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This infinite product converges uniformly on compact subsets of the complex plane. Pair the factors at $iz$ and $-iz$ in the Weierstrass product for the reciprocal gamma function, and use the Gamma reflection formula together with $\Gamma (1+w)=w\Gamma (w)$. This gives $[\Gamma (1+iz)\Gamma (1-iz){]}^{-1}=\sinh (\pi z)/(\pi z)$ and the displayed product. The removable value at $z=0$ is one. Matching zeros alone would not exclude multiplication by a nonvanishing entire function; the normalized product fixes that ambiguity.

Take a directed graph with vertices $v_0,v_1,\dots$ and $n+1$ parallel bonds from $v_n$ to $v_{n+1}$. This is a locally finite graph, although the incoming degrees are unbounded. For bond percolation, ${\theta }_b(p)={\prod }_{n\ge 0}(1-(1-p{)}^{n+1})>0$ whenever $p>0$: the failure probabilities are summable, so the infinite product is positive. For site percolation, survival requires every vertex of the ray to be open, which has probability zero for $p<1$. Thus the survival and susceptibility thresholds for directed percolation satisfy $p_H^b=0$ and $p_H^s=1$.

For a submartingale $(Z_n)$ and $a<b$, let $U_N(a,b)$ count the completed upcrossings of $[a,b]$ by time $N$. The Doob upcrossing inequality givesFor a martingale $(X_n)$, a useful sharper form isHere $x^{+}=\max (x,0)$ and $x^{-}=\max (-x,0)$. To see the martingale form, use a bounded predictable process $H_k\in \{0,1\}$ that holds one unit after an observation at or below $a$ and releases it on the next observation at or above $b$. Its gain $G_N={\sum }_{k=1}^N{H}_k(X_k-X_{k-1})$ is at least $(b-a)U_N-(X_N-a{)}^{-}$: completed trades earn at least $b-a$, and any unfinished trade loses at most $(X_N-a{)}^{-}$. Since $\mathbb{E}{G}_N=0$, the bound follows. For the submartingale version, apply the same strategy to $(Z_n-a{)}^{+}$, itself a submartingale. Each purchase occurs at value zero, so its gain is at least $(b-a)U_N$. The complementary predictable strategy has nonnegative expected gain, bounding this strategy's expected gain by $\mathbb{E}(Z_N-a{)}^{+}-\mathbb{E}(Z_0-a{)}^{+}$, and in particular by the stated right side.

The almost-sure martingale convergence theorem says that if $C={\sup }_n\mathbb{E}|X_n|<\infty$, then $X_n$ converges almost surely to a finite, integrable random variable $Y$, with $\mathbb{E}|Y|\le C$. Indeed, the Doob upcrossing inequality bounds the expected total number of upcrossings of every rational interval by $(C+|a|)/(b-a)$. By monotone convergence, each such total count is finite almost surely; intersect these probability-one events over all rational $a<b$. If the lower and upper limits of a sample path differed, some rational interval strictly between them would be crossed infinitely often. Thus each remaining path has a limit in the extended real line. The Fatou lemma givesso the limit is finite almost surely and integrable. Almost-sure convergence by itself neither preserves expectations nor supplies this uniform first-moment bound.

For the loss of expectation, let $({\eta }_k)$ be independent variables with the fair Bernoulli distribution and setConditional on the past, survival doubles the value with probability $1/2$ and otherwise sets it to zero, so this is a nonnegative martingale. An infinite string of successes has probability zero. The process therefore eventually vanishes almost surely, giving yes: $X_0=1$ and $Y=0$ almost surely are possible. Its expectations remain one at every finite time; the process is not uniformly integrable.

A nonintegrable limit is also possible for a signed martingale. Take independent ${\xi }_k$ withand let $M_0=1$, $M_n=1+{\sum }_{k=1}^n{\xi }_k$. Every increment has mean zero and a finite absolute expectation, so each $M_n$ is integrable and $(M_n)$ is a martingale. Since ${\sum }_k\mathbb{P}({\xi }_k\ne 0)<\infty$, the Borel-Cantelli lemma makes only finitely many increments nonzero almost surely; hence the finite limit $Y=1+{\sum }_k{\xi }_k$ exists almost surely. To compute its absolute expectation, let $A_k$ be the event that only the $k$th increment is nonzero. Independence givesThe infinite product is positive because its tail logarithms are bounded in absolute value by a constant times the summable series ${\sum }_j{2}^{-j}$. The events $A_k$ are disjoint, and on $A_k$, $|Y|\ge 4^k-1$. ThereforeThus yes: a finite almost-sure martingale limit can have infinite absolute expectation, even with deterministic $M_0=1$. This is almost-sure martingale convergence to a nonintegrable limit; the missing hypothesis is the uniform first-moment bound. For a nonnegative martingale, in contrast, the Fatou lemma would force the limit to be integrable.

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.

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.

The infinite product converges locally uniformly in $\mathbb{H}$, because ${\sum }_{n\ge 1}|q{|}^n<\infty$ uniformly on compact subsets. None of its factors vanishes there, and the convergent logarithm of its tail shows that the product is nonzero. Logarithmic differentiation, justified by locally uniform convergence, givesThe middle equality groups the absolutely convergent terms $n{q}^{nr}$ by $m=nr$. Thus $F^{\prime }/F=2\pi i{E}_2$.

Let $H(\tau )={\tau }^{-12}F(-1/\tau )/F(\tau )$. It is nonzero and holomorphic on the simply connected upper half-plane. Using the printed transformation $E_2(-1/\tau )={\tau }^2{E}_2(\tau )+6\tau /(\pi i)$, its logarithmic derivative isSo $H$ is constant. At $\tau =i$ we have $-1/i=i$ and $i^{-12}=1$, giving $H(i)=1$. Therefore $F(-1/\tau )={\tau }^{12}F(\tau )$. Also $F(\tau +1)=F(\tau )$ directly from $q=e^{2\pi i\tau }$. Since these two transformations generate $S{L}_2(\mathbb{Z})$, $F$ has the weight-twelve modular form transformation law.

At the unique modular cusp its expansion is $F=q+O(q^2)$, so it extends holomorphically with zero value. HenceThis is the modular discriminant. The transformed argument contributes $6\tau /(\pi i)$; the converted TeX's numerator $67$ is a transcription error, not the printed relation.

No. The parallel-bond ray with different site and bond thresholds supplies a locally finite graph counterexample. Take vertices $v_0,v_1,\dots$ and $n+1$ parallel directed edges from $v_n$ to $v_{n+1}$. Its underlying multigraph is connected and every individual degree is finite, but there is no uniform incoming-degree bound.

For bond percolation rooted at $v_0$, survival means that every successive group of parallel bonds contains at least one open bond. Independence givesFor $p>0$, the sum of the failure probabilities is ${\sum }_{n\ge 0}(1-p{)}^{n+1}=(1-p)/p<\infty$. Eventually these terms are at most $1/2$, and $\log (1-t)\ge -2t$ for $0\le t\le 1/2$. The tail of the infinite product is therefore bounded below by a positive exponential; its finitely many initial factors are positive as well. Thus ${\theta }_b(p)>0$ for every $p>0$.

For site percolation, however, every vertex of the unique underlying ray must be open. For $p<1$, the probability that its first $N$ vertices are all open is $p^N\to 0$. At $p=1$ the ray is open. Consequently

For infinite products, a nonzero limiting product requires that its factors tend to one. A useful sufficient condition isAfter finitely many factors, $|u_j|<1/2$ and the principal holomorphic logarithm satisfies $|\log (1+u_j)|\le 2|u_j|$. Therefore the sum of logarithms converges, and exponentiating it gives a finite nonzero product. The same argument on compact sets proves infinite product convergence from logarithmic tails: a locally uniformly absolutely convergent tail of holomorphic logarithms gives a holomorphic nonvanishing tail product. Finite factors then determine all zeros and their orders.

The Weierstrass elementary factors areFor $|w|<1$,Their only zero is a simple zero at $w=1$.

Work with entire functions on $\mathbb{C}$. As usual for prescribed exact zero orders, the distinct zero locations must have consistent multiplicities. The literal statement allows repeated locations with conflicting orders; that cannot be true, for example if the same point is prescribed order one and order two. Remove consistent repetitions rather than adding their orders, and separate a possible zero at the origin.

List the distinct nonzero locations as $a_j$, with prescribed orders $n_j$, and write $m$ for the prescribed order at zero, or zero if the origin is not prescribed. Choose $p_j$ large enough thatThen the Weierstrass factorization theorem construction isOn every compact set, $|z/a_j|\le 1/2$ for all sufficiently large $j$ because $|a_j|\to \infty$. Its logarithmic tail is bounded byThus the product converges locally uniformly, is entire, and has exactly the specified zeros with exactly their orders. At a prescribed zero, only its own finite factor vanishes; all other finite factors and the tail product are nonzero.

Without escape to infinity the conclusion fails in general. Distinct proposed zeros $1/j$ accumulate at zero. The identity theorem would force an entire function with those zeros to vanish identically, which does not have precisely the prescribed isolated zeros. A finite zero set can of course be realized by a polynomial.

For entire functions with the same zero divisor, the quotient $q=f/g$ extends through every common zero by cancelling equal local powers. It is entire and nowhere zero. Hence $q^{\prime }/q$ is entire and has a primitive on the simply connected plane. Choose $h(0)$ with $e^{h(0)}=q(0)$ and putThe derivative of $q{e}^{-h}$ vanishes, and its value at zero is one. ThereforeIf also $f=e^{k}g$, then $e^{h-k}=1$ everywhere. The continuous difference $h-k$ takes values in the discrete set $2\pi i\mathbb{Z}$ and so is constant on the connected plane:The whole-plane hypothesis matters: a zero-free holomorphic quotient on a multiply connected domain need not possess a global holomorphic logarithm.

PutThe infinite product converges locally uniformly and never vanishes on the complex upper half-plane. With $D=qd/dq=(2\pi i{)}^{-1}d/d\tau$, logarithmic differentiation gives

The product is unchanged by $T:\tau \mapsto \tau +1$. To study $S:\tau \mapsto -1/\tau$, defineUsing the transformation law for the Eisenstein series of weight two,Thus $R$ is constant. At the fixed point $\tau =i$, one has $i^{12}=1$, so $R(i)=1$. HenceThe transformations under $S$ and $T$, which generate the modular group, show that $P$ is a weight-twelve modular form. Its Fourier expansion begins $q+O(q^2)$, so it is a cusp form. The normalized element of $S_{12}(\Gamma (1))$ is unique by part (a), and therefore

The specialization of the Jacobi triple product to the Jacobi theta function isSince $\tau \in \mathfrak{h}$, one has $|q|<1$, so every displayed factor is nonzero. MoreoverThe standard convergence criterion for infinite products therefore shows that the product converges to a nonzero value. This proves the nonvanishing of the Jacobi theta function throughout $\mathfrak{h}$.

Every closed discrete subset of a plane domain is the zero set of a holomorphic function, with assigned finite positive multiplicities if desired. Exhaust the domain by relatively filled compact sets. For a prescribed point outside an earlier compact, choose a rational factor with that zero and its possible pole outside the domain in the same complementary component. A slit joining zero and pole allows a holomorphic logarithm near the earlier compact. Approximate that logarithm by Runge theorem and multiply the rational factor by the negative exponential of the approximant. The resulting factors tend to one with summable errors on each compact. Their infinite product has the prescribed zeros and no additional zeros.

// Destination: analysis.bigb