Complex functions bounded in the upper half plane Given a complex function $G(z)$, $z=x+iy, x,y\in \mathbb R$ which is analytic and bounded in the upper half-plane, i.e.,
$$|G(z)|\leq C, \forall z\in \mathbb C^{+}$$
does it imply that $G(x)$ is bounded on the real line (if we know that $G(x)$ has no poles on $\mathbb R$), i.e., 
$$|G(x)|\leq C_{o}, \forall x\in \mathbb R$$
EDIT: $\mathbb C^{+}=\{z\in \mathbb C, \text{Im}(z)>0\}$.
 A: THE ANSWER IS NO.
Page numbers are from Boundary Behavior of Conformal Maps by Christian Pommerenke
The Pommerenke book concentrates on the unit disk, so here is some terminology for the theorem:
(page 1): he writes $\mathbb D$ for the open unit disk, then $\bar{\mathbb D}$ for the closed unit disk, finally $\mathbb T$ for the unit circle. 
(page 30): A bounded simply connected domain $G$ in the complex plane comes equipped with a set called its prime ends, the set denoted $P(G).$ 
(page 32): For a prime end $p \in P(G),$ there is a set $I(p) \subseteq \mathbb C$ called the impression of $p.$ The impression of a prime end is a non-empty compact connected set. 
(page 30): Theorem 2.15. 
Let f map $\mathbb D$ conformally onto the bounded simply connected domain $G,$ by the Riemann Mapping Theorem. There is an induced  bijection $\hat{f}$ with
$$ \hat{f} : \mathbb T \rightarrow P(G).   $$
(page 35): Corollary 2.17(ii) with $\zeta \in \mathbb T,$ the mapping $f$ extends continuously to $\zeta,$ with limit $a,$ if and only if 
$$  I( \hat{f}(\zeta)) = \{a\},    $$ the singleton set consisting of the point $a.$
(page 35): Exercise 2.5.2. Let $$ G = \mathbb D \backslash \{ (1 - e^{-t})e^{it} : 0 \leq t < \infty      \}.   $$
Then $G$ has a prime end $p$ with $I(p) = \mathbb T.$ That is, $I(p)$ is not a single point. 
Thus, there is some point $\zeta \in \mathbb T,$ with $\zeta = \hat{f}^{-1}(p), $ at which $f$ does not extend continuously. Put another way, $f(\zeta)$ cannot be defined in a  way that makes $f$ continuous at $\zeta.$
To return to the upper half plane, let us  call it $\mathbb U.$ Take a Möbius transformation $m$ that maps the real line to $\mathbb T$ minus one point of course, does NOT map $\infty$ to the point $\zeta$ where $f$ is discontinuous, and maps  $\mathbb U$ conformally onto 
$\mathbb D.$  Then the function composition
$$ f \circ m : \mathbb U \rightarrow G  $$ and is thus bounded, but cannot be extended continuously at $m^{-1}(\zeta)$ on the real line.
A: I borrowed some books, the most important being Boundary Behavior of Conformal Maps by Christian Pommerenke. Much of this subject goes back to Constantin Caratheodory.
The main thing I want to emphasize is that there is no guarantee that a bounded function on the open upper half plane extends continuously to the real line. The Pommerenke book concentrates on the unit disk, so here is some terminology for the theorem: he writes $\mathbb D$ for the open unit disk, then $\bar{\mathbb D}$ for the closed unit disk, finally $\mathbb T$ for the unit circle. We have Theorem 2.1 on page 20:
Let f map $\mathbb D$ conformally onto the bounded domain $G.$ Then the following four conditions are equivalent:
(i) $f$ has a continuous extension to $\bar{\mathbb D}$
(ii) $\partial G$ is a curve, that is $\partial G = \{\varphi(\zeta): \zeta \in \mathbb T   \}$ with continuous $\varphi$
(iii) $\partial G$ is locally connected
(iv) $\mathbb C \backslash G$ is locally connected.
What this says to me is that there is no guarantee of a continuous extension to the boundary, but examples would be fairly difficult to describe, and it is possible that nobody has ever written one down. I would be very happy to learn of a reference where somebody gave a single example.
A: The boundedness is not an issue, if it extent continuously, since your bound extent continuously.
Often function extent to the boundary under certain conditions in a certain measure theoretic sense, (Fatou's theorem and similar stuff for $H^p$ spaces) 
More over Phragmen-Lindeloeff gives you a necessary condition, for the function to extent continuously. 
Note that this is the important question, not the boundedness.
