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Use curvature-minus-one Hyperbolic distance in the Poincare disc. Each Blaschke factor has modulus $\tanh (\rho (w,a)/2)=(1-t)/(1+t)$, with $t=e^{-\rho (w,a)}$. For $0\le t<1$, differentiating shows $\log [(1+t)/(1-t)]\ge 2t$. Sum this logarithmic inequality over the zero factors and pass to the Blaschke product by locally uniform convergence. At a prescribed zero, its modulus vanishes and the bound is immediate. All sums count multiplicities.

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.

Let $f$ be a nonzero meromorphic function in a neighborhood of the closed disk $|z|\le R$, with no boundary zero or pole. List its interior zeros $a_j$ with multiplicities $m_j$ and its interior poles $b_k$ with orders $n_k$. Define the disk Green function and Poisson kernel byAt every interior point that is neither a zero nor a pole, the Poisson-Jensen formula isIn particular, at $z=0$ when $f(0)$ is finite and nonzero, this becomes Jensen's formula, with each zero contributing $-m_j\log (R/|a_j|)$ and each pole contributing $n_k\log (R/|b_k|)$.

Here is a proof that also explains the signs. The disk Blaschke factorshave a simple zero at $a$, no poles in the disk and boundary modulus one. Thereforeextends to a zero-free holomorphic function throughout the disk. Its logarithmic modulus is a harmonic function and agrees with $\log |f|$ on the boundary. The Poisson integral represents $\log |F(z)|$ by that boundary data. Since $\log |B_a|=-G_R(\cdot ,a)$, substitution gives the displayed Poisson-Jensen formula. Boundary zeros or poles can be handled by taking limiting radii: their logarithmic singularities are integrable. When zero is itself a zero or pole, factor out its signed power of $z$ before evaluating Jensen's formula there.

Write a Blaschke product as $B(z)=\lambda z^m{\prod }_n{b}_{a_n}(z)$, where $a_n\ne 0$, $|\lambda |=1$, and the zeros are counted with multiplicity. With normalized angular measure $dm=d\theta /(2\pi )$, Jensen's formula for each Blaschke factor givesIndeed, the mean of $\log |a-r{e}^{i\theta }|$ is $\log \max (r,|a|)$, while $\log |1-\overline{a}r{e}^{i\theta }|$ has mean zero by the mean value property for harmonic functions. Each factor has modulus at most one. Applying the monotone convergence theorem to the negative logarithms of the finite products yields the radial logarithmic mean of a Blaschke product:The Blaschke condition implies ${\sum }_n-\log |a_n|<\infty$: only finitely many zeros have modulus below $1/2$, and $-\log t\le 2(1-t)$ for $1/2\le t<1$. Every summand tends to zero and is bounded in absolute value by $-\log |a_n|$. The dominated convergence theorem for this series, together with $m\log r\to 0$, therefore givesThis also covers finite products and constant unimodular products. Zeros on a circle create integrable logarithmic singularities and do not change the conclusion.

Use the curvature $-1$ hyperbolic metric $ds=2|dz|/(1-|z{|}^2)$. For the disk Möbius transformation ${\varphi }_w(z)=(z-w)/(1-\overline{w}z)$, set $r_w(z)=|{\varphi }_w(z)|$. The hyperbolic distance satisfiesThe triangle inequality givesThus convergence of the exponential sum is independent of the base point. Taking the base point zero, it is equivalent to the Blaschke condition ${\sum }_n(1-|z_n|)<\infty$. This condition implies that only finitely many terms lie in any compact subdisk, including finitely many repetitions of any point.

To construct the required holomorphic function, for $a\ne 0$ define the Blaschke factorFor $|z|\le R<1$, writing $a=|a|e^{i\alpha }$ givesConsequently, after the finitely many factors corresponding to small $|a|$ are removed, the sum of $|1-b_a|$ converges uniformly on each compact subdisk. Those factors are uniformly close to one, so their logarithms form an absolutely convergent series there. Their product tail is holomorphic and nonvanishing. Including the finitely many removed factors gives the Blaschke productwith exactly the requested zero multiplicities. Each finite product has modulus at most one in the disk, so the limit does too. Hence $f=B/2$ maps into the open disk and has exactly the prescribed zeros, proving the implication from the summability condition.

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.

For a nonzero holomorphic function, isolated boundary zeros produce only integrable logarithmic singularities: locally a zero of multiplicity $m$ contributes $m\log |\theta -{\theta }_0|$ plus a bounded term. Thus the circle means below are well-defined. The identically zero function may be assigned $J=-\infty$.

1. Multiplication. The pointwise identity $\log |F_1{F}_2|=\log |F_1|+\log |F_2|$ gives $J(F_1{F}_2,R)=J(F_1,R)+J(F_2,R)$ by integration, with the same extended interpretation if a factor is identically zero.

2. Zero-free normalized functions. Compactness lets us choose a slightly larger open disc on which $F$ is holomorphic and zero-free. On this simply connected disc, $F^{\prime }/F$ has a holomorphic primitive $L$ normalized by $L(0)=0$. Differentiating $F{e}^{-L}$ shows it is constant and equal to $F(0)=1$, so $F=e^L$ and $L$ is a holomorphic logarithm of $F$. The Cauchy integral formula givesTaking real parts, using $\Re L=\log |F|$, proves $J(F,R)=0$. For a zero-free function with arbitrary nonzero value at zero, applying this to $F/F(0)$ gives $J(F,R)=\log |F(0)|$.

3. Disc factors. For $|z|=R$,as follows by expansion and $|z{|}^2=R^2$. The denominator's possible zero is at $R^2/\overline{w}$, outside the closed disc; when $w=0$ it is constant. Thus the given Blaschke factor is holomorphic on a neighbourhood of the disc, has modulus one on its boundary and satisfies $J({\psi }_{w,R},R)=0$.

For the general formula, a nonzero entire function has finitely many zeros in the compact disc: an infinite set would have an accumulation point in a holomorphic neighbourhood and the identity theorem would make the function zero identically. List its interior zeros with multiplicities. When $F(0)\ne 0$, divide by their factors ${\psi }_{w_j,R}$ to obtain a holomorphic function $G$ zero-free on a neighbourhood of the disc. The preceding parts give $J(F,R)=J(G,R)=\log |G(0)|$, while ${\psi }_{w_j,R}(0)=-w_j/R$. Hence Jensen's formula isThe source does not explicitly rule out a zero at zero in this general step. If its order there is $m$ and $F(z)=z^m{F}_{\ast }(z)$ with $F_{\ast }(0)\ne 0$, the correct version isThus a formula using $\log |F(0)|$ alone requires the indicated nonzero-centre assumption.

Finally, the lattice hypothesis forces quadratic exponential growth. For $R\ge 8$, all integer pairs with $|j|,|k|\le \lfloor R/4\rfloor$ lie inside the disc of radius $R/2$. Excluding zero, there are at least $R^2/16$ such points, and each contributes at least $\log 2$ to Jensen's sum. Thereforefor boundary-zero-free radii, with the additional nonnegative $m\log R$ term if needed. For all sufficiently large such radii, this is at least $(\log 2)R^2/32$. Choose $R_j\to \infty$ avoiding zero moduli and choose $z_j$ on each circle where $|F|$ is maximal. The logarithm of this circle maximum is at least the logarithmic circle mean, soThis proves lattice zeros force quadratic exponential growth; the smaller constant makes the required inequality strict.

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.

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.