3
$\begingroup$

In Walter Rudin's Real and Complex Analysis, second edition, on page 213, two definitions are stated. One of them says the derivative of $f$ at $z_0$ is

$$f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}.\tag 1$$

The other says $f$ is holomorphic in an open set if $f'(z_0)$ exists at every value of $z_0$ in that set.

By that definition, holomorphic doesn't just mean differentiable; it means differentiable at every point in some open set.

He never defines “holomorphic at” a point, but only “holomorphic in” an open set.

Before opening the book this afternoon, just remembering from years ago, I thought what it said was

  • $f$ is differentiable at $z_0$ if $(1)$ holds; and
  • $f$ is holomorphic at $z_0$ if the derivative exists at every point in some open neighborhood of $z_0$.

By that definition $f(z)=|z|^2$ would be differentiable at $0$ but not holomorphic at $0$.

Thus differentiability at an isolated point would be weaker than holomorphy at that point.

QUESTION: Do both conventions exist?

I take “$f$ is analytic at $z_0$” to mean $f$ has a convergent power series expansion at $z_0$, which actually converges to the right thing, the value of $f$, in the interior of its circle of convergence. Differentiability at every point in some open neighborhood of $z_0$ is enough to prove analyticity, but differentiability at $z_0$ is not. By one of the conventions above, one can say a function holomorphic at a point is analytic at that point; by the other one cannot.

$\endgroup$
  • 1
    $\begingroup$ For what it's worth, I only saw holomorphic/analytic applied to functions in open (connected) sets. $\endgroup$ – Jonathan Y. Sep 4 '13 at 21:44
  • $\begingroup$ @PeterTamaroff : As I said, $z\mapsto|z|^2$ is differentiable only at one point, a fortiori isolated. Apply the definition and you'll see that. $\endgroup$ – Michael Hardy Sep 4 '13 at 22:08
  • 1
    $\begingroup$ Doesn't "differentiable at an isolated point" convey that information? $\endgroup$ – Michael Hardy Sep 4 '13 at 22:16
  • $\begingroup$ @MichaelHardy Maybe the first read felt weird. I'll clean up. $\endgroup$ – Pedro Tamaroff Sep 4 '13 at 22:26
4
$\begingroup$

There is almost never any reason to talk about functions that are (complex-) differentiable at isolated points (other than as exercises to show the students that the derivative may exist at some exceptional points, and to show that Cauchy-Riemann's equations don't quite tell the full story).

I can't think of any book or text that uses "holomorphic at a point", at least not meaning that $f'$ exists at that particular point. If I would see the term, I would interpret it as synonymous with "analytic at a point", i.e. holomorphic on some open neighbourhood of the point.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Indeed! In_practice, functions "holomorphic at a point" do not arise. This issue would only conceivably be contrived in a textbook to make a (rather pedantic) point. Thus, to avoid this pedantry, one might (semi-mysteriously) insist on differentiability on a non-empty open set in the "definition". In real life, typically one has differentiability on a (non-empty) open set, anyway, so this "extra requirement" is not a "burden", but simply a thing that naturally arises, and we "have". It is not good to overload language with delicate distinctions whose origins or purposes are suppressed. $\endgroup$ – paul garrett Sep 4 '13 at 22:23
  • $\begingroup$ John Bush, an expert in fluid dynamics who has been affiliated with Harvard and MIT at various times, taught a course to prepare engineering graduate students with topics in differential equations and in complex variables that were to be used in fluid dynamics. He included an example of a fairly simply defined function that is differentiable everywhere on a straight line in the complex plane but nowhere else in the plane. It seems safe to predict they won't see that very often in fluid dynamics. $\endgroup$ – Michael Hardy May 15 at 21:59
2
$\begingroup$

I am writing this like a answer, but only because for me to add a comment here I have t have 50 reputation (but I don't have still). So, Michael Hardy, I was with a similar doubt. Because in Churchill's Complex variables and its applications, there's a section where he define an analytical function just like a function which derivative exists in a point $z_{0}$ and in the points that are a neighborhood of the point $z_{0}$.

In the case of the function $f(z)={\vert}z{\vert}^{2}$, this function has a derivative defined in the point $z=0$, but in the $z=0$ neighborhood we cannot define the derivative (because the lateral limit does not exist, so the derivative does not exists too).

But there is another case: the function $g(z)={\frac{1}{z}}$. It's derivative is not defined for the point $z=0$, but it exists to the neighbor points of $z=0$. In this case, $z=0$ is a singular point of the function.

Note: I would like to write this like an comment (in fact, this is not the answer!), but my S.E. points are not enough to post a comment.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Yes, both conventions exist, and are, as far as I'm aware, standard. Differentiability is a pointwise property (in the sense that differentiablity at $x$ does not imply differentiability at any other point), and holomorphy (or analyticity) a local property (in the sense that holomorphy at $z_0$ implies holomorphy in a neighbourhood of $z_0$).

Fischer/Lieb, A Course in Complex Analysis, define

A complex-valued function $f$ defined on an open set $U \subset \mathbb{C}$ is (complex) differentiable at $z_0 \in \mathbb{C}$ if there exists a function $\Delta$ on $U$, continuous at $z_0$, such that $$f(z) = f(z_0) + \Delta(z)(z-z_0)\tag{1}$$ holds for all $z \in U$. If $f$ is complex differentiable at all points $z_0 \in U$, then we say that $f$ is holomorphic on $U$. We say that $f$ is holomorphic at $z_0$ if there exists an open neighbourhood of $z_0$ on which $f$ is holomorphic.

(The $z_0 \in \mathbb{C}$ is a typo, should of course read $z_0 \in U$.)

Ahlfors also differentiates between the derivative existing at a point and analyticity/holomorphy:

The class of analytic functions is formed by the complex functions of a complex variable which possess a derivative wherever the function is defined. The term holomorphic function is used with identical meaning.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Whether a function $f$ is differentiable at a point $z_0$ depends on the behavior of $f$ in some neighborhood of $z_0$. So I'd have guessed you'd call it a "local" property. Can you tell me what definition of "pointwise" is consistent with this? $\endgroup$ – Michael Hardy Sep 4 '13 at 22:14
  • $\begingroup$ Well, yes, of course the existence of the derivative at a point is local in the sense that it depends on the values in a neighbourhood of that point. What I meant, and don't know how to succinctly express, is that to say that "$f$ is differentiable at $z_0$" does not imply the existence of the derivative at any other point, while "$f$ is holomorphic at $z_0$" does imply the existence of the derivative in a full neighbourhood of $z_0$. If you (or anybody) can suggest a good wording, I'd be grateful. $\endgroup$ – Daniel Fischer Sep 4 '13 at 22:20
  • $\begingroup$ If there's no difference in meaning between "holomorphic" and "analytic", then what does "real-analytic" mean? I thought it meant being equal to its power-series expansion. $\endgroup$ – Michael Hardy Sep 4 '13 at 23:32
  • 2
    $\begingroup$ "Real-analytic" means representable by a power series in the real coordinates (or, in the case of domains in $\mathbb{C}^n$, equivalently, power series in the $z_i$ and $\overline{z}_i$). "Analytic" sans "real" is - in complex analysis - reserved for holomorphic functions; historically, as far as I remember, people called functions analytic that were representable as power series (in $z$, resp the $z_i$, only), and holomorphic when they had a complex derivative/partial complex derivatives in an open set. It turned out, both conditions are equivalent, but that was not obvious from the start. $\endgroup$ – Daniel Fischer Sep 4 '13 at 23:38
1
$\begingroup$

According to Wikipedia, a function $f$ is said to be holomorphic at a point $z_0$ if it is holomorphic in some neighbourhood of $z_0$. It goes on to say that $f$ is defined to be holomorphic on a non-empty set $A$ if it is holomorphic on some open neighbourhood of $A$.

I don't know if any other reference uses this terminology though.

| cite | improve this answer | |
$\endgroup$

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.