Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 9 (The Value Distribution of Analytic Functions)

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.

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

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.

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 $p>0$.

The printed proximity integral is missing a factor $1/(2\pi )$. Denote that literal integral by $m_{\mathrm{print}}$ and write $\overline{m}=m_{\mathrm{print}}/(2\pi )$. The correct Annular spherical Jensen formula, with the printed definitions of $N$ and $T$, isThe following derivation proves this identity and also isolates the normalization issue.

For the chordal metric put $u_a(z)=-\log k(f(z),a)$ and let $M_a(s)=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_a(s{e}^{i\theta })d\theta$. In a finite target coordinate,For $a=\infty$ use $u_{\infty }=\frac{1}{2}\log (1+|f{|}^2)$. Differentiating away from the preimages of $a$ gives half the squared spherical derivative of a meromorphic function. At an $a$-point of local degree $d$, $u_a=-d\log |z-z_0|+O(1)$; the second distributional derivatives give the Laplacian of $\log |z-z_0|$ is $2\pi {\delta }_{z_0}$. Thus, including poles by changing spherical coordinates,Because $a$ is off the image of $|z|=1$, $u_a$ is smooth near that circle and $I(a)=M_a^{\prime }(1)$ is finite and independent of $r$.

Apply Green's second identity to the harmonic weight $w(z)=\log (|z|/r)$ on $r<|z|<1$. On the outer boundary $w=\log (1/r)$ and ${\partial }_{n}w=1$; on the inner boundary $w=0$ and ${\partial }_{n}w=-1/r$. HenceThe distributional formula makes the left side $2\pi T(r)-2\pi N(r;a)$, and $M_a(r)-M_a(1)=\overline{m}(r;a)$, proving the corrected Annular spherical Jensen formula. Initially take an inner circle avoiding $a$-points. Logarithmic singularities have continuous angular means, and a point on the inner circle has zero logarithmic counting weight, so continuity extends the identity to every $0<r\le 1$.

There is no way to absorb the missing factor into a different constant $I(a)$. For the holomorphic function $f(z)=z$ and $a=\infty$, direct integration givesThe value of $(T-m_{\mathrm{print}})/\log (1/r)$ tends to $1/2$ as $r\downarrow 0$ and to $\pi$ as $r\uparrow 1$. Therefore the literal printed identity has no constant $I(\infty )$, whereas the corrected identity has $I(\infty )=1/2$.

For the singularity criterion, define the nondecreasing normalized spherical areaTonelli's theorem applied to the logarithmic weight gives $T(e^{-l})={\int }_0^{l}s(t)dt$. Averages of a nondecreasing function converge to its supremum: for fixed $t_0<l$, $(1-t_0/l)s(t_0)\le T(e^{-l})/l\le s(l)$. Thus the limit, not just the lower limit, exists in $[0,\infty ]$ and equalsIf $f$ has a removable singularity or pole at zero, it extends as a holomorphic map to the Riemann sphere; the pulled-back spherical density is smooth and bounded on the closed disk, hence its area is finite. Conversely an essential singularity gives, by the meromorphic version of the Great Picard theorem, infinitely many preimages of every spherical target with at most two exceptions in each punctured neighborhood. The area formula givesThe formula follows by applying change of variables on inverse-function neighborhoods away from the discrete critical points, then summing with multiplicity; the countable critical values have area zero. This proves the finite spherical area criterion for an isolated singularity and the requested equivalence between a finite lower limit and a removable singularity or pole.

A covering-surface formulation of the Ahlfors second fundamental theorem is as follows. For a regular finite bordered Riemann surface $X$ mapped holomorphically to a fixed metric bordered Riemann surface $Y$, let $S=A_f(X)/A(Y)$ be the average sheet number and $L$ the pulled-back length of the relative boundary, namely the boundary mapped into the interior of $Y$. With the modern sign convention for the Euler characteristic,where $h$ depends on $Y$ and its metric, not on $X$ or the map. The associated area comparison is $|S_D-S|\le h_{D}L$ for each fixed regular target region $D$. These formulations and the sign convention are explained in www.math.purdue.edu/~eremenko/dvi/ahl.pdf .

The island version of the Ahlfors second fundamental theorem for a disk source, which is the useful form here, says that for $q\ge 3$ fixed Jordan domains $D_j$ with disjoint closures on the Riemann sphere,Here $n_j$ counts simply connected islands of a meromorphic function over $D_j$, once each without weighting by mapping degree, andAn island of a meromorphic function is a relatively compact inverse-image component mapped properly onto its target domain; components reaching the source boundary are not counted. The simply connected island estimate is recorded, for spherical disks, in Lemma 2 of pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf . Regular target regions give the same estimate with a region-dependent constant. General Jordan domains can be enclosed in slightly larger regular Jordan domains still having disjoint closures. A simple island over an enlarged domain restricts to one over the original domain, so it suffices to prove the conclusion for regular domains.

Suppose a nonconstant meromorphic function on the plane has no simple islands over five such domains. On every counted simply connected island its proper map has integer degree at least two. The area formula therefore givesTogether with area comparison and the Ahlfors second fundamental theorem, this yieldsThe area comparison can also be seen directly. The two-form $[1_D/A(D)-1/(4\pi )]dA$ on the target sphere has zero integral and a bounded one-form primitive. One construction solves its Poisson equation: the gradient of the spherical Green function has an integrable $1/\mathrm{distance}$ singularity, so convolution against the bounded density gives a bounded primitive. Pulling back and applying Stokes theorem, or smooth approximation at the domain boundary, bounds the integral by a constant times $L(R)$.

We now derive the needed length-area exhaustion of the complex plane. The Cauchy-Schwarz inequality on a circle givesIf $L(R)/S(R)\ge \epsilon >0$ for all sufficiently large $R$, then $S(R)>0$ andIntegration makes $1/S(R)$ negative, which is impossible. Hence there are radii $R_k\to \infty$ with $L(R_k)/S(R_k)\to 0$. This contradicts $S/2\le CL$ and proves the Ahlfors five islands theorem: one of the five target domains has a bounded simply connected inverse-image component mapped biholomorphically onto it.

Four domains do not suffice. Take the Weierstrass elliptic function $\wp$ for the square lattice $\mathbb{Z}+i\mathbb{Z}$. The equationshows the three distinct finite branch values $e_1,e_2,e_3$; the fourth is infinity, because all poles are double. Equivalently, the degree-two map from the elliptic torus to the Riemann sphere has these four branch values, with all their preimages of local degree two. Choose four small disjoint Jordan domains, each containing one of these four totally ramified Weierstrass values. A degree-one island of a meromorphic function over any one would contain a preimage of its central value of local degree one, contradicting total ramification. Thus $\wp$ provides four domains with no simple island.

Write $s=A(S)=4\pi$, $L=L(\partial U)$, and $a=\min \{A(U),A(S\setminus U)\}$. We first prove a constant-version spherical isoperimetric inequality directly, including regions with several boundary components.

If $L\ge 1$, the arithmetic bound $A(U)A(S\setminus U)\le s^2/4=4{\pi }^2$ already gives $A(U)A(S\setminus U)\le 4{\pi }^2{L}^2$. If $L=0$, a region with smooth boundary is empty or the whole connected sphere up to boundary-null changes, and its area product is zero. For $0<L<1$, write the smooth boundary as finitely many disjoint simple closed curves ${\gamma }_j$ of lengths $l_j$. Fix a point $p_j$ on each curve. The shorter boundary arc from $p_j$ to any other point has length at most $l_j/2$, so ${\gamma }_j$ is contained in the closed geodesic ball of radius $l_j/2$ about $p_j$.

The complement of that small ball is connected and lies entirely on one side of ${\gamma }_j$. The Jordan curve theorem therefore gives a disk $E_j$ bounded by ${\gamma }_j$ contained in the ball. Its area satisfiesThe indicator function of $U$ and the sum ${\sum }_j{1}_{E_j}$ have the same jumps modulo two across every boundary curve. Their difference modulo two is consequently constant throughout the sphere off the boundary. Thus either $U$ or its complement lies in ${\bigcup }_j{E}_j$, up to its area-zero boundary. This also handles nested curves and disconnected regions. It follows thatCombining the cases, one valid universal choice is $\kappa =4{\pi }^2$. Sharpness is not required.

For the inclusion estimate put $x=A(V)$, $d=A(D)>0$ and $y=A(V\cap D)$. Splitting $D$ into its portions inside and outside $V$ givesBecause $0\le A(D\setminus V)\le s-x$ and $0\le A(D\cap V)\le x$, this expression lies between $-x(s-x)$ and $x(s-x)$. Apply the preceding spherical isoperimetric inequality to $V$ and divide by $sd$:Thus the constant in the spherical inclusion area discrepancy is independent of the nonempty open set $D$.

Let $g$ and $\tilde{g}$ be the old and new smooth Riemannian metrics. Their area forms satisfyThe determinant ratio is invariant under changes of coordinates, so $\theta$ is a globally defined smooth function. The layer cake representation follows from $\theta (x)={\int }_0^{\infty }{1}_{\{\theta (x)>t\}}dt$ and Tonelli's theorem:Smoothness makes every $D_t$ open. Compactness of the sphere gives $0<m=\min \theta \le M=\max \theta <\infty$. Multiply the inclusion bound by $A(D_t)$ before integrating; for empty $D_t$ the resulting bound is trivially zero. For $t<m$, $D_t=S$ and the discrepancy is exactly zero; for $t>M$, $D_t$ is empty. ThereforeDividing by $\tilde{A}(S)>0$ provesThis constant depends only on the two metrics. If $\theta$ is constant the normalized areas agree exactly, as the bound also shows.