Talk:Morera's theorem

message: The proof provided here is unclear, the function F(z) should be defined by picking any point z_0 in D and letting F(z) be the path integral of f(z) along any path from z_0 to z.  The arguments presented in the given proof show why this is well defined.  I would fix it myself but. . . i'm not too hot with the math editing. —Preceding unsigned comment added by 66.75.254.51 (talk • contribs) 
:There should be a link to the article about Morera somewhere in this article (or at least his name if he isn't notable enough for an article). -- Obradović Goran  (talk 19:22, 21 June 2007 (UTC)
replies: text: I would prefer to go back to the older statement, with the integral. Charles Matthews 07:01, 14 August 2007 (UTC)

text: they ain't the same though. the old version says that if f has an anti-derivative everywhere on its domain (or, equivalently, has zero integral along every closed curve in the domain), then it is holomorphic. but that is trivial and, far as i know, not Morera's theorem. Mct mht 15:04, 14 August 2007 (UTC)
replies: text: well, it's getting reverted to the previous version. one might wanna revert further back though, as right now the article is not quite coherent. there's really nothing complex-analytic about this formulation of the theorem, the argument would be a verbatim copy from the real variable case. Mct mht 22:28, 13 September 2007 (UTC)
replies: text: The version of Morera's theorem now in the article is the same version stated in Rudin's book, in Ahlfors' book, and in Conway's book.  There's nothing trivial about it: "holomorphic" means "complex differentiable", and it's a relatively deep result that a complex function that locally possesses an anti-derivative should have a derivative too.  The corresponding statement in the real case would be "any vector field on R2 whose line integral around every closed loop is zero must be differentiable", and this is completely false.  The proof in the article makes the theorem seem somewhat easy, but only because it assumes that holomorphic functions are analytic. Jim 19:51, 15 September 2007 (UTC)
replies: text: "the integral of a continuous function is differentiable", that part of the claim is a word by word copy of a similar fact for functions on the real line and is entirely trivial. of course one would then invoke differentiability implies analyticity in the complex-variable case. the point ought to be made that this formulation of the theorem does not use the full strength of the fact that holomorphy means analyticity. having zero integral along a every closed path, i.e. possessing an anti-derivative in the complex variable sense everywhere on its domain, is a far more restrictive condition than being holomorphic. Mct mht 23:56, 15 September 2007 (UTC)
replies: text: You seem to be confused about something, though maybe I'm misinterpreting.  The theorem states that a function which can be integrated is differentiable.  It is not part of the hypothesis of the theorem that f is the integral of a continuous function. Jim 08:30, 21 September 2007 (UTC)
replies: text: yes, you are misinterpreting. Mct mht 19:43, 21 September 2007 (UTC)
thread_title: Untitled

message: I've attempted to add some detail to the proof, however, there is still something to be desired. It might be good to say a little bit more about how the derivative of F is f. Additionally, the TeX didn't come out quite the way I expected. Perhaps somebody to look into that to enhance readability. 
Coolkid70 (talk) 04:22, 29 October 2010 (UTC)
replies: text: Thanks, looks great! One more thing I'm not seeing in the current version is why F should be holomorphic. Perhaps a word about that would be a good idea. Hilbert Space Shuttle (talk) 23:36, 30 December 2022 (UTC)
thread_title:  Revising the proof 

message: The statement that the hypothesis can be weakened is not cited and provided without proof. Please at least add a source.
136.152.179.213 (talk) 20:48, 25 July 2011 (UTC)
thread_title: Weakening of Hypothesis

message: Why is the domain D of definition assumed to be connected? This seems to be completely superflous (simply apply the "connected" theorem on each component).  — Preceding unsigned comment added by 91.61.122.126 (talk) 20:57, 29 September 2014 (UTC)
thread_title: Assumption of connectedness

message: Substituted at 02:21, 5 May 2016 (UTC)
thread_title: Assessment comment

message: Morera's theorem requires that the integral of the function be 0 only around every triangle in the region. The conclusion is still that the function is analytic in the region.
This is much stronger than what is stated in the article, since triangles form only a tiny subset of "all piecewise C1 curves".  
The article should be modified to reflect this.Daqu (talk) 21:03, 11 June 2016 (UTC)
thread_title:  Theorem as stated is a lot weaker than Morera's theorem