 integral converges at infinity and since zeta hat of zero is zero and zeta is smooth, then at least zeta hat grows at worst linearly when t is less than one, okay? And so that kills the singularity in this integral convergence. And then it's just a matter of renormalizing to make that the integral equal one, okay? All right. So then in that case this operator qt that we defined to be convolution with zeta t, then qt satisfies the so-called calderon reproducing formula, which is that the integral from zero to infinity of qt squared dt over t is the identity operator in what sense? In a strong operator topology, the space of bounded operators on L2. I'll explain what that means when we prove this. So this thing is called the calderon reproducing formula, right? It's a resolution of the identity, right? Okay. So here's the proof. It's going to be very easy. So what we're saying here, what we need to show in other words is that for f in L2 that integral from let's say epsilon to 1 over epsilon of qt squared f dt over t converges to f in L2 as epsilon goes to zero. All right. So this is an easy exercise in a planche-Rels theorem or in the use of planche-Rels theorem. Okay. So the planche-Rels we have the following. Take the L2 norm of the difference. Planche-Rels theorem tells us that the L2 norm of this expression is the same as the L2 norm of its Fourier transform. So this is the same as integral from epsilon to 1 over epsilon. When we take the Fourier transform, the Fourier transform turns its convolutions into multiplications. And the thing we're multiplying by is the Fourier transform of zeta t, which becomes, I'm going to use the sort of sloppy notation, but I think you know what I mean. It's going to be zeta hat of t times c, and I'm going to abuse the notation since this is radial and right to modulus of c. Since there are two of them, this gets squared since zeta was real and radial, so was zeta hat. Then we have dt over t minus 1 times f hat in L2. Okay. All right. But now you just here make the change of variable t goes to t divided by mod c, and then let epsilon go to zero and use dominated convergence along with this fact. Okay. And then you're done. So now, associated to such a qt operator, we're going to define the vertical square function, which is also known as sometimes little wood-paley g function. Well, the original little wood-paley g function was for a very particular qt, one based on derivatives of the Poisson extension, but this is sort of a generalization of that. Okay. Let's define this follows. g of f of x, which for a given zeta, maybe I'll write it as g zeta of f at x. This is going to be the L2 norm on the half line of qt f of x squared dt over t. Okay. Where qt, again, is say the t star f. Okay. So next fact is the following for zeta as above. And in fact, we can even weaken this to simply insisting that zeta hat of c, well, doesn't even really need to be radial for this part. Z hat of c is less than or equal to the minimum, well, up to a constant, the minimum of c to the alpha, c to the minus alpha for some alpha positive. Okay. Because if c hat has satisfies these bounds, then the analog of this integral is going to converge. Okay. All right. So in other words, integral 0 to infinity of zeta hat of t c squared dt over t, which, okay, this is going to be less than or equal to say some bound mu of zeta uniformly in c. Okay. So then what we're going to say is that the conclusion is that maybe let me do it like this. That the L2 norm of g zeta is less than or equal to mu of zeta times the L2 norm of f. And I'm going to omit the proof. It's actually in the notes, but this is again just an exercise in Pleitrall's theorem. Okay. And use this fact. Mu of zeta is a constant. Yeah. We're saying that uniformly in c, this is easiest to see if this guy is radial. Because if this guy is radial, then you make the change of variable t goes to t divided by length of c. And this isn't even there. And so it really is a constant. But otherwise, with these bounds, it's going to be bounded by a constant. Okay. A uniform constant. Depending on alpha. Okay. Right. So, of course, I told you I was going to deal with non-convolution operators. And the only thing we've done so far is to talk about convolution ones. But here's where we want to make a generalization. All right. So we're going to define the notion of a little wood-paley kernel L, L-p stands for little wood-paley. Okay. So we're going to be talking about a family of operators or family of kernels, psi t of x, y, indexed by some positive parameter t. Psi t is going to map rn cross rn into the complex numbers. And we're going to assume that we have the following two conditions. There's a little wood-paley size condition, which is that... So we're going to assume that there exists some positive alpha and some finite constant c such that we have this. The size condition is that we have this upper bound for the modulus of psi t. And we're going to have a smoothest condition, which is basically a local helicotnuity condition in the y variable only. And you might wonder about the x variable. For L2 theory, we don't need it in the x variable. We just need it in the y variable. To do Lp theory, you want it at least for p less than one. Sorry, for p bigger than one. For p bigger than two, I should say. For p bigger than two, you want smoothest in the x variable as well. But we're not going to worry about that. So the bound should be c times length of h to the alpha over t plus x minus y to the n plus alpha. And this is to hold provided that length of h is less than or equal to t. So a kernel with those bounds satisfies the little wood-paley size and smoothest condition. Or generally, a little wood-paley kernel condition. So of course, these nice zeta t's that were c0 and 3 in complexly supported satisfy this property, but other things do as well. For example, if you take t times a derivative of the classical Poisson kernel, it's going to satisfy those conditions. Actually, the Poisson kernel itself does. But in a moment, we're going to want some kind of cancellation condition, which does not apply to the Poisson kernel, but does apply to its derivatives. This is Poisson kernel in the half-space. I should say. Yes, psi. Yeah. Yeah. Sorry, I think like handwriting is starting to shrink as I go along. So if it starts getting too small, you can't read it. Please let me know. Okay. All right. And now associated to these guys, we're going to define an operator theta t f of x is going to be defined to be integral in our n against the kernel psi t. And I apologize for this because when I start writing rapidly, my thetas and my qs start to morph into each other. I will try to reserve the qt for the convolution guys and theta for the nonconvolution. Hopefully, you'll be able to read the difference. Okay. But in fact, that's not a complete coincidence. The theta t's are going to be generalizations to the nonconvolution setting of these qt's. Okay. All right. So just a remark, note that soup over t bigger than zero of theta t f is controlled point wise by the Herdy-Littlewood Maxwell function of f. And that's just because this kernel, the size condition of the kernel, it's controlled by something that's a rescaling of a radial decreasing L1 kernel. Okay. So this lemma 2.1 applies here, for example. Okay. So in particular, we have uniform L2 bounds, uniform in t for the theta t's. The operator norm, and for me, operator norm means L2 to L2 operator norm. The operator norm of theta t is bottom by some uniform constant, uniformly in t. Okay. All right. So let's see, can I do this in five minutes? All right. So we're going to assume that we have a family of operators that belong to this Littlewood-Paley class. Theta t is going to be defined this way. And we're going to assume further that theta t of one is identically zero. Which means, in other words, that if you integrate this expression with f set equal to one, this is going to equal zero for all x and r in. Okay. All right. Then in that case, the Littlewood-Paley g function associated to theta, and I'll write it out explicitly in a moment, is bounded in L2. Okay. So g theta just means, well, it's this thing but with qt replaced by theta t. And of course, with the constants in the L2 bound, depending only on dimension and these Littlewood-Paley kernel conditions. Okay. So the proof, well, I'll do at least part of the proof here. Okay. So I'm going to set for future reference some expression I'll call i. Well, to be the square of the L2 norm of g theta of f. And the square of the L2 norm of g theta of f, if you think about it, you're integrating an Rn and squaring that thing. That's this guy. You should be inverted the way I wrote this, but you know what I mean. Okay. So at this point, we're going to use this Coulter-Unreproducing formula. Okay. So let zeta be C0 infinity, I'll give it ball, radial, integral of zeta equals 0. And satisfying the Coulter-Unreproducing formula. Which we can always do if this is non-trivial after normalization. Okay. So then this is the same as integral 0 to infinity, integral of Rn. I expand f as the integral of qs squared f integrated from 0 to infinity ds over s. And the integral interchanges with this L2-bondered operator. It interchanges with L2-bondered operators because this convergence is in the strong operator topology on B of L2. Okay. So this is integral from 0 to infinity of theta t of qs squared of f ds over s squared dx dt over t. Okay. And now I'm going to multiply and divide by a certain factor and then use Cauchy-Schwarz in the s integration. All right. So this is less than or equal to integral 0 to infinity, integral on Rn. I have an integral from 0 to infinity of the minimum of s over t, t over s, to some positive power beta to be chosen in a moment. All right. ds over s. And then I have similarly, well, the reciprocal of this will give me the maximum of those guys. And then I'll have this expression squared. And then I have here ds over s and then dx dt over t. Okay. Well, notice that this is just some constant that depends on beta. Right? That's integrable with bound independent of t. Right? Just make a change of variable. Okay. And so what things boil down to is the following fact, which is sometimes called quasi-orthogonality, which is that the operator norm here is, that is the L2-L2 operator norm of theta t composed with Qs. It's less than or equal to a constant times the minimum of s over t, t over s, to the power beta. What I should say is that there exists a beta positive such that this is true. Okay. So I'm running out of time. I basically have run out of time. So I won't have time to prove that. I'll leave it to you to look in the notes and see how it's done. There's also an exercise associated to this. But once you have that, then notice that if I integrate an RN here first using Fubini, I pick up the square of the operator norm there, which gives me this factor but with the minimum and with power two beta. And that beats the maximum, and then this just converges. Okay. And then the result follows. All right. And since I'm out of time, I won't write down the details, but again, it's in the notes. Okay. So at the limit of the time, so let me stop. Thanks for your attention this morning. Are there any questions? You can save your questions for Zui. All right. Thank you.