What is the condition for two real analytic functions to be identically equal? We know that there is a nice condition (Identity Theorem) for holomorphic function to check if they are the same. What is its version for real analytic functions?

6If $U$ is a domain, and $f,g$ are two realanalytic 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 onedimensional (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 $U$. – Daniel Fischer Apr 04 '14 at 12:09

@DanielFischer Thanks, I also think it should be like this way, but would you give me some references about it? – Yui Apr 04 '14 at 12:12

1Can't list a reference, but it's proved exactly like the holomorphic case. For the nonempty open $V$ case (without loss of generality $g\equiv 0$), you consider the set $W = \{ x : f\lvert_N \equiv 0 \text{ for some neighbourhood } N \text{ of } x\}$, and show $W$ is open (by definition) and closed. For the $M$ has an accumulation point case, consider the power series expansion in the accumulation point, derive all coefficients are $0$, and you're in the first case. – Daniel Fischer Apr 04 '14 at 12:19

@DanielFischer Right,thank you. – Yui Apr 04 '14 at 12:25
1 Answers
As a reference, I suggest A Primer of Real Analytic Functions by Krantz and Parks.
The two versions of the Identity Theorem stated by Daniel Fisher can be unified at the expense of more complicated statement.
If $U$ is a domain, and $f,g$ are two realanalytic 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 onedimensional (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 $U$.
Claim. If $f,g$ are real analytic and there is a point $p\in \mathbb R^n$ such that the set of all limits $$ \left\{\lim_{n\to \infty} \frac{x_np}{x_np} : \qquad f(x_n)=g(x_n),\ x_n\to p, \ x_n\ne p\right\} \tag{1}$$ has an interior point in the topology of the sphere $S^{n1}$, then $f\equiv g$.
Indeed, suppose $fg$ is not identically zero. Express its Taylor series at $p$ as the sum of homogeneous polynomials $P_d$. Let $d$ be the smallest degree for which $P_d$ is not identically zero. Then the set defined by (1) can be shown to be precisely $$ S^{n1} \cap \{ P_d =0\} \tag{2}$$ Since the zero set of a polynomial has empty interior in $\mathbb R^n$, it follows that (2) has empty interior in $S^{n1}$. $\quad \Box$
When $n=1$, the sphere $S^0$ is a twopoint set, so any nonempty subset of it has nonempty interior. We thus recover the onedimensional result.
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1@Yui The part beginning with "Indeed" is meant to be a proof, at least in an outline. – user127096 Apr 08 '14 at 05:05

I know this is an old post. But does this conclusion hold in $\mathbb{R}^n$? – Boby Apr 28 '17 at 20:19

I think there is something wrong with the claim. In $\mathbb{R}^2$ the analytic function $\sin(2\pi (x^2+y^2))$ vanishes on the unit circle. What am I missing? – Kosh Aug 11 '20 at 18:07

2@Kosh, the set points where that function vanishes in R^2 has an empty interior. You need this to hold over an entire area in R^2. – Joel Apr 10 '21 at 19:33