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Let $f(x)$ be a real-valued analytic bounded function. Can I find a open set $U\subset\mathbb{C}$, containing $\mathbb{R}$, and a holomorphic complex-valued function $\bar{f}(z)$ such that $\bar{f}(z)$ is bounded on $U$ and $\bar{f}(x)=f(x)$ for every $x \in \mathbb{R}$? Is there any class of real-valued function with such properties? I just remark that the open set $U$ is not necessarily a strip.

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Yes always; let $M$ be a bound for $f$; since $f$ is analytic at $x$ there is a small plane disc $U_x$ centered at $x$ for which $f$ extends to a complex analytic function $f_x$ on $U_x$ and by continuity at $x$ we can shrink it so $|f_x| \le 2M$ there; by analytic continuation $f_x=f_y$ whenever $U_x \cap U_y$ non empty since the intersection is connected (and contains the real segment joining $x,y$) as $U_x, U_y$ are discs centered at $x,y$. In particular this means that we can define a global complex analytic function $F$ on $U=\cup U_x$ and $F=f$ on the reals, while $|F| \le 2M$ on $U$

($U$ is not a strip in general unless $f$ is the quasi analytic class $1/n!$ which has one of the equivalency definition precisely strip extension; there are bounded analytic functions on the reals that are not in there - see my answer HERE)

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  • $\begingroup$ Thank you very much! $\endgroup$ – Canjioh Feb 17 at 17:00
  • $\begingroup$ Happy to be of help $\endgroup$ – Conrad Feb 17 at 17:12

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