3
$\begingroup$

In my complex analysis class, I learned that a complex function $f(z)$ is said to be analytic at $z_0$ if there is a neighborhood around $z_0$ in which $f$ is differentiable. Then, I learned various properties of analytic functions, but I never quite understood why the notion of pointwise differentiability is not enough and we in fact need differentiability "in nearby neighborhoods".

What is the importance/role of "neighborhoods" in this context? What goes wrong if one replaces analyticity with differentiability? I thought that when one talks about pointwise differentiability, one would like to work in open sets and since analyticity and differentiablity are the same in open sets, I do not see what is important about defining analyticity the way it is defined

$\endgroup$
1
  • 1
    $\begingroup$ It's a contraction of language: instead of saying "analytic on a neighborhood of $z_0$" all the time, we declare the shorter phrase "analytic at $z_0$" to have that broader meaning. The notion of a derivative is a natural pointwise concept, so saying "differentiable at $z_0$" already means something and thus is not available as a term to describe behavior in an open set around $z_0$. The function $f(z) = z|z|$ is differentiable at the origin ($f(h)/h \rightarrow 0$ as $h \rightarrow 0$ in $\mathbf C$), but is not expressible as a complex power series on a neighborhood of the origin. $\endgroup$
    – KCd
    Feb 22, 2017 at 4:45

1 Answer 1

Reset to default
2
$\begingroup$

Because while the definition "differentiable in a neighborhood" is a suitably-equivalent way of saying "analytic" in the context of complex analysis, it is not necessarily the most informative way.

The most informative definition of analytic is just the extension of the one from real analysis: a complex $f(z)$ is analytic at $z_0$ iff its Taylor series expansion at $z_0$ converges to $f(z)$ in a neighborhood of $z_0$.

In complex analysis, this is equivalent to the statement that it be differentiable in a neighborhood. In real analysis, it is not equivalent.

In fact, in complex analysis the following equivalences hold: all (complex-)differentiable functions are (complex-)smooth and all smooth functions are analytic. But the definitions of these terms need not be the same, even though we could do so because of these equivalences. IMO, it's better to use the regular definitions and then prove the equivalence as a useful theorem. To go from Real to Complex with a change of definition of "analytic" from "converging Taylor series" to "differentiable in a neighborhood" would be rather jarring, and leave one scratching one's head for reasons, imo. And it obscures some of the beauty, which is precisely that in the complex plane the added structure is far more restrictive on the behavior of the function than in real-number analysis and the reasons for it being so (essentially, that a function which is locally complex-linear (and non-constant) preserves the shape of vanishing objects at (almost) every point, while more general 2D real $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ functions do not.).

$\endgroup$
6
  • $\begingroup$ So, there is a real analytic function which is differentiable only at the point of expansion, and equals to its Taylor series, yet it is not differentiable in a neighborhood? $\endgroup$
    – nan
    Feb 22, 2017 at 5:29
  • $\begingroup$ @momonao: The Taylor series must converge in a neighborhood to the function, this implies the function will be differentiable at other points than just the point of expansion. Any Taylor series will converge at its point of expansion, of course. $\endgroup$
    – The_Sympathizer
    Feb 22, 2017 at 5:31
  • $\begingroup$ Then, what do you mean by "in real analysis, it is not"? $\endgroup$
    – nan
    Feb 22, 2017 at 5:32
  • $\begingroup$ @momonao: Ah. I see. Not equivalent, that is (to being differentiable in a neighborhood). $\endgroup$
    – The_Sympathizer
    Feb 22, 2017 at 5:34
  • $\begingroup$ Is one stronger than the other? or how are they related? I mean, in real analytic setting $\endgroup$
    – nan
    Feb 22, 2017 at 5:38

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.