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Real analytic functions are defined as functions on the Euclidean spaces with convergent power series at each point.

My question is that, is there some kind of identity theorem for real analytic functions? My book (John Lee's smooth manifolds) says on p.46 that a real-analytic function defined on a connected domain and vanishes on an open set is identically zero.

But I have the impression that this kind of fact holds for holomorphic functions only. Am I missing something?

Guy Fsone
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Keith
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1 Answers1

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Note that the notion of $C^\infty-$differentiability and analyticity are defined separately

your book is totally, right since a function $f$ being analytic mean that at each point $a$ of the domain there exists $r=r(a)>0$ and a sequence $(a_n)_n$ such that $$f(x) =\sum_{n=1}^{\infty} a_n(x-a)^n~~~~~|x-a|<r $$

and this implies the $C^\infty-$differentiability. The converse is not true for real-function but it is only true for holomorphic functions.

Indeed, you may presumably reconsider your definition within the following function

$$f(x)=e^{-\frac{1}{x^2}}~~~~~f(0)=0$$

This function has a pathology that is $C^\infty(\Bbb R)$ but we have, $$f^{(n)}(0)=0~~~~\forall ~~n\ge0$$

Hence, for all $x\in \Bbb R$$$\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!}x^n =0$$

this function is $C^\infty(\Bbb R)$ non-constant but not analytical near $x=0$.

We can conclude that being real-analytic implies the $C^\infty$ differentiable but the converse is not true.

Guy Fsone
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  • Yes that is what I think. – Keith Jan 20 '18 at 09:20
  • @MartinR Well, he asks to OP to reconsider his definition. And explicitly states that his example is not analytical. – Thomas Jan 20 '18 at 09:30
  • The statement "it is only true for holomorphic functions" is still wrong. It is true for real-analytic functions (if the correct definition is used). – Martin R Jan 20 '18 at 09:37
  • @MartinR you are right that was senseless I meant the converse is only true for holomorphic functions. – Guy Fsone Jan 20 '18 at 09:39