There is a theorem that states that if a function $h$ is harmonic on a simply connected domain, there exists a holomorphic function $f$ such that $h = Re f$.

Now, I am having a problem with the statement of this exercise:

Let $h$ be a function harmonic on $\{z\in\mathbb{C}: \rho_1 < |z| < \rho_2\}$. Using the fact that $h_x - ih_y$ is holomorphic, prove that...

What we need to prove is not important, I've already done that. The question is about the fact mentioned in the statement: that $h_x - ih_y$ is holomorphic.

The domain is not simply connected. Normally, we need a simply connected domain to prove that $h_x - ih_y$ is holomorphic, because we use the path-independence of an integral of the form $h(z_0) + \int_{z_0}^z (h_x - ih_y)(w)dw$.

Is there another proof that $h_x - ih_y$ is holomorphic in a non-simply connected domain? Or did I misread the statement and this is just an additional condition on our function $h$?

  • $\begingroup$ Remember that Morera's Theorem is local. As long as you have a continuous function, you're set to go. $\endgroup$ May 4 '15 at 13:26
  • $\begingroup$ That one, I don't understand. If my function $h$ is such that $g(z) = h_x - ih_y$ has a pole in zero, then to solve $f'(z) = g(z)$, I will need to take an integral over some path from $z_0$ to $z$, but it will be path-dependent... $\endgroup$ May 4 '15 at 13:44
  • $\begingroup$ To show $g$ is holomorphic, you only need a local primitive $f$. You're right that globally $f$ may have periods, but that doesn't stop $f'=g$ from being holomorphic. Try the example of $h(z)=\log |z|$. Then we get, indeed, $g(z)=1/z$. This is holomorphic on the annulus, even though it does not have a global primitive. $\endgroup$ May 4 '15 at 14:13
  • $\begingroup$ Thanks, Ted! I think I understand why the exercise statement is correct. $\endgroup$ May 4 '15 at 14:24

No need for simple connectedness here. The function $$ h'_x - ih'_y $$ (is $C^1$ and) satisfies Cauchy-Riemann's equations.

With $u = h'_x$ and $v = -h'_y$, we have \begin{align} u'_x &= h''_{xx} & v'_y &= -h''_{yy} \\ u'_y &= h''_{xy} & v'_x &= -h''_{yx}. \end{align} Hence $u'_x = v'_y$ (since $h$ is harmonic) and $u'_y = -v'_x$ (since $h$ is $C^2$ and the mixed partials are the same).

  • $\begingroup$ So, basically, we don't need the simply connectedness for the fact that $g = h_x - ih_y$ is holomorphic, right? And if I want to get $f'(z) = g(z) = h_x(z) - ih_y(z)$, then I need that path integral to define $f(z)$, so I can't use it in my proof, I guess? $\endgroup$ May 4 '15 at 13:31
  • $\begingroup$ Right, $g=h'_x-ih'_y$ is holomorphic, but doesn't necessarily have an anti-derivative. $\endgroup$
    – mrf
    May 4 '15 at 15:11

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.