 Welcome to the second module of point set topology course part 1, this is an NOC course of NPTEL. So today's topic is normed linear spaces. Recall that we use this notation K for either a set of real numbers or a set of complex numbers. Amongst all properties of the modulus function on K, you know, you talk about modulus of z which is equal to what? x plus x square plus y square and then we take the square root. So there are modulus functions on both r and c. So I am trying to handle both of them together. So what is the striking properties of this modulus function? I would like to list three of them. The first thing is that modulus of any non-negative modulus is always a non-negative real number. Whether x is a real number or a complex number, mod x is a non-negative real number. It is 0 if and only if x is 0. So that is the first property. Second property is if you multiply by some number r times x, then the modulus of that is modulus of r into modulus of x. I deliberately write r and x instead of x and y because the multiplication this r is supposed to be scalar multiplication. It is not a real number, it is real or complex, no problem. But just to indicate that they have different roles here. So that is why I am writing r here, that is all. r, even if you write modulus of x, y is less than or equal to mod x, mod y, it is correct, no problem. But do not take r as a real number, it is both for real and complex number. And the third property, modulus of x plus y is always a similar mod x plus mod y. So these things we have been using all the time, that is why they are the most important ones. So these properties will be now taken and made into an axiom. So that is what we are going to do. So I already told you that what I am going to do if it is not real, k is not real but k is c and c can be taken as r cross r. Namely, every complex number can be written as real part comma imaginary part, x comma y. Then modulus is nothing but square root of x square plus y square. So this looks like I have the same modulus function, same property wherein k would be r or r square. But no, you can do it for all r n. So that is the whole idea. Easily generalized to all r n. How? If you just take a vector now, an n vector x equal to x1, x2, xn, all x1, xn, xn are real numbers. Just like when n was 2, this was real part and imaginary part of a complex number. So in more general case, you have n coordinate real numbers here, the coordinates of x, they are all real. Then I denote it for, in a future reference also, this norm x2, that 2 indicates that I am going to take squares here and then square root here. Take the squares of each xi, sum it up, then take the square root. This is what is called as Euclidean norm. So we have just generalized the complex number of modulus into any, you know, you can call it as a modulus. So just to confuse, if you construct, you call it as modulus, do not get confused. So I am having a different notation here that is all. So the same properties can be checked, namely this norm now, norm x is non-negaterial number. It is 0 if and only if x is 0. Now x is 0 means what? Each x1, x2, xn are all 0. Similarly now it comes down. I cannot take r as x1, x2, xn, r1, r2, rn and x as x1, xn, no. x will be taken as x1, x2, xn, r will be now taken only from k, either to a complex number or it is just a real number. Now I want to take real number because all xi's I have taken real number. So this is the second property. Third property is again x plus y is the same thing, x1, xn plus y1, y2, yn here will be x1 plus y1, x2 plus y2, xn plus yn. The norm of that instead of modular is less than or equal to norm of x plus norm of yn. So these are the properties which will be, now these three properties which will be carried out for these normal function arms is number 2. So many of you have seen this one, I am just listing it. This x2 norm is non-negative and it is 0 if and only if all the coordinates are 0. So r times norm x is modulus of r times norm x. So r is inside now real number and x is r. Norm of x plus y, namely this 2 norm is less than equal to norm of x plus norm of y. So this was all right with the easy step. What we want to take the first step is something more substantial that is our first step of generalization. What is this? This normed linear spaces is what we are going to define. So what instead of taking Rn what we want to take is any vector space over k. So I can take c cross c cross c also. So in that case this r will become a complex number. So this makes sense. That is what we want to do. So take any vector space over k. k is a field. Whether k is r or c is a field, we have to take a vector space over k. Need not be even finite dimensional? It could be infinite dimensional also. Take any vector space. So by a norm function on v, we mean a function. This is a somewhat you know this notation is not functional notation. But this is the way we would like to write namely this is the slot for the function. So norm is determined from this vector space into what? Non-negatorial numbers. The closed interval 0 to open infinity. This entire ray here. Where is the codomain? So a function which satisfies these three hypotheses n1, n2, n3, n for norm. What are they? They are exactly the same thing as what we have seen here. Same guess what we have seen for the modulus function here. Very easy thing that we have done. Same property we write here. Only we have to take care here namely the x is a vector. x is always inside v. But r there whatever r appears, I have written deliberate you know alpha, which alpha is a scalar in the field whether k is either r or c. The first one is positive definiteness is a name. Norm x is 0 if we don't leave x is 0. Positive part corresponds to that it is taking non-negative numbers here. It is 0 if we don't leave x is 0. It is definiteness. This is a standard terminology here. The homothecy says that scalar multiplication then take the norm is the same thing as multiplying by the modulus of the scalar into the norm of x. This happens for every alpha in scalar and every vector. The third one is again a copy of the corresponding third property there. It says a name triangle inequality which is derived from the two-dimensional you know school geometry. If you take a triangle some of the two sides you know always slightly bigger than or equal to the third side. So that is what this triangle inequality means here. Norm of x plus y is less than or equal to norm of x plus norm of y. These x and y are now arbitrary vectors inside this vector space. When you have such a function it is called a norm. Together with such a function the vector space v will be called a normed linear space. This is a standard terminology now. Nobody else speaks about anything else. Earlier in the development of these things there were different names and so on. Let us have some examples here. What are called as LP norms. See now why I denoted this x with a norm x here with a 2 that will be clear here. Instead of these two I would like to write any positive real number. I would like to but I cannot. I will have to restrict myself namely p greater than or equal to 1. So suppose it is 1 then what is the meaning of this? This is summation modulus of xi I have to take. Don't take xi with as they are. This square was okay. Real numbers square are already positive, non-negative. So you would have to take modulus. But the correct thing is to take always modulus. Then it will work for complex numbers also. So this thing which is very simple has to be correctly generalised and correctly modified. So instead of 2 I can write p here. Then modulus of xi raised to p. Then instead of square root what should I do? I should put a pth root. Raise to the p here and then take the pth root of that. That is what I am doing here. Come here. Modulus of xi raised to p. Take the summation. Take the 1 by pth root and pth root of that of this total summation. So that to define again a norm is the claim here. So what you have to verify? You have to verify that if I put this whole thing to 0 then all xi's must be 0. That is easy thing. Some power raised to 1 by p is 0 means the summation itself is 0. But the summation is all non-negative number. So each xi raised to p must be 0. Therefore each xi must be 0. So the first part is totally obvious. The second thing is if I multiply this one by scalar alpha, all the xi's will be multiplied by scalar alpha. This is just scalar x comma y raised to p modulus. Modulus of x comma y first of all is modulus of x to the modulus. Raise to p is alpha raised to p into xi raised to p. Then alpha raised to p will come out here. When you take 1 by p, pth root of that it will just modulus of alpha. So that is the second one. So the first two n1 and n2 are straightforward. Computation, little more analysis is needed to prove the third one namely triangle inequality. Triangle inequality now becomes for p what it becomes modulus of xi plus yi raised to p summation and then take the pth root. It is less than or equal to individually you do the same thing for both x and y namely xi raised to p summation then raised to 1 by p plus yi raised to p summation and then 1 by p. So this is what you have to verify. This has a name. This is called Minkowski's inequality. It is not very straightforward but it is not very difficult also. There are theories here. So these things are very nicely done in elementary real analysis books. So what I want you to do is if you have not seen it please read it from that book. I am assuming that you know already a bit of real analysis. Real analysis courses will do have this kind of things. There are many books. If you do not know you can approach me then I will tell you exactly some books also no problem. So we shall leave it to you to read the proof of this Minkowski inequality from any elementary analysis book. Of particular interest for us is the cases where p equal to 1 and p equal to 2. When you put p equal to 1 what is it? You have to take modules of xi raised to 1. So there is no way here. You do not have to write. Take the summation and then take the 1th root. 1th root is same thing. So it is just summation modulus of xi. The second one which we already started with what is called as Euclidean norm. Started with the observation that how it happens, what it happens in r cos r namely in the plane. Summation xi square. If it is a real number then you do not need to put modules but complex numbers also it is valid. So you can take z 1 z 2 then mod z 1 you have to take mod z 1 square plus mod z 2 square you have to take then take the square root. So that is Euclidean norm. The first one this is you would call it as L 1 norm is 1 2 3 etcetera or p L p that is the what we have L p spaces L p norm. So this is L 2 norm this is L 1 norm. This L 1 norm as a different name it is called taxicab norm. This was actually introduced by Jordan. Before that people were usually thinking of this one single norm. This was something contribution from Jordan okay very simple minded thing. There were now many other norms which you will have to study. So these examples generalize immediately and effortlessly to the case when v is an infinite direct sum of copies of k. If I write a direct sum of copies of k v SWAT it is say z 1 comma z 2 comma z n then 0 0 0. So where you want to stop nobody tells you. So if it is only k power n then you have to stop it z 1 z 2 z n n to pulse. So that is the infinite direct sum. It could be uncountable also then you cannot write it as z 1 z 2 z n you have to write z alpha where alpha is an indexing set for them and so on. So but at each vector only finitely many entries will be non-zero. So you can take you can add you can take the modulus of them you can take the raise to p then you can add then you can all these things make sense immediately and everything works because each time you have to restrict it to only finite limit. So if you have proved Minkov's inequality other things are just easy or now right now you think of that you have proved it then you can apply it. Now there is another interesting case namely it is not direct sum or anything it is Lp is infinite sequences of real or complex number such that when you take this sum you see if it is finite sum there is no problem it is infinite sum what you have to do you have to say it is convergent. Now it becomes very crucial that I assume p is between 1 and infinity otherwise Minkov's inequality will not be true. Other things are okay but Minkov's inequality will not be true. So p belong to 1 to infinity I have taken then you get all this Lp that is a notation now L upper p remember L lower p was for the norm L upper p was a space here all sequences are exist okay infinite sequences all x i's are inside k okay summation modulus is absolute convergence modulus raised to p pth power that must be finite take such a thing to take two of them the sum will be also finite Minkov's inequality it will prove to prove that this is a Lp this is a vector space alpha times this is finite is of obvious okay so it is vector in proving this is vector space itself you can use Minkov's inequality okay so all that you have to do is you replace this n here by infinity to justify what you have to do you have to show that these are convergent and okay once you made it convergent you have made it as a hypothesis here you have put this hypothesis so finally what you have these why this one is true for when you have infinity you take just the limit it is true for each n okay it is true for partial sums so you can take the limit then for the limit also it will prove okay you do not have to prove Minkov's inequality separately for the convergent things that will automatically follow all right so we have all these L upper p spaces what are they they are convergent sequence convergence is with respect to this p modulus of summation x i raised to p summation x sorry modulus x i raised to p summation so those systems to converge okay there is one more interesting thing don't worry about convergence but put another condition just a weaker condition namely all those sequences which are bounded by some number see when you have infinite sequence you do not know make a one two three four five why it can keep going you do not want bounded okay take real or complex number no problem take only bounded sequences then you can talk about supremum remember supremum is not not the same thing as maximum because we are taking infinite sets here okay supremum will be also some finite number because it is bounded you put that supremum mass is norm x infinity but it is just a symbol here I am not taking p first and then taking the limit okay in some sense that is also true if you understand the geometry but this is just a symbol here for the supremum okay now verifying n1 and n2 is easy as before verifying n3 is also easier here you think about it if you haven't done it these are not difficult for the supremum you don't need lot of analysis it's elementary you know inequalities will do so that will give you minkowski something corresponding minkowski neutrality which I have called it as triangle inequality in this case okay for norm x infinity so n3 will be also easy here so this is called l infinity norm or sometimes sup norm okay yeah so if you take a vector subspace of vector space which has a norm then everything works for this vector subspace also if you just restrict the function norm function to the subspace that way you will get another norm near space okay so for example you can take this kn itself sitting inside all this l infinity space how kn is what kn is just z1 z2 zn then put zeros so a finite sequence can be made into infinite sequence you know extending by putting extra zeros okay so then take the maximum norm supremum norm various things all old minks are there already lp norms will be lp norms on kn but now I get one extra thing which I had not done namely the supremum norm becomes maximum the maximum of modulus of x okay so that is called norm x infinity is the same symbol because she is restricted but now it has a different name on the right hand side because she is maximum supremum becomes maximum okay only when these things are finite dimensional vector space so here is a remark about these lp spaces which just for getting because these are examples that I am giving you I am telling but you should be learning these things in analysis but let me just explain this point because I may sometimes use this one namely if p is less than q then lp is contained inside leq what is the meaning of that if a if a sequence if a series summation xn is convergent after raising power p you know absolute convergence is to power p then it will be convergent if you take a few hours why this is just a comparison test see a sequence which is convergent like this I am taking modulus here you can always assume they are real number positive real numbers okay so it is absolute convergence positive real number so any positive real number summation convergence means after certain stage okay these things must be small in fact limit of xn as a n tends to infinity must be 0 so they are smaller than 1 therefore when you take a higher power it will be smaller therefore each xn raise to q after certain stage will be smaller than xn raise to p therefore if this is convergent this will be also convergent so that is the comparison test here okay so this is elementary analysis so I have given you full explanation here okay why if you take a sequence like this it will be this one will also convergent convergence of suggesting one divided by n raised to s for example okay you have to use why this convergent if and only if s is bigger than 1 you take this sequence okay 1 by n raised to s between where s is between 1 by q and 1 by p this sequence will be inside lq but not inside lp that will give you that lp is contained inside lq but there are points in lq which are not in lp so containment is strict here okay one single example wherein s say you have to choose correctly okay so 1 divided by n raised to s this is a beautiful example this this series itself is very important okay so containment is fine but there are some other kind of relations also close relations between lp norms so now I am coming to the geometric aspect of this one look at just the numbers 1 2 3 p and then infinity don't worry about the in between real numbers there only take the integer okay just to concentrate on what is going on not for the sake of you know logical statement just for getting some ideas okay you can take one and a half and so on that may be more difficult to imagine what is happening so just take these numbers look at this picture this is in r2 okay because I am drawing a picture means it has to be yeah the plane right so plane is r3 r4 at a difficult to imagine okay so what is this picture this squared here this is set of all points x y such that modulus of x plus modulus of y is less than or equal to 1 this is x plus y less than or equal to 1 you can say this is 0 which is exactly y axis right so what is this one this is the circle what is this last thing this is the maximum of x comma y modulus okay modulus of x comma y is less than or equal to 1 so all these things are unit disks inside corresponding LP spaces this is L1 this is L2 the standard Euclidean this is L3 or L2 or L5 I don't know I have only drawn one of them then this last one is L infinity so you see now if you keep on taking bigger and bigger this line becomes flatter and flatter like this keeps going so in the limiting case it will become the outer square in that sense L infinity is actually the limiting of these things that is what I meant earlier okay so these are d2 for you know n equal to 2 p for p norm or x around r2 that norm of x is less than or equal to 1 so various disks I have shown okay so when p is smaller the the corresponding disk is also smaller this is L1 this is smallest one is contained in all of them as p increases they become increasing you know bigger and bigger and finally it will become a square like this so that is the picture in r2 similar pictures you can see in all n okay so that I can't draw but I can make this conclusion if p is less than q norm of xq is always less than equal to same x is same okay norm of xp the qth norm you know lq norm is less than lp norm so this is easy to verify okay argument is similar what we have done this kind of argument you have to use so this is the play for today there is a little more deeper relations between these lp spaces interrelation so that we will investigate a little later okay thank you