Can every smooth $(C^{\infty}(\mathbb{R}))$ Function $f$ be extended to a holomorphic function? Considering the Identity Theorem:
Let $G \subset \mathbb{C}$ be a non empty, connected open subset of $\mathbb{C}. For f,g: G \rightarrow \mathbb{C}$ the following are equivalent:
i) $f \equiv g$
ii) The set ${w \in G : f(w)=g(w)}$ has a limit point (in $G$)
iii) There exists a point $c \in G$, such that $f^{(n)}(c)=g^{(n)}(c)$ for all $n \in \mathbb{N}$
Now the following Question came up:
Can every smooth $(C^{\infty}(\mathbb{R}))$ Function f be extended to a holomorphic function, i.e is there a open set $U \subseteq \mathbb{C}$ with $\mathbb{R} \subseteq U$ and a holomorphic function $g: U \rightarrow \mathbb{C}$ such that $g_{| \mathbb{R}}=f$
My thoughts:
As far as I know, in complex analysis $sin$, $cos$, and $exp$ got extended to a holomorphic function using their power series. Because of the Identity Theorem iii) there is only one way to expand these ($sin$,$cos$, $exp$) function holomorphically.
But I also remember that the Function
$h: \mathbb{R} \rightarrow \mathbb{R}$,
$h:=exp(-x^{-2})$ for $x \neq 0$
$h(0):=0$ for $x=0$
can't be developed into a power series around $x=0$.
Because of this my answer would tend to "no", but I am not sure if there is another method to do this expansion.
Would be nice if someone cloud explain me the answer to this question.
 A: A holomorphic function cannot have a point of accumulation of zeros. A smooth function can. So, the answer is no
A: Your idea is (nearly) correct. If$$h(x)=\begin{cases}e^{-1/x^2}&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}$$then you cannot extend it to a holomorphic function $H\colon U\longrightarrow\Bbb C$, for some open subset $U$ of $\Bbb C$ such that $U\supset\Bbb R$. In fact, we would have $(\forall n\in\Bbb Z_+):H^{(n)}(0)=0$, and therefore near $0$ we would have$$H(z)=\sum_{n=0}^\infty\frac{H^{(n)}(0)}{n!}z^n=0.\tag1$$
In fact, if $U$ turned out to be connected, then it would follow from $(1)$ that $H$ would have to be the null function, by the identity theorem.
A: There is no other method, because if $f$ is a holomorphic function in an open disk $B(w,r)$, then $f$ can be expanded into a Taylor series around $f(z)=\sum_{n\ge 0} \frac{f^{(n)}}{n!} (z-w)^n$ that converges in that disk. In particular, $f(x)=\exp(-x^{-2})$ which is in $C^\infty(\mathbb R)$, cannot be extended to a holomorphic function in any open  set that contains $0$.
