First fundamental Theorem of Nevanlinna Theory MAT 205B

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1 First fundamental Theorem of Nevanlinna Theory Ojesh Koul June 0, 208

2 The theory aims to describe the value distribution of meromorphic functions by looking at various formulae connecting the values of meromorphic functions with the distribution of its zeros and poles. In the beginning we will derive some of these formulae and later use them to get some connections with main characteristics of meromorphic functions. In the next section we will describe some preliminary results which will prove useful in our later endeavors. Preliminary Results Let D be a bounded region with boundary Γ consisting of piecewise analytic curves, denote the differentiation along inwardly directed normal vector to Γ, and be the 2-D Laplacian operator. The following formula is called the second Green s formula u v v u)dσ = u v v u ) dl ) D The above formula can be derived from the regular Green s formula by applying it to u v, u v ) y x u v + u v D x x + u ) v dσ = u v y y Γ dl Reversing u and v and subtracting we get the second Green s formula. Next we need the concept of Green s function. On a domain D the Green s function is a function Gζ, z) defined for ζ, z D, ζ z, satisfying the following. For each z D Gζ, z) = ζ z + h z ζ) where h z is a harmonic function in D and continuous in D. 2. If ζ Γ, z D or vice-versa Gζ, z) = 0 The Green s function is unique which can be seen by considering two Green s functions G and H. Using the properties of Green s function the function vζ) = Gζ, z) Hζ, z) is harmonic and continuous in D. Also it is zero on the boundary of D. By the maximum and minimum modulus principle of harmonic functions, v has to be zero everywhere. Hence G=H. Using the max and min principle we can also show that Green s function is greater than 0 for ζ, z D. For simply connected domains the existence can be proved by using Riemann mapping theorem. This gives us a conformal function from D onto unit disc wζ) = f z ζ) such that f z z) = 0. This function is continuous on D and has modulus on the boundary. Then the Green s function is given by Gζ, z) = 2) f z ζ) 2 Γ

3 Example... For D = { ζ < R}. Then Gζ, z) = R 2 ζ z Rζ z) 3) 2. For D = { ζ < R, Imζ > 0}. Then Gζ, z) = R 2 ζ z Rζ z Rζ z) R 2 ζz 4) Suppose D be a simply connected region. Let γ, γ 2,..., γ p be the analytic curves forming Γ. Let A l, l=,2,...,p be the common end points of these curves, let πα l 0 < α l < 2) be the angles between γ l and γ l+ γ p+ = γ ). Then f z has an extension to a domain containing D\{A,...A p }. We wil assume that in a sufficiently small neighborhood U l of A l, f z has a representation f z ζ) = ζ A l ) α l φ l ζ) + w l 5) with φ l analytic in U l, φ l A l ) 0 and w l =. For our purpose it is enough to see that that this is true for the two domains in example.). Since Gζ, z) > 0 for ζ, z D and G is zero on the boundary, G > 0. Along the curve we have f zζ) dζ = i G f z ζ dlζ A l) 6) This can be proved by taking the logarithmic derivative of f z and using the fact that f z ζ) = on the boundary. Then using Cauchy Riemann equations we can get the above equation. Theorem.2. Let D be a simply connected domain with a piecewise analytic boundary Γ, and let uz) be a twice continuously differentiable function in some domain containing D, excluding a finite set of points {c, c 2,...c q } D. In a neighborhood of these points u has the form uz) = d k z c k + u k z) 7) where d k are constants and u k is a twice continuously differentiable function in a neighborhood of the point c k. Then uz) + Gζ, z) uζ)dσ = uζ) G 2π D 2π Γ ds d k Gc k, z) 8) c k D Proof. Let us exclude from D the discs of radius ɛ centered at c,..., c q, z, A,..A p. We construct the discs such that there is no overlap between the discs and if the center of the discs is in D then the whole disc is in D. Let the domain obtained be D ɛ and the part of Γ not in any of the discs by Γ ɛ. By Cɛ, a) we denote the intersection of {z : z a = ɛ} and D. 3

4 Letting u=uζ) and v=gζ, z) in the second Green s formula and observing that v = 0 we get q p ) v udσ = u v D ɛ Γ ɛ Cɛ,z) Cɛ,c k ) Cɛ,A l ) v u ) ds 9) k= If any of A l matches with c k we consider it only once. Let us find the limits as ɛ 0. Since v=0 on Γ we have lim u ɛ 0 Γ v ɛ v u ) ds = uζ) G Γ By using mean value theorem we have for a D Cɛ,a) u v v u ) ds = length of Cɛ, a)) u v v u ds) ζ ζ being a point on Cɛ, a). Length of Cɛ, z) is 2πɛ and we have the following estimates as ɛ 0 v = v + O), = u + O), u = uz) + o), = O) ɛ ɛ then, u v v u ) ds = 2πuz) 0) lim ɛ o Cɛ,a) If c k D, then on Cɛ, c k ), we have the following estimates u = d k ɛ + O), u = d k ɛ + O), v = Gc k, z) + o), v = O) Therefore for c k D we have lim ɛ o Cɛ,c k ) l= u v v u ) ds = 2πd k Gc k, z) If a is one of the points c k belonging to the boundary or one of the points A,..A p, the length of Cɛ, a) < 2πɛ and the following estimates are valid u = O ɛ ), u = O v ), v = o), = O ɛ ɛ ) The last of these comes from equation 5. Hence we have lim ɛ o Cɛ,c k ) u v v u ) ds = 0 Now taking limit ɛ 0 in equation 9 we get the statement of the theorem. 2 Poisson Jensen formulae In this and the next sections we will prove a bunch of formulae connecting the behavior of meromorphic function to their zeros and poles. Most of these are special cases of 8. 4

5 Theorem 2.. Let D be a simply connected region with a piecewise analytic boundary Γ and fz) 0 meromorphic function in D. Then fz) = fζ) G 2π dl Ga m, z) + Gb n, z) ) Γ a m D where a m and b n are the zeros and poles of fz) respectively. Proof. We apply Theorem.2 with uz)= fz). If z is neither a pole nor a zero then u is zero. Hence the integral on the right is zero. If c k is a zeropole) of fz) of order χ k then d k = χ k d k = χ k ). Hence the right hand side of equation 8 is equal to the right hand side of equation One of the most important special case of the above case is when D is the disc {z : z < R}. Theorem 2.2. Let fz) 0 be meromorphic on the disc {z : z < R}. Then the following formula known as Poisson Jensen formula holds fz) = 2π fre iθ ) Re Reıθ+z 2π Re i θ z dθ R 2 a m z Rz a m ) + R 2 b n z Rz b n ) 0 a m <R where a m and b n are the zeros and poles of fz) respectively. b n D Proof. The theorem follows from entering the explicit expression for the Green s function in Theorem 2. and using equation 6 to get G. Theorem 2.3. Let fz) 0 be meromorphic on the disc {z : z < R} and let 2) fz) = c λ z λ + c λ+ z λ+ +..., c λ 0 3) be the laurent expansion. Then the following known as the Jensen formula holds c λ = 2π fre iθ) dθ 2π 0 0< a m <R R a m + where a m and b n are the zeros and poles of fz) respectively. 0< R λ R 4) b n Proof. Let z 0 in the Poisson Jensen formula. For the sum on the right we get a m <R + and on the left hand side we get = 0< a m <R = 0< a m <R + 0< R a m + Now taking z 0 the Jensen formula follows. +λ z R 0< R b n + λ z R + o) fz) = λ z + c λ + o) 5) 5

6 3 Shimizu Ahlfors formula This is again a special case of Theorem.2. Theorem 3.. Let fz) 0 be meromorphic on the disc {z : z < R}. If f0), then the following formula known as Shimizu Ahlfors formula holds R ) f z) 2 π z R z + fz) 2 ) dσ = 2π + fre 2 2π iθ 2 dθ 0 + f0) 2 + R 6) b n If f0)=, then the right hand is replace by 2π + fre 2π iθ 2 dθ c λ λ R + R 0 b n here c λ is the same as in Theorem 2.3. Proof. We use Theorem.2 with uz) = + fz) 2, D the disc with radius R. Let A=Refz) and B=Imfz). We compute uz). Using the Cauchy Riemann equation and the fact that A and B are harmonic we find u = 2 f z) 2 + fz) 2 ) 2 Observing that in a neighborhood of a pole b k of f of order χ k, c k in Theorem.2 is b k and d k = χ k, we get + fz) 2 + R 2 ζz f z) 2 π ζ R Rz ζ) + fz) 2 ) dσζ) 2 = 2π R 2 z 2 2π 0 Re iθ z + fre iθ ) 2 dθ + R 2 b n z 7) 2 Rz b n ) If f0) taking z to 0 we get the formula. In case f0)= we take into account when z 0. Taking z 0 we get the formulae. + fz) 2 = λ z R + c λ + o) R 2 b n z z Rz b n ) = λ R + R + o) b n 0< 6

7 4 Nevanlinna Characteristics. The first fundamental theorem We will define several real functions defined on [0, ) which characterize the behavior of a meromorphic function fz). These functions are called the Nevanlinna characteristics of fz). The number of poles of fz) in the disc { z R} is denoted by nr,f). It is an integer valued function, non decreasing and right semi continuous. Any point r, is a point of discontinuity if f has poles on the circle of radius r. The jump value is equal to the number of poles on the circle. Since the poles of f can t have limit point, nr, f) is piecewise constant on every [a, b] [0, ). Let r nt, f) n0, f) Nr, f) = dt + n0, f) r 8) t 0 This characteristic describes the location of poles of fz).the function Nr, f) is continuous, non decreasing on 0, ). Also, as a function of r, this is a convex function. It is noteworthy that the characteristic takes into account only the absolute value of the poles. The Jensen formula can be succinctly written in terms of Nr,f) in the following way N r, ) Nr, f) = 2π fre iθ ) dθ c λ 9) f 2π 0 This can be seen from the following equalities r = Nr, f) n0, f) r b n 0< b n <r r a m = N r, ) n 0, ) r f f 0< a m <r n0, f) n 0, ) = λ f Let us now define the function + x which is useful in defining another characteristic. + x = max x, 0) We have the following relations for this function x = + x + x, x = + x + + x, + x = x 20) where x 0,x = maxx, ). We will also require the following inequalities 7

8 n n + x m + x m 2) n + x m n + x m + n 22) + x + x 2 x x 2 23) + x + x 2 + x x ) Let mr, f) = 2π + fre iθ ) dθ 2π 0 This function gives ) us information about the growth of the function fz).when a the function m r, tells us about the proximity of fz) to a on the circle { z = R} and the f a ) function n characterize the modulii of the roots of the equation fz) = a. When r, f a the function ) f we are) dealing with is clear, we write nf, a), mf, a) and so on instead of n, m. Introduce a new characteristic r, f a r, f a T r, f) = mr, f) + Nr, f) 25) Now we are in a position to prove the first fundamental theorem of Nevanlinna theory. Theorem 4.. Let fz) be a non constant meromorphic function. Then if a where ɛr, a) = O) as r mr, a) + Nr, a) = T r, f) + ɛr, a) 26) Proof. We use the form of Jensen formula as given in equation 9. We can do this since the function f a is non constant. Since the poles of f a coincide with the poles of f we have ) N r, Nr, f) = 2π fre iθ ) a dθ c λ a) 27) f a 2π 0 where c λ a) is the first non zero term in the Laurent expansion of f a in the neighborhood of 0. By the first equality in 20 we have 2π ) fre iθ ) a dθ = mr, f a) m r, 28) 2π 0 f a 8

9 and we can write equation 27 as. Nr, a) + mr, a) = Nr, f) + mr, f a) c λ a) 29) Now we use the inequality 24 to get + f a + f + a + 2 mr, f a) mr, f) + a + 2 Defining ɛr, a) = mr, f a) mr, f) c λ a) we get equation 26. To complete the proof observe that ɛr, a) + a c λ a) = O) as r Intuitively the theorem states that if fz) takes on the value a more often than it takes the value b, then it approaches a more slowly than it approaches b. Now we show that the T r, f) tends to as r, so the contribution of the term ɛr, a) can be ignored. Let a = f0), then n0, a) > 0 and Nr, a) = r since mr, a) is non negative 0 nt, a) n0, a) dt + n0, a) r n0, a) r t T r, f) Nr, a) + O) as r 30) 5 Further topics in Nevanlinna theory Writing the theorem as first fundamental theorem surely means that we have a second fundamental theorem. The second theorem deals with Nr, f) defined in the same way as Nr, f) but without taking multiplicity of poles into account and gives bound on k mr, a i, f) where a i are distinct values on the Riemann sphere. One of the application of the second fundamental theorem is to prove the Picard s theorem. An interesting topic to read parallel to Nevanlinna theory is the Ahlfors theory which is the geometric conterpart of Nevanlinna theory. Lars Ahlfors was awarded the Fields medal for his work on this in

10 References [] Anatoly A. Goldberg, Iossif V. Ostrovskii. Value distribution of Meromorphic Functions, Translations of Mathematical Monographs Volume 236. [2] W.K Hayman. Meromorphic Functions [3], theory 0