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An announcing sequence consists of stopping times increasing to $T$ almost surely, strictly smaller than $T$ on $\{T>0\}$. Such a lifetime can be approached through stopped intervals on which a locally defined stochastic process is an ordinary adapted process. For the usual maximal local solution of a stochastic differential equation, the exit times ${\tau }_n$ from a nested compact exhaustion of the open domain, capped as $T_n={\tau }_n\wedge n$, supply this sequence. The existence of such a sequence is an extra lifetime convention; arbitrary stopping times need not admit one.

A normal family is a family such that every sequence has a subsequence converging locally uniformly in the spherical metric to a meromorphic limit, possibly the constant infinity. For a family uniformly bounded in modulus, the possible limits are finite holomorphic functions, and ordinary locally uniform convergence suffices. The limit need not take values in the open unit disc; a sequence of constants can approach its boundary.

Here is a direct compactness proof for disc-valued maps. For each compact subset $K\subset \Omega$, choose a slightly larger compact neighborhood contained in $\Omega$. The Cauchy estimate on a fixed-radius disk around each point bounds $|f^{\prime }|$ independently of $f$ because $|f|<1$. A finite covering of $K$ yields uniform boundedness and equicontinuity there. The Arzelà-Ascoli theorem, applied successively to a compact exhaustion and followed by a diagonal subsequence, gives a limit uniformly on every compact. The Cauchy integral formula passes to that limit on small circles, proving it is a holomorphic function. This proves the required normal family assertion rather than just invoking Montel theorem.

Every $g_n(z)=g(z/n)$ is a holomorphic function on the complex upper half-plane with values in the unit disc, so the same argument makes $(g_n)$ a normal family. If a subsequence converges locally uniformly to $h$, then for every $t>0$,The identity theorem for holomorphic functions gives $h\equiv \ell$. Since every convergent subsequence has this same limit, the whole sequence converges locally uniformly to $\ell$: otherwise a compact set and a subsequence with a fixed positive discrepancy would have a further locally uniformly convergent subsequence, a contradiction.

To obtain the nontangential limit, take any $z_j=x_j+i{y}_j\to 0$ with $|x_j|\le A{y}_j$ and $y_j>0$. Put $n_j=\lfloor 1/y_j\rfloor$. Then $n_j\to \infty$, $n_j{y}_j\to 1$, and the points ${\zeta }_j=n_j{z}_j$ lie, for large $j$, in a fixed compact rectangle inside the half-plane. ThereforeThis is the normal-family proof of angular boundary convergence; using integer rescalings requires this step because the points need not lie exactly on one fixed scaled ray.

For a tangential counterexample, useIf $z=x+iy$ with $y>0$, then $\mathrm{Re}(-i/z)=-y/(x^2+y^2)<0$, so $g$ is a holomorphic function and $|g|<1$. Also $g(iy)=e^{-1/y}\to 0$. ButTheir ratio $|\mathrm{Re}{z}_n|/\mathrm{Im}{z}_n=2\pi n^2$ tends to infinity. Thus the nontangential limit is zero while the unrestricted boundary limit does not exist.

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.

Use complex-linear distributions, so the pairing contains no complex conjugation. The space of smooth functions is $\mathcal{E}(X)=C^{\infty }(X)$, with seminormsThus $f_j\to f$ means uniform convergence of every derivative on every compact subset of $X$. A cofinal compact exhaustion with $K_j\subset \mathrm{int}{K}_{j+1}$ makes $p_j=p_{K_j,j}$ an increasing defining family. Such an exhaustion exists for every open set; for example, use bounded sets staying a positive distance from its complement and enlarge them slightly. This is a Fréchet space.

The compactly supported distribution space is the continuous dual space ${\mathcal{E}}^{\prime }(X)$. We use its weak-star topology: $u_j\to u$ precisely when $⟨u_j,f⟩\to ⟨u,f⟩$ for every $f\in \mathcal{E}(X)$. A stronger standard choice is the strong dual topology, which requires uniform convergence on bounded subsets of $\mathcal{E}(X)$; specifying the weak convention avoids conflating these definitions.

A continuous linear functional necessarily takes null sequences to zero. Conversely, suppose a linear functional $u$ fails to be continuous. For each $j$, it is unbounded on $\{f:p_j(f)\le 1\}$; otherwise scaling would give a continuity estimate. We can therefore choose $f_j$ withFor each fixed $l$, $p_l(f_j)\le p_j(f_j)\to 0$ once $j\ge l$. This contradicts the assumed null-sequence property. Hence sequential continuity characterizes the continuous dual:This is the sequential continuity criterion in a metrizable vector space applied to a linear functional.

Continuity also yields one compact $K⋐X$, an integer $m$ and a constant $C$ such that $|u(f)|\le C{p}_{K,m}(f)$. Thus $u$ has compact support in $K$ and finite order of a distribution. Conversely, a compactly supported distribution extends to $\mathcal{E}(X)$ by $u(f)=u(\eta f)$, where a cutoff function $\eta$ equals one near its support of a distribution. The local finite-order estimate makes this extension continuous and independent of $\eta$. This identifies the dual definition with compactly supported distributions.

Here is an explicit compact continuous-derivative representation of a distribution. Extend $u$ by zero to ${\mathbb{R}}^n$ using a cutoff function inside $X$, retaining finite order $m$. Put $r=m+2$ andThis locally $C^m$ function satisfies ${\partial }^{\gamma }E={\delta }_0$. Indeed, each one-dimensional factor has $r$th distributional derivative equal to the Dirac delta distribution. The convolution $F=u\ast E$ is continuous: the finite-order estimate extends $u$ to $C^m$ functions near its compact support, and translated $E$ varies continuously in their $C^m$ norms. Equivalently, mollify $E$ and use the estimate to obtain local uniform convergence of the convolved functions. Distributional differentiation gives ${\partial }^{\gamma }F=u$.

Choose $\chi \in C_c^{\infty }(X)$ equal to one near $\mathrm{supp}u$. Since $\chi u=u$, repeated Leibniz rule givesEvery coefficient function on the right is continuous and compactly supported in $X$. Thus the required representation is finite:The equality first holds on test functions and then on all smooth functions after inserting a common cutoff, so it holds in ${\mathcal{E}}^{\prime }(X)$.

A general distribution need not admit one finite such sum, even if compact support is not required of the continuous functions. On $X=\mathbb{R}$, considerThe sum is locally finite, so it defines a distribution. If it were a finite sum ${\sum }_{l=0}^M{\partial }^l{g}_l$ with all $g_l$ continuous, then on every fixed compact set its action would be bounded by test derivatives through order $M$. Near an integer $j>M$, however, it is exactly ${\delta }_j^{(j)}$, whose order is $j$. To see the contradiction directly, choose a test function $\psi$ with ${\psi }^{(j)}(0)\ne 0$ and put ${\psi }_{\epsilon }(x)={\epsilon }^j\psi ((x-j)/\epsilon )$. All derivative norms through order $M$ tend to zero, while $⟨{\delta }_j^{(j)},{\psi }_{\epsilon }⟩=(-1{)}^j{\psi }^{(j)}(0)$. This is a distribution of unbounded order. In dimensions $n>1$, the same example uses point masses at $(j,0,\dots ,0)$ and derivatives in the first coordinate.

A locally defined stochastic process is a pair $(X,T)$, where $T$ is a lifetime stopping time and $X_t(\omega )$ is defined for $0\le t<T(\omega )$, that is, on a stochastic interval. For continuous local problems one uses stopping times $T_n\uparrow T$ with $T_n<T$ on $\{T>0\}$, so the process stopped at each $T_n$ has an ordinary continuous adapted process version. This is an announcing sequence for a stopping time; no value at $T$ or after $T$ is implicit in the pair.

A local solution of a stochastic differential equation on an open domain $U$ takes its values in $U$ before $T$ and, on every such stopped interval, satisfieswith the local integrability needed for the ordinary and Itô integrals. A maximal local solution of a stochastic differential equation has no extension to a strictly later lifetime that agrees with it before $T$. For locally Lipschitz continuous coefficients on $U$, take a nested compact exhaustion $K_n$ of $U$ and the successive exit times: their increasing limit is the maximal lifetime, and on $\{T<\infty \}$ the solution eventually leaves every compact subset of $U$. To obtain the announcing sequence with finite stops even when the path never exits a compact set, cap these exit times by $n$; the capped times still increase to the maximal lifetime. Thus a finite maximal lifetime means exit from the domain or explosion; it need not mean divergence to infinity. For $U=(0,\infty )$, reaching the boundary zero terminates the local solution even if an absorbing extension could be defined for a different domain.