If $\mu\in\mathcal{M}$ with $\mathrm{Ind}(\mu)>1$ then $\phi_\mu$ has a continuous extension to $\mathbb{C}^+\cup\mathbb{R}$.

Measures with mean zero and finite variance
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Recall that for $\mu\in\mathcal{M}$ with mean zero and unit variance, $\Phi(\mu)$ is the unique measure in $\mathcal{M}$ satisfying the Eq. (\[EMaa\]) with $\sigma^2=1$, i.e.,$
E_\mu=G_{\Phi(\mu)}$. In general, a measure $\mu$ has mean $m$ and finite variance $\sigma^2$ if and only if $\mu\boxplus\delta_{-m}$ has mean zero and variance $\sigma^2$ because $d(\mu\boxplus\delta_{-m})(s)=d\mu(s+m)$, and hence $E_{\mu\boxplus\delta_{-m}}=\sigma^2G_{\Phi((\mu\boxplus\delta_{-m})^{\uplus1/\sigma^2})}$. Since $\mathrm{Ind}(\mu)=\mathrm{Ind}(\mu\boxplus\delta_a)$ for any $a\in\mathbb{R}$ by \[\[Japan\], Proposition 3.7\], in what follows we only consider measures with mean zero and finite variance.

Recall that the free Brownian motion started at $\nu\in\mathcal{M}$ is the process $\{\nu\boxplus\gamma_t:t\geq0\}$. The connection among this process, the map $\mathbb{B}_t$, and the subordination function of the $\boxplus$-convolution powers is described in the following theorem, which was proved in \[\[BN1\]\] and \[\[Biane1\]\]. For the completeness, we provide its statement and proof.

\[Brownian\] If $\mu\in\mathcal{M}$ has mean zero and variance $\sigma^2$, and $\nu=\Phi(\mu^{\uplus1/\sigma^2})$ then $$\label{Gmotion}
G_{\nu\boxplus\gamma_{t\sigma^2}}(z)=
G_\nu(F_{\mathbb{B}_{t+1,t/(t+1)}(\mu)}(z))=\frac{E_{\mathbb{B}_{t+1,t/(t+1)}(\mu)}(z)}{t\sigma^2},\;\;\;\;\;
z\in\mathbb{C}^+\cup\mathbb{R},$$ where $t>0$. Consequently, we have $\phi_{\mathbb{B}_{t+1,t/(t+1)}(\mu)}=t\sigma^2G_\nu$ and $$\label{motion}
E_{\mathbb{B}_t(\mu)}=\sigma^2G_{\nu\boxplus\gamma_{t\sigma^2}}.$$

Let $p=t+1>1$. Since $E_\mu=\sigma^2G_\nu$, it follows that $$H_p(z):=pz+(1-p)F_\mu(z)=z+(p-1)\sigma^2G_\nu(z),\;\;\;\;\;z\in\mathbb{C}^+.$$ If $\omega_p$ is the right inverse of $H_p$ then \[\[Biane1\], Proposition 2\] shows that $$G_{\nu\boxplus\gamma_{(p-1)\sigma^2}}(z)=G_\nu(\omega_p(z))=\frac{z-\omega_p(z)}{(p-1)\sigma^2},
\;\;\;\;\;z\in\mathbb{C}^+\cup\mathbb{R}.$$ Since $\omega_p=F_{\mathbb{B}_{p,1/p^*}(\mu)}$ by Proposition \[3.1\], 