I have a function $F(x,y)$ which is continuous and analytic on the complement of a certain function $x(y)$. Is it possible to use Morera's theorem to show that it is analytic everywhere? Clearly, this approach can be used if it fails to be analytic on some linear function $x = a y + b$, for example (as far as I can guess, we could complexify $z = x + iy$ and use Morera's theorem by approximating the integrals over circles passing through the line until we showed analyticity everywhere). If $x(y)$ is also continuous (at least on a certain range of interest), could we use a similar method? If I'm wrong, can anyone think of convincing counterexamples?!

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    $\begingroup$ Your terminology is confusing. $F$ is a function on $\mathbb{R}^2$; do you mean to have it be a function on $\mathbb{C}$? What do you mean by the complement of the function $x(y)$? $\endgroup$
    – robjohn
    Apr 9, 2013 at 8:54

1 Answer 1


If I understand correctly, "the complement of a function $x(y)$" means the complement of its graph, that is, the set $\{x+iy:x\ne x(y)\}$. You assume that $F$ is continuous in some domain $\Omega$ (possibly $\mathbb C$) and is holomorphic (=analytic) on $\Omega\setminus \Gamma$, where $\Gamma$ is the graph of some continuous function. The question is: does it follow that $F$ is holomorphic in $\Omega$?

(Stated more succinctly: is $\Gamma$ removable for continuous holomorphic functions?)

In general, the answer is no. Here is why.

Let $x(y)$ be a continuous function on a closed interval whose graph $\Gamma$ has Hausdorff dimension $D>1$. Such functions exist: for example, see reference 6 in the Wikipedia article on the Weierstrass function. Pick $1<d<D$. By Frostman's lemma, there exists a finite positive measure $\mu$ with support contained in $\Gamma$ and with the growth bound $\mu(B(z,r))\le r^{d}$ for all $z\in \mathbb C$ and $r>0$. Define $$F(z)=\int_{\mathbb C}\frac{1}{z-\zeta}d\mu(\zeta)$$ By construction, the function $F$ is holomorphic on $\mathbb C\setminus \Gamma$. It also tends to zero as $z\to\infty$. Since $F$ is not identically zero, it cannot be extended to a holomorphic function on $\mathbb C$; such an extension would contradict Liouville's theorem.

It remains to show that $F$ is continuous at every point $p\in \Gamma$. Pick a small $\delta>0$ and write $$F(z)=\int_{B(p,\delta)}\frac{1}{z-\zeta}d\mu(\zeta)+\int_{B(p,\delta)^c}\frac{1}{z-\zeta}d\mu(\zeta)=:F_1(z)+F_2(z) $$ The function $F_2$ is holomorphic in $B(p,\delta)$, and therefore continuous there. For $z\in B(p,\delta)$ the first term can be estimated from above by switching to polar coordinates and integrating by parts: $$ |F_1(z)|\le \int_{B(p,\delta)}\frac{1}{|z-\zeta|}d\mu(\zeta) = \int_0^\infty \mu(B(z,\rho)\cap B(p,\delta))\,\frac{d\rho}{\rho^2} \\ \le \int_0^\infty \min(\rho,\delta)^d \,\frac{d\rho}{\rho^2} = \int_0^\delta \rho^d \,\frac{d\rho}{\rho^2}+\int_\delta^\infty \delta^d \,\frac{d\rho}{\rho^2}=C\delta^{d-1} $$ Since $\delta^{d-1}\to 0$ as $\delta\to 0$, the function $F_1$ is continuous at $p$. In fact, the estimate shows that it is Hölder continuous with exponent $d-1$. This relation between Hölder exponent and Hausdorff dimension is not accidental: see this thread and references quoted there.

However, the answer is yes if $x(y)$ is a Lipschitz function. That is, Lipschitz graphs are removable for continuous holomorphic functions. More generally, curves of finite length are removable for continuous holomorphic functions: this a theorem of Painlevé. See Analytic capacity and measure by Garnett (Springer LNM series, 297).


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