 Hi, I'm Zor. Welcome to Unisor Education. I would like to present another lecture part of the course of advanced mathematics on Unisor.com. Now, this lecture is about certain property of continuous functions. It's called extreme value theorem. I do suggest you to watch this lecture from the website because it contains very nice notes for each lecture. So before or after you listen to the lecture, you can actually read the notes, plus the website contains certain educational functionality like taking exams for example. And the site is free by the way, so you can use it as many times as you want. Okay, extreme value theorem. This particular theorem is definitely based on something which I have proven before. I would also encourage you to, after certain introduction which I'm going to make right now, just to stop watching the lecture and try to continue proof by yourself because I will give certain ideas how I want to approach this. And again, this course is not about jamming facts about certain mathematical objects into your head. It's about basically teaching how to think, how to prove, how to use the logic, how to creatively use certain analytics, etc. It's very important if you try to prove this theorem just by yourself after whatever introduction I'm going to make right now. Alright, let's go for it. Extreme value theorem. First of all, it's about continuous functions which are defined on segment AB with both ends included. So you have to really understand what continuous means. Secondly, you have to really think about this particular segment from A to B. It's a segment which means the end points are included. That's very important too. Now, in the previous lecture we were talking about the theorem which is called boundedness theorem, that such a continuous function defined on a segment is bounded from above and from below. It cannot go to infinity because both ends are supposed to be fixed and it's continuous. So this is fixed, this is fixed, this is function of A, this is function of B. And in between you have a continuous curve which means it cannot go to infinity. That's on an intuitive level, but we have proven it relatively rigorously. Now, it's bounded, that's fine, but now let's just think about this. We know the axiom of completeness of the real numbers which means, let me just remind you, if you have certain set of real numbers and it's bounded from above, then there is something which is called the least upper bound, which means it's an upper bound, but it's less than or equal to any other upper bound. So all the upper bounds are from here. Now the least upper bound probably for this function would be this level. Now similarly there is the bounding from below. So if a set of real numbers, any set of real numbers is bounded from below, like in this case we are bounded by any level here, then there is something which is called the highest lower bound, which in this case is probably this level. Now the theorem, which is called extreme value theorem, is about attaining by this function its highest lower bound and the least upper bound. So in both cases if this function is continuous and therefore bounded, as we know from the previous lecture, then there are certain points where the value of the function is exactly equal to, in this particular case, the highest lower bound and in this case it's the least upper bound. Now these two points are actually maximum and minimum of this function on the segment AB. These least upper bound and highest lower bound are called supremum and infimum. So what basically this theorem states is that the maximum of the function is equal to its supremum and the minimum of the function is equal to its infimum. That's what it means that the function actually attains its least upper and highest lower bounds. Attains means there are certain points where the function is exactly equal to these and if these are these points, in this case it's this one and this one, then obviously the value of the function, which is actually the maximum, is the same as supremum and the value of the function in this point, which is minimum, is actually an infimum. Now it's not always maximum and minimum are equal to these kind of values. For instance, if the function, if this function is defined not on a segment AB, but on open interval AB, which means n's are not included, the x is equal to A is not part of the definition of the function and x is equal to B is not part of the definition. Then as we see, supremum is still this value of the function, but we cannot say that this is the maximum because at point B it's not really defined. So supremum is still this, this is equal to supremum of f of x, this is the level, but it's not attained by the function if it's not defined at point B. So similarly, if this point L is not part of the definition of the function, then we have infimum, which is this level, but it's not, there are little errors here because this particular point is not the point where the function is defined. So this is all about this theorem. So we have to basically prove that if this is a closed segment, closed in terms of every points including the end points are included and this is a continuous function, so there are no gaps here. Then the maximum is equal to supremum and minimum is equal to infimum. So there is a point in this case L where function attains its infimum and that's why it's called the minimum because it's a real function. Minimum and maximum are real values of the function. So supremum and infimum are limits where the function is approaching its value, either the least upper or highest lower value bound, but not necessarily equal. But in this case these values are attained and that's why we can call them maximum and minimum. So maximum and minimum are always real, they are always existent values of the function. So supremum and infimum might or might not be. In case of continuous function on a segment, yes they are. Now how can I approach the proof of this function? Well obviously, as it was mentioned already, the previous lecture was very important where I proved that this type of a function always bounded from the above and from below. Well, which means that there is a least upper bound and there is a highest lower bound. Now we have to basically prove that these type of equalities by actually finding the point, exactly the point where the minimum is attained or the maximum of this function is attained. We have to really physically find these points constructively. And that's how we will prove that this point is actually the maximum and this point is the minimum. Now I will do it for the minimum, for infimum. While in the notes I have the same for maximum function, the attained supremum value. Just to be different basically, so you will have both ways. Lecture is from the below and the notes are from the above. Okay, so from below let's just think about this way. So we have to find this point. Now how can we find this point? The useful theorem about this is that if I can find a sequence of points which are actually going somewhere to some limit and the values at these points, this point, this point, this point and the values are approaching my infimum, then the limit of the points would be probably, I still have to prove it but will probably be exactly the point where my infimum is really attained which is the minimum of the function. Now this is probably a good place to stop if you are watching this lecture and start thinking just yourself. How can I get to this particular sequence of points which are approaching our point where infimum is really attained? Okay, so if you do it, just stop it now and now start thinking and I will continue basically for completeness of this lecture. Here is the way how I propose actually to find this sequence of points which are leading to my very important point where the minimum is equal to infimum. First of all, the function is bounded from below as has been stated before which means according to the completeness axiom this function on this segment has the highest lower bound which is this one, m. So m is the highest lower bound. So any lower bound from below its highest level is this. It exists according to completeness theorem. So I know that m exists actually. Okay, fine. Now what I'm going to do, I'm going to construct the sequence of arguments with such values of functions which are actually approaching m. How can I do it? Here is the way. Let's consider a set. Let's say s1 is my segment AB. Now s2 would be set of all arguments x such that f of x is greater or equal to m plus one-half. So this is m, this is m plus one-half. Now this is the area from this to this where the value of the function is below m plus one-half. So this is my area s2. Now obviously I didn't specify it here but obviously f of x would be less than or equal to m because m is the lower bound. It's the highest lower bound but it's still lower bound, right? Now, here is my first statement. This set of these points x is not empty. So there are points in this particular set. Why? Well, if there are no points here, it means that m plus one-half would be a better lower bound than m. So if there are no points x where the function value is in between, it means function value is all above, right? Because m is definitely a lower bound. Now I assume that this is the highest lower bound. Now that's why since it's a highest lower bound, m plus one-half cannot be the highest lower bound, right? But if the function values are above, I think I make a mistake here. If the function value are above this value m plus one-half, then obviously m plus one-half is the lowest bound. So all argument values between these two are those where the function is lower than m plus one-half. And it's not empty because if it's empty, then m plus one-half would be the highest lower bound. Okay, now let's just reduce this. I will put m plus one-third. Now this is slightly closer to the m, right? So my error should be narrower. So s3 is narrower than s2. s2 is narrower than s1, which is the whole segment AB. So again, this is not an empty set because if it is, if there are no points where the function is less than m plus one-third, then m plus one-third is the highest lower bound, which we have assumed that m is the highest, right? m is the highest, which means there is no better bound, higher bound than m. Okay, now what's important here is that s2, which contains certain restriction on the function f at x, is definitely part of the s1. Now s3 has a more strict restriction than s2. So there are less points or equal points in this area than in this. So this is part of s2. Now obviously, as you understand, I can go to m plus one-fourths as m plus one-fifths, etc. sn is x where f of x is less than or equal to m plus one-n. So this is sn. And again, it's a subset of the previous one. So all these sets, they are actually nested into each other, right? So s1 includes s2, includes s3, etc., includes sn, includes, etc. Okay, so we have certain sequence of sets. And that's exactly where I'm getting my points, which will produce me a sequence, which is approaching l. From each set, s1, s2, s3, etc., I will choose one single point, any point. So x1 belongs here, x2 belongs from s2, x3 is a point in s3, xn a point in sn. And they form a sequence, right? Now, every sequence in the segment from a to b, according to a Bolzano-Werstrass theorem, has a converging subsequence. So it's a limited space. There are infinite number of points here. And if you remember the proof I just divided by two, it should be an infinite point in half. Divide by two again should be infinite number of points in a quarter. And they're all getting closer and closer. So that's how I am getting a converging subsequence. Same thing here. From each set, I have a point. And from this set of points, I obviously can have a converging subsequence. So let's say I have chosen yk. It's converging and the converging point, the limit point is, I assume it's some kind of L. So far, this L and this L, obviously they will be the same point, but so far we don't know that. So this is a converging point for yk. So, but now let's just think about it. We are talking about continuous function. If it's a continuous function and if arguments are going to certain point L, which belongs to a segment AB, obviously, because all yk belongs to this segment. And that's why their limit belongs to this segment as well, because it includes both ends, right? So we know that f of yk would go to f of L. Because this is the continuousness, continuity of function f of x, right? Now, let's just think. I would like actually to prove that f of L is equal to m. How can I prove this? Well, let's just think about it. If my yk belongs to point 1k is a point in sk, which is a set of those x, where f of x is less than or equal m plus 1k, right? Now, I didn't specify it, but it's kind of obvious that f of x is greater than m, because m is a lower bound, right? So, yk belongs to the set of those x with this type of inequality between them. Now, so what does it mean? Well, basically it means that in particular f of yk, yk is one of those, right? So f of yk is from m to m plus 1k, right? That's what basically it means. If yk is a point in the set of all those x with this inequality to be true, then obviously it's true for yk. Now, what does it mean? As k goes to infinity, these two are going to m, obviously. I mean, this is m constant and this is going to m, which means that this also goes to m by the theorem about two policemen and drunk men. So this guy, I know that this guy is going to m. So what we have here, that this sequence converges to m because of this and to f of l because function f is a continuous function. Well, the sequence cannot go to two different limits, obviously, right? So that means that these two guys are equal to each other. So at point l, which is a limit point for yk, which exists because this is a closed segment, at point l, the value of the function is equal to the highest lower bound, which means that this infimum is really attained at point l. And that's exactly what we wanted to prove. We wanted to prove that if it's a continuous function, there is a point somewhere where the value of the function is minimum and it's equal to its highest lower bound. Similarly, there is another point which is maximum and equal to the least upper bound. Well, that's it. That's basically the theorem. What's important is that you understand the logic of the whole thing. What exactly we were using and how we have concluded whatever we needed. So the first we were using, obviously, completeness theorem. We were using the theorem about continuous function to be bounded on a segment. We used the Bolzano-Berstrasse theorem that any bounded sequence of points has a converging subsequence on this segment, right? So these are all kind of things which we used. And then we made certain logical conclusions to come to this. This is a very important diagram, if you wish. So the values of the function on one hand, according to the continuity of the function, go to this one. On another hand, it goes to this minimum because that's exactly how we chose yk's. And that's why they are equal. I do suggest you to read again the notes. Notes are about maximum being equal to supremum. I specifically decided to do it on a minimum, so we'll have both sides. And I do suggest you to sign on to the website and take exams wherever it's available. It's not for every subject available, but for many available. That's basically it. Thank you very much and good luck.