I wonder if my statements below are correct.

Let $V$ be an open domain in $\mathbb{R}^d$, and $U$ an open domain in $\mathbb{C}^d$ with $V=\operatorname{Re}U:=\{\operatorname{Re}z:z\in U\}$.

I want to define $\mathcal{H}(V,U)$ to be the space of complex analytic functions on $U$ which has a continuous extension to $\overline{U}$ and is real analytic when restricted to $V$. Let $\|f\|:=\sup_{z\in U}|f(z)|$. I think $\mathcal{H}(V, U)$ is a Banach space?

I am thinking, the space of real analytic functions on $V$ which has a complex analytic extension to $U$ and further continuous extension to $\overline{U}$ should be an equivalent definition for $\mathcal{H}(V,U)$? (Because I think if a real analytic function $f$ on $V$ can be analytically extended to $U$, this extension is unique.)

Any help is much appreciated! Thanks.

  • $\begingroup$ To be a normed space you needd $||f||\in \mathbb{R}$ $\forall f \in \mathcal{H}(V,U)$, so $U$ has to be bounded set or consider the subset of bounded functions in $U$. Note: if $U=\mathbb{C}$ then, in this case, the elements of $\mathcal{H}(U,V)$ are the real constants. From here it's easy to show that is Banach. $\endgroup$ – Rafael Gonzalez Lopez Jul 18 '18 at 11:35

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