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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.

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 $e^{-\rho (0,a)}=(1-|a|)/(1+|a|)$.

Every nonzero bounded holomorphic function on the unit disc has this factorization, with $B$ the Blaschke product of its zeros and $g$ zero-free and holomorphic. The Blaschke condition follows from Jensen's formula. Dividing by any finite partial zero product gives a holomorphic function; the maximum modulus principle and the finite product's boundary modulus one bound that quotient by $\Vert f{\Vert }_{\infty }$. Locally uniform convergence of the partial products gives the same bound for $g$, with removable extensions at their zeros. This factorization alone does not make $g$ constant or a full inner-outer factorization.

A sequence in the unit disc, counted with multiplicity, is a Blaschke sequence when it satisfies the Blaschke condition. The zeros of every nonzero bounded holomorphic function form such a sequence: Jensen's formula bounds ${\sum }_{|a_n|<r}\log (r/|a_n|)$ uniformly as $r\uparrow 1$, and $1-|a|\le -\log |a|$ gives the assertion. There are only finitely many zeros at the origin, which can be factored out first.

For nonzero prescribed zeros $a_n$ with $|a_n|\to 1$, the Weierstrass elementary factors of increasing order give the displayed holomorphic function on the unit disc; $m$ handles finitely many zeros at zero. On any compact disc, eventually $|z/a_n|\le q<1$, and $|\log E_n(z/a_n)|\le q^{n+1}/[(n+1)(1-q)]$. The infinite product convergence from logarithmic tails gives a nonvanishing tail. The finite initial factors supply exactly the prescribed zeros and multiplicities. No Blaschke condition is required unless boundedness is imposed.

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

Use the Nevanlinna proximity function $m(r,f)=(2\pi {)}^{-1}\int {\log }^{+}|f(r{e}^{i\theta })|d\theta$ and the Nevanlinna integrated counting function $N(r,f)$ for poles, counting multiplicity. If the pole order at zero is ${\nu }_0$, thenBoundedness of the Nevanlinna characteristic refers to $r\uparrow 1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when ${\nu }_0>0$.

The Nevanlinna first main theorem states, for every fixed finite value $a$ and every nonconstant meromorphic function on the disk,For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while Jensen's formula gives $T(r,h)-T(r,1/h)=\log |c|$ when $h(z)=c{z}^{\nu }(1+O(z))$ near zero; these observations explain the first theorem.

The Nevanlinna second main theorem in the unit disc states that for $q\ge 3$ distinct values $a_j$ on the Riemann sphere,as $r\uparrow 1$ outside an exceptional set $E$ satisfying ${\int }_{E}dr/(1-r)<\infty$. The truncated Nevanlinna counting function $\overline{N}$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge 0$, its poles satisfy the Blaschke condition, counting their orders: let $r\uparrow 1$ in their integrated count and use $1-|b|\le -\log |b|$. Let $B$ be their Blaschke product, including the origin factor if necessary. Then $g=Bf$ extends to a holomorphic function on the whole disk, and $|B|\le 1$ givesHere the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge {\nu }_0\log r_0$ on $r\ge r_0$.

The nonnegative subharmonic function ${\log }^{+}|g|$ has a finite harmonic majorant. Indeed its Poisson integrals $H_R$ on expanding disks dominate it by the maximum principle for subharmonic functions, increase with $R$, and have $H_R(0)\le C$. The Harnack inequality for harmonic functions bounds them on every compact subdisk, so their increasing limit is a finite harmonic function $H\ge {\log }^{+}|g|$. Since the disk is simply connected, $H$ has a harmonic conjugate, giving a holomorphic function $A$ with $\mathrm{Re}A=H$. ConsequentlyThis proves the constructive half of the quotient characterization of bounded characteristic.

Conversely, if $f=u/v$ with $u,v$ bounded holomorphic functions and $v\not\equiv 0$, each pole of $f$ is a zero of $v$ of at least the same order. The elementary inequality ${\log }^{+}|u/v|\le {\log }^{+}|u|+{\log }^{+}|1/v|$ and the Nevanlinna first main theorem giveAny cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus bounded characteristic is exactly a quotient of two bounded analytic functions.

An unbounded example with bounded characteristic isIt is a ratio of the bounded holomorphic functions $1$ and $1-z$. More explicitly, Jensen's formula gives the mean of $\log |1-r{e}^{i\theta }|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le \log 2$, although $g(r)\to \infty$.

Finally, set $Y=(1+|g{|}^2{)}^{p/2}\ge 1$. Pointwise ${\log }^{+}|g|\le p^{-1}\log Y$. Jensen's inequality for the concave logarithm givesThere is no pole term because $g$ is holomorphic. Hence the stated integral bound implies bounded characteristic. This argument in fact works for every $p>0$.

Let $m$ be the multiplicity at zero, and write $f(z)=z^{m}h(z)$ with $h(0)\ne 0$. The proof in solution gives $|f|\le 1$. For radii without boundary zeros, Jensen's formula saysThe absolute logarithmic hypothesis makes the left side tend to zero. The nonnegative zero sum increases to ${\sum }_n\log (1/|a_n|)$ by the monotone convergence theorem, so this series is finite. Hence ${\sum }_n(1-|a_n|)<\infty$, and the Blaschke condition supplies a Blaschke product $B$ with exactly these zeros and their multiplicities.

The quotient $g=f/B$ extends holomorphically across every zero because the multiplicities agree, and the extensions have nonzero values. Thus $f=Bg$ with $g$ zero-free and holomorphic. We will also need the stronger Blaschke factorization of a bounded holomorphic function, namely $|g|\le 1$. Let $B_N$ contain the factor $z^m$ and the first $N$ other zero factors. The quotient $f/B_N$ is holomorphic. On $|z|=R$ enclosing those finitely many zeros, its modulus is at most $1/{\min }_{|z|=R}|B_N(z)|$. The maximum modulus principle transfers this bound to the interior. Since the finite Blaschke product has boundary modulus one continuously, letting $R\uparrow 1$ gives $|f/B_N|\le 1$. Now let $N\to \infty$ using locally uniform convergence of the Blaschke product, first off the zeros and then by continuity at them. This proves $|g|\le 1$.

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.

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.

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.

Let $f$ be the nonzero bounded holomorphic function in the zero-set condition. Choose $v$ with $f(v)\ne 0$ and compose with a disk Möbius transformation carrying zero to $v$. Call the resulting function $g$ and its zeros $a_n$, counted with multiplicity. Then $g(0)\ne 0$ and $|g|\le 1$.

For a radius $r$ with no zero on its circle, Jensen's formula givesTo see the formula directly, divide $g$ by its finitely many interior zero factors. The logarithm of the modulus of the resulting zero-free holomorphic function is harmonic near the closed disk and satisfies the mean value property. The circle average of $\log |r{e}^{i\theta }-a|$ for $|a|<r$ is $\log r$, obtained by expanding $\log (1-(a/r)e^{-i\theta })$. Subtracting its value at zero gives exactly the displayed zero sum.

Let $r\uparrow 1$ through zero-free circles. Increasing limits of the nonnegative summands giveSince $1-|a_n|\le -\log |a_n|$, the Blaschke condition follows for the transformed sequence. The hyperbolic metric identity and the base-point comparison proved above giveThis proves the reverse implication and completes the equivalence.

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.

Write ${\log }^{+}x=\max (0,\log x)$. The Nevanlinna proximity function and Nevanlinna integrated counting function areThe pole sum counts multiplicity; it is finite for each $r$ because $f(0)$ is finite and nonzero and poles of a meromorphic function are isolated. Equivalently $N(r,f)={\int }_0^{r}n(t,f)dt/t$, where $n(t,f)$ counts poles in the disc. The Nevanlinna characteristic isFor a general nonzero meromorphic function $g$ with signed order $\nu$ at zero, use $g(z)=c{z}^{\nu }(1+O(z))$, $c\ne 0$. The general counting convention includes a pole at zero:where $n(0,g)=\max (0,-\nu )$. This convention will also cover $g=f-a$ when the target $a=f(0)$.

Here is a direct proof of the exact reciprocal identity underlying the Nevanlinna first main theorem. Choose a circle with no zero or pole of $g$ on it, and a slightly larger disc with the same interior zeros and poles. Divide $g$ by $z^{\nu }$ and its nonzero zero factors and multiply by its nonzero pole factors, obtaining a holomorphic, nowhere-zero function $H$ on that disc. For any factor $z-a$ with $|a|<r$,Indeed, factor out $r{e}^{i\theta }$; the remaining logarithm has real part equal to that of $-{\sum }_{k\ge 1}(a/r{)}^k{e}^{-ik\theta }/k$, whose angular mean is zero. Also, $\log |H|$ is harmonic, so the mean value property for harmonic functions equates its circle mean to $\log |H(0)|$. Putting back the zero and pole factors provesThe identity ${\log }^{+}x-{\log }^{+}(1/x)=\log x$ now yieldsZeros or poles on the circle cause only integrable logarithmic singularities; the same identities hold at those radii by limits. A boundary zero or pole contributes $\log (r/r)=0$ to its counting term.

For a finite target $a$, set $g=f-a$. Its finite poles and their orders equal those of $f$. The elementary estimatesgive $|T(r,f-a)-T(r,f)|\le \log (1+|a|)$. Combining this with the reciprocal identity proves the first main theorem:The error is bounded independently of $r$, with an explicit bound $\log (1+|a|)+|\log |c_a||$, where $c_a$ is the first nonzero local coefficient of $f-a$ at zero. The count on the right records the occurrences of the value $a$ with multiplicity. The target infinity is the defining identity $T=m+N$. If $f$ is a constant equal to $a$, the reciprocal expression is undefined; the value-distribution assertion assumes $f-a\not\equiv 0$, as it automatically does for nonconstant $f$. The exact reciprocal identity remains valid for every nonzero constant $f$.

Taking $g=f$, whose signed order is zero and whose local coefficient is $f(0)$, rearranges that exact identity into Jensen's formula:Thus the zero and pole terms, their signs and the initial-value constant follow from the same proved identity, rather than from a theorem citation without proof.

Finally let $f$ be a nonzero bounded holomorphic function on the unit disc, with $|f|\le M$. If its order at zero is $m$, write $f(z)=z^{m}h(z)$, $h(0)\ne 0$. The factorization argument above works on every compact subdisc and gives, for its nonzero zeros,For $1/2\le r<1$ this is a uniform finite upper bound. As $r\uparrow 1$, monotone convergence givesSince $1-u\le -\log u$ for $0<u<1$, and there are only $m$ zeros at zero, we obtainThis is precisely the Blaschke condition, so the zeros, counted with multiplicity, form a Blaschke sequence.

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.

For $B(z)=\lambda z^m{\prod }_n{b}_{a_n}(z)$ with nonzero zeros $a_n$ counted with multiplicity, Jensen's formula gives the displayed identity. The negative logarithms of partial products increase, so the monotone convergence theorem permits termwise integration. The Blaschke condition implies ${\sum }_n-\log |a_n|<\infty$, which dominates the absolute summands. The dominated convergence theorem therefore gives a limit of zero as $r\uparrow 1$. Since $|B|\le 1$, the mean of $|\log |B||$ tends to zero as well.