physics2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Derivation

Analytic Continuation & the Identity Theorem

D-187 Home PU-205 Threads symmetry Depends on Taylor Series of Analytic Functions
Statement

Let \(f\) and \(g\) be holomorphic on a connected open set (domain) \(D\subseteq\mathbb{C}\). If \(f=g\) on a subset \(S\subseteq D\) that possesses at least one limit point \(z_0\in D\), then \(f=g\) throughout \(D\). Equivalently, the set on which two analytic functions agree, if it accumulates anywhere inside \(D\), is all of \(D\). Consequently an analytic function is determined on the whole of \(D\) by its germ at a single interior point, and any function agreeing with \(f\) on such a set is the unique analytic continuation of \(f\).

Why it matters

Analytic functions are absurdly rigid: fixing them on a tiny convergent sample — a short arc, a sequence \(1/n\), even a single point together with all derivatives — pins them everywhere their domain reaches. This rigidity is what licenses analytic continuation: a formula valid on a small disc (a power series, an integral like \(\Gamma\) or \(\zeta\)) extends to a unique function on a far larger set, and the extension cannot depend on how it was constructed.

In physics the theorem underwrites the entire machinery of continuing scattering amplitudes, propagators, and partition functions off the real axis into the complex plane. Dispersion relations, the crossing symmetry linking distinct physical processes, Wick rotation from Euclidean to Lorentzian signature, and Regge theory all rest on the fact that the continuation is unique: there is one analytic object, not a family, so a relation proven in one regime holds by continuation in another.

Assumptions
Holomorphy on all of \(D\).If \(f\) is merely smooth (real-differentiable), agreement on a set with a limit point says nothing globally — a bump function vanishes on a half-line yet is nonzero elsewhere. Complex differentiability is essential.
\(D\) is connected.On a disconnected domain the conclusion holds only on the component containing the accumulation point. Two components can carry independent analytic functions, so agreement on one says nothing about the other.
The limit point lies inside \(D\).If \(S\) accumulates only at a boundary point of \(D\), rigidity fails: \(\sin(\pi/z)\) vanishes at \(z=1/n\), which accumulate at \(0\notin D=\mathbb{C}\setminus\{0\}\), yet \(\sin(\pi/z)\not\equiv 0\). The limit point must be an interior point where the function is holomorphic.
\(S\) has a genuine limit point, not merely infinitely many points.An infinite discrete set with no accumulation point in \(D\) (e.g. the integers in \(\mathbb{C}\)) does not force agreement — the zeros of \(\sin(\pi z)\) are exactly \(\mathbb{Z}\) and it is not identically zero.
Derivation

Set \(h=f-g\). Then \(h\) is holomorphic on \(D\), and \(h=0\) on \(S\). By continuity \(h(z_0)=0\) at the limit point \(z_0\). We show \(h\equiv 0\).

1
\[ h=f-g,\qquad h\ \text{holomorphic on }D,\qquad h\big|_{S}=0 \]
Difference of holomorphic functions is holomorphic; reduce "\(f=g\)" to "\(h=0\)". A
2
\[ h(z)=\sum_{n=0}^{\infty} a_n\,(z-z_0)^n,\qquad a_n=\frac{h^{(n)}(z_0)}{n!},\quad |z-z_0|<R \]
Holomorphy at the interior point \(z_0\) gives a Taylor expansion converging on a disc \(D_R\subseteq D\) (assumed prior result: analytic ⇒ Taylor series). A
3
\[ \text{Claim: } a_n=0\ \text{for all } n\ge 0. \]
Suppose not, and let \(m\) be the smallest index with \(a_m\neq 0\); we derive a contradiction with the limit point. B
4
\[ h(z)=(z-z_0)^m\,\varphi(z),\qquad \varphi(z)=\sum_{k=0}^{\infty} a_{m+k}\,(z-z_0)^k,\qquad \varphi(z_0)=a_m\neq 0 \]
Factor out the lowest nonvanishing power. The remaining series \(\varphi\) is holomorphic on \(D_R\) (same radius) and \(\varphi(z_0)=a_m\neq0\). B
5
\[ \varphi\text{ continuous},\ \varphi(z_0)\neq0 \ \Rightarrow\ \exists\,\rho>0:\ \varphi(z)\neq 0\ \text{on } 0<|z-z_0|\le\rho \]
Continuity of \(\varphi\) gives a punctured neighbourhood where \(\varphi\) stays near \(a_m\neq0\), hence nonzero. There \((z-z_0)^m\neq0\) too, so \(h(z)\neq0\). B
6
\[ z_0=\lim_{k\to\infty} s_k,\quad s_k\in S\setminus\{z_0\}\ \Rightarrow\ h(s_k)=0\ \text{with } s_k\to z_0 \]
\(z_0\) is a limit point of \(S\), so points of \(S\) (where \(h=0\)) approach \(z_0\); infinitely many lie in the punctured disc \(0<|z-z_0|\le\rho\). B
7
\[ h(s_k)=0\ \text{yet } h\neq0 \text{ on } 0<|z-z_0|\le\rho\ \Rightarrow\ \Rightarrow\!\Leftarrow \]
The vanishing points \(s_k\) fall inside the zero-free punctured disc — contradiction. Hence all \(a_n=0\) and \(h\equiv0\) on \(D_R\). B
8
\[ U=\{z\in D:\ h^{(n)}(z)=0\ \forall n\ge0\}=\operatorname{int}\{h=0\} \]
Define \(U\), the set of points around which \(h\) vanishes identically. Step 7 shows \(z_0\in U\), so \(U\neq\varnothing\). C
9
\[ U\ \text{open}\quad\text{and}\quad U\ \text{closed in } D \]
Open: on a disc where \(h\equiv0\) every derivative is \(0\), so neighbours lie in \(U\). Closed: if \(z_j\to w\in D\) with \(z_j\in U\), continuity gives \(h^{(n)}(w)=0\) for all \(n\), so its Taylor series vanishes and \(w\in U\). C
10
\[ D\ \text{connected},\ U\neq\varnothing,\ U\ \text{clopen}\ \Rightarrow\ U=D\ \Rightarrow\ h\equiv0\ \Rightarrow\ f\equiv g \]
A nonempty subset that is both open and closed in a connected space is the whole space. Therefore \(h=0\) on all of \(D\). C
Result
\[ f,g\ \text{holomorphic on connected }D,\quad f=g\ \text{on } S\subseteq D\ \text{with a limit point in }D \ \Longrightarrow\ f\equiv g\ \text{on } D \]

Reading. Two analytic functions that coincide on any set thick enough to accumulate at an interior point are the same function everywhere their common domain reaches. In particular the germ at one point — the value and all derivatives, equivalently one Taylor series — determines \(f\) globally, and analytic continuation is unique whenever the target domain is connected.

Units check. The statement is dimensionless and structural: \(z\), \(f\), \(g\) are pure complex numbers (or carry whatever units the modelled quantity has, identically on both sides). The relation \(f=g\) preserves units trivially; no dimensional balance is at stake because the content is the equality of two functions, not a physical equation.

Limiting cases
  • If \(S\) is an open subdisc, the hypothesis is trivially met (every interior point is a limit point) — recovers the familiar "agree on a small disc ⇒ agree everywhere."
  • If \(S\) is a convergent sequence \(z_n\to z_0\in D\), agreement at those countably many points already forces \(f\equiv g\); no interval needed.
  • If \(g\equiv0\), the theorem becomes the identity theorem for zeros: the zeros of a nonzero analytic function are isolated, with no interior accumulation point.
  • For a real-analytic function on an interval, the same proof (real Taylor series) gives uniqueness of continuation along the real line before complexifying.
Breaks when
  • Only smooth, not holomorphic. \(f(z)=e^{-1/x^2}\) (with \(f=0\) for \(x\le0\)) is \(C^\infty\) on \(\mathbb{R}\), vanishes on a whole half-line — a set with limit points — yet is not identically zero. Real smoothness carries none of this rigidity.
  • Limit point on the boundary, not the interior. \(f(z)=\sin(\pi/z)\) on \(D=\mathbb{C}\setminus\{0\}\) vanishes at \(z=1/n\), accumulating at \(0\), which is not in \(D\). \(f\not\equiv0\): the accumulation point must be interior.
  • Disconnected domain. On \(D=D_1\sqcup D_2\), set \(f=0\) on \(D_1\) and \(f=1\) on \(D_2\). It is holomorphic and vanishes on the limit-point-rich \(D_1\), but not on \(D_2\). Connectivity is indispensable.
  • Zero set discrete with no interior accumulation. \(\sin(\pi z)\) vanishes exactly on \(\mathbb{Z}\), an infinite set, yet the integers have no limit point in \(\mathbb{C}\); the function is nonzero. Infinitely many agreements are not enough — they must cluster.
Failure modes
  • "Infinitely many shared points ⇒ equal." False without accumulation: \(\sin(\pi z)\) and \(0\) agree on all of \(\mathbb{Z}\). The set needs a limit point inside \(D\).
  • Confusing "limit point" with "point of \(S\)." The limit point \(z_0\) need not itself lie in \(S\); it must lie in \(D\). Students often demand \(z_0\in S\), which is unnecessary.
  • Applying it across a natural boundary. Assuming continuation always exists. The theorem gives uniqueness of a continuation, not existence; \(\sum z^{2^n}\) cannot be continued past \(|z|=1\) at all.
  • Ignoring connectivity. Concluding global agreement on a domain with several components from data on one component.
  • Dropping to real analysis. Invoking the theorem for a merely \(C^\infty\) function; the whole argument uses the Taylor series equalling the function, which needs holomorphy.
  • Multivaluedness overreach. Treating \(\sqrt{z}\) or \(\log z\) continued around a branch point as single-valued; uniqueness holds only along simply-connected paths, else monodromy intervenes.
Discussion

The identity theorem expresses that holomorphy is an enormously overdetermined condition. The Cauchy–Riemann equations couple the real and imaginary parts so tightly that the Taylor coefficients at one point encode the function everywhere the domain of convergence — extended by chaining discs — can reach. This is qualitatively unlike real smooth functions, which have infinitely many local degrees of freedom (bump functions live in disjoint regions). Analytic functions have, in effect, only countably many degrees of freedom globally: one Taylor series.

Uniqueness of continuation is the theorem's working face. The Gamma function defined by \(\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt\) converges only for \(\operatorname{Re} z>0\), yet the functional equation \(\Gamma(z+1)=z\,\Gamma(z)\) extends it to a meromorphic function on \(\mathbb{C}\); the identity theorem guarantees this extension is the only analytic one, so all its properties are unambiguous. The same logic makes the Riemann \(\zeta\)-function's continuation past \(\operatorname{Re} z=1\) well-defined, and it is what allows physicists to Wick-rotate a Euclidean correlator into Lorentzian signature and know they have recovered the physical amplitude.

The proof is a template for the "propagation of local information" theorems in complex analysis: an open-and-closed (clopen) argument on a connected set. The vanishing of all derivatives at one point spreads to an open set (via the power series), and continuity makes the maximal such set closed; connectedness then forces it to be everything. The reflection principle, the monodromy theorem, and the edge-of-the-wedge theorem in quantum field theory are elaborations of this same rigidity applied to boundary values and overlapping domains.

Sheaf-theoretically, the collection of analytic germs forms a sheaf whose sections over a connected base are rigid: the restriction map from global sections to the germ at a point is injective. The identity theorem is precisely this injectivity. Continuation is then the study of sections of the sheaf over the maximal domain — the Riemann surface of \(f\) — where multivaluedness (\(\log\), \(\sqrt{\,}\)) reappears as nontrivial monodromy: the germ propagated around a loop encircling a branch point returns transformed, so global single-valuedness holds only on simply-connected subdomains. Uniqueness is a statement about germs along a fixed path; the failure of global uniqueness is entirely topological, encoded in \(\pi_1\) of the punctured domain.

Common misconceptions. The theorem does not assert that a continuation always exists (natural boundaries can block it), nor that it is single-valued globally (branch points spoil this). It asserts only that if a continuation exists along a route through a connected domain, it is unique. And the required data is astonishingly small: one convergent sequence of agreements, not a whole region.

Worked examples
1
\[ f\ \text{entire},\quad f\!\left(\tfrac{1}{n}\right)=\frac{1}{n^2}\ \ (n=1,2,3,\dots)\ ;\quad \text{find } f. \]
Candidate \(g(z)=z^2\) is entire and matches at the sample points. A
2
\[ g(z)=z^2:\quad g\!\left(\tfrac1n\right)=\frac{1}{n^2}=f\!\left(\tfrac1n\right) \]
\(f\) and \(g\) agree on \(S=\{1/n\}\), which accumulates at \(z_0=0\), an interior point of \(D=\mathbb{C}\). A
3
\[ S\ \text{has limit point } 0\in\mathbb{C},\quad f,g\ \text{holomorphic}\ \Rightarrow\ f\equiv g \]
Identity theorem applies: the two entire functions coincide everywhere. A
\[ f(z)=z^2\quad\text{(unique such entire function)} \]

Reading. Values on the single convergent sequence \(1/n\to0\) already determine the entire function; no other holomorphic \(f\) fits. Units: if \(z\) carried a length, \(f\) would carry length\(^2\), consistently on both sides.

1
\[ f(z)=\sum_{n=0}^{\infty} z^n\quad\text{on } D_1=\{|z|<1\};\quad \text{continue to } z=2i. \]
The geometric series converges only on the unit disc, but sums to a closed form there. A
2
\[ f(z)=\frac{1}{1-z},\qquad |z|<1 \]
Closed form on \(D_1\); this rational function is holomorphic on all of \(\mathbb{C}\setminus\{1\}\). A
3
\[ g(z)=\frac{1}{1-z}\ \text{on } D_2=\mathbb{C}\setminus\{1\}\supset D_1,\quad g\big|_{D_1}=f \]
\(g\) agrees with \(f\) on the whole open disc \(D_1\) (a set with interior limit points), so it is the continuation. B
4
\[ g(2i)=\frac{1}{1-2i}=\frac{1+2i}{(1-2i)(1+2i)}=\frac{1+2i}{5} \]
Evaluate the unique continuation at the target point, rationalising the denominator. A
\[ f(2i)=\frac{1+2i}{5}=0.2+0.4\,i \]

Reading. Although the defining series \(\sum z^n\) diverges wildly at \(z=2i\), the function it represents has a unique value there, \(0.2+0.4i\), fixed by its restriction to the unit disc. Units: dimensionless throughout (\(z\) and \(f\) are pure numbers). The continuation is unique because \(\mathbb{C}\setminus\{1\}\) is connected.

Problems
  1. An entire function satisfies \(f(1/n)=0\) for all positive integers \(n\). What is \(f\)?
    Solution The zeros \(1/n\) accumulate at \(0\in\mathbb{C}\), an interior point. \(f\) is holomorphic there, so by the identity theorem (with \(g\equiv0\)) \(f\equiv0\). The only entire function vanishing on \(\{1/n\}\) is the zero function. (Contrast \(\sin(\pi z)\), which vanishes on \(\mathbb{Z}\) — no interior limit point — and is nonzero.)
  2. Does there exist an entire \(f\) with \(f(1/n)=(-1)^n\) for all \(n\ge1\)?
    Solution No. Take the even subsequence: \(f(1/(2k))=1\to\) forces, by continuity, \(f(0)=\lim f(1/(2k))=1\). The odd subsequence gives \(f(1/(2k+1))=-1\to f(0)=-1\). A continuous function cannot have \(f(0)=1\) and \(f(0)=-1\). No such (even merely continuous) function exists, let alone analytic.
  3. Two functions holomorphic on the connected domain \(D=\mathbb{C}\setminus\{0\}\) agree on the real segment \((1,2)\). Must they agree on all of \(D\)?
    Solution Yes. The segment \((1,2)\) has interior limit points lying in \(D\) (e.g. \(1.5\)), and \(D\) is connected. The identity theorem gives \(f\equiv g\) on all of \(D=\mathbb{C}\setminus\{0\}\). (The puncture at \(0\) does not disconnect \(\mathbb{C}\), so connectivity holds.)
  4. The function \(f(z)=\sum_{n=0}^\infty z^{2^n}=z+z^2+z^4+z^8+\cdots\) is holomorphic on \(|z|<1\). Explain why it cannot be analytically continued to any larger connected domain, and reconcile this with the uniqueness theorem.
    Solution Every root of unity \(z=e^{2\pi i p/2^q}\) is a singularity: as \(r\to1^-\) along that ray the partial sums blow up, because \(z^{2^n}\to1\) for all large \(n\). These singular points are dense on \(|z|=1\), forming a natural boundary; no disc straddling the circle can be a domain of holomorphy. The identity theorem guarantees only that if a continuation existed it would be unique — it never promises existence. Here existence fails entirely, so there is nothing to be unique about.
  5. Let \(f\) be holomorphic on the unit disc with \(f(0)=1\) and \(f'(z)=f(z)\) for all \(z\). Identify \(f\), and state the largest domain to which it uniquely continues.
    Solution Seek \(f(z)=\sum a_n z^n\). The ODE \(f'=f\) gives \((n+1)a_{n+1}=a_n\), so \(a_{n+1}=a_n/(n+1)\); with \(a_0=f(0)=1\) this yields \(a_n=1/n!\). Hence \(f(z)=\sum_{n\ge0} z^n/n! = e^z\), which converges on all of \(\mathbb{C}\). The candidate \(g(z)=e^z\) is entire and agrees with \(f\) on the disc (a set with interior limit points), so by the identity theorem the unique continuation is \(f(z)=e^z\) on the whole connected plane \(\mathbb{C}\). Numerically, e.g. \(f(1)=e\approx2.71828\).