 d bar s, and suppose f of s is compactly supported between minus 1 and 1. If this is an L1 function on the interval from minus 1 to 1, this behaves like 1 over z, which does not belong to L1. So it's obvious. It does belong to weak L1, but let me not go there. OK. Now, the next thing is, as I mentioned, and as we draw on the board, we're not just interested in Riemann-Hilbert problems where the contour is the line. We're interested in Riemann-Hilbert problems on contours, which have points of self-intersection. There can be many of them. They can be disconnected, and so on. So we would certainly want something like the following. Suppose this is the contour, the part of the contour here, which is sigma 1. And this is the part of the contour, which is sigma 2. What we'd like to have, so the integration is taking place over sigma, we would like, for example, let's suppose that this vector f here, this function f here, is just supported on sigma 1. And let's take z now to belong to sigma 2. So I've got this picture here, like this. z is sitting here, and f is supported here. If this operator there is going to be bounded in LP on this contour, we would certainly need to know that if I look at f of z on sigma 2, this should be bounded, or this is cf on sigma 2, should be bounded by f on sigma 1. That would be a consequence of my hope that the operator c is in fact bounded on the whole contour. Now there's a very nice calculation which one can make again using Fourier theory, but now using Fourier theory of the multiplicative group on the positive numbers, which is the same thing as the Mellon transform. And so it's a very nice exercise to show the following, just to give you a flavor of it. Take the line 0 infinity, and take a line here going off at an angle theta. And we have theta f at r is integral 0 to infinity f of s d by s upon s minus r z hat, where z hat is e to the i theta. So I'm just computing this object where sigma 1 is the real line, the positive real line, and sigma 2 is the real half line going off at an angle theta. Then you treat this as a group, a multiplicative group, with a Fourier transform which is given by the Mellon transform. That will diagonalize this operator here, just as the ordinary Fourier transform diagonalized because you operate on the line. That will turn your operator here into a multiplication operator. And what you find is that c theta f in L2 is less than or d r, some constant in theta if an L2 of s. And again, you can turn this into instead of 2, you can put an LP here by other kinds of arguments. But I just wanted to give you this as more or less as an exercise and use Mellon transform. This gives you a little bit of the flavor. Now why are how the many ways that this is useful? So let me give you an application of this which begins to take us towards the question of in what sense do we achieve boundary values? And when will those boundary values, when will the function not just have boundary values but when will it be continuous up to the boundary? So observe the following. Suppose f belongs to h1 of the real line. So that is f in L2, f prime in L2. Well, this is the distribution derivative of f. And suppose I take the derivative of d by dz of cr of f of z. So it's d by dz integral over r f of s upon s minus z d by s. So I can put the derivative on the z. I can transfer it over to the s. I can integrate by parts. And this comes out to be f prime of s d by s upon s minus z. So this is just cr of f prime. Now why is this useful? Consider the following. Let me take, and what we know then, suppose I've got two points, z prime and z double prime in the upper half plane. Then I can draw a line through them going through some point to which I'm going to call the point 0. So now I've got an angle theta here and an angle pi minus theta over here. And suppose I want to look at cf at z prime minus cf at z double prime. So that will be the integral from 0 to infinity of f of s ds of s minus z. Or let me first of all take it like this. OK, so let me do this calculation better. So this will be the integral over this line from z double prime to z prime of d by dz of cf of z dz just by integrating along. But this object here, if I now take absolute values, will be bounded by z prime to z double prime to the half times the integral along this entire line from 0 to infinity, say, of d by dz of cf squared dz to the power half by the Schwarz inequality. But this object here is going to be bounded double prime of 1 half f prime on L2 of r because of this fact here. If I look at the derivative of this object here, it's the Fourier transform of an L2 function. So what that means is that this object over here on this line given by there is going to be bounded by the L2 norm of this function here. So I've got my function defined here and everywhere in here. It has an h1. It's an h1. It has a derivative. So that means that if I take cf of that function, it will give me a function over here which has a derivative in L2. So I get that bound. And so what's the conclusion from here? Is that a priori, a priori, the function cf in the upper half line is uniformly and globally holder half. In particular, it has boundary values in the continuous boundary values. OK, so that's a little bit about how boundary values arise. Now, so I have gone through this to just give you some sense of what you can pick up from a first course in complex analysis or theory of bounded analytic function. So now we begin to get to the real stuff. So a composed curve is a finite union of arcs in the closed in the Riemann sphere, which can intersect only at their end points, things like this. And an arc, let's call arc gamma, is phi of t, where t runs from, say, a to b, and is homeomorphism on its from the interval onto the arc. Now, we allow for arcs to go through in infinity, and you use the usual topology on the sphere to control that. And it is curves, a composed curve are going to be the contours on which you do Riemann Hilbert theory. So we always are going to be look at a composed curve. Because they are given in this particular fashion, they are automatically oriented. They come with an automatic orientation, which is prescribed by the parameterization. Now, again, in this picture, we think of a circle, say, as being a composed curve, because we always think of it as a union of two arcs. That's convenient. So the first question having said what we have, so these are going to be the curves we do Riemann Hilbert theory on. So what do we mean by LP? It's the first question. So what we do is we introduce arc length. Suppose I've got some simple contours, a simple arc sigma. And I've got a succession of points, C0, Z1, all the way up to Zn. These points have an ordering which is induced by the parameterization. And we look at the quantity of Z0, Zn. For any choice, any partition of the curve as being the supremum over all partitions of the sum of Zi plus 1 minus Zi from 1 up to n. And if this quantity is finite, we say that the curve is rectifiable. And we're going to be interested in composed curves which are locally rectifiable, which means that if I take my contour sigma and intersect it with any ball, say Z is less than r, then the intersection of sigma with r must be rectifiable. So that means you could have, here's my ball. And I look at the curve going around like this with different pieces like this. When I intersect with the ball, I'm going to get a countable number of arcs. And I want each of those arcs to be rectifiable and their sum to be finite in any ball. So the first definition is we're going to look at Riemann-Hilbert problems on contours which are composed curves and which are locally rectifiable. Now, for a rectifiable or locally rectifiable curve, we can introduce arc lengths. For example, mu i of this interval is just to be the length from alpha to beta. Curve goes from L. Now, things are very similar to just the construction of the Bayek measure. And such sets, semi-open intervals form a semi-algebra. And hence, by the usual extension theory, you obtain a complete measure on a sigma algebra which contains the Borel subsets of that contour. In that way, you get a measure theory and you get a measure which enables you to integrate on each one of these curves. So then you've got an LP theory. So you get this immediately implies. We have semi-algebra, which implies a measure on a sigma algebra, and algebra, et cetera. So all of this is a standard construction. And then you get LP theory, which would be measurable functions with respect to that measure, with respect to that sigma algebra. And you take the LP integral of that finite and so on. It's absolutely analogous to what's going on. Just to remark, this is a little exercise. d mu is equivalent to a Hausdorff one measure. If you were thinking about it, so you can also put measure on the curves by regarding and by defining a Hausdorff one measure, it's exactly the same quantity that you get. OK. Now, we come to, and once you have this, you can define CH on your contour of Z as being integral of f of s ds upon s minus z over the contour. And the reason that you can do this now is that you take your parametrization of the curve, so s of sum t, you substitute it in here, and then you end up with an integral with respect to the arc measure. So this is the usual way in calculus where you have line integrals, which are expressed in terms of some parametrized integration. So all the natural things you can do with respect to these curves because you have a measure on these curves, you can integrate, you can have line integrals. Now comes to the most important thing of all. We want to be sure that this operator here has got all the properties that we saw that the Cauchy transform had when we're just looking at the Cauchy transform on the real line. Do we need to assume anything else about the contour sigma in order to conclude that? Well, we'll have that C plus minus C plus f at Z equals C minus C f at Z is equal to f of Z. Can't turn any f belonging to Lp, where p now is bigger than 1 and less than 1. Notice we include 1. This still remains true. And similarly for the Hilbert transform. But what's missing so far, although that exists as a point-wise limit, so let me just take a step back, skipped over something. Just on the assumption that the curve is rectifiable or locally rectifiable, you know the following. That because it's rectifiable, at almost every point there'll be a tangent vector at almost every point, which means you have a normal and which means you have a cones. And this object here will have boundary values almost everywhere, which are non-tangential. And this is true for all p bigger or equal to 1 and less than infinity. So that part of the theory, so we have non-tangential boundary values, boundary values almost everywhere for f belonging to Lp, p bigger than 1 and less than infinity, as just as long as the contour sigma is locally rectifiable. So this much of the theory goes through completely in this way. And you get C plus minus C minus. What you don't know at this point is that the operator C plus and C minus are bounded in any function space. Now, there is an absolutely remarkable theorem in analysis, which in many people's view is one of the major achievements of the latter part of the 19th, of the 20th century in analysis. And that is there are necessary and sufficient conditions on a rectifiable contour in order that C plus and C minus and hence the Hilbert transform. Again, you'll have C plus. Minus will be plus or minus a half f plus i times the Hilbert transform of f. But again, hf at z is i 1 over pi times the limit as epsilon goes down to 0 of s minus z bigger than epsilon f of s d bar s upon s minus z. And this isn't the integral over the contour, so it's exactly the analog of what you'd expect. But now let me repeat that again. What we don't know is that the Hilbert transform of C plus or C minus is a bounded operator in LP. And there is a theorem which begins with Corke-Colderon in the 1950s, 1960s, then was developed by many people, Kofman, Mayer and McIntosh, and then eventually put in its final form by Guy David, which gives necessary and sufficient conditions, necessary plus sufficient conditions on C plus minus or h to be in the bounded operators in LP. So what are those conditions? So you take your contour. You take a point z. And you take a ball of radius r. And you intersect the ball of radius r upon z with the contour. So let me expand it out a bit. And here's the point z. Here's the ball. And your contour goes around like this and comes out like this. So you take the ball of radius r and you intersect it with the contour. And you compute the arc length of what there is in there. And you divide by r. And you take a soup of r positive and all z in the contour. And you call that the quantity c gamma. And the theorem is this remarkable theorem is that C plus minus or h are in LP, bounded operators in LP, if and only if c gamma is finite. Remarkable theorem. Moreover, you can show that c gamma has a form looking like this, phi p of the bound. Suppose we have, let me write it like this, that hf in p is bounded by cp times f in LP. And cp will have the form phi p of c gamma, where phi p is a function which goes to zero and some continuous function like this, which is independent of all contours. It's an a priori function. So this is a marvelously useful formula, because as we hinted at, early on, the steepest ascent method operates by deforming contours. So what's going to change as you change the contours is going to be this. But as long as these numbers remain finite, you're going to retain this bound. Theorem is this, if c gamma is finite, then the operator h, for example, is bounded in L2. And then by some kind of method, we saw interpolation and things like this, you know it's going to be true for all LP bigger than 1. Conversely, if the Hilbert transform is bounded in L2 or in any fixed LP, it's bounded in all LP bigger than 1, and c gamma is finite. So let me just finish with the following remark. Here's an example. Take the contour y equals x squared, 0. This goes to 1. This goes to the point 1, 1. Take such a contour. As an exercise, I suggest it's obvious from here that c gamma is finite. And therefore, h belongs to the bounded operator, say, in L2. Prove that the operator is a bounded operator in L2 directly without this theorem. And you see that the effect of this cuspia is very non-trivial from the analytic point of view. So if you want to get something of the depth of the theorem, just work out this one example. Finish off with the following comment. A Riemann-Hilbert problem is going to be something which takes place on an oriented, composed contour. Each of whose arcs have a finite c gamma, and such arcs are called carloson curves. So Riemann-Hilbert problems take place naturally on composed curves. Each of whose arcs is a carloson curve. OK, thank you. Any quick question for Percy? Yes, no? No. So let's thank Percy again. Bravo. Thank you for this question. Sure.