Understanding a version of the Phragmén-Lindelöf Principle

Question

I am reading Ransford's "Potential Theory in the Complex Plane", and I am having trouble understading the proof Theorem 2.3.5 p. 31. The proof is also to be found in these notes http://www.dm.unibo.it/~arcozzi/subharmonic.pdf that originate from Ransford's book (see thm. 9).

My problem is understanding why $v(z)=\cos(\beta x)\cosh(\beta y) >0$. Of course $\cosh$ is always positive, but why is $\cos(\beta x)>0$ on $S$? I don't see why it is not a problem that $\beta$ can be negative? Wouldn't that allow $\cos(\beta x)$ to be negative as well?

Answer 1

The theorem in question is:

Theorem 9 (Phragmén-Lindelöf in its original form.) Let $\gamma > 0$ and consider the strip $S = \{ z : |\operatorname{Re}| < \frac{\pi}{2 \gamma} \}$. Let $u$ be subharmonic in $S$ such that, for some $A > 0$ and $\alpha < \gamma$, $$ u(x+iy) \le A e^{\alpha |y|} \, . $$ If $\limsup_{z \to \zeta} u(z) \le 0$ for all $\zeta \ne \infty $ in $\partial S$, then $u \le 0 $ on $S$.

If $\alpha < 0$ then we can replace it with any $\alpha' \in (0, \gamma)$, this only weakens the condition: $$ u(x+ix) \le A e^{\alpha |y|} \le A e^{\alpha' |y|} \, . $$ (Also for $\alpha < 0$ it is easy to see that $u$ has boundary values $\le 0$ everywhere, so that $u \le 0$ follows immediately.)

Therefore we can assume without loss of generality that $\alpha \ge 0$, and proceed as in the given proof: If $\alpha < \beta < \gamma$ and $z = x+iy \in S$ then $$ |\beta x| = \beta |x| < \gamma \frac{\pi}{2 \gamma} = \frac{\pi}{2} \Longrightarrow \cos (\beta x) > 0 \, . $$