 So, we're gonna finish up these lectures today. So, it'll be a little bit less heavy-handed than yesterday. So, I mean, from a nicer point of view, it would have been good to switch the two lectures, but I kind of wanted you to see one thing before I did it at the end of today, because even if you kind of only vaguely understood the neck region of structures, I mean, I can at least pull back to something so we can sort of discuss morally what's gonna be happening. So what I wanna do today is discuss a little bit of the proof of Reifenberg itself. This is where I'm gonna start, even just the classical case. And for that matter, just for the classical case, what I'm gonna do first, actually, is simply write down again this example that we were discussing the first couple of days, and we'll prove it carefully there that we actually have a bi-holder equivalence between this wiggling example and an interval. But we're gonna do this not the way I kind of outlined it on the first day, but we're gonna do this in a somewhat different way, because I'm gonna do it in a way where literally, once we build the right structures, the proof of the classical Reifenberg will be verbatim with the verbatim estimates. It will be no different. So we'll get to seeing a nice clean example, how to actually, you know, how this looks and what it looks like, and then we can wave our hands through through what constructions are missing in the general case. So let's begin by recalling what this example looks like. So what we did was the following. So we had off our start with an interval and I did it from minus two to two. I really don't know why I picked two off hand. I did that so long ago and I've been writing with it for a while, I'm just kind of stuck with it. And then this is what we call our sort of S naught. And we started to build a sequence of guys, right? And then the first thing we did was we said, okay, we're gonna fix some background epsilon. And for us here, it's just some fixed background. Something, we're gonna build an assassin's triangle. Say this guy goes up, so if the length of this guy is the length of the single interval that sort of connects from from minus two to two, of course that's four, then this guy here is supposed to be epsilon times the length of the interval of the opposite side. And that was S one, which was itself a union of two intervals, right? So S one was union of, in general, SI, right, was a union of L, AI, J, and AI, J plus one. We'll recall by definition, these were points in R two and this was the segment connecting the two points by definition. And then the next step was that, okay, now we start to alternate. We went down by epsilon times the length of whatever each of these intervals are. And then we went up the next guy and this built us our nice S two, I guess, right? S zero, S one, S two, and so forth and so on. I mean, we kept alternating like this. And what you get out of this construction is a nice sort of Reifenberg set in the end and we sort of suggested why this thing was actually bi-holder to an interval and what I'd like to see is a little bit more carefully how we can construct this map to see that this thing is bi-holder to the unit interval. And the way I wanna do it now is the following. So the way I'd suggested before recall was that in the one dimensional case we can cheat and say that even though the lengths of all these things are going to infinity, we can parametrize each of them by arc length and try to start keeping track of what's happening with those mappings. But this is not at all how I want to do this. Instead of what I wanna do is the following. Let me make sure I have all my information recorded. So recall from our notation before, so our SI was our piecewise linear guy and the length of all these guys was the same because we always cut it up and their length was, because I want this down for later, the length of any interval in the ith segments was, so four was the original one and then every time it was what happened. It was the square root of one plus epsilon squared over two and then we multiply by ih time. All right, because what happens, we have the length and then we came out and did a nice little Pythagorean theorem. Okay, now here's how I wanna build the mapping. So here's our R as zero, here's R as one, here's R as two. Note the following and let me just actually draw the first two steps or one can view this as being any step somewhere in the middle. Maybe this is one of our LI, so this becomes LI plus one of length LI plus one. Then I wanna look at the projection map, pi, what's my notation? I, from SI plus one into SI. So there's a nice orthogonal projection here. So I can take any point here and look at the closest point, I just sort of boom projected, look at the closest point projection down. So if I'm here, that's literally just there. If I'm here, I'm literally just there. And I wanna build a map but what I wanna do is make the following sort of interesting observation. We're trying to eventually build a mapping from let's just say some SI. I want uniform estimates on Enei so I can pass it to the limit to get our wild sets into an interval. And what's one way of identifying the interval? The interval was actually our original S naught. I like that observation is sort of silly sounding but quite useful because I'm gonna try to build a mapping from SI and to the interval S naught by composing these guys. So the goal is the following. I'm gonna let, what's my notation? Those pies look exactly the same. I need different notation. Let's do phi. On the right up it's lower case and up a little bit who can tell here. I'm gonna get a mapping from SI. So after we wiggle on the scale two to the minus I roughly speaking into the interval which is just S naught so minus two to two by just composing. The goal is to show that this mapping is gonna be uniformly byholder independent of I. So in the end when we take our same limits to get our crazy set S we can just limit these maps right along with it. So we wanna get some nice estimates on this guy here. Okay so in order to do that all I want actually are two dumb observations. We're just gonna play with these observations over and over again. So if we look at this mapping here which maps SI plus one to SI it always looks like this. I mean you've always got some interval on SI and then that's connected by two intervals in SI plus one and therefore the projection map from SI plus one to SI is you just look at that interval and you plop it down. So if you look at that this tells you two things right away. So let's call these estimates one. Minus pi I of Y on the one hand is actually precisely equal in this case but you can think less than or equal to for the context of the standard Reifenberg obviously. The square root of one plus epsilon squared times X minus Y. So Y is that. I mean note if I take two points even if they're far away here right I have a point X here and a point Y over here then the well what's the bilipsic constant of this mapping? If I take a unit curve here then it always or yeah we're not here. If I take a unit curve here then it always this length gets divided by the square root of one plus epsilon squared here. So if I take my unit curve connecting these two to get the length then they just get projected right in there and we actually get the precise quality up here. So that is to say these maps are bilipsics right? No bi-holder. There's a uniform bilipsic constant. If I only move one step right? You can't induct on that obviously. If I tried to use that forever you'd never get anything about your end map because all you'd be saying is you know after you do I steps to get down here you'd have a bilipsic constant that's one plus epsilon squared to the I which is useless information right? So clearly we need more than this but this is one thing we'll use. It holds for all X and Y. Well okay one should be a little careful about this you're right. If I'm using the intrinsic metric it's a quality. If I'm using actually Euclidean which is of course what I'm doing. Let's say that. It's actually less than or equal. I do not. I mean one step exactly. Yes. Cause if this was true for capital fee not only would we have bi-holder we'd have bilipsics right? So we definitely don't have bilipsics. So just one step nothing goes too wrong we have a nice bilipsics map. And the other thing I want to point out is the following. So what if I take some point and I just take it here and I project it down and I look how far does that point get moved in total? Now that point gets moved basically epsilon times the length of your side right? Cause that's the most distance it can be. So this is bounded by or we can say bounded by epsilon times the length of Li. Now the longer the two sides doesn't really matter. That's the farthest it can move is epsilon times this length right here cause by definition that there was epsilon times Li. Everything else is less. Again that's just one step one movement. Okay good. So the claim now though is that okay so bilipsics for one step not moving very much for one step. Actually if you combine these two in a clever enough fashion that's by holder. That's the claim. So this is worth absorbing a little bit cause this is the how does one prove by holder estimates? This is how one tries to prove by holder estimates. So the way it's gonna work is the following. So let's take our Si. Actually let's take three. Here's S naught our eventual goal. Now let's say we're on some Si which is what we're trying to study here to get our projection map down. So let me just wiggle this around. Who the heck knows what this thing looks like. It's really really small. Here's our Si. We've wiggled a lot at this point maybe. And what I wanna do is take two points X and Y over here. And I wanna introduce so we have two scales in front of us right. We have our I which is where we're at. We have the not which is where we're heading. And in fact we have an in between scale which is roughly the distance of these two points which may not be sort of comparable either to Li or L naught. So what we wanna look at is let's let X minus Y be approximately LJ. So let's find the J so that if we look at SJ the amount each segment is moving is roughly on the same order as the distance between X minus Y. And the precise way of saying this is just let's find the largest J say for which this is less than or equal to that. So that is more precisely we can find a J with the following properties. So it's less than LJ but maybe bigger than LJ plus one is the point so we find just that point. Okay great and let's look at that guy too. Maybe it looks something like that. Here's our SJ. Okay now here's what we wanna do. We wanna take these two points in SI right. We wanna move them all the way down to S naught. So I'm gonna break this into two steps. So let's define it in between map here which is I'll call it phi IJ that maps SI and do SJ basically by just stopping the composition at some point. So two that J or J plus one J great that's a zero. And likewise just by switching I and J it's the exact same definition but let's go ahead and point it out. I can let J naught be defined as J minus one composed all the way down to K naught. I'm gonna say Euclidean here and say Euclidean yeah. Oops, did I do those backwards? Do I have to compose? What do I have to compose these in? I'm sorry. My left and right are totally off right. You gotta do this map first and then you end with this map. You gotta do this map first and then you end with this map. We really need to write compositions the other direction. You think about moving in that direction unless you're Chinese, right? So you think about moving that way and it's actually moving this way, right? So I think I confused everybody except our Chinese students. Okay, great. So then what I wanna do now is simply write phi I as being the composition of these two guys. Let's make sure I do this one right. So that would be phi J naught composed with phi IJ. Okay, great. Yes. So, okay, goal is have a picture for what these estimates are supposed to look like. So my basic claim is that we're gonna use one of these to control one of these maps and the other one to control the other map, right? So we can try to end up with our bi-holder. So in some sense, J here basically just depends on the distance between X and Y. So what we don't wanna do just to get a hint of sort of what we might expect to be doing here. What we don't wanna do is apply this infinitely often, right? So in fact, we're going to be applying this to this map where we know we're only applying it a certain number of times based on the distance apart. That's almost the moral of what uniform continuity is. And if we check the actual constants, this is what bi-holder will be for us in the end. And what we can apply infinitely often is this because if we're applying this guy for all K in between I and J, it's like a geometric series. It's gonna end up not being that big of a deal to apply it infinitely often. So this is what we wanna think about. So let me talk about this map first for just a second. So let's write a goal, goal one I wanna show. So I'm interested in looking at, right? So what are we trying to show? Let's do our series of goals to each keep getting more and more refined. So our map was defined here, right? So let's just say, let's show one direction of the bi-holder estimate. The other is verbatim. Hold on to this board before it jumps away. So I wanna show there's a beta and we're gonna be able to take this beta explicitly and it'll go to one as epsilon goes to zero so that phi i of x minus phi i of y is bounded by something small. I don't care, something close to one, I don't care. Sometimes the distance of x and y to the power of beta, so this is our ultimate goal of what we wanna show. I wanna start with a weaker goal over here for this map. Basically I wanna say this map actually doesn't move a fixed point very much relative to the distance of x and y. C for me, by the way, always basically means some universal constant here because it's all one dimensional. If you were doing a standard right from Burguet we mean a dimensional constant, the same for y. So my first claim is that this map basically doesn't matter. When we do end estimates here, the amount that's gonna move the point x is gonna be small relative to the distance in x and y anyway. So in the end it won't count for anything in terms of bad estimates. So this is gonna be a good thing because this one definitely will count for bad stuff in the end. So we wanna see nothing moves much. And how do we see this? So let's just do it inductively. So let's let x k be phi of i k of x so I want k here to go from i to so phi i i is just x itself, right, nothing's happened and then I want k to go from i to j. And all I wanna do is the absolute dumbest thing possible which is to estimate. So I wanna estimate x of j. I know that i i is x. I'm just gonna do a stupid triangle and a quality to try to sum up all those distances in between. This is gonna be the goal. And here we're gonna use this estimate here which is gonna be less than or equal to epsilon times lk I guess, k plus one, I don't care. And lk itself, well I don't care about lk, I care about lj, right? lj is the distance between x and y by definition essentially, right? So if I compare this, so k is gonna be potentially much smaller than than much bigger than than than j. So this is gonna be much smaller than l sub j and the amount smaller it'll be is I mean this is equals epsilon times square root of one plus epsilon squared over two, say to the j minus i, right? lj. So in other words, this is something summable is all you care about, right? If k is gonna be, yes, that's less than one. So if k is gonna be less than j, right? This sums just like two, right? As we move on, something a little bit bigger than one. Which means that if we want to prove this, that phi ij of x minus x, this is less than or equals to the sum by just sort of definition of what these things are, right? Of, I'm likely gonna get all my i's and j's messed up here. So I can do the silly thing where I can just sort of add and subtract in all the in-between steps and just do a triangle inequality, plug all these guys in. They're all epsilon times l sub j, but something summable. All right, so what I get is c, which is maybe two even, compute times l sub j, and this is like x minus y. So as claimed below, well, once you start wiggling more than the distance between them, you don't care, things don't move around much. So this map ends up not counting for much. I've got five pictures, I can erase one. So what have we done? We've taken our point on, let's not draw it that way, let's draw it this way, I don't know why I'm intersecting the two, I shouldn't be. Here's y, right? So I've got my two points on si, and I've now projected them up to sj and nothing much has changed, right? They've moved here. So here's my phi ij of x, and this has moved over here somewhere. I mean it's bounced around, but it's gotten up around here, my phi ij of y. So comparing these two distance is now basically equivalent to comparing these two distances, it doesn't matter. So now how do I finish the game and prove this? So let's just look at this for a second. Sorry about the two lines. So by definition, the phi i, which is the composition of all those maps, I can get it by first doing the first i minus j of them and then doing the last j of them, and this is the map we just studied. You get from here to here. Now, what do we know about this mapping here in each of these? Well, this thing guy here is a composition of j of our single projection maps from point to point. If we look at this estimate up here, we know the distance, every time we do this, the distance multiplies by a factor, so in fact, it's close to one, right? And it'll most now multiply by this guy j times. So in fact, all this guy does is multiply distances by at most the square root of one plus epsilon squared to the power j, right? So this is less than or equal to one plus epsilon squared to the j over two times the original distance of these two points, the ij of x minus phi ij. Y, okay. Now, phi ij of x and phi ij of y, what do we know about them? This hasn't moved x by more than some small factor times the distance. This hasn't moved y by more than some small factor times this distance. So this here is less than or equal to, in fact, maybe I'll, let me just do it for you. I'll do it over here so I don't have to mess up my computation. So what happens when we do this twice? I mean, that picture should be telling you the answer. So how far away is phi ij of x minus phi ij of y in comparison to x minus y? We'll just do a triangle inequality. Let's bound this by by, I have a small blackboard, I'm afraid, the sum of that minus that plus that minus that. You can do this by a triangle inequality. So it's just twice this, right? So this is another c, I don't care. I'm an analyst, my c is never the same from line to line. So this is also just bound to by c epsilon times x minus y, which is to say, this distance is up to one plus c epsilon, the same as this distance. Because this didn't move x much and this didn't move y much relative to the distance of x and y. That's why we chose to scale j. Okay, now what? All we actually really have to do at this point is fiddle around with the definition of this and what this is for a minute. So let's just write like four lines and we'll have a bi-holder estimate. We're gonna have to erase another board. I guess that one's gonna go. So this is, so one of the definition of x minus y was that it was roughly equivalent to lj in particular less than or equal to lj. So let me put that back. And I'm basically gonna turn this into lj, manipulate all my j's and turn them back into x minus y just because we can see it cleaner that way. And let me stick straight in the definition of lj. So this is bounded by lj, which is itself precisely equal to what? Square root of one plus epsilon squared. That looks familiar. Over two to the power of j. This is equal to one plus c epsilon. I just wanna collect together my two square root of one plus epsilon squared because why not for j? And now here we're sort of the nifty definition comes from. So let's define, let me try to do this right. Oh, sorry. There's actually a four here, right? When I put in the definition of l sub j, it's four times this. Doesn't matter. It gets it right back in just by definition it's four times that because I start on the interval of like four and then I keep multiplying by this. This is gonna be equal, now equal mind you. So I'm gonna be giving you a definition in one second. I don't like this one plus epsilon squared. I prefer square root of one plus epsilon squared because that gets me back to l j. So what I'm gonna do is simply write this as being the square root of one plus epsilon squared over two to the power of beta. And let's observe something, right? So this is less than one, right? So this is a little bit bigger than this is but this is less than one which means I need to raise to a power it's a little bit less than one. So beta is less than one here. I mean this equality here is my definition of beta, right? It's less than one but clearly as epsilon goes to zero, beta goes to one. What's that? This? The last one, the very last one. Yeah, this line here only holds though when the distance between x minus y is equivalent to l sub j, right? That was the point. So this distance here ended up being so you got this line here, right? Which is always true but that only became x minus y on this level. So I can't let j go to zero. That's the point. I can only do that when the wiggling is more than the distance is less than the distance between x and y. That's the point. Okay, where am I? I'm over here. I just wanna switch beta and j. Well, yes. I don't care, I'm gonna do either. So this is, I wanna do a less than or equal to so I can shove a beta here. And then I'll just, I'll rewrite it because I don't wanna do too much in one line. Two, I'm gonna stick a beta there and a j there. I'm just gonna switch beta and j. And then, so all I did was raise this to a power so I could absorb this into here now. One plus C epsilon times four times the square root of one plus epsilon squared over two to the j to the beta. And this is l sub j by definition, right? So we can get back to that. I might have to make this a little bit bigger. I didn't actually keep track of which one made it less than or equal to. This is a little bit less than one, which actually makes this a little bit smaller but only by an amount that's very, very close to one. Right, so I made this C worse. That's how I made it less than or equal to. L sub j to the beta. And now one more time we turn that x minus y and we're done. And the way I'm doing this, I guess I'm gonna shove a factor that's close to two. You can actually be more careful in that, but I don't care. So, two, whatever. And I may as well just drop the one plus C epsilon and I'm gonna put two there. L sub j x minus y to the beta. The two was because of this guy here. You can be more careful in that, but I didn't. That's it, right? So, starting up here, it's saying we have a holder condition on how these are and the opposite direction is basically exactly the same. You just gotta trace the inequalities the other direction. So, good. So, what's aggravating about this is that it's aggravating. However, it is a comparatively simple situation, right? You sit down for two hours and this becomes clear. It's just a projection from line to line to line to line and you trace through. The clever point you really wanna keep track of here is that when you're doing your estimate, you cannot, you've gotta have both these guys. If you do this too much, you lose all control. If you do this too much, well, actually, you might be able to preserve like room of how or how's worth close, but you'll never get a byholder condition, right? So, you need to preserve the distance between the two. Okay. Good.