Let $H\doteqdot\{z\in\mathbb{C}\colon \operatorname{Re}z>0\}$ denotes the half plane and $f\colon H\rightarrow\mathbb{C}$ be a holomorphic function, with the property that $$f(n)=0\qquad\forall n\in\mathbb{N}\,.$$
It is then clear that the function $$\frac{f(z)}{\sin(\pi z)}\qquad z\in H$$ is also a holomorphic function, taking into account that this function has removable singularity at the points which the denominator vanishes, which are precisely the zeros of $f$. It is then true, by an elementary Theorem that concerns subharmonic functions, that $$u(z)\doteqdot\log\left|\frac{f(z)}{\sin(\pi z)}\right|$$ is a subharmonic function defined on the half plane.
Assume now that $f$ has is of an exponential type $\gamma<\pi$, meaning that for some constant $C>0$, $$|f(z)|\leq C\cdot e^{\gamma|z|}\qquad\forall z\in H.$$ If $u$ is given by $$u(z)\doteqdot\log\left|\frac{f(z)}{2C\sin(\pi z)}\right|$$ then of course this function is still subharmonic, by the reasoning stated above.
We want to prove the following two inequalities.
- $u(z)\leq A+B|z|,$ where $A,B$ are constants (not necessarily positive).
- $\limsup_{z\to\zeta}u(z)\leq -(\pi-\gamma)|\zeta|$, where $\zeta\in\{z\in\mathbb{C}\colon \operatorname{Re}\zeta=0\}$, in particular $\zeta$ lies on the boundary of $H$ and it is not $\{\infty\}$.
Both inequalities are not clear why they hold. This is exercise 2.3.6 of Ransford Potential Theory of the Complex Plane, so they should be true. My main problem lies to proving an upper bound for the sin function in (i) and I expect similar reasoning will also yield the second inequality.
Appreciate any help or comments in this exercise.