Suppose $u$ is a real valued continuous function on $\overline{\mathbb D}$, harmonic on ${\mathbb D}$\ $\{0\}$ and $u=0$ on $\partial\mathbb D$, show $\mathbb u$ is constant in $\mathbb D$.

I'm going through old exams and I spent quite some time on this one without success. I think of using log$|z|$ first as an example. But it fails since $u$ is continuous on $\overline{\mathbb D}$. If we are working with holomorphic functions $f$: $\mathbb{C} \to \mathbb{C}$, $0$ is a removable singularity and everything works out. But I don't remember any similar technic to work with harmonic functions. Any help will be appreciated!

  • 3
    $\begingroup$ The conclusion here is helpful. $\endgroup$
    – 23rd
    May 19, 2013 at 4:13

1 Answer 1


Hint: use the maximum principle for harmonic functions.

For the details, it depends a little on what tools you have available. Let me assume that you know the following theorem about removable singularities for subharmonic functions:

Theorem Assume that $\Omega$ is a domain in $\mathbb{C}$ and that $E \subset \Omega$ is polar. If $u$ is an upper bounded subharmonic function on $\Omega \setminus E$, then $u$ extends (uniquely) to a subharmonic function on $\Omega$.

From the theorem, it follows that $u$ is (can be extended to be) subharmonic on $\mathbb{D}$, and so can $-u$. Hence $u$ is in fact harmonic on $\mathbb{D}$, so by the maximum principle $u = 0$ on $\mathbb{D}$.

Just in case, here is a proof of the removable singularities theorem:

Let $\phi$ be a negative subharmonic function such that $\phi = -\infty$ on $E$. (In your particular case, you can take $\phi(z) = \log|z|$.) Let $u_\epsilon = \max \{ u, \epsilon \phi \}$. Then $u_\epsilon$ is subharmonic for every $\epsilon$, since $u_\epsilon = \phi$ near $E$. Furthermore $u_\epsilon$ increases to $u$ on $\Omega \setminus E$ as $\epsilon \to 0^+$, and the function $$v(z) = \limsup_{\zeta \to z} \left(\sup_{\epsilon > 0} u_\epsilon(z)\right)$$ is subharmonic on $\Omega$. By continuity of $u$, $v(z) = u(z)$ on $\Omega \setminus E$, and we have our subharmonic extension of $u$. (In fact, you can check that $v(z) = \limsup_{\zeta \to z} u(\zeta)$ for $z \in E$.)

  • $\begingroup$ Sorry for the late reply. I know eventually I will need to use the maximum principle, but I'm still stuck on u(0). Could you explain more? Thank you very much! $\endgroup$
    – BigTree
    May 19, 2013 at 23:42
  • $\begingroup$ @BigTree Better? $\endgroup$
    – mrf
    May 20, 2013 at 5:49
  • $\begingroup$ I have never seen this theorem, and it seems to be a very useful tool to have. Thank you very much! $\endgroup$
    – BigTree
    May 20, 2013 at 13:09

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