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?

  • $\begingroup$ 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

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 $\Omega$.


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