 Now what we are going to do now is move ahead to you know try to give a proof of the very important Riemann mapping theorem okay and for this I will need to know I mean we will have to look at other things in as a preparation, so the first things that we will be looking at are the so-called harmonic functions about which you would have studied in a first course in complex analysis okay, so what I am going to do is try to recall harmonic functions the so-called mean value property then the maximum principle and then the Schott's lemma which is which is the fundamental lemma that we need in the context of the that is the simplest lemma that we want in the context of Riemann mapping theorem. So much of this is something that you would have seen in a first probably you would have seen in a first course in complex analysis but nevertheless it is important, so this will help you to refresh your memory if you have seen it once and if you have not seen it this is an opportunity to learn it, so we are looking at harmonic functions, so basically you know you take D or rather U in the complex plane a domain and f, so let me use a small U from capital U to R, so it is a real valued function okay is called harmonic if it is continuous and has partial derivatives up to order less than or equal to 2 which are continuous satisfies Laplace's equation Laplace's equation which is del U is equal to 0 where this where this that is delta U is equal to 0 where delta is the Laplace operator it is dou square by dou x square plus dou square by dou y square okay. So of course in this definition so what I have done in this definition is I have defined real valued harmonic function on a domain in the complex plane and I am assuming the function is continuous and it has partial derivatives up to order less than or equal to 2 of course you know partial derivatives means I am taking partial derivatives with respect to x and y okay with respect to x and y, so this means that you know dou U by dou x the first partial derivative with respect to x then dou U by dou y and then for the second partial derivatives you have the pure derivatives dou square U by dou x square dou square U by dou y square and you have also the mixed partial derivative dou square U by dou x dou y and dou square U by dou y dou x and we assume that all these partial derivatives exist and they are continuous okay and I mean the point is that somehow you know the at least in the definition it we insist only up to the existence of derivatives of order up to 2 okay but the fact is that you know it is rather amazing the fact is that you put this condition and then the partial derivatives of all orders will exist okay so the requirement of the partial derivatives of order partial derivatives of order up to 2 existing is just so that this equation can be written down okay and of course to write this equation I do not need the mixed partial derivative dou square U by dou x dou y or dou square U by dou y dou x okay but the point is normally continuity of the function and of the of all these partial derivatives is assumed but the big theorem the big theorem that you get from complex analysis is that you take such a harmonic function then it is infinitely differentiable okay that is partial derivatives of all orders exist okay and they are all continuous it is a very deep theorem and why it is why it is so deep is because the you are getting infinite differentiable you are getting existence of partial derivatives of any order okay and you know let me tell you a few words about this I mean you would have seen in a first course in complex analysis that probably we will revisit that again that you know if you you know that if you take an analytic function okay then the real and imaginary parts of an analytic function are harmonic functions okay and we and we call the imaginary part harmonic conjugate of the real part okay and conversely if you give me a harmonic function alright if the domain is simply connected okay then it will always have a harmonic conjugate that means so that means that you give me a harmonic function it will be the real part of an analytic function at least on a small disc okay and the moment it is real part of an analytic function you know analytic functions are infinitely differentiable because analyticity the beautiful property about analyticity is that you would assume differentiability once with respect to the complex variable z and you get infinite differentiability and the moment you know that the moment you know that it follows that the real and imaginary parts of an analytic function are also infinitely differentiable okay therefore the fact that you take harmonic function which has derivatives only up to order less than or equal to 2 and satisfies Laplace equation okay actually it is a very it is weaker when compared to the result that the function you will actually have derivatives of all orders okay and it is a and all the derivatives of all possible orders all mixed partial derivatives will exist and they will all be continuous though the reason is because even you is locally the I mean you locally is the real part of an analytic function and an analytic function is infinitely differentiable that is the reason okay. So what you must remember is that this condition that we have put that the u satisfies Laplace equation and all these partial derivatives exist up to order 2 and they are continuous is rather weak condition okay. Now you see now for such harmonic functions they have a very important property that is called the mean value property okay so I will explain what this mean value property is so maybe I can save some space and rub this off and write partial derivatives here with respect to x and y okay. So let me define this mean value property so well I can of course you know and now I can extend this before I do that I can extend this definition to a complex valued function being harmonic so you know if f is now a function from u to c when will I say f is harmonic if it is real and imaginary parts are harmonic so let me write that down also that is just an extension of this definition we say f from u to c is harmonic if real part of f imaginary part of f from u to r are harmonic so complex valued function is harmonic if and only if it is real and imaginary parts of harmonic alright. Now what is this business about mean value okay see suppose you have suppose you have some domain u and you have a function f defined on u taking values in the complex plane and you take this point take a point z0 inside u right then what you do is you take a circle centered at z0 radius r right then well of course I am taking the circle inside u so that even on the on the circle at every point on the circle f is defined okay. Now what I do is you know I define f mean value okay this is the mean value of f at over the circle so over this circle centered at z0 radius r I define this mean value this is to be and this is a function of r okay it will change if I change the radius of the circle so I am looking at circles centered at the point z0 okay of various radii small r okay and for given one such circle with radius small r I am defining this mean value okay of f with respect to that circle and what is it is just you know the mean value is defined like this you simply calculate f of z you take f of z that is the value of f at a point z on the circle and then you and then you integrate with respect to mod dz because you know integrating with respect to mod dz will actually give you the arc length okay integrate integrating mod dz over an arc will give you the length of the arc alright so and what is the path of integration is the circle it is mod z-z0 equal to okay I do this and then this is you know this is to be thought of as summing all the values of f as I move across the circle okay and this mod dz should be thought of as the arc length right so this is the sum of all the values of f as you take the as you move the point around the circle and then if you want the average value I have to divide by the arc length of the circle which is 2 pi r okay so this is the mean value so the mean value is the sum of all you know how a mean value is defined it is an average so what I am doing is I am taking I am summing up all the values of the function on the on the boundary circle that is what the numerator gives that is what the integral gives and then I am dividing by the length of the circle that the circumference circle which is 2 pi r okay and this is called the mean value of f for that circle alright now you know if you put z is equal to z0 you know you can parameterize the circle as z equal to z0 plus r e power i theta where theta varies from 0 to 2 pi okay you can you can parameterize this circle and if you do that you know how what will this integral change to see dz will be r into e power i theta into id theta okay mind you r is fixed theta is varying alright and if I differentiate e to the i theta with respect to theta I will get e power i theta times i right and if I calculate mod dz I will end up with r d theta okay in fact I will get r mod d theta if you want and I can write as r d theta because if I take theta to be increasing then d theta is a change in theta is also positive okay. So if you put it in this uhh if you substitute if you for this if you if you change this integral in z into this integral into an integral in theta what you will get is the mean value of f over r is well its integral from 0 to 2 pi so theta will vary from 0 to 2 pi f of z0 plus r e power i theta and mod dz is going to give me r d theta and of course I will have I will have a 2 pi r. So it it also has this expression you can also write this integral from 0 to 2 pi f of z0 plus r e to the i theta into d theta by 2 pi so this is another expression for the mean value okay of f over the circle. Now what is uhh so this is this is how you define this this is just the mean value and the only thing you must notice is that the mean value is taken on the on the circular arc the mean value of is taken on the circular arc and the fact that you are doing it on the circular arc is reflected by integrating with respect to mod dz okay because integral with respect to mod dz over an arc gives you the length of the arc okay. So this mod dz comes for that reason because you are adding up values of the function on the arc right and of course you divide by the length of the arc which is 2 pi r in this case the arc is a circle the whole circle. Now what are the properties of this mean value of r okay the first property is that this so you know I have cooked up a new function I have cooked up a new function uhh for sufficiently small r I have to take sufficiently small r starting from well starting from r equal to 0 alright and I am going to look at sufficiently small r all r below a certain value so that all these circles are inside my domain alright and I am getting a function based on that r I am getting values I am getting mean values of f based on this r okay. Now what is the property of so you know so f is f sub mean value is defined from well it is defined on 0 rho if you want well maybe I can even put so the way I have done it it is 0, rho to C where you know rho is if you want the maximum of all r such that the circle mod z-z0 less than r lies in u I am actually taking the largest possible circle alright in fact let me put equal to r I want the circle take the you take the maximum over all r okay such that mod z-z0 equal to r lies in u and sometimes probably this maximum may not exist okay so I think it is safer to put supremum here if you want can put supremum so if I want to write it like this should write a supremum over all r such that mod z-z0 equal to r is in u in fact I want the whole I want the whole I want the whole disc inside u so you know it is very important that I do not want any holes in between okay I do not want any holes in between so I have this mean value function alright now the claim is that this mean value function is continuous okay it is a continuous function okay and so f sub mv is continuous that is a that is a first observation why is it continuous because you see is it continuous in what it is continuous in the variable r which lies in 0 rho okay well why is that so because you know it is pretty easy actually it is because the continuity of f the continuity of f sub mv the mean value of f as a function of r is a result of the continuity of f as a function of z of course here I have not I have not mentioned that so you assume so here let me add that if f is continuous on u so if I take a continuous complex valued function then the mean value function that it defines is also continuous why is that so it is because you see if given epsilon greater than 0 okay if you estimate the difference between f mean value of let us say some r0 and f mean value of r okay then this turns out to be I mean it is going to be modulus of if I use this I will get integral from 0 to 2 pi f of z0 plus r0 e power i theta minus f of z0 plus r e power i theta the whole into d theta by 2 pi this is what I will get by the definition of the mean value and but you know I can use this you know this fact modulus of the integral is less than or equal to integral of the modulus so this is less than or equal to integral from 0 to 2 pi of mod of the integrand which is z0 plus r0 e power i theta minus f of z0 plus r e power i theta into mod d theta by 2 pi which again continues to be d theta by 2 pi right and well this can be you know this can be made less than integral from 0 to 2 pi epsilon d theta by 2 pi which is equal to epsilon okay if we choose delta greater than 0 which exists by continuity of f such that mod of f of z done let me write z prime minus f of z can be made less than epsilon if mod z minus z prime is less than delta you see f is continuous therefore you know if you give me any z and z prime then I can make the values of f at z and z prime as close as I want if I choose z and z prime close enough and of course here I am taking z prime to be any point of this type z0 plus r0 e power i theta namely a point on the circle with radius r0 and I am taking z to be a point on the circle with radius r okay and I can do this just because of continuity and what does this calculation tell you it tells you that it tells you that you know z is lying on circle of radius r z prime is lying on circle of radius r0 alright and the fact that this distance can be made less than delta means that you are you know bringing r0 and r close okay. So the moral of the story is that if you bring r0 and r close okay then f r0 and f mv r0 and f mv r come close so that will tell you that if r tends to r0 okay then f mv r tends to f mv r0 which means that f mv is a continuous function of r okay so this tells you that f sub mv the mean value function it is a continuous function of r alright. So you know so the diagram is like this I mean I am having this z0 I am having this r0 and I am having this circle and this is where I am taking my z prime okay which is this argument and then I have this bigger circle which is bigger or smaller it does not matter both ways. So this is r and I am having and I am having point z there okay and if I bring if I bring r close to r0 I am actually bringing z and z prime close okay and if z and z prime come close then f of z and f of z prime come close and therefore the difference between f mv r0 and f mv r0 f mv r becomes small enough okay. So this is the proof of the fact that the mean value function defined by a complex value to continuous function is a continuous function of r alright then the further thing is actually this mean value function actually is even defined at the origin and it takes the value f of z0 okay. So further limit r tends to 0 f mv of r is actually f of z0 the mean values the mean value function tends to fz0 as r goes to 0 or you must understand that as r goes to 0 the circle is becoming smaller and smaller and smaller circle centre at z0 and you know it is natural to expect you know as I make this circle smaller and smaller the function values are also going to come close to fz0 therefore you should expect the average also to be fz0 I mean if all the function values are close enough to fz0 then the average should also become close enough to fz0 okay and then therefore if you take the limit you should get only fz0 okay. So this is intuitively correct but then you can rigorously prove it by the same method what you can do is you can calculate the value fmvr-f of z0 if you calculate this what you will get is the same kind of you will get modulus the same kind of estimation as here integral is 0 to 2pi I will get f of ez0 plus re power i theta minus fz0 d theta by 2pi so and again modulus of the integral is less than or equal to integral of the modulus so I will get this is less than or equal to integral from 0 to 2pi mod of f of ez0 plus re power i theta minus f of z0 well into d theta by 2pi and this can be made lesser than integral 0 to 2pi epsilon d theta by 2pi which is equal to epsilon I can make this less than epsilon because I can make this less than epsilon and that is because of continuity at z0 of f if I make r sufficiently small okay this is less than epsilon if r is sufficiently small due to continuity of f at z0 after all f is continuous also at the centre alright. So the same so what this tells you is as r tends to 0 fmvr tends to fz0 so actually you know you have this the mean value function actually extends to 0 so 0, rho to c with f of mean value at 0 is fz0 okay so what you have done is you have defined a mean value function which at 0 the mean value is fz0 which is just the value of f at z0 and for any r you get the mean of the values of f along the circle centre at z0 radius r alright. Now this is all about the mean value function now when does the function have a mean value property if the function is said to have a mean value property if the mean values of the function all they are all equal to the value at the centre for sufficiently small for sufficiently small discs okay so here is a definition so this is the definition of mean value property the continuous function f from u to c is said to have the mean value property I will abbreviate it as mvp it has a mean value property at z0 at z0 if f mean value of r is actually equal to fz0 which is just f mean value at 0 for all r sufficiently small so that is for all r belonging to 0 epsilon okay for some epsilon greater than 0 okay in other words the function a complex continuous complex value function has a mean value property at a point if you take sufficiently small circle surrounding that point and you take the mean value of the function you get exactly the value of the function at the centre okay this is the mean value property. So you see this mean value property is some it is a completely it is a kind of integral condition right because after all the mean value is defined by an integral mean value is defined by this integral here or here okay and the integral of a continuous function always exists right so it is very easy to define and now comes the big theorem really big theorem so the big theorem is function is harmonic if and only if it has a mean value property a continuous function is harmonic if and only if it has a mean value property at every point it is a terrific theorem because you see one part of the theorem says that if it is harmonic it has a mean value property that is more or less easy to prove okay and you would have seen a proof of that in the first course in complex analysis by for example taking the function to be the real part of an analytic function okay or you and if it is a complex value function you can even take I mean you can do it using analytic functions you can use analytic functions for example and show that analytic functions have the mean value property okay in fact the fact the in fact the analytic functions having the mean value property is exactly Cauchy's integral formula in a way alright and on the other hand the striking thing is the other the implication in the other direction that you start with the continuous function which has only mean value property and what you get is that it is harmonic and why it is so powerful is because the condition you have put is an integral condition I mean you have this mean value function you have the mean value associated to that function which is defined by an integral okay and it is defined on just a you know interval to the right of 0 and of course you can include 0 also the fact that this is for sufficiently small value is equal to the function value at z0 okay it seems to be rather simple condition but that condition if you put at every z0 what it results in is that the function is harmonic and what is harmonicity that is a great deal as I told you saying that the function is harmonic means you are saying it is infinitely differentiable okay at least it is I have told you that even though in the definition of harmonic functions we only require that it is derivatives of order to exist and are continuous but it is actually infinitely differentiable and it is also satisfies Laplace's equation so it is rather amazing that you know you put this simple condition on a continuous function okay it results in the function becoming harmonic it makes the it makes a function see infinity it makes both the real and imaginary parts of the function infinitely differentiable with respect to both variables and that is amazing and all the derivatives all mixed partial derivatives of all orders exist they are all continuous and on top of that the function also satisfies the Laplace's equation okay so you see this condition it is a really a remarkable property so here is a theorem the theorem is for a function for a continuous function f from u to c the following are equivalent so what are they number 1 is f is harmonic number 2 f has the mean value property at each point of so it is an amazing it is an amazing equivalence okay see 1 implies 2 is easier okay it is easier provided you use some Cauchy theory which you would have covered in a first course in complex analysis okay but 2 implies 1 is serious harder that if you have you can obviously expected to be harder because the condition 2 seems to be very simple condition you are just saying that some integral is you calculate some integral for small values of r and all the values should coincide with the value of the function at the center okay the mean values of the function around small enough circles should give you the value at the center of the circle that is the condition 2 so relatively simple condition but then from that you are trying to conclude that f is harmonic okay which means effectively you are getting infinite differentiability of f which is very very serious alright so 2 implies 1 is harder and 1 proof of 2 implies 1 involves the so-called Dirichlet problem okay it involves the Poisson kernel the Poisson integral formula using which you can solve the Dirichlet problem for the disc okay and then you can get 2 implies 1 okay I will try to see if we can cover that we have enough time to do that in the series of lectures but I can for the moment the easier part 1 implies 2 is something that we can easily check okay so well for 1 implies 2 of course I am going to use some complex analysis to do it alright so what I am going to do is you know put u equal to real part of f okay mind you f is I am assuming f is harmonic alright and I take u to be real part of f okay and what I am going to do is I am just going to look at a sufficiently small disc small disc mod z minus z naught less than rho inside u for given z naught in u okay. Now mind you u is real part of f and f is harmonic and you know f is mind you f is only a complex valued harmonic function okay I am not saying f is analytic I am not saying f is analytic I am just saying f is a complex valued harmonic function therefore the definition is that both the real part and the imaginary part of f are real valued harmonic functions that is all I have so I am taking u to be real part of f okay it is harmonic is given to me now I am going to use this very powerful theorem that if you have a harmonic function and if it is defined on a simply connected domain okay then it has a harmonic conjugate okay namely I can find another function such that if you put that as imaginary part of a new function then that with this as real part then you will get an analytic function okay so I am going to use that which is this is a fact from complex analysis okay so I am going to use that so yeah so since mod z-z naught less than rho is simply connected there exists g analytic that is holomorphic in mod z-z naught less than rho with real part of g is equal to u okay and imaginary part of g will be a harmonic conjugate of u. So imaginary part of g is a harmonic conjugate of u and you would have seen in first course in complex analysis that you can get different harmonic conjugates but they will only differ by a constant right. So of course you know you can prove one implies two also directly appealing to some version of Green's theorem right that is also another way of proving it but I am trying to circumvent I am trying to avoid all that and I am trying to give a an elegant proof which anyway uses some powerful results but my main idea is to you know give you an indication of one line of argument that will convince you that at least statement one implies statement two alright. So and now you see now you know now I am in the following situation so you know I have this I have this z naught I have if I take an r such that r is less than rho so this is inside u and I have f I have this g defined here with real part of g is equal to u okay. Now you know you apply the Cauchy integral formula okay by the Cauchy integral formula you know that if I calculate the integral over mod z minus z naught equal to r of course whenever I am calculating such integrals I am taking the positive orientation every the anticlockwise orientation. Suppose I calculate f of z by z minus z naught okay you know Cauchy integral formula tells sorry not f g you know that I will get I think maybe I have to put a 1 by 2 pi i you know that I will get g of I will simply get g of z naught okay this is the Cauchy integral formula alright okay and you know if you write out the integral in terms of theta on the right side then this is the same as 1 by 2 pi i integral from theta equal to 0 to 2 pi g of z is r z naught plus r e power i theta I am just parameterizing this circle with radius small r centers z naught as z equal to z naught plus r e power i theta where r small r is fixed and theta is varying from 0 to 2 pi. So then I will get this of course I have forgotten d z here I should put that d z there is a variable of integration and well you know if I write out what the d z is I have already written it out here it is r e to the i theta i d theta divided by z minus z naught is r e to the i theta and what I will get is I get 1 by 2 pi integral 0 to 2 pi g of z naught plus r e to the i theta so you know I will let me write it in this form d theta by 2 pi this is what I get okay and what is this you know if you this g is real part of g plus imaginary part of g plus i times imaginary part of g so you know so if I write it like that this is integral if I write g as real part of g plus i times imaginary part of g what I will get is I will get real part of g mean value at r plus i times imaginary part of g sub mean value at r this is what I get and that is equal to on the left side g z naught which is real part of g z naught plus imaginary part i times imaginary part of g z naught this is what I get okay but then what is real part of g my real part of g is u real part of g is u so what I will get is if I compare real parts I will get u at z naught is the mean value of u at r for 0 less than or equal to r less than rho and this is the same as saying that you know u has the mean value property okay so I mean what I have done is the same proof actually is the usual proof that you would have seen a first course in complex analysis that tells you that the real and imaginary parts of a real and imaginary parts of an analytic function have the mean value property okay so what I am doing is you give me a harmonic function I am using this rather powerful theorem that if you take a sufficiently small disc which is simply connected then it has a I mean if I take a real valued harmonic function then this is real part of an analytic function okay and then I am using the fact that the real and imaginary parts of an analytic function have the mean value property okay and I am saying I am just trying to deduce it using Cauchy integral formula so this proves this is this is one way of showing one implies two okay for two implies one we need lot of machinery so I would not get into that but my aim is to tell you that you know harmonic functions have the mean value property alright which in some sense you should have seen in a first course in complex analysis. Now now I come to the so called maximum principle I come to the so called maximum principle and what is this maximum principle so you would have you would have again loosely learnt this in a first course in complex analysis as if you take an analytic function on a domain with boundary bounded okay a bounded domain so the domain is bounded so its boundary is also bounded then the analytic function if you take the modulus of the analytic function that will attain its maximum only on the boundary and not in the interior and if it attains a maximum the maximum of the interior then it has to be constant so a non constant analytic function will attain its maximum only on the boundary okay now this is actually a property of this is actually a property of harmonic functions actually so the maximum modulus being attained at the boundary is a property of the harmonic functions okay and that is why analytic functions have that property because analytic function is harmonic the analytic function has both real and imaginary parts which are harmonic okay and the mean value property for I mean the maximum the maximum modulus principle for harmonic functions gives you the maximum modulus principle for analytic functions in fact it also gives you maximum modulus principle for complex valued harmonic functions okay. So let me write this maximum principle so let me write a few cases so this is a maximum principle for the real case and what is a maximum principle for the real case you have u from capital U to R harmonic on domain u in the complex plane so u is a real valued harmonic function alright and so this is let suppose M is an upper bound for u for small u in capital U then unless u is a is constant on u M is a strict upper bound I should say strict upper bound on u okay. So this is the real this is the maximum principle for real valued harmonic functions so you know the situation is that I have this domain u in the complex plane and I have a harmonic function on that right and suppose all the values of u are bounded by bounded above by a real number M okay so that is if so you know if I if I write it in symbols if u of z is less than or equal to M for all z belonging to u okay and u is not constant on u then u of z is strictly less than M for all z in so this is the so every upper bound is a strict upper bound that is the maximum principle. In other words another way of saying it is that if u attains this upper bound that is if the upper bound is not strict and if that upper bound is attained in at some point then it has to be constant so another way of saying it is another way of saying it is if u of z not is equal to M for some z not in u then u is equal to constant is equal to M on capital alright this is the strict version of the maximum principle right this is the maximum principle for the real case and then you have the same statement as a maximum principle for the complex case and then you have a maximum principle for also for analytic functions okay so so let me write out the other case also maximum principle complex case so here I take f from u to r u to c complex harmonic function so it is a complex valued harmonic function on a domain u and c and so same statement suppose M is an upper bound for mod f okay now because it is complex valued by a bound for f we actually mean a bound for mod f okay in u then unless f is constant on u on u M is a strict upper bound for mod f on u so this is a statement okay the only thing is when you take a complex valued function you have to and when you talk about bounds you know you have to take mod f because you know I cannot write complex number one complex number lesser than another complex number because complex numbers are not ordered okay and the ordering of the real numbers does not exist extend to complex numbers right so I cannot write a statement such as M is an upper bound for u small u here u is real valued I can write that okay but I cannot write a M is an upper bound for f does not make sense because f is complex value I should only write M is an upper bound for mod f okay in general whenever we say f is bounded we always mean there is a bound for the modulus of f okay and then the rest of the statement is the same thing so that is if so let me write this if f of mod fz is lesser than m for all z in u and f is not constant on u then mod fz is strictly less than m for all z in u so it is a strict upper bound and there is and of course the other way of writing it is also like this if f of mod fz is equal to m for some z not in u then f is constant on u okay and therefore you know this applies also for mind you here f is harmonic complex value but need not be analytic okay but then even if f is analytic this applies because after all if f is analytic then both the real and imaginary parts of f are harmonic therefore f is also harmonic analytic function is always harmonic because it satisfies Laplace equation both the real and imaginary parts are harmonic but harmonic complex value function need not be analytic okay. So this also applies to analytic functions right and the usual version that we often use is the is the the counter positive of this which is that if your domain is bounded okay then f attains its in the in the real case the maximum is attained on the boundary in the complex case the modulus of the the modulus maximum modulus is attained on the boundary. So let me write that also if u is bounded and u extends continuously to u union dou u dou u is a boundary of u then u attains a maximum on the boundary. So this is the version that you are all familiar with which is actually which is also equivalent to that. So let me write the same thing here if u is bounded and f extends to a continuous function to a continuous function on u union dou u mind you in these cases u union dou u is compact because it is closed because I have added the boundary dou u to u and it is bounded. So it is closed and bounded so it is compact okay then mod f attains a maximum on dou u and and only on dou u okay you cannot get a maximum in the interior okay I am not so I should say in fact instead of saying a maximum I will say all maximum only on the boundary and here also I should say all maximum only on the boundary you cannot get a maximum in the interior unless it is constant okay. So of course unless u is constant so here also unless f is a constant okay. So the either the function is a constant in which case it has the same value throughout if it is real value and it has the same modulus throughout including the boundary if it is complex value or if it is not constant then the maximum is only on the boundary for the real value function and for the modulus of the complex value function. So this is a maximum prop this is a maximum principle for harmonic functions okay. I will continue in the next lecture and I will give you a proof of this I think it just will use the mean value property very easily and then so I am making use of fact that harmonic functions have the mean value property which I more or less sketched a proof of right and then my aim of doing all this is to get to the so called Schwartz lemma okay and that is required in a preliminary discussion of the Riemann mapping theorem okay. So I will continue in the next lecture.