Analytic Continuation

This is precisely the Identity Theorem and Zeros: let $h=f-g$, holomorphic on the connected open set $D$. Since $h\equiv 0$ on the open set $U$, every point of $U$ is a non-isolated zero of $h$ (in fact an accumulation point of zeros, being an interior point). The Identity Theorem states that a holomorphic function on a domain with a zero set having an accumulation point in the domain is identically zero on the domain. Hence $h \equiv 0$ on $D$, i.e. $f=g$ on $D$. $\blacksquare$

Technique: connectedness of $[0,1]$ via the "closed and open" argument. Let $S = \{t \in [0,1] : [f_t]_{\gamma(t)} = [g_t]_{\gamma(t)}\}$. We show $S$ is nonempty, open, and closed in $[0,1]$; since $[0,1]$ is connected, this forces $S = [0,1]$. Nonempty. By hypothesis $(f_0,D_0) = (g_0, E_0)$ as the initial element (both continuations start from the same germ at $\gamma(0)$), so $0 \in S$. Open. Let $t_0 \in S$, so $[f_{t_0}]_{\gamma(t_0)} = [g_{t_0}]_{\gamma(t_0)}$, meaning $f_{t_0} = g_{t_0}$ on some open neighborhood $V$ of $\gamma(t_0)$ contained in $D_{t_0}\cap E_{t_0}$. By definition of continuation along a path (condition 2), there is $\varepsilon>0$ such that for $|t - t_0|<\varepsilon$: $\gamma(t) \in D_{t_0}\cap E_{t_0}$, $f_t = f_{t_0}$ near $\gamma(t)$, and $g_t = g_{t_0}$ near $\gamma(t)$ (shrinking $\varepsilon$ if necessary to satisfy both continuation conditions simultaneously — possible since each condition individually supplies such an $\varepsilon$, and we take the minimum). Since $f_{t_0}=g_{t_0}$ throughout the neighborhood $V$ (by definition of $t_0\in S$), and both $f_t,g_t$ agree with $f_{t_0}, g_{t_0}$ respectively near $\gamma(t)$, shrinking further to lie inside $V$ gives $f_t = g_t$ near $\gamma(t)$. Hence $t \in S$ for all $t$ with $|t-t_0|<\varepsilon$ (intersected with $[0,1]$), so $S$ is open in $[0,1]$. Closed. Let $t_0 \in [0,1]$ be a limit point of $S$; we show $t_0 \in S$. By the continuation condition applied at $t_0$, there is $\varepsilon>0$ such that for $|t-t_0|<\varepsilon$, $\gamma(t) \in D_{t_0}\cap E_{t_0}$ and $f_t=f_{t_0}$, $g_t=g_{t_0}$ near $\gamma(t)$. Since $t_0$ is a limit point of $S$, choose $t_1 \in S$ with $|t_1 - t_0| < \varepsilon$, $t_1 \neq t_0$. Then $[f_{t_1}]_{\gamma(t_1)} = [g_{t_1}]_{\gamma(t_1)}$, i.e. $f_{t_1}=g_{t_1}$ on a neighborhood $W$ of $\gamma(t_1)$. But also $f_{t_1}=f_{t_0}$ and $g_{t_1}=g_{t_0}$ near $\gamma(t_1)$ (by the choice of $\varepsilon$ applied to $t_1$). So $f_{t_0} = f_{t_1} = g_{t_1} = g_{t_0}$ on a common neighborhood of $\gamma(t_1)$ contained in $D_{t_0}\cap E_{t_0}$ (all these domains being open sets containing $\gamma(t_1)$, their intersection is open and nonempty). Since $D_{t_0}, E_{t_0}$ are domains (connected open sets) and $f_{t_0}, g_{t_0}$ agree on a nonempty open subset of $D_{t_0}\cap E_{t_0}$'s component containing $\gamma(t_1)$, and $\gamma(t_0)$ lies in the same domain — we now apply the Local persistence of equality Lemma on the connected component $C$ of $D_{t_0}\cap E_{t_0}$ containing $\gamma(t_1)$: $f_{t_0}=g_{t_0}$ on all of $C$. Since continuation requires $\gamma(t)\in D_{t_0}$ for $t$ near $t_0$ and by continuity of $\gamma$, $\gamma(t_0)$ and $\gamma(t_1)$ lie in the same connected component $C$ (as $\gamma$ is continuous and connects them through nearby values of $t$, all mapping into $D_{t_0}\cap E_{t_0}$ by the openness condition — more directly, $\gamma(t_0) \in D_{t_0}\cap E_{t_0}$ trivially and is in the same component as $\gamma(t_1)$ because both are limits along the connected path-image for $t$ between $t_0,t_1$, which lies in $D_{t_0}\cap E_{t_0}$ by the $\varepsilon$-condition). Hence $f_{t_0}=g_{t_0}$ near $\gamma(t_0)$, i.e. $t_0 \in S$. So $S$ is closed. Since $S$ is nonempty, open, and closed in the connected space $[0,1]$, $S=[0,1]$. This proves part (A). $\blacksquare$

Technique: compactness of $[0,1]^2$ and a grid subdivision argument, combined with part (A). Let $H : [0,1]\times[0,1] \to \Omega$ be a homotopy rel endpoints with $H(s,0)=\gamma_0(s)$, $H(s,1) = \gamma_1(s)$ wait — we set conventions: let $H(0,t)=z_0$, $H(1,t)=z_1$ for all $t$, $H(s,0) = \gamma_0(s)$, $H(s,1)=\gamma_1(s)$, and $H$ continuous on $[0,1]^2$ with values in $\Omega$. For each $t \in [0,1]$, let $\gamma_t(s) := H(s,t)$, a path from $z_0$ to $z_1$ in $\Omega$. By hypothesis, $(f_0,D_0)$ continues along every path in $\Omega$ from $z_0$; in particular it continues along each $\gamma_t$. By part (A), for each fixed $t$ the germ $\Phi(t) := [\,\text{continuation of } f_0 \text{ along } \gamma_t\,]_{z_1}$ at the endpoint $z_1$ is unambiguous — the notation $\Phi(t)$ is well defined. We must show $\Phi(t)$ is locally constant in $t$, hence constant on the connected interval $[0,1]$ (again by connectedness), giving $\Phi(0)=\Phi(1)$, i.e. continuation along $\gamma_0$ and $\gamma_1$ agree. Claim: for each $t_0$, there is $\delta>0$ such that $|t-t_0|<\delta \implies \Phi(t)=\Phi(t_0)$. Fix $t_0$. Let $\{(f_s^{t_0}, D_s^{t_0})\}_{s\in[0,1]}$ denote a continuation of $(f_0,D_0)$ along $\gamma_{t_0}$. For each $s \in [0,1]$, $D_s^{t_0}$ is an open set containing $\gamma_{t_0}(s) = H(s,t_0)$; let $r(s) > 0$ be such that the open disk $B(H(s,t_0), r(s)) \subset D_s^{t_0}$. Since $[0,1]$ is compact and $s\mapsto H(s,t_0)$ is continuous, and since the continuation family also satisfies a local-persistence property, we may — by compactness of $[0,1]$ — choose a finite partition

and radii $r_0,\dots,r_n>0$ so that:

• $\overline{B(H(s_k,t_0), r_k)} \subset D_{s_k}^{t_0}$ for each $k$;
• $H(s,t_0) \in B(H(s_k,t_0), r_k)$ for all $s \in [s_k, s_{k+1}]$ (i.e. the disk covers the path segment) — such a partition exists because $\{s : H(s,t_0)\in D_u^{t_0} \text{ for } u \text{ near } s\}$ covers $[0,1]$ and Lebesgue's number lemma applied to the compact metric space $[0,1]$ with the open cover by preimages of the disks gives a uniform partition mesh.

By continuity of $H$ on the compact square $[0,1]^2$, $H$ is uniformly continuous. Hence there is $\delta>0$ such that $|t-t_0|<\delta$ implies

and moreover $H(s,t) \in B(H(s_k,t_0), r_k)$ whenever $s\in[s_k,s_{k+1}]$ (shrinking $\delta$ further using uniform continuity of $H$ so that nearby-in-$t$ paths stay within the same chain of disks — this is possible since uniform continuity controls $|H(s,t)-H(s,t_0)|$ for all $s$ simultaneously once $|t-t_0|<\delta$). Now fix such $t$ with $|t-t_0|<\delta$. We claim the continuation of $f_0$ along $\gamma_t$ can be carried out using the same chain of function elements $(f_{s_0}^{t_0}, D_{s_0}^{t_0}), \dots, (f_{s_n}^{t_0}, D_{s_n}^{t_0})$ (restricted appropriately), because the path $\gamma_t$ restricted to $[s_k,s_{k+1}]$ stays inside the disk $B(H(s_k,t_0),r_k) \subset D_{s_k}^{t_0}$, on which the germ is already $f_{s_k}^{t_0}$. Concretely: define a continuation along $\gamma_t$ by setting, for $s\in[s_k,s_{k+1}]$, the function element to be the germ of $f_{s_k}^{t_0}$ at $\gamma_t(s)$ (well-defined since $\gamma_t(s) \in D_{s_k}^{t_0}$). At the junction points $s=s_{k+1}$, both the element inherited from the $k$-th disk and from the $(k+1)$-th disk are restrictions of the original continuation-along-$\gamma_{t_0}$ chain, which are consistent by construction of a valid continuation (they were direct analytic continuations of each other along $\gamma_{t_0}$, and since $\gamma_t(s_{k+1})$ also lies in the overlap of the two disks — using that the disks have radius large enough to contain a neighborhood of both $H(s_{k+1},t_0)$ and the nearby $H(s_{k+1},t)$ — the same equality on overlap propagates). Hence this defines a valid continuation of $(f_0,D_0)$ along $\gamma_t$, terminating at $s=1$ with the germ $[f_{s_n}^{t_0}]_{z_1}$ — the same germ as the continuation along $\gamma_{t_0}$, because $\gamma_t(1) = z_1 = \gamma_{t_0}(1)$ and the last function element in the chain is literally $f_{s_n}^{t_0}$ (unchanged, since $s_n=1$ is a fixed endpoint of the homotopy, $H(1,t)=z_1$ for all $t$). By part (A) (uniqueness of continuation along a fixed path $\gamma_t$), any continuation of $(f_0,D_0)$ along $\gamma_t$ — in particular the one we are given by hypothesis — must yield the same terminal germ as this explicitly constructed one. Therefore $\Phi(t) = [f_{s_n}^{t_0}]_{z_1} = \Phi(t_0)$. This proves the claim: $\Phi$ is locally constant on $[0,1]$. A locally constant function on the connected space $[0,1]$ is constant, so $\Phi(0) = \Phi(1)$, i.e. continuation along $\gamma_0$ and continuation along $\gamma_1$ produce the same germ at $z_1$. This proves part (B). For the final statement: if $\Omega$ is simply connected, any two paths in $\Omega$ from $z_0$ to any point $z_1$ are homotopic rel endpoints (definition of simple connectivity), so $\Phi$ depends only on the endpoint $z_1$, not on the path. Defining $F(z_1) := \Phi(\text{any path from } z_0 \text{ to } z_1)$ (as a germ, hence a well-defined holomorphic function value near $z_1$) gives a well-defined function on $\Omega$, holomorphic since it agrees locally with each continued function element, and it extends $f_0$ since $\Phi$ restricted to the constant path at $z_0$ recovers $[f_0]_{z_0}$. $\blacksquare$