MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 5.. The Arzela-Ascoli Theorem.. The Riemann mapping theorem Let X be a metric space, and let F be a family of continuous complex-valued functions on X. We have the following definitions: () The family F is equicontinuous on a subset Y X if for every ɛ > 0 there exists δ > 0 so that for all p, q Y and all f F, if d X (p, q) < δ, then f(p) f(q) < ɛ. (2) The family F is normal on a subset Y X if for every sequence {f n } F, there exists a subsequence {f nj } which converges uniformly on every compact subset of Y. Also recall that a metric space X is separable if there is a countable subset {p j } X which is dense. Theorem. (Arzela-Ascoli). Let X be a separable metric space and suppose there are compact subsets K n X such that K n K n+, and n K n = X. Let F be a family of continuous complexvalued function on X. Then F is normal if and only if: (i) the family F is equicontinuous on every compact subset of X; (ii) for any p X, there is a constant C p > 0 so that f(p) C p for all f F. We shall actualy only need the sufficiency of the conditions (i) and (ii). Proof of necessity. To see the necessity of (i), we argue by contradiction. Suppose that F is a normal family, and that F is not equicontinuous on some compact subset K X. Then there exists ɛ > 0 so that for every n there are points p n, q n K and a function f n F such that d X (p n, q n ) < n and f n (p n ) f n (q n ) ɛ. Since F is normal, there is a subsequence {f nj } which converges uniformly on K to a limit f 0, which must then be continuous. We can select further subsequences (which we again call {n j } so that p nj p 0 and q nj q 0. But since d X (p nj, q nj ) < n j 0, it follows that p 0 = q 0. However, since f 0 is continuous which is a contradiction. ɛ lim sup f nj (p nj ) f jj (q nj ) = f 0 (p 0 ) f 0 (q 0 ) = 0, j To see the necessity of (ii), suppose there exists p X so that the set of values {f(p)} for f F is unbounded. Then for every n there exists a function f n F with f n (p) > n. But this clearly contradicts the hypothesis that there is a subsequence {f nj } which converges on the (compact) set {p} X. This completes the demonstration of the necessity of (i) and (ii). Date: April 26, 2009.
COMPLEX ANALYSIS 2 Proof of sufficiency. Now suppose that a family of functions F satisfies (i) and (ii), and let {f n } F be a sequence. Let {p n } be a countable dense subset of X. Since the set of values { f n (p ) } is contained in the compact set {ζ C ζ C p }, we can find a first subsequence n, < n,2 < < n,k < such that the sequence {f n,k (p )} converges, say to the value f 0 (p ). Next, since the set of values { f n,k (p 2 ) } is contained in the compact set {ζ C ζ C p2 }, we can find a second subsequence n 2, < n 2,2 < < n 2,k < such that the sequence {f n2,k (p 2 )} converges, say to the value f 0 (p 2 ), and the second subsequence is a subset of the first subsequence. Continuing in this way, we can find an infinite sequence of subsequences so that n, < n,2 < < n,k < n 2, < n 2,2 < < n 2,k < n 3, < n 3,2 < < n 3,k <.. n j, < n j,2 < < n j,k <.. (i) the limit as k of the sequence {f nj,k (p j )} exists (and equals f 0 (p j )); (ii) each subsequence {n j,k } is a subset of the pervious subsequence {n j,k }. It follows that for any fixed j, the elements of the diagonal sequence {n l,l } are a subset of the sequence {n j,k } for l j. Thus for the sequence of functions {F l = f nl,l }, we have lim l F l (p j ) = f 0 (p j ) for all j. We claim that this sequence {F l } converges uniformly on any compact subset K X. Since the sequence is equicontinuous on K, for any ɛ > 0 there exists δ > 0 so that d(p, q) < δ implies F l (p) F l (q) < ɛ. Now K q K B X(q, δ), and since K is compact, there is a finite set {q,..., q N } so that K N j= B X(q j, δ). Since the sequence {p k } is dense in X, for each j N, there is a point p jr B X (q j, δ). Since lim l F l (p jr ) = f 0 (p jr ), and since we are dealing with only a finite set of points, we can find M so that l, l 2 M implies F l (p jr ) F l2 (p jr ) < ɛ. Now let p K. Then p B X (p jr, δ) for some j r. We have F l (p) F l2 (p) F l (p) F l (p jr ) + F l (p jr ) F l2 (p jr ) + F l2 (p jr ) F l2 (p) ɛ + ɛ + ɛ. This shows that the sequence {F l } is uniformly Cauchy in the supremum norm on K, and it follows that the sequence converges uniformly on K. This completes the proof....2. Montel s theorem. 4/20/09 Theorem.2 (Montel). Let Ω C be open and connected, and let F be a family of holomorphic functions on Ω. Suppose that for every compact subset K Omega, there is a constant C K > 0 so that f(w) C z for every f F and all z K. Then the family F is normal on Ω. Proof. It follows from the Arzela-Ascoli theorem that it suffices to show that the family F is equicontinuous on every compact subset K Ω. For this it suffices to observe that the family of holomorphic functions bounded by M on a disk of radius R is equicontinuous on every smaller disk.
COMPLEX ANALYSIS 3 Let r < R, and choose ρ with r < ρ < R. If f is holomorphic on D(a, R) and bounded in absolute value by M, and if z, w D(a, r), then f(z) f(w) = 2πi ζ a =ρ ζ z dζ 2πi ζ a =ρ ζ w dζ z w = 2πi (ζ z)(ζ w) dζ This completes the proof. ζ a =ρ 2π z w M 2π z w 0 Mρ (ρ r) 2. ρ dt ρe it + a z ρe it + a w.3. Simply connected domains. Definition.3. A topological space X is simply connected if for every continuous function γ : [0, ] X with γ(0) = γ(), there exists a continuous mapping Γ : [0, ] [0, ] X such that () Γ(t, 0) = γ(t) for 0 t ; (2) Γ(0, s) = Γ(, s) = Γ(0, 0) for 0 s ; (3) Γ(t, ) = Γ(0, ) = Γ(, ) for 0 t. In other words, X is simply connected if and only if every continuous mapping of the circle into X passing through a given point x 0 X can be continuously deformed to the constant mapping whose image is {x 0 }. We are concerned primarily with open subsets of C. In this situation, we have Proposition.4. A connected open set Ω C is simply connected if and only if its complement in the Riemann sphere Ĉ has at most one connected component. Equivalently, the complement of Ω has no compact components. The topological condition of being simply connected has the following analytic consequence: Proposition.5. Suppose that Ω C is a connected, simply-connected open set. If f is holomorphic on Ω and f(z) 0 for all z Ω, then there is a holomorphic branch of log(f) defined on Ω. That is, there exists a holomorphic function g defined on Ω so that f(z) = exp(g(z)) for all z Ω. In particular, if n is any positive integer, the function h n (z) = exp [ n g(z)] is holomorphic and satisfies h n (z) n = f(z) for all z Ω..4. The Riemann Mapping Theorem. Proposition.6. Suppose that F : D D is a biholomorphic mapping of the unit diks such that F (0) = 0, and F (0) > 0. Then F (z) = z for all z D. Proof. Let G be the inverse mapping to F so that F (G(z)) = z and G(F (w)) = w for all z, w D. It follows from the chain rule that F (G(z)) G (z) =, and so F (0) G (0) =. On the other hand, F (z) <, G(z) < for all z D, and F (0) = G(0) = 0, so it follows from the Schwarz lemma that F (0) and G (0). Thus we must have F (0) =, and again by the Schwarz lemma, F (z) = e iθ z. But since F (0) > 0 we must have θ = 0, and so F (z) = z. 4/22/09
Proposition.7. Let a C with a <, and define COMPLEX ANALYSIS 4 φ a (z) = a z āz. Then φ a is a fractional linear transformation, and has the following properties: (i) φ a (0) = a and φ a (a) = 0; (ii) φ a φ a (z) = z; (iii) φ a : D D is a biholomorphic mapping. Proof. Statements (i) and (ii) are simple computations. We know that every fractional linear transformation is a one-to-one and surjective mapping of the Riemann sphere to itself. If we can show that φ a carries the unit circle to itself, statement (iii) will follow. But if z =, then φ a (z) 2 = a z āz and so φ a (z) =, which completes the proof. a z āz = a 2 āz a z + z 2 āz a z + a 2 z 2 = a 2 āz a z + āz a z + a 2 =, Theorem.8 (Riemann Mapping Theorem). Let Ω C be a non-empty, connected, simplyconnected open subset of C which is not all of C. Let z 0 Ω. Then there exists a unique biholomorphic mapping F : Ω D, the open unit disk, such that F (z 0 ) = 0 and F (z 0 ) > 0. Uniqueness follows from Proposition.6. In fact, if F and F 2 are two biholomorphic mappings from Ω to D carrying z 0 to zero, then F 2 F is a biholomorphic mapping of the unit disk which carries 0 to 0. The proof of the existence of a biholomorphic mapping proceeds in three steps. Let F denote the set of all holomorphic functions f : Ω D such that f is one-to-one, f(z 0 ) = 0, and f (z 0 ) > 0. Step : The set F is not empty. Since Ω is not the entire complex plane, there exists a C \ Ω. Then the function z a is holomorphic and never zero on Ω, and since Ω is simply connected, there exists a holomorphic function g on Ω such that g(z) 2 = z a. The function g is has the following properties: (i) g is one-to-one. In fact, if g(z ) = g(z 2 ), then z a = g(z ) 2 = g(z 2 ) 2 = z 2 a, and so z = z 2. (ii) The image of g does not contain any pair of points {w, w}. In fact, if g(z ) = g(z 2 ), then z a = g(z ) 2 = ( g(z 2 )) 2 = g(z 2 ) 2 = z 2 a, and so z = z 2, which yields g(z ) = g(z ). This implies g(z ) = 0, and hence z = a, which is impossible. (iii) The image of g contains some closed disk {w C w g(z0 ) δ} where δ > 0. It now follows from (ii) and (iii) that the range of the function g does not contain any point in the closed disk centered at g(z 0 ) of radius δ. Hence for every z Ω, we have Let g(z) + g(z 0 ) > δ. h(z) =: δ g(z) + g(z 0 ).
COMPLEX ANALYSIS 5 Then: (a) h is holomorphic on Ω; (b) for all z Ω, we have h(z) < ; (c) since the mapping w δ (w + g(z 0 )) is one-to-one, h is also one-to-one; (d) h(z 0 ) = δ (2g(z 0 )) = a D. We now have h : Ω D is holomorphic and one-to-one. We now compose the function h with the fractional linear transformation φ a and define F (z) =: φ a h(z) = a h(z) + āh(z). Then F (z 0 ) = φ a (a) = 0. If we multiply F by a suitable constant of absolute value, we can insure that the derivative at zero is positive. Thus F. Step 2: There exists F F such that F (z 0 ) = sup f (z 0 ). f F Let A = sup f F f (z 0 ), so that A (0, + ]. Choose a sequence {f n } in F so that lim n f n(z 0 ) = A. Since the family F is uniformly bounded on Ω, it is normal, and hence there is a subsequence {f nk } which converges uniformly on compact subsets of Ω to a limit F. Since F is the uniform limit of holomorphic functions, F is also holomorphic. We have F (z 0 ) = lim k f nk (z 0 ) = 0, and F (z 0 ) = lim k f n k (z 0 ) = A. (It follows that A < +. Also, if z Ω, F (z) = lim k f nk (z). But then it follows from the maximum modulus theorem that we actually have F (z) < for all z Ω. It remains to show that F is one-to-one. Suppose F (z ) = F (z 2 ) = w. The functions {f nk (z) w} converges uniformly on Ω to the function F (z) w, which by assumption has two zeros. It follows that for k large enough, f nk also has zeros near z and z 2, which implies z = z 2. Thus F F. 4/24/09 Step 3: The range of the function F is all of D. Suppose there exists a D such that F (z) a for all z Ω. Then the function z φ a F (z) is one-to-one, maps Ω to D, is never zero, and φ a F (z 0 ) = a. Since Ω is simply connected, there is a function g, defined and holomorphic on Ω, so that g(z) 2 = φ a F (z). Then g : Ω D, and as in the proof of Step, it follows that g is one-to-one. Moreover g(z 0 ) = b with b 2 = a. If we let S(w) = w 2, we have constructed a holomorphic function g on Ω so that S g = φ a F. Put G = φ b g. Then G is one-to-one, maps Ω to D, and satisfies G(z 0 ) = 0. We can write g = φ b G = φ b G, and we have or, finally, It follows that φ a F = S g = S φ b G, F = φ a S G = φ a S φ b G. F (0) = [ φ a S φ b ] (0) G (0). It remains to compute [ ] (0) φ a S φ b = φ a (a)s (b) φ b (0) and show that its absolute value is strictly less than. We have φ a(z) = ( āz)( ) (a z)( ā) ( āz) 2 = + āz + a 2 āz ( āz) 2 = a 2 ( āz) 2.
Thus and so φ a(a) = COMPLEX ANALYSIS 6 a 2 = b 4, φ b (0) = b 2, S (b) = 2b φ a(a)s (b) φ b (0) = 2b + b 2 which has absolute value strictly less than. It follows that F (z 0 ) < G (z 0 ), which contradicts the assumed maximality of F (z 0 ). This completes the proof. 2. Polynomial and Rational Approximation We address the following question; let K C be a compact set, and let Ω be an open neighborhood of K. If f is holomorphic on Ω, can f be approximated uniformly on K by rational functions or polynomials? Theorem 2. (Runge). Let K be a compact set, and let C\K = j U j where each U j is a connected open set. For each j let a j U j. Let R K denote the algebra of rational functions whose poles belong to the set {a,...}. () If Ω is any open neighborhood of K, f is any holomorphic function on Ω, and ɛ is any positive real number, there exists R R K such that f(z) R(z) < ɛ for all z K. (2) If C \ K is connected, if Ω is any open neighborhood of K, f is any holomorphic function on Ω, and ɛ is any positive real number, there exists a polynomial P so that f(z) P (z) < ɛ for all z K. 4/27/09 2.. Construction of a contour. Lemma 2.2. Let K C be compact, and let Ω K be an open neighborhood. There exists a finite collection {γ,..., γ N } of parameterized curves, each of which is either a horizontal or a vertical straight line segment, and an open set V Ω so that: (a) each γ j Ω \ L; (b) K V ; (c) if f is holomorphic on Ω and z V, then N f(z) = j= 2πi γ j ζ z dζ. Proof. The complement of Ω is a closed set which is disjoint from K, and so there exists δ > 0 so that z K and w / Ω implies z w > δ 2. We can tile the complex plane C (or actually R 2 ) with a collection S of closed squares with side length δ. For example, we can take all closed squares whose four vertices are points of the form {δ(m + in), δ(m + + in), δ(m + (n + )i), δ((m + ) + i(n + ))} where m, n Z. Any two of these squares are either disjoint, or intersect along a single edge, or intersect only at a vertex. Denote by S K the set of squares in S which intersect K. Since K is compact, S K consists of a finite collection of squares {Q,..., Q M }. From our construction, for each j we have Q j K, and Q j (C \ Ω) = so that Q j Ω. If z belongs to the interior of one of the squares Q k, it follows from the Cauchy integral representation that 2πi Q j ζ z dζ = { f(z) if j = k, 0 if j k.
Summing over all j, we obtain M f(z) = j= 2πi Q j COMPLEX ANALYSIS 7 M ζ z dζ = 4 j= i= 2πi Q i j dζ (2.) ζ z for any z in the interior of some square belonging to S K, where { Q i j, i 4} are the four edges of the square Q j. If any of the edges in (2.) appears twice, it occurs with opposite orientation, and thus the two corresponding integrals cancel. In particular, if any one of the edges Q i j has non-empty intersection with K, then that edge is actually an edge for two squares belonging to S K. Thus after canceling repeated edges, each of the remaining edges occurs exactly once and lies in Ω but does not intersect K. Let {γ,..., γ N } be an enumeration of these remaining edges. We then have f(z) = N j= 2πi γ j dζ (2.2) ζ z for any z belonging to the interior of any of the original squares in S K. By continuity, formula (2.2) continues to hold for any z V =: M j= Q j \ N k= γ k. This completes the proof. We now start on the proof of Theorem 2.. We begin with the following: Lemma 2.3. Let K C be a compact set, let Ω K be an open neighborhood, and let f be holomorphic on Ω. Then we can uniformly approximate f on K by rational functions whose poles lie in the complement of K. Proof. Let {γ j } be the line segments constructed in Lemma 2.2. For all z K we have N N f(z) = 2πi ζ z dζ = f j (z) where f j (z) = 2πi ζ z dζ. j= γ j j= Let γ j have the parameterization γ j (t) for 0 t, so that M k= f j (z) = 2πi 0 f(γ j (t)) γ j (t) z γ j(t) dt. The function F j (t, z) = f(γ j (t)) 2πi γ j (t) z γ j(t) is jointly continuous on [0, ] K, and so M ( ) k lim M F j M, z = F j (t, z) dt = f j (z) and, what is critical for us, the convergence is uniform on K. But M M M F j (k/m, z) = 2πiM f(γ j (k/m)) γ j (k/m) (γ j (k/m) z) k= k= is a rational function whose poles lie on the curve γ j which is contained in the complement of K. Since we can uniformly approximate each f j, we can also uniformly approximate the finite sum f, and this completes the proof. We can now move the location of the poles of the approximating rational functions using the following. Lemma 2.4. Let K be a compact set, and let γ : [0, ] C be a piecewise smooth curve whose image is in the complement of K. For any n, the function (z γ(0)) n can be uniformly approximate on K by rational functions whose only poles are at the point γ(). 0 γ j
COMPLEX ANALYSIS 8 Proof. Since K and the image of γ are both compact and are disjoint, there exists η > 0 so that if z K and w γ, then z w > 3η. Let a γ, let b C with a b < η, and let z K. Then z b = z a + a b z a a b > 3η η = 2η. Thus (z a) = ( (z b) (a b) ) = (z b) = (z b) ( ) a b k = z b [ a b ] z b (a b) k (z b) k+, and this series converges uniformly on K. In particular, given any ɛ > 0 there exists N 0 depending on ɛ but independent of a and b such that if N N 0 and z K, N (z (a b) k a) (z b) k+ < ɛ. It follows that if a, b γ and if a b < η, then rational functions with poles at a can be uniformly approximated on K by rational functions with poles at b. But then we can cover γ by a finite number of disks of radius η 2, and the Lemma follows. Finally, to show that if the complement of K has only one component, we can approximate by polynomials, we use the following. Lemma 2.5. Let K C be a compact set. Suppose that if z K then z M. Then if a > 2M, the function (z a) can be uniformly approximated on K by polynomials. Proof. Let z K. Then (z a) = a [ z a ] = a ( z k = a) a k+ zk, and this series converges uniformly on K. But then (z a) can be uniformly approximated on K by finite partial sums, which are polynomials. This completes the proof.