Suppose $a,b>0$ are contants and $F$ is a non-constant function such that $F(z+a)=F(z)$ and $F(z+ib)=F(z)$. Prove $F$ is not analytic in the rectangle $0\leq a \leq b$ and $0\leq y \leq b$
I don't see how I'd apply Liouville (requirements are bounded+analytic).
I know bounded+analytic imply constant, but does non-constant imply unbounded+non-analytic?
Since there is still some confusion after the comment of @carmichael561. Suppose $F$ is analytic for all $z=x+iy$ with $0\leq x \leq a$ and $0\leq y \leq b$. Since $F$ is analytic it is also continous so $\vert F(z) \vert$ will obtain it's maximum on $z_0$ in this rectangle. Say $\vert F(z_0) \vert=M$. Now for arbitrary $z \in \mathbb{C}$ you find $m,n \in \mathbb{Z}$, s.t. $z=x+iy=x'+an+(y'+mb)i$ with $0\leq x'\leq a$ and $0\leq y'\leq b$.
Now by the functional equation that is satisfied by $F$ you get
$$ \vert F(x+iy)\vert=\vert F(x'+iy')\vert\leq M $$
so $F$ is bounded. Now use Liouville.