 So welcome to this morning's session. We are the third lecture by lecture by Percy Dived Okay Okay, so Let me just give you a quick reminder what we've done so far we have Discussed the following that a remand hillbill problem Sigma V Involves an oriented contour our convention Plus side is on your left as you go along and V is a jump matrix, which is a invertible matrix at each point of the Contour such that the matrix V and its in inverse are bounded over the whole Contour the kind of contours we look at are going to be what we call composed contours or contours Composed curves, which are finite unions of arcs. In other words, homeomorphic images of the interval 01 and They had additional properties these each arc was rectifiable or locally rectifiable That enabled us to introduce on any contour Sigma We could introduce C come to H of Z say to be the integral Of H of s d bar s upon s minus z and this is over the contour D bar means DS over to 2 pi i So because we have rectifiable con contours we have a measure on each of the arcs We have a measure theory. We can integrate and Just with these assumptions we know that We can introduce the operators C plus minus which are the boundary values as You approach the point Z From the plus side or the minus side and that is equal to one-half of H plus i over 2 times a Hilbert transform on H and We always have that C plus minus C minus is the identity and This is true for all Lp Where P is bigger or equal to one and less than infinity But if you want these operators C plus minus or H to be bounded in Lp now P will only run from bigger than one and less in Infinity you need another property, which is called the Which is the property that the C gamma little little call it C gamma Which is equal to the supremum over all Z belonging to the contour and all numbers are which are positive Times the ball of radius R around Z intersect with the contour divided by R This quantity is finite. So You take any ball around any point Z So you could take a point like this you take a ball and You intersect with the con contour you compute its arc length Divided by R and the supremum over Z and R should be fun finite Such a curve is called a Carl Carlson curve is called the Carl Carlson constant This is finite if and only if H or C plus minus Belong to the bounded operators and Lp Also, whenever you have just this over here these bound boundary values are non-tangential Limits Okay, so now we can begin Speaking concretely about what we mean by a Riemann Hilbert problem and So we always have Composite curve of these kinds are not going to say anything more more about it now we say a pair of Lp Z functions F plus minus belongs to and This is just notation Boundary C Lp if and only if F is equal to C plus minus of some H Where H belongs to Lp, so that's a definition so that means If we define F of Z to be CH At Z For Z belonging to C take away the contour. This is called naturally the extension of F plus minus off the contour so those extension of F plus minus of You notice that this H is necessarily unique because What we have to have yeah, this is So F plus minus right equals that So if I have this pair here It's given by C plus C minus so because of the property that C plus minus C minus is the identity I'll keep putting the pluses Because of that you see that H is just F plus minus F minus So H is unique Okay, now I'm going to introduce two kinds of Riemann-Hilbert problems and both are used for different purposes, but they are equivalent to to each other So let me just give the following So we fix P less than one unless bigger than one less than finding and We're given Sigma V and a given Sigma V and a Measurable function F say We say M plus minus belongs to F plus Boundary C of LP So this means if I take M plus minus minus F It can be written as C plus minus of some function H. There's some function H Solves the Inhomogeneous Riemann-Hilbert problem of type 1 and LP So it's inhomogeneous Riemann-Hilbert problem of type 1 and LP if M plus equals M minus times V on the contour So the function has this property if you take away this function F for just a measurable function It's not in any LP space or anything, but just M plus minus minus F is equal to C plus minus of some function H and it has this relation almost everywhere on the contour okay, so We say so given Some function now which belongs to LP of Sigma we say M plus minus which belongs to the boundary and LP solves The inhomogeneous Riemann-Hilbert problem of type 2 and LP if M plus equals M minus times V plus F almost everywhere one second So you have these two different things now you notice the big difference between them is that M plus minus is something which automatically has an analytic continuation above the off Sigma and M plus minus might not because you don't know anything about F Now these are useful in many different ways Okay Okay now at V equal V minus inverse V plus the any Point wise almost everywhere factorization of V V plus minus V plus minus inverse belongs to L infinity now This is just a point-wise factorization. There's no Analysti attached to it now you should think about it in in the following way you should think of yourself in the context of Suter differential operators where you're obtaining an a priori Picture of what V is going to look like which will then be useful for further analysis So in other words you have this freedom to factor it point-wise and Achieve different goals in how you choose all the results that we're going to be speaking about going to be Independent of how you make this choice and I'll make this clear We write V plus because our density plus W plus and V minus as being equal to one Identity minus W minus so these are going to be bounded Okay, and we introduce the following operator CW we let W be the pair W plus W minus CWH is just C plus H of W minus C minus H of W plus and this belongs to the bounded operators in LP Okay, you've got a matrix valued function everything is of size k k bar k You take your function H you multiply it From the left on to W minus Then you take C plus of that C plus of course is applied when I write C of H I mean if H is a vector if H is a matrix and you just mean C acting on H I J each component And you do the same thing here now because this is bounded W minus is bounded W Plus is bounded and C plus and C minus of bounded this makes for a bounded operator and As I said right in the very beginning That as a analytical problem a Riemann-Hilbert problem is all about Solving singular integral equations on the contour so we're getting closer to that point Okay Now I just want to draw a little Important Diagram now these are going to be equivalent and You use one equation or the one point of view or the other depending on what you want to achieve This is going to be Connected with trying to solve this equation this on the other hand Is going to be useful for deformation theory? see over here in plus and minus always have automatic analytical continuations so Depending on what you want to do you use one version of Riemann-Hilbert theory or the audience here is a theorem Here's a theorem so suppose F times V minus one belongs to LP and P is less than one and bigger than infinity then M plus M plus plus if solves In the homogeneous Riemann-Hilbert problem of type Okay Of type one if M Let me put it like this Okay. All right, so M plus minus. Let me put it like that Okay, am I saying this correctly? Okay If M plus minus solves the inner homogeneous Riemann-Hilbert problem of type two and LP with F equals F times V minus the identity Conversely If F belongs to L2 F belongs to LP then M plus equals M plus plus F and M minus equals M minus Solves in homogeneous Riemann-Hilbert problem of type two P if This object here solves the inhomogeneous Riemann-Hilbert problem of type one and LP with F equal to C minus of F So the first part is Really very very trivial the thing I want you to pick up from this is the obvious point if you solve the one Riemann-Hilbert problem you solve the other and vice versa I'm going to leave this proof as the full proof as an exercise, but the first the first part is really very trivial as so if M plus minus belongs to LP And M plus Equals M minus V plus F Where F is equal to F times V minus 1 Then this is M minus V Plus F V minus 1 so that means M plus plus F Equals M minus plus F V So if you define this to be little M plus and This here to be little M minus you've got M plus equals M minus times V and M plus minus minus F belongs to the boundary So that's a triviality To go the other way around is strange enough a little bit more subtle Because you can't in general reverse this to go if I know a little F I know capital F, but if I know a capital F, I don't know little F was this might not be in Invertible, so I'll leave it for you to just Check that through Now Okay, now the connection To a normalized Riemann-Hilbert problem is simply the following we say M plus minus Solves the normalized Riemann-Hilbert problem sigma V If it solves the In homogeneous Riemann-Hilbert problem of type 1 and P with F just to be the identity function It's another words. I'm saying that M plus equals M minus times V and M plus equals the identity plus minus plus boundary You notice that with this definition you obviously see that if you take the extension now of You define MZ to be 1 plus C so this means it's equal to Identity plus C plus months of some H, so you just define it to be this Then this function here solves what we called formally a Riemann-Hilbert problem It satisfies this jump condition that goes to the identity as Z goes to infinity In a trivial way So this is the precise meaning of what we define to be Formally a solution of the Riemann-Hilbert problem. All right now Where does this operator come into the picture? There's a very nice and cute Calculation so let F belong to LP and M plus minus Equals F plus C plus minus H and M plus Because M minus times B. So in other words M plus minus solves the inhomogeneous Riemann-Hilbert of type 1 with this function F. So we've got M plus Because M minus times V minus inverse times V plus for any factorization pointwise almost everywhere factorization We define mu to be M plus Times V plus inverse which is also equal to M minus times V minus inverse This is bounded And Now we make this definition H of Z by definition is just C of mu W plus plus W minus So mu is a matrix valued function on the Condole you multiply on the right by this and you take the Cauchy transform of this object now here comes Perhaps the main calculation of the whole theory very simple, but extremely important if I look now at H plus That's going to be C plus of mu W plus plus mu W minus and That's equal to C plus mu W plus Plus C plus mu W minus and now we use the fact that C plus minus C minus is the identity So that means that H plus now it's going to be C plus mu W minus Plus C minus mu W plus plus mu W plus using C plus equals C minus times times that this object now this object is what we called CW mu And then we've got here plus mu W plus That's CW minus the identity mu plus the identity Plus W plus mu Which is C mu minus one mu plus V plus Me and we've got Sorry mu I'm putting it on the wrong side mu V mu now This is what we the CW minus one Mu And this is what we called This is how we defined mu. It's equal to M plus So we've got this Simple calculation, but that's what we get and we get similarly H minus is going to be CW minus one mu Plus M minus H minus So we get that So that means if we look at this combination M plus minus one minus F minus H plus minus that's equal to one minus CW mu Minus F. So the minus F I've added to both sides and I've just put the I've taken this put it on this side Put the one minus C minus on the other side. I've got this identity Now by assumption this object here belongs to the boundary of C and LP and the same thing is true for that So that means that the whole left-hand side belongs to This quantity here. So that means that we have that M plus minus minus F minus H plus minus is equal to C plus minus of some function H But by this relationship, this is independent on the right-hand side of plus or minus So that means that C plus H is equal to C minus H But we know that C plus H minus C minus H is just H So it must be that H is zero Which means that The left-hand side there is zero which implies That one minus CW mu equals F So there you have the fundamental connection between the Riemann-Hilbert problem and a particular singular integral operator equation on the contour. This is the fundamental connection Keep losing this. Now conversely So let me just summarize quickly. We've shown that the inhomogeneous Riemann-Hilbert problem of type 1 is equivalent to knowing how to solve that of type 2 Now we begin to make the connection to this operator 1 minus C is CW And we see that if we can solve the inhomogeneous Riemann-Hilbert problem of type 1 We can solve equations of the form 1 minus CW mu F Now the converse is true that if 1 minus CW mu equals F for some mu belonging to LP F is given then if we define H C mu W plus plus W minus if then if we make this definition we have That H plus minus equals F minus F Plus mu V plus minus Mu is here mu is the solution of this equation before we define mu Over there So if I solve the inhomogeneous Riemann-Hilbert problem of type 1 and I define mu in this way Then that mu Is going to solve this equation Solve an inhomogeneous problem of type 1 or 2 and I get a solution of that equation now If I have a solution of that equation I'm saying I reverse this little calculation here, and I see that this function here H has these properties and if we now set M plus minus to be mu V plus minus We see that M plus Equals M minus times V by the definition of these things and from here we see that M plus minus is equal to F plus the boundary Operator or C plus minus of some function H So if we solve this problem We see we obtain a solution of the inhomogeneous Riemann-Hilbert of type 1 now So Not only are these true algebraically, but they also true and analytically So you see I'm not going to write this fact out. It's obvious That if we solve this problem here And we obtain a solution here Where we've got a bound on M plus minus in terms of the norm of F Okay, so let me just write Write this out Let me call it a theorem that F belonging to P of sigma at M plus minus solve problem of type 1 with P F given M plus minus solve to P With F given by F times V minus the identity then one minus CW inverse on F Is bound by C times M plus minus P one minus CW inverse on F Is bounded by some other constant Plus minus That's F Such a relationship follows very easily from the algebraic facts because mu is expressed in terms of M plus minus So all of this we dealing with three equivalences if you can solve type 1 or type 2 Or you can invert the operator 1 or 1 minus C If anyone of those three three things is known The other two immediately follow their equivalent Now what you see here is they not only algebraic facts But you have these analytical facts which follow directly from the algebra and if I would know for example That the solution of the type 1 problem Was bounded in the natural sense for example of M plus minus NLP is bounded by some constant times norm F Similar thing to that I would know immediately from here that I have an a priori bound