Phragmen-Lindelof principle on a half-plane or strip

Prove Lindelöf's theorem: if $\left|f(z)\right| \leqslant C$ on the edges of the region $-a < x < a$, and inside this region $\left|f(z)\right| < e^{\lambda \left|y\right|}$, where $\lambda$ is some constant, then also inside the region $\left|f(z)\right| \leqslant C$, and moreover equality at an interior point is possible only when $f(z) \equiv C$.

The function $f(z)$ is assumed to be regular in the half-plane $\mathrm{Re}\left(z\right)\geq 0$ and to satisfy the following conditions:

1. there exist two constants $A$ and $B$, $A>0$, $B>0$ such that in the entire half-plane

   $$\left|f(z)\right| \leq Ae^{B\left|z\right|};$$

2. we have for $r\geq 0$

   $$\left|f(ir)\right| \leq 1, \quad \left|f(-ir)\right| \leq 1;$$
3. $$$$

\limsup _{r\to +\infty } \frac{\log \left|f(r)\right|}{r} \leq 0.

\left|f(z)\right| \leq 1 \quad \text{for} \quad \mathrm{Re}\left(z\right)\geq 0.

Let the function $f(z)$ be regular in the half-plane $\mathrm{Re}\left(z\right)\geq 0$ and satisfy the following conditions:

1. there exist two constants $A$ and $B$, $A>0$, $B>0$, such that in the entire half-plane

   $$\left|f(z)\right| < Ae^{B\left|z\right|};$$

2. $f(z)$ is bounded on the imaginary axis,

   $$\left|f(ir)\right| \leq 1, \quad \left|f(-ir)\right| \leq 1, \quad r\geq 0;$$

3. there exists an angle $\alpha$, $-\dfrac {\pi }{2}<\alpha <\dfrac {\pi }{2}$ such that

   $$\lim _{r\to +\infty } \frac{\log \left|f(re^{i\alpha })\right|}{r} = -\infty .$$

Such a function must vanish identically. [Examine the function $e^{\omega z}f(z)$, $\omega >0$.]

The function $f(z)$ is supposed to be regular in the half-plane $\mathrm{Re}\left(z\right)\geq 0$ and to satisfy the conditions:

1. there exist two constants $A$ and $B$, $A>0$, $B>0$, such that on the entire half-plane

   $$\left|f(z)\right| < Ae^{B\left|z\right|};$$

2. there exist two constants $C$ and $\gamma$, $C>0$, $\gamma >0$ such that for $r\geq 0$

   $$\left|f(\pm ir)\right| \leq Ce^{-\gamma r}.$$

A function satisfying these conditions must vanish identically. [Examine the function $f(z)e^{-\beta z\log (z+1)}$.]

Suppose that the function $f(z)$ is not a constant and that it is regular in the half-strip $\mathfrak {G}$ defined by the inequalities

If there exist two constants $A$ and $a$, $A>0$, $0<a<1$, such that in $\mathfrak {G}$

and if

on the boundary of $\mathfrak {G}$ (i.e.for $x=0$, $-\dfrac {\pi }{2}\leq y\leq \dfrac {\pi }{2}$ and for $x\geq 0$, $y=\pm \dfrac {\pi }{2}$) then $f(z)$ satisfies the strict inequality

in the interior of $\mathfrak {G}$. [The comparison function is of the type $e^{e^{bz}}$.]

Let $\omega (x)$ have the same properties as in problem 329. Every function $f(z)$ that is regular in the half-strip

must satisfy the inequality

at least at one point $z=x+iy$ of the half-strip.

Let $\Pi$ be the open right half plane. Suppose that $f \in H(\Pi )$, that $\left|f(z)\right| < 1$ for all $z \in \Pi$, and that there exists $\alpha ,-\pi / 2 < \alpha < \pi / 2$, such that

Prove that $f = 0$. Hint: Put $g_{n}(z) = f(z) e^{n z}, n = 1,2,3, \ldots$. Apply Exercise 9 to the two angular regions defined by $-\pi / 2 < \theta < \alpha , \alpha < \theta < \pi / 2$. Conclude that each $g_{n}$ is bounded in $\Pi$, and hence that $\left|g_{n}\right| < 1$ in $\Pi$, for all $n$.

Suppose $\Gamma$ is the boundary of an unbounded region $\Omega , f \in H(\Omega ), f$ is continuous on $\Omega \cup \Gamma$, and there are constants $B < \infty$ and $M < \infty$ such that $\left|f\right| \leq M$ on $\Gamma$ and $\left|f\right| \leq B$ in $\Omega$. Prove that we then actually have $\left|f\right| \leq M$ in $\Omega$.

Suggestion: Show that it involves no loss of generality to assume that $U \cap \Omega = \varnothing$. Fix $z_{0} \in \Omega$, let $n$ be a large integer, let $V$ be a large disc with center at 0, and apply the maximum modulus theorem to the function $f^{n}(z) / z$ in the component of $V \cap \Omega$ which contains $z_{0}$.