 dy dx, OK? And by the anti-symmetry, this becomes minus the same thing if we replace k by its transpose, k of yx. And now we're going to do something totally stupid. We're going to use the fact that y and x are dummy variables, so we can just interchange their roles. Because this truncation is symmetric with respect to x and y. So this is the same as minus integral where x minus y is bigger than epsilon, k of xy now. But now we have f of x and g of y, OK? And now this is equal to this, which means it's equal to the average of the two, OK? So this is the same as 1 half integral where x minus y is bigger than epsilon k of xy times f of y g of x minus f of x g of y dy dx, OK? And now I'm going to take absolute values, all right? I'm going to bring absolute value all the way in. And when I do that, I'm just going to crudely integrate over all of rn cross rn. And I'm going to use the cauldron-sigmund-size condition. So up to a constant, this is 1 over x minus y to the n. And then here, I'm going to subtract off and add back in f of x times g of x, OK? So this first pairing, what will I pick up? I pick up this first pairing. I pick up y minus x times the l infinity norm of grad f times the l infinity norm of g. And then from this second pair, I pick up the same thing, but with the roles of f and g interchange. Except actually before I do that, I want to do one other thing. Each of these is supported in a ball centered at x naught of radius r, OK? So when I integrate in, let's say, x, I have x naught minus x less than r. And I also have, just by the triangle inequality, x minus y less than 2r, OK? And I put all that together. All of this stuff pulls out. I notice what happens that I've weakened the singularity. By order 1 now, it becomes an integrable singularity. And when you integrate in y, you pick up a factor of r. When you integrate it in x, you pick up r to the n. So this is less than or equal to r to the n times r times grad f infinity g infinity plus the same thing with the roles of f and g interchanged, OK? And that's weak-bonedness, OK? So weak-bonedness is just, should think of it actually as a cancellation condition. There's some weak cancellation implicit in this anti-symmetry. In fact, if we had a convolution kernel, that anti-symmetry would be in and of itself enough cancellation to give you L2-bonedness. But we're in the non-convolution case, so we need a little bit more. OK. So maybe one more comment is in order. Remember the examples that I gave you, the Cauchy integrable and Lipschitz curve, the color on commutators. What you notice is that they actually do satisfy this anti-symmetry condition, OK? If you just interchange the rules of x and y, all you do is introduce a minus sign. So that's worth remarking, right? The Cauchy integrable and Lipschitz curve, the color on commutators associated to Lipschitz function, have anti-symmetric kernels, OK? So that we can make sense of their principal values in this way. And in fact, the associated principal value operator satisfies the weak-bonedness property, OK? About that, maybe I should make a comment. Why does this give you existence of the principal values? It's just dominated convergence, right? Because you could put absolute values in. It says the limit exists. OK. So now another theorem, which in a way can be thought of as kind of a converse to the T of 1 theorem, a strong converse, but actually much easier and preceded it by a number of years. There's a theorem due independently to Petrie, to Span, and to Stein. They all found this theorem independently around the same time. It says the following thing. Suppose that we have a Kolder-Ohn-Ziegmann operator that we know is bounded on NL2. Then T maps L infinity to BMO, OK? Kolder-Ohn-Ziegmann operators are typically not bounded at the endpoints, right? They're bounded on, I mean, what you typically hope to find anyway is that once you get L2 boundedness, you get LP boundedness. But you don't get L infinity boundedness, nor do you get L1 boundedness. You get weak type 1, 1. And this is the naturally associated endpoint result in infinity, OK? And just a comment, of course, just basic Hilbert space theory tells you that if T is bounded on L2, then so is T star, and so then also T star maps L infinity to BMO, OK? So let's see why this is true. All right, so there's one sort of technical issue is for Kolder-Ohn-Ziegmann operators, right, we originally have defined them as mappings from test functions to distribution. So they're well-defined on C's or infinity. Once you know it's bounded on L2, then it makes the operator make sense on elements of L2, but how do we make sense of it on L infinity? So we have to interpret, OK, so for f and L infinity, we have to interpret T of f modulo constants, OK? And this makes sense, after all, because we're trying to show their maps into BMO. BMO is only defined modulo constants, so this is natural, OK, OK? All right, so we need to show that the BMO norm of T of f is bounded, all right? So let f be an L infinity, and we fix a cube, Q, OK? And we need to consider this, OK, at least formally, T f of x minus C Q, OK, where, all right, I've kind of glossed over the definition of this, but it'll kind of become apparent as we go through the arguments, OK, what we're doing here, all right? We need to show that for, is that there exists a constant, depending on Q, such that this is bounded by some uniform constant times the L infinity norm of f, OK? All right, OK, so then we're going to split f into two pieces. F is going to be f1 plus f2, where f1 is f times the indicator of, let's say, 5 times the Q. That's the concentric dilate. And so f2 is obviously living on the complement of 5Q, all right? And so the mean value on Q of T f of x minus C Q is going to be less than or equal to the mean value of Q of T of f1 plus mean value of Q T f2 minus C Q. Now, this is perfectly well-defined, because T is a minor operator in L2, and once we've truncated this to have compact support, f1 is in L2. So this makes sense. This will need to make sense of. All right, all right, so what about the first term? Let's call these a name, 1 plus 2, OK? Term 1's going to be easy. We're going to just use L2 boundedness, which we know by hypothesis. So term 1, by Cauchy-Schwarz, this is less than or equal to the average, the L2 average. And now we use that T is bounded on L2. At this point, just integrate over all of our n. This will be less than or equal to the L2 operator norm of T. We'll have 1 over the measure of Q. And then we'll pick up the L2 norm of f1. But f1, remember, is truncated to live in 5Q. So up to a constant, we're just taking an average of this L infinity function. Of course, that's controlled by the L infinity norm, OK, which is the bound we want for that term, all right? So what about the bound for term 2? For term 2, now this is where we have to make sense of what we mean here. So what we mean is that, OK, so first I have to tell you what CQ is. So CQ is going to be T of f2 evaluated at xQ, which is the center of Q. And of course, that doesn't necessarily really make sense either. But in conjunction, we make sense of them, all right? OK. So then for x in Q, T f2 of x minus T f2 of xQ, what we mean by this is that this is really, we think of this way, this is really the kernel for T, K of xy evaluated at x and y minus the value at xQ and y. And then that's integrated against f2. And this makes sense because of the Coulomb-Zigman-Smuthers condition, all right? This is going to be less than we bring absolute values in, use the Coulomb-Zigman-Smuthers condition. We're going to have x minus xQ to some positive power alpha divided by x minus y to the power n plus alpha. And here, then, we're integrating the picture is this, x is in the QQ, y is outside 5 times Q. So the distance from x to y is always bigger than some constant times the length of Q. So we're integrating where x minus y because of the truncation of f2. We're integrating where x minus y is bigger than a constant times the length of Q. On the other hand, x is in Q, so x minus xQ is bounded by a constant times the length of Q. And then we just pull out the L infinity norm, all right? And this just integrates to give you some constant that depends on n and alpha, OK? So that's actually a point-wise bound for this thing. And then you integrate it, you take its average over Q. Since it's point-wise bounded, then the average is also bound by the same thing that you're done, OK? All right, and I guess that's a good place to stop right of time. Thanks for your attention again. Any questions? Any questions today? OK, we'll have fun at the problem session. See you Thursday.