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The Blaschke condition for a sequence in the unit disc is ${\sum }_n(1-|a_n|)<\infty$. It characterizes the zero sequences, counted with multiplicity, of nonzero bounded holomorphic functions. Sufficiency follows by constructing the Blaschke product. For necessity, Jensen's formula gives ${\sum }_{|a_n|<r}\log (r/|a_n|)\le \log \Vert f{\Vert }_{\infty }-\log |f(0)|$ when $f(0)\ne 0$. Letting $r\uparrow 1$ and using $-\log |a|\ge 1-|a|$ proves the condition. A finite zero at zero is factored out first. Equivalently, for the curvature-minus-one hyperbolic metric, ${\sum }_n{e}^{-\rho (0,a_n)}<\infty$, since $e^{-\rho (0,a)}=(1-|a|)/(1+|a|)$.

The Cayley transform gives a conformal bijection between an unbounded half-plane and the unit disc. Two conventions are useful. For the right half-plane, useFor the upper half-plane, useRotating the upper half-plane onto the right half-plane by $z\mapsto -iz$ explains the relationship $C(z)=R(-iz)$. The factors of $i$ distinguish which half-plane is mapped onto the unit disc. ^[ref]

In particular, $C(i)=0$, $C(2i)=1/3$ and $C(1+i)=(1-2i)/5$. The whole positive imaginary axis becomes the real diameter: $C(iy)=(y-1)/(y+1)$ for $y>0$. Horizontal grid lines become circles tangent to the unit circle at $1$ when completed by their endpoint at infinity; they are not hyperbolic geodesics.

On the Riemann sphere, $C$ maps the extended real line onto the unit circle:Every finite real $x$ has $|C(x)|=1$; the remaining point $w=1$ corresponds to infinity. The pole $z=-i$ is outside the upper half-plane, and the pole of $C^{-1}$ at $w=1$ is on the boundary of the unit disc. Neither pole obstructs the interior conformal bijection.

For a short verification, subtract the squared moduli:These identities put each interior image in the required domain; solving $w(z+i)=z-i$ gives the stated inverse function. Both maps are holomorphic, with $C^{\prime }(z)=2i/(z+i{)}^2\ne 0$, so the conformal bijection preserves Euclidean angles between tangent directions.

Euclidean distance is not preserved: $i$ and $2i$ are one unit apart, whereas $0$ and $1/3$ are only one third of a unit apart. The striking feature is that $C$ does preserve length for the hyperbolic metrics of the Poincaré half-plane model and Poincaré disk model:Consequently it is an isometry for these hyperbolic metrics, including their induced hyperbolic distances. For the same example, the corresponding hyperbolic geodesic segments have equal length:Thus the Cayley transform lets us choose the most convenient picture: a straight boundary for calculations, or a bounded unit disc for visualization. It also transports automorphisms of the unit disk into upper half-plane automorphisms by function composition $C^{-1}\circ g\circ C$.

This logarithmic area integral measures the accumulated pullback hyperbolic metric area of a holomorphic disc map. Its nonnegative kernel makes it increasing in $R$. Schwarz-Pick theorem bounds $H_g(z)$ by $2/(1-|z{|}^2)$, which controls its characteristic through a critical-point Jensen identity.

For a nonconstant holomorphic map into the unit disc, the displayed density pulls the target hyperbolic metric back to the source Euclidean coordinate. It is not the operator norm between two hyperbolic tangent spaces, which additionally divides by the source metric density. Away from ramification points $\Delta \log H_g=H_g^2$; a ramification point of multiplicity $m$ adds $2\pi m$ times its point mass to the distributional Laplacian.

Hyperbolic distance is the Riemannian distance induced by the hyperbolic metric.

The universal covering map $\tau \mapsto e^{i\tau }$ from the complex upper half-plane to the punctured unit disc pulls back the metric $|dz|/(|z|\log (1/|z|))$ to $|d\tau |/\mathrm{Im}\tau$. This is its complete curvature-minus-one hyperbolic metric, which is different from restricting the ordinary disk metric to the punctured disk.

// Destination: analysis.bigb

The infimum of nonconstant closed geodesic lengths in curvature $-1$, with value $+\infty$ if there are none. On a closed surface with a hyperbolic metric this is a positive attained minimum, and a minimizing geodesic is simple. For fixed closed genus $g\ge 2$, a common positive lower bound gives relative compactness in the unmarked moduli space of Riemann surfaces by Mumford's compactness theorem.

In the upper half-plane model, hyperbolic lines are vertical Euclidean rays and semicircles centred on the real axis, equivalently circle arcs meeting the real boundary orthogonally. A semicircle with endpoints $r<s$ is mapped to the imaginary axis byIts determinant is one, and its real endpoints map to $0,\infty$. A vertical line $x=a$ is mapped to the imaginary axis by $z\mapsto z-a$. Thus every hyperbolic line can be carried to a fixed one; composing one such map with the inverse of another proves transitivity of $\mathrm{PSL}(2,\mathbb{R})$ on the lines.

Reflection in $x=a$ is $R_a(z)=2a-\overline{z}$. Reflection in the unit semicircle is $R_{\circ }(z)=1/\overline{z}$. The latter fixes the semicircle pointwise, is an involution, and preserves the hyperbolic metric because $|dw|=|dz|/|z{|}^2$ and $\mathrm{Im}w=\mathrm{Im}z/|z{|}^2$. The vertical reflection similarly preserves $|dz|/\mathrm{Im}z$. Their composition in the specified order is

A simply connected plane domain carrying a hyperbolic metric is proper. By the Riemann mapping theorem, choose a biholomorphism $\varphi :\Omega \to \mathbb{D}$ sending a basepoint to zero. The same distance formula now givesThus the Blaschke condition holds. In particular the $a_n$ have no interior accumulation point: near any such point the quantities $1-|a_n|$ would stay bounded away from zero.

Construct the Blaschke productHere $m=1$ precisely when the sequence contains zero. For $|\zeta |\le r<1$, its normalized Blaschke factor $b_a$ satisfiesThe Blaschke condition therefore makes the infinite product converge uniformly on every compact subdisk. Away from the prescribed zeros its tail factors tend uniformly to one with summable errors, so their logarithms converge and the product is nonzero. It has exactly the prescribed zeros, and the maximum modulus principle bounds every finite partial product by one, hence $|B|\le 1$. Thus $f=B\circ \varphi$ is the required bounded analytic function, with no extra zeros.

Simple connectivity cannot be dropped. On the punctured disk chooseThe hyperbolic metric on the punctured disk is $|dz|/(|z|\log (1/|z|))$. Its universal covering map $\tau \mapsto e^{i\tau }$ pulls it back to the curvature-minus-one upper-half-plane metric. The imaginary axis is a geodesic; the lift from $i$ to $i{n}^2$ has length $\log n^2$, and any other lift differs by a real deck translation and has at least that distance. ThereforeThe triangle inequality again makes the sum finite for every basepoint. However, any bounded holomorphic function on the punctured disk extends over zero by the Riemann removable singularity theorem. These zeros accumulate at zero, so the identity theorem makes the extension identically zero. It cannot have exactly the prescribed zero set. This example of hyperbolic zero sequences on a punctured disk satisfies the distance condition but not the analytic zero-set condition.

We use the complete hyperbolic metric of curvature $-1$; on the unit disc its length element is $2|dz|/(1-|z{|}^2)$. This normalization matters in the exponential summability criterion. A nonzero bounded holomorphic function on the entire plane is constant by Liouville's theorem, so the prescribed infinite zero set forces $\Omega \ne \mathbb{C}$. The Riemann mapping theorem supplies a biholomorphism $\varphi :\Omega \to \mathbb{D}$, and transports the disk hyperbolic metric to $\Omega$.

Fix $w\in \Omega$ and choose $\varphi (w)=0$. Put $a_n=\varphi (z_n)$ and $F=f\circ {\varphi }^{-1}$. First suppose $F(0)\ne 0$. For radii avoiding its zeros, Jensen's formula givesThe monotone convergence theorem as $r\uparrow 1$ gives ${\sum }_n-\log |a_n|<\infty$, hence ${\sum }_n(1-|a_n|)<\infty$. If $F$ has a zero at zero, factor off its finite order first; the resulting holomorphic function is still bounded, since division by that power is bounded away from zero and extends near zero. Thus the Blaschke condition holds in either case, with the possible origin zero treated as one additional finite term.

The Hyperbolic distance in the Poincare disc satisfiesConsequently the exponential distances are summable. This holds for every basepoint: the triangle inequality bounds the ratio of $e^{-\rho (w,z_n)}$ and $e^{-\rho (w_0,z_n)}$ between $e^{-\rho (w,w_0)}$ and $e^{\rho (w,w_0)}$.

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 $f(z)=Bz$.

A normal family of holomorphic functions on a domain is one from which every sequence has a subsequence converging locally uniformly to a holomorphic function, or locally uniformly to infinity in the extended convention. For a bounded family, only finite limits occur.

For functions mapping a plane domain $D$ into the unit disc, the Cauchy integral formula supplies uniform derivative bounds on each compact subset. More explicitly, if a closed disc of radius $r$ around $z$ lies in $D$, the Cauchy estimate gives $|f^{\prime }(z)|\le 1/r$. A slightly larger compact neighborhood of any given compact set therefore gives uniform boundedness and equicontinuity there. Apply the Arzelà-Ascoli theorem on a compact exhaustion and take a diagonal subsequence. Its limit is locally uniform; the Cauchy integral formula shows that the limit is holomorphic. This proves normality rather than merely invoking it.

Now let $|g|\le M$ on the complex upper half-plane. Every dilation $g_t(w)=g(tw)$ has the same bound, so the argument just given, after scaling the bound if necessary, proves that $\{g_t:t\ge 1\}$ is a normal family. For any sequence $t_j\to \infty$, every locally uniform subsequential limit $G$ satisfiesThe identity theorem for holomorphic functions forces $G\equiv \ell$. It follows that $g_t\to \ell$ locally uniformly as $t\to \infty$: otherwise a sequence violating convergence on one compact set would have a convergent subsequence with this same constant limit, a contradiction.

For $0<\epsilon <\pi /2$, the arc $K_{\epsilon }=\{e^{i\theta }:\epsilon \le \theta \le \pi -\epsilon \}$ is a compact subset of the complex upper half-plane. Write $z=|z|w$ with $w\in K_{\epsilon }$. Local uniform convergence of the dilations givesFor $\epsilon \ge \pi /2$ the specified sector is empty. This is the same compactness mechanism as the normal-family proof of angular boundary convergence.

To transfer the result to the unit disc, use the Cayley transform between the half-plane and diskIt sends $r\omega$ to $iy$ with $y=(1+r)/(1-r)$, so $g(iy)\to \ell$ as $y\to \infty$. Choose the curvature-minus-one normalization of the hyperbolic metric, for which the disc line element is $2|dz|/(1-|z{|}^2)$ and the half-plane line element is $|dw|/\mathrm{Im}w$. The Cayley transform between the half-plane and disk preserves these metrics. For $w\in \mathbb{H}$ and $y>0$,Thus ${\rho }_{\mathbb{D}}(z,r\omega )<k$ implies $\mathrm{Im}w/|w|>\mathrm{sech}k$. This is the hyperbolic tube around a radial segment estimate. It places $w$ in a fixed sector bounded away from the real axis. Moreover $z\to \omega$ implies $|C_{\omega }(z)|\to \infty$. Applying the sector result givesA different constant normalization of the hyperbolic metric only rescales $k$ and gives the same conclusion.

In the upper half-plane model, a hyperbolic line is either a vertical line or a Euclidean semicircle perpendicular to the real axis. For the vertical line $x=a$, set $R_L(z)=2a-\bar{z}$. For the circle $|z-a|=r$ with real $a$, setThe vertical formula preserves $y$ and Euclidean speed. For the circular formula, $\mathrm{Im}{R}_L(z)=r^2\mathrm{Im}z/|z-a{|}^2$, while its local Euclidean length multiplier is $r^2/|z-a{|}^2$. Thus both preserve $|dz|/\mathrm{Im}z$, the hyperbolic metric, and consequently all curve lengths and distances. Each is an involution preserving the half-plane; each fixes its specified line pointwise and moves points off that line. This establishes the required hyperbolic reflection explicitly.

To factor an arbitrary isometry $g$, first recall why a bisector reflection exchanges any distinct points $P,Q$. The hyperbolic distance formula isEquating the distances to $P$ and $Q$ yields a vertical line or a circle with centre on the real axis, hence a hyperbolic line. Substitution in the reflection formulas above shows that its reflection exchanges $P$ and $Q$. Choose a base point $P$. If $gP\ne P$, compose $g$ on the left with their bisector reflection, obtaining $h$ that fixes $P$; otherwise take $h=g$.

Choose $Q\ne P$. The points $hQ,Q$ have the same distance from $P$, so their bisector contains $P$. If they differ, its reflection corrects $hQ$ while retaining $P$. We obtain an isometry $k$ fixing both $P,Q$ after at most two reflections.

An isometry fixing two points is either the identity or reflection in their joining line. To see this directly, a point with prescribed distances from $P,Q$ lies at the intersection of two hyperbolic circles; these are Euclidean circles in this model and have at most two intersections. When there are two, reflection in the line $PQ$ interchanges them. Pick a third point off that line. Either $k$ fixes it, or composing $k$ with that line reflection fixes it. An isometry fixing three noncollinear points is the identity: the first two distances leave only the reflected pair, and the distance to the third point distinguishes that pair. Thus $k$ is the identity or one additional reflection. Reversing the compositions proves every isometry is a product of at most three hyperbolic reflections.

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.

In the upper half-plane model $H=\{(x,y):y>0\}$ the hyperbolic metric isThe length of a differentiable curve is $\int \sqrt{{\dot{x}}^2+{\dot{y}}^2}dt/y$. The area element is the square root of the metric determinant times $dxdy$, so the hyperbolic area of a measurable region $S$ is ${\int }_S{y}^{-2}dxdy$.

The Gauss-Bonnet theorem for a geodesic hyperbolic triangle of angles $\alpha ,\beta ,\gamma$ in curvature $-1$ gives area $\pi -\alpha -\beta -\gamma$; an ideal vertex has angle zero. The horizontal edge in the present region is not a geodesic, so it is simpler to use the area element directly:The open boundary does not affect the integral because its area is zero.

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.