TITLE: About B. Ya Levin's proof that $|f(x)| \leq M$ implies $|f(x+iy)| \leq Me^{\sigma y}$
QUESTION [8 upvotes]: This question is about Theorems 1 through 3 on pages 37-38 of B. Ya Levin's Lectures on Entire Functions, available on Google Books.
If you can't access the Google Books link there is also a screenshot of the relevant portion of the book available here: http://i.stack.imgur.com/8f9kV.jpg
I'm trying to understand the proofs of these three theorems for the case where the hypotheses of Theorem 3 are satisfied.
In Theorem 3, $f$ is a function analytic in $\operatorname{Im} z \geq 0$ which is bounded on the real axis by a constant, $M$.  Further, there is a positive number $\sigma$ such that, for any $\epsilon > 0$, the bound
$$
|f(z)| < e^{(\sigma + \epsilon)|z|}
$$
holds for $|z|$ large enough with $\operatorname{Im} z > 0$.  The theorem states that, for such an $f$,
$$
|f(x+iy)| \leq M e^{\sigma y}
$$
for all $x+iy \in \mathbb C$ with $y \geq 0$.
To prove Theorem 3, the author recommends applying Theorem 2 with $\alpha = \pi/2$ and $\rho = 1$ to the rotated function $f(iz)$.  (Actually the author suggests using $f(-iz)$ instead, but I believe this is a typo.  This function is not analytic for $\operatorname{Re} z > 0$, which is a requirement of Theorem 2.)
Now in Theorem 2, the author defines
$$
D = \left\{z : |\arg z| < \alpha = \frac{\pi}{2\rho}\right\} = \{z:\operatorname{Re}z > 0\}
$$
and
$$
\varphi_\epsilon(z) = f(iz)e^{-(\sigma + \epsilon)z}.
$$
The function $\varphi_\epsilon$ is bounded by $M$ on the boundary of $D$, since
$$
|\varphi_\epsilon(ix)| = \left|f(-x) e^{i(\sigma + \epsilon)x}\right| \leq M.
$$
Further, it tends to zero along the positive real axis.  Indeed, for $|z|$ large enough we have $|f(z)| < e^{(\sigma + \epsilon/2)|z|}$, so for $x > 0$ large enough we see that
$$
|\varphi_\epsilon(x)| < e^{(\sigma+\epsilon/2)x}e^{-(\sigma+\epsilon)x} = e^{-x\epsilon/2}.
$$
Thus there is a constant $C_\epsilon$ such that
$$
|\varphi_\epsilon(x)| \leq C_\epsilon
$$
for all $x \geq 0$.  I believe this is what the author means by the first sentence of the proof,

The function $\varphi_\epsilon(z)$ is bounded on a positive ray and on the boundary of $D$.

I can understand the next line,

According to the previous theorem [Theorem 1], it is bounded by a constant in each angle $D_+ = \{z:0<\arg z<\pi/2\}$ and $D_- = \{z:-\pi/2<\arg z<0\}$.

Here the author seems to have applied Theorem 1 with $\rho = 1$ and $\lambda = 2$ to $\varphi_\epsilon(z)$ in each of the angles $D_+$ and $D_-$.  It seems that the $M$ in Theorem 1 was taken to be $\max\{M,C_\epsilon\}$.  This is then the constant bound being referred to.
I do not understand the next line,

Applying the previous theorem once more, we obtain $|\varphi_\epsilon(z)| \leq M$ for $z \in D$.

How was the previous theorem applied this time?  How does he conclude that $|\varphi_\epsilon(z)| \leq M$ when we apparently only have
$$
|\varphi_\epsilon(z)| \leq \max\{M,C_\epsilon\}?
$$

REPLY [3 votes]: The first application of Theorem 1 upgrades the upper bound on $M_\varphi$ from some power of $r$ to a constant. Now we can take $\rho $ to be $0$ (or arbitrarily close to $0$) in Theorem 1 and apply it to $\varphi$ on the whole angle $D$. Since $|\varphi|\le M$ on the sides of the angle, we get $|\varphi|\le M$ inside. 
On other words, the angle $D$ was too big for Theorem 1 to chew at once, so  it ate it in two bites, regurgitated, and rechewed.