The following is an old qual problem I came across.

If $h$ is harmonic on $D-\{0\}$, where $D$ is the unit disk, show that

$h(z) = \Re(f(z)) + c \log|z|$ for where $f$ is analytic on $D- \{0\}$.

This is obvious (with $c=0$) if $h$ extends to a harmonic function on $D$ but I don't how to treat the case where $h$ is not extendable.


1 Answer 1


You know that a harmonic function has a conjugate in a simply connected domain. But $D\setminus \{0\}$ is not simply connected. Solution: use its universal cover, which is conveniently realized as the left half-plane $H$. Namely, $\exp:H\to D\setminus\{0\}$ is a holomorphic covering map.

The composition $u=h\circ \exp$ is a harmonic $2\pi i $-periodic function in $H$. Since $H$ is simply-connected, has a harmonic conjugate $v$. If $v$ is also $2\pi i$-periodic, we can grab $u+iv$ and jump right back to the disk. But in general it's not. In general, from $u(z)=u(z+2\pi i)$ we get $\nabla v(z)=\nabla v(z+2\pi i)$ which implies that $v(z+2\pi i)-v(z)$ is a constant. Let $c$ be this constant.

Define $g(z) = u(z)+iv(z)-\frac{c}{2\pi }z$; this is constructed so that $g$ is $2\pi i$-periodic. Therefore, $g\circ \log $ is a well-defined holomorphic function in $D\setminus\{0\}$. This is $f$ that you are looking for. The difference between $\mathrm{Re}\,f$ and $h$ comes from $\mathrm{Re}\,\left(\frac{c}{2\pi }\log z\right)$, and this is the multiple of $\log|z|$ that you see in the statement of the problem.

  • $\begingroup$ @user75064 can you please explain why we do ask periodicity from $g$? Also, which function is $\overline v$ and how do you derive that $Re(f)-h=Re(\frac{c}{2\pi}\log z)$? Thank you! $\endgroup$
    – Dimitris
    Apr 30, 2013 at 8:54
  • $\begingroup$ I believe periodicity is needed to ensure that $g$ composed with $\log$ is well defined regardless of the branch of $\log$ choosen. $\endgroup$
    – Mykie
    Apr 30, 2013 at 23:34
  • 1
    $\begingroup$ @DimitrisDallas (a) what jennifer said; (b) $\tilde v$ should not have been there, I meant $g\circ \log$. Corrected now. And (c), observe that the real part of $g$ is $u-(c/2\pi)\mathrm{Re}z$. Composing this with $\log$, you get $h-(c/2\pi)\mathrm{Re}\log z$. $\endgroup$
    – 75064
    Apr 30, 2013 at 23:58

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