 Okay, so let us continue with our discussion of harmonic functions and the maximum principle. So if you recall we have this maximum principle, so this is the real version. So you have U defined on capital U with values in R, U is a domain in the complex plane and U is real valued function and in fact you assume U is a function which is actually harmonic okay, U is harmonic on capital U okay and suppose U of z is less than or equal to m for all z in U then either U is a constant, U is constant on capital U or the bound is inequality is strict or U of z is strictly less than m for all z in U okay. So in other words a harmonic function and which is a non-constant which is non-constant namely a non-constant harmonic function it can never attain a maximum in the in an open set okay in an open connected set. So and of course the point about an open set is that every point is an interior point right, a set is open in topology if it only every point is an interior point. So what you are saying is that a harmonic function cannot attain a maximum at an interior point and therefore you know if you can extend the harmonic function to the boundary to a continuous function then you should expect the maximum to be attained on the boundary okay. So let me state that also in other words so if U extends continuously to the boundary of U and capital U is bounded then unless then U attains its maximum on the boundary okay. So this is the real version of the maximum principle so in fact the fact is that the condition that U is harmonic can be replaced by the condition that by the equivalent condition that U is continuous and satisfies the mean value property has the mean value property at each point okay. So instead of assuming U is harmonic in U is equivalent to assuming U is continuous in capital U and has the mean value property on U okay. So this is something that I told you is a theorem that for a continuous function harmonicity on a domain is equivalent to its having the mean value property at each point and I think and we saw I gave you a proof in one direction namely if the function is harmonic then it has a mean value property I use the Cauchy integral formula to prove that last time and the harder part is to show that a continuous function which has a mean value property is harmonic okay that is the statement that I did not prove and I am not going to give the proof of that okay but I am going to use that fact that harmony is harmonicity is equivalent to continuity with the mean value property okay. So in other words I am just trying to say that maximum principle applies to any continuous function which has a mean value property because this is the same as harmonic okay. So what is the proof of this so the proof of this is pretty simple it of course uses a mean value property. So you know so the idea is that you know see so you know you have so if let me draw a diagram so you have so here is the complex plane and you have some domain U well I am always drawing a bounded domain it cannot be bounded but and then there is a point I take a point Z0 in the domain alright and well the fact is that so what is given to me is that U is real value small U is real valued and it is bounded all the values of small U on this open set U which is the interior of this amoeba like region that I have drawn that is where the function U is defined small U is defined. So there is a function like this taking values in the real line and all the values of small U are bounded above by M okay and what do I have to show I have to show that if U is not constant then U never attains the value M in the in that open set in that open connected set it is a domain it is both open and connected as you will see the connectedness is very important it will be used in the proof. So the other way of proving that statement is trying to show that if U assumes the value capital M at a point inside I will try to show that U is constant okay so which so the counter positive of that will be that if U is not constant then it cannot assume the value M at any point inside so at every point inside the values of U will always be strictly less than M which is the assertion of the statement of the assertion in the statement okay. So suppose so let me write this let us assume that Z let Z not belong to U with U small U at Z not equal to capital M let us assume this okay and now you know the fact is that small U is a harmonic function but more importantly what I want is that it has the mean value property okay so a harmonic function is a continuous function which has the mean value property at every point this is a this is an equivalence okay by the way you know that harmonic means that the original straight definition of a harmonic function is that it should satisfy Laplace's equation it should have continuous derivatives of orders up to 2 and it should be continuous okay and it should satisfy Laplace's equation but let me again repeat it is an important fact that a function which is harmonic has the mean value property and conversely a function which is continuous just continuous and has the mean value property is harmonic and the beautiful thing about harmonic functions is how are they related to complex analysis they are related in the sense that harmonic functions are always on small disc like neighbourhoods namely some on simply connected open sets harmonic functions are always real value I mean real parts of analytic functions and since analytic functions are infinitely differentiable it follows that the harmonic functions are infinitely differentiable okay and that is great because when you define a harmonic function you only want the derivatives to occur up to order 2 and that they should be continuous but then it they turn out to be infinitely differentiable okay that is because of the fact that harmonic functions are locally real parts of analytic functions are infinitely differentiable okay. Of course what is more amazing is the statement that you assume nothing about the derivative at all if you assume the mean value property you are just assuming that the function is continuous and it has the mean value property which is you take the mean of all the values on a circle surrounding a point sufficiently small circle surrounding a point then the the mean value average value you get is the value at the at that point at the centre of the circle and this holds for all sufficiently small circles that is the mean value property it is a property which is defined by an integral and it only and for the integral to be defined you only need continuity so you know if you have a function which is just continuous and which has a mean value property the up short is that it is harmonic and as a result is infinitely differentiable so it is really amazing okay. So anyway so see at at the point z0 you will have the small you will have the mean value property so it means that you know if I take a small disc surrounding z0 inside your domain so you know so I will have the following thing u u has mean value property at z0 implies for all so implies there exists an r0 greater than 0 with the value the average value or mean value of f on mod z-z0 equal to rho okay the average value on the circle is actually equal to the not f so it is u is equal to u at the centre of the circle so this is the this is the mean value property okay and this is for all rho greater than 0 rho less than r0 so for sufficiently small circles centre at z0 the average value of the function on the circle is equal to the value at the centre that is the mean value property but and how is this defined this is defined as 1 by so it is defined by integrate you integrate from 0 to 2 pi the function values u of z0 plus rho e to the i theta that is how a point on this circle looks like as theta varies from 0 to 2 pi you get the whole circle centre at z0 radius rho and then you integrate with respect to d theta by 2 pi this is the mean value and this is equal to u z okay but of course you know I have assumed that u z0 is m okay so you know I can actually write this as integral 0 to 2 pi u z0 minus u of z0 plus rho e to the i theta the whole into d theta by 2 pi equal to 0 I can write it like this that is because you know u z0 is a constant if the first integral will just give me u z0 because if I integrate d theta over 2 pi from 0 to 2 pi I will simply get 1 okay so now what you must understand is that this is the integrand this integrand is a you know this is always greater than or equal to 0 okay that is because u z0 is m u of z0 is m and all other u values are less than or equal to m that is already given to me okay so the difference is always non-negative so I have a non-negative so you know I have this fact I have this situation where I have a integral of a integral of a real value function on a close interval okay which is non-negative and the integral is 0 okay but then that means the integrand has to be identically 0 okay this is something that you know so this implies that u of z0 has to be equal to u of z0 plus rho e to the i theta for all rho with 0 less than rho less than r0 okay and of course so what have I proved I proved the following thing I proved the following thing I proved that if u attains the value m this maximum value m at an interior point z0 then there is a whole disc surrounding z0 where u is constantly equal to that value m okay so on this whole disc okay as I increase as I allow and of course you know it is very important that theta of course is varying from 0 to pi okay so if I fix a rho and let theta vary from 0 to pi I will get the circle of radius rho and then if I make rho small then I will get the whole disc I will get I will get all points in the disc centre at z0 and radius rho r0 okay so what all this tells you is that so if u of z0 is m then there exist r0 greater than 0 such that u of z is m for all z with mod z-z0 less than r0 this is what this is what we have proved okay if u attains the value m at the point z0 then there is a whole disc centre at z0 contained in your domain with radius r0 okay where u attains u where u is equal to where u attains the same maximum m okay so what this tells you is that so this so I have basically I have used the mean value property at z0 I am applying the mean value property at the point where u has attained the maximum okay and I am using the fact that the integral of a non-negative if the integral of a non-negative function is 0 okay then that non-negative function has to be 0 itself okay so so the moral story is what does this tell you this tells you this tells you the following thing the if you take the set of all z in u capital u such that u of z is equal to m it tells you that this is open in u is an open set because whatever I proved whenever z0 is in this set I proved that there is a whole disc centre at z0 which is in this set so every point of this set is an interior point of this set and that means that this set is open okay on the other hand you also have the set of all points of u where u is at strictly less than m this is also open in u okay this is also open why because you see why is that true because you see u of z the set of all points in u where u of z strictly less than m is just u inverse of minus infinity, m it is inverse image of the this interval minus infinity m on the real line I mean this is precisely all those points where u takes a value less strictly less than m and you see u is continuous function u is a continuous function and this is an open set and you know one of the characterization of continuity is that the inverse image of an open set is open under a continuous map so therefore so this set is open but then what is the union of these two open sets the union of these two open sets is all of u okay so you get u is equal to the set of all points where small u attains the value capital m and disjoint union with the set of the other points where small u is strictly less than m because anyway it is given that all the values of u are less than or equal to m so the those points where you takes the value m form one piece and those who take the value strictly less value strictly less than m form another piece and the union of these two is u but what of we just now seen we have seen this also open this also open these are two open sets okay and u is union of two open sets but u is a domain capital u is a domain it is connected and a connected set cannot be written as a disjoint union of two open sets two proper open sets because any set that is written as a disjoint union of two non-empty open sets is already disconnected writing a set as a union of two pieces which are disjoint from each other and which are open it is already disconnecting the set okay so so the connectedness of u will tell you that one of these has to be empty okay so but you know is it not life here so this set is non-empty therefore this has to be empty so it will mean u is equal to this which means u is constant okay and that will be the end of the proof so now you connected because it is a domain a domain is an open connected set implies u is exactly the set of all points where u small u is equal to m this other set is an empty okay this implies that u is equal to m on u so that finishes the proof that if u attains the value m at an interior point then it has to be constant okay alright now there is one more statement the other statement is if u extends continuously to the boundary and u is bounded okay then u attains its maximum on the boundary okay and what is the proof for that the proof is very simple you know if you take if capital U is bounded okay then its boundary is also bounded so if you take u union the boundary it is a bounded and closed set it is closed because you added the boundary to it so it is compact any subset of Euclidean space which is both closed and bounded is compact so that is why I am using the boundedness of u okay the boundedness of u is also going to give you the boundedness of dou u okay so it gives you the boundedness of u union dou u which is a closed set so which therefore becomes compact and you know a continuous function on a compact set is uniformly continuous and attains its bounds okay therefore u has to attain its maximum on u union dou u but we have already proved that it will not attain its maximum in the interior unless it is constant therefore it has to attain its maximum only on the boundary and that is the proof for the second part of the statement okay so the second part of the statement just follows from the first part of the statement okay so I will just write that down if u is bounded then so is then so is then so are dou u and u union dou u which is and so yeah let me write that if u is bounded then so are dou u and u union dou u further u u union dou u is compact and u necessarily attains its maximum on u union dou u and hence on dou u because it see it cannot attain a maximum in the interior the only case when it attains a maximum in the interior it is when it is constant but if it is constant then its extension to the boundary will also be constant by continuity. So it will attain that maximum that constant value the constant value will be also a maximum value it will be the maximum value and that will also be attained on the boundary okay so on the other hand if it is not constant then certainly it cannot attain the maximum value in the interior so it will attain the value on the boundary alright fine so that finishes the real version of the so I mean the point I wanted to remember is that a continuous function which has the mean value property on a domain has the maximum principle okay a continuous function which has a mean value property on a domain satisfies the maximum principle that is what you must understand it cannot attain its maximum in the domain it has to attain the maximum only on the boundary provided you can extend it to the boundary okay and of course the boundary should be bounded right. So alright so this is the real version now let me go to the complex version so here so the statement is the same except that instead of taking a real valued function you take a complex valued function okay so f from U to C U inside C a domain and you assume f is harmonic on capital U and of course saying that f is harmonic on U is equivalent to saying both the real and imaginary parts of f are harmonic on capital U okay so a complex valued harmonic function is by definition something for which both the real and imaginary parts which are real valued functions they are harmonic okay so I have the same statement only thing is that since it is complex valued I have to write I cannot write U is I cannot write f is less than or equal to m because it is complex valued I have to use the modulus of f so let me write that down if mod f of z is less than or equal to m on U then an either f is constant on U or mod fz is strictly less than m for all z in U that is the maximum the modulus of f will not attain its maximum in the interior okay and basically that means that if you find if you take any point with a certain and take the modulus of the function at that point you can always find another point where the modulus of function is bigger right when you say function does not attain its maximum it means that given its value at any point I can find another point where its value is bigger that is when that is what it means to say a function does not attain its maximum okay so it exceeds its function values at every point by function values at some other point suitable point okay. So if so let me write the other part as in that case if U is bounded and f extends continuously to the boundary then f attains its maximum in mod f attains its maximum on the boundary okay. So this is the complex version so what is the proof the proof is that we just use we apply the real version so the proof is well so again proof is the same thing you assume that it there is a there is an interior point where mod f attains the value m and show that f is constant okay. So suppose z0 is a point of view with mod f z0 is equal to m okay then I have to show that f is constant right so this means f z0 is e to the i alpha m okay so see mod f z0 is equal to m and you assume that of course I am assuming that m is not 0 okay yeah so I mean even if m is 0 I could have taken alpha to be 0 so f of z0 is a complex number with mod less m alright then I can write f of z0 as e to the i alpha times m okay for suitable alpha right. Of course if m is not 0 then I can divide by m and I will get f of z0 by m as mod less 1 so it is a unimodular complex number so it is of the form e to the i alpha so I will get f z by m f z0 by m is equal to e to the i alpha and well if f z0 is 0 then I just take alpha equal to 0 okay so in any case this equation holds for suitable alpha right. Now of course if mod f z0 is 0 it means f z0 itself is 0 okay fine so I can write this now what I am going to do is you look at look at e to the minus i alpha f of z look at this function okay now this function is just f of z multiplied by a constant e to the minus i alpha is some constant it is a unimodular complex number it is a complex number of modulus 1. So you have multiplied a function by a constant and a harmonic function multiplied by a constant is again a harmonic function because after all function is harmonic if it is for example if it satisfies Laplace's equation and multiplying a function that is harmonic by a constant it is going to keep it if it is already harmonic it is going to give rise to again a harmonic function because the differential operator the constant will come out of the differential operator okay. So this is also this is also harmonic alright so look at this which is also harmonic harmonic okay and if you look at so in particular real part of that is also harmonic is harmonic because you know our definition of a complex function being harmonic is that both the real and imaginary parts should be harmonic so the real part of this is also harmonic okay. Now you see if I look at the real part of e to the minus i alpha f z this is less than or equal to modulus of real part of e to the minus i alpha f z okay because any real number is less than or equal to modulus okay and this is certainly less than or equal to modulus of e to the minus i alpha f z because you know for any complex number the modulus of its real part is less than or equal to its modulus okay mod real part of w is less than or equal to mod w okay for any complex number w but this is the same as mod f z because modulus of e to the minus i alpha is 1 alright and mod f z is always less than or equal to m okay. So what I have got is that I have got that this is a real valued harmonic function which is bounded by m and what is its value at z0 at z0 its value is exactly m okay real part of e to the minus i alpha f z0 is actually real part of m which is equal to m okay. So I have a harmonic function which has values less than or equal to m and it has attained the value m at an interior point therefore by the previous case namely the real version of the maximum principle what I can conclude is that this real part of e to the minus i alpha f has to be constant okay so by the real version of the maximum principle real part of e to the minus i alpha f z is equal to m it has to be a constant okay. So real version of the maximum principle says that whenever you have a real valued harmonic function on a domain if it is if it attains a maximum value in the interior it has to be constant so this function the real part of e to the minus i alpha f z is a harmonic function on the domain capital U it is bounded by m all its values are bounded by m and it attains the value m at a point z0 inside the domain therefore it has to be constant so it has to be also and that constant value has to be the same as the constant value at z0 which is m so this function is exactly m okay. Now look at e to the minus i alpha f z this is actually real part of e to the minus i alpha f z plus i times imaginary part of e to the minus i alpha f z okay that is just expressing a complex number as the real part plus i times its imaginary part and then but this is equal to m plus i times imaginary part of e to the minus i alpha f z okay. So I have this but on the other hand what is the modulus of this the modulus of e to the minus i alpha f z cannot exceed the modulus is same as mod f z because mod of e to the minus i alpha is actually 1 and that is less than or equal to m okay what do these two equations tell you see I have a complex number whose real part is m and its modulus cannot exceed m the only way is that the imaginary part has to be 0. So this will tell you that the imaginary part of e to the minus i alpha f z is 0 okay because a modulus is square root of m squared plus the square of this imaginary part and if and that can never be less than or equal to m unless it can only be equal to m and in that case the imaginary part should vanish okay therefore you get this and the moment the imaginary part is 0 the function is equal to its real part and that is equal to m so you will get e to the minus i alpha f z equal to m and this will tell you that f z is a constant it is just m e to the minus i alpha e to the i alpha so that completes it okay and again as far as the latter statement is concerned the proof again uses the fact that if capital u is bounded then the boundary of u del u that is also bounded and u union del u will become both bounded and close it will be compact and mod f which will be a continuous function on this compact set will attain its bounds and therefore you know that if f is not constant then mod f is always strictly does not attain its bound in the interior so it has to attain its bound on the boundary okay so that finishes the second part the second part just follows as the real case okay so in particular what you must understand is that if you take an analytic function okay this is what you would have seen in a first course in complex analysis you take an analytic function analytic function also satisfies the maximum principle okay namely if you take an analytic function which is all mind you analytic function is also harmonic function because the real and imaginary parts of an analytic function are harmonic okay but the only extra condition is that the imaginary part is a harmonic conjugate of the real part that is what makes it analytic if I simply take two real harmonic functions and write them as u plus iv that will not give me an analytic function unless the v is a harmonic conjugate of u okay so even for analytic functions the maximum principle applies and that is what you would have seen in an earlier course in complex analysis but what I want you to appreciate is the important fact that at the base of all this is the mean value property I mean it is just if a function has the mean value property okay then if it is continuous and has the mean value property then it automatically has the maximum it satisfies the maximum principle the maximum can only be attained on the boundary not in the interior okay so now I need to come to so called Schwartz lemma which I need to use later on so here is the Schwartz lemma so what is this Schwartz lemma so you know Schwartz lemma is well so delta is the unit disc okay and you have and of course delta closure is its closure the close unit disc and you are looking at analytic functions defined on delta and ticking values in delta closure okay so f from delta to delta closure analytic okay so that means what you are saying is f is analytic in mod z less than 1 which means f is defined in delta on delta and it takes value in delta closure so mod fz is less than or equal to 1 I mean if you want to state this in words you will say let f be an analytic function on the unit disc with modulus less than or equal to 1 okay you want to state it geometrically you are looking at an analytic function which is defined on unit disc and taking values inside the closed unit disc okay and let us also assume that 0 goes to 0 f of 0 equal to 0 okay so assume this right then this Schwartz lemma is lemma that compares the modulus of fz and the modulus of z okay so this lemma is mod fz is always less than or equal to mod z okay for all mod z less than 1 okay so in some sense what it means is that in terms of lengths it is a contraction it is a contraction map in the sense that if I start with the complex number z in the unit disc okay its length is mod z okay but if I take its image fz its length will become smaller so which means that the mapping is kind of contractive okay but it is not entirely that only in terms of length it is contracting but there is also twist okay so the question is the other part of the Schwartz lemma is the case when equality occurs so the next part of the Schwartz lemma tells that if you get equality even for a single point in the unit disc then f is a rotation okay further mod fz0 is equal to mod z0 for and z0 with mod z0 less than 1 if and only if f is a rotation that is f of z is equal to e power i alpha z okay so this is Schwartz lemma in other words any mapping from the unit disc taking values in the closed unit disc if it is not a rotation I mean you take an analytic map you take an analytic function defined on the unit disc and taking values in the closed unit disc and assume 0 goes to 0 map that preserves 0 preserves the origin okay if it is not a rotation then mod fz will always be strictly less than mod z it will be strictly contractive okay the only case when it is not contractive is when it is a rotation and if it is a rotation then you know mod fz will be equal to z for all z because mod of e to the i alpha will be 1 so if equality occurs for a single z0 it will occur for all z with mod z less than 1 so let me write that okay this is the Schwartz lemma okay and it is a powerful lemma which is used it is a proof is rather simple because it uses the maximum principle but it is a very powerful lemma and finds lot of use in conformal mapping and many other situations so what you must understand is in the case when f of z is a rotation it is a rotation about the origin then f becomes a 1 to 1 conformal map namely it becomes a holomorphic isomorphism of the you know unit disc on to itself okay and in fact you know and in fact these are the only holomorphic isomorphism of the unit disc on to itself the only holomorphic automorphism of the unit disc which preserves the origin or the rotations that is like a that seems like a converse but actually it is also a corollary of the Schwartz lemma so let me write that also let me write this corollary any automorphism any holomorphic automorphism of the unit disc that fixes the origin so automorphism is a self isomorphism okay automorphism means self isomorphism it is an it is a holomorphic isomorphism from the unit disc back it back to itself and it should fix the origin okay is a rotation. So this is the corollary of Schwarz's lemma okay the only conformal automorphisms of the unit disc are rotations those that fix the origin are rotations okay so we will look at a proof of this in the next lecture.