 So, let 1 bigger than p less than infinity. Look at Lp. Now, you realize why I was using 1 and 2 and so on. It is all sequences such that sigma mod x i in L1, we just took the sum. In L2, we squared them and now we take the pth power and take the pth root is finite. So, and define for every x belonging to Lp, p to be equal to sigma i equal to 1 to n mod x i to the power p raise to power 1 by p. And then you can define dp x, the distance equal to x y belonging to Lp. Here is something which we would like to prove, claim dp is a metric. So, let us try to prove it is a metric. So, to prove this, the first step will be trying to prove something like Cauchy-Schwarz inequality and then using that prove triangle inequality. The main problem is the triangle inequality. All other two properties are okay, they do not cause any problem. So, only thing is the triangle inequality. Even for 2, p equal to 2, we use Cauchy-Schwarz inequality and then use that to prove that it is a norm, it is a metric on it. So, same root we will follow. So, let us, so step one corresponding, what should be Cauchy-Schwarz inequality? What should be Cauchy-Schwarz inequality? So, that is called, so let me write that is called, it is called Holder's inequality. It is called Holder's inequality. And what does it say? So, Holder's inequality says the following. So, let us first state it in Rn and then do it for general thing. So, let 1 less than p less than infinity, 1 less than q less than infinity be such that 1 over p plus 1 over q is equal to 1. Then, what is n? For every a equal to say a 1, a 2, a n belong a n. Let us write another part, a vector b is b 1, b 2, b n belonging to Rn. If I look at product a i b i, sigma i equal to 1 to n, that is less than or equal to summation mod a i raise to power p, raise to power 1 by p n tau equal to 1 to n, mod b i raise to power q, raise to power 1 over q. That is what we say that the inequality says. Now, let us see why it is a generalization of Cauchy-Schwarz inequality. If you take p equal to 2, if p is equal to 2, 1 over p plus 1 over q equal to 1 should give you, sorry, p equal to 2. What is q? That also should be 2. So, what is this? 1 to n 2 raise to power 1 over 2, that is precisely Cauchy-Schwarz inequality. So, then for 2 is Cauchy-Schwarz inequality. It is precisely Cauchy-Schwarz inequality. We were able to prove Cauchy-Schwarz inequality very easily by looking at the linear combination of the vector and norm. A more general thing for this, another route of proving this is why I am looking at arithmetic geometric mean, the relation between them. What is the relation between the two numbers a and b? What is the relation between the arithmetic mean and the geometric mean? Arithmetic mean is always bigger than the geometric mean. But you think of in the arithmetic, in the geometric mean, you take the square root kind of. So, instead of square root, if you think of p, 1 over p, you want to generalize that inequality arithmetic mean, geometric mean. So, let me state that and then prove that and then come to this thing. So, here is the generalize. If a and b are positive real numbers and p is bigger than 1 and q is a number such that 1 over p plus 1 over q is equal to 1, then a raised to power 1 over p into b raised to power 1 over q, that is the geometric mean. If p is equal to 2, q is equal to 2 is less than a by p plus b by q. That is the right hand side is the arithmetic mean. So, it is a generalization of that, once you want to go beyond powers of 2. So, we will be, one can prove it. So, very nicely easily. So, probably I think I will go through the proof of this and then using this, one proves Holder's inequality, that is generalization of Cauchy-Schwarz inequality and then we will prove what is called the triangle inequality. So, let us see what is the proof of this. So, here things involved are p, q, a and b. So, the idea of the proof is first step, try to bring it to a equation in one variable kind of a thing. So, here is the, how do you do it? So, here is the idea of the proof. So, this is what you want to, proving everything on one side, you want to show it is less than or equal to 0. Now, write everything in terms of, instead of a, b, write in terms of a by b. a and b are two numbers given, positive. If I write everything in terms of a by b, then one variable only comes, a and b together do not come. So, that is the idea. So, let us write. So, this a raise to power p, b raise to power 1 over q, what is 1 over q? Keep in mind 1 over p plus 1 over q is equal to 1. So, 1 over q is 1 minus. So, write that is less than or equal to 1 and p q is equal to 1. So, this is same as this equation, the given equation is same as this equation in a by b. Now, q is also gone, only p is there. So, only two variables, p is fixed for us. a by b, you can call it as a variable t. So, what is this? If I call a by b as t, are you following what I am saying? The given thing less than or equal to 0, I divide by b. So, that everything becomes divided by b and 1 over q write in terms of 1 minus 1 by p. So, that gives me a by b. If you call it as t, t raise to power 1 over p minus 1 over p into t plus 1 over p minus 1 is less than or equal to 0. For every t bigger than a by b, you are called as t. So, if you can show it for every t bigger than 0, this holds, then we are through. What we are saying is, for every value of t, the function f of t, this is the value of the function f of t is less than or equal to 0. That means what? If I want to show f of t is less than or equal to 0, that means what? The minimum value of the function should be 0. So, let us and that calculus now comes to calculus. You have got a function which is differentiable derivative equal to 0. That gives you what is the point of minimum maximum analyze. So, it is a calculus proof. So, let us look at the proof. So, we want to show f of t is less than or equal to f of 1, which is equal to 0. At 1, the value is 0. At 1, what is the value at the point 1? That is 0. So, we want to show f is a function defined by this quantity and this function has the minimum value 0 at the point x t equal to 1. So, calculus problem now. So, how do you solve that problem in calculus? Take the derivative. Derivative equal to 0 and analyze. Either you can analyze the second derivative at that point or just look at left and right. So, you do that. So, one variable calculus. So, let us go back. What happened? So, this is one variable proof calculus only. No problem. Using this, we will prove our holders inequality. So, what is holders inequality? So, let us go to holders inequality. So, that says that if p and q are between 1 and infinity, then the dot product is less than or equal to this p raise to power 1 over p into q raise to power 1 over q. Generalization of the usual Cauchy-Schwarz inequality. For p equal to 2, q equal to 2, it is a usual Cauchy-Schwarz inequality. The proof is very simple. If you look at the idea, the same idea that if all x i's and all y i's are 0, there is a typo here. This q is the power. I will correct it. If all x i and all y i are 0, then both sides are equal, nothing to prove. So, assume that the vector x with component x 1, x 2, x n and the vector y with components y is not 0, divide by that if required. Then you get vectors with components x is equal to 1, y equal to 1 and you have to show that this inequality holds. So, let us look at a proof of that. So, that is the basic idea. So, divide alpha is norm. Here, the norm is to the p raise to power 1 over p. So, divide the vector by each component by alpha, each component by beta, call this as a i, call this as b i. Apply that lemma a raise to power p into b raise to power q is less than or equal to 1. So, apply that with this each alpha i. So, you get, apply for each alpha i, you get this component a raise to power p plus b raise to power q. So, a i raise to power x i to the power p. So, if you apply to each, you get this summation. Now, this is for every i. So, for each component I am applying that previous inequality. Now, sum it both sides over i. So, when you sum it up, i equal to 1 to n, i equal to 1 to n plus i equal to 1 to n. What is sigma x i to the power p? What is this quantity? What is this quantity? That is precisely alpha to the power p. Alpha was the norm and this is beta to the power p. So, they cancel out. So, it is 1 over p plus 1 over q. That is equal to 1. So, it says this is less than or equal to 1. So, that means sigma mod alpha i by i is less than, almost the same proof. Only instead of using that arithmetic mean is less than, is bigger than the geometric mean, we are using generalized inequality. a raise to power p into b raise to power q is less than or equal to a p divided by. So, that, so that is, so now I am going to, I think, let me do some writing also. So, we have got the holders inequality. Using this, we want to prove that what is called Minkowski's inequality. Do not go by, get scared by the name. It precisely says for x and y belonging to L p, norm of x plus norm of y is less than or equal to the pth. That triangle inequality for this, so what was norm? So, that is defined as sigma mod x i to the power p i equal to 1 to n raise to power 1 by p. That is the definition that this is equal to this. That goes by the name Minkowski's inequality. So, let us give a proof of that. So, proof is reasonably simple. So, let us look at the left hand side. So, that is norm of x plus norm of y p. What is that quantity? So, that is equal to sigma mod x i plus y i raise to power p 1 to n. Whole thing raise to power 1 over p. So, let me raise the power p, so that I do not have to write every time. So, left hand side, I have raised the power. So, that is equal to this. Now, here is how one thinks of a proof from this product. This is a product, which is the left hand side. I want to split it into two parts. Some, right hand side is a sum, left hand side is a product. So, I have to split this left hand side into a sum. Somehow, two terms. Otherwise, I cannot reach the target. So, how I can split it? The idea is very simple. So, let us look at that. So, write this equal to summation i equal to 1 to n mod x i plus y i raise to power p minus 1 into, right? Is that okay? Why I am doing that? Because this gives me immediately that this is less than or equal to, if I look at this quantity, that is less than or equal to mod x i plus mod y i. So, let us write that summation i equal to 1 to n x i plus y i raise to power p minus 1 into mod x i plus mod y i, right? Using triangle inequality. So, that is less. Now, let us split the two terms. So, less than or equal to summation i equal to 1 to n x i plus y i raise to power p minus 1 into mod x i plus sigma x i plus y i raise to power p minus 1 into mod of y i. Now, I have got two terms at least. I am nearing the target. Now, idea is somehow make these two each sum, right? What do you want? We want this quantity to be less than or equal to norm of x i plus, right? So, how do we do that? Now, this product, now let us use our holders inequality in this. Remember, improving triangle inequality for two, we use Cauchy Schwarz inequality. Same thing we are doing here. Now, at this step, we will use holders inequality. This is one term. This is another term. Product of these two, sigma mod, should be less than or equal to, what is holders inequality? Here is holders inequality. Product of a i b i is less than a i to the power p raise to power 1 over p into b i raise to power q 1 over q, right? So, here let us use holders inequality. So, less than or equal to. So, sigma i equal to 1 to n x i minus y i raise to power p minus 1. So, this one divided by raise to power p, right? So, this is my a i, that is b i. So, p minus 1 raise to power p raise to power 1 by p. Let me write because x i, I will write q here and instead of p, let me use q for this and p for, because x i belong to 1 power p. So, q raise to power 1 over q, sigma mod x i to the power p i equal to 1 to n raise to power 1 by p. Is it okay? I have called a i b i. So, a i is x i, b i are the other ones. Similarly, plus the second term. So, second term will be sigma i equal to 1 to n x i. Sorry, it is plus or minus, I do not know. So, it was plus, plus y i raise to power p minus 1 raise to power q raise to power 1 over q into b i raise to power 2 mod of summation mod y i raise to power p 1 to n raise to power 1 by p. And what was on the left hand side? Keep in mind, on the left hand side was mod x plus y p raise to power p. So, that is less than or equal to this. Now, one uses the effect that 1 over p plus 1 over q equal to 1. So, use that. So, what is this? So, this is p minus 1. I am just working out roughly here, p minus 1 times q. So, what is that equal to? Using this effect. So, if you want q times, so 1 over q is equal to 1 over q is equal to 1 over q is equal to 1 minus 1 by p. So, that is equal to p minus 1 divided by p. So, if I put q here, is it okay? 1 over p plus 1 over q is 1. So, what is 1 over q? It is 1 minus 1 over p. And that is p minus 1 divided by p. I take it there. So, this quantity is equal to 1. So, this quantity is 1. So, this quantity is q times. So, what is it? Divided by p. So, this is equal to times p. So, this quantity comes out to be. So, the right hand side. So, let me write. It is just an mathematical simplification of this powers. Nothing more than that. x plus mod y p to the power p is less than or equal to this quantity. That is same in both. So, let me write it outside that quantity. So, it is summation x i plus y i summation raise to power 1 over q of p summation of x i plus y i raise to power p i equal to 1 to n. So, that is equal to p raise to power 1 over q into norm of x. And that second term is norm of… Can you see some relation between this and this? This is norm of… So, this quantity is norm of x plus y. p raise to power 1 over p is the norm. So, it is just norm of p raise to power q 1 over q. Now, you can transport it on the left hand side. So, this, shift it on this side. So, what do you get? So, norm of x plus y raise to power p minus 1 over and on this side, divide by both sides by this term. So, when you divide, what you will get? This is less than or equal to… Is that okay? This is to the power p. Yes? This is to the power p. And this is to the power p raise to power p by q. Sorry, not the same. So, this is… This is… So, p minus… Sorry, anyway I will be getting problem with this if I do not write it correctly. p minus p by q. Because this thing is p by q, mod norm of x plus y raise to power p by q. And that is power p. So, shift to the other side. And what is this quantity left hand side? What is p minus p by q? 1 over p plus 1 over q is equal to 1. So, that is the quantity p into 1 minus 1 over q, if you like. So, that is 1 over p p by p that is equal to 1. So, this quantity is just 1. So, you get the required thing that this is less than or equal to p plus norm y. The basic idea is of the proof is going parallel to one variable, parallel to L2 case. You first generalize Cauchy Schwarz inequality to… Instead of p equal to q equal to 2, you generalize it to Cauchy Schwarz inequality called Holder's inequality, where a product is less than or equal to a sum. Basically, that is what the AMGM is. A product is less than or equal to sum. Arithmetic mean is bigger than the geometric mean. So, generalize that quantity. And that gives you Holder's inequality. And the proof from how do you use Holder's inequality in Minkowski's inequality is precisely from the product you want to go to the sum. So, this is a crucial step. Get the power into two parts, that into minus 1 absolute value, this minus 1 into this. And this gives you less than or equal to norm. So, splits into two terms. And then is only the arithmetic jargon. You simplify the powers and all that. So, that is Minkowski's inequality. So, that says on Rn for every 1 less than p, actually we have got 10 to the power 1 also. You can define on Rn, you look at what is called, because it is finite. So, no problem. So, define norm of x to p to be equal to sigma mod x i, i equal to 1 to n to the power p, raised to the power 1 over p is a norm. So, what does norm mean? It is something like absolute value. We give it as name as norm and dp of xy equal to x minus yp is a metric. This is a metric on Rn. So, on the same set Rn, you have got infinite number of matrix. For every p, p equal to 1 also, we have done it. p equal to infinity also, we have done it. Every power in between, p bigger than 1 and less than infinity, also we are getting a metric. In the next thing, what we will show is not only a metric on Rn, it is a metric on R infinity also. It is a metric on, but that is a minor step, because these are partial sums already, 1 to n kind of thing, we have it. From there, we will go over to what should be the next step? R infinity, what should be the next thing? R1, Rn, R infinity, what should be the next thing? Anybody has a guess? Is there anything beyond R infinity? R to the power infinity? So, think about it till the next lecture.