6
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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 '13 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. $\endgroup$ – Digital Gal Apr 30 '13 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 '13 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.