1

I'm struggling a little bit around three theorem of complex analysis, here are the statements

Theorem 1: Suppose $f$ is a continuous function on an open connected set $G \subset \Bbb C$, the following are equivalent:

  1. Integrals are path-independent;
  2. Integrals around closed curves are $0$;
  3. There is a global antiderivative for $f$ on $G$.

Theorem 2: If $f$ is analytic on a simply connected region $G$ and $\gamma$ is a closed curve in $G$ then $$\int_\gamma f = 0$$

Theorem 3: Suppose $R$ is a rectangular path with sides parallel to the axes and $f$ is an analytic function on an open set $G$ containing $R$ and its interior then $$\int_R f =0$$

Aside for the second theorem where we add the topological specification, which since I'm a little bit rusty I think it refers to a holesless region, to me they are saying the same thing. Here's my reasoning, in the first theorem we state that the continuity of $f$ on an open set (whatever form the region has) we have the three equivalence. In the second theorem $G$ is a connected region so it could also be closed and I can appreciate this difference, so we are in a slightly different case from the first one. So in this case is path independence lost? As well antiderivative?

The third one honestly looks to me as the third implication of the first theorem. Analytic $\implies$ continuous so we have an open region and a closed curve. Isn't the same hypothesis?

I think maybe problem arises in my assumption that analytic implies continuity, I'm catching information from different texts so maybe I'm getting confused between analytic and holomorphic... Any help is appreciated!

Turquoise Tilt
  • 2,571
  • 3
    Analytic most definitely implies continuity. – Chris Sep 06 '24 at 14:03
  • I’d say that pretty clearly Theorem 1 $\Rightarrow$ Theorem 2 $\Rightarrow$ Theorem 3. I don’t think it’s correct to say that Theorem 3 implies Theorem 1, because the hypothesis is stronger. – Chris Sep 06 '24 at 14:09
  • @Chris I totally agree with you, but when the text provides proofs for theorem 2,3 are presented with some "extra work". Personally I think that they follow almost directly from theorem 1. So it's a little mysterious to me why they aren't present as theorem 1's corollaries... – Turquoise Tilt Sep 06 '24 at 14:16
  • Which text do you use? And which definition of "simply connected" is used? – Kritiker der Elche Sep 06 '24 at 15:25
  • @KritikerderElche text are some notes professor gave, but I think also "Basic Complex Analysis" - J. E. Marsden, M. J. Hoffman follows a pretty similar approach. For simply connected: A space $X$ is s.c. iff its fundamental group is trivial, in other terms the space could be contracted to a point. – Turquoise Tilt Sep 06 '24 at 16:54

1 Answers1

2

Theorem 1 $\implies$ Theorem 3 definitely holds with the reasoning given by you. Analytic functions are continuous.

In Theorem 2, I noticed the word "region". This term is not consistent among literature on Complex Analysis, I would guess that in your case it refers to open and connected sets. In that case there would be no topological difference to Theorem 1. See here for a discussion on the term.

I'm a little confused that your textbooks work with the concept "analytic" instead of "holomorphic", I guess it's only later shown that they are equivalent. Once equivalence is proven, both Theorem 2 and 3 are consequences of Cauchy's Integral Theorem, which you may or may not know yet.

David Birkenmayer
  • 361
  • Since my reasoning is valid why do we need the urge to make a statement about rectangular path, when we have a more general results? – Turquoise Tilt Sep 06 '24 at 16:57
  • 1
    Maybe this theorem (which is more of a corollary) is at some point used in the proof of a version of Cauchy's Integral Theorem, which makes statements about paths which are images of rectangles of $C^1$ functions. It involves splitting the path on the rectangle into paths on arbirarily many smaller rectangles and making a limit argument. – David Birkenmayer Sep 06 '24 at 17:22