Suppose we know the values of a complex analytic function $f$ at all $x+iy$, for $x,y\in\mathbb{Z}$. Can we uniquely determine $f$?
More generally, are there examples of nowhere-dense sets $E$ s.t. $f\mid_E$ determines $f$?
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Equivalently: Are there nowhere-dense sets $E$ such that if $f$ vanishes on $E$ it vanishes on $\mathbb C$? – Milo Brandt Jul 09 '15 at 02:47
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No to your first question. The Weierstrass factorization theorem asserts that for any sequence $\{a_n\}$ of nonzero complex numbers with $|a_n| \to \infty$, there is a nontrivial analytic function $f$ whose zero set is precisely $\{a_n\}$. Clearly we can enumerate all the nonzero points $a_n$ of the integer lattice in such a way that $|a_n| \to \infty$, so there is a nontrivial analytic $f$ vanishing at all those points. Then the function $z f(z)$ is also not the zero function, but vanishes at all those points and 0 as well.
Yes to your second question. The zeros of a nontrivial analytic function must be isolated. So take $E$ to be any nowhere dense set that has a limit point (for instance, the set $\{1/n : n = 1, 2, \dots\} \cup \{0\}$). If $f,g$ are analytic functions which agree on $E$, then $f-g$ is an analytic function whose zero set has a limit point, hence $f-g=0$.
Nate Eldredge
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+1: So (correct me if I'm wrong), one might say that $f\mid_E$ fails to uniquely determine $f$ if and only if $E$ is a discrete set. Does that seem to be a fair summary? – Cameron Buie Jul 09 '15 at 03:20
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@CameronBuie: No, careful. The set $E = {1/n : n = 1,2,\dots}$ is discrete (every point is isolated), but it does uniquely determine $f$ (since it has a limit point, namely 0). I think it holds iff $E$ is a closed discrete set, or equivalently, has no limit points. You have to show that such $E$ is countable and can be enumerated as a sequence ${a_n}$ with $|a_n| \to \infty$, but I believe this follows by considering a sequence of balls $B(0,m)$; by compactness, each such ball contains only finitely many points of $E$. – Nate Eldredge Jul 09 '15 at 03:26
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@NateEldredge: Thanks. I misstated the second question. I really meant, are there sets with no accumulation points that can determine $f$. I forgot that sets like ${1/n}$ count as nowhere-dense. – TorsionSquid Jul 09 '15 at 03:34
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@TorsionSquid: Ok, then the answer is no, as explained in my previous comment. – Nate Eldredge Jul 09 '15 at 03:38