Analytic Continuation

Transcription

1 Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for all z 1 ad is i fact differetiable, ad thus aalytic for all such z. Also H (z) = h (z) for all z < 1, whe we say that H is a aalytic cotiuatio of h. Ufortuately this example is too good i that a aalytic cotiuatio to all of C \ {1} was foud so easily. But we will cotiue with this example to fid the aalytic cotiuatio usig Taylor Series. Retur to the example h (z) = 1 + z + z 2 + z which coverges i z < 1. Take ay a D (0, 1), a 0. For the Taylor Series for h (z) aroud a, we ca use the fact that close if z is sufficietly close to a the z < 1 ad we have h (z) = 1/ (1 z) so h () (a) =! (1 a) +1. Thus the Taylor Series for h (z) aroud a is (z a) +1. (1) (1 a) This was calculated subject to z beig sufficietly close to a, but it is a geometric series which therefore coverges for z a 1 a < 1, i.e. z a < 1 a. (The radius of covergece is the distace of a to the sigularity of h at z = 1.) So (1) is a aalytic cotiuatio of h (z) to D (a, 1 a ). This process is oly of iterest if the ew disc cotais poits ot i the origial. If Re a < 0 say, the 1 a 2 = (Re (1 a)) 2 + (Im (1 a)) 2 > 1. I which case D (a, 1 a ) will cotai poits ot i D (0, 1). To cotiue assume z 0 C \ D (0, 1) ad cosider two cases. I the first assume z Choose a path γ from 0 to z 0 such that all poits z γ satisfy z 1 1. Cover the path by a sequece of ope discs D (a i, 1) such that a i γ, 1 i, a 1 = 3/4, a = z 0 ad 1

2 a i+1 D (a i, 1). Apply the idea step-by-step. So at the ed of the i-th step you will a Taylor Series defiig a fuctio f i o D (a i, 1). I priciple (i.e. it may be hard i practice) you ca calculate f () i (a i+1 ), costruct f i+1 which will the coverge o D (a i+1, 1) (if ot a larger disc). The coclusio is that at the ed of the sequece get a cotiuatio to D (z 0, 1), ad i particular z 0. I the secod case z 0 1 < 1 with z 0 1 whe there exists z 1 : z ad z 0 z 1 < 1. I particular, if z 0 = δe iθ + 1 for some 0 < δ < 1 choose z 1 = (1 + δ/2) e iθ + 1. Repeat the method above to fid a cotiuatio to D (z 1, 1). Sice z 0 D (z 1, 1) we have a cotiuatio to z 0. Combiig both cases we see that we have a cotiuatio to all poits of C \ {1}. The questio is, is the cotiuatio at each step uique? Lemma 1 Assume F (z) is aalytic i a domai D cotaiig the overlappig ope discs D (z 0, r 0 ) ad D (z 1, r 1 ) where z 1 D (z 0, r 0 ). If F is idetically zero o D (z 0, r 0 ) the it is idetically zero o D (z 1, r 1 ). Proof Write the Taylor Series of F about z 0 ad z 1 respectively, as F (z) = (z z 0) m ad F (z) = r=0 b r r! (z z 1) r. The Taylor series of a holomorphic fuctio f at a poit a has radius of covergece equal to the distace from a to the earest sigularity of f. Sice F is aalytic i D ay sigularities of F lie outside of D ad thus outside of both D (z 0, r 0 ) ad D (z 1, r 1 ). Hece these series have radii of covergece respectively r 0 ad r 1. For z D (z 0, r 0 ) D (z 1, r 1 ) write z z 0 = (z z 1 ) + (z 1 z 0 ) so Hece (z z 0 ) m = m ( ) m (z z 1 ) (z 1 z 0 ) m. 2

3 (z z 0) m = = m ( ( ) m (z z 1 ) (z 1 z 0 ) m ( ) ) m (z 1 z 0 ) m (z z 1 ), havig iterchaged summatios. Thus i D (z 0, r 0 ) D (z 1, r 1 ) ( ( ) ) m (z 1 z 0 ) m (z z 1 ) b ad! (z z 1) represet the same fuctio, F (z). Our fial assumptio is that z 1 D (z 0, r 0 ) D (z 1, r 1 ) i which case the exists d > 0 such that Hece for all 0. Fially, b = 1 2πi C d D (z 1, d) D (z 0, r 0 ) D (z 1, r 1 ). F (z) +1 dz = (z z 1 ) = F (m) (z 0 ) = 1 2πi C δ ( ) m (z 1 z 0 ) m, (2) F (w) (w z 0 ) m+1 dw where C δ is ay circle cetered z 0 lyig withi the regio i which F is aalytic. If we choose δ < r the F (w) = 0 for all w C δ i which case = 0 for all m 0. By (2) this implies b = 0 for all 0 ad thus F (z) 0 o D (z 1, r 1 ). We ca repeat this lemma alog a sequece of discs. Ulike the earlier example we do t split ito two cases but choose a radius for the discs differet to 1. Theorem 2 Assume F (z) is aalytic i a path coected domai D cotaiig the ope disc D (z 0, r) o which F is idetically zero. The F is idetically zero o D. 3

4 Proof Let z D. Sice D is path coected there exists a path γ from z 0 to z. Further, D is a ope set so there is iimum distace, δ say, from γ to the boudary of D. Cover γ with a sequece of discs D (c i, γ/2), c i γ, 0 i, c 0 = z 0, c = z ad c i+1 D (c i, γ/2). Repeatedly apply the lemma to deduce that F is idetically zero o D (z, γ/2) ad i particular at z. Sice z was arbitrary F is idetically zero o D. Corollary 3 Assume f (z) ad g (z) are aalytic i a path coected domai D cotaiig the ope disc D (z 0, r). Assume f (z) = g (z) for all z D (z 0, r). The f (z) = g (z) for all z D. Proof Apply Theorem to F (z) = f (z) g (z). Example 4 ζ (s) = 1 s s (s + 1) 2 is the uique aalytic cotiuatio of to Re s > 1, s 1. ζ (s) = =1 1 ψ 2 (u) du, (3) us+2 1 s, (4) Corollary 3 says that if two aalytic fuctios agree o a disc they agree i a larger path-coected domai o which they are both defied. Far more tha this is satisfied by the series ad itegral form of the Riema zeta fuctio, (4) ad (3). They agree o a half-plae Re s > 1. Yet a disc is a far larger set tha is ecessary. Theorem 5 Assume that F (z) is aalytic i a path coected domai D. Assume there exists a coverget sequece {z i } i 1 of poits of D with limit poit z l also i D for which F (z i ) = 0 for all i 1 ad F (z l ) = 0. The F (z) = 0 for all z D. Proof. It suffices to show that F is zero o some disc cetred o z l for the result will the follow from Theorem 2. For cotradictio assume that F is ot idetically zero i ay disc aroud z l. This meas that the Taylor 4

5 Series aroud z l is ot idetically zero. Let m be the smallest iteger with F (m) (z l ) 0. Thus ) (z z l ) +... F (z) = (z z l ) m ( F (m) (z l ) = (z z l ) m h (z), + F (m+1) (z l ) (m + 1)! say, where h (z) is aalytic at z = z l ad o-zero at z l. Sice h is aalytic at z l it is cotiuous at z l. Choosig ε = F (m) (z l ) /2 i the defiitio of cotiuity there exists δ > 0 such that if 0 < z z l < δ the h (z) F (m) (z l ) < F (m) (z l ) 2. By the triagle iequality h (z) > F (m) (z l ) 2. I particular h (z) > 0 i the puctured disc 0 < z z l < δ. Yet z l is the limit poit of the sequece {z i } i 1 ad, from the defiitio of covergece, there exists such that 0 < z z l < δ. sice F (z ) = 0 we get the required cotradictio. Corollary 6 Assume that F (z) ad G (z) are aalytic i a path coected domai D. Assume there exists a coverget sequece {z i } i 1 of poits of D with limit poit z l also i D for which F (z i ) = G (z i ) for all i 1 ad F (z l ) = G (z l ). The F (z) = G (z) for all z D. Note I the above I have made two assumptios without proof. i. The Taylor Series for a fuctio f aalytic at c has radius of covergece equal to the distace of c from the earest sigularity of f. ii. If γ is alie betwee two poits lyig wholly withi a ope set the the earest distace from γ to the boudary of D is o-zero. That is mi { z w : z γ, w D c } > 0. This is true sice γ is a closed ad bouded set (i.e. compact), D c is closed ad γ D c = φ. 5