POWER SERIES AND ANALYTIC CONTINUATION
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1 POWER SERIES AND ANALYTIC CONTINUATION 1. Analytic functions Definition 1.1. A function f : Ω C C is complex-analytic if for each z 0 Ω there exists a power series f z0 (z) := a n (z z 0 ) n which converges to f(z) for all z in some disc D r (z 0 ) Ω. Real-analytic functions are defined analogously, except the domain Ω R is a subset of the reals and the two-dimensional disc D r (z 0 ) is replaced by 1-dimensional open interval in Ω. Unless stated otherwise, we consider functions of a complex variable and use the term analytic to mean complex-analytic. Theorem 1.1. Holomorphic functions are analytic. More precisely, if f : Ω C C is holomorphic (complex-differentiable) and D r (z 0 ) Ω is a closed disc contained in Ω, then f has a power series expansion f (n) (z 0 ) (z z 0 ) n for all z D r (z 0 ). In particular, if f : C C is entire, its power series expansion converges to f for all z C, uniformly on compact subsets of C. Remark. The theorem is false for real differentiable functions; in HW6 you encounter an example of a smooth (infinitely-differentiable) function of a real variable which is not real analytic. Proof. By Cauchy s integral formula for the closed disc D r (z 0 ), for each z D r (z 0 ) we have 1 2πi D r(z 0) ζ z dζ = 1 2πi D r(z 0) (ζ z 0 )(1 z z0 ζ z 0 ) dζ = 1 ( z z0 ) n dζ. 2πi ζ z 0 ζ z 0 D r(z 0) Since z z 0 < ζ z 0 = r, the geometric series converges uniformly in ζ D r (z 0 ), so we may interchange the sum and integral to obtain (z z 0 ) n[ 1 ] 2πi (ζ z 0 ) n+1 dζ f (n) (z 0 ) = (z z 0 ) n, D r(z 0) where we used the CIF for derivatives in the last step. 1
2 2 POWER SERIES AND ANALYTIC CONTINUATION Remark. Note that in the above theorem, the same power series f (n) (z 0 ) (z z 0 ) n converges to f(z) on every disc D r (z 0 ) Ω. In particular, its radius of convergence is at least sup{r > 0 : D r (z 0 ) Ω}. However it could conceivably converge on a larger disc D R (z 0 ) not entirely contained in Ω. In this case, however, there is no guarantee that it will converge to f(z) for all z D R (z 0 ) Ω. In the next section we shall characterize more precisely its radius of convergence. Definition 1.2. We say f has a zero of order N at z 0 if f (k) (z 0 ) = for all 0 k n 1 but f (n) (z 0 ) 0. Lemma 1.2. Assume f : Ω C is analytic. If z 0 Ω is a zero of order N, then there exists a open disc D r (z 0 ) Ω and a nonvanishing analytic function g : D r (z 0 ) C such that (z z 0 ) N g(z) for all z D r (z 0 ). Corollary 1.3. If f has a zero of order N at z 0, then there exists a disc D r (z 0 ) which contains no other zeros of f. Proof. Indeed, with D r (z 0 ) as in the lemma above, both factors in the product are nonzero when z D r (z 0 ) \ {z 0 }. Theorem 1.4. Let Ω C be open and connected, and let f : Ω C be analytic. If f 1 (0) has a limit point in Ω, then 0 for all z Ω. Proof. We begin with a Lemma. Let U C be any open set and f : U C be analytic. Then the set is both open and closed relative to U. K := {z U : f (n) (z) = 0 for all n}. Proof of lemma. Since f is locally represented by a power series, for each z 0 K there exists a disc D r (z 0 ) U such that f (n) (z 0 ) (z z 0 ) n = 0 forall z D r (z 0 ). Therefore f (n) (z) = 0 for all z D r (z 0 ), so D r (z 0 ) K and K is open. On the other hand, if z 0 U \ K, by definition there exists some integer n such that f (n) (z 0 ) 0. Since f (n) is holomorphic and in particular continuous, there exists a small ball D ε (z 0 ) U such that that f (n) (z) 0 for all z D ε (z 0 ). In other words D ε (z 0 ) U \ K so U \ K is open, and K is closed relative to U. Now if we could show that K is nonempty, the connectedness assumption would imply that K = Ω so that f is identically zero. Let z 0 be a limit point of f 1 ({0}); we claim that z 0 K. For if not, there exists an integer N such that z 0 is a zero of order N. But the above Corollary then implies that there exists a disc centered at z 0 which contains no other zeros of f, contradicting the choice of z 0 as a limit point of f 1 ({0}). This completes the proof of the theorem.
3 POWER SERIES AND ANALYTIC CONTINUATION 3 Corollary 1.5. Let Ω C be open and connected. Suppose f 1, f 2 : Ω C are analytic and that {z Ω : f 1 (z) = f 2 (z)} has a limit point in Ω. Then f 1 (z) = f 2 (z) for all z Ω. Proof. Apply the previous theorem with f = f 1 f Analytic Continuation Definition 2.1. Let Ω 0 C be any set and f : Ω 0 C a function. We say a function f : Ω C C is an analytic continuation or analytic extension of f if Ω 0 Ω, f is analytic, and f Ω0 = f (that is, f(z) for all z Ω 0 ). Lemma 2.1. If Ω is open and connected, Ω 0 Ω, and Ω 0 has a limit point in Ω, then f : Ω 0 C has at most one analytic continuation to Ω. Proof. If f, g : Ω C are both analytic extensions of the same f, then by continuity they agree on the closure Ω 0 Ω which contains a limit point, so the claim follows from the last Corollary in the previous section. Examples. The complex exponential exp : C C exp(x + iy) := e x (cos(y) + i sin(y)) is the unique analytic extension of the real exponential function x e x, x R to the complex plane. On the other hand, if Ω C is an open set containing R which is not connected, then there exist multiple analytic continuations of the real exponential function x e x to Ω. Let f(x) = a n (x x 0 ) n be a real power series with radius of convergence R = (lim sup n a n 1/n ) 1 > 0; thus it converges to a smooth function f : (x 0 R, x 0 + R) R R. Then, replacing x with the complex variable z yields a complex power series which converges to a holomorphic function on the complex open disc D R (z 0 ) = {z C : z x 0 < R}. Thus F : D R (x 0 ) C, F (z) = is an analytic extension of f, and is the unique analytic extension of f to the disc D R (x 0 ) as D R (x 0 ) is connected. Using the technique of the last example, we can quickly write down power series expansions of some analytic functions. For instance, we know that the real exponential function x e x has power series expansion e x = x n, for all x R.
4 4 POWER SERIES AND ANALYTIC CONTINUATION Replacing the left side with the complex exponential exp(z) and the right side with the complex power series zn, we conclude that z n exp(z) = for all z C, since both sides are analytic on C and agree on the real line. Definition 2.2. Let Ω 0 C be a set and f : Ω 0 C a function. For z 0 Ω 0, we say an analytic function f : Ω C is a (local) analytic extension/continuation of f at z 0 if there exists an open disc D r (z 0 ) Ω 0 Ω such that f Dr(z 0) = f Dr(z 0); in other words, f is an analytic continuation of the restriction of f Dr(z 0) to some neighborhood D r (z 0 ) of z 0. Remarks. Note that Ω is not required to contain all of Ω 0 ; we merely require that Ω 0 Ω contain an open disc. Note that by the uniqueness theorem for analytic continuations, we know that f(z) for all z in the connected component of Ω 0 Ω containing z 0. However, there is no guarantee that f agrees with f on other components of Ω 0 Ω. Example 1. Define f : D 1 (0) (C \ D 1 (0)) C by { 0, z < 1 z, z > 1. Then f is analytic. The function 0, z C is an analytic continuation of f at z = 0; indeed f(z) for all z < 1. However f(z) f(z) for z > 1. Example 2. Let Log : C \ (, 0] C be the principal logarithm Log(z) := log z + i arg ( π,π) (z), and let log (0,2π) : C \ [0, ) C be the branch of log defined by log (0,2π) (z) := log z + i arg (0,2π) (z). then log (0,2π) is an analytic continuation of Log at any point z 0 with Im(z 0 ) > 0. Indeed log (0,2π) (z) = Log(z) for all such z. However log (0,2π) (z) Log(z) for any z in the lower half plane. We return to the radius of convergence of power series for analytic functions. Lemma 2.2. Let f : Ω C be analytic and for z 0 Ω, let f f(z) (n) (z 0 ) := (z z 0 ) n be the power series expansion of f at z 0. Let R z0 denote the radius of convergence of this power series. Then f : D Rz0 (z 0 ) C is an analytic continuation of f at z 0 (but need not be a (global) analytic continuation of f : Ω C). Moreover, R z0 = sup{r > 0 : F : D R (z 0 ) C an analytic continuation of f at z 0 }.
5 POWER SERIES AND ANALYTIC CONTINUATION 5 Proof. The first bullet point was already proved in the Theorem at the beginning of these notes. To provve the second bullet point, note that follows immediately from the first bullet point. To prove, let F : D R (z 0 ) C be an analytic continuation of f at z 0. By definition there exists some disc D r (z 0 ) Ω such that F Dr(z 0) = f Dr(z 0). In particular, F (n) (z 0 ) = f (n) (z 0 ) for all n. Now as F is itself analytic, by the Theorem at the beginning of these notes we have F (z) = F (n) (z 0 ) (z z 0 ) n = f (n) (z 0 ) (z z 0 ) n for all z D R (z 0 ). Therefore the power series for f has radius of convergence at least R. Example 3. Let 1 z 2 +1 for z Ω := C \ {±i}. Then its power series at z 0 = 0 has radius of convergence 1. To see this, note that f : Ω C is a local analytic continuation of itself at 0, and D 1 (0) Ω. On the other hand, f admits no analytic extension to a larger disc D r (0) for r > 1. Indeed, if f : D r (0) C were such an extension, then f would agree with f on D r (0) \ {±i} by the uniqueness of analytic extensions on connected domains. But f would be bounded near z = ±i, while lim z ±i f(z) =.
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