Suppose \(D\) is a domain and \(f, g\) are analytic on \(D\).
Prove that if \(fg = 0\) on \(D\), then either \(f \equiv 0\) or \(g\equiv 0\) on \(D\).
Solution
Goal: Prove that if \(f, g\) are analytic on a domain \(D\) and \(fg \equiv 0\) on \(D\), then either \(f \equiv 0\) or \(g \equiv 0\) on \(D\).
1. Assume \(f \not\equiv 0\); it suffices to show \(g \equiv 0\).
By symmetry between \(f\) and \(g\).
2. The zero set \(Z \definedas \theset{z \in D \suchthat f(z) = 0}\) is a closed subset of \(D\) with no accumulation point in \(D\).
2.1. \(Z\) is closed in \(D\).
\(f\) is continuous, and \(Z\) is the preimage of the closed set \(\theset{0}\).
2.2. \(Z\) has empty interior.
If \(Z\) contained a nonempty open set \(U\), then \(f \equiv 0\) on \(U\), and by the identity theorem \(f \equiv 0\) on \(D\), contradicting 1.
3. \(D \setminus Z\) is nonempty and open in \(D\).
\(Z \neq D\) (else \(f \equiv 0\), contradicting 1) and \(Z\) is closed by 2.1.
4. \(g \equiv 0\) on \(D \setminus Z\).
For \(z \in D \setminus Z\), \(f(z) \neq 0\), so \(fg(z) = 0\) forces \(g(z) = 0\).
5. \(g \equiv 0\) on \(D\).
3 gives a nonempty open set (e.g. a disk contained in \(D \setminus Z\)) on which \(g\) vanishes; by the identity theorem, \(g\) vanishes on all of \(D\).
6. Q.E.D.
<1>1–<1>5 show that if \(f \not\equiv 0\) then \(g \equiv 0\); the alternative case is symmetric.