 Yeah so what I want to do is you know try to tell you about some other important theorems which are connected with normal families okay. So the theorem that we are going to look at today is called Royden's theorem okay and the whole point about the theorem is that you know whether you can decide the normality of a family if you have a growth condition on the derivatives of the functions in the family okay. So remember that normality is connected with the derivatives okay Montel's theorem will tell you that you know if for a family of analytic functions normality is the same as saying that the you know the original functions are themselves normally uniformly bounded okay and the first course the Cauchy integral formula will then tell you that the derivatives will also be you know normally uniformly bounded and the normal uniform boundedness of the derivatives will give rise to equicontinuity okay and then you are in an Arzela Ascoli kind of situation and you will get normal sequential compactness okay and then the same kind of thing there is the same kind of philosophy with Marty's theorem as well because in that case you are looking at Meromorphic functions and then the theorem says that the normality of a family of Meromorphic functions is directly the same as the normal boundedness of the spherical derivatives okay. So you have to take derivatives with respect to the spherical matrix alright and then so but mind you normally whenever you have a family of functions that whose derivatives for example if you have a family of functions which satisfy Lipschitz condition okay that is the difference in the function values is are bounded by a constant times the difference in the variable values okay this is the kind of condition that you will get if for example the derivatives are bounded okay. So basically if you have Lipschitz type condition which is how you must think of a condition where your derivatives are bounded whenever you have Lipschitz type condition then you are actually getting equicontinuity and then you can apply Arzela Ascoli theorem to get to get normal sequential compactness okay so here is what we are going to look at today is Roiden's theorem which says that if you have a family of functions okay and assume that the derivatives of the family. So I am looking at either I can look at analytic functions or I can look at Meromorphic functions the only thing is that if I look at analytic functions I mean I must consider the derivatives where they defined okay so if they are Meromorphic functions I should not worry about the derivatives at the poles and I am not in principle I am not looking at the spherical derivatives okay I am looking at the ordinary derivatives at all points other than poles okay so if the derivatives grow as at most like an increasing function of the original functions okay then the family is normal this is Roiden's theorem okay so let me write this down so Roiden's theorem suppose script F is a family of analytic functions on a domain D the complex plane such that there exists a strictly increasing function so this is the function psi from non-negative real numbers to non-negative real numbers so it is strictly increasing function such that the modulus of the derivatives of the functions in the family they are bounded by psi of the modulus of the functions okay so this is the condition in Roiden's theorem the condition is that the derivatives of your functions in a family the way they grow is at most like an increasing function of the growth of the original functions so what is there on the right side is psi of mod F okay mod F indicates the growth of F okay in modulus and psi of mod F is psi of the growth of it says that the psi of mod F will be an increasing function of mod F because psi is an increasing function okay so what you are saying is that the derivatives grow as an increasing function of the original functions okay then the family F is normal okay and the significance of this is that you know because of Montel's theorem this will tell you that the family is going to be normally uniformly bounded okay it also tells you that the derivatives will also be normally uniformly bounded okay it is a very powerful condition alright the point is that when you look at this condition it looks as if you know the derivatives are growing pretty fast okay it looks as if the derivatives are growing see what you want is the derivatives to be bounded on compact subsets okay you want derivatives to be normally uniformly bounded that means you should you want them to be uniformly bounded on compact subsets which means that on a compact subsets you want a uniform bound okay you know but the point is that what this says is the derivatives are growing but the growth is at most like an increasing function of the modulus of the original functions okay so it looks it directly does not look as if this is going to lead to normality but the theorem is that it does lead to normality and the reason is that this is also a kind of boundedness of derivatives with respect to a different kind of metric on the Riemann sphere which can be defined using psi so that is the whole idea okay so let me write this let me write the proof so here is the proof probably I will use proof firstly assume that psi of t is you know less than or equal to 1 plus t square by 2 okay suppose you assume this suppose you assume this alright because I want to then consider the case when it is greater than or equal to 1 plus t square by 2 suppose I assume this alright then you see if you calculate the spherical derivative of any function in the family what I will get is f hash of z is what it is going to be by definition 2 times mod f dash of z divided by 1 plus mod f of z the whole square this is what it is alright this is the spherical derivative mind you these are all analytic functions but I can also consider them as meromorphic functions and spherical derivatives are defined okay even for meromorphic functions. So now you see the condition is that mod f dash mod f dash is supposed to be less than or equal to psi of mod f okay this is the condition in the theorem so I can write this is less than or equal to 2 psi of mod f divided by 1 plus mod f the whole square alright but then 2 psi of mod f by 1 plus mod of the whole square is less than or equal to 1 that is because of this okay so if I assume psi of t is less than or equal to 1 plus t square by 2 then all the spherical derivatives are bounded on the whole domain and of course you know if the spherical derivatives are bounded on the whole domain then Marty's theorem tells you that this is equivalent to the normality of the family okay. So this is equal to normality of the family considered as a family of meromorphic functions but that is also the same as a normality of the family considered as a family of analytic functions only the only thing is that you should allow the possibility that you can have normal convergence to the function which is identically infinity okay when you consider the spherical measure okay. So you know the case when psi of t is less than or equal to 1 plus t square by 2 is trivial alright it is just because of Marty's theorem then by Marty's theorem so let me write this down f is normal okay so assume the other possibility that psi of t is greater than or equal to 1 plus t square by 2 for beyond certain stage t greater than or equal to t not okay I can replace the function psi of t by another function for which the same condition holds with t greater than or equal to 0 that means I am saying that I can without loss of generality can assume t greater than t not equal to 0 and why is that so because you just have to put psi 1 of t to be equal to psi of t plus t not if you do this alright then you know psi of t plus t not is certainly going to be certainly going to be equal to is going to be greater than or equal to 1 plus t plus t not the whole squared by 2 that is if t is greater than or equal to 0 okay so the mind you psi of t is greater than or equal to 1 plus t square by 2 provided the t is greater than t not so I need t plus t not greater than t not and that is the same as t greater than or equal to 0 okay so I will get this but then of course this is greater than or equal to 1 plus t square by 2 okay and so the so you know if psi of t is greater than or equal to 1 plus t square by 2 for t greater than or equal to t not I can replace psi of t by psi 1 of t with psi 1 of t greater than or equal to 1 plus t square by 2 for all t greater than or equal to 0 okay and the point is that psi 1 of t is certainly greater than psi of t because psi 1 of t is psi of t plus something and psi is increasing so psi 1 of t will be also greater than or equal to psi of t this will also be true okay and the point is that this is of course because psi is increasing okay and therefore what will happen is that you would have got mod f dash is going to be less than or equal to psi of t which is less than or equal to psi 1 of t okay so all these considerations tell you that without loss of generality can assume psi of t to be greater than or equal to 1 plus t square by 2 for t greater than or equal to 0 okay so without loss of generality we may assume that psi of t is greater than or equal to 1 plus t square by 2 for t greater than or equal to 0 okay you can do that now namely if not you replace psi by psi 1 alright and call psi 1 as psi if you want alright again so then the next reduction I am going to make is that I am going to assume that psi is also continuously differentiable function of t that is because you know psi is a monotonic function okay and I can always approximate it to buy continuously differentiable function okay so we can also assume that psi dash of t exists and is continuous this is the same as saying psi is c 1 okay it is continuously differentiable function you can do that because after all you can replace psi by any bigger function okay and you can replace psi by a bigger function which is smooth okay you can always do that and well now you know what one is going to do you are going to use this psi to give a metric on the Riemann sphere okay and look at the induced metric on the domain in your in the complex plane so what you do is for z 1, z 2 in the domain okay define d psi of z 1, z 2 to be equal to okay you see you take integral from z 1 to z 2 okay along any smooth path or along any even piecewise smooth path even a contour okay and what you integrate is that so this is you integrate mod d zeta by psi of mod zeta okay see notice and of course you know you have to take the infimum of all this so this is this integral is over gamma so infimum over all gamma where gamma is a smooth or piecewise smooth path from z 1 to z 2 you take all possible paths contours from z 1 to z 2 integrate along that this quantity mod d zeta by psi of mod zeta you will do this okay and notice that you know the point is psi psi is see try to understand psi is greater than psi of t is greater than 1 plus t squared by 2 so 1 by psi will be bounded by 2 by 1 plus t squared 1 by psi of t will be less than or equal to 2 by 1 plus t squared okay and 2 by 1 plus t squared is you know that is the that is the form of the integrand that you will have to put when you want the spherical metric okay and 2 by 1 plus t squared is bounded as t goes to infinity if you want right so it is a so the integral is always nicely defined alright so you and you take this infimum alright then the point is that this is a this gives you a metric on the this d sub psi the way I have written it gives a metric on the on the Riemann sphere okay or you can think of it also as a metric on d d considered as d identified with its image in the Riemann sphere if you want okay but the point is that this gives a metric okay and you know in particular the an infimum of a set of quantities is always you know less than or equal to each of those quantities so if for gamma I had taken the straight line path from z 1 to z 2 this is also less than or equal to integral from z 1 to z 2 okay is straight line path straight line segment from z 1 to z 2 of this quantity mod d zeta by psi of mod zeta okay and the point is that see but the point is that this is this is bounded see 1 by psi zeta is bounded by 2 by 1 plus mod zeta the whole square because that is exactly that is this condition right we have psi of we have assumed psi of t is greater than or equal to 1 plus t squared by 2 okay so 1 by psi of zeta psi of mod zeta is going to be this is going to be bounded by 2 by 1 plus mod zeta the whole square okay you have this alright so this is bounded by integral from z 1 to z 2 mod d zeta 2 mod z d zeta by 1 plus mod zeta the whole square okay but what is this is actually the spherical length from z 1 to z 2 so this is just d spherical of z 1, z 2 so what you have proved is that the d sub psi is a metric which is bounded above by the spherical metric okay but the point is that for this for the spherical metric the Riemann sphere itself is compact okay it is compact with respect to the spherical metric because you know the spherical distance maximum spherical distance is there is a maximum to it is you know it is the it is going to be just the half the circumference of a sphere of radius 1 okay it is going to be namely it is going to be just pi that is the maximum distance you can get on the Riemann sphere the Riemann sphere is the sphere of radius 1 okay then any great circle on it will have radius 2 pi I mean we will have circumference 2 pi okay and the maximum distance you can get is from the for example from a point to its antipodal point for example from the north pole to the south pole north pole representing infinity south pole representing 0 and the maximum distance you can get is pi so see it is a space with finite diameter and it is compact okay and the point is that for a compact space okay if one metric is bounded above by another metric then these two metrics are uniformly equivalent okay so the point is that since the Riemann sphere so the Riemann sphere sphere or you can you can just consider c union infinity or the extended complex plane is compact with respect to the spherical metric okay we have that these two metrics d psi and ds are strongly equivalent. So at this point let me very quickly tell you about the strong equivalence see if you have two metrics on a space we generally say that these two metrics are equivalent if they induce the same topology okay this is called weak equivalence okay now what is strong equivalence is a condition that each metric is bounded by the other metric by up to a constant an absolute constant so if you have two metrics d1 and d2 on a space x we have to say that x to say that d1 and d2 d2 are equivalent for x means to say that the topology induced by d1 is the same as topology induced by d2 okay one of the sufficient conditions is that every ball in d1 contains a ball in d2 and every ball in d2 contains a ball in d1 okay the nesting of balls condition as it is called okay. So this is just to say that the two metrics are topologically equivalent but there is something called strong equivalence. Strong equivalence is that the two metrics each metric is bounded above by the other metric up to multiplication by an absolute constant. So if the two metrics are d1 and d2 you should get d1 less than or equal to lambda d2 and you should get d2 less than or equal to mu d1 you should be able to find such absolute constants if you are able to find such absolute constants these metrics are said to be strongly equivalent. Now of course strongly equivalent means equivalent but what is the beauty about the strong equivalence it is the following see if you on a space suppose you are considering functions continuous functions okay then if you change the metric to an equivalent metric that is you change it up to weak equivalence okay that is you change the metric by another metric which gives the same topology continuity will not be affected because after all you have not changed the topology continuity that just depends on the topology but the problem is uniform continuity will become a problem okay if you have a uniformly continuous function okay on a subset suppose you have a function which is continuously uniformly continuous on subset with respect to one metric if you replace that metric by an equivalent metric the uniform continuity may not be preserved if you want also the uniform continuity to be preserved you should replace it the metric necessarily by a strongly equivalent metric not just by any other metric which will give the same topology. So what will happen is that if you change the metric by just another equivalent metric namely you do not that you are only worried about the topology what will happen is that a function which is continuous uniformly with respect to one metric may fail to be uniformly continuous with respect to the other metric but if you want to preserve uniform continuity you have to replace the metric only by a strongly equivalent metric okay and the point is that you see if you have two metrics on a space and such that one metric bounds is an upper bound for the other metric and the bounding metric with respect to the bounding metric the space is compact then it also means that the some multiple of the smaller metric some multiple of the smaller metric also bounds the larger metric that can be proved okay because of compactness therefore just the bounding of one metric by another on a compact space where the bigger metric where the space is compact with respect to the bigger metric will tell you that they are strongly equivalent okay. Now you know why I am saying all this I am just saying all this to tell you that you know if you are looking at a family of functions you know to decide uniform continuity with respect to the spherical metric I can as well decide uniform continuity with respect to the metric any metric that is strongly equivalent to the spherical metric okay and so this applies to uniform continuity it applies to equicontinuity and things like that right. So we are now more or less done see now what you do is that now you look at the following thing so if you calculate for small f in script f what happens is that if I calculate the under the new metric d sub psi if I calculate fz1 fz2 okay what will I get this is going to be infimum over all contours from z1 to z2 integral from z1 to z2 along that contour gamma of mod d zeta by psi of mod zeta this is what it is with you know the substitution zeta equal to f of z okay this is the definition right and you well what is this going to give me see notice that this is being an infimum this is certainly less than or equal to you know the integral over the straight line segment okay. So if I take this integral over z1 to z2 and I took take the straight line segment what I am going to get is mod d zeta by psi of mod zeta and of course I have to put psi zeta equal to fz so if I do that I am going to get integral from z1 to z2 along the straight line segment of I will get mod f dash of z into mod d z because that is what mod d zeta will be okay and I will get mod divided by psi of mod fz okay but then what is my what is the what is the hypothesis in the theorem the hypothesis in Royden's theorem is that the numerator mod the numerator of the integral mod f dash of z is bounded by psi of mod fz so it means that this integrand is less than or equal to 1 so you know what it means is that I will get this is less than or equal to integral from z1 to z2 over the straight line segment of you know mod d z but you know integrating mod d z will give you just Euclidean metric okay so this will be see it will simply be mod z1-z2 this is all that I am going to get okay so alright so what have we proved we proved therefore that what is there on the left side on the left side I have d psi of fz1 z1, fz is you have proved that this d psi of fz1, fz2 is bounded by is mod z1-z2 that is a Lipschitz condition that is a Lipschitz condition on f with respect to the new metric d psi on the Riemann sphere but the moment you have a Lipschitz condition it implies equicontinuity so it means that the family f is equicontinuous okay this family script f is equicontinuous with respect to the metric d psi but then d psi is strongly equivalent to ds therefore the family script f is also equicontinuous with respect to the spherical metric but if it is equicontinuous with respect to the spherical metric I am in the I am I can use as in the proof of Marty's theorem I can use the Arzela Ascoli theorem and I can use a diagonalization argument to conclude that the family is normal and that is the proof okay. So thus script f is equicontinuous on d with respect to the spherical metric but since the spherical metric I mean sorry with respect to the script f is equanimous on d with respect to d psi but since d psi is strongly equivalent to the spherical metric ds script f is equicontinuous on d with respect to the spherical metric. Now by the Arzela Ascoli theorem and the diagonalization argument we conclude script f is normal it is a normal family okay so that is the proof of Royden's theorem. So the whole idea is that this Royden's condition is actually the Royden's condition translates to Ellipschitz condition when you use different metric on the Riemann sphere okay. So what is the advantage of this? The advantage of this is we can apply it to an example like this let d be a domain and script f be the set of all functions f analytic on d such that you know mod f dash of z is bounded by e power mod f z look at functions like this mod f dash is bounded by e power mod f okay now then this family is normal because here e power mod f is the function e power t and e power t is of course a smooth it is an increasing function of t for t greater than equal to 0 and you can apply Royden's theorem okay. So you know you get this nice condition that you know if you are looking at a family of analytic functions whose derivatives grow at most as the grow exponentially as the functions okay the derivatives are f dash the derivatives of the functions are given by f dash okay and they are mod f dash that is the rate at which f dash grow and mod f dash is bounded by is at most like e power mod f that is a condition but the so the function involved is e power t which is an increasing function of t alright and you know it mind you this is for any domain for example if you are taking the domain to be the whole plane okay e power mod f okay could e power mod f is a kind of exponential growth okay and it seems to be even if you put mod f dash is equal to e power mod f okay it looks like the derivatives are growing exponentially as the original functions okay but so it looks as if you know the derivatives they seem it seems as if the derivatives cannot be normally uniformly bounded it looks because whenever there is something growing exponentially you are worried about boundedness okay. So if you take a compact subset of the domain alright one is worried whether you know whether the derivatives will be bounded but the truth is they will be and it is not obvious it is not it is not an obvious result what you have is the derivatives are growing exponentially okay as the original functions and from that trying to conclude that the derivatives are bounded uniformly on any compact subset is a great thing. So you see Royden's theorem we can will now apply it will tell you that the that this one family is normal but if this family is normal Montel's theorem will tell you again or you can directly even use Marty's theorem Marty's theorem will tell you the spherical derivatives are bounded okay and if you want use you can use Montel's theorem which will tell you that the original functions are themselves normally uniformly bounded okay so if you take a compact subset of the domain then the functions are themselves uniformly bounded and the Cauchy integral formulas will then tell you that the derivatives of the functions are also normally uniformly bounded that they are bounded on every compact subset which is not at all obvious because the derivatives seem to be growing exponentially as the functions. So that is the significance of this theorem okay and it has got to do with the study of normal families and that is the reason why I wanted to present it here okay. So I will stop here so let me complete the sentence I will just say that then by Royden's theorem script f is normal in particular the derivatives f belonging to script f are uniformly bounded on compact subsets of D which is quite surprising given that the original condition was just that the derivatives are growing exponentially with respect to the functions.