 Let me just remind you what's involved in a Riemann-Hilbert problem. There's a contour, sigma, which is oriented, and by convention, the plus sign that lies on your left as you follow the arrow. And there's also a jump matrix, which goes from the contour to the invertible k by k matrices, such that v and v inverse are an infinity of the contour. And we say m, which is a function of z, solves the Riemann-Hilbert problem, or is a solution of the Riemann-Hilbert problem. If m of z is analytic, in c, take away the contour. And secondly, m plus of z equals m minus of z times v of z. This is for z belonging to the contour. And in the case where m of z goes to the identity, as z goes to infinity, this is what we mean by saying that m solves. Can people hear me? Yeah. OK. Took it off. That's what we mean by saying that m solves the normalized Riemann-Hilbert problem. Now, there are many, many, many technical issues connected with a Riemann-Hilbert problem. I wrote on a list of nine or 10 of them. I just mentioned some of them. The most obvious ones is what you mean by m plus and m minus. I remind you that m plus means, if you have a point z over here, m plus of z is the boundary value of m of z. As you approach z from the plus side, m minus is the boundary value from the left. In what sense are those limits taken? When m goes to the identity, in what senses are taking place? Different problem is how do you know that there exists such a solution? Is the solution unique? What kind of contours are you going to look at? Many, many problems. I'm not going to address this in a systematic fashion at all. We're just going to go along and face some of the issues as they come up. Now, a key observation which has been apparent in many of the recent lectures is the role of the Cauchy operator, or sometimes people call it the Cauchy transform. That had the effect, for example, we saw in Los Los Talk that it draws the analysis from the upper half plane down to the axis. In exactly the same way, the Cauchy operator is going to take the problem from the complement of the contour onto the contours. So the most important operator in the whole business is the Cauchy operator. And it looks like this. We have a contour, an oriented contour. It acts in the following fashion, f of s, d bar s upon s minus z, so the sigma, whenever I d bar s, I mean ds over 2 pi i. And this is for z, allowing to the complex take away the contour. This is the Cauchy operator. And what we'll be studying in this lecture is the properties of some of these properties, some of the properties of this operator, because as I say, it's going to play the key role in understanding Riemann-Hilbert problems. So, just as an example, I wanted to just take the case where sigma is just the real line oriented from Meijers-Vietu plus infinity. So here I'll be just reviewing for you some of the properties of this operator, which you all meet when you have a basic course in complex analysis. So this first part of this talk is just going to be that kind of review and to just move you on ahead, the remarkable thing is that these properties for the Cauchy operator just on the line are now going to be shown to be true in a much more general situation where we have a contour of that kind. So here, let me just write c of f of z as the integral of f of s d bar s upon s minus z, and now the integration takes place over R. If I end up writing too small, can somebody at the back just give me a word, okay? All right. Now, what we have, let's do a very simple computation. Let's take z equal to x plus i epsilon, epsilon positive, and x real. So then we've got cf at x plus i epsilon integral over R f of s d bar s upon s minus x minus i epsilon. And you write that out in terms of the real and imaginary parts, and you get the following that this is equal to f of s times i epsilon d bar s upon s minus x squared plus epsilon squared plus f of s d bar s, we've got over here, s minus x, s minus x squared, epsilon squared d bar s. And so let's call that 1 plus 2, okay? So now, if we take 1 and we change variables, this will take the form one-half of f of x plus epsilon u du over pi times u squared plus 1. So that's the first thing. So let me assume, first of all, that f belongs to the short space. So it immediately means we can let epsilon go down to zero, and this is going to go to one-half, one over two pi times f of x by dominated convergence du over u squared plus 1, which is just f of x over 2. If we look at the second part, let me call it 2 less than epsilon and 2 bigger than epsilon, where 2 less than epsilon is the integral of 1 over 2 pi i integral x minus s is less than epsilon f of s ds upon x minus s squared plus epsilon squared s minus x ds. And the part where 2 is bigger than epsilon, I just put here x minus s is bigger than epsilon. Okay, now comes the key fact about these operators. And one of the reasons that these operators are so important in analysis is that they are a prime example of what are called pseudo-differential operators, or particularly here, called they're on Ziegman operators, and there's a fundamental cancellation which is built into them which you're about to see now. And that is, I can write 2 less than epsilon as being the integral of 1 over 2 pi i times s minus x upon s minus x squared plus epsilon squared s minus x is less than epsilon, and I could put f of s minus f of x ds. I can put this in for free because this is an odd function and I'm integrating over an even interval, so I can just put that in for free. And this is this cancellation which is absolutely essential for what comes. We find that this then here tells us that pi less than epsilon, this quantity, is then less than or equal to 1 over 2 pi times f prime in the infinity norm times the integral of s minus x squared, s minus x squared plus epsilon squared ds, and this is over s minus x is less than epsilon. This number is less than 1, so this is less than 2 epsilon upon 2 pi f prime in the infinity and that goes to 0 as epsilon goes down to 0, essential thing. Now if we do this bigger than epsilon, and I can write it in the following way, 1 upon 2 pi i s minus x is bigger than epsilon upon s minus x squared plus epsilon squared minus 1 over s minus x times f of s ds plus 1 over 2 pi i times the integral s minus x bigger than epsilon, sorry about this, times 1 upon s minus x f of s ds, and let me call that 3 and 4, so this thing here is 3 plus 4, so that's 3 and that's 4. Now, so 3 is, oh 4, okay, so 3 here is going to equal, going to be bounded, 1 over 2 pi times integral s minus x bigger than epsilon, and we've got epsilon squared upon s minus x squared epsilon squared f of s ds quantity, and we change variables, this is 1 over 2 pi, and we've got f of x plus epsilon u, d u left something off, there's an s minus x here. This changes to u bigger than 1 of u times u squared plus 1, and again by dominated convergence, this goes as epsilon to 1 over 2 pi times the integral of d u upon u times u squared plus 1 mod u bigger than 1 times f of x. And again, this is an odd function, so this integral is 0, so this limit here is 0, so what we left with is a part 4 which is just the integral, so this is equal to 1 over 2 pi times part 4, which is just the, okay, so the limit as epsilon goes down to 0 of 1 upon pi integral of f of s, we've got your x minus s is bigger than epsilon f of s upon s minus s ds. So that's what we left with that, let me put up what I want to say, and this is the key fact, c plus f at a point x, which is the limit, goes down to 0, cf on the real, it's 1 half f of x plus i over pi, cf of x is 1 over pi x minus s, yes, and this is what's called the Hilbert transform. So that is the key fact, if I come from the bottom, I let epsilon be negative and go up to 0, everything stays the same, instead of having plus here I have a minus, the big one so far, as I said I'm just sticking with f in short space. Okay, now the key things to observe is that c plus of f minus c minus of f is just f, and c plus of f plus c minus of f is just i times the Hilbert transform. Now one of the things is the fact is we looked at this limit, epsilon goes down to 0, you can replace it with what's called a non-tangential. So instead of just taking the limit as you come down there, you can, here's your point x, you can put any cone, say an opening angle to theta over here, and you can take the limit not just as you come down but anywhere as you come down. In general you can't go if this was a parabola, that wouldn't work. But so all the limits are going to be point-wise limits which are non-tangential. So far we're just working on short space functions. So now to get a handle on this as an operator we use the Fourier transform f hat of z, which is the limit in the mean, or the limit, so r f of s e to the minus i xz e to the i sz ds upon root 2 pi. And it's inverse transform which will be the same thing as this except you put e to the i sz ds over root 2 pi f of s, this was dz now. You can get this at s now. And again it's a limit in the mean from minus r to r, you just flip the sign. Now it's a straightforward calculation in residue theory to establish the following. Fix epsilon positive and define c plus, let me call it cx or c epsilon, f at x to be c, the transform at the point x plus i epsilon. So think of epsilon as being fixed now. It's a calculation, a straightforward calculation using residues and you get the following. That if you want to compute c epsilon f, you take your function, you take its Fourier transform, you multiply by the function chi bigger than zero e to the minus dot epsilon and you take the inverse Fourier transform and okay that's what it is. So what it is to evaluate this operator here which is now a function of x, epsilon is fixed, x is real. You take the Fourier transform of the given function, you multiply it by this function, zero of s e to the minus epsilon s, for s positive and zero otherwise and that's what you have. In particular it tells you that this operator, initially you do all these calculations on short space and you immediately see from here because of the properties of the Fourier transform, I remind you the Fourier transform is a bijection, an isomorphism between l2 and itself and the inverse Fourier transform is the inverse transform. So this takes a function f onto function f hat with the same l2 norm. You're multiplying by a bounded function so it stays in l2. Then f inverse brings it back to the original l2 with the same norm. So this is a bounded operator along to the bounded operators in l2. And moreover we see that as epsilon goes down to zero, c epsilon minus c plus in l2 goes to zero. Now c plus is clearly our original operator. It's the value of this quantity when epsilon goes down to zero. On the other hand we see if we look at this object here when epsilon goes down to zero, this goes just down to a particular operator. What I do with, so we've got that c epsilon goes down to the operator c plus which is just the inverse Fourier transform multiplied by this bigger than this and then the Fourier transform and this operator is necessarily the Cauchy transform limit as epsilon goes down and down from the top. So you've established a number of things over here which are very important in application. First of all you see that the Cauchy operator which we wrote down, bb4 for points above the line converges to the operator c plus. That operator c plus is necessarily a bounded operator. It's just multiplication by the characteristic function of the right hand line in Fourier space. Moreover, c plus is the limit in l2 of the functions defined on lines of height epsilon above the real axis. You just look at the definition. It's an integral only over the real line. The complex comes in only in the variable z. This is purely real. Fourier transform goes from the real line to the real line. So here, cf is defined, let me remind you, cf at z is defined for complex number z which is the integral of f of s. Have you got it? Yes. Okay. So let me just re-interpret these are absolutely critical facts. The operator c plus is a bounded operator from l2 to l2. That follows from the calculation it is on short space and the density of short space in l2. Moreover, if you look at the values of this function for fixed epsilon in l2 of x, it will converge in l2 to a function on the line which is given by c plus of f. Okay. Now, if we use this fact that c plus and similarly we would find that c of epsilon converges to c minus as epsilon goes up to zero in a similar calculation. And there, if you write out so a similar calculation will give you c minus but if you now write out c plus is one-half, c plus on f, plus i hf over 2 and you take the Fourier transform of both sides, so you get one-half fh plus plus i over 2 and then you take the inverse Fourier transform you'll see that this implies that hf is a Fourier transform times i of the signum function times the inverse Fourier transform. So instead of just being the characteristic function of the right-half interval, it's now the signum function which is plus one on the right and minus one on the left. So this is just a little piece of algebra where you plug in this fact over here, this fact that we just plug it in over here. Okay. Now, there's a beautiful argument due to Ries and he tells us the following. If you take c f at z, z belonging to the upper half plane which we know of f of s upon s minus z d by s on the real line. And let's assume first of all that f belongs to c zero infinity or complexly, compactly supported infinitely differentiable functions. Then this function here clearly has an analytic continuation or analytic in c plus and is of order one over z as z goes to infinity. This means we can integrate c f of z d z over a contour where mod z is equal to r. And we can raise this to any power we want. Let's raise it to the fourth power. Then because this decays like one over z, this goes like one over z to the fourth, so it looks like one over r to the fourth. This will give you the length of the contours r. So this behaves like one over r cubed, which goes to zero as zero as you go. But then by Cauchy's theorem, the integral around here is just the integral around here. So this tells you that the integral of c plus of f to the power four dx minus infinity to infinity is zero. This is great because you take this formula here and you just plug it in there. So let's also assume for simplicity that f is real. So you got this integral here. Now you write out the real and imaginary parts and you get an f to the fourth minus, I think it is four, f squared f hf squared plus hf to the fourth integral must be zero, taking the real part of that. But now you see what's going on. You've got an h to the fourth here. You've got an f to the fourth here and you've got this thing over here, which by Schwartz's inequality is controlled by this object and by that object. So a little application of Schwartz tells you hf to the fourth. So you immediately see a beautiful argument that not only is the Hilbert transform bounded in L2, it's bounded in L4. And I can repeat this argument for any even integer just using Schwartz over here. So then you can do it for every even integer. So by interpolation you then have it for any positive integer bigger than two and then by duality you get it for all p bigger than one. So this implies together with interpolation and duality to the p. So that's the first big piece of information that the Hilbert transform is a bounded operator in Lp for any p bigger than one and less than infinity. It's not a bounded operator in L1 as you can see easily because suppose I've got some function of s minus z.