large enough. This implies that the Harnack inequality for $X$ is not possible.

Notice that for $1<\alpha<2$, we have $$\begin{aligned}
  \frac{u_n(0)}{u_n(w_0)}\geq c(\alpha) \cdot \frac{1}{a_n} \cdot \frac{1}{b_n},
 \end{aligned}$$ where $$\begin{aligned}
 \frac{1}{b_n}=(1-2y_n)^\alpha \cdot \frac{y_n}{\delta_n+x_n} \cdot y_n^{\alpha-1}.\end{aligned}$$ Notice that from here, we have $$\begin{aligned}
\lim_{n\rightarrow \infty} \frac{1}{b_n}=0,\end{aligned}$$ therefore the proof breaks down.

In connection with the article [@bg-sz-1], there it was shown that as in our case ($d=2$), for $1<\alpha<2$ Harnack inequality holds (see Corollary 13 of the aforementioned article).

Weak Harnack Inequality
=======================

*The weak Harnack inequality* for a symmetric $\alpha$-stable Lévy processes $X$ holds if there is a constant $C\geq 1$ such that for every non-negative function $u\colon \mathbb{R}^d\rightarrow \mathbb{R}$, which is harmonic in $B_1$ with respect to $X$, the inequality $$\begin{aligned}
 \|u\|_{L^{1}(B_{1/2})} \leq C\cdot \inf_{B_{1/4}} u\end{aligned}$$ holds.

We continue with the definitions regarding the spherical part of the Lévy measure (see [@bg-sz-2] (cf. [@priola])).

A measure $\lambda$ on $\mathbb{R}^d$ is called *degenerate* if there is a proper linear subspace $M$ of $\mathbb{R}^d$ such that $Spt(\lambda)\subset M$, where $Spt(\lambda)$ denotes the support of the measure $\mu$.

A measure $\lambda$ is called *non-degenerate* if it is not degenerate.

\[def\_ned\] A measure $\mu$ on $\mathbb{S}^{d-1}$ is called a *spectral measure* if it is positive, finite, non-degenerate and symmetric.

For the equivalence of the non-degeneracy of the measure $\mu$ and the condition $$\begin{aligned}
\label{nondeg}
\Phi(u)\geq c\cdot |u|^\alpha, \ u\in \mathbb{R}^d,\end{aligned}$$ where $c=c(\alpha)$ is a positive constant and $\Phi$ is as in , see [@priola].

[\[Weak Harnack\]]{} Let $X$ be a symmetric $\alpha$-stable Lévy process in $\mathbb{R}^d$, $d\geq 2$, with index of stability $\alpha \in (0,2)$ and the characteristic function of the form $$\begin{aligned}
\mathbb{E}^0 e^ {iu\cdot X_t} =  e^{-t \Phi(u)}, \ \ u \in
\mathbb{R}^d,\ t\geq 0,\end{aligned}$$ where the characteristic exponent is given by $$\begin{aligned}
\Phi(u)=\int_{\mathbb{S}^{d-1}} |u\cdot \xi|^\alpha \mu(\mathrm{d}\xi),\end{aligned}$$ and $\mu$ is a spectral measure. Furthermore, let $\mu$ be absolutely continuous with respect to the uniform measure $\sigma$ on the sphere $\mathbb{S}^{d-1}$ and denote by $f_\mu$ its density. Assume that there is a positive constant $m$ such that $$\begin{aligned}
0\leq f_{\mu}(\xi)\leq m,\  \xi \in \mathbb{S}^{d-1}.\end{aligned}$$ Then the weak Harnack inequality for $X$ holds.

\[prva\_gornja\] Let $\alpha\in (0,2), d\geq 2$. There is a constant $c_1=c_1(\alpha,d)$ such that for every $|w|<3/4$, the inequality $$\begin{aligned}
   \int_{B_{1/2}} G_{B_{3/4}}(x,w)\ \mathrm{d}x\leq c_1
  \end{aligned}$$ holds.

By the inequality $p_D\leq p$ and the estimate of the transition density $p$ for small times (see e.g. [@sz-1], Theorem 1), it follows:$$\begin{aligned}
&\int_{B_{1/2}} G_{B_{3/4}}(x,w)\ \mathrm{d}x\\
&= \int_{B_{1/2}} \bigg[\int_0 ^\infty p_{B_{3/4}}(t,x,w)\ \mathrm{d}t\bigg]\ \mathrm{d}x \\
&\leq\int_{B_{1/2}} \bigg[\int_0 ^\infty p(t,x,w)\ \mathrm{d}t\bigg]\ \mathrm{d}x\\
&\leq c(\alpha,d) \int_{B_{1/2}} \bigg[\int_0 ^1 \bigg(t^{-d/\alpha} \wedge \frac{t}{|x-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}x\\
&\text{\;\;\;\;\;}+\int_{B_{1/2}} \bigg[\int_1 ^\infty t^{-d/\alpha}\cdot p(1, t^{-1/\alpha}x, t^{-1/\alpha}w)\ \mathrm{d}t\bigg]\ \mathrm{d}x\\
&\leq c(\alpha,d) \int_{B_{1/2}} \bigg[\int_0 ^{|x-w|^\alpha\wedge 1} \frac{t}{|x-w|^{\alpha+d}}\ \mathrm{d}t\bigg]\ \mathrm{d}x\\
&\text{\;\;\;\;\;} + c(\alpha,d) \int_{B_{1/2}} \bigg[\int_{|x-w|^\alpha\wedge 1}^1 t^{-d/\alpha}\ \mathrm{d}t\bigg]\ \mathrm{d}x\\
&\text{\;\;\;\;\;}+c(\alpha,d) \bigg[\int_{B_{1/2}} \int_1 ^\infty t^{-d/\alpha}\ \mathrm{d}t\bigg]\ \mathrm{d}x\\
&\leq c(\alpha,d) \Bigg[ \int_{B_{1/2}} |x-w|^{\alpha-d}\ \mathrm{d}x\\
&\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}+\frac{\alpha}{d-\alpha}\int_{B_{1/2}}\bigg[\big(|x-w|^\alpha\wedge 1\big)^{\frac{\alpha-d}{\alpha}}-1\bigg]\ \mathrm{d}x\Bigg]\\
&\text{\;\;\;\;\;}+c(\alpha,d)\ \frac{\alpha}{d-\alpha}|B_{1/2}|\\
&\leq c(\alpha,d).\end{aligned}$$

\[prva\_donja\_1\] Let $\alpha \in (0,2), d\geq 2$. There exist $\delta_1=\delta_1(\alpha,d)>0$ and $c_2=c_2(\alpha,d,\delta_1)$ such that for every $|\bar{x}|<1/4$ and every $w\in B(\bar{x},\delta_1)$ the inequality $$\begin{aligned}
G_{B_{3/4}}(\bar{x},w)\geq c_2\end{aligned}$$ holds.

Using $$\begin{aligned}
G_{B_{3/4}}(\bar{x},w)=G(\bar{x},w)-\mathbb{E}^{\bar{x}}[G(X_{\tau_{B_{3/4}}},w)],\label{formula}\end{aligned}$$ in order to prove the lemma, we compute the estimates for $G(\bar{x},w)$ from below and $\mathbb{E}^{\bar{x}}[G(X_{\tau_{B_{3/4}}},w)]$ from above. Using the heat kernel estimates for small times ([@sz-1], Theorem 1), we obtain $$\begin{aligned}
&\mathbb{E}^{\bar{x}}[G(X_{\tau_{B_{3/4}}},w)]\notag\\
&=\int_{{B_{3/4}}^c}\ G(u,w)P_{B_{3/4}}(\bar{x},u)\ \mathrm{d}u\notag\\
&= \int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{0}^{\infty} p(t,u,w)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&= \int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{0}^{1} p(t,u,w)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&\ + \int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{1}^{\infty} p(t,u,w)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&\leq c(\alpha,d)\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u) \bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&+ \int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{1}^{\infty} t^{-d/\alpha}\cdot p(1, t^{-1/\alpha}u, t^{-1/\alpha}w)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&\leq c(\alpha,d)\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[ \int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&+ c(\alpha, d)\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{1}^{\infty} t^{-d/\alpha}\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&\leq c(\alpha,d)\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&+ c(\alpha, d)\label{int}.\end{aligned}$$

Examining the integral in more closely, we obtain: $$\begin{aligned}
&\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&=\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|<1\}}(u)\bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&+\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|>1\}}(u)\bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
&=I_1+I_2.\label{skupa}\end{aligned}$$

$$\begin{aligned}
 I_1&=\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|<1\}}(u)\notag\\
 &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
 &=\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|<1\}}(u)|w-u|^{-\alpha-d}\bigg[\int_{0}^{|w-u|^\alpha}t\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
 &+\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|<1\}}(u)\bigg[\int_{|w-u|^\alpha}^{1}t^{-d/\alpha}\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
 &=c_1 \int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|<1\}}(u)|w-u|^{\alpha-d}\ \mathrm{d}u\notag\\
 &+c_2\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|<1\}}(u)\frac{\alpha}{d-\alpha}\big(|w-u|^{\alpha-d}-1\big)\ \mathrm{d}u\notag\\
 &\leq c_1(\alpha,d,\bar{\delta}_1),\label{prvi_int}\end{aligned}$$

where in the last inequality we have used $w \in B(\bar{x},\bar{\delta}_1)$, for $\bar{\delta}_1>0$ small enough.

$$\begin{aligned}
 I_2&=\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|>1\}}(u)\notag\\
   &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\bigg[\int_{0}^{1} \bigg(t^{-d/\alpha} \wedge \frac{t}{|u-w|^{\alpha+d}}\bigg)\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
   &= \int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|>1\}}(u)|w-u|^{-\alpha-d}\bigg[\int_{0}^{1} t\ \mathrm{d}t\bigg]\ \mathrm{d}u\notag\\
   &= c\int_{{B_{3/4}}^c}\ P_{B_{3/4}}(\bar{x},u)\mathbf{1}_{\{|w-u|>1\}}(u)|w-u|^{-\alpha-d}\ \mathrm{d}u\notag\\
      &\leq c_2.\label{drugi_int}\end{aligned}$$

In conclusion, by , , and , we obtain $$\begin{aligned}
\label{gornja}
 \mathbb{E}^{\bar{x}}[G(X_{\tau_{B_{3/4}}},w)]\leq \bar{c}_1(\alpha, d, \bar{\delta}_1),\end{aligned}$$ for all $w \in B(\bar{x},\bar{\delta}_1)$ and $\bar{\delta}_1>0$ small enough.

To estimate $G(\bar{x},w)$ from below, we use the continuity of the potential density (see [@taylor]).

Due to $$\begin{aligned}
 p(1,0)\geq c>0,\end{aligned}$$ by continuity of $p(1, \cdot)$ in $x=0$, there is $R>0$ such that $p(1,x)>\frac{1}{2}\cdot p(1,0)$, for all $|x|<R$.

Furthermore, for $|\xi|=1$, since: $$\begin{aligned}
 G(0,\xi)&\geq\int_{R^{-\alpha}}^{\infty} p(t,\xi)\ \mathrm{d}t\\
       &= \int_{R^{-\alpha}}^{\infty}t^{-d/\alpha}\cdot p\Big(1,\frac{\xi}{t^{1/\alpha}}\Big)\ \mathrm{d}t > \frac{1}{2}\cdot \int_{R^{-\alpha}}^{\infty}t^{-d/\alpha}\cdot p(1,0) \ \mathrm{d}t\\
    &=c_1(\alpha,d)>0,\end{aligned}$$ we obtain $$\begin{aligned}
\label{ocjena}
 G(0,\xi)\geq c_1(\alpha,d).\end{aligned}$$

For $|x|\neq 0$, by scaling and $$\begin{aligned}
 G(0,x)=|x|^{\alpha-d}\cdot G\bigg(0,\frac{x}{|x|}\bigg)\geq c_1 \cdot |x|^{\alpha-d}.\end{aligned}$$

Therefore, $$\begin{aligned}
\label{donja}
 G(\bar{x},w) \geq \bar{c}_2 \cdot |\bar{x}-w|^{\alpha-d}.\end{aligned}$$

Now, choose $\delta_1$ such that $\delta_1 < (\bar{c}_2/(\bar{c}_1+c))^{\frac{1}{d-\alpha}}\wedge \bar{\delta}_1$, where $c>0$. Then for every $|\bar{x}|<1/4$ and $w\in B(\bar{x},\delta_1)$, combining , and , we obtain $$\begin{aligned}
 G_{B_{3/4}}(\bar{x},w)\geq \bar{c}_2\cdot |\bar{x}-w|^{\alpha-d}-\bar{c}_1\geq c>0.\end{aligned}$$ Define $c_2=c$ and now the statement follows.

\[nap\] Notice that, according to the Lemma \[prva\_gornja\] and Lemma \[prva\_donja\_1\], there are $\tilde{c}=\tilde{c}(\alpha,d)$ and $\delta_1=\delta_1(\alpha,d)$ such that for every $\bar{x}\in B_{1/4}$ and for every $w\in B(\bar{x},\delta_1)$ the inequality $$\begin{aligned}
  \int_{B_{1/2}}\ G_{B_{3/4}}(x,w)\ \mathrm{d}x \leq \tilde{c}\cdot G_{B_{3/4}}(\bar{x},w)\end{aligned}$$ holds.

\[treca\_donja\_1\] For $\alpha \in (0,2)$ and $d\geq 2$, let $\delta_1>0$ be as in Lemma \[prva\_donja\_1\]. There is a constant $c_3=c_3(\alpha,d)$ such that for every $\bar{x}\in B_{1/4}$ and every $\tilde{u}\in B(0,3/4)\setminus B(\bar{x},\delta_1)$ the inequality $$\begin{aligned}
\int_{B_{1/2}}\ G_{B_{3/4}} (x,\tilde{u})\ \mathrm{d}x \leq c_3 \cdot G_{B_{3/4}} (\bar{x}, \tilde{u})
$$ holds.

The proof relies on the maximum principle (cf. [@landkof]). We use the fact that $G_D (\bar{x}, \cdot)$ is regular harmonic in $D \setminus B(\bar{x}, \varepsilon)$ with respect to $X$ for every $\varepsilon > 0$ (cf. [@bg-1]). $$\begin{aligned}
\int_{B_{1/2}} \ G_{B_{3/4}}(x,\tilde{u})\ \mathrm{d}x &= \int_{B_{1/2}} \mathbb{E}^{\tilde{u}}[G_{B_{3/4}}(x, X_{\tau_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}})]\ \mathrm{d}x\\
                             &= \int_{B_{1/2}} \bigg[\ \int_{\mathbb{R}^2\setminus ({B_{3/4}}\setminus B(\bar{x},\delta_1))}\\
                             &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}G_{B_{3/4}}(x,z)\ P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}z\bigg]\ \mathrm{d}x\\
                             &= \int_{\mathbb{R}^2\setminus ({B_{3/4}}\setminus B(\bar{x},\delta_1))} \ \bigg[\int_{B_{1/2}} \\
                             &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}G_{B_{3/4}}(x,z)\ P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}x\bigg]\ \mathrm{d}z\\
                             &= \int_{B(\bar{x},\delta_1)}\bigg[\int_{B_{1/2}} \\
                             &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}G_{B_{3/4}}(x,z)\ P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}x\bigg]\ \mathrm{d}z\\
                             &\text{\;\;}+ \int_{\mathbb{R}^2\setminus B_{3/4}} \bigg[\int_{B_{1/2}} \\
                             &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}G_{B_{3/4}}(x,z)\ P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}x\bigg]\ \mathrm{d}z\\
                             &=\int_{B(\bar{x},\delta_1)}  \bigg[\int_{B_{1/2}}\\
                             &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}G_{B_{3/4}}(x,z)\ P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}x\bigg]\ \mathrm{d}z\\     
                                                         &\overset{Rem. \ref{nap}}{\leq}\tilde{c}\int_{B(\bar{x},\delta_1)} \ G_{B_{3/4}}(\bar{x},z)P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}z\\
                             &= \tilde{c} \int_{\mathbb{R}^2\setminus ({B_{3/4}}\setminus B(\bar{x},\delta_1))}
                             G_{B_{3/4}}(\bar{x},z)P_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}(\tilde{u},z)\ \mathrm{d}z\\
                             &= \tilde{c} \cdot \mathbb{E}^{\tilde{u}}[G_{B_{3/4}}(\bar{x}, X_{\tau_{{B_{3/4}}\setminus B(\bar{x},\delta_1)}})]\\
                             &= \tilde{c} \cdot G_{B_{3/4}} (\bar{x}, \tilde{u}).\end{aligned}$$

Define $c_3=\tilde{c}$ and the lemma follows.

$$\begin{aligned}
&\|u\|_{L^{1}(B_{1/2})}\notag\\
&=\int_{B_{1/2}} u(x)\ \mathrm{d}x\notag\\
&=\int_{B_{1/2}} \mathbb{E}^{x}[u(X_{\tau_{B_{3/4}}})]\ \mathrm{d}x\notag \\
&=\int_{B_{1/2}} \bigg[\int_{(\bar{B}_{3/4})^{c}} \ u(y)P_{B_{3/4}} (x,y)\ \mathrm{d}y\bigg]\mathrm{d}x\notag \\ &=\int_{B_{1/2}} \Bigg[\int_{(\bar{B}_{3/4})^{c}} u(y)\ \bigg[\int _{B(y,3/4)} f_\nu(z) \ G_{B_{3/4}}(x,y-z)\ \mathrm{d}z\bigg]\ \mathrm{d}y\Bigg]\ \mathrm{d}x\notag \\
&= \int_{(\bar{B}_{3/4})^{c}} u(y) \Bigg[\int _{B(y,3/4)} f_\nu(z)\ \bigg[\int_{B_{1/2}}  G_{B_{3/4}}(x,y-z)\ \mathrm{d}x\bigg]\ \mathrm{d}z\Bigg]\ \mathrm{d}y\label{zad}\end{aligned}$$

By Remark \[nap\] and Lemma \[treca\_donja\_1\] there is a constant $c=c(\alpha,d)$ such that for every $\bar{x}\in B_{1/4}$ $$\begin{aligned}
  \eqref{zad}\ &\leq\ c\cdot \int_{(\bar{B}_{3/4})^{c}}  u(y)\bigg[\int _{B(y,3/4)} \mathbf{1}_{B(\bar{x},\delta_1)}(y-z)\\
  &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}f_\nu(z)\cdot G_{B_{3/4}}(\bar{x},y-z)\ \mathrm{d}z\bigg]\ \mathrm{d}y\\
  &\text{\;\;}+ c\cdot \int_{(\bar{B}_{3/4})^{c}} u(y)\bigg[\int _{B(y,3/4)} \mathbf{1}_{B(\bar{x},\delta_1)^c}(y-z)\\
  &\text{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}f_\nu(z)\cdot G_{B_{3/4}}(\bar{x},y-z)\ \mathrm{d}z\bigg]\ \mathrm{d}y\\
  &= c\cdot \int_{(\bar{B}_{3/4})^{c}} u(y)\bigg[\int _{B(y,3/4)} f_\nu(z)\ G_{B_{3/4}}(\bar{x},y-z)\ \mathrm{d}z\bigg]\ \mathrm{d}y\\
  &=c\cdot \int_{(\bar{B}_{3/4})^{c}} u(y)P_{B_{3/4}} (\bar{x},y)\ \mathrm{d}y\\
  &=c\cdot \mathbb{E}^{\bar{x}}[u(X_{\tau_{{B_{3/4}}}})]\\
  &=c\cdot u(\bar{x}).\end{aligned}$$

Since the inequality $$\begin{aligned}
 \|u\|_{L^{1}(B_{1/2})}\leq c\cdot u(\bar{x}),\end{aligned}$$ holds for any $\bar{x}\in B_{1/4}$, the proof is finished.

Appendix {#Ap}
========

[\[lema\_pr\]]{} Let:

1.  $\alpha_k = 2^{-k}, \ \beta_k = \frac{1}{k(k+1)}$,

2.  $\alpha_k = a^{-k}, \ \beta_k = b^{-k}$,   $1 < b \leq a$,

3.  $\alpha_k =  \frac{1}{k(k+1)} , \ \beta_k = k^{-1-\delta},\ \delta > 0$,

4.  $\alpha_k = 2^{-(k^3)}, \ \beta_k = 2^{-(k^2)}$\
    $\alpha_k = a^{-(k^3)}, \ \beta_k = b^{-(k^2)}$,$a>1,b>1$.

Then none of these sequences does not satisfy the property from the Lemma \[lema\].

Notice that $$\begin{aligned}
\lim_{n \to \infty}\frac{\tan(A_n + B_{n+1})}{\tan(A_n + B_n)}=\infty\end{aligned}$$ implies $$\begin{aligned}
\label{limes1}
\lim_{n \to \infty}\frac{\cos(A_n + B_{n+1})}{\cos(A_n + B_n)}=0.\end{aligned}$$

1.  Set $$\begin{aligned}
     \alpha_k = 2^{-k}, \qquad \beta_k = \frac{1}{k(k+1)}.
     \end{aligned}$$

    Since $A_n = 1-2^{-n}$, $B_n=\frac{n}{n+1}$ and $B_{n+1}=B_n+\beta_{n+1}$, we have: $$\begin{aligned}
     \lim_{n \to \infty}\frac{\cos(A_n + B_{n+1})}{\cos(A_n + B_n)}&=\lim_{n \to \infty}\frac{\frac{d}{dn}(A_n+B_{n+1})}{\frac{d}{dn}(A_n+B_n)}\\
     &=1+\lim_{n \to \infty}\frac{\frac{d}{dn}\beta_{n+1}}{\frac{d}{dn}(A_n+B_n)}\\
     &=1+ \lim_{n \to \infty} \frac{\frac{-2n-3}{(n+1)^2(n+2)^2}}{\ln2\cdot 2^{-n}+(n+1)^{-2}}\\
     &=1+0=1.
     \end{aligned}$$

    This means that the condition (\[limes1\]) is not fulfilled.

2.  Set $$\begin{aligned}
     \alpha_k = a^{-k}, \ \beta_k =  b^{-k},\quad 1<b\leq a.
     \end{aligned}$$

    Since $A_n = \frac{1-a^{-n}}{a-1}$, $B_n=\frac{1-b^{-n}}{b-1}$, we have: $$\begin{aligned}
     \lim_{n \to \infty}\frac{\cos(A_n + B_{n+1})}{\cos(A_n + B_n)}&=\lim_{n \to \infty}\frac{\frac{d}{dn}(A_n+B_{n+1})}{\frac{d}{dn}(A_n+B_n)}\\
     &=1+ \lim_{n \to \infty}\frac{\frac{d}{dn}\beta_{n+1}}{\frac{d}{dn}(A_n+B_n)}\\
     &=1+ \lim_{n \to \infty} \frac{-\ln b\cdot b^{-(n+1)}}{\frac{\ln a}{a-1}a^{-n}+ \frac{\ln b}{b-1}b^{-n}}\\
     &=1+ \lim_{n \to \infty} \frac{-\ln b\cdot b^{-1}}{\frac{\ln a}{a-1}(\frac{b}{a})^{n}+ \frac{\ln b}{b-1}}.
     \end{aligned}$$

    If $a=b$, the last expression is equal to

    $$\begin{aligned}
     1- \frac{\ln a\cdot a^{-1}}{2\frac{\ln a}{a-1}}&=1-\frac{a-1}{2a}\\
     &=\frac{a+1}{2a}.\end{aligned}$$

    If $b<a$, we obtain

    $$\begin{aligned}
     1- \frac{\ln b\cdot b^{-1}}{\frac{\ln b}{b-1}}&=1-\frac{b-1}{b}\\
     &=\frac{1}{b}.\end{aligned}$$

    In conclusion, the condition (\[limes1\]) is not fulfilled.

    In the examples that follow, we will proceed in the following way.

    We use the equality $$\begin{aligned}
    \frac{\cos(A_n + B_{n+1})}{\cos(A_n + B_n)}
    =
    \frac{\sin(\frac{\pi}{2}-A_n - B_{n+1})}{\sin(\frac{\pi}{2}-A_n - B_n)},\end{aligned}$$ and the estimates $$\begin{aligned}
    c_1\cdot \frac{\frac{\pi}{2}-A_n - B_{n+1}}{\frac{\pi}{2}-A_n - B_n}\leq\frac{\sin(\frac{\pi}{2}-A_n - B_{n+1})}{\sin(\frac{\pi}{2}-A_n - B_n)}\label{oc_1}\\
    \frac{\sin(\frac{\pi}{2}-A_n - B_{n+1})}{\sin(\frac{\pi}{2}-A_n - B_n)}\leq c_2\cdot \frac{\frac{\pi}{2}-A_n - B_{n+1}}{\frac{\pi}{2}-A_n - B_n}\label{oc_2},\end{aligned}$$ for some constants $c_1, c_2>0$.

    Examining the lower (or upper) bound in the expression (or ), we see $$\begin{aligned}
    \frac{\frac{\pi}{2}-A_n - B_n-\beta_{n+1}}{\frac{\pi}{2}-A_n - B_n}&=1-\frac{\beta_{n+1}}{\frac{\pi}{2}-A_n - B_n}\notag\\
    &=1-\frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k}\label{limes2}.\end{aligned}$$

    Now, we can use integral test for convergence to estimate series

    $$\sum_{k=n+1}^\infty \alpha_k$$ and $$\sum_{k=n+1}^\infty \beta_k.$$

3.  Let $$\begin{aligned}
     \alpha_k =  \frac{1}{k(k+1)} , \qquad \beta_k = k^{-1-\delta},\end{aligned}$$ for some $\delta > 0$. In this example we will use integral test.

    Notice, $$\begin{aligned}
     \sum_{k=n+1}^\infty \alpha_k = \frac{1}{n+1}.  \end{aligned}$$

    The integral test implies $$\begin{aligned}
     \int_{n+1}^\infty \beta(x)dx \leq \sum_{k=n+1}^\infty \beta_k \leq \beta_{n+1} + \int_{n+1}^\infty \beta(x)dx,
     \end{aligned}$$ which is equivalent to $$\begin{aligned}
      \sum_{k=n+1}^\infty \alpha_k + \int_{n+1}^\infty \beta(x)dx &\leq  \sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k,\notag\\
      \sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k &\leq \sum_{k=n+1}^\infty \alpha_k + \beta_{n+1} + \int_{n+1}^\infty \beta(x)dx.
     \end{aligned}$$ These inequalities yield $$\begin{aligned}
      \frac{1}{\sum_{k=n+1}^\infty \alpha_k + \beta_{n+1} + \int_{n+1}^\infty \beta(x)dx} &\leq \frac{1}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k},\notag\\
      \frac{1}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k} &\leq \frac{1}{\sum_{k=n+1}^\infty \alpha_k + \int_{n+1}^\infty \beta(x)dx}.
     \end{aligned}$$

    Multiplying the two inequalities by $\beta_{n+1}$, we obtain the lower and the upper bound for the expression in (\[limes2\]),

    $$\begin{aligned}
     \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \beta_{n+1} + \int_{n+1}^\infty \beta(x)dx}
     \leq \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k}
     \end{aligned}$$

    and

    $$\begin{aligned}
     \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k}
      \leq \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \int_{n+1}^\infty \beta(x)dx}.
     \end{aligned}$$

    Since $$\begin{aligned}
     \lim_{n \to \infty} \frac{\sum_{k=n+1}^\infty \alpha_k}{\beta_{n+1}}=\infty,  \end{aligned}$$ it follows

    $$\begin{aligned}
     \lim_{n \to \infty} \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \int_{n+1}^\infty \beta(x)dx}=0
     \end{aligned}$$

    and consequently $$\begin{aligned}
     \lim_{n \to \infty}\frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k}=0.
     \end{aligned}$$ This means, by ,

    $$\begin{aligned}
     \lim_{n \to \infty} \frac{\frac{\pi}{2}-A_n - B_{n+1}}{\frac{\pi}{2}-A_n -
    B_n}=1-0=1,
     \end{aligned}$$

    which implies the condition is not fulfilled.

4.  Set $$\begin{aligned}
      \alpha_k = 2^{-(k^3)}, \ \ \beta_k = 2^{-(k^2)}.
      \end{aligned}$$ As in the second example, we use the integral test. Similarly, it implies $$\begin{aligned}
      &\int_{n+1}^\infty \alpha(x)dx +  \int_{n+1}^\infty \beta(x)dx \leq \sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k,\\
      &\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k \leq \alpha_{n+1}+\int_{n+1}^\infty \alpha(x)dx + \beta_{n+1}+\int_{n+1}^\infty \beta(x)dx,
     \end{aligned}$$ which yields $$\begin{aligned}
     &\frac{1}{\alpha_{n+1}+\int_{n+1}^\infty \alpha(x)dx + \beta_{n+1}+\int_{n+1}^\infty \beta(x)dx}\notag \\
     &\leq \frac{1}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k},\notag\\
     &\frac{1}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k} \leq \frac{1}{\int_{n+1}^\infty \alpha(x)dx +  \int_{n+1}^\infty \beta(x)dx}.\label{njd}
     \end{aligned}$$

    We will only observe the inequality . Multiplying (\[njd\]) by $\beta_{n+1}$, it follows $$\begin{aligned}
     \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k}
      \leq \frac{\beta_{n+1}}{\int_{n+1}^\infty \alpha(x)dx +  \int_{n+1}^\infty \beta(x)dx}.
     \end{aligned}$$ Since $$\begin{aligned}
     \frac{\beta_{n+1}}{\int_{n+1}^\infty \alpha(x)dx +  \int_{n+1}^\infty \beta(x)dx} \leq \frac{\beta_{n+1}}{\int_{n+1}^\infty \beta(x)dx},
     \end{aligned}$$ it is enough to show $$\label{uvjet_pr}
     \lim_{n \to \infty}\frac{\beta_{n+1}}{\int_{n+1}^\infty \beta(x)dx}=0.$$

    Notice that then $$\begin{aligned}
     \lim_{n \to \infty} \frac{\beta_{n+1}}{\sum_{k=n+1}^\infty \alpha_k + \sum_{k=n+1}^\infty \beta_k}=0,
     \end{aligned}$$ which implies the condition is not fulfilled.

    Using polar coordinates we obtain $$\begin{aligned}
     \int_{n+1}^\infty \beta(x)dx &=\int_{n+1}^\infty 2^{-(x^2)}dx\\
     &=\frac{1}{2}\sqrt{\frac{\pi}{\ln2}}\cdot 2^{-\frac{1}{2}(n+1)^2},
     \end{aligned}$$ which implies $$\begin{aligned}
     \lim_{n \to \infty} \frac{\beta_{n+1}}{\int_{n+1}^\infty \beta(x)dx}&=\lim_{n \to \infty}\frac{2^{-(n+1)^2}}{\frac{1}{2}\sqrt{\frac{\pi}{\ln2}}\cdot 2^{-\frac{1}{2}(n+1)^2}}\\
     &=\lim_{n \to \infty}\frac{2^{-\frac{1}{2}(n+1)^2}}{\frac{1}{2}\sqrt{\frac{\pi}{\ln2}}}\\
     &=0,
     \end{aligned}$$ and that is what we wanted to prove.

Notice that only in the very end the explicit formula for $\beta_n$ has been used. This means that if $$\begin{aligned}
 \beta_k = b^{-(k^2)},\ b>1,
 \end{aligned}$$ the same reasoning applies.

In this case, we would have $$\begin{aligned}
 \int_{n+1}^\infty \beta(x)dx &=\int_{n+1}^\infty b^{-(x^2)}dx\\
 &=\frac{1}{2}\sqrt{\frac{\pi}{\ln b}}\cdot b^{-\frac{1}{2}(n+1)^2}
 \end{aligned}$$ and $$\begin{aligned}
 \lim_{n \to \infty} \frac{\beta_{n+1}}{\int_{n+1}^\infty \beta(x)dx}&=\lim_{n \to \infty}\frac{b^{-(n+1)^2}}{\frac{1}{2}\sqrt{\frac{\pi}{\ln b}}\cdot b^{-\frac{1}{2}(n+1)^2}}\\
 &=\lim_{n \to \infty}\frac{b^{-\frac{1}{2}(n+1)^2}}{\frac{1}{2}\sqrt{\frac{\pi}{\ln b}}}\\
 &=0.
 \end{aligned}$$

[\[lema4\]]{} Let the quantities $\delta_n$, $I_n$, $P_n$, $P_n'$, $\delta_n$, $x_n$, $y_n$, $c_n$ and $d_n$ be as in the Proposition \[prop4\]. Then:

1.  $d_n=(\delta_n + x_n)\cdot \tan(A_n+B_{n+1})-1$;

2.  all quantities above are positive and $c_n<d_n$.

<!-- -->

1.  By the definition , $$\begin{aligned}
    &\delta_n=\frac{I_n(1+\tan(A_n+B_n))\cdot(1+\tan(A_{n+1}+B_{n+1}))}{P_n^2(1+\tan(A_n+B_{n+1}))+P_n'I_n(1+\tan(A_{n+1}+B_{n+1}))}.\end{aligned}$$ This expression is equivalent to $$\begin{aligned}
     &I_n(1+\tan(A_n+B_n))\cdot(1+\tan(A_{n+1}+B_{n+1}))\\
     &=\delta_n P_n'I_n\cdot(1+\tan(A_{n+1}+B_{n+1}))+\delta_n P_n^2\cdot(1+\tan(A_n+B_{n+1})),\end{aligned}$$ from where $$\begin{aligned}
     &I_n(1+\tan(A_n+B_n))\cdot(1+\tan(A_{n+1}+B_{n+1}))\\
     &=\delta_n P_n'I_n\cdot(1+\tan(A_{n+1}+B_{n+1}))\\
     &+\delta_nP_n(1+\tan(A_n+B_{n+1}))\cdot(I_n-2(1+\tan(A_n+B_{n+1})))\end{aligned}$$ follows. Therefore, $$\begin{aligned}
     &2\delta_n P_n\cdot(1+\tan(A_n+B_{n+1}))^2 \\
     &=\delta_n P_n I_n\cdot(1+\tan(A_n+B_{n+1}))\\
     &+I_n(1+\tan(A_{n+1}+B_{n+1}))\cdot (\delta_n P_n'-\tan(A_n+B_n)-1),\end{aligned}$$ and eventually $$\begin{aligned}
     &2\delta_n\frac{P_n}{I_n}\cdot(1+\tan(A_n+B_{n+1}))=\\
     &=\delta_n P_n +\frac{1+\tan(A_{n+1}+B_{n+1})}{1+\tan(A_n+B_{n+1})}\cdot (\delta_n P_n'-\tan(A_n+B_n)-1).\end{aligned}$$ By the definition of $x_n$ , the last equality yields $$\begin{aligned}
    &x_n\cdot\tan(A_n+B_{n+1})-1=\\
     &=\delta_n P_n +\frac{1+\tan(A_{n+1}+B_{n+1})}{1+\tan(A_n+B_{n+1})}\cdot (\delta_nP_n'-\tan(A_n+B_n)-1).\end{aligned}$$ Therefore, $$\begin{aligned}
     \frac{1+\tan(A_n+B_n)-\delta_n P_n'}{1+\tan(A_n+B_{n+1})}=\frac{\delta_n P_n-x_n\cdot\tan(A_n+B_{n+1})+1}{1+\tan(A_{n+1}+B_{n+1})}.\end{aligned}$$ Notice that by the left-hand side is $y_n$, hence $$\begin{aligned}
     y_n=\frac{\delta_n P_n-x_n\cdot\tan(A_n+B_{n+1})+1}{1+\tan(A_{n+1}+B_{n+1})}.\end{aligned}$$ From here, by the definition of $P_n$ , $$\begin{aligned}
    &(\delta_n-y_n)\cdot\tan(A_{n+1}+B_{n+1})-y_n\\
    &=(\delta_n + x_n)\cdot \tan(A_n+B_{n+1})-1.\end{aligned}$$ Consequently, by the definition of $d_n$ , the last equality implies $$\begin{aligned}
     d_n=(\delta_n + x_n)\cdot \tan(A_n+B_{n+1})-1,
     \end{aligned}$$ and hence the first part of the Lemma.

2.  Using the definition of $\delta_n$ one sees that $y_n>0$. Namely, the nominator of $y_n$ is positive if and only if $$\begin{aligned}
     1+\tan(A_n+B_n)-\delta_n\cdot P_n'>0,\end{aligned}$$ which is equivalent to $$\begin{aligned}
     \frac{1+\tan(A_n+B_n)}{P_n'}>\delta_n.\end{aligned}$$ By the definition of $\delta_n$, the condition above is fulfilled if and only if $$\begin{aligned}
      &\frac{1+\tan(A_n+B_n)}{P_n'}\\
      &>\frac{I_n\cdot (1+\tan(A_n+B_n))\cdot (1+\tan(A_{n+1}+B_{n+1}))}{H_n},\end{aligned}$$ which is equivalent to $$\begin{aligned}
     H_n>P_n'\cdot I_n\cdot (1+\tan(A_{n+1}+B_{n+1})).\end{aligned}$$ Using , we see that it holds.

    Consequently, $c_n$ is also positive.

    Next we show that $c_n<d_n$, which also implies that $d_n>0$. Since $$\begin{aligned}
     I_n>0,\end{aligned}$$ by the definition of $P_n$ , the inequality $$\begin{aligned}
     2\cdot (1+\tan(A_{n+1}+B_{n+1}))>P_n\end{aligned}$$ holds.

    Therefore, $$\begin{aligned}
    &2\cdot P_n (1+\tan(A_n+B_{n+1}))\cdot(1+\tan(A_{n+1}+B_{n+1}))\\
    &>P_n^2(1+\tan(A_n+B_{n+1})).\end{aligned}$$

    Using the expression for $H_n$ , $$\begin{aligned}
     &2\cdot P_n (1+\tan(A_n+B_{n+1}))\cdot(1+\tan(A_{n+1}+B_{n+1}))\\
     &+P_n'I_n\cdot(1+\tan(A_{n+1}+B_{n+1}))> H_n,\end{aligned}$$ which, by the definition of $\delta_n$ yields $$\begin{aligned}
     \delta_n>\frac{I_n\cdot(1+\tan(A_n+B_n))}{2\cdot P_n (1+\tan(A_n+B_{n+1}))+P_n'I_n}.\end{aligned}$$

    Therefore, $$\begin{aligned}
     2\delta_n\cdot P_n (1+\tan(A_n+B_{n+1}))> (1+\tan(A_n+B_n)-\delta_n P_n')\cdot I_n,\end{aligned}$$ which implies $$\begin{aligned}
     2\delta_n\cdot\frac{1+\tan(A_n+B_{n+1})}{\tan(A_n+B_{n+1})}\cdot \frac{P_n}{I_n}>\frac{1+\tan(A_n+B_n)-\delta_n P_n'}{\tan(A_n+B_{n+1})}.\end{aligned}$$

    Consequently, by the definition of $x_n$ $$\begin{aligned}
    x_n>\frac{2+\tan(A_n+B_n)-\delta_n P_n'}{\tan(A_n+B_{n+1})}.\end{aligned}$$

    Hence, using the definition for $y_n$ , $$\begin{aligned}
     \frac{x_n\cdot\tan(A_n+B_{n+1})-1}{1+\tan(A_n+B_{n+1})}>y_n.\end{aligned}$$

    Therefore, $$\begin{aligned}
    (\delta_n + x_n)\cdot \tan(A_n+B_{n+1})-1> (\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n.\end{aligned}$$

    By the first part of the Lemma and the definition of $c_n$ , the equality $$\begin{aligned}
    d_n>c_n,\end{aligned}$$ holds.

    Hence $d_n>0$ and the proof is finished.

[\[lema5\]]{} Let $I_n$, $J_n$, $P_n$, $P_n'$, $\delta_n$, $x_n$, $y_n$, $c_n$, $d_n$, $U_n$, $\delta_n'$, $S_n=(x_{S_n}, y_{S_n})$ and $r_n$ be as in the Proposition \[prop5\]. Then:

1.  $B(S_n, r_n)\subset Q_n:=(\delta_n', \delta_n)\times (c_n,c_n+(\delta_n-\delta_n'))$;

2.  the inequalities $$\begin{aligned}
     c_n &\geq(\varepsilon+1)\cdot\tan(A_n+B_n)+1,\\
     c_n+(\delta_n-\delta_n') &\leq(\varepsilon+x_n)\cdot\tan(A_n+B_{n+1})-1,\end{aligned}$$ hold true, for some $\varepsilon>0$ such that $\delta_n'<\varepsilon<\delta_n$.

<!-- -->

1.  Let $z=(z_1, z_2)\in B(S_n, r_n)$. Clearly, $$\begin{aligned}
    |z_1-x_{S_n}|<r_n,\\
    |z_2-y_{S_n}|<r_n.\end{aligned}$$

    If we show: $$\begin{aligned}
    \label{prvo}
    x_{S_n}-r_n=\delta_n'\end{aligned}$$ $$\begin{aligned}
    \label{drugo}
    x_{S_n}+r_n=\delta_n\end{aligned}$$ $$\begin{aligned}
    \label{trece}
    y_{S_n}-r_n=c_n\end{aligned}$$ $$\begin{aligned}
    \label{cetvrto}
    y_{S_n}+r_n=c_n+(\delta_n-\delta_n'),\end{aligned}$$ then the proof of the first part of the Lemma is finished.

    Let us first check . The definitions of $x_{S_n}$ , $r_n$ , $I_n$ , $P_n$ and $\delta_n'$ imply $$\begin{aligned}
     x_{S_n}-r_n&=\frac{\delta_n+y_n}{2}\cdot\frac{I_n}{1+\tan(A_{n+1}+B_{n+1})}\\
    &\text{\;\;\;\;}-\frac{1}{2}\cdot\frac{\delta_n  P_n-y_n  I_n}{1+\tan(A_{n+1}+B_{n+1})}\\
      &=\frac{\delta_n(I_n-P_n)+2y_nI_n}{2(1+\tan(A_{n+1}+B_{n+1}))}\\
      &=\frac{\delta_n(1+\tan(A_n+B_{n+1}))+y_nI_n}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=\delta_n'.\end{aligned}$$ Let us show . Similarly, the definitions of $x_{S_n}$ , $r_n$ , $I_n$ , $P_n$ and $\delta_n$ imply $$\begin{aligned}
     x_{S_n}+r_n&=\frac{\delta_n+y_n}{2}\cdot\frac{I_n}{1+\tan(A_{n+1}+B_{n+1})}\\
    &\text{\;\;\;\;}+\frac{1}{2}\cdot\frac{\delta_n  P_n-y_n  I_n}{1+\tan(A_{n+1}+B_{n+1})}\\
      &=\frac{\delta_n(I_n+P_n)}{2(1+\tan(A_{n+1}+B_{n+1}))}\\
     &=\frac{2\delta_n(1+\tan(A_{n+1}+B_{n+1}))}{2(1+\tan(A_{n+1}+B_{n+1}))}\\
     &=\delta_n.\end{aligned}$$ Now, we check . Likewise, the definitions of $y_{S_n}$ , $r_n$ , $I_n$ , $J_n$ , $P_n$ and $c_n$ yield $$\begin{aligned}
     y_{S_n}-r_n&=\frac{\delta_n+y_n}{2}\cdot\frac{J_n}{1+\tan(A_{n+1}+B_{n+1})}\\
    &\text{\;\;\;\;}-\frac{1}{2}\cdot\frac{\delta_n P_n-y_n I_n}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=\frac{\delta_n(J_n-P_n)+y_n(I_n+J_n)}{2(1+\tan(A_{n+1}+B_{n+1}))}\\
     &=\delta_n\tan(A_n+B_{n+1})+y_n\tan(A_n+B_{n+1})+y_n\\
     &=(\delta_n+y_n)\cdot\tan(A_n+B_{n+1})+y_n\\
     &=c_n.\end{aligned}$$ Lastly, we check . The definitions of $y_{S_n}$ , $r_n$ , $I_n$ , $J_n$ and $P_n$ imply $$\begin{aligned}
     y_{S_n}+r_n&=\frac{\delta_n+y_n}{2}\cdot\frac{J_n}{1+\tan(A_{n+1}+B_{n+1})}\\
    &\text{\;\;\;\;}+\frac{1}{2}\cdot\frac{\delta_n\cdot P_n-y_n\cdot I_n}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=\frac{\delta_n\tan(A_{n+1}+B_{n+1})(1+\tan(A_n+B_{n+1}))}{1+\tan(A_{n+1}+B_{n+1})}\\
     &+\frac{y_n(\tan(A_n+B_{n+1})\tan(A_{n+1}+B_{n+1})-1)}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=\frac{\delta_n\tan(A_{n+1}+B_{n+1})(1+\tan(A_n+B_{n+1}))}{1+\tan(A_{n+1}+B_{n+1})}\\
     &+\frac{y_n((1+\tan(A_n+B_{n+1}))(1+\tan(A_{n+1}+B_{n+1}))-I_n)}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=\frac{(\delta_n+y_n)(1+\tan(A_n+B_{n+1}))(1+\tan(A_{n+1}+B_{n+1}))}{1+\tan(A_{n+1}+B_{n+1})}\\
     &+\frac{-\delta_n(1+\tan(A_n+B_{n+1}))-y_nI_n}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=(\delta_n+y_n)(1+\tan(A_n+B_{n+1}))\\
     &+\frac{-\delta_n(1+\tan(A_n+B_{n+1}))-y_nI_n}{1+\tan(A_{n+1}+B_{n+1})}\\
     &=c_n+(\delta_n-\delta_n'),\end{aligned}$$ where in the last equality we use the definitions of $c_n$ and $\delta_n'$ .

2.  Recall, $$\begin{aligned}
    \Delta_{n}^{\varepsilon}=\{(y_1,y_2): &y_1=\varepsilon,
    (\varepsilon+1)\cdot\tan(A_n+B_n)+1<\\
    &<y_2<(\varepsilon+x_n)\cdot\tan(A_n+B_{n+1})-1\}.\end{aligned}$$

    We start by proving the inequality $$\begin{aligned}
    (\varepsilon+1)\cdot\tan(A_n+B_n)+1\leq c_n,\end{aligned}$$ for some $\delta_n'<\varepsilon<\delta_n$.

    Let $\delta_n'<\varepsilon<\delta_n$. Then the inequality $$\begin{aligned}
    (\varepsilon+1)\cdot\tan(A_n+B_n)+1\leq (\delta_n+1) \tan(A_n+B_n)+1\end{aligned}$$ holds. Therefore, the definition of $P_n'$ yields $$\begin{aligned}
    &(\varepsilon+1)\cdot\tan(A_n+B_n)+1\\
    &\leq \delta_n\cdot\tan(A_n+B_{n+1})+(1+\tan(A_n+B_n)-\delta_n P_n').\end{aligned}$$

    The definition of $y_n$ implies $$\begin{aligned}
    (\varepsilon+1)\cdot\tan(A_n+B_n)+1\leq (\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n.\end{aligned}$$

    Eventually, the definition of $c_n$ implies $$\begin{aligned}
    (\varepsilon+1)\cdot\tan(A_n+B_n)+1\leq c_n,\end{aligned}$$ and hence the inequality.

    In order to prove the inequality $$\begin{aligned}
    c_n+(\delta_n-\delta_n')\leq(\varepsilon+x_n)\cdot\tan(A_n+B_{n+1})-1,\end{aligned}$$ it is enough to show it for $\varepsilon=\delta_n'$.

    We also use two relations: $$\begin{aligned}
     &\delta_n-\delta_n'=\delta_n'\cdot\tan(A_{n+1}+B_{n+1})-\delta_n\cdot\tan(A_n+B_{n+1})-y_n\cdot I_n,\label{del}\\ &y_n=\frac{\delta_n P_n-x_n\tan(A_n+B_{n+1})+1}{1+\tan(A_{n+1}+B_{n+1})}.\label{ips}$$ Notice that these statements follow directly from the definition of $\delta_n'$ and the equality $$\begin{aligned}
    d_n=(\delta_n+x_n)\cdot\tan(A_n+B_{n+1})-1\end{aligned}$$ from the Lemma \[lema4\], respectively.

    Due to $$\begin{aligned}
    &\delta_n'<\delta_n,\end{aligned}$$ and the definition of $P_n$ , the inequality $$\begin{aligned}
    &-\delta_n P_n + x_n\cdot\tan(A_n+B_{n+1})-1+\delta_n'\cdot\tan(A_{n+1}+B_{n+1})\\
    & \leq(\delta_n'+x_n)\cdot\tan(A_n+B_{n+1})-1\end{aligned}$$ holds. Therefore, the equality implies $$\begin{aligned}
    &\delta_n'\cdot\tan(A_{n+1}+B_{n+1})-y_n\cdot(1+\tan(A_{n+1}+B_{n+1}))\\
     &\leq (\delta_n'+x_n)\cdot\tan(A_n+B_{n+1})-1.\end{aligned}$$ From here, using , the inequality $$\begin{aligned}
    &(\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n+(\delta_n-\delta_n')\\
    &\leq(\delta_n'+x_n)\cdot\tan(A_n+B_{n+1})-1\end{aligned}$$ follows. The definition of $c_n$ yields, $$\begin{aligned}
    &(\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n+(\delta_n-\delta_n')\\
     &\leq(\delta_n'+x_n)\cdot\tan(A_n+B_{n+1})-1,\end{aligned}$$ and the inequality is proved.

\[har1\] Let $I_n$, $P_n$, $P_n'$, $\delta_n$, $x_n$, $y_n$, $c_n$, $d_n$, $U_n$, $\delta_n'$, $S_n=(x_{S_n}, y_{S_n})$, $r_n$, $y_{P_1}$ and $ y_{P_2}$ be as in the Proposition \[prop5\]. Then

1.  $y_{P_1}\leq c_n$,

2.  $c_n + (\delta_n-\delta_n')\leq y_{P_2}$.

<!-- -->

1.  The inequality $$\begin{aligned}
    \varepsilon<\delta_n,\end{aligned}$$ implies $$\begin{aligned}
    (\varepsilon + y_n)\cdot \tan(A_n+B_{n+1})+y_n \leq (\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n.\end{aligned}$$ Therefore, together with and , the inequality $$\begin{aligned}
    y_{P_1}\leq c_n\end{aligned}$$ follows.

2.  To prove the second inequality, we use $\delta_n'<\varepsilon$ and .

    Namely, $$\begin{aligned}
    \delta_n'\leq \varepsilon\end{aligned}$$ implies $$\begin{aligned}
    &\delta_n'\tan(A_{n+1}+B_{n+1}))-y_n(1+\tan(A_{n+1}+B_{n+1}))\\
    &\leq \varepsilon\cdot\tan(A_{n+1}+B_{n+1})-y_n(1+\tan(A_{n+1}+B_{n+1})).\end{aligned}$$ From here we obtain $$\begin{aligned}
    &(\delta_n + y_n)\cdot\tan(A_n+B_{n+1})+y_n +\\
    &+\delta_n'\cdot\tan(A_{n+1}+B_{n+1})-\delta_n\cdot \tan(A_n+B_{n+1})-y_n\cdot I_n\leq\\
    &\leq (\varepsilon-y_n)\cdot\tan(A_{n+1}+B_{n+1})-y_n\end{aligned}$$ Using , and it follows $$\begin{aligned}
    c_n + (\delta_n-\delta_n')\leq y_{P_2},\end{aligned}$$ and hence the Lemma.

\[har\] Let $I_n$, $P_n$, $P_n'$, $\delta_n$, $y_n$, $b_n$ and $M$ be as in the proof of Theorem \[har\_thm\_mat\], i.e. from Proposition \[prop4\]. Then: $$\begin{aligned}
\limsup_{n\rightarrow \infty} \frac{M}{y_n}<\infty.\end{aligned}$$

By the definition of $\delta_n$ , it follows: $$\begin{aligned}
 y_n&=\frac{1+\tan(A_n+B_n)-\delta_n\cdot P_n'}{1+\tan(A_n+B_{n+1})}\notag\\
 &=\frac{P_n^2\cdot(1+\tan(A_n+B_n))}{H_n},\label{y_n_g}\end{aligned}$$ where $H_n$ is like in the proof of the Proposition \[prop4\].

From here and it follows $$\begin{aligned}
\label{komb1}
 &\frac{\delta_n}{y_n}=\frac{ I_n\cdot(1+\tan(A_{n+1}+B_{n+1}))}{P_n^2}.\end{aligned}$$

Also, by and the definitions of $\delta_n$ and $x_n$ , it follows $$\begin{aligned}
\label{komb2}
 \frac{x_n}{y_n}=R_1+R_2+R_3,\end{aligned}$$ where $$\begin{aligned}
 R_1&=2\cdot\frac{1+\tan(A_n+B_{n+1})}{\tan(A_n+B_{n+1})}\cdot\frac{1+\tan(A_{n+1}+B_{n+1})}{P_n},\\
 R_2&=\frac{1+\tan(A_n+B_{n+1})}{(1+\tan(A_n+B_n))\cdot\tan(A_n+B_{n+1})},\\
 R_3&=\frac{I_n\cdot(1+\tan(A_{n+1}+B_{n+1}))}{P_n^2} \frac{P_n'}{\tan(A_n+B_{n+1})}\frac{1}{1+\tan(A_n+B_n)}.\end{aligned}$$

Combining (\[komb1\]) and (\[komb2\]), the conditions (\[uvjet\]) and (\[uvjet\_d\]) imply $$\begin{aligned}
 \limsup_{n\rightarrow \infty} \frac{M}{y_n}<\infty,\end{aligned}$$ and hence the proposition.

[10]{}

K. Bogdan, *The boundary Harnack principle for the fractional Laplacian*, Studia Math. 123 (1997), no.1, 43–80. MR 1438304 (98g:31005)

K. Bogdan and P. Sztonyk, *Harnack’s inequality for stable [L]{}évy processes*, Potential Anal. 22 (2005), no.2, 133–150. MR 2137058 (2005m:31015)

K. Bogdan and P. Sztonyk, *Estimates of the potential kernel and [H]{}arnack’s inequality for the anisotropic fractional [L]{}aplacian*, Studia Math. 181 (2007), no.2, 101–123. MR 2320691 (2008k:31015)

Z.-Q. Chen, *Lecture Notes: Potential Theory for Jump Processes; Summer School 2012 on Nonlocal Operators*, 2012.

N. Ikeda and S. Watanabe, *On some relations between the harmonic measure and the [L]{}évy measure for a certain class of [M]{}arkov processes*, J. Math. Kyoto Univ. 2 (1962), 79–95. MR 0142153 (25 \#5546)

T. Jakubowski, *On Harnack inequality for $\alpha$-stable Ornstein-Uhlenbeck processes*, Math. Z. 258 (2008), no. 3, 609–628. MR 2369047 (2009b:60242)

N. S. Landkof, *Foundations of modern potential theory*, Die Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen mit besonderer Berücksichtigung der Anwendungsgebiete ; 180, vol. 180, Springer, Berlin \[u.a.\], 1972.

E. Priola, *Pathwise uniqueness for singular [SDE]{}s driven by stable processes*, Osaka J. Math. 49 (2012), no. 2, 421–447. MR 2945756

K. Sato, *Lévy processes and infinitely divisible distributions*, digital print. ed., Cambridge studies in advanced mathematics; 68, vol. 68, Cambridge Univ. Press, Cambridge \[u.a.\], 2007 (eng).

P. Sztonyk, *Estimates of tempered stable densities*, J. Theoret. Probab. 23 (2010), no. 1, 127–147. MR 2591907 (2011b:60183)

S. J. Taylor, *Sample path properties of a transient stable process*, J. Math. Mech. 16 (1967), 1229–1246. MR 0208684 (34 \#8493)

[^1]: The author appreciates the support of the International Graduate College “Stochastics and Real World Models”, Universität Bielefeld.