If $\Omega$ is a bounded open set and $u$ is harmonic function in $\Omega$ with $u=0$ on the boundary $\partial\Omega$

Question

If $\Omega$ is a bounded open set and $u$ is harmonic function in $\Omega$ with $u=0$ on the boundary $\partial\Omega$, then $u=0$ inside $\Omega$ as well. Show that boundedness is required for this theorem to be true by using counterexample on the upper half-plane.

I can't think of a nice funtion that will do the job.

Answer 1

How about $u(x,y)=y$? Then $u$ is harmonic (obviously) and $u(x,0)=0$, so $u$ is zero on the boundary of the upper half-plane (i.e. the line $y=0$).

Answer 2

The theorem follows from the maximum principle.

For the counterexample, consider a degree one polynomial.