 I hope this is a stable arrangement here. All right, so thank you all for coming. Thank you all for showing up in such numbers at 8.30 in the morning. Thanks to both the organizers and the staff here at PCMI for inviting me to do this. It's a pleasure to do so. So, as Tatiana said, this talk is gonna be in some sense of an introductory nature. It has to do with finding criteria for to establish the L2 boundedness of singular integrals and square functions that are not of convolution type. In the convolution case, there's a standard tool for establishing L2 boundedness. You just use the Fourier transform in Planche-Roll's theorem. All right, that's not available in the nonconvolution case, so other techniques had to be developed. So we're gonna talk about some of those criteria and if I move expeditiously enough, then we'll show some extra applications as well. By the way, the prerequisites for the course in principle should be the basics of harmonic analysis, things like knowing about the Fourier transform in Planche-Roll's theorem, knowing about the Hardy-Littlewood maximal theorem, the basics of the Calderon-Ziegman singular integral theory in the convolution case. If you're not familiar with those things, that's not a disaster, but it means probably there are some things that you'll need to kind of take for granted and fill in the gaps later, okay? So first, let me start with some examples, okay? So the first example, sort of a fundamental example, is the Cauchy integral on a Lipschitz curve. Is that big enough? Can everyone see even in the back? It's okay? Good, good, okay. So the Cauchy integral is defined as follows. So we're gonna have a curve gamma, which is gonna be a Lipschitz curve in the complex plane. So it'll be the set of X plus IA of X, X on the line, and A is Lipschitz, okay? And then the Cauchy integral operator is defined to be C gamma F of Z for Z on the curve. It's gonna be the principal value. I'll say what that is in a second. There's a normalizing constant, one over two pi I, and we're integrating on the curve. And we have F of V Z minus V DV, okay? Sometimes that's normalized with a minus sign. It doesn't matter too much, all right? What does this principal value mean? Well, we'll say a little bit more about that a little bit later. It basically just means though that you're interpreting this in some kind of limiting sense, sort of like an improper integral from first year calculus. The reason being that you'll notice that the order of the singularity here is not absolutely integrable, right? It fails to be absolutely integrable. So that's one reason it's called a singular integral. Ah, I guess that's the main reason it's called a singular integral. So you can't just write this down as an absolute conversion integral. You have to say what you mean. We'll say what we mean in a little bit. But just for now, think of this as being some kind of improper limiting operation, okay? Now, if we write this in the, if we parameterize in these graph coordinates, we have that c gamma f at the point x plus i a of x is gonna equal, filling up my board here. Okay, let me write the integrand over here. So the integrand is gonna become one over x minus y plus i times a of x minus a of y. And then we're integrating f of y plus i a of y. And then the complex measure dv becomes one plus i a prime of y, d y. And we're gonna give this object a name too, this parametric thing. We're gonna call this, let's say, c sub a of g of x, where g of x is, well, let me say it this way, g of y, is gonna be this stuff with this factor absorbed in. And this is a harmless factor because a is Lipschitz, therefore, a prime is bounded. And because this factor is harmless, that means that L2 boundedness of this operator on the curve gamma with respect to the arc length measure on the curve is equivalent to L2 boundedness of this operator on the line, okay? Of course, there are many reasons that this might be of interest. Of course, if you've had a course in complex analysis, you know, several reasons why the Cauchy integrals of interest. It's also of interest for other reasons, which we probably won't have too much time to get into. But it turns out that once you have the L2 boundedness of this guy, there are transference techniques to let you get L2 boundedness for a huge class of singular integral operators that are not of convolution type. And in particular, there are transference methods that allow you to extend up into higher dimensions. There's some, I'm not gonna take time this morning to mention more examples along that line, but in the type lecture notes, which have been posted, there are additional examples. All right, so related to these are the so-called cauldron commutators, okay? So these are gonna be, again, A is gonna be this Lipschitz function. This is defined to be the principal value, one over two pi I, integral of the line. And again, there are higher dimensional analogs, but we won't say anything about that this morning. I'm not budgeting my space on the board very well. And then, of course, then we have F of y dy, okay? So for example, notice that the first commutator, you might wonder why do they call it a commutator? Well, C1A is at least formally one can make sense of this with some normalizing constant, which I guess should be minus one over two I. This is the commutator of the Hilbert transform with multiplication by the function, ah, not the Hilbert transform. The derivative composed with the Hilbert transform, take the commutator of that with multiplication by the function, a, and then apply it to the function, f. Well, this is the operator, okay? Culdron was interested in this particular object because it was the fundamental building block for his work on building algebras of singular integral operators in order to build sort of a pseudo-differential calculus to treat partial differential operators that had minimally smooth coefficients, say, Lipschitz coefficients. The standard pseudo-differential calculus treats see infinity coefficients, but Culdron was able to build a sort of pseudo-differential calculus to treat the case of minimally smooth coefficients and the L2 boundedness of this was, of this guy was sort of the fundamental tool that he needed in order to do that, okay? So let me point out also a connection between these operators and this Cauchy integral operator, okay, notice that if we look at this kernel here, the kernel for this operator that I'm calling C sub a, okay? All right, so what you notice is that one over x minus y plus i times a of x minus a of y, if we factor out one over x minus y, then we've got what's left is one over one plus i times a of x minus a of y over x minus y, which at least formally, at least if the Lipschitz constant say less than one, you can expand in a power series and write this as one over x minus y times the sum, k runs from zero to infinity of minus i times a of x minus a of y over x minus y to the k, which means that this operator, this operator Ca can be expanded in a power series in terms of these guys, because notice that, well, ignoring this harmless factor of negative i to the k power, this is exactly the kernel that we have for these cultural commutators, okay? Yep, they can be, you can interpret them as iterated commutators, yeah. If you wanna see the exact formula, it's actually in the notes, that's the published notes, or in the posted notes. Yeah, a good question, but yeah, in fact, yes. Okay, so this leads me to a couple of definitions. We're gonna give a definition of a class of operators that includes these operators as special cases and of course many more. So first of all, a Calderon-Zigman kernel, and for short, I'm gonna abbreviate that Cz, is a function k of x, y, defined on the product rn cross rn minus the diagonal. You'll see why we need to take away the diagonal in a second, such that these two conditions hold. There's a size condition, which is that k of x, y is bounded in modulus, oh, and we're, let's say, complex value. Okay, so this is an rn, so there's some uniform constant, such that this is bounded by one over length of x minus y to the n power, and then there's a smoothest condition, which is that k of x plus h, y minus k of x, y, and the same for y plus h, right? So there's smoothest at x and y. The size of this is less than or equal to uniform constant, length of h to the alpha over x minus y to the n plus alpha. Here alpha's some positive number between strictly positive and less than or equal to one, and this is to hold whenever h in modulus is less than or equal to, let's say, well, two times length of h is less than or equal to length of x minus y. By the way, there's a typo in the notes. The length of the vector was omitted, that doesn't make sense, okay? Okay, so you can see that in the case n equals one, these kernels that we're looking at have this property, right? Because a is Lipschitz and it's easy to check that this kernel and these kernels here satisfies those conditions with n equals one, and with alpha equals one because, again, a is Lipschitz, okay? And then we say that, here we're defining the notion of a generalized Kaldron-Ziegman operator, CZO for short, and I'll sometimes also call these SIOs for singular integral operator, is a mapping T from test functions to distributions. So here, remember, D is the space C zero infinity of Rn and D prime is the, that should be a D prime there. D prime is its dual, okay? Which is associated to a CZ kernel in the following sense. In the sense that for all Fg in C zero infinity with disjoint supports, the pairing of Tf with g, which makes sense, g is a test function, T maps test functions to distribution, so this is a distribution. So this just means the pairing of an element in D with its dual space, okay? All right, so this is going to be equal to the integral of this kernel integrated against F of y integrated against g of x, dy of x, okay? And the point here is that if these guys have disjoint supports, then this integral makes sense because that keeps us away from the diagonal where this kernel has a singularity, okay? So notice formally what we're saying Tf of x is equal to the integral in Rn k of x y f of y dy, all right? And this actually makes sense if x is not in the supported app, okay? So what we're going to be striving for among other things are boundedness in this course and we're going to be bounding this criteria to deal with operators of this type, not necessarily a convolution type, but which satisfy these kind of standard Calderon-Zigman size and smoothness conditions, just like the kernels of the Koshy integral operator, for example, okay? All right, so let me say one more thing on this topic before we move on to the next topic, which is let's recall the classical theorem of Calderon-Zigman getting back to 1952, which is that if T is a Calderon-Zigman operator, of course they didn't formulate it with that language. And if T is bounded on L2, what do we mean by boundedness on L2 when initially this is only defined on test functions? It means that this pairing in absolute value is controlled by a uniform constant times the L2 norm of F times the L2 norm of G. And since C0 or infinity is dense in L2, then you make an extension by company way. All right, so the theorem is that if the signal integral is bounded on L2, then it is bounded or extends to a bounded operator on LP, one less than P less than infinity, and it's also of weak type one one. So of course in 1952, Calderon-Zigman were only looking at convolution type operators, but in fact their proof works the same way. It's the same proof. Convolution is really only an extra help in proving the L2 boundedness when you can use plantarole. Once you have L2 boundedness, LP bounds work the same way whether it's convolution type or not. It's just an artifact of these Calderon-Zigman kernel conditions plus the Calderon-Zigman method. Okay? All right. So the fundamental tool for us as we go through this course is gonna be little wood-Paley theory, by which I mean the analysis of square functions and the use of square functions to infer boundedness of other operators. Okay? So for the rest of today, we're gonna talk about little wood-Paley theory. In fact, I'll maybe call this section generalized little wood-Paley theory because at some point, I want to be talking about little wood-Paley square functions that are induced that are not of convolution type. Okay? The classical little wood-Paley theory, of course, is with convolution operators. Okay? So first, let me recall a couple of standard facts from basic harmonic analysis. Okay? We're gonna need a couple of tools from approximate identities. So here's the setting. We're gonna let, I suppose we have a function phi that's in L1 of our N, radial and decreasing. Meaning radially decreasing, decreasing as you move away from the origin. Okay? All right? And we're gonna set phi sub t of x. This is in our N. So the scaling will be t to the minus N phi of x over t. And let me recall also the hardy little wood-maximal function, which is the following thing. It's enough of x is defined to be the soup over all t bigger than zero of the average over the ball of radius t centered at x of the absolute value of f of y dy. Okay? And of course, as you probably know, we can also use cubes in place of balls. And we can also work with non-centered balls or non-centered cubes. They all give equivalent operators. Okay? All right? So then I wanna state a lemma whose proof is gonna be left to you as an exercise. And I'm gonna try to use the numerology that's actually in the type notes. So this is a limit 2.1. All right? So you can find it in the notes. Okay? So let phi be as above, okay? So let's call one radial decreasing. And suppose that g is some other measurable function which is whose absolute value or modulus is bounded by phi point-wise. Okay? All right? Then in that case, if we convolve gt, gt is defined the same way as phi t. If we convolve gt with a function f, this is gonna be point-wise bounded by a dimensional constant times the l one norm of phi times the hardly loaded maximal function of f at x point-wise. Okay? And the proof, this is safe for all f and lp, one less than or equal to p less than or equal to a. All right? And the proof is an exercise. And there's a hint in the notes to do the exercise. Let me just make one brief comment. This is actually true with constant one here. The argument is a little more subtle. For our purposes, having a purely dimensional constant doesn't bother us at all. And in that case, you can prove this by sort of elementary methods. Just making dyadic angular decomposition and just estimating things and adding them up. Okay? All right, and then a second related lemma. All right, so let phi and g be as in the previous lemma. Okay? And let's also suppose that phi of zero is finite and that, let's say g is continuous. Don't really need that, but it makes life slightly simpler. Okay? So then for all f and lp, one less than or equal to p less than or equal to infinity, we have that the non-tangential maximal function, which I'll explain to you in a second what that is, of gt star f at a point x. This is by definition, this soup over y, should be bigger, y and t such that x minus y is less than t. And there's nothing special about making the aperture to the comb be one. You could have any fixed aperture if you like. Okay, and then the constant's gonna depend on the aperture. And then we look at gt star f of y. This is gonna be bounded also by a dimensional constant times phi of zero plus the l one norm of phi times the hardly little maximal function of f. Again, I think I will, well, not exactly leave it as an exercise, sort of an informal exercise, because actually the proof, the proof of this is in the type notes. So you can refer to that if you like, but it's basically sort of an elementary thing. So I'm not gonna prove it. Yes. No, we're taking the, it's a non-tangential maximal function. So think of a point x on the boundary and then we're looking at arbitrary y and t, taking a supremum over all y and t in the cone. Okay? So, so right, if the point y and t satisfies this inequality, then it's in this cone, right? That's what we're saying. That's why it's called a non-tangential maximal function. Okay? Okay. Let me remind you of the definition of the 48 transform, let's say for, initially for f and l1, f hat of c is defined to be integral on our n, e to the two pi i c dot x. Some people normalize it with a minus sign, it doesn't really matter, of f of x dx. And there's a famous theorem of Plancherelle, which I've already mentioned. The 48 transform extends to not just a bounded operator, but an isometry on l2. Okay? All right. So the first thing that I'm actually gonna prove for you, in the notes, this is proposition 2.4. This is gonna be easy. So let's let zeta be in C0 infinity on the uniball in our n. We're gonna define an operator qt of f is zeta t star f. Zeta t defined the same way we defined phi t earlier. It's t to the minus n, zeta of x over t. And here, besides being C0 infinity, we're gonna assume that zeta is real and radial and the integral of zeta is zero. Hence the integral of each of the zeta t's is zero. Okay? And we're gonna assume that, oh, I should have said it's also non-trivial. Non-trivial so that therefore, zeta hat is not zero, not identically zero. And so if it's non-trivial, we can normalize it so that the integral from zero to infinity of zeta hat of t squared. What am I doing here? I'm abusing notation because if zeta is real and radial and invariance properties of the Fourier transform tell us that also zeta hat is real and radial. And since it's a radial function, I'm kind of abusing notation here and identifying it with a naturally associated function of the real line. So we're gonna normalize so that this is one. Okay? All right, oh, why do we know that this is finite? Well, because, okay, since the integral of zeta is zero, this says that zeta hat of zero equals zero if you just think of the definition of the Fourier transform. And if zeta sees your infinity, then zeta hat is a Schwarz function. So infinitely smooth and rapidly decaying. Since it's rapidly decaying, this