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Addison Beer
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Date Created: 09/09/15
printed on 2172009 Complex Analysis Math 535 Winter 2009 Donald E Marshall VII Series and Products 1 Mittag Le ier s Theorem So far we7ve learned a little bit about polynomials rational functions exponentials and logarithms NOW Well look at more complex functions Suppose f is meromorphic in a region Q With a pole at b E 9 Then recalling the Laurent expansion fz07 c 1b 27 Zibwil a0a1ziba2zib2 for 2 near b The sum of the first 11 terms Sbzmmzjb is called the singular part of f at b If f is rational then by a partial fraction expansion HZ Z Sbk Z 192 k1 Where p is a polynomial If f is meromorphic in a region Q With only finitely many poles then M Z Sm 92 k1 Where g is analytic in 9 This follows because fz 7 55 is analytic at each bk and therefore in all of Q In this section well find a similar expansion for meromorphic functions in Q With infinitely many poles We say that an infinite sequence bk 6 Q a 89 as k a 00 if each compact K C 9 contains only finitely many bk Theorem 11 Mittag Le ier Suppose bk 6 Q a 89 Set 71k CV k S 17 V Z 7 1 3 4 VII Series and Products where each nk is a positive integer and ch E C Then there is a function meromorphic in Q with singular parts Sk at bk k 12 and no other singular parts in 9 Proof Let Kn z E Q distzz9 2 2 and S 1 n Then Kn is a compact subset of 9 such that each component of C Kn contains a point of C Q and Kn C Kn By Runge7s Theorem lV34 we can find a rational function fn with poles in C 9 so that i 2 SM 7 Mz lt 2 bkEKn2Knl for all z E Kn Then 612 M2 7127 bkEKn2Kn1 converges uniformly on Km to an analytic function on Km by the Weierstrass M test and Weierstrass7s Theorem lV39 Thus fltzgtZSkltzgtZ Z Skltzgtef ltzgt 11 bk K2 711 bkEKn2Knl is the desired function E A more constructive proof of Mittag Lef er7s Theorem is useful in some circumstances We will illustrate the technique in some special cases and leave the more general case to the reader Suppose we7d like to find a function meromorphic in C with singular part akz 7 bk at bk where bkl a 00 Then 00 1k Zzibk k1 will work provided the sum converges When does it converge lf S R lt 00 write 0 1k 1k 1k leJr bk 13 2 bk 2 Z Z Z lt gt k1 kzibkigza kzibkpm Mittag Lef er7s Theorem 5 The first sum has only nitely many terms For the second sum S R lt lbkl2 so that Thus if M lt 00 14 then the second sum in 13 converges uniformly on S R to an analytic function by Weierstrass again The right side of 13 then is meromorphic in lt R with singular part akz 7 bk at bk lbkl lt R Since R is arbitrary the sum in 13 is meromorphic in C What if lbkl tends to 00 more slowly If we examine the proof we just need the tail of the series to converge for each R Mittag Lef er7s idea was to subtract a polynomial from each term so that the result converges As we have seen before 1 7 1 7 1 z j Z 7 bk bk17 bk 160 bk So it is natural to subtract a few terms of the expansion to make it smaller 1 1 i z j 7 1 f z j 2 bk bk 1 bk T bk 1 bk 10 Jw1 z WH 1 E lil l 1 bk provided lt For example the following proposition holds Proposition 12 If bk a 00 and if for some n lt 00 Ml lt 00 2 2 k lbkl 6 V11 Series and Products then V 0 77 J ak Lk 3 fltzgtizzibk bkZltbk k1 30 is meromorphic in C with Singular part Sbk kabk at bk k 1 2 and no other poles To prove Proposition 12 if lt R split the sum into two pieces a nite sum of the terms with lbkl 3 2B and a convergent sum of the terms with lbkl gt 2R Example 13 7r2 0 1 m 7oo 27112 17 The right side of 17 converges uniformly on compact subsets of C and the limit is meromorphic with singular parts 5712 12 7 n2 at z n and has no other poles To see this follow the proof of Proposition 12 fix R lt 00 and split the sum into two pieces a finite sum of terms with lt 2B and the remaining infinite sum 1n the second sum when S R the nth term is uniformly bounded by 7R2 3 4112 Thus the second in nite sum converges uniformly and absolutely on S R to an analytic function by Weierstrass7s Theorem The function 7rz sin 7rz has a removable singularity at z 0 and is an even function so that 1 022 near 0 By squaring and dividing by 22 we conclude that the the singular part of 7r2 sin2 7rz 2 7rz 7 n sin2 7w we conclude that the singular part of the at z 0 is 122 Since sin left side of 17 is the same as the singular part of the right side of 17 at z n for each 11 Set 7r2 0 1 f 7 sin 7rz Then F is entire and Fz 1 Write z x iy and suppose 0 S x S 1 We claim that 1F2l 0 L8 Mittag Lef er7s Theorem 7 as lyl 7 00 If so F is bounded in the strip 0 S x S 1 and since Fz 1 Fz the function F is bounded in C By Liouville7s Theorem F is constant and by 18 F E 0 proving the Example To see 18 first observe that lsin7rzl 6 yi 7 e ry imlZ 7 oo as lyl 7 00 Thus the left side of 17 tends to 0 as lyl 7 00 With 0 S x S 1 Likewise in the strip 0 S x S 1 the right side of 17 is dominated by 1 lnl12y2 Z Which also tends to 0 as lyl 7 00 by comparison With the integral flee dab2 242 This proves 18 and the Example D Example 14 1 1 1 t 7 E 7 7rco 7rz 2 lt 19 7170 2 7 n 11 First observe that 2 7r 1 7 t 7 7 7 d2 7r CO 7w sin2 7rz Z 2 7 112 But 1 7rcot7rz7 Z because the latter sum does not converge The dif culty is that for behaves like 111 However 2quot S R the nth term 1 1 z 1 7 2711 n T 2 7 nn N W for large n and Zn 2 lt 00 To prove Example 14 suppose S R and split the sum on the right side of 19 into the sum of the terms With 3 2B and the sum of the terms With gt 2B The first sum is finite and the second sum has terms satisfying 1 1 Z R 2R 7 73 i 2711 n 27nn n2 8 V11 Series and Products Since Z2Rn2 lt 00 the right side of 19 is meromorphic in S R with poles only at the integers and prescribed singular parts 12 7 n for each R lt 00 Furthermore by Weierstrass7s Theorem 1V39 and Example 13 the right side of 19 has derivative 72712 7m 7rcot7rz Thus the two sides of 19 differ by a constant The convergence of the right side of 19 is absolute so that we can rearrange the terms and the sum will converge to the same limit This allows us to conclude that the right side of 19 is an odd function Since the left side of 19 is also odd the difference between the left and right sides is odd and constant and hence is identically 0 D We will use the function 7r cot 7rz in the section on Calculus of Residues Example 15 Weierstrass 73 Function Suppose 11111112 E C with wlwg not real In other words ml and 1112 are not on the same line through the origin There are no entire function f satisfying fz 1 ml fz 1 mg fz for all 2 by Liouville7s Theorem but there are meromorphic functions with this property The Weierstrass 73 function is defined by 1 1 1 732 7 7 110 m nmo o 2 T m T quotW2 mm my where the sum is taken over all pairs of integers except 00 To prove convergence of this sum we first observe that there is a 6 gt 0 so that lmwl nwgl 2 6 unless m n 0 for if lmjwl 1 nngl a 0 then 13 le 0 contradicting the assumption that wlwg is not real Thus Cmm mwl nw2 m n E Z forms a lattice of points in C with no two points closer than 6 If we place a disk of radius 62 centered at each point of the lattice then the disks are disjoint The area of the annulus k S Kquot S k 1 is 2k 1 17r so there are at most Ck lattice points in this annulus for some constant 0 depending on 6 Weierstrass Pro ducts 9 For lt R we split the sum in 110 into a nite sum of terms with Km 3 2B and the sum of terms with Kmml gt 2R Note that if lt R and id gt 2R then i 2702 C2 22C 7 22 22702 Rlt21C1R 1412147212 1413 We conclude that 1 1 Z WW 3w 1 1 Z Cmn2 3m 7 lcmnlgt2K k2K kltlcmmlgk1 S f 016 lt oo k2K By Weierstrass7s Theorem the Weierstrass 73 function is meromorphic in C with singular part 52 12 7 mwl 7 nw22 at mwl 711112 and no other poles Next we show that 732 717 ml By Weierstrass7s Theorem 7 2 2 P 2 73 By the same estimate this series converges absolutely so that 732w1 732 and hence 73z w1773z is a constant But the series for 73 is even so 73z w1773z 0 when 2 7w12 and thus 73zw1 A similar argument shows that 73zw2 The Weierstrass 73 function is a basic example of an elliptic function Elliptic functions are important in the application of complex analysis to number theory 2 Weierstrass Products Just as infinite sums are used to create meromorphic functions with prescribed poles infinite products are used to create analytic functions with prescribed zeros 1f fj are analytic and f 0 then Hy fj vanishes at 2122 zn If we want a function to vanish at infinitely many 2 it is natural to try 77 lim H fj 71700 13971 10 VII Series and Products However this limit may not exist Moreover it is possible for the product of infinitely many non zero numbers to converge to zero which might create additional zeros First we will treat the convergence of infinite products of complex numbers Proposition 21 Suppose p E C Then Hydpj converges to a non zero complex number P as n a 00 if and only if 210ng j1 converges to a complex number S where logpj is de ned so that 77139 lt arg p 3 7r More over if convergence holds then P es and limp 1 Proof Let 5 21logpj and P7 11371193 lf 5 a S then P7 65 a es 6 C0 Conversely if P7 a P E C 0 then by altering 191 if necessary we may suppose P gt 0 Then log P7 a log P where the logarithm is defined so that 77r lt arg P7 3 7139 Note that anialn i 1i an Pnili a 0 so that 1 a l and argpn a 0 But 77 log P7 Zlogpj 27rikn j1 for some integers k7 and since lngn a 0 and loan 7 log P71 a 0 we must have kn km for no suf ciently large and n gt no Thus limn 21logpj exists and the Proposition is proved B One possibility for a definition of absolute convergence of an infinite product Hp would be the convergence of the partial product Hyd lpjl However Hyd fij eimnil does not converge even though the product of the absolute values is l The notion of absolute convergence was useful for infinite sums because it implies that the terms in the sum can be rearranged without affecting convergence De nition 22 lfpj are non zero complex numbers then we say that pj converges absolutely if E i log pjl converges Weierstrass Pro ducts 11 It is not hard to see that if p are nonzero complex numbers such that 1 con verges absolutely then any rearrangement of the terms in the in nite product will not affect lim Hy 1 The next Lemma is useful for determining absolute convergence Lemma 23 lfpj are nonzero complex numbers then p converges absolutely if and only if Z My 1 j1 converges Proof logz i d 1 71 so that if p 7 1 then for j suf ciently large 1 v 1 amp 1 lt and thus 1 3 5le i 1l S llogpjl S 5le i 1la and the Lemma follows E Lemma 23 says that p converges absolutely if and only if p 71 converges absolutely The analogous statement for convergence is false If p e 1j then 2 logpj converges by the alternating series test and hence Hp converges but Zj p 7 l diverges Similarly if p l then 21 7 l converges but 2 logpj diverges and hence Hp diverges The proof of these statements is an exercise for the reader As mentioned at the start of this section we would like to consider infinite products of analytic functions H fjz but we would like to allow the f to have zeros For that reason we make the following definition De nition 24 Suppose fj are analytic on a region Q We say that f con verges on Q iffor every compact K C 9 there is a jo lt 00 so that f 31 0 for all z E K 12 VII Series and Products and allj 2 jg and if Hugo H 222 jjo converges uniformly on K to a non zero function g If the product converges then x jail n H m2 2 H W W113 H W 11 11 jjo is analytic on Q The next theorem says that no extra zeros are created in an in nite pro duct Theorem 24 Hurwitz Suppose 97 are analytic and gnz 31 0 for all z in a region Q If 97 converges uniformly to g on compact subsets of Q then either 9 is identically zero in Q or g is non zero in 9 Proof The limit function g is analytic on Q by Weierstrass7s Theorem In particular if g is not identically zero then the zeros of g are isolated If y N 0 is a simple curve on which 9 31 0 then by Weierstrass7s Theorem 9 converges to g and hence gggn converges to gg uniformly on 7 By the argument principle for n suf ciently large the number of zeros of g enclosed by y is the same as the number of zeros of 97 enclosed by y and since 97 is never zero the theorem follows E Hurwitz7s Theorem shows that the zero set of an infinite product U f is the union of the zero sets of the f As an aside we point out the following Corollary to Hurwitz7s Theorem which is useful for studying conformal maps Corollary 25 If 97 are analytic and one to one in a region 9 such that 97 converges uniformly on compact subsets of Q to 9 then either 9 is one to one or g is constant Proof Apply Hurwitz7s Theorem to gnz 7 gn2 0 in Q 20 D The next result is the analog of Mittag Lef er7s Theorem for products Weierstrass Products 13 Theorem 26 Weierstrass Ifbj C 9 With bj 7 89 and if nj are positive integers then there eXists an analytic function f on 9 such that f has a zero of order exactly nj at bj forj 12 and no other zeros Proof We first suppose that 89 is nonempty and bounded We include the case where 00 is an interior point of 9 As in the proof of Mittag Lef er7s Theorem let Kn z E 9 distz89 2 1 n Then Kn is a compact subset of 9 such that each component of 3 Kn contains a point of 3 9 and Kn C Kn If 00 is an interior point of 9 then Kn is viewed as a compact subset of the Riemann sphere 3 Choose aj E 89 so that distbj 89 bj 7aji We can de ne logz 7 bjz7 1 so as to be analytic in 3 ab bj Where ab bj denotes the line segment from aj to bj In particular if bj Kn then logz 7 bjz 7 1 is analytic on K By Runge7s Theorem lV34 we can find a rational function fn With poles in 3 9 so that Z 110g7fn2 lt 2 21 bk EKn2Kn1 for all z E Kn Then 1 nj loglt7 a7fnz 712m bk EKn2Knl converges uniformly on Km to an analytic function on Km by Weierstrass7s Theorem lV39 Thus i 27 W 00 27bit W fnz m H W H H 7 e bk K2 1 bk EKn2Knl is the desired function Note that if 00 is an interior point of 9 then f is analytic at 00 If 89 is non empty but unbounded then choose b E 9 With b 31 bk for all k Apply the result above on the image of the region 9 by the map Tz lz 7 b then compose 14 V11 Series and Products the resulting function with T l The only remaining case to consider is Q C In this case we set Kn z S n and instead of 21 we find polynomials fn so that nj logz 7 bj 7 fnz lt 2 71 bkEKn2Kn1 for all z E K Otherwise the proof is identical D Here is a more constructive version in the plane Theorem 27 Suppose an C C 0 and suppose g is a nonnegative integer such that 1 Z lt 00 711 Then 0 9 z 39 Hlt1i igteZF1 1a 22 an 711 converges and represents an entire function with zeros at an and no other zeros The number 9 in Theorem 27 is called the genus of the infinite product In the case when 9 0 we interpret the sum as equal to 0 The an in Theorem 27 do not need to be distinct so that the Theorem can be used to construct functions with zeros of order more than one Proof By integrating the series for 11 7 w we have j 1 log17w7 fw HMS J where the sum converges and therefore is analytic for lwl lt 1 1f lt R and lal gt 2R 2 g 1 2 339 91 00 1 1 17971 logll alfgiial 2 35 2 jg1 then R 91 1 R S NEXT TIME JUST USE THAT log17 c 31 314739 VG is analytic on the unit disk and therefore bounded on the disk of radius 12 It is easier to remember and apply Weierstrass Products 15 Thus if S R then 9 j z 1 z 1 lt17 i E 7 i og an j1jltan So by Weierstrass7s Theorem and Hurwitz7s Theorem since lanl a 00 the function given 2 lanlgt2R l S 2Rg1 7 lt oo laEZR lanlg1 by 22 is analytic in lt R and is zero at each 1 with lanl lt R and no other zeros in 2 lt R Since R is arbitrary the theorem follows E Example 28 The function 0 2 z 1 a m converges since Z n 2 lt 00 g l and vanishes precisely at the positive integers Thus we can write 0 2 z sin7rz 269Z H 1 geg 71700 where g is entire lndeed sin 7rz has a simple zero at each integer and no other zeros so if we divide by 2 times the product the resulting function is entire and never equal to 0 in C and hence has an analytic logarithm To find 9 we take the logarithmic derivative ff 1 27 l l t 7 E 7 7rco 7rz Zgz 71 n7 0 But then by 19 we must have gz E 0 and hence g is constant Since sin 7rz lim 7 7r ZHO 2 we must have 690 7r and we obtain 0 Z 1 sin7rz7rz H 177 6 n 71700 We now give a few corollaries of the Weierstrass Product Theorem 26 16 VII Series and Products Corollary 29 H9 is a region then there is a function f analytic on 9 such that f does not extend to be analytic in any larger region Proof Take a sequence 17 7 89 such that 89 C an By the Weierstrass Product Theorem Theorem 26 we can find f analytic on 9 with an 0 but f not identically zero If f extends to be analytic in a neighborhood of b E 89 then the zeros of the extended function would not be isolated D In several complex variables a similar result is not true If Br 2111 M2 11 lt r2 then any function which is analytic on 32 Bl extends to be analytic on Bg Corollary 210 Suppose 9 is a region and an 7 89 and suppose on are complex numbers then there exists f analytic on 9 such that fltan Cn Results like this corollary are usually called interpolation theorems Proof By the Weierstrass Product Theorem we can find G analytic on 9 with a simple zero at each an Let G 1 lim 7 Gan Z7117 z 7 7 Since the zero of G at an is simple 1 31 0 By Mittag Lef e s Theorem we can find F meromorphic on 9 with singular part cndn 5712 27 a at an and no other poles in C Then fz is analytic on 9 an and C2 c7 2 7 1 d7 0 lim Jim 2 7 anFz z7gtan Thus the singularity of f at each an is removable and f extends to be entire with an on D Blaschke Gamma and Zeta 17 Corollary 211 If f is meromorphic in Q then there are functions 9 and h analytic on 9 such that Proof Let an be the poles of f where the list is written such that a pole of order k occurs k times in this list By the Weierstrass Product Theorem there is a function h analytic on Q with zeros an and no other zeros Then 9 t is analytic on Q 17 with removable singularity at each an Thus 9 extends to be analytic in Q and f gh D The idea behind the proof of Corollary 211 can be used to prove the analog of Mittag Lef er7s Theorem for Taylor polynomials Corollary 212 Suppose 1 is a polynomial of degree 1 for n 1 2 and suppose an 7 89 Then there is a function f analytic on 9 such that for each n 1 2 7 pnz 7 an has a zero of order 1 1 at an In other words the Taylor polynomial for f at an of degree 1 is the prescribed function 1 z 7 17 Proof By the Weierstrass Product Theorem we can find h analytic on Q with zeros an of order dn1 Let 5 be the singular part ofpz 7 So 5 712 7 anhz is analytic in a neighborhood of an By Mittag Lef er7s Theorem we can find 9 meromorphic on Q with singular part 542 at an Then 92 7 pnz 7 anhz extends to be analytic near an and so 7p z 7 an is analytic near 1 with a zero of order at least 1 1 at an This proves the corollary with f gh D 3 Blaschke Gamma and Zeta Weierstrass7s Theorem shows the zero set of an analytic function on the unit disc can be any sequence tending to the unit circle More information on the growth of the analytic 18 VII Series and Products function will give us more information on how quickly the zeros approach the circle This is the content of Jensen7s formula Theorem 31 Jensen Suppose f is meromorphic on S R with zeros a1 an and poles b1 bn Suppose also that 0 is not a zero or a pole off Then 12quot t R R 7 log f ReZ dtlog f 0 logi 7 logi 27f 0 l lt M l M Z W Z W Proof Replacing fz by fR2 we may assume R 1 First suppose that f has no poles or zeros on 1 Write and where g is analytic on 3 land has no zeros Then lf0l lajlHk lbkllg0 log lfeitl log lgeitl Re loggeit where loggz is analytic on S 1 Thus 123 ifltitgtidt7 123 lltitgtldt 271390 0g 6 27r0 Ogge loglg0gtl 7 loglf0l 7 Zloglajl Zloglbkl j k To prove Theorem 31 in the case where f has zeros or poles on 1 it suf ces to show that 27r I I log 6 7 eZtOldt 0 31 0 by factoring out the zeros and poles on the circle Furthermore we can suppose to 0 by the change of variables 5 t 7150 Since logl 7 r2 is analytic on S 1 when 7 lt l we have f027r log ireit 7 lldt 0 For 6 gt 0 write 27r I 27r logl62t7lldt log 0 0 log ltllt5 6 re 7 1 it 71 Eif dt log W 1 M25 71 dt cit 7 1 re 7 l dt 32 Blaschke Gamma and Zeta 19 1f 1151 2 6 then 1 61171 0 o 7 7 g reZt71 uniformly int 1f 1151 lt 6 then 621571 2 1 l 7 ltl 7 l 7 lt21 7 0g re 7111 0g re 711 0g ems 711 0g cltl since reit 7 11 2 lsintl 2 if 6 S Thus lo ltllt5 as 6 7 0 Equation 31 with to 0 now follows from 32 D g it 71 1 1 67 dtg 2log7dt3026log7 70 re 71 MG cltl 6 Corollary 32 Suppose f is analytic in D f f 0 with zeros an and suppose 27r sup logl reit dt lt oo rlt1 0 Then 2171an1 lt 00 710 Proof Since f has only nitely many zeros at 0 we may divide them out and suppose f0 31 0 Then by Jensen7s formula 1 Zlog 7 lt 00 710 1an1 But limmml 1 so that Zlog lt 00 if and only if 21 7 lanl lt 00 D We remark that log fwt 0p freitPdt for some constant Op depending on p gt 0 So if lfreitl 6 LP then the hypothesis of Corollary 32 holds As a partial converse we have the following Theorem 20 VII Series and Products Theorem 33 lfan C D such that 21 7 lanl lt 00 then 0 lanl Zia71 32 H n 0 7 fan li z converges uniformly and absolutely on compact subsets of D Where we de ne the conver gence factor lagfan to be equal to 1 if an 0 The function B is analytic on D is bounded by 1 and has zero set exactly equal to an The function in Theorem 33 is called a Blaschke product Proof Without loss of generality an 31 0 for all n Then for S 7 lanl Z an 71 17 lawllllanlz l anl lt 217 lanl fan li z lanli zl 7 inflaanir By Weierstrass7s M test B converges uniformly and absolutely on S 7 Note that the partial products for B are all bounded by l and analytic on D D Theorem 33 gives the following factorization result Theorem 34 lff is bounded and analytic on D With zero set an counting multiplicity and if B is the Blaschke product With zero set an then 2 BZ69Z for some analytic function g With suplegwl suplf2l ZED ZED Proof Without loss of generality l S l on D Write N an 27 an 711 fan 1 7amp2 Then by Schwarz7s Lemma S 1 Choose rn a 1 so that B 31 0 on Tn Then fBN converges uniformly to fB on Tn and by the Maximum Principle the convergence is uniform on S Tn Now let rn a 1 Thus h fB is bounded by l on The Gamma and Zeta Functions 21 D and non vanishing This implies g log h can be de ned as an analytic function on D and Theorem 33 follows D It is dif cult to overestimate the importance of Blaschke products in function theory on D 4 The Gamma and Zeta Functions The simplest entire function With zeros at the integers is sin 7w By the same conver gence proof 0 2 a E 1 7 7 lt2 ll ngte is the simplest entire function With zeros only at the negative integers Moreover 1 2G2G72 7 s1n7rz 7r Note that Cz 7 has no zeros and hence 67Z2G2 Cz 71 41 42 43 Where 72 is entire By Weierstrass7s Theorem and the uniform convergence of the product 41 we can take logarithmic derivatives and 1 0 1 1 70 11 WltZgtzg zn n T 271n n 711 1 1 1 771 n z zn n1 1 1 1 1 1 771 77 77 z zn n1 711 Thus yo 71 7 10 n1 22 V11 Series and Products and y is a constant By 41 and 43 m 1 1 G0 any lirn H 1 Ee 1 lirn m15 1 mace 711 mace Thus 1 1 1 y lirn 17iilogm1 m057722 mace 2 3 m The proof above shows that this limit called Euler s constant exists The Gamma Function is de ned by 1 1 1 P i39yzi 7 zn Z 2Gze Z 6 2 711114 2116 Then T is a non vanishing meromorphic function With poles at the non positive integers and no other poles Moreover by 43 Pz 1 2Tz 44 and by 42 P2P1 a z 45 It follows from 41 and 42 that P1 1 and by induction Pn n 7 1 By 45 at z 12 we also obtain 7r Theorem 41 If Rez gt 0 then m Awtz le tdt 46 Proof The integral converges uniformly and absolutely on compact subsets of Rez gt 0 because ltz ll lthez l is integrable near 15 0 and because lthez le t 3 ft2 for 15 large By Morera7s Theorern the right side of 46 is analytic for Rez gt 0 r742 Ontz11andt Set The Gamma and Zeta Functions 23 Then we claim that for Rez x gt 0 lim PM tm le tdt 47 71700 0 1 1 139 7 7 ninio PM W lf 47 and 48 hold then Theorem 41 follows from the Uniqueness Theorem 48 Lemma 42 KM lt 00 then t n 15 17 7 6 increases to 1 n uniformly for 0 S t S M Proof The slope of the secant line to the graph of logu with one end at u 1 7 1511 and the other end at u 1 decreases to 1 as n 7 00 since the graph of logu is concave Thus 15 log17 3 7 log1 l 1 7 7 t 7 1 7t 0 49 noglt ngt 17 gt7 lt gt uniformly for 0 S t S M Lemma 42 now follows since 715 S 0 and 65 is an increasing function of s D By Lemma 42 ma 3 131435 tm le tdt 0 and 47 holds for x gt 0 Substitute t ms in the definition of Pnz and obtain 1 new 37117 was 2 Mg 410 0 Note that f0z 12 and fn1z fnz 7 fnz 1 so that by induction n fnz zz1zn39 Thus 1 7 n z n z 7 Z H 110 n 1 711 Z2 17z 17e Zkezk1k g 7 Pnz E k E k PltZgt H 24 V11 Series and Products as n a 00 proving 48 and Theorem 41 The Riemann Zeta Function is de ned by 00 Zniza 711 Where n z e ZIOg The zeta function is important because of its connection With the prime numbers as illustrated in Theorem 43 below By the integral test the series for 2 converges uniformly and absolutely on compact subsets of Re2 gt 1 and hence is analytic in Re2 gt 1 The zeta function can be extended to Re2 gt 0 by comparing the sum to the appropriate integral For Re2 gt 1 n1 n Z 7 tizdt n1 t 257Z 1ds 411 m A 2 V l 2 l H H M 8 3 l l H 8 1 n The estimate n1 t Zsizil ds iniRezi1 77 77 2 and the Weierstrass M test show that 411 converges uniformly and absolutely on com pact subsets of Re2 gt 0 to an analytic function Thus 2 extends to be meromorphic in Re2 gt 0 With a simple pole at 2 1 and no other poles The Riemann Hypothesis perhaps the most famous problem in mathematics is that 2 31 0 for Re2 gt The next Theorem proves this statement for Re2 gt 1 and gives the connection between 2 and the prime numbers The Gamma and Zeta Functions 25 Figure VII1 C6 lg 740 S y S 40 Theorem 43 Let pn be the primes With 191 2 p2 3 For Rez gt 1 1117157 412 In particular 2 31 0 for Rez gt 1 Proof The in nite product in Theorem 43 converges uniformly and absolutely because forRezgtx0gtl mm 1 S if so that by the Weierstrass M test the right side of 412 is analytic in Rez gt 1 Because the series is absolutely convergent we can rearrange the terms and obtain Czl 7 2 Z Zn z 7 22n z Z 771 m odd Similarly Cz1i 217 3 7 Z miz 7 Z 3mz 7 Z miz m odd m odd m not divisible b 2 07 3 By induction k C2H1ipif Z m zlp1 l l H n m not div by p17pk 26 VII Series and Products Since pk1 a 00 and since 2 in zi converges the product converges and cltzgtHlt1epZgt1 H 3 for Rez gt 1 D As one might expect from 410 there is a connection between N2 and By the same substitution used to establish 410 Hz nz szile mds 0 and 00 00 Zn zmz szile mds 711 711 0 sZ 1 67 ds 0 176 5 Thus oo szil 2 1 0 ds 413 for Rez gt 1 Contour Integration and Residues 27 VIII Calculus of Residues 1 Contour Integration and Residues In this chapter we will learn one of the main applications of complex analysis to engineering problems but well concentrate on the math not the engineering Cauchy7s Theorem says that if f is analytic in a region Q and if y is a closed curve in 9 which is homologous to 0 then f7 fzdz 0 What happens if f has an isolated singularity at a E 9 but otherwise is analytic Expanding f in its Laurent series about a we have 27a dz new n71 So if A is a disc centered at a and contained in Q then by the Fundamental Theorem of Calculus fzdz b1 dz 27mm 11 8A 8A 2 i 1 provided 8A is oriented in the positive or counterclockwise direction De nition Iff is analytic in 0 lt 27 ai lt 6 for some 6 gt 0 then the residue of f at a written Resaf is the coef cient of z 7 a 1 in the Laurent expansion off about 2 1 Theorem 11 Residue Theorem Suppose f is analytic in 9 except for isolated singularities at 11 an If y is a closed curve in Q with y N 0 then fzdz 27rd Z n y a Resak f 397 n 28 VIII Calculus of Residues Usually the Residue Theorem will be applied to curves y such that n y 1k 0 or 1 so that the sum on the right is 27rd times the sum of the residues of f at points enclosed by 7 Proof Let Ak be a disc centered at 1k k 12 n such that Km Zk Q if m 31 k Orient BAk in the positive or counterclockwise direction Then 7 ZDW akWAk N 0 k in the region Q 11 an By Cauchy7s Theorem if le inm aw Ah fem 0 k1 Then Theorem 11 follows from 11 D Here are some examples illustrating several useful techniques for computing residues CSZ 1 g W has a simple pole at z 2 and hence 6 Resgf llH12Z 7 2fz 672 The residue at z 4 can be calculated similarly CSZ 2 92 We expand egz in a series expansion about 2 2 Z 6663z72 66 0 37 66 366 7 7 2 n 7 7 92 2722 2722gmiz 2722272 so that R8829 366 In this case limz 7 2292 is not the coef cient of z 7 2V1 and limz 7 2gz is infinite Of course the full series for egz was not necessary to compute the residue A g V Contour Integration and Residues 29 We can find the appropriate coef cient in the series expansion for CSZ by computing the derivative of egz More generally if Cz is analytic at z a then Cz 2 7 a i Gn71a R a 7 es n 71 Another trick that can be used With simple poles When pole is not already written as a factor of the denominator is illustrated by the example CazZ4 1 Then h has simple poles at the fourth roots of 71 If w4 71 then z 7 w 6 ea Reswh lim 4 Zoo 24 1 139 z 1 lmzgtw Z Note that the denominator is the limit of difference quotients for the derivative of 24 l at z w and hence aw aw 6 we Caz 241 473 4 Resw Another method using series is illustrated by the example 7rcot 7rz M2 72 2 To compute the residue of k at z 0 note that cot 7rz has a simple pole at z 0 and hence k has a pole of order 3 so that b2 b1 7rcot7rz b4 77727b0m z z 22 23 Then 7rcos7rz b4 by b1z2 bog3 z lnserting the series expansions for cos and sin we obtain 2 3 7r 7r 7rli 22 7T7 322b3b2zb122 Equating coef cients 3 7T igbii Wbila 30 V111 Calculus of Residues and Resok 7 Examples a If y is the circle centered at 0 With radius 3 then CSZ d 2 66 z 7 m7 7272274 2 by the Residue Theorem and Example 1 above dac b 7 ltgtlw 1 We construct the contour y consisting of the interval 71 R followed by the semi circle in H of radius R With R gt 1 OR Figure VIII1 Half disc contour By the Residue Theorem R fzdz fzdz 27riReszlf ResZQf 12 7R CR Where 21 and 22 are the roots of 24 1 0 in the upper half plane llll Note that lt 2 Rda 0 7 0 R4 i 1 as R a 00 Since the integral fRac4 1 1dac is absolutely convergent it equals limRnoo ffRx4 1 1dac so that by 12 and Example 3 above 1 d7727r2 277r mx41 4 1 27 fzdz CR Contour Integration and Residues 31 The technique in Example b can be used to compute the integral of any rational function with no poles on R if the degree of the denominator is at least 2 plus the degree of the numerator This latter condition is needed for the absolute convergence of the integral 27r 1 fd 0 3 sin 6 Set 2 629 Then 27r d 2d 0 3sm6l Z 1226w7 l The roots of 22 612 7 1 occur at 2122 i73 i Only i73 lies inside 1 dz lzl13 2 712 i2 2 1 By the Residue Theorem and method for computing residues in Example 1 or Example 3 27139 27r l 2 ids 2 R 7 0 3 sin l m eSZ 3 22 6n 7 1 The technique in Example c can be used to compute 27r Rcos 6 sin 6d6 0 where Rcos 6 sin 6 is a rational function of sin l and cos 6 with no poles on the unit circle An integral on the circle as in Example c can be converted to an integral on the line using the Cayley transform 2 7 w of the upper half plane onto the disc It is interesting to note that you obtain the substitution at tang which you might have learned in calculus cosx d 00x21 A first guess might be to write cosz 6 e iZ2 but if y lmz then cosz elyllle for large This won7t allow us to find a closed contour where the part off the real line has small contribution to the integral Instead we use elf22 1 then 32 V V111 Calculus of Residues take real parts of the resulting integral Using the same contour as in Example b we have the estimate 6 6 9 7TB 7d lt 7 d lt 7 0 CR221Z CRR271 ZquotR271T as R a 00 where y lmz gt 0 By the Residue Theorem and the method in Example 3 ix 239 6 ooxhrl In this particular case we did not have to take real parts because the integral itself 62quot 2i 6 e dab 27m 1 Ego fiesta271 27m is real since sin 2 1 is odd The technique in Example d can be used to compute 00 f6igtmd 13 for A gt 0 provided f is meromorphic in the closed upper half plane H U R with no poles on R and S Klzll5 for some 8 gt 0 and all large with lmz gt 0 1f the integral 13 is desired for all real A then for negative A use a contour in the lower half plane provided is meromorphic and S Vizll in lmz lt 0 As the reader can deduce if f is meromorphic in C satisfies this inequality for all large then f is rational and the degree of the denominator is at least 2 plus the degree of the numerator The integral in 13 is called the Fourier Transform of f as a function of A f Example e cannot be done by the method in Example d because the integrand x sin Ax 00 x21 does not decay fast enough to prove fCR a 0 where fz 2 tiigt Z2 2 1 Indeed it is not even clear that the integral in Example e exists Letvv1v2vsv4wherev1liAaBlm2Biy20yAB 73zxiABBsziAand y427AiyzABZy20 where y is oriented as indicated in Figure V1112 Contour Integration and Residues 33 13 B m1 B Y4 Y2 7A 71 B Figure VIII2 Rectangle contour To prove convergence of the integral in Example e we will let A and B tend to 00 independently and use the estimate 1 3 71 S when gt 1 For A and B large iAz 226 dz quot3 Z 1 asABaoo Also 26m AB 2 2 1767AAB d lt 7 Md lt7 0 2221Z0 B6 y B A R as B a 00 A similar estimate holds on 74 as A a 00 By the Residue Theorem B 7AAB 2 7AAB 25 lt lt 7714 6 dz AB 0 B iAoc iAz 7A 2 lim zeidac 27rd ResiL w i7re 14 143400 Aac 1 221 21 Indeed by our estimates the integrals over y2 V9 and 74 tend to 0 as A and B tend to 00 so that the limit on the left side of 14 exists and 14 holds Example e follows from Proposition 12 by taking the imaginary parts The technique in Example e can be used to compute 13 with the weaker as sumption S for large sinac f 00 de The main difference between Example e and Example f is that the function fz e 2 has a simple pole on R The function sin acac is integrable near 0 since sin acac a 1 as x a 0 but fac is not integrable However the imaginary part of f is odd so that 75 1 eim lim idx 540 1 5 x exists 34 VIII Calculus of Residues De nition If f is continuous on y a Where y is a smooth curve y is continuous then the Cauchy Principal Value of the integral of f along y is de ned by PVfzdz lim fzdz w 50 W zealz provided the limit exists Note that we have deleted points in a ball centered at a If the limit exists for all balls containing a then the usual integral of f exists For example 1 cosx PV 7dr 0 1 because the integrand is odd but the integral itself does not exist Proposition 12 Suppose f is meromorphic in lmz 2 0 such that K WM 3 7 M When lmz 2 0 and gt R lfA gt 0 then PV 00 fxe mdx 27rd I Ego Resaemz z 27rd I 0 Resaemzfz 15 Part of the conclusion of Proposition 12 is that the integral exists Proof Note that f has at most finitely many poles in lmz 2 0 because S so that both sums in the statement of Proposition 12 are finite Construct a contour similar to the rectangle in Figure Vlll2 but avoiding the poles on R using small semicircles Oj of radius 6 gt 0 centered at each pole aj E R See Figure VIII3 73 BiAB Y4 0139 n 7 A a Y 1 B Figure VIII3 Contour avoiding poles on R Contour Integration and Residues 35 The integral of f9 along the top and sides of the contour tend to 0 as A B a 00 as in the previous example The semi circle Oj centered at aj can be parameterized by z aj 669 7r gt 6 gt 0 so that if M i W Z 7 lj where g is analytic in an neighborhood of aj then 0 b I fzdz iewd gzd2 Cj 7r 552 Cj Because gj is continuous at aj and the length of Oj is 6 we have lim fzdz imbj 0339 5 0 By the Residue Theorem PV fzeigt zdz 7 i7r Z bj 27rd Z Resafzeigt z 1R g lmagt0 and 15 holds B One way to remember the conclusion of Proposition 12 is to think of the real line as cutting the pole at each aj in half The integral contributes half of the residue of f at aj In our example PV dac m 16 00 x 00 smac dac 7139 00 at Note that sin acx a l as x a 0 so that fsinacx exists as an ordinary Riemann integral so that by taking imaginary parts For this reason we can drop the PV in front of the integral 0 g Mellin Transforms0 mdx where 0 lt 04 lt 1 Here we define 20 calogz in C 0 00 where 0 lt argz lt 27139 and set fz 122 1 Part of the dif culty here is constructing a closed contour that will give 36 VIII Calculus of Residues the desired integral Consider the keyhole contour y consisting of a portion of a large circle OR of radius R and a portion of a small circle 05 of radius 6 both circles centered at 0 along With two line segments between 05 and OR at heights is oriented as indicated in Figure V1114 Figure VIII4 Keyhole contour By the Residue Theorem for R large and 6 small zo fzdz 27riResZzafz Resizafz Calogi Calogii Ma Is a 2 T 7 1T 17 mlt 2 72 7re e We Will first let 5 a 0 then R a 00 and 6 a 0 Even though the integrals along the horizontal lines are in opposite directions they do not cancel as 8 a 0 For 8 gt 0 11135 ie0 fx is 6mg WWW lin1ac 7 i6o fx 7 is eaaog lml27rigtfltgta because of our definition of log 2 Thus the integral over the horizontal line segments tends to R A 17 62quotaafxd 18 For B large 27r Ra Zafzdz S R2 7 1Rd a 0 19 CR 0 as R a 00 Similarly 2 Zafzdz S 1i625d9 a 0 110 C 0 Contour Integration and Residues 37 as 6 a 0 By 17 18 19 and 110 2397ra 37ra 0 I e 2 7 62 2 27139 M 7r 0 2 1 17 627ml cos our2 This line of reasoning works for meromorphic f satisfying S 0121 2 for large and with at worst a simple pole at 0 The function 20 can be replaced by other functions which are not continuous across R such as log 2 In this case real parts of the integrals along 0 00 will cancel but the imaginary parts will not Mellin transforms are used in applications to signal processing image filtering stress analysis and other areas h Inverse Laplace Transforms The Laplace Transform of a function F defined on 0 00 is given by MW macaw 0 As we saw last quarter the Laplace Transform can be used to solve ordinary differential equations The Laplace Transform converts a differential equation into an algebra problem roughly converting differentiation into multiplication The algebra problem is generally easy to solve but to find the solution to the original differential equation we need find the inverse of the Laplace of the solution to the algebra problem In this example we will find an integral equation for the inverse Laplace Transform In other words given a function F Wed like to find f so that f F Theorem 13 Suppose F is analytic in Rez gt a and satis es 0 1F 21 S 1 121 when Rez gt a for some constant C and some 1 gt 1 For b gt 1 de ne 1 0 7 Mugt15 m 2 00Fbtye dy Then the de nition of f does not depend on b lftl S Olebt where 01 is a constant depending only on O and p and 77 7d ltfgtltcgt 2rm Hm0 e e y 27 Re bzic 2 where the integral along Rez b is in the upward direction By the assumptions on the growth of if OR is the serni circle in the right half plane with radius R centered at b then F 2 AH as R a 00 Applying the Residue Theorern to the integral over the closed contour con C ldzl S EWR a 0 sisting of OR together with the portion of Rez b between the endpoints of OR and letting R a 00 we obtain We can turn this game around Instead of using sums to compute integrals we can use integrals to compute surns Set fz and consider the rnerornorphic function fz7r cot 7rz Then 7r cot 7rz has a simple pole at z 0 with residue 1 and since cot 7rz 7 n cot 7rz for any integer n it has a simple pole with residue 1 at each integer Since the poles are simple fz7r cot 7rz has a simple pole with residue n at z n We consider Contour Integration and Residues 39 the contour integral of fz7r cot 7rz around the square SN with vertices N il i The function 7r cot 7rz is uniformly bounded on SN independent of N To see this write 627m39z 1 cot 7rz 1 The linear fractional transformation C lC 71 is bounded in the region K7 1 gt 6 and 62 Z 7 1 gt 1 7 C Z VN on SN as can be seen by considering the horizontal and vertical segments separately Since S Olzl Z we have fz7r cot 7rde 7 0 Sn By the Residue Theorem 0 Resifz7r cot 7rz ResZfz7r cot 7rz Z n and hence 0 1 7r 6quote 277 71 n2l 2 673976 7 710 This techique can be used to compute Z An 71700 provided f is meromorphic with 3 GM for large If some of the poles of f occur at integers then the residue calculation at those poles is slightly more complicated because the pole of fz7r cot 7rz will not have order 1 If only the weaker estimate S Olzl l holds then f has a removable singularity at 00 and so 92 fz f7z satisfies S GM for large Applying the techique to 9 we can find the symmetric limit 40 IX Normal Families IX Normal Families Our next major goal is to prove the Riemann mapping theorem that every simply connected plane domain except the plane itself can be mapped one to one and analyt ically onto the unit disk The original approach to this problem due to Riemann was to minimize a certain integral with side conditions over a collection of functions This method had the advantage of a physical interpretation but it assumed that there was a minimizer Weierstrass then showed that there are similar minimization problems with no minimizer Existence proofs of this and other related problems dominated mathematical thinking in analysis for the next 50 or so years Around 1900 Hilbert finally patched up the dif culties See Courant Dirichlet Principle and J Grey n1 for a discussion of some of the history It is possible to base a proof on constructive methods such as the Zipper algorithm presented last quarter but here we will use a different approach that is rather elegant and which is a useful technique for solving other problems It is based on a notion of convergence of analytic functions called normal convergence 1 Normality and Equicontinuity De nition 11 A collection or family f of continuous functions on a region Q C C is said to be normal on 9 provided every sequence fn C f contains a subsequence which converges uniformly on compact subsets of Q A normal family f is not necessarily countable a limit function is continuous but not necessarily in f and if f is a sequence which is normal it does not necessarily converge For example f is normal on D but does not converge uniformly on D The only limit function is the zero function which is not in f A simple example of a normal sequence which does not converge is f 97 where 927 E l and 927 E 0 The first lemma says that normality is local Normality and Equicontinuity 41 Lemma 12 A family of continuous function f is normal on a region 9 if and only if is normal on each disk A whose closure is contained in 9 Proof If f is normal on 9 it is normal on each disk A C Q by de nition Suppose f is normal on each disk A with A C 9 Write Q U lAj where C 9 Suppose fn C f Then there is a subsequence an C fn such that lm converges uniformly on A1 Likewise there is a sequence f9 C sh which converges uniformly on A2 and indeed there is a sequence aw C gkim which converges uniformly on Ak Then the diagonal sequence 229 converges uniformly on each Aj since it is a subsequence of y for each k Since each compact set K C Q can be covered by nitely many Aj the diagonal sequence converges uniformly on K D The proof of the above lemma shows that if for each compact subset there is a subsequence which converges uniformly then there is a single subsequence which converges uniformly on all compact subsets We can de ne a metric on the space 09 of continuous functions on a region Q as follows Write Q Ug iAg a where each Aj is a disk with C Q If g 6 09 set 33039 Zefj 1 792 7 pltf79gt Z 2Tijltfag 31 Then p is a metric If pfg 0 then f g on each Aj and hence f g on Q Mfg My f for 9 E 09 p089 S p08 h Way for 189 h 6 09 42 IX Normal Families The triangle inequality follows from the observation that if a and b are non negative num bers then a b a b 1 b S m T m Proposition 13 A sequence fn C O converges uniformly on compact subsets of Q to f 6 09 if and only if lignpUmf 0 In other words the space 09 with the topology of uniform convergence on compact subsets is a metric space Proof If p03 f 7 0 then pjfn f 7 0 for each j This implies f7 converges uniformly to f on A for each j Since each compact set K C Q can be convered by finitely many A the sequence f7 converges uniformly on K Conversely if f7 converges to f uniformly on each A then given 8 gt 0 choose 11 and m so that 8 lf z 7 lt i for n 2 n on A forj l23 and 2 3 lt 7 Z 2 Jm Then for m 2 maxn1 nm1 M71 00 5 Pfnf z2 2 7lte D 13971 Fm Normality is related to a strong concept of continuity called equicontinuity De nition 14 A family of functions f is called equicontinuous on a set E if for all 8 gt 0 there exist a 6 gt 0 so that iflz7 wl lt 6 then 7 fwl lt 8 for all f E 7 For example if fM is the family of analytic functions on D such that MW 3 M lt 00 for all z E D then writing f as the integral of its derivative we obtain lfz7fwl S Mlz7wl Thus 7 fwl lt 8 whenever l2 7 ml lt 6 eM Continuity uniform continuity and equicontinuity are all related to the statement Normality and Equicontinuity 43 l2 7 ml lt 6 implies 7 fwl lt 8 For continuity 6 is allowed to depend on 8 the function f and the point w For uniform continuity 6 depends on 8 and the function f but works for all w For equicontinuity 6 depends on 8 but works for all w and for all f E f Theorem 15 Arzela Ascoli A family f ofcontinuous functions is normal on a region Q C C if and only if i f is equicontinuous on every compact subset of Q and ii there is a 20 E 9 so that the collection f20 f E f is a bounded subset ofC Proof Suppose f is normal If f is not equicontinuous on a compact set K C Q then there is an 8 gt 0 and 2mm 6 K and f7 6 f such that 27 7 wnl lt 111 and C Z 8 By normality we may suppose f7 7 f uniformly on K Thus For suf ciently large n the first and third terms on the right side of 11 are less than 53 by uniform convergence on K and the second term is less than 53 because the limit function f is uniformly continuous on K This contradiction shows that f is equicontinuous on K If 20 E Q then S f20 f E f is a bounded set by the normality of f arguing by contradiction Conversely suppose and hold We first show that holds at each 2 E 9 By equicontinuity each w E Q is contained in a disk Aw C 9 such that lfz 7 fwl lt l for all z E Aw and all f E f 12 If 21 E Q then we can find an arc y connecting 20 to 21 Cover y by finitely many Aw Since y is connected we can order these disks so that Aw intersects j71 Uk0Awk 44 IX Normal Families for j l n and where wo 20 and 111 21 By 12 and f21 is a bounded set Now suppose fn C 7 Let be a countable dense subset of 9 Since f21 f E f is a bounded set we can find a subsequence an such that f7121 converges Likewise we can find a sequence g C gm such that f7222 converges and indeed there is a sequence 0 C Akin such that converges Then as in the proof of Lemma 12 the sequence M converges on 27 for each k We may relabel this sequence as By i if 8 gt 0 then we can find disks A7 8 centered at 2 such that W2 f2wl lt 83 13 for all f E f and all z E Aw Moreover we can find a n so that lfMZj fp2jl lt 83 14 whenever 7 2 11 forj 12 lf K is compact then we can cover K by finitely many disks A j 1M Choosing N maxn1n2nM we have that for z E A j 1M andmpgtN we 7 mm s we 7 mm we 7 WM w 7 new 15 The first and third term on the right side of 15 are at most 53 by 13 and the middle term is at most 53 by 14 Thus for 7 gt N we 7 w lt e on K Since 8 gt 0 is arbitrary this proves that fn is a Cauchy sequence in the uniform topology on K Thus fn converges uniformly on compact subsets of Q and hence f is normal D De nition 16 A family f of continuous functions is said to be locally bounded on 9 if there exist disks A C C Q UA and M lt 00 j 12 such that S M for allz E A and aHf E f Normality and Equicontinuity 45 An equivalent de nition is that f is locally bounded on 9 if and only if for each compact K C Q the set f E fz E K is bounded Theorem 17 A family f of analytic functions on a region Q is normal if and only if f is locally bounded on 9 Proof Suppose f is normal In the course of the proof of the Arzela Ascoli Theorem we showed that holds for each 2 E Q by covering a curve with a finite number of disks AW Similarly if K is any closed disk contained in 9 we can cover K with finitely many such disks Thus the same argument shows that f E f z E K is bounded and so f is locally bounded Conversely if f is locally bounded on Q and if A C K C Q is a disk then by Cauchy7s g 1 odc BAC Z Theorem If B C A is a smaller disk then for 2111 6 B i m2 mm M fC C M 7 w 27 where O Mlength8A27rdist8AB2 So if 2 7 ml lt 80 then 7 fwl lt 8 S Olziwl for all f E f We can cover any compact subset of Q by finitely many of these smaller disks and so it follows that f is equicontinuous on compact subsets of Q and so by the Arzela Ascoli Theorem f is normal D In the discussion above we used the Euclidean distance lai l to measure the distance between 04 fz and We could similarly consider families of continuous functions with values in a complete metric space For example we could consider continuous functions with values in the Riemann sphere using the chordal distance between any two points on the sphere In this case we must allow the north pole 0 0 l as a possible value of a function Equivalently we can consider functions with values in the extended plane 3 C U 00 with the metric 2la7 l Oz 6 C 1a2132 2 7 if oo xl al XWB 46 IX Normal Families Then to say that a function f is continuous at 20 With f20 00 means that for all R lt 00 there is a 6 gt 0 so that gt R for all 2 With l2 720 lt 6 The Euclidean metric is usually used for families of analytic functions because by Weierstrass7s Theorem if fn is a sequence of analytic functions on a region Q converging to f uniformly on compact subsets of Q then the limit function f is analytic The chordal metric is used for families of meromorphic functions Proposition 18 If fn is a sequence of meromorphic functions Which converges uni formly on compact subsets of a region Q in the chordal metric then the limit function is either meromorphic or identically equal to 00 If fn is a sequence of analytic functions Which converges uniformly on compact subsets of Q in the chordal metric then the limit function is either analytic or identically equal to 00 Proof Note that 1 1 WW M3 3 16 If fn is a sequence of meromorphic functions Which converges uniformly on compact subsets of Q in the X metric then the limit function f is continuous as a map into the extended plane Cquot If f20 00 then lfn is bounded in a neighborhood of 20 for n suf ciently large and hence analytic By Weierstrass7s Theorem lf is analytic in a neighborhood of 20 and hence f is meromorphic in a neighborhood of 20 or lf is identically 0 in a neighborhood of 20 Since 9 is connected either lf E 0 or f is meromorphic in Q If the sequence fn consists of analytic functions and if the limit function f satisfies f20 00 then applying HurWitZ7s Theorem to lfn in a neighborhood of 20 we conclude lf E 0 in this neighborhood Since 9 is connected lf E 0 in 9 Thus if f is not identically equal to 00 then f is analytic D m is a normal family in C in the X metric With limit function 12 For example by 16 One of the exercises is to prove that a family f of analytic functions is normal on a region 9 if and only if the family is bounded at one point of Q and the family of derivatives Riemann Mapping Theorem 47 is normal To make a similar statement about meromorphic functions we need to use the spherical derivative Theorem 19 Marty A family of meromorphic functions is normal in the chordal metric if and only if the expressions 2W2 f 2 1 WW are locally bounded The quantity fff is the limit 7 we 1 and is sometimes called the spherical derivative Proof Suppose f is normal in the chordal metric on a region Q and suppose f7 is a sequence in f which converges to f uniformly on compact subsets of Q If f is analytic at 20 then f7 converges uniformly to f in a neighborhood of 20 in the Euclidean metric By Weierstrass7s theorem converges to f uniformly in a neighborhood of 20 and hence ff converges to fff uniformly in a neighborhood of 20 If lf is analytic at 20 then may converges to lfy in a neighborhood of 20 and hence ff lfn converges to lf fff It follows that ff cannot be unbounded in a neighborhood of 20 and so the spherical derivatives are locally bounded Conversely if the spherical derivatives are locally bounded then the family f is equicontinuous we 1 igf fCldCl MXZw where the infimum is taken over all curves y C 9 connecting z to w This can be seen by considering Riemann sums for the integral using the remark just after the statement of Marty7s Theorem Condition in the Arzela Ascoli Theorem always holds for the chordal metric because the chordal distance is always bounded by 2 Thus by the Arzela Ascoli Theorem f is normal D 48 IX Normal Families 2 Riemann Mapping Theorem Theorem 21 Riemann Mapping Theorem Suppose Q C C is simply connected and Q 31 C Then there exist a one to one analytic map f on onto D z lt 1 If 20 E Q then there is a unique such map With f20 0 and fzo gt 0 Proof Fix 20 E Q The idea of the proof is to show that there is conformal map of 9 into D vanishing at 20 With largest possible derivative at 20 This forces the image region to be as large as possible and hence all of D First we prove that there is a one to one map into D By assumption there is a point 21 E C 9 Since 9 is simply connected we can define logz 7 21 so as to be analytic on 9 Set f1z xz 7 21 6 10gz zl lf f1 f1w then by squaring 2721 111721 and hence z w so that fl is one to one Note that fz covers a neighborhood of f20 and hence fz omits a neighborhood of 7f20 since fz and 7fz have the same square Thus 0 f2ltzgt f12 f12 0 is analytic and one to one on Q and bounded by l for O suf ciently small Set f f f is one to one analytic and lt l on Q With f20 0 The function f 2 i f 2 1 flt20f2 is the composition of f With a linear fractional transformation and hence one to one With 920 0 Thus f is non empty By Theorem 17 f is normal Let fn C f such that W113 mew M supmzo f e r 21 Zalcman Montel and Picard 49 Since f is normal we may suppose that fn converges uniformly on compact subsets By Weierstrass7s Theorem the limit function f is analytic Moreover g is also normal and lf20l M In particular f is not constant By Hurwitz7s Theorem see Corollary Vll25 f is one to one Finally we show that f must map 9 onto D Suppose not Then there is a point 0 E D such that f 31 0 on Q But then M 912 7 were is a non vanishing function on the simply connected domain 9 and hence it has an analytic square root Set 922 912 and Then 92 6 f To compute g20l first observe that 1 27a ililalZ dz 1762 Tili le39 By the chain rule the derivative of a composition is the product of the derivatives at the appropriate points so that 1lC0l2 1 1l9220l2 l1 C3f20l2 2MM 1 l9220l22 1ofzo lgZo lfZo 1lC0l 2VlC0l Thus lg20l gt lf20l M contradicting 21 and hence f maps 9 onto D lfZo Replacing the map f by Af where W l we may suppose that fzo gt 0 If fl is another such map then f1 o f 1 is a one to one analytic map of D onto D with positive derivative at 0 and so by the maximum principle or equality in Schwarz7s lemma flof 1zzand so f1f D 3 Zalcman Montel and Picard 50 IX Normal Families In this section we Will prove Zalcman7s clever lemma characterizing non normal fam ilies and use it to prove far reaching extensions of Liouville7s Theorem and Riemann7s Theorem on removable singularities Theorem 31 Zalcman s Lemma A family f of meromorphic functions on a region Q is not normal in the chordal metric if and only if there exists a sequence converging to 200 E Q a sequence of positive numbers p7 converging to 0 and a sequence fn C f such that gnC fn2n MC 31 converges uniformly on compact subsets of C to a nonconstant meromorphic function g With the property that We 3 W0 1 32 for all C E C For example the family f is not normal on 2D z lt 2 The lack of normality comes from the lack of equicontinuity in a neighborhood of any point on the unit circle So set 27 l and p7 ln Then for C E C flt2npncgt 1 g a e4 as n a 00 as can be seen by taking logarithms It is easy to check that the spherical derivative of 6C satisfies 32 and therefore is not constant the main point of the Theorem Proof One direction of Zalcman7s Lemma is easy If f is normal then any sequence fn contains a convergent subsequence Which we can relabel as fn and call the limit function f If 2 a 200 E Q and p7 a 0 then for each C E C as n a 00 since the family fn is equicontinuous Thus the limit of 97 is constant Zalcman Montel and Picard 51 Conversely suppose f is not normal By Marty7s Theorem there exists 2 a 200 E Q and fn E f such that a 00 Without loss of generality we may suppose zoo 0 and S 7 C 9 Then Mn mam lzl z 7 lwnl wn lzl r for some lwnl lt 7 since ff is continuous Note that Mn a 00 since 2 a 0 Then fwltwn is de ned on 3 Mn since lwnCfwnl S lwnlMnffwn lwnHrilwnl 7 Fix C E C Then for Mn gt Kl gm mmm 9 0 fit W Mn 7 7 lwnl wmm m m r lwnl lt 7 rilwnlilClf w 1 1 lewdM as n a 00 Since the spherical derivatives of 9 are locally bounded the family 9 contains a convergent subsequence by Marty7s Theorem Relabeling the subsequence 31 holds With p71 lffwn The limit function 9 satisfies 32 and is meromorphic by Proposition 18 Since g0 l g is nonconstant D The first consequence of Zalcman7s Lemma is an improvement of the locally bounded condition for normality 52 1X Normal Families Theorem 32 Montel A family f of meromorphic functions on a region 9 that omits three xed values a b c is normal in the chordal metric Proof Normality is local so we may assume 9 D Composing with a linear fractional transformation we may suppose a 0 b 1 and c 00 Without loss of generality we may assume that f is the family of all analytic functions on D which omit the values 0 and 1 Set fm f analytic on D f 31 0 and f 31 62Wik27mk 1 2m Then ff0f13f23 If f 6 fm then f is analytic and f 31 0 so that we can de ne f so as to be analytic Moreover f E fm1 If f is not normal then there exists a sequence fn C f with no convergent subsequence Moreover f is then a sequence in 71 with no convergent subsequence By induction each fm is not normal Thus for each m we can construct a limit function hm as in Zalcman7s lemma The functions hm are entire by Proposition 18 and nonconstant since 1 By Marty7s Theorem hm is a normal family If h is a limit of a subsequence uniformly on compact subsets of C then h is entire by Proposition 18 and nonconstant since h0 1 By Hurwitz7s Theorem h omits the 2m roots of 1 for each m These points are dense in the unit circle Since MC is connected and open either MC C D or MC C C D Thus either h or 1h is bounded By Liouville7s Theorem h must be constant which is a contraction D We remark that the families fm in the proof of Montel7s Theorem are not closed Indeed constant functions are in the closure For example 2 z omits 0 and 1 in D and tends to 2 in D as n a 00 However Zalcman7s lemma yields limit functions which are not constant and therefore by Hurwitz7s Theorem in the family fm The next consequence of Zalcman7s Lemma can be viewed an improved version of Riemann7s Theorem on removable singularities Zalcman Montel and Picard 53 Theorem 33 Picard s Great Theorem If f is meromorphic in Q z 0 lt 127201 lt 6 and if f omits three distinct values in C then f extends to be meromorphic in Q The reader can verify that an equivalent formulation of Picard7s great theorem is that an analytic function omits at most one complex number in every neighborhood of an essential singularity The function fz elZ does omit the value 0 in every neighbhood of the essential singularity 0 so that Picard7s theorem is the strongest possible statement in terms of the range of an analytic function The weaker statement that an entire function can omit at most one complex number is usually called Picard7s Little Theorem and can be viewed as an extension of Liouville7s Theorem Proof As before we may assume a 0 b 1 c 00 and 20 0 Let 6 a 0 If f omits 0 and 1 in a neighborhood of 0 then by Montel7s Theorem the family end is normal on compact subsets of C Relabelling a subsequence we may suppose enz converges uniformly on compact subsets of C0 to a function g analytic on C 0 or to g E 00 lfg is analytic then S M lt 00 on 1 and hence l S M1 on 6 for n 2 no But then by the maximum principle l S M 1 on 5 lt lt 6 for n 2 no Thus l S M 1 on 0 lt lt 670 so that by Riemann7s Theorem on removable singularities f extends to be analytic in a neighborhood of 0 If g E 00 we can apply a similar argument to 1fenz to conclude that 1f extends to be analytic at 0 and hence f is meromorphic in a neighborhood of 0 D Picard7s Little Theorem follows from exactly the same proof by letting 5 a 00 using Liouville7s Theorem instead of Riemann7s Theorem to conclude that f is constant Zalcman7s Lemma has been used in many situations with the heuristic principle that if 73 is a property of meromorphic functions then f meromorphic on Q f has 73 is normal on 9 if and only if no meromorphic function on C has 73 For example 73 could be omits three values 54 IX Normal Families We end this chapter with a comment about normal families and some open problems Normal families can be used to prove results like see the exercises Koebe s Theorem There is a K gt 0 so that iff is analytic and one to one on D with f0 0 and fO 1 then fD D 2 lt Landau s Theorem There is a constant L gt 0 so that iff is analytic on D with fO 1 then fD contains a disk of radius L Bloch s Theorem There is a constant B gt 0 so that iff is analytic on D with fO 1 then there is a region Q C D such that f is one to one and analytic on Q and f 2 is a disk of radius B In each theorem we already know that there is a constant for each f Normal families are use to prove that the constants can be taken to be independent of the functions f but this method of proof does not give any information about what the constants are The largest K is known to be 14 The conformal map of D onto C 14 00 is an example where 14 is the best possible The largest L is called Landau7s constant and it is known that 5 ltL L0 M 0544 1 We It is conjectured that Landau7s constant is L0 The largest B is called Bloch7s constant and it is known that i 1 PG 7 0433 lt B S BO and it is conjectured that B B0


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