Picard’s Theorem

In this post, we will deal with the proof of Picard’s Great and Little theorem via the method of differential geometry.

Ahlfors-Schwarz Lemma

Assume \Omega_1 and \Omega_2 are 2 domains in \mathbb{C}, and f: \Omega_1\to \Omega_2 is an analytic function. If \rho is a metric on \Omega_2 and f'\neq 0, then the pull-back of \rho, defined on \Omega_1, is f_*\rho=(\rho\circ f)|f'|. Recall that the definition of Curvature with respect to metric \rho is K(z,\rho):=-\dfrac{\Delta ln \rho(z)}{\rho^2(z)}, so

Lemma 0: If notations as above, then K(z,f_*\rho)=K(f(z),\rho)

Proof: This follows from a directly computation.

\begin{aligned}K(z,f_*\rho)&=-\dfrac{|f'(z)|^2(\Delta_fln \rho)\circ f(z)}{(\rho\circ f(z))^2|f'(z)|^2}\\ &=-\dfrac{(\Delta_fln\rho)\circ f(z)}{(\rho \circ f(z))^2}\\ &=K(f(z),\rho)\end{aligned} \Box

Theorem 1(Ahlfors-Schwartz Lemma) f(z) is a holomorphic function on D, f maps D to U. If there is a metric \rho on U, that is, ds_{\rho}^2=(\rho(z))^2|dz|^2, such that its curvature is not bigger than -1, then f_*\rho(z)\leq \lambda(z), where \lambda(z)=\dfrac{2}{1-|z|^2}. In another word, ds_{\rho}^2\leq ds_{\lambda}^2

Proof: fix any r\in (0,1) and define a metric on D(0,r) as \lambda_r=\dfrac{2r}{r^2-|z|^2}. Obviously, for any point z\in D(0,r), its curvature is -1. Define a function v(z)=\frac{f_*\rho(z)}{\lambda_r} on D(0,r). It is easy to notice that v is bounded on \bar{D(0,r)} and it is 0 on the boundry. So v must attain its maximum M at an interior point \tau. So all we are left to do is prove M\leq 1

1)f_*\rho(\tau)=0, then v=0, there’s nothing to say.

2) f_*\rho(\tau)>0, then K(\tau,f_*\rho) is defined, by the definition K\leq-1. Since \tau is the maximal point, then

\begin{aligned} 0 & \geq \Delta ln\ v(\tau)=\Delta ln f_*\rho(\tau)-\Delta ln \lambda_r(\tau) \\ &= -K(\tau,f_*\rho)\cdot (f_*\rho(\tau))^2+K(\tau,\lambda_r)(\lambda_r)^2\\ &\geq (f_*\rho)^2-(\lambda_r)^2\end{aligned}

In another word, this is \dfrac{f_*\rho(\tau)}{\lambda_r(\tau)}\leq 1, or M\leq 1. \Box

In the above theorem, if we choose U=D(0,1) and \rho=\lambda, it agrees with the Schwartz-Pick Lemma.

Further, if we define \lambda_{\alpha}^A=\dfrac{2\alpha}{\sqrt{A}(\alpha^2-|z|^2)} on D(0,\alpha), here A>0,\alpha >0, then this metric has curvature  -A at any point in D(0,\alpha).

Theorem 2(General form of Ahlfors-Schwartz Lemma) f(z) is a holomorphic function on D(0,\alpha)f maps D(0,\alpha) to U. If there is a metric \rho on U, that is, ds_{\rho}^2=(\rho(z))^2|dz|^2, such that its curvature is not bigger than -B, then f_*\rho(z)\leq \dfrac{\sqrt{A}}{\sqrt{B}}\lambda_{\alpha}^A(z) for each z\in D(0,\alpha) and B>0 \Box

Generalized Liouville Theorem

In classical Complex Analysis, we know that a bounded entire function must be constant. Now by the Ahlfors-Schwartz Lemma, there is a corresponding theorem with respect to curvature.

Theorem 3(Generalized Liouville Theorem) Assume f(z):\mathbb{C}\to U to be entire, and \rho be a metric on U, such that for any point z\in U, K(z,\rho)\leq -B<0, for some positive constant B. Then f(z) must be a constant.

Proof: for any \alpha>0, f(z) maps D(0,\alpha) to U. By our assumption, and general form of Ahlfors-Schwartz Lemma, we have

f_*\rho(z)\leq \dfrac{\sqrt{A}}{\sqrt{B}}\lambda_{\alpha}^A(z)

Since \lambda_{\alpha}^A\to 0, \alpha \to \infty, then we may conclude that f_*\rho \leq 0. Thus f_*\rho(z)=0, this force f to be a constant.\Box

Deduction of Liouville thoerem: If f is bounded by M, then $g=\frac{1}{M}f$ maps \mathbb{C} \to D. If we choose a metric \lambda on D such that its curvature is -1. So choose B=1 in the above theorem, then g is a constant, so is f. \Box

Picard’s Little Theorem

With the theorem above, we may prove Picard’s little theorem.

Lemma 4: If U\subset \mathbb{C} is an open set, with at least 2 point in \mathbb{C}\backslash U, then we can introduce a metric \mu such that K(z,\mu)\leq -B<0 for each z\in U, where B is some positive constant.

Proof: Choose a linear transform to map 2 fixed point in \mathbb{C}\backslash U to \{0,1\}, denote E=\mathbb{C}\backslash\{0,1\}, and introduce a metric \mu \text{ on }E by

\mu(z)=\dfrac{(1+|z|^{1/3})^{1/2}}{|z|^{5/6}}\cdot \dfrac {(1+|z-1|^{1/3})^{1/2}}{|z-1|^{5/6}}

By directly computation we have

K(z,\mu)=-\dfrac{1}{18}\big{[}\dfrac{|z-1|^{5/3}}{(1+|z|^{1/3})^3(1+|1-z|^{1/3})}+\dfrac{|z|^{5/3}}{(1+|1-z|^{1/3})^3(1+|z|^{1/3})}\big{]}

In this case, we know

  1. \lim_{z\to 0}K=\lim_{z\to 1}K=\frac{-1}{36}
  2. \lim_{z\to \infty}=-\infty

So K is bounded \Box

Theorem 5(Picard’s Little Theorem): If entire function w=f(z) maps \mathbb{C} to U, and \mathbb{C}\backslash U contains at least 2 points, then f is a constant.

Proof: Directly by theorem 1 and lemma 2. \Box

 

Picard’s Great Theorem

Def 6.1 If g_j is a sequence of functions defined on \Omega,it is called normally convergence if for any \epsilon>0 and K\subset \Omega a compact subset, there must exists a natural number J,(depends on \epsilon and K), such that for all j>J we have |g_j(z)-g(z)|<\epsilon. That is if g_j inner-closed uniformly convergence, then it is normally convergence.

And we call it compactly divergence, if for any compact subset K\subset\Omega and L\subset \mathbb{C}, there exists a natural number J,(depends on K and L), such that for all j>J we have g(z)\notin L,z\in K. That is if g_j uniformly diverge to \infty on any compact subset, then it is compactly divergent.

Def 6.2 If \mathcal{F} is a familly complex function defined on \Omega such that any sequence in \mathcal{F} has a subsequence normally convergent or compactly divergent, then we call \mathcal{F} a normal family.

Def 6.3 If \mathcal{F} is a familly meromorphic function defined on \Omega\subset \bar{mathbb{C}} normally convergent with respect to the sphere metric, is called a normal family.

Theorem 7(Marty): If \mathcal{F} is a family of meromorphic function defined on \Omega, then \mathcal{F} is normal iff the set \{f_*\sigma:f\in \mathcal{F}\} are uniformly bounded on any compact set, where \sigma is the sphere metric. In another word, there exists a constant M_K for each compact set K such that \dfrac{2|f'(z)|}{1+|f(z)|^2}\leq M_K.

Proof: If uniformly bounded, then

\begin{aligned} d(f(z_1),f(z_2))&= \inf_{\gamma}\int_{\gamma}ds\\ &=\inf_{\gamma'}\int_{\gamma'}\dfrac{2|f'(z)|}{1+|f(z)|^2} |dz|\\ &\leq \int_{\gamma_0}\dfrac{2|f'(z)|}{1+|f(z)|^2} |dz| \\ &\leq M_K|z_1-z_2|\end{aligned}

where \gamma is a curve connecting f(z_i) and contained in K, \gamma'=f^{-1}(\gamma), and \gamma_0 is the strait line connecting z_i. In this case, the sphere metric is bounded and d under this metric is uniformly continuous, so by A-A lemma, \mathcal{F} is normal.

On the other hand, If \mathcal{F} is normal, and assume the boundedness is not true, then there is a compact set E and a sequence \{f_n\}\subset \mathcal{F} such that \max_{z\to E}(f_n)_*\sigma(z) is unbounded on E. Then by assumption, there is a subsequence f_{n_k} uniformly converging to f on E.

At each point of E, there can be a closed circle \bar{D}\subset \Omega, with either f or \frac{1}{f} is holomorphic on \bar{D}. For the first case, f is bounded on \bar{D}. Since f_{n_k} converges under sphere metric, so when n_k large enough, f_{n_k} has no poles on \bar{D}. So (f_{n_k})_*\sigma converges uniformly to f_*\sigma on a slightly smaller circle that D. Since f_*\sigma is continuous, then (f_{n_k})_*\sigma are bounded on the slightly smaller circle. The same argument can show that (\dfrac{1}{f_{n_k}})_*\sigma are bounded, and by the fact (\dfrac{1}{f_{n_k}})_*\sigma=(f_{n_k})_*\sigma. Cover E by finitely many such circles, we thus deduced a contradiction that \{(f_{n_k})_*\sigma\} are bounded on E. \Box

Theorem 8(Montel): If \mathcal{F} is a family of meromorphic function on \Omega, P,Q,R are three different point. If each function in \mathcal{F} missed the three points, then the family is normal.

Proof: Use linear transformation to transform P,Q,R to 0,1,\infty. So we just need to show All holomorphic function families with values in E=\mathbb{C}\backslash\{0,1\} are normal. This suffices to prove \mathcal{F} is normal for any D(z_0,\alpha)\subset \Omega, and W.L.O.G. we may assume z_0=0.

Let \mu(z)=\dfrac{(1+|z|^{1/3})^{1/2}}{|z|^{5/6}}\cdot \dfrac {(1+|z-1|^{1/3})^{1/2}}{|z-1|^{5/6}}, multiply it by a constant such that the upper bound of its curvature is -1, and still denote it by \mu. By General Ahlfors-Schwartz Lemma, for any f\in \mathcal{F}, we have f_*\mu\leq \dfrac{2\alpha}{\sqrt{A}(\alpha^2-|z|^2)}, that is for each z\in D(0,\alpha)

\mu(f(z))|\dfrac{df}{dz}|\leq \dfrac{2\alpha}{\sqrt{A}(\alpha^2-|z|^2)}

compare the sphere metric \sigma(w) with \mu(w) in E. It is easy to conclude that( directly computation)

\dfrac{\sigma(w)}{\mu(w)}=\dfrac{\dfrac{2}{1+|z|^2}}{\dfrac{a(1+|w|^{1/3})^{1/2}}{|w|^{5/6}}\cdot \dfrac {(1+|w-1|^{1/3})^{1/2}}{|w-1|^{5/6}}}\to 0, z\to 0,1,\infty

So there is a positive constant M such that \sigma(w)\leq M\mu(w), so for each z\in D(0,\alpha) we know

\begin{aligned} f_*\sigma(z)&= \sigma(f(z))|\dfrac{df}{dz}|\\&\leq M\mu(f(z))|\dfrac{df}{dz}|\leq Mf_*\mu(z)\\&\leq M\lambda_{\alpha}^A(z)=\frac{2\alpha M}{\sqrt{A}(\alpha^2-|z|^2)}\end{aligned}

This means that f_*\sigma is bounded on any compact subset of D(0,\alpha) uniformly(not dependent of the choice of f), by Marty’s theorem, we proved this theorem. \Box

Theorem 8′(Montel) If \mathcal{F} is a holomorphic familly on \Omega, such that each function in \mathcal{F} miss 2 fixed point in \mathbb{C}, then \mathcal{F} is normal familly.

Proof: In fact, this is what we proved in theorem 8.\Box

Theorem 9(Picard’s Great Thoerem): If f(z) is holomorphic on D'(0,r):=D(0,r)\backslash\{0\}, and z=0 is its essential sigular point, then \mathbb{C}\backslash f(U) contains at most one point for any neighborhood of 0.

Proof: Otherwise, assume f is holomorphic on D'(0,1) and f(D') do not contain 0,1. Claim $z=0$ is its removable sigular point or polar point.

Define f_n(z):=f(\dfrac{z}{n}),(0<|z|<1). Construct holomorphic family \mathcal{F}:=\{f_n\}, with image contained in E=\mathbb{C}\backslash\{0,1\}. By theorem 8′ we know it is normal family, so there is a subsequence f_{n_k}, which is normally convergent or compactly divergent.

  1. If normally convergent, then f_{n_k} are inner-closed uniformly convergent, thus is bounded on any compact subset of D'. In partucularly, f_{n_k} are bounded on \{z:|z|=\frac{1}{2}\} by M, that is f is bounded by M on \{z:|z|=\frac{1}{n_k}\}. By the maximal mudulus theorem, f is bounded by M on 0<|z|<\frac{1}{2}. So z=0 is removable, a contradiction.
  2. If compactly divergent, consider g_k=\dfrac{1}{f_{n_k}}, we can prove that \dfrac{1}{f}\to 0,z\to 0, then z=0 is a polar point, a contradiction. \Box
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Hadamard’s Three Circles Theorem and Three Lines Lemma

Today I will record 2 lemmas I met.

Hadamard three-lines theoremLet f(z) be a bounded function of z=x+iy defined on the strip \{x+iy:a\leq x\leq b\}holomorphic in the interior of the strip and continuous on the whole strip. If M(x)=sup_y |f(x+iy)|then logM(x) is a convex function on [a,b]. More precisely, if x=ta+(1-t)b  with 0\leq t\leq 1, then M(t)\leq M(a)^tM(b)^{1-t}

Proof: WLOG we may assume a=0,b=1, then define F(z) by F(z)=f(z) M(a)^{\frac{z-b}{b-a}} M(b)^{\frac{z-a}{a-b}}, and define F_n(z):=F(z)e^{\frac{z^2}{n}}e^{\frac{-1}{n}}use the maximum modulus principle to conclude that F(z)\leq 1. \Box

Hadamard three-circle theorem: Let f(z) be a holomorphic function on the annulus r_1\leq|z|\leq r_3. Define M(r):=\sup_{|z|=r}|f(z)| . Then, logM(r) is a convex function of the logarithms log(r). Moreover, if f(z) is not of the form cz^{\lambda} for some constant \lambda and c, then M(r) is strictly convex as a function of log(r). In another word, \log \left({\frac {r_{3}}{r_{1}}}\right)\log M(r_{2})\leq \log \left({\frac {r_{3}}{r_{2}}}\right)\log M(r_{1})+\log \left({\frac {r_{2}}{r_{1}}}\right)\log M(r_{3}) for any three cocentered circle of radii 0<r_1<r_2<r_3

Proof: It follows from the fact that for any real a, the function Re log(z^af(z)) is harmonic between two circles, and therefore takes its maximum value on one of the circles. Take a such that the two value on the boundry agrees,  then the theorem comes from the equality Re(log(z^af(z))_{|z|=r_2}\leq \max_{|z|=r_1}Re(z^af(z)) \Box

Riemann’s Mapping Theorem

The following are outlines of the proof of Riemann’s mapping theorem, which is the most important and profund theorem in the Riemann conformal mapping thoery.

Theorem:(Riemann) : If \Omega \subset \mathbb{C} is a simply connected domain, not equal to \mathbb{C}z_0\in \Omega, \theta_0 is a real number (0\leq \theta_0<2\pi), then there exists a unique function $latexf(z)$ such that : w=f(z) is monodromy and analytic in \Omega, f(z) maps \Omega to the unit disc D conformally and  f(z_0)=0, e^{i\theta_0}f'(z_0)>0

Proof: To prove this, we may assume that \theta_0 = 0, and then we proceed in 4 steps:

1)Consider the familly

 \mathcal{M}=\{g:\text{ g is monodromy and analytic in } \Omega,g(z_0)=0,g'(z_0)>0\}

we will show by constructing a specific function \mathcal{M} is non-empty in this step .

2)let \lambda:=sup_{g\in \mathcal{M}}g'(z_0) and restrict g to a circle |z-z_0|<r_0 and use Schwartz Lemma to show that 0<g'(z_0)\leq \frac{1}{r_0}. This implies that \lambda <\infty, so by the definition, we may find a sequence g_n such that g_n\in \mathcal{M}, \lim g'_n(z_0)=\lambda. Since g_n is uniformlly bounded in \Omega, according to Motel Theorem, there is a subsequence g_{n_k} converges uniformly to a f(z). We can claim that f(z) is monodromy and analytic in \Omega, f\in \mathcal{M} and f'(z_0)=\lambda (Use Weierstrass theorem, Hurwitz theorem and Montel thoerem)

Theorem:(Montel) If f_n are analytic in D, and inner-closed uniformly bounded in D, then there must be a sequence f_{n_k} inner-closed uniformly converge.

3)Claim: w=f(z) maps \Omega conformally to $latex

Lemma: If \Omega is a simply connected subset of D,  not equaling to D, with 0\in \Omega, then there exists a monodromy analytic function h:\Omega\to D such that h(0)=0,h'(0)>1

Proof: choose a\notin \Omega, and define \varphi_a(z)=\frac{z-a}{1-\bar{a}{z}}, choose a monodromy branch as g(z)=\sqrt{\varphi_a(z)}, let b=g(0)\in D, and thus define h(z) to be e^{iargb}\varphi_b'[g(z)]:\Omega\to D, h(0)=0 \Box

If otherwise, by the above lemma, there exists a h(w) with respect f(\Omega)\subset D, define \tilde f(z)=h\circ f(z), we have \tilde{f}'(z_0)>f'(z_0), a contradiction.

4)By Swartz Lemma, we can show such f is unique.\Box