Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p<\infty$, as the class of functions $f$ that are holomorphic on $\Omega$ such that $|f|^p$ has a harmonic majorant on $\Omega$, i.e. there is a function $u$ harmonic on $\Omega$ such that $$|f(z)|^p \leq u(z) $$ for all $z \in \Omega$.

For $p=\infty$, $H^\infty(\Omega)$ is the class of bounded holomorphic functions on $\Omega$.

I'm interested in cases when $H^p(\Omega)$ consist only of constant functions. For example, this is the case when $\Omega$ is the whole plane, because positive harmonic functions on $\mathbb{C}$ are constant.

I came upon the following question :

Let $E$ be a compact subset of the real line, and suppose that $E$ has zero length. Let $\Omega$ be the complement of $E$. Does $H^p(\Omega)$ consist only of the constant functions?

For $p=\infty$, the answer is yes : one can use Cauchy's formula to extend any bounded holomorphic function on $\Omega$ to a bounded holomorphic function on $\mathbb{C}$, and that function is now constant by Liouville's theorem.

For $1 \leq p<\infty$, I am pretty sure the answer is also yes. However, I can't seem to find a way to extend $f$ or the harmonic majorant of $|f|^p$ to the whole plane. Is there any way to do so?

Thank you, Malik

  • 1
    $\begingroup$ For the benefit of other readers, I'll point out the answer given at MathOverflow. It would be interesting to have a non-probabilistic proof, though. $\endgroup$
    – user31373
    Jul 2, 2012 at 16:57

1 Answer 1


An answer by Georges Lowther has been given at Math Overflow.


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