Talk:De Branges' theorem

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Now a theorem, so a conjecture category could mislead.

Charles Matthews 11:28, 9 Oct 2004 (UTC)

Both cases a0 and a1 are simply consequences of normalization. The case a2 is elementary (originally due to Bieberbach, see Hille's book on analytic functions, vol II, page 350). The first deep case is a3, and was settled by Loewner (then at Stanford, i believe). He sat on the result for a year, hoping that it could be extended. His method is in Ahlfors' little green book on geometric function theory (not the standard Ahlfors' text), and probably in Peter Duren's book and elsewhere. (See also http://mathworld.wolfram.com/BieberbachConjecture.html which has dates.) Schiffer had a big hand in a4 and a5 (the even coefficients are easier, iirc), and i heard him one time remark at a talk that somebody (whose name escapes me) said that the Bieberbach conjecture was perfect for a mathematician, because it was guaranteed employment (you'd do one coefficient at a time, because there would never be a general proof). I believe Schiffer's work used his method of interior variation, instead of Loewner's method of boundary variation. In fact, Loewner's method yields a PDE which in principle could be studied to derive all the coefficients, and that's what de Branges did, in the context of his own theory.

[edit] Most popular name?

So the article says it was formerly called the Bieberbach conjecture. I found that odd as I've always thought of it as Bieberbach conjecture, and heard it often referred that way. Do specialists really call it de Branges' theorem? A preliminary look through MathSciNet, seems to indicate that "the de Branges theorem" actually refers to a more general theorem that implies (among other important stuff) the Bieberbach conjecture. --Chan-Ho (Talk) 07:36, 15 April 2006 (UTC)

[edit] Demonstration of non-injectivity in the sufficiency counterexample f(z):=z+z^2

The article states "it is holomorphic on the unit disc and satisfies [...], but it is not injective since f '(-1/2) = 0." Surely this is not complete without also noting f''(-1/2) > 0, correct? 06:44, 15 May 2006 (UTC)

No, you're thinking of real valued functions defined on the real axis. A holomorphic function f defined in the complex plane is injective if and only if its derivative is non-zero. As soon as you have a point z₀ with f '(z₀) = 0, you know that f cannot be injective near z₀. The intuitive reason is that such a function, near z₀, has approximately the form g(z) = a + b(z-z₀)ⁿ for some n≥2 (Taylor series), and the function g wraps the circle around z₀ n-times around the point a, so is not injective. AxelBoldt 15:14, 16 May 2006 (UTC)