1
$\begingroup$

Let $h$ be an harmonic function on $\Omega\subset\mathbb{C}$. Let $A\neq\emptyset$ an open subset of $\Omega$ such that $h\mid_A\equiv 0$. Prove $h\mid_\Omega\equiv 0$.

I thought on taking a point on the boundary of $A$ and apply the maximum principle for harmonic functions but on the other hand, we don't know neither that $A$ is bounded nor $h$ continuous on $\partial A$. How can we overcome it and apply the maximum principle getting $h\mid_\Omega$=0?

$\endgroup$
2
  • $\begingroup$ First show that $h$ and all of its derivatives vanishes in $A^{c}\cap\Omega$, where $A^c$ is the closure of $A$. Conclude that $h$ vanishes on an open neighborhood of $A^{c}\cap\Omega$. Use connectedness. $\endgroup$ – Disintegrating By Parts Jan 12 '16 at 1:38
  • $\begingroup$ $\Omega$ needs to be connected; please edit. $\endgroup$ – zhw. Jan 12 '16 at 4:59
0
$\begingroup$

Consider the function $f=\frac{\partial h}{\partial x}-i \frac{\partial h}{\partial y}$. It's easy to check, using Cauchy-Riemann equations, that f is holomorphic on $\Omega$ (and yes, $\Omega$ must be connected) and identically zero on $A$. So, by Identity Principle, we obtain that $f \equiv 0$ on $\Omega$.

$\endgroup$

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.