# product of harmonic functions

"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.

• I tried this: Suppose u is not identically zero. Then $u(z_1) \neq 0$ for some $z_1$. But this forces $v(z_1)=0$. As $u$ being cts., there exists a ball around $z_1$ s.t. $u \neq 0$. So $v$ is zero on this ball. Hence $v$ is identically zero on $G$. Does this work? – wqr Mar 21 '13 at 4:45
• I did not assume any zero is 'isolated' ? – wqr Mar 21 '13 at 4:50
• @Potato It's good. It shows $v$ is zero on an open ball. So $v$ is constant equal to $0$ on $G$ by the fact I mentioned. – Julien Mar 21 '13 at 4:51
• @julien Ah yes, I misread that. Sorry. Good work wqr. – Potato Mar 21 '13 at 4:51
• thanks! but I used your idea Potato, thx again. – wqr Mar 21 '13 at 4:53

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.

• from potato's argument, I tried this: Suppose u is not identically zero. Then $u(z_1) \neq 0$ for some $z_1$. But this forces $v(z_1)=0$. As $u$ being cts., there exists a ball around $z_1$ s.t. $u \neq 0$. So $v$ is zero on this ball. Hence $v$ is identically zero on $G$. Does this work? – wqr Mar 21 '13 at 4:46
• @wqr Oh boy! Of course. No Baire...Thanks! – Julien Mar 21 '13 at 4:48

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 $G$.

• In your answer "As u being cts." what does cts. means? – Kavita Mar 28 '17 at 14:28