 So, applying Akutani's fixed y n theorem, we are going to get to Nash's theorem. So, this is Nash 1. There exists a y 1 star to y n star such that j i of y 1 star to y n star is less than equal to j i of y i comma y minus i star for all y i in capital y i and for all i n n. This is what we are going to prove. So, the proof idea is going to be to simply apply Akutani to R. So, what did we had just establish that y is a Nash equilibrium if and only if y sorry y star is a Nash equilibrium if and only if y star belongs to R of y star. That means if and only if y star is a fixed point of R. So, to show that there is at least one Nash equilibrium, all I need to do is show that R has a fixed point. So, and now to show that what how will we show that R has a fixed point, we will show that we will apply Akutani's fixed point theorem which means that we are going to show that if I take phi as R, all these hypotheses are satisfied. So, this is that is that is going to be the proof idea. So, we need to check a few things. So, let us go through them one by one. So, in order to apply Akutani's fixed point theorem to R, we need to check that the domain of R is close convex and bounded. We need to check that R actually maps that set the domain to subsets of itself. We need to check that R is convex valued and we need to check that R has a close graph. Once these once these are checked Akutani will then give us that there is always a fixed point and hence there is always a Nash equilibrium. Alright. So, firstly what is the domain of R? So, domain of R remember was of R is y. So, R maps R maps y to subsets of itself and so R is in fact mapping from a set to subsets of itself. But what but is this domain closed and close convex and bounded. So, what was y remember y was simply the product of these yi's right. y was the product of these yi's. Each yi itself was a set of probability distributions on a finite set. So, probability distribution that means basically it is every component is between 0 and 1. So, the entire each yi is basically in some box of dimension 1 lying in some box of dimension 1 in its own space. So, y being a product of all of that is in a hyper box of higher dimensions, but at the end of the day it is in some box or in some bounded set in its own in its space right. So, therefore so since y so since yi's is a set of is the set of probability distributions the set of probability distributions on Si this is in fact. So, this set is therefore yi therefore is bounded it is also close because you take the take a sequence of probability distributions it will converge to and if it converges then the limit will also be a probability distribution why because you can see that if I take vectors yi say yi k these belong to y capital yi and if I take their limit as k tends to infinity sorry take the limit of k as k tends to infinity as yi suppose they converge to some yi then each of these belonging to yi means that this equals 1 for all k it also means that this is greater than equal to 0 for all k and therefore if I take the limit as k tends to infinity the limit should also be should also have this property that one transpose that limit would be equal to 1 and that limit would be greater than equal to 0 which means which effectively implies that that one transpose yi will be 1 and yi will also be greater than equal to 0. So, in other words therefore so yi is bounded because it is these are probability distributions and they are all this it is closed for the same reason and then what about convexity you take two probability distributions and take a convex combination of those two probability distributions that is another probability distribution right you mix two probability distributions you get another probability distribution. So, which means that it is its convex another way of looking at this is that remember yi is simply defined by linear equations and inequalities. So, therefore it is a convex set and it is also therefore a closed set but because these they distribute the what it defines is a probability is the set of all probability distributions it automatically is bounded as well. So, in other words capital yi is closed convex and bounded ok alright. So, once capital yi is closed convex and bounded therefore yi is also being a product of such sets is closed convex and bounded. So, with this then what we have ticked off is this we have ticked off the first part. So, we this one we have a we have a function that maps a set to subsets of itself and it is a domain to subsets of that domain and that domain is a closed convex and bounded set alright. So, the first this is done now let us go to the next one. So, now we are we are looking at this one we need to ask is the function is their set valued map convex valued ok. That means what we are asking is R convex valued ok. Convexity I just argue I just said that it follows because you take convex combination of two distributions it is a convex it is a probability distribution ok. Alternatively you can see that y here you see yi is just defined by linear inequalities and equations. So, therefore necessarily for you know yi is a convex set alright ok. So, now we will be showing that R is we will show that R is convex valued it follows trivially actually that if A and B this is a standard property that if A and B are convex or they have A and B are convex or closed or bounded the product will also have inherit that property trivially actually. So, I just write I just quickly write the proof here. So, take any two sets A comma B or convex and take now you take a point and take a point take two points x1 x2 in 8 cross B this one you let us write this as x11 x12 and this guy let us write it as x21 x22. So, now if you take a convex combination of x1 and x2 you take alpha times x1 plus 1 minus alpha times x2 where alpha is in is in 0 1 ok. So, this is therefore now alpha times the first you know you get a convex combination of the first component. So, this here would reduce to alpha x11 plus 1 minus alpha x21 and alpha x12 plus 1 minus alpha x22. So, this is in A this the lower one is in B the A is because A is convex and B is convex and so, therefore, the product is also in is in A is in A cross B no, no, no. So, the multilinearity. So, this is the set of all probability distributions the set of all probability distributions is convex if you constrain the set of probability distributions like through a Markov chain or something then you do not then it is not necessarily convex alright. So, now we want to show that R is convex valued. So, for this let us recall go back and see what R was. So, R was the product of R R i's and each R i is the set of best responses of player i ok. So, R i of y minus i is the set of best responses of player i to y minus i ok. Now this what we want to so, it is to show that R is convex valued what we will show is that each R i is convex each R i is convex valued in other words for each for if you fix a y if you fix a y then for every i and if R i of y minus i is going to be is a convex set ok. Now, why is this a convex set is why is the set of best responses always a convex set ok. So, for that we need to just carefully look at what this what is actually happening here ok. So, and this is once we understand this the next part will also follow very easily. So, if you look at the cost of player i that is listed here that is written here it is it is this is it is this expectation right. Now what we are doing is when we are evaluating the best response what we are basically saying asking is we are fixing all the y minus i's ok. So, we are fixing the mixed strategy of the other player and asking what is the best that this guy can do. So, if I fix everything other than y i here all. So, fix all the y j's other than y i then what sort of expression is this in y i it is linear in y i right. So, if you look at this expression as a if you see it is as a function of y 1 to y n it is actually it has this product of all the y's involved ok. So, it will in fact be a polynomial in y 1 to y n right because it will be a polynomial each term involving a product of n n monomials right ok. So, this is going to be a polynomial, but if I fix the y minus i then it is linear in y i clear. So, which means that is what are this these best responses the best responses are actually solutions of a linear program they are the ones that minimize this linear function over all the y's right. So, in other words this here. So, now let me go back here r i of y minus i is simply the solution set of this linear program right it is r min of g i of this over y i in capital y i ok and this here this chap here this guy is linear in y i ok. Now, if this is linear the objective is linear in y i what about the constraints which is which is this capital y i capital y i itself is defined by two linear constraints is defined by the requirement at y 1 transpose y i is equal to 1 and y i is greater than equal to 0 right. So, if you put together this these two facts the objective is linear in y i the constraints are also linear in y i which means then that r i of y minus i is simply a solutions is the solution set of a linear program or a linear optimization problem. Now, if this is a solution set of a linear program then what is this what do we know about the solution set of a linear program we know that the solution set is always going to be if there are two solutions to a linear program then every segment the entire line segment joining them is also going to be a solution right which means that if you have two points in r i of y minus i then the entire line segment joining them is also in r i of y minus i right which means what which means that if you have two points in r i of y minus i the convex every convex combination of those two points is going to be in r i of y minus i which means r i of y minus i is actually convex right. So, if you take two points in it the entire segment joining it is also in that set which by definition makes the set convex. So, which means that the r i of y minus i is actually a convex set is convex and it is convex for each value of y minus i for each y minus i. So, you fix y minus i you get a convex set. So, what does this mean then therefore, then r itself r of y was itself r 1 of y minus 1 till r n of y minus n and each of these guys is a convex set each of these factors in this product is a convex set. So, therefore, this r of y is also a convex set and this is exactly what we wanted to show we wanted to show that r is convex valued right which means that for every y r of y is a convex set means that r is convex valued ok. So, let us go back to Cacutani again and let us say what he let us see what he has to say. So, we have now knocked off this we have shown that r satisfies this is that it is convex it is convex value.