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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: $\mathbb C^{+}=\{z\in \mathbb C, \text{Im}(z)>0\}$.

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    Is $\mathbb{C}^+$ the set for which $\text{Re}(z) > 0$?2012-03-21
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    Sorry, I meant to type $\text{Im}(z) > 0$, of course. Ok.2012-03-21
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    Yeah, me too!! I fixed it above. Thanks2012-03-21
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    Strange way to ask this. If your function extends continuously to the real line, then of course it is bounded there, by continuity. Now, there are "lacunary functions" such as $$ \sum z^{2^k} $$ which cannot be analytically continued past a natural boundary, in this case the unit circle. Also, in this case, doubtless no continuous extension to the closed unit disk is possible, by the maximum principle (no finite value at $1$ for a clear instance). However, you are asking about bounded in the interior, and it may be possible to extend continuously only. Not sure yet.2012-03-21

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