 that T satisfies the weak bond in this property and T of one and T star of one belong to BMO. So for now, once again, the convention T star is just a transpose. It's not Hermitian adjoint. There's no complex conjugation at this point, okay? And the second thing is that T is bounded on L2. All right? So a couple of remarks are in order. So let's try to put a little perspective on this theorem. Remember that we have this theorem of Petri, Span, and Stein. I don't remember the numerology of it now, but there was only one theorem with those three guys. All right, which said that if T is bounded on L2, then T maps L infinity into BMO. And in fact, of course, just by basic Hilbert space theory, if T is bounded on L2, so is T star, and therefore also T star maps L infinity into BMO, okay? On the other hand, suppose that you don't necessarily know a priori that T is bounded on L2, but you do know for some reason that T and T star map L infinity into BMO. Well, dualizing the estimate for T star, then you also get that T maps Hardy space H1 into L1. And then interpolating, there's a version of interpolation due to Phefermann and Stein where BMO can substitute for L infinity as an endpoint and H1 can substitute for L1, and you pick up all the LPs in between. In particular, you pick up L2. So then you get that T maps L2 to L2. Okay? So what's the T of one theorem saying? It's saying that you don't have to test T and T star on all of L infinity. You only have to test them on a very, very special function, which is the constant function of one, okay? All right. Well, except with one additional, but as we've seen sort of mild hypothesis, which is the weak-boundness property, right? Remember, we saw that weak-boundness automatically holds if you're looking at the principal value operator associated to an anti-symmetric kernel, which our examples often happen to be. In fact, for the examples that we've discussed, Cauchy-Energy operator and the Color-owned commutators, they are anti-symmetric. Okay? All right. So the second remark is that the direction two implies one, right? That the L2-boundness implies the first thing. We already know this, okay? Because Petrie-Spanstein tells us that T and T star map L and 30 to BMO. So in particular, they map one into BMO because one is an L and 30. And we've also seen that L2-boundness trivially gives you weak-boundness property just by Cauchy-Schwarz, okay? Okay? So the real content of the theorem is to prove that one implies two, all right? So last comment before we go to the proof. There's an issue. If you don't already know boundness from L infinity to BMO, then what do we mean by T of one? Well, this is one of the exercises, but I'll just say a brief word now, that you define T of one as an element of the distribution's modular constants, which means, in other words, that it's in the dual space of what I'll call D zero. D zero is gonna be the set of psi and C zero infinity with mean value zero, okay? And you can make sense then by sort of mimicking when you test T of one against such a psi, that means because psi has mean value zero, you're gonna be able to subtract an appropriate constant, okay? And then you sort of mimic the proof of the Petrie-Spanstein theorem, all right? And that's how you prove it. And there's just a little bit of work, but not much to verify that the thing is actually well-defined, right? The idea with Petrie-Spanstein was to split your function F into a local part plus a far away part. And somehow you need to get well-defined in this, you need to verify that really what the thing that you're defining is independent of the particular splitting that you make, but that's not hard, okay? And that's the exercise, all right? So let's proceed to the proof, all right? So, we're gonna use Littlewood-Paley theory, all right? So we're gonna introduce our usual sort of nice Littlewood-Paley operators that give us a color on reproducing formula. So let QSF equals Zeta S star F, Zeta S as usual as to the minus N Zeta of X over S. I'm not sure why I'm using S. I'm gonna be switching to T in a moment, but I hope that's not confusing. We wanna have Zeta in C0 infinity of, well, just for a slight amount of convenience, I'll choose it to be sporting the ball of radius a half rather than one that doesn't really matter. And with mean value zero, and we're gonna choose Zeta to be real and radial. And non-trivial, of course. And then as we've seen in those circumstances after a normalization, we get the color on reproducing formula, all right? Okay, for that family of QSs, okay? All right, so now I'm gonna define an operator PT, which is gonna be this guy, all right? And it turns out, well, first of all, notice that as T goes to zero, this guy is approximating the identity, at least in L2, because that's the meaning of this statement, okay? But one can actually say more. I'm gonna omit the proof here. It's in the notes. It's just a sort of elementary real variable argument with a, based on some changes of variables, and using the fact that you're dealing with a radial guy, okay? What you can show is that PT, in fact, is a standard modifier, a nice approximate identity, okay? So in fact, PTf is Ft star f, Ft defined in the usual way, with phi in C0 infinity, real, radial, supported in the unit ball. That's why I made this support a half, because then this was in the unit ball, all right? If we made that one in the unit ball, this would be the ball radius two. Doesn't really matter. And the integral of phi is one, okay? So you get a very nice standard modifier, nice approximate identity, okay? And as I say, I'll omit the proof here to save time, but you can find it in the notes and it's not hard, okay? So a couple of further sort of easy remarks about this PT. Of course, we can see because PT is a standard modifier, we know that it converges to the identity in strong operator topology on B of L2, okay? We also have that PTf converges to F point-wise that PTf converges to F point-wise almost everywhere for F in L2, say. And in fact, everywhere for F continuous. But moreover, we also, and again, this is as T goes to zero, but also that PT of F converges to F in D for F in C0 infinity. That's routine to check. And one more thing, which is that the limit as R goes to infinity of T of PR of F, PR of F, PR of G, I should say, is zero for F and G in C0 infinity. And this is a consequence of the weak boundaries property and the fact that PR, this operator PR, is mapping L1 to L infinity with norm R to the minus N, okay? You can buy that fact with the weak minus property in. Yeah, you get it, okay? Again, the details, well, I don't know. Maybe I didn't put the details in the notes, but it's not hard. I guess it's an informal exercise. It's not on the official exercise list, okay? So if we combine these facts together, what does that say? It says that, okay. So let F and G be in C0 infinity. And our goal is to show that if we make this pairing, initially defined in the sense of distributions, that there's some uniform constant depending on the dimension and the color and Ziegman's estimates for the kernel and the BLMO norm of T1 and T star of one and the constant of the weak minus property, you know, all that usual stuff, then they should be bounded by the L2 norm of F times the L2 norm of G. Because if you can do that for all F and G in this dense subspace of L2, then you would just make an extension by continuity and that T extends to a bounded operator in L2, okay? So let's look at this. Because of what we've said here, we can obtain this thing as a limit as epsilon goes to zero of the integral from epsilon to one over epsilon of, let me say it this way, d by dt of pt t pt f g dt, right? Because you just integrate out and use these facts. Oh, and one other fact that the conditions on pt make it self-adjoint. So putting the pt here is no different than putting the pt here, okay? Right? Oh, I think I'm missing a minus sign, right? There should be a minus sign. Because we get the identity on this end and zero on this end. So now we take this derivative and notice what we get from differentiating this integral here. We get that this equals, limit as epsilon goes to zero, integral from epsilon to one over epsilon of qt squared t pt f g dt. Dt over t plus the same thing, let's see, the same thing with a pt here, t qt squared f g. And again, the measure is dt over t, okay? All right? Now I claim that these two terms are essentially of the same form. Okay, so let's give these a name. Let's call it one epsilon of f and g plus two epsilon of f and g, okay? So for example, let's look at two epsilon of f and g. All right, as we said, the pt is self-adjoint. So you can bring it over here. You can bring t over, it becomes t star, and you can bring one of the qt's over. The qt's are also self-adjoint, they're real and radial, okay? And so this becomes integral epsilon to one over epsilon qt f qt of t star of pt of g dt over t. But notice that if we brought the qt over here, this is the same, exactly of the same form as this term, only interchanging the roles of f and g and the roles of t and t star. But everything is symmetric here. So if you can handle one, you can handle the other, okay? All right, yep, yeah, yeah, it's not hard. Yeah, it's because, yeah, it's in the distribution sense because the pt is also given by a C0 infinity thing, right? And so you're still, and whether you have pt or the derivative of pt, everything that you have here is always of the form t applied to a thing in d, paired with a thing in d, and it's easy to justify using that, okay? So therefore it's enough to treat the term one, okay? All right, so it's enough to treat the term one epsilon, and in fact, it's enough to show the bound that we're after for one epsilon as long as it's independent of epsilon, okay? So our goal now at this point is to show that the sup over epsilon bigger than zero of one epsilon of fg is less than or equal to some permitted constant times the L2 norm of f times the L2 norm of g, okay? So then let's look at that guy. Again, we're gonna move one of the qt's over onto the g and then notice that this pairing is no longer merely a distributional pairing. It's actually, it can actually be understood as an integral, okay? So that's because right when you modify a distribution, you actually have a function, okay? So one epsilon of fg is equal to integral from epsilon to one over epsilon integral on rn of qt, pt of f of x integrated against qt g of x, dx dt over t. And we're gonna bring absolute values all the way in at this point and then use Cauchy-Schwarz. And when we use Cauchy-Schwarz, we might as well, once we bring the absolute values in, we might as well integrate from zero to infinity. So for sure, any bounds we get will be independent of epsilon. And here we have qt t ptf squared dx dt over t to the one half times the same thing with qt g of x squared. And let's give these a name, let's call them a times b, okay? And so notice that term b by the basic Littlewood-Paley theorem for very nice convolution type qt's. I guess that was what? Proposition 2.6, that will check me on the numerology, but I think that's right. We get that b is bounded by the L2 norm of g, okay? Right, that's immediate. So what we've boiled things down to is that a is less than or equal to the L2 norm of f. So notice what we've got here. What we've got is a square function. And if I call this thing qt t pt, if I call this thing theta t, it is sort of a familiar appearance to what we've been working with. All right, so in other words, we need to show this where theta t is qt t pt, just the composition of those three operators. So at this point, you can probably guess where this is going. We're gonna prove this by using the t of one theorem for square functions. All right, maybe I should give this a name, let me call this thing star, okay? So to prove star by the t of one theorem for square functions, this was, I think, again, not super confident about the numerology, but I think it was 3.1. What we need to show are two things. First, that this guy has a kernel, which satisfies the standard little wood Paley kernel conditions, okay? So it's enough to show that, one, theta t has a kernel, psi t of x, y, in the little wood Paley class, all right? So we have a little wood Paley family, I mean, all right? And second, that the measure d mu of xt, given by theta t of one of x, squared dx dt over t is a Carlson measure, right? Those were the two hypotheses of that t of one theorem for square functions. Once we have these two things, we're done. All right, now the second one is essentially immediate, because recall that pt of one equals one, again, because the kernel pt had integral one, okay? And also, t of one is in BMO, all right? So therefore, qt of t of pt, so let me write it this way, theta t of one is qt t pt of one, pt of one is one, so this is qt of t of one, but this is in BMO, okay? So therefore, the Carlson measure estimate is immediate now from the Pfeffermin-Stein lemma that you proved as an exercise, okay? And I think, if I remember the numerology, there's 222 maybe in the notes, okay? So that part's easy. Part one is longer and more complicated, I think because time is short, I won't go through the details of that, but they are in the notes, but let me just make a couple of comments, okay? So what you want to do, okay, maybe I should tell you what the kernel looks like, all right? So you split into two cases. First, where the distance between x and y is say at least 100 times t. And second, where the distance from x to y is less than 100 times t, okay? Now in this case, we have separation and we can write out explicitly what the kernel looks like as an absolute conversion integral, okay? So this kernel is gonna be, well, it's a convolution of qt with t of pt, all right? So this will be zeta t of x minus z, say. Then integral of k of let's say z and v, the kernel for the singular integral, and then the kernel for pt, which is ft of v minus y, and then we have dv and then dz, okay? Right, because if you apply this to a function f integrating in y, this will give you pt of v, applying this gives you t of pt of, sorry, pt of f. Applying this, integrating here gives you t of pt of f and integrating here gives you qt of t of pt of f. And this is legitimate because notice that if x minus y is bigger than or equal to 100t, then z minus v is bigger than or equal to, let's say 98t, right? By the triangle inequality since each of these guys, zeta t and vt in particular support in a ball of radius one, or ball of radius t, okay? All right, and so we have separation here and this is an absolute conversion integral, right? The kernel is away from its singularity. Okay, and then what do you do? Okay, you do a thing that we've done a few times now, use the cancellation here, which allows you to subtract off so we can replace k of zv by k of zv minus k of xv, all right? And then use the Kolder-Ohn-Zigman condition and you go through. The smoothest condition in the little, that'll give you the little wibbilly size condition. The smoothest condition, you need to only verify it in the y variable and so then you just work with the gradient of phi t instead of phi, and then, and again, you get things rather easily. Okay, again, the details are in the notes, but that's the idea. For case two, the idea is somehow to use the weak-bounderness property, right? You're not, in case two, you don't get to write this out as an absolutely conversion integral, right? Because you don't have separation, you're not away from the diagonal there, okay? So I'll just say briefly that in case two, use weak-bounderness, okay? And again, the details here are in the notes, all right? And that's the proof of the theorem, okay? So except for this kind of, you know, not deep and difficult but technically messy business of verifying the kernel conditions, it actually turned out to be a f-