Holomorphic versus harmonic functions Is it true that any holomorphic function on domain $D$ is of the form $u_x-iu_y$ for some harmonic function $u$?
Motivation for this question is the problem of existence of harmonic conjugates.
 A: I'm going to rephrase everything in terms of forms. Given a holomorphic function $f$, rewrite it as $u-iv$, where $u$ and $v$ are real. Suppose you had a function $w$ ($u$ in your original question) with $w_x = u$, $w_y = v$. Then $w$ is automatically harmonic by the Cauchy-Riemann equations. Now, the existence of such a $w$ is exactly the same statement as that $udx+vdy$ is an exact 1-form - it's automatically closed by the Cauchy-Riemann equations again.
So this is now a question of simple connectedness. If your domain is simply connected, every closed form is exact, as desired. If it's not, I can find some closed form that's not exact (because for domains in the plane, $\pi_1 \neq 0$ means that the de Rham cohomology $H^1(D;\Bbb R) \neq 0$). Now a harmonic 1-form is one such that $d\omega = d*\omega = 0$, where $*$ is the Hodge star; to phrase that in terms of $\omega = udx+vdy$, we say precisely the Cauchy-Riemann equations for $u-iv$: $v_y - u_x = 0$ (the condition that $d\omega = 0$) and $v_x + u_y = 0$ (the condition that $d*\omega = 0$).
Now, lastly, what we need to know is that every cohomology class has a harmonic representative. I have not checked this carefully, but I believe this still follows from Hodge theory and the elliptic operator theory; what you lose when you don't assume compactness is uniqueness of harmonic representative. So the answer to your question is "Yes, when $\Omega$ is simply connected; and no, when $\Omega$ is not."
