The Hartogs Phenomena

Fix an integer $n$. By [L1], for $\mathrm{\mid }\zeta \mathrm{\mid }=\rho$ and $\mathrm{\mid }w\mathrm{\mid }<1$ the point $(\zeta ,w)$ lies in $H(r,s)$, so the integrand $(\zeta ,w)\mapsto f(\zeta ,w){\zeta }^{-n-1}$ is continuous on the circle times the unit disc and, for fixed $\zeta$, holomorphic in $w$. Therefore [L2] makes $a_n$ holomorphic on $\{\mathrm{\mid }w\mathrm{\mid }<1\}$.

Fix $w$ with $\mathrm{\mid }w\mathrm{\mid }<1$. Then the slice $z\mapsto f(z,w)$ is holomorphic on the annulus $r<\mathrm{\mid }z\mathrm{\mid }<1$, again by [L1]. Applying [L3] to that one-variable slice on the circle $\mathrm{\mid }z\mathrm{\mid }=\rho$ shows that its Laurent coefficients are exactly the numbers $a_n(w)$ defined above.

The Laurent expansion from step 1.2 is therefore $f(z,w)={\sum }_{n\in \mathbb{Z}}{a}_n(w)z^n$ for $r<\mathrm{\mid }z\mathrm{\mid }<1$, and step 1.1 gives the holomorphic dependence of every coefficient on $w$.

Fix $w$ with $\mathrm{\mid }w\mathrm{\mid }<s$. Then $(z,w)\in H(r,s)$ for every $\mathrm{\mid }z\mathrm{\mid }<1$, so the slice $z\mapsto f(z,w)$ is holomorphic on the whole unit disc. By [L1], the coefficients $a_n(w)$ are the Laurent coefficients of that slice on $r<\mathrm{\mid }z\mathrm{\mid }<1$, and [L2] therefore forces $a_n(w)=0$ for every $n<0$.

For each fixed $n<0$, step 1.1 shows that the holomorphic function $a_n$ is zero on the nonempty open disc $\{\mathrm{\mid }w\mathrm{\mid }<s\}$. The domain $\{\mathrm{\mid }w\mathrm{\mid }<1\}$ is connected, so [L3] gives $a_n\equiv 0$ there.

As $n<0$ was arbitrary, every negative Laurent coefficient vanishes on the whole parameter disc $\{\mathrm{\mid }w\mathrm{\mid }<1\}$.

Fix $\rho$ with $r<\rho <1$ and define $F_{\rho }$ by the Cauchy integral in [L2]. Since $\mathrm{\mid }\zeta \mathrm{\mid }=\rho$ and $\mathrm{\mid }w\mathrm{\mid }<1$ place $(\zeta ,w)$ inside $H(r,s)$ by [L1], the integral is well defined. For fixed $z$ with $\mathrm{\mid }z\mathrm{\mid }<\rho$, [L2] applied to the one complex parameter $w$ makes $w\mapsto F_{\rho }(z,w)$ holomorphic on $\mathrm{\mid }w\mathrm{\mid }<1$. For fixed $w$, the same theorem applied to the one complex parameter $z$ makes $z\mapsto F_{\rho }(z,w)$ holomorphic on $\mathrm{\mid }z\mathrm{\mid }<\rho$. The ML estimate gives local bounds on compact subsets, so [L6] upgrades $F_{\rho }$ to a jointly holomorphic function on $\{\mathrm{\mid }z\mathrm{\mid }<\rho ,\mathrm{\mid }w\mathrm{\mid }<1\}$.

If $\mathrm{\mid }w\mathrm{\mid }<s$, then the slice $z\mapsto f(z,w)$ is holomorphic on $\mathrm{\mid }z\mathrm{\mid }<1$. Therefore [L3] gives $F_{\rho }(z,w)=f(z,w)$ whenever $\mathrm{\mid }z\mathrm{\mid }<\rho$ and $\mathrm{\mid }w\mathrm{\mid }<s$. So $F_{\rho }$ extends $f$ across the missing core over that open overlap.

If $r<{\rho }_1<{\rho }_2<1$, then both $F_{{\rho }_1}$ and $F_{{\rho }_2}$ are holomorphic on $\{\mathrm{\mid }z\mathrm{\mid }<{\rho }_1,\mathrm{\mid }w\mathrm{\mid }<1\}$ by step 1.1, and step 2.1 shows that they agree on the nonempty open subset $\{\mathrm{\mid }z\mathrm{\mid }<{\rho }_1,\mathrm{\mid }w\mathrm{\mid }<s\}$. Hence [L4] forces $F_{{\rho }_1}=F_{{\rho }_2}$ on the whole connected domain $\{\mathrm{\mid }z\mathrm{\mid }<{\rho }_1,\mathrm{\mid }w\mathrm{\mid }<1\}$.

For each point $(z,w)\in \widehat{H}(r,s)$ choose any $\rho$ with $\max \{r,\mathrm{\mid }z\mathrm{\mid }\}<\rho <1$, and set $F(z,w):=F_{\rho }(z,w)$. Step 3.1 shows that this does not depend on the chosen $\rho$, so $F$ is well defined and holomorphic locally, hence holomorphic on all of $\widehat{H}(r,s)$.

Step 2.1 gives $F=f$ on the open set $\{\mathrm{\mid }w\mathrm{\mid }<s\}\subseteq H(r,s)$, so $F$ is a holomorphic extension of $f$ to the bidisc. If $G$ is another such extension, then $F$ and $G$ are holomorphic on the connected bidisc and agree on the same nonempty open overlap with $f$; [L4] gives $F=G$. Thus the extension is unique.

The Laurent-coefficient view in [L5] is compatible with the integral construction above: the Cauchy kernel removes the vanished negative part and rebuilds the same holomorphic continuation.

Let $f\in \mathcal{O}(\mathrm{\Omega })$. Its restriction to the open subset $H(r,s)$ is holomorphic, so [L2] gives a holomorphic function $F_f\in \mathcal{O}(\widehat{H}(r,s))$ with $F_f=f$ on $H(r,s)$.

The domain $\widehat{H}(r,s)$ is not contained in $\mathrm{\Omega }$ by hypothesis, and $H(r,s)$ is a nonempty open subset of $\mathrm{\Omega }\cap \widehat{H}(r,s)$. Thus the same open overlap works for every holomorphic function on $\mathrm{\Omega }$, namely the fixed set $U_1:=H(r,s)$ and the larger domain $U_2:=\widehat{H}(r,s)$.

By the definition in [L1], the existence of that common pair $U_1,U_2$ shows that $\mathrm{\Omega }$ is not a domain of holomorphy.

By [L1], choose radii ${\rho }_1>\rho >0$ and $\tau >0$ with ${\mathrm{\Delta }}_{{\rho }_1}(a_1)\times {\mathrm{\Delta }}_{\tau }(a^{\mathrm{\prime }})\subseteq \mathrm{\Omega }$. For $(z_1,z^{\mathrm{\prime }})$ with $\mathrm{\mid }{z}_1-a_1\mathrm{\mid }<\rho$ and $\mathrm{\mid }{z}^{\mathrm{\prime }}-a^{\mathrm{\prime }}\mathrm{\mid }<\tau$, define $F(z_1,z^{\mathrm{\prime }}):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta -a_1\mathrm{\mid }=\rho }\frac{f(\zeta ,z^{\mathrm{\prime }})}{\zeta -z_1}d\zeta \mathrm{.}$ When $\mathrm{\mid }\zeta -a_1\mathrm{\mid }=\rho$ and $\mathrm{\mid }{z}^{\mathrm{\prime }}-a^{\mathrm{\prime }}\mathrm{\mid }<\tau$, the point $(\zeta ,z^{\mathrm{\prime }})$ lies in ${\mathrm{\Delta }}_{{\rho }_1}(a_1)\times {\mathrm{\Delta }}_{\tau }(a^{\mathrm{\prime }})\setminus \{a\}$, so the integrand is well defined and continuous there.

Fix all variables except one. If the free variable is $z_1$, the Cauchy kernel makes $z_1\mapsto F(z_1,z^{\mathrm{\prime }})$ holomorphic on $\mathrm{\mid }{z}_1-a_1\mathrm{\mid }<\rho$. If the free variable is one coordinate of $z^{\mathrm{\prime }}$, then [L2] applies to that single complex parameter. So $F$ is separately holomorphic on ${\mathrm{\Delta }}_{\rho }(a_1)\times {\mathrm{\Delta }}_{\tau }(a^{\mathrm{\prime }})$. The same integral formula gives local bounds on compact subsets, so [L5] upgrades $F$ to a jointly holomorphic function there.

If $z^{\mathrm{\prime }}\mathrm{\ne }{a}^{\mathrm{\prime }}$, then the slice $\zeta \mapsto f(\zeta ,z^{\mathrm{\prime }})$ is holomorphic on the full disc $\mathrm{\mid }\zeta -a_1\mathrm{\mid }<\rho$, because the deleted point $a$ does not lie on that slice. Hence [L3] gives $F(z_1,z^{\mathrm{\prime }})=f(z_1,z^{\mathrm{\prime }})$ for every $\mathrm{\mid }{z}_1-a_1\mathrm{\mid }<\rho$.

The set $\{(z_1,z^{\mathrm{\prime }})\in P:z^{\mathrm{\prime }}\mathrm{\ne }{a}^{\mathrm{\prime }}\}$ is a nonempty open subset of $P\setminus \{a\}$, and step 2.2 shows that $F$ and $f$ agree there. Both are holomorphic on the connected punctured polydisc $P\setminus \{a\}$, so [L4] forces $F=f$ on all of $P\setminus \{a\}$.

Thus $F$ is a holomorphic extension of $f$ across $a$ on the neighborhood ${\mathrm{\Delta }}_{\rho }(a_1)\times {\mathrm{\Delta }}_{\tau }(a^{\mathrm{\prime }})$. Repeating the same construction at each puncture point gives a local extension, and uniqueness on overlaps again follows from [L4]. Therefore the local extensions glue to a unique holomorphic function on all of $\mathrm{\Omega }$.

Apply [L1] to the punctured domain $\mathrm{\Omega }\setminus \{a\}$. It produces a holomorphic extension across $a$.

So the deleted point cannot support a nonremovable isolated singularity.

Fix $z^{\mathrm{\prime }}\in U$. The slice $w\mapsto f(z^{\mathrm{\prime }},w)$ is holomorphic on the punctured disc $0<\mathrm{\mid }w\mathrm{\mid }<R$ and bounded on $0<\mathrm{\mid }w\mathrm{\mid }<\eta$. Therefore [L1] gives a holomorphic extension $g_{z^{\mathrm{\prime }}}$ to the full disc $\mathrm{\mid }w\mathrm{\mid }<R$.

Fix one coordinate of $z^{\mathrm{\prime }}$ and hold the others fixed. For $\mathrm{\mid }\zeta \mathrm{\mid }=\eta$ the integrand $(z^{\mathrm{\prime }},\zeta )\mapsto f(z^{\mathrm{\prime }},\zeta )\mathrm{/}\zeta$ is jointly continuous in that parameter and $\zeta$, and for fixed $\zeta$ it is holomorphic in the chosen coordinate. So [L2] makes $F$ separately holomorphic on $U$. The same circle formula gives local bounds on compact subsets of $U$, so $F$ is locally bounded there.

Applying [L3] to the holomorphic function $g_{z^{\mathrm{\prime }}}$ on the circle $\mathrm{\mid }\zeta \mathrm{\mid }=\eta$ gives $g_{z^{\mathrm{\prime }}}(0)=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=\eta }\frac{f(z^{\mathrm{\prime }},\zeta )}{\zeta }d\zeta =F(z^{\mathrm{\prime }})\mathrm{.}$ So $F(z^{\mathrm{\prime }})$ is exactly the removable value of the slice at $w=0$.

Step 2.1 identifies $F(z^{\mathrm{\prime }})$ with the extension value at $w=0$ for every $z^{\mathrm{\prime }}$. Step 1.2 makes $F$ separately holomorphic and locally bounded, so [L4] upgrades it to a holomorphic function on $U$.

Let $p=(p^{\mathrm{\prime }},0)\in H$. By local boundedness, choose a product neighborhood $U\times \{\mathrm{\mid }w\mathrm{\mid }<R\}\subseteq \mathrm{\Omega }$ of $p$ on which $f$ is bounded whenever $0<\mathrm{\mid }w\mathrm{\mid }<R$, after translating coordinates so that $p_m=0$. Applying [L1] on this product neighborhood gives a holomorphic function $F_p:U\times \{\mathrm{\mid }w\mathrm{\mid }<R\}\to \mathbb{C}$ extending $f$ across the slice $U\times \{0\}$.

The functions $F_p$ and $f$ agree on the nonempty open overlap $U\times \{0<\mathrm{\mid }w\mathrm{\mid }<R\}$, so each $F_p$ is a local holomorphic extension in the sense of [L3].

If two such neighborhoods overlap, their local extensions agree on the nonempty open subset of the overlap where $w\mathrm{\ne }0$, because both equal $f$ there. The overlap is connected after shrinking if necessary, so [L2] makes the two local extensions equal on the whole overlap.

The local extensions therefore glue to a single holomorphic function on $\mathrm{\Omega }$ that agrees with $f$ off $H$. Uniqueness follows from [L2], since two global extensions agree on the nonempty open set $\mathrm{\Omega }\setminus H$.

We prove a stronger local claim by induction on $m$: every separately holomorphic function on ${\mathrm{\Delta }}_R^m$ is holomorphic on a neighborhood of each point of ${\mathrm{\Delta }}_R^m$. Once that is known, the displayed boundedness follows, because the compact set ${\bar{\mathrm{\Delta }}}_r^m$ is covered by finitely many such holomorphic neighborhoods and each holomorphic function is bounded on a smaller closed polydisc inside its neighborhood.

Base case $m=1$: separate holomorphy is ordinary one-variable holomorphy by [L1], so the local claim and the boundedness statement are immediate. Assume now that the local claim is known in dimension $m-1$, and prove it in dimension $m\ge 2$.

Fix $p\in {\mathrm{\Delta }}_R^m$. Choose $\rho >0$ with $0<\rho <R$ and ${\bar{\mathrm{\Delta }}}_{2\rho }^m(p)\subseteq {\mathrm{\Delta }}_R^m$. After translating and rescaling each coordinate disc, it is enough to prove that a separately holomorphic function on ${\mathrm{\Delta }}_2^m$ is holomorphic in a neighborhood of the origin. Write $z=(z^{\mathrm{\prime }},w)$ with $z^{\mathrm{\prime }}\in {\mathbb{C}}^{m-1}$ and $w\in \mathbb{C}$.

Box claim. If a closed real box $K=I_1\times \cdots \times I_d\subseteq {\mathbb{R}}^d$ is covered by countably many closed sets $F_n$, then some $F_n$ contains a smaller closed real box with nondegenerate sides. We prove this by induction on $d$. For $d=1$ it is [L2]. Assume the claim in dimension $d-1$. Write $K=I\times K^{\mathrm{\prime }}$. Enumerate the closed subboxes of $K^{\mathrm{\prime }}$ with rational endpoints in the coordinates of $K^{\mathrm{\prime }}$ as $Q_1,Q_2,\dots$. For each pair $(n,j)$ let $E_{n,j}:=\{x\in I:\{x\}\times Q_j\subseteq F_n\}\mathrm{.}$ Each $E_{n,j}$ is closed. Fix $x\in I$. The sections $F_n(x):=\{y\in K^{\mathrm{\prime }}:(x,y)\in F_n\}$ are closed and cover $K^{\mathrm{\prime }}$, so the induction hypothesis in dimension $d-1$ gives some $n$ such that $F_n(x)$ contains a smaller closed box; shrinking slightly if needed, that box contains a rational-endpoint subbox $Q_j$. Hence $x\in E_{n,j}$. So the countable family $E_{n,j}$ covers $I$, and [L2] gives one pair $(n,j)$ for which $E_{n,j}$ contains a nondegenerate closed subinterval $J$. Then $J\times Q_j\subseteq F_n$, proving the claim.

Whenever $0\le d<1<\eta$, one can choose radii $d<s<r_1<1<r_2<\eta$.

For each positive integer $B$, define ${\mathrm{\Omega }}_B:=\{z^{\mathrm{\prime }}\in {\bar{\mathrm{\Delta }}}_1^{m-1}:\mathrm{\mid }f(z^{\mathrm{\prime }},w)\mathrm{\mid }\le B\text{ for every }\mathrm{\mid }w\mathrm{\mid }\le 1\}\mathrm{.}$ For fixed $w$ with $\mathrm{\mid }w\mathrm{\mid }\le 1$, the induction hypothesis applied to $z^{\mathrm{\prime }}\mapsto f(z^{\mathrm{\prime }},w)$ makes that function holomorphic, hence continuous, on ${\mathrm{\Delta }}_2^{m-1}$. Therefore each set $\{z^{\mathrm{\prime }}:\mathrm{\mid }f(z^{\mathrm{\prime }},w)\mathrm{\mid }\le B\}$ is closed, and so every ${\mathrm{\Omega }}_B$ is closed. Also ${\bigcup }_{B\ge 1}{\mathrm{\Omega }}_B={\bar{\mathrm{\Delta }}}_1^{m-1}$, because for fixed $z^{\mathrm{\prime }}$ the slice $w\mapsto f(z^{\mathrm{\prime }},w)$ is holomorphic on ${\mathrm{\Delta }}_2$ and therefore bounded on $\mathrm{\mid }w\mathrm{\mid }\le 1$.

Apply the box claim to the real box $[-1\mathrm{/}\sqrt{2},1\mathrm{/}\sqrt{2}{]}^{2m-2}\subseteq {\bar{\mathrm{\Delta }}}_1^{m-1}$ and the closed cover $({\mathrm{\Omega }}_B{)}_{B\ge 1}$. We obtain some $B_0$ and a nondegenerate closed real box contained in ${\mathrm{\Omega }}_{B_0}$. Inside its relative interior choose a closed complex polydisc ${\bar{\mathrm{\Delta }}}_{2\epsilon }^{m-1}(a^{\mathrm{\prime }})$ for some $a^{\mathrm{\prime }}$ and some $\epsilon >0$. Its centre satisfies $d:={\max }_{1\le j<m}\mathrm{\mid }{a}_j^{\mathrm{\prime }}\mathrm{\mid }<1,$ and $\mathrm{\mid }f(z^{\mathrm{\prime }},w)\mathrm{\mid }\le B_0\text{for }{z}^{\mathrm{\prime }}\in {\bar{\mathrm{\Delta }}}_{2\epsilon }^{m-1}(a^{\mathrm{\prime }}),\mathrm{\mid }w\mathrm{\mid }\le 1.$ Hence $f$ is bounded on the product polydisc $E:={\mathrm{\Delta }}_{2\epsilon }^{m-1}(a^{\mathrm{\prime }})\times {\mathrm{\Delta }}_1$.

By [L3], the separately holomorphic and bounded function $f$ is jointly holomorphic on $E$. Put $\zeta =z^{\mathrm{\prime }}-a^{\mathrm{\prime }}$ and retain the notation $f(\zeta ,w)$ after this translation. The original first-variable domain contains the centred polydisc ${\mathrm{\Delta }}_{\eta }^{m-1}$, where $\eta :=2-d>1$, while the original target $z^{\mathrm{\prime }}=0$ now has coordinate $\zeta =-a^{\mathrm{\prime }}$. Thus $f$ is separately holomorphic on ${\mathrm{\Delta }}_{\eta }^{m-1}\times {\mathrm{\Delta }}_2$ and jointly holomorphic on ${\mathrm{\Delta }}_{2\epsilon }^{m-1}\times {\mathrm{\Delta }}_1$. Choose radii $d<s<r_1<1<r_2<\eta$, which is possible because $d<1<\eta$.

For each multi-index $\alpha \in {\mathbb{N}}^{m-1}$, define $c_{\alpha }(w):=\frac{1}{\alpha !}{\mathrm{\partial }}_{\zeta }^{\alpha }f(0,w)(\mathrm{\mid }w\mathrm{\mid }<1),$ using the jointly holomorphic function from step 4.1. By [L4], every $c_{\alpha }$ is holomorphic on ${\mathrm{\Delta }}_1$. For fixed $w$ with $\mathrm{\mid }w\mathrm{\mid }<1$, the induction hypothesis makes $\zeta \mapsto f(\zeta ,w)$ holomorphic on ${\mathrm{\Delta }}_{\eta }^{m-1}$, so these $c_{\alpha }(w)$ are exactly its Taylor coefficients at $0$. Since $f$ is bounded by $B_0$ on the larger product ${\mathrm{\Delta }}_{2\epsilon }^{m-1}\times {\mathrm{\Delta }}_1$, [L5] applied at the strictly smaller radius $\epsilon$ gives $\mathrm{\mid }{c}_{\alpha }(w)\mathrm{\mid }\le B_0{\epsilon }^{-\mathrm{\mid }\alpha \mathrm{\mid }}(\mathrm{\mid }w\mathrm{\mid }<1)\mathrm{.}$ For each nonzero multi-index $\alpha$ with $c_{\alpha }\equiv ̸0$, define $u_{\alpha }(w):=\mathrm{\mid }\alpha {\mathrm{\mid }}^{-1}\log \mathrm{\mid }{c}_{\alpha }(w)\mathrm{\mid }$. By [L6], each such $u_{\alpha }$ is subharmonic on ${\mathrm{\Delta }}_1$. If $c_{\alpha }\equiv 0$, then the term $c_{\alpha }(w){\zeta }^{\alpha }$ vanishes identically and is already harmless for the later power-series tail estimate. The displayed estimate gives a uniform upper bound for the whole family $(u_{\alpha }{)}_{\alpha \mathrm{\ne }0,c_{\alpha }\equiv ̸0}$.

Fix $w\in {\mathrm{\Delta }}_1$. The Cauchy estimates [L5] applied to the holomorphic function $\zeta \mapsto f(\zeta ,w)$ on ${\mathrm{\Delta }}_{\eta }^{m-1}$, with the strictly smaller radius $r_2$, show that $\mathrm{\mid }{c}_{\alpha }(w)\mathrm{\mid }\le M(w)r_2^{-\mathrm{\mid }\alpha \mathrm{\mid }}$ for some finite constant $M(w)$ and every nonzero multi-index $\alpha$. Therefore ${\mathrm{lim\hspace{0.17em}sup}}_{\begin{array}{c}\mathrm{\mid }\alpha \mathrm{\mid }\to \mathrm{\infty }\\ c_{\alpha }\equiv ̸0\end{array}}{u}_{\alpha }(w)\le -\log r_2\mathrm{.}$

Let $S:=\{\alpha \in {\mathbb{N}}^{m-1}\setminus \{0\}:c_{\alpha }\equiv ̸0\}\mathrm{.}$ If $S$ is finite, the required tail estimate is immediate. Otherwise enumerate it as $(\alpha (j){)}_{j\ge 1}$ with nondecreasing degrees and put $v_j=u_{\alpha (j)}$. By steps 5.1 and 6.1, the subharmonic functions $v_j$ have a common upper bound $A$ on ${\mathrm{\Delta }}_1$ and satisfy ${\mathrm{lim\hspace{0.17em}sup}}_j{v}_j(w)\le C:=-\log r_2$ pointwise.

We claim that for every compact $K⋐{\mathrm{\Delta }}_1$ and every $\delta >0$, $v_j\le C+\delta$ on $K$ for all sufficiently large $j$. Otherwise choose $j_k\uparrow \mathrm{\infty }$ and $q_k\in K$ with $v_{j_k}(q_k)>C+\delta$, and pass to a subsequence with $q_k\to q\in K$. Choose $t>0$ with $\overline{D(q,t)}\subseteq {\mathrm{\Delta }}_1$, discard finitely many terms so $q_k\in D(q,t)$, and put $P_k(\theta ):=P((q_k-q)\mathrm{/}t,e^{i\theta })$. By the definition of $S$ and [L9], no selected coefficient can vanish on a nonempty open subset of ${\mathrm{\Delta }}_1$. More specifically for the boundary argument, $v_{j_k}$ cannot be identically $-\mathrm{\infty }$ on the circle: if it were, every constant harmonic function $-n$ would majorize its boundary values, so [L10] would give $v_{j_k}(q_k)\le -n$ for every $n$, contradicting the finite strict lower bound just chosen. For $\zeta$ on the circle define ${\varphi }_{k,n}(\zeta ):={\sup }_{\eta \in \mathrm{\partial }D(q,t)}(v_{j_k}(\eta )-n\mathrm{\mid }\zeta -\eta \mathrm{\mid })\mathrm{.}$ Compactness and the common upper bound make each ${\varphi }_{k,n}$ finite and continuous, and upper semicontinuity gives ${\varphi }_{k,n}\downarrow v_{j_k}$ pointwise. Harmonic comparison, their Poisson extensions, and monotone convergence in [L7] therefore give $v_{j_k}(q_k)\le \frac{1}{2\pi }{\int }_0^{2\pi }{P}_k(\theta )v_{j_k}(q+t{e}^{i\theta })d\theta \mathrm{.}$ The kernels $P_k$ tend uniformly to $1$. The functions $P_k(\theta )(A-v_{j_k}(q+t{e}^{i\theta }))$ are nonnegative and measurable, so Fatou's lemma [L7] and the pointwise limsup bound make the lower limit of their normalized integrals at least $A-C$. Since each $P_k$ has normalized integral $1$, the preceding inequality gives ${\mathrm{lim\hspace{0.17em}sup}}_k{v}_{j_k}(q_k)\le C$, a contradiction.

Fix $0<\sigma <1$. Apply step 8.1 to $K={\bar{\mathrm{\Delta }}}_{\sigma }$ and $\delta =\log (r_2\mathrm{/}{r}_1)>0$. For all sufficiently large $j$, $v_j(w)\le -\log r_1(\mathrm{\mid }w\mathrm{\mid }\le \sigma ),$ or equivalently $\mathrm{\mid }{c}_{\alpha (j)}(w)\mathrm{\mid }{r}_1^{\mathrm{\mid }\alpha (j)\mathrm{\mid }}\le 1$.

Fix such a $\sigma$. If $S$ is finite, then ${\sum }_{\alpha }{c}_{\alpha }(w){\zeta }^{\alpha }$ is a finite sum in $\zeta$. Otherwise step 9.1 dominates its tail on ${\bar{\mathrm{\Delta }}}_s^{m-1}\times {\bar{\mathrm{\Delta }}}_{\sigma }$ by the convergent product-geometric majorant ${\sum }_{\alpha }(s\mathrm{/}{r}_1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}$; the finitely many low-degree terms are harmless and the coefficients outside $S$ vanish. Thus the series converges locally uniformly. Every partial sum is holomorphic, so [L8] gives a jointly holomorphic limit $G(\zeta ,w)$ on ${\mathrm{\Delta }}_s^{m-1}\times {\mathrm{\Delta }}_{\sigma }$.

On the open set ${\mathrm{\Delta }}_{\min (\epsilon ,s)}^{m-1}\times {\mathrm{\Delta }}_{\sigma }$, the Taylor expansion of the jointly holomorphic function from step 4.1 is exactly the series defining $G$. Thus $G=f$ on that nonempty open set. For fixed $w\in {\mathrm{\Delta }}_{\sigma }$, both $\zeta \mapsto G(\zeta ,w)$ and $\zeta \mapsto f(\zeta ,w)$ are holomorphic on ${\mathrm{\Delta }}_s^{m-1}$ and agree on a nonempty open subset, so [L9] gives equality on all of ${\mathrm{\Delta }}_s^{m-1}$. Hence $f=G$ on ${\mathrm{\Delta }}_s^{m-1}\times {\mathrm{\Delta }}_{\sigma }$, and $f$ is jointly holomorphic there.

Because ${\max }_j\mathrm{\mid }-a_j^{\mathrm{\prime }}\mathrm{\mid }=d<s$ and $0<\sigma$, the translated coordinates of the original target, $(-a^{\mathrm{\prime }},0)$, lie in the product from step 11.1. Thus that step proves the required local holomorphicity at the original origin in dimension $m$. By the reductions in steps 1.1 and 1.3, every point of ${\mathrm{\Delta }}_R^m$ has a holomorphic neighborhood. Therefore $f$ is locally bounded on ${\mathrm{\Delta }}_R^m$, and in particular bounded on every smaller closed polydisc ${\bar{\mathrm{\Delta }}}_r^m$. This closes the induction.

Fix $p\in U$. Because $U$ is open, there is a polydisc neighborhood $P\subseteq U$ of $p$. The restriction $f{\mathrm{\mid }}_P$ is still separately holomorphic by [L1], so [L2] makes it locally bounded near $p$.

Applying [L3] to that same restriction shows that $f$ is holomorphic on a neighborhood of $p$. As $p$ was arbitrary, $f$ is holomorphic on all of $U$.