"Let $u, v$ be harmonic functions on a region G. Prove that if the product $uv$ is identically zero, then either $u$ or $v$ must be identically zero." Could sb. give me a hint for this question? I tried all the properties and theorems on harmonic functions (definition, poisson integral, max. principle, passed a holomorphic function,..), but could go nowhere.
Thanks.
Suppose $u$ is not identically zero. Then $u(z_1)≠0$ for some $z_1$. But this forces $v(z_1)=0$ since their product is zero. As $u$ being cts., there exists a ball around $z_1$ s.t. $u≠0$. So $v$ is zero on this ball. Hence, by the identity theorem for harmonic functions, $v$ is identically zero on $G$.
Fact 1: $u$ and $v$ are real analytic on $G$.
Fact 2: if $w$ is real analytic on $G$ and if $w^{-1}(0)$ has non empty interior, then $w=0$ on $G$.
Reference: Theorems 1.27 and 1.28 here. Note that the zeros of a harmonic function are never isolated (nice application of the mean value property, and the intermediate value theorem). So a sequence strategy does not work.
Edit: As pointed out by @wqr, one really does not need Baire (see the generalization below for something which really requires Baire). If $u$ is not identically $0$, then it is nonzero on some open ball by continuity. Therefore $v$ is zero on this same open ball. By Fact 2, it follows that $v=0$ on $G$. Likewise $u=0$ on $G$ if $v\neq 0$.
Generalization: If $(u_n)$ is a sequence of harmonic functions on $G$ such that for every $z\in G$, there exists $n$ such that $u_n(z)=0$, then there exists one of these functions which is identically $0$ on $G$. Indeed, writing $$ G=\bigcup u_n^{-1}(0) $$ we see that one of the closed (in $G$) sets $u_n^{-1}(0)$ must have nonempty interior (in $G$) by Baire property of the Baire space $G$ (every open subspace of a Baire space is a Baire space). Finally, since $G$ is open, the interiors relative to $G$ and to $\mathbb{C}$ coincide. So it only remains to apply Fact 2.
