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Let a group $G$ of Möbius transformations of the unit disk have an orbit satisfying the Blaschke condition, counted once per distinct orbit point. For its Blaschke product $B$, invariance of the pseudohyperbolic distance and permutation of the orbit give $|B(Tz)|=|B(z)|$. The ratio $B(Tz)/B(z)$ extends across its zeros and has modulus one, so the open mapping theorem makes it a constant $\chi (T)$. Composition gives a group homomorphism $\chi :G\to \{\lambda :|\lambda |=1\}$. It is essential to compare moduli of the products: two arbitrary bounded holomorphic functions with identical zeros need not have constant quotient.

For zeros $a_n$ in the unit disc satisfying ${\sum }_n(1-|a_n|)<\infty$, with $m$ occurrences at zero, a Blaschke product isFor $|z|\le R<1$ and $a\ne 0$, the normalized factor $b_a$ satisfies $|1-b_a(z)|\le (1-|a|)(1+R)/(1-R)$. The product therefore converges locally uniformly; outside its prescribed zeros the logarithms of its tail factors converge absolutely, so it has no additional zeros. Every finite product has modulus at most one in the disc, and the limit does too. Since $|b_a(w)|$ is pseudohyperbolic distance from $w$ to $a$, its modulus can be estimated geometrically.

First construct a holomorphic function without requiring boundedness. Separate the finitely many zeros at zero into $z^m$, and enumerate the remaining prescribed zeros as $a_n$. Since $|a_n|\to 1$, there is no interior accumulation point and every prescribed point has finite multiplicity. Using the Weierstrass elementary factors, setFor a compact disc $|z|\le r<1$, choose $q$ with $r<q<1$. Eventually $|z/a_n|\le q$. The power series identitygives an absolutely summable uniform bound. The infinite product convergence from logarithmic tails therefore makes the tail holomorphic and nonvanishing on this compact set. The finitely many initial factors have exactly their prescribed zeros, with the prescribed multiplicities. Hence $F$ has exactly the required zero sequence. This is a direct canonical product construction for zeros escaping to the disk boundary.

For a nonzero bounded holomorphic function, Jensen's formula bounds ${\sum }_n\log (1/|a_n|)$ after separating the zero at zero. Thus the Blaschke condition is necessary; the Blaschke product construction makes it sufficient. Use the curvature-minus-one convention for Hyperbolic distance in the Poincare disc:The latter lies between $(1-|z|)/2$ and $1-|z|$. Consequently the precise bounded-function criterion isThis counts multiplicities; the finitely many occurrences at zero cause no problem.

For the requested hyperbolic zero estimate for a Blaschke product, the pseudohyperbolic distance identity isIf $w$ is not a zero, $0<t<1$. The real function $L(t)=\log [(1+t)/(1-t)]-2t$ has $L(0)=0$ and $L^{\prime }(t)=2t^2/(1-t^2)\ge 0$, so $L(t)\ge 0$. Thus $\log |b_a(w)|\le -2e^{-\rho (w,a)}$. Apply this to each factor, including factors at zero, and then pass from finite products to the Blaschke product by locally uniform convergence:At a zero of $B$ the left side is zero, so the inequality is immediate; the logarithmic calculation at $t=1$ is unnecessary. The unimodular constant contributes nothing. Away from the zeros the product is nonzero, so the same inequality also bounds the sum and shows it is finite there.

The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.

Zeros are counted with their prescribed finite multiplicities, and the function is not identically zero. The criterion for zero sets in the unit disc is local finiteness: no point of the disc is an accumulation point of the zero sequence, and no point has infinite multiplicity. Necessity follows from the identity theorem for holomorphic functions. For sufficiency, factor a finite number of zeros at zero as $z^m$. For an infinite sequence of remaining zeros $a_n$, local finiteness implies $|a_n|\to 1$. With Weierstrass elementary factorsuse $f(z)=z^m{\prod }_n{E}_n(z/a_n)$. On each compact subdisc, eventually $|z/a_n|\le q<1$ and the logarithm of the tail factor is $-{\sum }_{j>n}(z/a_n{)}^j/j$, bounded in modulus by $q^{n+1}/((n+1)(1-q))$. These bounds are summable, so infinite product convergence from logarithmic tails gives locally uniform convergence and nonvanishing away from the listed zeros. Its finite initial factors give exactly their multiplicities. For a finite sequence an ordinary polynomial suffices.

Use the curvature-minus-one hyperbolic metric, with length element $2|dz|/(1-|z{|}^2)$. Integrating radially givesTherefore the printed sum condition is equivalent to the Blaschke condition ${\sum }_n(1-|a_n|)<\infty$.

To prove necessity for bounded $f$, first suppose $f(0)\ne 0$ and choose a radius $r<1$ with no zero on its circle. Remove the finitely many factors $(z-a_n)$ for $|a_n|<r$. The logarithm of the modulus of the remaining zero-free function is harmonic on the closed disc. Its mean value property, together with the mean of $\log |r{e}^{i\theta }-a|$ being $\log r$ for $|a|<r$ (expand the logarithm in powers of $a/(r{e}^{i\theta })$), proves Jensen's formulaLet $r\uparrow 1$ through such radii. Each summand increases to $-\log |a_n|$, and $-\log t\ge 1-t$ for $0<t\le 1$ by integration of $1/t\ge 1$. Consequently ${\sum }_n(1-|a_n|)<\infty$. If $f$ has order $m$ at zero, apply the argument to $f/z^m$, which is still bounded: outside $|z|=1/2$ it is bounded by $2^m\Vert f{\Vert }_{\infty }$, and inside use the maximum modulus principle. Add the finite contribution of the zeros at zero.

For sufficiency use a Blaschke product. For $a\ne 0$ its normalized Blaschke factor isOn $|z|\le R<1$, $|1-b_a(z)|\le (1-|a|)(1+R)/(1-R)$. The Blaschke condition thus makes the product converge locally uniformly; outside the specified zeros, the tail logarithms converge absolutely and the product is nonzero. Each factor has modulus at most one in the disc, so the limit is bounded by one. A factor $z^m$ supplies any finite multiplicity at zero. This proves

For the elementary inequality, put $H(t)=\log ((1+t)/(1-t))-2t$. Then $H(0)=0$ andso $\boxed{\log ((1+t)/(1-t))\ge 2t}$ for $0\le t<1$.

For any point $w$ not among the zeros, the pseudohyperbolic distance formula givesApply the proved inequality with $t=e^{-\rho (w,a)}$ to obtain $\log |b_a(w)|\le -2e^{-\rho (w,a)}$. Sum over finite products and pass to the limit, including factors at zero. This givesThe sum is finite: the triangle inequality gives $e^{-\rho (w,a_n)}\le e^{\rho (0,w)}{e}^{-\rho (0,a_n)}$. If $w$ is itself a zero, $B(w)=0$ and the same bound is immediate, so no substitution at the excluded value $t=1$ is needed. A unimodular constant in $B$ has no effect.

If “zeros at the points” is interpreted as a set without assigned multiplicities, first delete repeated entries throughout. Otherwise infinitely repeating a single zero would invalidate the claimed necessity of the series condition; the standard zero-sequence convention counts multiplicities.

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 $L=1$.

Represent a transformation byMultiplication preserves this form and determinant, and the inverse has the same form with parameters $(\overline{u},-v)$. To check the congruence condition, complex conjugation is the identity modulo $2\mathbb{Z}[i]$. Thus $u+v\equiv 1$ says exactly that $M$ fixes the column $(1,1{)}^T$ modulo two. Products and inverses preserve this property. The identity belongs to the set, and changing the common sign of $(u,v)$ does not change either the transformation or its congruence. Hence these transformations form a group.

Because $|u|>|v|$, the denominator has no zero in the disc. Direct calculation givesThe inverse also maps the disc into itself, so $g$ is a disc automorphism, and these identities show that it preserves the hyperbolic metric $2|dz|/(1-|z{|}^2)$.

The action is properly discontinuous. If $C\subset \mathbb{D}$ is compact, put $R={\max }_{z\in C}\rho (0,z)$. If $gC\cap C\ne \varnothing$, choose $z,w\in C$ with $w=gz$. ThenSince $g0=v/\overline{u}$ and $|u{|}^2-|v{|}^2=1$, its distance bound gives $|v/u|\le \tanh R<1$ and therefore $|u{|}^2=1/(1-|v/u{|}^2)\le {\cosh }^{2}R$. There are only finitely many Gaussian integers $u,v$ with these bounds, so only finitely many $g$ satisfy $gC\cap C\ne \varnothing$. This proves the required properly discontinuous group action directly.

If $g0=0$, then $v=0$ and $u$ is one of the Gaussian units $1,-1,i,-i$. The congruence $u-1\in 2\mathbb{Z}[i]$ leaves only $u=\pm 1$. Both give the identity transformation. Thus the point stabilizer of zero is trivial.

The closed Dirichlet region centered at zero is the set of $z$ with $\rho (z,0)\le \rho (z,g0)$ for every $g$. For a nonidentity element, $v\ne 0$. WriteThe distance inequality is equivalent, by pseudohyperbolic distance, to $|z|\le |(z-q)/(1-\overline{q}z)|$. Squaring and subtracting yieldsSince $|z|<1$ and $|q{|}^2=N/(N+1)$, the bisector half-plane is exactlyFor $u=1+i$ or $1-i$ and $v=i$ or $-i$, all the group conditions hold and the products $uv$ are the four numbers $\pm 1\pm i$, with $N=1$. Their four inequalities forceConversely, for any group element $m^2+n^2=|uv{|}^2=N(N+1)<(N+1{)}^2$. As $m,n$ are integers, $|m|,|n|\le N$. Thus every point satisfying those four inequalities satisfies every other one:This proves, with no missing bisectors, the Gaussian-integer disc group with an ideal-square Dirichlet domain formulaEquivalently it is the part of the unit disc outside all four open circles of radius one centered at $1+i,1-i,-1+i,-1-i$. Its sides are the inward arcs of those circles. They meet the unit circle orthogonally, and their ideal vertices are $\boxed{1,i,-1,-i}$; these vertices themselves lie outside the open disc.

The origin has trivial point stabilizer and the group orbit is locally finite by the discontinuity proof, so this closed bisector intersection has translates covering the disc with disjoint interiors, as in the definition of a Dirichlet domain. Side identifications are allowed on its boundary.

On the unit disc, the pseudohyperbolic distance isIt is invariant under disc Möbius transformations. For the curvature-minus-one hyperbolic metric with length element $2|dz|/(1-|z{|}^2)$, $\delta (z,w)=\tanh (\rho (z,w)/2)$ and $e^{-\rho (z,w)}=(1-\delta )/(1+\delta )$. The Schwarz-Pick theorem says that holomorphic functions from the disc into itself contract this distance. A Blaschke factor has modulus $\delta (z,a)$, so the distance also controls the size of a product with prescribed zeros.