 Before starting, I would like to fix some notations that I will be using quite a lot. So a path, x is just a continuous function. A continuous function from the interval zero one to rd. It could be from any interval, but just, let's keep it simple. I will use the notation x of t for x of t. So it's quite common in probabilities. And xst will denote the increment xt minus xs for s smaller than t. And although everything we will do is, we'll hold for any dimension to keep the notation simple. I will always use a notation for d equal one. So we will write everything as if d is equal to one. So the second definition should be quite familiar. So I will say that x is c alpha or alpha in one third one. If the following quantity is finite, this is the usual Holder semi-norm. So it's the supremum of the increment of the paths over t minus s to the alpha. And basically I take t and s in zero one and of course different. And finally I will say, yes, xst, yeah, you can put a comma, I don't know if that's the problem. Yeah, yeah, yeah. So the one third, of course, we can define this for alpha smaller than one third, but that's the setting we will be working with. And finally, I will say that x is in c alpha minus. If x is in c beta, or all beta smaller than alpha. And the very last thing is, Brownian motion is a random t one-half minus path. That's all we need to know about Brownian motion. Okay, so today I want to talk about two problems that are related actually. So problem one is the following. I want to define the integral of x, the ys against the x for x and y in c alpha and alpha between one third and one. Second problem is the following. I want to give sense of the following differential equation. dy t equal f yt dx t, some initial condition for x in c alpha. But I want to do it in a nice way. That would mean at least existence and uniqueness, but also the continuity of the solution map in such a way that the map that takes the initial condition and the path x and gives you solution is continuous. So that would allow you to use approximations, to treat approximated problems and say that the solutions are similar. What we will do now, I will spend some time to tell you why this is not trivial and why the usual tools don't work. And then I will give you, as I was saying, the title of Glynx, that's what can be done. I will talk about the specific theory which is the theory of control roughbats. So this was introduced by Massiminello Gubinelli back in 2004 and it has become quite relevant these times. It has been one of the fundamental tools used by Martin Heijer to build a theory of existence and uniqueness of some problematic stochastic PDs in some topology. So in particular we will see that the uniform topology does not work with a concrete example. So an observation first is that these two problems are quite equivalent. So P1 can be written in the language of P2 as solving the differential equation, dZT, equal y of t dx t with initial condition z of zero equals zero. So that would be a particular case. And P2 is not exactly equivalent to P1 but can be written in the language of P1 as y of t equal the initial condition plus the integral of the right-hand side. In order to solve P2 in particular we have to give meaning of the integral of the right-hand side which is of the same kind of the one above. Okay, so let's point out what are some of the problems we would encounter to answer these questions. The first one is that Riemann-Steelge's integration does not work. So I remind you what the Riemann-Steelge integral is. You define this to be some kind of Riemann sum. I hope the notation is clear. So here I have a partition of the interval zero one. UV would be the extreme points of each interval of the partition and just build this Riemann sums where you belong to UV. So that'd be a tricky part. And when a classical result says that if G is a bounded variation then this converts for any continuous F and any choice of U bar in the intervals of the partition. Verges or F continues if G is a bounded variation. Now if the function is C alpha and only C alpha it is not a bounded variation. So say if it's C one-half then the size of these increments would be one over square root of N. So if you take the regular partition then if you are summing N terms of size one over square root of N it will blow up unless you have some miraculous cancellation. So it seems that the Riemann-Steelge's integration is doomed to fail if you want to apply to solve these problems. You can think well it's just that maybe we are clumsy and we are not able to prove such a result for the wider family of functions. And actually there is a theorem that says that you cannot do better. The Riemann is the following. So let's define an operator on continuous functions as follows where I take the partition to be the regular partition. So Tk is K over N. Then if Sn of F converges for all continuous F then G has bounded variation. So it's not that you can define Riemann-Steelge's integral for function of bounded variation best to best you can do. And the proof of this result is quite simple. It just uses a bit of functional analysis and it's very short. So the first thing you have to notice is that Sn is in fact a linear operator from the space of continuous functions to the real numbers. So we can use all the tools of functional analysis to deal with this. So I will do the following thing that will look mysterious at the beginning is that I will define a function Fn continuous function with infinity norm equal to one and such that, yes? Yes, there is nothing. So I will define a certain test function. So I will prescribe its value at the left points at the left points of the partition. So this will take the value minus one or plus one according to the sign of the increment of G. It will be the sign of G Tk plus one minus G Tk. So Sn of Fn is equal to the following. So if this increment is positive, I'm multiplying it by plus one. If it's negative, I'm multiplying it by minus one. So in any case, I get the absolute value of the increment and this is related to two things. So first, this is related to the total variation of G. When I take the measure of the partition to go to zero, this will converge to the total variation. And on the other side, I'm testing Sn against a function which has norm one. So it's also related to the norm of Sn as an operator. This is smaller or equal to the norm of Sn as an operator, okay? So this means that these quantities will be smaller than the supremum of all these norms. And now I can also take the limit in here when n goes to infinity and I will get the total variation of G. Now if I can show that this is bounded, I will get that the total variation is bounded. And that's where I assume that this thing converges. So if Sn of f converges for all f in C01, then this sequence is bounded in R for all f. It means that the family of operators Sn is point-wise bounded. And I can use the Banach-Steinhaus theorem which says that then it's bounded in norm. And this gives us an upper bounds on the total variation of G. So there is no hope to use Riemann-Steins integration to go beyond the total bounded variation case. So we still learn something from this proof beside the fact that this will be useless. Is that, so in the core of the proof is the fact that I can engineer a bad function, a bad test function, okay? So maybe I'm being too ambitious and I want to build the integral against anything. And anything means that I can build this pathological test functions. But maybe I don't want to integrate against anything. And that's in fact true. So we don't want to integrate anything against anything. For instance, let's go to problem one. In problem one, remember that we had dyT equal f of yT, dxT. So it means that on small scales, I should have that the increment of y is that f of yS times the increment of X. So if I want to solve this problem too, sorry. If I want to solve problem two, basically I'm interested to integrate Y to integrate only functions that's locally look like X or see it like X. So I'm not integrating anything against X, only things that are quite related. So the second observation is that there is one classical theory that goes beyond the bounded variation case. So, and that's the young integral is well defined for f in C alpha and G in C beta as long as A plus beta is bigger than one. So that's good, but that's not good enough. For instance, it doesn't cover the Brownian motion case. So it does not cover, let's say the integral of B against itself. Why remember that B was C one half minus? So when I add up the regularities of B and the integrator, I get something which is slightly below one. So we are below this threshold. So let me take a second to show you that some very exotic thing happened when you go actually to bad exponents. So let's talk about some pathologies of Brownian motion. There is a very funny fact that you can say you want to build the Riemann sums of Brownian motion and try to decide if they convert to something. Say I take B of T, something I will call Tk epsilon times the increment. Tk epsilon is something that interpolates between Tk minus one and Tk. Well, if this was Riemann-Steelge integral, this would converge no matter what the epsilon is. But actually in the case of Brownian motion, this converges to what you would expect plus something that depends on epsilon. And there is more. So this is, B is the Brownian motion evaluated at time one. It's a Brownian motion, it's the same B, the square. Yeah, so if it was a usual function, you would stop here. So let's see in a heuristic why such thing happened. So let's be even more general. So let's try to see what would happen if it's a fundamental theorem of calculus. So say take F smooth and let's look at F of B one minus F of B zero. So we do the usual thing. We split this big increment into smaller increments and then we Taylor expand. We get F prime of B Tk minus one times the increment and I will continue up to order two. Now if B was smooth, say differentiable with the bounded derivative, this increment would be of order one over N which is the size of the increment to the square. So this is one over N square. You are summing N of those terms. So this whole sum as order one over N, it vanishes. Then you can just stop there to obtain the usual fundamental theorem of calculus. Now this guy says C one half, a bit less. So this should be of size square root of the increment, one over square root of N. So the square is one over N which compensates exactly with this N. So I cannot throw this sum away. So in the limit, what happens is the following that F of B one equal to F of B zero plus the integral of the derivative, whatever this means, plus a sum on order term. What you have to put here is actually the S. So this is something called Ito's formula, okay? So yeah, so things can get very exotic when you try to integrate against things that are quite irregular. So we talked about the difficulties we encounter when trying to define these integrals but let's talk a little bit about the problem of the continuity of the solution map. So that problem too. Let's look at the following example. So let's say we want to solve this differential equation. So I have two one-dimensional paths, so two one-toedomentional paths. Let's say, well, let's give an initial condition. So this can be solved explicitly you just use an integrating factor. The solution is here and I will take a very specific sequence of paths. We'll see what happens, okay? So this is what we will consider, take. So in particular, this path goes to zero uniformly. So let's have a look at what happens with the solutions. So the solution in this case looks like this, yt of n, just writing this in this specific case. So it will be where this y and t is the integral. So this is not something we can compute but at least we can estimate it. Okay, so let's say what this looks like. So ytn is, I have the exponential of minus xn1, which is these things there. And then I have the dx2, which is n times sine of cosine, sorry, n squared t dt. Okay, so I do the obvious change of variables which would be s equal n squared t and then s will go between zero and n squared t. I have cosine of s here, the sine of s ds. And then I have an extra one over n squared which here, one over n. So this is correct. Now let's integrate by part. So say this is the derivative of something, this is just something. So we exchange the rule. So let's not care about the boundary terms. I will get 12 between zero and n squared t of e minus one n, the sine of s. And what we'll pop out actually is a sine square of s ds. And an extra one over n which comes from this derivative. So let's evaluate this. So this will not matter because here I have something that goes to zero uniformly, exponential of this, just goes to one. No, I have something which is positive over an interval of order n squared. I'm just taking the average of that. So the integral of sine square usually is about one half the length of the interval. It's true if you do it between zero and two pi, the integral is pi. So this should converge t over two. That's actually true. Hence the solution that we call yn converges to t over two plus y zero. Again, these exponentials just disappear. So the map that takes y zero, x one, x two and gives you y is not continuous in the uniform topology. So if you want a continuous solution map, you have to do something else. And it's actually, it's not just that you have to change the topology on these two paths. You have to do something more. And the nice thing is that this example tells you exactly what you have to do. So we will approach it in a different way. Zero or the initial condition, say y zero, yeah. Yeah, because the, right. So let's do something else. So let's rearrange things a bit. So let's extract this guy. This is one. And now very naively, we will tell or expand. I mean, I will tell or expand inside this integral. So the first order term will just be one. And in the second order term, I will see this increment, et cetera. Now this will go to zero. This is just the increment of x and two. And we know that it goes uniformly to zero. You can, having a look at the, the higher order terms, you will see that they converge to zero as well. But not this one. So let's have a look at that one. So you see, this is an iterated integral. So this is minus integral between zero and t, integral between zero and s, dx and one, u, dx and two, s. You know, it's something you can compute. You just write down explicitly these two things. And I have the answer over here. So after doing all the simplification, you will get that this is, except for boundary terms that will go to zero, this is what you get. And this again converges to t over two. So all the problem was apparently in this iterated integral. So some, let's summarize everything we have learned on these examples. So the relevant things, of course, that remain still disintegration, it's not enough. Uniform topology is not suitable. So that's the pessimistic side of what we have done up to now. But there is some hope. This term seems to matter. So what this example says, that whatever topology, I'm trying to use, to have continuity of this solution map, it has to include this iterated integral. And that's actually what the theory of rough paths does. Everything clears up to this point. So I apologize in advance, because I won't be completely rigorous here. I will give mostly intuitions and then I will write down theorems which will not be completely correct. So there will be some bounds that will, there will be things missing, like initial condition and stuff like that. But at least the relevant part, in my opinion, will be there. So we will talk about control rough paths. So that's not the genesis of rough paths actually. It's quite older than that. It goes back to maybe the 80s of the 90s. It's by words by Terry Lyons, 80s maybe. What happens is that at some point, Guvinelli's reformulated the theory in a different way. And that reformulation was the one that has been used a lot in stochastic PDs. So I will actually base this lecture on some lecture notes of Hayer and Fritz lecture notes. So they present Guvinelli's theory, but maybe in a more accessible way than the original paper. Let's start with the definition. So let's take one alpha between one third and one. The rough paths is a pair such that. So X is C alpha. This strange X goes from the interval zero one square to R and it's such that it also has some kind of holder norm. We have to be a little careful here because now this is a function of two variables. So this st are actually two different variables. It's not an increment. Okay, so say it's a pair of a holder alpha function and some kind of holder two alpha function. And then I have to impose something more. So they cannot be completely independent one from the other. So we have the following relation. So actually this has a name. It's called a chance relation. It's the integrated integral here. Yeah. So this relation is not completely arbitrary, but it's exactly what would happen if we take X ST to be the iterated integral of X against itself. Now this identity is tricky because if I'm taking X to be a very irregular path, so far I don't know how to define this. It doesn't matter. If I choose the point is that if I choose this X double bar, satisfying this algebraic condition and this analytic condition, then it gives a satisfactory notion of an iterated integral. We'll have all the properties that I would expect from an iterated integral. And I will take that as a definition. So to work with this, I have, so the expert used the name postulates. I have to postulate the value of the iterated integral. And once I have that, I can continue with the theory. So of course I still have to do something reasonable there. Usually these objects one has to construct them by hand. They say when you deal with a probabilistic problem, what you do is that you use this theory which is completely deterministic. But the place where you have to do probabilities is to build this additional object. Okay, so an observation. Okay, so we denote the alpha, the space of such growth paths. And we'll use the short hand notation X underline. So sometime we will just say X belongs to the alpha. It's just an abuse of notation. So we say that I have to provide the iterated integral. And that's tricky because I lose uniqueness. So let's take a rough path and a function F smooth. Then I can build a different, and this will satisfy a chance relation. So given a path X, this is no unique way to construct this double integral. Okay, so let's define some norms. Just one, Charlie. And not really a norm, but a semi norm. The important thing is that it gives a notion of distance in the space of growth paths. Okay, so back to problem one. Let's see how this could be of any use for what we have in mind. So remember that Y looks like X on small scales. Say for S close to T. So for short, let's write Y as T more or less Y prime of S, X as T. So let's naively try to integrate this thing and see if we can guess what could be done here to define an integral. So let's integrate this approximation from S to T. So I will have here S R, V X R. Now S is fixed, so this goes outside of the integral and here I have X S R, V X R. So now, so here I have an increment. Only R will see the integrator, Y S is a constant. So I get integral of Y R, V X R minus Y S increment of X, more or less. And here what we see is the iterated integral. So it would seem that the correct notion of integral should satisfy the following. And you see, that's not Riemann integration or Riemann-Sidius integration. In Riemann integration you would stop here. What do I mean by that is that when I build Riemann sums, I just look at sums of these objects and they converge to what I call the integral. But now there is something here that seems to matter. It's different. So it seems that when we take Y of this form, there is a natural way to build this integral. And this leads to the concept of controlled rough paths. Go there. So again, I take alpha between one third and one. Believe me, in a few minutes you will see why we have to restrict to alpha bigger than one third. Okay, so let X be a rough path. So we say that the paths Y is controlled by X. Either exists the paths Y alpha, alpha and again a function of two variables. Like I say, from zero one square R, which has this two alpha non-finite. Such that the increment of Y is exactly Y prime of S times the increment of X plus a reminder. And the point is that the reminder has is higher order than everything else. I don't both face here. No, no, it's the increment. Yeah, yeah, the both. Yeah, yeah. Yeah, the iterated integral does not enter explicitly here. Okay, so an example, this is not too crazy. Yeah, so first let's see how we will quantify the norms of these things. So for, so we denote by the alpha X, the space of control rough paths. And we will define a norm. So technically a control rough path is not just a path. It's a, there is more information there to say the Y prime. We define the norm to be the norm of Y prime plus the norm of the remainder. And you have some choice here. There are three objects which are not X in this, in this equality, they're not independent. So you have to pick two of them to quantify the size of the others. Okay, so an observation is that what happens with the alpha norm of Y. So actually that's something you can estimate quite easily using this decomposition. So this will be one equal to the infinity norm of Y prime, alpha norm of X plus the two alpha norm of this. You could put an alpha norm, but it doesn't matter. And now you could put these two things together to get the norm in the sense of control rough paths. And then you could even put more here to get the norm of X as a rough path. We can bound this by the rough paths norm of X times the control rough paths norm of Y. So an example, this actually covers, this covers a case which motivated us to introduce a notion of control rough paths. I have until four. I said one hour and a half. It says, here it says until four. Sorry? Oh, they asked me one hour, one hour and a half. Sorry about that. So examples of control rough paths. If F is, let's say C2, then F of X is controlled by X. And it's the reactive is interested. And you can even estimate the norm. So here you have something that depends on the function. It's C2 norm actually. And also you can look at the composition of a Y controlled by X. And it will be also controlled by X. And the derivative now, the prime of these paths will be this, it's sort of a chain rule. Okay, so back to integrals. So let's repeat this de-heuristic with it before, but now with equalities instead. So let's again integrate naively on both sides. Remember that this was just iterated integral. So now I have the following equality. So it means that a priori, if I want to define this, I should build a Riemann sum that involves these three terms. That's not very satisfactory because a few minutes ago we said, well, these two terms should matter and not this one. And that's actually the case because you see this increment there is of order t minus s to the two alpha. The increment of X is of order t minus s to the power alpha. So this gives us code for t minus s power three alpha, but now we choose alpha bigger than one third. So this is little o of t minus s. So when I build Riemann sums with these guys, this will completely disappear. And that's the main theorem. So take again alpha between one third and one X of parts and Y controlled by X. Then I can define an integral via Riemann sums, but now instead of just putting the increments I have to put more information. And in addition here is incomplete because it involves this term here which is not uniquely defined. So this is really an integral against the roughness is well defined. So furthermore, we have some bounds. So when we started the heuristic, we say that these integrals should be a good approximation of this increment plus this second order increment. And it's actually the case. This is bounded by, so here's where it's not really exact. But the important thing is that here there is the t minus l minus s to the three alpha which is related to the heuristic I gave there. And also an interesting fact is that if I define the parts z t to be this integral, then z is again controlled by X and the derivative is Y. Is what we would expect. So of course one should turn the heuristic into a proof. You would have to do several things. So first that show that this remand sense that I wrote somewhere converge that what you get at the end is additive and stuff like that. So it's quite a bit of work but I think the heuristics gives you the general idea. So now we have all the tools we need to solve a problem too. By that I mean that we have the correct spaces where we can use fixed point arguments to get unique solutions. Okay, so it was something like that. So what I will do here is to define a certain map M from the space of control paths to itself. And now here I will not just take Y but Y together with a Y prime. It's a control path and I will map it to Y zero plus the integral F of Y is the XS. Now it's the rough integral. So to get the map from control of paths to control of paths I also have to give a prime for this guy and the theorem above tells us that the right notion of the relative if whatever I have inside the integral. So this is indeed the control of paths. So let's say of course there are some analytical bounds but they are provided by the theorem above. So if Y, Y prime is a fixed point what we would have is that Y of T is equal to Y zero plus this integral and Y prime is exactly what we want. So what we have to do, everything we have, what we have to do is to set a fixed point argument. There are two things we have to do is to see that M maps balls into balls and that M is a contraction. So we can give a fast sketch of proof that M leaves some ball invariant and there is actually a very funny idea in there. So let's see. So I will define a ball which is suitable for this problem. So I will fix the initial condition but see if I fix the initial condition I'm also fixing the initial condition for the derivative which is F of Y. So Y prime of zero should be F of Y zero and I will impose that Y, Y prime is known smaller or equal to one. Then what I have to do is I have to take Y, Y prime in this ball and show that what I have here is also in that ball. So I have to control the norm of the first entry and the second entry. So let's have a look at the second entry. So it's something I wrote as an example. This is bounded by F, a certain norm for F. So norm of Y. Now you see what we have to control the norm of this thing and show that it's small enough and here it seems hopeless because these are fixed quantities. There is no way we can make them small and there is one where we have to use a certain trick and it's the following. So let's try to estimate this norm. So let's estimate the increments. So remember that this can be decomposed in this way. Now this is smaller or equal to say Y prime and infinity. A certain older norm of X times T minus S to the same exponent plus the two alpha norm of the reminder, T minus S to the two alpha. So what do I want to do? I want to show that this is small, at least the norm. So I will have to divide by T minus S to the alpha and show that that will remain small. So here I'm good because when I divide by T minus S to the alpha I keep something which will go to zero when T minus S goes to zero. But now here the natural thing would be to put alpha and alpha. So when I divide by T minus S to the alpha again I will remain with something that cannot be small. And that's where you have to go a little above alpha for X. Now if you do that, then when you divide by T minus S to the alpha and you take the supremum you get that the alpha norm of Y is bounded by point infinity. Then you will have a factor B minus alpha there. And if you take T and S in an interval zero T this will be bounded by T minus alpha and there P to the alpha. Okay I'm just bounding this increments by the total size of the interval. Now this is part of the norm of Y, Y prime. So it will not matter because it's smaller than one. This also can be bounded above by the norm of Y as a control path. This will be bounded by the norm of X. So overall this is bounded by the norm of Y norm of X theta, T to the beta minus alpha. This is smaller than one because we fixed it that way. This is the data, this is axis fixed. So this is order, it's a beta minus alpha. So if you take T small enough then this will be small and at least for the second entry you will be in this ball. And the first entry can be treated similarly. It's a standard fixed point argument. So with the same ideas you can show that it's a contraction, it's a bit more involved and also then from there it's quite easy to show that it will be continuous in this rough path. You really have to use this as a rough path not just as one entry. Okay that's all I wanted to tell you today. Thank you very much.