Okay, so now we come to the main things. There are desperate to say orally and then go through it one by one. So number one is, if I have a function in HKU, I want to extend it to HK aren't going to pour some very simple questions which we'll see. It's surprisingly they're even, those are different ones are number one. Number two, if I have a function in the HKU, I want to take us restriction to the boundary because when I solve PDEs and bring boundary value problems, my solutions are going to be in HKU. So what is the value of u on the boundary, right? I mean these are support the functions and L2, which are only defined almost everywhere. The boundary has measure zero. So does it even make sense to talk of the restrictions? But now I'm not saying L2. L2 does not have restrictions on the boundary, but h k does read at least one derivative, each one at least. Then it does. We'll see that that's extension theorem, restriction theorem. Third thing, if a function is an HKU, H2, H3, you, Does it have any real parameter? Is the function even contend with this? These are L2 functions with derivatives. Is the function even continuous? Sometimes, you know, H2 is good to prove theory, but you actually want to get your hands on that when you're doing Estimates, error estimate. You want to approximate one function by another. Hotkey L2 error is good. But very, uh, computing you're doing pointwise things. Is the value of the function at this point really close to that? So pointwise you need that is you need the continuous norm, the l infinity norm, not the L2 norm. Which makes the question if a function is in HKU, is it actually continuous? Does it make sense to talk of its value at a point? Correct? And we'll see that if k is large enough, the theorem is k is greater than n over two. Then you can talk of its pointwise value. It's actually a continuous function. And the next question you ask is, does it have derivatives? Yes, if k is large enough. So the gap between k and n over two. So look at K minus N over two. That's roughly how many actual derivatives it has. Alright? Okay, then that's going to be, another thing I'm going to prove is what our cow compacts subsets of HKU. That's going to be very important for us. And I'm not going to characterize all, but the very important result will be. So you have HKU, which is a classical functions with k, l two derivatives. And then you have h k plus one k plus one derivatives. So h k plus one u is sitting inside HKU, right? And that is a compact subset of HKU. So H1 U is a compact subset of L2 you, okay? So if you have any box sequence in H1 U which is bounded, it will have a convergent subsequence converging in L2. You guaranteed. This is going to be important for us when we proved theorems about existence of PDEs. Okay? So several things just let me first is extension theorem. So you go from HKU to HK RN trace theorem. You go from HKU and you don't go to HK of partial, you lose one derivative. That is why you cannot have a L2. We cannot have a trace. You need at least one derivative to have a trace on the boundary. Then the third was what I was saying was, yeah, if k is greater than n over two, Then you have a map from HKU to CJEU bar, or any j strictly less than K minus N over two. That many derivatives you gain. In particular, it's going to be continuous. Sorry. You're working in R n j 0. Okay? Right, Jade derivatives. Okay? And there'll be an estimate. So all of these are bounded linear maps. Meaning so there'll be a norm of that is less than the norm of that times the constant. Okay? So the log of the bounded linear maps. So the way you will prove them as you start working with a dense subset C infinity u bar, right? All of these things make sense for C infinity u bar. The important thing is to prove that the normal estimates are there. Then you have the unique extension, the unique linear extension, right? Bounded linear operator defined on a dense vector space. That homework problem that I gave you. So all of these terms, the way they approved as I've just proved, C infinity u bar is dense in HKU. So all of these rights infinity u bar. How do you extend it? Okay, so we'll see that there is a way to extend it. So that's not so clear. This one is clear, right? You have a function on new bar, it has a limit on the boundary C infinity U bar. So the important thing there is to prove the estimate than the mapping is already there. But you don't know whether that map is bounded if you restrict to see infinity u bar. Same thing here, right? If you have a function which is in C infinity u bar, well certainly it's CJ new barn. But the important thing is, is there a norm estimate that goes with it? Okay, And then the fourth thing was the compactness. But basically each k plus one is setting image K plus HKU. So h three is sitting inside. H1 is three u, three big derivatives. What's the H1? Only one repair it whatever required. So it is three. You are H2, you, they're all sitting in each one. And let's say that those subsets are compact in the sense that every bounded sequence has a not, well, it's that map is compact, which means bounded sequence has a convergent sub-sequence. So these are the four important theorems. And once we're done with that, then we can get to pedis. Alright? This one, I'm not going to use so much in my PDEs. In the PDS. In fact, I'll not use it at all, but it's good to know. This will be very important for us. When I'll try to prove existence of some, some solution or something, I will construct a sequence and then I'll have to show it has a limit and all that stuff. So this is very useful that H1 is dense and it's called Math is compact. Then this of course, would be important. And number one is really hard to say if you want to prove something on HKU, this is another way you make it in at each k RN and then prove it on our end. This is another way to do it. In fact, number one, I'm going to use a number three. Alright? Okay. So let's look at first. First thing is the extension theorem. Is that okay? Do you need to look at that or right, for things, extension, trace, actual derivatives and compact subsets or map HKU is not compact but bounded subsets of h k plus one u and that is compact or pre-compiled mean close this compound. Okay, so first thing is the extension theorem. So let me ask you this right here is a very elementary question. Suppose you have a map from you, from U to R, Okay? You uniformly continuous on you. Okay? So therefore, we have a map. Still call you. We have a continuous function actually, right? I have a continuous function on u bar. I want an extension of this to a continuous function on RN. Forget, so we just continuity. How do I do that? I gave you a continuous function on, Let's take a ball or even like maybe a rectangle that are easier to work with. So I gave you my clothes rectangle now, it's continuous on that. How do I extend it to a continuous function outside? Right? Now, whatever the continuous function, basically two, you have to define you outside. And all it should do is, you know, really match up on the boundary. That's all it needs to do. Okay? So rectangle is particularly actually easy because it's close to a half-plane. That's what I'm going to do. Right? So let's, let's ask this question. Suppose I gave you a function on here, x n equal to zero. This is x direction. So I've given you a function here and aren't pleasant. Closure is continuous. Can you give me an extension of that function in R n, which is continuous? And it's very simple. It's not zero. You can't because you need zero on the boundary for that to have zero. No. Because then you would have to be zero at x equal to zero. For that to be a continuous extension, which does not in general, right? I'm giving you any function, continuous controller on RN plus Br. And I'd say extended. Go ahead. I was just extended, correct? Right. Okay. That's one. Okay. Now, I want to so basically it was like this, right? Okay, good. So let's just do it in 1D. So suppose you have a function on some function like this, right? So this is x, y, y equals fx. So it was a continuous function. You said to make it continues all the way. Just that's not what I had in mind when you give the answer, but this is actually even simpler than what I hadn't planned. Okay. But, okay, so I did that and that's continuous. Now I want to see one extension, meaning derivatives should match up. So you know, whatever you defined on the left, its derivative at the origin should match up with the derivative from the right. So let's get the derivative at that point was, what does it look like? Minus one, right? So what could you do? Want to extend it to a C1 function? Sorry, yeah, You take my, take any function, right? So not minus x, but you'd have to go through that point to figure out the height also. So you make a line and make it go like that. Correct? If I say two derivatives, you make a parabola fits it like that, and a third to a polynomial. Correct? So that is good. But there's one problem. Remember, I'm trying to do an extension from h one plus h one are okay, and I want this map to be continuous. So which means that the H1 norm of the extension should be C times the H1 norm of the original. I want a continuous exchange like the bounded linear operator. So arbitrarily thing this polynomial business, everything is just based on this point, right? And that's pointwise stuff rather than what was whatever was here on the other side. This kind of extension is not going to be good, right? If you did an extension like this, you will not have a bounded linear operator from H one out. So the, whatever you construct this H1 norm is not going to be controlled by the H1 norm on that side. It's just going to be pointwise and you know, pointwise things are not controlled very well by h. So we'll let things. So I need a different way of extending things. Alright? Okay, so now let's see. We've tried to find ways of extending it, which would sort of in some sense use the whole graph. Correct? So let's see how that could be done. So I have some function here like this. Maybe it's dying out later. It's in each one, so maybe dies out. Okay? So now let's see to match derivative continuity ZZ. Well, okay, let's see, I want to do a continuous extension now. But I want a norm on the left to be controlled by the norm on the right. The L2 norm, Let's say. So can you think of any other extension? I want it to be continuous. That's it. The whole function should be continuous. So still it has to match the value at the origin, correct? But after that, how you define it has to be in such a way so that it's controlled by what's on the right. Okay. I shifted button that will not make it continuous to the reflection. Okay? So you do reflection. Okay? But then now you, then you can see the control, right. The left is controlled by the right. Say it again. No, it will not be right. So I need to do something else. So the question is, how do we do this, right? So someone came with it. It's very elementary trick. When he say, Yeah, of course, it should work, right? We will see that how to do it? Well, you will see that it's just slightly not so it's a mixture of these things. So here is the theorem which I'm going to write. Okay? So then I will do it on our end ultimately. Okay? So here's the theorem. Let me stick to the state when I wrote down. Suppose u is a bounded subset of R n with a smooth boundary, and K is a non-negative integer. There is a bounded linear map from HKU to HK ARN such that EU is equal to u. Sorry I wrote that wrong to fix my notes. Right? Eu is an extension so they should agree on. Okay, and this extension map is a bounded linear map. So the norm of the extension, they age care. So what this means is if you take the norm of the extension on our end is less than the norm of the original view. Okay, So what you put on the left-hand side is so-called. Its norm has to be controlled by it. Okay? Now, so how are we going to do this, right? Again, they're going to be our n plus. How is it going to be this part? I'm just going to talk about it. It's in the notes written carefully, but it's the same way we've done it, right? So here is my u, right? Then you can cover it with things be at the boundary. You can do these balls. Are these open sets, right? You have the UIs. And then you need one in the inside, which should sort of covered this like that. Okay, That's another one. Let's call it U is zero. Okay? And then, you know, you, basically we've done this before, like so then u is the sum of UIs I going from one to some capital N with support of UI contained in the US? Correct. Okay. So I'm I'm just stating it quickly. So if your u is support, the UI is supported here. At zero outside. Then let's just take the zero extension. There's nothing under norm is till norm of u, norm of u. Okay? So the unusual cases are when your u is supported on these things. So actually, sorry. So your u is inside that, but actually, but you just defined here, right? To draw another picture. Your user only defined here. But I can flatten these things. And then they will go to things which are defined here. And they are zero. The compactly supported here, meaning actually basically there's zero outside this. So it's things that are like this. Okay? Firstly, they are defined only above it. And they are supported in some ball like this. You don't know what's happening here. Right? It's in C infinity aren't bar. Now we're in RM plus br because we proved that. Okay, but they're compact we support. So if you know how to do extensions for this with the norm preserving norm of continuity will know how to do it here. Just use the change of variables to map it back. Okay? So here we don't worry about it shifts because it's embedded in the correct. So we're already working with C infinity RM plus bar. We don't need to worry about shifts anymore. Things all the way to the art on x n equal to zero all the way there. It's nice. The problem is how to go beyond it. Right? This theorem will follow from the following proposition for our in-class. Okay? This will be standard every time I will say if supportive view is in the interior of you, you can, something easy will happen. It's only when you're near the boundary, then you have to worry about, okay. So here's the proposition. And this is where the real thing is. It's a simple calculus trick. Given a non-negative integer k, There is a bounded linear map E from h k r plus r n h k with e u equal to u on the plus. Okay? It will also do some other things. Either way the extension is not unique. It same thing is true for that. Okay? Sorry. Okay. Okay. So what it will do is further, functions supported in B are zero, are mapped functions, right? So what I'm saying is, you know, you have this is the origin. You have your ball of radius r. Your original function is zero outside this. It may be, it will be zero here. But of course, you know, whatever it is on the boundary on x n equal to zero. It doesn't matter, right? So when I extend it, it will be supported in that is going to be based on reflection. So you can see how the support is going to be managed. But you saw that even in the one dimensional case, just pure reflection will not do it, right? Continuity is there, but the derivative's don't match up. So how did we do this? Right here is the trick. So by the densities for your take you to be C infinity R n plus Br with support of your contained in B are zero. So take a function like that. This is dense in H K RN. Okay? So if you just prove it for that, define the operator on this. Then by that linear extension thing, I'll have it on h k are in place. Right? So how do I extend it? Okay, So by the way, I'm going to assume that I also have a definition. So it's hidden from the RM plus. They are defined on R n plus. But we've seen the uniform continuity. You can extend it to RM plus bar. Okay? So we've seen that if it's uniformly continuous, you can actually go all the way to the boundary. So I'm assuming that, okay, so here is my extension. So how about how do I define U bar x? Okay? If it's x, n is greater than or equal to zero, you just define it, UX. Same, right? You don't do anything that's supposed to be the extension. Here is the clever trick. If x n is less than zero. So I'm going to stick to j going from one to, remember I'm doing k, h k. So it's u of x prime. Right? Now this would be, if I just wrote this, that would be the reflection. Correct? But we saw that just then. I don't need to write any sum. Okay? So, but the reflection we know is not differentiable at the intersection. So you have to change this a little bit. I'm going to put some constants in front. And I'm going to reflect. But not just reflect, but multiplied by J. J is some constant from one to k plus one. So I'm taking combinations of reflection then some stretching or shrinking. Okay? With Cj's tilde to be chosen. Okay? That's what I'm going to do. So okay, so here is the picture. You have this, you have you here. And then you have the u bar, the extension. Okay? Now for it to be, I'm going to show that this u bar is actually not in C infinity. I'm going to show it's in CK. So then it'll be in HK, RN. Okay? So I just need to match up derivatives on x equal to zero. Now, which derivatives do I? Firstly, is it continuous? What happens at x is equal to zero, the top and the bottom. So you put x n equal to zero, so you just get zero here. Okay? And that's with the same value except they're all the Cj's. Need to choose my CEJ. So that comes out so that seniors will have to be chosen carefully. So that the sum of those is one. So what do match up, mash them up here, and then matched all the derivatives up to k. Okay? So I need to match continuity. That's the zeroeth derivative all the way to k derivatives. So there'll be k, k plus one equations. I need k plus one constants, but things to match up. The c k plus one will be the unknowns. And I'll get a system of k plus one equations and k plus one of nodes. So let me write this down and then we'll see, okay, By the way, what do I need to value of the two things to match up on the thing that's continuity. What about derivatives, right? Which derivatives have to match up? All derivatives, right? We need to match up all the derivate, well, up to k on x n equal to zero. So let's say that we'll be doing two variables. Let's say this is x and this is X1 and X2. Suppose the function matches up on x n equal to zero. What about the X1 derivatives? Do they match up? Do I need to impose any conditions to make separately to make the x1 derivatives match up? No, right, because the values are the same, so it's just differentiating horizontally. It's not, nothing is changing. So X1 derivatives don't really match up the values on the boundary. We X1 Vx prime derivative match up automatically. You really need to work on the x n derivatives. Okay? So all I need to do is make sure that the values of the function match-up on set equal to zero. They accept derivatives up to order k match up. Once I do that, I have a seek a functional manner and then it's going h k. Okay? And that'll be the extension. Does manipulations. Because otherwise he was gonna give us one. Right? Right now there are two other thing by the way, now, you know, but maybe I need to choose each seizure depends on which point on the boundary you work with. Maybe the seizure depends on the bottom point on x and you understand, right? But the way that choice has been made as it's independent of which boundary point you're working with. Okay, you will see that it comes out in that automatically. Second thing is notice that the extension, the norm on the FDA extension, the lower part is bounded by the UK, is controlled by the bottom and the top part is given in terms of you. So you can see that the norms will be controlled also. Okay? So now let me write this. So I'm going to erase this. So here it goes. So we choose C k, so that delta n minus j. Let's call it a tie now, of u bar at x prime zero is equal to delta n plus i of u bar at x prime zero. I going from zero to I just need to match up the left, the up and down derivatives. Okay? What is the, if you saw the down derivatives are coming from this the lower part, right? If you differentiate this part with respect to x and what will you get? It'll pull out the minus j is correct. Are you okay with that? What I'm saying, right? So you want sigma, sigma j going from one to k plus one CJ. And you differentiate this I times. So you get minus j to the power i and delta i u x prime. And now you plug in x, x n equal to zero. So you will just get zero. That should match up the derivative of the top in the positive x and direction. Are you with me on that? Right? For the devil is in the negative x direction. You use this for the derivate is from the positive you use that. So that should match up. I going from zero to k. That's what we want. So notice that I have this delta x and zero, same thing. So all I basically want is that this quantity should be one. That comes out. It doesn't even depend on j. Okay? Therefore, we need to choose c j. That's what I need to do. A system of equations and k plus one equations. Notice it doesn't depend on x prime. That is the point I was trying to make. Did not depend on which point on x n equal to zero you working with. Initially were scared the CDF might depend on the x prime, but it does not. So someone did this cleverly. Correct. Okay, Now, does this system have a solution where we have to look at the matrix? What is the matrix? So matrix is okay, so let's look at the i equal to zero equation. Everything is 11111. Then you look at the I equals one equation. It's minus j minus one, minus two, all the way to k, k plus one. Okay, can think about it a little, but that's the second equation. For each die, you get one equation. If you do the second equation, I equals two, you'll get this. And the last equation will be That's your matrix. Now if you've spent time with matrices, you will recognize this as a kind of a matrix which is well known. It's called the Vandermonde matrix. Okay, is basically determined by k. N by n matrix is done by n different numbers made up of powers of those numbers. Okay? There's an example of a Vandermonde matrix. And its determinant is basically the differences of these minus one through m minus k. Take pairs of products, like you know, you have. So suppose this was x1, x2, XK plus one, then its products of xy minus x j, the determinant, all the various products and stuff. While the combinations you can obtain, because all the numbers are different. So inward. So therefore you bought belongs to seek ARN. Okay? Mashed up all the derivatives. Okay, What about the support? So there is UX. And if you look at that, say let's say, you know, you hear this are okay. And then you looked at U of I minus j X, right? So, when is it non-zero? Or excellent, less than zero. U of x prime minus J x n is non-zero. It's easier to do it with rectangles rather than the square. Let's just do it with rectangles. It's easier, like, like that. What I'm saying is correct AQI also for their butt. Let's hit this height was five. So how will the J, Let's say I get here. This is each. If I looked at minus phi of x n, how low is this going to go? The support, right? The, the Phi of x n. So you know, five times absolute x n has to be less than h. So x and x b is less than h over five. In fact, actually it's closer. Alright? So rather than expanding it downwards, it's actually moving it closer. So that's the same thing you can see. It's going to be true for the ball. If it is enclosed in the upper hemisphere like that, then the lower part is actually even closer inside. It doesn't shift in the x prime direction at all. But the vertical direction, it's moving closer to the x-axis, the support. Okay? So I am not writing this down very carefully, but you can see this argument here that actually the height, therefore, if support you contained in B are zero then also, okay, I've written it. I haven't written it so carefully. It's all done in the notes. It's written carefully. Okay? Alright, and I've come to the last bit, which is, what about the h k norm of u bar? Correct? How is that controller? Last bit? Again, now, I can write it down, but let's just look at this, this part. So here is the at the top part, we know what's happening. There is nothing there. So it's the lower part. So this is u x prime minus Jx and that's what it is. Sorry. Okay? So if I take the derivative in the x prime direction, if you're differentiating the x prime direction, nothing happens. It just falls on this directly. If you're doing the Zen directions and minus j is pop-up, right? So delta alpha u bar, there are no x and derivative that is that. Okay? And so you know, you can just do just these things here, okay? And if you have x and derivatives. Okay, it's complicated to write this, but it's straightforward. We'll see I just pulled out some of the minus j's. Okay, just pulls out that. But these are just some constants that they're not. So it's not changing very much. Only thing which is changing is that I have these minus j x ands. How is that? How's that going to affect the norm? That's the only point you see, right? So it's, nothing else is changing. It's just a minus J exons. How's that going to affect the norm? Right? Then? But this is the point here, right? If you take any, you know, if you take delta n i u x prime and you take Jx and right, and you do it on the RN plus. You had minus here, which is an R in R n minus. Okay? Drinker to an RN minus. Okay? And you take it any square root and stuff. What happens, right? How does that change anything? Well, you flip it to our N plus. Correct. And then you have to make a change of variables. So you know Jx and you played by y and you get dx n is one over J y and d y n. So it's basically, it's one over J. So that J thing actually reduces the normal. Not that it matters, but it's not changing anything. So you can see that the norm, h k norm of the lower part is controlled or dominated by the norm of the upper part up to a constant. It's written more carefully here, but you understand the idea here. Okay, so she's writings, but the important thing is to make it c k on x n equal to zero. And then from the definition itself, Okay, you can see that the h k norm on the lower part is controlled by the H1 norm on the upper part, we have an extension which is CK. So it's NHK, RN. Okay? And the extension is a bounded linear maps. So I'm not doing by the way, notice it's not a scene. I did not tell sense C infinity to infinity extension. I only stopped at k because I had to match k derivatives. With k plus one coefficients. I cannot match how derivatives. There's not a C infinity extension. But it's an h k extension. Correct? And notice that these coefficients were determined by k. So suppose you did an extension which is an h5. Say, okay, I want to do any age seven extension. Can you elaborate two different stages? Right? Because the seizures are obtained by solving this system of equations. The system will change. If instead of phage five-year doing a needs statement. Okay. By the way, this Accenture and I chose this extension, but there could be other extensions. In fact, once you've extended after that, what you do, you could cut it off by a smooth function. That's telling an extension which is an age can control by norm. Okay. There is, in the notes I say something about there is a way to construct an extension which works for every k. You have to modify this a little and I give a reference to it if you want to look. But the way we've done it and this is easier is it depends on k. So now we know how to extend half-plane. But then halfway, you can go to the boundary for Delta U and extended there. It's complicated. He's tried to do it directly. So easy that okay, I'm sorry, the ideas of simple here, it just takes a while to write it. Okay? So this is the crucial thing here. What you see. Okay? Alright. Stopped.
Extension theorem Math 836 Spring 2023
From Rakesh Rakesh March 23, 2023
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