1 00:00:18,198 --> 00:00:20,240 YUFEI ZHAO: So we've been discussing graph limits 2 00:00:20,240 --> 00:00:21,790 for a couple of lecturers now. 3 00:00:21,790 --> 00:00:23,970 In the first lecture on graph limits, 4 00:00:23,970 --> 00:00:28,340 so two lectures ago, I stated a number of main theorems. 5 00:00:28,340 --> 00:00:30,830 And today, we will prove these theorems 6 00:00:30,830 --> 00:00:34,310 using some of the tools that we developed last time, namely 7 00:00:34,310 --> 00:00:35,720 the regularity lemma. 8 00:00:35,720 --> 00:00:40,130 And also we proved this Martingale convergence theorem, 9 00:00:40,130 --> 00:00:42,800 which will also come in. 10 00:00:42,800 --> 00:00:45,920 So let me recall what were the three main theorems that we 11 00:00:45,920 --> 00:00:49,160 stated at the end of two lectures ago. 12 00:00:49,160 --> 00:00:53,150 So one of them was the equivalence of convergence. 13 00:00:53,150 --> 00:00:56,570 On one hand, we defined a notion of convergence 14 00:00:56,570 --> 00:01:04,190 where we say that Wn approaches W, by definition, 15 00:01:04,190 --> 00:01:09,340 if the F densities converge. 16 00:01:15,390 --> 00:01:17,780 We can say convergence even without a limit in mind, 17 00:01:17,780 --> 00:01:22,640 where we see a sequence converges if all of these F 18 00:01:22,640 --> 00:01:25,400 densities converge. 19 00:01:25,400 --> 00:01:28,040 So the first main theorem was that the two notions 20 00:01:28,040 --> 00:01:34,280 of convergence are equivalent, so one notion being convergence 21 00:01:34,280 --> 00:01:38,120 in terms of F densities, and the second notion 22 00:01:38,120 --> 00:01:43,720 being convergence in the sense of the cut norm, the cut 23 00:01:43,720 --> 00:01:46,500 distance. 24 00:01:46,500 --> 00:01:48,130 There was a second term that tells us 25 00:01:48,130 --> 00:01:51,700 that limits always exist. 26 00:01:51,700 --> 00:01:53,650 If you have a convergent sequence, 27 00:01:53,650 --> 00:01:58,730 then you can represent a limit by a graphon. 28 00:01:58,730 --> 00:02:01,300 And the third statement was about compactness 29 00:02:01,300 --> 00:02:03,452 of the space of graphons. 30 00:02:03,452 --> 00:02:05,410 So we're actually going to prove these theorems 31 00:02:05,410 --> 00:02:06,380 in reverse order. 32 00:02:06,380 --> 00:02:09,590 We're going to start with the compactness and work backwards. 33 00:02:09,590 --> 00:02:12,100 So this is not how these theorems were originally 34 00:02:12,100 --> 00:02:15,610 proved, or not in that order, but it will be helpful 35 00:02:15,610 --> 00:02:18,640 for us to do this by first considering 36 00:02:18,640 --> 00:02:20,950 the compactness statement. 37 00:02:20,950 --> 00:02:23,480 So remember we're compactness says. 38 00:02:23,480 --> 00:02:31,310 I start with this space W tilda, which 39 00:02:31,310 --> 00:02:39,730 is the space of graphons, where I identify 40 00:02:39,730 --> 00:02:41,530 graphons that have distance 0. 41 00:02:51,610 --> 00:02:55,780 So if they have cut distance 0, then I refer to them 42 00:02:55,780 --> 00:02:56,680 as the same point. 43 00:02:56,680 --> 00:02:59,360 So this is now a metric space. 44 00:02:59,360 --> 00:03:02,990 And the theorem is that this space is compact. 45 00:03:13,890 --> 00:03:15,975 So I think this, it's a really nice theorem. 46 00:03:15,975 --> 00:03:18,480 It's a beautiful theorem that encapsulates 47 00:03:18,480 --> 00:03:21,720 a lot of what we've been talking about so far with similarities, 48 00:03:21,720 --> 00:03:27,150 regularity, and what not, in a qualitatively succinct way, 49 00:03:27,150 --> 00:03:31,080 just that this space of graphons is compact. 50 00:03:31,080 --> 00:03:34,410 You may not have some intuition about what the space looks 51 00:03:34,410 --> 00:03:36,480 like at the moment, but we'll see the proof 52 00:03:36,480 --> 00:03:39,090 and hopefully that will give you some more intuition 53 00:03:39,090 --> 00:03:41,040 I first learned about this theorem 54 00:03:41,040 --> 00:03:45,240 when Laszlo Lovasz, who was one of the pioneers in the subject, 55 00:03:45,240 --> 00:03:47,250 when he came to MIT to give a talk 56 00:03:47,250 --> 00:03:48,840 when I was a graduate student. 57 00:03:48,840 --> 00:03:52,290 And he said that analysts thought 58 00:03:52,290 --> 00:03:55,110 that they pretty much knew all the naturally occurring 59 00:03:55,110 --> 00:03:57,140 compact spaces out there. 60 00:03:57,140 --> 00:03:59,940 So there are lots of spaces that occur in analysis 61 00:03:59,940 --> 00:04:01,530 and topology that are compact. 62 00:04:01,530 --> 00:04:05,910 I mean, the first one you learn in analysis undergraduate 63 00:04:05,910 --> 00:04:08,160 is probably that an interval is compact. 64 00:04:08,160 --> 00:04:10,590 But there are also many other spaces. 65 00:04:10,590 --> 00:04:13,290 But this one here doesn't seem to be 66 00:04:13,290 --> 00:04:16,089 any of these classical notions of compactness. 67 00:04:16,089 --> 00:04:22,480 So it's, in some sense, a new compact space. 68 00:04:22,480 --> 00:04:23,880 So let's see how the proof goes. 69 00:04:27,520 --> 00:04:32,620 Now, because we are working in a metric space, 70 00:04:32,620 --> 00:04:42,330 it suffices to show, due to the equivalence between compactness 71 00:04:42,330 --> 00:04:45,340 in the sense of finite open covers 72 00:04:45,340 --> 00:04:54,280 and sequential compactness in the metric space, 73 00:04:54,280 --> 00:04:57,340 so it suffices to show sequential compactness, 74 00:04:57,340 --> 00:05:28,190 that every sequence of graphons has a convergent subsequence 75 00:05:28,190 --> 00:05:32,000 with respect to this cut metric and also will 76 00:05:32,000 --> 00:05:40,000 produce a limit as a convergence subsequence with a limit point. 77 00:05:43,390 --> 00:05:44,620 So that's what we'll do. 78 00:05:44,620 --> 00:05:46,720 I give you an arbitrary sequence of graphons. 79 00:05:46,720 --> 00:05:49,390 I want to construct by taking subsequences a convergent 80 00:05:49,390 --> 00:05:49,890 sequence. 81 00:05:49,890 --> 00:05:53,150 And I will tell you what that limit is. 82 00:05:53,150 --> 00:05:57,280 So here is what we're going to do, given this sequence. 83 00:05:57,280 --> 00:06:00,190 As I hinted before, it has to do with the regularity lemma. 84 00:06:00,190 --> 00:06:01,960 So we're going to apply the regularity 85 00:06:01,960 --> 00:06:05,600 lemma in the above form, which we did last time. 86 00:06:05,600 --> 00:06:16,100 So apply the weak regularity lemma, 87 00:06:16,100 --> 00:06:24,300 which will tell us that for each Wn, 88 00:06:24,300 --> 00:06:31,190 there exists a partition, in fact, a sequence of partitions, 89 00:06:31,190 --> 00:06:33,560 each one refining the next. 90 00:06:33,560 --> 00:06:37,400 So what's going to happen is, I'm going to start with Wn 91 00:06:37,400 --> 00:06:42,020 and starting with a trivial partition, apply that lemma, 92 00:06:42,020 --> 00:06:48,460 and obtain a partition P sub n, 1. 93 00:06:48,460 --> 00:06:51,820 And then starting with that as my P0, 94 00:06:51,820 --> 00:06:56,020 I'm going to apply regularity lemma again and obtain 95 00:06:56,020 --> 00:06:56,800 a refinement. 96 00:07:02,010 --> 00:07:05,530 I will have this sequence of partitions, 97 00:07:05,530 --> 00:07:11,470 each one refining the next. 98 00:07:17,640 --> 00:07:19,590 So all of these are going to be partitions 99 00:07:19,590 --> 00:07:21,690 of the 0, 1 interval. 100 00:07:21,690 --> 00:07:23,882 And as I mentioned last time, everything's 101 00:07:23,882 --> 00:07:24,840 going to be measurable. 102 00:07:24,840 --> 00:07:27,070 I'm not going to even mention measurability. 103 00:07:27,070 --> 00:07:31,610 Everything will be measurable such 104 00:07:31,610 --> 00:07:35,640 that they satisfy the following conditions. 105 00:07:35,640 --> 00:07:38,750 So the first one is what I mentioned 106 00:07:38,750 --> 00:07:41,670 earlier, is that you have a sequence of refinements. 107 00:07:41,670 --> 00:07:51,400 So each P sub n k plus 1 refines the previous one for all n 108 00:07:51,400 --> 00:07:51,900 and k. 109 00:07:55,560 --> 00:07:59,780 And the second condition, as given by the regularity lemma, 110 00:07:59,780 --> 00:08:03,410 you get to control the number of parts. 111 00:08:03,410 --> 00:08:07,500 So I will say in the third part what the error of approximation 112 00:08:07,500 --> 00:08:08,000 is. 113 00:08:08,000 --> 00:08:10,280 But you get to control the number of parts. 114 00:08:10,280 --> 00:08:16,760 So in particular, I can make sure that this number here, 115 00:08:16,760 --> 00:08:22,580 the number of parts in the k'th partition depends only on k. 116 00:08:32,299 --> 00:08:36,039 Now, you might complain somewhat, 117 00:08:36,039 --> 00:08:40,059 because the regularity lemma only tells you an upper bound 118 00:08:40,059 --> 00:08:42,150 on the number of parts. 119 00:08:42,150 --> 00:08:45,030 But that's OK. 120 00:08:45,030 --> 00:08:46,470 I can allow empty parts. 121 00:08:50,990 --> 00:08:58,620 So now I make sure that the k'th partition has exactly nk parts. 122 00:08:58,620 --> 00:09:04,380 And the third one has to do with the area of approximation. 123 00:09:07,110 --> 00:09:17,100 OK, so suppose we write W sub nk as the graphon obtained 124 00:09:17,100 --> 00:09:18,870 by applying the stepping operator. 125 00:09:18,870 --> 00:09:24,210 So this is the averaging operator on this partition, 126 00:09:24,210 --> 00:09:25,740 corresponding to the k'th partition. 127 00:09:25,740 --> 00:09:30,060 I apply that partition, do a stepping averaging operator 128 00:09:30,060 --> 00:09:32,080 on the n'th graphon. 129 00:09:32,080 --> 00:09:36,080 I get W sub nk. 130 00:09:36,080 --> 00:09:44,380 The third condition is that the k'th partition approximates-- 131 00:09:44,380 --> 00:09:51,780 it's a good approximation in the cut norm up to error 1 over k. 132 00:09:51,780 --> 00:09:54,950 So 1 over k is some arbitrary sequence going to 0 133 00:09:54,950 --> 00:09:57,138 as k goes to infinity. 134 00:10:12,100 --> 00:10:14,520 So I obtained a sequence of partitions, 135 00:10:14,520 --> 00:10:17,390 so by applying the regularity lemma 136 00:10:17,390 --> 00:10:21,480 to each to each graphon in the sequence. 137 00:10:24,300 --> 00:10:27,530 Now, these graphons, I mean, they 138 00:10:27,530 --> 00:10:29,260 each have their own vertex set. 139 00:10:29,260 --> 00:10:32,040 And so far, they're not related to each other. 140 00:10:32,040 --> 00:10:37,300 But to make the visualization easier and also in order 141 00:10:37,300 --> 00:10:39,620 to do the next step in the proof, 142 00:10:39,620 --> 00:10:46,440 I am going to do some measure-preserving bisection So 143 00:10:46,440 --> 00:10:49,960 think of this as permuting the vertex labels. 144 00:10:49,960 --> 00:11:04,760 So by replacing each Wn by some W sub 145 00:11:04,760 --> 00:11:15,350 n of phi, where phi is a measure-preserving bisection, 146 00:11:15,350 --> 00:11:24,080 we can assume that all these partitions 147 00:11:24,080 --> 00:11:27,940 are partitions into intervals. 148 00:11:36,660 --> 00:11:38,670 So initially, you might have a partition 149 00:11:38,670 --> 00:11:41,640 into arbitrary measurable sets. 150 00:11:41,640 --> 00:11:47,120 Well, what I can do is to push over the first set to the left, 151 00:11:47,120 --> 00:11:50,400 and so on, so do a measure-preserving bisection 152 00:11:50,400 --> 00:11:55,740 in a way so that I can maintain that all the partitions are 153 00:11:55,740 --> 00:11:58,189 visually chopping up into intervals. 154 00:11:58,189 --> 00:11:58,689 Yeah? 155 00:11:58,689 --> 00:12:01,179 AUDIENCE: So at some point, we need just one measure 156 00:12:01,179 --> 00:12:04,920 for the projection, like all of them be in k? 157 00:12:04,920 --> 00:12:07,110 YUFEI ZHAO: OK, so the question is, 158 00:12:07,110 --> 00:12:11,510 it may be the case that, for a given k, 159 00:12:11,510 --> 00:12:15,640 I can do this arrangement, but it's not 160 00:12:15,640 --> 00:12:18,040 clear to you at the moment why you can 161 00:12:18,040 --> 00:12:21,790 do this uniformly for all k. 162 00:12:21,790 --> 00:12:24,610 So one way to get around this is, for now, 163 00:12:24,610 --> 00:12:26,500 just think of for each given k. 164 00:12:26,500 --> 00:12:29,290 And then you'll see at the end that that's already enough. 165 00:12:32,898 --> 00:12:34,512 OK, any more questions? 166 00:12:38,170 --> 00:12:41,940 So now assume all of these P sub nk's are intervals. 167 00:12:41,940 --> 00:12:45,210 So in fact, what you said may be a better way to go. 168 00:12:45,210 --> 00:12:48,430 But to make our life a little bit easier, 169 00:12:48,430 --> 00:12:50,430 let's just assume for now that you can do this. 170 00:12:53,670 --> 00:12:56,190 OK, and what's going to happen next 171 00:12:56,190 --> 00:12:59,490 is some kind of a diagonalization argument. 172 00:12:59,490 --> 00:13:01,290 We're going to be picking subsequences. 173 00:13:05,760 --> 00:13:10,740 So I'm going to be picking subsequences 174 00:13:10,740 --> 00:13:12,810 so that they are going to have very 175 00:13:12,810 --> 00:13:14,850 nice convergence properties. 176 00:13:14,850 --> 00:13:18,150 And so I'm going to repeatedly throw out 177 00:13:18,150 --> 00:13:19,930 a lot of the sequence. 178 00:13:19,930 --> 00:13:21,930 So this is a diagonalization argument. 179 00:13:21,930 --> 00:13:25,390 And basically what happens is that, 180 00:13:25,390 --> 00:13:27,370 by passing two subsequences-- 181 00:13:31,564 --> 00:13:35,450 and we're going to do this repeatedly, many times-- 182 00:13:35,450 --> 00:13:51,420 we can assume, first, that the end points of P sub n1, 183 00:13:51,420 --> 00:13:59,070 they converge as n goes to infinity. 184 00:13:59,070 --> 00:14:14,250 So each P sub n1 is some partition of interval 185 00:14:14,250 --> 00:14:17,370 into some fixed number of parts. 186 00:14:17,370 --> 00:14:19,320 So by passing to a subsequence, I 187 00:14:19,320 --> 00:14:22,005 make sure that the division points all converge. 188 00:14:28,850 --> 00:14:34,420 And now, by passing one more time, 189 00:14:34,420 --> 00:14:36,860 so by passing to subsequence one more time, 190 00:14:36,860 --> 00:14:46,020 let's assume that also, W sub n1 converges to some function, 191 00:14:46,020 --> 00:14:49,080 some graphon u1, point-wise. 192 00:15:00,580 --> 00:15:12,100 So initially, I have these graphons. 193 00:15:12,100 --> 00:15:15,710 Each one of them is an m by n block. 194 00:15:15,710 --> 00:15:19,060 They have various division points. 195 00:15:19,060 --> 00:15:21,590 By passing to a subsequence, I assume 196 00:15:21,590 --> 00:15:24,560 that the points of division, they converge. 197 00:15:24,560 --> 00:15:27,130 And now by passing to an additional subsequence, 198 00:15:27,130 --> 00:15:32,600 I can make sure the individual values, they converge. 199 00:15:32,600 --> 00:15:37,700 So as a result, W sub n, 1 converges to W1-- 200 00:15:37,700 --> 00:15:42,382 converges to some graphon, u1, point-wise, almost everywhere. 201 00:15:46,030 --> 00:15:54,990 And we repeat for W sub nk for each k. 202 00:15:59,970 --> 00:16:01,520 So do this sequentially. 203 00:16:01,520 --> 00:16:03,490 So we just did it for k equals to 1. 204 00:16:03,490 --> 00:16:05,890 Now do it for 2, 3, 4, and so on. 205 00:16:08,537 --> 00:16:10,120 So this is a diagonalization argument. 206 00:16:10,120 --> 00:16:12,670 We do this countably many times. 207 00:16:12,670 --> 00:16:15,310 At the end, what do we get? 208 00:16:15,310 --> 00:16:20,970 We pass down to the following subsequence. 209 00:16:26,850 --> 00:16:29,820 And just to make my life a bit more convenient, 210 00:16:29,820 --> 00:16:34,900 instead of labeling the indices of the subsequence, 211 00:16:34,900 --> 00:16:37,380 I'm going to relabel the sequence 212 00:16:37,380 --> 00:16:40,560 so that it's still labeled by 1, 2, 3, 4, and so on. 213 00:16:40,560 --> 00:16:47,320 So we now pass to a sequence W1, W2, W3, 214 00:16:47,320 --> 00:16:54,630 and so on, such that if you look at the first partition, 215 00:16:54,630 --> 00:16:59,620 the first weak regularity partition, they produce W1,1, 216 00:16:59,620 --> 00:17:05,079 W2,1, W3,1, and so on. 217 00:17:05,079 --> 00:17:09,981 And these guys, they converge to u1, point-wise. 218 00:17:15,849 --> 00:17:19,695 The second level, W2,1-- 219 00:17:19,695 --> 00:17:32,310 sorry, W1,2, W2,2, W3,2, they converge to u2, point-wise, 220 00:17:32,310 --> 00:17:32,810 and so on. 221 00:17:46,820 --> 00:17:48,893 OK, so far so good? 222 00:17:48,893 --> 00:17:49,800 Question? 223 00:17:49,800 --> 00:17:52,310 AUDIENCE: Sorry, earlier, why did that 224 00:17:52,310 --> 00:17:54,110 converge to u1 point-wise? 225 00:17:54,110 --> 00:17:56,620 YUFEI ZHAO: OK, so the question is, why is this true? 226 00:17:56,620 --> 00:18:03,490 Why is Wn,1 converge to u1 point-wise? 227 00:18:03,490 --> 00:18:05,200 Initially, it might not. 228 00:18:05,200 --> 00:18:07,900 But what I'm saying is, you can pass to a subsequence. 229 00:18:07,900 --> 00:18:08,663 AUDIENCE: Yes. 230 00:18:08,663 --> 00:18:10,330 YUFEI ZHAO: You can pass to subsequence, 231 00:18:10,330 --> 00:18:15,210 because there are only n1 parts. 232 00:18:17,800 --> 00:18:24,500 So it's an n1 by m1 matrix of real numbers. 233 00:18:24,500 --> 00:18:27,720 And so you only have finite bounded many of them. 234 00:18:27,720 --> 00:18:30,080 So you can pick a subsequence so that they converge. 235 00:18:30,080 --> 00:18:30,580 Yeah? 236 00:18:30,580 --> 00:18:31,955 AUDIENCE: So how do you make sure 237 00:18:31,955 --> 00:18:35,420 that your subsequence is not empty at the end-- like, 238 00:18:35,420 --> 00:18:37,800 could you fix the first k? 239 00:18:37,800 --> 00:18:39,760 YUFEI ZHAO: OK, so you're asking, 240 00:18:39,760 --> 00:18:44,080 if we do this slightly not so carefully, 241 00:18:44,080 --> 00:18:46,180 we might end up with an empty sequence. 242 00:18:46,180 --> 00:18:48,430 So this is why I say, you have to do a diagonalization 243 00:18:48,430 --> 00:18:49,240 argument. 244 00:18:49,240 --> 00:18:53,170 Each step, you keep the first term, the sequence, 245 00:18:53,170 --> 00:18:56,440 so that you always maintain some sequence. 246 00:18:56,440 --> 00:19:00,870 You have to be slightly careful with diagonalization. 247 00:19:00,870 --> 00:19:01,976 Any more questions? 248 00:19:06,080 --> 00:19:07,900 So by passing to a subsequence, we 249 00:19:07,900 --> 00:19:11,680 obtain this very nice sequence, this nice subsequence, 250 00:19:11,680 --> 00:19:16,960 such that each row corresponding to each level of regularization 251 00:19:16,960 --> 00:19:20,740 converges point-wise to some u. 252 00:19:20,740 --> 00:19:23,370 So what do this u's look like? 253 00:19:23,370 --> 00:19:24,800 So they are step graphons. 254 00:19:27,460 --> 00:19:29,740 So let's explore the structure of u a bit more. 255 00:19:36,050 --> 00:19:43,880 OK, so since we have that each-- 256 00:19:43,880 --> 00:19:47,070 OK, so we have the property that each partition 257 00:19:47,070 --> 00:19:49,190 refines the previous partition. 258 00:19:53,190 --> 00:20:00,680 And as a result, if you look at the k plus 1'th stepping, 259 00:20:00,680 --> 00:20:07,490 and I step it by the previous partition in the sequence, 260 00:20:07,490 --> 00:20:14,460 I should get back, I should go back one in the sequence. 261 00:20:14,460 --> 00:20:19,490 So this was this graphon obtained by averaging over 262 00:20:19,490 --> 00:20:20,870 the k'th partition. 263 00:20:20,870 --> 00:20:26,000 And this is the graphon obtained by averaging over the k'th plus 264 00:20:26,000 --> 00:20:28,070 1st partition. 265 00:20:28,070 --> 00:20:33,000 So if I go back one more, I should go back in the sequence. 266 00:20:35,640 --> 00:20:41,530 And since the u's are the point-wise limit of these 267 00:20:41,530 --> 00:20:50,530 W's, the same relationships should also hold 268 00:20:50,530 --> 00:21:06,820 for the u's, namely that u sub k should equal to u sub k plus 1 269 00:21:06,820 --> 00:21:17,210 if I step it with Pk, where Pk is the-- 270 00:21:20,020 --> 00:21:22,870 so if you look at, all these endpoints converge. 271 00:21:22,870 --> 00:21:28,550 And these partitions, they converge to P1. 272 00:21:28,550 --> 00:21:36,760 So if you look at the partitions that correspond to P1,1, P2,1, 273 00:21:36,760 --> 00:21:41,480 and so on, I want these partitions to converge to P1. 274 00:21:41,480 --> 00:21:51,460 I want these partitions to converge to P2. 275 00:21:51,460 --> 00:21:55,313 So all these partitions, they are partitions into intervals. 276 00:21:55,313 --> 00:21:56,980 So I'm just saying, if you look at where 277 00:21:56,980 --> 00:21:59,350 the intervals, where the divisions of intervals go, 278 00:21:59,350 --> 00:22:00,250 they converge. 279 00:22:00,250 --> 00:22:02,830 And then I'm calling the limit partition P sub k. 280 00:22:07,810 --> 00:22:16,240 And here we're using that P sub k plus 1 refines P sub k, 281 00:22:16,240 --> 00:22:18,712 because the same is true for each end. 282 00:22:18,712 --> 00:22:20,670 So in the limit, the same must be true as well. 283 00:22:25,080 --> 00:22:27,620 So you have this column of u's. 284 00:22:27,620 --> 00:22:32,852 So let me draw you a picture of what these u's could look like. 285 00:22:32,852 --> 00:22:34,810 So here is an illustration that may be helpful. 286 00:22:40,980 --> 00:22:43,830 So what could these u's look like? 287 00:22:43,830 --> 00:22:51,390 Each one of them is represented by values on the unit square. 288 00:22:51,390 --> 00:22:53,520 And I write this in matrix notation 289 00:22:53,520 --> 00:22:55,480 so that inversion is in the top left corner. 290 00:23:00,180 --> 00:23:05,400 Well, maybe P1 is just the trivial partition, 291 00:23:05,400 --> 00:23:08,910 in which case u1 is going to be a constant graphon. 292 00:23:08,910 --> 00:23:14,630 Let's say it has value 0.5. 293 00:23:14,630 --> 00:23:19,300 u2 came from u1 by some partitioning. 294 00:23:19,300 --> 00:23:21,290 And suppose just for the sake of illustration, 295 00:23:21,290 --> 00:23:23,210 there was only a partitioning into two parts. 296 00:23:26,950 --> 00:23:30,123 And OK, so it doesn't have to be at the origin. 297 00:23:30,123 --> 00:23:31,540 It doesn't have to be at midpoint. 298 00:23:31,540 --> 00:23:33,550 But just for illustration, suppose the division 299 00:23:33,550 --> 00:23:36,800 were at the midpoint. 300 00:23:36,800 --> 00:23:40,520 Because u1 needs to have-- 301 00:23:40,520 --> 00:23:44,220 so this 0.5 value should be the average value 302 00:23:44,220 --> 00:23:46,530 in all of these four squares. 303 00:23:46,530 --> 00:23:59,020 So for instance, the points may be like 0.6, 0.6, 0.4, 0.4, 304 00:23:59,020 --> 00:24:01,900 so for example. 305 00:24:01,900 --> 00:24:08,820 And in u3, the partition, the P3 partition-- so here 306 00:24:08,820 --> 00:24:10,120 are the partition is P1. 307 00:24:10,120 --> 00:24:11,740 The partition is P2. 308 00:24:11,740 --> 00:24:13,070 There's two parts. 309 00:24:13,070 --> 00:24:18,142 And suppose P3 three now has four parts. 310 00:24:18,142 --> 00:24:19,600 And again, for illustration's sake, 311 00:24:19,600 --> 00:24:22,120 suppose it is equally dividing the interval 312 00:24:22,120 --> 00:24:25,000 into four intervals. 313 00:24:25,000 --> 00:24:28,960 It could be that now each of these parts 314 00:24:28,960 --> 00:24:32,950 is split up into four different values in a way 315 00:24:32,950 --> 00:24:37,471 that so you can obtain the original numbers by averaging. 316 00:24:40,560 --> 00:24:44,010 So that's one possible example. 317 00:24:44,010 --> 00:24:50,514 Likewise, you can have something like that. 318 00:24:54,888 --> 00:25:07,220 Sorry, 4, 7-- so and I should maintain symmetry 319 00:25:07,220 --> 00:25:08,990 in this matrix. 320 00:25:08,990 --> 00:25:11,520 And the last one, I'm just going to be lazy and say 321 00:25:11,520 --> 00:25:15,960 that it's still 0.4 throughout. 322 00:25:15,960 --> 00:25:20,080 OK, so this is what the sequence of u's are going to look like. 323 00:25:20,080 --> 00:25:25,660 Each one of them splits up a box in the previous u in such way 324 00:25:25,660 --> 00:25:31,700 that the local averages, the step averages are preserved. 325 00:25:31,700 --> 00:25:32,590 Any questions so far? 326 00:25:35,410 --> 00:25:41,270 All right, so now we get to Martingales. 327 00:25:41,270 --> 00:25:45,870 So I claim that this is basically a Martingale. 328 00:25:45,870 --> 00:25:56,550 So and suppose you let x, y be a uniform point in the unit 329 00:25:56,550 --> 00:25:57,050 square. 330 00:25:59,930 --> 00:26:05,220 And consider this sequence. 331 00:26:05,220 --> 00:26:07,190 So this is now a random sequence, 332 00:26:07,190 --> 00:26:09,290 because x, y are random. 333 00:26:12,030 --> 00:26:17,752 I evaluate these u's on this uniform random point, x, y. 334 00:26:17,752 --> 00:26:18,960 So this is a random sequence. 335 00:26:23,800 --> 00:26:27,279 And the main observation is that this is a Martingale. 336 00:26:34,320 --> 00:26:38,450 So remember the definition of a Martingale from last time. 337 00:26:38,450 --> 00:26:42,990 Martingale is one where, if you look 338 00:26:42,990 --> 00:26:47,430 at the value of u sub k conditioned 339 00:26:47,430 --> 00:26:53,730 on the previous values, the expectation 340 00:26:53,730 --> 00:26:56,190 is just the previous term. 341 00:26:56,190 --> 00:26:58,493 And I claim this is true for the sequence, 342 00:26:58,493 --> 00:27:00,035 because of the way we constructed it, 343 00:27:00,035 --> 00:27:03,750 it's splitting up each box in an averaging-preserving way. 344 00:27:07,380 --> 00:27:09,177 A different way to see this, and for those 345 00:27:09,177 --> 00:27:11,010 of you who actually know what the definition 346 00:27:11,010 --> 00:27:13,920 of a random variable is in the sense of probability theory, 347 00:27:13,920 --> 00:27:21,160 is that you should view this 0, 1 squared as the probability 348 00:27:21,160 --> 00:27:27,115 space, in which case u itself is the random variable. 349 00:27:30,050 --> 00:27:34,480 And this partitioning gives you a filtration of the space. 350 00:27:34,480 --> 00:27:37,450 It's a sequence of sigma algebras dividing up the space 351 00:27:37,450 --> 00:27:39,590 into finer and finer pieces. 352 00:27:39,590 --> 00:27:43,280 So this is really what a martingale is. 353 00:27:43,280 --> 00:27:45,710 So we have a Martingale. 354 00:27:45,710 --> 00:27:56,770 It's bounded because the values take place in 0, 1. 355 00:27:56,770 --> 00:28:00,340 So by the Martingale convergence theorem, 356 00:28:00,340 --> 00:28:15,370 which we proved last time, we find that this sequence 357 00:28:15,370 --> 00:28:18,760 must converge to some limit. 358 00:28:21,550 --> 00:28:24,970 So this sequence of Martingale converges, 359 00:28:24,970 --> 00:28:29,490 which means, so if you think about the interpretation 360 00:28:29,490 --> 00:28:33,480 up there, so there exists a u which 361 00:28:33,480 --> 00:28:45,210 is a graphon such that uk converges to u point-wise 362 00:28:45,210 --> 00:28:51,086 almost everywhere as k goes to infinity. 363 00:28:57,410 --> 00:28:59,000 That's the limit. 364 00:28:59,000 --> 00:29:00,897 So this is the limit. 365 00:29:00,897 --> 00:29:02,980 And we're going to show that it is indeed a limit. 366 00:29:02,980 --> 00:29:05,110 But you see, this is a construction of the limit, 367 00:29:05,110 --> 00:29:08,950 where we took regularity, got all these nice pieces, 368 00:29:08,950 --> 00:29:11,140 found convergent subsequences, and then 369 00:29:11,140 --> 00:29:14,140 applied the martingale convergence 370 00:29:14,140 --> 00:29:17,290 theorem to produce for us this candidate 371 00:29:17,290 --> 00:29:19,170 for the limit, this u. 372 00:29:19,170 --> 00:29:28,520 So now let us show that it is indeed the limit that we're 373 00:29:28,520 --> 00:29:33,650 looking for in the subsequence. 374 00:29:33,650 --> 00:29:37,500 So again, I've tossed out all the terms 375 00:29:37,500 --> 00:29:41,510 which we removed in passing to subsequences. 376 00:29:41,510 --> 00:29:43,290 So in the remaining subsequence, I 377 00:29:43,290 --> 00:29:48,920 want to show that the Wn's indeed converge to u. 378 00:29:48,920 --> 00:29:52,130 And this is now a fairly straightforward three epsilons 379 00:29:52,130 --> 00:29:55,910 argument, the standard analysis type argument. 380 00:29:55,910 --> 00:29:57,630 But OK, so let's carry it through. 381 00:29:57,630 --> 00:30:01,370 So for every epsilon bigger than 0, 382 00:30:01,370 --> 00:30:08,360 suppose you pick a sufficiently large k. 383 00:30:08,360 --> 00:30:11,170 There exists a sufficiently large k. 384 00:30:11,170 --> 00:30:17,010 And we make sure k is large enough 385 00:30:17,010 --> 00:30:22,880 such that u differs from u sub k in l1 norm 386 00:30:22,880 --> 00:30:28,400 by, at most, epsilon over 3, because the uk's, 387 00:30:28,400 --> 00:30:32,090 they converge to u point-wise almost everywhere. 388 00:30:32,090 --> 00:30:33,150 So we find this k. 389 00:30:33,150 --> 00:30:34,060 So let's fix this k. 390 00:30:39,220 --> 00:30:47,080 Then there exists an n0 such that if you look at this u sub 391 00:30:47,080 --> 00:30:51,930 k, it does not-- 392 00:30:51,930 --> 00:31:03,540 it is very close to W sub nk for all n 393 00:31:03,540 --> 00:31:13,150 large enough because of what happened up there. 394 00:31:18,300 --> 00:31:25,435 So we can now compute the difference between-- 395 00:31:27,950 --> 00:31:30,770 in fact, let's do it this way. 396 00:31:35,400 --> 00:31:42,585 So now let's compute the difference, 397 00:31:42,585 --> 00:31:45,810 the cut norm of the difference between the term 398 00:31:45,810 --> 00:31:48,060 in the sequence W sub n and u. 399 00:31:48,060 --> 00:31:53,510 So by triangle inequality, we have 400 00:31:53,510 --> 00:31:54,900 that the following is true. 401 00:32:11,060 --> 00:32:15,940 The cut norm is upperbounded by the l1 norm. 402 00:32:15,940 --> 00:32:17,332 Look at the definitions. 403 00:32:22,420 --> 00:32:25,710 So I'm going to replace the first couple of these cut norms 404 00:32:25,710 --> 00:32:28,290 by l1 norms and leave the last one in tact. 405 00:32:35,500 --> 00:32:41,000 The first term, I claim is, at most, epsilon over 3, 406 00:32:41,000 --> 00:32:44,810 because up there. 407 00:32:44,810 --> 00:32:47,690 The second term is going to be at least epsilon 408 00:32:47,690 --> 00:32:52,040 over 3, because over here. 409 00:32:52,040 --> 00:32:56,510 And the third term is going to be also, at most, epsilon 410 00:32:56,510 --> 00:33:02,660 over 3, because well, from the regularity approximation, 411 00:33:02,660 --> 00:33:07,010 I know that it is, at most, 1 over k. 412 00:33:07,010 --> 00:33:09,970 And I chose k large enough so that there is also, 413 00:33:09,970 --> 00:33:12,410 at most, epsilon over 3. 414 00:33:12,410 --> 00:33:17,285 Put everything together, we find that these two are-- 415 00:33:17,285 --> 00:33:21,770 they different by, at most, epsilon if n is large enough. 416 00:33:21,770 --> 00:33:30,380 But now, since epsilon can be arbitrarily small, 417 00:33:30,380 --> 00:33:38,460 we find that you indeed have convergence, as claimed. 418 00:33:41,750 --> 00:33:46,160 And this finishes the proof of compactness. 419 00:33:46,160 --> 00:33:47,700 So there are a few components. 420 00:33:47,700 --> 00:33:48,910 One is passing to-- 421 00:33:48,910 --> 00:33:53,970 so applying regularity, passing to subsequences, 422 00:33:53,970 --> 00:33:58,630 and obtaining this limit from the regularity approximations, 423 00:33:58,630 --> 00:34:00,140 these u's. 424 00:34:00,140 --> 00:34:03,840 And then we observe that these u's, they form a Martingale. 425 00:34:03,840 --> 00:34:08,340 So we can apply the Martingale convergence theorem to get 426 00:34:08,340 --> 00:34:12,280 us a candidate for the limit. 427 00:34:12,280 --> 00:34:14,449 And then the rest is fairly straightforward, 428 00:34:14,449 --> 00:34:17,074 because all the steps are good approximations. 429 00:34:17,074 --> 00:34:19,960 You put them together, you prove the limit. 430 00:34:23,090 --> 00:34:24,163 Any questions? 431 00:34:27,790 --> 00:34:30,980 All right, so you may ask, well, now we have compactness. 432 00:34:30,980 --> 00:34:33,179 What is compactness good for? 433 00:34:33,179 --> 00:34:36,022 So it may seem like a somewhat abstract concept. 434 00:34:36,022 --> 00:34:37,730 So in the second half of today's lecture, 435 00:34:37,730 --> 00:34:41,870 I want to show you how to use this compactness claim combined 436 00:34:41,870 --> 00:34:46,790 with the first definition of compactness that you've seen, 437 00:34:46,790 --> 00:34:51,110 namely every open cover contains a finite sub cover, 438 00:34:51,110 --> 00:34:56,179 and to use that to prove many consequences 439 00:34:56,179 --> 00:34:58,460 about the space of graphons. 440 00:34:58,460 --> 00:35:02,030 And some things that we had to work a bit hard at, 441 00:35:02,030 --> 00:35:06,790 but they turn out to fall from the compactness statement. 442 00:35:06,790 --> 00:35:09,910 So let's take a quick break. 443 00:35:09,910 --> 00:35:11,420 In the first part of this lecture, 444 00:35:11,420 --> 00:35:14,270 we proved that a space of graphons is compact. 445 00:35:14,270 --> 00:35:17,840 So now let me show you what we can reap as consequences 446 00:35:17,840 --> 00:35:20,450 from the compactness result. So I 447 00:35:20,450 --> 00:35:24,140 want to show you how to apply compactness 448 00:35:24,140 --> 00:35:25,820 and prove some consequences. 449 00:35:35,540 --> 00:35:38,450 As I mentioned earlier, the compactness result 450 00:35:38,450 --> 00:35:40,550 is related to regularity. 451 00:35:40,550 --> 00:35:43,040 And in fact, many of the results I'm going to state, 452 00:35:43,040 --> 00:35:47,870 you can prove maybe with some more work using the regularity 453 00:35:47,870 --> 00:35:49,010 lemma. 454 00:35:49,010 --> 00:35:51,320 But I also want to show you how to deduce them directly 455 00:35:51,320 --> 00:35:52,190 from compactness. 456 00:35:52,190 --> 00:35:54,650 In fact, we'll deduce the regularity lemma 457 00:35:54,650 --> 00:35:55,850 from compactness. 458 00:35:55,850 --> 00:35:58,550 So these two ideas, compactness and regularity, 459 00:35:58,550 --> 00:36:01,830 they go hand-in-hand. 460 00:36:01,830 --> 00:36:05,390 So first, more so as a warm up, but as also an interesting 461 00:36:05,390 --> 00:36:07,880 result, statement, an interesting statement 462 00:36:07,880 --> 00:36:11,548 on its own, so let me prove the following. 463 00:36:14,970 --> 00:36:21,580 So here is a statement that we can deduce from compactness. 464 00:36:21,580 --> 00:36:26,720 So for every epsilon, there exists some number N 465 00:36:26,720 --> 00:36:36,530 which depend only on epsilon, such that for every W graphon, 466 00:36:36,530 --> 00:36:49,440 there exists a graph G with N vertices, 467 00:36:49,440 --> 00:37:02,770 such that the cut distance between G and W 468 00:37:02,770 --> 00:37:05,100 is, at most, epsilon. 469 00:37:08,440 --> 00:37:10,540 So think about what this says. 470 00:37:10,540 --> 00:37:12,930 So for every epsilon, there is some 471 00:37:12,930 --> 00:37:17,100 bound N such that every graphon-- so a graphon is 472 00:37:17,100 --> 00:37:22,900 some real-value function, so taking values between 0 and 1. 473 00:37:22,900 --> 00:37:24,990 You can approximate it in the distance 474 00:37:24,990 --> 00:37:28,560 that we care about by a graph with a bounded number 475 00:37:28,560 --> 00:37:31,180 of vertices. 476 00:37:31,180 --> 00:37:33,700 This is kind of like regularity lemma. 477 00:37:33,700 --> 00:37:38,210 If you are allowed edge weights on this graph G, 478 00:37:38,210 --> 00:37:43,220 then it immediately follows from the weak regularity lemma 479 00:37:43,220 --> 00:37:44,570 that we already proved. 480 00:37:48,500 --> 00:37:50,930 And from that weak regularity lemma 481 00:37:50,930 --> 00:37:55,370 which allows you to get some G with edge weights, 482 00:37:55,370 --> 00:37:57,440 you can think about how you might 483 00:37:57,440 --> 00:38:03,100 turn an edge-weighted graph into an unweighted graph. 484 00:38:03,100 --> 00:38:04,968 So that can also be done. 485 00:38:04,968 --> 00:38:07,010 But I want to show you a completely different way 486 00:38:07,010 --> 00:38:10,700 of proving this result that follows from compactness. 487 00:38:13,570 --> 00:38:16,930 And so I say it's a warm up, because it's really 488 00:38:16,930 --> 00:38:19,918 a warm up for the next thing we're going to do. 489 00:38:19,918 --> 00:38:21,460 This is an easier example showing you 490 00:38:21,460 --> 00:38:24,690 how to use compactness. 491 00:38:24,690 --> 00:38:26,440 So the idea is, I have this compact space. 492 00:38:26,440 --> 00:38:32,380 I'm going to cover this space by open sets, by open balls. 493 00:38:32,380 --> 00:38:38,230 So the open balls are going to be this B sub epsilon 494 00:38:38,230 --> 00:38:41,200 G. So for each graph, G, I'm going 495 00:38:41,200 --> 00:38:53,005 to consider the set of graphons that are within epsilon of G. 496 00:38:53,005 --> 00:38:54,755 So this is you have some topological space 497 00:38:54,755 --> 00:38:55,810 or some metric space. 498 00:38:55,810 --> 00:38:58,260 I have a point G. And I look at its open ball. 499 00:39:07,930 --> 00:39:09,580 This is the ball. 500 00:39:09,580 --> 00:39:22,753 So I claim that these open balls, they form an open cover, 501 00:39:22,753 --> 00:39:25,620 of the space. 502 00:39:30,300 --> 00:39:31,410 Where is that? 503 00:39:31,410 --> 00:39:35,010 So I want to show every point W is covered. 504 00:39:39,220 --> 00:39:55,930 So this follows from the claim that every W is the limit 505 00:39:55,930 --> 00:39:57,340 of some sequence of graphs. 506 00:40:04,940 --> 00:40:08,600 So we didn't technically actually prove this claim. 507 00:40:08,600 --> 00:40:11,895 I said that if you take W random graphs, you get this. 508 00:40:11,895 --> 00:40:15,320 So we didn't technically prove that. 509 00:40:15,320 --> 00:40:16,870 But OK, so it turns out to be true. 510 00:40:16,870 --> 00:40:18,780 There are easier ways to establish it as well 511 00:40:18,780 --> 00:40:20,790 by taking l1 approximations. 512 00:40:20,790 --> 00:40:24,780 But the point is that, if you use this claim here, 513 00:40:24,780 --> 00:40:28,460 you do not get a bound on the number of vertices. 514 00:40:28,460 --> 00:40:32,660 It could be that for very bizarre-looking W's, you 515 00:40:32,660 --> 00:40:37,690 might require much more number of vertices. 516 00:40:37,690 --> 00:40:40,320 And a priori, you do not know that it is bounded 517 00:40:40,320 --> 00:40:43,510 as a function of epsilon. 518 00:40:43,510 --> 00:40:47,110 But now we have this open cover. 519 00:40:47,110 --> 00:40:55,870 So by compactness of this space of graphons, 520 00:40:55,870 --> 00:41:03,340 we can find an open cover using a finite subset 521 00:41:03,340 --> 00:41:10,510 of these graphs, so G1 to Gk, so a finite subset 522 00:41:10,510 --> 00:41:13,740 to do an open cover. 523 00:41:13,740 --> 00:41:22,330 And now we let N to be the least-common multiple of all 524 00:41:22,330 --> 00:41:25,105 of these vertex set sizes. 525 00:41:49,130 --> 00:41:59,140 So all of these graphs, they are within-- 526 00:42:03,010 --> 00:42:05,270 so for each of these graphs, I can replace it 527 00:42:05,270 --> 00:42:07,550 by a graph on exactly N vertices. 528 00:42:16,300 --> 00:42:28,440 There exists a graph Gi prime of exactly N vertices, 529 00:42:28,440 --> 00:42:34,208 such that they represent the same point 530 00:42:34,208 --> 00:42:35,250 in the space of graphons. 531 00:42:38,436 --> 00:42:40,740 So why is this? 532 00:42:40,740 --> 00:42:43,950 Think about the representation of a graphon using 533 00:42:43,950 --> 00:42:45,630 from a graph. 534 00:42:45,630 --> 00:43:00,650 If I start with G and I blow up each vertex into some k 535 00:43:00,650 --> 00:43:04,530 vertices, then it turns out-- 536 00:43:04,530 --> 00:43:06,530 I mean, you should think about why this is true. 537 00:43:06,530 --> 00:43:09,580 But it's really not hard to see if you draw the picture. 538 00:43:09,580 --> 00:43:12,662 So remember, this black and white picture, that actually, 539 00:43:12,662 --> 00:43:13,620 they're the same point. 540 00:43:13,620 --> 00:43:17,360 They are represented by the same graphon. 541 00:43:17,360 --> 00:43:18,890 OK, and that's it. 542 00:43:18,890 --> 00:43:23,790 So we found these G's. 543 00:43:23,790 --> 00:43:27,540 All of them have exactly N vertices such 544 00:43:27,540 --> 00:43:32,700 that their epsilon open balls form an open cover 545 00:43:32,700 --> 00:43:33,870 of the space of graphons. 546 00:43:33,870 --> 00:43:36,000 So every graphon can be approximated 547 00:43:36,000 --> 00:43:37,172 by one of these graphs. 548 00:43:40,070 --> 00:43:44,030 So you get that from compactness. 549 00:43:44,030 --> 00:43:46,760 The statement says, for every epsilon, their exists an N. 550 00:43:46,760 --> 00:43:50,650 So N is a function of epsilon.l What's the function? 551 00:43:54,590 --> 00:43:57,090 This proof doesn't tell you anything about that. 552 00:43:57,090 --> 00:44:06,860 So this proof gives no information 553 00:44:06,860 --> 00:44:18,750 about the dependence of N on epsilon. 554 00:44:18,750 --> 00:44:21,420 So in some sense, it's even worse than some of the things 555 00:44:21,420 --> 00:44:23,700 we've seen in the earlier discussion 556 00:44:23,700 --> 00:44:25,950 on Szemerédi's regularity lemma where there were tower 557 00:44:25,950 --> 00:44:27,030 or Wowzer-types. 558 00:44:27,030 --> 00:44:29,340 Here there is no information, because it comes 559 00:44:29,340 --> 00:44:32,850 from a compactness statement. 560 00:44:32,850 --> 00:44:39,316 So you just know there exists a finite open cover, no bounds. 561 00:44:39,316 --> 00:44:43,970 OK, any questions about this warm-up application? 562 00:44:43,970 --> 00:44:45,160 So it feels a bit magical. 563 00:44:45,160 --> 00:44:46,160 So you have compactness. 564 00:44:46,160 --> 00:44:51,216 And then you have all of these consequences. 565 00:44:51,216 --> 00:44:54,080 So now let me show you how you can deduce the regularity 566 00:44:54,080 --> 00:44:56,900 lemma itself from compactness. 567 00:44:56,900 --> 00:45:03,010 In fact, in the proof of the existence, 568 00:45:03,010 --> 00:45:06,840 in the proof of compactness, we only used weak regularity. 569 00:45:06,840 --> 00:45:09,590 And now let me show you how you can use the weak regularity 570 00:45:09,590 --> 00:45:11,780 consequence of namely compactness 571 00:45:11,780 --> 00:45:16,650 to bootstrap itself to strong regularity. 572 00:45:16,650 --> 00:45:20,150 So we saw a version of strong regularity in the earlier 573 00:45:20,150 --> 00:45:23,510 chapter when we discussed Szemerédi's regularity lemma. 574 00:45:23,510 --> 00:45:27,770 So let me state it in a somewhat different-looking form, 575 00:45:27,770 --> 00:45:32,640 but that turns out to be morally equivalent. 576 00:45:32,640 --> 00:45:35,590 Suppose I have a vector of epsilons. 577 00:45:41,740 --> 00:45:44,090 So all of these are positive real numbers. 578 00:45:49,680 --> 00:45:55,470 The claim is that there exists an M which 579 00:45:55,470 --> 00:46:10,310 only depends on this vector such that for every graphon W, 580 00:46:10,310 --> 00:46:10,950 one can-- 581 00:46:10,950 --> 00:46:25,010 so every graphon W can be written as the following, 582 00:46:25,010 --> 00:46:26,260 decomposing the following way. 583 00:46:26,260 --> 00:46:31,970 We write W as a sum of a structured part, 584 00:46:31,970 --> 00:46:38,540 a pseudo-random part, and a small part, 585 00:46:38,540 --> 00:46:51,060 where the structured part is a step function with k parts, 586 00:46:51,060 --> 00:46:57,450 but k is, at most, M, this claimed bound M. 587 00:46:57,450 --> 00:47:03,640 The pseudo-random part has a very small cut norm, 588 00:47:03,640 --> 00:47:08,520 so its cut norm, very small, even compared 589 00:47:08,520 --> 00:47:10,870 to the number of parts. 590 00:47:10,870 --> 00:47:22,117 And finally, the small part has l1 norm bounded by epsilon 1. 591 00:47:22,117 --> 00:47:22,950 So that's the claim. 592 00:47:22,950 --> 00:47:25,530 You can always-- there exists some 593 00:47:25,530 --> 00:47:28,200 bound M in terms of these error parameters 594 00:47:28,200 --> 00:47:30,667 so that you have this decomposition. 595 00:47:30,667 --> 00:47:32,250 So we saw some version of this earlier 596 00:47:32,250 --> 00:47:35,280 when we discussed the spectral proof of regularity lemma. 597 00:47:35,280 --> 00:47:38,520 And I don't want to go into details of how these two 598 00:47:38,520 --> 00:47:41,880 things are related, but just to comment that depending 599 00:47:41,880 --> 00:47:44,010 on your choice of the epsilon parameters, 600 00:47:44,010 --> 00:47:46,110 it relates to some of the different versions 601 00:47:46,110 --> 00:47:48,820 of regularity lemma that we've seen before. 602 00:47:48,820 --> 00:47:54,600 So for example, if epsilon k is roughly epsilon, 603 00:47:54,600 --> 00:47:57,240 some fixed epsilon over k squared, 604 00:47:57,240 --> 00:48:08,400 then this is basically the same as Szemerédi's regularity 605 00:48:08,400 --> 00:48:16,920 lemma, whereas if all the k's are the same epsilon, 606 00:48:16,920 --> 00:48:20,236 then this is roughly the same as the weak regularity lemma. 607 00:48:27,540 --> 00:48:30,820 All right, so how to prove this claim? 608 00:48:30,820 --> 00:48:32,430 We're going to use compactness again. 609 00:48:38,070 --> 00:48:43,670 So first, there always exists an l1 approximation 610 00:48:43,670 --> 00:48:55,870 so that every W has some step function u associated 611 00:48:55,870 --> 00:49:02,010 to it such that the l1 distance between W and u 612 00:49:02,010 --> 00:49:06,060 is, at most, epsilon 1. 613 00:49:06,060 --> 00:49:08,730 So again, this is one of these more measured theoretic 614 00:49:08,730 --> 00:49:10,610 technicalities I don't want to get into, 615 00:49:10,610 --> 00:49:12,567 but so it's not hard to prove. 616 00:49:12,567 --> 00:49:14,400 So roughly speaking, you have some function. 617 00:49:14,400 --> 00:49:16,820 You can approximate it using steps. 618 00:49:26,860 --> 00:49:29,950 So similar to what we did just now, 619 00:49:29,950 --> 00:49:33,220 if you just do that, the number of steps 620 00:49:33,220 --> 00:49:36,840 might not be a function of epsilon, 621 00:49:36,840 --> 00:49:40,480 so you might need much more steps just doing 622 00:49:40,480 --> 00:49:45,570 that if your W looks more pathological. 623 00:49:45,570 --> 00:49:48,120 So now what we're going to do is consider 624 00:49:48,120 --> 00:49:52,050 the following function, k of W. And I 625 00:49:52,050 --> 00:49:57,390 define it to be the minimum k such 626 00:49:57,390 --> 00:50:10,660 that there exists a k step graphon u such that u you 627 00:50:10,660 --> 00:50:20,290 minus W is, at most, epsilon 1. 628 00:50:20,290 --> 00:50:26,030 So among all the step function approximations, 629 00:50:26,030 --> 00:50:29,330 pick one that has the minimum number of steps 630 00:50:29,330 --> 00:50:34,755 and call the number of steps k of W. So now, 631 00:50:34,755 --> 00:50:36,130 as before, we're going to come up 632 00:50:36,130 --> 00:50:38,780 with an open cover of the space of graphons. 633 00:50:41,380 --> 00:50:44,770 So the open cover is going to be consisting 634 00:50:44,770 --> 00:50:50,722 of the cut norm balls of-- 635 00:50:50,722 --> 00:50:52,750 actually, what notation did I use over there? 636 00:51:01,200 --> 00:51:03,990 So this is a ball centered around W 637 00:51:03,990 --> 00:51:08,660 with radius epsilon sub kW. 638 00:51:08,660 --> 00:51:15,972 This is an open cover of the space of graphons 639 00:51:15,972 --> 00:51:19,490 as W ranges over all graphons. 640 00:51:23,980 --> 00:51:26,490 So I'm literally looking at every point in the space 641 00:51:26,490 --> 00:51:28,020 and putting an open ball around it. 642 00:51:28,020 --> 00:51:31,750 So obviously, this is an open cover. 643 00:51:31,750 --> 00:51:33,850 And because of compactness, there 644 00:51:33,850 --> 00:51:35,340 exists a finite sub cover. 645 00:51:41,080 --> 00:51:46,910 So there exists a finite set, we write 646 00:51:46,910 --> 00:52:06,940 curly s, of graphons such that these balls, as I range over W 647 00:52:06,940 --> 00:52:13,010 and curly s, they cover the space of graphons. 648 00:52:22,550 --> 00:52:25,430 Now the goal is, given the W, I want 649 00:52:25,430 --> 00:52:26,870 to approximate it in some way. 650 00:52:26,870 --> 00:52:31,580 So having a finite set of things to work with 651 00:52:31,580 --> 00:52:36,010 allows us to do some kind of approximations. 652 00:52:36,010 --> 00:52:45,950 So thus, for every W graphon, there exists a W prime in s 653 00:52:45,950 --> 00:52:53,420 whose ball in that collection covers the point W, such that W 654 00:52:53,420 --> 00:53:06,400 is contained in this ball. 655 00:53:06,400 --> 00:53:16,670 And OK, so given this W prime, because of this definition 656 00:53:16,670 --> 00:53:22,460 over here, so there exists a u which 657 00:53:22,460 --> 00:53:35,570 is a k step graphon with k, at most, 658 00:53:35,570 --> 00:53:48,450 the maximum over all such possible number of steps, 659 00:53:48,450 --> 00:53:59,830 such that W and W prime, they are close in cut norm 660 00:53:59,830 --> 00:54:03,250 because you have this open cover. 661 00:54:03,250 --> 00:54:11,380 And furthermore, W prime is close to a graphon 662 00:54:11,380 --> 00:54:13,180 with a small number of steps. 663 00:54:21,970 --> 00:54:32,860 So suppose we now write W as u plus W minus W prime and then 664 00:54:32,860 --> 00:54:37,870 plus W prime minus u. 665 00:54:37,870 --> 00:54:39,850 We find that this is the decomposition that we 666 00:54:39,850 --> 00:54:44,290 are looking for because the u-- 667 00:54:44,290 --> 00:54:46,380 so this is the structural component-- 668 00:54:46,380 --> 00:54:58,635 has k steps, where k is less than this quantity here. 669 00:54:58,635 --> 00:55:03,890 And that quantity there is just some function of epsilons. 670 00:55:03,890 --> 00:55:05,900 So it's, at most, some function of the epsilons. 671 00:55:12,030 --> 00:55:18,700 It doesn't depend on the specific choice of W. 672 00:55:18,700 --> 00:55:22,570 The second term, this is this pseudo-random piece, 673 00:55:22,570 --> 00:55:35,586 because it's cut norm is small, so what we have here. 674 00:55:35,586 --> 00:55:38,508 Yeah, so this entire thing should be subscript. 675 00:55:45,690 --> 00:55:52,620 And finally, the third term here is the small term, 676 00:55:52,620 --> 00:55:54,810 because it's l1 norm is small. 677 00:56:01,307 --> 00:56:03,557 So putting them together, we get the regularity lemma. 678 00:56:06,480 --> 00:56:10,140 So again, the proof gives you no information whatsoever 679 00:56:10,140 --> 00:56:13,230 about the bound M as a function of the input 680 00:56:13,230 --> 00:56:16,180 parameters, the epsilons. 681 00:56:16,180 --> 00:56:18,890 So it turns out you can use a different method 682 00:56:18,890 --> 00:56:20,030 to get the bounds. 683 00:56:20,030 --> 00:56:22,700 Namely, we actually more or less did this proof 684 00:56:22,700 --> 00:56:25,830 when we discussed regularity lemma, the strong regularity 685 00:56:25,830 --> 00:56:26,330 lemma. 686 00:56:26,330 --> 00:56:29,120 So we did a different proof where we iterated an energy 687 00:56:29,120 --> 00:56:30,780 increment argument. 688 00:56:30,780 --> 00:56:33,740 And that gave you some concrete bounds, some bounds which 689 00:56:33,740 --> 00:56:36,020 iterates on these epsilons. 690 00:56:36,020 --> 00:56:37,565 But here is a different proof. 691 00:56:37,565 --> 00:56:41,150 It gives you less information, but it elegantly 692 00:56:41,150 --> 00:56:43,690 uses this compactness feature of the space of graphons. 693 00:56:47,440 --> 00:56:48,660 Any questions? 694 00:56:54,310 --> 00:56:56,040 OK, so we proved compactness. 695 00:56:56,040 --> 00:56:59,130 So now let's go on to the other two claims, 696 00:56:59,130 --> 00:57:01,210 namely the existence of the limit 697 00:57:01,210 --> 00:57:04,080 and that equivalences of convergence. 698 00:57:04,080 --> 00:57:05,800 The existence of the limit more or less 699 00:57:05,800 --> 00:57:07,410 is a consequence of compactness. 700 00:57:15,380 --> 00:57:23,690 So you have this sequence of graphons, W1, W2, and so on. 701 00:57:23,690 --> 00:57:34,260 And the claim is that, if this sequence of F densities 702 00:57:34,260 --> 00:57:47,750 converges for each F, then there exists 703 00:57:47,750 --> 00:57:57,680 some limit W such that all of these sequences of F densities 704 00:57:57,680 --> 00:58:01,140 converge to the limit density. 705 00:58:01,140 --> 00:58:05,810 So that was the claim, so nothing about cut norms 706 00:58:05,810 --> 00:58:10,040 in at least as far as the statement goes. 707 00:58:10,040 --> 00:58:15,540 Well OK, from compactness, we know 708 00:58:15,540 --> 00:58:19,680 that you can produce always a subsequential limit. 709 00:58:19,680 --> 00:58:28,030 So by compactness or sequential compactness, 710 00:58:28,030 --> 00:58:36,810 there exists some limit point which we call W. 711 00:58:36,810 --> 00:58:40,950 And this W has the property that, 712 00:58:40,950 --> 00:58:49,230 for some subsequence, the cut distance 713 00:58:49,230 --> 00:58:54,120 from the subsequence converges to W. 714 00:58:54,120 --> 00:59:07,660 So for some subsequence n0 as ni going to infinity. 715 00:59:07,660 --> 00:59:20,785 But now, by the counting lemma, the sequence of F densities-- 716 00:59:20,785 --> 00:59:22,410 so the counting Lemma tells you, if you 717 00:59:22,410 --> 00:59:26,040 have cut distance going to 0, then the F density 718 00:59:26,040 --> 00:59:28,120 should also go to 0. 719 00:59:28,120 --> 00:59:29,620 So indeed, that's what we have here. 720 00:59:38,450 --> 00:59:40,840 So this is so far just for the subsequence. 721 00:59:40,840 --> 00:59:45,340 But we assumed already that the entire sequence 722 00:59:45,340 --> 00:59:49,550 converges in respect to every F densities. 723 00:59:56,940 --> 00:59:59,310 So it must be the same limit. 724 01:00:04,130 --> 01:00:08,660 And that finishes the proof of convergence, 725 01:00:08,660 --> 01:00:11,830 so proof of the existence of the limit. 726 01:00:11,830 --> 01:00:16,260 So we obtain this limit from compactness. 727 01:00:16,260 --> 01:00:19,590 Next, let's prove the equivalence of convergence. 728 01:00:19,590 --> 01:00:21,310 And this one is somewhat trickier. 729 01:00:21,310 --> 01:00:24,150 So what happens here is that we would 730 01:00:24,150 --> 01:00:27,840 like to show that these two notions of convergence, 731 01:00:27,840 --> 01:00:30,750 one having to do with F densities 732 01:00:30,750 --> 01:00:33,320 and another having to do with cut distance, 733 01:00:33,320 --> 01:00:36,100 that these two notions are equivalent to each other. 734 01:00:41,250 --> 01:00:51,700 So the goal here is to show that this F density convergence is 735 01:00:51,700 --> 01:01:01,390 equivalent to the statement that W sub n is Cauchy with respect 736 01:01:01,390 --> 01:01:04,630 to the cut distance. 737 01:01:07,410 --> 01:01:11,360 All right, claim one of the directions is easy. 738 01:01:11,360 --> 01:01:12,360 Which direction is that? 739 01:01:22,070 --> 01:01:24,298 So which direction is the easy direction? 740 01:01:28,590 --> 01:01:31,230 So which way, left, going left, going right? 741 01:01:34,520 --> 01:01:36,240 OK, so going left? 742 01:01:36,240 --> 01:01:40,620 So I claim that this is easy, because it 743 01:01:40,620 --> 01:01:41,830 follows from counting lemma. 744 01:01:50,633 --> 01:01:53,050 Counting lemma, remember the spirit of the counting lemma, 745 01:01:53,050 --> 01:01:56,130 at least qualitatively, is that if you have two graphons that 746 01:01:56,130 --> 01:01:59,760 are close in cut distance, then they are close in F densities. 747 01:01:59,760 --> 01:02:03,030 So if you have Cauchy with respect to cut distance, 748 01:02:03,030 --> 01:02:06,764 then they are Cauchy, and hence, convergent in F densities. 749 01:02:09,370 --> 01:02:12,760 And it's the other direction that will require some work. 750 01:02:12,760 --> 01:02:16,130 And this one is actually genuinely tricky. 751 01:02:16,130 --> 01:02:21,160 So and it's almost kind of a miraculous statement, 752 01:02:21,160 --> 01:02:25,630 that somehow if you only knew the F densities-- 753 01:02:25,630 --> 01:02:28,480 so somebody gives you this very large sequence of graphs 754 01:02:28,480 --> 01:02:30,760 and only tells you that the triangle densities, 755 01:02:30,760 --> 01:02:34,090 the C4 densities, all of these graph densities, they converge. 756 01:02:34,090 --> 01:02:37,600 Somehow from these small statistics, 757 01:02:37,600 --> 01:02:41,140 you conclude that the graphs globally look 758 01:02:41,140 --> 01:02:43,370 very similar to each other. 759 01:02:43,370 --> 01:02:44,870 That's actually, if you think about, 760 01:02:44,870 --> 01:02:49,035 this is an amazing statement. 761 01:02:49,035 --> 01:02:50,160 OK, so let's see the proof. 762 01:02:54,030 --> 01:02:56,100 The proof method here is somewhat 763 01:02:56,100 --> 01:03:00,450 representative of these graph-limit-type arguments. 764 01:03:00,450 --> 01:03:03,870 So it's worth paying attention to see how this one goes. 765 01:03:03,870 --> 01:03:07,380 So by compactness, if-- 766 01:03:07,380 --> 01:03:11,550 OK, we're going to set up by contradiction. 767 01:03:11,550 --> 01:03:24,290 If the sequence is not Cauchy, then there 768 01:03:24,290 --> 01:03:26,380 exists two limit points. 769 01:03:31,930 --> 01:03:34,830 So there exists at least two distinct limit points. 770 01:03:34,830 --> 01:03:40,200 And call them u and W, and such that-- 771 01:03:46,370 --> 01:03:50,110 so because you have two separate limit points, 772 01:03:50,110 --> 01:03:54,590 you must have that this sequence, 773 01:03:54,590 --> 01:03:57,990 at least along a subsequence that converges to W, 774 01:03:57,990 --> 01:04:05,970 converges in F densities to W. So initially, this 775 01:04:05,970 --> 01:04:07,260 is true along subsequence. 776 01:04:07,260 --> 01:04:09,160 But the left-hand side is convergent, 777 01:04:09,160 --> 01:04:12,790 so this is true along the sequence. 778 01:04:12,790 --> 01:04:15,370 But u is also a limit point. 779 01:04:15,370 --> 01:04:21,990 So the same is true for u. 780 01:04:26,337 --> 01:04:34,670 And therefore, the F density in W must equal to the F density 781 01:04:34,670 --> 01:04:45,670 in u for all F. 782 01:04:45,670 --> 01:04:49,420 So we would be done if we can prove that the F densities, 783 01:04:49,420 --> 01:04:51,670 the collection of all these F densities, 784 01:04:51,670 --> 01:04:53,705 they determine the graphon. 785 01:04:53,705 --> 01:04:54,830 And that's indeed the case. 786 01:04:54,830 --> 01:04:56,540 And so this is the next claim. 787 01:04:56,540 --> 01:05:00,140 So it's what I will call a moment 788 01:05:00,140 --> 01:05:17,450 lemma, is that if u and W are graphons such that the F 789 01:05:17,450 --> 01:05:27,380 densities agree for all F, then the cut distance between u 790 01:05:27,380 --> 01:05:31,910 and W is equal to 0. 791 01:05:31,910 --> 01:05:34,880 Somehow the local statistics tells you globally 792 01:05:34,880 --> 01:05:39,420 that these two graphons must agree with each other. 793 01:05:39,420 --> 01:05:42,360 Does anyone know why I call it a moment lemma? 794 01:05:42,360 --> 01:05:46,730 There is something else which this should remind you of. 795 01:05:46,730 --> 01:05:50,040 So there are some classical results in probability that 796 01:05:50,040 --> 01:05:53,240 tells you, if you have two probability distributions, 797 01:05:53,240 --> 01:05:57,830 both, assume are nice enough, then if they have the same 798 01:05:57,830 --> 01:06:01,490 k'th moment for every k, so first moment, second moment, 799 01:06:01,490 --> 01:06:03,770 third moment, if all the moments agree, 800 01:06:03,770 --> 01:06:07,260 that these two probability distributions should agree. 801 01:06:07,260 --> 01:06:10,130 And this is some graphical version of that. 802 01:06:10,130 --> 01:06:12,600 So instead of looking at the probability distribution, 803 01:06:12,600 --> 01:06:15,590 we're looking at graphons, which are two-dimensional. 804 01:06:15,590 --> 01:06:17,135 These are two-dimensional objects. 805 01:06:17,135 --> 01:06:19,760 And this moments lemma tells you that in these two-dimensional, 806 01:06:19,760 --> 01:06:21,927 in the corresponding two-dimensional moments, namely 807 01:06:21,927 --> 01:06:25,040 these F moments, if they agree, then 808 01:06:25,040 --> 01:06:27,380 the two graphons must agree. 809 01:06:30,240 --> 01:06:33,810 So it's the analog of the probability theory 810 01:06:33,810 --> 01:06:37,860 statement about moments. 811 01:06:37,860 --> 01:06:39,540 The proof is actually somewhat tricky. 812 01:06:39,540 --> 01:06:41,190 So I'm only going to give you a sketch. 813 01:06:44,880 --> 01:06:49,590 And the key here is to consider the W random graph, which 814 01:06:49,590 --> 01:06:51,090 we saw last lecture. 815 01:06:57,326 --> 01:07:00,320 So this is W random graph with k vertices 816 01:07:00,320 --> 01:07:06,070 sampled using the graphon W. 817 01:07:06,070 --> 01:07:09,010 So a key observation here is that, 818 01:07:09,010 --> 01:07:19,140 for every F, the probability that the sampled W 819 01:07:19,140 --> 01:07:26,560 random graph agrees with F-- 820 01:07:26,560 --> 01:07:29,960 and here, there is a bit of a technicality. 821 01:07:29,960 --> 01:07:31,910 I want them to agree as labeled graphs. 822 01:07:35,880 --> 01:07:39,340 So the vertices of W are a priori labeled 1 through k. 823 01:07:39,340 --> 01:07:43,330 And this kW random graph is generated with vertices 824 01:07:43,330 --> 01:07:44,890 labeled 1 through k. 825 01:07:44,890 --> 01:07:48,640 They agree with some probability that is completely 826 01:07:48,640 --> 01:07:54,170 determined by the F densities. 827 01:07:54,170 --> 01:07:54,670 Yes? 828 01:07:54,670 --> 01:07:56,740 AUDIENCE: Is k the number of vertices of F? 829 01:07:56,740 --> 01:08:02,390 PROFESSOR: Yeah, so k is the number of vertices of F. 830 01:08:02,390 --> 01:08:04,867 And the specific formula is not so important. 831 01:08:04,867 --> 01:08:05,950 Let me just write it down. 832 01:08:05,950 --> 01:08:09,470 But the point is that, if you know all the F densities, then 833 01:08:09,470 --> 01:08:12,500 you have all the information about the distribution 834 01:08:12,500 --> 01:08:15,110 of this W random graph. 835 01:08:15,110 --> 01:08:18,810 And the way you can calculate the actual probability 836 01:08:18,810 --> 01:08:21,173 is via an inclusion exclusion. 837 01:08:23,840 --> 01:08:26,090 And the reason we have to do this inclusion exclusion 838 01:08:26,090 --> 01:08:28,340 is just because this is more like counting 839 01:08:28,340 --> 01:08:29,760 induced subgraphs. 840 01:08:29,760 --> 01:08:31,939 And this is counting actual subgraphs. 841 01:08:31,939 --> 01:08:33,770 So there is an extra step. 842 01:08:33,770 --> 01:08:36,880 But the point is that, if you knew this data, the moment's 843 01:08:36,880 --> 01:08:40,026 data then you immediately know the distribution of the W 844 01:08:40,026 --> 01:08:40,609 random graphs. 845 01:08:51,800 --> 01:08:57,520 OK, so if I have two graphons for which I know that their F 846 01:08:57,520 --> 01:09:00,350 densities agree, then I should be 847 01:09:00,350 --> 01:09:05,660 able to conclude that the corresponding W 848 01:09:05,660 --> 01:09:19,580 random graphs also have the same distribution, in particular, 849 01:09:19,580 --> 01:09:25,340 this, the W random graph and the u random graph 850 01:09:25,340 --> 01:09:27,324 have the same distribution. 851 01:09:34,010 --> 01:09:37,939 I am going to create a variant of the W random graph 852 01:09:37,939 --> 01:09:42,979 which is something called an H random graph. 853 01:09:42,979 --> 01:09:44,720 It's kind of like the W random graph 854 01:09:44,720 --> 01:09:46,800 except I forget the very last step. 855 01:09:46,800 --> 01:09:51,319 So I only keep a weighted-- 856 01:09:51,319 --> 01:10:06,310 think of it as, so think this is an edge-weighted graph where 857 01:10:06,310 --> 01:10:16,860 you sample x1 through xk uniformly between 0 and 1. 858 01:10:16,860 --> 01:10:26,680 And I put edge weight between i and j to be W of xi xj. 859 01:10:26,680 --> 01:10:29,580 So the difference between this H version and the G version 860 01:10:29,580 --> 01:10:31,110 is that the G version is obtained 861 01:10:31,110 --> 01:10:34,120 by turning this weight into an actual edge 862 01:10:34,120 --> 01:10:37,320 with that probability, but if I don't do the last step, 863 01:10:37,320 --> 01:10:41,150 I obtain this intermediate object. 864 01:10:41,150 --> 01:10:46,920 So the following are true. 865 01:10:46,920 --> 01:10:49,430 And this is where I'm going to skip the proofs. 866 01:10:52,700 --> 01:11:02,430 If I look at this H random graph and the G random graph, 867 01:11:02,430 --> 01:11:05,663 they are very close in cut distance. 868 01:11:05,663 --> 01:11:07,080 You can think of this as the claim 869 01:11:07,080 --> 01:11:11,250 that G and P is very close to G in cut distance. 870 01:11:11,250 --> 01:11:12,930 So they are very close in cut distance. 871 01:11:18,160 --> 01:11:23,550 As k going to infinity with probability 1-- 872 01:11:23,550 --> 01:11:25,050 so now I'm going to do the proof. 873 01:11:25,050 --> 01:11:27,720 But it's some kind of a concentration argument. 874 01:11:33,990 --> 01:11:41,930 And the second claim is that the H random graph is actually 875 01:11:41,930 --> 01:11:46,660 very close to the original graphon W, as well. 876 01:11:51,260 --> 01:11:57,770 This is also little l1 in distance as k goes to infinity. 877 01:11:57,770 --> 01:12:01,180 So this one is, again, not so obvious. 878 01:12:01,180 --> 01:12:08,870 But it's easier in the case when W 879 01:12:08,870 --> 01:12:16,520 itself is a step function, in which case, the produced 880 01:12:16,520 --> 01:12:20,180 H is almost the same as W, except the boundaries are 881 01:12:20,180 --> 01:12:24,150 slightly shifted, perhaps. 882 01:12:24,150 --> 01:12:30,890 And so you first approximate W by a step function, 883 01:12:30,890 --> 01:12:33,530 and prove this up to an epsilon approximation, 884 01:12:33,530 --> 01:12:37,470 and then let the steps go to infinity. 885 01:12:37,470 --> 01:12:41,060 So if you have these two claims, so then 886 01:12:41,060 --> 01:12:58,640 we see that this one here is identically distributed as ku. 887 01:12:58,640 --> 01:13:06,200 So it should follow that the corresponding H random graph 888 01:13:06,200 --> 01:13:14,765 for u, if you place the same inequalities by the u versions, 889 01:13:14,765 --> 01:13:16,550 it should also be true. 890 01:13:16,550 --> 01:13:21,730 So because these two are the same distribution, 891 01:13:21,730 --> 01:13:24,220 if you follow this chain, you obtain 892 01:13:24,220 --> 01:13:30,590 that the cut distance between u and w is equal to 0. 893 01:13:38,410 --> 01:13:45,790 I want to close by mentioning, in some sense-- 894 01:13:45,790 --> 01:13:50,580 so here, you have two graphons that have exactly the same F 895 01:13:50,580 --> 01:13:51,770 moments. 896 01:13:51,770 --> 01:13:54,790 But what if I give you two graphons which have very 897 01:13:54,790 --> 01:13:56,500 similar moments to each other? 898 01:13:56,500 --> 01:13:59,140 Can you conclude that the two graphons 899 01:13:59,140 --> 01:14:01,370 are close to each other? 900 01:14:01,370 --> 01:14:04,510 And that will be some kind of an inverse counting lemma. 901 01:14:04,510 --> 01:14:06,670 And in fact, it does follow as a corollary. 902 01:14:17,070 --> 01:14:20,340 And the statement is that, for every epsilon, 903 01:14:20,340 --> 01:14:32,330 there exists k and eta such that if the two graphons u and W are 904 01:14:32,330 --> 01:14:42,080 such that the F densities do not differ by more than eta 905 01:14:42,080 --> 01:14:51,460 for every F on, at most, k vertices, 906 01:14:51,460 --> 01:15:01,120 then the cut distance between u and W is, at most, epsilon. 907 01:15:01,120 --> 01:15:03,910 So the counting lemma tells you, if the cut distance is small, 908 01:15:03,910 --> 01:15:07,120 then all the F moments are close to each other. 909 01:15:07,120 --> 01:15:09,430 And the inverse tells you this converse. 910 01:15:09,430 --> 01:15:13,090 So it tells you this, if you have similar F moments, 911 01:15:13,090 --> 01:15:18,690 up to a certain point, then this is small. 912 01:15:18,690 --> 01:15:20,720 You can deduce the inverse counting lemma 913 01:15:20,720 --> 01:15:23,730 from the moments lemma via a compactness argument 914 01:15:23,730 --> 01:15:27,420 similar to the one that we did in class today. 915 01:15:27,420 --> 01:15:30,240 And I want to give you a chance to practice with that argument. 916 01:15:30,240 --> 01:15:33,120 So this will be on the homework, for the next homework. 917 01:15:33,120 --> 01:15:34,860 I'll give you some practice with using 918 01:15:34,860 --> 01:15:37,780 these compactness arguments. 919 01:15:37,780 --> 01:15:42,670 But you see, just with the other compactness statements, 920 01:15:42,670 --> 01:15:46,860 it doesn't tell you anything about the k and the epsilon 921 01:15:46,860 --> 01:15:48,210 as a function of-- 922 01:15:48,210 --> 01:15:51,600 the k and eta as a function of epsilon. 923 01:15:51,600 --> 01:15:54,400 So there are other proofs that gives you concrete bounds, 924 01:15:54,400 --> 01:15:58,270 but this proof here is much simpler 925 01:15:58,270 --> 01:16:04,370 if you assume the corresponding results about compactness. 926 01:16:04,370 --> 01:16:11,780 And finally, I want to mention that in the moments lemma, 927 01:16:11,780 --> 01:16:15,980 in order to deduce that u and w have the same-- 928 01:16:15,980 --> 01:16:18,320 that they are basically the same graphon, 929 01:16:18,320 --> 01:16:23,060 we need to consider F moments for all F's. 930 01:16:23,060 --> 01:16:25,640 So you might ask, could it be the case 931 01:16:25,640 --> 01:16:31,800 that we only need some finite set of F's to deduce-- 932 01:16:31,800 --> 01:16:34,220 to recover the graphon? 933 01:16:34,220 --> 01:16:40,400 Is it the case that you can recover W 934 01:16:40,400 --> 01:16:44,410 from only a finite number of F moments? 935 01:16:44,410 --> 01:16:47,000 And this is, it's actually a very interesting problem 936 01:16:47,000 --> 01:16:50,660 for which we already saw one instance. 937 01:16:50,660 --> 01:16:54,710 Namely, when we discussed quasi-random graphs, 938 01:16:54,710 --> 01:17:01,910 we saw that if you know that the k2 moment is p 939 01:17:01,910 --> 01:17:12,860 and also the C4 moment is p to the 4, then 940 01:17:12,860 --> 01:17:16,880 we can deduce that the graphon must 941 01:17:16,880 --> 01:17:18,942 be the constant graphon, p. 942 01:17:18,942 --> 01:17:20,930 OK, so we didn't do it in this language, 943 01:17:20,930 --> 01:17:23,060 but that's what the proof does. 944 01:17:23,060 --> 01:17:27,050 And likewise, you can use this to deduce a qualitative version 945 01:17:27,050 --> 01:17:40,670 where you have an extra slack and an extra slack over here. 946 01:17:40,670 --> 01:17:45,410 So you might ask, except for the constant graphons, 947 01:17:45,410 --> 01:17:48,050 are there other graphons for which you can similarly 948 01:17:48,050 --> 01:17:49,880 deduce-- 949 01:17:49,880 --> 01:17:53,840 recover this graphon from just a finite amount of moments data? 950 01:17:53,840 --> 01:17:56,510 And such graphons are known as finitely forcible. 951 01:18:03,350 --> 01:18:08,000 So finitely forcible graphons W such 952 01:18:08,000 --> 01:18:20,690 that a finite number of moments can uniquely recover-- 953 01:18:25,350 --> 01:18:28,620 can uniquely identify this graphon, W. 954 01:18:28,620 --> 01:18:30,510 And a very interesting question is, 955 01:18:30,510 --> 01:18:34,130 what is the set of all finitely forcible graphons? 956 01:18:34,130 --> 01:18:36,290 And it turns out, this is not at all obvious. 957 01:18:36,290 --> 01:18:38,790 And let me just give you some examples, highly non-trivial, 958 01:18:38,790 --> 01:18:42,240 that turned out to be finitely forcible. 959 01:18:42,240 --> 01:18:47,040 For example, anything which is a step graphon 960 01:18:47,040 --> 01:18:47,915 is finitely forcible. 961 01:18:52,120 --> 01:18:55,350 The half graphon which corresponds 962 01:18:55,350 --> 01:18:58,870 to the limit of a sequence of half graphs 963 01:18:58,870 --> 01:19:00,450 is finitely forcible. 964 01:19:00,450 --> 01:19:03,630 I mean, already, I think neither of these two examples 965 01:19:03,630 --> 01:19:06,570 are easy at all. 966 01:19:06,570 --> 01:19:10,140 And this example here can be generalized where 967 01:19:10,140 --> 01:19:12,000 you have any polynomial curve. 968 01:19:20,820 --> 01:19:22,905 I think this has to be-- 969 01:19:22,905 --> 01:19:28,660 so if it's a polynomial curve, it's also finitely forcible. 970 01:19:28,660 --> 01:19:30,370 But turns out finitely forcible graphons 971 01:19:30,370 --> 01:19:32,920 can get quite complicated. 972 01:19:32,920 --> 01:19:38,150 And there is still rather quite a bit of mystery around them. 973 01:19:38,150 --> 01:19:43,810 OK, so next time, I want to discuss some inequalities that 974 01:19:43,810 --> 01:19:45,640 come out of-- 975 01:19:45,640 --> 01:19:49,034 you can state between different F densities. 976 01:19:49,034 --> 01:19:49,940 OK, great. 977 01:19:49,940 --> 01:19:51,820 That's all for today.