But11NN+1diverges to infinity(that is,thepartial sum →oo),so we can choosep,such that11>2eNN+pNextwechoosesocloseto1 tahtrN+p>.Then2N+p11N+pNNN+1N+pVAconstractionisobtainedHowever,notethat f(z)is still continuous onAbecauseit is continuousat each z,sinceeach zlies in some Ar on which wehaveuniformconvergence.FCV&IT13/50Uni.ofSci&Tech)August 19,2019Ming Li (C)
Seqences of complex numbers But 1 N + 1 N + 1 + · · · diverges to infinity (that is, the partial sum → ∞), so we can choose p, such that 1 N + · · · + 1 N + p > 2ε Next we choose x so close to 1 taht x N+p > 1 2 . Then x N N + · · · + x N+p N + p > xN+p 1 N + 1 N + 1 + · · · + 1 N + p > ε A constraction is obtained. However, note that f(z) is still continuous on A because it is continuous at each z, since each z lies in some Ar on which we have uniform convergence. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 13 / 50
niimherTheorem (4.1.6)If fn are continuous function defined on A and f(z)=k- fi(z)converges uniformly on A,f is continuous on A.Proof.We wish to show that for zo E A, given > 0, there is a >0 such thatz-zo<implesthatf(z)-f(zo)/<e.O f(z) =k=1 fi(z) converges uniformly on A → N, forSn(z) =k=1 fk(z) such that [Sn(z) - f(z)/ <e/3 for all zE A.S(z)iscontinuous=thereisa>Osuchthat[S(z) -Sn(z0)[<e/3 if |z - z0|< .OIf(z) - f(z0)/< If(z) - S(z)/+ |SN(z) - S(z0)/ +|S(z0) - f(z0eee=E33+3口FCV&ITAugust 19.201914/50Uni.ofSci&Tech)GLI
Seqences of complex numbers Theorem (4.1.6) If fn are continuous function defined on A and f(z) = P∞ k=1 fk(z) converges uniformly on A, f is continuous on A. Proof. We wish to show that for z0 ∈ A, given ε > 0, there is a δ > 0 such that |z − z0| < δ imples that |f(z) − f(z0)| < ε. 1 f(z) = P∞ k=1 fk(z) converges uniformly on A ⇒ ∃N, for SN (z) = PN k=1 fk(z) such that |SN (z) − f(z)| < ε/3 for all z ∈ A. 2 SN (z) is continuous ⇒ there is a δ > 0 such that |SN (z) − SN (z0)| < ε/3 if |z − z0| < δ. 3 |f(z) − f(z0)| < |f(z) − SN (z)| + |SN (z) − SN (z0)| + |SN (z0) − f(z0)| < ε 3 + ε 3 + ε 3 = ε Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 14 / 50
nmherTheorem (4.1.7)Let C: Ja,b]→Abe a curve in a domain A and let fn be a sequence ofcontinuous functions defined on C converging uniformly to f on C.Then( fn(a)dz → / f(a)dzSimilarly,if,gk(z)converges uniformly on C,then(2g(2) dz=2 /,g(2)d2Proof.Given >O.wecan chooseN,suchthatn>N,implesthat[fn(z)-f(z)/<eforall zon C.Thenfn(2)dz - [ f(2)dz≤ / 1fn(2) - f(2)]ds<eL.August19,201915/50FCV&IT&Tech)
Seqences of complex numbers Theorem (4.1.7) Let C : [a, b] → A be a curve in a domain A and let fn be a sequence of continuous functions defined on C converging uniformly to f on C. Then Z C fn(z)dz → Z C f(z)dz Similarly, if P∞ k=1 gk(z) converges uniformly on C, then Z C X∞ k=1 gk(z) ! dz = X∞ k=1 Z C gk(z)dz Proof. Given ε > 0, we can choose N, such that n > N, imples that |fn(z) − f(z)| < ε for all z on C. Then Z C fn(z)dz − Z C f(z)dz ≤ Z C |fn(z) − f(z)|ds < εL. Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 15 / 50
numhersTheorem (4.1.8)() Let D bea domain in C and let fn be a sequence of analyticfunctionsdefined on D.If fn→f isuniformon everycloseddiskcontained in D, then f isanalytic.Furthermore,f→f'ispointwiseonDanduniformon everyclosed disk in D(i) If gn are analytic functions on D and g(z) =k=1 gk(z) convergesuniformly on every closed disk in C, then g is analytic on D andg'(z)=k=1gi(z) is pointwise on D and also uniform on everycloseddiskcontainedonDProof.(1)Letzo EDand letB=(z:|z- zol ≤r) CD.Wewish toshowthat f isanalyticon D(zo,r)={z:z-zol<rl.Itis clearthatfn→fisuniformonD(zo,r)Theorem4.1.6=f is continuousonD(zo,r) Let C be a closed curve in D(z0, r), we have fc fn = 0.Theorem 4.1.7=→ fc fn → fc f. so fc f =0.By Morera theorem, f is analytic on D(zo,r).FCV&ITAugust 19, 201916/50ngLi(Changsha Uni.of Sci&Tech)
Seqences of complex numbers Theorem (4.1.8) (i) Let D be a domain in C and let fn be a sequence of analytic functions defined on D. If fn → f is uniform on every closed disk contained in D, then f is analytic. Furthermore, f 0 n → f 0 is pointwise on D and uniform on every closed disk in D. (ii) If gn are analytic functions on D and g(z) = P∞ k=1 gk(z) converges uniformly on every closed disk in C, then g is analytic on D and g 0 (z) = P∞ k=1 g 0 k (z) is pointwise on D and also uniform on every closed disk contained on D. Proof. (1) Let z0 ∈ D and let B = {z : |z − z0| ≤ r} ⊂ D. We wish to show that f is analytic on D(z0, r) = {z : |z − z0| < r}. It is clear that fn → f is uniform on D(z0, r). Theorem 4.1.6 ⇒ f is continuous on D(z0, r). Let C be a closed curve in D(z0, r), we have H C fn = 0. Theorem 4.1.7 ⇒ H C fn → H C f, so H C f = 0. By Morera theorem, f is analytic on D(z0, r). Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 16 / 50
ForanyzEB=[z:z-zol≤r] CD,by Cauchy integral formula1Tfn(S)1fn(S)1()=2 Je(-d(a)=2 J(-dBy hypothesis, fn f is uniform on {z:z- zol≤ p] C D, here p > r.Then,given >O,wepick N,such that n≥N,imples thatIfn(z) -f(z)| < e, for all z in the disk. Since C : [z : [z -zol = p) is theboundary of this disk, n ≥ N impliesfn()-f(S)l< on C. By Cauchyintegral formula fn(C) -f(C)dIfn(z) - f(z)/ =(C - z)22元iJcandobservethatforSonCandzEB,IS-z≥p-r.Hencen≥NimpliesthatepE1%(2)-F(2)≤2元(p-· L=(p-r)2FCV&ITAugust 19,201917/50MineLileUni.ofSci&Tech)
Seqences of complex numbers For any z ∈ B = {z : |z − z0| ≤ r} ⊂ D, by Cauchy integral formula f 0 n (z) = 1 2πi Z C fn(ζ) (ζ − z) 2 dζ f0 (z) = 1 2πi Z C fn(ζ) (ζ − z) 2 dζ By hypothesis, fn → f is uniform on {z : |z − z0| ≤ ρ} ⊂ D, here ρ > r. Then, given ε > 0, we pick N, such that n ≥ N, imples that |fn(z) − f(z)| < ε, for all z in the disk. Since C : {z : |z − z0| = ρ} is the boundary of this disk, n ≥ N implies |fn(ζ) − f(ζ)| < ε on C. By Cauchy integral formula |f 0 n (z) − f 0 (z)| = 1 2πi Z C fn(ζ) − f(ζ) (ζ − z) 2 dζ and observe that for ζ on C and z ∈ B, |ζ − z| ≥ ρ − r. Hence n ≥ N implies that |f 0 n (z) − f 0 (z)| ≤ 1 2π · ε (ρ − r) 2 · L = ερ (ρ − r) 2 Ming Li (Changsha Uni. of Sci & Tech) FCV & IT August 19, 2019 17 / 50