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Data leaders are updating their playbook with an offense that focuses on increasing business value with data and AI. Take Sol Rashidi, Chief Analytics Officer of Estée Lauder. She holds seven patents and was recently recognized as one of the 50 Most Powerful Women in Tech. In this episode, Sol dives into how she's taking charge of product development to add business value, all while overcoming organizational challenges and becoming a change agent.
Sol Rashidi
Chief Analytics Officer
Sol is a subject matter expert in the areas of data, analytics, data science, automation, product development and UI/UX experience with a proven ability to drive innovative solutions to meet consumer challenges. She was the Chief Data and Analytics Officer for Merck. Prior to that, Sol was EVP, Chief Data Officer for Sony Music, the Chief Data and AI Officer for Royal Caribbean, and she helped IBM launch Watson, democratizing AI into the global marketplace.
Sol has seven patents granted with 21 patents filed, and she has received numerous awards and accolades over the years for her leadership in the tech, data and analytics space. Recently, she was named to the Top 50 Technologists, New York's Top 21 CDOs (#4), and the 2020 list of Global Data Power Women in CDO Magazine. Sol graduated from UC Berkeley where she played NCAA water polo, holds an MBA from Pepperdine University, and played rugby on the Women's National Team for many years.
In her spare time, Sol is an avid runner, a master cocktail maker, music fanatic, roasts her own coffee beans, and runs after her 3 and 5 year olds.
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Speaker 1:
Welcome Champions of Data and AI. Brought to you by Databricks. Thousands of data leaders rely on Databricks to simplify data and AI so data teams can innovate faster and solve the world's toughest problems. Visit Databricks.com to learn how data leaders are unlocking the true potential of all their data.
In each episode, we salute champions of data and AI. The change agents who are shaking up the status quo. These mavericks are rethinking how data and AI can enhance the human experience. We'll dive into their challenges and celebrate their successes all while getting to know these leaders a little more personally.
Chris D'Agostino:
Welcome to Champions of Data and AI. I am your host, Chris D'Agostino, global principle technologist at Databricks. Data leaders are updating their playbook with an offensive strategy that focuses on increasing business value with data and AI. Take Sol Rashidi, Chief Analytics Officer for Estee Lauder.
She holds seven patents and is recognized as one of the most innovative women in tech. Today, Sol dives into how she's taking charge on creating data products that add business value, and we'll discuss her approach to working with business stakeholders to bring about organizational change.
Sol, welcome and it's great to have you with us today.
Sol Rashidi:
Thank you, Chris.
So, before we get started, let's go back to your time at UC Berkeley. Most people would imagine that the role of a CDAO, or a CDO, is really technical and focuses on the technology and computer science aspects of things, but you're bringing a broader dimension than that to the role that you're in and the career that you've got. So would love to hear a little bit more about your background at UC Berkeley. I know, from the last time that we talked, you were a professional athlete at one point in your career. And so, would love to hear how your studies at school, plus your professional athlete career, combined and shaped who you are now and helped you get to where you are today.
Yeah. No, I appreciate that. To be honest with you, the whole thing happened by accident. No one ever grows up and says, "Hey, I want to be a data expert or I want to give analytics or insights." That stuff just wasn't cool in the 90s or the early 2000s. And so, the whole thing happened by accident. I was a chemistry major at Berkeley, and not because I loved chemistry. I literally didn't know what else to study. And my favorite teacher, Ms. Spring, from high school, studied chemistry at Berkeley. And I was like, "Okay. It makes sense." But I actually went to Cal to play water polo. I was competing NCAAs, I got recruited, and so I was like, "Okay, it's great. Best of both worlds. I get to play the sport I love and major in a major that inspired me because of a teacher, not because I was good at it." But, like any other has been athlete, you could be phenomenal in high school and be barely mediocre at the collegiate level, and that was just the informative truth.
I was second string, and I was training five, six hours a day, and I never got a chance to play unless it was a team that just wasn't as good and they were, quite frankly, my coach was like, "All right. It's fine. We'll throw in the second string." But at some point in time, it was sophomore year, my coach and I had to have a conversation. She's just like, "I love your passion and you show up every day committed and I love your work ethic, but you're just not as tall and you're not as fast as the rest of the girls. You're only 5'3″-and-a-half on a good day, and everyone else is 5'10."
It's like me trying out for the basketball team.
But that's exactly it. And it wasn't a issue for me because I could make up for it in power and height in high school. But you get to a certain level, professionally, where just your skills and you're physical dimensions just don't align to be able to compete at the caliber that they want. And so, it was a sad truth of, "Okay, maybe sports isn't going to be in my works." And I remember, that day, I was walking down Bancroft Way, and another woman tapped me on the shoulder and she's like, "What sport do you play?" And I was like, "Well, up until a few hours ago, I played water polo." And she's like, "I think you're going to be perfect for something. Do you want to try out for another sport?" And I said, "What is it?" She's like, "Rugby." I'm like, "What on earth is rugby?" And she's like, "Well, American's call it football without pads." But I'm like, "Why would anyone play that? Why would a woman play that?"
The whole ironic thing is what was a deficiency for me in water polo, turned out to be a phenomenal strength for me in rugby. I tried out. I was a natural because I knew how to handle a ball, but because I was 5'3," I had a low center of gravity and no one could tackle me. And it turned out to be a tremendous strength of mine. And it was the first lesson of some cultures, some environments, aren't going to bring the best you out. And others thrive off of what you bring to the table. And it's not a matter of no, but more a matter of how and when. And it was just the lesson learned, so I thought I was going to play professional sports for the rest of my life and the odd thing about that is I graduated, played on the women's national team, and then you get a point, in your twenties, where you're like, "Yeah, I wonder what my shelf life is going to be." And, to be honest with you, I took the one job that no one wanted and it was combining these massive Excel spreadsheets. Remember the 63,000 row limitations in creating these magnanimous pivot tables? And then, taking a screenshot and sending it in email to the business leaders and saying, "Hey, here's what I'm noticing and here's what I'm finding."
Literally, no one wanted to do that and I took the job no one wanted to do. And it was my transition from, "I'm no longer going to be a professional athlete for the rest of my life. I should probably become a professional. I don't know what I want to do." So I took a job that no one wanted to do and that's just how the whole thing started.
Yeah, that's great. That's great. And you're probably a master at Excel, right?
I can do a VLOOKUP statement like nobody's business, but to be honest with you, I'm not finger-to-keyboard like I used to be. If I don't do another VLOOKUP statement for God knows when, I'll be a happy camper. I've had my fair share.
So you did the athletic side of things and you learned finding the right sport, if you will, for what you're best at and most suited for, but there's a life span to that. You've carried that over into a longer career in data and analytics. And tell me a little bit about what aspects of what you learned in school, what you learned from team sports, to bring forward into the role that you've had doing data.
There are themes. Not everyone's juiced, or excited, or have that competitive spirit that I may have. I, fundamentally, love what I do. And if I don't love what I do, I change the circumstances. And not many people are in that same gear. Some folks are just okay doing their 9-5 jobs and a steady paycheck. It's not bad, but it's not great. It's fine. And they get comfortable. I get uncomfortable feeling comfortable because I know I'm not growing, I'm not being challenged. And so, for me, I've learned that everyone's motivations are very different. They may not be aligned to mine, and that's okay. There's room for everyone, but I also stay true to myself and know that I'm not your support and maintenance person. I'm a builder. I love to create things. I love to go in with an empty canvas, hear the business problems and the symptoms of what the company's are feeling and coming up with a grand vision, selling that vision, getting alignment on that vision, and building a team to be able to deliver on that vision.
It sounds simple in theory, but it's a lot of social science and very little data science, to be honest with you. But that's the first lesson. Everyone's motivated differently and there's room for a ton of people, but if you're in a position of building and creating, you're always better positioned if others are also in that mode.
The second is know your shelf life. Just like with sports, at some point in time, I had to make a judgment call. I'm like, "I'm 24-years-old and I'm getting tackled left and right. And my recovery is just taking a lot longer. It used to be days, now it's weeks. Now, I sprain an ankle and it's a month-and-a-half and I still feel it." The same applies to work. Know your shelf life. You're not meant to be somewhere forever, and if you are, that's okay. Goes back to point one, but for me, it's really important to know, "Have I done what I needed to do here and has my shelf life expired? And if so, because I'm not growing or the company's changed visions or they're intent on transformation really meant, 'We want incremental steps forward and not exponential steps forward.' Know when to call the game."
Yeah, it's interesting. Back to that comment about being comfortable, I had a software company for a number of years and one of my colleagues had a phrase that I thought was just fantastic. And this is way back when, so I don't know if he coined it or if he just adopted it from somewhere else, but he said, "Get comfortable with being uncomfortable." So, this idea that is technology is evolving so fast and new programming languages, new frameworks are coming out. Don't sit still. Make sure that you're constantly evolving, learning something new, and if you are uncomfortable, that's a good sign.
So that leads into both: either taking the very calculated risk averse approach to work and almost guaranteed for success versus being willing to risk it a bit and hope to achieve some great outcomes but, along the way, you might come across a few failures. Can you give us some sense of the big accomplishments? Things that anchor you back to the role that you've got and your passion for it? And some of the challenges you faced where, maybe, things didn't work out quite as you would hope they would?
It's a double-edge sword. I would say some of my greatest accomplishments were also the most epic failures at first. Businesses talk about wanting to derive a commercial strategy off of the insights that our teams are fundamentally responsible for providing. And that sounds great, in theory, but there's a few legs in the stool that create an unstable environment for the theory.
The first one is what if the insights doesn't align with the way they've been running their business?
You're essentially, one, you're exposing yourself to just be told you're wrong. Or the team's exposing themselves to be told they're wrong. You're putting credibility at risk and, potentially, if the business is not of a mindset of, "We need to evolve and, therefore, really need to take this seriously. And let's work on this together because this completely conflicts with the way we've been running the business," there could be a detrimental impact. Like, "How dare you? Based on my experience, my intuition, how I've grown the business the past 10 or 15, 20 years, is because of X, Y, and Z. And now you're coming in with A, B, and C and saying, 'Well, those levers weren't actually the things that helped you grow the business. It was a whole different series of components that came together that helped create this elevation, but let's make sure the credit is given where it deserves.'"
So, insights is great and dandy, but you've got to be careful and you got to know who you're presenting it to because if it conflicts with the leader's way of running the business, it's never a good position to be in. And so, you've got to be able to avoid a maneuver around those land mines very carefully.
And I had an experience, way back when, where… Remember the notion of building that 360 view of the customer? Well, yeah. We pulled in data sources from multiple databases into a lake. We created aggregated layers and then we consumed it via the app that we had built. And in that 360 view, because we brought in all pieces of information, we connected MDM data with call center data with revenue management data, you name it. And what we discovered was is who we thought were our highest margin customers, actually, ended up being our lowest margin income customers because when we pulled in customer service data and information, we found out that because the systems weren't integrated, and the data wasn't integrated, we were comping this consumer class so much that, actually, we weren't making double-digit margins. We were making single-digit margins and even, sometimes, losing margin on them because we were comping their relatives or their sons and daughters or anyone who was traveling with them. And when you aggregate it all together, the margins weren't nearly were we thought they were.
We talk about this a lot, internally, at Databricks. The things that got you to where you are today likely are not the same techniques, processes, and approaches that are going to get you to the next level. And we often talk about, "Let the data decide." And so, it's interesting that you're in what would otherwise be viewed as a very successful organization, you start analyzing the data, you're seeing that there are areas where, maybe, the success isn't as profound as you think it might be and data is suggesting that, perhaps, there's a different way to think about the business. And then you've got that cultural shift and, really, that willingness to embrace data, the willingness to look at the data and the insights and start changing how the business functions as a result.
That's precisely it. So now, just think about that, though. The role of a CDO, in 2020, there's a reason why [inaudible 00:14:00] released a publication that says, "The average life span's 18 months." And the ones that last more than 18 months tend to be CDO's that play the defensive playbook. The one's that bring you down to the average of 18 months are the ones that tend to play the offensive playbook because the defensive playbook is really around putting in place the data ecosystems. It's the ETLs, the data integrations, the quality, the cleansing, the government, what I call the 'defensive playbook.' It's all around the backend ecosystem. The stuff that, quite frankly, if you had a discussion with around the presidents of brands or whatever they may be, their eyes would just roll and they're like, "Yeah, that's geek talk. Just step away."
That doesn't get you in trouble. The stuff that gets you in trouble is when you're having the conversations with the business leaders because you've a genuine interest in establishing a strong, commercial presence, either to maintain the current positioning or to continue to create a competitive advantage or the competitive landscape has changed. You've got to know their business as well as they know their business, but they're still going to question you because you're not the one running the P&O. And, you're trying to bring in new insights and you're trying to get them to change the [inaudible 00:15:09] in which they approach things. You're completely exposed 100% of the time. And I always say, in this position, you need a backbone and not a wishbone because you're not going to be popular all the time. And you need to be okay, sometimes, with not being liked because people are going to talk behind your back and go, "She doesn't know. She doesn't know our business yet."
It doesn't matter if you spent nothing but four months understanding that business inside and out and all the KPIs that drive it. You're still never one of them. So, you're now, if you're a CDO that plays both the offensive and defensive playbook, you're not IT, you're not business, you're that weird overlap in the Venn diagram that's orphaned and doesn't have a home. But your job is to operate as a proxy between the two teams and make them work together.
Yeah. It's extra challenging in that you need to understand the data, you need to understand the technology and what it's capable of doing with the data. But building those relationships and getting the different C-suite members and leads for the P&O to trust you and, maybe, take that chance of adjusting the way in which they're approaching the business based on the feedback that you're giving them. Like you said, it's easy to just say, "I'm going keep the lights running and I'll be in this job forever and it's a great job that pays well. The company is a great company to work for, but if you're really trying to make a difference, you're going to be pushing the envelope. And that envelope, sometimes, is going to come along with some paper cuts from people that don't like the response.
You're going to kick up some dust. And some organizations, when they say 'transformation,' they expect the dust and they're going to support it. Other organizations talk transformation, but as soon as dust is kicked up, it's amazing. What ends up happening is they just don't create the runway that's needed for you to continue to drive the capabilities that you want establish and impact the commercial strategy based on the insights that you've derived. And if they don't create a runway for you, then you automatically know this is a word and a term, but there is a genuine interest to truly do this.
And that's why you've always got to make a judgment call. Like, "Am I in the right environment? Are they sincere about the transformation? Or am I not?" Because you're going to get paper cuts along the way and you need a strong leadership team that says, "This is part of the job, but change isn't easy because, if it was, we would have already been done it. We would've already done it by now."
Yeah. And you've taken a bit of an interesting approach where you focused a bit more on product definition, like data product definition and data product management, as a tangible asset for the organization. And we talk, often, about data native companies, companies that are really looking to think about data in a different way, make it part of the fabric of the culture of the company. And so, can you shed some light for people listening in on how do you approach your job from a product management standpoint and the assets that you create for the business?
Yeah. So, I was fortunate enough, I had a really great… I don't think I ever told him he was a mentor of mine, but I always watched from the side, and a part of his leadership meetings, and he taught me how to develop digital products. Whether it's a mobile app or a web-based application, design matters. User experience matters. The grace and elegance associated with using an iPhone versus a Blackberry matters. Pay attention to the details because the adoption and the utilization of whatever capability you're trying to create, 60% of it comes from the way a person feels when you're using it. And so, for me, I've actually pivoted from back office, or defensive playbook, which is where my world was, to evolving into an offensive playbook, but it's not just about analytics and KPIs. It's now evolved into, really, product development and developing assets.
Data is just a means to get to the end. And the end state is really around producing insights and presenting it in a way that the business can understand it. So, here's an example. Going from an environment that's Excel Spreadsheet and PowerPoint based to understand the health of your business to a mobile app that, at any point in time, you can click on a button and take a look and see what gross sales were. If actual sales is aligning with what forecasted sales were, are we over indexing or under indexing? What's the demographic of the consumers that are using our product? Get out of Excel and PowerPoint and go into apps because the technology's there. The data ecosystem's are there. We just got to be a little bit creative about it. And so, for me, it doesn't matter what forum that we present this information, I just choose not to do it in PowerPoint. I want to be able to develop apps and I want to be able to develop an interface that resonates with the business.
And businesses remember tangible products. They don't remember slide ware that you showed them about all the pipes that you plugged in to a data lake because the question is, "So, what does that do for me?"
Okay. So, Sol, that's really interesting. You've talked about this idea of moving away from Excel Spreadsheets as a way to analyze the business. You want to move it more towards real-time, presumably, with these mobile apps and you can see the health of the business at a glance. How do you approach, when you're in the C-suite meetings, how do you approach convincing these stakeholders that you should take the time, the budget, the people power to go out and build these solutions? And then, once you get approval, what do you do to go motivate the teams that report into you, or who you partner with, in order to focus on that and prioritize it?
I wish the process was clean. I wished it was consistent. It's not. It's messy. One, you never know when inspiration's going to strike. One example I had was… I remember asking, he was the president of one of the brands, and I said, "Can I sit in on your executive leadership meetings? I'd really like to learn the questions you ask. And I'd really like to learn how you gauge how things are going. More importantly, I just want to understand the language." And some presidents are like, "Yeah. Absolutely. No problem." And they'll just forward me the invites and tell their EAs and, all of a sudden, I'm in these biweekly leadership meetings. And others are like, "Well, we're having one in three months. Why don't we get over this product launch and then I'll invite you to the next one and we'll figure it out." I was like, "Okay."
You can always tell which presidents you're going to build alliances with and which ones… It's going to take some time. But in the cases where it's like, "Yeah, not a problem. It's great. Here. Here's the invite," I'll listen. And oddly enough, there's always a common theme. The common theme is there'll be questions and the president will ask the SVP, who will then ask the VP, who then will ask the senior director, who then asks the senior manager, who then asks the analyst. "So-and-so's looking for this. Can you put this information together?" The analyst, or the data scientist, will go and cobble bits and pieces of information together. Sometimes it's a pivot table, an Excel spreadsheet, and you send a screenshot in an email that then gets forwarded, that then gets forwarded, that then gets forwarded. In a good case, it takes a couple days. More often than not, it takes a couple weeks.
And I remember this clear example. I'm like, "They're asking the same questions week after week. So, while the analyst knows every two week that they're going to have to present this information, that's literally all the analyst is doing their entire career, is supporting what the president wants to know on a weekly basis. And, by the way, the questions are consistent, the datasets are consistent. Where's the creativity in this? In this one particular case, where things went really, really well, I remember I ran back to my desk and I'm like, "What if we automate this whole thing? What if we actually just build an app, gave them a prototype in TestFlight? Let him click on it. If he liked it, awesome. We can build a full blown production app."
And I started sketching wire frames. I'm old school. If I have a white board, I'll do white board. If it's piece of paper, markers, I'll just start sketching it. I literally gave it to a vendor that I work with. I'm like, "Here's my idea. Can you make this come to life?" And she'd use InVision and we created clickable buttons and created a screen. And then, I learned giving something on paper doesn't work the creatives to physically, tangibly develop it into an app. So, we literally took dummy data, supported the clickable app with some dummy data, converted it into an app. Then we used TestFlight and then I asked his assistant. I said, "Are you in charge of his phone? Or is he in charge of his phone?" She's like, "I do everything for him." I'm like, "Great. Can we download this?"
We downloaded it and, in my next meeting, they literally went through the same questions. And afterwards, I said, "Hey, I created something for you. It's on your phone. It's not live, but would something like this help? Could you take your phone out?" And he took his phone out and he opened the app. And he clicked on the app. And it was literally all the answers to the questions he was asking on a consistent basis. And he looked at me. First, he cursed. He's like, "Is this definitely possible?" I said, "Of course it's possible. We're just taking it from Excel Spreadsheets and pivot tables and emails and putting into a digital format, an app on your phone, and know you don't have to ask the questions. It's brought to you, but more importantly, here's some things that I want to suggest." You always ask actual sales against forecast. What if we were to send you push notifications anytime sales exceeded forecasts by 10%? And we use anomaly detection to show you in which markets, or countries, you're exceeding expectations and which markets, or countries, you're not exceeding expectations? He's like, "Is that possible?" I said, "Of course it's possible." He says, "I want everyone on my leadership team to use this app and I want to start doing it tomorrow."
And I was just like, "Whoa, whoa, whoa. Hold on. So this is dummy data, this isn't real. I've got to connect all the pipes. I've got to build the data ecosystem. I've got to get the app to really make this production ready. And we've got to go through our infosec protocols. My pace, I can build it for you in six months, but knowing our culture, and how I've got to get everyone aligned, it's probably going to take a year."
And, luckily, he said, "That's fine. Let's do this." And he was the person-
He wasn't freaked out that you, in a stealth manner, got an app installed on his phone?
His assistant does everything. So, luckily, as long as I got her blessing, and that was my first lesson learned. Always make friends with the executive assistant. They control everything. And because of that, [crosstalk 00:25:42] to get a pass for it.
You might have a future career in the intelligence communities, so we can talk about that on another episode.
Well, you just got to get creative at finding ways of just getting things done. And that's one of them. They're fundamentally awesome people to know, and they know everything about the organization, but they're also the biggest gatekeepers. So, if you show a little bit of appreciation, because they manage everything, it's amazing how far it goes.
But it's also an indication of, especially in data and AI as we transition the talk a little bit into the AI space, the quick wins. And what define 'quick,' it goes organization by organization, but the notion of, "What questions need to be answered routinely? How can you automate the answer in and of those questions? How can you surface those insights more quickly?" So oftentimes, especially in data engineering functions, where I ran one for quite some time, you want to boil the ocean. And sometimes, keeping it really simple, answering some very critical business questions in the moment, at a rate that the executives haven't experience before, it goes a long way.
So, how do you take the excitement that the executives see that, maybe, the developer community thinks, "Well, this is pretty straightforward," and "Wow, is this all we're going to do?" But just getting those quick, early wins and building the traction and momentum that you need to get the buy-in support funding to go big beyond that?
There is no clear recipe. You go in and all you do is talk to as many people as possible. Find out where they're hurting the most or what their vision is or what they think the initiative or transformation's around. And you're literally on a fact finding and discovery zone for a solid three to four months. And you're just nonstop meeting people, asking questions, listening to them, and I don't know if you guys ever saw A Beautiful Mind with Russell Crowe, and this is the analogy, and I tell my team, I'm getting information from all over the place. But my job is, legitimately, to connect the dots of the pain points and the symptoms that people are feeling and find where the root sources and causes are.
And oftentimes with big bets and quick wins, every company calls it different, low hanging fruit, there's a lot of fruit on the ground. You don't even need to reach up. There's just stuff all over the orchard and it's on the ground and we're stepping over them because we always want to do the sexy things. We always want to do the innovative things, but the fact of the matter is 80% of our problems can be solved without using data science and machine learning techniques. Low hanging fruit and high hanging fruit sounds glamorous, but there's a lot of fruit on the ground that's ready to be picked if we just listen.
And when you start pulling all these together, you realize that it always comes down to three things: your impact, commercial impact, so top-line growth, cost-reduction, so it's an efficacy play, or convenience. When you can categorize into those three C's, those three buckets, every leader's going to be after something different. If you're in supply chain and manufacturing, the name of the game is really around how efficient can you be. And it's mostly a cost-reduction play. How can we do more with less? And that's where you get into automation and machine learning and it's a fun playground, but know your place. Now what the value drivers are.
Well, if you're in the commercial division of a business, it's all around top-line revenue and competitive stance and positioning. Not losing market share, gaining market share. And it's around the insights that support that. Know your leaders and know what matters to them most. But what's interesting is, oftentimes, you got to pay attention to the undercurrent because it's never just the presidents. You got to know who the presidents rely on, who their number two, number three, and number four is. And the ones that are very vocal. Yes, top-line matters for them. Yes, cost matters for them. But they're so busy running around supporting the business and supporting the president, that convenience is what matters for them the most. And whether it's taking away 10 spreadsheets and PowerPoints and creating an app, whether it's consolidating all the data sources into a single dashboard and building canned reports, which just doesn't sound sexy or glamorous by any means, but in a way that's self-service-oriented, and they can access it on a tablet or a computer or a mobile app. It's all about convenience for them, and that matters. It matters a lot.
And so, you've got to know your audience. You've got to know what matters to them and you've got to gear it towards that.
So, Sol, you talked about just the importance of understanding the role of the presidents within each of the P&Os and who his or her direct reports are and what are the key leaders. And, of course, that ties back into the concept of that EA network and understanding the support function and how the EAs, in many large organizations, really understand who the key players are, how do you get on to their calendars and making sure that you get time. So, certainly an important set of relationships to build.
One of the things that you mentioned, in terms of data and surfacing data, and then you referenced the Russell Crowe movie, A Beautiful Mind, and I'm noticing the painting behind you which is, in some ways, very abstract, but in other ways, actually surfaces… The circles become very well-defined and well-shaped and they're connected. I wanted to know is that painting something that you draw inspiration from, in terms of how you think about data and connecting data?
Very much so. So I live in Miami, still. And I moved from New York City about almost four years ago. And we moved to Miami, fell in love with it. We haven't left yet. And so, this is actually made by a local artist, her name is Natalia, and she's got an obsession with cognitive services and neural networks and the connection of information. And I've borderline got a little bit of an obsession, if you will, with data and information and how certain things connect and triggers neurons and signals. And you'll notice that there's a little bit of color as well because when things collide, yes, they get dismissed, but when things collide and it creates a signal. And the passage of information's [inaudible 00:32:18], it lights up. Your brain, if you will, and if you've ever seen those scans, you can see parts of your brain that lights up. And so, this is, actually, a really creative depiction of information passed through neurons, connecting signals, lighting up when things work.
And so, our space is a very creative space. And oftentimes, folks think that we're talking geek talk, but it's not. We're legitimately building assets and, sometimes, my brain resembles the painting behind me. When things light up, it's magic. So…
Awesome. Well, let's hope the fuzzy parts don't [crosstalk 00:32:51].
[crosstalk 00:32:51] a lot of fuzzy parts, for sure. But eventually, something will connect and something will light up and something will work. And that fuels you for the next two months of frustration.
Yeah. That's awesome. So, we talked the concept of low hanging fruit and fruit being on the ground, things that you don't have to reach too far in order to help automate and drive some new efficiency into the organization, the cultural component of making sure that the organization is ready for someone in the CDO, CDAO position, and that you build those relationships. We've talked about building applications that give insights. And those applications, obviously, require an underlying architecture that's really designed to support that. And, as you probably know, with Databricks, we've coined this term called 'lakehouse,' which is this desire to combine the best of both worlds. The worlds being the traditional enterprise data warehouse, which gives you a lot of great data management capabilities, good concurrency for serving a large number of users, but it structures the data, oftentimes, in ways that maybe limits the types of questions you can ask.
The data lake, in contrast, talks about using low-cost object stores to be able to put as much data, and different data types, as possible, and then run analytics against that. But then, you've got this divide where you're making copies of data into two different environments. You're not using the same view of the data. You might have synchronization issues between data that's loaded from the data lake into the EDW and it's no longer in sync and up-to-date.
So I wanted to just get your thoughts on architectures and bringing forward the ability to do AI at scale in a company the size of Merck, for example. There's tons of data. If you want to bring new insights into the business, how do you view the architecture as part of that equation?
I fundamentally believe that there's multiple ways to solve the problem. And so, what you've got to take into consideration is the maturity of the organization and where they are in the spectrum of technology. There are companies that have always been forward-leaning on technology. The early adopters, they can experiment. They've got enough margin to support an environment of experimentation and business continuity. And that has it's own home, when you can test and try and push the envelope. There are other companies whose success is predominantly based off of the brands marketing, sales reps. And so technology isn't necessarily front and center. It's more of a creative than it is a technical art. And so, you've got to understand what position, or what place technology has, in enabling these capabilities.
And I always say, "Those environments are the trickiest because you have to introduce technology as an enabler and foundational, to all the things that they talk about, without boring them gory details, but doing it fast enough where you still have their sponsorship and their support." So it's like flying the plane while building it at the same time and you're always pivoting and you're always gauging. My job is to bend, but not break the environment. And so, if I'm talking about AI and machine learning and developing automated forecasts models in an environment where they're like, "No, we're not ready for that." I can't introduce that stuff. I've got to go back and really, again, it comes down to the three C's that I was sharing, is the commercial impact, the convenience impact, or the cost-reduction impact.
Know which of the three C's matters most and align an architecture that's as simple as possible that gets you some quick wins and some credibilities, out the door, that aligns with those three C's. And, by the way, depending on the business stakeholder that you're working with, or the business stakeholders that you have to serve, the answer will be fundamentally different. So then you got to look at the architecture and go, "I can provide convenience for this business leader and build some credibility there. I can reduce cost-reduction in this particular area, or this business unit by doing this. Is there an architecture that's going to support both endeavors versus having to create different architectures?"
And so, architecture's always top of mine. It's always top of discussion. What's interesting is, if you get 10 engineers into a room, you get 10 architects in a room, you're probably going to get 100 different answers of how to approach something. And that's the hardest part about the architecture. There's multiple ways of solving a problem. You can't let the team that's discussing architecture lose sight of why we're discussing the architecture and you've always got to remind them that, more than likely, the simplest approach is going to be the best one for us right now. We can't lean in with all the sexy terms and tools and technology that's in place. If we're still crawling, then we're not ready to run a marathon yet. So there's also an element of maturity and speed that comes into play.
But architecture's probably one of the most important areas that gets people into trouble, very quickly, if you don't simplify and if you don't constantly reinstate why we're even having this discussion because everyone wants to play with the coolest things, but the coolest things don't always solve the quickest problems.
Yeah. It's a great point about the architecture and reminding the engineering teams, and the architecture teams, that the technology solution is a business enabler. Right?
Especially in organizations like Merck, for example. You're not in the business of building out infrastructure. You're in the business of delivering pharmaceuticals to consumers and making a difference in people's lives and their health. The faster you can generate the insights, the better you're going to be in terms of providing that service. And so, figuring out how to build an architecture that's elegant and simple and gets the job done as quickly as possible is something that, at least when I was at Capital One and running data engineering function, you try to remind the leaders that, "As much as it would be cool to build out this really complex, cool thing, it's going to take 12 months and the business can't afford to wait." And, at the end of the day, the business leaders, as much as they can appreciate the challenge of building out good software and a good architecture, they're more interested in supporting their P&O and getting their job done. And so, so long as the environment provides the support they need, supports their use cases as reliable and secure, they probably don't care that much what the under pennies are.
They don't care at all. And you've only got their attention span for, maybe, nine months.
So, yeah. The undercurrent of architecture is so fundamental, my only advice is don't complicate things because the attention span of the stakeholders that you have to serve is limited and minimal. And they don't care how you do it. They just want you to do it.
And this isn't specific to Merck by any means, this is any organization I've come into. You can get as fancy as you want, but that just kicks the can down the road and you're going to lose their attention span. So your job is to create impact. Build something tangible that they remember. Get their trust so that they can take a leap of faith with something more complicated that you want to do.
Sounds great. Well, we're almost at time and wanted to just ask a couple of final questions. You've obviously had a great career. A lot of success. You're in a role that is new, in a lot of industries, and you've talked about the positives, some of the challenges that come along with that. Especially as a woman in tech, what advice would you give to other people who are aspiring to have your type of career path and aiming for your type of role? And, for women in particular, are there things that you would give advice to, say, your younger self of what to consider and what to look out for?
Get comfortable with not being perfect. Get comfortable with criticism. Get comfortable that the ratio of successes is going to equal the ratio of failures. When I started out, I tended to overcompensate because I thought everything had to be perfect in order for it to be deemed, or perceived, as successful. And I had to work twice as hard and this and this and that. I actually made more impact when I took a more casual approach towards things, but I also realize that sometimes it doesn't matter how much of a vision I have or how much effort I put into something or how much it's going to help the company. It just doesn't stick and it doesn't go anywhere. And it's sweat, blood, and tears of, not only yours, but your teams as well, and it's not a home run. And it gets shelved. And, sometimes, failures turn into successes. And, sometimes, your successes turn into failures. And, again, the proportion's equal.
You got to get uncomfortable with knowing not everything's going to be successful and you're going to have just as many failures. And you've got to get comfortable with knowing that you're going to have just as many naysayers as you're going to have promoters and allies. And you've got to be okay with that because if you fundamentally take the posture of being a change agent, if you fundamentally chose to be a leader where you're going into an environment because they haven't been able to do something before in the past and they need someone with your skill sets, your personality, your tenacity, your relentlessness, know that it's going to be an uphill battle constantly. And be okay with it.
Go in knowing that you're going to have more bad days than good days. But that one good day you have, it juices you and it powers you up for the bad days that are about to come because a job of a change agent, the job of a CDO, the job of a CAO, doesn't matter what title your given, if you're fundamentally there to generate insights to create a deeper, stronger, more penetrable commercial strategy, you're going to rock the boat. And that's part of your job. And if you can get comfortable with that, you'll be okay.
The other advice I would give is, it goes back to shelf life, know your shelf life. Know the environment that's going to create the best version of you. Not all environments will. And don't stick with it just for the sake of sticking with it. If the subconscious biases are kicking in, and they're so subtle, but you feel it, pay attention. If you want to go at a pace and the environment's just not ready you yet, pay attention. If you're walking on eggshells because you feel like anytime you open your mouth, you're going to be criticized, or your point-of-view isn't being heard, pay attention to the signals. It's not an accident. And if it's an environment where when all those things come together, and you're okay and you're comfortable with, then that's okay. But if it's an environment where, day-to-day, you're just not and you're taking away a piece of you to accommodate, don't do that.
We talk about our careers and what kind of legacy we want to leave behind. And when people view our careers and describe us to other peers and other people in the industry, it sounds to me like from the moment you recognized that water polo wasn't the perfect sport for you, but you could be really great at something that, like you said, "What is rugby? What are women doing in rugby?" Breaking a mold and willing to take a chance on something based on just, maybe, some gut instinct and raw talent. Understanding how long that shelf life is, whether you outgrow the sport or the sport outgrows you in some way. But, like in an organization, if you come in as a change agent and you're doing a lot of these massive transformation initiatives, it could very well be that you go through all those steps that you talked about in terms of being willing to take the criticism and having the successes and failures. At some point, you will have evolved the organization.
And if your true nature is to be a change agent, there may not be enough change to be done anymore and you may have outgrown the organization and outgrown your role there. And now it's time to move on to something different. And probably telling viewers that that's an actual fine career path, that you don't need to get to a point where you're there for 25 years and you get the gold watch when you retire. That the ability to make an impact, know when your time is up, and move on to the next challenge.
It's true. You got to spend time with yourself and know who you are and be genuine about it. I'm not a support and maintenance person. At this stage in my life, I have too much energy and I want to make an impact. And I love the hustle and I love the aptitude that's needed to constantly just shift and pivot, and the successes and the failures that come with it. I'm comfortable in that state right now. Now in 15 years, and with my two young kids, it may be a totally different story. But that's not my story right now. And so, you got to get-
They will wear you down.
Oh, they've been wearing me down. They're starting to. I have a three and a five-year-old. And I love the rambunctiousness of both of them, but yes. They will wear you down.
Wait until they're 15 and 17 like mines. So… Yeah.
I know. I know.
Awesome. Well, thank you so much for your time today. And just want to, again, appreciate the insights that you brought to the discussion and I'm sure a lot of people watching this will really appreciate your career path and, especially, your role in such a large organization, a woman in this role, and the challenges of a chief data analytics officer. So, Sol Rashidi, thank you very much and look forward to talking in the future.
Thank you for joining this episode of Champions of Data and AI. Brought to you by Databricks. Thousands of data leaders rely on Databricks to simplify data and AI so data teams can innovate faster and solve the world's toughest problems. Visit Databricks.com to learn how data leaders are unlocking the true potential of all their data.
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Realistic drawing is difficult, but luckily there are other, more forgiving styles out there. For example, the chibi style has clear rules, and you can use it to draw literally everything.
In this tutorial, I will show you how to draw five simple chibi animals, and you'll learn how to create more just by applying the same rules to your own drawings.
The smaller the drawings, the easier it will be for you to keep the correct proportions. Draw all the animals on one sheet and treat them like small practice sketches rather than artworks on their own.
The illustrations in this tutorial are clear so that you see what you're supposed to draw—yours don't need to and even shouldn't be so perfectly drawn.
Don't try to draw the shapes with a single continuous line. Watch the videos to see how I build the shapes out of curves, connecting them only when I see what I was trying to draw.
The base for each drawing should be drawn with a pencil, with subtle lines, without pushing the tool too hard. This way the lines will not be visible when you finish the drawing with dark outlines. Watch the videos to see both phases of drawing.
Watch the video first to see the exact technique of drawing.
Start with an oval that will be the body. Remember: draw it lightly, don't press too hard. This is just a sketch!
Let's add the head now. It should be round and about twice as big as the body.
Add the legs. The forelegs can be drawn with simple straight lines, but in the hind legs draw the lines in the feet only. Add simple "paws" at the end of each line.
Add some volume to the forelegs and the feet. They should be quite puffed.
Time for the thigh. Make it very round, connecting directly with the foot.
Now we're going to create a cute face for our wolf. Draw an oval muzzle at the bottom of the head. It must be much, much smaller than the head to achieve a proper "cute" feeling.
Cross the muzzle with a wide smile.
Time for the eyes. Draw the lower part right next to the muzzle.
Make the eyes big, but leave a lot of space for the forehead—a huge forehead will make it look younger and cuter.
The ears can be either huge or tiny—both extremes will add to the cute feeling.
... and finish with a triangle.
Take your darker tool now and finish the details. This time you're allowed to press harder, but be careful—these are the final lines! You don't need to copy me—just look at your base and add some wolf features to it. It will look cute no matter what you do!
Let's start with an oval again. Remember to draw it subtly!
Add the round head. It must be above the body, because we'll add the neck later.
Draw the legs. They must be longer than in the wolf, but still use the same rule: straight lines for the forelegs, shorter lines for the feet.
Time for the eyes. Draw them on the sides of the muzzle again.
Because the legs are so long, we need to add some details to them to keep them interesting. Add a circle to each to create an outline of the joint.
Lead a line from each side of the circle to the hoof.
Finish the legs, attaching these new parts to the circles as well.
Add a gentle neck, making it an extension of the body.
Add the huge ears—exaggeration is very important in this style!
Add the tail, drawing a teardrop shape.
Time for "inking" again! Feel free to add little stubs of antlers if your deer is male. They should be as small and cute as the rest of the body.
Start with the oval body once again—nothing new here.
Add a huge head, just like in the wolf.
Add the legs. As you can see, you can fully control the position just by adding these four lines.
The mouth is open, so place the oval muzzle somewhere in the middle, and the lower part of it at the bottom.
Add a pair of lips—one for each part of the mouth.
Add some facial features. Normal, cute eyes wouldn't fit here, because the tiger wants to look dangerous. Keep them cutely closed instead.
... and finish the hind legs.
Connect both parts of mouth with subtly curved lines.
Our "dangerous" tiger is still cute, so don't be afraid to add that huge ear!
... and then copy the inner line on the sides.
Let's try something different this time. Draw the body as a half of a circle—an oval flattened by the weight of the body.
Now the head—big and "levitating", because the neck will be long.
Add the muzzle. Make it round for a change.
Give your little giraffe a big smile.
Don't forget about the huge ears!
The neck is not very long for a real animal, but long enough for a chibi!
Add the details: the little horns and the tail.
All these animals were fluffy mammals, let's try something different this time. Sharks can be cute as well! This time, start with the head.
Make the smile toothed and sinister.
Sketch the tail—look how short it is in comparison to the head!
Connect the tail with the rest of the body.
Let's add the fins now. Treat them as reversed legs.
Make the muzzle sharp to mix "cute" with "dangerous".
Start the eye the same way as usual.
To make the shark look dangerous, keep its eye half closed with a wrinkle of anger.
Only then add the pupil inside.
And if you're thinking seriously about learning how to draw, try this article—it will tell you everything you need to know!


text:
\section{Introduction}

Let $X=(X_t,\P_x)_{t\geq 0,x\in{\ensuremath{\mathbb{R}}}^d}$ be a  L\' evy process in ${\ensuremath{\mathbb{R}}}^d$ ($d\geq 3$) with the characteristic exponent
\[
	\Phi(\xi)=\frac{|\xi|^2}{\ln(1+|\xi|^2)}-1.
\]

Let us give some motivation for the process $X$. It is known that the  variance gamma process can be obtained as a subordinate Brownian motion, where the corresponding subordinator is the gamma subordinator, i. e. a L\' evy process  whose Laplace exponent (cf. (\ref{eq:sub})) is given by
\[
	\ln(1+\lambda).
\]
 It belongs to the class of the geometric stable processes (cf. \cite{SSV2}). The process $X$ is also a subordinate Brownian motion with subordinator that is a conjugate of the gamma subordinator. Namely, we take a subordinator with the Laplace exponent
\begin{equation}\label{eq:chexp1}
	\frac{\lambda}{\ln(1+\lambda)}-1.
\end{equation}
To avoid killing we subtract $1$ in (\ref{eq:chexp1}). Therefore we can say that  the process $X$ is (almost) conjugate  to the variance gamma process and so they are on the 'opposite sides'.

Another interesting property of  this process is that it is closer to the Brownian motion than any stable process. This can be argumented as follows.
Consider the potential operator $G$ defined by
\[
	Gf(x)=\ensuremath{\mathbb{E}}_x\left[\int_0^\infty f(X_t)\,dt\right].
\]
This operator has a density, i.e. there exists a function $G(x,y)$, usually called the \emph{Green function}, with the following asymptotical properties (cf. Proposition \ref{prop:green_asymp})
\begin{equation}\label{eq:introt01}
	G(x,y)\sim \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}|x-y|^{2-d}\ln\frac{1}{|x-y|^2}\ \ \textrm{as }\ |x-y|\to 0+
\end{equation}
and
\begin{equation}\label{eq:introt02}
	G(x,y)\sim \frac{\Gamma(\frac{d}{2}-1)}{2\pi^{d/2}}|x-y|^{2-d}\ \ \textrm{as }\ |x-y|\to \infty.
\end{equation}
Comparing  (\ref{eq:introt01}) and (\ref{eq:introt02}) with  the Green function of the rotationally invariant $\alpha$-stable process  ($0<\alpha<2$):
\[
	G^{(\alpha)}(x,y)=\frac{\Gamma(\frac{d}{2}-\frac{\alpha}{2})}{2^\alpha\pi^{d/2}\Gamma(\frac{\alpha}{2})}|x-y|^{\alpha-d}
\]
and the Green function of the Brownian motion in ${\ensuremath{\mathbb{R}}}^d$:
\[
	G^{(2)}(x,y)=\frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}|x-y|^{2-d}.
\]
we see that the Green function of $X$ is "between" the Green functions of  Brownian motion and any stable process. We remark that $X$ is still a pure jump L\' evy process.

The aim of this paper is to investigate some potential-theoretic notions of the process $X$. To be more precise, we investigate asymptotic behavior of the Green function and the L\' evy density of this process. Furthermore, we prove the scale invariant Harnack inequality and apriori H\" older estimates for the corresponding harmonic functions.

Although the origin of potential theory  is in the theory of differential equations,  it has also a probabilistic counterpart. The reason is that many local and non-local operators can be considered as infinitesimal generators of some Markov processes.

Recently, probabilistic methods turned out to be very successful (cf. \cite{BL1}) in providing some steps in the proofs of the Harnack inequality and regularity estimates. Extensions to certain classes of L\' evy processes were obtained in \cite{SV1,BS,RSV,SSV2,KS,Mi}. More general jump processes were treated in \cite{BKa,BKa2,CK2}.

Before stating our results precisely, let us give a few comments on the main ingredient in the proof: a Krylov-Safonov type estimate. This kind of estimate for  jump processes appeared first in \cite{BL1}. In \cite{SV1} it was extended to some Markov processes.

In these papers the following Krylov-Safonov type estimate was used:
\begin{equation}\label{eq:intro t05}
	\P_y(T_A<\tau_{B(0,r)})\geq c\,\frac{|A|}{|B(0,r)|},
\end{equation}
for all $r\in (0,1/2)$, $\ y\in B(0,r/2)$ and closed $A\subset B(0,r/2)$.
Here $T_A$ and $\tau_{B(x_0,r)}$ denote the first hitting time of $A$ and the first exit time from the ball $B(0,r)$, $|\cdot|$ denotes the Lebesgue measure in ${\ensuremath{\mathbb{R}}}^d$ and $c>0$ is a constant that does not depend on $r\in (0,1/2)$.

Applying the same techniques to our case would lead to the  following estimate:
\begin{equation}\label{eq:intro t06}
	\P_y(T_A<\tau_{B(0,r)})\geq \frac{c}{\ln\frac{1}{r}}\,\frac{|A|}{|B(0,r)|}.
\end{equation}

The main difference between estimates  (\ref{eq:intro t05}) and (\ref{eq:intro t06}) is that the second one is not scale invariant. If we replace Lebesgue measure by some other set function, it is still possible  to obtain a scale invariant estimate of the similar type. More precisely, we have the following estimate (cf. Proposition \ref{heat:prop.1001})
\begin{equation}
\P_y(T_A<\tau_{B(x_0,r)})\geq c\,\frac{\ensuremath{\mathrm{Cap}}(A)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,r)}\right)},
\end{equation}
where $\ensuremath{\mathrm{Cap}}$ denotes the $0$-order capacity with respect to the process $X$ (cf. Section \ref{sect:prep}).

This idea appeared first in \cite{SSV2} and \cite{RSV}. We mention that in  the case of the geometric stable process considered in \cite{SSV2}, the known techniques lead to  the Krylov-Safonov type estimates that are not scale invariant neither with capacity nor with the Lebesgue measure.

In Section  \ref{sect:prep} we will see that the process $X$ is a purely discontinuous L\' evy process. Thus, the characteristic exponent $\Phi$ of $X$  is of the form
\begin{equation}\label{eq:chexp}
\Phi(\xi)=\int_{{\ensuremath{\mathbb{R}}}^d\setminus \{0\}}\left(e^{i\xi\cdot y}-1-i\xi\cdot y1_{\{|y|<1\}}\right)\Pi(dy).
\end{equation}
Here $\Pi$ denotes the \emph{L\' evy measure}, i.e. a measure on ${\ensuremath{\mathbb{R}}}^d\setminus \{0\}$ which satisfies the following integrability condition
\begin{equation}\label{eq:levycond}
\int_{{\ensuremath{\mathbb{R}}}^d\setminus\{0\}}(1\wedge |y|^2)\Pi(dy).
\end{equation}
Moreover, in our case $\Pi$ is absolutely continuous with respect to the Lebesgue measure:
\[
	\Pi(dy)=J(y)\,dy.
\]
The function $J$ is called  the \emph{L\' evy density}. At this point it is interesting to mention asymtpotical properties of $J$ (cf. Proposition \ref{prop:jump_asymp}):
\begin{equation}\label{eq:intt10}
	J(y)\sim \frac{4\Gamma(\frac{d}{2}+1)}{\pi^{d/2}}\cdot\frac{1}{|y|^{d+2}\left(\ln\frac{1}{|y|^2}\right)^2} \ \ \textrm{as }\ |y|\to 0.
\end{equation}
Comparing this with the L\' evy density of the rotationally invariant $\alpha$-stable process
\[
	J^{(\alpha)}(y)=\frac{\alpha2^{\alpha-1}\Gamma(\frac{d}{2}+\frac{\alpha}{2})}{\pi^{d/2}\Gamma(1-\frac{\alpha}{2})}\cdot\frac{1}{|y|^{d+\alpha}}
\]
we see that small jumps of the process $X$ are more intensive than the corresponding small jumps of any stable process. Using (\ref{eq:intt10}) we can see that the integrability condition of $\Pi$ given in (\ref{eq:levycond}) is barely satisfied.

We say that a function $h\colon {\ensuremath{\mathbb{R}}}^d\rightarrow [0,\infty)$ is \emph{harmonic} in an open set $D\subset {\ensuremath{\mathbb{R}}}^d$ with respect to the process $X$ if for any open set $B\subset D$ such that $B\subset \overline{B}\subset D$ the following is true
\[
	h(x)=\ensuremath{\mathbb{E}}_x[h(X_{\tau_B})1_{\{\tau_B<\infty\}}] \ \ \ \textrm{ for all }\  x\in B.
\]
Here $\tau_B=\inf\{t>0:X_t\not \in B\}$ is the \emph{first exit time} from $B$.
Denote by $B(x_0,r)$ the open ball in ${\ensuremath{\mathbb{R}}}^d$ with center $x_0\in {\ensuremath{\mathbb{R}}}^d$ and radius $r>0$.

The first result of this paper is the scale invariant Harnack inequality.
\begin{Thm}[Harnack inequality]\label{tm:harnack}
There exist $R>0$ and $L_1>0$ such that for any $x_0\in{\ensuremath{\mathbb{R}}}^d$ and $r\in (0,R)$ and any non-negative bounded function $h$ on ${\ensuremath{\mathbb{R}}}^d$ which is harmonic with respect to $X$ in $B(x_0,6r)$,
\[
 h(x)\leq L_1\,h(y)\ \textrm{ for all } x,y\in B(x_0,r).
\]
\end{Thm}

This type of Harnack inequality does not imply H\" older continuity directly via Moser's method of oscillation reduction. The relation of this two properties is currently investigated (cf. \cite{Kass}).
Next result shows that harmonic functions locally satisfy uniform H\" older estimates.
\begin{Thm}[H\" older continuity]\label{tm:hoelder}
There exists $R'>0$, $\beta>0$ and $L_2>0$ such that for all $a\in {\ensuremath{\mathbb{R}}}^d$, $r\in (0,R')$ and any bounded function $h$ on ${\ensuremath{\mathbb{R}}}^d$ which is harmonic in $B(x_0,r)$ we have
\[
	|h(x)-h(y)|\leq L_2\|h\|_\infty r^{-\beta}|x-y|^\beta\ \textrm{ for all }\ x,y\in B(x_0,r/4).
\]
\end{Thm}

The paper is organized as follows. In Section \ref{sect:prep} we show that the process $X$ can be obtained as a subordinate Brownian motion and show some asymptotic properties of the L\' evy density and the Green function. We also prove a Krylov-Safonov type estimate. In Sections \ref{sect:hi} and \ref{sect:reg} we prove main results of the paper: Theorem \ref{tm:harnack} and Theorem \ref{tm:hoelder}.

\section{Preparatory results}\label{sect:prep}

A function $\phi\colon (0,\infty)\rightarrow (0,\infty)$ is called a \emph{Bernstein function}  if $\phi\in C^\infty((0,\infty))$ and
\[
(-1)^{n-1} \phi^{(n)}(\lambda)\geq 0\ \textrm{ for all } \ n\in {\ensuremath{\mathbb{N}}} \ \textrm{ and }\  \lambda>0.
\]
We say that  $\phi\colon (0,\infty)\rightarrow (0,\infty)$ is a \emph{completely monotone function} if $\phi\in C^\infty((0,\infty))$ and
\[
(-1)^{n} \phi^{(n)}(\lambda)\geq 0\ \textrm{ for all } \ n\in {\ensuremath{\mathbb{N}}}\cup \{0\} \ \textrm{ and }\  \lambda>0.
\]

A \emph{subordinator} $S=(S_t)_{t\geq 0}$ is a L\' evy process taking values in $[0,\infty)$ and starting at 0. The Laplace transform of $S_t$ is given by
\[
	\ensuremath{\mathbb{E}} e^{-\lambda S_t}=e^{-t\phi(\lambda)},\ \lambda>0,
\]
where  $\phi$ is called the \emph{Laplace exponent} and  it is of the form (cf. \cite[p. 72]{Be})
\begin{equation}\label{eq:sub}
\phi(\lambda)=d\lambda+\int_{(0,\infty)}(1-e^{-\lambda t})\mu(dt).
\end{equation}
Here $d\geq 0$ and $\mu$ is a measure on $(0,\infty)$ (called a \emph{L\' evy measure})  satisfying
\[
\int_{(0,\infty)}(1\wedge t)\mu(dt)<\infty.
\]

Using \cite[Theorem 3.2]{SSV} we conclude that the Laplace exponent $\phi$ of $S$ is a Bernstein function. Conversely, if $\phi$ is a Bernstein function such that $\lim_{\lambda\to 0+}\phi(\lambda)=0$, then there exists a subordinator $S$ with the Laplace exponent $\phi$ (cf. \cite[Theorem I.1]{Be}).

We say that $f\colon (0,\infty)\rightarrow (0,\infty)$ is a \emph{ complete Bernstein function} if it has representation (\ref{eq:sub}) such that the L\' evy measure has a completely monotone density (with respect to the Lebesgue measure). It follows from \cite[Proposition 7.1]{SSV} that $f^*\colon (0,\infty)\rightarrow (0,\infty)$ defined by
\[
f^*(\lambda)=\frac{\lambda}{f(\lambda)}
\] is also a complete Bernstein function.

The \emph{potential measure} $U$ of the subordinator $S$ is defined by
\[
U(A)=\ensuremath{\mathbb{E}} \left[\int_0^\infty 1_{\{S_t\in A\}}\, dt\right],\ A\subset [0,\infty).
\]
The Laplace transform of $U$ is then
\begin{equation}\label{eq:lapl}
\mathcal{L}U(\lambda)=\int_{(0,\infty)}e^{-\lambda t}\,U(dt)=\frac{1}{\phi(\lambda)},\ \lambda>0.
\end{equation}

Let $B=(B_t,\P_x)_{t\geq 0, x\in {\ensuremath{\mathbb{R}}}^d}$ be a Brownian motion in ${\ensuremath{\mathbb{R}}}^d$ independent of the subordinator $S$. The process $X=(X_t,\P_x)_{t\geq 0,\ x\in {\ensuremath{\mathbb{R}}}^d}$ defined by
\[
X_t=B_{S_t},\ t\geq 0
\]
is called the \emph{subordinate Brownian motion}. By \cite[Theorem 30.1]{S} we conclude that $X$ is a L\' evy process and
\[
\ensuremath{\mathbb{E}}_x\left[e^{i\xi\cdot (X_t-X_0)}\right]=e^{-t\Phi(\xi)},\ \xi\in {\ensuremath{\mathbb{R}}}^d
\]
with $\Phi(\xi)=\phi(|\xi|^2)$. Moreover, we can rewrite $\Phi$ in the following way
\[
\Phi(\xi)=\int_{{\ensuremath{\mathbb{R}}}^d\setminus \{0\}}\left(e^{i\xi\cdot y}-1-i\xi\cdot y1_{\{|y|\leq 1\}}\right)j(|y|)\,dy
\]
where $j\colon (0,\infty)\rightarrow (0,\infty)$ is given by
\[
j(r)=(4\pi)^{-d/2}\int_{(0,\infty)}t^{-d/2}e^{-r^2/4t}\,\mu(dt),\ r>0.
\]
Therefore, the L\' evy measure of $X$ has density given by $J(y)=j(|y|)$.
Note that $j$ is a non-increasing function.

From now on we denote by $S=(S_t)_{t\geq 0}$ the subordinator with the Laplace exponent
\[
\phi(\lambda)=\frac{\lambda}{\ln(1+\lambda)}-1
\]
and by $X$ the correspoding subordinate Brownian motion.

Note that
\[
\ell(\lambda)=\ln(1+\lambda)=\int_0^\infty (1-e^{-\lambda t})\,t^{-1}e^{-t}\,dt
\]
and thus $\ell$ is a complete Bernstein function. Therefore, $\phi$ is also a complete Bernstein function and so the L\' evy measure of the subordinator $S$ has a completely monotone density $\mu(t)$.

Let $T=(T_t)_{t\geq 0}$ be the subordinator with the Laplace exponent $\ell$. This process is known as the \emph{gamma subordinator}.
 It follows from \cite[Corollary 10.7 and Corrolary 10.8]{SSV} that the potential measure $V$ of $T$ has a non-increasing density $v(t)$ and the following is true
 \begin{equation}\label{eq:connection}
 v(t)=1+\int_t^\infty \mu(s)\,ds,\ t>0.
 \end{equation}
By \cite[Theorem 2.2]{SSV2} we get the following asymptotic behavior of $v$
\begin{equation}\label{eq:v_asym}
v(t)\sim t^{-1}\left(\ln\frac 1 t\right)^{-2}\ \textrm{ as } t\to 0+.
\end{equation}

Now we can prove the asymptotic behavior of the jumping function $J$. The proof of the following proposition is basically the proof of \cite[Lemma 3.1]{SSV2} but with the use of Potter's theorem (cf. \cite[Theorem 1.5.6 (ii)]{BGT}), which was also done in \cite[Lemma 5.1]{KSV2}.

\begin{Prop}\label{prop:jump_asymp}
The following asymptotic behavior of the function $j$ holds
  \[
   	j(r)\sim \frac{4\Gamma(\frac{d}{2}+1)}{\pi^{d/2}}\cdot \frac{1}{r^{d+2}\left(\ln\,\frac{1}{r^2}\right)^2}\ \textrm{ as } \ r\to 0+.
  \]
\end{Prop}
\noindent\textbf{Proof.}\
Using (\ref{eq:connection}) and (\ref{eq:v_asym}) we get
\[
 \int_t^\infty \mu(s)\,ds\sim \frac{1}{t(\ln{t})^2}\ \textrm{ as } \ t\to \,0+
\]
and thus by the Karamata's monotone density theorem (see \cite[Theorem 1.7.2]{BGT}) we have
\begin{equation}\label{eq:mu_asymp}
 \mu(t)\sim \frac{1}{t^2(\ln\,t)^2}\ \textrm{ as } \ t\to 0+.
\end{equation}
By change of variable we get
\begin{align}
j(r)&=(4\pi)^{-d/2}\int_0^\infty t^{-d/2}e^{-\frac{r^2}{4t}}\mu(t)\,dt\nonumber\\&=4^{-1}\pi^{-d/2}r^{-d+2}\int_0^\infty t^{d/2-2}e^{-t}\mu\left(\frac{r^2}{4t}\right)\,dt\nonumber\\
&=4^{-1}\pi^{-d/2}r^{-d+2}\mu(r^2)\int_0^\infty t^{d/2-2}e^{-t}\frac{\mu\left(\frac{|h|^2}{4t}\right)}{\mu(r^2)}\,dt.\label{eq:tmp101}
\end{align}
By Potter's theorem (cf.  \cite[Theorem 1.5.6 (iii)]{BGT}) we see that there is a constant $c_1>0$ such that
\[
\frac{\mu\left(\frac{r^2}{4t}\right)}{\mu(r^2)}\leq c_1 (t^{2-1/2}\vee t^{2+1/2})\ \textrm{ for all } t>0 \textrm{ and } r>0.
\]
Therefore we can apply the dominated convergence theorem in (\ref{eq:tmp101}) and to get
\begin{equation}\label{eq:tmpj10}
\lim_{r\to 0+}\frac{j(r)}{4\pi^{-d/2}r^{-d+2}\mu(r^2)}=\Gamma\left(\frac{d}{2}+1\right).
\end{equation}
Combining (\ref{eq:mu_asymp}) and (\ref{eq:tmpj10}) we finish the proof.
\hfill\ensuremath{\Box}\\

If $d\geq 3$, then by  \cite[Corollary 37.6]{S} and Proposition \ref{prop:jump_asymp} we conclude that $X$ is transient. Thus we can define a measure $G(x,\cdot)$ in the following way
\begin{equation}\label{eq:greenm}
G(x,A)=\ensuremath{\mathbb{E}}_x\left[\int_0^\infty 1_{\{X_t\in A\}}\,dt\right]=\int_0^\infty \P_x(X_t\in A)\,dt.
\end{equation}
Using \cite[Proposition 28.1]{S} we deduce that $X$ has a transition density $p(t,x,y)$ and the measure $G(x,\cdot)$ defined by (\ref{eq:greenm}) has a density which we denote by
\[
G(x,y)=\int_0^\infty p(t,x,y)\,dt
\]
and call the \emph{Green function} of $X$. Using \cite[Theorem 30.1]{S} we see that $G(x,y)=g(|x-y|)$, with
\begin{equation}\label{eq:greenj}
g(r)=(4\pi)^{-d/2}\int_{(0,\infty)} t^{-d/2}e^{-r^2/4t}\,U(dt),
\end{equation}
where $U$ is the potential measure of the subordinator $S$.
Combinig  \cite[Corollary 10.7 and Corollary 10.8]{SSV} we see that $U$ has a completely monotone density $u(t)$.

\begin{Lem}\label{prop:pot_asymp}
  We have the following asymptotics of $u$
\begin{align*}
 u(t)&\sim \ln\frac{1}{t}\ \textrm{ as } \ t\to 0+,&
u(t)&\to 2\ \textrm{ as } \ t\to \infty.
\end{align*}
\end{Lem}
\noindent\textbf{Proof.}\
We can readily check that
\begin{align*}
 	\phi(\lambda)&\sim \frac{\lambda}{2}\ \textrm{ as } \ \lambda\to 0+,&
 	\phi(\lambda)&\sim \frac{\lambda}{\ln\lambda}\ \textrm{ as } \ \lambda\to\infty,
\end{align*}
and thus by (\ref{eq:lapl}) we deduce
\begin{align*}
 	\mathcal{L}U(\lambda)&\sim \frac{2}{\lambda}\ \textrm{ as } \ \lambda\to 0+, &
 	\mathcal{L}U(\lambda)&\sim \frac{\ln\lambda}{\lambda}\ \textrm{ as } \ \lambda\to\infty.
\end{align*}
By the Karamata's Tauberian theorem (cf.  \cite[Theorem 1.7.1]{BGT}) we conclude that
\begin{align*}
 	U(t)&\sim 2t\ \textrm{ as } \ t\to \infty, &
 	U(t)&\sim t\ln\frac{1}{t}\ \textrm{ as } \ t\to 0+.
\end{align*}
Finally, using Karamata's monotone density theorem (cf. \cite[Theorem 1.7.2]{BGT}) we get
\begin{align*}
 	u(t)&\sim 2\ \textrm{ as } \ t\to \infty,
&
 	u(t)&\sim \ln\frac{1}{t}\ \textrm{ as } \ t\to 0+.
\end{align*}
\hfill\ensuremath{\Box}\\
\begin{Prop}\label{prop:green_asymp}
The following is true
	\begin{align*}
		g(r)&\sim \frac{\Gamma(\frac{d}{2}-1)}{2\pi^{d/2}}r^{2-d}\ln\frac{1}{r} \ \textrm{ as } \  r\to 0+, &
		g(r)&\sim \frac{\Gamma(\frac{d}{2}-1)}{2\pi^{d/2}}r^{2-d} \ \textrm{ as } \  r\to \infty.\\
	\end{align*}
\end{Prop}
\noindent\textbf{Proof.}\
Using (\ref{eq:greenj}) and changing variable we have
\begin{align}
g(r)&=4^{-1}\pi^{-d/2}r^{-d+2}\int_0^\infty t^{d/2-2}e^{-t}u\left(\frac{r^2}{4t}\right)\,dt\nonumber\\
&=4^{-1}\pi^{-d/2}r^{-d+2}u(r^2)\int_0^\infty t^{d/2-2}e^{-t}\frac{u\left(\frac{r^2}{4t}\right)}{u(r^2)}\,dt.\label{eq:tmp102}
\end{align}
From Potter's theorem (cf.  \cite[Theorem 1.5.6 (ii)]{BGT}) we deduce that there is a constant $c_1>0$ such that
\[
\frac{u\left(\frac{r^2}{4t}\right)}{u(r^2)}\leq c_1(t^{1/2}\vee t^{-1/2})\ \textrm{ for all } t>0 \textrm{ and } x\in {\ensuremath{\mathbb{R}}}^d,\ x\not= 0.
\]
Therefore we can use the dominated convergence theorem in (\ref{eq:tmp102}) to get
\begin{equation}\label{eq:tmpgj01}
\lim_{r\to 0+}\frac{g(r)}{4^{-1}\pi^{-d/2}r^{-d+2}u(r^2)}=\Gamma\left(\frac{d}{2}-1\right).
\end{equation}
Using (\ref{eq:tmpgj01}) and Lemma \ref{prop:pot_asymp} we obtain the asymptotics ti $0$. The other asymptotical formula is obtained simliarly.
\hfill\ensuremath{\Box}\\

We will need the following technical lemma later.
\begin{Lem}\label{heat:lem.100}
\begin{itemize}
\item[(a)]
 Let $f\colon (0,1)\rightarrow{\ensuremath{\mathbb{R}}}$ be defined by
\[
 f(t)=\frac{t^{d-2}}{\ln\frac 1 t}.
\]
Then $f$ is stricly increasing and
\[
 f^{-1}(t)\sim (d-2)^{-\frac{1}{d-2}}\, t^{\frac{1}{d-2}}\left(\ln \frac 1 t\right)^{\frac{1}{d-2}}\ \textrm{ as } \ t\to 0+.
\]
\item[(b)] The following is true
\[
\int_0^rs^{d-1}g(s)\,ds \sim \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}r^2\ln\frac 1 r\ \textrm{ as }\ r\to 0+.
\]
\end{itemize}
\end{Lem}
\noindent\textbf{Proof.}\
\begin{itemize}
\item[(a)]
It is easy to see that $f$ is a strictly increasing function. Define $h\colon (0,1)\rightarrow {\ensuremath{\mathbb{R}}}$ by
\[
 h(t)=t^{\frac{1}{d-2}}\left(\ln \frac 1 t\right)^{\frac{1}{d-2}}.
\]
Then
\begin{align*}
 \lim_{t\to 0+}\frac{f^{-1}(t)}{h(t)}&=\lim_{t\to 0+}\frac{f^{-1}(f(t))}{h(f(t))}=\lim_{t\to 0+}\frac{t}{\frac{t}{\left(\ln\frac 1 t\right)^\frac{1}{d-2}}\left(\frac{t^{d-2}}{\ln\frac 1 t}\right)^\frac{1}{d-2}}\\
&=\lim_{t\to 0+}\left(\frac{\ln\frac 1 t}{(d-2)\ln\frac 1 t+\ln\ln\frac{1}{t}}\right)^\frac{1}{d-2}=(d-2)^{-\frac{1}{d-2}}.
\end{align*}
\item[(b)] By applying Karamata's theorem (cf. \cite[Proposition 1.5.8]{BGT}) we get
\begin{align*}
 \int_0^r s^{d-1}g(s)\,ds&\sim \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}\int_0^r s\ln\frac{1}{s^2}\,ds\\&\sim \frac{\Gamma(\frac{d}{2}-1)}{4\pi^{d/2}}r^2\ln \frac{1}{r}\ \textrm{ as } \ r\to 0+.
 \end{align*}
 \end{itemize}
\hfill\ensuremath{\Box}\\

Let $D\subset {\ensuremath{\mathbb{R}}}^d$ be an open set. We define the \emph{killed process} $X^D=(X^D_t)_{t\geq 0}$ by killing process $X$ upon exiting set $D$, i.e.
\[
X^D_t=\left\{
\begin{array}{cl}
X_t, & t<\tau_D\\
\partial,& t\geq \tau_D.
\end{array}
\right.
\]
Here $\partial$ is an extra point adjoined to $D$. In this case the killed process also has a transition density and it is given (cf. proof of \cite[Theorem 2.4]{CZ}) by
\begin{equation}\label{eq:tranD}
p_D(t,x,y)=p(t,x,y)-\ensuremath{\mathbb{E}}_x \left[p(t-\tau_D,X_{\tau_D},y);\tau_D<t\right].
\end{equation}
The Green function of $X^D$ also exists and it is given by
\begin{equation}\label{eq:greenD}
G_D(x,y)=\int_0^\infty p_D(t,x,y)\,dt=G(x,y)-\ensuremath{\mathbb{E}}_x\left[G(X_{\tau_D},y)\right]\ \textrm{ for } x,y\in D.
\end{equation}
 Since $\P_x(X_{\tau_{B(x_0,r)}}\in \partial B(x_0,r))=0$ (cf. \cite{SZ}) for $x_0\in {\ensuremath{\mathbb{R}}}^d$ and $r>0$, it follows from Theorem 1 in \cite{IW} that for any non-negative function $h\colon{\ensuremath{\mathbb{R}}}^d\rightarrow [0,\infty)$ we have
 \begin{equation}\label{eq:tmp120}
 \ensuremath{\mathbb{E}}_x[h(X_{\tau_{B(x_0,r)}})]=\int_{\overline{B(x_0,r)}^c}\int_{B(x_0,r)} G_{B(x_0,r)}(x,y)j(|z-y|)h(z)\,dy\,dz.
 \end{equation}
 If we define \emph{Poisson kernel} $K_{B(x_0,r)}\colon B(x_0,r)\times \overline{B(x_0,r)}^c\rightarrow [0,\infty)$ by
  \[
 K_{B(x_0,r)}(x,z)=\int_{B(x_0,r)}G_{B(x_0,r)}(x,y)j(|z-y|)\,dy\ \textrm{ for } x\in B(x_0,r),\ z\in \overline{B(x_0,r)}^c,
 \]
from (\ref{eq:tmp120}) we get
 \begin{equation}\label{eq:Poisson1}
 \ensuremath{\mathbb{E}}_x[h(X_{\tau_{B(x_0,r)}})]=\int_{\overline{B(x_0,r)}^c} K_{B(x_0,r)}(x,z)h(z)\,dz.
 \end{equation}

 \begin{Prop}\label{heat:prop.80}
 There exist constants  $R_0\in(0,1/6)$ and $C_1>0$ such that for  any  $r\leq R_0$ and $x_0\in{\ensuremath{\mathbb{R}}}^d$,
\begin{equation}
G_{B(x_0,4r)}(x,y)\geq C_1r^{2-d}\ln\frac{1}{r}\  \textrm{ for all }  x,y\in B(x_0,r).
\end{equation}
\end{Prop}
\noindent\textbf{Proof.}\
Choose $0<c_1<1<c_2$ such that \[c_1^2\left(\frac{1}{2}\right)^{d-2}-c_2^2\left(\frac{1}{3}\right)^{d-2}> 0.\]
Using Proposition \ref{prop:green_asymp}  we can choose $R_0\in(0,1/6)$ such that for $r\leq 3R_0$ we have
\begin{equation}\label{est_1}
 c_1c_3r^{2-d}\ln\frac{1}{r}\leq g(r)\leq c_2c_3r^{2-d}\ln\frac{1}{r},\ \ \ c_1\leq \frac{\ln\frac{1}{2r}}{\ln\frac{1}{r}}\leq c_2,\ \ \  c_1\leq \frac{\ln\frac{1}{3r}}{\ln\frac{1}{r}}\leq c_2,
\end{equation}
where $c_3=\frac{\Gamma(d/2-1)}{2\pi^{d/2}}$.
Let $r\leq R_0$, $x_0\in{\ensuremath{\mathbb{R}}}^d$ and  $x,y\in B(x_0,r)$. By (\ref{est_1}) and monotonicity of $g$ we get
\begin{align*}
 G_{B(x_0,4r)}(x,y)&=G(x,y)-\ensuremath{\mathbb{E}}_x[G(Y_{\tau_{B(x_0,4r)}},y)]=g(|x-y|)-\ensuremath{\mathbb{E}}_x[g(|Y_{\tau_{B(x_0,4r)}}-y|)]\\
&\geq g(2r)-g(3r)\geq c_3\left( c_1(2r)^{2-d}\ln\frac{1}{2r}-c_2\left(3r\right)^{2-d}\ln\frac{1}{3r}\right)\\
&=c_3r^{2-d}\ln{\frac{1}{r}}\left(c_1\,\left(\frac{1}{2}\right)^{d-2}\frac{\ln{\frac{1}{2r}}}{\ln{\frac{1}{r}}}-c_2\,\left(\frac{1}{3}\right)^{d-2}\frac{\ln{\frac{1}{3r}}}{\ln{\frac{1}{r}}}\right)\\&\geq c_3r^{2-d}\ln{\frac{1}{r}}\left(c_1^2\left(\frac{1}{2}\right)^{d-2}-c_2^2\left(\frac{1}{3}\right)^{d-2}\right).
\end{align*}
Hence we may take
\[C_1=c_3\left(c_1^2\left(\frac{1}{2}\right)^{d-2}-c_2^2\left(\frac{1}{3}\right)^{d-2}\right)>0.\]
\hfill\ensuremath{\Box}\\

\begin{Prop}\label{prop:Poisson1}
 There exist $R_1\in(0,R_0]$ and  a constant $C_2>0$ such that for  any $r\leq R_1$ and $x_0\in{\ensuremath{\mathbb{R}}}^d$,
\begin{equation}
 K_{B(x_0,r)}(x,z)\leq C_2\,K_{B(x_0,4r)}(y,z)\ \textrm{ for all } x,y\in B(x_0,r/2),\,z\in B(x_0,4r)^c.
\end{equation}
\end{Prop}
\noindent\textbf{Proof.}\
Take $r\leq R_0$, $x_0\in{\ensuremath{\mathbb{R}}}^d$, $x,y\in B(x_0,r/2)$ and $z\in B(a,3r)^c$. Using Proposition \ref{heat:prop.80} we get
\begin{align}\nonumber
 K_{B(x_0,4r)}(y,z)&=\int_{B(x_0,4r)}G_{B(x_0,4r)}(y,u)\,j(|z-u|)\,du\\&\geq\int_{B(x_0,r)}G_{B(x_0,4r)}(y,u)\,j(|z-u|)\,du\\\label{heat:eq.34}
&\geq C_1r^{2-d}\ln\frac{1}{r}\int_{B(x_0,r)}j(|z-u|)\,du.
\end{align}
On the other side, applying  \cite[Lemma 2.7]{Mi} to $j$, and then using Proposition \ref{prop:green_asymp}
 and Lemma \ref{heat:lem.100} (b)  we see that there exist constants  $R_1\in(0,R_0]$ and
$c_1,c_2,c_3>0$ such that\\
\begin{align*}
 K_{B(x_0,r)}(x,z)=&\,\int_{B(x_0,3r/4)}G_{B(x_0,r)}(x,v)j(|z-v|)\,dv\\&+\int_{B(a,
r)\setminus B(x_0,3r/4)}G_{B(x_0,r)}(x,v)j(|z-v|)\,dv\\
\leq\,&
c_1r^{-d}\int_{B(x_0,r)}j(|z-u|)\,du\int_{B{(a,3r/4)}}g(|x-v|)\,dv\,\\&+c_2r^{2-d}
\ln\frac{1}{r}\int_{B(x_0,r)\setminus B(x_0,3r/4)}j(|z-u|)\,du\\
\leq\,& c_3\,r^{2-d}\ln\frac{1}{r}\int_{B(x_0,r)}j(|z-u|)\,du+c_2r^{2-d}\ln\frac{1}{r}\int_{B(x_0,r)}j(|z-u|)\,du\\
=&(c_2+c_3)r^{2-d}\ln\frac{1}{r}\int_{B(x_0,r)}j(|z-u|)\,du,
\end{align*}
where in the first term in the last inequality we have used
\[
 \int_{B(x_0,3r/4)}g(|x-v|)\,dv\leq \int_{B(x,2r)}g(|x-v|)\,dv\leq c_1'r^2\ln\frac{1}{r}.
\]
Therefore
\[
 K_{B(x_0,r)}(x,z)\leq \frac{c_2+c_3}{C_1}K_{B(x_0,4r)}(y,z).
\]
\hfill\ensuremath{\Box}\\

 For a measure $\rho$ on ${\ensuremath{\mathbb{R}}}^d$ we define its \emph{potential} by
 \[
 G\rho(x)=\int_{{\ensuremath{\mathbb{R}}}^d}G(x,y)\,\rho(dx).
 \]
Denote by  $\ensuremath{\mathrm{Cap}}$ the (0-order) \emph{capacity} with respect to $X$ (cf.  \cite[Section II.2]{Be}). It is proved in \cite[Corollary II.8]{Be} that for any compact set $K\subset {\ensuremath{\mathbb{R}}}^d$ there exists a measure $\rho_K$, called the  \emph{equilibrium measure}, which is supported by $K$ and satisfies
\begin{equation}\label{eq:equilibrium1}
G\rho_K(x)=\P_x(T_K<\infty)\ \textrm{ for a.e. } x\in {\ensuremath{\mathbb{R}}}^d.
\end{equation}
Moreover, the following is true
\[
\frac{1}{\ensuremath{\mathrm{Cap}} (K)}=\inf\left\{\int_{{\ensuremath{\mathbb{R}}}^d} G\rho(x)\,\rho(dx)\colon \rho \textrm{ is a probability measure supported by}\ K\right\}
\]
and the infimum is attained at the equilibrium measure $\rho_K$. If we combine Lemma \ref{heat:lem.100} (b) and  \cite[Proposition 5.3]{SSV2} we conclude that there exist constants $C_3, C_{4}>0$ such that
\begin{equation}\label{eq:est_cap}
C_3\frac{r^{d-2}}{\ln\frac{1}{r}} \leq \ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,r)}\right)\leq C_{4}\frac{r^{d-2}}{\ln\frac{1}{r}}\ \textrm{ for } x_0\in{\ensuremath{\mathbb{R}}}^d, \ 0<r\leq 1/2.
\end{equation}
Now we can prove a Krylov-Safonov-type estimate.
\begin{Prop}\label{heat:prop.1001}
 There exists a constant $C_{5}>0$  such that for any $x_0\in{\ensuremath{\mathbb{R}}}^d$, $r\leq R_1$, closed subset A of $B(x_0,r)$ and $y\in B(x_0,r)$,
\begin{equation*}
 \P_y(T_A<\tau_{B(x_0,4r)})\geq C_{5}\frac{\ensuremath{\mathrm{Cap}}(A)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,4r)}\right)}.
\end{equation*}
\end{Prop}
\noindent\textbf{Proof.}\
Let $x_0\in{\ensuremath{\mathbb{R}}}^d$, $r\leq R_1$ and let $A\subset B(x_0,r)$ be a closed subset. We may assume that $\ensuremath{\mathrm{Cap}}(A)>0$. Let $\rho_A$ be the equilibrium measure of $A$. If $G_{B(x_0,4r)}$ is the Green function of the process $X$ killed upon exiting from $B(x_0,4r)$, then  for $y\in B(x_0,r)$ we have
\begin{align}
 G_{B(x_0,4r)}\rho_A(y)&=\ensuremath{\mathbb{E}}_y[G_{B(x_0,4r)}\rho_A(Y_{T_A});T_A<\tau_{B(x_0,4r)}]\nonumber\\
&\leq \sup_{z\in{\ensuremath{\mathbb{R}}}^d}G_{B(x_0,4r)}\rho_A(z)\P_y(T_A<\tau_{B(x_0,4r)})\nonumber\\
&\leq \P_y(T_A<\tau_{B(x_0,4r)}),\label{heat:eq.40}
\end{align}
since $G_{B(x_0,4r)}\rho_A(z)\leq G\rho_A(z)\leq 1$ by (\ref{eq:equilibrium1}).
Also, for any $y\in B(x_0,r)$ we have
\begin{align}
 G_{B(x_0,4r)}\rho_A(y)&=\int_{{\ensuremath{\mathbb{R}}}^d}G_{B(x_0,4r)}(y,z)\rho_A(dz)\geq
\rho_A({\ensuremath{\mathbb{R}}}^d)\inf_{z\in B(x_0,r)}G_{B(x_0,4r)}(y,z)\nonumber\\
&=\ensuremath{\mathrm{Cap}}(A)\inf_{z\in B(x_0,r)}G_{B(x_0,4r)}(y,z).\label{heat:eq.41}
\end{align}
Using (\ref{heat:eq.40}), (\ref{heat:eq.41}) and Proposition \ref{heat:prop.80} we  obtain
\[
 \P_y(T_A<\tau_{B(x_0,4r)})\geq C_1\ensuremath{\mathrm{Cap}}(A)r^{2-d}\ln\frac{1}{r}.
\]
By (\ref{eq:est_cap}) we see that
\[
 \P_y(T_A<\tau_{B(x_0,4r)})\geq C_1C_3\frac{\ensuremath{\mathrm{Cap}}(A)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,4r)}\right)}
\]
for $r\leq R_1$.
\hfill\ensuremath{\Box}\\

\section{Harnack inequality}\label{sect:hi}

{\bf Proof of Theorem \ref{tm:harnack}.} Define $f(t)= \frac{t^{d-2}}{\ln\frac{1}{t}}$. By Lemma \ref{heat:lem.100} (a) we can choose $R\leq \frac{R_1}{4}\wedge\frac{1}{16}$ such that
\begin{equation}\label{heat:tmp}
 f^{-1}(r)\leq c_0r^{\frac{1}{d-2}}\left(\ln\frac 1 r\right)^{\frac{1}{d-2}}\ \textrm{ for all } \ r\leq R
\end{equation}
and for some constant $c_0\geq 1$.

Let $x_0\in{\ensuremath{\mathbb{R}}}^d$ and  $r\leq R$.
	Without loss of generality we may suppose
	\[
	 	\inf_{z\in B(x_0,r)} h(z)=\frac{1}{2}
	\]
Let $z_0\in B(x_0,r)$ be such that $h(z_0)\leq 1$. It is enough to show that $h$ is bounded from above by some constant independent of $h$. By Proposition \ref{heat:prop.1001}  there exists $c_1>0$ such that
\begin{equation}\label{eq:har1}
 	\P_x(T_F<\tau_{B(x,s)})\geq c_1,
\end{equation}
for any $s\in (0,R_1)$, $x\in {\ensuremath{\mathbb{R}}}^d$ and a compact subset $F\subset B(x,s/4)$ such that
\[
 \frac{\ensuremath{\mathrm{Cap}}(F)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x,s/4)}\right)}\geq \frac{1}{3}.
\]
Put
\begin{equation}\label{eq:eta}
 \eta=\frac{c_1}{3},\ \ \zeta=\frac{\eta}{3}\wedge \frac{\eta}{C_2}.
\end{equation}
Suppose that there exists $x\in B(x_0,r)$ such that $h(x)=K$ for
\begin{equation}\label{eq:K_1}
 K>c_0^{d-1}4^{3(d-2)}\,\left(\frac{2C_4}{C_3C_5\zeta}\right)^{1/2}.
\end{equation}
It is possible to choose a unique $s>0$ such that
\[
f(s/4)=\frac{2\,C_4}{C_3\,C_5\,\zeta\,K}f(4r),
\]
since $f$ is strictly increasing and continuous on $(0,1)$ and $c_2:=\frac{2\,C_4}{C_3\,C_5\,\zeta\,K}<\frac{1}{4^3}$.

Using inequality
\[
 p\ln\frac 1 p<2\sqrt{p}\ \textrm{ for } \ p\in (0,1)
\]
and (\ref{heat:tmp}) we get
\begin{align}
 \frac s 4&\leq c_0\left(c_2f(4r)\right)^{\frac{1}{d-2}}\left(\ln\frac{1}{c_2f(4r)}\right)^{\frac{1}{d-2}}\nonumber\\
 &\leq c_0^{1+\frac{1}{d-2}}c_2^{\frac{1}{d-2}}\left(\frac{\ln\frac{1}{c_2}}{\ln\frac{1}{4r}}+d-2-\frac{\ln\ln\frac{1}{4r}}{\ln\frac{1}{4r}}\right)^{\frac{1}{d-2}}4r\nonumber\\
 &\leq  c_0^{1+\frac{1}{d-2}}\left(c_2\ln\frac{1}{c_2}+(d-2)c_2\right)^{\frac{1}{d-2}}4r\nonumber\\
 &\leq  c_0^{1+\frac{1}{d-2}}\left(2\sqrt{c_2}+(d-2)c_2\right)^{\frac{1}{d-2}}4r\nonumber\\
 &\leq c_0^{1+\frac{1}{d-2}}(d\sqrt{c_2})^{\frac{1}{d-2}}4r\label{eq:tmpharn01}
 \end{align}
and thus $s\leq \frac{r}{4}$ by (\ref{eq:K_1}).

By (\ref{eq:est_cap}) we obtain
\[
 \ensuremath{\mathrm{Cap}}\left(\overline{B(x,s/4)}\right)\geq C_3f(s/4)\geq \frac{2}{C_5\,\zeta\,K}\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,4r)}\right).
\]
Let $A$ be a compact subset of
\[
 A'=\{t\in B(x,s/4)\colon h(t)\geq \zeta K \}.
\]
By the optional stopping theorem  we have
\begin{align}
 1&\geq h(z_0)\geq E_{z_0}[h({X_{T_A\wedge \tau_{B(x_0,4r)}}});T_A< \tau_{B(x_0,4r)}]\nonumber\\
&\geq \zeta K\P_{z_0}(T_A< \tau_{B(x_0,4r)})\geq C_5\zeta K\frac{\ensuremath{\mathrm{Cap}}(A)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,4r)}\right)},\label{eq:tmpharn02}
\end{align}
where in the last inequality we have used Proposition \ref{heat:prop.1001}. Therefore, from (\ref{eq:tmpharn01}) and (\ref{eq:tmpharn02}) we conclude
\[
 \frac{\ensuremath{\mathrm{Cap}}(A')}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x,s/4)}\right)}= \frac{\ensuremath{\mathrm{Cap}}(A')}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,4r)}\right)}\cdot  \frac{\ensuremath{\mathrm{Cap}}\left(\overline{B(x_0,4r)}\right)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x,s/4)}\right)}\leq \frac{1}{2}
\]
and thus, by subadditivity of capacity, there exists a compact set \[F\subset \overline{B(x,s/4)}\setminus A'\] such that
\begin{equation}\label{eq:f}
 \frac{\ensuremath{\mathrm{Cap}}(F)}{\ensuremath{\mathrm{Cap}}\left(\overline{B(x,s/4)}\right)}\geq \frac{1}{3}.
\end{equation}
 Next we prove that
\[
 \ensuremath{\mathbb{E}}_x[h(X_{\tau_{B(x,s)}}); X_{\tau_{B(x,s)}}\not\in B(x,4s)]\leq \eta K.
\]
If the latter is not true, then \[\ensuremath{\mathbb{E}}_x[h(X_{\tau_{B(x,s)}}); X_{\tau_{B(x,s)}}\not\in B(x,4s)]> \eta K\] and by Proposition \ref{prop:Poisson1} and (\ref{eq:Poisson1}) for any $y\in B(x,s/4)$ we have
\begin{align*}
 h(y)&=\ensuremath{\mathbb{E}}_y[h(X_{\tau_{B(x,4s)}})]=\ensuremath{\mathbb{E}}_y[h(X_{\tau_{B(x,4s)}});X_{\tau_{B(x,4s)}}\not\in B(x,4s)]\\
&=\int_{\overline{B(x,4s)}^c} K_{B(x,4s)}(y,z)h(z)\,dz\geq C_2^{-1}\int_{\overline{B(x,4s)}^c} K_{B(x,s)}(y,z)h(z)\,dz\\
&=C_2^{-1}\ensuremath{\mathbb{E}}_y[h(X_{\tau_{B(x,s)}});X_{\tau_{B(x,s)}}\not\in B(x,4s)]\geq C_2^{-1}\eta K\geq \zeta K,
\end{align*}
which is a contradiction with (\ref{eq:f}) and the definition of the set $A'$.

Set
\[
 M=\sup_{B(x,4s)}\,h.
\]
We have
\begin{align*}
 K&=h(x)=\ensuremath{\mathbb{E}}_x[h(Y_{\tau_{B(x,s)}})]\\
&=\ensuremath{\mathbb{E}}_x[h(Y_{T_F});T_F<\tau_{B(x,s)}]+\ensuremath{\mathbb{E}}^x[h(Y_{\tau_{B(x,s)}});\tau_{B(x,s)}<T_F,X_{\tau_{B(x,s)}}\in B(x,4s)]\\
&+\ensuremath{\mathbb{E}}_x[h(Y_{\tau_{B(x,s)}});\tau_{B(x,s)}<T_F,X_{\tau_{B(x,s)}}\not\in B(x,4s)]\\
&\leq \zeta K\P_x(\tau_{B(x,s)}<T_F)+M\P_x(\tau_{B(x,s)}<T_F)+\eta K\\
&= \zeta K\P_x(\tau_{B(x,s)}<T_F)+M(1-\P_x(\tau_{B(x,s)}<T_F))+\eta K
\end{align*}
and thus
\[
 \frac{M}{K}\geq \frac{1-\eta-\zeta \P^x(\tau_{B(x,s)}<T_F)}{1-\P^x(\tau_{B(x,s)}<T_F)}\geq 1+2\beta,
\]
for some $\beta>0$. It follows that there exists $x'\in B(x,4s)$ such that $h(x')\geq K(1+\beta)$. Repeating this procedure, we get a sequence $(x_n)$ such that $h(x_n)\geq K(1+\beta)^{n-1}$ and
\[
|x_{n+1}-x_n|\leq \left(4^3c_0\left(dc_0\sqrt{c_2}\right)^{\frac{1}{d-2}}\right)^nr=\left(c_3K^{-\frac{1}{d-2}}\right)^n\,r
\]
Therefore,
\[
 \sum_{n=1}^\infty |x_{n+1}-x_n|\leq c_4K^{-\frac{1}{d-2}}r.
\]

If $K>c_4^{d-2}$ we can find sequence $(x_n)$ in $B(x_0,2r)$ such that $h(x_n)\to \infty$ which is a contradiction to $h$ being bounded and so
\[
 \sup_{x\in B(x_0,r)}h(x)\leq c_4^{d-2}.
\]
\hfill\ensuremath{\Box}\\

\section{Regularity}\label{sect:reg}
 \begin{Lem}\label{lem:tmp200}
 There exists a  constant $C_6>0$ such that for all $r\in (0,1/8)$, $s\in [4r,1/2)$ and $x_0\in {\ensuremath{\mathbb{R}}}^d$ we have
 \[
 \P_x(X_{\tau_{B(x_0,r)}}\not \in B(x_0,s))\leq C_6\frac{r^2\ln\frac 1 r}{s^2\left(\ln\frac 1 s\right)^2}\ \textrm{ for all }\ x\in B(x_0,r/2).
 \]
 \end{Lem}
 \noindent\textbf{Proof.}\  Let $r\in (0,1/8)$, $s\in [4r,1/2)$ and $x_0\in {\ensuremath{\mathbb{R}}}^d$.
 Using (\ref{eq:tmp120}) with $h=1_{B(x_0,s)^c}$ for $x\in B(x_0,r/2)$ we have
 \begin{align*}
 \P_x(X_{\tau_{B(x_0,r)}}\not \in B(x_0,s))&=\int_{B(x_0,r)}G_{B(x_0,r)}(x,u)\int_{B(x_0,s)^c}j(|z-u|)\,dz\,du\\
 &\leq \int_{B(x_0,r)}g(|u-x|)\int_{B(u,s/2)^c}j(|z-u|)\,dz\,du\\
 &=\int_{B(x-x_0,r)}g(|u|)\,du\cdot\int_{B(0,s/2)^c}j(|z|)\,dz\\
 &\leq \int_{B(0,2r)}g(|u|)\,du\cdot\int_{B(0,s/2)^c}j(|z|)\,dz,
 \end{align*}
 where in the second inequality we have used the facts that $B(u,s/2)\subset B(x_0,s)$ and $G_{B(x_0,r)}(y,u)\leq g(|u-y|)$, while in the last inequality we have used $B(x-x_0,r)\subset B(0,2r)$. Now the conclusion follows from Proposition \ref{prop:jump_asymp} and Proposition \ref{prop:green_asymp}.
 \hfill\ensuremath{\Box}\\

 {\bf Proof of Theorem \ref{tm:hoelder}.} Let $r\in (0,1/6)$, $x_0\in {\ensuremath{\mathbb{R}}}^d$ and let  $h\colon {\ensuremath{\mathbb{R}}}^d\rightarrow [0,\infty)$ be bounded by $M>0$ and harmonic in $B(x_0,r)$.

 Let $z_0\in B(x_0,r/4)$. Define
 \[
 	r_n=\gamma_1 \,4^{-n},
 \]
 where we choose $\gamma_1>0$ small enough so that $B(x_0,4r_1)\subset B(z_0,r/4)$. Set $B_n=B(z_0,r_n)$ and $\tau_n=\tau_{B_n}$ and
 \[
 s_n= b^{-n},
 \]
 where constant $b>1$ will be chosen later.
 Let
 \[
 m_n=\inf_{x\in B_n}h(x)\ \ \ \textrm{ and }\ \ \  M_n=\sup_{x\in B_n}h(x).
 \]
 It is enough to prove that
 \begin{equation}\label{eq:ind}
 M_k-m_k\leq s_k
 \end{equation}
 for all $k\geq n_0$ for some $n_0\in {\ensuremath{\mathbb{N}}}$.
 We prove this by induction. Assume that (\ref{eq:ind}) holds for $k\in \{n_0,n_0+1,\ldots,n\}$, where $n\geq n_0$.

 Let $\varepsilon>0$ and let $x,y\in B_{n+1}$ such that $h(x)\leq m_{n+1}+\varepsilon$ and $h(y)\geq M_{n+1}-\varepsilon$. Our aim is to show that $h(y)-h(x)\leq s_{n+1}$. Then we have
 \[
 M_{n+1}-m_{n+1}\leq s_{n+1}+2\varepsilon
 \]
 and since $\varepsilon>0$ is arbitrary, we get (\ref{eq:ind}) for $k=n+1$.

 Let $A=\left\{x\in B_{n+1}\colon h(x)\leq \frac{m_n+M_n}{2}\right\}$ and assume that
 \[
 	\frac{\ensuremath{\mathrm{Cap}}(A)}{\ensuremath{\mathrm{Cap}}(B_{n+1})}\geq \frac{1}{2}
 \]
 (if this is not true, then we consider function $M-h$ and use the subadditivity of capacity). By Choquet's theorem $A$ is capacitable and therefore there exists a compact subset $K\subset A$ such that
 \[
	 \frac{\ensuremath{\mathrm{Cap}}(K)}{\ensuremath{\mathrm{Cap}}(B_{n+1})}\geq \frac{1}{3}.
 \]

 By the optional stopping theorem, we have
 \begin{align*}
 h(x)-h(y)=&\ \ensuremath{\mathbb{E}}_x[h(X_{\tau_n\wedge T_K})-h(y)]\\
 =&\ \ensuremath{\mathbb{E}}_x[h(X_{\tau_n\wedge T_K})-h(y);T_K<\tau_n,X_{\tau_n}\in B_{n-1}\setminus B_n]\\ &+\ensuremath{\mathbb{E}}_x[h(X_{\tau_n\wedge T_K})-h(y); T_K>\tau_n, X_{\tau_n}\in B_{n-1}\setminus B_n]\\
 &+\sum_{i=1}^{n-2}\ensuremath{\mathbb{E}}_x[h(X_{\tau_n\wedge T_K})-h(y);X_{\tau_n}\in B_{n-i-1}\setminus B_{n-i}]\\
 &+\ensuremath{\mathbb{E}}_x[h(X_{\tau_n\wedge T_K})-h(y); X_{\tau_n}\not\in B_1]\\
 \leq & \left(\frac{m_n+M_n}{2}-m_n\right)\P_x(T_K<\tau_n)+(M_{n-1}-m_{n-1})\P_x(T_K>\tau_n)\\
 &\ + \sum_{i=1}^{n-2}(M_{n-i-1}-m_{n-i-1})\P_x(X_{\tau_n}\not \in B_{n-i-1})+2M\P_x(X_{\tau_n}\not\in B_1).
  \end{align*}

  It follows from Proposition \ref{heat:prop.1001} that there is a constant $c_1>0$ such that
  \begin{equation}\label{eq:tmphold01}
  p_n:=\P_x(T_K<\tau_n)\geq c_1.
  \end{equation}
By Lemma \ref{lem:tmp200} we obtain
\[
h(x)-h(y)\leq \frac{1}{2}s_n\,p_n+s_{n-1}(1-p_n)+C_6\sum_{i=1}^{n-2} s_{n-i-1}\frac{r_n^2\ln\frac{1}{r_n}}{r_{n-i}^2\left(\ln\frac{1}{r_{n-i}}\right)^2}+2MC_6\frac{r_n^2\ln\frac{1}{r_n}}{r_{1}^2\left(\ln\frac{1}{r_{1}}\right)^2}.
\]
Then there exist constants $c_2,c_3>0$ such that
\begin{align}\nonumber
&h(x)-h(y)\leq\nonumber\\&\leq s_{n+1}\left[\frac{b}{2}\,p_n+b^2(1-p_n)+c_2\,n\left(4^{2}b^{-1}\right)^{-n}\,b^2\sum_{i=2}^{n-1}\frac{(4^2b^{-1})^{-i}}{i^2}+c_3Mb \,n\,(4^2b^{-1})^{-n}\right]\nonumber\\
&\leq s_{n+1}\left[\frac{b}{2}\,p_n+b^2(1-p_n)+c_2\,n\left(4^{2}b^{-1}\right)^{-n}\,b^2\sum_{i=2}^{\infty}\frac{(4^2b^{-1})^{-i}}{i^2}+c_3Mb \,n\,(4^2b^{-1})^{-n}\right].\label{eq:tmp234}
\end{align}
Choose  $b\in (1,2)$ such that $(1-c_1)b^2+c_1\,\frac{b}{2}<1$. By (\ref{eq:tmphold01}) we conclude
\[
	\frac{1}{2}b\,p_n+b^2(1-p_n)=b^2-p_n\left(b^2-\frac{b}{2}\right)\leq b^2-c_1\left(b^2-\frac{b}{2}\right)<1.
\]
Since $4^2b^{-1}>1$, we see that the last two terms in the parenthesis in (\ref{eq:tmp234}) can be made arbitrary small for $n$ large enough. Thus there is $n_0\in{\ensuremath{\mathbb{N}}}$ such that for any $n\geq n_0$ we have
\[
h(x)-h(y)\leq s_{n+1},
\]
which was to be proved.
\hfill\ensuremath{\Box}\\

\textbf{Acknowledgements.} The author would like to thank Z. Vondra\v{c}ek and M. Kassmann for many helpful suggestions and comments.

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