Let $f:\mathbb C\to \mathbb C$ be a continuous function and $A,B\subseteq \mathbb C$ two open connected sets with $\overline{A \cup B}=\mathbb C$. Further, we know that $f\mid_A$ and $f\mid_B$ are holomorphic. Does this also imply holomorphy of $f$? If not can you give precise conditions on when we get holomorphy?

I think in the case where $A\cap B\ne \emptyset$ we should get holomorphy.

I'm in particular interested in the case where $A$ is the upper half plane $\mathbb H$ and $B$ is the lower half plane.

  • 2
    $\begingroup$ For the half-planes, $f$ is holomorphic. That can easily be seen using Morera's theorem. The general case is non-obvious, since the sets can be pretty wild. $\endgroup$ May 1, 2015 at 21:25

1 Answer 1


In general, $f$ need not be holomorphic on the entire plane. Let $K\subset \mathbb{C}$ be a compact set with empty interior and positive Lebesgue measure, such that its complement is connected, e.g. a product of two fat Cantor sets. Define

$$f(z) = \int_K \frac{d\overline{\zeta}\wedge d\zeta}{\zeta-z}.$$

Then $f$ is holomorphic on $\mathbb{C}\setminus K$, and continuous on all of $\mathbb{C}$.

But, if $C$ is a circle containing $K$ in its interior, then we have

\begin{align} \int_C f(z)\,dz &= \int_C \int_K \frac{d\overline{\zeta}\wedge d\zeta}{\zeta-z}\,dz\tag{Fubini}\\ &= \int_K \left(\int_C \frac{dz}{\zeta-z}\right) d\overline{\zeta}\wedge d\zeta\\ &= \int_K -2\pi i\, d\overline{\zeta}\wedge d\zeta\\ &= 4\pi \lambda(K), \end{align}

where $\lambda$ is the Lebesgue measure. Thus $\int_C f(z)\,dz \neq 0$, and by Cauchy's integral theorem it follows that $f$ is not holomorphic on all of $\mathbb{C}$.

Now pick e.g. $A = B = \mathbb{C}\setminus K$, or $A,B$ such that $A\cup B = \mathbb{C}\setminus (K \cup L)$, where $L$ is a straight line, if you want $A$ and $B$ different, even disjoint.

In the case of particular interest, where $A$ is the upper half-plane, and $B$ the lower half-plane, however, it follows that $f$ is entire.

That is most conveniently seen by using Morera's theorem.

  • $\begingroup$ Thank you very much for your answer. I don't know much about distributions so I can't fully understand the construction of your counterexample. Which book do you recommend? But your example shows me that even in case where A and B are not disjoint evil things may happen. I somehow thought this case is not far apart from the case where we have isolated points, but it turns out to be. $\endgroup$ May 2, 2015 at 8:34
  • $\begingroup$ Now let's talk about the case with the two half-planes. Using Morera's theorem we have to show the line integral over a triangle is 0. In case the triangle is in one of the two half-planes this is clearly the case. Otherwise we can write the integral as sum of two line integral where we integrate over an interval of the real line in both directions. Now each of those integrals can be written as limit of integrals which are completely in the respective half-planes. Is that the approach you had in mind? $\endgroup$ May 2, 2015 at 8:35
  • $\begingroup$ It was late yesterday evening, I overlooked the simple way to show that the constructed $f$ is not entire, I'm going to replace that part in a moment with something not mentioning distributions. Concerning the other comment: Yes, that's precisely what I had in mind (the standard argument). $\endgroup$ May 2, 2015 at 9:05

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