 fairly easy consequence of the theorem of Christian Jornet that we'd already proved the T of 1 theorem for square functions. You get the singular intervals from the square functions. OK? All right. So now, from almost immediately in the sense that you're going to do it as an exercise, I hope it's not too bad. So some corollaries. OK? Let's recall the definition of the color on commutators. All right? We had these operators c, a, k, f of x is defined to be, I'm going to run out of room there, so let me write it here, 1 over 2 pi i, principal value, integral on the line of 1 over x minus y, a of x minus a of y over x minus y to the k, f of y dy with a-lipschitz. OK? Notice that we can also consider the case k equals 0. In that case, we just have a Hilbert transform with a slightly different constant. The usual Hilbert transform constant is 1 over pi. OK? All right? So the theorem is, the numerology is 325. This is originally due to Calderon. There exists some constant c1, a universal constant. All right? Such that these Calderon commutators are bounded on L2 with L2 operator norm, less than or equal to c1 to the k, times, of course, the Lipschitz constant to the k. Of course, that's just homogeneity, right? You have to pick up the Lipschitz constant to the k power. All right? And that's an exercise. You prove it using the T of 1 theorem. It's an induction argument. This is one of your exercises. OK? So maybe I'll just give you a heuristic. Let's take the k equals 1 case. OK? So remember that the first commutator really is a commutator. It's the derivative, at least formally. It's the derivative operator composed with h. And then you take the commutator with the operator of multiplication with the Lipschitz function a. OK? So if we apply this to the constant function 1, this becomes what? This being by the definition of the commutator, this is d by dx h of a times 1 minus a times d by dx of h of 1. But that's 0, all right? But also, the derivative commutes with h. Well, except I don't know. I have a constant. Sorry. Let me do it this way. I think 2i. OK? And then this is a true statement. So the derivative commutes with the Hilbert transform. And this is just h of a prime. But we know the Hilbert transform is bounded at line 2. A is Lipschitz, so a prime is in L infinity. So this is in BMO by Petrie-Spandstein. And then you're done because this is an anti-symmetric kernel. So we bounded this as immediate. And t star is just negative of t. So if t of 1 is in BMO, then you have anti-symmetry, then so is t star of 1. OK? So then it's just an immediate application of the t of 1 theorem. Well, except that you need to justify these manipulations. All right? OK? And so the way you justify the manipulations, one way to justify the manipulations is to smoothly truncate the kernel here, introduce a smooth truncation, integrate by parts. You get bounds independent of the truncation. And then you play an induction game. Reducing the k plus 1 order of commutator applied to 1 to the kth order of commutator applied to a prime. OK? Modulo some errors that you get from the truncation. All right? You prove a bound independent of the truncation. So that's how the exercise works. OK. And as an immediate corollary of that, we have this is in the notes. This is called theorem 326, which is that if the Lipschitz constant of a is strictly less than 1 over c1, where c1 is that c1, then the Cauchy integral operator, written to graph coordinate parameterized version, CA, is bounded in L2. OK? And the proof of this is just using the fact, as we observed earlier on the first day, that CA is the sum k runs from 0 to infinity of negative 1 times CA, negative 1 to the k times CAk. All right? And now you just sum the Neumann series because you get convergence here if the Lipschitz constant is less than this, strictly less than this constant c1. OK? Now, it was interesting that Calderon actually, he was working on this problem well before the advent of the T of 1 theorem. So he didn't have this technology available. And his argument actually reverses the order. He proves this by hand. And then from that, you get the other thing. OK? But T of 1 makes things a little easier. In fact, it makes it a lot easier. OK. But now, of course, we'd like to do more than just have to be restricted to this small Lipschitz constant case. And in fact, the theorem is true in general, for an arbitrary Lipschitz function. The Cauchy integral operator is bounded on L2 for any Lipschitz graph A, or for any Lipschitz function A. OK? That's a theorem of, well, a very famous theorem of Corifian Macintosh in my air. We will at least, we will prove it, maybe, modulo a little bit of waving my hands and leaving you to check the notes. OK? But the first, the way we're going to approach that theorem is via an extension of the T of 1 theorem called the T of B theorem. OK? So where somehow you can increase, you can generalize the class of testing functions that you work with, not just the constant function 1, but more general functions that have some suitable non-degeneracy property. OK? So we need a definition. So we say that a function B in L infinity of Rn is a creative if there exists some positive constant delta such that the real part of B is always bounded from below by delta. OK? So it's bounded above because it's L infinity, but the real part is bounded below. All right? And similarly, we have that B in L infinity function B is pseudo-accretive if there exists a standard modifier pt and a constant delta such that the modulus of pt of B is always bigger than or equal to delta. Yes, this is uniform in T. Yes, thank you. Uniform in T. And finally, we say that B is dyadic pseudo-accretive if, again, uniformly in T, At of B is bigger than or equal to delta in modulus, where At is the dyadic averaging operator, meaning what? Meaning that At f of x is, by definition, the average over a cube q of xt, I'll explain whether it is in a second, of f, where q of xt is the unique half-open dyadic cube containing x with side length at the minimal one with side length at least t, in other words, with length of q of xt over 2 less than t less than or equal to length of q of xt. So it's the unique minimum-sized dyadic cube with side length at least t, which contains x. So in other words, we're averaging at a scale, at a dyadic scale, that's comparable to t. All right? By the way, I haven't apologized for this notation. It's not my fault. It's just part of the literature. We use queues two different ways. Queues are cubes, and queues are also these little wood-paley operators. Yeah, it's just convention in this subject. I hope that's not too confusing. The subscript t is always going to be the operator. All right? Otherwise, it's going to be a cube. OK? Notice that the point-wise acrativity implies either pseudo-acrativity or dyadic pseudo-acrativity and also dyadic pseudo-acrativity. For example, what's because the absolute value of pt of b is bigger than or equal to the real part of pt of b. But pt has a real kernel. OK? The convolution kernel here is a real valued kernel. So this is the same as pt of the real part of b, which is bounded below by delta, and the integral of pt is 1. So I should emphasize that we're here taking the kernel to be non-negative, with pt bigger than or equal to 0. OK? All right, similarly for at. On the other hand, maybe it's also worth mentioning that it's not enough to simply assume that the absolute value is bounded from below. Absolute value or the modulus of b is bounded from below. Think of a complex exponential where the modulus is 1. But when you average it out, you're getting a lot of cancellation. OK? All right, so somehow when you're talking about point-wise acrativity, the lower bound on the real part is saying something stronger than just lower bound on the modulus. OK? All right, so now that brings us to the next theorem, which is a so-called t of b theorem for square functions. And this is due to SEMS. OK, it's a t of b theorem because, as you might guess, instead of testing on the constant function 1, we're going to test on some b, and the b is going to be a credit or say pseudo-credit. OK? So the theorem says the following. Let theta t f of x be defined to be integral of psi t of xy f of y dy, where psi t, again, is a little with Paley family. All right? And suppose that there exists at least a pseudo-accretive function such that the measure, the n plus 1 dimensional measure d mu of xt defined to be theta t of b of x squared dx dt over t is a Carlson measure. OK? Then the conclusion is that, once again, the theta t's give rise to a bounder square function. OK? And the constant c, of course, is going to depend on the allowable stuff. OK? So where c is going to depend on dimension, the little with Paley constants for psi t, of course, the Carlson norm of mu and the L infinity norm of mu, and also the constant delta in the acrativity condition, or pseudo-acrativity condition. All right? So maybe just before we prove it, just a brief remark. Of course, the natural thing is to compare to theorem, I think, 3.1, which is the t of 1 theorem for square functions, the Christian A theorem. So this actually turns out to be a special case of this one. Right? If you take b to be 1, of course, that's accretive. OK? So this is actually a special case of this theorem. So before we prove it, maybe one more remark also. We proved t of 1 theorems for both square functions and for singular integrals. And in fact, it is indeed true that there is also a t of b theorem for singular integrals. That's harder to prove. We don't have time to do it. There's some comments about it in the notes, but no details. For our purposes, this is going to be enough. And this is actually not going to be a hard theorem, given what we've already done. OK? This is due to David, Jornay, and Semes. But we won't take time to talk about this. All right? OK. So how do we prove this t of b theorem for square functions? All right? Well, the proof is going to go via reduction to the t of 1 theorem for square functions. We're going to reduce matters to theorem 3.1. In other words, what do we need to show? We already know that we're given a Littlewood-Paley family of kernels. So it's enough to show that theta t of 1 of x squared dx dt over t is a Carlson measure, right? Because then the conditions of the t of 1 theorem over square functions are verified. And we just get the square function bound from that. OK? So somehow the idea is to control this guy by this thing that we know is a Carlson measure. We're going to control the Carlson norm of that by this Carlson norm. OK? Somehow. OK? All right? And this is where we use the pseudo-recreativity. All right? Since pt of b, the real part, or sorry, the modulus, is bigger than or equal to delta, therefore theta t of 1 is point-wise bounded by 1 over delta times theta t of 1 of x times pt of b of x. OK? OK. So therefore, if we fix a cube q and try to verify the Carlson condition on that cube and 1 over the measure of q integral from 0 to length of q, integral on q, theta t of 1 of x squared dx dt over t, is less than or equal to 1 over delta squared, 1 over the measure of q, integral from 0 to length of q, integral of q of theta t of 1 times pt of b squared dx dt over t. All right? And we have two minutes, and I think that will be enough. OK? So now we use a familiar trick, this idea of Carlson-Meier. OK? So we write theta t of 1 times pt of b as theta t of 1 times pt minus theta t, all applied to b, plus theta t of b. So now the contribution of this guy is OK by hypothesis. Right? Because we're assuming that theta t of b gives us a Carlson measure. All right? And now what about this part? Let's call this RT. And what you notice about RT, because pt of 1 is 1, this is kind of a familiar thing by now, we note that RT of 1 is 0. And also, the kernel of RT also satisfies the Lew-Wood-Paley conditions. OK, so now you just invoke the Phefermann-Steinlemma, which you did as an exercise. And you get that RT of b as a Carlson measure. OK? In fact, you can even control in terms of the BMO norm of b, but you don't even need something that strong. You just need the L of 30 norm of b. OK, so then you conclude RT b squared dx dt over t as a Carlson measure by the Phefermann-Steinlemma, which I think was theorem 222, I believe. OK, and then we're done. OK, so that's a good place to stop. Any questions? No? OK. Thanks. We've got one more tomorrow. See you there.