Let $\mu$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ and $\varphi_\mu$ denote the characteristic function of $\mu$.
Remember that $\mu$ is called infinitely divisible if for all $k\in\mathbb N$, there is a probability measure $\mu_k$ with$^1$ $$\mu=\mu_k^{\ast k}\tag1.$$
I wonder whether we can uniquely define the "$k$th convolution root $\mu^{\ast\frac1k}$ of $\mu$ by setting $$\mu^{\ast\frac1k}:=\mu_k.\tag2$$
To elaborate on that, let $\varphi_\nu$ denote the characteristic function of a finite measure $\nu$ on $(\mathbb R,\mathcal B(\mathbb R)$ and note that, for all $k\in\mathbb N$, $(1)$ is equivalent to $$\varphi_\mu=\varphi_{\mu_k}^k\tag{1'}.$$
Now I've read the following argument: If $$\varphi_\mu(t)\ne0\;\;\;\text{for all }t\in\mathbb R\tag3,$$ then $\ln\left(\varphi_\mu\right)$ is a well-defined continuous function and hence there is a unique continuous function $$\varphi_\mu^{\frac1k}=\exp\left(\frac{\ln\left(\varphi_\mu\right)}k\right)\tag4$$
Question 1: Neither I understand the argument, nor do I understand whether I need to understand $\varphi_\mu^{\frac1k}$ on the left-hand side as the function $$\mathbb R\ni t\mapsto\varphi_\mu(t)^{\frac1k}\tag5$$ or $\varphi_\mu^{\frac1k}$ is supposed to be a symbol defined by $(4)$. I guess this is a actually a rather trivial argumentation, but I'm quite confused and it would help a lot if someone could explain how we need to understand this.
Question 2: How do we verify $(3)$? I've read that we can utilize Lévy’s continuity theorem to show that $$\varphi_{\mu_k}\xrightarrow{k\to\infty}1\tag6,$$ where the convergence is not only pointwise, but compactly, and we could deduce $(3)$ from that.
However, I guess I'm missing something crucial, since $(6)$ seems to me to be a trivial consequence of $(1')$, since $(1')$ shouly yield $$\varphi_{\mu_k}=\varphi_\mu^{\frac1k}\xrightarrow{k\to\infty}\varphi_\mu^0=1\tag7,$$ where I understand $\varphi_\mu^{\frac1k}$ as being $(5)$ and the convergence should clearly be compactly. At which stage do we need Lévy’s continuity theorem?
$^1$ If $\nu$ is a finite measure on $(\mathbb R,\mathcal B(\mathbb R))$ and $k\in\mathbb N_0$, then $\nu^{\ast k}$ is the $k$-fold convolution of $\nu$ with itself, i.e. $\nu^{\ast 0}:=\delta_0$ is the Dirac measure on $(\mathbb R,\mathcal B(\mathbb R))$ at $0$ and if $k\in\mathbb N$, then $\nu^{\ast k}$ is the pushforward $\tau_k\left(\nu^{\otimes k}\right)$ of the $k$-fold product measure $\nu^{\otimes k}$ of $\nu$ with itself under the map $$\tau_k:\mathbb R^k\to\mathbb R\;,\;\;\;x\mapsto x_1+\cdots+x_k.$$