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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 Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $b$ is algebraic irrational, every number $e^{b\lambda }$ with $e^{\lambda }=a$ is transcendental. Thus the statement includes every determination of the power, not merely a chosen principal branch.

Here is an auxiliary-function outline with the arithmetic and analytic steps specified. Assume $v=e^{b\lambda }$ were algebraic, and put $K=\mathbb{Q}(a,b,v)$, $d=[K:\mathbb{Q}]$. For a large integer $T$, choose degrees $L_1\asymp T\log T$ and $L_2\asymp T/\log T$, with constants large enough that $(L_1+1)(L_2+1)>2d(T+1{)}^2$. Apply Siegel lemma to imposeon a nonzero polynomial $P(X,Y)\in \mathbb{Z}[X,Y]$ of these two degrees. Each field equation gives at most $d$ rational equations. Clearing denominators and bounding conjugates gives equation-coefficient logarithmic size $O(L_1\log T+L_{2}T)$; the surplus of variables therefore permits coefficient height $H$ with

The entire function $F(z)=P(z,e^{\lambda z})$ is nonzero: its polynomial-exponential terms have distinct exponents $j\lambda$, and their independence follows by the differential-operator argument in question 2. Here $\lambda \ne 0$. Its maximum modulus satisfies $\log M_F(R)=O(\log H+L_1\log R+L_{2}R)$. The grid points $s+tb$ are distinct because $b$ is irrational, and lie in a disk of radius $(1+|b|)T$.

Suppose the vanishing grid has been extended to $0\le s,t\le B$, for some $B\ge T$. Divide $F$ by the product of its $(B+1{)}^2$ known linear zero factors and use the maximum principle on a disk of radius $CB$, with $C$ large and fixed. At a point of the doubled grid, the quotient estimate givesif the value is nonzero. But this value equals the algebraic number $P(s+tb,a^s{v}^t)$. Clearing its denominators and taking its norm gives the opposite boundFor sufficiently large initial $T$, the terms on the right are $o(B^2)$ uniformly for every $B\ge T$: $L_2/B=O(1/\log T)$ and $L_1\log B/B^2=O((\log T{)}^2/T)$. Thus every doubled-grid value must be zero. Iterating doubles the grid indefinitely. This supplies quadratically many distinct zeros in disks of linearly growing radius, whereas Jensen's formula and the exponential-polynomial growth bound allow only $O(R)$ zeros for a nonzero $F$. The contradiction proves the theorem. This is the Schneider lattice proof of the Gelfond–Schneider theorem.

Now fix any logarithms of algebraic $\alpha ,\beta \ne 0,1$. Their values are nonzero. If $q=\log \alpha /\log \beta$ were algebraic irrational, then $e^{q\log \beta }=\alpha$ would be algebraic, contradicting the theorem with base $\beta$. HenceFor arbitrarily many logarithms the qualitative Baker theorem on algebraic linear combinations of logarithms states that logarithms of algebraic numbers which are linearly independent over $\mathbb{Q}$ remain independent over $\overline{\mathbb{Q}}$, even after adjoining one to the list. Equivalently a nonzero homogeneous algebraic-coefficient linear combination of such logarithms is transcendental. Rational relations are therefore the only source of homogeneous algebraic relations among the chosen logarithms.

For a nonconstant meromorphic map, write $T_f(R)=m(R,f)+N(R;\infty )$ in the usual normalization, where $m=(2\pi {)}^{-1}\int {\log }^{+}|f(R{e}^{i\theta })|d\theta$. Let $N(R;a)$ count $a$-points with their local degrees, and let $\overline{N}(R;a)$ count each distinct point once, using the same integrated logarithmic weights. At infinity the local degree is the pole order. The Nevanlinna deficiency and Nevanlinna ramification index areThe latter is an asymptotic excess-multiplicity index, not the integer local ramification degree. The Nevanlinna first main theorem gives $N(R;a)\le T_f(R)+O(1)$ and identifies deficiency with the lower limiting normalized proximity. The target $a$ ranges over the Riemann sphere, not just the real line.

The exponential is entire and has no poles. ConsequentlyThus $c=1/\pi$ in this normalization.

To obtain the rational-composition growth, represent the nonconstant degree-$d$ sphere map by relatively prime homogeneous polynomials $(P,Q)$ of degree $d$. Their joint norm on the unit sphere in ${\mathbb{C}}^2$ has a positive minimum and finite maximum: simultaneous vanishing would contradict relative primeness on the projective line. Homogeneity therefore gives constants $0<A\le B<\infty$ withApply this to the entire nonvanishing lift $(e^z,1)$. The logarithmic joint norm differs from $d\log \Vert (e^z,1)\Vert$ by a bounded function. Its circular mean minus the value at zero is the spherical characteristic. It differs from the ordinary characteristic by $O(1)$: Jensen's formula applied to the entire denominator counts its zeros, whilecontrols the proximity replacement. If the denominator vanishes at zero, factor its zero and use its leading coefficient in Jensen's formula; the discrepancy is still a basepoint-dependent constant. This proves the rational composition law for the Nevanlinna characteristic here and givesIf $U$ is constant, $d=0$ and its characteristic is bounded instead.

For a simultaneous nonzero deficiency and ramification example choosea degree-three polynomial composed with the exponential. Its characteristic is $3R/\pi +O(1)$. Its only zeros are $2\pi in$, $n\in \mathbb{Z}$, all of multiplicity two. The number of distinct such points in a large disc is $R/\pi +O(1)$, so integration of their count givesThe origin contribution is handled by its usual logarithmic counting term. ThereforeThis is simultaneous deficiency and ramification for an exponential polynomial: the missing exponential value contributes a deficit, while the attained value is attained with double multiplicity.

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.

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.

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.

For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

The Riemann xi function is the entire functionwith $\xi (s)=\xi (1-s)$. Its Hadamard factorization iswhere the Nontrivial zeros of the Riemann zeta function are repeated by multiplicity and the factors are canonical genus-one factors.

Here is the growth estimate needed for the Jensen zero-count bound. For $|s|\le 2T$, functional symmetry reduces to $\Re s\ge 1/2$. Euler summation truncated at $T^2$ bounds $(s-1)\zeta (s)$ by a fixed power of $T$, uniformly in that region; the multiplication cancels the pole at one. The logarithmic gamma estimate bounds $\log |\Gamma (s/2)|$ by $O(T\log T)$, including the bounded small-$s$ part separately. The remaining elementary factors obey the same bound. ThusFor a zero with $|\rho |\le T$, its contribution in Jensen's formula on radius $2T$ is at least $\log 2$. ConsequentlyIf a zero lies on the integration circle, use nearby radii and continuity of the zero-count estimate. Hence $n_{\xi }(T)=O(T\log T)$ for $T>2$. The growth also gives order at most one and justifies the stated Hadamard factorization; the zero-count bound gives convergence of its genus-one factors.

The printed logarithmic Stirling hint drops the term $-\frac{1}{2}\log z$. The correct expansion is $(z-\frac{1}{2})\log z-z+\frac{1}{2}\log (2\pi )+O(|z{|}^{-1})$ in a fixed sector. Its consequence $\log |\Gamma (z)|=O(|z|\log (2+|z|))$ is all that the argument needs.

For a meromorphic function with no zero or pole on $|w|=R$, set $G_R(z,a)=\log |(R^2-\overline{a}z)/(R(z-a))|$. At source points that are neither zeros nor poles, the displayed identity expresses its logarithmic modulus as the Poisson integral of its boundary values, minus its zero Green potentials plus its pole Green potentials. The Poisson kernel is $(R^2-|z{|}^2)/|R{e}^{i\theta }-z{|}^2$ with normalized angular measure. Subtracting the corresponding logarithmic singularities leaves a harmonic function, so the harmonic Poisson formula proves the identity. At zero it reduces to Jensen's formula.

For a polynomial $p(z)=c{\prod }_j(z-a_j)$ with $c\ne 0$, no polynomial root at zero, and $R>0$ different from every $|a_j|$, Jensen's formula gives the displayed expression. The polynomial roots are repeated according to their multiplicity of a zero: an exterior root contributes $\log |a_j|$, whereas an interior root contributes $\log R$.

For an explicit example, take the polynomialIts two simple zeros are $1/4$ and $-1/2$. The circle log-mean of a holomorphic function is thereforeAt $R=0.2$ the open disc contains no zeros of a function, at $R=0.4$ it contains $1/4$ only, and at $R=0.75$ it contains both. These three circles avoid the polynomial roots; the radii $1/4$ and $1/2$ are excluded from the displayed version.

The corresponding values of the circle log-mean of a holomorphic function and the circle geometric mean of a holomorphic function areFor example, at $R=0.4$ the contribution of the sole interior zero of a function is $\log (0.4/0.25)=\log (8/5)$; adding it to $\log (3/8)$ gives $\log (3/5)$. The second polynomial root still affects $f(0)$, but contributes no additional interior-zero term at this radius.

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.

For a degree-$d$ rational function $U$ of the Riemann sphere, choose relatively prime homogeneous polynomials $(P,Q)$ of degree $d$. Their joint norm on the unit three-sphere has positive lower and finite upper bounds. Thus $\log \Vert (P,Q)(v)\Vert -d\log \Vert v\Vert$ is bounded. Averaging over a circle using a nonvanishing entire homogeneous lift of a meromorphic function proves the displayed law for the spherical characteristic. Jensen's formula and $0\le \frac{1}{2}\log (1+x^2)-{\log }^{+}x\le \frac{1}{2}\log 2$ show that it also holds for the ordinary Nevanlinna characteristic.