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 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.

What about the case $p < \infty$? Any reference would be quite useful.

Thank you, Malik

  • $\begingroup$ Are you looking for an analogue of Analytic Capacity? I think that Guy David's proof of the Vitushkin Conjecture should answer your question. Also, a good book to check out is "Analytic Capacity, Rectifiability, Menger Curvature and the Cauchy Integral." by Hervé Pajot. $\endgroup$ Apr 15, 2011 at 17:43
  • $\begingroup$ Analytic Capacity would be the case $p = \infty$ of my question, which I already know the answer. Thanks for the reference though! $\endgroup$ Apr 19, 2011 at 19:44

1 Answer 1


Yes, it is true that $H^p(\Omega)$ consists only of the constant functions for all $1\le p\le\infty$. This is because all nonnegative harmonic functions $f\colon\Omega\to\mathbb{R}$ are constant.

You can prove this using the properties of Brownian motion. If $\Omega$ is a connected open subset of the plane and $B_t$ is a two dimensional Brownian motion with $B_0\in\Omega$ then either (i) $B_t\in\Omega$ for all times $t$, with probability one, or (ii) $B_t\in\mathbb{R}^2\setminus\Omega$ for some time $t$, again with probability one. Also, which case holds depends on the domain $\Omega$ but not on the initial distribution of $B$. If case (ii) holds then $\Omega$ is called a Greenian domain. As I'll show below, the $\Omega$ you are considering is not a Greenian domain, which implies that all nonnegative harmonic functions on $\Omega$ are constant (so $H^p(\Omega)$ consists of the constant functions). For this terminology, I'm going from Foundations of Modern Probability, Second Edition, Olav Kallenberg, Chapter 24.

The idea is that a twice continuously differentiable function $f\colon\Omega\to\mathbb{R}$ is harmonic if and only if $f(B_t)$ is a local martingale when run up until the first time at which $B$ exits $\Omega$. This follows from Ito's formula $$ f(B_t)=f(B_0)+\sum_{i=1}^2\int_0^tf_{,i}(B_s)\\,dB^i_s+\frac12\int_0^t\Delta f(B_s)\\,ds $$

The first two terms on the right hand side are local martingales, and the third term will be a local martingale if and only if $\Delta f=0$, so that $f$ is harmonic. Now, if $\Omega$ is not a Greenian domain and $f$ is a nonnegative harmonic function then $f(B_t)$ is defined at all times and is a nonnegative local martingale. By martingale convergence (see my blog post on martingale convergence, or any textbook on martingale theory), this implies that $f(B_t)$ converges to a limit as $t\to\infty$ with probability one. However, two dimensional Brownian motion is recurrent, so that it keeps getting arbitrarily close to any given point. Then, $f(B_t)$ can only converge for continuous $f$ if $f$ is constant.

We can prove the following theorem so that, by your observation that bounded holomorphic functions on $\Omega$ are constant (similar reasoning shows that bounded harmonic functions on $\Omega$ are constant), it follows that $\Omega$ is not a Greenian domain and that all nonnegative harmonic functions on $\Omega$ are constant.

Theorem: If $\Omega$ is a connected open subset of $\mathbb{R}^2$ then the following are equivalent.

  1. Every bounded harmonic function $f\colon\Omega\to\mathbb{R}$ is constant.
  2. Every nonnegative harmonic function $f\colon\Omega\to\mathbb{R}$ is constant.
  3. Every harmonic function $f\colon\Omega\to\mathbb{R}$ which is bounded below is also constant.
  4. $\Omega$ is not a Greenian domain.

As argued above, statement 4 imples 2. Also, statements 2 and 3 are equivalent by adding a constant to $f$ to make it nonnegative. Also, 3 clearly implies 1. So, it just remains to show that 1 implies 4. I'll prove the contrapositive, so supposing that $\Omega$ is Greenian, we just need to construct a non-constant bounded harmonic function on $\Omega$.

Let $\tau$ be the first time at which $B_t$ exits $\Omega$ which, as we are assuming it is Greenian, is finite with probability one. Also, for each $x\in\Omega$, let $\mathbb{P}\_x$ be the probability measure for the Brownian motion started at $B_0=x$, and $\mathbb{E}\_x$ denotes expectation with respect to $\mathbb{P}\_x$. Then, for any bounded measurable function $g\colon\mathbb{R}^2\setminus\Omega\to\mathbb{R}$ we can define $$ \begin{align} &f\colon\Omega\to\mathbb{R},\\\\ &f(x)=\mathbb{E}\_x\left[g(B_\tau)\right]. \end{align} $$ Then $f$ is a bounded smooth function satisfying $f(B_t)=\mathbb{E}[g(B_\tau)\mid\mathcal{F}_t]$ (whenever $t < \tau$). So, $f(B_t)$ is a martingale run up until time $\tau$. It follows that $f$ is harmonic. Furthermore, by martingale convergence, $f(B_t)\to g(B_\tau)$ as $t\uparrow\uparrow\tau$ (with probability one). To ensure that $f$ is non-constant, we just have to make sure that $g(B_\tau)$ is not equal to a constant with probability one. However, $B_\tau$ cannot be almost surely constant (because two dimensional Brownian motion has zero probability of hitting any given fixed point). So, $B^i_\tau$ is not almost-surely constant for at least one of $i\in\{1,2\}$ and we can take $g(x)=1_{\{\vert x\vert\le K\}}x_i$ for large enough $K$, making $f$ a bounded non-constant harmonic function on $\Omega$ and contradicting property 1 above.

Update: I just want to add some comments on the use of Brownian motion in this answer. We have a domain $\Omega$ on which we know that all bounded harmonic functions are constant, and want to show that all nonnegative harmonic functions are constant. Or, looking at the contrapositive, given a nonnegative and non-constant harmonic function $f\colon\Omega\to\mathbb{R}$, we want to construct a non-constant bounded harmonic function $g\colon\Omega\to\mathbb{R}$. One way you might think of doing this is to construct $g$ directly from $f$. Say, by capping $f$ at some positive value $K$. You could try looking at $\min(f,K)$ and then choose $g$ to be the maximal harmonic function with $g\le\min(f,K)$. However, this does not work. There can exist positive harmonic functions $f$ on a domain for which every bounded harmonic function $g\le f$ is non-positive everywhere (this is related to the existence of positive martingales which tend to zero). So, this approach would give you $g\equiv0$, which is not very helpful. For example, let $S\subseteq\mathbb{C}$ consist of the negative real axis $S=\{x\in\mathbb{R}\colon x\le0\}$ and $\Omega=\mathbb{C}\setminus S$. Then, $f(z)=\Re[\sqrt{z}]$ defines a nonnegative harmonic function on $\Omega$ (taking the square root with positive real part). However, $f$ vanishes on $S$, so any bounded harmonic function $g\le f$ must be nonnegative on $S$. This implies that $g\le0$ everywhere. So, constructing $g$ directly from $f$ in this way is tricky (maybe there is some way around this). Instead, the approach I took was to use the existence of a nonnegative harmonic function to imply some property of the domain, which then allows us to construct bounded harmonic functions. Taking the expected value of a function of a Brownian motion when it first exits the domain provides one way of doing this, so it is natural to try to show that Brownian motion must exit the domain at some time. In the example just mentioned, there do exist bounded harmonic functions -- e.g., $g(z)=\Re[(1+\sqrt{z})^{-1}]$, which tends to $1/(1-x)$ for $x\in S$, so $g(z)=\mathbb{E}\_z[1/(1-B_\tau)]$.

Also, the result I stated in the answer relating the existence of nonnegative harmonic functions to the existence of bounded harmonic functions is intimately related to the behaviour of 2 dimensional Brownian motion. In particular, it fails in $n > 2$ dimensions, for which Brownian motion is not recurrent. Consider, for example, $\Omega=\mathbb{R}^n\setminus\{0\}$. If $f$ is a nonnegative harmonic function on $\Omega$ and $B$ is a Brownian motion in $\mathbb{R}^n$ then it is true that $f(B_t)$ will converge as $t\to\infty$. However, as $B$ is not recurrent ($\Vert B_t\Vert\to\infty$) in more than 2 dimensions, this does not allow us to conclude that $f$ is constant. In fact, there do exist non-constant and positive harmonic functions on $\Omega$, such as $f(x)=\Vert x\Vert^{2-n}$. However, all bounded harmonic functions are constant. You can prove this by extending any such bounded harmonic $g$ to all of $\mathbb{R}^n$ and applying Liouville's theorem. Or martingale convergence can be used to show that $g(B_t)=\mathbb{E}[X\mid\mathcal{F}\_t]$, where $X=\lim\_{t\to\infty}g(B_t)$ is in the tail $\sigma$-algebra of $B$ which, by Kolmogorov's zero-one law, is almost surely constant. In any case, the result I stated in the answer fails in $n > 2$ dimensions in which Brownian motion is not recurrent.

  • $\begingroup$ Thank you very much for your answer! Your approach to this kind of problem using Brownian motion is very interesting, to say the least. +1! $\endgroup$ Apr 19, 2011 at 19:43
  • $\begingroup$ @George: Nice answer. Are you a stochastic analyst? $\endgroup$
    – JT_NL
    May 8, 2011 at 22:09
  • $\begingroup$ @Jonas: Well, I'm not in academia. I have a PhD in which I specialized in stochastic processes and have a blog on which I've been posting stochastic calculus notes, but am not a professional mathematician. $\endgroup$ May 10, 2011 at 19:30
  • $\begingroup$ maybe you can anso help out for this other question too? mathoverflow.net/questions/62347/… I would be curious about that answer too. But I'm not sure that he uses the same notation. $\endgroup$
    – Mircea
    Jan 18, 2012 at 16:52

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