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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.

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 $e^{\rho (w,w_0)}$.

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 Poincare disc model isThis Riemannian metric has constant Gaussian curvature $-1$.

For $a\in D$ and $|\eta |=1$, the Möbius transformationis an isometry of the disc, and every orientation-preserving disc isometry has this form. In particular, $T_{z_1,1}$ sends $z_1$ to zero and $z_2$ to a point of modulusRotations are also isometries, so it remains only to integrate the metric along a radial geodesic from $0$ to $\rho$:Thus the Hyperbolic distance in the Poincare disc is