2
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Looks like a duplicate of The Identity Theorem for real analytic functions to me. $\endgroup$ – Martin R Jan 20 '18 at 9:07
  • $\begingroup$ The same proof of analytic continuation for holomorphic functions works for real analytic functions. What is more interesting is that other facts from complex analysis also carry over to real analytic functions since real analytic functions can be extended to holomorphic functions. Indeed, around each point there is a nonzero radius of convergence, so the function can be extended to an open disk around that point in $\mathbb{C}$. This provides a proof of facts such as the composite of two real analytic functions is analytic. $\endgroup$ – Ishan Levy Jan 20 '18 at 9:17
  • $\begingroup$ Possible duplicate of The Identity Theorem for real analytic functions $\endgroup$ – Martin R Jan 20 '18 at 9:21
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Yes that is what I think. $\endgroup$ – Keith Jan 20 '18 at 9:20
  • $\begingroup$ @MartinR Well, he asks to OP to reconsider his definition. And explicitly states that his example is not analytical. $\endgroup$ – Thomas Jan 20 '18 at 9:30
  • $\begingroup$ 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). $\endgroup$ – Martin R Jan 20 '18 at 9:37
  • $\begingroup$ @MartinR you are right that was senseless I meant the converse is only true for holomorphic functions. $\endgroup$ – Guy Fsone Jan 20 '18 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.