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Math 246A
Homework 6 Solutions
Due Thursday May 17
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.
19 Prove the maximum principle for harmonic functions, that is:
(a) If u is a non-constant real-valued harmonic function in a region Ω, then u cannot attain a maximum (or a minimum) in Ω. Following the hint, let f be holomorphic in a neighborhood of z 0 with f = u + iv for some v. Then f is not constant and must be open. It follows exp(f) is also open. But |exp(f(z))| = exp(u(z)) and since exp is monotone increasing it follows if u(z 0 ) is a local maximum or minimum then so is exp(u(z 0 )), which contradicts exp(f(z)) being open. You can also do this without using exp, which is what most people did. (b) Suppose that Ω is a region with compact closure Ω. If¯ u is harmonic in Ω and continuous in Ω,¯ then sup z∈Ω
|u(z)| ≤ sup z∈ Ω¯−Ω
|u(z)|.
As Ω¯ is compact |u(z)| has a maximum value for z ∈ Ω¯. By part (a) this maximum cannot be for z ∈ Ω.
[Hint: To prove the first part, assume that u attains a local maximum at z 0. Let f be holomorphic near z 0 with u = Re (f), and show that f is not open. The second part follows directly from the first.]
20 This exercise shows how the mean square convergence dominates the uniform convergence of analytic functions. If U is an open subset of C we use the notation
‖f‖L (^2) (U ) =
U
|f(z)|^2 dx dy
for the mean square norm, and ‖f‖L∞(U ) = sup z∈U
|f(z)|
for the sup norm.
(a) If f is holomorphic in a neighborhood of the disc Dr (z 0 ), show that for any 0 < s < r there exists a constant C > 0 (which depends on s and r) such that ‖f‖L∞(Ds (z 0 )) ≤ C ‖f‖L (^2) (Dr (z 0 )).
By the maximum modulus principle, we know that ‖f‖L∞(Ds(z 0 )) ≤ supz∈γ |f(z)| where γ is the boundary of D(r+s)/ 2 (z 0 ). Let z 1 ∈ γ be a point at which |f(z 1 )| = supz∈γ |f(z)|. Now |f(z 1 )| ≤ (^21) π
∫ (^2) π 0 |f(z^1 +^ te
iθ (^) )|dθ where 0 < t < r−s
- using Cauchy-Schwarz this is less than 1 2 π (
∫ (^2) π 0 |f(z^1 +^ te
iθ (^) )| (^2) dθ) 1 / (^2) (∫^2 π 0 dθ)
1 / (^2). So |f(z 1 )|
2 π
∫ (^2) π 0 |f(z^1 +^ te
iθ (^) )| (^2) dθ. Thus
|f(z 1 )|2 (r−s)
2 8 =^
∫ r− 2 s 0 |f(z^1 )|
(^2) tdt ≤ 1 2 π
∫ r− 2 s 0
∫ (^2) π 0 |f(z^1 +^ te
iθ (^) )| (^2) dθtdt = 1 2 π
|z−z 1 |≤ r− 2 s^ |f(z)|
(^2) dxdy ≤ 1 2 π
|z−z 0 |≤r |f(z)|
(^2) dxdy = 1 2 π ‖f‖
2 L^2 (Dr(z 0 )) This implies ‖f‖L∞ (^) (Ds(z 0 )) ≤ |f(z 1 )| ≤ √π(^2 r−s) ‖f‖L (^2) (Dr (z 0 )). (b) Prove that if {fn} is a Cauchy sequence of holomorphic functions in the mean square norm ‖·‖L (^2) (U ), then the sequence {fn} converges uniformly on every compact subset of U to a holomorphic function. Since {fn} for any there exists some N such that ‖fm − fn‖L (^2) (U ) < for m, n ≥ N. Given a compact subset Ω ⊂ U , let Dri (zi) be a finite cover of Ω such that Dsi (zi) also covers Ω where 0 < si < ri for all i. Then by part (a), ‖fm − fn‖L∞(Dsi (zi)) ≤ Ci ‖fm − fn‖L (^2) (Dri (zi)) < C for some Ci where C = max {Ci}. The result follows by Theorem 5.2 of Chapter 2 since L∞^ is complete.
[Hint: Use the mean-value property.]
21 Certain sets have geometric properties that guarantee they are simply connected.
(a) An open set Ω ∈ C is convex if for any two points in Ω, the straight line segment between them is contained in Ω. Prove that a convex open set is simply connected. The homotopy of Example 1 of Stein and Shakarchi, γs(t) = (1 − s)γ 0 (t) + sγ 1 (t) works for any convex open set. (b) More generally, an open set Ω ∈ C is star-shaped if there exists a point z 0 ∈ Ω such that for any z ∈ Ω, the straight line segment between z and z 0 is contained in Ω. Prove that a star-shaped open set is simply connected. Conclude that the slit plane C − {(−∞, 0]} (and more generally any sector, convex or not) is simply connected. We discussed this problem in some detail in section. Basically the idea is to contract all the points into z 0 but design it so that the endpoints are fixed. Note its enough to show each curve is homotopic to the curve consisting of a straight line from γ 0 (0) to z 0 followed by another straight line from z 0 to γ 0 (1) and then to use that being homotopic is an equivalence class. Create a homotopy αs(t) by making αs = γ 0 ∗ (1 − s) + z 0 ∗ s with straight lines connecting the endpoints back to γ 0 (0) and γ 0 (1) (Note I’m not saying explicitly how to parameterize the curves αs). This is clearly a homotopy proving the claim.
22 Show that there is no holomorphic function f in the unit disc D that extends continuously to ∂D such that f(z) = 1/z for z ∈ ∂D. Say that such a function f exists. For 0 < r ≤ 1 , let Fr (z) =
Cr f(ζ)^ dζ, where^ Cr^ is the circle^ |z|^ =^ r. Then Fr =
∫ (^2) π 0 gr^ (t)^ dt^ where^ gr^ (t) =^ f(re
it (^) )ireit. Since f is continuous on ¯D it follows that for all > 0 there exists δ > 0 such that 1 − r < δ implies |g 1 (t) − gr (t)| < , which holds for all t since t ∈ [0, 2 π], a compact set. It follows limr→ 1 Fr (z) =
∫ (^2) π 0 f(e
it (^) )ieit (^) dt. However since f is holomorphic in D it follows Fr = 0 for all r < 1 , whereas
∫ (^2) π 0 f(e
it (^) )ieit (^) dt = 2πi, a contradiction.
P2 Let u be a harmonic function in the unit disc that is continuous on its closure. Deduce Poisson’s integral formula
u(z 0 ) =
2 π
∫ (^2) π
0
1 − |z 0 |^2 |eiθ^ − z 0 |^2
u(eiθ) dθ for |z 0 | < 1
from the special case z 0 = 0 (the mean value theorem). Show that if z 0 = reiϕ, then
1 − |z 0 |^2 |eiθ^ − z 0 |^2
1 − r^2 1 − 2 r cos(θ − ϕ) + r^2
= Pr (θ − ϕ),
and we recover the expression for the Poisson kernel derived in the exercises of the previous chapter. [Hint: Set u 0 (z) = u(T (z)) where T (z) = 1 z−^0 −z 0 zz. Prove that u 0 is harmonic. Then apply the mean value theorem to u 0 , and make a change of variables in the integral.] T is a conformal self-map of D that maps the boundary to the boundary, and has itself as an inverse. Now let f = u + iv in D; then f ◦ T is also holomorphic in D with real part u 0 (z) = u(T (z)) which is then harmonic. By the mean-value theorem we have u(z 0 ) = u 0 (0) = (^21) π
∫ (^2) π 0 u^0 (re
iω (^) ) dω for 0 < r < 1 , and since u and thus u 0 is continuous on D¯ we have u(z 0 ) = (^21) π
∫ (^2) π 0 u^0 (e
iω (^) ) dω.
Let eiθ^ = T (eiω^ ) = e
iω (^) −z 0 z 0 eiω^ − 1 ; then since^ T^ is its own inverse we also have^ e
iω (^) = T (eiθ (^) ) = eiθ^ −z 0 z 0 eiθ^ − 1. It follows eiω^ dω = e
iθ (^) (|z 0 | (^2) −1) eiθ^ z¯ 0 −1)^2 from which it follows^ dω^ =^
1 −|z 0 |^2 |eiθ−z 0 |^2 dθ, and noting that the limits of integration remain the same the result follows. The calculation of Pr (θ − ϕ) is basic algebra.
- A holomorphic mapping f : U → V is a local bijection on U if for every z ∈ U there exists an open disc D ⊂ U centered at z, so that f : D → f(D) is a bijection. Prove that a holomorphic map f : U → V is a local bijection on U if and only if f′(z) 6 = 0 for all z ∈ U. [Hint: Use Rouch´e’s theorem as in the proof of Proposition 1.1.]
