Let $\Omega = R \times(-1,1) \subset R^2$. Assume $u:\Omega \rightarrow R$ Is harmonic and continuous on $\overline{\Omega}$ and u is periodic with period 2.
If u = 0 on boundary, I want to prove that u=0 on $\overline{\Omega}$.
My attempt: I am trying to prove with strong maximum principle. Since it needs to be on the bounded domain, I restrict the domain to be $(0,2)\times (-1,1)$
Prove by contradiction,
Assume that there exists a max in the interior, then by the mean value theorem of harmonic function, the center equals the average of boundary, which is 0. Therefore, the max is on the boundary and equals 0. However, I am not sure how to continue and prove that u is constant on the whole domain.