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If a holomorphic function $F$ on the unit disc has positive real part, the Cayley transform between the half-plane and disk $\varphi =(F-F(0))/(F+\overline{F(0)})$ maps the disc into itself and vanishes at zero. The Schwarz lemma gives the displayed derivative bound. Equality holds exactly when $\varphi (z)=\omega z$ with $|\omega |=1$, equivalently $F(z)=i\mathrm{Im}F(0)+\mathrm{Re}F(0)(1+\omega z)/(1-\omega z)$.

Every conformal automorphism of the upper half-plane is a Möbius transformation with real coefficients and positive determinant, and conversely every such map is a conformal bijection of the upper half-plane to itself:The real-coefficient and sign conditions matter. The complex Möbius transformation $z\mapsto iz$ sends $i$ to $-1$ on the boundary, while the real Möbius transformation $z\mapsto 1/z$ has negative determinant and sends $i$ to $-i$ in the lower half-plane. ^[ref]

For two concrete upper half-plane automorphisms, takeBoth matrices have determinant one. Here $T(i)=1+i$, $S(i)=i$, and $S(1+i)=(-1+i)/2$. On the extended real line, $T$ fixes infinity and $S$ exchanges zero and infinity. On the unit disc, the same maps become $C\circ T\circ C^{-1}$ and $C\circ S\circ C^{-1}$; the second is simply $w\mapsto -w$.

Multiplying all four coefficients by a nonzero real scalar leaves $f$ unchanged. Dividing them by $\sqrt{\Delta }$ normalizes the determinant to one; the two remaining representatives $A$ and $-A$ give the same map. Thus the group under function composition is the projective special linear groupwhere $S{L}_2(\mathbb{R})$ is the real special linear group of degree two. Composition of the Möbius transformations corresponds to multiplication of their matrices. ^[ref]

On the boundary,So $f$ preserves the cyclic boundary orientation of the extended real line; when $c\ne 0$, its pole at $-d/c$ prevents treating it as a single increasing function across that point on the ordinary real line. Three distinct boundary images determine a Möbius transformation uniquely, but an upper half-plane automorphism can realize only triples with matching cyclic boundary orientation.

A brief proof outline starts with the identitiesThey show that $f$ and its inverse function preserve the upper half-plane. For completeness of the classification, conjugate an arbitrary conformal bijection by the Cayley transform. Normalize its image of zero with an automorphism of the unit disk, then apply the Schwarz lemma to the resulting map and its inverse function; both inequalities force a rotation. Conjugating back gives the real Möbius transformation above.

Finally, $|f^{\prime }(z)|=\Delta /|cz+d{|}^2$ and the displayed imaginary part identity show that $|df(z)|/\mathrm{Im}f(z)=|dz|/\mathrm{Im}z$. Every upper half-plane automorphism is therefore also an isometry of the Poincaré half-plane model. The same formulas describe both holomorphic coordinate symmetries and motions of the hyperbolic plane.

Let $L>0$ and consider holomorphic functions $f$ from the unit disc into $e^{-L}<|w|<e^L$, normalized by $f(0)=1$. Their largest possible derivative modulus is $4L/\pi$. Lift $f$ to its holomorphic logarithm $h$ with $h(0)=0$, so $|\mathrm{Re}h|<L$. The function $\tan (\pi h/(4L))$ maps the disc into itself and fixes zero. The Schwarz lemma gives $|h^{\prime }(0)|\le 4L/\pi$. Equality holds precisely forHere $\arctan$ is its holomorphic branch vanishing at zero; $(4L/\pi )\arctan z$ is a conformal map from the disc onto the vertical strip. Thus the maximal derivative modulus is attained, with a circle of distinct extremizers rather than a unique map.

Exhaust a simply connected hyperbolic Riemann surface in potential theory by compact bordered Jordan domains $R_n$. Their Green functions on a Riemann surface $G_n(\cdot ,p)$ have logarithmic poles and zero boundary values. Exponentiating their harmonic conjugates produces proper degree-one maps $f_n:R_n\to \mathbb{D}$ with $f_n(p)=0$. If $G_n=-\log |\zeta |+c_n+o(1)$, normalize $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity and a global positive Green function bound $c_n$ increasingly from above. Thus Montel theorem gives a nonconstant limit $F$, which is injective by the locally uniform limit of univalent functions theorem. The Schwarz lemma makes $F^{\prime }(p)$ extremal among normalized disk-valued maps. If $F$ omitted $a\ne 0$, taking a global holomorphic square root of a disk automorphism applied to $F$, then renormalizing at $p$, would multiply its derivative by $(1+|a|)/(2\sqrt{|a|})>1$. Thus $F$ is onto and gives the desired biholomorphism.

Either equality condition says that the holomorphic function $h$ from part (i) has modulus one at an interior point: at zero in the derivative case, or at the given nonzero point in the other case. Since $|h|\le 1$ everywhere, the maximum modulus principle makes $h$ a constant $w$ with $|w|=1$. ThereforeThese maps are rotations of the unit disk and do achieve both equalities, so this also identifies all equality cases of the Schwarz lemma.

Because $f(0)=0$, the Taylor series of $f$ at zero shows thatis holomorphic throughout the unit disk. Fix $0<r<1$. On $|z|=r$ we have $|h(z)|<1/r$, and the maximum modulus principle on the closed disk of radius $r$ gives $|h(z)|\le 1/r$ for $|z|\le r$. For any fixed $z$ inside the unit disk we may let $r$ increase to one, obtaining $|h(z)|\le 1$. Hence the Schwarz lemma inequalities areThe derivative bound is the same estimate at $z=0$, using the holomorphic extension of the quotient.

Since $f(0)=0$, the function $g(z)=f(z)/z$ extends analytically to zero by $g(0)=f^{\prime }(0)$. It is bounded on the disk: outside $|z|<1/2$, $|g(z)|\le 2$, while on the closed half-radius disk its analytic extension is continuous and hence bounded.

Fix $z\ne 0$ and choose $|z|<r<1$. Let $T_r$ be the first exit time of Brownian motion started at $z$ from the radius-$r$ disk. The bounded analytic martingale identity in part (a) givesOn this exit circle, $|g(Z_{T_r})|=|f(Z_{T_r})|/r\le 1/r$. The triangle inequality gives $|g(z)|\le 1/r$. Letting $r\uparrow 1$ proves $|g(z)|\le 1$, and thereforeAt zero the claim holds by the assumed zero value. This is the Schwarz lemma, obtained from Brownian optional stopping rather than an analytic maximum principle.

For $a\in \mathbb{D}$, the automorphism of the unit disk $T_a(z)=(z-a)/(1-\bar{a}z)$ sends $a$ to zero. The holomorphic function $F=T_{f(a)}\circ f\circ T_a^{-1}$ maps the unit disc into itself and satisfies $F(0)=0$. The Schwarz lemma therefore gives $|F(w)|\le |w|$. Evaluate at $w=T_a(b)$ to obtain the Schwarz-Pick theorem:The denominators do not vanish because both their products have modulus less than one.

For a holomorphic function $h:\mathbb{D}\to \mathbb{D}$, the Schwarz-Pick lemma statesEquality in the first inequality for distinct points, or in the derivative inequality at one point, occurs exactly when $h$ is an automorphism of the unit disk.

Here is a proof, including the underlying Schwarz lemma. If $u:\mathbb{D}\to \mathbb{D}$ is holomorphic and $u(0)=0$, then $u(z)/z$ has a removable singularity at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle on this circle, followed by $r\uparrow 1$, gives $|u(z)|\le |z|$ and $|u^{\prime }(0)|\le 1$. Equality at a nonzero point, or $|u^{\prime }(0)|=1$, forces $u(z)/z$ to be a constant of modulus one by the same maximum modulus principle.

Set ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$, an automorphism of the unit disk. The composite ${\varphi }_{h(a)}\circ h\circ {\varphi }_a^{-1}$ fixes zero. The Schwarz lemma gives the first inequality, while its derivative at zero gives the second. The equality conclusion follows because a composite attaining equality is a rotation. Conversely, every automorphism of the unit disk attains equality. This also proves contraction of the hyperbolic metric.

The boundary lower bound first guarantees a zero inside the unit circle. If $f(0)=0$, there is nothing to prove about existence. Otherwise, if $f$ had no zeros on the closed unit disc, $1/f$ would be holomorphic on a neighborhood of it and satisfy $|1/f|\le 1/A$ on the boundary. The maximum modulus principle would give $|f(0)|\ge A$, contradicting the hypothesis. The boundary lower bound excludes boundary zeros, so there is a zero $a$ with $|a|<1$.

The upper boundary bound and the maximum modulus principle give $|f|\le B$ on the unit circle and its interior. The function is nonconstant under the stated hypotheses, so $f/B$ takes its interior values in the open unit disc. Apply the Schwarz-Pick lemma to the points $a$ and $0$:This is the zero location bound from Schwarz-Pick, and it applies to every zero inside the unit circle.

The bound is sharp. Given $0<a<1$ and $B>0$, chooseThis holomorphic function has no pole in ${\mathbb{D}}_R$, has constant modulus $B$ on the unit circle, and satisfies $|f(0)|=Ba<B$. With $A=B$, its zero at $a$ gives equality. The case $a=0$ is also attained by $f(z)=Bz$.

Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.

For a plane domain, the following are equivalent descriptions of a simply connected domain: every based loop contracts in the domain with its basepoint fixed; any two paths with the same endpoints are homotopic while those endpoints stay fixed; its complement in the Riemann sphere is connected; every holomorphic function has a single-valued primitive; and every nowhere-zero holomorphic function has a holomorphic logarithm. The complement must be taken in the extended plane, including infinity. An equivalent winding number formulation is that every closed curve has index zero about every point outside the domain.

Here is a direct equivalence proof for the first two topological formulations. If all paths with given endpoints are homotopic relative to those endpoints, apply this to a loop based at $a$ and the constant path at $a$, obtaining a contraction. Conversely, let ${\gamma }_0,{\gamma }_1$ join $a$ to $b$. The loop ${\gamma }_0\ast {\gamma }_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with ${\gamma }_1$ produces a homotopy from ${\gamma }_0\ast {\gamma }_1^{-1}\ast {\gamma }_1$ to ${\gamma }_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a homotopy from that first path to ${\gamma }_0$. Hence ${\gamma }_0$ and ${\gamma }_1$ are homotopic relative to their endpoints. Planar domains are path-connected, so no extra path-component condition is needed.

The analytic descriptions follow from path independence of the holomorphic integral. Local primitives exist on discs by integrating power series; a homotopy can be subdivided into pieces lying in such discs, and local path integrals cancel along its common edges. Thus integrals over homotopic paths agree, giving a global primitive. Applied to $h^{\prime }/h$ for a nowhere-zero function, a primitive $H$ makes $h{e}^{-H}$ constant and hence supplies a holomorphic logarithm. These observations are also the tools used in the mapping proof below.

The Riemann mapping theorem states that every nonempty simply connected proper domain $D⊊\mathbb{C}$ admits a conformal bijection onto the unit disc. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f^{\prime }(z_0)>0$. The exclusion of the whole plane is necessary: an entire map into the disc would be constant by Liouville's theorem.

An extremal proof has the following concrete steps. First construct an injective bounded holomorphic map. Choose $a\notin D$ and a holomorphic square root $h$ of $z-a$. It is injective, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is injective and bounded. Scaling it and applying an automorphism of the unit disk gives a map into the disc taking $z_0$ to zero; a rotation makes its derivative positive.

Let $\mathcal{U}$ consist of all injective maps $D\to \mathbb{D}$ with this normalization. It is nonempty. The derivative at $z_0$ is bounded above by the Cauchy integral formula on a small disc compactly contained in $D$. Choose $f_n\in \mathcal{U}$ with derivatives tending to their positive supremum $L$. Boundedness, the Cauchy estimates and a diagonal Arzelà-Ascoli theorem argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the Montel theorem. Then $f(z_0)=0$, $f^{\prime }(z_0)=L>0$, and the maximum modulus principle puts its image in the open disc. The limit is injective: if distinct $z,w$ had $f(z)=f(w)$, apply Hurwitz's theorem near $w$ to $f_n(\cdot )-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f^{\prime }(z_0)>0$.

Finally suppose $a\in \mathbb{D}$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline{a}w)$ and choose a holomorphic square root $h$ of $T_a\circ f$. It is injective and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a rotation. The new map's derivative magnitude at $z_0$ isThis square-root improvement of a normalized conformal map contradicts extremality. Thus $f$ is onto. Two normalized bijections differ by a disc automorphism fixing zero; Schwarz lemma applied to it and its inverse makes it a rotation, and positivity of the normalized derivatives forces that rotation to be the identity. This completes the requested proof outline.

An explicit disk map of a slit half disk makes the two boundary approaches visible without assuming an extension theorem for the boundary. We construct the map as a succession of bijections and then determine its limits.

First $w=z^2$ maps the upper half-disc of radius two bijectively onto $|w|<4$ slit along $[0,4)$. Indeed, its inverse chooses the square root whose argument lies in $(0,\pi )$. The removed segment on the imaginary axis becomes $[-1,0)$, so the image of $D$ is preciselyThe Möbius transformation $v=4(w+1)/(w+16)$ maps the radius-four disc bijectively onto the unit disc, taking $-1$ to zero and $4$ to one. This follows also by writing it as $(w/4+1/4)/(1+w/16)$, an automorphism of the unit disk after the scaling. It takes the removed real interval to $[0,1)$, so the image is the unit disc with this radial slit removed. Choosing $0<\arg v<2\pi$ defines a single-valued square root $u=\sqrt{v}$ and maps that slit disc bijectively ontoAll these maps have nonzero derivatives on their respective domains, since their potentially problematic zero or branch point is on the removed slit.

Next the Joukowski map $J(u)=u+1/u$ is a conformal bijection from $U$ onto the lower half-plane. For $u\in U$,If $J(u)=J(t)$, then $(u-t)(1-1/(ut))=0$, and $|ut|<1$ rules out the second factor vanishing. To prove surjectivity, take $q$ in the lower half-plane and consider $u^2-qu+1=0$. Neither root lies on the unit circle, where $u+1/u$ is real. Their product is one, so exactly one has modulus less than one. For that root the displayed imaginary-part identity forces $\mathrm{Im}u>0$. Thus it lies in $U$. Also $J^{\prime }(u)=1-u^{-2}$ cannot vanish there. Finally $C(q)=(q+i)/(q-i)$ maps the lower half-plane bijectively onto the unit disc: the inequality $|q+i|<|q-i|$ is exactly $\mathrm{Im}q<0$, and its inverse is $q=i(\zeta +1)/(\zeta -1)$. Consequentlyis a conformal bijection of $D$ onto the disc.

At the slit point $z_0=i/2$ the continuous rational expression for $v$ tends to $4/21$. In a sufficiently small neighborhood of $z_0$, points of $D$ have nonzero real part. Since $\mathrm{Im}(z^2)=2\mathrm{Re}z\mathrm{Im}z$ and the real-coefficient Möbius map $w\mapsto v$ preserves the sign of the imaginary part, right-side approach gives $\arg v\to 0$, while left-side approach gives $\arg v\to 2\pi$. HenceThe two limits of $F$ are thereforeThey have modulus one and are distinct, since $L>0$ and the Möbius map $C$ is injective on the extended real line. The two-sided cluster set at an interior slit point is a geometric consequence of opening the slit into two boundary intervals.

For the given map $f$, the composition $B=f\circ F^{-1}$ is an automorphism of the unit disk. To see its precise boundary behavior, let $a=B^{-1}(0)$ and compose $B$ with the inverse of $T_a(\zeta )=(\zeta -a)/(1-\overline{a}\zeta )$. This gives a disc automorphism fixing zero. Applying Schwarz lemma to it and its inverse shows it is a rotation. ThusIts denominator never vanishes on the closed disc, and its inverse has the same property; it therefore extends to a continuous bijection of the unit circle. Put $\alpha =B({\gamma }_{+})$ and $\beta =B({\gamma }_{-})$. These are distinct points of that circle, independent of the approaching sequence.

For any sequence approaching $z_0$, every infinite right-side subsequence has image tending to $\alpha$, and every infinite left-side subsequence has image tending to $\beta$. At least one side occurs infinitely often. Conversely, any convergent image subsequence has a further subsequence confined to one side, so its limit must be the corresponding one of these two points. The cluster set is therefore exactly $\{\alpha \}$, $\{\beta \}$ or $\{\alpha ,\beta \}$, according to which sides occur infinitely often. Both sides really occur in the domain, for example along $i/2\pm {\epsilon }_n$ with positive ${\epsilon }_n\downarrow 0$.

Use the curvature-minus-one normalization of the hyperbolic metric:The associated distance is $d(z,w)=2\mathrm{artanh}|(z-w)/(1-\overline{w}z)|$. A connected Riemann surface has its complete canonical Poincare metric on a Riemann surface precisely when its universal cover is conformally the unit disc. By the uniformization theorem, the other simply connected covering types are the complex plane and Riemann sphere. Disc automorphisms preserve $\lambda (z)|dz|$, as direct differentiation of $(z-a)/(1-\overline{a}z)$ shows; therefore the metric descends through the deck group. This is the uniformization notion of hyperbolicity, not the different potential-theoretic Green-function classification.

First prove the needed Schwarz lemma. If $F:\mathbb{D}\to \mathbb{D}$ is holomorphic and $F(0)=0$, then $F(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ give $|F(z)|\le |z|$ and $|F^{\prime }(0)|\le 1$.

Write ${\psi }_a(z)=(z-a)/(1-\overline{a}z)$. Conjugating $f$ by the source and target disc automorphisms, ${\psi }_{f(a)}\circ f\circ {\psi }_a^{-1}$ fixes zero. Applying the derivative bound givesIntegrate along any piecewise smooth path and then take the infimum of lengths: $d(f(z),f(w))\le d(z,w)$. This proves the Schwarz-Pick theorem. Contraction means nonexpansion, not a uniform Lipschitz constant strictly smaller than one; automorphisms preserve all distances.

If both $R$ and $S$ have disc universal covers, lift $g\circ {\pi }_R$ to $\tilde{g}:\mathbb{D}\to \mathbb{D}$ through ${\pi }_S$. The disc is simply connected, so the lift exists, and it is holomorphic because the covering maps are local biholomorphisms. Both covering maps are local hyperbolic isometries. The disc contraction therefore descends to $g^{\ast }d{s}_S\le d{s}_R$ and, by the path-length argument, $d_S(g(p),g(q))\le d_R(p,q)$. If only $S$ is known to have a hyperbolic metric, a nonconstant $g$ forces $R$ to have one too: otherwise lift from the plane or sphere universal cover of $R$ to the disc cover of $S$. A bounded entire function is constant by the Liouville theorem, and a holomorphic map from the compact sphere into the disc is constant by the maximum modulus principle. Thus a nonconstant map into a hyperbolic surface is distance-nonincreasing between the canonical hyperbolic metrics; a nonhyperbolic source admits only constant such maps.

For the prescribed point, Schwarz-Pick theorem gives $|{\psi }_{w_0}(f(z))|\le |{\psi }_{z_0}(z)|$. Thereforeis holomorphic after filling the removable singularity at $z_0$, and $|h|\le 1$. Inverting the target automorphism provesConversely every holomorphic $h$ with this bound gives an admissible disc map: $|{\psi }_{z_0}(z)h(z)|<1$, its inverse target automorphism remains in the disc, and the denominator cannot vanish.

Fix $z$ and put $r=|{\psi }_{z_0}(z)|<1$. The possible intermediate values fill $|v|\le r$, since constant choices $h\equiv c$, $|c|\le 1$, attain them all. Thus the one-point value region of a holomorphic disc map is $|{\psi }_{w_0}(w)|\le r$. Completing the square in $|w-w_0{|}^2\le r^2|1-{\overline{w}}_{0}w{|}^2$ gives its Euclidean centre and radius:This closed disc lies strictly inside the unit disc because $r<1$. For $z=z_0$ it degenerates to the single point $w_0$. Its boundary is attainable by unimodular constant $h$; requiring $f$ to map into the open disc does not remove that boundary.

Write $Y={\mathbb{P}}^1\setminus \{0,1,\infty \}$. One concrete uniformization uses the image in ${\mathrm{PSL}}_2(\mathbb{R})$ of the level-two principal congruence subgroupIt acts on the complex upper half-plane by Möbius transformations. Integer matrix entries make the group discrete, and the action is properly discontinuous. There are no nonidentity elliptic stabilizers: an elliptic element would have integer even trace with absolute value less than two, hence trace zero. But then $d=-a$ gives $bc=-(a^2+1)\equiv 2(\mathrm{mod}4)$, contradicting the evenness of $b,c$. Thus the quotient is an ordinary Riemann surface, with no orbifold points.

A fundamental region is the ideal quadrilateralIts vertical sides are identified by $T(\tau )=\tau +2$, and its semicircular sides by $U(\tau )=\tau /(1-2\tau )$. The ideal vertices are $-1,0,1,\infty$; $-1$ and $1$ represent one puncture, while zero and infinity give two others. Filling these three punctures gives genus zero: the identified quadrilateral has three vertices, two paired edges and one face, so its compactification has Euler characteristic $3-2+1=2$. Its complex compactification is a sphere, and a Möbius transformation puts the punctures at $0,1,\infty$. HenceThis is the thrice-punctured sphere as a modular quotient. Equivalently, the uniformization theorem applied to its universal covering map produces the upper-half-plane model: the sphere cannot cover this noncompact surface, and an entire covering from $\mathbb{C}$ would contradict Little Picard theorem by omitting both zero and one.

The curvature-minus-one metric $d{s}_{\mathbb{H}}=|d\tau |/\mathrm{Im}\tau$ is invariant under every real Möbius transformation preserving $\mathbb{H}$. Indeed, for determinant-one coefficients,Thus it descends to the hyperbolic metric on $Y$. This defines its distance by the infimum of path lengths; a choice of local lift does not affect lengths.

For the Schwarz-Pick lemma, first prove the Schwarz lemma. If $h:\mathbb{D}\to \mathbb{D}$ is a holomorphic function and $h(0)=0$, the function $h(z)/z$ extends as a holomorphic function across zero. On $|z|=r$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ yield $|h(z)|\le |z|$ and $|h^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative makes $h(z)/z$ a constant of modulus one, again by the maximum modulus principle, so $h$ is a rotation.

For a general holomorphic function $F:\mathbb{D}\to \mathbb{D}$, conjugate by the automorphisms of the unit disk $A_a(z)=(z-a)/(1-\overline{a}z)$, putting $a$ and $F(a)$ at zero. Applying the proved Schwarz lemma givesThe disk hyperbolic metric is $2|dz|/(1-|z{|}^2)$, and its distance is twice the inverse hyperbolic tangent of the displayed pseudohyperbolic distance. Both forms of Schwarz-Pick lemma therefore express metric contraction. Equality in the derivative at one point makes $F$ a disk automorphism by the equality case already proved.

For $X=\mathbb{D}\setminus \{0\}$ and a holomorphic function $f:X\to Y$, compose with a universal covering map of $X$ and lift to a universal covering map of $Y$. The simply connected source of the lift is a disk or half-plane. Schwarz-Pick lemma there bounds the lifted metric derivative, and the covering maps are local isometries, soThe second inequality follows by applying the length inequality to every joining path and taking the infimum. The hyperbolic metric on the punctured disk is its own intrinsic metric,obtained from the covering $z=e^{i\tau }$. It is not merely the ordinary disk metric restricted to the punctured disk.

Yes, equality can occur for distinct points. The cyclic subgroup $⟨T⟩$ lies in $\Gamma (2)$. Its quotient is conformally the punctured disk, using $z=e^{\pi i\tau }$, because this identifies precisely $\tau \sim \tau +2$. The subgroup inclusion induces a covering map that is a holomorphic functionIt is a local isometry. Choose a sufficiently small geodesically convex ball around any target point that is evenly covered. Two distinct sufficiently close points in one lift of that ball are joined by the lift of its minimizing target geodesic. The lift has the same length, giving $d_X\le d_Y$ for this pair, while contraction gives the reverse inequality. Thus equality holds.

For context, if the lifted map is not an automorphism, the derivative inequality is strict at every point; integrating along a minimizing geodesic in the complete punctured-disk metric gives a strict distance inequality for every distinct pair. Equality forces the covering-map case, but even a covering map need not preserve all distances globally because shorter paths can pass through other sheets. This is the distinction captured by equality in hyperbolic Schwarz-Pick contraction.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

The Schwarz-Pick lemma says that every holomorphic self-map $F$ of the unit disc contracts pseudohyperbolic distance and therefore hyperbolic distance:Its infinitesimal form isEquality for distinct points in the distance inequality, or at one point in the derivative inequality, holds exactly when $F$ is a disc Möbius transformation.

For completeness, first prove the Schwarz lemma. If $H:\mathbb{D}\to \mathbb{D}$ is holomorphic and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle, followed by $r\uparrow 1$, gives $|H(z)|\le |z|$ and $|H^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative forces that quotient to be a constant of modulus one, again by the maximum modulus principle; hence $H(z)=e^{i\theta }z$.

Now let ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$ and apply this result to $H={\varphi }_{F(w)}\circ F\circ {\varphi }_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho =\log ((1+\delta )/(1-\delta ))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|{\varphi }_a^{\prime }(a)|=(1-|a{|}^2{)}^{-1}$ and $|({\varphi }_w^{-1}{)}^{\prime }(0)|=1-|w{|}^2$, so the chain rule proves the derivative inequality. The equality statement follows from the rotation case of the Schwarz lemma after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.

There is a maximum of the derivative modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and Cauchy estimates bound their derivatives at zero. Take a sequence whose derivative moduli approach the finite supremum. By Montel theorem it has a subsequence converging locally uniformly to a holomorphic function $f$, with $f(0)=1$ and $e^{-1}\le |f|\le e$. The upper equality cannot occur at an interior point, because the maximum modulus principle would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is holomorphic because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. Locally uniform convergence gives convergence of derivatives by the Cauchy integral formula, so this limit attains the supremum.

To compute it and all extremizers, the nonvanishing function $f$ on the simply connected disc has a holomorphic logarithm $h$ with $h(0)=0$. One can construct it by integrating $f^{\prime }/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\mathrm{Re}h<1$. The mapis a biholomorphism from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi /2,\pi /2)$. Thereforehas real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the tangent, so these are inverse maps.

Applying the Schwarz lemma to $T\circ h$ gives $(\pi /4)|h^{\prime }(0)|\le 1$. Since $f^{\prime }(0)=h^{\prime }(0)$, the exact maximum isIt is attained by $f(z)=\exp ((4/\pi )\arctan z)$. The equality case requires $T(h(z))=e^{i\theta }z$, and hence all maximizing functions areTheir derivatives are $(4/\pi )e^{i\theta }$, so the extremizer is not unique: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the extremal derivative of a disc map into a symmetric annulus for $L=1$.

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.