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Math 246A
Homework 8 Solutions
Due Thursday, June 7
Note the solutions are often just an outline, but I have tried to address all of the key points and you should be able to fill in the details. If you have questions about them feel free to ask me; also if there are any errors in my solutions please let me know. Of course there are usually many other correct ways to solve each problem as well.
8 Find a harmonic function u in the open first quadrant that extends continuously up to the boundary except at the points 0 and 1, and that takes on the following boundary values: u(x, y) = 1 on the half-lines {y = 0, x > 1 } and {x = 0, y > 0 }, and u(x, y) = 0 on the segment { 0 < x < 1 , y = 0}. [Hint: Find conformal maps F 1 , F 2 ,... , F 5 indicated in Figure 11. (See page 250 of Stein and Shakarchi. If you don’t have the text ask a classmate or the TA to look at their copy.) Note that 1 π arg(z) is harmonic on the upper half-plane, equals 0 on the positive real axis, and 1 on the negative real axis. Check that the following conformal maps work, including mapping the boundaries properly. Then let u be the real part of (^) πi^1 log ◦F 5 ◦ F 4 ◦ F 3 ◦ F 2 ◦ F 1. All logarithms are the principal branch.
F 1 =
z − 1 z + 1 F 2 = log z F 3 = −iz − π/ 2 F 4 = sin z F 5 = z − 1
9 Prove the function defined by
u(x, y) = Re (
i + z i − z
) and u(0, 1) = 0
is harmonic in the unit disc and vanishes on the boundary. Note that u is not bounded in D. Clearly i i+−zz is holomorphic on D (its only pole in C is at i which is not in D) and thus being the real part of a holomorphic function it is harmonic. Looking at u(0, t) with 1 − < t < 1 shows that u is unbounded, now lets consider the points on the boundary and show they vanish. Note we can assume x + iy 6 = i by hypothesis, so assume w = ieiθ^ and 0 < θ < 2 π, then i i+−ww = 1+e
iθ 1 −eiθ^ =^
eiθ/^2 +e−iθ/^2 eiθ/^2 −e−iθ/^2 =^
i cos(θ/2) sin(θ/2) which is well-defined and totally imaginary for appropriate θ showing u vanishes on the boundary.
11 Show that if f : D(0, R) → C is holomorphic, with |f(z)| ≤ M for some M > 0, the
f(z) − f(0) M 2 − f(0)f(z)
|z| M R
[Hint: Use the Schwarz lemma] Consider g(w) = f^ ( MRw )a holomorphic function from the disc into the closed disc D. Using the open mapping theorem we can in fact see that it must map into the open disc D. Let ψα(z) = 1 α−−αzz , which is an automorphism of the disc, then ψg(0) ◦ g maps D to D and takes 0 to 0 so we can apply Schwarz to find | (^) Mf 2 (−Rwf (0))−ff (^ (0)Rw) | ≤ | Mw| for all w ∈ D. Letting z = Rw gives the result.
Problem 7 Applying the ideas of Carathe´odory, Koebe gave a proof of the Riemann mapping theorem by constructing (more explicitly) a sequence of function that converges to the desired conformal map. Starting with a Koebe domain, that is, a simply connected domain K 0 ⊂ D that is not all of D, and which contains the origin,the strategy is to find an injective function f 0 such that f 0 (K 0 ) = K 1 is a Koebe domain ”larger” than K 0. Then one iterates this process, finally obtaining functions Fn = fn ◦... ◦ f 0 : K 0 → D such that Fn(K 0 ) = Kn+1 and lim Fn = F is a conformal map from K 0 to D. The inner radius of a region K ⊂ D that contains the origin is defined by rK = sup{ρ ≥ 0 : D(0, ρ) ⊂ K}. Also, a holomorphic injection f : K → D is said to be an expansion if f(0) = 0 and |f(z)| > |z| for all z ∈ K − { 0 }.
(a) Prove that if f is an expansion, then rf (K) ≥ rK and |f′(0)| > 1. [Hint: Write f(z) = zg(z) and use the maximum principle to prove that |f′(0)| = g(0) > 1.] Following the hint, note g is holomorphic since f(0) = 0 and g(0) = f′^ (0). Furthermore |g(z)| = |f(z)/z| > 1 for z 6 = 0 since f is an expansion so in particular |g(0)| ≥ 1 and g is non-vanishing on K. If we assume |g(0)| = 1 and apply the maximum principle to 1 /g (which is holomorphic) then g must have a maximum at 0 which implies g is a constant of modulus 1 which contradicts that f is an expansion, so |g(0)| = |f′(0)| > 1. To show rf (K) ≥ rK its enough to show D(0, r) ⊂ f(D(0, r)) for any r < rf (K). This follows since |f(z)| > |z| and f is injective since then f(∂D(0, r)) is a simple closed curve with all points of absolute value greater than r and since f(D(0, r)) is simply connected and f(0) = 0 this implies that f(D(0, r)) is the interior of f(∂D(0, r)) which contains D(0, r).
Suppose we begin with a Koebe domain K 0 and a sequence of expansions {f 0 , f 1 ,... , fn,.. .}, so that Kn+1 = fn(Kn) are also Koebe domains. We then define holomorphic maps Fn : K 0 → D by Fn = fn ◦... ◦ f 0.
(b) Prove that for each n, the function Fn is an expansion. Moreover, F (^) n′(0) =
∏n k=0 f ′ k (0), and conclude that limn→∞ |f n′(0)| = 1. [Hint: Prove that the sequence {|F (^) n′(0)|} has a limit by showing that it is bounded above and monotone increasing. Use the Schwarz lemma.] An easy induction argument shows Fn is a holomorphic injection, Fn(0) = 0 and |Fn(z)| > |z| so Fn is an expansion. Furthermore F (^) n′(0) =
∏n k=0 f ′ k (0)^ follows from the chain rule. Since the absolute value of each term in the product is greater than one the sequence {|F (^) n′(0)|} is monotone increasing. Since K 0 is a Koebe domain rK 0 > 0 so for any 0 < r < rK 0 Fn|D(0,r)(rz) satisfies the conditions for the Schwarz lemma so |F (^) n′(0)| < 1 /r showing the sequence is bounded above thus limn→∞
∏n k=0 |f
′ k(0)|^ exists which implies^ limn→∞^ f
′ n(0) = 1. (c) Show that if the sequence is osculating, that is, rKn → 1 as n → ∞, then {Fn} converges uniformly on compact subsets of K 0 to a conformal map F : K 0 → D. [Hint if rK 0 ≥ 1 then F is surjective.] Note this problem is false as stated, as a counter example consider fn : D(0, 2
n+1 (^) − 1 2 n+1^ )^ →^ D(0,^
2 n+2− 1 2 n+2^ )^ defined by^ f(z) =^ e
2 πiα 2 n+2− 1 2 n+1− 1 z. Then clearly^ fn^ satisfies all the necessary conditions for the problem for all α in R, so Fn : D(0, 1 /2) → D(0, 2
n+2− 1 2 n+2^ ), Fn(z) =^ e
2 π1(N +1)α 2 N+2^ − 1 2 N+1^ z, but this sequence does not converge uniformly on compact subsets unless α ∈ Z since FN (1/4) = e^2 πiα(N^ +1) 1 22
N+2− 1 2 N+2^ doesn’t converge for instance. So we change the problem to require that we only to show that a subsequence of {Fn} converges uniformly on compact subsets to the desired conformal map. Note |Fn(z)| < 1 for all n, the Fn’s are uniformly bounded on compact subsets of K 0. Thus the sequence {Fn} forms a normal family by Montel’s Theorem and there is a subsequence {Fnj } which converges uniformly on compact subsets to some function F. Since rKn → 1 we know that rF (K 0 ) = 1 which implies F is surjective onto D. Clearly F is holomorphic on K 0 (Theorem 2.5.2), injective (F is not constant since it surjects) (Proposition 8.3.5). So we have the desired conformal map.
To construct the desired osculating sequence we shall use the automorphisms ψα(z) = 1 α−−αzz.
