A set $M$ is called a set of uniqueness for functions of a class $\mathcal{F}$ if any function $f \in \mathcal{F}$, equal to $0$ on $M$, is identically equal to $0$. Prove that the following sets are sets of uniqueness for functions holomorphic on $\mathbb{C}^2$:

(a) a real hyperplane in $\mathbb{C}^2$;

(b) the real two-dimensional plane $\{z_{1} = \bar{z_2} \}$;

(c) the arc $\{z_2 = \bar{z_1}, y_1 = x_1 \text{ sin }(1/x_1)\}$

Any help with these would be great. I was wondering specifically if there is a systematic way to address each of these proofs. If so, some help with part (a) would be great, and I can attempt (b) and (c). Thanks!

  • $\begingroup$ Isn't sufficient to check that these sets are non-discrete? Holomorphic non-constant maps should have discrete sets of zeroes (if I am not wrong this is true in several complex variables as in one variable). $\endgroup$ – Crostul Mar 20 '16 at 11:16
  • $\begingroup$ Not in SCV. The zero set of a holomorphic function on a domain $D \subset \mathbb{C}^n$, $n \geq 2$ contains no isolated points. $\endgroup$ – K.Reeves Mar 20 '16 at 11:31
  • $\begingroup$ @Crostul Note $f(z,w) = z$ is zero on $\{0\}\times \mathbb C.$ $\endgroup$ – zhw. Mar 20 '16 at 23:08

Sets of uniqueness for (subclasses of) holomorphic functions in several variables is fairly tricky, and not completely understood. For example there is no known simple, general characterisation of sets of uniqueness for bounded holomorphic functions in several variables.

For your specific question, for a) you can assume that the hyperplane is $\operatorname{Im} z_2 = 0$. Assume that $f = 0$ on the hyperplane and look at functions $$ g(z) = f(c,z) $$ for a fixed $c$. Then $g$ is entire and vanishes for $\operatorname{Im} z = 0$, hence everywhere. Can you take it from there?

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  • $\begingroup$ Not really...I don't really see how $g = 0$ in this hyperplane helps. $\endgroup$ – K.Reeves Mar 20 '16 at 23:53
  • $\begingroup$ For each $c$, $g(z) = 0$ for all $z$. But this implies that $f=0$ everywhere. $\endgroup$ – mrf Mar 21 '16 at 13:43
  • $\begingroup$ Yeh I think I get it. Very subtle... but makes sense. Thanks! $\endgroup$ – K.Reeves Mar 22 '16 at 6:37
  • $\begingroup$ @mrf How do you characterise this hyperplane. Shebat's text defines a real hyperplane as the set of all vectors in $\mathbb{C}^n$ that are real orthogonal to some vector $z^0$. How have you used that here? $\endgroup$ – user412674 Mar 10 '17 at 1:07
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    $\begingroup$ @mrf Also, Shabat states that if a function $f \in \mathcal{O}(D)$ vanishes in a real neighbourhood of a point $a \in D$, then $f \equiv 0$ in $D$. Would we not then simply assert that for every point on a real hyperplane, there is a real neighbourhood about that point? Similarly for (b)? $\endgroup$ – user412674 Mar 10 '17 at 1:15

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