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Let $\Omega$ be a domain in $\mathbb C$, and let $\mathscr X$ be some class of functions from $\Omega$ to $\mathbb C$. A set $E\subset \Omega$ is called removable for holomorphic functions of class $\mathscr X$ if the following holds: every function $f\in\mathscr X$ that is holomorphic on $\Omega\setminus E$ is actually holomorphic on $\Omega$ (possibly, after being redefined on $E$).

It is clear that the larger $\mathscr X$ is, the smaller is the class of removable sets. In the extreme case, if $\mathscr X$ contains all functions $\Omega\to\mathbb C$, there are no nonempty removable sets. Indeed, if $a\in E$, then $f(z)=\frac{1}{z-a}$ (arbitrarily defined at $z=a$) is holomorphic on $\Omega\setminus E$ but has no holomorphic extension to $\Omega$.

The problem of describing removable sets is nontrivial in many classes $\mathscr X$ such as

  • $L^{\infty}(\Omega)$, bounded functions
  • $C(\Omega)$, continuous functions
  • $C^{\alpha}(\Omega)$, Hölder continuous functions
  • $\mathrm{Lip}(\Omega)$, Lipschitz functions

Which sets are removable for holomorphic functions in these classes?

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1 Answer

The case $\mathscr X=C^{\alpha}(\Omega)$, $0\le \alpha<1$, was settled by E. P. Dolzhenko in 1963: a compact set $E$ is removable if and only if its $(1+\alpha)$-dimensional Hausdorff measure is zero.

As often happens, the integer Hölder exponent turned out to be harder than fractional. It waited under 1979, when Nguyen Xuan Uy proved that a compact set $E$ is removable for holomorphic functions of class $\mathrm{Lip}(\Omega)$ if and only if $E$ has 2-dimensional measure zero.

The two remaining cases, $L^\infty$ and $C$, are much more involved. The removable sets for these classes admit no simple characterization in terms of Hausdorff measures. They are studied via the concepts of analytic capacity $\gamma$ and continuous analytic capacity $\alpha$ (the latter is defined similarly to $\gamma$ but using $C$ instead of $L^{\infty}$.) Removability is easily shown to be equivalent to the vanishing of the appropriate capacity. One then proceeds to investigate necessary and sufficient conditions for the vanishing of analytic capacity, its invariance under classes of maps, continuity under nested unions/intersections, etc... The webpage of Xavier Tolsa has a wealth of information on this topic.

Here is a proof of an elementary result: If $L$ is a line, then $L\cap \Omega$ is removable for $\mathscr X=C(\Omega)$. Let $f\in C(\Omega)$ be holomorphic on $\Omega\setminus L$. By Morera's theorem it suffices to prove that the integral of $f$ along the boundary of every triangle $T\subset \Omega$ is zero. Consider three cases:

  • $T$ does not meet $L$. This case is clear.
  • $T$ has a side that lies on $L$. Let $T_n=T+\delta_n$ where $\delta_n$ are complex numbers such that $\delta_n\to 0$ and $T_n\cap L=\varnothing$. In other words, we translate $T$ by a small amount to move it off the line $L$. Using the uniform continuity of $f$ on compact subsets of $\Omega$, we find that $$\int_{\partial T}f(z)\,dz=\lim_{n\to\infty}\int_{\partial T_n}f(z)\,dz=0$$
  • $T$ has points on both sides of $L$. Then $L$ divides $T$ into a triangle and another polygon (triangle or quadrangle). The quadrangle can be further divided into two triangles. The integral of $f$ over the boundary of each resulting triangle is $0$ by the above. Add these integrals together to obtain $\int_{\partial T}f(z)\,dz=0$. QED

To complement the above result, I'll add a simple example: a line segment is not removable for $\mathscr X=L^{\infty}(\mathbb C)$. Indeed, the function $f=z+z^{-1}$ maps the unit disk $\mathbb D$ bijectively onto $\mathbb C\setminus [-2,2]$. The inverse $g=f^{-1}$ is holomorphic in $\mathbb C\setminus [-2,2]$ and is bounded by $1$, but has no holomorphic extension to $\mathbb C$. (For one thing, $g(z)$ approaches both $i$ and $-i$ as $z\to 0$. For another, a holomorphic extension would have to be constant by Liouville's theorem.)

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