# Identity theorem for $\mathbb{R}^n$

This question is follow up to an interesting question I found here.

The results of this question states the following:

If $U$ is a domain, and $f,g$ are two real-analytic functions defined on $U$, and if $V\subset U$ is a nonempty open set with $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$. If the domain is one-dimensional (an interval in $\mathbb{R}$), then it suffices that $f\lvert_M \equiv g\lvert_M$ for some $M\subset U$ that has an accumulation point in $U$.

1. I was looking for a reference for this. However, in the previously pointed out reference A Primer of Real Analytic Functions by Krantz and Parks, I was not able to locate this theorem. I would appriciate if someone could point me to proper reference of this theorem.
2. My main question, is about the second part of the theorem. Specifically, I am interested in why there is such a difference going from $\mathbb{R}$ to $\mathbb{R}^n$. That is in one-dimension we can assume that $M$ is just a set with an accumulation point, but in $\mathbb{R}^n$ we have to assume that $M$ is an open set. I would really like see a counter example that demonstraes that assuming that $M$ is a set with accumulation points is not sufficient in $\mathbb{R}^n$.
3. I would really like for you to speculate or suggest an extra assumptions on functions $f$ and $g$ such that it suffices to consider $M$ to be only a set with an accumulation point.
• As for your question (2): You could take a set $M$ with an accumulation such that every point of $M$ is colinear. What could this possibly tell you about how the function behaves in a perpendicular dimension? You could add multiples of $y, y^2$ (say), and not affect $f|_M$. – Joppy May 1 '17 at 14:44

$M$ being open is not a necessary condition for $M$ to be a uniqueness set in higher dimensions. For example, if $f,g$ are real analytic on $\mathbb R^2$ and $f = g$ on $\cup_{k=1}^\infty \{(x,y):x=1/k\},$ then $f\equiv g.$ Also any set of positive measure, in any dimension, will give the result.
• You mentioned that $f$ and $g$ agree on a set of a positive measure then they agree everywhere. Does this set have to be connected? For example, if $f$ and $g$ agree on a fat Cantor set which has a positive measure do they agree everywhere? – Boby May 4 '17 at 12:56
• @Boby I don't really know of a reference. For $\mathbb R$ it's simple: Suppose $E\subset \mathbb R$ has positive measure. Then $E$ is uncountable. Therefore $E\cap [-a,a]$ is infinite for some $a>0.$ Hence $E$ has a limit point. By the usual result in one variable, $E$ is a uniqueness set for real analytic functions. continued below ... – zhw. May 4 '17 at 16:21
• Suppose now $E\subset \mathbb R^2$ has positive two-dimensional Lebegue measure. By Fubini, $$m_2(E) = \int_{\mathbb R} \int_{\mathbb R} m_1(E\cap ( \{x\}\times \mathbb R))\, dm_1(y) \, dm_1(x).$$See if you can get the two dimensional result from this and the one-dimensional result. This points the way to a general inductive proof based on the dimension. – zhw. May 4 '17 at 16:22
For 2 and 3: the functions $0$ and $x$ are a counterexample if all you want is an accumulation, and since a general analytic function will have an $n-1$ dimensional $0$ set, the best you can do is what Mark van Leeuwen suggests here.
• Ok. I see now how these are counter examples. I see that the reference that you pointed out is related. But can I explain more in the context of this question. Specifically, commenting on the $\mathbb{R}^n$. – Boby May 1 '17 at 14:45