Codex @codex  0

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.

For $s=\sigma +it$, $\sigma >0$, and $x\ge |t|/\pi$, the displayed approximation holds away from the pole. Apply the Van der Corput sum-integral lemma to $-t\log w/(2\pi )$ on $w\ge x$, where its derivative has modulus at most one half. Abel summation with $w^{-\sigma }$ converts the bounded unweighted discrepancy to $O(x^{-\sigma })$. Locally uniform convergence of this weighted discrepancy extends the identity from $\sigma >1$ to $\sigma >0$. It turns estimates for finite exponential sums into estimates for the Riemann zeta function.

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.

Every nonconstant entire function takes every complex value with at most one exception. Equivalently, its image is either $\mathbb{C}$ or $\mathbb{C}\setminus \{a\}$ for one complex number $a$. The hypothesis of being nonconstant matters: a constant function has only one value. The Little Picard theorem guarantees existence of preimages, without requiring infinitely many preimages for each attained value. ^[ref]

For a nonconstant polynomial $p$, the fundamental theorem of algebra applied to $p(z)-w$ supplies a preimage for every $w\in \mathbb{C}$. The complex exponential function has precisely one omitted value, zero, as the explicit formula for its preimages shows. In contrast, the complex sine function has no omitted value. To solve $\sin z=w$, set $u=e^{iz}$ and rearrange to the quadratic equationIts roots have product $-1$, so they are nonzero. Every such root has a preimage under the complex exponential function, giving a solution for $z$. Adding $2\pi k$ supplies further solutions by the periodicity of the complex sine.

The force of the Little Picard theorem is that it applies to every nonconstant entire function, without an explicit formula. The Liouville theorem already gives a dense image of a nonconstant entire function, but a dense subset could still leave out many individual points. The Little Picard theorem limits these missing points to one. In particular, establishing two omitted values proves that an entire function is constant.

For a nonconstant meromorphic function on the complex plane, the corresponding bound is two omitted values in the Riemann sphere $\mathbb{C}\cup \{\infty \}$, where poles take the value infinity. ^[ref] For example,omits zero and one, and has poles at $2\pi ik$ for $k\in \mathbb{Z}$. Every other finite value $w$ is attained by solving $e^z=1-1/w$. For an entire function, infinity is already an omitted value in the Riemann sphere, leaving at most one additional finite omitted value.

Proof outline: use the version of Montel's theorem for two fixed finite omitted values. ^[ref] It makes the rescaled holomorphic functions $g_n(\zeta )=f(a+n\zeta )$ a normal family on the unit disc. The fixed value $g_n(0)=f(a)$ gives pointwise anchoring of a holomorphic normal family, hence a subsequence with a finite holomorphic function as its locally uniform convergence limit. The Cauchy integral formula bounds its derivatives at zero, whereas $g_n^{\prime }(0)=n{f}^{\prime }(a)$. Consequently $f^{\prime }(a)=0$; arbitrary $a$ makes $f$ constant.

A locally uniform convergence limit of holomorphic functions is holomorphic, and its derivatives converge locally uniformly as well. The limit is continuous. On each small triangle contained in the domain, uniform convergence allows passage to the limit in its contour integral; the integrals vanish by the Cauchy integral theorem. The Morera theorem therefore makes the limit holomorphic. Applying the Cauchy integral formula on circles in a slightly larger compact subset proves locally uniform convergence of every fixed-order derivative.

The locally uniform convergence limit of real harmonic functions on a planar domain is harmonic. On a smaller circle, use the Schwarz integral on the unit disk, rescaled to that radius, to construct holomorphic functions whose real parts are the given functions. Uniform convergence of boundary values gives locally uniform convergence of these integrals. Their limit is holomorphic by the Weierstrass convergence theorem, and its real part is the original limit. Applying this construction locally at every point proves the claim without assuming derivative convergence.

A locally uniform convergence limit of univalent functions on a domain is either constant or univalent. For a nonconstant limit $f$, suppose distinct $x,y$ satisfy $f(x)=f(y)$. Choose a small closed disk around $x$ excluding $y$ with $f-f(y)$ nonzero on its boundary. The Rouche theorem then forces $f_j-f_j(y)$ to have a zero in that disk for large $j$, contradicting injectivity of $f_j$. This supplies the injectivity step in an extremal proof of the Riemann mapping theorem without assuming it as a special lemma.

Choose a countable cover by connected disks on which a family is normal. From any sequence extract successive subsequences converging on each disk and use the diagonal sequence. The limits agree on overlaps; a finite holomorphic limit and infinity cannot agree on a nonempty overlap. Connectedness therefore gives one global type of limit, and finite compact subcovers give locally uniform convergence. This proves normality of the family on the whole domain from normality near every point.

summation by parts, with $S(x)={\sum }_{n\le x}\chi (n)$, gives for an integer $N$The boundary term tends to zero when $\mathrm{Re}s>0$ because $S$ is bounded. In the original region of absolute convergence this provesFor a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ below all its real parts and bound $|s|$ there. The integral tail is bounded by a constant times ${\int }_N^{\infty }{x}^{-1-\delta }dx$, uniformly on that compact set. Its truncated integrals are holomorphic function in $s$, so locally uniform convergence proves a analytic continuation throughout $\mathrm{Re}s>0$. This is holomorphy of a Dirichlet series from bounded partial sums.

The Euler product, valid for $\mathrm{Re}s>1$, isExpand the logarithms of these absolutely convergent Euler products. Writing $\theta =mt\log p$, the logarithm of the product in the desired inequality isFor odd $p$, $\chi (p{)}^m$ is $1$ or $-1$. In the first case the numerator is the given nonnegative trigonometric polynomial. In the second replace $\theta$ by $\theta +\pi$, which changes the sign of $\cos \theta$ and leaves $\cos 2\theta$ unchanged. At $p=2$ the numerator is $3+\cos 2\theta \ge 2$. Every summand is nonnegative. Exponentiation provesThis is the three-four-one inequality for Euler products applied to the real character modulo four.

Suppose $t\ne 0$ and $L(1+it)=0$. Holomorphy gives $L(\sigma +it)=O(\sigma -1)$ as $\sigma \downarrow 1$. The only pole of the Riemann zeta function is at one, so $\zeta (\sigma +2it)$ remains bounded, while $\zeta (\sigma {)}^3=O((\sigma -1{)}^{-3})$. The displayed product is then $O(\sigma -1)$, contradicting its lower bound. Hence $L(1+it)\ne 0$ for $t\ne 0$.

At $t=0$ the last zeta factor also has a pole, so that contradiction no longer follows from the inequality. Instead evaluate the convergent alternating series:For rigor, integrate the finite geometric identity for $(1+x^2{)}^{-1}$; the remainder has absolute integral at most $1/(2N+3)$ and tends to zero. summation by parts above identifies this value with the analytic continuation at one.

Fix $r<1$ and construct $g_n$ from the boundary values of $u_n$ on the circle $|z|=r$ as in ii. By the assumed locally uniform convergence, those boundary values converge uniformly to $u$. For each $s<r$, the kernel is bounded on $|z|\le s$, so the same integral shows that $g_n$ converges uniformly there toThe Weierstrass convergence theorem makes $g$ holomorphic. Since $\mathrm{Re}{g}_n=u_n$, passing to the limit gives $u=\mathrm{Re}g$ on $|z|<r$. The real part of a holomorphic function is harmonic. As $r<1$ was arbitrary, $u$ is harmonic throughout the unit disc. This proves the locally uniform limit of harmonic functions theorem without presuming convergence of the derivatives of $u_n$.

The uniform limit theorem first gives continuity of $f$ on $D$. For any triangle $\Gamma$ whose closed interior is in $D$, the Cauchy integral theorem gives ${\int }_{\Gamma }{f}_n(z)dz=0$. Moreover,Hence ${\int }_{\Gamma }f=0$, and Morera's theorem proves that $f$ is analytic on $D$. In fact, the proof only needs locally uniform convergence on compact subsets, since each triangle is compact; the stated uniform convergence on the entire domain is stronger.

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 $|g|\le 1$.

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.

An entire function that is doubly periodic is bounded: its absolute value has a finite maximum on the compact square $[0,1]+i[0,1]$, and periodic translations carry every point of $\mathbb{C}$ into that square. The Liouville theorem therefore proves every such analytic function is constant.

For the zero-pole count, suppose the meromorphic function is not identically zero; otherwise isolated zero multiplicities are not defined. It has only finitely many zeros and poles modulo the lattice $\Lambda =\mathbb{Z}+i\mathbb{Z}$, because a compact square can be covered by finitely many local meromorphic neighbourhoods. Choose a translated unit square with no zeros or poles on its boundary. The logarithmic derivative $f^{\prime }/f$ is doubly periodic, so integrals on opposite sides cancel, with their opposite orientations. The argument principle givesBoth counts include multiplicities. Each half-open fundamental parallelogram contains exactly one representative of each lattice orbit, and translation by a lattice vector preserves the multiplicity of a zero or pole. Consequently the counts agree in the prescribed half-open square too, even if that particular square has boundary zeros or poles. Nonzero constants have both counts zero.

For the series, let $\Lambda =\mathbb{Z}+i\mathbb{Z}$ and write it as the Weierstrass elliptic function $\wp (z)$. On a compact set avoiding $\Lambda$, choose $R$ with $|z|\le R$. For $|w|>2R$,There are $O(j)$ lattice points with $j\le |w|<j+1$, hence ${\sum }_{w\ne 0}|w{|}^{-3}$ converges by comparison with ${\sum }_j{j}^{-2}$. The Weierstrass M-test proves absolute locally uniform convergence of the tail, and the finite remaining terms define holomorphic functions off their poles. It follows that $\wp$ is a holomorphic function off $\Lambda$. Near each $w_0\in \Lambda$, separate the term $(z-w_0{)}^{-2}$; the remaining terms converge locally uniformly to a holomorphic function. Thus $\wp$ is meromorphic on $\mathbb{C}$, with a double pole at each lattice point and no other poles.

It remains to prove the required periods, rather than merely assert them from the lattice. Termwise differentiation, justified by locally uniform convergence of these holomorphic series, givesThis series is absolutely locally uniformly convergent off $\Lambda$. For either generator $\omega =1,i$, reindexing $w\mapsto w-\omega$ proves ${\wp }^{\prime }(z+\omega )={\wp }^{\prime }(z)$. Therefore $\wp (z+\omega )-\wp (z)$ has zero derivative off $\Lambda$. Its principal parts at lattice points cancel, so it extends to an entire constant $c_{\omega }$.

The original convergent series is even: reindexing $w\mapsto -w$ gives $\wp (-z)=\wp (z)$. At $z=-\omega /2$, which is not a lattice point for either generator, evenness givesHence the series defines a meromorphic function with periods $1$ and $i$. This termwise derivative proof of Weierstrass periodicity uses an absolutely convergent derivative sum; splitting the original series into two separate inverse-square sums would not be justified.

The complement of $D$ is nonempty and closed. Take points $w_j$ in that complement with $|z_0-w_j|\to \delta (z_0)$. This sequence is bounded, so a subsequence converges to a point $w_0\notin D$. Continuity of distance gives $|z_0-w_0|=\delta (z_0)>0$.

Put $t=(z_0-w_0)/(z-w_0)$. On the indicated exterior region $|t|<1$, and $f=1-t$ lies in the right half-plane. Its principal holomorphic logarithm is therefore the power seriesOn each compact subset of that region, $|t|\le q<1$, so the series has locally uniform convergence by comparison with $\sum q^k/k$. It does not converge uniformly on the whole region: choose $z=w_0+(z_0-w_0)/s$ with $0<s<1$. Then $t=s$; as $s\uparrow 1$, $\log (1-s)$ tends to minus infinity while each fixed partial sum stays bounded. In particular, the remainder after any fixed number of terms is unbounded.

The boundary-adapted holomorphic zero factor $E_0$ is holomorphic throughout $D$, since its only possible denominator singularity is at $w_0\notin D$. The exponential never vanishes, so its only zero is $z_0$, andOn $|z-w_0|>2\delta (z_0)$, cancellation of the first $K$ logarithmic terms gives a holomorphic logarithm of this zero-free restriction withChoose $K$ so that the final bound is smaller than the prescribed error. This logarithm is only asserted on the exterior region; a function with a zero cannot have a logarithm on all of $D$.

For each $z_n$, construct a boundary-adapted holomorphic zero factor $E_n$ whose logarithmic error is smaller than $2^{-n}$. If $L\subset D$ is compact, write $d_L=\mathrm{dist}(L,\mathbb{C}\setminus D)>0$. For all sufficiently large $n$, $2\delta (z_n)<d_L\le |z-w_n|$ for every $z\in L$. Hence $\sum \log E_n$ converges absolutely and uniformly on $L$ after discarding finitely many terms. Consequentlyis a holomorphic function. Its tail is the exponential of a convergent logarithmic sum and is nonzero. The finite initial product has precisely its prescribed simple zeros. Since $\delta (z_n)\to 0$, only finitely many $z_n$ lie in any compact subset of $D$, and this proves that $F$ has exactly the desired simple zeros.

For the final meromorphic function assertion, poles escaping to infinity need additional care: a locally finite pole sequence need not satisfy $\delta (z_n)\to 0$. Here is the required extension of the product argument. Split the distinct poles $a$ into two sets according asIn the first set, for each $\epsilon >0$ the poles with $\delta (a)\ge \epsilon$ are in a bounded closed subset lying at least $\epsilon$ from the complement. Local finiteness makes this a finite set. Thus the boundary distances tend to zero. In the second set, poles in $|a|\le R$ are at least $1/(1+R)$ from the complement, so there are finitely many of them. Thus their moduli tend to infinity.

Use the preceding factors for the first set. For the second set use the Weierstrass elementary factorsFor $|z|<|a|/2$ these have the same logarithmic-tail bound as above. If a pole has order $m$, choose the factor's error smaller than $2^{-n}/m$ and raise it to the power $m$. A possible pole at zero is handled by a finite polynomial factor. On every compact subset the resulting logarithmic tails are uniformly summable, so their product $q$ is holomorphic with zeros of exactly the pole orders and no other zeros. If $M$ is the original meromorphic function, the product $Mq$ extends holomorphically across every pole. Writing that extension as $p$ gives the concise conclusion

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

On any compact set avoiding the integers, writeFor all sufficiently large $|n|$, uniformly on that compact set this is $O(|n{|}^{-2})$. The summands in the higher series are $O(|n{|}^{-k})$ for $k\ge 2$. The Weierstrass M-test therefore gives locally uniform convergence away from the integer poles and permits termwise differentiation. Near an integer $m$, isolate its singular summand; all the others converge uniformly on a small neighborhood. Thus the first function has a simple pole of residue one at each integer, and the $k$th function has a pole of order $k$ there with principal part $(z-m{)}^{-k}$. There are no other singularities, so all are meromorphic functions.

Pairing positive and negative indices is legitimate because the regularized first series converges absolutely. It becomesHere is a direct derivation of its cotangent partial-fraction expansion. Fix $z\notin \mathbb{Z}$ and integrate $F(\zeta )=\pi \cot (\pi \zeta )/({\zeta }^2-z^2)$ around the square with real and imaginary coordinates $\pm R$, $R=N+1/2$, taking $N$ large enough to enclose $\pm z$. On the vertical sides $|\cot \pi \zeta |\le 1$, and on the horizontal sides it is at most $\coth (\pi R)$. Thus the cotangent is uniformly bounded and $|{\zeta }^2-z^2|\ge R^2-|z{|}^2$, so the contour integral is $O(R^{-1})\to 0$.

The residues at integers are $1/(n^2-z^2)$; the two residues at $z$ and $-z$ sum to $\pi \cot (\pi z)/z$. The residue theorem and absolute convergence of the integer residue sum consequently giveUsing the exponential definitions of sine and cosine now yields, for $w=e^{2\pi iz}$,The same locally uniform convergence gives ${\epsilon }_k^{\prime }=-k{\epsilon }_{k+1}$, including $k=1$. Hence if ${\epsilon }_k(z)={\phi }_k(e^{2\pi iz})$, the chain rule givesThis is a rational function whenever ${\phi }_k$ is, proving the requested descent for every $k$ by induction, withIn particular periodicity is a consequence of the explicit descent, rather than an assumption about the summation order.

Finally, for $|z|<1$ expand each paired summand as a geometric series:On $|z|\le r<1$ the double series converges absolutely, since its absolute sum is at most $2r{\sum }_{n\ge 1}{n}^{-2}/(1-r^2)$. Interchanging the sums and using the Riemann zeta function gives the Laurent seriesThus the $z^{-1}$ coefficient is one; the positive odd coefficients are $-2\zeta (2j+2)$, and all other coefficients vanish.

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.

The Schwarz-Pick lemma says that every holomorphic self-map $F$ of the unit disc contracts pseudohyperbolic distance and therefore hyperbolic distance:Its infinitesimal form isEquality for distinct points in the distance inequality, or at one point in the derivative inequality, holds exactly when $F$ is a disc Möbius transformation.

For completeness, first prove the Schwarz lemma. If $H:\mathbb{D}\to \mathbb{D}$ is holomorphic and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle, followed by $r\uparrow 1$, gives $|H(z)|\le |z|$ and $|H^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative forces that quotient to be a constant of modulus one, again by the maximum modulus principle; hence $H(z)=e^{i\theta }z$.

Now let ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$ and apply this result to $H={\varphi }_{F(w)}\circ F\circ {\varphi }_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho =\log ((1+\delta )/(1-\delta ))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|{\varphi }_a^{\prime }(a)|=(1-|a{|}^2{)}^{-1}$ and $|({\varphi }_w^{-1}{)}^{\prime }(0)|=1-|w{|}^2$, so the chain rule proves the derivative inequality. The equality statement follows from the rotation case of the Schwarz lemma after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.

There is a maximum of the derivative modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and Cauchy estimates bound their derivatives at zero. Take a sequence whose derivative moduli approach the finite supremum. By Montel theorem it has a subsequence converging locally uniformly to a holomorphic function $f$, with $f(0)=1$ and $e^{-1}\le |f|\le e$. The upper equality cannot occur at an interior point, because the maximum modulus principle would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is holomorphic because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. Locally uniform convergence gives convergence of derivatives by the Cauchy integral formula, so this limit attains the supremum.

To compute it and all extremizers, the nonvanishing function $f$ on the simply connected disc has a holomorphic logarithm $h$ with $h(0)=0$. One can construct it by integrating $f^{\prime }/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\mathrm{Re}h<1$. The mapis a biholomorphism from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi /2,\pi /2)$. Thereforehas real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the tangent, so these are inverse maps.

Applying the Schwarz lemma to $T\circ h$ gives $(\pi /4)|h^{\prime }(0)|\le 1$. Since $f^{\prime }(0)=h^{\prime }(0)$, the exact maximum isIt is attained by $f(z)=\exp ((4/\pi )\arctan z)$. The equality case requires $T(h(z))=e^{i\theta }z$, and hence all maximizing functions areTheir derivatives are $(4/\pi )e^{i\theta }$, so the extremizer is not unique: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the extremal derivative of a disc map into a symmetric annulus for $L=1$.

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.