was that if you have a sober life space, HKU, first thing I want to do is what was that? The extension theorem. So you can take things in HKU and make it into h k RN. And I showed you how to do that in a continuous way. Meaning I created a bounded operator from HKU to HK are rendered. It doesn't extension. Okay? Then the second thing you're trying to do now is the trace. So if you have a region, you okay, I'm trying to take the trace on the boundary of you. If you was an L2 function, right? So if use an L2 function is only defined almost everywhere the boundary has measure zero actually. So you could make anything on the boundary of you. And so it cannot have a trace. There is no way to give a. So what I want to do with, and I want to take a map from some HKU. Who's something H s on the boundary, right? Which it should be that if you want a smooth function all the way up to the boundary, then the trace of you should be just I'm not writing it carefully. I'm just trying to talk right now. Okay? All this is very clearly written in the notes. So if you as a smooth function on your right, if u belongs to, let's erase this. If u belongs to C infinity u bar, okay? Then, which means then a uniformly continuous implies u has an extension, has a continuous extension u bar, and call it u bar. That was one of the homework problems. You saw that in the half-plane. I think it's in homework before perhaps, or if it is in C infinity R n plus bar, it has an extension all the way to x n equal to zero, which is continuous. And then it's question too, I think that it also will have derivatives in the horizontal direction and so on. Okay? So if you have something which is uniformly continuous on you, then it has an extension all the way to the boundary and that'll be continuous and that's called u bar. So I have this map from C infinity u bar to a function on the boundary, right? You take u and you take the restriction of u bar to delta U. I have this trace map. Right? Now, C infinity u bar is dense in HKU. That is my question. If the function was, if the function is uniformly continuous on you, I know it has the extension. I can take its trace, no problem. But I have something in HKU. Can I take its trace from the boundary? I know LTU does not work. Okay? Ltu does not work. Okay? Now, to put this extension to exist, and I don't I'm not even saying which S. On delta u. I'm not even talking of that which has, but for this extension, we exist. What I need to prove is that this map is a bounded linear map under these norms. Because then the extension will exist by that standard work. Right? Whenever you have a dense set. So C infinity you will be NHS. So you already have a map from C infinity u bar to hs delta U, whatever it is. If you want the extension to HKU, you have to prove it's a bounded linear map and then you're done. So I need to, for this extension to exist, we need to prove an estimate that for all u and That's what I need to show. Okay? If I can prove such an estimate for all, take the U's, which are C infinity nu bar. Okay? They all have traces on the boundary. This u bar makes sense, is restriction to delta U make sense. If you want to prove that this trace map has this extension to HKU to some hs delta U. I need, I need to, I need to prove such an estimate just for you and smooth functions. That the trace of a smooth function on the boundary can be bounded by the h k norm of u inside. Think about that. That the NHS norm of the bound of u on the boundary can be bounded by the HK norm of u inside. You're estimating boundary values by interior values. Correct? That's what I need to do if I'm going to such an extension to will exist. Okay. And we'll show that it's possible. But I didn't tell you anything about the S. So make sure to show that as an extension trees and as we'll see in H, Correct? Correct. This is sitting in some hs of delta. You get that the normal curve. And then we do like on Barnard. Don't even want Monarch because remember, this is a dead set on monarch is needed harm by not only applies to functionals, when the range only goes to R or C, like when this is R or C. Only then can you apply our RAM or CM only then can you apply on buttock? On the right-hand side? Probably comes from there from the HCl. Does not involve the insurance. Yeah. Right. If whenever you have a V2, why we then sin x? And if you have norm d v is less than, this is the y norm is less than norm. Read the x norm. If this is true, then there is a unique extension. Right? So that's what I'm applying. The question is, what is this such an estimate true? How is it that you can estimate on the boundary by interior values? All right, and what is this? S, S naught, which is, is it? The answer is actually, we'll show that S has to be k minus one. You lose one derivative when you take the trace. Actually the correct answer is you lose half a derivative. So if you are, there are several questions which are, which I haven't. Suppose you are an H1 here, right? So which is the correct answer is you lose half a derivative. Well, I haven't even defined h half on our end. What does HR from the boundary? So I'm not going to define it because it requires a different way of defining the Sobolev spaces, which is through Fourier transforms. Okay? Which I haven't done that. So basically, you can define HIS of you by first defining Hs of RN by using Fourier Transform. I'll maybe I'll, when I have time, I'll tell you it's not very long, but you define Hs of RAM for any S. And because you can take Fourier transforms, transforms are only four. Functions are not red. They're not for functions are. You do HIS RN, you define things with Fourier transform. What is that? And then you restrict those things to you. And those RHS you. So you can talk of fractional. There are there ways of, Yeah, what is h as far and I'll just tell you why you can do this. So these are all you on our end whose Fourier transform where one-plus is finite. Notice the S. So S can actually be negative. There are negative slope, so Sobolev spaces but negative when you wonder, yeah, what do you mean by taking negative derivatives? There's meetings, but this is what a cesarean is. All functions. There's more here. I can say we've got this integral has to exist. You can have two will take Fourier transformed and stuff. So these are all distributions for which all this happens and so on. Okay, so there's a definition and then HSE would just take this restriction to you. Objects on a cesarean, these are distributions. Take the restriction to you, That's HSC. So you can define it even for half. The theorem is actually this half. You lose half a derivative when you take the trace. Okay? But we don't know how to do half derivatives. So I'm going to show that you lose one derivative. So the theorem will be okay. So you can see that, right? E.g. if an k is zero, you can't have L2. L2 because I know L2 things don't have traces. You need H1, then you traces and L2. So if you're in H1, you, the trace on the boundary is an L2 of the boundary. So that's, so here is, I'm going to state the theorem and there are lots of issues even with the statement. I haven't yet told you. How do you define Sobolev spaces on the boundary? What does it take to define? What does this mean? E.g. let's say you're doing it in three dimensions, so you as the sphere or less to a two-dimension, it's even easier. Or you as the circle delta U is this boundary. What does it mean to be h one of the circle? So this one is not so hard because if you are on the circle, then you really have a periodic function of zero to two pi, right? It's, it's based on an angle. So you do, so you can think of it as a periodic function on the real line. Then you run the real line. It's just a periodic function. So you can talk of each one of ours. So that's easy, right? Because you could parametrize it like a line. But what do you do for any arbitrary thing? Curve C, this is you and that's the boundary. Well, every piece is like, you can straighten it out like aligned. So on every piece you can talk of H1 of that. So you do partition of unity, you break it up into pieces which can be strengthened by aligned. So that's how you define Sobolev spaces on that same thing if you're on a surface. So you're on this, you do partition of unity and you break it up by a boundary. Smooth boundary means you can flatten it. That's what it means. So you know how to define each one of that piece. It's like a really flat piece by change of variables and you patch it up together. That's how you would define h one on the bottom. So to define on the boundary, you need to do, need to be able to integrate on the boundary. You need to take derivatives on the boundary. How would you differentiate on the boundary, right? I said you just basically the way you do it as you straighten it out. So let me let me do something. I'm not going to go into that. Let me state the parts which are the most important things. And then we'll talk a little bit more time. I spend more time about this. How do you work on the boundary? Alright? So here is the theorem that I'm going to prove or well, I'm going to actually just state. So this is all in a very organized fashion in the notes. I'm I'm updating the notes and I'm posting it and stuff so you will see it there. And all the question about how to integrate on the boundary, hard to differentiate on the boundary, it's in the very beginning. Go and look, you'll see that. So suppose, and I'll talk about it also. Suppose u is a bounded open. Subset of R, n with a smooth boundary. And k is a positive integer, you need care at least, greater than or equal to one. Okay? Let me stick to the statement I have. Yeah. There is bounded linear map f from HKU h k minus one delta u, such that d of u is the u bar restricted to delta U. Forearm. You belong to C infinity u bar, where u bar is the continuous extension of you. Let me explain what I had said. I said this before, but I'm going to repeat this. U is a bounded open sets smooth boundary case non-negative, positive integer. Then there is a map, right? Whenever you are in C infinity u bar the EU. So it's a function on you which is uniformly continuous. So it has an extension, you bar to u bar, so it has a trace. So for C infinity u bar function, that is what that map is. And this, this map has a unique extension as a bounded linear map from HKU to h k minus one. Yeah, right? In particular. So that's the general statement. And as I said, you know, actually remark. So one loses one derivative on the boundary. Actually it's only half are deliberate. You can see book by Sobolev space by Adams. You have to do it with, define it differently. So there's only a loss of half a derivative. And you need to have, you don't need H1, you need any HHS as greater than a half, then you lose half a deliberate. Okay, that's the correct theorem. Okay? So it's the usual trick which I will not go right down to, which is you cut out the boundary into pieces and you go and do stuff like that. So let's see, to really understand it, we have to do it when you as the half-plane. You understand how to do it. Why this theorem is true in the half-plane. It just change your variables and comes up, okay? As a consequence. The following proposition for you isn't that? How much time do I have? Minutes? And the proof is so trivial. But it also tells you why you lose half, why you lose one derivative, at least in this case, the Fourier transform proof showed you a different way. Okay? So here is the statement of the proposition. So proposition says the following. I'm just going to repeat. Okay? There is a unique bounded linear map. Sends each k of the half-plane. What is the boundary? X n equal to zero. So it looks like R n minus one. The boundary of the half-plane is just the, so you lose one disk cleanly. Wrong you, and let's try to understand this statement. So what I'm saying is, take any function in C infinity R n plus Br is uniformly continuous. It has an extension all the way to x n equal to zero. So I call that extension nu bar. And it's, what you're doing is you're taking the value and x n equal to zero. Remember my style, I always write X as X prime comma X and correct I'm convention where x prime is n minus one, that's in red. So that's what I'm doing here. So the trace of that is just take x n equal to zero, That's your map is well-defined if you're smooth and UNHCR. And then I claim that this map, this map has a unique extension. Okay? Because these things are dense there as a map from h k to the h k minus one of the boundary. Okay? And the proof of that, okay? So to prove that, I basically need to show. So there's a theorem that if you take functions which are in scene from the RM plus bar, but the support is bounded inside some ball, then they are actually dense in h k or n plus one. Okay? So since functions in C infinity of n plus bar with support in a ball are dense in each k, R and plus. It is enough to prove this thing. For all u in skin. Trna plus bar, support a few in some ball. This was it. That's all you need to do. You need to prove this estimate and everything follows the extension theorem. Alright, let's make sure we understand what this statement is saying. It says takeoff function, which is c infinity, our infamous part means infinitely differentiable on the half-plane. And the derivatives are uniformly continuous. Okay? So they all have tracers for the boundary. Okay? But I also impose that the support of q is in some ball. That's one of the homework problems I think I gave you that it's dense in HDR. And plus, if you prove this estimate or a data set, you have a bounded linear operator on a dense set to h k minus one as a unique extension. That's what I need to do. Okay? Other thing is this. So let's say we agree that all I need to put this estimate, okay? And now the h k minus one norm is, it involves taking derivatives of u bar, but only in the x prime direction, the horizontal directions. So if I can prove this for k equals one, then applying it to derivatives of you, you just don't get it for every, okay, it's not a complicated thing. Prove it. For k equals one. Then you apply to Dell file for you and it'll just come out. You can think about it. It's nothing complicated there. You just need to prove it because it just shows you need to lose one. Think about that. It's not very hard. Just need to show. Are all U and C infinity support a few in some model. That's what I want to do. Right? So on the left-hand side is you take the value of u bar, take the value of u on the boundary and over the boundary, and you take the L2 norm on the right-hand side, I'm taking the value of u in the interior and also its derivatives in the interior. Actually, all I need is the NF derivative, the derivative direction perpendicular. That's all I need. Do we know any relation which connects the interior values to the boundary value in the relation. But I'm allowed to use a derivative, right? Because I have H1 and I have no derivative there. So the value at one point on the boundary are in place in terms of maybe you and its derivatives in the interior. Do I have any relation? Mean value theorem means you have two points and then which two points? Okay? So here, let's draw this picture here. So I have x prime zero here, right? And I want to take the value of u at this point and express it in terms of the value of u in the interior. And I'm allowed to use the derivative of u also because I'm H1. So actually all you need to do is just work on this line. Here is x prime zero. Take the vertical line all the way to infinity. But remember by you is supported in a ball. So it doesn't matter what happens. And just use this fundamental theorem of calculus. If you integrate that, you will get the value of q square at infinity or x n equals infinity. Right? Let me write this again. You take u square, the function u squared not you. Because I need to do the integral of u squared not you write L2 norms. So look at the value at any x prime and x, and it's a function. So think of it as a function of x and variable X prime is fixed. So I can differentiate it with respect to x n. Correct? Now I integrate it from x and going from zero to infinity, I'm integrating on that line. Correct? So you will get the value of u square, the fundamental theorem of calculus. At infinity and zero, infinity, u is zero anyway after some time. So it's just negative u square minus two, right? The boundary value in terms of the inferior. Okay? I'll just take 2 min and then we'll just put the minus here because it doesn't matter. And then this is zero to infinity. Who you? X prime X n and delta n, U, X prime X and dx. That's the derivative will be u squared. So therefore, absolute u squared x prime zero is less than or equal to absolute value of follow that. Notice the integral is only an x and direction. Okay? And this is less than or equal to two a b is less than a squared plus b squared. So all I did was this very simple relation that the value there is, you express it in terms of the interior by doing differentiation in the Zen direction. Okay, and you're done. This is the proof. So now I want the L2 norm of u squared on the horizontal plane. So I integrate this with respect to x prime. So this is already zero to infinity. So what do you get on the right-hand side? You already had a dx and now you get into dx prime. Correct? So this is the square of the l2 norm of the trace. And what does that? It's used squared and the derivative of u integrated over the full half space. Okay? And this is less than or equal to the H1 norm. The proof is very elementary that this has to take it. That's how you lose one derivative because he needed the derivative to go inside and apply the fundamental theorem of calculus. Okay, there's nothing I only did was this just this fundamental theorem of calculus. This is how you relate the boundary value to the interior value. And then this, right, then you do that. You can see why it is going to be half and not one. You need. When you do this here. It's related to that, that you need only half a derivative when you thinking Fourier transform spaces. Second thing is notice you needed only the nth derivative and you need the horizontal, first-order horizontal derivatives to prove this. But this was the proof, the trace tariff. Now I'm then of course, you jazz it up with the boundary and change the variables. Can you get your theorem one delta u? But this is the main thing. This is why some people, you just tried to get to the essential idea by doing it in R plus end. And then you, whatever you need to do, change it and do that. But this is not complicated or right? I've already taken up a lot of the time. I'll just stop.
Trace Theorem I, Math 836 Spring 2023
From Rakesh Rakesh April 04, 2023
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