The Riemann Mapping Theorem

Because $\mathrm{\Omega }⊊\mathbb{C}$, choose $a\in \mathbb{C}\setminus \mathrm{\Omega }$. The function $z\mapsto z-a$ is holomorphic and nowhere zero on $\mathrm{\Omega }$, so [L1] gives a holomorphic $q:\mathrm{\Omega }\to \mathbb{C}$ with $q(z{)}^2=z-a$.

If $q(z_1)=q(z_2)$ then $z_1-a=q(z_1{)}^2=q(z_2{)}^2=z_2-a$, so $z_1=z_2$; thus $q$ is injective. Also $0\mathrm{\notin }q(\mathrm{\Omega })$, and $q(\mathrm{\Omega })\cap (-q(\mathrm{\Omega }))=\mathrm{\varnothing }$ because $q(z_1)=-q(z_2)$ would again force $z_1=z_2$, hence $q(z_1)=0$, impossible.

Put $w_0:=q(z_0)$. Since $q$ is nonconstant, [L3] makes $q(\mathrm{\Omega })$ open, so choose $\rho >0$ with $D(w_0,\rho )\subseteq q(\mathrm{\Omega })$. Step 2.1 gives $q(\mathrm{\Omega })\cap (-q(\mathrm{\Omega }))=\mathrm{\varnothing }$, hence $D(-w_0,\rho )=-D(w_0,\rho )\subseteq -q(\mathrm{\Omega })$ is disjoint from $q(\mathrm{\Omega })$. Therefore $\mathrm{\mid }q(z)+w_0\mathrm{\mid }\ge \rho$ for every $z\in \mathrm{\Omega }$.

Define $h(z):=\frac{\rho }{2(q(z)+w_0)}\mathrm{.}$ Step 3.1 gives $\mathrm{\mid }h(z)\mathrm{\mid }\le 1\mathrm{/}2$, so $h(\mathrm{\Omega })\subseteq \mathbb{D}$. The reciprocal affine map is injective away from $-w_0$, and step 2.1 makes $q$ injective, so $h$ is holomorphic and injective on $\mathrm{\Omega }$.

Let $b=h(z_0)\in \mathbb{D}$. By [L2], $g:={\phi }_b\circ h$ is holomorphic and injective from $\mathrm{\Omega }$ into $\mathbb{D}$, and $g(z_0)=0$. Differentiating $q^2=z-a$ gives $q^{\mathrm{\prime }}(z_0)=1\mathrm{/}(2w_0)\mathrm{\ne }0$, so $g^{\mathrm{\prime }}(z_0)\mathrm{\ne }0$. Multiplying by the unimodular constant $\overline{g^{\mathrm{\prime }}(z_0)}\mathrm{/}\mathrm{\mid }{g}^{\mathrm{\prime }}(z_0)\mathrm{\mid }$ makes the derivative at $z_0$ positive. The resulting map lies in $\mathcal{F}(\mathrm{\Omega },z_0)$.

Fact [L1] gives at least one map in $\mathcal{F}(\mathrm{\Omega },z_0)$, so the derivative set $E$ is nonempty. Fact [L2] makes every element of $E$ strictly positive.

Choose $\rho >0$ with $\overline{D(z_0,\rho )}\subset \mathrm{\Omega }$. If $f\in \mathcal{F}(\mathrm{\Omega },z_0)$, then $\mathrm{\mid }f\mathrm{\mid }\le 1$ on $D(z_0,\rho )$ because $f(\mathrm{\Omega })\subset \mathbb{D}$, so [L3] gives $\mathrm{\mid }{f}^{\mathrm{\prime }}(z_0)\mathrm{\mid }\le 1\mathrm{/}\rho$. Since [L2] makes $f^{\mathrm{\prime }}(z_0)$ positive real, this is the same as $f^{\mathrm{\prime }}(z_0)\le 1\mathrm{/}\rho$.

Therefore $E\subset (0,1\mathrm{/}\rho ]$, so $E$ has a finite supremum.

By [A1] and [L1], choose a sequence $(f_n)$ in $\mathcal{F}(\mathrm{\Omega },z_0)$ with $f_n^{\mathrm{\prime }}(z_0)\to M$. Because every $f_n$ maps $\mathrm{\Omega }$ into $\mathbb{D}$, the family is locally bounded, so [L2] gives a locally uniformly convergent subsequence, still denoted $(f_n)$, with holomorphic limit $f$ on $\mathrm{\Omega }$.

For every $n$, one has $f_n(z_0)=0$, so the locally uniform convergence of step 1.1 gives $f(z_0)=0$. Fact [L3] gives $f_n^{\mathrm{\prime }}(z_0)\to f^{\mathrm{\prime }}(z_0)$, hence $f^{\mathrm{\prime }}(z_0)=M$.

Because $M>0$ by [L1], step 2.1 makes $f$ nonconstant. Also $\mathrm{\mid }f\mathrm{\mid }\le 1$ on $\mathrm{\Omega }$ as a locally uniform limit of disc-valued maps. If $\mathrm{\mid }f(a)\mathrm{\mid }=1$ at some $a\in \mathrm{\Omega }$, then $f(\mathrm{\Omega })$ would be an open subset of the closed unit disc by [L4], impossible. Hence $f(\mathrm{\Omega })\subset \mathbb{D}$.

The map $f$ therefore has the required normalization and extremal derivative.

Fix $a\in \mathrm{\Omega }$. For each $n$, define $h_n(z):=\frac{f_n(z)-f_n(a)}{z-a}(z\in \mathrm{\Omega }\setminus \{a\})\mathrm{.}$ Since each $f_n$ is injective by [L1], the function $h_n$ has no zeros on $\mathrm{\Omega }\setminus \{a\}$. The removable singularity at $a$ is filled by $h_n(a):=f_n^{\mathrm{\prime }}(a)$, so each $h_n$ is holomorphic and nowhere zero on $\mathrm{\Omega }$.

The functions $h_n$ converge locally uniformly to $h(z):=\{\begin{array}{ll}{\displaystyle \frac{f(z)-f(a)}{z-a}},& z\mathrm{\ne }a,\\ f^{\mathrm{\prime }}(a),& z=a,\end{array}$ because $f_n\to f$ locally uniformly and derivatives converge locally uniformly as well. Fact [L2] therefore makes $h$ either identically zero or nowhere zero.

Since $f$ is nonconstant, the function $h$ is not identically zero. Hence step 2.1 makes $h$ nowhere zero. If $f(z)=f(a)$, then $h(z)=0$ unless $z=a$, so necessarily $z=a$. As $a$ was arbitrary, $f$ is injective and therefore univalent by [L1].

By [L1], the derivative of $f$ at the basepoint is positive, so $f$ is nonconstant.

The maximizing sequence consists of univalent maps, so [L2] applies to the local uniform convergence in the given data. Together with step 1.1 it yields that $f$ is univalent.

Assume toward a contradiction that $f(\mathrm{\Omega })⊊\mathbb{D}$. Choose $c\in \mathbb{D}\setminus f(\mathrm{\Omega })$. Since $f(z_0)=0$, one has $c\mathrm{\ne }0$. Put $\beta :={\phi }_c\circ f$. Then [L2] makes $\beta$ holomorphic and injective into $\mathbb{D}$, with $\beta (\mathrm{\Omega })\subset \mathbb{D}\setminus \{0\}$.

By [L3], the nowhere-zero holomorphic function $\beta$ has a holomorphic square root $q$ on $\mathrm{\Omega }$ with $q^2=\beta$. If $q(z_1)=q(z_2)$ then $\beta (z_1)=\beta (z_2)$, so injectivity of $\beta$ gives $z_1=z_2$. If $q(z_1)=-q(z_2)$ then again $\beta (z_1)=\beta (z_2)$, so $z_1=z_2$ and then $q(z_1)=0$, impossible because $\beta$ never vanishes. Hence $q$ is injective.

Let $a:=q(z_0)$, so $a^2=\beta (z_0)={\phi }_c(0)=c$. Define $g:={\phi }_a\circ q$. Then [L2] makes $g$ holomorphic and injective from $\mathrm{\Omega }$ into $\mathbb{D}$, with $g(z_0)=0$. Thus after multiplying by a unimodular constant if needed, $g$ is another competitor in $\mathcal{F}(\mathrm{\Omega },z_0)$.

Differentiate $q^2=\beta$ at $z_0$ to obtain $2a{q}^{\mathrm{\prime }}(z_0)={\beta }^{\mathrm{\prime }}(z_0)={\phi }_c^{\mathrm{\prime }}(0)f^{\mathrm{\prime }}(z_0)=-(1-\mathrm{\mid }c{\mathrm{\mid }}^2)M$. Since ${\phi }_a^{\mathrm{\prime }}(a)=-1\mathrm{/}(1-\mathrm{\mid }a{\mathrm{\mid }}^2)$ and $\mathrm{\mid }a{\mathrm{\mid }}^2=\mathrm{\mid }c\mathrm{\mid }$, one gets $\mathrm{\mid }{g}^{\mathrm{\prime }}(z_0)\mathrm{\mid }=\frac{1-\mathrm{\mid }c{\mathrm{\mid }}^2}{2\mathrm{\mid }a\mathrm{\mid }(1-\mathrm{\mid }a{\mathrm{\mid }}^2)}M=\frac{1+\mathrm{\mid }c\mathrm{\mid }}{2\sqrt{\mathrm{\mid }c\mathrm{\mid }}}M>M,$ because $(1+\mathrm{\mid }c\mathrm{\mid })\mathrm{/}2>\sqrt{\mathrm{\mid }c\mathrm{\mid }}$ for $0<\mathrm{\mid }c\mathrm{\mid }<1$. This contradicts the extremal definition of $M$.

Therefore the assumption of step 1.1 is false, so $f(\mathrm{\Omega })=\mathbb{D}$.

By [A1] and [L1], choose a holomorphic map $f:\mathrm{\Omega }\to \mathbb{D}$ with $f(z_0)=0$ and extremal derivative $f^{\mathrm{\prime }}(z_0)>0$.

Fact [L2] makes this map injective and gives $f(\mathrm{\Omega })=\mathbb{D}$. Since $\mathrm{\Omega }$ is a complex domain by the given data, [L3] makes $f$ biholomorphic onto its image, which is exactly $\mathbb{D}$.

The map of step 2.1 has the required normalization, so it is the desired conformal equivalence.

The composite $h:=g\circ f^{-1}$ is a biholomorphic self-map of $\mathbb{D}$, and $h(0)=0$ because both maps send $z_0$ to $0$. Thus [L2] gives $h(\zeta )=e^{i\theta }\zeta$ for some real $\theta$.

Differentiate the identity $g=h\circ f$ at $z_0$. By [L3], $g^{\mathrm{\prime }}(z_0)=h^{\mathrm{\prime }}(0)f^{\mathrm{\prime }}(z_0)=e^{i\theta }{f}^{\mathrm{\prime }}(z_0)\mathrm{.}$ Since both displayed derivatives in the statement are positive real numbers, $e^{i\theta }=1$.

Hence $h$ is the identity on $\mathbb{D}$, so $g=h\circ f=f$.

Fix $0<r<1$ and put ${\gamma }_r(t):=g(r{e}^{it})$. By [L1], ${\gamma }_r$ is simple on $[0,2\pi )$, and by [L2], ${\gamma }_r^{\mathrm{\prime }}(t)=ir{e}^{it}{g}^{\mathrm{\prime }}(r{e}^{it})\mathrm{\ne }0.$ Thus ${\mathrm{\Gamma }}_r:={\gamma }_r([0,2\pi ])$ is a regular real-analytic simple closed curve. Since $g(z)=z^{-1}+O(z)$ near zero, the image $g(\{0<\mathrm{\mid }z\mathrm{\mid }<r\})$ is the unbounded side of ${\mathrm{\Gamma }}_r$; write $E_r$ for the bounded side. The parametrization ${\gamma }_r$ is clockwise relative to $E_r$.

The real and imaginary coordinate functions of ${\gamma }_r$ and their derivatives are real analytic. Neither coordinate derivative is identically zero, since a regular simple closed curve cannot lie in one vertical or horizontal line. By [L3], the zeros of each derivative are isolated; periodic real analyticity and compactness of the parameter circle make both zero sets finite. Subdivide at those finitely many critical parameters and at the finitely many intersections with their horizontal and vertical critical lines. The nonintersecting coordinate-monotone arcs then bound finitely many pieces, each describable both between two piecewise-$C^1$ graphs in $x$ and between two such graphs in $y$. These pieces have disjoint interiors and share complete oppositely oriented arcs, so they supply $E_r$ with a finite elementary Green decomposition in the sense of [L4].

Apply [L5] to the decomposition in step 2.1 and reverse the clockwise orientation from step 1.1. Writing $w=g(z)$ and using $dw=g^{\mathrm{\prime }}(z)dz$ gives $\mathrm{Area}(E_r)=-\frac{1}{2i}{\int }_{\mathrm{\mid }z\mathrm{\mid }=r}\overline{g(z)}{g}^{\mathrm{\prime }}(z)dz\mathrm{.}$

On $\mathrm{\mid }z\mathrm{\mid }=r$, one has $\bar{z}=r^2\mathrm{/}z$, so $\overline{g(z)}=\frac{z}{r^2}+{\sum }_{n\ge 1}\overline{b_n}\frac{r^{2n}}{z^n}\mathrm{.}$ Multiplying by $g^{\mathrm{\prime }}(z)=-\frac{1}{z^2}+{\sum }_{n\ge 1}n{b}_n{z}^{n-1}$ and taking the contour integral leaves only the $z^{-1}$ coefficient. Hence $\frac{1}{2i}{\int }_{\mathrm{\mid }z\mathrm{\mid }=r}\overline{g(z)}{g}^{\mathrm{\prime }}(z)dz=\pi \text{\hspace{0.17em}⁣}({\sum }_{n\ge 1}n\mathrm{\mid }{b}_n{\mathrm{\mid }}^2{r}^{2n}-\frac{1}{r^2})\mathrm{.}$

Combining steps 3.1 and 4.1 with $\mathrm{Area}(E_r)\ge 0$ gives ${\sum }_{n\ge 1}n\mathrm{\mid }{b}_n{\mathrm{\mid }}^2{r}^{2n}\le \frac{1}{r^2}\mathrm{.}$ For every $N$, discard the nonnegative terms with $n>N$ and let $r\uparrow 1$ to obtain ${\sum }_{n=1}^{N}n\mathrm{\mid }{b}_n{\mathrm{\mid }}^2\le 1$. Letting $N\to \mathrm{\infty }$ proves the asserted inequality.

Since $f$ is injective and $f(0)=0$, the only zero of $f$ in $\mathbb{D}$ is $0$. Hence $F(z):=\frac{f(z^2)}{z^2}=1+a_2{z}^2+a_3{z}^4+\cdots$ extends holomorphically and nowhere vanishingly to $\mathbb{D}$. By [L2] and [L3], choose a holomorphic square root $q$ on $\mathbb{D}$ with $q(z{)}^2=F(z)$ and $q(0)=1$.

Put $h(z):=zq(z)$. Then $h(z{)}^2=f(z^2)$. The function $h$ is odd and univalent: if $h(z_1)=h(z_2)$ then $f(z_1^2)=f(z_2^2)$, so $z_1^2=z_2^2$; if $z_1=-z_2\mathrm{\ne }0$, oddness gives $h(z_1)=-h(z_2)$, contradiction. Thus $h(z)=z+\frac{a_2}{2}{z}^3+\cdots \mathrm{.}$

Define $G(\zeta ):=\frac{1}{h(\zeta )}=\frac{1}{\zeta }-\frac{a_2}{2}\zeta +\cdots (0<\mathrm{\mid }\zeta \mathrm{\mid }<1)\mathrm{.}$ Since $h$ is injective on $\mathbb{D}$ and vanishes only at $0$, the map $G$ is holomorphic and injective on the punctured disc. Fact [L1] therefore applies and yields ${\mid \frac{a_2}{2}\mid }^2\le 1.$

Therefore $\mathrm{\mid }{a}_2\mathrm{\mid }\le 2$.

Assume toward a contradiction that $w\in \mathbb{C}$ with $0<\mathrm{\mid }w\mathrm{\mid }<1\mathrm{/}4$ is omitted by $f(\mathbb{D})$. Define $g(z):=\frac{wf(z)}{w-f(z)}\mathrm{.}$ Then $g$ is holomorphic and univalent on $\mathbb{D}$, $g(0)=0$, and $g^{\mathrm{\prime }}(0)=1,g(z)=z+(a_2+\frac{1}{w})z^2+\cdots \mathrm{.}$

Since $g\in \mathcal{S}$, fact [L1] gives $\mid a_2+\frac{1}{w}\mid \le 2.$ Applying [L1] again to $f$ yields $\mathrm{\mid }{a}_2\mathrm{\mid }\le 2$, so $\frac{1}{\mathrm{\mid }w\mathrm{\mid }}\le \mid a_2+\frac{1}{w}\mid +\mathrm{\mid }{a}_2\mathrm{\mid }\le 4.$ Thus $\mathrm{\mid }w\mathrm{\mid }\ge 1\mathrm{/}4$, contradicting step 1.1.

Therefore no value of modulus less than $1\mathrm{/}4$ is omitted, so $D(0,1\mathrm{/}4)\subset f(\mathbb{D})$.

By rotating the source and target, it is enough to treat the case $z=r\in [0,1)$. Define ${\psi }_r(\zeta ):=\frac{\zeta +r}{1+r\zeta },g(\zeta ):=\frac{f({\psi }_r(\zeta ))-f(r)}{(1-r^2)f^{\mathrm{\prime }}(r)}\mathrm{.}$ Fact [L1] makes ${\psi }_r$ an automorphism of $\mathbb{D}$, so $g\in \mathcal{S}$.

Differentiate twice at $0$. Since ${\psi }_r^{\mathrm{\prime }}(0)=1-r^2$ and ${\psi }_r^{\mathrm{\prime }\mathrm{\prime }}(0)=-2r(1-r^2)$, one gets $g^{\mathrm{\prime }\mathrm{\prime }}(0)=(1-r^2)\frac{f^{\mathrm{\prime }\mathrm{\prime }}(r)}{f^{\mathrm{\prime }}(r)}-2r\mathrm{.}$ Because $g\in \mathcal{S}$, [L2] gives $\mathrm{\mid }{g}^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{\mid }\le 4$. Therefore $\mid \frac{f^{\mathrm{\prime }\mathrm{\prime }}(r)}{f^{\mathrm{\prime }}(r)}-\frac{2r}{1-r^2}\mid \le \frac{4}{1-r^2}\mathrm{.}$

Put $H(r):=\log \text{\hspace{0.17em}⁣}((1-r^2)f^{\mathrm{\prime }}(r)),$ choosing a continuous branch along $[0,r]$ since $f^{\mathrm{\prime }}$ never vanishes on $\mathbb{D}$ for univalent $f$. Then step 2.1 yields $\mathrm{\mid }{H}^{\mathrm{\prime }}(t)\mathrm{\mid }=\mid \frac{f^{\mathrm{\prime }\mathrm{\prime }}(t)}{f^{\mathrm{\prime }}(t)}-\frac{2t}{1-t^2}\mid \le \frac{4}{1-t^2}(0\le t<1)\mathrm{.}$

Integrating step 3.1 from $0$ to $r$ and using $f^{\mathrm{\prime }}(0)=1$ gives $\mathrm{\mid }H(r)\mathrm{\mid }\le {\int }_0^r\frac{4}{1-t^2}dt=2\log \frac{1+r}{1-r}\mathrm{.}$ Exponentiating and dividing by $1-r^2$ yields $\frac{1-r}{(1+r{)}^3}\le \mathrm{\mid }{f}^{\mathrm{\prime }}(r)\mathrm{\mid }\le \frac{1+r}{(1-r{)}^3}\mathrm{.}$

The argument of steps 1.1 through 4.1 applies after rotation to every point of modulus $r$, so the same bounds hold for the original $z$.

Write $z=r{e}^{i\theta }$. Since $f(0)=0$, $f(z)={\int }_0^r{e}^{i\theta }{f}^{\mathrm{\prime }}(t{e}^{i\theta })dt\mathrm{.}$ Therefore $\mathrm{\mid }f(z)\mathrm{\mid }\le {\int }_0^r\mathrm{\mid }{f}^{\mathrm{\prime }}(t{e}^{i\theta })\mathrm{\mid }dt\mathrm{.}$

For the lower bound, choose $z_0$ on $\mathrm{\mid }z\mathrm{\mid }=r$ for which $\mathrm{\mid }f(z_0)\mathrm{\mid }$ is minimal. The segment from $0$ to $f(z_0)$ lies in $f(\overline{D(0,r)})$: otherwise its first exit point from $f(D(0,r))$ would be an image of the circle $\mathrm{\mid }z\mathrm{\mid }=r$ having modulus strictly smaller than $\mathrm{\mid }f(z_0)\mathrm{\mid }$. Since $f$ is univalent, this segment has a lift $\gamma$ from $0$ to $z_0$.

Applying [L1] inside the integral gives $\mathrm{\mid }f(z)\mathrm{\mid }\le {\int }_0^r\frac{1+t}{(1-t{)}^3}dt=\frac{r}{(1-r{)}^2}\mathrm{.}$

The image of $\gamma$ is a straight segment, so [L1] gives $\mathrm{\mid }f(z_0)\mathrm{\mid }={\int }_{\gamma }\mathrm{\mid }{f}^{\mathrm{\prime }}(\zeta )\mathrm{\mid }\mathrm{\mid }d\zeta \mathrm{\mid }\ge {\int }_{\gamma }\frac{1-\mathrm{\mid }\zeta \mathrm{\mid }}{(1+\mathrm{\mid }\zeta \mathrm{\mid }{)}^3}\mathrm{\mid }d\zeta \mathrm{\mid }\ge {\int }_0^r\frac{1-t}{(1+t{)}^3}dt=\frac{r}{(1+r{)}^2}\mathrm{.}$ The penultimate inequality follows because $\gamma$ joins radius $0$ to radius $r$, while the integrand is positive and depends only on the radius.

Minimality of $z_0$ now gives $\mathrm{\mid }f(z)\mathrm{\mid }\ge \mathrm{\mid }f(z_0)\mathrm{\mid }\ge r\mathrm{/}(1+r{)}^2$ for every $\mathrm{\mid }z\mathrm{\mid }=r$. Together with step 2.1 this proves both bounds.

Define $g(\zeta ):=\frac{f({\phi }_a(\zeta ))-f(a)}{-(1-\mathrm{\mid }a{\mathrm{\mid }}^2)f^{\mathrm{\prime }}(a)}\mathrm{.}$ By [L1], the map ${\phi }_a$ is an automorphism of $\mathbb{D}$ with ${\phi }_a(0)=a$ and ${\phi }_a^{\mathrm{\prime }}(0)=-(1-\mathrm{\mid }a{\mathrm{\mid }}^2)$, so $g$ is holomorphic and univalent on $\mathbb{D}$, with $g(0)=0$ and $g^{\mathrm{\prime }}(0)=1$.

Thus $g\in \mathcal{S}$, and [L2] gives $D(0,1\mathrm{/}4)\subset g(\mathbb{D})$. Multiplying by the affine factor from step 1.1 and translating back yields $D\text{\hspace{0.17em}⁣}(f(a),\frac{(1-\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{\mid }{f}^{\mathrm{\prime }}(a)\mathrm{\mid }}{4})\subset f(\mathbb{D})\mathrm{.}$