Continuous + holomorphic on a dense open => holomorphic? - MathOverflow

Continuous + holomorphic on a dense open => holomorphic?

2010-08-06T13:16:07Z

Let D ⊂ ℂ be the closed unit disc in the complex plane, and let C be a continuously embedded path in D between the points -1 and 1. The curve C splits D into two halfs $D_1$ and $D_2$.

Let f : D→ℂ be a continuous function that is holomorphic on the interiors of $D_1$ and $D_2$.
Is f then necessarily holomorphic?

PS: If he path C is sufficiently smooth (so that ∫_C f(z) dz makes sense), then f is necessarily holomorphic, as it is then given by Cauchy's formula $f(w)=1/(2\pi i)\int _ {\partial D} f(z) /(z-w) dz$.

Answer by Andrea Ferretti for Continuous + holomorphic on a dense open => holomorphic?

2010-08-06T13:27:26Z

If the image of the curve has measure 0, it seems true. Indeed it is enough to prove that both the real and imaginary part are harmonic. This will be true if they satisfy the mean value identity on small balls.

The mean value identity is clear outside of $C$; on points of $C$ it follows by continuity and the fact that the integral of $f$ on $C$ with respect to the $2$-dimensional Lebesgue measure is $0$.

EDIT: I'm sorry, according to Mohan Ramachandran comment below, this answer is wrong.

Answer by Tom Boardman for Continuous + holomorphic on a dense open => holomorphic?

2010-08-06T14:49:53Z

We can go via Morrera's Theorem as well:

Let T be a triangle in the disc: partition the preimage of your curve $\cap T$ into n pieces (in a suitably 'going to be dense when $n \to \infty$' sort of way), now barycentrically subdivide T until we have a simplicial approximation of the curve. Since $f$ is continuous (and so has no singularities) and C has measure zero, as $n \to \infty$ the intergral around the subtriangles containing the curve must tend to $0$, the remainder are zero by Cauchy.

In steps Morrera, and the day is saved.

Answer by Mohan Ramachandran for Continuous + holomorphic on a dense open => holomorphic?

2010-08-07T15:51:54Z

If the curve C is rectifiable then the answer is yes.Under the assumption that C is rectifiable your question is known as Painleve's Theorem.It follows from the strong version of Cauchy's theorem which is stated as follows: If C is a simple closed rectifiable curve in the plane and f is holomorphic in the interior and continuous in the closed bdd region enclosed by C then the integral of f over C is zero.See for example the book by Behnke and Sommer page 119 (the book is in German).You can also find a proof of the strong form of Cauchy's theorem in the book titled:Elements of the topology of plane sets of points by M H A Newman,2nd edn page 187. If the jordan arc has positive area the answer to the question is no.See pages 122- 123 of the article in the amer math monthly vol 81 no 2 pages 115-137 year 1974.The paper is by Lawrence Zalcman who has other papers on this topic.

Answer by Zen Harper for Continuous + holomorphic on a dense open => holomorphic?

2010-08-08T13:31:38Z

Not an answer, but too big for a comment and useful for many similar problems.

From Chapter VI of Theory of the Integral by Stanislaw Saks, p.197, available online at http://banach.univ.gda.pl/pdf/saks/

For a domain $G$:

Theorem (Attributed to Besicovitch) Let $f$ be continuous on $G$, satisfying:

(i): $f'(z)$ exists for each $z \in G \setminus E_1$,

(ii): $$ \limsup_{h \to 0} \left| \frac{f(z+h) - f(z)}{h} \right| < \infty $$
for all $z \in G \setminus E_2$.

(iii): $E_1$ has Lebesgue measure zero.

(iv): $E_2$ is a countable union of finite-length curves.

Then: $f$ is holomorphic on $G$.

Answer by Piero D'Ancona for Continuous + holomorphic on a dense open => holomorphic?

2010-08-08T23:15:43Z

Denjoy makes a detailed study of this question, and in particular constructs counterexamples where the curve C is the graph of a continuous function. Apparently, the construction works for curves which are 'very' non rectifiable, i.e., the local variation is infinite of a suitably high order at each point.