1 Space of Analytic Functions
The space of continuous functions C(G, Ω) Definition : If G is an open set in C and (Ω, d) is a complete metric space the set of all continuous functions from G to Ω, designated by C(G, Ω) is called space of continuous functions. Compact set of a metric space Definition : A subset K of a metric space X is compact if for every collection
ς = {G : G is open;G ⊂ X} of open sets in X, ....(1)
i.e. there is a finite number of sets G1,G2 ,...,Gn in ς such that
Cover of a compact set ςK ⊂ UG{GG: G ∈....ς} G {}xn 1 U 2 U U n A collection of set ς satisfying (1) is called cover of the compact set K. If each member of is an open set then is called open cover of K. e.g. empty set and all finite sets are compact.
Cauchy sequence : A sequence is called a Cauchy sequence if
for every ε>0 there is an integer N such that d (xn , xm ) < ε, ∀n,m ≥ N
Complete metric space : A metric space (X,d) is called complete metric space if each Cauchy sequence has a limit in X.
Let G ⊂ C, G is an open subset of C and H(G) the set of all analytic functioins defined on G, i.e. H(G) = {f : f is analytic function on G} then H(G) be a subset of space of continuous functons from G to C. 2
H (G) ⊂ C(G,C) We denotes the set of analytic function on G by H(G) rather than A(G) because g continuous function that are analytic in G}
Thus A(G) ≠ H (G)
Theorem: If be a sequence in H(G) and f ∈C(G,C) such that
(k ) (k ) then f is analytic and fn → f ∀ integer k ≥1 Proof : To prove that f is analytic, we use Morera’s theorem. If T be a triangle contained inside a disk D ⊂ G, then T is compact, and the sequence { fn } converges to f uniformly over T. Hence by Morera’s theorem
∫ f = lim ∫ f = 0 ...(1) T T
since each f n is analytic.
(k ) (k ) − {AfTnf((Gk→)R})→=f {f g :(k, )G ⊂ Ck;! | f (w) − f (w) | a fn n (z) − f (z) ≤ n | dw| r n D ∫ k+1 2π γ | w− z | γ G
Thus f must be analytic in every disk D ⊂ G . Now we show that
Let D = B(a; r) ⊂ G; then there is a number R > r such that B(a, R) ⊂G.
If we take a circle γ ≡| z − a |= R then by Cauchy’s Integral formula
k! ⎡ f (w) f (w) ⎤ f (k) (z) − f (k) (z) = n − dw n ∫ ⎢ k+1 k+1 ⎥ ∀ z ∈ D 2πi γ ⎣(w− z) (w− z) ⎦
⇒ ...(2)
Since fn → f and f n are continuous in C then ∃ M n > 0 where
M n = sup{| fn (w) − f (w) |: | w − a |= R} such that 3
| fn (w) − f (w) |≤ M n , then from the equation (2) k! M f (k) (z) − f (k) (z) ≤ n | dw| ∴ n ∫ k+1 2π γ (R − r) k!M k!M = n | dw| = n .2πR k+1 ∫ k+1 2π(R − r) γ 2π(R − r)
k!M n R = k+1 for |z-a| Since fn → f , and limMn =0 then (3) gives (k ) (k ) ⇒ f n → f uniformly on B(a;r). Now if K is an arbitrary compact subset of G and distance of each element of K from any of the boundary point of G is greater than r, i.e. 0 < r < d(K,∂G) then in K such that n K ⊂ B(a ;r) j K Uj=1 ∂G (k ) (k ) {δ|∃nfz(≥=:−az|1N)infza,≠a−|=20{|a,....,R|f=(Rza)}n|:| z − a |= R} > 0 Since f n → f uniformly on each B(a j ;r) (k) (k) ⇒ f n converges uniformly to f on K Theorem : Hurwitz’s Theorem Let G be a region and suppose the sequence {fn} converges to f in H(G). If f ≡/ 0, B(a; R) ⊂ G and f (z) ≠ 0 for | z − a |= R then there is an integer N such that for n ≥ N , f and fn have the same number of zeros in B(a;R). Proof : Since for | z − a |= R , therefore we can define a positive number δ as But fn converges uniformly to f on . Therefore integer N such that if and then δ | f (z) − f (z) |< <| f (z) | n 2 ≤| f (z) | + | f n (z) | 4 Thus from above it follows that f and fn satisfies the condition of Rouche’s theorem. Thus f and fn have the same number of zeros in B(a;R). Theorem: Let be a sequence in M(G) and suppose f n → f in C(G,C∞ ) then either f is meromorphic or f ≡ ∞ . If each f n is analytic then either f is analytic or f ≡ ∞ Proof : Suppose there is a point ‘a’ in G with f (a) ≠ ∞ and let | f (a) |= M a fa() G C∞ Now therefore we can find a number ρ > 0 such that ....(1) {|Bfzf(−,a((})ffza∈)(,|a<||Cf−);≤r,....}|ρ| f)f(⊂a()zB)| (−≤f |(faf();a()Mz|)+)−| ff ((a))|| But f n → f so there is an integer n| 0f∞n suchnn(z)1| ≤ that2|∞fn(zn)− f (a)|+|nf(a)|≤2M ∀ z∈B(a; r) and n≥n0 ρ d( f (a); f (a)) < , ∀n ≥ n n 2 0 Also the set is compact inC(G,C∞ ) ⇒ it is equicontinuous i.e. ∃ an r > 0 such that ρ ⇒ d( fn (z), f (z)) < 2 ρ ⇒ d( f (z); f (a)) < n n 2 This gives that d ()f n (z), f (a) ≤ ρ for | z − a |≤ r and for n ≥ n0 Now ⇒ ...(2) In view of the ρ chosen in (1); the expression (2) can be written as .....(3) 5 2 | f ( z ) − f ( z ) | d ()f ( z ) , f ( z ) = n But n 2 2 1 / 2 {}[][]1+ | f n ( z ) | 1+ | f ( z ) | ∀ z ∈ B(a; r) and n ≥ n0 2 | f (z) − f (z) | ≥ n 1/ 2 {}(1+ 4M 2 )(1+ 4M 2 ) ∀ z ∈ B(a; r) and n ≥ n0 Since d{ f n (z), f (z)} → 0 uniformly for z ∈ B(a; r) this gives that | f n (z) − f (z) |→ 0 uniformly for z ∈ B(a; r) From (3) {fn} is bounded on B(a; r) ⇒ fn has no poles and must analytic near z =a, ∀ n ≥ n0 . ⇒ f is analytic in a disc about a...... (4) Now suppose there is a point ‘a’ in G with . Then, if g ∈C(G,C∞ ) ; define 1/g as follows: f (a) = ∞ 1 1 (z) = g g(z) if g(z) ≠ 0 or ∞; 1 (z) = 0 g if g(z) = ∞; and 1 (z) = ∞ g if g(z) = 0 1 Then it follows that ∈C(G,C∞ ) g Also, since f n → f inC(G,C∞ ) then by the property of metric over non zero complex number z1 and z2 , i.e. ⎛ 1 1 ⎞ 1 ⎜ ⎟ ⎛ ⎞ d(z1, z2 ) = d⎜ , ⎟ and d(z,0) = d⎜ ,∞⎟ ⎝ z1 z2 ⎠ ⎝ z ⎠ 1 1 → It follows that in C(G,C∞ ) fn f 1 Now each is meromorphic on G; f n 6 1 1 So, ∃ r > 0 and an integer n0 such that and are analytic on B(a; r) for f f n n ≥ n0 1 1 ⇒ → uniformly on B(a; r). fn f 1 1 From Hurwitz theorem, either ≡ 0 or has isolated zeros in B(a; r). f f 1 ∴ if f ≡/ ∞ then ≡/ 0 and f must be meromorphic in B(a; r) f Combining this with the result (4) that f is meromorphic in G if f is not identically infinite. 1 If each fn is analytic then has no zeros in B(a; r). Then [ corollary : {fn} ⊂ fn H(G) converges to f in H(G) and each fn never vanishes on G then either f ≡ 0 or 1 1 ≡ 0 never vanishes] then either f or f never vanishes. 1 But since f (a) = ∞ we have that has at least one zero; thus f ≡ ∞ in B(a;r) f |sup∀Fz −f{|a∈|f Locally bounded set of analytic functions Definition : A set F ⊂ H (G) is locally bounded if ∀ a ∈G, there are constants M and r >0, such that | f (z) |≤ M, for Alternately F is locally bounded if there is an r >0 such that Montel’s Theorem A family F in H(G) is normal iff is locally bounded Proof : To prove that is normal i.e. each sequence in has a subsequence 7 which converges to a function f in H(G). Necessary : Suppose is normal but fails to be locally bounded. then there is a compact set K ⊂ G such that ....(1) [contradiction of locally bounded set] i.e. there exists a sequence { f n } in F such that Since F is normal therefore a function and a subsequence such that f → f nk sup{| f (z) − f (z)|:z ∈ K} → 0 ⇒ nR as k → ∞ If | f (z) |≤ M for z ∈ K n ≤ sup{| f (z) − f (z)|:z ∈ K} + M k nk sup{|supmHFf ′∈(≥Gz{|(1G))ffn)({z∞) ||}::zz∈∈KK},≥fn∈F} = ∞ {fn } U when nk → ∞ right hand side convergesk to M which cannot be true ( ∞ ≤ M ) Hence the assumption taken at start is wrong. Therefore if is normal it is locally bounded. Corollary : is closed in C(G,C∞ ) . For this we prove the normality of M(G). For this let us introduce the quality 2 | f ′(z) | 2 ....(1) 1+ | f (z) | for each meromorphic function. However if z is a pole of f then has no meaning because derivative increases the order of the pole. Therefore (1) is meaningless. This can be rectify by taking limit of (1) as z approaches the pole. Now we show that the limit of (1) when z tends to pole. Let ‘a’ be a pole of f of order ; then 8 f(z) can be expressed as A A f (z) = g(z) + m + .... + 1 (z − a) m (z − a) for z in some disk about a and g(z) analytic in that disk. For z ≠ a mA A 2 m + .... + 1 − g′(z) 2 | f ′(z) | (z − a)m+1 (z − a) 2 = 2 2 ∴ 1+ | f (z) | A A 1 + m + .... + 1 + g(z) (z − a)m (z − a) 2 | z − a |m−1| mA + ..... + A (z − a) m−1 − g′(z)(z − a) m+1 | = m 1 2m m−1 m 2 | z − a | + | Am + ..... + A1 (z − a) + g(z)(z − a) | Thus if m ≥ 2 2 | f ′(z) | lim = 0 2 z→a 1+ | f (z) | ⎡ mA A ⎤ f ′(z) = g′(z) − m + .... + 1 ⎢ m+1 2 ⎥ If m = 1 then ⎣(z − a) (z − a) ⎦ A 2 1 − g′(z) 2 | f ′(z) | (z − a)2 lim lim 2 = 2 z→a 1+ | f (z) | z→a A 1+ 1 + g(z) (z − a) 1 2 A − (z − a)2 g′(z) | z − a |2 1 = lim z→a 1 []| z − a |2 + | A + g(z)(z − a) |2 | z − a |2 1 2 | A | 2 = 1 = 2 | A1 | | A1 | 2 | f ′(z) | which shows that for the limit of exits. m ≥ 1 1+ | f (z) |2 9 Theorem : Riemann Mapping Theorem Let G be a simply connected region which is not the whole plane and let Then there is a unique analytic function having the properties. f D f(G) G C (a) f(a) = 0 and (b) f is one-one (c) f (G) ={z :| z |<1} Proof : First we prove uniqueness of f, let g be a function having the same properties like f, i.e. g(a) =0 and g'(a) >0, g is one-one and |afc∈′:(|G=aG)1→= C0 g(G) = {z :| z |<1} g :G → D and D ={z :| z |<1} then 1−1 g :G → D onto 1−1 and fog−1:D →D and analytic. onto Also fog −1(0) = f (g −1(0)) = f (a) = 0 Then by the theorem- 1−1 Let f :G → D analytic and f (a) = 0 ∃ a complex number c with | c |= 1 onto z −α such that , ϕα (z) = is Mobious transformation. f =cϕα 1−α z Implies that ∃ a constant c; and fog−1(z) =cz ∀z ⇒ f (z) = cg(z) 10 ⇒ Since and cg′(a) > 0 ⇒ c =1 ⇒ f = g Hence f is unique. For the existence of f, consider the family F of all analytic function f having properties (a) and (b) and satisfying for z in G. The idea is to choose a member of F having property (c). Suppose {Kn} is a sequence of compact subsets of G such that and a ∈ K n ∀ n Then { f (K n )} is sequence of compact subsets of D ={z :| z |<1} . Also, as n becomes larger { f (K n )} becomes larger and larger and tries to fill out the disk. ∞ Choosing f ∈ F with the largest possible|F0gff′′=(<(aaz{) )derivativef>|><∈′01(0aH)(G=) atc: g fa′, (iswea ); onechoose− one, the functionf (a) = 0, f ′(a) > 0, f (G) ⊂ D} Kf (nK=nG) = D nU=1 which “starts out the fastest” at z =a. Thus Lemma : Let G be a region which is not the whole plane and such that every non vanishing function on G has an analytic square root. If a ∈G then there is an analytic function f on G such that : (a) f(a) = 0 and (b) f is one-one (c) f (G) = D ={z :| z |<1} Proof : Let we define F as Since f (G) ⊂ D then 11 sup{| f (z) |:z ∈G} ≤1 for f ∈F By Montel theorem F is normal if it is non-empty. Thus first of all it is to be proved that ....(1) − and F = F ∪{}0 ...(2) Suppose (1) and (2) hold and consider the function f → f ′(a) of H(G) → C This is a continuous function and since F− is compact there is a f in with F . Since (empty set) therefore f ∈F. f(G) = D is remains to shows now. Fwff≠′∉−(zaφ)f) −(≥Gwg)′(a) ∀g ∈ Suppose w ∈ D such that , Fthen the function 1− wf (z) is analytic in G and never vanishes. then by the hypothesis there is an analytic function h : G → C which is equal to the square root of a analytic function, i.e. f (z) − w [h(z)]2 = 1− wf (z) ....(3) Since the Mobious transformation ς − w Tς = maps D onto D, h(G) ⊂ D. 1− wς Define g :G → C by | h′(a) | h(z) − h(a) g(z) = . ....(4) h′(a) 1 − h(a) h(z) Then | g(G) |≤ 1 ⇒ g(G) ⊂ D, 12 which is obvious from (4) and g is one-one, as f and hence h both are one-one. Also | h′(a) | h′(a)[1− | h(a) |2 ] g′(a) = . h′(a) [1− | h(a) |2 ]2 | h′(a) | = 1− | h(a) |2 But f (a) − w 0− w | h(a) |2 = = =| −w|=| w| 1− wf (a) 1−0 and derivative of (3) gives f ′(z)(1− w f (z)) − ( f (z) − w) (−w f ′(z)) 2h(z)h′(z) = []1− w f (z) 2 f ′(a) − w w f ′(a) 2h(a)h′(a) = 1 2h(a)h′(a) = f ′(a)(1− | w|2 ) egfh((az)))=≠=0f0(z) f ′(a)(1− | w |2 ) 1 ∴ g′(a) = . 2 | w | 1− | w | ⎛ (1+ | w|)⎞ ⎡ 1+ | w | ⎤ = f ′(a)⎜ ⎟ > 1 ⎜ ⎟ ⎢Q ⎥ ⎝ 2 | w| ⎠ ⎣⎢ 2 | w | ⎦⎥ > f ′(a) This contradict that g is in F and the choice of f. Thus f(G) = D. Theorem : Let f(z) be analytic in a simply connected region R and suppose that f(z) has no zeros in R. Then there is a analytic function h(z) such that for all z ∈ R . Proof : Since in R f ′(z) then f (z) is also analytic in R and therefore for any two points a and z in R the 13 integral ....(1) defines an analytic function of z in R Let α be an argument of f(a) then f(a) can be expressed as f (a) = r eiα and set β = log| f (a) | +iα eβ = elog|f (a)|+iα = e log| f ( a )| e iα = e log r e iα = r e iα Then eβ = f (a) ...(2) and if we consider z f ′(ς ) h(z) = β + ∫ dς ....(3) a f (ς ) h(a) = β z h(z) We get , then from (2) we haveF(′(zz))=f e(′(fzς)′()−z)F(z) f ′(z) = 0 hI ′=(z) = dς h(a) β ∫ f (z) e = e = f (a) a f (ς ) ...(4) ⇒ h(z) is analytic in R for all z ∈ R from (3) Let F(z) = e h(z) f ′(z) F′(z) = eh(z) h′(z) = F(z) then f (z) F ′(z) f ′(z) = ⇒ F(z) f (z) ⇒ F ′(z) f (z) − F(z) f ′(z) ⇒ = 0 [F(z)]2 Q be non vanishing 14 d ⎛ f (z) ⎞ ⇒ ⎜ ⎟ = 0 dz ⎝ F(z) ⎠ f (z) = κ ⇒ F(z) (a constant ) for all z ∈ R Putting z =a, we obtain ∴ f (z) = F(z) = e h(z) The analytic function h(z) defined as above is called logarithm of f(z) in R and we write h(z) = log f (z) . Clearly, if h(z) is a logarithm of f(z), then (m an integer) is also logarithm of f(z). Convergence of logarithm series : ∞ zn z2 z3 log(1+ z) = (−1)n−1 = z − h(+z) +f (2a−m)...... πi f (a) f (a) ∑ n 2κ = 3 = = =1 n=1 F(a) eh(a) f (a) with radius of convergence 1. If |z| < 1 then log(1 + z) ⎛ z z 2 ⎞ ⎜ ⎟ 1 − = 1 − ⎜1 − + − ...⎟ z ⎝ 2 3 ⎠ z 1 ⎛ 1 1 ⎞ = − z 2 + .... ≤ ⎜ | z | + | z |2 +....⎟ 2 3 ⎝ 2 3 ⎠ 1 ≤ (| z | + | z |2 +.....) 2 1 | z | ≤ ....(1) 2 1− | z | 1 Further if | z |< then 2 log(1+ z) 1 1− ≤ z 2 15 z − log(1+ z) 1 ⇒ ≤ z 2 | z | ⇒ | log(1 + z) | − | z |≤ 2 3 ⇒ | log(1 + z) |≤ | z | 2 1 Since log (1+z) is converges to 1, then for | z |< , we have 2 | z | 3 ≤| log(1+ z) |≤ | z | ....(2) 2 2 Proposition : ∞ ∞ z log z Re zn > 0 ∀n ≥1. Then ∏ n converges to a non-zero number iff.∏ n n=1 n=1 converges. Proof : Let pn = (z1 ⋅z2 ⋅ z3 ⋅.... ⋅ zn ) iθ mkzlim=, nre=p≥nkn=0 z=− πk < θ < π n→m∞ n and ln( pn ) = log | pn | +iθ n , θ − π < θn < θ + π If S n = log z1 + log z2 + ..... + log z n then exp(S n ) = exp[log z1 + log z 2 + ..... + log zn ] = exp[log( z1.z2 ....zn )] = pn ∴ S n = ln( pn ) + 2πik n for some integer kn. Now suppose , then S n − S n−1 = log zn → 0 Also ln( pn ) − ln( pn−1 ) → 0 Hence (k n − k n−1 ) → 0 as n → ∞ Since kn is an integer, there exists integers n0 and k such that for 16 So S n → ln(z) + 2πiK Corollary : If Re z n > 0 then the product. ∏ zn converges absolutely iff the series ∑ (zn −1) converges absolutely. Lemma : Let X be a set and f1, f2, f3 ... be functions from X into C such that fn (x) → f (x) uniformly for x in X. If there is a constant ‘a’ such that Re f (x) ≤ a ∀x∈ X then exp f n (x) → exp f (x) uniformly for x in X. Proof : For given ε > 0 choose such that whenever | z |< δ ....(1) Now choose n0 such that {δng≥>z}0n −a −−aa | exp[expn 0ffnn((xx))−−expf (xf)](−x)1|<|<εεee | exp f (x) |≤ ε | f n (x) − f (x) |< δ ∀x ∈ X |whenevere −1|< εne≥ n0 Thus, from (1) exp f (x) n −1 < εe −a ⇒ exp f (x) −a ⇒ | exp f n (x) − exp f (x) |< εe | exp f (x) | It follow that for any x ∈ X and ⇒ exp f n (x) → exp f (x) uniformly for x ∈ X . Lemma : Let (X, d) be a compact metric space and let be a sequence of continuous function from X into C such that ∑ g n (x) converges absolutely and uniformly for x in 17 X. ∞ Then the product f (x) = ∏ (1 + g n (x)) converges absolutely and uniformly for n=1 x in X. Also there is an integer n0 such that f (x) = 0 iff gn (x) = −1 for some n, 1 ≤ n ≤ n0 Proof : Since ∑ g n (x) converges uniformly for x ∈ X therefor their exists an integer n0 such that and n ≥ n0 3 ⇒ and | log(1+ g (x))|≤ | g (x)|; ∀n≥ n and x∈X n 2 n 0 ∞ Thus h(x) = ∑ log(1+ g n (x)) converges uniformly for x ∈ X n=n0 +1 Now ∴ is also continuous. Hence h(x) is continuous. Q X is compact ⇒ h(x) must be bounded. log(expRe[Regf n(x(h11x)h(+)(=+x∈x)[gg)1 ⎡ ∞ ⎤ exp h(x) = exp⎢ ∑ log(1 + g n (x))⎥ ⎣n=n0 +1 ⎦ ∞ = ∏(1+ gn (x)), converges uniformly for x ∈ X and n=n0+1 for any x ∈ X . Now if we take then ∞ f (x) =∏(1+gn(x)), n=1 18 converges uniformly for x ∈ X . Definition : An Elementary factor is one of the following function or E(z; p) for p = 0, 1, 2, 3,... defined as E(z; 0) or E0 (z) = 1 − z ⎛ z 2 z p ⎞ ⎜ ⎟ E(z; p) or E p (z) = (1 − z) exp⎜ z + + .... + ⎟, p ≥1 ⎝ 2 p ⎠ 2 p z ⎛ z ⎞ ⎛ z ⎞ ⎛ z 1 ⎛ z ⎞ 1 ⎛ z ⎞ ⎞ E ( ; p ) or E ⎜ ⎟ = ⎜1 − ⎟exp⎜ + ⎜ ⎟ + ...... ⎜ ⎟ ⎟, p ≥1 a p ⎜ ⎟ ⎝ a ⎠ ⎝ a ⎠ ⎝ a 2 ⎝ a ⎠ p ⎝ a ⎠ ⎠ ⎛ z ⎞ ⇒ E p ⎜ ⎟ has a simple zero at z =a and no other zero. ⎝ a ⎠ Also if b is a point in C−G then 2 p ⎛ a − b ⎞ ⎛ a − b ⎞ ⎡a − b 1 ⎛ a − b ⎞ 1 ⎛ a − b ⎞ ⎤ E = 1 − exp + + .... + p ⎜ ⎟ ⎜ ⎟ ⎢ E (z) ⎜ ⎟ ⎜ ⎟ ⎥ ⎝ z − b ⎠ ⎝ z − b ⎠ ⎣⎢ z −pb 2 ⎝ z − b ⎠ p ⎝ z − b ⎠ ⎦⎥ has a simple zero at z =a, and is analytic in G. Lemma p+1 If | z |≤1 and p ≥ 0 then |1− E p (z) |≤| z | Proof : We restrict for p ≥1. For a fixed p let the power series expansion of E p (z) about z =0 is ∞ k E p (z) =1 + ∑ ak z ....(1) k =1 differentiating (1), we have ∞ ′ k −1 E p (z) = 1 + ∑ k ak z k =1 From the definition of Ep (z) 19 ⎛ z 2 z p ⎞ ⎛ z 2 z p ⎞ ′ ⎜ ⎟ ⎜ ⎟ 2 p−1 E p (z) = −1exp⎜ z + + .... + ⎟ + (1 − z) exp⎜ z + + .... + ⎟(1 + z + z + .... + z ) ⎝ 2 p ⎠ ⎝ 2 p ⎠ ⎛ z 2 z p ⎞ ⎜ ⎟ 2 p−1 = exp⎜ z + + .... + ⎟[]−1 + (1 − z)(1 + z + z + .....z ) ⎝ 2 p ⎠ On simplifying expressiion in the square bracket only z p remains and all other terms will cancelled out. ∞ ⎛ z 2 z p ⎞ k−1 p ⎜ ⎟ ∴ ∑ k ak z = −z exp⎜ z + + ..... + ⎟ k=1 ⎝ 2 p ⎠ On comparing coefficients of like power of z, we find that a1 = a2 = .... = a p = 0 ⎛ z 2 z p ⎞ ⎜ ⎟ Also the coefficient of the expression of exp⎜ z + + ..... + ⎟ are all positive ⎝ 2 p ⎠ 2 p k ≥ p +1 p ⎛ z z ⎞ , therefore ak ≤ 0 for ′ ⎜ ⎟ Ep = −z exp⎜z + +....+ ⎟ ⎝ 2 p ⎠ Thus | ak |= −ak for k ≥ p +1 ∞ ⇒ E p (1) = 0 = 1 + ∑ a n k = p+1 ∞ ∞ or ∑| ak | = − ∑ ak =1 k = p+1 k = p+1 Hence for | z |≤1 ∞ k | E p (z) −1|= ∑ ak z k= p+1 ∞ p+1 k − p−1 =| z | ∑ ak z k= p+1 ∞ p+1 ≤ | z | ∑ | ak | k=p+1 =| z | p+1 20 Proposition Let Re z n > −1 then the series ∑ log(1 + zn ) converges absolutely iff ∑ zn converges absolutely. Proof : If ∑| zn | converges then zn → 0 1 ⇒ | z |< n 2 1 3 1 Since | z | ≤ | log(1+ z) |≤ | z | for | z |< 2 2 2 1 Therefore for | z |< the series | log(1+ z ) | is convergent. n 2 ∑ n Conversely if, ∑| log(1+ zn ) | converges then 1 | z |< n 2 ∴ ∑| zn | is converges Absolute convergence of an infinite product. ∏ | z n | converges ⇒/ ∏ z n converges ∏ zn Example: Let Z n = −1 ∀ n then | z n |=1 ∀ n ⇒ ∏| zn | converges to 1. n However ∏ zk is ±1 depending on whether n is even or odd, k=1 ⇒ does not converges. Definition : If Re z n > 0 , ∀n then the infinite product ∏ zn is said to converges absolutely if the series ∑ log zn converges absolutely. Theorem : Weirestrass factorization theorem Given an infinite sequence of complex numbers a0 = 0, a1 , a2 ,...., an with no finite point of accumulation, the most general entire function having zeros at those points only (a zero an n≥1 of multiplicity α being repeated α times in the sequence) is given 21 by ∞ ⎛ z ⎞ F(z) = eh(z) z m E⎜ ;k ⎟ ∏ ⎜ n ⎟ ...(1) n=1 ⎝ an ⎠ where h(z) is an arbitrary entire function, m ≥ 0 is the order of multiplicity of a0 =0, and the kn are non negative integers such that the series k +1 ∞ z n ∑ ...(2) n=1 an Converges for each finite value of z. Proof : Arranging the given zeros in increasing order of modulus, so that 0 <| a1 |≤| a2 |≤| a3 |≤ .... with | an |→ ∞ The sequence {k n } of non-negative integers can always be found such that the series (2) is convergent for each z. We may take k n+1 = n , since n z 1 z 1 ≤ n or ≤ ...(3) an 2 an 2 as soon as | an |≥ 2 | z | which for any given z holds for sufficiently large value of n , say n>N ⇒ N depends on |z| ⇒ pointwise convergence. Since we are not interested in uniform convergence, only point wise convergence. From the already proved result that 1 if | z |≤ <1 then p p k+1 logE(z;k) ≤ | z | ...(4) p −1 then for p =2 log E(z; k ) ≤ 2 | z |k +1 provided 22 1 | z |≤ 2 We have y k +1 ⎛ z ⎞ z an log E⎜ ; k ⎟ ≤ 2 ⎝ an ⎠ an provided Z x z 1 O R 2R ≤ an 2 ⇒ | a n |≥ 2 | z | Let R be an arbitrary positive number and consider two circles with centres at the origin and radii R and 2R. If we take |z| < R and n large enough so that , we have| an |> 2 | z | Hence the series | an |> 2R ⎛ z ⎞ is absolutely and uniformly convergent for all z, suchlog E that⎜ |z|; ⎛ z ⎞ E ⎜ ; k ⎟ ∏ ⎜ a n ⎟ |a n |> 2 R ⎝ n ⎠ is also absolutely and uniformly convergent for |z| < R. Since the product ⎛ z ⎞ E ⎜ ; k ⎟ ∏ ⎜ a n ⎟ |a n |≤ 2 R ⎝ n ⎠ contain a finite number of factors each of which is an analytic function, it follows that ⎛ z ⎞ f ( z ) = E ⎜ ; k ⎟ 1 ∏ ⎜ a n ⎟ |a n |≤ 2 R ⎝ n ⎠ is analytic in |z| < R and vanishes in the disc only at those points of the sequence a1, a2, ... which lie in |z| < R. Then from the theorem : If fk (z); k = 1,2,... are analytic in open set 23 ∞ A⊂C, ∑| fk (z)| is uniformly convergent on every compact set K ⊂ A , then n=1 converges absolutely in A to F(z) which is analytic in A. We have ⎛ z ⎞ f ( z) = E ⎜ ; k ⎟ 2 ∏ ⎜ a n ⎟ |a n |> 2 R ⎝ n ⎠ is analytic and different from zero in |z| < R. Hence the product ∞ ⎛ z ⎞ ⎜ ⎟ f (z) = f1(z) f2 (z) = ∏ E⎜ ;kn ⎟ n=1 ⎝ an ⎠ is also analytic in |z| ⇒ It is an entire function and has less zero precisely at the points a1 , a2 ...., an ... ∴ F (z) = e h( z ) z m f (z) is most general entire function with the prescribed zeros. Factorization of sinπz : sin πz is an entire function with simple zeros at 0, ±1, ± 2,... Then by Weierstrass factorization theorem the most general form of this entire function be ∴ ∞ g ( z ) ⎡⎛ z ⎞ − z / n ⎛ z ⎞ z / n ⎤ = e z.∏ ⎢⎜1− ⎟e ⎜1+ ⎟e ⎥ n=1 ⎣⎝ n ⎠ ⎝ n ⎠ ⎦ 24 ∞ ⎛ z 2 ⎞ = e g ( z ) z ⎜1 − ⎟ ∏ ⎜ 2 ⎟ ...(1) n =1 ⎝ n ⎠ Taking logarithmic differentiation of (1) ∞ d ⎡ 2 2 2 ⎤ π cot πz = ⎢g(z) + log z + ∏ ()log(n − z ) − log n ⎥ dz ⎣ n=1 ⎦ 1 ∞ ⎛ 2z ⎞ = h′(z) + + ∏ ⎜ 2 2 ⎟ z ≠ ±n ...(2) z n=1 ⎝ z − n ⎠ But ...(3) Hence h′(z) = 0 and it follows that h(z) = C (a constant) C Let e = C1 then (1) gives ∞ ⎛ z 2 ⎞ sin πz = C z ⎜1 − ⎟ 1 ∏ ⎜ 2 ⎟ n =1 ⎝ n ⎠ sinπz C ∞ ⎛ z2 ⎞ = 1 ⎜1− ⎟; z ≠ 0 ∏⎜ 2 ⎟ πz π n=1 ⎝ n ⎠ 1 ∞ ⎛ 2z ⎞ π cotπz = + Taking z → 0 ∏⎜ 2 2 ⎟ z n=1 ⎝ z − n ⎠ C n ⎛ k 2 ⎞ 1 = 1 lim lim ⎜1− ⎟ z→0 n→∞ ∏ ⎜ 2 ⎟ π k =1 ⎝ n ⎠ ∞ ⎛ z 2 ⎞ ⎜1− ⎟ Since ∏⎜ 2 ⎟ is uniformly converges, therefore the order of limit can be n=1 ⎝ n ⎠ changed C n ⎛ z 2 ⎞ = 1 lim lim ⎜1− ⎟ n→∞ z→0 ∏ ⎜ 2 ⎟ π k =1 ⎝ k ⎠ C n C ∴ 1 = 1 lim ∏ (1) = 1 π n→∞ K =1 π ⇒ C1 = π ∞ ⎛ z 2 ⎞ sin πz = πz ⎜1 − ⎟ ⇒ ∏ ⎜ 2 ⎟ n =1 ⎝ n ⎠ 25 Gamma function According to Weierstrass (the weierstrass fact. th) the Γ function can be defined as the reciprocal of a particular entire function with simple zeros at the points 0, -1, - 2,.. namely ∞ ⎛ z ⎞ F (z) = eγ z z∏ ⎜1+ ⎟e−z / n n=1 ⎝ n ⎠ −1 1 e−γ z ∞ ⎛ z⎞ ⇒ =Γ(z) = ∏⎜1+ ⎟ ez/n ...(1) F(z) z n=1 ⎝ n⎠ where the constant γ (Euler or Mascheroni constant) is chosen so that Γ(1) =1 From (1) we see that is a meromorphic function on C with simple poles at z =0, -1, -2,... Existence of γ Now we show that there exists such γ. Γ∞((1z))=∞1 −1log C−1 ∞−1 −1 γ ⎛ e⎛ z ⎞1 ⎞ ⎛1/ n 1 ⎞ 1/ n eΓ(1=)⎜=∏1+⎜1 +⎟ .⎟e∏z /e⎜n1 + ⎟ e ∏ n=1 ⎝ 1 n ⎠n=1 ⎝ n ⎠ substituting z =1 in the infinite productn= 1 ⎝ n ⎠ , we get −1 ∞ ⎛ 1 ⎞ ∏⎜1 + ⎟ e1/ n = C a finite positive number ...(2) n=1 ⎝ n ⎠ Let γ = logC then on substituting z =1 in (1), we get 1 = .C =1 C ∴ From (2), we see that the constant γ satisfies the equation ...(3) Hence the existance of γ is acertained. 26 Lemma : 1 1 1 Let H =1+ + + ... then γ = lim(H n − log n) n 2 3 n n→∞ Proof : Taking logarithm on both sides of −1 ∞ ⎛ 1 ⎞ e γ = ∏ ⎜1 + ⎟ e1/ k k =1 ⎝ k ⎠ We have −1 ∞ ⎡⎛ 1 ⎞ ⎤ γ = ∑ log⎢⎜1 + ⎟ e1/ k ⎥ k=1 ⎣⎢⎝ k ⎠ ⎦⎥ ∞ ⎡⎛ k ⎞ 1/ k ⎤ = ∑ log⎢⎜ ⎟e ⎥ k=1 ⎣⎝ k + 1⎠ ⎦ n ⎡ 1 ⎤ = lim log k − log(k +1) + n→∞ ∑ ⎢ ⎥ k =1 ⎣ k ⎦ ⎡ ⎤ ⎢ ⎥ n n 1 = lim⎢ {}log k − log(k +1) + ⎥ n!.n z n→∞⎢∑ Γ(z) = lim∑ ⎥ k =1 14244 4344 n→∞ kz=(1z k+1)(z + 2).....( z + n) ⎢ ⎪⎧log1=0 and all int ermediate ⎪⎫ ⎥ ⎨terms will cancelled out except ⎬ ⎣⎢ ⎩⎪log(n+1) ⎭⎪ ⎦⎥ ⎡ ⎛ 1 1 ⎞⎤ = lim ⎢− log(n + 1) + ⎜1 + + ... + ⎟⎥ n→∞ ⎣ ⎝ 2 n ⎠⎦ ⎡ ⎛ 1 1 ⎞ ⎤ = lim ⎢log n − log(n + 1) + ⎜1 + + ... + ⎟ − log n⎥ n→∞ ⎣ ⎝ 2 n ⎠ ⎦ ⎡⎛ 1 1 ⎞ ⎛ (n +1) ⎞⎤ = lim ⎢⎜1+ + ... + ⎟ − log n − log⎜ ⎟⎥ n→∞ ⎣⎝ 2 n ⎠ ⎝ n ⎠⎦ ⎡⎛ 1 1 ⎞ ⎤ ⎡ ⎛ (n +1) ⎞ ⎤ ⇒ γ = lim ⎢⎜1+ + ... + ⎟ − log n − 0⎥ ⎢Q lim log⎜ ⎟ = 0⎥ n→∞ ⎣⎝ 2 n ⎠ ⎦ ⎣ n→∞ ⎝ n ⎠ ⎦ γ = lim(H n − log n) ⇒ n→∞ Euler’s formula for Γ(z) or Gauss formula From the definition of Γ(z) 27 −1 e−γ z n ⎛ z ⎞ = lim ⎜1+ ⎟ ez /k n→∞ ∏ z k=1 ⎝ k ⎠ e−γ z n kez /k = lim n→∞ ∏ z k=1 (z + k) ⎛ z z z ⎞ e−γ z (1.2.3....n).exp⎜ z + + + .... + ⎟ 2 3 n = lim ⎝ ⎠ n→∞ z(z +1)(z + 2).....( z + n) −z(H −logn) ⎧ ⎛ 1 1 1 ⎞⎫ e n .n!.exp⎨z⎜1+ + +....+ ⎟⎬ ⎝ 2 3 n ⎠ = lim ⎩ ⎭ n→∞ z(z +1)(z + 2).....(z + n) n!.e−zHn .e−zlogn.ezHn = lim n→∞ z(z +1)(z + 2).....(z + n) n!.n z −γ z ∞ −1z+1 Γ(z) = lim e ⎛ nz!⎞.n z / n ...(5) n→∞ z(z +1)(z + 2).....(Γz(+z)n=) lim ⎜1+ ⎟ e n→∞ ∏ z (zn+=11⎝)(z +n2⎠).....(z + n +1) Functional equation Γ(z+1)=zΓ(z) Replacing z by (z+1) in the equation (5) n! nz n z = lim n→∞ z(z +1)(z +2).....(z +n) (z+n+1) n! nz z = lim lim n→∞ n→∞ z +1 z(z +1)(z + 2).....(z + n) ( +1) n = Γ ( z ) . z 28 Reflection formula Proof : Using Euler’s formula ⎡ (n!) 2 .n ⎤ = lim ⎢ 2 2 2 2 2 ⎥ n→∞ ⎣ z(1− z )(2 − z ).....( n − z )(n +1− z) ⎦ ⎡ ⎤ ⎢ ⎥ 1 = lim⎢ ⎥ n→∞ 2 2 ⎢ 2 ⎛ z ⎞ ⎛ z ⎞⎛ n +1− z ⎞⎥ (1− )⎜1− ⎟.....⎜1− ⎟ ⎢ z z ⎜ 2 ⎟ ⎜ 2 ⎟⎜ ⎟⎥ ⎣ ⎝ 2 ⎠ ⎝ n ⎠⎝ n ⎠⎦ 1 = ⎧ ⎛ z 2 ⎞ ⎛ z 2 ⎞⎫ (1− 2 )⎜1− ⎟.....⎜1− ⎟ z lim⎨ z ⎜ 2 ⎟ ⎜ 2 ⎟⎬.1 n→∞⎩ ⎝ 2 ⎠ ⎝ n ⎠⎭ 1 π z Γ(z)=Γe(1−−s logz)n = ⎡; z ≠ 0, ±n1,!.±n2,... ⎤ ⎡ n!.n 1 Γ(s z) Γ(1− z) = lim .lim = n sinπz ⎢ ⎥ ⎢ 2 n→∞ z(z +1)(z + 2).....(z + n) n→∞ (1− z)(2 − z ∞ ⎛ z ⎞ ⎣ ⎦ ⎣ ⎜1− ⎟ z∏⎜ 2 ⎟ k =1 ⎝ n ⎠ 1 = sin πz z. πz π = sin πz Ordinary Dirichlet Series Definition : The series of the form ∞ an ∑ s ....(1) n=1 n called ordinary Dirichlet series, where the an are given constants, s = σ + it is a complex variable and . 29 Zeta function of Riemann If an =1, ∀n then the series (1) becomes ∞ 1 ς(s) = , Re s =σ >1 ∑ s n=1 n represents Zeta function of Riemann. ∞ −s Theorem : The series ∑ n converges absolutely and uniformly for σ ≥1 + ε n=1 (ε > 0 arbitrary) Proof : We have 1 1 1 1 1 = = = ≤ ns | nσ +it | | nσ || nit | nσ n1+ε ∞ 1 and the series ∑ 1+ε converges, by Weiestrass M-test n=1 n Theorem : For Re s > 1 ∞ t s−1 Γ(s) ς (s) = dt ∫ t 0 e −1 ∞ t s−1 dt Proof : The integral ∫ t converges at both the lower and upper limits 0 e −1 whenever Re s > 1 Since t lim =1 t→0+ et −1 so that by the definition of limit there is a δ > 0 such that for 0 < t ≤ δ the inequality t 1 −1 < ε = et −1 2 t 1 ⇒ − 1 < e t − 1 2 30 t 3 ⇒ < holds. et −1 2 Hence for 0 < δ1 < δ and σ = Re s >1 δ t 3 δ 3 1 tσ −2dt ≤ tσ −2dt = ()δ σ −1 − δ σ −1 ∫ et −1 2 ∫ 2 σ −1 1 δ1 δ1 δ t σ −2 3 δσ−1 t t dt → as δ → 0 ⇒ ∫ e −1 1 δ1 2 σ −1 on the other hand, we have t m lim = 0 t→+∞ et −1 t m 1 So that there is a b , such that for t ≥ b the ineuqality < is satisfied 1 1 et −1 2 Hence for 0 < b1 < b and choosing m > σ we get b m ∞ b t sσ−m−−1nu s−1 σ−m−1 1 1 σ−m σ−m Γ(s) = nt e dtu≤ dut dt = ()b − b1 1 1 σ−m ∫ t ∫ 2 ∫ 2 σ − m → b as b → ∞ b1 e −1 0 b1 2 m − σ 1 ∞ −t s−1 Now in Γ(s) = ∫ e t dt 0 which is valid for σ > 0 , we make the substitution t =nu to obtain ⎛ ⎞ ∞ ⎜ ⎟ N 1 N Γ(s) = ⎜ e−nu ⎟u s−1du and ∑ n s ∫ ⎜ ∑ ⎟ n=1 0 ⎜ 1n=123 ⎟ ⎝ (sum of G.P) ⎠ ∞ ⎛ e −u (1 − e −Nu ) ⎞ = ⎜ ⎟u s−1du ∫ ⎜ −u ⎟ 0 ⎝ (1 − e ⎠ ∞ u s −1 ∞ u s −1e − Nu = du − du ∫ u ∫ u 0 e − 1 0 e − 1 31 Now it is required to prove that ∞ u s−1e−Nu du → 0 as N → ∞ ∫ u 0 e −1 ∞ us−1e−Nu δ us−1e−Nu ∞ us−1e−Nu du= du+ du; δ >0 ∫ u ∫ u ∫ u 0 e −1 0 e −1 δ e −1 ∞ u s−1e− Nu δ uσ −1 ∞ uσ −1 du < du + e− Nδ du ⇒ ∫ u ∫ u ∫ u 0 e −1 0 e −1 δ e −1 For a given ε >0, we choose δ sufficiently small so as to make the first integral ε on right hand side less than . 2 Fixing that δ we can now take N large enough so as to make the second term on the right less than ∴ for N → ∞ ∞ u s−1 Γ(s) ς (s) = du ∫ u valid for Re s >1 0 e −1 Note:B(E) denotes a closed algebra of C(K, D) that contains every rational U|rb∃ab,ε∈−=V0aC<|<−r Lemma If a∈C−k then (z − a)−1 ∈ B(E) Proof : Case I : ∞ ∉ E Let and let V ={a∈C : (z − a)−1∈ B(E)} so E ⊂ V ⊂ U If and then ...(1) ⇒ V is open ⇒ a number such that |b−a| But −1 −1 ⎡ b − a ⎤ = (Z −1) ⎢1 − ⎥ ...(3) ⎣ Z − a ⎦ From (2) |b−a| |z−a|−1 −1 n ⎡ b − a ⎤ ∞ ⎛ b − a ⎞ ⎢1 − ⎥ = ∑ ⎜ ⎟ ...(4) ⎣ z − a ⎦ n=0 ⎝ z − a ⎠ By the Weirestrass M-Test series on right hand side of (4) converges uniformly on K. k n ⎛ b − a ⎞ If Qn (z) = ∑ ⎜ ⎟ then k =0 ⎝ z − a ⎠ −1 (z − a) Qn (z) ∈ B(E) Since a∈V and B(E) is an algebra (3) shows that B(E) is closed and uniformly −1 −1 {ba∈n }∂V convergence of (4) implies that (z−b)−1 ∈B⎡(zE−) a) − (b − a) ⎤ −1 (z − b) = ⎢ ⎥ (z − a) ⎣ z − a ⎦ ⇒ b∈V Then (1) implies that V is open If then let be a sequence in V with b = lim a n n → ∞ Since b∈/ V then from (1) it follows that | b − an |≥ d(an , K) then for n →∞; an →b ; we get 0 = d (b, K ) or b ∈ K Thus ∂V I U = φ (null set ) If H is a component of U = C − K then H IK ≠φ So H I V ≠ φ 33 ∴ H ⊂ V But H was arbitrary so or V = U Case 2 : Let d = the metric on Let a0 is in the unbounded components of U = C − K such that 1 d(a ,∞) ≤ d(∞, K) 0 2 and | a0 |> 2max{| z |: z ∈ K} ...(5) Let so E0 meets each components of −1 If a∈C−K then case-I gives that (z − a) ∈ B(E0 ) 1 Now | z / a |≤ ; ∀z ∈K [from (5)] 0 2 Therefore 1 1 1 ∞ = = − (z / a ) n z − a − a (1− z / a ) a ∑ 0 0 0 0 U0 n⊂=0V C∞QE0n∞∈∈=−E(BKE(E−){∞}) U {a0 } Converges uniformly on K. Then n 1 k Qn (z) = − ∑ (z / a0 ) is a polynomial a0 k =0 −1 and (z − a0 ) = u − lim Qn on K. Since Qn has its only pole at ∞ , −1 Thus (z − a0 ) ∈ B(E) ⇒ B(E0 ) ⊂ B(E) ⇒ (z − a)−1 ∈ B(E) for each a ∈ C − K Definition : Function element is a pair (f,G) where G is a region and f is an analytic function on G. Definition : Germ of f at a : [ f ]a . For a given function element (f, G) the germ of f at a, denoted by [ f ]a , is the 34 collection of all function elements (g, D) such that a ∈ D and in a neighbourhood of a. ⇒ [ f ]a = collection of function elements and it is not a function element itself. Note: The equivalence of two germs of f i.e. [ f ]a = [ f ]b is meaning less until a = b. Definition : Analytic continuation along a path. Let γ :[0,1] → C be a path and suppose that for each t in [0,1] there is a function element ( ft , Dt ) such that (a) γ(t)∈ D (b) for each t in [0,1] there is a δ > 0 such that | s − t |< δ [ f ] = [ f ] ⇒ γ(s)∈ Dt and s γ(s) t γ(s) (i.e. germ of fs and ft at γ(s) are equal) Then (f, D) is the analytic continuous of ( f0 , D0 ) along the path γ; or ( f1, D1) is obtained from ( f0 , D0 ) by analytic continuation along γ. Power series method of analytic continuation Rf (tz))≡=∞g(z) ∀ z Lemma : Let γ :[0,1] → C be a path and let {( f t , Dt ):0 ≤ t ≤1} be an analytic continuation along γ. For 0 ≤ t ≤1 let R(t) be the radius of convergence of the power series expansion of ft about z = γ(t). Then either or R:[0,1] → (0,∞) is continuous. Proof : If R(t) = ∞ for some value of t then it is possible to extend ft to an entire function. Since f s (z) = ft (z) ∀ z ∈ Ds ⇒ R(s) = ∞ ∀ s ∈[0,1] i.e. R(s) ≡ ∞ Suppose that R(t) < ∞ ∀t For a particular t ∈[0,1] let τ = γ(t) 35 let be the power series expansion of ft about τ. Let δ1 > 0 be such that | s − t |< δ1 ⇒ γ(s)∈ Dt I B (τ;R(t)) and [ f s ]γ(s) = [ ft ]γ(s) Now let σ = γ(s) for a fix s; Now ft can be extended to an analytic function on B(τ;R(t)) . Since fs agrees with ft on a neighbourhood of σ (by definition germs), fs can be extended so that it is also analytic on . If fs has a power series expansion ∞ n fs (z) = ∑σn (z − σ) about z = σ n=0 then the radius of convergence R(s) must be at least as big as the distance from |RRz(−(ss)τ)≥−|−=dRRR∞((σ(t(t)t,))≤{||z≤γ:|(|tzγ)−(−t)τγ−|(=sγ)R(|s()t)})| ≥ R(t)− | τ − σ | |Bs(−τ;tR|<(tδ))1U Ds n σ to the circle ft (z) = ∑τn (z − τ) n=0 i.e. ⇒ R(t) − R(s) ≤| γ(t) − γ(s) | Similarly it can be shown that Hence for Since γ:[0,1] → C is continuous. ⇒ R must be continuous at t. Rung’s Theorem : Let K be a compact subset of C and let E be a subset of C∞ − K that meets each component of C∞ − K . If f is analytic in an open set containing K and ε>0 then there is a rational function R(Z) whose only poles lie in E and such that | f (z) − R(z) |< ε ∀ z ∈ K 36 Proof : By the fact that if K be a compact subset of the region G, then there are straight line segments γ1 , γ 2 ,.....γ n in G – Ksuch that for every function f in H(G). n 1 f (w) f (z) = ∑ ∫ dw ∀z ∈ K K =1 2πi w − z γk The line segments form a finite number of closed polygons. Also If γ be a rectifiable curve and let K be compact set such that K I {γ} = φ . If f is continuous function on {γ} and ε>0 then there is a rational function R(z) having all its poles on and such that Definition : Unrestricted analytic continuation : Let (f, D) be a function element and let G be a region which contains D: then (f, D) admits unrestricted analytic continuation in G if for any path γ in G with initial point in D there is an analytic continuation of (f, D) along γ. {γ}f (w) ∫ dw − R(z); ∀z ∈ K γ w − z Fixed-End-Point (FEP) homotopic Definition : If γ 0 , γ1 :[0,1] → G are two rectifiable course in G such that γ 0 (0) = γ1 (0) = 0 and γ 0 (0) = γ1 (0) = b then γ 0 and γ1 are FEP homotopic if there is a continuous map Γ:I 2 → G such that Γ(s,0) = γ 0 (s) Γ(s,1) = γ1 (s) Γ(0,t) = a Γ(1,t) = b for 0 ≤ s, t ≤1 I 2 = [0,1]×[0,1] 37 Theorem : Monodromy : Let (f, D) be a function element and let G be a region containing D such that (f, D) having unrestricted continuation in G. Let a ∈ D,b ∈ G and let γ 0 and γ1 be paths in G from a to b; let {( f t , Dt ) : 0 ≤ t ≤1} and {(g t , Dt ) : 0 ≤ t ≤1}be analytic continuation of ( f, D) along γ 0 and γ1 respectively. If γ 0 and γ1 are fixed-end-point homotrpic in G then [ f1 ]b = [g1 ]b Proof : Since γ 0 and γ1 are FEP homotopic in G ⇒ there is a continuous function Γ:[0,1]×[0,1] → G such that Γ(t,0) = γ 0 (t) Γ(t,1) = γ1 (t) Γ(0,u) =a; Γ(1,u) =b for all t and u in [0, 1] [hf11,0]b]b==[h[1h,10,1]b] p Let γ u (t) = Γ(t,u) for a fix u, 0 ≤ u ≤1, from a to b. ....(1) By hypothesis there is an analytic continuation {(ht,u ,Dt,u ):0 ≤ t ≤1} of (f, D) along γ u By the result that if γ:[0.1] → C be a path from a to b and {( f t , Dt ):0 ≤ t ≤1} and {(g t , Bt ):0 ≤ t ≤1} be analytic continuations along γ such that [ f 0 ]a = [g 0 ]a . Then [ f1 ]b = [g1 ]b Then it follows that [g1 ]b = [h1,1 ]b and Now it is sufficient to show that ...(2) To show this Let U = {u ∈[0,1]:[h1,u ]b = [h1,0 ]b} ...(3) and we show that U is non-empty open and closed subset of [0,1]. 38 Since Now it is remains to show that U is both open and closed. Let us consider for u ∈[0,1] ∃ a δ > 0 such that if | u − v |≤ δ then [h1,u ]b = [h1,v ]b ....(4) For a fixed there is an ε > 0 such that if σ is any path from a to b with|γ u (t)−σ(t)|<ε , ∀ t and if to any continuation of (f,D) along σ, then [h1,u ]b = [K1 ]b ....(5) Now Γ is a uniformly continuous function, so there is δ >0 such that if then | γ u (t) − γ v (t) |=|Γ(t,u) − Γ(t, v) | < ε ∀ t. ....(6) Suppose u ∈U and let δ > 0 be the number given in (4). By definition of U, u|v0u∈:G−[U0v→,|1− ∃ a such that But form (4) [h1,u ]b = [h1,v ]b ; and since ⇒ ⇒ U is closed Mean Value Theorem: Let be a harmonic function and let B(a; r) be a closed disk contained in G. If γ is the circle centred at a and of radius r i.e. γ :| z − a |= r then 1 2π u(a) = ∫ u(a + reiθ )dθ 2π 0 Proof : Let D be a disk such that 39 B(a;r) ⊂ D ⊂ G, and Let f be analytic on D such that u = Ref Then by Cauchy integral formula 1 f (z) f (a) = ∫ dz 2πi γ z − a | z − a |= r z = a + reiθ dz = ir iθ 1 2π f (a + reiθ ) dθ = ∫ iθ 2π 0 re 1 2π = ∫ f (a + reiθ )dθ 2π 0 1 2π f (a) = ∫ f (a + reiθ )dθ 2π 0 2π 1 iθ Then Re f (a) = ∫ Re f (a + re )dθ 2π 0 A ={Z ∈G:u(z) = u(a)} 2π 1 iθ ⇒ u(a) = ∫ u(a + re )dθ 2π 0 Mean Value Property (MVP) Definition : A continuous function u:G → R has the MVP if whenever B(a;r) ⊂ G 1 2π u(a) = ∫ u(a + reiθ )dθ 2π 0 Maximum Principle : Let G be a region and let u is a continuous real valued function on G with the MVP. If there is a point a in G such that u(a) ≥ u(z) ∀z ∈G, then u is constant function. Proof : Let the set A be defined by Since u is continuous, the set A is closed in G. 40 If let r be chosen such that B(Z 0 ; r) ⊂ G . Suppose b∈ B(z0 ;r) such that u(b) ≠ u(a) ; then . ∴ By continuity in the neighbouhood of b iθ If | z0 − b |= ρ and b = z0 + ρe 0 ≤ θ ≤ 2π then there is a proper interval I of [0, 2π] such that θ∈ I and Hence by MVP 2π 1 iφ u(z0 ) = ∫u(z0 + ρe )dφ So B(z0 ;r) ⊂ A and A is also open. by connectedness of G, A = G. Harmonic function on a disk : iφ iθ 2 uz0(bz∈) Let z = reiθ , 0 ≤ r <1 then 1+ reiθ 1+ z ⇒ = = (1+ z)(1− z)−1 1− reiθ 1− z = (1+ z)(1+ z + z 2 + ....) ∞ =1+ 2∑ z n n=1 41 ∞ =1 + 2∑ r n einθ n=1 ⎛1+ reiθ ⎞ ∞ Re⎜ ⎟ =1+ 2 r n cosnθ ⇒ ⎜ iθ ⎟ ∑ ⎝1− re ⎠ n=1 ∞ (einθ + e−inθ ) =1+ 2∑r n n=1 2 ∞ = ∑ r |n|einθ n=−∞ = Pr (θ) ....(1) Also 1 + reiθ (1 + reiθ )(1 − re −iθ ) 1 + reiθ − re −iθ − r 2 = = 1 − reiθ |1 − reiθ |2 |1 − reiθ |2 (1 − r 2 ) + r(i2sin θ) = |1 − reiθ |2 ⎛1 + reiθ ⎞ 1 − r 2 Re⎜ ⎟ = ⇒ ⎜ iθ ⎟ 2 ....(2) ⎝1 − re ⎠ 1 − 2r cos θ + r iθ 2 2 [Q|1− re | =1− 2r cosθ + r ] Combining (1) and (2), we get the result. Prop. 2.3 1 π (a) ∫ Pr (θ)dθ =1 2π −π (b) Pr (θ) > 0 ∀θ, Pr (−θ) = Pr (θ) and Pr is periodic in θ with period 2π. Theorem : Let D = {z:| z |<1} and suppose that f :∂D → R is a continuous function. Then there is a continuous function u:D − → R such that (a) u(z) = f (z) ∀ z ∈∂D (b) u is harmonic in D. Moreover u is unique and is defined by the formula π iθ 1 it u(re ) = ∫ Pr (θ − t) f (e )dt for 0 ≤ r <1, 0 ≤ θ ≤ 2π 2π −π 42 Proof : Define u:D → R as π iθ 1 it u(re ) = ∫ Pr (θ − t) f (e )dt for 0 ≤ r <1 ...(1) 2π −π and let u(eiθ ) = f (eiθ ) ...(2) Then u satisfies part (a). Now we have to show that u is continuous on D and harmonic in D. (i) Proving u is harmonic in D. If 0 ≤ r <1 then from (1) 1 π ⎡1+ rei(θ−t) ⎤ u(reiθ ) = Re f (eit )dt ∫ ⎢ i(θ−t) ⎥ 2π −π ⎣1− re ⎦ 1424 434 by definition of Pr (θ) ⎪⎧ 1 π ⎡1+ rei(θ−t) ⎤ ⎪⎫ = Re f (eit ) dt ⎨ ∫ ⎢ i(θ−t) ⎥ ⎬ ⎩⎪2π −π ⎣1− re ⎦ ⎭⎪ ⎪⎧ 1 π ⎡eit + reiθ ⎤ ⎪⎫ = Re f (eit ) dt ⎨ ∫ ⎢ it iθ ⎥ ⎬ ...(3) ⎩⎪2π −π ⎣e e∂ε−Di>αre0 ∃⎦ a⎭⎪ ρ, 0 < ρ <1 Let us define g : D → C by 1 π ⎡eit + z ⎤ g(z) = f (eit ) dt ∫ ⎢ it ⎥ ...(4) 2π −π ⎣e − z ⎦ Then g is analytic and Re g = u ...(5) ∴ ∇ 2u = 0 ⇒ u is harmonic. (ii) Continuity of u on D– Since u is harmonic on D therefore it remains to prove that u is continuous at each point of the boundary of D. If α ∈[−π, π] and and an arc A of about such that for ρ < r <1 and eiθ in A | u(reiθ − f (eiα ) |< ε ....(6) 43 for particular α =0 we show that (6) holds. Since f is continuous at z =1 ∃ a δ > 0 such that ε | f (eiθ ) − f (1) |< if | θ |< δ ...(7) 3 Let M = max{| f (eiθ ) | :| θ |≤ π} Then from the result on Poisson kernel Pr (θ) < Pr (δ) if 0 < δ <| θ |≤ π There exists a ρ, 0 < ρ <1 such that ...(8) δ for ρ < r <1 and | θ |≥ 2 ⎧ iθ δ⎫ Let A be the arc ⎨e :| θ |< ⎬ then if ⎩ 2⎭ eiθ ∈ A and ρ < r <1 π iθ 1 it u(re ) − f (1) = ∫ Pr (θ − t) f (e )dt − f (1) 2π −π ε P (θ) < r 3M 1 it 1 it = ∫ Pr (θ − t)[ f (e ) − f (1)]dt + ∫ Pr (θ − t)[ f (e ) − f (1)]dt 2π |t|<δ 2π |t|≥δ δ δ if | t |≥ δ and | θ |≤ then | t − θ |≥ 2 2 iθ 1 it 1 it | u(re ) − f (1) |≤ ∫ Pr (θ − t) | f (e ) − f (1) | dt + ∫ Pr (θ − t) | f (e ) − f (1) | dt 2π |t|<δ 2π |t|≥δ ε ε ≤ + 2M. [from (7) and (8)] 3 3M ⇒ | u(reiθ ) − f (1) |< ε Hence for general value of α. | u(reiθ ) − f (eiα ) |< ε ...(9) Show that u is continuous on D–. (iii) For u is unique 44 Suppose v ∈ D − which is harmonic on D and v(eiθ ) = f (eiθ ) ∀ θ . then (u-v) is harmonic is D and (u − v)(z) = 0. ∀z ∈∂D ⇒ u - v = 0 ⇒ u is unique. Harnack inequality If u:B(a; R) → R is continuous, harmonic in B(a; R) and u ≥ 0 then for 0 ≤ r < R and all θ R − r R + r u(a) ≤ u(a + reiθ ) ≤ u(a) R + r R − r Harnack’s theorem : Let G be a region with (a) The metric space Har(G) is complete. (b) If {u n } is a sequence in Har(G) such that u1 ≤ u2 ≤ .... then either un (z) → ∞ uniformly on compact subset of G or {u n } converges in Har(G) to a harmonic function. Proof : (a) Har(G) is complete Let {u n } be a sequence in Har(G) such that un → u in C(G, R) Then u has the MVP ⇒ u is harmonic [by the theorem if u:G → R which has MVP then u is harmonic.] (b) Assuming u1 ≥ 0 Let u(z) = sup{(u n (z):n ≥1}; ∀ z ∈G ⇒ Either u(z) = ∞ or u(z)∈R and un (z) → u(z), ∀ z ∈G Define A ={z ∈G:u(z) = ∞} ....(1) 45 B ={z ∈G:u(z) < ∞} ...(2) Then A U B = G and A I B = φ Now we show that both A and B are open If a∈G, and R be chosen such that Then by Harnack’s inequality R− | z − a | R+ | z − a | u (a) ≤ u (z) ≤ u (a); ∀ z ∈ B(a; R) and ∀ n ≥1. R+ | z − a | n n R− | z − a | n ...(3) If a ∈ A then so that from (3) un (z) → ∞, ∀ z ∈ B(a; R) [left half of (3)] ⇒ B(a; R) ⊂ A ⇒ A is open. Similarly : If a ∈ B then right half of (3) gives that uB(n((za()a;;<)R→)∞) ⊂∞GB for | z − a |< R ⇒ un (z) < ∞ ∀ z ∈ B(a; R) ⇒ ⇒ B is open Since G is connected, either A =G or B =G Suppose A = G; that is u ≡ ∞ If and 0 < ρ < R then (R − ρ) M = > 0 and R + P (3) gives that Mun (a) ≤ un (z) for | z − a |≤ ρ ⇒ un (z) → ∞ uniformly for z ∈ B(a;ρ) ⇒ ∀a∈G ∃ a ρ > 0 such that un (z) → ∞ uniformly for | z − a |≤ ρ 46 ⇒ un (z) → ∞ uniformly for z in any compact set. Now suppose B= G i.e. u(z) < ∞ ∀ z ∈G If ρ < R then there is a constant N, which depends only on a and ρ such that for | z − a |≤ ρ and ∀ n So if ≤C[un (a) − um (a)] for some constant C. ⇒ {un (z)} is a uniformly Cauchy sequence on B(a;ρ) ⇒ {u n } is a Cauchy sequence in Har(G) and must converge to a harmonic function [by part (a)] Since un (z) → u(z) m ≤ n ⇒ u is the required harmonic function.ϕ0MuB≤:(aGnu;(nr→a())z⊂)≤R−Guu,n (mz()z≤) ≤NNuun n(a(a)) − Mu m (a) Definition : Subrarmonic function- If ϕ be a continuous function and if such that 1 2π ϕ(a) ≤ ∫ϕ(a + reiθ )dθ 2π 0 Definition: Superharmonic- If ϕ be a continuous function and if such that 1 2π ϕ(a) ≥ ∫ϕ(a + reiθ )dθ 2π 0 47 Definition : If G is a region and f be a continuous function f :∂ ∞G → R then the Parron family denoted by Ρ( f ,G) consists of all subharmonic function ϕ:G→R such that Dirichlet Problem : It consists in determining all regions G such that for any continuous function there is a continuous function u:G− → R such that u(z) = f (z) for and u is harmonic in G. Definition : Barrier for G at a Let G be a region and . A barrier for G at a is a family {ψ r :r > 0} of functions such that (a) ψ r is defined and superharmonic on G(a; r) with 0 ≤ ψ r (z) ≤ 1; wψazˆf ∈(∈:a∂G∂G) =−→G0{BwR(a:;|rw) −a|= r} limψ r (z) = 0 limz ∈r ψsup∂∞GrI(ϕz()z=) 1≤ f (a) ∀a ∈∂∞G (b) z→a z→aw (c) for If we define ψˆ r by letting ψˆ r = ψ r on G(a; r) ψˆ r (z) = 1 for then is superharmonic. ⇒ ψˆ r approaches the function which is one everywhere but at z=a, ψˆ r is zero. Theorem : Let G be a region and let a ∈∂ ∞G such that there is a barrier for G at a. If f :∂∞G → R is continuous and u is the Perron function associated with f then lim u(z) = f (a) z→a Proof : Let {ψ r :r > 0} be a barrier for G at a. Assuming a ≠ ∞ and 48 (otherwise consider the function f- f(a)} Let ε > 0 and choose δ > 0 such that | f (w) |< ε whenever w ∈ ∂ ∞ G and | w − a |< 2δ ; Let ψ = ψ δ Let ψˆ : G → R defined by ψˆ (z) =ψ (z) for z ∈G(a;δ ) and for z ∈G − B(a;δ ) ...(1) Then is superharmonic. If | f (w)|≤ M for all w in ∂ ∞G ...(2) then − Mψˆ − ε is subharmonic. Consider − Mψˆ − ε in Ρ( f ,G) ....(3) ψ∀ϕ∂ψ−ˆˆM(ϕGz∈ψ)ˆ=≥(Ρz1(0)f−,Gε)≤ u(z) ; ∀ z ∈G If w∈ ∂ ∞G − B(a;δ ) then from (1) and∞ (2) lim sup[−Mψˆ (z) − ε ] = −M − ε < f (w) z→w ...(4) Because limsup[−Mψˆ(z) − ε ] ≤ −ε ∀ w∈∂∞G ⇒ z→w ....(5) If particular if w ∈ ∂ ∞G I B(a;δ ) then limsup[−Mψˆ(z) −ε ] ≤ −ε < f (w) z→w by the choice of δ. ⇒ the consideration (3) is valid. Hence ...(6) Similarly liminf[Mψˆ(z)+ε] ≥ limsup(ϕ(z)) z→w z→w and w in . By the Maximum principle for subharmonic and superharmonic 49 we have ∀ϕ in and z ∈ G Hence ....(7) (6) and (7) ⇒ − Mψˆ (z) − ε ≤ u(z) ≤ Mψˆ (z) + ε ∀z ∈ G ....(8) But ( is Parron function) Since ε is arbitrary (8) implies limu(z) = 0 = f (a) Z →a Harmonic function Definition : Let G be an open set and G⊂C, u:G→R is harmonic if u has continuous partial derivatives and satisfying the Laplace equation, i.e. ∂ 2u ∂ 2u 2 ˆ ˆ + = 0. or ∇ u = 0 ϕulimvψΡ((=(zzf)ψI)sup,ˆm≤G(fMzϕ))ψ=≤ˆ(limz) +ψinfε(;ψz) ∂x 2 ∂y 2 zz→wa zz→→wa Harmonic conjugate Definition : Let f be an analytic funcion defined as f :G → C then u = Re f and are called harmonic conjugates. 1 p | z |≤ <1 | log E(z;k) |≤ | z |k +1 Theorem : If p then p −1 . Proof : Let k > 0 ⎛ z 2 z k ⎞ log E(z;k) = Log (1− z) + ⎜ z + + ..... + ⎟ ⎝ 2 k ⎠ z 2 z k 1 Log(1− z) = −z − − − z k +!.... 2 k k +1 for | z |<1 1 1 1 ∴ log E(z;k) = − z k+1 − z k +2 − ..... for | z |≤ <1 k +1 k + 2 p 50 k+1 ⎛ 1 1 ⎞ ∴ | Log E(z;k) |≤| log E(z;k) |≤| z | ⎜ + | z | +...⎟ ⎝ k +1 k + 2 ⎠ k+1 2 ≤| z | (1+ | z | + | z | +.....) [Qk > 0] ⎛ 1 1 ⎞ ≤| z |k+1 ⎜1 + + + ...⎟ ⎜ 2 ⎟ ⎝ p p ⎠ p = | z |k +1 p −1 for k =0 ⎛ | z | ⎞ | log E(z;0) |=| log(1− z) |≤| z | ⎜1+ + ...⎟ ⎝ 2 ⎠ ⎛ 1 1 ⎞ ≤| z | ⎜1+ + + ...⎟ ⎜ 2 ⎟ ⎝ p p ⎠ p | log E(z;0) |= | z | p −1 for p =2 2 | Log E(z;k) |≤ | z |k+1= 2 | z |k+1 2 −1 1 provided | z |≤ 2 Example : construct an entire function with simple zeros at the point 0,1, 2p , 3p , ... , n p , ...(p >1) Solution : We may take kn = 0 for every n. The series ∞ z ∞ 1 ∑ p =| z | ∑ p n=1 n n=1 n converges for ∀z when p >1 ∴ by Weierstrass factorization Theorem 51 1 m=1 (simple zero at z=0), k = 0, a = n n np ∞ ⎛ z ⎞ F(z) = eh( z) z 1− ⇒ ∏ ⎜ p ⎟ n=1 ⎝ n ⎠ Example : Express sin πz as an infinite product. Solution : since sin πz is an entire function with simple zeros at 0,±1,±2,±3,.... Then by Weierstrass factorization Theorem ....(1) k + 1 ∞ z n provided ∑ converges for ∀z where an , n ≥1are zeros. n =1 a n ∞ ∞ kn +1 ⎛ ⎛z z ⎞ ⎞ ∞ sin πz = he(zh)( z )mz1 ⎜E⎜ ; k⎟ ⎟ r F(z) = e z ∏∏E⎜ ⎜ ; kn ⎟n ⎟ n=1 a a since ∑ < ∞; n ≠ 0, ∀r > 0 n=1 ⎝ ⎝n n ⎠ ⎠ n=−∞ an Then it is sufficient to choose k n =1, ∀ n, 2 ∞ z ⎛ 1 1 1 1 ⎞ =| z |2 ⎜ + + + + .... ⎟ then ∑ 2 2 2 2 n = −∞ an ⎝ 1 1 2 2 ⎠ n ≠ 0 ∞ 1 = 2 | z |2 ∑ 2 that conveyes for each z n=1 n ∴ in (1) we may put an = ±n, kn =1 to get ∞ ⎛ z ⎞ sinπz = eh(z)z∏E⎜ ;1⎟ n=1 ⎝ ±n ⎠ ∞ h(z) ⎡⎛ z ⎞⎛ z ⎞ ⎛ z ⎞ ⎛ z ⎞⎤ sinπz = e z∏⎢⎜1+ ⎟⎜1− ⎟ exp⎜ ⎟exp⎜− ⎟⎥ n=1 ⎣⎝ n⎠⎝ n⎠ ⎝ n⎠ ⎝ n⎠⎦ 52 ∞ ⎛ z 2 ⎞ sinπz = eh(z) z ⎜1− ⎟ ∏⎜ 2 ⎟ n=1 ⎝ n ⎠ Example : Construct an entire function with simple zero at the point 0, -1, -2, ...., -n. Solution : Given that an = -n then the series k +1 ∞ z n ∑ be convergent if kn =1 n =1 a n k +1 ∞ z n ∞ 1 ∞ 1 =| z |2 =| z |2 i.e. ∑ ∑ 2 ∑ 2 n=1 an n=1 | −n | n=1 n Then Weierstrass factorization Theorem gives z ∞ − h(z) ⎛ z ⎞ n F(z) = e z∏⎜1 − ⎟e n=1 ⎝ (−n) ⎠ z ∞ ⎛ z ⎞ − = e h(z) z∏⎜1 + ⎟e n n=1 ⎝ n ⎠ Definition : Rank of infinite product ∞ 1 The smallest non negative integer k for which the series ∑ k +1 converges n=1 | an | ∞ ⎛ z ⎞ ⎜ ⎟ is said to be rank of the infinite product. ∏ E⎜ ;kn ⎟ n=1 ⎝ an ⎠ Definition Canonical (or regular) product : If k denotes the rank of infinite product and if we take kn = k ∀ n , then the infinite product ∞ ⎛ z ⎞ ∏ E⎜ ;k ⎟ is called canonical product. n=1 ⎝ an ⎠ If no such integer k exists, the infinite product is said to be of infinite rank. Definition : Exponent of convegence of zeros of an entire function 53 If F(z) = eh(z).zmP(z) ...(1) where P(z) is canonical product of rank k and A a non-empty set of non negative real members α such that converges. Then the number ρ = glb A is called the exponent of convegence of the zeros of the entire function given by (1) Theorem : Theorem for the computation of the exponent of convegence. Let | an |= rn (0 < r1 ≤ r2 ≤ ... ≤ rn ≤ ....;rn → ∞) then the exponent of convergence is given by 1 logr = lim n n→∞ ρc logn 1 n Example : If (i) 2 (ii) | a |=| a | , | a |>1 then find ρ | an |= n logn n c. Ans. (i) 2 (ii) 0 2 F∞(z)1= z 2e z ∑ α Definition : Genus and Exponential ndegree=1 | an | of an entire function : If F(z) = e h(z) z m P(z) is an entire function with canonical product P(z) of rank k and h(z) is a polynomial of degre q ≥ 0 then non-negative integer p = max(k, q) is called the genus of the entire function and q is said to be the exponential degree of F. If P(z) is not of finite rank k, or if h(z) is not a polynomial, then F is said be of infinite genus. Example : Find genus of the entire function 2 F(z) = e z z 2 P(z) h(z) = z 2 is a polynomial of degree 2. P(z) =1 ∴ k n = k = 0 ∴ max(k, q) = max(0,2) = 2 is the genus of F(z). 54 Definition : Order of an entire function If F(z) be an entire function and A is a positive constant such that A max | F(z) |= M (r) < e r |z|=r for all sufficiently large value of r = |z|, then F(z) is caled an entire function of finite order. Alternatively F(z) is of finite orer if ∃ A > 0 such that A | F(z) |= 0 (e r ) A or | F(z) |< K e r ; K>0 Definition : Order ρ of an entire function of finite order : r A Let S = {A:| F (z) |< e , r > r0} then the order ρ of an entire function F of finite A ρ order is defined by ρ = inf{A :| f (z) |< e r }M (r) < e Br ∀ If there is not a positive constant A such that A | F(z) |< e r ∀ r large enough F(z) is said be of infinite order, ρ = ∞ . Definition : Type of entire function: If F(z) is an entire function of finite order ρ and there exists a constant B >0 such that r large enough then F(z) is said to be of finite type and ρ σ = inf{B:M(r) < eBr , r > R} is called the type of F. If σ > 0 , F is said to be of normal type. If σ = 0, F is called minimum type. 55 ρ If there is no B such that M (r) < eBr , then F is called of infinite type (or maximum type). exponential type σ : An entire function F is said to be of expontial type σ(σ < ∞) if either the function is of order ρ =1 and type , or the function is of order less then 1. Theorem : The order ρ of an entire function is given by the formula Proof : Suppose ρ < ∞ then ∀ r large enough ...(1) then for given ε >0 we have ρ+ε M (r) < e r ∀ r large enough ...(2) Also, there are some z with arbitrary modulus for which r ρ−ε M (r) < e ≤logσ log Mlogr(ρ rlog) M (r) ..(3) ρM=(rlim) < e < ρ + ε (r =| z |) Then (2) implies rlog→∞ r log r log log M (r) < (ρ + ε) log r i.e. ∀r < Re ...(4) and (3) implies loglog M (r) > ρ − ε log r ...(5) log log M (r) ρ − ε < < ρ + ε ∴ log r loglog M (r) ⇒ ρ = lim r→∞ log r Theorem : The type σ of an entire function of finite order ρ is given by 56 logM(r) σ = lim r→∞ r p Proof : Since the entire function f(z) of finite order ρ is of type σ Therefore ρ M (r) < e σr ...(1) then for given ε > 0, we have ρ M (r) < e (σ+ε)r ∀ r large enough ...(2) ρ and M (r) > e (σ−ε)r ...(3) for infinity many values of r Then (2) implies log M (r) < σ + ε ...(4) r p and (3) implies log M (r) > σ − ε ....(5) r p | z |≤ R log M (r) ⇒ σ − ε < < σ + ε r p log M (r) ⇒ σ = lim r→∞ r p Theorem : Jensen Formula If f(z) is analytic in the disc | z |≤ R and if ak ≠ 0 (1≤ k ≤ n) are the zero of f(z) in those zeros being repeated according to their multiplicities, then 1 2π n R ∫ log | f (Re iθ ) | dθ = log | f (0) | +∑ log 2π 0 k=1 | ak | Proof : Since f(z) is analytic in | z |≤ R, ∃ an open disc | z |< R′ = R + δ (δ > 0) , where f(z) is analytic and has no other zero than the ak Then if the function a1a2 ...an f (z) g(z) = ...(1) (a1 − z)(a2 − z)...(an − z) f (0) 57 is dfined at the point a1 , a2 ,..., an then it becomes analytic in | z |< R′ and does not vanish analytic in this disc. Then there exists a function h(z) analytic in R'such that eh(z) = g(z) an analytic branch h(z) of log g(z) in the disc R'. From (1) g(0) =1, we may coose h(0) =0 [from (2)] h(z) Then can be made analytic in | z |< R′ z If we consider C:z = Reiθ , 0 ≤ θ ≤ 2π then by the Cauchy-Gousrat theorem 1 h(z) 1 2π h(Re iθ ) dz = Rie iθ dθ = 0 ∫ ∫ iθ 2πi C + z 2πi 0 Re 2π 1 iθ = ∫ h(Re )dθ = 0 ...(3) 2π 0 But Re h(z) = log | g(z) | Taking real part of (3) 1 2π a .a ...... a f (Reiθ ) log 1 2 n dθ = 0 ∫ iθ iθ 2π 2π 2π 0 (a1 − Re n)...(2π an − Re ) f (0) 1 iθ 1 1 ⇒ ∫ log | f (Re ) | dθ − ∫ log | f (0) | dθ + ∑ ∫ log | ak | dθ 2π 0 2π 0 2π k=1 0 n 2π 1 iθ − ∑ ∫ log | ak − Re | dθ = 0 2π k =1 0 2π n 1 iθ 1 ⇒ ∫ log | f (Re ) | dθ − log | f (0) | + ∑ log | ak |×2π 2π 0 2π k=1 1 n 2π 1 + log dθ = 0 ∑ ∫ iθ 2π k =1 0 | ak − Re | 2π n 1 iθ 1 ⇒ ∫ log | f (Re ) | dθ − log | f (0) | + ∑ log | ak |×2π 2π 0 2π k=1 58 ⎛ ⎞ ⎜ ⎟ 1 n 2π ⎜ 1 ⎟ + log dθ =0 ∑ ⎜ −iθ ⎟ 2π ∫ a e k=1 0 ⎜ iθ k ⎟ ⎜| Re | −1 ⎟ ⎝ R ⎠ 2π n 1 iθ 1 ⇒ ∫ log | f (Re ) | dθ − log | f (0) | + ∑ log | ak |×2π 2π 0 2π k=1 1 n 2π ⎡ a e−iθ ⎤ + ∑ ∫ ⎢−log| R | −log1− k ⎥dθ = 0 2π k=1 0 ⎣ R ⎦ 2π n 1 iθ 1 ⇒ ∫ log | f (Re ) | dθ − log | f (0) | + ∑ log | ak |×2π 2π 0 2π k=1 1 n 1 n 2π a e−iθ + ∑2π(−logR) − ∑∫log1− k dθ = 0 2π k=1 2π k=1 0 R 2π n 1 iθ 1 ⇒ ∫ log | f (Re ) | dθ − log | f (0) | + ∑ log | ak |×2π 2π 0 2π k=1 1 n 2π a e−iθ + (−n)log R − ∑ ∫ log1− k dθ = 0 2π k=1 R 0 1424 434 (=0 when |ak / R|≤1) 1 2π n n log | f (Reiθ ) | dθ = log | f (0) | − log | a | + log R ⇒ ∫ ∑ k ∑ 2π 0 k =1 k =1 n ⎛ R ⎞ = log | f (0) | +∑log⎜ ⎟ k=1 ⎝ | ak | ⎠ Corollary : If n(r) denotes the number of zeros of the entire function F(z) in the disc | z |≤ r , and F(0) ≠ 0, then R n(t) ∫ dt ≤ log M (R) − log | F(0) | 0 t Proof : Let a1 , a2 ,...., an are zeros of F(z) in |z| n ⎛ R ⎞ n ⎜ ⎟ ∑log⎜ ⎟ = nlog R − ∑log | ak | k =1 ⎝ | ak | ⎠ k =1 59 n−1 = n log R + ∑ k[log | ak +1 | − log | ak |] − n log | an | k=1 n−1 |ak +1| 1 = ∑ ∫ k dt + n(log R − log | an |) k =1 t |ak | n−1|ak+1| k R 1 = ∑ ∫ dt + n ∫ dt ....(1) k =1 t t |ak | |an | n(t) = 0 for 0 ≤ t ≤| a1 | n(t) = k for | ak |≤ t <| ak +1 |, k =1,2,....,(n −1) n(t) = n when | an |≤ t < R We have (from 1) n R |a2 | 1 |a3| 2 |a4 | 3 |an | (n −1) R 1 ∑ log = ∫ dt + ∫ dt + ∫ dt + ... + ∫ dt + n ∫ dt k =1 | a | t t t t t k |a1| |a2 | |a3| |an−1| |an | R n(t) = ∫ dt 0 t ρ+ε / 3 Then by Jesens formula Mρc(2r) < exp{(2r) } R n(t) 1 2π ∫ dt = ∫ log | F(Reiθ )dθ − log | F(0) | 0 t 2π 0 ≤ log M (R) − log F(0) Theorem : (Hadamard ) The exponent of convergence of the zeros of an entire function of finite order is no greater then ρ. i.e. ρc ≤ ρ ∞ Proof : {an } sequence of zeros of F(z), | ak |≤| ak +1 | and real fucntion n(t) is monotonic increasing with t 2r n(t) 2r n(t) 2r dt ∴ ∫ dt ≥ ∫ dt ≥ n(r) ∫ = n(r) log 2 0 t r t r t 1 2r n(t) ⇒ n(r) ≤ ∫ dt ....(1) log 2 0 t For given ∈> 0 we have . For r large enough 60 or log M (2r) < (2r)ρ+ε / 3 < r ρ+2ε / 3 ...(2) Replacing R by 2r in last result of the corollary and using (2) 2r n(t) 2 / 3 ∫ dt < r ρ+ ε − log | F(0) |< r ρ+ε ...(3) 0 t (1) and (3) 1 ρ +ε n(r) < r ...(4) log 2 ⇒ n(r) =o(rρ+ε ) as r → ∞ Now we show that converges whatever be δ > 0 Let ε be such that and r = rn 1 ρ+ε n = n(rn ) < rn ....(5) log 2 ∞ ρ0 c<≤ε1ρ< δ ∑ ρ+ε for all r large enough n=1 rn from (5) 1 1 1 < ρ+ε rn n log 2 (ρ +δ )/(ρ +ε ) 1 1 ⎛ 1 ⎞ < A and ρ+ε (ρ+δ )/(ρ+ε ) A = ⎜ ⎟ rn n ⎝ log 2 ⎠ (ρ + δ) ∞ 1 >1 , the series ∑ (ρ+δ )/(ρ+ε ) converges (ρ + ε) n=1 n ∞ −(ρ+ε) Hence ∑ rn n=1 ⇒ Exponent of convergence ρ c of zeros is ρ + δ ⇒ Theorem : Suppose that about each zero an ,| an |>1, of a cononical product 61 1 P(z), a dix of radius p is described where rn =| a n | and = order P(z). Then in rn the region R complementary to the union of all those discs, the inequality holds an infinitely many circles of radii arbitrarily large. ∞ 1 2 ∑ Proof : It is obivious that the sum p infinite as ( p > ρ = ρc ) the union of n=1 rn − p − p [r r , r r ] − p the intervals n − n n + n on the real axis each of length 2r does not cover Un n the entire positive real axis. There are infinitely many circles with center at the origin and radii arbitrarily large which lie in R. k = rank of P(z), then ⎛ z ⎞ ⎛ z ⎞ | P(z) |= E⎜ ; k ⎟ E⎜ ; k ⎟ ∏ ⎜ a ⎟ ∏ ⎜ a ⎟ ....(1) rn ≤2r ⎝ n ⎠ rn >2r ⎝ n ⎠ p > ρ ρ+ε | P(z) |> e−r (ε > 0) ⎛ z ⎞ ⎛ z ⎞ log | P(z) | = log E⎜ ;k ⎟ + log E⎜ ;k ⎟ ⇒ ∑ ⎜ a ⎟ ∑ ⎜ a ⎟ rn ≤2r ⎝ n ⎠ rn >2r ⎝ n ⎠ k z z ⎛ z ⎞ 1 ⎛ z ⎞ ≥ log1− − + .... + ⎜ ⎟ . − log E⎜ ;k ⎟ ∑ a ∑ a ⎜ a ⎟ k ∑ ⎜ a ⎟ ....(2) rn ≤2r n rn ≤2r n ⎝ n ⎠ rn >2r ⎝ n ⎠ [on using triangle inequally | z1 + z2 |≥| z1 | − | z 2 | ] For |z| =r and rn ≤ 2r , we obtain 2 k k ⎧ k −1 k −1 ⎫ z 1 ⎛ z ⎞ 1 ⎛ z ⎞ ⎛ z ⎞ ⎪1 1 ⎛ an ⎞ ⎛ an ⎞ ⎪ + ⎜ ⎟ + ..... + ⎜ ⎟ ≤ ⎜ ⎟ ⎨ + ⎜ ⎟ + .... + ⎜ ⎟ ⎬ an 2 ⎝ an ⎠ k ⎝ an ⎠ ⎝ an ⎠ ⎩⎪k k −1⎝ z ⎠ ⎝ z ⎠ ⎭⎪ k ⎧ k −1 ⎫ ⎛ r ⎞ ⎪ 1 an ⎪ ≤ ⎜ ⎟ ⎨ + .... + ⎬ ⎝ rn ⎠ ⎩⎪ k z ⎭⎪ 1424 434 ( A1 which does not depend on r ) k ⎛ r ⎞ ⎜ ⎟ = A1 ⎜ ⎟ ....(3) ⎝ rn ⎠ 62 k k z 1 ⎛ z ⎞ ⎛ r ⎞ ∑ + ...... + ⎜ ⎟ ≤ A1 ∑ ⎜ ⎟ ∴ a k ⎜ a ⎟ ⎜ r ⎟ rn ≤2r n ⎝ n ⎠ rn ≤2r ⎝ n ⎠ 2ρ+ε / 2−k .r ρ+ε / 2+k = A 1 ∑ ρ+ε / 2−k ρ+ε / 2 k rn ≤2r 2 .r (rn ) 2ρ+ε / 2−k .r ρ+ε / 2 = A 1 ∑ ρ+ε / 2−k k rn ≤2r (2r) .(rn ) 1 = A 2ρ+ε / 2−k .r ρ+ε / 2 1 ∑ ρ+ε / 2−k k rn ≤2r (rn ) .(rn ) ⎛ 1 ⎞ = r ρ +ε / 2 ⎜ A 2 ρ +ε / 2−k ⎟ ⎜ 1 ∑ ρ +ε / 2 ⎟ ⎝ n=1 rn ⎠ 14244 4344 ( A2 which does not depend on r ) ρ+ε / 2 = A2 r ....(4) − p If k =0 the sum in (4) does not appear in (2). For z outside every circle | z − an |= rn with rn ≤ 2r . We have − p z | an − z | rn − p−1 − p−1 1− = = = rn ≥ (2r) an | an | rn Hence for all circles |z| =r in the region R, with r sufficiently large, ρ+ε log | P(z) |> −r ρ+ε ⇒| P(z) |> e −r 63 Hadanard’s Factorization theorem If F(z) is an entire function of finite order ρ, then the factorization F(z) =eh(z)zmP(z) is always possible where h(z) is a polynomial of degree ≤ρ, m≥o is the multiplicity of z =0 and P(z) is a canonical product of rank . Proof : According to Weierstrass factorization theorem an entire function F(z) can be factorize in the following from ...(1) When h(z) is an entire function and P(z) a product which may or may not be canonical. By the previous theorem ρc ≤ ρ, so that P(z) is of finite rank k ≤ ρ . ρ+ε Also since | P(z) |> e−r replacing ε by we have ρ+ε / 2 |≤FεPρ((zz))=|>eeh(−zr)zmP(z) on infinitely many circles |z| =r of arbitrarily2 large radius. ρ+ε/ 2 Aslo | F(z)| rρ+ε / 2 Reh(z) h(z) | F(z) | e e =| e |= < ρ+ε / 2 | z m P(z) | e−r ρ+ε / 2 ρ+ε = e2r < er on circles of arbitrarily large radius. Hence on such circles Reh(z) < rρ+ε ⇒ h(z) is a polynomial of degree not greater then ρ. ∞ ⎛ z 2 ⎞ sinπz = πz ⎜1− ⎟ Example : Show ∏⎜ 2 ⎟ by Hadaward Factorization Theorem. n=1 ⎝ n ⎠ ∞ 2n n sinπ z n π z Solution : Let F(z) = = ∑(−1) ....(1) π z n=0 (2n +1)! 64 π 2 z =1 − + ..... 3! ∞ n when F(z) = ∑cn z then by the formula n=0 1 log()1/ | cn | = lim ...(2) ρ n→∞ n log n 1 1 = 2 i.e. ρ = ...(3) ρ 2 Since zeros of F(z) are z =1, 4, ..., n2, ... we have sin π z ∞ ⎛ z ⎞ F(z) = = eh( z) 1− ∏⎜ 2 ⎟ ...(4) π z n=1 ⎝ n ⎠ But according to Hadamard Factorization theorem, h(z) must be a polynomial of degee ≤ order of F(z) i.e. degree of ⇒ h(z) = C (constant) 1 ∞ ⎛ z ⎞ h(z) ≤ F(z) eC 1 2 ∴ = ∏⎜ − 2 ⎟ n=1 ⎝ n ⎠ Since F(0) =1 ⇒ C =0 sinπ z ∞ ⎛ z ⎞ = F(z) = 1− ∴ ∏⎜ 2 ⎟ π z n=1 ⎝ n ⎠ Replacing z by z2 sin πz ∞ ⎛ z 2 ⎞ = ⎜1 − ⎟ ∏⎜ 2 ⎟ πz n=1 ⎝ n ⎠ 65