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For a plane domain, the following are equivalent descriptions of a simply connected domain: every based loop contracts in the domain with its basepoint fixed; any two paths with the same endpoints are homotopic while those endpoints stay fixed; its complement in the Riemann sphere is connected; every holomorphic function has a single-valued primitive; and every nowhere-zero holomorphic function has a holomorphic logarithm. The complement must be taken in the extended plane, including infinity. An equivalent winding number formulation is that every closed curve has index zero about every point outside the domain.

Here is a direct equivalence proof for the first two topological formulations. If all paths with given endpoints are homotopic relative to those endpoints, apply this to a loop based at $a$ and the constant path at $a$, obtaining a contraction. Conversely, let ${\gamma }_0,{\gamma }_1$ join $a$ to $b$. The loop ${\gamma }_0\ast {\gamma }_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with ${\gamma }_1$ produces a homotopy from ${\gamma }_0\ast {\gamma }_1^{-1}\ast {\gamma }_1$ to ${\gamma }_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a homotopy from that first path to ${\gamma }_0$. Hence ${\gamma }_0$ and ${\gamma }_1$ are homotopic relative to their endpoints. Planar domains are path-connected, so no extra path-component condition is needed.

The analytic descriptions follow from path independence of the holomorphic integral. Local primitives exist on discs by integrating power series; a homotopy can be subdivided into pieces lying in such discs, and local path integrals cancel along its common edges. Thus integrals over homotopic paths agree, giving a global primitive. Applied to $h^{\prime }/h$ for a nowhere-zero function, a primitive $H$ makes $h{e}^{-H}$ constant and hence supplies a holomorphic logarithm. These observations are also the tools used in the mapping proof below.

The Riemann mapping theorem states that every nonempty simply connected proper domain $D⊊\mathbb{C}$ admits a conformal bijection onto the unit disc. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f^{\prime }(z_0)>0$. The exclusion of the whole plane is necessary: an entire map into the disc would be constant by Liouville's theorem.

An extremal proof has the following concrete steps. First construct an injective bounded holomorphic map. Choose $a\notin D$ and a holomorphic square root $h$ of $z-a$. It is injective, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is injective and bounded. Scaling it and applying an automorphism of the unit disk gives a map into the disc taking $z_0$ to zero; a rotation makes its derivative positive.

Let $\mathcal{U}$ consist of all injective maps $D\to \mathbb{D}$ with this normalization. It is nonempty. The derivative at $z_0$ is bounded above by the Cauchy integral formula on a small disc compactly contained in $D$. Choose $f_n\in \mathcal{U}$ with derivatives tending to their positive supremum $L$. Boundedness, the Cauchy estimates and a diagonal Arzelà-Ascoli theorem argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the Montel theorem. Then $f(z_0)=0$, $f^{\prime }(z_0)=L>0$, and the maximum modulus principle puts its image in the open disc. The limit is injective: if distinct $z,w$ had $f(z)=f(w)$, apply Hurwitz's theorem near $w$ to $f_n(\cdot )-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f^{\prime }(z_0)>0$.

Finally suppose $a\in \mathbb{D}$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline{a}w)$ and choose a holomorphic square root $h$ of $T_a\circ f$. It is injective and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a rotation. The new map's derivative magnitude at $z_0$ isThis square-root improvement of a normalized conformal map contradicts extremality. Thus $f$ is onto. Two normalized bijections differ by a disc automorphism fixing zero; Schwarz lemma applied to it and its inverse makes it a rotation, and positivity of the normalized derivatives forces that rotation to be the identity. This completes the requested proof outline.

On the right half-plane a single-valued holomorphic logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$, with $-\pi /2<\mathrm{Arg}z<\pi /2$, and its derivative is $1/z$. ThusThis proves the conclusion directly from the integral definition of the winding number, without appealing to its homotopy invariance.

The zero-prescription claim requires a closed discrete subset, meaning no accumulation in $\Omega$. If “discrete” means only that each listed point is isolated, it is false: $\{1/n:n\ge 1\}\subset \mathbb{C}$ is a discrete subset, but any holomorphic function vanishing there is identically zero by the identity theorem for holomorphic functions, and then has additional zeros. In the usual zero-set interpretation, discreteness includes local finiteness. We prove the claim with this necessary interpretation.

First choose an increasing compact exhaustion $K_j$ of $\Omega$ by relatively filled sets: every component of $\hat{\mathbb{C}}\setminus K_j$ meets $\hat{\mathbb{C}}\setminus \Omega$. Such an exhaustion is obtained by exhausting with finite unions of small closed squares lying in $\Omega$, then adding all complementary components relatively compact in $\Omega$. The filled sets remain compact in $\Omega$: a complementary component containing a point within distance $\mathrm{dist}(K_j,\mathbb{C}\setminus \Omega )$ of the complement could join that complement through a ball disjoint from $K_j$, and hence would not be filled. Their boundedness follows by using a disk containing the initial compact. Nesting can be ensured by enlarging each stage first. This is also the geometric description of the holomorphic convex hull proved below.

A finite prescribed set is realized by a finite product of linear factors. For an infinite closed discrete subset, enumerate the distinct points as $a_n$. Every compact contains only finitely many of them, so there are indices $j(n)\to \infty$ with $a_n\notin K_{j(n)}$. Finitely many initial factors can be handled without an approximation requirement. For the remaining factors, let $U_n$ be the complementary component containing $a_n$. It also contains a point $c_n\notin \Omega$ or infinity. Join $a_n$ to $c_n$ by an arc in that component, or to infinity by a ray-like arc. DefineEach $p_n$ is a holomorphic function on $\Omega$, has exactly one simple zero, at $a_n$, and no other zeros there.

The arc is disjoint from $K_{j(n)}$, and $p_n$ has a holomorphic logarithm $L_n$ near that compact. To justify the logarithm, any closed curve avoiding the arc has equal winding numbers about its finite endpoints, so the integral of $p_n^{\prime }/p_n$ around it is zero. For a ray to infinity the winding number about $a_n$ is zero. A primitive of $p_n^{\prime }/p_n$, adjusted by a constant on each component, supplies the logarithm.

The Runge theorem proved in Question 4 approximates $L_n$ uniformly on $K_{j(n)}$ by $h_n\in \mathcal{O}(\Omega )$: all needed finite poles can be chosen outside $\Omega$ because every complementary component meets its complement. Choose the error small enough thatEach $F_n$ still has exactly the simple zero $a_n$. On every fixed compact, all sufficiently late factors satisfy the summable bound above. Hence their infinite product converges locally uniformly. It has a nonzero tail because, once $|F_n-1|<1/2$, its logarithmic tails satisfy $|\log F_n|\le 2|F_n-1|$ and converge absolutely. Thus multiplying by the finitely many earlier factors gives a holomorphic function with precisely the prescribed zeros, and no additional zeros. Repeating a factor to any prescribed finite positive multiplicity gives the corresponding multiplicity version. This is prescribed zeros in a plane domain via Runge approximation, not an invocation of the existence theorem being asked for.

For the specified infinite product, put $a_k=q^{-2k-1}$. On any compact annulus $0<r\le |z|\le R$,The logarithmic-tail estimate therefore proves locally uniform convergence on ${\mathbb{C}}^{\ast }$ and nonvanishing wherever no individual factor vanishes. At a zero of one factor, the product is zero and the remaining product converges to a nonzero value. ConsequentlyAll these zeros are simple and distinct: their moduli distinguish the two strings and their indices. At $z=0$ the reciprocal factors are undefined, so the given expression has no convergence value there. The infinite sequence of zeros tending to zero also rules out continuation across that point as a holomorphic function.

Set $P(z)={\prod }_{k\ge 1}(1-q^{-2k}z)(1-q^{-2k}{z}^{-1})$. Absolute local convergence allows reindexing:Since $1-z=-z(1-z^{-1})$, $\sigma (qz)=-z\sigma (q^{-1}z)$ for every $z\ne 0$. No division by a vanishing factor is involved, so the identity also holds at all the zeros.

Choose centred charts so the coordinate expression $F$ satisfies $F(0)=0$. On a component where the map is nonconstant, the identity theorem makes $F$ nonconstant. Its Taylor expansion factors as $F(z)=z^{n}h(z)$, with $n\ge 1$ and $h(0)\ne 0$. In a sufficiently small disc, choose a holomorphic logarithm of $h$, and hence a holomorphic $n$th root $q$ with $q^n=h$. The function $zq(z)$ has nonzero derivative at zero. The holomorphic inverse function theorem, which says that a holomorphic function with nonzero derivative has a holomorphic local inverse, makes it a valid new source coordinate. In that coordinate,If the map is bijective, local injectivity forces $n=1$: for $n>1$, a small nonzero value has $n$ distinct preimages in this local disc. The inverse is therefore holomorphic in a neighbourhood of every point, and the local inverses agree, proving the map is biholomorphic. The nonconstant hypothesis must apply to the component containing $p$ if disconnected Riemann surfaces are allowed; otherwise a constant component is an exception to the proposed local form.

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

For the unrestricted zero sequence, introduce the Weierstrass elementary factorsFor $|u|\le 1/2$, cancellation of the first $p$ terms of the logarithm power series gives a holomorphic logarithm of the factor:Define the canonical productFix any compact disc $|z|\le R$. Since $|z_n|\to \infty$, eventually $|z/z_n|\le 1/2$ uniformly on this disc. The tail logarithms are then bounded by $2(1/2{)}^{n+1}$, a summable bound independent of $z$. Their sum is holomorphic by locally uniform convergence of holomorphic functions, and its exponential is nowhere zero. Multiplication by the finite initial product therefore gives an entire function. Each factor has a simple zero exactly at its specified $z_n$, the exponential factor has none, and a tail beginning beyond any given compact disc is nonvanishing there. Thus $f$ has precisely the prescribed zeros and no others, each simple, and $f(0)=1$. This proves the required existence without an assumption on ${\sum }_n|z_n{|}^{-1}$.

For the final, unheaded product request, suppose now that ${\sum }_n|z_n{|}^{-1}<\infty$. On $|z|\le R$, all sufficiently late factors have $|z/z_n|\le 1/2$, so part (a) bounds their principal complex logarithms by $2R/|z_n|$. The logarithms converge absolutely and uniformly; infinite product convergence from logarithmic tails proves thatis entire and again has exactly the prescribed simple zeros. For its finite partial products,Passing to the limit provesIn particular $P$ is an entire function of exponential type. This argument uses a valid modulus estimate and does not require the false logarithm assertion in part (c).

For a positive integer $N$, a Dirichlet character $\psi$ modulo $N$ is a group homomorphism $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }$, extended periodically to the integers and set equal to zero on integers not coprime to $N$. Its nonzero values are roots of unity. For $\mathrm{Re}s>1$, its Dirichlet L-function is the absolutely convergent Dirichlet seriesComplete multiplicativity of $\psi$ proves the Euler product by expanding each factor as a geometric series and applying unique integer factorization; absolute convergence justifies these rearrangements.

Suppose $\psi$ is a nonprincipal Dirichlet character. Choose a unit $u$ with $\psi (u)\ne 1$. Multiplication by $u$ permutes a complete residue system, so ${\sum }_{a=1}^N\psi (a)=\psi (u){\sum }_{a=1}^N\psi (a)$, and the sum is zero. Consequently $A(x)={\sum }_{n\le x}\psi (n)$ is bounded: complete periods contribute zero and the remaining incomplete period has at most $N$ terms. The Abel summation formula gives, initially for $\mathrm{Re}s>1$,For $\mathrm{Re}s>0$ the integral converges, locally uniformly together with its derivatives in $s$, because on a compact set with $\mathrm{Re}s\ge \delta >0$ the differentiated integrands are bounded by constants times $x^{-1-\delta }(1+\log x{)}^k$. Thus holomorphy of a Dirichlet series from bounded partial sums proves

For a nonreal $\psi$, the character ${\psi }^2$ is also nonprincipal. The supplied logarithmic Euler product yields, for real $\sigma >1$,At a prime dividing $N$ the numerator is $3$. At every other prime put $\psi (p{)}^k=e^{i\theta }$; then the numerator is $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. This Euler product positivity for L-function nonvanishing proves that the product inside the logarithm is at least one. If $L_N(\psi ,1)=0$, its holomorphy would give $L_N(\psi ,\sigma )=O(\sigma -1)$. Meanwhile $L_N({\psi }^2,\sigma )$ is bounded near one and the Riemann zeta function has a simple pole there. The product would therefore be $O(\sigma -1)$ and tend to zero, a contradiction.

For a real nonprincipal $\psi$, suppose instead that $L_N(\psi ,1)=0$. The product $H(s)=\zeta (s)L_N(\psi ,s)$ would then be holomorphic on the whole half-plane $\mathrm{Re}s>0$, because the zero cancels the only zeta pole there. In $\mathrm{Re}s>1$ it has coefficientsThese are nonnegative zeta-times-real-L coefficients. They are multiplicative, and their prime-power values are $k+1$ if $\psi (p)=1$, one or zero according as $k$ is even or odd if $\psi (p)=-1$, and one if $\psi (p)=0$. In particular $b_n\ge 0$ and $b_{m^2}\ge 1$ for every $m$.

The Taylor series of the assumed holomorphic $H$ about $2$ converges in the disk $|s-2|<2$. Termwise differentiation of its absolutely convergent Dirichlet series at $2$ gives $(-1{)}^k{H}^{(k)}(2)={\sum }_n{b}_n(\log n{)}^k/n^2$. Evaluate the Taylor series at $s=1/2$ and apply the Tonelli theorem to the resulting nonnegative terms:This contradicts holomorphy at $1/2$. The argument is the positive-coefficient mechanism of the Landau theorem for a Dirichlet series with nonnegative coefficients, with the needed step proved explicitly. Combining the real and nonreal cases gives the nonvanishing of a nonprincipal Dirichlet L-function at one:

Now let $a$ be coprime to $N$. The Dirichlet theorem on primes in arithmetic progressions asserts that infinitely many prime numbers satisfy $p\equiv a(\mathrm{mod}N)$. To prove it, separate the first-power terms in the given logarithmic Euler product:For $\sigma \ge 1$, the remainder is bounded uniformly byFor a nonprincipal character, holomorphy and nonvanishing near one give a local holomorphic logarithm. Along a short real interval to the right of one, this logarithm and the supplied continuous logarithm differ by a fixed integer multiple of $2\pi i$. Hence the prime-character sum near one is bounded:For the principal Dirichlet character ${\psi }_0$,The latter follows from the simple zeta pole and the bounded remainder. Finally Orthogonality of Dirichlet characters givesHere $\phi$ is the Euler totient function. The right-hand side tends to infinity as $\sigma \downarrow 1$, whereas a finite set of primes would give a bounded left-hand side. Therefore

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.

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 $L=1$.

For $\Re s>1$, the Euler products areThe Dirichlet character is completely multiplicative, and unique prime factorization expands each finite product into the terms of the corresponding Dirichlet series supported on those primes. Absolute convergence permits passage to all primes. More strongly, the absolutely convergent Euler product logarithm givessince the sum of absolute values is bounded by ${\sum }_{n\ge 2}{\sum }_{m\ge 1}{n}^{-m\Re s}/m<\infty$.

We prove nonvanishing of a nonprincipal Dirichlet L-function at one, including real Dirichlet characters. If $\chi$ is nonprincipal modulo $q$, periodicity and the given Orthogonality of Dirichlet characters bound $A_{\chi }(x)={\sum }_{n\le x}\chi (n)$ by $q$. Partial summation giveswhich is holomorphic for $\Re s>0$ by locally uniform convergence, including after differentiation. The principal Dirichlet character instead haswith a simple pole at one.

For a nonreal Dirichlet character, consider $P(s)={\prod }_{\chi \mathrm{mod}q}L(s,\chi )$. For real $\sigma >1$, its Euler product logarithm isThe given character inversion with $a=1$ makes each inner sum either $\phi (q)$ or zero, so $P(\sigma )\ge 1$. If $L(1,\chi )=0$ for a nonreal Dirichlet character, then $L(1,\overline{\chi })=0$ as well, since their values at real $s$ are conjugate. These are two distinct factors. All nonprincipal factors are holomorphic at one, and only the principal factor has a simple pole. Their two zeros would force $P(\sigma )\to 0$, a contradiction.

For a real nonprincipal Dirichlet character, suppose $L(1,\chi )=0$. Then $G(s)=\zeta (s)L(s,\chi )$ is holomorphic throughout $\Re s>0$: the supplied continuation of the Riemann zeta function has no other pole there, and the zero cancels the pole at one. On $\Re s>1$ its Dirichlet series has coefficientsTo check the sign explicitly, for a prime power $p^j$ the local sum is $j+1$ if $\chi (p)=1$, is $1$ for even $j$ and $0$ for odd $j$ if $\chi (p)=-1$, and is $1$ if $\chi (p)=0$. Multiplication over prime factors proves nonnegativity and $g_{m^2}\ge 1$. Also $g_n\le \tau (n)$, so termwise derivatives at $s=2$ are justified by absolute convergence:The disc $|s-2|<2$ lies in $\Re s>0$. Its Taylor series therefore converges at $s=1/2$, and Tonelli theorem givesThis contradicts holomorphy at $1/2$. It is the positive-coefficient argument behind the Landau theorem for a Dirichlet series with nonnegative coefficients, proved here rather than invoked. Thus $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

It remains to detect the primes. For real $\sigma >1$, the Euler product logarithm satisfiesuniformly as $\sigma \downarrow 1$: terms with $m\ge 2$ are bounded in absolute value by ${\sum }_{n\ge 2}{\sum }_{m\ge 2}{n}^{-m}/m<\infty$. For a nonprincipal Dirichlet character, a holomorphic logarithm of $L$ exists near one by the nonvanishing just proved. On a short real interval it differs from the Euler product logarithm by a fixed multiple of $2\pi i$, so the prime sum stays bounded. For the principal Dirichlet character, the simple pole instead givesApplying the supplied Orthogonality of Dirichlet characters for a reduced residue class $a$ now yieldsThe right side tends to infinity, whereas a finite set of primes would give a bounded left side. Therefore every residue class coprime to $q$ contains infinitely many primes, proving the Dirichlet theorem on primes in arithmetic progressions.

We use the Dirichlet density form of the Chebotarev density theorem, the version directly proved by the permitted information at $s=1$. If $L/K$ is a finite Galois extension of number fields with group $G$ and $C$ is a conjugacy class, let $S_C$ be the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. ThenThe same conclusion holds for a union of conjugacy classes, replacing $|C|$ by the size of that union. The denominator may equivalently be the sum over all primes of $K$. Here and below $s>1$ is real. This specifies the density being proved; an unweighted prime-counting asymptotic is a stronger formulation and is not being inferred merely from a limit at one.