Are absolutely continuous functions analytic? I am asking for a proof that any absolutely continuous function with absolutely continuous derivatives is analytic, once I am studying a function with the first property and 'd like to obtain the second one.
I know that exist smooth functions that are not analytical, but and if the derivatives are absolutely continuous?
If I am wrong, a counterexample is really welcome.
Thank you.
 A: You are asking whether any absolutely continuous function, all of whose derivatives exist and are absolutely continuous, is analytic.  This is false.  Counterexample: any smooth, nonanalytic function.
Indeed, any smooth function on an interval $(a,b)$ is absolutely continuous (on any closed subinterval $[c,d]$ of $(a,b)$), and all its derivatives exist and are also absolutely continuous.
This is because if $f: [c,d] \rightarrow \mathbb R$ is continuous and differentiable on $(c,d)$, with continuous derivative, then $f$ is already absolutely continuous.
A: A famous counterexamples can be obtained by studying the function
$$f(x)=\begin{cases} e^{-\frac{1}{x}} & \text{ when }x>0, \\ 0 &\text{ when }x\leq 0.\end{cases}$$
This function happens to be $\mathcal{C}^\infty(\mathbb{R})$ (this requires some work but nothing beyond calculus)
Then, the function $\phi (x)=f(1-x)f(x+1)$ is also in $\mathcal{C}^\infty(\mathbb{R})$, is nonnegative and vanishes outside $(-1,1)$. Hence, $\phi$ along with all its derivatives, is absolutely continuous. This function however is not analytic!
This kind of functions (some variations of them rather) are called smooth mollifiers. This is also a related Wikipedia article.
A: Posting the comment above as answer, as requested.
The function
$$g(x) = \frac{1}{2} x|x|$$
is differentiable everywhere, with
$$g'(x) = |x|$$
absolutely continuous. However, $g$ is not analytic: if $g$ is analytic, then $g'$ is differentiable. So this $g$ is a counterexample.
An absolutely continuous function (which might not be differentiable) in general is far from analytic. Even when a function is infinitely differentiable, it might not be analytic as the other answer shows.
