equivalence between (2) and (4) is proved as follows. Let $g(z):=G_\mu(\frac{1}{z})$ for $z \in \operatorname*{\mathbb{C}_{+}}$. We have the identity $$\frac{1}{g^{-1} (z)} - \frac{1}{z} = F^{-1}_\mu\Big(\frac{1}{z}\Big) - \frac{1}{z}$$ for $z \in \Omega^{-1}$; this enables us to prove the equivalence. More precisely, we can show that (2) is equivalent to (2’) which states that $g^{-1}(z)$ defined in $\Omega^{-1}$ has analytic continuation to $\{z \in \operatorname*{\mathbb{C}}; |z| < r \}$ for some $r > 0$. Using this we can prove the equivalence between (2) and (4).

**4.** We prove the equivalence between (1) and (5). If (1) holds, the first statement in (5) follows from Proposition \[prop2\]. The second follows from the estimation in Proposition \[prop13\].

Conversely we assume that (5) holds. Let $c_n:= \frac{d^n \widetilde{F}}{d \xi^n}(0)$ for $n \geq 0$; in particular $c_0 = 1$. Then for each $A >R$, there exists $C$ depending on $A$ such that $|c_n| \leq CA^n$ for all $n$. This is proved as follows. The Cauchy’s integral formula implies that $|c_n| \leq \frac{n! e^x R^n}{x^n}$ for any $x > 0$ and $n$. Since $\frac{e^x}{x^n}$ takes its minimum value at $n$, we take $x=n$. If $A > R$, we can prove that $|c_n| < A^n$ for large $n$ 