5
$\begingroup$

"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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – wqr Mar 21 '13 at 4:45
  • $\begingroup$ I did not assume any zero is 'isolated' ? $\endgroup$ – wqr Mar 21 '13 at 4:50
  • $\begingroup$ @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. $\endgroup$ – Julien Mar 21 '13 at 4:51
  • 1
    $\begingroup$ @julien Ah yes, I misread that. Sorry. Good work wqr. $\endgroup$ – Potato Mar 21 '13 at 4:51
  • $\begingroup$ thanks! but I used your idea Potato, thx again. $\endgroup$ – wqr Mar 21 '13 at 4:53
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – wqr Mar 21 '13 at 4:46
  • 2
    $\begingroup$ @wqr Oh boy! Of course. No Baire...Thanks! $\endgroup$ – Julien Mar 21 '13 at 4:48
2
$\begingroup$

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 $G$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.