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Let $D$ be a domain and $f,g:D\to\mathbb{C}$ holomorphic functions. Show that $f\cdot g\equiv 0$ on $D$ implies either $f\equiv 0$ or $g\equiv 0$.

I have a vague idea for a proof: Let's assume $g\neq 0$ for some $z\in D$, i.e. $z_0$. It follows that $f(z_0)=0$. Now if $z_0$ is a limit point, the identity theorem implies that $f\equiv 0$ on $D$.

Empty
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user424862
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3 Answers3

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You would need to conclude that with $g(z_0)\ne 0$ then by continuity also $g(z)\ne 0$ for all $z\in B(z_0,r)$ for some small $r>0$. Then $f(z)=0$ on an open set, but all roots of non-zero holomorphic functions are isolated.

Lutz Lehmann
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As $f\cdot g\equiv0$, therefore $D = Z_f \cup Z_g$ where $Z_f, Z_g$ are zero sets of $f$ and $g$ respectively.

Suppose $f \not \equiv 0$ and $g \not \equiv 0$. Then $Z_f$ and $Z_g$ are isolated sets. It follows that $D$ is an isolated set. But we know that a connected set with at least two points is not isolated.

Sahiba Arora
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Let $K$ be a compact subset of $D$. Then for every point $z$ of $K$, either $f(z) = 0$ or $g(z) = 0$. So at least one of $f$ or $g$ has infinitely many roots in $K$. Since $K$ is compact, these infinitely many roots will have a limit point in $K$, so that $f$ or $g$ must be identically $0$.

mathworker21
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  • So my proof would be correct if I restrict it to a compact subset? I guess there must be some theorem stating that on a compact set there will be a limit point? Which is it though? – user424862 Jun 26 '17 at 19:51
  • The theorem I'm using is that any infinite subset of a compact set has a limit point. This theorem is not true for open sets: consider $1-\frac{1}{n}$ in the open unit disk. – mathworker21 Jun 26 '17 at 19:52
  • I only know limit points in the context of sequences. It seems the identity theorem takes another definition of limit points. Could you explain that as well? – user424862 Jun 26 '17 at 19:53
  • I don't think it takes another version. So a limit point of a set $E$ is a point $z$ (that might be in $E$ or not) so that every nbhd of $z$ intersects $E\setminus{z}$. A limit point of a sequence $(z_1,z_2,\dots)$ is the same as a limit point of the set ${z_1,z_2,\dots}$. So the theorem I mentioned says that, in particular, any sequence in a compact set has a limit point in that compact set. – mathworker21 Jun 26 '17 at 19:55
  • Ah, I see! Thanks for clarifying! – user424862 Jun 26 '17 at 19:58
  • It is not necessary that $K$ has to be compact..! – Empty Jun 26 '17 at 20:01
  • Nobody said it was necessary.... but it is sufficient – mathworker21 Jun 26 '17 at 20:28