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. $\Box$

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

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| and use Schwartz Lemma to show that $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.$\Box$