# Extension of real bounded analytic function to a holomorphic bounded function.

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.

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)