1
$\begingroup$

I'm using the power series representation definition of analyticity, and I thought that this definition, together with the Term-By-Term Differentiation Theorem for power series implied that all the derivatives and antiderivatives of an analytic function are analytic, since if a function can be given by a convergent power series then all of its derivatives and antiderivatives can also be given by power series and the radius of convergence is preserved.

My book actually mentions what I just said, but only for derivatives, that is, that an analytic function have infinite derivatives and all of them are analytic. It also mentions that an analytic function have infinite antiderivatives but it said nothing about them being analytic. Then I was doing some problems and I came with the following:

A set $E\subseteq C$ is starlike with respect to $a\in E$ (or just starlike) if $[a, z]\subseteq E$ for all $z\in E \setminus \{a\}$. Show that if $Ω\subseteq C$ is a starlike domain and $f\in H(Ω)$, then there is $F\in H(Ω)$ such that $F' = f$.

Where $H(\Omega)$ is the set of all analytic functions over the set $\Omega$. If my previous reasoning was correct then the proof would be trivial and the hypothesis that $\Omega$ is starlike wouldn't be necessary. I don't know what I am missing. Is analyticity actually not preserved by antidifferentiating a function and why? Thanks in advance.

$\endgroup$
1
  • $\begingroup$ what you are missing is that while antiderivatives are indeed analytic they may not exist in the full domain unless it is simply connected and starlike is indeed such, as for example $\frac{1}{z}$ has no antiderivative in the punctured unit disc and to get one you need to cut a segment (or a Jordan arc) going from zero to the boundary $\endgroup$
    – Conrad
    Commented Feb 12, 2020 at 14:37

2 Answers 2

4
$\begingroup$

You are missing .... topology.

Consider the function $\dfrac{1}{z}$ on $\mathbb{C}\setminus \{0\}$. This function is holomorphic on the domain.

Formally its antiderivative should be $\ln(z)$. However, we have a problem actually defining this function, since if you take a small circle $\gamma$ that encapsulates the origin and integrate $\dfrac{1}{z}$ around $\gamma$, you get a non-zero value.

(In other words, is is not the regularity that kills you, but the fact that you can't even necessarily define a global anti-derivative. I suspect when your book talks about analytic functions having anti-derivatives they are taking this from a very local point of view.)

$\endgroup$
1
  • $\begingroup$ I checked and you are right about the local point of view, and the book actually says that the function has infinite antiderivatives on a neighborhood of each point, it but my brain totally choose to ignore that part. Thanks a lot! $\endgroup$
    – Kerd
    Commented Feb 12, 2020 at 14:55
0
$\begingroup$

Yes, an antiderivative of an analytic function is always analytic. The problem here is not about the analyticity of the antiderivative. Instead, it is about its existence.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .