Bloch, Schottky, and the Picard Theorems

The closed disc $\mathrm{\mid }z\mathrm{\mid }\le R$ is compact, and $z\mapsto \mathrm{\mid }{f}^{\mathrm{\prime }}(z)\mathrm{\mid }$ is continuous, so [L1] justifies the maximizing point $z_0$. Since $\mathrm{\mid }{z}_0\mathrm{\mid }+r=(R+\mathrm{\mid }{z}_0\mathrm{\mid })\mathrm{/}2<R$, the affine disc $z_0+r\mathbb{D}$ lies in $\mathbb{D}$, so $g$ is holomorphic on $\mathbb{D}$ and direct differentiation gives $g(0)=0$, $g^{\mathrm{\prime }}(0)=1$.

If $\mathrm{\mid }w\mathrm{\mid }<1$, then $\mathrm{\mid }{z}_0+rw\mathrm{\mid }\le \mathrm{\mid }{z}_0\mathrm{\mid }+r<R$. Maximality of $z_0$ yields $(R-\mathrm{\mid }{z}_0+rw\mathrm{\mid })\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0+rw)\mathrm{\mid }\le (R-\mathrm{\mid }{z}_0\mathrm{\mid })\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }=2r\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }$. Because $R-\mathrm{\mid }{z}_0+rw\mathrm{\mid }\ge R-\mathrm{\mid }{z}_0\mathrm{\mid }-r=r$, dividing gives $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }=\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0+rw)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\le 2$.

Since $0$ lies in the maximizing disc, maximality also gives $R\mathrm{\mid }{f}^{\mathrm{\prime }}(0)\mathrm{\mid }\le (R-\mathrm{\mid }{z}_0\mathrm{\mid })\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }=2r\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }$, which is the claimed lower bound.

Define $h(w):=(g^{\mathrm{\prime }}(w)-1)\mathrm{/}3$. Then $h$ is holomorphic on $\mathbb{D}$, $h(0)=0$, and $\mathrm{\mid }h(w)\mathrm{\mid }\le (\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }+1)\mathrm{/}3\le 1$. Hence [L1] gives $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)-1\mathrm{\mid }\le 3\mathrm{\mid }w\mathrm{\mid }$ for every $w\in \mathbb{D}$.

If $\mathrm{\mid }w\mathrm{\mid }\le 1\mathrm{/}6$, step 1.1 gives $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)-1\mathrm{\mid }\le 1\mathrm{/}2$. For $w_1,w_2\in D(0,1\mathrm{/}6)$, one has $g(w_1)-g(w_2)={\int }_{[w_2,w_1]}{g}^{\mathrm{\prime }}(\zeta )d\zeta$, so $\mathrm{\mid }g(w_1)-g(w_2)-(w_1-w_2)\mathrm{\mid }\le \mathrm{\mid }{w}_1-w_2\mathrm{\mid }\mathrm{/}2$. Therefore $\mathrm{\mid }g(w_1)-g(w_2)\mathrm{\mid }\ge \mathrm{\mid }{w}_1-w_2\mathrm{\mid }\mathrm{/}2$, which shows injectivity on $D(0,1\mathrm{/}6)$.

On $\mathrm{\mid }w\mathrm{\mid }=1\mathrm{/}6$, one has $\mathrm{\mid }g(w)-w\mathrm{\mid }\le {\int }_0^1\mathrm{\mid }{g}^{\mathrm{\prime }}(tw)-1\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }dt\le 3\mathrm{\mid }w{\mathrm{\mid }}^2\mathrm{/}2=1\mathrm{/}24<1\mathrm{/}6=\mathrm{\mid }w\mathrm{\mid }$. Fix $\xi$ with $\mathrm{\mid }\xi \mathrm{\mid }<1\mathrm{/}12$. Then on $\mathrm{\mid }w\mathrm{\mid }=1\mathrm{/}6$, $\mathrm{\mid }(g(w)-\xi )-w\mathrm{\mid }\le \mathrm{\mid }g(w)-w\mathrm{\mid }+\mathrm{\mid }\xi \mathrm{\mid }<1\mathrm{/}24+1\mathrm{/}12=1\mathrm{/}8<1\mathrm{/}6=\mathrm{\mid }w\mathrm{\mid }$. Rouche [L2] gives the same zero count for $g(w)-\xi$ and $w$, so $g(w)=\xi$ has exactly one solution in $D(0,1\mathrm{/}6)$.

Step 2.2 shows every $\mathrm{\mid }\xi \mathrm{\mid }<1\mathrm{/}12$ lies in $g(D(0,1\mathrm{/}6))$, and step 2.1 shows the restriction there is univalent.

Apply [L1] with $R=1\mathrm{/}2$ to obtain $z_0\in D(0,1\mathrm{/}2)$, a radius $r>0$, and a normalized rescaling $g(w)=(f(z_0+rw)-f(z_0))\mathrm{/}(r{f}^{\mathrm{\prime }}(z_0))$ with $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }\le 2$ on $\mathbb{D}$ and $r\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\ge 1\mathrm{/}4$.

By [L2], the restriction of $g$ to $D(0,1\mathrm{/}6)$ is univalent and its image contains $D(0,1\mathrm{/}12)$. Take the inverse image of that round disc under this univalent restriction and then undo the affine source and target normalizations. This gives a subdomain on which $f$ maps univalently onto $D\text{\hspace{0.17em}⁣}(f(z_0),r\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\mathrm{/}12)$. Since step 1.1 gives $r\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\ge 1\mathrm{/}4$, this radius is at least $1\mathrm{/}48$.

Thus $\beta (f)\ge 1\mathrm{/}48$. Taking the infimum over all normalized $f$ gives $\mathcal{B}\ge 1\mathrm{/}48>0$.

By [L1], some subdomain of $\mathbb{D}$ is mapped by $f$ univalently onto a round disc of radius at least $1\mathrm{/}48$. That round disc is contained in the full image $f(\mathbb{D})$, so $\lambda (f)\ge 1\mathrm{/}48$.

Since every schlicht disc counted by $\beta (f)$ is also a disc inside $f(\mathbb{D})$, one has $\lambda (f)\ge \beta (f)$ for each normalized $f$. Taking infima and using [L1] gives $\mathcal{L}\ge \mathcal{B}>0$.

Both $f$ and $1-f$ are holomorphic and nowhere zero on $\mathbb{D}$. Fact [L1] makes $\mathbb{D}$ homologically simply connected.

Applying [L2] first to $f$ and then to $1-f$ gives holomorphic logarithms $F$ and $G$ with $e^F=f$ and $e^G=1-f$.

By [L1], choose $h$ with $e^{2\pi ih}=f$. Since $f$ omits $1$, the function $h$ omits every integer. In particular $h$ and $h-1$ are nowhere zero, so [L2] gives holomorphic $u,v$ on $\mathbb{D}$ with $u^2=h$ and $v^2=h-1$. Then $(u-v)(u+v)=1$, so $u-v$ is nowhere zero, and [L2] gives a holomorphic $g$ with $e^g=u-v$.

Using [L4] and $(u-v)(u+v)=1$, one gets $u+v=e^{-g}$ and hence $2u=e^g+e^{-g}=2\cosh g$. Therefore $u=\cosh g$, $h=u^2={\cosh }^{2}g=(1+\cosh (2g))\mathrm{/}2$, and $f=e^{2\pi ih}=-e^{\pi i\cosh (2g)}$.

Put $R_0:=\max \{2,R\}$. First suppose $R_0^{-1}\le \mathrm{\mid }f(0)\mathrm{\mid }\le R_0$. In step 1.1 choose the logarithm $h$ so that $\mathrm{\mid }\mathrm{Re}h(0)\mathrm{\mid }\le 1\mathrm{/}2$, which is possible because $h$ is determined up to an integer. Since $\mathrm{\mid }f(0)\mathrm{\mid }=e^{-2\pi \mathrm{Im}h(0)}$, this also gives $\mathrm{\mid }\mathrm{Im}h(0)\mathrm{\mid }\le (\log R_0)\mathrm{/}(2\pi )$ and hence $\mathrm{\mid }h(0)\mathrm{\mid }\le H(R_0)$ for a fixed bound $H(R_0)$. Now $\mathrm{\mid }u(0)\mathrm{\mid }=\sqrt{\mathrm{\mid }h(0)\mathrm{\mid }}$ and $\mathrm{\mid }v(0)\mathrm{\mid }=\sqrt{\mathrm{\mid }h(0)-1\mathrm{\mid }}$, so for $P(R_0):=\sqrt{H(R_0)}+\sqrt{H(R_0)+1}$ one has $\mathrm{\mid }u(0)-v(0)\mathrm{\mid }\le P(R_0)$. Because $(u-v)(u+v)=1$ and $\mathrm{\mid }u(0)+v(0)\mathrm{\mid }\le P(R_0)$, one also has $\mathrm{\mid }u(0)-v(0)\mathrm{\mid }\ge P(R_0{)}^{-1}$. Finally choose the logarithm $g$ of $u-v$ so that $\mathrm{\mid }\mathrm{Im}g(0)\mathrm{\mid }\le \pi$; then $\mathrm{\mid }\mathrm{Re}g(0)\mathrm{\mid }\le \log P(R_0)$ and therefore $\mathrm{\mid }g(0)\mathrm{\mid }\le A(R_0):=\log P(R_0)+\pi$.

Let ${\alpha }_n:=\frac{1}{2}\mathrm{arcosh}(2n+1)$ for $n\ge 0$, and set $E:=\{\pm {\alpha }_n+m\pi i:n\ge 0,m\in \mathbb{Z}\}\cup \{\pm {\alpha }_n+(m+\frac{1}{2})\pi i:n\ge 0,m\in \mathbb{Z}\}$. If $g(z)=\zeta \in E$, then $\cosh (2\zeta )$ is an odd integer, so step 2.1 gives $f(z)=1$, impossible. Hence $g(\mathbb{D})\cap E=\mathrm{\varnothing }$. The horizontal gaps between consecutive ${\alpha }_n$ are less than $1$, and the two vertical translates reduce the vertical gap to $\pi \mathrm{/}2$, so every open Euclidean disc of radius $2$ in $\mathbb{C}$ meets $E$.

Fix $w\in \mathbb{D}$ and rescale $g$ from the disc $D(w,1-\mathrm{\mid }w\mathrm{\mid })$ to the unit disc. If $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }>2\mathrm{/}(b(1-\mathrm{\mid }w\mathrm{\mid }))$, then [L3] would produce a schlicht disc of radius greater than $2$ inside $g(\mathbb{D})$, contradicting step 3.1. Therefore $\mathrm{\mid }{g}^{\mathrm{\prime }}(w)\mathrm{\mid }\le 2\mathrm{/}(b(1-\mathrm{\mid }w\mathrm{\mid }))$ for every $w\in \mathbb{D}$.

If $\mathrm{\mid }f(0)\mathrm{\mid }\ge R_0^{-1}$, integrate step 4.1 radially and use step 2.2 to get $\mathrm{\mid }g(z)\mathrm{\mid }\le A(R_0)+(2\mathrm{/}b)\log \text{\hspace{0.17em}⁣}(1\mathrm{/}(1-r))$ for $\mathrm{\mid }z\mathrm{\mid }\le r$. The formula in step 2.1 then bounds $\mathrm{\mid }f(z)\mathrm{\mid }$ by a constant depending only on $R$ and $r$. If $\mathrm{\mid }f(0)\mathrm{\mid }<R_0^{-1}\le 1\mathrm{/}2$, apply the same construction to $1-f$: its center value lies between $1\mathrm{/}2$ and $3\mathrm{/}2$, so the preceding case with center parameter $2$ uniformly bounds $\mathrm{\mid }1-f(z)\mathrm{\mid }$ and hence $\mathrm{\mid }f(z)\mathrm{\mid }\le 1+\mathrm{\mid }1-f(z)\mathrm{\mid }$. Taking the larger of the two bounds gives the required $C(R,r)$.

If $\mathcal{F}$ is empty, it is chordally normal vacuously. Otherwise fix $a\in \mathrm{\Omega }$ and choose $\rho >0$ with $\overline{D(a,2\rho )}\subset \mathrm{\Omega }$. For each $f\in \mathcal{F}$, at least one of $\mathrm{\mid }f(a)\mathrm{\mid }$, $\mathrm{\mid }1\mathrm{/}f(a)\mathrm{\mid }$, or $\mathrm{\mid }1\mathrm{/}(1-f(a))\mathrm{\mid }$ is at most $2$: if both $\mathrm{\mid }f(a)\mathrm{\mid }<1\mathrm{/}2$ and $\mathrm{\mid }1-f(a)\mathrm{\mid }<1\mathrm{/}2$ held, the triangle inequality would fail. Thus one of the three transforms $T_1(z)=z$, $T_2(z)=1\mathrm{/}z$, $T_3(z)=1\mathrm{/}(1-z)$ has center value of modulus at most $2$.

Each $T_j\circ f$ omits $0$ and $1$, so [L1] applied after rescaling $D(a,2\rho )$ to $\mathbb{D}$ gives a bound $\mathrm{\mid }{T}_j(f(z))\mathrm{\mid }\le M(\rho )$ on $D(a,\rho )$ for the transform selected in step 1.1. Hence every member of the transformed family is locally bounded there. Fact [L2] makes each transformed subfamily locally equicontinuous, and because there are only three fixed inverse transforms, the original family is chordally locally equicontinuous on $D(a,\rho )$.

The target $\widehat{\mathbb{C}}$ is compact, so pointwise relative compactness is automatic. Therefore [A1] and [L3] apply on each $D(a,\rho )$, giving chordal normality there. As $a$ was arbitrary, $\mathcal{F}$ is chordally normal on $\mathrm{\Omega }$.

Assume toward a contradiction that $f$ omits two finite values. After an affine change of target, we may suppose those values are $0$ and $1$. For every $R>0$, apply [L1] with the fixed inner radius $1\mathrm{/}2$ to the map $f_R(z):=f(Rz)$ on $\mathbb{D}$. Since $f_R(0)=f(0)$, the resulting bound is independent of $R$ and gives $\mathrm{\mid }f(w)\mathrm{\mid }\le C(\mathrm{\mid }f(0)\mathrm{\mid },1\mathrm{/}2)$ whenever $\mathrm{\mid }w\mathrm{\mid }\le R\mathrm{/}2$.

Since $R$ in step 1.1 is arbitrary, those discs exhaust $\mathbb{C}$ while the same constant bounds all of them. Thus $f$ is bounded on $\mathbb{C}$. Fact [L2] then makes $f$ constant, contradicting the assumption.

Therefore a nonconstant entire function omits at most one finite value.

After an affine change of target, we may assume the omitted values are $0$ and $1$. Choose radii ${\rho }_n\downarrow 0$ with $2{\rho }_1<R$, and define $f_n(\zeta ):=f(a+{\rho }_n\zeta )$ on the fixed annulus $A:=\{1\mathrm{/}2<\mathrm{\mid }\zeta \mathrm{\mid }<2\}$. Each $f_n$ omits $0$ and $1$, so [A1] and [L1] give a chordally locally uniformly convergent subsequence on $A$; relabel it again as $(f_n)$, with the corresponding radii still written $({\rho }_n)$.

By [L2], the limit of that subsequence is either holomorphic on $A$ or identically $\mathrm{\infty }$. In the first case, chordal local uniform convergence to a finite holomorphic limit is Euclidean local uniform convergence on the unit circle, so there are $M>0$ and $N$ with $\mathrm{\mid }f(a+{\rho }_n\zeta )\mathrm{\mid }\le M$ for every $n\ge N$ and $\mathrm{\mid }\zeta \mathrm{\mid }=1$. In the second case, the same argument applied to the infinity chart gives $M>0$ and $N$ with $\mathrm{\mid }1\mathrm{/}f(a+{\rho }_n\zeta )\mathrm{\mid }\le M$ for every $n\ge N$ and $\mathrm{\mid }\zeta \mathrm{\mid }=1$.

In the first case, fix $n\ge N$ and apply [L3] to the bounded annulus ${\mathrm{\Omega }}_n:=\{z:{\rho }_{n+1}\le \mathrm{\mid }z-a\mathrm{\mid }\le {\rho }_n\}$. Step 2.1 bounds $\mathrm{\mid }f\mathrm{\mid }$ by $M$ on both boundary circles of ${\mathrm{\Omega }}_n$, so $\mathrm{\mid }f(z)\mathrm{\mid }\le M$ throughout ${\mathrm{\Omega }}_n$. As this holds for every $n\ge N$, the function $f$ is bounded on $0<\mathrm{\mid }z-a\mathrm{\mid }\le {\rho }_N$. Fact [L4] then makes $a$ removable.

In the second case, apply the same annulus argument to $1\mathrm{/}f$. Step 2.1 bounds $\mathrm{\mid }1\mathrm{/}f\mathrm{\mid }$ by $M$ on both boundary circles of each ${\mathrm{\Omega }}_n$ for $n\ge N$, hence throughout every such annulus. Therefore [L4] extends $1\mathrm{/}f$ holomorphically across $a$. If the extension is nonzero at $a$, then its reciprocal extends $f$, so $a$ is removable for $f$. If the extension vanishes at $a$, [L6] makes $a$ a pole of $f$.

Steps 3.1 and 3.2 show that only the removable and pole branches of [L5] can occur, so $a$ is not an essential singularity.

Suppose two distinct finite values $w_1,w_2$ each failed to occur infinitely often in some punctured neighborhood of $a$. After passing to the smaller of those neighborhoods, each equation $f(z)=w_j$ would have only finitely many solutions there. Shrink once more past all those finitely many points. The resulting punctured disc omits both $w_1$ and $w_2$, so [L1] would make the singularity removable or a pole, contradicting the hypothesis that it is essential.

Step 1.1 shows that at most one finite value can fail the asserted infinitely-often property. Every other finite value is therefore assumed infinitely often in every punctured neighborhood of $a$.

This is exactly the Great Picard conclusion for finite values.

Assume toward a contradiction that $f$ omits three distinct sphere values. By [L1], choose a Möbius transformation $M$ sending them to $0$, $1$, and $\mathrm{\infty }$. Then $g:=M\circ f$ is meromorphic by [L2], omits $0$, $1$, and $\mathrm{\infty }$, and therefore is actually entire.

Fact [L3] makes an entire function omitting $0$ and $1$ constant, so $g$ is constant. Since $M$ is biholomorphic by [L2], $f=M^{-1}\circ g$ is constant as well, contradicting the hypothesis.

Therefore a nonconstant meromorphic function on the plane omits at most two sphere values.

Suppose three distinct sphere values each failed to occur infinitely often in some punctured neighborhood of $a$. After passing to a common smaller neighborhood and then shrinking past their finitely many preimages, all three values are omitted. By [L2], choose a Möbius transformation $M$ sending them to $0$, $1$, and $\mathrm{\infty }$. Then $g:=M\circ f$ is holomorphic on that smaller punctured disc and still has an essential singularity at $a$, because a biholomorphic target change cannot turn an essential singularity into a removable singularity or pole.

The function $g$ omits the finite values $0$ and $1$, so [L1] gives a contradiction. Thus at most two sphere values can fail the infinitely-often property, and every other sphere value occurs infinitely often in every punctured neighborhood.

This is the meromorphic Great Picard conclusion.