Complex Analysis II, Final Reports

Complex Analysis II, Final Reports 王⾦金龍

2015 Spring semester, NTU Week I

[1] June 9 ⿈黃哲宏 Big Picard Theorem

[2] June 11 李昱陞 Modular Forms and Moduli Problem [3] June 11 林肱慶 (Confluent) Hypergeometric Functions

Week II

[4] June 16 ⿈黃庭瀚 Sum of Squares

[5] June 16 古晉丞 Fundamental Groups and Covering Spaces [6] June 18 ⾼高尉庭 Topological Classification of Compact Surfaces

[7] June 18 李⾃自然 Frobenius Method for ODE with Regular Singularities [8] June 19 陳學儀 Hecke Operators on Modular Forms

[9] June 19 江泓 Asymptotic of Airy Function

Week III

[10] June 23 唐爾晨 Mandelbrot Sets and Julia Sets

[11] June 23 林東成 Asymptotic of Partition Function

[12] June 25 廖偉恩 Dirichlet Theorem with Density

[13] June 25 李龍欣 Dirichlet Principle

Dirichlet’s Principle

李龍欣

2015.06.25

Notation. Let A, B be subsets of a topological space. We say A ⊂⊂ B if A, the closure of A, is contained in ˚B, the interior of B.

Let (Ω, z) be a coordinate patch of a Riemann surface S. Then for some a ∈

C

and r > 0, if B(a; r) ⊂⊂ z(Ω), then we call Bz(a; r) := z⁻¹(B(a; r)) a z-disk.

1 Das Dirichletsche Integral

Notation (p.107). Let S denote a connected (oriented) Riemann surface. Any- thing related to “K” denotes a z-disk for some z. In particular, we arbitrarily fix a point p₀ ∈ S, a coordinate map z₀ with z₀(p₀) = 0, and some appropriate 0 < R₀ < R⁰₀. Then we call K₀ := B_z0(p₀; R₀) the hole, call K₀⁰ := B_z0(p₀; R⁰₀) the lid, call K₀⁰\K₀ the lock-ring, and call S\K₀ the punched surface.

Recall (p93, p72). For η = (η1dx + η2dy) and ξ = (ξ1dx + ξ2dy) being two 1- forms, we define [η, ξ] := η ∧ (^∗ξ) = (η₁ξ₁+ η₂ξ₂)(dx ∧ dy), which is symmetric and bilinear on the two inputs.

Definition (p.97). Let A ⊆ S be a region, and v, w ∈ C¹(A). The Dirichlet integral is defined to be D_A(v, w) :=

R

A[dv, dw]. If v = w, we denote the integral by D_A(v) := D_A(v, v) ≥ 0. The set of admissible functions is defined to be

M

(A) := {v ∈ C¹( ˚A) ∩ C⁰(A) : D_A(v) < ∞}

Notation (p.114). For v ∈

M

(K), define v to be the harmonic function on K that agrees with v on ∂K (which may be derived from Poisson’s integration formula).

Lemma 1 (p.97). ∀v ∈

M

(K), D_K(v) − D_K(v) = D_K(v − v) ≥ 0.

(hint: D_K(v, v − v) = 0)

Theorem 2 (p.106). Let Φ be a harmonic function on the lid which is regular in the lock-ring, and satisfies ^∂Φ_∂n = 0 along ∂K₀. There exists a harmonic function U such that U is regular in S\K₀ and that U − Φ is regular in K₀.

1

Definition (p.108). The set of competing functions is defined to be F := {(v, v^∗) : v ∈

M

(S\K₀), v^∗ ∈

M

(K₀⁰), v ≡ v^∗+ Φ in K₀⁰\K₀}

Whenever there is no ambiguity, we tend to use v in place of (v, v^∗). We define the potential to be D(v) := D_S\K

0(v) + D_K0(v^∗).

Remark (p.108). The potential can be also derived by the following process: Let a smoothing function λ be fixed, which is identically 1 in the hole, and vanishes outside the lid. We define the 2-forms Ψ = (1 − λ)[dv, dv] + λ[dv^∗, dv^∗] over S, and that Ψ⁰ = λ ([dv, dv] − [dv^∗, dv^∗]) over K₀⁰\K₀. Then D(v) can be given by the sum of D_λ(v) :=

R

SΨ and D_λ⁰(v) :=

R

K₀⁰\K0Ψ⁰. Fact 3 (pp.108–109).

1. ∀v ∈ F , 0 ≤ D(v) < ∞.

2. If U exists, then (u, u^∗) := (U |_S\K

0, U |_K⁰

0 − Φ) ∈F .

3. If Φ can be extended on an open disk K that contains the closure of the lid, then there exists a cut-off function λ such that λ|_K⁰

0 ≡ 1 and λ|_S\K ≡ 0.

Therefore the pair (v₀, v₀^∗) which is defined by v₀^∗ ≡ 0, v₀ ≡ λΦ on K\K₀, and v0 ≡ 0 on S\K is a competing function.

In summary, we are free to assume F 6=

∅

4. Let K be contained in the lid or the punched surface. Suppose that v1, v2∈ F coincide outside of K. That is, v1 ≡ v2 and v₁^∗≡ v^∗₂ respectively on each of their domains except on K. Then

D(v₁) − D(v₂) =

(

D_K(v₁) − D_K(v₂) whenever K ⊆ S\K₀ D_K(v₁^∗) − D_K(v^∗₂) whenever K ⊆ K₀⁰ (hint: for the second case, apply Green’s theorem)

Observation 4 (p.110). F = v0+

M

(S) in the following senses:

First, for all v₁, v₂ ∈ F , v1 − v₂ and v^∗₁ − v^∗₂ agree on the lock-ring, so they define an admissible function on S. Conversely, for all v ∈ F and w ∈

M

(S), (v + w, v^∗ + w) lies in F . Therefore for a fixed member v0 ∈ F , there is a one-to-one correspondenceF ↔

M

(S), v 7→ v − v0

Second, define T := K₀+ (S\K₀) to be the direct sum of spaces, which may be identified with S\∂K₀ sometimes. We identify v ∈ F with the corresponding function in C¹(T ), which is defined by

p 7→

(

v(p) if p ∈ S\K₀ v^∗(p) if p ∈ K₀

2

and satisfies D_T(v) = D(v) < ∞. Thus v ∈

M

(T ).

Finally, notice that (^{M(T )}/∼, D_T(·, ·) ) is a inner-product space over

R

^{, where} the equivalence relation ∼ presents “v₁ ∼ v₂ ⇔ v₁ − v₂ = const.” In addition,

M

(S), which is included in

M

(T ) by restriction, is a subspace. Therefore we can handle the problem as a problem of orthogonal projection: find v_//= w ∈

M

(S) so that the norm of v_⊥= u = v − w is minimized.

Proposition 5 (p.110, due to Beppo Levi). Define d := inf{D(v) : v ∈ F }.

Then for all v₁, v₂ ∈F ,

p

D_S(v₁− v₂) ≤

p

D(v₁) − d +

p

D(v₂) − d

Proof. As mentioned, we identify F as a subset of

M

(T ).

Let λ ∈

R

. If λ 6= −1, then ^λv_λ+1^1+v2 ∈F . Hence DT(^λv_λ+1^1+v2) = D(^λv_λ+1^1+v2) ≥ d, so D_T(λv₁+ v₂) ≥ (λ + 1)²d. The last inequality remains valid when λ = −1.

In summary, the quadratic function on λ

λ²(D_T(v1) − d) + 2λ(D_T(v1, v2) − d) + (D_T(v2) − d) is always ≥ 0. Hence we have the discriminant

(D_T(v1, v2) − d)²− (D_T(v1) − d)(D_T(v2) − d) ≤ 0 It follows that

0 ≤ D_T(v1− v2)

= D_T(v₁) − 2D_T(v₁, v₂) + D_T(v₂)

= (D_T(v₁) − d) + (D_T(v₂) − d) − 2(D_T(v₁, v₂) − d)

≤ (D_T(v₁) − d) + (D_T(v₂) − d) + 2

p

(D_T(v1) − d)(D_T(v2) − d)

=

p

D_T(v1) − d +

p

D_T(v2) − d



2

⇒

p

D_T(v₁− v₂) ≤

p

D_T(v₁) − d +

p

D_T(v₂) − d

⇒

p

D_S(v₁− v₂) ≤

p

D(v₁) − d +

p

D(v₂) − d

Corollary (p.111). If a minimizing function exists, it is unique up to an additive constant.

Notation (p.111). lim

v means the limitation taken as D(v) → d among those v ∈F⁰, whereF⁰:=

n

v ∈F :

R

∂K0v^∗ds = 0

o

.

3

2 Fourierreihe

Let K = B_z(0; R) be a fixed z-disk, and z = x + iy = re^iθ. For all v, w ∈

M

(K), define J_z,K(v, w) :=

RR

z(K)v(z)w(z) dx dy, and that J_z,K(v) := J_z,K(v, v).

Let u = v be the harmonic function on K that agree with v ∈

M

(K) on ∂K.

Then u is the real part of an analytic function f (z) =

P

∞

n=0c_nzⁿ. Hence u(z) = Re(f (z)) =

∞

X

n=0

(Re(c_n)Re(zⁿ) − Im(c_n)Im(zⁿ))

= a₀+

∞

X

n=1

(a_nrⁿcos(nθ) + b_nrⁿsin(nθ))

where a_n = Re(c_n) and b_n = −Im(c_n). Notice that

R

2π

0 f (re^iθ)e^−niθdθ = 2πrⁿc_n for n ≥ 0, and = 0 for n < 0. Hence for all n > 0,

a_n = 1 2πrⁿRe

Z

2π 0

f (re^iθ)e^−niθdθ



= 1

2πrⁿRe

Z

2π 0

f (re^iθ)(e^−niθ+ e^niθ) dθ



= 1

2πrⁿ

Z

2π

0

Re f (re^iθ)(2 cos(nθ))

dθ

= 1 πrⁿ

Z

2π 0

u(re^iθ) cos(nθ) dθ , and similarly,

b_n = 1 πrⁿ

Z

2π 0

u(re^iθ) sin(nθ) dθ Note that a₀ = _2π¹

R

2π

0 u(re^iθ) dθ

Define P_n = Re(zⁿ) = rⁿcos(nθ), Q_n = Im(zⁿ) = rⁿsin(nθ) ∈

M

(K). Ob- serve that dP_n =^∗dQ_n, so that by Green’s formula,

D_K(v, P_n) =

Z

K

dv ∧ dQ_n =

Z

∂K

v dQ_n

= nRⁿ

Z

2π

0

v(Re^iθ) cos(nθ) dθ

= nRⁿ

Z

2π

0

u(Re^iθ) cos(nθ) dθ

= πnR²ⁿa_n , and similarly,

D_K(v, Q_n) = πnR²ⁿb_n

By setting u = v = P_n or Q_n, we have the orthogonality relations



 

 

D_K(P_m, Q_n) = 0 without exception D_K(P_m, P_n) = D_K(Q_m, Q_n) = 0 if m 6= n

D_K(P_n) = D_K(Q_n) = πnR²ⁿ without exception

4

Also, by integrating under the polar coordinate, we have



 

 

 

 

J_z,K(P_m, Q_n) = 0 without exception J_z,K(P_m, P_n) = J_z,K(Q_m, Q_n) = 0 if m 6= n

J_z,K(P_n) = J_z,K(Q_n) = _2n+2^π R²ⁿ⁺² if n > 0 J_z,K(P₀) = πR²

Since u(z) = a₀ +

P

∞

n=1(a_nP_n+ b_nQ_n) converges uniformly, the orthogonality relation of D_K provides that

D_K(v) ≥ D_K(u) =

∞

X

n=1

πnR²ⁿ(a²_n+ b²_n) Similarly,

J_z,K(u) = πR²a²₀+

∞

X

n=1

π

2n + 2R²ⁿ⁺²(a²_n+ b²_n)

Lemma 6 (p.103). For all v ∈

M

(K), ∃a ∈

R

such that J_z,K(v−a) ≤ const.D_K(v) Proof. On one hand, take a = a₀ with respect to u = v, then

J_z,K(u − a₀) =

∞

X

n=1

π

2n + 2R²ⁿ⁺²(a²_n+ b²_n) ≤ R² 4

∞

X

n=1

πnR²ⁿ(a²_n+ b²_n)

= R²

4 D_K(u)

On the other hand, for w = v − u, which vanishes on ∂K, w(ρe^iθ) =

Z

ρ R

∂w(z)

∂r dr By Schwartz’s inequality,

w(ρe^iθ)² =

Z

ρ R



∂w(z)

∂r

√r

 

1

√r



dr



²

≤

Z

ρ

R



∂w(z)

∂r



² r dr

Z

ρ R

1 rdr

=

Z

R

ρ



∂w

∂x cos θ + ∂w

∂x sin θ



²

r dr(log R − log ρ)

=

Z

R

ρ

2

"



∂w

∂x



² +



∂w

∂y



²

#

r dr(log R − log ρ) Next, integrate the previous equation in order to yield that

J_z,K(w) ≤

Z

R

0

Z

2π 0

Z

R ρ

2

"



∂w

∂x



2

+



∂w

∂x



2

#

r(log R − log ρ)ρ dr dθ dρ

=

Z

R

0

2

Z

ρ≤|z|≤R

[dw, dw]



(log R − log ρ)ρ dρ

≤ 2D_K(w)

Z

R

0

(log R − log ρ)ρ dρ = R²

4 D_K(w) 5

Finally,

J_z,K(v − a₀) = J_z,K((u − a₀) + w) ≤ 2(J_z,K(u − a₀) + J_z,K(w))

≤ R²

2 [D_K(u) + D_K(w)] = R²

2 D_K(v)

Proposition 7 (p.112). For all K = B_z(0; R), there is a constant C so that for every w ∈

M

(S) that satisfies

Z

∂K0

w ds = R₀

Z

2π

0

w z₀⁻¹ R₀e^iθ

dθ = 0 we have J_z,K(w) ≤ CD_S(w).

Proof. Recall that K₀ is the hole. Let each 1 ≤ j ≤ n be corresponded with K_j, which is a z_j-disk with radius R_j, such that K_n = K, z_n = z, and that ∀1 ≤ j ≤ n, K_j−1∩ K_j 6=

∅

. Set the constants c_j so that

R

∂Kj(w − c_j) = 0 ds, where c₀= 0.

We prove by induction. If n = 0, i.e., K = K0, we take C = ^R₂²⁰ by Lemma 6.

It suffices to prove that if our claim holds on K_n−1, then it holds on K_n. Let k ⊂⊂ K_n−1 ∩ K_n be a z_n-disk with radius tR_n, where 0 < t < 1. Let m be an upper bound for

   

dz_n dz_n−1

   

on k. By the inductive hypothesis, there is a constant C⁰ which only depends on K_n−1 such that

J_zn,k(w) ≤ m²J_zn−1,k(w) ≤ m²C⁰D_S(w) In addition, by Lemma 6, we have

J_zn,k(w − c_n) ≤ J_zn,Kn(w − c_n) ≤ 1

2R_n²D_k(w) ≤ 1

2R²_nD_S(w) It follows that

πc²_nt²R²_n = J_zn,k(c_n) ≤ 2(J_zn,k(w) + J_zn,k(w − c_n))

≤ (2m²C⁰+ R²_n)D_S(w) Finally, we have

J_z,K(w) ≤ 2(J_zn,Kn(w − cn) + J_zn,Kn(cn))

≤ 2



₁

2R_n²D_K(w) + πc²_nR_n²



≤ 2



1

2R²_nD_S(w) + 2m²C⁰+ R²_n

t² D_S(w)



=



R²_n+4m²C⁰+ 2R²_n t²



D_S(w)

6

3 Die Mittelwertfunktion

Recall. Let z = x + iy be a local coordinate map and K = B_z(0; R) be a open disk with “center” p = z⁻¹(0). If v is harmonic, then

v(p) = 1 πR²

Z Z

K

v(x + iy) dx dy

Notation (p.113). From now on, let a point p ∈ S, a coordinate map z at p be fixed. In addition, let K = B_z(0; R) be contained in the punched surface or the lid. Define a map M_z,K :

M

(K) →

R

, which is abbreviated to M, as following:

M_z,K(w) = 1 πR²

Z Z

K

v(x + iy) dx dy = 1 πR²

Z

R 0

Z

2π 0

v(re^iθ) r dθ dr

If K is contained in the punched surface, one yields from Schwarz’s inequality, and the Propositions 5 and 7 that for all v₁, v₂∈F⁰,

(M(v1) − M(v2))² =



1 πR²

Z Z

K

(v1− v2) dx dy



2

≤ 1

πR²

Z Z

K

(v₁− v₂)²dx dy = 1

πR²J_z,K(v₁− v₂)

≤ C

πR²

p

D(v₁) − d +

p

D(v₂) − d



² That is,

|M(v₁) − M(v₂)| ≤ 1 R

r

C π

p

D(v₁) − d +

p

D(v₂) − d



(1) Therefore lim

v M(v) exists. We denote the limit by u(p). Then by the previous estimation,

|M(v) − u(p)| ≤ 1 R

r

C π

p

D(v) − d (2)

For all q ∈ K, let M_q denote M_z,kq, where the disk k_q := B_z(z(q); R − |z(q)|) is contained in K. Since we have an estimation which is similar to (1), the limit u(q) := lim

v M_q(v) exists. Moreover, in place of (2),

|M_q(v) − u(q)| ≤ 1 R − |z(q)|

r

C π

p

D(v) − d

It follows that M_q(v) converges uniformly to u(q) on q ∈ k, where k ⊂⊂ K is a disk (concentric with K).

Remark (p.114). If K is contained in the lid, we can compute u^∗(p) := lim

v M(v^∗), which existence and estimations are given in a similar way. In particular, if K is contained in the lock-ring, u = u^∗+ Φ because Φ is harmonic.

7

Proposition 8 (p.114). u : K →

R

^{or u∗ : K →}

R

is harmonic (whenever any one of which is defined).

Proof. For simplicity, we suppose that K ⊆ S\K₀ and consider v ∈F⁰. A similar argument holds for K ⊂ K₀⁰ and v^∗.

Recall that v ∈

M

(K) is harmonic. We define

e

^{v ∈}F by applying a smoothing process so that

e

v coincides with v outside of K, but with v in k = B_z(0; r), where 0 < r < R. Let the smoothing be well chosen so that D_K(

e

^{v) → DK}(v) as r → R⁻. By Lemma 1, D_K(v) ≤ D_K(v), and it takes “=” if and only if v is harmonic, namely v = v =

e

v. Therefore for sufficiently large r, we have D_K(

e

^{v) ≤ DK(v).}

Notice that v =

e

v, so that D_K(v) ≤ D_K(

e

v). Hence D_K(v) ≤ D_K(

e

^{v) ≤ DK(v).}

By Fact 3.4, D(

e

^{v) ≤ D(v).}

We replace v₂ with ^v

e

² in Levi’s inequality to yield that

p

D_K(v₁−^v

e

^{2) ≤}

p

D_S(v₁−^v

e

²⁾

≤

p

D(v₁) − d +

p

D(^v

e

^{2) − d}

≤

p

D(v1) − d +

p

D(v2) − d Take r → R⁻. Thus

p

D_K(v₁− v₂) ≤

p

D(v₁) − d +

p

D(v₂) − d (3)

Similarly,

p

D_K(v₁− v₂) ≤

p

D(v₁) − d +

p

D(v₂) − d Repeat the argument for (1). So lim

v Mq(v) = u(q). Note that Mq(v) = v(q) because v is harmonic. Hence in place of (2),

|v(q) − u(q)| ≤ 1 R − |z(q)|

r

C π

p

D(v) − d

As a result, lim

v v(q) = u(q) uniformly on q ∈ k for any k ⊂⊂ K. Therefore u is also harmonic.

Lemma 9 (p.115). For all v ∈F⁰, we have

ˆ DK(v − v) ≤ 4(D(v) − d)

ˆ Jz,K(v − v) ≤ R²(D(v) − d)

Proof. First, take v₁ = v₂ = v in (3) to get the first estimation. Next, since (v − v) vanishes on ∂K, J_z,K(v − v) ≤ ^R₄²D_K(v − v) ≤ R²(D(v) − d) by the inequality for w in Lemma 6.

8

In order to make u an ansatz, we need one more step:

Claim (p.114). u(p) := lim

v M_z,K(v) (or u^∗, resp.) does not depend on z nor K.

Proof. Let z⁰ = x⁰+ iy⁰ be another coordinate, and K⁰ = Bz⁰(0; R⁰) be a z⁰-disk with center p⁰. Observe that it suffices to prove for K⁰⊂⊂ K and p = p⁰.

Note that

   

dz dz⁰

   

has an lower bound _m¹ > 0 on K⁰. Therefore

(M_z⁰_,K⁰(v) − M_z⁰_,K⁰(v))²≤ 1 πR⁰²

Z Z

K⁰

(v − v)²dx⁰dy⁰

≤ m² πR⁰²

Z Z

K

(v − v)²dx dy

= m²

πR⁰²J_z,K(v − v)

≤ m²R²

πR⁰² (D(v) − d)

Because v is harmonic on K⁰, we have M_z⁰_,K⁰(v) = v(p). Hence u⁰(p) := lim

v M_z⁰_,K⁰(v) = lim

v v(p) = u(p)

Proof of Theorem 2. We claim that (u, u^∗) minimizes D(·).

First, observe that for B, a smaller z-disk concentric with K (the radius of B is smaller than the radius of K), lim

v D_B(v − v) = 0 follows from Lemma 9, and lim

v D_B(v − u) = 0 follows from the fact that the derivatives of v converge uniformly to those of u on B. Therefore lim

v D_B(v − u) = 0 follows from the triangle inequality.

Next, associate each point p with a local coordinate z, a z-disk K = K(p), and a smaller z-disk B = B(p) such that p ∈ B(p) ⊂⊂ K(p). Since {B(p)}_p∈S covers S, there is a countable subcover {B(p_i)}^∞_i=1 (by Lindel¨of’s covering theorem).

Next, we construct Diudonn´e factors µ_i by {K(p_i)} and {B(p_i)} such that

P

iµ_i ≡ 1 with each µ_i ∈ C¹(S, [0, 1]), and vanishes outside K(p_i). (See p.74) The conclusions above lead to

limv

Z

S

µ_i[d(v − u), d(v − u)] ≤ lim

v

Z

K(pi)

[d(v − u), d(v − u)] = 0

⇒limv

Z

S

µi[d(v − u), d(v − u)] = 0 (4)

In the statements above, v − u ∈ C¹(S). Naturally, for all v1, v2∈F , we define D_i(v₁, v₂) =

Z

S\K0

µ_i[dv₁, dv₂] +

Z

K0

µ_i[dv₁^∗, dv₂^∗]

9

Observe that the triangle inequality of

p

D_i(·) holds. Hence



 

p

D_i(v) −

p

D_i(u)

  

^≤

p

D_i(v − u)

Combine this with (4). It follows that lim

v

P

n

i=1D_i(v) =

P

n

i=1D_i(u). Observe that for all v,

P

∞

i=1D_i(v) increases to D(v). Therefore D(u) = lim

n→∞

n

X

i=1

D_i(u) = lim

n→∞lim

v n

X

i=1

D_i(v) ≤ lim

n→∞lim

v D(v) = d

By the definition of d, D(u) ≥ d, so D(u) = d. As a result, for all w ∈

M

(S) and ε ∈

R

, (u + εw) ∈F implies D(u + εw) ≥ D(u), so D(u, w) = 0.

Finally, we claim that the function U , given by u on the punched surface and u^∗+ Φ on the lid, minimizes D_S(·). It suffices to take any w ∈

M

(S) that vanishes in some neighborhood of every singularity of Φ, and check that D_S(U, w) = 0. We derive from the equation D(u, w) = 0 that

0 = D(u, w) =

Z

S\K0

[du, dw] +

Z

K0

[du^∗, dw]

=

Z

S\K0

[dU, dw] +

Z

K0

[d(U − Φ), dw]

=

Z

S

[dU, dw] −

Z

K0

[dΦ, dw]

= D_S(U, w) −

Z

K0

[dΦ, dw]

= D_S(U, w) −

Z

∂K0

w∂Φ

∂nds

= D_S(U, w) because ^∂Φ_∂n = 0 along ∂K₀.

References

[1] Hermann Weyl, The Concept of a Riemann Surface, 3rd ed., Dover edition, translated by Gerald R. MacLane, Dover, Mineola, N.Y., 2009. pp.73–74, 93–118.

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