the restriction of an entire function $\widetilde{\mathcal{F}}_\mu(z)$ of order one and of finite type.

Moreover, we define $r_\mu(k)$ to be the supremum of the radius $r > 0$ in the statement ($k$) for $k = 2,~3$ and $\widetilde{R}_\mu:=\inf \{R > 0; \sup_{z \in \operatorname*{\mathbb{C}}} \frac{|\widetilde{\mathcal{F}}_\mu(z)|}{e^{R|z|}} < \infty \}$. Then $r_\mu(k) = R_\mu^{-1}$ and $\widetilde{R}_\mu = R_\mu$. The second relation implies that the type of $\widetilde{\mathcal{F}}_\mu(z)$ is $R_\mu$.

**1.** We show the equivalence between (1) and (2). With Proposition \[prop1\], (1) implies (2). Conversely, we assume (2). Then we have the expansion $$G_\mu \Big(\frac{1}{z}\Big) = \sum_{n=0}^\infty d_n z^{n+1}$$ for $|z| < r$. $d_0 = 1$ since $z G_\mu(z) \to 1$ in the non tangential limit $z \to \infty$, $z \in {\mathbb{C}_{-}}$. We take any $R > r^{-1}$. Clearly $\frac{1}{\pi}{\text{\normalfont Im~}}G_\mu (x - iy)$ converges to $p(x):=\sum_{n=1}^\infty \frac{{\text{\normalfont Im~}}d_n}{x^{n+1}}$ locally uniformly on $|x| \geq R$ as $y \searrow 0$. Then $\mu$ has an absolutely continuous density $p(x)$ on $|x| \geq R$ by using the Stieltjes-Perron inversion formula [@Akh]. Therefore, $\mu$ is of the form $\mu = \mu|_{[-R, R]} + \lambda$, where $\lambda = p(x)dx$ on $|x| \geq R$.

**2.** The equivalence between (2) and (3) is immediate from simple computation.

**3.** The 