Let $U$ be a non-empty open convex of $\mathbb{R}^n$, and :
$$\begin{array}{ccccc}
\lambda & : & (\mathbb{C^{\times}})^n & \to & \mathbb{R}^n \\
& & z=(z_1, \dots, z_n) & \mapsto & (\log(|z_1|), \dots , \log(|z_n|)) \\
\end{array}$$
Some notations : if $f$ is a Laurent series, $C_f = \{z \in (\mathbb{C^{\times}})^n \; | \; f(z)$ is absolutely convergent $\}$, $U_f = \overset\circ{C_f}$ the interior of $C_f$.
I have to show that it exists $f$ a Laurent series such that $U_f = \lambda^{-1}(U)$.
I have treat the case $n=1$ which is only a consequence of the fact that a convex set of $\mathbb{R}$ is an interval (bounded of not).
I tried to proceed via a similar way to the case $n \geq 2$, but I had a problem.
Actually, I think I have also treat the case where $U$ is a $n$-dimensional cube, i.e $U=\prod_{i \in \{1, \dots, n \}} ]a_i;b_i[$. But now, I want to extend this to the general case where $U$ is only an open convex set, and there is where I have a problem.
Someone would have an idea ?
Thank you !
Best Answer
Given a half-space $H_{w,r}=\{ x\in \Bbb{R}^n, \langle x,w\rangle < r\}$ with $w\in \Bbb{Z}^n,r\in \Bbb{R}$ then $$F_{w,r}(z)=\sum_{m=0}^\infty (z^w)^m e^{-rm},\qquad z\in \Bbb{C}^n$$ is a Laurent series which converges for $|z^w|< e^r$ ie. for $\langle \lambda(z),w\rangle < r$ and diverges for $\langle \lambda(z),w\rangle > r$.
Since $\overline{U}$ is convex closed, it is the intersection of countably many such half-spaces $$\overline{U}=\bigcap_k H_{w_k,r_k}$$ thus for $c_k$ decreasing fast enough we'll have that $$\sum_k c_k F_{w_k,r_k}(z)$$ converges for $\lambda(z)$ in the interior of $\overline{U}$ and diverges on its exterior.