 So you see I have this situation where I have a path beginning at so this is the path which is given by continuous function gamma from the closed interval a, b on the real line to complex plane and of course this point is z0 which is gamma of a and this terminal point of the path is z1 which is gamma of b and uhhh what we defined as uhhh an indirect penalty continuation uhhh in terms of power series is an expression uhhh it is a uhhh it is an expression involving uhhh power series which continuously vary with respect to the parameter t as t varies on the uhhh on this closed interval on the real line okay. So uhhh for every uhhh t in a, b uhhh we have gamma uhhh we have we have the function f t of z which is given by this power series it is a convergent power series uhhh centred at gamma of t uhhh so it is an expansion in integral powers of z-gamma of t and with coefficients a n which are again functions of t uhhh n is relevant to 0 and uhhh of course uhhh with uhhh radius of convergence uhhh r of t and disc of convergence uhhh mod z-gamma t is less than r of t. So, this is a so this is a one parameter family of power series okay the parameter is t alright and we want this we want a as t moves uhhh uhhh which means you are thinking of the point gamma of t moving on this path so t is moving on this interval okay from a to b gamma of t traces this path okay and we want the power series to move uhhh to be be continuous in the sense that uhhh we want successively uhhh the uhhh power series in a neighbourhood uhhh being direct analytic continuations uhhh uhhh and then that is a condition so the condition is uhhh for uhhh t uhhh t prime near t uhhh f t prime uhhh is equal to f t this is the condition we have okay. So, so in other words uhhh so so so for each each uhhh t in a b there exists an epsilon uhhh there exist an epsilon of t uhhh greater than 0 such that uhhh uhhh uhhh f t prime is the same as f t uhhh whenever t prime is in this epsilon neighbourhood of t so t prime belongs to uhhh uhhh t-epsilon t t plus epsilon t intersection of course a b this is the condition this is the condition that we put to say that uhhh uhhh so if you have come a t here uhhh corresponding to a point t so if I draw if I draw the real line uhhh here and this is a and this is b and I have a t then I have a an open neighbourhood an open interval centred at t uhhh given by t-epsilon t t plus epsilon t such that for every t prime in that neighbourhood uhhh the uhhh the power series at t prime and the power series at t represent the same analytic function okay. So, uhhh uhhh of course when I say the power series are equal I mean that the analytic functions they represent are equal in the in the uhhh of course in the intersection of the uhhh uhhh uhhh it will be in it will be in a neighbourhood of uhhh it will be in this neighbourhood okay. So, so what is happening is here is gamma of t and then I have gamma of t prime and you know there is a gamma of t uhhh has it is a power series with certain radius of convergence and uhhh what I want is that uhhh uhhh if I draw if I draw the power series uhhh centred at gamma t prime what I want is that in this uhhh in this intersection uhhh I want uhhh uhhh the the power series at gamma t and gamma t prime to represent same analytic function that is what I want okay. So, uhhh so, so at gamma t this is the uhhh this is the disc of convergence with radius equal to uhhh uhhh radius of convergence equal to rt and at gamma t prime I have another disc of convergence uhhh with uhhh radius of convergence rt prime okay and then I in this disc of convergence I have ft uhhh which is a analytic function of z with this power series expansion it is a Taylor series expansion of ft and here I have in this disc I have the uhhh the power series ft prime okay and what I want is that in this intersection of these two discs I want ft and ft prime to be the same alright. So, what this essentially means and I want this to happen for all t prime in uhhh all t prime sufficiently close to t. So, in other words uhhh if you take any such uhhh t prime uhhh then uhhh what is happening is that uhhh ft prime and ft are direct analytic continuations of each other. In fact, they are one in the same function on the intersection. So, they are direct analytic continuations of each other. So, uhhh thus thus for t prime close to its close to t uhhh ft prime uhhh come up. So, if I take the pair which corresponds to uhhh uhhh the disc of convergence for ft prime uhhh I mean for ft and for ft prime they these two pairs are direct analytic continuations of one another. So, uhhh so so if I take mod z-gamma t less than rt this is the domain the disc of convergence of ft and uhhh the the other one is z-gamma t prime less than rt prime f of t prime are direct analytic continuations continuations of one another. So, this is what I have and the claim is that this uhhh this definition uhhh uhhh so the claim is that this definition of uhhh thinking of uhhh a one parameter family of power series okay uhhh such that close by power series are the same. That means uhhh they represent same analytic function uhhh this is I claim this gives a an equivalent definition of uhhh an indirect analytic continuation. So, the claim is uhhh that uhhh uhhh the power series the power series uhhh at B is the is an indirect analytic continuation of the power series at A. So, so the claim is claim is that the the the power series uhhh claim is mod z-gamma B uhhh gamma A uhhh rather gamma B uhhh less than r of B uhhh which is uhhh B is gamma uhhh B is just a z1. So, this is z1 gamma the power series at B uhhh which is gamma uhhh uhhh uhhh which is a power series at gamma B which is a power series at this point z1 uhhh is uhhh uhhh an indirect analytic continuation analytic continuation continuation of the initial uhhh the initial pair this is the final pair this is the final pair of the initial pair uhhh which is at the starting point gamma A is uhhh z0 okay. So, this is the uhhh this is the claim and to prove this claim I will have to show that you know our definition of indirect analytic continuation the original definition of indirect analytic continuation is that there are finitely many uhhh uhhh you know uhhh uhhh direct it is a chain of finitely many direct analytic continuation that is our definition of indirect analytic continuation that is what I will have to prove. So, what I will have to show is that this uhhh is obtained from this by uhhh chain of direct analytic continuation okay. Now uhhh see uhhh this is what we have to prove it is uhhh more or less intuitively obvious but then it needs a proof okay because it uses some topology. So, uhhh so the so the point is that while the definition insist that nearby uhhh nearby power series uhhh are equal okay which means that you are saying as T varies a uhhh in a small neighbourhood uhhh you are getting the same function okay it is clear that nearby power the the functions that you get nearby are one in the same. So, they are direct analytic continuation but if uhhh but it does not it does not imply that uhhh this and the starting one the the power series and the starting point the power series at the ending point the analytic functions that they represent are uhhh indirect analytic continuation of each other that is not imply okay. The the the definition only says nearby uhhh uhhh power series uhhh uhhh if you take 2 nearby points uhhh then the corresponding power series are direct analytic continuation okay. But it does not tell you uhhh uhhh that for example uhhh anything is an indirect analytic continuation of something before it okay. So, uhhh that is the power series at any point is an indirect analytic continuation of a power series at a point before okay that needs to be proved. So, uhhh so how does one prove this well one essentially uses uhhh uhhh some topology. So, the first thing is what you do is uhhh start with start with uhhh T not equal to a let epsilon not uhhh epsilon not be uhhh say the maximum uhhh uhhh such that uhhh f T prime is equal to f T not for all T prime in uhhh uhhh T not equal to a to T not as epsilon okay epsilon not right. So, choose uhhh uhhh you choose a maximum epsilon not see the definition says that you give me any point then there is a then there is a small neighbourhood uhhh uhhh of the parameter corresponding to that point where the power series are all equal alright. So, I start with T equal to T not then I will get an epsilon neighbourhood of that T equal to T not where all the power series are equal okay. Then what I do is so you know in this way uhhh so my uhhh inductively I do the following thing. So, I have this is a this is b on the real line okay and well and uhhh a is T not and what I do is now I have now I have uhhh I have this uhhh epsilon not uhhh plus T not and for all T in this uhhh interval for all T prime in this interval f T is equal to f T prime I have that okay. Now what I will do is I will start with epsilon plus T not I will call that I will call that as T one okay. If you call that as T one uhhh then there exists an epsilon one uhhh maximum such that uhhh uhhh uhhh f T f T uhhh T one is equal to uhhh f T prime is equal to f T one for all T prime in uhhh uhhh uhhh well this epsilon one neighbourhood about T one alright. So, uhhh T one T one minus epsilon one T one plus epsilon one intersection a b okay. This is again just this condition uhhh that the locally the power series uhhh for sufficiently closed values of the parameter are the same I am just supplying that condition here okay. And of course uhhh ice if if you know uhhh uhhh if this interval already contains a b I am done okay if if uhhh if the interval a, T not plus uhhh uhhh a, a plus epsilon epsilon not contains if it contains a b uhhh we are done because in that case you are saying that you are just saying that the uhhh you are just saying that the power series at b is same as a power series at a alright uhhh and uhhh uhhh I mean what you are saying is that uhhh the uhhh uhhh not the power series but the function represented by the power series at b namely f b is the same as the function representing the power series at a alright that is what you are saying and we are done. And which means that uhhh uhhh the this function is actually the same as this function and there is a whole neighbourhood uhhh there is a whole open set which contains this whole arc where uhhh the the f a and f b define same function okay. So, it it means that this is actually an extension it is a direct extension of uhhh f b is a direct analytic extension of f a okay. And of course is a direct analytic extension is also an indirect analytic extension okay it is a chain consisting of just one pair uhhh uhhh I mean uhhh just uhhh couple of pairs okay. So, uhhh so if we stop here but if it is not okay we continue okay so what what we do is if if not if not we do the following thing we put t1 is this and then you again uhhh try to look for a neighbourhood surrounding so surrounding t1 where uhhh ft1 where ft prime is the analytic function represented by ft prime is same as the analytic function represented by ft1 okay and uhhh uhhh clearly uhhh f t1 uhhh f t1 is a direct analytic continuation of f t0 which is f a okay this is because is because for t prime close to t1 and t prime less than t1 we have uhhh ft prime is equal to ft1 and ft prime is also equal to ft0 okay. Because ft prime is equal to ft0 the moment t prime goes lesser than t1 alright that is how t0 uhhh that is how epsilon0 was chosen that is how t1 was chosen alright. So, you have both uhhh which implies that uhhh f t1 is equal to ft0 okay. So, uhhh so in this way what has happened is that uhhh you know I have I have I have this point I have I have chosen this point t1 such that ft1 is uhhh direct analytic continuation of ft0 okay and uhhh and you know I now I continue this process I now continue this process until I reach the other uhhh the other end point uhhh uhhh until I until the my parameter t reaches b alright. So, you will have to show the parameter t uhhh if I do this process I have to show that I have to hit b at some point. Now again if uhhh well uhhh this this interval uhhh uhhh uhhh b is less than or equal to if b is strictly less than uhhh t1 plus epsilon1 is equal to t2 we are done okay we will be done uhhh and if not and if not we set uhhh uhhh uhhh we set t2 is equal to t1 plus epsilon1 and choose epsilon2 maximum so that uhhh ft prime is equal to ft2 uhhh for all t prime in uhhh t2 minus epsilon2 t2 plus epsilon2 intersection a b okay. So, you so you continue this process we continue by induction by induction getting an infinite chain uhhh chain of uhhh direct successive direct analytic continuations continuations. So, what you get is well I have first uhhh uhhh I want write the domains. So, I have ft0 which is fa and then uhhh ft1 ft2 and so on and the point is that uhhh and this corresponds to choosing points on the on the on the parameterizing interval. So, this is t0 then I get t1 then I get t2 and it goes on like this okay and every ftj is a direct analytic continuation of ftj minus 1 okay and I go on like this. Now the the point I want to say is that uhhh my my claim is that these t's okay there are 2 possibilities to the sequence of t's one is I am I might uhhh this sequence of t's could could could just stop okay. If the sequence of t's uhhh stops at a finite stage then I am done okay because then I would have then this would be a finite chain of direct analytic continuation starting from fa and ending at fb which will tell you that fb is an indirect analytic continuation of fa okay. See the only problem is that I need a finite chain for for fb to show that fb is an indirect analytic continuation of fa I need only a the only thing is I need is a is a finite chain okay it is a finiteness that is bothering me alright. So, if if the sequence uhhh t0 t0 t1 uhhh stops at a finite stage finite stage we are done we are done because you got a you got a finite chain of successive direct analytic continuation starts with fa and ends with fb and that will tell you that fb is an indirect analytic continuation of fa okay. So, this case can be ruled out okay this this case is done okay what is the other case the other case is that the sequence is an infinite sequence okay and if it is an infinite sequence there are 2 possibilities okay it is an infinite sequence which is uhhh which is an increasing sequence okay. So, it has to it is bounded above by the monotone convergence it has to converge okay. So, if it so there are again 2 possibilities if it converges to b also I am done okay the only problematic situation is when it converges to a point before b and in that case you get an uhhh that case will not happen because you will get a contradiction okay. So, uhhh if not the sequence uhhh ti tn n to the 0 is infinite and uhhh and it certainly converges to uhhh say uhhh t infinity uhhh in a b okay. So, uhhh so you are having a so I am using the monotone convergence theorem here I have a sequence of uhhh real numbers which is bounded above and which is increasing then it has to converge okay and the con uhhh whatever the limit is I am calling that limit as t infinity and it has to it has to belong to this interval close interval a b because it is closed a closed set will always contain the limits of its sequences okay. So, this t infinity is in a b and if this t infinity is b then I am done okay. We claim we claim t infinity is equal to b okay we claim t infinity is is the point b alright. If not if not uhhh uhhh then t infinity is a point strictly less than b okay. And uhhh uhhh that will give us a contradiction because you know see the situation is that uhhh so you have a which is t0 and you have t1 and so on. Suppose it converges to t infinity and this t infinity is strictly less than b uhhh will give a contradiction as follows what will happen is well uhhh you see the the condition the this condition that the power series f t you know uhhh it locally uhhh should match with the power series in nearby points also applies at t infinity okay. So, uhhh uhhh at at uhhh by by by the definition uhhh there exists a epsilon uhhh of t infinity greater than 0 such that uhhh f t prime is equal to f t infinity for all t prime in uhhh an epsilon t infinity neighbourhood of t infinity. So, it is so I should write this t infinity minus epsilon t infinity uhhh t infinity plus epsilon t infinity uhhh intersect a b I will get this okay and now this will give me a contradiction because you see uhhh if you choose a if you choose a uhhh uhhh so you know I I get this neighbourhood uhhh like this and this left end point is t infinity minus epsilon of t infinity the right end point is t infinity plus epsilon of t infinity of course I am not worried about uhhh the right end point it could go uhhh beyond b also okay. So, I am not worried about that but I am but what is important for me is the points before t infinity. So, if you if you choose a point if you choose a point t prime there then you will be in trouble uhhh choose choosing a point for m sufficiently large because after all uhhh uhhh every neighbourhood of the uhhh limit point contains points uhhh as close as you want uhhh of the sequence. So, I can find a t m there alright and then uhhh but what is the contradiction is uhhh we have that f t prime is equal to f uhhh t m for all t prime in uhhh uhhh t m minus epsilon m t m plus epsilon m intersection a b. I mean this is how you are choosing see what you should understand is that this t m plus epsilon m which is mind you this is t this is t m plus 1 this t m plus epsilon m is t m plus 1 alright and this t m plus 1 is to the left of t infinity okay this t m plus 1 is to the left of t infinity. So, I have this uhhh so, I have t infinity here I have on this on this side I have t infinity minus epsilon of t infinity okay and I have uhhh and on the other hand I have chosen t m here okay I have chosen t m here and then I have uhhh uhhh then there is a uhhh there is a neighbourhood alright uhhh t m minus epsilon m t m plus epsilon m alright and uhhh t m plus epsilon m is is to the left of t infinity okay. So, this is t m plus 1 this is t m minus plus epsilon m which is t m plus 1 alright. But the point is I get a contradiction here okay uhhh I will get a contradiction here in the sense that uhhh if I take any point here at a point here for a t there if I call this is this t st prime I will have that f t prime is equal to f t m okay I will have that and on the other hand I will also have f t prime is also equal to f uhhh t infinity okay and that means that you know uhhh uhhh uhhh so, it means that uhhh this t infinity has to be strictly less than epsilon uhhh the distance of t infinity from t m has to be strictly less than epsilon m which is not correct okay. So, so this implies uhhh distance of t infinity from t m should be lesser than epsilon m which is which is which is not correct this is what you get. See please try to understand I have this uhhh left side of the neighbourhood of t infinity with length of epsilon of t infinity such that for every t prime there the power series at f t prime is the same as the power series at f t infinity alright. In that neighbourhood I choose for m sufficiently large at t m okay and for that t m I have an epsilon m which is a maximum uhhh uhhh uhhh interval around t m such that f t prime is equal to f t m alright. But I have found a point outside uhhh that interval okay uhhh so, you know I have so, you know I have I have points uhhh I have this point t infinity which is outside that interval and such that uhhh uhhh uhhh yeah the the analytic function representing representing f t infinity is the same as analytic function representing f t m that is a contradiction to the definition of epsilon m okay and that contradiction tells me that this case will not happen okay. So, uhhh uhhh thus uhhh thus we end up with uhhh thus t infinity is actually equal to b. So, so I will just have to say that uhhh uhhh now since I know that t infinity is b uhhh then I use the compactness of the arc to get you know finitely many points uhhh in the sequence uhhh with the corresponding discs of convergence you know covering the uhhh covering the arc alright. And then it will follow that uhhh f b is uhhh uhhh in a uhhh continuation of f a right. So, uhhh now the discs the discs of convergence uhhh mod z-gamma ti this is an rti uhhh I grade I greater than or equal to 0 uhhh uhhh cover uhhh the arc traced by gamma uhhh which is compact which is compact and uhhh uhhh thus uhhh finitely many of these discs also also cover uhhh they also cover gamma this uhhh and uhhh uhhh so this shows then uhhh f b uhhh becomes an uhhh indirect analytic continuation continuation. So, let us stop here.