3
$\begingroup$

The following is the textbook definition of holomorphic functions:

Let $f:\mathbb{C}\to\mathbb{C}$ be a complex function and $U$ an open subset of $\mathbb{C}$. We say $f$ is holomorphic on $U$ if $f$ is differentiable at each point of $U$.

Here is my question:

Do we have an example of $f$ such that it is holomorphic on some proper closed subset $U\subset\mathbb{C}$ but not holomorphic on $\mathbb{C}\setminus U$?


If one only consider the real variable case, there is an example of a function which is differentiable at a single point, and discontinuous everywhere else: $$ f= \begin{cases} x^2,&x\in\mathbb{Q}\\ -x^2,&x\not\in\mathbb{Q}. \end{cases}$$ But I don't know about the complex case.

$\endgroup$
  • 2
    $\begingroup$ There is a very subtle distinction between a function being holomorphic and it being (complex) differentiable. Holomorphic is specifically meant for open sets so it does not make sense to ask about being holomorphic on a closed set; differentiability can occur on any kind of set. For such a function, consider $f(z) = |z|^2$. This is differentiable only at the origin. $\endgroup$ – Cameron Williams Jan 28 '16 at 3:42
  • $\begingroup$ @CameronWilliams: That's a very good point. Would you bother writing the comment to an answer? $\endgroup$ – Jack Jan 28 '16 at 4:06
7
$\begingroup$

There is a very subtle distinction between a function being holomorphic and it being complex differentiable. Holomorphic is specifically meant for open sets so it does not make sense to ask about being holomorphic on a closed set; differentiability can occur on any kind of set.

For an example of a function that is complex differentiable on a (proper) closed set but not holomorphic anywhere on the complement, consider $f(z)=|z|^2$. This is differentiable only at the origin.

$\endgroup$
1
$\begingroup$

No.

A function $f$ is considered holomorphic on some set $U$ if and only if, for every point $u\in U$, $f$ is differentiable on some neighbourhood of $u$.

That means that there needs to be some "space" (in every direction) around $u$ for which $f$ is differentiable.

Closed sets have boundary points included, so if $U$ was closed, it would contain [boundary] points that have "space" around them not contained in $U$. So if $f$ was not differentiable on any point outside of $U$, then every neighbourhood of a boundary point would contain some points outside of $U$, i.e. points that $f$ is not differentiable on, i.e. there would be no neighbourhood of $u$ for which $f$ is differentiable on.

This means that $f$ cannot be holomorphic at every point in a closed set $U$ if $f$ is not differentiable on any point in the complement $U\setminus\mathbb{C}$.

In summary, a function $f$ can only be holomorphic on a set $U$ if there exists some superset $V: U\subseteq\mathbb{C}\setminus\overline{\mathbb{C}\setminus V}$ (i.e. $U\subseteq interior(V)$) that $f$ is differentiable on.

$\endgroup$
0
$\begingroup$

Sure. Let $U=\{0\}.$ For $z\in \mathbb C,$ define $f(z) = z^2\chi_{\mathbb Q\times \mathbb Q}(z).$ Then $f'(0) = 0,$ but $f'(z)$ fails to exist for all $z\in \mathbb C \setminus \{0\}.$

$\endgroup$
  • $\begingroup$ Which means that the function is nowhere holomorphic. $\endgroup$ – N. S. May 11 '16 at 23:35
  • $\begingroup$ Fine. But then your issue is with the OP, who opened the door to an expanded notion of "holomorphic" in the question. $\endgroup$ – zhw. May 11 '16 at 23:45

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.