 So, last time we were looking at properties of continuous functions and we proved that as a consequence of intermediate value property that if i is an interval, the domain is an interval then the range of the function is also an interval. We have started looking at the special case when i is a close bounded interval what can you say about the range. So, let us look at that. So, f is defined on i and taking values in R, let us say i is ab close bounded and f continuous. So, claim is that f of i is also a close bounded interval. We already know one, we know f of i is a interval. So, let us and we had proved last time that it is also bounded. So, we also proved f of i is also bounded and the basic idea was that if it is not bounded then there is a sequence in the range which is goes to absolute value of that goes to infinity. Look at the images, pre-images that being in i which is close bounded should have a convergent subsequence, but that is not possible because if it has a convergent subsequence then the image must be convergent, but that is unbounded. So, that cannot be converged. So, that was the idea. We had proved that. So, let us prove the third that namely f of i is closed. So, let us look at y, y be such that there exists a sequence y n belonging to the range, there is a sequence y n belonging to f of i such that y n converges to y. So, what is to be shown? To show that f of i is closed we have to take a sequence y n which is convergent to some point y. So, to show that y belongs to f of i that will prove that f of i is a closed set. So, let us start looking at because what is given is i is closed bounded. So, we have to shift everything to the domain. So, y n belongs to f of i implies y n must be equal to f of x n for some x n belonging to i. For every n, this is happening. So, implies x n is a sequence in the interval i which is given to be closed and bounded. So, it must have a convergent subsequence. So, implies because i compact or closed bounded, there exists a subsequence say x n k of x n such that x n k converges to a point x belonging to the interval i. So, that is compactness. Every sequence has got a subsequence which is convergent in the set. But then f continues implies f of x n k converges to f of x. But what is f of x n k? f of x n k is equal to y n k which converges where does y n k converge? Because the sequence y n is convergent to y. So, that is what is given to us. y n converges to y. Every subsequence must also converge to y. So, now look at this statement 1, look at the statement 2. So, 1 and 2 limit being unique implies f of x is equal to y. So, the idea is from the range shift everything to the domain, analyze and come back. So, that proves that the subsequence f of i is also closed. So, what we have proved is if i is a closed bounded interval, then f of i also is a closed bounded interval. So, let us write this as it is slightly more elaborate way. So, always we had i is a b, f of i is a closed bounded interval. So, it has some n points. It is closed, it is bounded. So, let us say this is small m and capital M. So, let us write corollary of double. Then what does it imply? What does it mean? That the range f of i is the closed bounded interval small m and to capital M. That means small m is in the range of the function. So, implies for every x belonging to i, m is less than or equal to f of x is less than or equal to capital M. Range is equal to small m to capital M. That means what? Small m is the smallest value of the function. Small m is the smallest value of the function and capital M is the largest value of the function on the interval a, b. Not only that, small m is in the range. So, small m must be equal to image of some point. So, implies there exist some, let us call x min and some x, x belonging to a, b such that the value at x point x min is equal to small m less than or equal to value at every point is less than or equal to is less than or equal to capital M, which is the value at the point x min. Is that clear? I am just re-interpreting that result. For the, if the domain is a closed bounded interval a, b, we just now proved range must be also a closed bounded interval. So, if domain is a, b, range is small m to capital M. What does it mean? That range is equal to this. That means every value of the function is between small m and capital M. That means function is bounded. Small m is the minimum value, capital M is the maximum value and small m and capital M being in the range, they are the values taken at some point. So, what we are saying is, so you can this corollary can be rewritten as, so rewrite. If you like, you can write it as a theorem. It is due to v stress. So, the theorem says every continuous function f from a, b to r, I am just rewriting, doing nothing more than that, is bounded and attains its maximum. There are points where the maximum value of the function is, the range is bounded. So, by least upper bound property, it must have greatest over bound and least upper bound. What we are saying is, this theorem says that not only they exist by the completeness property, they are actually attained at some points in the domain. So, this is very useful in proving when we analyze maximum, minima of functions and optimization problems. So, that is, so we showed, in fact, if you look at the proof, the proof does not use the fact that other than the fact that you are in interval. If you do not want to claim that the range is a interval, you can just assume domain is a compact set, then the range is a also compact set. It is closed bounded. That is all. Only because of intermediate value property, we get intervals get mapped into intervals. Otherwise, the same proof works by saying that if you have domain to be a compact set, then the range is also a compact. Range is also closed and bounded. So, we did not use the fact that the domain was an interval in proving that the range is a closed bounded set. Only in interval, we use intermediate value property and those things. So, for a connected set, continuity preserves connectedness. Image of a connected set is connected. Image of a compact set is compact. Let us look at one or two more properties of continuous functions before we go over to something else. So, let us look at, right, there are some, you must have already come across, but let me define it. What is called a monotone function? A function f in a domain d contained in r to r is only for real valued functions upon reals is called monotonically increasing f for every x1, x2 belonging to the domain. The name is indicative of what we are looking at. x1 less than or equal to x2 should imply the image f of x1 is less than or equal to f of x2, right, increasing. As you move from left to right, your function is going up and up, geometrically. So, this is monotonically increasing and similarly you can define what is monotonically. So, one, two monotonically decreasing if x1, x2 belonging to the domain, x1 less than x2 less than or equal to the point x2 implies f of x1 is bigger than or equal to f of x2. The graph is going down and down as you move from left to right. So, that is decreasing. Keep in mind, we are saying less than or equal to monotonically increasing. I am not saying this is a strict inequality. If we want to say the function, if we want to indicate that f of x1 is strictly less than, then we will add the word strictly monotonically increasing and strictly monotonically decreasing. We will qualify what is increasing and decreasing if you want to say strict. So, let us say a function is called monotone if it is either monotonically increasing or monotonically decreasing. If the function is either increasing or decreasing, we are not very particular whether it is increasing or decreasing. We are only interested in knowing that it is increasing or it is decreasing everywhere. See, increasing and decreasing are not properties of the function at a point. These are properties of the function over the domain. By limit, continuity are properties of the function at a point. So, that is the difference. This is the property of the function over a domain, over a subset of the domain at least. So, monotone if it is either increasing or decreasing. So, let us observe our observations or properties. Let us look at one. Suppose f in a domain D to R is monotone is strictly monotonically increasing or let us say decreasing. Either it is strictly increasing or strictly decreasing. It is given to us. Then this implies the obvious property that f is 1, 1. f is to be 1, 1. Is that okay? The function is either strictly increasing or strictly decreasing. It should be 1, 1. Is it clear? Everybody why is so? Because if not, then what will happen? At two different points, f of x 1 is equal to f of x 2. But either x 1 will be less than x 2 or x 2 is bigger than x 1 either of it. So, either of it will contradict the fact that it is monotonically increasing or decreasing. Let us look at what about the converse of this statement. Suppose we are given given f is strictly monotone. Can we say, given if function is 1, 1 sorry the other way round. So, converse will be if f is given f is strictly, we are looking at the converse of it. Suppose f is 1, 1. Can we say f is strictly increasing or decreasing? Obviously not. Many examples. Let us say y equal to x square for example, if we can look at or any function could be anything 1, 1. Graph could be a weird function. So, this is the graph of the function say on the interval a to b. It is 1, 1. Every horizontal line cuts the graph only once. So, it is 1, 1. If we just want to look at the geometrical thing, but it is not increasing or decreasing. It is clear or if you want you can make it more complicated. You can just change the graph to something like anything is possible 1, 1, but still it is not monotonically increasing or decreasing. So, when can we say it is given to be 1, 1, but still ok. When can we say a 1, 1 function? Oh, sorry. Wait, did I there is not 1, 1 sorry. I gave a very wrong graph kind of a thing. I want example of a function which is 1, 1, but it is not strictly increasing or decreasing. So, can you think of a graph of such a function? A function which is 1, 1, which is 1, 1, but is not increasing or decreasing. Quite simple again. Let me just change this to what we want? We want a graph of function which is 1, 1. Can you think of drawing a picture of such a thing? I want a function which is 1, 1, but it is not monotonically increasing or decreasing. Just a picture. I do not want a formula. So, it seems it is not possible to draw it. Looks like you may not, but let us try to draw. If you want to draw it what should happen? So, this is a point A and this is a point B. I start here and I keep on increasing. It is 1, 1, but I want to break the property. It is increasing, but I still want to keep the property. It is monotone. So, I cannot go from here, but I can go like this. Is this a function? It goes up to here and then it starts here at this point. This is the point C. It goes up to here. On the left side, A to C, it is monotonically increasing. On the right side, it is monotonically decreasing and it is 1, 1. This function is not 1, 1. Shall we make it 1, 1? Let us make it 1, 1. Still it is not very good. Let me make it 1, 1. So, how do I make it 1, 1? Let me take this. Now it is 1, 1. So, that is how we experiment in mathematics. So, I have got a function which is 1, 1, but it is not monotone in the interval in the domain A to B. Why it is not happening? Because there is a break in the graph of the function. So, probably it is true if the function is continuous and 1, 1, then it should be monotone. So, let us write that as a theorem. So, when the function is 1, 1, what can you say if a function is 1, 1 and continuous? So, let us try to analyze a theorem F from on an interval I. Let us keep it on an interval I to is continuous and 1, 1. Then F is monotone. If it is continuous and 1, 1, that should be 1, 1. So, essentially what we are saying is we start at the point A and you decide whether you want to go up or you want to go down. So, when you start drawing, you start going up. You cannot come down because 1, 1 will be contradicted. So, the function should be monotonically increasing. So, let us write a proof of that. So, proof. Let us discuss the proof first. So, let us say, if let us assume F is not. So, this is also a good idea of not 1, 1, not monotone. If the function is not monotone, then what should happen? Mathematically, what should I say? There are, say for example, then I should happen, I should have three points A, B and C. So, then implies there exists A less than B less than C below in I, but what should happen? F of A, so either F of A, what should happen? If it is monotone, F of A will be less than F of B, is bigger than F of B and F of B is less than F of C. So, but either this or other way, something, other inequality should happen. So, what is happening? Here is a F of A or what should happen? F of A or let me write that also. If it is less than F of B is bigger than F of C, one of these things should happen. One of the things should be contradicted. So, now let us say F of, let us visualize F of B is somewhere here. So, let me write F of B and F of, it is bigger than F of A. So, probably here is F of A. It is bigger than F of A. Are you able to see or shall I draw a bigger picture? Maybe I think a bigger picture will be a better idea. So, let me draw somewhere here. So, here is A, here is B and here is my, sorry, here is B and here is my C, three points we have got. So, here is F of B, this is F of B and probably here is F of A because we are looking at the case, F of B is bigger than F of A as well as bigger than F of C. So, F of B and F of C either is here or it is in between somewhere here. So, two possibilities, either it is F of C is less than F of B, but it can be bigger than F of A. We do not know what is or this could be somewhere in between here. So, let us say this is one of the cases. Everything will be similar. Now, what is happening? So, look at this. This value F of A is between C and B. This value F of A is between F of C and F of, so F at B is less than F of A is less than F of C. Function is given to be continuous. So, F of A must be attained, F of A must be attained at some value in between B and C by intermediate value property. This value is between these two values. So, implies by intermediate value property, there exists a point X1 such that F of X1 is equal to A and where is X1? X1 is between B and C. Now, F of X1 is A, F of X1 is not F of A, sorry not F of A. In between values F of A. So, there is a point where the value F of A is taken and where is X1? Here is X1. X1 is between B and C. So, F of A is taken at A as well as at X1, but X1 cannot be equal to A. So, this contradicts one-oneness of the function F. Is that for everybody clear? Yes? If F of C was up and F of A was down, then we would have seen C is taken somewhere else also between the point A and again there will be a contradiction. So, this implies F is not. So, we analyze this case with a property that it is F of B is bigger than F of A. So, we specialize it, say one of the cases is this. If it is other way around A and C, then again there will be a contradiction. Similarly, you can analyze this one also. So, basic idea is the one-one property and continuity implies it has to be monotone. If not, it will contradict intermediate value property somewhere. So, write all other cases. Other cases are similar. So, write down the other possible cases. We looked at when this is in between. Other cases, F of C is in between F of B less than F of C less than F of A. So, that gives you two and similarly two other cases. So, a function which is continuous and one-one has to be monotone. So, this is the theorem that we just know. One-one continuous, it has to be monotone. Let us look at monotone function stately more...