10.1 Uniqueness of the Continuation
Everything in this chapter rests on the following, which is the Identity Theorem of Chapter 6 read in a new way.
Theorem 10.4 (uniqueness of analytic continuation). Let \(D_1\) and \(D_2\) be domains whose intersection \(D_1 \cap D_2\) is non-empty and connected. Let \(f_1\) be analytic on \(D_1\) and \(f_2\) analytic on \(D_2\). If \(f_1 = f_2\) on some subset of \(D_1 \cap D_2\) possessing a limit point in \(D_1 \cap D_2\), then \(f_1 = f_2\) throughout \(D_1 \cap D_2\).
Proof. Put \(g = f_1 - f_2\) on the domain \(\Omega = D_1 \cap D_2\). Then \(g\) is analytic on \(\Omega \) and its zero set contains a set with a limit point in \(\Omega \). By the Identity Theorem \(g \equiv 0\) on \(\Omega \), since \(\Omega \) is connected. Hence \(f_1 = f_2\) on \(\Omega \). \(\blacksquare \)
The hypothesis that \(D_1\cap D_2\) be connected is not decoration. If the intersection falls into two pieces, agreement on one piece says nothing about the other, and this is exactly the door through which multi-valuedness enters later in the chapter. □
Corollary 10.5. An analytic function on a domain \(D\) is determined by its values on any subset of \(D\) having a limit point in \(D\) — on any small arc, or on any convergent sequence of distinct points together with its limit. In particular a continuation of a given element to a given domain, if one exists, is unique.
This is a statement with no counterpart in real analysis. A real function that is infinitely differentiable on \(\mathbb {R}\) may vanish on a whole interval and be non-zero elsewhere; \(e^{-1/x^2}\) patched to \(0\) is the standard example. Analyticity is a far heavier constraint than infinite differentiability, and uniqueness of continuation is the sharpest expression of it.
Proof. This is the identity theorem restated. If \(f\) and \(g\) are analytic on \(D\) and agree on a set \(S\) with a limit point in \(D\), then \(f-g\) vanishes on \(S\) and hence, by the identity theorem, vanishes identically on \(D\).
For the uniqueness of continuation, suppose \(F_1\) and \(F_2\) are analytic on a domain \(D\) and both restrict to the given element on some subdomain. That subdomain certainly has a limit point in \(D\), so \(F_1\equiv F_2\).
The hypothesis cannot be weakened to ”agree on an infinite set”. The set \(\left \{\frac {1}{n}\right \}\) has a limit point at \(0\), and \(\sin \frac {\pi }{z}\) vanishes on it, yet the function is not identically zero — because \(0\) is not a point of the domain. The limit point must lie in \(D\). □
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.