# The Brown measure of a sum of two free random variables, one of which is
R-diagonal

HARI BERCOVICI AND PING ZHONG
Hari Bercovici, Department of Mathematics, Indiana University, Bloomington, IN
47401, USA. Email<EMAIL_ADDRESS>Ping Zhong, Department of Mathematics
and Statistics, University of Wyoming, Laramie, WY 82072, USA. Email:
<EMAIL_ADDRESS>
(Date: September 23, 2022. Preliminary version.)

###### Abstract.

We obtain a formula for the Brown measure of the sum of an $R$-diagonal
operator and a free random variables using subordination functions. This
allows us to provide a unified approach to the Brown measures of many known
examples which include $R$-diagonal operators, the sum with a free circular
operator and the sum with a free Cauchy operator. This answers a question of
Biane and Lehner.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries: free additive convolution and $R$-diagonal operators
  3. 3 The free convolution of symmetric probability measures
  4. 4 The sum of an $R$-diagonal operator and a free random variable
  5. 5 Addition with $R$-diagonal operators: examples

##  1\. Introduction

Let ${\mathcal{M}}$ be a finite von Neumann algebra equipped with the faithful
normal tracial state $\phi$. We denote by $\widetilde{\mathcal{M}}$ the set of
closed, densely defined operators affiliated with ${\mathcal{M}}$. Given
$T\in\widetilde{\mathcal{M}}$, the _Fuglede-Kadison determinant_ of $T$,
$\Delta(T)$, is defined by

$\Delta(T)=\exp\Big{(}\int_{0}^{\infty}\log t\,{\rm d}\mu_{|T|}(t)\Big{)}.$

Let ${\mathcal{M}}^{\Delta}$ be the set of operators
$T\in\widetilde{\mathcal{M}}$ satisfying the condition

$\phi(\log^{+}|T|)=\int_{0}^{\infty}\log^{+}(t)\,{\rm d}\mu_{|T|}(t)<\infty.$

It is known that the set ${\mathcal{M}}^{\Delta}$ is a subspace of
$\widetilde{\mathcal{M}}$. In particular, for $T\in{\mathcal{M}}^{\Delta}$ and
$\lambda\in\mathbb{C}$, $T-\lambda{\bf 1}\in{\mathcal{M}}^{\Delta}$. For
$T\in{\mathcal{M}}^{\Delta}$, the function $\lambda\mapsto
f(\lambda):=\log\Delta(T-\lambda{\bf 1})$ taking values in $[-\infty,\infty)$
is subharmonic on $\mathbb{C}$. The _Riesz measure_ associated with this
subharmonic function is called the _Brown measure_ [10, 14] of $T$ and we
denote it by $\mu_{T}$. More precisely,

${\rm d}\mu_{T}=\frac{1}{2\pi}\nabla^{2}f\,{\rm d}\lambda$

taken in the distribution sense. Moreover, the Brown measure $\mu_{T}$
satisfies

$\quad\log\Delta(T-\lambda{\bf 1})=\int_{\mathbb{C}}\log|\lambda-z|\,{\rm
d}\mu_{T}(z),\qquad\forall\lambda\in{\mathbb{C}}.$

In a breakthrough paper [22], Voiculescu discovered that the mixed moments of
a family of independent Gaussian random matrices of large dimension are
approximately equal to the mixed moments of a family of free random variables.
This spectacular result suggests that the Brown measure of an operator in
${\mathcal{M}}$ arising from free probability is a natural candidate of
empirical spectral distribution of some suitable random matrix model. Let
$U,V$ be $N\times N$ independent Haar random unitary and let
$\Sigma=\Sigma_{N}$ be a sequence of diagonal matrices of size $N$ that
converges in moments to some operator in ${\mathcal{M}}$. It is known as the
single ring theorem [12, 21] that the empirical spectral distribution of
random matrix model $U\Sigma V$ converges to the Brown measure of an
$R$-diagonal operator [19]. The Brown measure of an $R$-diagonal operator $T$
is supported in a single ring where the inner and outer radii, and the density
formula can be expressed by the spectral measure of $|T|$, which were
calculated by Haagerup–Larsen–Schultz [13, 14].

Let $x_{0}\in{\mathcal{M}}^{\Delta}$ be a random variable with an arbitrary
distribution, $*$-free from the $R$-diagonal operator $T$. The purpose of this
article is to calculate the Brown measure of $T+x_{0}$. This question was left
open in Biane–Lehner’s work [9], where they obtained Brown measure formula for
such operators in some special cases. In a work by Ho and the second author,
the Brown measure of $T+x_{0}$ is shown to be the limit of the empirical
spectral distribution of deformed single ring random matrix model [18]. See
Figure 1 for a simulation of eigenvalue distribution of random matrix model.

######  Theorem 1.1 (See Theorem 4.1 and Theorem 4.13).

Let $T\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal and
$x_{0}\in{\mathcal{M}}^{\Delta}$ be be a operator that is $*$-free from $T$.
Then the support of the Brown measure $\mu_{T+x_{0}}$ is contained in the
closure of following set

$\Omega(T,x_{0})=\bigg{\\{}\lambda\in\mathbb{C}:||(x_{0}-\lambda)^{-1}||_{2}||T||_{2}>1\,\,\text{and}\,\,||x_{0}-\lambda||_{2}||T^{-1}||_{2}>1\bigg{\\}}.$

The Brown measure $\mu_{T+x_{0}}$ is absolutely continuous with respect to
Lebesgue measure in $\Omega(T,x_{0})$ except possibly at finitely many points.
Moreover, the density formula can be expressed explicitly in terms of some
subordination functions parametrized by $\lambda$.

(a) Brown($x_{0}+u$)

(b) Brown($x_{0}+c$)

Figure 1. Brown measure simulation for sum of Haar unitary/circular operator
and $x_{0}$ distributed as
$\mu_{x_{0}}=0.25\delta_{-1}+0.5\delta_{0}+0.25\delta_{1}$

Fix an operator $x_{0}$, then the boundary of $\Omega(T,x_{0})$ is completely
determined by the inner and outer radii of the single ring associated with the
$R$-diagonal operator $T$. See Figure 1 for the comparison of the addition
with a Haar unitary and a circular operator. The complement of
$\Omega(T,x_{0})$ in $\mathbb{C}$ naturally splits to the union of two
disjoint closed sets. If $x_{0}$ is selfadjoint, then the set
$\Omega(T,x_{0})$ is symmetric with respect to the $x$-axis. Moreover, the
complement $\mathbb{C}\backslash\Omega(T,x_{0})$ has at most one bounded
component, and the bounded component is a disk if $x_{0}$ is selfadjoint, as
shown in Figure 1. By choosing $x_{0}=0$ in Theorem 1.1, we see immediately
that the Brown measure of an $R$-diagonal operator $T$ is supported in a
single ring

${\text{\rm
supp}}(\mu_{T})=\\{\lambda\in\mathbb{C}:1/||T^{-1}||_{2}<|\lambda|<||T||_{2}\\}.$

Our method provides a new proof of the Haagerup-Larsen-Schultz’s formula for
Brown measure of $R$-diagonal operators. This was done by the second author
[25], where the key ingredient in that work was to study the free convolution
of a symmetric probability measure with a Bernoulli distribution. In this
article, the main tools are subordination functions in free convolution of two
symmetric probability measures with arbitrarily distributions.

We show that the density of the absolutely continuous part of $\mu_{T+x_{0}}$
is strictly positive in $\Omega(T,x_{0})$ for the following cases: (1) $T$ is
a Haar unitary operator, (2) $T$ is a circular operator, (3) $T$ is a circular
Cauchy operator. The case when $T$ is a circular operator generalizes some
results in [24], where the second author calculated the Brown measure of the
sum of a circular operator and a bounded operator in ${\mathcal{M}}$ (see also
[17] for earlier results). The circular Cauchy operator [14] can be expressed
as $xy^{-1}$, where $\\{x,y\\}$ is a circular family. The Brown measure of
$xy^{-1}+x_{0}$ plays a crucial role in the work of Haagerup–Schultz’s work
[15] on invariant subspace problem in type II1 factors. We provide an
alternative proof for the Brown measure of $xy^{-1}+x_{0}$ and our proof works
for unbounded operator $x_{0}$. The extension to unbounded operator is useful
for generalization of Haagerup–Schultz’s main results to unbounded operators
[11].

Let us outline the strategy of the proof. Let $U$ be a Haar unitary operator
that is $*$-free from $\\{T,x_{0}\\}$. We observe that the following
identification of distributions

$|x_{0}+T-\lambda{\bf 1}|\underset{\ast{\mathcal{D}}}{\sim}|(x_{0}-\lambda{\bf
1})+U^{*}T|\underset{\ast{\mathcal{D}}}{\sim}|U(x_{0}-\lambda{\bf 1})+T|.$

The operator $U(x_{0}-\lambda{\bf 1})$ is $R$-diagonal that is $*$-free from
$T$. It is well-known [14, 20] (see Proposition 2.5) that the addition
$U(x_{0}-\lambda{\bf 1})+T$ of two $R$-diagonal operators is also an
$R$-diagonal operator, and

$\tilde{\mu}_{|U(x_{0}-\lambda{\bf 1})+T|}=\tilde{\mu}_{|x_{0}-\lambda{\bf
1}|}\boxplus\tilde{\mu}_{|T|},$

where $\tilde{\mu}$ denote the symmetrization of $\mu$. That is, $\tilde{\mu}$
is defined by by

$\tilde{\mu}(B)=\textstyle{\frac{1}{2}}(\mu(B)+\mu(-B)),$

where $B$ is any Borel measurable set of real numbers. By studying the free
convolution of symmetric probability measures, we obtain a formula for the
Fuglede-Kadison determinant $\Delta(x_{0}+T-\lambda{\bf 1})$ using
subordination functions parameterized by $\lambda$. As subordination functions
satisfy fixed point equation, it allows us to take implicit differentiation.
The calculation of Brown measure is then reduced to questions about
subordination functions.

The paper is organized as follows. In Section 2, we review some relevant
results in free convolutions and $R$-diagonal operators. In Section 3, we
study the free convolution of two symmetric probability measures. In Section
4, we calculate the Fuglede-Kadison determinant formula and Brown measure
formula for the sum of an $R$-diagonal operator and a free random variable. In
Section 5, we study explicit formulas for three important cases.

##  2\. Preliminaries: free additive convolution and $R$-diagonal operators

###  2.1. Free additive convolution

Given a selfadjoint operator $x\in\mathcal{M}$, let $\mu_{x}$ be the
probability distribution of $x$ in the sense that $\mu_{x}(B)=\phi(E_{x}(B))$,
for any Boreal measurable set $B$, where $E_{x}$ denotes the spectral measure
of $x$. For a probability measure $\mu$ on $\mathbb{R}$, its Cauchy transform
is defined as

$G_{\mu}(z)=\int_{\mathbb{R}}\frac{1}{z-x}d\mu(x),\qquad z\in\mathbb{C}^{+};$

and the $F$-transform is its reciprocal, $F_{\mu}(z)=1/G_{\mu}(z)$. The
$R$-transform of $\mu$ is defined as
$R_{\mu}(z)=G_{\mu}^{\langle-1\rangle}(z)-1/z$ in some domain of the form
$\Gamma_{\alpha,\beta}=\\{z=x+iy:y>\beta,|x/y|<\alpha\\}$ for some
$\alpha,\beta>0$ (see [5]). Given two selfadjoint random variables $x,y$,
freely independent in $(\mathcal{M},\phi)$, denote $\mu_{1}=\mu_{x}$ and
$\mu_{2}=\mu_{y}$. The free additive convolution $\mu_{1}\boxplus\mu_{2}$ of
$\mu_{1}$ and $\mu_{2}$ is defined as $\mu_{1}\boxplus\mu_{2}=\mu_{x+y}$. The
free additive convolution is a binary convolution on probability measures on
the real line $\mathbb{R}$, which is the free analogue of usual convolution in
classic probability. The $R$-transform linearizes the free additive
convolution in the sense that
$R_{\mu_{1}\boxplus\mu_{2}}(z)=R_{\mu_{1}}(z)+R_{\mu_{2}}(z)$ in some domain
where all these functions are defined. It is known [8, 22, 23] that there are
two analytic functions
$\omega_{1},\omega_{2}:\mathbb{C}^{+}\rightarrow\mathbb{C}^{+}$ such that

(2.1)
$F_{\mu_{1}\boxplus\mu_{2}}(z)=F_{\mu_{1}}(\omega_{1}(z))=F_{\mu_{2}}(\omega_{2}(z)),\qquad
z\in\mathbb{C}^{+}.$

One can show that

(2.2) $\omega_{1}(z)+\omega_{2}(z)=F_{\mu_{1}\boxplus\mu_{2}}(z)+z,\qquad
z\in\mathbb{C}^{+}.$

It turns out that (2.1) and (2.2) determine $\omega_{1},\omega_{2}$ uniquely.
The subordination functions are defined everywhere in $\mathbb{C}^{+}$, while
$R$-transforms are defined only in some subsets. Hence, it is convenient to
use subordination functions solely to calculate free additive convolution. To
be more precise, the subordination functions are unique solutions of certain
fixed point equation [2]. Let $H_{1}(z)=F_{{\mu}_{1}}(z)-z$ and
$H_{2}(z)=F_{{\mu}_{2}}(z)-z$. Then $\omega_{1},\omega_{2}$ satisfy the
following fixed point equations respectively

$\omega_{1}(z)=z+H_{2}(z+H_{1}(\omega_{1}(z))),\qquad\omega_{2}(z)=z+H_{1}(z+H_{2}(\omega_{2}(z))).$

The above fixed point equations play an important role in our study.

The subordination functions and free convolution obey some nice regularity
properties. It is shown [1, 3] that the subordination functions
$\omega_{1},\omega_{2}$ extends as a continuous function
$\omega_{1},\omega_{2}:\mathbb{C}^{+}\cup\mathbb{R}\rightarrow\mathbb{C}^{+}\cup\mathbb{R}\cup\\{\infty\\}$.
By using (2.2), we see that $F_{\mu_{1}\boxplus\mu_{2}}$ has a continuous
extension to $\mathbb{C}^{+}\cup\mathbb{R}$. This extension satisfies
$\omega_{i}(\infty)=\infty$ for $i=1,2$, viewed as nontangential limits. We
summarize some relevant results as follows.

######  Theorem 2.1.

[1, 6] Let $\mu_{1},\mu_{2}$ be two probability measures on $\mathbb{R}$ and
assume that neither of them is a single delta mass. Then
$\mu_{1}\boxplus\mu_{2}$ has no singular continuous part. Its absolutely
continuous part of $\mu_{1}\boxplus\mu_{2}$ is always nonzero and its density
is analytic whenever positive and finite.

Moreover, the point $\gamma\in\mathbb{R}$ is an atom of the measure
$\mu_{1}\boxplus\mu_{2}$ if and only if there exists
$\alpha,\beta\in\mathbb{R}$ such that $\gamma=\alpha+\beta$ and
$\mu_{1}(\\{\alpha\\})+\mu_{2}(\\{\beta\\})>1$. In addition,
$\alpha=\omega_{1}(\gamma),\beta=\omega_{2}(\gamma)$ and
$(\mu_{1}\boxplus\mu_{2})(\\{\gamma\\})=\mu_{1}(\\{\alpha\\})+\mu_{2}(\\{\beta\\})-1$.

######  Proposition 2.2.

Let $\mu_{1},\mu_{2}$ be two probability measures on $\mathbb{R}$ and assume
that neither of them is a single delta mass. For $\gamma\in\mathbb{R}$,

  1. (1)

$\omega_{1}(\gamma)\in\mathbb{C}^{+}$ if and only if
$\omega_{2}(\gamma)\in\mathbb{C}^{+}$.

  2. (2)

if $F_{\mu_{1}\boxplus\mu_{2}}(\gamma)=0$, then

(2.3) $\mu_{1}(\\{\alpha\\})+\mu_{2}(\\{\beta\\})\geq 1.$

where $\alpha=\omega_{1}(\gamma),\beta=\omega_{2}(\gamma)$.

###### Proof.

Given $z\in\mathbb{C}^{+}$, we note that $\Im F_{\mu_{i}}(z)>\Im z$ for
$i=1,2$. If $\omega_{1}(\gamma)\in\mathbb{C}^{+}$, by using (2.2), we have

$\Im(\omega_{2}(\gamma))=\Im(F_{\mu_{1}}(\omega_{1}(\gamma))-\omega_{1}(\gamma))>0.$

Similarly, if $\omega_{2}(\gamma)\in\mathbb{C}^{+}$, then
$\omega_{1}(\gamma)\in\mathbb{C}^{+}$. This established item (1).

If $F_{\mu_{1}\boxplus\mu_{2}}(\gamma)=0$, the condition (2.3) was proved in
the proof of Part 3 of [1, Theorem 4.1] (see Pages 147-149). ∎

###  2.2. Unbounded $R$-diagonal operators

The $R$-diagonal operators, introduced by Nica-Speicher [19], satisfy a number
of remarkable symmetric properties. In [14], the class of $R$-diagonal
operators was extended to unbounded case. Recall that [19, 14]
$T\in\widetilde{\mathcal{M}}$ is $R$-diagonal if there exist a von Neumann
algebra ${\mathcal{N}}$, with a faithful normal tracial state, and $*$-free
elements $U,H\in{\mathcal{N}}$ with $U$ Haar unitary, $H\geq{0}$, and such
that $T$ has the same $*$-distribution as $UH$. If
$T\in\widetilde{\mathcal{M}}$ is $R$-diagonal with $ker(T)=0$, let the polar
decomposition of $T$ be $V|T|$, where $|T|=(T^{*}T)^{1/2}$, then $V$ is a Haar
unitary which is $*$-free from $|T|$.

######  Lemma 2.3.

[14] Let $\mu$ be a probability measure on $[0,\infty)$ or a symmetric
probability measure on $\mathbb{R}$, and put

$\lambda_{1}(\mu)=\Bigg{(}\int_{\mathbb{R}}\frac{1}{u^{2}}\,{\rm
d}\mu(u)\Bigg{)}^{-\frac{1}{2}}\quad{\rm
and}\quad\lambda_{2}(\mu)=\Bigg{(}\int_{\mathbb{R}}u^{2}\,{\rm
d}\mu(u)\Bigg{)}^{\frac{1}{2}},$

with the convention that $\infty^{-\frac{1}{2}}=0$. Then
$0\leq\lambda_{1}(\mu)\leq\lambda_{2}(\mu)\leq\infty$. If its symmetrization
$\tilde{\mu}$ is supported at least three points, then
$\lambda_{1}(\mu)<\lambda_{2}(\mu)$.

For a probability measure $\mu$ on $[0,\infty)$ that is not a delta mass at
zero, define $\psi_{\mu}:\mathbb{C}\backslash(0,\infty)\rightarrow\mathbb{C}$
by

$\psi_{\mu}(z)=\int_{[0,\infty)}\frac{zt}{1-zt}\,{\rm d}\mu(t),\qquad
z\in{\mathbb{C}}\setminus(0,\infty).$

One can check that: (i) $\psi_{\mu}^{\prime}(x)>0,x\in(-\infty,0)$; (ii)
$\psi_{\mu}(x)\rightarrow\delta-1$ as $x\rightarrow-\infty$ where
$\delta=\mu(\\{0\\})$; (iii) $\psi_{\mu}(x)\rightarrow 0$ as $x\rightarrow 0$
(see [5]). Hence, in a neighborhood ${\mathcal{V}}_{\mu}$ of $(\delta-1,0)$,
the inverse $\chi_{\mu}$ of $\psi_{\mu}$ exists:
$\chi_{\mu}(z)=\psi_{\mu}^{\langle-1\rangle}(z),z\in{\mathcal{V}}_{\mu}$. Then
Voiculescu’s $S$-transform is defined by

$S_{\mu}(z)=\frac{z+1}{z}\chi_{\mu}(z),\quad z\in{\mathcal{V}}_{\mu}.$

Given two probability measures $\mu,\nu$ on $[0,\infty)$, the $S$-transform
satisfies $S_{\mu\boxtimes\nu}(z)=S_{\mu}(z)S_{\nu}(z)$ in a region where all
these functions are defined.

The Brown measure of bounded $R$-diagonal operators was calculated by
Haagerup-Larsen [13] and the result was further extend by Haaagerup-Schultz
[14] for all $R$-diagonal operators in ${\mathcal{M}}^{\Delta}$. See [4] for a
new proof based on operator-valued free probability and [25] for another proof
based on free convolution of symmetric measures.

######  Theorem 2.4.

Let $T$ be an $R$–diagonal element in ${\mathcal{M}}^{\Delta}$ with Brown
measure $\mu_{T}$, and suppose $\mu_{T}$ is not a Dirac measure. The Brown
measure $\mu_{|T|}$ is the unique rotationally invariant probability measure
such that

$\displaystyle\mu_{|T|}\\{\lambda\in\mathbb{C}:|\lambda|\leq
r\\}=\begin{cases}0,\quad&\text{for}\quad r<\lambda_{1}(\mu_{|T|});\\\
1+S^{\langle-1\rangle}_{\mu_{\mu_{T^{*}T}}}(r^{-2}),\quad&\text{for}\quad\lambda_{1}(\mu_{|T|})<r<\lambda_{2}(\mu_{|T|});\\\
1,\quad&\text{for}\quad r\geq\lambda_{2}(\mu_{|T|}).\end{cases}$

When $\mu_{|T|}(\\{0\\})=\delta>0$, the Brown measure $\mu_{T}$ has a mass at
the origin with the mass $\mu_{T}(\\{0\\})=\delta$.

The following result is proved in [14, Proposition 3.5] (see also [20] for the
original proof for bounded case). It is a powerful tool to calculate the
distribution of the sum of $R$-diagonal operators.

######  Proposition 2.5.

Let $S,T\in\widetilde{\mathcal{M}}$ be $\ast$–free $R$–diagonal elements. Then
$S+T\in\widetilde{\mathcal{M}}$ is $R$-diagonal and

(2.4) $\tilde{\mu}_{|S+T|}=\tilde{\mu}_{|S|}\boxplus\tilde{\mu}_{|T|}.$

##  3\. The free convolution of symmetric probability measures

In this section, we will study the subordination functions appearing in the
free convolution of symmetric probability measures on $\mathbb{R}$. Those
results will be useful in later sections. Throughout this section, symmetric
probability measures are not the single delta mass at zero. For such a
symmetric probability measure $\mu$ on $\mathbb{R}$, we set

(3.1) $h_{\mu}(s)=\int_{\mathbb{R}}\frac{s}{s^{2}+u^{2}}\,{\rm d}\mu(u),\qquad
s>0$

and

(3.2) $f_{\mu}(s)=\frac{1}{h_{\mu}(s)}-s,\qquad s>0.$

By symmetry of ${\mu}$, for $s>0$, we have

(3.3)
$G_{{\mu}}(is)=\int_{\mathbb{R}}\frac{1}{is-x}d{\mu}(x)=\int_{\mathbb{R}}\frac{-is}{s^{2}+x^{2}}d{\mu}(x)=-ih_{\mu}(s),$

and consequently,

(3.4) $f_{\mu}(s)=\Im(F_{\mu}(is)-is),\qquad s>0.$

Let $\mu_{1},\mu_{2}$ be symmetric probability measures on $\mathbb{R}$,
neither of them is a single delta mass at zero. Let $\omega_{1},\omega_{2}$ be
subrodination functions such that

(3.5)
$F_{{\mu}_{1}\boxplus{\mu}_{2}}(z)=F_{{\mu}_{1}}(\omega_{1}(z))=F_{\mu_{2}}(\omega_{2}(z)),\qquad
z\in\mathbb{C}^{+}.$

######  Proposition 3.1.

For any $t>0$, the functions $\omega_{1}(it),\omega_{2}(it)$ are pure
imaginary. Set $\omega_{1}(it)=is_{1}(t)$ and $\omega_{2}(it)=is_{2}(t)$. We
have $s_{1}(t)>t,s_{2}(t)>t$, and

$h_{{\mu}_{1}\boxplus{\mu}_{2}}(t)=h_{\mu_{1}}(s_{1}(t))=h_{\mu_{2}}(s_{2}(t)).$

Moreover,

(3.6) $s_{1}(t)+s_{2}(t)=t+\frac{1}{h_{{\mu}_{1}\boxplus{\mu}_{2}}(t)}.$

###### Proof.

In some domain of the form
$\Gamma_{\alpha,\beta}=\\{z=x+iy:y>\beta,|x/y|<\alpha\\}$, we have

$\omega_{1}(z)=F^{\langle-1\rangle}_{{\mu}_{1}}\left(F_{{\mu}_{1}\boxplus{\mu}_{2}}(z)\right),\qquad
z\in\Gamma_{\alpha,\beta}.$

Hence, by (3.3) and (3.5), $\omega_{1}(it)$ is purely imaginary for $t>\beta$
and therefore $\omega_{1}(it)$ is purely imaginary for all $t>0$ by analytic
extension. Let $s_{1}(t)=\Im\omega_{1}(it)$, we have
$s_{1}(t)=\Im\omega_{1}(it)>\Im(it)=t$ (the strict inequality holds because
${\mu}_{1}$ is a non-trivial probability measure). Finally,

$h_{{\mu}_{1}\boxplus{\mu}_{2}}(t)=-\Im
G_{{\mu}_{1}\boxplus{\mu}_{2}}(it)=-\Im
G_{{\mu}_{1}}(\omega_{1}(it))=h_{{\mu}_{1}}(s_{1}(t)).$

Same argument works for $\omega_{2}$.

From free convolution of ${\mu}_{1}$ and ${\mu}_{2}$, we have

(3.7) $\omega_{1}(z)+\omega_{2}(z)=z+F_{{\mu}_{1}\boxplus{\mu}_{2}}(z),\quad
z\in\mathbb{C}^{+}.$

It follows that (3.6) holds by letting $z=it$. ∎

We have

(3.8) $sf_{\mu}(s)=\frac{s}{h_{\mu}(s)}-s^{2}=\frac{n_{\mu}(s)}{d_{\mu}(s)}$

where

$n_{\mu}(s)=\int_{\mathbb{R}}\frac{u^{2}}{s^{2}+u^{2}}d\mu(u)\qquad\text{and}\qquad
d_{\mu}(s)=\int_{\mathbb{R}}\frac{1}{s^{2}+u^{2}}d\mu(u).$

The following result is taken from the proof of [14, Lemma 4.8] (see also
[25]). We provide a new proof for reader’s convenience.

######  Lemma 3.2.

Let $\mu$ be a symmetric probability measure on $\mathbb{R}$, and
$\lambda_{1}(\mu),\lambda_{2}(\mu)$ be as in Lemma 2.3, and let $f_{\mu}$ be
as in (3.2). Then the function $s\mapsto sf_{\mu}(s)$ is a strictly increasing
bijection of $(0,\infty)$ onto $(\lambda_{1}(\mu)^{2},\lambda_{2}(\mu)^{2})$.

###### Proof.

Sine $\mu$ is a symmetric probability measure, then $F_{\mu}$ has the
following Nevanlinna representation

$F_{\mu}(z)=z+\int_{\mathbb{R}}\frac{1+uz}{u-z}d\rho(u),$

where $\rho$ is a finite measure on $\mathbb{R}$. It is easy to see that
$\rho$ is also a symmetric measure. Hence, by (3.4), we have

$sf_{\mu}(s)=s\cdot\left(\Im\int_{\mathbb{R}}\frac{1+ius}{u-is}d\rho(u)\right)=\int_{\mathbb{R}}\frac{(1+u^{2})s^{2}}{u^{2}+s^{2}}d\rho(u).$

This implies that the function $s\mapsto sf_{\mu}(s)$ is strictly increasing
function on $(0,\infty)$. By (3.8), we can verify that

$\lim_{s\rightarrow
0}sf_{\mu}(s)=\frac{1}{\int_{\mathbb{R}}\frac{1}{u^{2}}d\mu(u)}=\lambda_{1}^{2}(\mu),$

and

$\lim_{s\rightarrow
0}sf_{\mu}(s)=\int_{\mathbb{R}}u^{2}d\mu(u)=\lambda_{2}^{2}(\mu).$

This completes the proof. ∎

Recall that $\omega_{1},\omega_{2}$ satisfy the following fixed point
equations respectively

$\omega_{1}(z)=z+H_{2}(z+H_{1}(\omega_{1}(z))),\qquad\omega_{2}(z)=z+H_{1}(z+H_{2}(\omega_{2}(z))),$

where $H_{1}(z)=F_{{\mu}_{1}}(z)-z$ and $H_{2}(z)=F_{{\mu}_{2}}(z)-z$. Then,
for $z=it$, we have
$H_{1}(it)=\frac{1}{G_{{\mu}_{1}}(it)}-it=i\left(\frac{1}{h_{\mu_{1}}(t)}-t\right)=if_{\mu_{1}}(t)$
and $H_{2}(it)=i\left(\frac{1}{h_{\mu_{2}}(t)}-t\right)=if_{\mu_{2}}(t)$.
Using Proposition 3.1, the above fixed point equations for subordination
functions are then rewritten as the following equations for $s_{1}(t)$ and
$s_{2}(t)$,

(3.9) $s_{1}(t)=t+f_{\mu_{2}}(t+f_{\mu_{1}}(s_{1}(t))),\qquad
s_{2}(t)=t+f_{\mu_{1}}(t+f_{\mu_{2}}(s_{2}(t))).$

Thanks to [2, Theorem 4.1], the above equation (3.9) determines unique
solutions $s_{1}(t),s_{2}(t)$ for any $t>0$.

Recall that the subordination functions $\omega_{1},\omega_{2}$ extends as
continuous functions on $\mathbb{C}^{+}\cup\mathbb{R}$ taking values in
$\mathbb{C}^{+}\cup\mathbb{R}\cup\\{\infty\\}$. Hence,
$\Im\omega_{i}(x)\in[0,\infty]$ and the function $s_{1}(t),s_{2}(t)$ extends
as continuous function on $[0,\infty)$ taking values in $[0,\infty]$. We set

$s_{1}(0)=\lim_{t\downarrow 0}s_{1}(t),\qquad s_{2}(0)=\lim_{t\downarrow
0}s_{2}(t).$

######  Lemma 3.3.

Assume that $\mu_{1},\mu_{2}$ are symmetric probability measures and neither
of them is a single delta mass at zero. Then, at least one of
$s_{1}(0),s_{2}(0)$ is finite. Let $f_{\mu_{1}\boxplus\mu_{2}}$ be the density
function of the absolutely continuous part of $\mu_{1}\boxplus\mu_{2}$. Then,

  1. (1)

$s_{1}(0)\in(0,\infty)$ if and only if $s_{2}(0)\in(0,\infty)$, and in this
case

$f_{\mu_{1}\boxplus\mu_{2}}(0)\in(0,\infty);$

  2. (2)

if $s_{1}(0)=0,s_{2}(0)=\infty$, then
$\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$ and
$f_{\mu_{1}\boxplus\mu_{2}}(0)=0$. Moreover,

(3.10)
$\lim_{t\downarrow{0}}\frac{t}{s_{1}(t)}=\frac{\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2})}{\lambda_{1}^{2}(\mu_{1})},$

and

(3.11) $\lim_{t\downarrow{0}}t\cdot
s_{2}(t)=\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2}).$

  3. (3)

if $s_{1}(0)=\infty,s_{2}(0)=0$, then
$\lambda_{1}(\mu_{2})\geq\lambda_{2}(\mu_{1})$ and
$f_{\mu_{1}\boxplus\mu_{2}}(0)=0$. Moreover,

(3.12)
$\lim_{t\downarrow{0}}\frac{t}{s_{2}(t)}=\frac{\lambda_{1}^{2}(\mu_{2})-\lambda_{2}^{2}(\mu_{1})}{\lambda_{1}^{2}(\mu_{2})},$

and

(3.13) $\lim_{t\downarrow{0}}t\cdot
s_{1}(t)=\lambda_{1}^{2}(\mu_{2})-\lambda_{2}^{2}(\mu_{1}).$

  4. (4)

if $s_{1}(0)=s_{2}(0)=0$, then

$\lim_{\varepsilon\rightarrow 0}\Im
G_{\mu_{1}\boxplus\mu_{2}}(i\varepsilon)=\infty.$

Moreover, $\lambda_{1}(\mu_{1})=\lambda_{1}(\mu_{2})=0$ and

(3.14) $\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})\geq 1.$

###### Proof.

Recall that $s_{i}(t)=\omega_{i}(t)$ for $t\in[0,\infty)$. We have

$s_{2}(t)=\frac{1}{h_{\mu_{1}}(s_{1}(t))}-s_{1}(t)+t=\Im(F_{\mu_{1}}(is_{1}(t)))-is_{1}(t))+t.$

Hence, if $s_{1}(0)=\infty$, then $s_{2}(0)=0$ by the asymptotics of
$F$-transform. Similarly, if $s_{2}(0)=\infty$, then $s_{1}(0)=0$.

By item (1) in Proposition 2.2, we see that $s_{1}(0)\in(0,\infty)$ if and
only if $s_{2}(0)\in(0,\infty)$. In this case,

$F_{\mu_{1}\boxplus\mu_{2}}(0)=i(s_{1}(0)+s_{2}(0)),$

and by Stieltjes inversion formula, we obtain

$f_{\mu_{1}\boxplus\mu_{2}}(0)=-\frac{1}{\pi}\Im
G_{\mu_{1}\boxplus\mu_{2}}(0)=\frac{1}{\pi}\frac{1}{s_{1}(0)+s_{2}(0)}>0.$

This proves part (1).

We multiply the factor $(t+f_{\mu_{1}}(s_{1}(t)))$ to the first equation in
(3.9) and then rewrite it as follows

(3.15)
$\displaystyle(t+f_{\mu_{1}}(s_{1}(t))(s_{1}(t)-t)=(t+f_{\mu_{1}}(s_{1}(t)))f_{\mu_{2}}(t+f_{\mu_{1}}(s_{1}(t))).$

We recall that for a symmetric measure $\mu$, we have

(3.16) $\lim_{s\downarrow
0}sf_{\mu}(s)=\lambda_{1}^{2}(\mu),\qquad\lim_{s\uparrow\infty}sf_{\mu}(s)=\lambda_{2}^{2}(\mu).$

Hence, if $s_{1}(0)=0$, then

(3.17) $\displaystyle\lim_{t\downarrow 0}(t+f_{\mu_{1}}(s_{1}(t))(s_{1}(t)-t)$
$\displaystyle=\lim_{t\downarrow{0}}\left(s_{1}(t)f_{\mu_{1}}(s_{1}(t))-tf_{\mu_{1}}(s_{1}(t))\right)$
$\displaystyle\leq\lim_{s\downarrow
0}sf_{\mu_{1}}(s)=\lambda_{1}^{2}(\mu_{1}).$

And if $s_{1}(0)=\infty$, then

(3.18)
$\lim_{t\downarrow{0}}f_{\mu_{1}}(s_{1}(t))=\lim_{s\uparrow\infty}f_{\mu_{1}}(s)=0,$

and hence

$\displaystyle\lim_{t\downarrow 0}(t+f_{\mu_{1}}(s_{1}(t))(s_{1}(t)-t)$
$\displaystyle=\lim_{t\downarrow 0}ts_{1}(t)+\lim_{t\downarrow
0}s_{1}(t)f_{\mu_{1}}(s_{1}(t))$ (3.19) $\displaystyle\geq\lim_{t\downarrow
0}s_{1}(t)f_{\mu_{1}}(s_{1}(t))=\lambda_{2}^{2}(\mu_{1}).$

We now proceed the proof case by case.

(2) Assume that $s_{1}(0)=0,s_{2}(0)=\infty$, then by (3.6), we have

$f_{\mu_{1}}(s_{1}(t))=\frac{1}{h_{\mu_{1}}(s_{1}(t)}-s_{1}(t)=s_{2}(t)-t.$

and hence $\lim_{t\downarrow 0}(t+f_{\mu_{1}}(s_{1}(t)))=\infty$. We then have

(3.20) $\lim_{t\downarrow
0}(t+f_{\mu_{1}}(s_{1}(t)))f_{\mu_{2}}(t+f_{\mu_{1}}(s_{1}(t)))=\lim_{s\uparrow\infty}sf_{\mu_{2}}(s)=\lambda_{2}^{2}(\mu_{2}).$

By combing (3.18), we deduce that
$\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$.

It follows that $\lambda_{1}(\mu_{1})>0$. Hence,

$\displaystyle\lim_{t\downarrow 0}(t+f_{\mu_{1}}(s_{1}(t))(s_{1}(t)-t)$
$\displaystyle=\lim_{t\downarrow{0}}\left(s_{1}(t)f_{\mu_{1}}(s_{1}(t))-tf_{\mu_{1}}(s_{1}(t))\right)$
$\displaystyle=\lim_{t\downarrow{0}}s_{1}(t)f_{\mu_{1}}(s_{1}(t))\left[1-\frac{t}{s_{1}(t)}\right]$
$\displaystyle=\left(\lim_{s\downarrow{0}}sf_{\mu_{1}}(s)\right)\cdot\left(\lim_{t\downarrow{0}}\left[1-\frac{t}{s_{1}(t)}\right]\right)$
$\displaystyle=\lambda_{1}^{2}(\mu_{1})\cdot\left(\lim_{t\downarrow{0}}\left[1-\frac{t}{s_{1}(t)}\right]\right),$

which implies that $\lim_{t\downarrow 0}t/s_{1}(t)$ exists. By comparing
(3.20), we obtain

$\lim_{t\downarrow{0}}\frac{t}{s_{1}(t)}=\frac{\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2})}{\lambda_{1}^{2}(\mu_{1})}.$

We multiply the factor $(t+f_{\mu_{2}}(s_{2}(t)))$ to the second equation in
(3.9) and then rewrite it as follows

(3.21)
$\displaystyle(t+f_{\mu_{2}}(s_{2}(t))(s_{2}(t)-t)=(t+f_{\mu_{2}}(s_{2}(t)))f_{\mu_{1}}(t+f_{\mu_{2}}(s_{2}(t))).$

Then,

$\lim_{t\downarrow{0}}f_{\mu_{2}}(s_{2}(t))=\lim_{s\uparrow\infty}f_{\mu_{2}}(s)=0,$

which yields

(3.22)
$\lim_{t\downarrow{0}}(t+f_{\mu_{2}}(s_{2}(t)))f_{\mu_{1}}(t+f_{\mu_{2}}(s_{2}(t)))=\lim_{s\downarrow{0}}sf_{\mu_{1}}(s)=\lambda_{1}^{2}(\mu_{1}).$

Taking the limit for the left hand side of (3.21) similar to (3) (by switching
roles of $\mu_{1}$ and $\mu_{2}$), we deduce that

(3.23)
$\displaystyle\lim_{t\downarrow{0}}(t+f_{\mu_{2}}(s_{2}(t))(s_{2}(t)-t)$
$\displaystyle=\lim_{t\downarrow{0}}t\cdot
s_{2}(t)+\lim_{s\uparrow\infty}sf_{\mu_{2}}(s)$
$\displaystyle=\lim_{t\downarrow{0}}t\cdot s_{2}(t)+\lambda_{2}^{2}(\mu_{2}).$

By comparing (3.22) with (3.23), we deduce that

$\lim_{t\downarrow{0}}t\cdot
s_{2}(t)=\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2}).$

(3) Assume that $s_{1}(0)=\infty,s_{2}(0)=0$, then we may switch roles of
$\mu_{1}$ and $\mu_{2}$ and results follow from Case (2).

(4) Assume that $s_{1}(0)=s_{2}(0)=0$, then by (3.6),

(3.24) $\lim_{t\downarrow 0}h_{\mu_{1}\boxplus\mu_{2}}(t)=\lim_{t\downarrow
0}\frac{1}{s_{1}(t)+s_{2}(t)+t}=\infty.$

This means $-\lim_{\varepsilon\rightarrow 0}\Im
G_{\mu_{1}\boxplus\mu_{2}}(i\varepsilon)=\infty$. Hence,

(3.25) $\lim_{t\rightarrow 0}\frac{1}{h_{\mu_{1}\boxplus\mu_{2}}(t)}=0,$

and equivalently, $\lim_{t\rightarrow 0}F_{\mu_{1}\boxplus\mu_{2}}(it)=0$. By
(3.6), we see that

$\lim_{t\rightarrow 0}(s_{1}(t)+s_{2}(t))=0.$

Hence, $\lim_{t\rightarrow 0}s_{j}(t)=0$ for $j=1,2$. Recall that
$\omega_{j}(i\varepsilon)$ is a pure imaginary number for $\varepsilon>0$ and
$j=1,2$. It folllows that $\omega_{j}(0)=\lim_{t\rightarrow 0}\omega(it)=0$.
By item (2) of Proposition 2.2, we have

$\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})\geq 1.$

Finally, (3.25) and $s_{1}(0)=0$ also imply that

$f_{\mu_{1}}(s_{1}(t))=\frac{1}{h_{\mu_{1}}(s_{1}(t)}-s_{1}(t)\longrightarrow
0$

as $t\rightarrow 0$. Hence,

$\lim_{t\downarrow
0}(t+f_{\mu_{1}}(s_{1}(t)))f_{\mu_{2}}(t+f_{\mu_{1}}(s_{1}(t)))=\lim_{s\uparrow
0}sf_{\mu_{2}}(s)=\lambda_{1}^{2}(\mu_{2}).$

On the other hand, we have

$\displaystyle\lim_{t\downarrow 0}(t+f_{\mu_{1}}(s_{1}(t))(s_{1}(t)-t)$
$\displaystyle=\lim_{t\downarrow{0}}\left(s_{1}(t)f_{\mu_{1}}(s_{1}(t))-tf_{\mu_{1}}(s_{1}(t))\right)$
$\displaystyle=\lim_{s\downarrow 0}sf_{\mu_{1}}(s)+\lim_{t\downarrow
0}tf_{\mu_{1}}(s_{1}(t))$ $\displaystyle=\lambda_{1}^{2}(\mu_{1}).$

It follows from (3.15) that $\lambda_{1}(\mu_{1})=\lambda_{1}(\mu_{2})$.

We next show that $\lambda_{1}(\mu_{1})=0$ by contradiction. If
$\lambda_{1}(\mu_{1})=\lambda_{1}(\mu_{2})=r\neq 0$, then the assumption
$s_{1}(0)=s_{2}(0)=0$ implies that

$\displaystyle\lim_{t\downarrow 0}h_{\mu_{1}}(s_{1}(t))$
$\displaystyle=\lim_{s\downarrow 0}h_{\mu_{1}}(s)$
$\displaystyle=\lim_{s\downarrow
0}\int_{\mathbb{R}}\frac{s}{s^{2}+u^{2}}d\mu_{1}(u)$
$\displaystyle\leq\lim_{s\downarrow
0}\int_{\mathbb{R}}\frac{s}{u^{2}}d\mu_{1}(u)=0.$

This contradicts to (3.24). Hence,
$\lambda_{1}(\mu_{1})=\lambda_{1}(\mu_{2})=0$ when $s_{1}(0)=s_{2}(0)=0$. ∎

######  Lemma 3.4.

Let $\mu_{1},\mu_{2}$ be two symmetric probability measures on $\mathbb{R}$,
neither of them is a single delta mass. If $\lim_{t\downarrow
0}s_{1}(t)\in(0,\infty)$, then the two open intervals
$(\lambda_{1}(\mu_{1}),\lambda_{2}(\mu_{1}))$ and
$(\lambda_{1}(\mu_{2}),\lambda_{2}(\mu_{2}))$ must have nonempty intersection:

$(\lambda_{1}(\mu_{1}),\lambda_{2}(\mu_{1}))\cap(\lambda_{1}(\mu_{2}),\lambda_{2}(\mu_{2}))\neq\varnothing.$

###### Proof.

We set $s=\lim_{t\downarrow 0}s_{1}(t)\in(0,\infty)$. Notice now
$f_{\mu_{1}}(s)=-i\left(\frac{1}{G_{{\mu}_{1}}(is)}-is\right)=\Im\left(F_{{\mu}_{1}}(is)-is\right)>0$
and also $f_{\mu_{1}}(s)<\infty$. Similarly, $0<f_{\mu_{2}}(s)<\infty$. By
taking the limit as $t$ goes to zero for both side of the fixed equation
(3.9), we obtain

$s=f_{\mu_{2}}(f_{\mu_{1}}(s))=\frac{1}{f_{\mu_{1}}(s)}{f_{\mu_{1}}(s)f_{\mu_{2}}(f_{\mu_{1}}(s))}=\frac{1}{f_{\mu_{1}}(s)}\frac{n_{\mu_{2}}(f_{\mu_{1}}(s))}{d_{\mu_{2}}(f_{\mu_{1}}(s))}.$

Hence, we must have

(3.26)
$sf_{\mu_{1}(s)}=\frac{n_{\mu_{1}}(s)}{d_{\mu_{1}}(s)}=\frac{n_{\mu_{2}}(f_{\mu_{1}}(s))}{d_{\mu_{2}}(f_{\mu_{1}}(s))}.$

If
$(\lambda_{1}(\mu_{1}),\lambda_{2}(\mu_{1}))\cap(\lambda_{1}(\mu_{2}),\lambda_{2}(\mu_{2}))=\varnothing$,
we have either $\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$ or
$\lambda_{1}(\mu_{2})\geq\lambda_{2}(\mu_{1})$. If
$\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$, by Lemma 3.2, we have

$\frac{n_{\mu_{1}}(s)}{d_{\mu_{2}}(s)}>\lim_{t\downarrow
0}\frac{n_{\mu_{1}}(t)}{d_{\mu_{1}}(t)}=\lambda_{1}^{2}(\mu_{1})\geq\lambda_{2}^{2}(\mu_{2})=\lim_{t\uparrow\infty}\frac{n_{\mu_{2}}(t)}{d_{\mu_{2}}(t)}>\frac{n_{\mu_{2}}(f_{\mu_{1}}(s))}{d_{\mu_{2}}(f_{\mu_{1}}(s))}$

which contradicts to (3.26). Similarly, if
$\lambda_{1}(\mu_{2})\geq\lambda_{2}(\mu_{1})$, we have

$\frac{n_{\mu_{1}}(s)}{d_{\mu_{2}}(s)}<\lim_{t\uparrow\infty}\frac{n_{\mu_{1}}(t)}{d_{\mu_{1}}(t)}\leq\lim_{t\downarrow
0}\frac{n_{\mu_{2}}(t)}{d_{\mu_{2}}(t)}<\frac{n_{\mu_{2}}(f_{\mu_{1}}(s))}{d_{\mu_{2}}(f_{\mu_{1}}(s))}$

which again contradicts to (3.26). Hence if $\lim_{t\downarrow
0}s_{1}(t)\in(0,\infty)$, the two intervals must have non-empty intersection.
∎

######  Corollary 3.5.

Let $\mu_{1},\mu_{2}$ be two symmetric probability measures on $\mathbb{R}$,
neither of them is a single delta mass at zero. If
$\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})=1$, then $s_{1}(0)=s_{2}(0)=0$.

###### Proof.

For a probability measur $\nu$ on $\mathbb{R}$. It is known (see [6] or [1,
Lemm 2.17]) that $\nu(\\{x\\})=\lim_{t\downarrow 0}(it)G_{\nu}(x+it)$. If
$\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})=1$, then $\mu_{j}(\\{0\\})>0$ for $j=1,2$.
Hence, $\lim_{t\downarrow 0}G_{\mu_{j}}(it)=\infty$ and $\lim_{t\downarrow
0}F_{\mu_{j}}(it)=0$. Then, by [1, Lemma 2.13], the Julia-Carathéodory
derivative at zero is given by

$\liminf_{z\rightarrow 0}\frac{\Im F_{\mu_{j}}(z)}{\Im z}=\lim_{t\downarrow
0}\frac{F_{\mu_{j}}(it)}{it}=\frac{1}{\mu_{j}(\\{0\\})}.$

Therefore the Julia-Carathéodory of $H_{j}$ at zero are given by

$H_{j}^{\prime}(0)=\frac{1-\mu_{j}(\\{0\\})}{\mu_{(0)}},\qquad j=1,2,$

where $H_{j}(z)=F_{\mu_{j}}(z)-z$, and $H_{j}(it)$ tends to zero
nontangentially as $t\downarrow 0$. Consequently, the function $H_{2}\circ
H_{1}$ has Julia-Carathéodory derivative

(3.27) $(H_{2}\circ
H_{1})^{\prime}(0)=\frac{1-\mu_{1}(\\{0\\})}{\mu_{1}(\\{0\\})}\times\frac{1-\mu_{2}(\\{0\\})}{\mu_{2}(\\{0\\})}=1.$

Let $g(z,w)=z+H_{2}(z+H_{1}(2))$. For any $z\in\mathbb{C}^{+}\cup\mathbb{R}$,
it is known [2, 3] that $\omega_{1}(z)$ is the Denjoy-Wolf point of the
function $w\rightarrow g(z,w)$. In particular, $\omega_{1}(0)$ is the Denjoy-
Wolff point of $H_{2}\circ H_{1}$. Because $\lim_{t\downarrow
0}H_{2}(H_{1}(it))=0$ and (3.27), we deduce that the Denjoy-Wolf point of
$H_{2}\circ H_{1}$ at zero is zero. Hence, $\omega_{1}(0)=0$. By using
$H_{1}\circ H_{2}$, the same argument shows $\omega_{2}(0)=0$. We conclude
that $s_{1}(0)=s_{2}(0)=0$. ∎

######  Theorem 3.6.

Let $\mu_{1},\mu_{2}$ be two symmetric probability measures on $\mathbb{R}$,
neither of them is a single delta mass, and let $s_{1},s_{2}$ be subordination
function defined in Proposition 3.1. Then we have

  1. (1)

If $\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$, then $\lim_{t\downarrow
0}s_{1}(t)=0$ and $\lim_{t\downarrow 0}s_{2}(t)=\infty$.

  2. (2)

If $\lambda_{1}(\mu_{2})\geq\lambda_{2}(\mu_{1})$, then $\lim_{t\downarrow
0}s_{1}(t)=\infty$ and $\lim_{t\downarrow 0}s_{2}(t)=0$.

  3. (3)

If
$(\lambda_{1}(\mu_{1}),\lambda_{2}(\mu_{1}))\cap(\lambda_{1}(\mu_{2}),\lambda_{2}(\mu_{2}))\neq\varnothing$,
and either $\lambda_{1}(\mu_{1})>0$ or $\lambda_{1}(\mu_{2})>0$, then
$s_{1}(0)=\lim_{t\downarrow 0}s_{1}(t)\in(0,\infty)$ and
$s_{2}(0)=\lim_{t\downarrow 0}s_{2}(t)\in(0,\infty)$. Moreover,

(3.28)
$s_{1}(0)+s_{2}(0)=\frac{1}{h_{\mu_{1}}(s_{1}(0))}=\frac{1}{h_{\mu_{2}}(s_{2}(0))}.$

  4. (4)

If $\lambda_{1}(\mu_{1})=\lambda_{1}(\mu_{2})=0$, then
$s_{1}(0),s_{2}(0)\in[0,\infty)$. Moreover, $s_{1}(0)=s_{2}(0)=0$ if and only
if $\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})\geq 1$.

###### Proof.

If $\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$ then the limit
$\lim_{t\downarrow 0}s_{1}(t)$ can not be finite due to Lemma 3.4. If
$s_{1}(0)=s_{2}(0)=0$, then by item (4) of Lemma 3.3, we have

$\lambda_{1}(\mu_{2})=\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})>\lambda_{1}(\mu_{2})$

which contradicts to the assumption that $\mu_{2}$ is not a single delta mass.
By comparing (2) and (3) in Lemma 3.3, we see that $\lim_{t\downarrow
0}s_{1}(t)=0$ and $\lim_{t\downarrow 0}s_{2}(t)=\infty$ which proves (1). Then
(2) follows in the same way.

Assume now
$(\lambda_{1}(\mu_{1}),\lambda_{2}(\mu_{1}))\cap(\lambda_{1}(\mu_{2}),\lambda_{2}(\mu_{2}))\neq\varnothing$.
By (1) in Lemma 3.3, at least one of $s_{1}(0),s_{2}(0)$ must be finite. If
$s_{1}(0)=0,s_{2}(0)=\infty$ or $s_{1}(0)=\infty,s_{2}(0)=0$, then we will
have $\lambda_{1}(\mu_{1})\geq\lambda_{2}(\mu_{2})$ or
$\lambda_{1}(\mu_{2})\geq\lambda_{2}(\mu_{1})$, which contradicts to our
assumption. By (4) in Lemma 3.3, $s_{1}(0),s_{2}(0)$ can not be both equal to
zero. We then conclude that $0<s_{1}(0),s_{2}(0)<\infty$. This proves (3).

Assume that $\lambda_{1}(\mu_{1})=\lambda_{1}(\mu_{2})=0$. If
$s_{1}(0)=\infty$, we deduce that $\lambda_{1}(\mu_{1})=\lambda_{2}(\mu_{1})$,
which contradicts to the assumption that $\mu_{1}$ is not a single delta mass.
Hence, $0\leq s_{1}(0)<\infty$. Similarly, $0\leq s_{2}(0)<\infty$. By (4) in
Lemma 3.3, $s_{1}(0)=s_{2}(0)=0$ implies that
$\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})\geq 1$.

For the necessary part of (4), if $\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})>1$, then,
by Theorem 2.1, we have

$(\mu_{1}\boxplus\mu_{2})(\\{0\\})=\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})-1>0.$

Hence, $F_{\mu_{1}\boxplus\mu_{2}}(0)=0$. By using
$\frac{1}{h_{\mu_{1}\boxplus\mu_{2}}(0)}=\Im F_{\mu_{1}\boxplus\mu_{2}}(0)=0$
and

$s_{1}(t)+s_{2}(t)=\frac{1}{h_{\mu_{1}\boxplus\mu_{2}}(t)}-t,$

we deduce that $s_{1}(0)=s_{2}(0)=0$. Finally, if
$\mu_{1}(\\{0\\})+\mu_{2}(\\{0\\})=1$, then $s_{1}(0)=s_{2}(0)=0$ thanks to
Corollary 3.5. ∎

We end this section with an application to the inner and outer radii of the
support of the sum of two $R$-diagonal operators.

######  Proposition 3.7.

Let $T_{1},T_{2}\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal that are freely
independent. Denote by $r_{1}=\lambda_{1}(\mu_{|T_{1}+T_{2}|})$ the inner
radii of the support of the Brown measure of $T_{1}+T_{2}$.

  1. (1)

If $0<\lambda_{2}(\mu_{|T_{2}|})\leq\lambda_{1}(\mu_{|T_{1}|})$, then
$r_{1}={\lambda_{1}(\mu_{|T_{1}|})^{2}-\lambda_{2}(\mu_{|T_{2}|})^{2}}$.

  2. (2)

If $0<\lambda_{2}(\mu_{|T_{1}|})\leq\lambda_{1}(\mu_{|T_{2}|})$, then
$r_{1}={\lambda_{1}(\mu_{|T_{2}|})^{2}-\lambda_{2}(\mu_{|T_{1}|})^{2}}$.

  3. (3)

If
$(\lambda_{1}(\mu_{|T_{1}|}),\lambda_{2}(\mu_{|T_{1}|}))\cap(\lambda_{1}(\mu_{|T_{2}|}),\lambda_{2}(\mu_{|T_{2}|}))\neq\varnothing$,
then $r_{1}=0$.

###### Proof.

With $\mu_{1}=\tilde{\mu}_{|T_{1}|}$ and $\mu_{2}=\tilde{\mu}_{|T_{2}|}$, let
the function $h_{\mu_{1}},h_{\mu_{2}}$ and the subordination function
$s_{1}(t),s_{2}(t)$ be as in Section 3. When
$0<\lambda_{2}(\mu_{|T_{2}|})\leq\lambda_{1}(\mu_{|T_{1}|})$, by Theorem 3.6,
$\lim_{t\rightarrow 0^{+}}s_{1}(t)=0$ and $\lim_{t\rightarrow
0^{+}}s_{1}(t)=\infty$. By item (2) in Lemma 3.3, we have

$\lim_{t\downarrow
0}\frac{t}{s_{1}(t)}=\frac{\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2})}{\lambda_{1}^{2}(\mu_{1})}.$

Therefore,

$\displaystyle\lambda_{1}(\mu_{|T_{1}+T_{2}|})$
$\displaystyle=\frac{1}{\phi\large([(T_{1}+T_{2})^{*}(T_{1}+T_{2})]^{-1}\large)}$
$\displaystyle=\lim_{t\rightarrow
0^{+}}\frac{1}{\phi\large([(T_{1}+T_{2})^{*}(T_{1}+T_{2})+t^{2}]^{-1}\large)}$
$\displaystyle=\lim_{t\rightarrow
0^{+}}\frac{t}{h_{{\mu}_{1}\boxplus{\mu}_{2}}(t)}$
$\displaystyle=\lim_{t\rightarrow
0^{+}}\frac{s_{1}(t)}{h_{\mu_{1}}(s_{1}(t))}\frac{t}{s_{1}(t)}$
$\displaystyle=\lim_{t\rightarrow
0^{+}}\frac{s_{1}(t)}{h_{\mu_{1}}(s_{1}(t))}\lim_{t\rightarrow
0^{+}}\frac{t}{s_{1}(t)}$
$\displaystyle=\lambda_{1}^{2}(\mu_{1})\frac{\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2})}{\lambda_{1}^{2}(\mu_{1})}$
$\displaystyle={\lambda_{1}^{2}(\mu_{1})-\lambda_{2}^{2}(\mu_{2}}).$

The case when $0<\lambda_{2}(\mu_{|T_{1}|})\leq\lambda_{1}(\mu_{|T_{2}|})$
follows by symmetric role of $T_{1}$ and $T_{2}$.

When
$(\lambda_{1}(\mu_{|T_{1}|}),\lambda_{2}(\mu_{|T_{1}|}))\cap(\lambda_{1}(\mu_{|T_{2}|}),\lambda_{2}(\mu_{|T_{2}|}))\neq\varnothing$,
by part (3) and (4) in Theorem 3.6, we have $s_{1}(0),s_{2}(0)\in[0,\infty)$.

  * •

If $s_{1}(0),s_{2}(0)\in(0,\infty)$, then

$\displaystyle\frac{1}{\phi\large([(T_{1}+T_{2})^{*}(T_{1}+T_{2})]^{-1}\large)}=\lim_{t\rightarrow
0^{+}}\frac{s_{1}(t)}{h_{\mu_{1}}(s_{1}(t))}\frac{t}{s_{1}(t)}=0.$

  * •

If $s_{1}(0)=s_{2}(0)=0$, then
$\lambda_{1}(\mu_{|T_{1}|})=\lambda_{1}(\mu_{|T_{2}|})=0$. We note that

$\lim_{s\downarrow 0}\frac{s}{h_{\mu_{1}}(s)}=\lim_{s\downarrow
0}\frac{1}{\int_{\mathbb{R}}\frac{1}{s^{2}+u^{2}}d\mu_{1}(u)}=\lambda_{1}^{2}(\mu_{1})=0.$

Since $t\leq s_{1}(t)$, we have

$\displaystyle\frac{1}{\phi\large([(T_{1}+T_{2})^{*}(T_{1}+T_{2})]^{-1}\large)}=\lim_{t\rightarrow
0^{+}}\frac{s_{1}(t)}{h_{\mu_{1}}(s_{1}(t))}\frac{t}{s_{1}(t)}=0.$

This finishes the proof. ∎

######  Proposition 3.8.

Let $T_{1},T_{2}\in{\mathcal{M}}$ be bounded $R$–diagonal that are freely
independent. We have

$\lambda_{2}^{2}(T_{1}+T_{2})=\lambda_{2}^{2}(T_{1})+\lambda_{2}^{2}(T_{2}).$

##  4\. The sum of an $R$-diagonal operator and a free random variable

In this section, we outline an approach to the calculation of the Brown
measure of the sum of an $R$-diagonal operator and a free random variable with
an arbitrary distribution using subordination functions. This answers some
questions asked by Biane–Lehner [9].

Let $T\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal and
$x_{0}\in{\mathcal{M}}^{\Delta}$ be be a operator that is freely independent
from $T$. Let $U$ be a Haar unitary that is freely independent from
$\\{x_{0},T\\}$. Then, for any $\lambda\in\mathbb{C}$, we have

(4.1) $|x_{0}+T-\lambda{\bf
1}|\underset{\ast{\mathcal{D}}}{\sim}|x_{0}-\lambda{\bf
1}+U^{*}T|\underset{\ast{\mathcal{D}}}{\sim}|U(x_{0}-\lambda{\bf 1})+T|.$

The operator $U(x_{0}-\lambda{\bf 1})$ is $R$-diagonal. Hence, for any
$\lambda$ fixed, the distribution of $|x_{0}+T-\lambda{\bf 1}|$ can be
calculated by studying the sum of two $R$-diagonal operators. This observation
appearing in [9] is the key for us to study the Brown measure of $x_{0}+T$. We
denote $\mu_{1}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}|}$ and
$\mu_{2}=\widetilde{\mu}_{|T|}$, and
$\mu=\widetilde{\mu}_{|x_{0}+T-\lambda{\bf 1}|}$. By Proposition 2.5, we have
$\mu=\mu_{1}\boxplus\mu_{2}$. That is,

$\widetilde{\mu}_{|x_{0}+T-\lambda{\bf
1}|}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}|}\boxplus\widetilde{\mu}_{|T|}.$

Let $x\in{\mathcal{M}}^{\Delta}$ and $\varepsilon>0$. The following
observation provides a link between Fuglede-Kadison determinant and the Cauchy
transform,

$\frac{1}{2}\frac{\partial\log\Delta(|x|^{2}+\varepsilon^{2})}{\partial\varepsilon}=\int_{0}^{\infty}\frac{\varepsilon}{u^{2}+\varepsilon^{2}}d\tilde{\mu}_{|x|}(u)=h_{\widetilde{\mu}_{|x|}}(\varepsilon).$

Recall we also have
$G_{\widetilde{\mu}_{|x|}}(i\varepsilon)=-ih_{\widetilde{\mu}_{|x|}}(\varepsilon)$.

We restate the main theorem 1.1 as follows.

######  Theorem 4.1.

Let $T\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal and
$x_{0}\in{\mathcal{M}}^{\Delta}$ be be a operator that is freely independent
from $T$. Then the support of the Brown measure $\mu_{T+x_{0}}$ is contained
in the closure of following open set

(4.2)
$\Omega(T,x_{0})=\big{\\{}\lambda\in\mathbb{C}:||(x_{0}-\lambda)^{-1}||_{2}||T||_{2}>1\,\,\text{and}\,\,||x_{0}-\lambda||_{2}||T^{-1}||_{2}>1\big{\\}}.$

Moreover, there is no atom in the complement of $\Omega(T,x_{0})$.

We can write the complement of $\Omega(T,x_{0})$ as

$\displaystyle\mathbb{C}\backslash\Omega(T,x_{0})$
$\displaystyle=\left\\{\lambda\in\mathbb{C}:\lambda_{1}(\mu_{|x_{0}-\lambda|})\geq\lambda_{2}(\mu_{|T|})\,\,\text{or}\,\,\lambda_{2}(\mu_{|x_{0}-\lambda|})\leq\lambda_{1}(\mu_{|T|})\right\\}$
$\displaystyle=\left\\{\lambda\in\mathbb{C}:\lambda_{1}(\mu_{|x_{0}-\lambda|})\geq\lambda_{2}(\mu_{|T|})\right\\}\cup\left\\{\lambda\in\mathbb{C}:\lambda_{2}(\mu_{|x_{0}-\lambda|})\leq\lambda_{1}(\mu_{|T|})\right\\}.$

By Lemma 2.3, the two sets on the right hand side in above identity are
disjoint. The function $\lambda\mapsto\lambda_{1}(\mu_{|x_{0}-\lambda|})$ is
lower semicontinuous (see the Proof for Theorem 4) and the function
$\lambda\mapsto\lambda_{2}(\mu_{|x_{0}-\lambda|})$ is continuous. Hence, both
sets are closed.

######  Example 4.2.

If $x_{0}$ is selfadjoint, then a direct calculation shows that the complement
can be decomposed as the union of two close sets
$\mathbb{C}\backslash\Omega(T,x_{0})=A\cup B$, where

$A=\bigg{\\{}\lambda=a+bi:\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+b^{2}}d\mu_{x_{0}}(u)\leq\frac{1}{\phi(|T|^{2})}\bigg{\\}}.$

and

$B=\left\\{\lambda=a+bi:(a-\phi(x_{0}))^{2}+b^{2}\leq\frac{1}{\phi(|T^{-1}|^{2})}-\text{var}({x_{0}})\right\\},$

where $\text{var}({x_{0}})=\phi(x_{0}^{2})-(\phi(x_{0}))^{2}$ is the variance
of $x_{0}$. Then $B$ is the empty set if and only if
$\lambda_{1}(\mu_{|T|})^{2}-\text{var}(x_{0})<0$. This explains that the
bounded component in Figure 1 is a disk, in which case $B$ is nonempty.

######  Remark 4.3.

Let $\mu_{1}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}|}$ and
$\mu_{2}=\widetilde{\mu}_{|T|}$, and let $s_{1},s_{2}$ be the subordination
functions as in Proposition 3.1. Then, by Theorem 3.6, the following
statements are equivalent:

  1. (1)

$\lambda\in\Omega(T,x_{0})$;

  2. (2)

$\lambda_{1}(\mu_{|x_{0}-\lambda|})<\lambda_{2}(\mu_{|T|})$ and
$\lambda_{1}(\mu_{|T|})<\lambda_{2}(\mu_{|x_{0}-\lambda|})$;

  3. (3)

$s_{1}(0),s_{2}(0)\in[0,\infty)$.

######  Lemma 4.4.

Let $T_{1},T_{2}\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal that are freely
independent. With $\mu_{1}=\tilde{\mu}_{|T_{1}|}$ and
$\mu_{2}=\tilde{\mu}_{|T_{2}|}$, the function $h_{\mu_{1}},h_{\mu_{2}}$ and
the subordination function $s_{1}(t),s_{2}(t)$ as in Section 3, we then have

(4.3) $\displaystyle\Delta\large((T_{1}+T_{2})^{*}(T_{1}+T_{2})+t^{2}{\bf
1}\large)=\frac{\Delta(T_{1}^{*}T_{1}+s_{1}(t)^{2})\Delta(T_{2}^{*}T_{2}+s_{2}(t)^{2})}{(s_{1}(t)+s_{2}(t)-t)^{2}}.$

###### Proof.

We set

$H_{T_{1}+T_{2}}(t)=\frac{1}{2}\log\Delta\large((T_{1}+T_{2})^{*}(T_{1}+T_{2})+t^{2}{\bf
1}\large)$

and for $i=1,2$,

$H_{T_{i}}(t)=\frac{1}{2}\log\Delta\large(T_{i}^{*}T_{i}+t^{2}{\bf 1}\large).$

With $\mu_{1}=\tilde{\mu}_{|T_{1}|}$ and $\mu_{2}=\tilde{\mu}_{|T_{2}|}$, set
$\mu=\tilde{\mu}_{|T_{1}+T_{2}|}=\mu_{1}\boxplus\mu_{2}$. We have

$H_{T_{1}+T_{2}}(t)=\frac{1}{2}\int_{0}^{\infty}\log(u^{2}+t^{2})d\mu(u),$

and

$H_{T_{i}}(t)=\frac{1}{2}\int_{0}^{\infty}\log(u^{2}+t^{2})d\mu_{i}(u).$

We have

(4.4) $\lim_{t\rightarrow\infty}(H_{T_{1}+T_{2}}(t)-\log
t)=\frac{1}{2}\lim_{t\rightarrow\infty}\log\left(1+\frac{u^{2}}{t^{2}}\right)d\mu(u)=0$

and similarly

(4.5) $\lim_{t\rightarrow\infty}(H_{T_{i}}(t)-\log t)=0.$

Fix $t_{0}>0$ and recall that

$s_{1}(t)+s_{2}(t)=t+\frac{1}{h_{\mu}(t)}.$

By definition, we have

$H_{T_{1}+T_{2}}^{\prime}(t)=h_{\mu}(t),\quad
H_{T_{i}}^{\prime}(t)=h_{\mu_{i}}(t).$

It follows that

$\displaystyle\int_{t_{0}}^{t}h_{\mu}$ $\displaystyle(v)dv$
$\displaystyle=\int_{t_{0}}^{t}h_{\mu}(v)\bigg{(}s_{1}(v)+s_{2}(v)-\frac{1}{h_{\mu}(v)}\bigg{)}^{\prime}dv+$
$\displaystyle=\int_{t_{0}}^{t}h_{\mu_{1}}(s_{1}(v))s_{1}^{\prime}(v)dv+\int_{t_{0}}^{t}h_{\mu_{2}}(s_{2}(v))s_{2}^{\prime}(v)dv+\int_{t_{0}}^{t}\frac{h_{\mu}^{\prime}(v)}{h_{\mu}(v)}dv$
$\displaystyle=H_{T_{1}}(s_{1}(t))-H_{T_{1}}(s_{1}(t_{0}))+H_{T_{2}}(s_{2}(t))$
$\displaystyle\qquad\qquad\qquad-H_{T_{2}}(s_{2}(t_{0}))+\log h_{\mu}(t)-\log
h_{\mu}(t_{0}).$

Hence, there exists a constant $C$ such that

(4.6) $H_{T_{1}+T_{2}}(t)=H_{T_{1}}(s_{1}(t))+H_{T_{2}}(s_{2}(t))+\log
h_{\mu}(t)+C.$

The asymptotics (4.4) and (4.5) as $t\rightarrow\infty$ implies that $C$ must
be zero. Finally, note

$h_{\mu}(t)=\frac{1}{s_{1}(t)+s_{2}(t)-t},$

we obtain

$\exp(2H_{T_{1}+T_{2}}(t))=\exp(2H_{T_{1}}(t))\exp(H_{T_{2}}(t))\frac{1}{(s_{1}(t)+s_{2}(t)-t)^{2}}$

which is (4.3). ∎

######  Theorem 4.5.

Let $T_{1},T_{2}\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal that are freely
independent.

  1. (1)

If $\lambda_{1}(\mu_{|T_{1}})\geq\lambda_{2}(\mu_{|T_{2}|})$, then
$\Delta(T_{1}+T_{2})=\Delta(T_{1})$.

  2. (2)

If $\lambda_{1}(\mu_{|T_{2}})\geq\lambda_{2}(\mu_{|T_{1}|})$, then
$\Delta(T_{1}+T_{2})=\Delta(T_{2})$.

  3. (3)

If
$(\lambda_{1}(\mu_{|T_{1}|}),\lambda_{2}(\mu_{|T_{1}|}))\cap(\lambda_{1}(\mu_{|T_{2}|}),\lambda_{2}(\mu_{|T_{2}|}))\neq\varnothing$
and $s_{1}(0),s_{2}(0)\in(0,\infty)$, then

(4.7)
$\Delta(T_{1}+T_{2})=\frac{\big{(}\Delta(T_{1}^{*}T_{1}+s_{1}(0)^{2})\Delta(T_{2}^{*}T_{2}+s_{2}(0)^{2})\big{)}^{\frac{1}{2}}}{s_{1}(0)+s_{2}(0)}.$

###### Proof.

By Theorem 3.6, when
$\lambda_{1}(\mu_{|T_{1}})\geq\lambda_{2}(\mu_{|T_{2}|})$, we have
$\lim_{t\rightarrow 0^{+}}s_{1}(t)=0$ and $\lim_{t\rightarrow
0^{+}}s_{2}(t)=\infty$. We rewrite the right hand side of (4.3) as

$\displaystyle\Delta(T_{1}^{*}T_{1}+s_{1}(t)^{2})\frac{\Delta(T_{2}^{*}T_{2}+s_{2}(t)^{2})}{s_{2}(t)^{2}}\frac{s_{2}(t)^{2}}{(s_{1}(t)+s_{2}(t)-t)^{2}},$

where the first factor converges to $\Delta(T_{1}^{*}T_{1})$,and both the
second and third factor converge to $1$ as $t\rightarrow 0^{+}$. Now (4.7)
follows from this as

$\Delta(T_{1}+T_{2})^{2}=\lim_{t\rightarrow
0^{+}}\Delta\large((T_{1}+T_{2})^{*}(T_{1}+T_{2})+t^{2}{\bf 1}\large).$

The case when $\lambda_{1}(\mu_{|T_{2}})\geq\lambda_{2}(\mu_{|T_{1}|})$ is
similar, in which case $\lim_{t\rightarrow 0^{+}}s_{1}(t)=\infty$ and
$\lim_{t\rightarrow 0^{+}}s_{2}(t)=0$.

Finally, if
$(\lambda_{1}(\mu_{|T_{1}|}),\lambda_{2}(\mu_{|T_{1}|}))\cap(\lambda_{1}(\mu_{|T_{2}|}),\lambda_{2}(\mu_{|T_{2}|}))\neq\varnothing$
and $s_{1}(0),s_{2}(0)\in(0,\infty)$, then (4.7) follows from (4.3) by letting
$t\rightarrow 0^{+}$. ∎

######  Remark 4.6.

If
$(\lambda_{1}(\mu_{|T_{1}|}),\lambda_{2}(\mu_{|T_{1}|}))\cap(\lambda_{1}(\mu_{|T_{2}|}),\lambda_{2}(\mu_{|T_{2}|}))\neq\varnothing$
and $s_{1}(0)=s_{2}(0)=0$, then by Lemma 3.3 and Theorem 3.6, we must have

$\lambda_{1}(\mu_{|T_{1}|})=\lambda_{1}(\mu_{|T_{2}|})=0,$

and $\mu_{|T_{1}|}(\\{0\\})+\mu_{|T_{2}|}(\\{0\\})\geq 1$.

######  Proposition 4.7 ([14]).

If $T\in{\mathcal{M}}^{\Delta}$ with

(4.8) $\int_{0}^{1}\log t\,{\rm d}\mu_{|T|}(t)>-\infty,$

then $\mu_{T}(\\{0\\})=\mu_{|T|}(\\{0\\})=0$, and $T$ has an inverse
$T^{-1}\in{\mathcal{M}}^{\Delta}$. Moreover, $\mu_{T^{-1}}$ is the push-
forward measure of $\mu_{T}$ via the map $z\mapsto z^{-1}$.

The following result was proved in [24, Section 4.2]. We provide a proof for
completeness.

######  Lemma 4.8.

Let $\lambda\in\mathbb{C}$ and $x_{0}\in{\mathcal{M}}^{\Delta}$. If
$\phi(|x_{0}-\lambda{\bf 1}|^{-2})<\infty$, then

$\int_{\mathbb{C}}\frac{1}{|z-\lambda|^{2}}d\mu_{x_{0}}(z)\leq\phi(|x_{0}-\lambda{\bf
1}|^{-2}).$

If $\phi(|x_{0}-\lambda{\bf 1}|^{-2})<L$ in some neighborhood $U$ of
$\lambda_{0}$ for some $L>0$, then $\mu_{x_{0}}(U)=0$. In particular,
$\lambda_{0}\notin{\text{\rm supp}}(\mu_{x_{0}})$.

###### Proof.

Observe that for any $t\in(0,1)$, we have $2\log t>-\frac{1}{t^{2}}$. Hence,

$\displaystyle 2\int_{0}^{1}\log td\mu_{|x_{0}-\lambda{\bf 1}|}(t)$
$\displaystyle>-\int_{0}^{1}\frac{1}{t^{2}}d\mu_{|x_{0}-\lambda{\bf 1}|}(t)$
$\displaystyle>-\int_{0}^{\infty}\frac{1}{t^{2}}d\mu_{|x_{0}-\lambda{\bf
1}|}(t)$ $\displaystyle=-\phi(|x_{0}-\lambda{\bf 1}|^{-2})>-\infty.$

By Proposition 4.7, the operator $x_{0}-\lambda{\bf 1}$ has an inverse
$(x_{0}-\lambda{\bf 1})^{-1}\in{\mathcal{M}}^{\Delta}$ and the Brown measure
$\mu_{(x_{0}-\lambda{\bf 1})^{-1}}$ of $(x_{0}-\lambda{\bf 1})^{-1}$ can be
defined. Moreover, $\mu_{(x_{0}-\lambda{\bf 1})^{-1}}$ is the push-forward
measure of $\mu_{x_{0}}$ via the map $z\mapsto(z-\lambda)^{-1}$. Hence, by
[14, Theorem 2.19], we obtain

(4.9) $\displaystyle\int_{\mathbb{C}}\frac{1}{|z-\lambda|^{2}}d\mu_{x_{0}}(z)$
$\displaystyle=\int_{\mathbb{C}}|z|^{2}d\mu_{(x_{0}-\lambda{\bf 1})^{-1}}(z)$
$\displaystyle\leq||(x_{0}-\lambda{\bf
1})^{-1}||_{2}^{2}=\phi(|x_{0}-\lambda{\bf 1}|^{-2}).$

If $\phi(|x_{0}-\lambda{\bf 1}|^{-2})<L$ in some neighborhood $U$ of
$\lambda_{0}$ for some $L>0$. It then follows that

$\int_{\mathbb{C}}\frac{1}{|z-\lambda|^{2}}d\mu_{x_{0}}(z)<L$

for $\lambda\in U$, which yields that $\mu_{x_{0}}(U)=0$ (see [24, Lemma 4.8]
for example). Hence, $\lambda_{0}\notin{\text{\rm supp}}(\mu_{x_{0}})$. ∎

We are now ready to prove the main result of this section

###### Proof of Theorem 4.1.

Observe that
$||(x_{0}-\lambda)^{-1}||_{2}=\phi(|x_{0}-\lambda|^{-2})=\lim_{\varepsilon\downarrow
0}\phi((|x_{0}-\lambda|^{2}+\varepsilon)^{-1})$. The function
$\\{\phi((|x_{0}-\lambda|^{2}+1/n)^{-1})\\}_{n}$ is a increasing family of
continuous functions of $\lambda$. Hence, the function
$\lambda\mapsto||(x_{0}-\lambda)^{-1}||_{2}$ is lower semicontinuous. The
other function $\lambda\mapsto||x_{0}-\lambda||_{2}$ is a continuous function.
Hence, the set $\Omega(T,x_{0})$ is open.

Let $U$ be a Haar unitary operator that is free from $\\{T,x_{0}\\}$. For any
$\lambda\in\mathbb{C}$, as we observed earlier that

$|x_{0}+T-\lambda{\bf
1}|\underset{\ast{\mathcal{D}}}{\sim}|U^{*}T+x_{0}-\lambda{\bf
1}|\underset{\ast{\mathcal{D}}}{\sim}|T+U(x_{0}-\lambda{\bf 1}|,$

which yields

$\Delta(x_{0}+T-\lambda{\bf 1})=\Delta(T+U(x_{0}-\lambda{\bf 1}).$

We now apply Theorem 4.5 to our setting (for $T_{1}=U(x_{0}-\lambda{\bf 1})$
and $T_{2}=T$). Then, for any $\lambda\in\mathbb{C}\backslash\Omega(T,x_{0})$,
we have either

$||(x_{0}-\lambda)^{-1}||_{2}||T||_{2}\leq 1$

or

$||x_{0}-\lambda||_{2}||T^{-1}||_{2}\leq 1.$

We note

$||(x_{0}-\lambda)^{-1}||_{2}||T||_{2}\leq
1\Longleftrightarrow\lambda_{2}(\mu_{|T|})\leq\lambda_{1}(\mu_{|x_{0}-\lambda{\bf
1}|}))$

and

$||x_{0}-\lambda||_{2}||T^{-1}||_{2}\leq
1\Longleftrightarrow\lambda_{2}(\mu_{|x_{0}-\lambda{\bf
1}|}))\leq\lambda_{1}(\mu_{|T|})).$

Hence, if $\lambda_{0}$ is in the interior of
$\mathbb{C}\backslash\Omega(T,x_{0})$ and $||(x_{0}-\lambda{\bf
1})^{-1}||_{2}||T||_{2}\leq 1$ in some neighborhood $U$ of $\lambda_{0}$. That
is,

(4.10) $\phi(|x_{0}-\lambda{\bf 1}|^{-2})\leq\frac{1}{\phi(|T|^{2})}$

for $\lambda\in U$. By Lemma 4.8, we have $\mu_{x_{0}}(U)=0$. By Theorem 4.5,
in this case, we also have

(4.11) $\Delta(T+U(x_{0}-\lambda{\bf
1}))=\Delta(U(x_{0}-\lambda))=\Delta(x_{0}-\lambda{\bf 1}),$

which yields that $\mu_{x_{0}+T}$ and $\mu_{x_{0}}$ coincides in $U$ by taking
the Laplacian. Hence, $\mu_{x_{0}+T}(U)=0$ and $\lambda_{0}$ is not in the
support of the $\mu_{x_{0}+T}$.

Moreover, if $||(x_{0}-\lambda{\bf 1})^{-1}||_{2}||T||_{2}\leq 1$, then (4.10)
implies that $\phi(|x_{0}-\lambda{\bf 1}|^{-2})<\infty$. Consequently, by
observing that $2\log t>-\frac{1}{t^{2}}$ for $0<t<1$, we have

$\displaystyle 2\int_{0}^{1}\log t\,d\mu_{|x_{0}-\lambda|}(t)$
$\displaystyle>-\int_{0}^{1}\frac{1}{t^{2}}\,d\mu_{|x_{0}-\lambda|}(t)$
$\displaystyle>-\int_{0}^{\infty}\frac{1}{t^{2}}\,d\mu_{|x_{0}-\lambda|}(t)=-\phi(|x_{0}-\lambda|^{-2})>-\infty.$

Then, this together (4.11) imply that

$\int_{0}^{1}\log t\,d\mu_{|x_{0}+T-\lambda{\bf 1}|}>-\infty.$

By [14, Proposition 2.16], it follows that $\mu_{x_{0}+T-\lambda}(\\{0\\})=0$.
Therefore, we deduce $\mu_{x_{0}+T}(\\{\lambda\\})=0$

Similarly, if $\lambda_{0}$ is in the interior of
$\mathbb{C}\backslash\Omega(T,x_{0})$ and
$||x_{0}-\lambda||_{2}||T^{-1}||_{2}<1$, we have

(4.12) $\Delta(x_{0}+T-\lambda{\bf 1})=\Delta(T).$

Hence, within some neighborhood $U$ of $\lambda_{0}$, we have

$\nabla^{2}(\Delta(x_{0}+T-\lambda{\bf 1}))=\nabla^{2}(\Delta(T))=0,$

which means that $\mu_{x_{0}+T}(U)=0$ and $\lambda_{0}$ is not in the support
of $\mu_{x_{0}+T}$. If $||x_{0}-\lambda||_{2}||T^{-1}||_{2}<1$, then

$\lambda_{1}(\mu_{|T|})=\frac{1}{||T^{-1}||_{2}}>||x_{0}-\lambda||_{2}>0,$

because $x_{0}$ is not a scalar of identity. Hence, $\mu_{T}(\\{0\\})=0$ and
$\Delta(T)>0$. Therefore, by (4.12), $\Delta(x_{0}+T-\lambda{\bf 1})>0$. By
[14, Proposition 2.16], it follows that $\mu_{x_{0}+T-\lambda}(\\{0\\})=0$.
Therefore, we deduce $\mu_{x_{0}+T}(\\{\lambda\\})=0$.

We then conclude that the support of the Brown measure of $x_{0}+T$ is
contained in the closure of $\Omega(T,x_{0})$, and there is no atom at
$\lambda\in\mathbb{C}\backslash\Omega(T,x_{0})$. ∎

######  Definition 4.9.

Let $T\in{\mathcal{M}}^{\Delta}$ be $R$–diagonal and
$x_{0}\in{\mathcal{M}}^{\Delta}$ be be a operator that is freely independent
from $T$. Denote

$S(T,x_{0})=\\{\lambda\in\mathbb{C}:\mu_{|T|}(\\{0\\})+\mu_{|x_{0}-\lambda|}(\\{0\\})\geq
1\\}.$

######  Remark 4.10.

We point out that the set $S(T,x_{0})$ consists of finitely many complex
numbers. If $\lambda\in S(T,x_{0})$, then $\mu_{|x_{0}-\lambda|}(\\{0\\})\geq
1-\mu_{|T|}(\\{0\\})>0$. Hence, $\lambda$ is an eigenvalue of $x_{0}$ and its
eigenspace $p_{\lambda}$ has trace
$\phi(p_{\lambda})=\mu_{|x_{0}-\lambda|}(\\{0\\})$ with positive value. It
follows that $S(T,x_{0})$ is a finite set. The points in $S(T,x_{0})$ are
candidates for the atoms of the Brown measure $\mu_{T+x_{0}}$.

######  Proposition 4.11.

Let $\mu_{1}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}|}$ and
$\mu_{2}=\widetilde{\mu}_{|T|}$, and let $s_{1},s_{2}$ be the subordination
functions as in Proposition 3.1. Note that $s_{1},s_{2}$ are functions of
$\lambda$. The following statements are equivalent:

  1. (1)

$\lambda\in\Omega(T,x_{0})\backslash S(T,x_{0})$;

  2. (2)

$s_{1}(0),s_{2}(0)\in(0,\infty)$;

  3. (3)

the density function $f_{\mu_{1}\boxplus\mu_{2}}$ of the absolutely continuous
part of $\mu_{1}\boxplus\mu_{2}$ is finite and strictly positive
$0<f_{\mu_{1}\boxplus\mu_{2}}(0)<\infty$.

###### Proof.

The equivalence of $(1),(2)$ and $(3)$ follows from Remark 4.3, Lemma 3.3 and
Theorem 3.6. ∎

######  Proposition 4.12.

Let $\mu_{1}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}}$ and
$\mu_{2}=\widetilde{\mu}_{|T|}$, and let $s_{1},s_{2}$ be the subordination
functions as in Proposition 3.1. Then for any $\lambda\in\Omega(T,x_{0})$, we
have

(4.13) $\displaystyle\phi((\lambda{\bf 1}-x_{0}-T)^{*}$
$\displaystyle(|\lambda{\bf 1}-x_{0}-T|^{2})^{-1})$
$\displaystyle=\phi((\lambda{\bf 1}-x_{0})^{*}(|\lambda{\bf
1}-x_{0}|^{2}+s_{1}(0)^{2})^{-1})$

###### Proof.

Under these conditions, we have (4.7), which is rewritten as

(4.14) $\Delta(x_{0}+T-\lambda{\bf 1})^{2}=\frac{\Delta(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2})\Delta(|T|^{2}+s_{2}(0)^{2})}{(s_{1}(0)+s_{2}(0))^{2}}.$

Taking the logarithmic derivative, we obtain

$\displaystyle\frac{\partial}{\partial\lambda}\log\Delta(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2})$
$\displaystyle=\frac{\partial}{\partial\lambda}\phi(\log(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2}))$ $\displaystyle=\phi((\lambda{\bf
1}-x_{0})^{*}(|x_{0}-\lambda{\bf 1}|^{2}+s_{1}(0)^{2})^{-1})$
$\displaystyle\qquad\qquad+\phi((|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2})^{-1})\cdot 2s_{1}(0)\frac{\partial
s_{1}(0)}{\partial\lambda},$

and

$\displaystyle\frac{\partial}{\partial\lambda}\log\Delta(|T|^{2}+s_{2}(0)^{2})$
$\displaystyle=\phi((|T|^{2}+s_{2}(0)^{2})^{-1})\cdot 2s_{2}(0)\frac{\partial
s_{2}(0)}{\partial\lambda},$

and

$\displaystyle\frac{\partial}{\partial\lambda}\log(s_{1}(0)+s_{2}(0))^{2}=\frac{2}{s_{1}(0)+s_{2}(0)}\left(\frac{\partial
s_{1}(0)}{\partial\lambda}+\frac{\partial s_{2}(0)}{\partial\lambda}\right).$

In addition, by subordination relations, we have

$\displaystyle\frac{1}{s_{1}(0)+s_{2}(0)}$
$\displaystyle=h_{\mu_{1}}(s_{1}(0))=h_{\mu_{2}}(s_{2}(0)$
$\displaystyle=s_{1}(0)\phi((|x_{0}-\lambda{\bf 1}|^{2}+s_{1}(0)^{2})^{-1})$
$\displaystyle=s_{2}(0)\phi((|T|^{2}+s_{2}(0)^{2})^{-1}).$

Putting these identities together, we obtain

$\displaystyle\frac{\partial}{\partial\lambda}\log\Delta(x_{0}+T-\lambda{\bf
1})^{2}$ $\displaystyle=\frac{\partial}{\partial\lambda}\log(\text{Right hand
side of}\,\eqref{eqn:Delta-T-1-2-t0-in-proof})$
$\displaystyle=\phi((\lambda{\bf 1}-x_{0})^{*}(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2})^{-1}),$

which is the identity claimed. ∎

######  Theorem 4.13.

The Brown measure of $x_{0}+T$ is absolutely continuous in the set
$\Omega(T,x_{0})\backslash S(T,x_{0})$. The density is real analytic whenever
it is strictly positive. Moreover, the density can be expressed as

(4.15)
$\frac{1}{\pi}s_{1}^{2}\phi(h^{-1}k^{-1})+\frac{1}{\pi}\frac{2s_{1}\big{(}s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))\big{)}\left|\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}}\right|^{2}}{\left[\frac{1}{s_{1}}p_{\mu_{1}}(s_{1})p^{\prime}_{\mu_{2}}(s_{2})+(s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))p^{\prime}_{\mu_{1}}(s_{1})\right]}$

where $h=h(\lambda,s_{1}(0))$ and $k=k(\lambda,s_{1}(0))$ for

$h(\lambda,w)=(\lambda{\bf 1}-x_{0})^{*}(\lambda{\bf 1}-x_{0})+w^{2}$

and

$k(\lambda,w)=(\lambda{\bf 1}-x_{0})(\lambda{\bf 1}-x_{0})^{*}+w^{2},$

and

$p_{\mu}(s)=\frac{n_{\mu}(s)}{d_{\mu}(s)}.$

###### Proof.

We put

$g(\lambda)=\log\Delta\big{(}(x_{0}+c_{t}-\lambda{\bf
1})^{*}(x_{0}+c_{t}-\lambda{\bf 1})\big{)}$

The Brown measure can be calculated as

$d\mu_{x_{0}+c_{t}}(\lambda)=\frac{1}{\pi}\frac{\partial^{2}}{\partial\overline{\lambda}\partial\lambda}g(\lambda).$

We rewrite (4.13) as

$\frac{\partial}{\partial\lambda}g(\lambda)=\phi\left(h^{-1}\frac{\partial
h}{\partial\lambda}\right)$

where $\frac{\partial h}{\partial\lambda}=(\lambda{\bf 1}-x_{0})^{*}$. By
applying [16, Lemma 3.2], we have

$\displaystyle\frac{\partial^{2}}{\partial\overline{\lambda}\partial\lambda}g(\lambda)$
$\displaystyle=\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}\frac{\partial
h}{\partial\lambda}\right)$ (4.16)
$\displaystyle=\phi\left(-h^{-1}\left(\frac{\partial
h}{\partial\overline{\lambda}}+2s_{1}\frac{\partial{s_{1}}}{\partial\overline{\lambda}}\right)h^{-1}\frac{\partial
h}{\partial\lambda}+h^{-1}\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}\right)$
(4.17) $\displaystyle=\phi\left(-h^{-1}\frac{\partial
h}{\partial\overline{\lambda}}h^{-1}\frac{\partial
h}{\partial\lambda}+h^{-1}\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}\right)-2s_{1}\frac{\partial{s_{1}}}{\partial\overline{\lambda}}\phi\left(\frac{\partial
h}{\partial{\lambda}}(h^{-1})^{2}\right)$

and

$\frac{\partial h}{\partial\overline{\lambda}}=\frac{\partial
h(\lambda,s_{1})}{\partial\overline{\lambda}}=\lambda{\bf 1}-x_{0};$

and

$\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}=\frac{\partial}{\partial\overline{\lambda}}(\lambda{\bf
1}-x_{0})^{*}=1.$

We now apply the identity $x(x^{*}x+\varepsilon{\bf
1})^{-1}=(xx^{*}+\varepsilon{\bf 1})^{-1}x$ to $x=\lambda{\bf 1}-x_{0}$ and
$\varepsilon=s_{1}^{2}$, we find that

$\displaystyle-\frac{\partial
h}{\partial\overline{\lambda}}h^{-1}\frac{\partial
h}{\partial\lambda}+\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}$
$\displaystyle=1-x(x^{*}x+s_{1}^{2}{\bf 1})^{-1}x^{*}$
$\displaystyle=1-(xx^{*}+s_{1}^{2}{\bf 1})^{-1}xx^{*}$
$\displaystyle=s_{1}^{2}(xx^{*}+s_{1}^{2}{\bf 1})^{-1}$
$\displaystyle=s_{1}^{2}k^{-1}.$

Hence the firs term in the right hand side of (4.17) is

(4.18) $\phi\left(-h^{-1}\frac{\partial
h}{\partial\overline{\lambda}}h^{-1}\frac{\partial
h}{\partial\lambda}+h^{-1}\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}\right)=s_{1}(0)^{2}\phi(h^{-1}k^{-1}).$

We now find a formula for $\frac{\partial s_{1}}{\partial\overline{\lambda}}$
by implicit differentiation. By adopting notations in Section 3 (in particular
Lemma 3.3) and (3.26), the function $s_{1}(0)$ is the unique solution of

(4.19)
$\frac{n_{\mu_{1}}(s)}{d_{\mu_{1}}(s)}=\frac{n_{\mu_{2}}(f_{\mu_{1}}(s))}{d_{\mu_{2}}(f_{\mu_{1}}(s))}$

for $\mu_{1}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}}$ and
$\mu_{2}=\widetilde{\mu}_{|T|}$. Set
$p_{\mu}(s)=\frac{n_{\mu}(s)}{d_{\mu}(s)}$ and recall that
$f_{\mu}(s)=\frac{1}{h_{\mu}(s)}-s$. Then,

(4.20) $p_{\mu_{1}}(s_{1}(0))=p_{\mu_{2}}(f_{\mu_{1}}(s_{1}(0))),$

where $f_{\mu_{1}}(s_{1}(0))$ can be expressed as

$f_{\mu_{1}}(s_{1}(0))=\frac{1}{s_{1}(0)\phi(h^{-1})}-s_{1}(0).$

Note that

$f_{\mu_{1}}(s_{1}(0))=\frac{1}{s_{1}(0)\phi(h^{-1})}-s_{1}(0)=s_{2}(0).$

Taking implicit differentiation of (4.20), we have

(4.21) $p_{\mu_{1}}^{\prime}(s_{1})\frac{\partial
s_{1}}{\partial\overline{\lambda}}+\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}}=p_{\mu_{2}}^{\prime}(f_{\mu_{1}}(s_{1}))\frac{\partial}{\partial\overline{\lambda}}f_{\mu_{1}}(s_{1}).$

We calculate

$\displaystyle\frac{\partial}{\partial\overline{\lambda}}f_{\mu_{1}}(s_{1})$
$\displaystyle=-\frac{1}{s_{1}^{2}\phi(h^{-1})}\frac{\partial
s_{1}}{\partial\overline{\lambda}}-\frac{1}{s_{1}}\frac{\phi\left(-h^{-1}\left(\frac{\partial
h}{\partial\overline{\lambda}}+2s_{1}\frac{\partial{s_{1}}}{\partial\overline{\lambda}}\right)h^{-1}\right)}{(\phi(h^{-1}))^{2}}-\frac{\partial
s_{1}}{\partial\overline{\lambda}}$ $\displaystyle=-\frac{\partial
s_{1}}{\partial\overline{\lambda}}\left[1+\frac{1}{s_{1}^{2}\phi(h^{-1})}-2\frac{\phi((h^{-1})^{2})}{(\phi(h^{-1}))^{2}}\right]+\frac{1}{s_{1}}\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}}.$

Alternatively, we have

$f_{\mu_{1}}(s_{1})=\frac{1}{s_{1}}p_{\mu_{1}}(s_{1}).$

Hence,

$\displaystyle\frac{\partial}{\partial\overline{\lambda}}f_{\mu_{1}}(s_{1})$
$\displaystyle=-\frac{1}{s_{1}^{2}}p_{\mu_{1}}(s_{1})\frac{\partial
s_{1}}{\partial\overline{\lambda}}+\frac{1}{s_{1}}p^{\prime}_{\mu_{1}}(s_{1})\frac{\partial
s_{1}}{\partial\overline{\lambda}}+\frac{1}{s_{1}}\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}}.$

Plugging it to (4.21), we have

$\displaystyle-\frac{\partial s_{1}}{\partial\overline{\lambda}}\times$
$\displaystyle\left[\frac{1}{s_{1}}p_{\mu_{1}}(s_{1})p^{\prime}_{\mu_{2}}(s_{2})+(s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))p^{\prime}_{\mu_{1}}(s_{1})\right]$
$\displaystyle=\big{(}s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))\big{)}\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}}.$

It follows that

(4.22) $-\frac{\partial
s_{1}}{\partial\overline{\lambda}}=\frac{\big{(}s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))\big{)}}{\left[\frac{1}{s_{1}}p_{\mu_{1}}(s_{1})p^{\prime}_{\mu_{2}}(s_{2})+(s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))p^{\prime}_{\mu_{1}}(s_{1})\right]}\times\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}},$

We note that

$\phi\left(\frac{\partial
h}{\partial{\lambda}}(h^{-1})^{2}\right)=\overline{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}.$

Plugging (4.18) and (4.22) to (4.17), then we have

$\frac{\partial^{2}}{\partial\overline{\lambda}\partial\lambda}g(\lambda)=s_{1}^{2}\phi(h^{-1}k^{-1})+\frac{2s_{1}\big{(}s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))\big{)}\left|\frac{\phi\left((h^{-1})^{2}\frac{\partial
h}{\partial\overline{\lambda}}\right)}{(\phi(h^{-1}))^{2}}\right|^{2}}{\left[\frac{1}{s_{1}}p_{\mu_{1}}(s_{1})p^{\prime}_{\mu_{2}}(s_{2})+(s_{1}-p^{\prime}_{\mu_{2}}(s_{2}))p^{\prime}_{\mu_{1}}(s_{1})\right]}$

This finishes the proof. ∎

######  Example 4.14.

Let $c$ be a standard circular operator and let $x_{0}$ be a selfadjoint
operator that is $*$-free from $c$. Note that $\lambda_{1}(\mu_{|c^{n}|})=0$
and $\lambda_{2}(\mu_{|c^{n}|})=1$ for any $n\in\mathbb{N}$. Then for any
$n\in\mathbb{N}$, the set $\Omega(c^{n},x_{0})=\Omega(c,x_{0})$ and is
expressed as

$\Omega(c,x_{0})=\bigg{\\{}\lambda=a+bi:\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+b^{2}}d\mu_{x_{0}}(u)>1\bigg{\\}}.$

See Figure 2 for simulations.

(a) $x_{0}+c$

(b) $x_{0}+c^{2}$

(c) $x_{0}+c^{3}$

(d) $x_{0}+c^{4}$

Figure 2. $2\,000\times 2\,000$ Matrix simulations for the Brown measure of
$x_{0}+c^{n}$ for $x_{0}$ distributed as
$\frac{1}{4}\delta_{-0.8}+\frac{3}{4}\delta_{0.8}$. The boundary of
$\Omega(c,x_{0})$ is red dashed on top and blue dashed on bottom.

######  Corollary 4.15.

If $S(T,x_{0})=\varnothing$, then the Brown measure of $T+x_{0}$ is absolutely
continuous in $\Omega(T,x_{0})$. In particular, if $\lambda_{1}(\mu_{|T|})>0$,
then $\mu_{|T|}(\\{0\\})=0$. Consequently, $S(T,x_{0})=\varnothing$ and the
Brown measure of $T+x_{0}$ is absolutely continuous in $\Omega(T,x_{0})$.

##  5\. Addition with $R$-diagonal operators: examples

###  5.1. The sum of a Haar unitary operator and a free random variable

Let $u$ be a Haar unitary operator and $x_{0}\in{\mathcal{M}}^{\Delta}$ be a
random variable that is $*$-free from $u$. In this section, we study the Brown
measure of $x_{0}+u$. For any $\lambda\in\mathbb{C}$, we denote
$\mu_{1}=\widetilde{\mu}_{|x_{0}-\lambda{\bf 1}|}$,
$\mu_{2}=\frac{1}{2}(\delta_{\gamma}+\delta_{-\gamma})$ for some $\gamma>0$,
and $\mu=\mu_{1}\boxplus\mu_{2}$. Because of $\mu_{2}$ is so simple, we can
obtain more explicit formula of subordination functions $s_{1}$ and $s_{2}$.

We adopt notations in Section 3 and set two $R$-diagonal operators
$T_{1}=V(x_{0}-\lambda{\bf 1})$ and $T_{2}=\gamma u$, where $V$ is a Haar
unitary operator that is $*$-free from $\\{x_{0},u\\}$. We shall review the
subordination function appearing in the free additive convolution of an
arbitrary symmetric distribution $\mu_{1}$ and a Bernoulli distribution
$\mu_{2}$. We provide a proof for completeness.

######  Lemma 5.1.

Using notations in Section 3, the subordination function $s_{1}(t)$ is the
unique solution $s\in(t,\infty)$ to the following equation

(5.1) $(s-t)^{2}-\frac{s-t}{h_{\mu_{1}}(s)}+\gamma^{2}=0.$

###### Proof.

We calculate that
$G_{\mu_{2}}(z)=\frac{1}{2}\left(\frac{1}{z-\gamma}+\frac{1}{z+\gamma}\right)$.
Hence $H_{2}(z)=F_{\mu_{2}}(z)-z=\frac{-\gamma^{2}}{z}$ and
$f_{\mu_{2}}(t)=\Im H_{2}(it)=\frac{\gamma^{2}}{t}$. Recall Equation (3.9)
satisfied by $s_{1}(z)$

$s_{1}(t)=t+f_{\mu_{2}}(t+f_{\mu_{1}}(s_{1}(t)))$

which is now expressed as

$s_{1}(t)=t+\frac{\gamma^{2}}{t-s_{1}+1/h_{\mu_{1}}(s_{1}(t))}.$

The above equation can be further reduced to

$(s_{1}(t)-t)^{2}-\frac{s_{1}(t)-t}{h_{\mu_{1}}(s)}+\gamma^{2}=0.$

The uniqueness of the solution for (3.9) implies that the function $s_{1}(t)$
must be the unique solution of the above equation. ∎

######  Theorem 5.2.

Let $u$ be a Haar unitary operator and $x_{0}\in{\mathcal{M}}^{\Delta}$ be a
random variable that is $*$-free from $u$. Given $\lambda\in\mathbb{C}$ and
$\gamma>0$, we have

  1. (1)

If $\lambda_{1}(\mu_{|x_{0}-\lambda{\bf
1}|})<\gamma<\lambda_{2}(\mu_{|x_{0}-\lambda{\bf 1}|})$, then

(5.2) $\Delta(x_{0}+\gamma u-\lambda{\bf
1})=\Bigg{(}\frac{\gamma^{2}}{\gamma^{2}+s_{1}(0)^{2}}\,\Delta(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2}{\bf 1})\Bigg{)}^{\frac{1}{2}},$

where $s_{1}(0)$ is a real analytic function of $\lambda$ which satisfies

(5.3) $s^{2}-\frac{s}{h_{\mu_{1}}(s)}+\gamma^{2}=0.$

  2. (2)

If $\gamma\leq\lambda_{1}(\mu_{|x_{0}-\lambda{\bf 1}|})$, then
$\Delta(x_{0}+\gamma u-\lambda{\bf 1})=\Delta(x_{0}-\lambda{\bf 1})$.

  3. (3)

If $\gamma\geq\lambda_{2}(\mu_{|x_{0}-\lambda{\bf 1}|})$, then
$\Delta(T-\lambda{\bf 1})=\gamma$.

###### Proof.

Following Theorem 4.5, we only need to verify part (i). If
$\lambda_{1}(\mu_{|x_{0}-\lambda{\bf
1}|})<\gamma<\lambda_{2}(\mu_{|x_{0}-\lambda{\bf 1}|})$, we know that
$s_{1}(0)$ is a real analytic function of $\lambda$ which satisfies (5.3). We
hence have

(5.4)
$s_{2}(0)=\frac{1}{h_{\mu_{1}}(s_{1}(0)}-s_{1}(0)=\frac{\gamma^{2}}{s_{1}(0)}.$

Divide (5.3) by $s_{1}^{2}$, we have

$1+\frac{\gamma^{2}}{s_{1}^{2}}=\frac{1}{s_{1}h_{\mu_{1}}(s_{1})}.$

Hence, and by (5.4),

(5.5)
$s_{1}+s_{2}=\frac{\gamma^{2}}{s_{1}}+s_{1}=\frac{1}{h_{\mu_{1}}(s_{1}(0)}=\frac{1}{h_{\mu_{1}}(s_{2}(0)}.$

Note that, by applying (5.4), we have

$\displaystyle\Delta(|T_{2}|^{2}+s_{2}(0)^{2})$
$\displaystyle=\gamma^{2}+s_{2}(0)^{2}=\gamma^{2}\frac{s_{1}(0)^{2}+\gamma^{2}}{s_{1}(0)^{2}}.$

We then deduce

$\displaystyle\frac{\Delta(|T_{2}|^{2}+s_{2}(0)^{2}}{(s_{1}(0)+s_{2}(0))^{2}}=\gamma^{2}\frac{s_{1}(0)^{2}+\gamma^{2}}{s_{1}(0)^{2}}\left(\frac{s_{1}^{2}}{\gamma^{2}+s_{1}^{2}}\right)^{2}=\frac{\gamma^{2}}{\gamma^{2}+s_{1}^{2}}.$

Hence (5.2) follows from (4.7). ∎

######  Remark 5.3.

Let $T\in{\mathcal{M}}^{\Delta}$ be a $R$-diagonal operator and
$\gamma\in\mathbb{C}$. Then

$|T-\gamma|\underset{\ast{\mathcal{D}}}{\sim}|T-\gamma
u|\underset{\ast{\mathcal{D}}}{\sim}|T+|\gamma|u|.$

Replacing $|x_{0}-\lambda{\bf 1}|$ by $T$, we see that Theorem 5.2 is
essentially equivalent to [14, Theorem 4.15].

In [9, Section 4], a Brown measure formula for $u+x_{0}$ was obtained for
bounded, normal operator $x_{0}$ (see [9, formula (4.8)]). The following
result generalizes their formula for $x_{0}\in{\mathcal{M}}^{\Delta}$ with an
arbitrary distribution.

######  Theorem 5.4.

The Brown measure of $x_{0}+\gamma u$ is absolutely continuous on the set

$\Omega(\gamma u,x_{0})=\\{\lambda:1/\phi(|x_{0}-\lambda{\bf
1}|^{-2})<\gamma^{2}<\phi(|x_{0}-\lambda{\bf 1}|^{2})\\}$

and the density is strictly positive. Moreover, the density is expressed as

$\frac{1}{\pi}\left(s_{1}(0)^{2}\phi(h^{-1}k^{-1})+\frac{|\phi\left(\frac{\partial
h}{\partial\overline{\lambda}}(h^{-1})^{2}|^{2}\right)}{\phi((h^{-1})^{2})-1/(s_{1}(0)+\gamma^{2})^{2}}\right)$

where

$h=h(\lambda,s_{1})=(\lambda{\bf 1}-x_{0})^{*}(\lambda{\bf
1}-x_{0})+s_{1}(0)^{2}$

and

$k=k(\lambda,s_{1})=(\lambda{\bf 1}-x_{0})(\lambda{\bf
1}-x_{0})^{*}+s_{1}(0)^{2}.$

######  Remark 5.5.

By the second identity in (5.5), we have

$\frac{1}{s_{1}^{2}+\gamma^{2}}=\phi(h^{-1}).$

Hence, by Cauchy-Schwartz inequality, we have

$\phi((h^{-1})^{2})-1/(s_{1}(0)+\gamma^{2})^{2}=\phi((h^{-1})^{2})-(\phi(h^{-1}))^{2}>0.$

###### Proof.

By Proposition 4.11, for $\lambda\in\Omega(u,x_{0})$, the limit values of
subordination functions $s_{1}(0),s_{2}(0)$ are strictly positive and are real
analytic function of $\lambda$. We put

$g(\lambda)=\log\Delta\big{(}(x_{0}+u-\lambda{\bf 1})^{*}(x_{0}+u-\lambda{\bf
1})\big{)}.$

By Proposition 4.12, we have

$\frac{\partial}{\partial\lambda}g(\lambda)=\phi\left(h^{-1}\frac{\partial
h}{\partial\lambda}\right)$

where $\frac{\partial h}{\partial\lambda}=(\lambda{\bf 1}-x_{0})^{*}$. We
adapt the calculation the proof of Theorem 4.13 (see also [24, Theorem 4.6]
and [14, Lemma 2.8]) to get the density formula. By [16, Lemma 3.2], since
$g(\lambda)$ is a real analytic function of $\lambda$, we have

(5.6)
$\displaystyle\frac{\partial^{2}}{\partial\overline{\lambda}\partial\lambda}g(\lambda)=\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}\frac{\partial
h}{\partial\lambda}\right)$
$\displaystyle=\phi\left(-h^{-1}\left(\frac{\partial
h}{\partial\overline{\lambda}}+2s_{1}(0)\frac{\partial{s_{1}(0)}}{\partial\overline{\lambda}}\right)h^{-1}\frac{\partial
h}{\partial\lambda}+h^{-1}\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}\right)$
$\displaystyle=\phi\left(-h^{-1}\frac{\partial
h}{\partial\overline{\lambda}}h^{-1}\frac{\partial
h}{\partial\lambda}+h^{-1}\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}\right)-2s_{1}(0)\frac{\partial{s_{1}(0)}}{\partial\overline{\lambda}}\phi\left(\frac{\partial
h}{\partial{\lambda}}(h^{-1})^{2}\right)$

where

$\frac{\partial h}{\partial\overline{\lambda}}=\frac{\partial
h(\lambda,s_{1})}{\partial\overline{\lambda}}=\lambda{\bf 1}-x_{0};$

and

$\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}=\frac{\partial}{\partial\overline{\lambda}}(\lambda{\bf
1}-x_{0})^{*}=1.$

We now apply the identity $x(x^{*}x+\varepsilon{\bf
1})^{-1}=(xx^{*}+\varepsilon{\bf 1})^{-1}x$ to $x=\lambda{\bf 1}-x_{0}$ and
$\varepsilon=s_{1}(0)^{2}$, we find that

(5.7) $\displaystyle-\frac{\partial
h}{\partial\overline{\lambda}}h^{-1}\frac{\partial
h}{\partial\lambda}+\frac{\partial^{2}h}{\partial\overline{\lambda}\partial\lambda}$
$\displaystyle=1-x(x^{*}x+s_{1}(0)^{2}{\bf 1})^{-1}x^{*}$
$\displaystyle=1-(xx^{*}+s_{1}(0)^{2}{\bf 1})^{-1}xx^{*}$
$\displaystyle=s_{1}(0)^{2}(xx^{*}+s_{1}(0)^{2}{\bf 1})^{-1}$
$\displaystyle=s_{1}(0)^{2}k^{-1}.$

For $\lambda\in\Omega(u,x_{0})$, the function $s_{1}(0)$ is determined by
(taking limit for (5.1) as $t$ goes to zero)

$s_{1}(0)^{2}-\frac{s_{1}(0)}{h_{\mu_{1}}(s_{1}(0))}+\gamma^{2}=0.$

which can be rewritten as

$\frac{1}{s_{1}(0)^{2}+\gamma^{2}}=\frac{h_{\mu_{1}}(s_{1}(0))}{s_{1}(0)}=\phi(h^{-1}).$

Taking implicit differentiation $\frac{\partial}{\partial\overline{\lambda}}$
and applying again [16, Lemma 3.2], we then obtain

$-\phi\left(h^{-1}\left(\frac{\partial
h}{\partial\overline{\lambda}}+2s_{1}(0)\frac{\partial{s_{1}(0)}}{\partial\overline{\lambda}}\right)h^{-1}\right)=-\frac{1}{(s_{1}(0)^{2}+\gamma^{2})^{2}}2s_{1}(0)\frac{\partial
s_{1}(0)}{\partial\overline{\lambda}},$

where $\frac{\partial h}{\partial\overline{\lambda}}=\lambda{\bf 1}-x_{0}$.
This implies

(5.8)
$-2s_{1}(0)\frac{\partial{s_{1}(0)}}{\partial\overline{\lambda}}=\frac{\phi\left(\frac{\partial
h}{\partial\overline{\lambda}}(h^{-1})^{2}\right)}{\phi((h^{-1})^{2})-1/(s_{1}(0)+\gamma^{2})^{2}}.$

By the tracial property, we observe that

$\phi\left(\frac{\partial
h}{\partial\overline{\lambda}}(h^{-1})^{2}\right)=\overline{\phi\left(\frac{\partial
h}{\partial{\lambda}}(h^{-1})^{2}\right)}.$

Hence, by plugging identities (5.7) and (5.8) into (LABEL:eqn:Laplacian-g-
lambda), we obtain

$\displaystyle d\mu_{x_{0}+u}$
$\displaystyle=\frac{1}{\pi}\frac{\partial^{2}}{\partial\overline{\lambda}\partial\lambda}g(\lambda)$
$\displaystyle=\frac{1}{\pi}\left(s_{1}(0)^{2}\phi(h^{-1}k^{-1})+\frac{|\phi\left(\frac{\partial
h}{\partial\overline{\lambda}}(h^{-1})^{2}|^{2}\right)}{\phi((h^{-1})^{2})-1/(s_{1}(0)+\gamma^{2})^{2}}\right)$
$\displaystyle=\frac{1}{\pi}\left(s_{1}(0)^{2}\phi(h^{-1}k^{-1})+\frac{|\phi\left(\frac{\partial
h}{\partial\overline{\lambda}}(h^{-1})^{2}|^{2}\right)}{\phi((h^{-1})^{2})-(\phi(h^{-1}))^{2}}\right).$

which is strictly positive, where we used

$\frac{1}{s_{1}^{2}+\gamma^{2}}=\phi(h^{-1})$

due to the second identity in (5.5). ∎

###  5.2. The sum of a circular operator and a free random variable

Denote by $c$ a standard free circular operator in ${\mathcal{M}}$, and by
$c_{\varepsilon}$ a free circular operator with variance
$\phi(c_{\varepsilon}^{*}c_{\varepsilon})=\varepsilon$. Let $x_{0}$ be a
random variable with an arbitrary distribution that is freely independent with
$c$ and $c_{\varepsilon}$. We give a new proof for a main result in [17] and
[24], where the case when $x_{0}$ is a bounded operator was studied. Our
approach allows us to describe the Brown measure of $x_{0}+c_{\varepsilon}$
with $x_{0}\in{\mathcal{M}}^{\Delta}$ having an arbitrary distribution.

Let $x_{0}\in{\mathcal{M}}^{\Delta}$ be a random variable with an arbitrary
distribution that is freely independent with
$\\{c_{\varepsilon}\\}_{\varepsilon\geq 0}$. Fix $\varepsilon>0$ and for any
$\lambda\in\mathbb{C}$, we denote $T_{1}=V(x_{0}-\lambda{\bf 1})$ and
$T_{2}=c_{\varepsilon}$, where $V$ is a Haar unitary $*$-free from $x_{0}$ and
$c_{\varepsilon}$. Denote $\mu_{1}=\tilde{\mu}_{|x_{0}-\lambda{\bf 1}|}$ and
$\mu_{2}=\tilde{\mu}_{|c_{\varepsilon}|}$ and
$\mu=\mu_{1}\boxplus\mu_{2}=\tilde{\mu}_{|T_{1}+T_{2}|}$.

######  Lemma 5.6.

The subordination function $s_{1}$ satisfies

$\varepsilon h_{\mu}(t)=s_{1}(t)-t,\qquad\forall t>0.$

###### Proof.

Note that $\mu_{2}=\mu_{|c_{\varepsilon}|}$ has semicircle distribution with
variance $\varepsilon$. Hence its $R$-transform is given by
$R_{\mu_{2}}(z)=\varepsilon z$. The additive of $R$-transform yields

$R_{\mu_{1}}(z)+R_{\mu_{2}}(z)=R_{\mu}(z)$

which yields

$G_{\mu_{1}}^{-1}(z)+R_{\mu_{2}}(z)=G_{\mu_{1}}^{-1}(z)+\varepsilon
z=G_{\mu}^{-1}(z).$

By replacing $z$ by $G_{\mu}(z)$, we obtain $\omega_{1}(z)=z-\varepsilon
G_{\mu}(z)$, which holds for any $z\in\mathbb{C}^{+}$ by analytic
continuation. In particular, when $z=it$, we have

$\omega_{1}(it)-it=-\varepsilon G_{\mu}(it).$

Recall that $G_{\mu}(it)=-ih_{\mu}(t)$ and $\omega_{1}(it)=is_{1}(t)$. The
result follows. ∎

######  Lemma 5.7.

For any $t>0$, we have

(5.9)
$\Delta(|c_{\varepsilon}|^{2}+s_{2}(t)^{2})(h_{\mu}(t))^{2}=\exp\left(-\varepsilon
h_{\mu}(t)^{2}\right)$

###### Proof.

The logarithmic derivative of (5.9) with respect to $t$ is given by

(5.10)
$\displaystyle\phi((|c_{\varepsilon}|^{2}+s_{2}(t)^{2})^{-1})s_{2}(t)s_{2}^{\prime}(t)+\frac{h_{\mu}^{\prime}(t)}{h_{\mu}(t)}=-\varepsilon
h_{\mu}(t)h_{\mu}^{\prime}(t).$

We will first verify that (5.10) holds. Using notations in Section 3, we have

$\phi((|c_{\varepsilon}|^{2}+s_{2}(t)^{2})^{-1})s_{2}(t)=h_{\mu_{2}}(s_{2}(t))=h_{\mu}(t),$

which allows us to rewrite (5.10) as

(5.11)
$h_{\mu}(t)s_{2}^{\prime}(t)+\frac{h_{\mu}^{\prime}(t)}{h_{\mu}(t)}=-\varepsilon
h_{\mu}(t)h_{\mu}^{\prime}(t).$

We also have $s_{1}(t)+s_{2}(t)=t+1/h_{\mu}(t)$. Hence,

$s_{2}^{\prime}(t)=(1/h_{\mu}(t)+t-s_{1}(t))^{\prime}=-\frac{h_{\mu}^{\prime}(t)}{h_{\mu}(t)^{2}}+1-s_{1}^{\prime}(t).$

Plugging this into (5.11), we see that to prove (5.10), it suffices to show
that

(5.12) $h_{\mu}(t)-h_{\mu}(t)s_{1}^{\prime}(t)=-\varepsilon
h_{\mu}(t)h_{\mu}^{\prime}(t).$

Lemma 5.6 tells that $s_{1}$ satisfies

(5.13) $\varepsilon h_{\mu}(t)=s_{1}(t)-t.$

We note that (5.13) actually implies (5.12) by plugging the derivative of
(5.13) to (5.12). Hence, (5.10) holds for all $t>0$.

We next observe that $\lim_{t\rightarrow\infty}h_{\mu}(t)=0$, and

$\lim_{t\rightarrow\infty}\log\Delta(|c_{\varepsilon}|^{2}+s_{2}(t)^{2})-2\log
t=0,$

and

$\lim_{t\rightarrow\infty}\log(h_{\mu}(t))+2\log t=0.$

Hence (5.9) holds for all $t>0$. ∎

######  Lemma 5.8.

We have, for any $t>0$,

(5.14) $\Delta(|x_{0}+c_{\varepsilon}-\lambda{\bf
1}|^{2}+t^{2})=\Delta(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(t)^{2})\exp\left(-\varepsilon h_{\mu}(t)^{2}\right)$

where $s_{1}(t)$ is the subordination as in Section 3 and

$h_{\mu}(t)=t\phi((|x_{0}+c_{\varepsilon}-\lambda{\bf 1}|^{2}+t^{2})^{-1}).$

###### Proof.

Note that $s_{1}(t)+s_{2}(t)-t=1/h_{\mu}(t)$. Then result then follows from
Theorem 4.5 and Lemma 5.7. ∎

The following result extends [24, Theorem 3.12] and [17, Theorem 3.10] to
unbounded operators in ${\mathcal{M}}^{\Delta}$.

######  Theorem 5.9.

Let $x_{0}\in{\mathcal{M}}^{\Delta}$ be a random variable freely independent
from the circular operator $c_{\varepsilon}$ with variance $\varepsilon>0$.

  1. (1)

If $||(x_{0}-\lambda{\bf 1})^{-1}||_{2}^{2}>\frac{1}{\varepsilon}$, then

(5.15) $\displaystyle\Delta(x_{0}+c_{\varepsilon}-\lambda{\bf
1})=\bigg{(}\Delta\big{(}(x_{0}-\lambda{\bf 1})^{*}(x_{0}-\lambda{\bf
1})+s_{1}(0)^{2}{\bf
1}\big{)}\bigg{)}^{\frac{1}{2}}\exp\left(-\frac{s_{1}(0)^{2}}{2\varepsilon}\right)$

where $s_{1}(0)$ is determined by

(5.16) $\phi\bigg{[}\big{(}(x_{0}-\lambda{\bf 1})^{*}(x_{0}-\lambda{\bf
1})+s_{1}(0)^{2}{\bf 1}\big{)}^{-1}\bigg{]}=\frac{1}{\varepsilon}.$

  2. (2)

If $||(x_{0}-\lambda{\bf 1})^{-1}||_{2}^{2}\leq\frac{1}{\varepsilon}$, then

(5.17) $\displaystyle\Delta(x_{0}+c_{\varepsilon}-\lambda{\bf
1})=\Delta(x_{0}-\lambda{\bf 1}).$

###### Proof.

We have
$\lambda_{2}(\mu_{2})=\lambda_{2}(\mu_{|c_{\varepsilon}|})=\varepsilon$ and
$\lambda_{1}(\mu_{1})=\frac{1}{||(x_{0}-\lambda{\bf 1})^{-1}||_{2}^{2}}$.
Hence

(5.18) $||(x_{0}-\lambda{\bf
1})^{-1}||_{2}^{2}>\frac{1}{\varepsilon}\Longleftrightarrow\lambda_{1}(\mu_{1})<\lambda_{2}(\mu_{2}).$

Since $\lambda_{1}(\mu_{2})=0$, when $||(x_{0}-\lambda{\bf
1})^{-1}||_{2}^{2}>\frac{1}{\varepsilon}$, Case 3 in Theorem 3.6 applies, in
which case $0<s_{1}(0)<\infty$. Hence, (5.15) follows from Lemma 5.8 by taking
the limit $t\rightarrow 0$. Now, as in Lemma 5.6, $s_{1}(t)$ satisfies

$h_{\mu_{1}}(s_{1}(t))=\frac{s_{1}(t)-t}{\varepsilon}.$

Hence, $s_{1}(0)\in(0,\infty)$ satisfies

$\phi\bigg{[}\big{(}(x_{0}-\lambda{\bf 1})^{*}(x_{0}-\lambda{\bf
1})+s_{1}(0)^{2}{\bf
1}\big{)}^{-1}\bigg{]}=\frac{h_{\mu_{1}}(s_{1}(0))}{s_{1}(0)}=\frac{1}{\varepsilon}.$

Finally, rewrite the condition $||(x_{0}-\lambda{\bf
1})^{-1}||_{2}^{2}\leq\frac{1}{\varepsilon}$ as
$\lambda_{2}(\mu_{2})\leq\lambda_{1}(\mu_{1})$. We apply Case 1 and Case 4 in
Theorem 3.6 and have $s_{1}(t)\rightarrow 0$ as $t\rightarrow 0$. Then (5.17)
from Lemma 5.8. ∎

The following result extends [24, Theorem 4.6] to all
$x_{0}\in{\mathcal{M}}^{\Delta}$.

######  Theorem 5.10.

Let $x_{0}\in{\mathcal{M}}^{\Delta}$ be a random variable freely independent
from the circular operator $c_{\varepsilon}$ with variance $\varepsilon>0$.
The Brown measure of $x_{0}+c_{\varepsilon}$ is supported in the closure of
the set

$\Xi_{\varepsilon}=\left\\{\lambda\in\mathbb{C}:\phi\bigg{[}\big{(}(x_{0}-\lambda{\bf
1})^{*}(x_{0}-\lambda{\bf 1})+s_{1}(0)^{2}{\bf
1}\big{)}^{-1}\bigg{]}>\frac{1}{\varepsilon}\right\\}.$

The Brown measure in this set is given by

(5.19) $\frac{1}{\pi}\left(\frac{|\phi((\lambda{\bf
1}-x_{0})(h^{-1})^{2})|^{2}}{\phi((h^{-1})^{2})}+s_{1}(0)^{2}\phi(h^{-1}k^{-1})\right)$

where $h=h(\lambda,s_{1})$ and $k=k(\lambda,s_{1})$ for

$h(\lambda,s_{1})=(\lambda{\bf 1}-x_{0})^{*}(\lambda{\bf
1}-x_{0})+s_{1}(0)^{2}$

and

$k(\lambda,s_{1})=(\lambda{\bf 1}-x_{0})(\lambda{\bf
1}-x_{0})^{*}+s_{1}(0)^{2}.$

In particular, the density of the Brown measure of $x_{0}+c_{\varepsilon}$ is
strictly positive within the domain.

###### Proof.

The set $\Xi_{\varepsilon}=\Omega(c_{\varepsilon},x_{0})$ as we observed in
the proof for Theorem 5.9, in particular (5.18). Hence, the Brown measure of
$x_{0}+c_{\varepsilon}$ is supported in the closure of $\Xi_{\varepsilon}$ by
Theorem 4.1. See [24, Theorem 4.6] for a proof for the density formula. ∎

The above result allows us to express the Brown measure of
$x_{0}+c_{\varepsilon}$ and recover results in [17, Section 3].

######  Remark 5.11.

For $x_{0}$ self-adjoint, let $\mu_{x}$ be the spectral measure of $\mu_{x}$
and $\lambda=a+bi$, then the set $\Xi_{\varepsilon}$ in Theorem 5.10 is
expressed as

(5.20)
$\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+b^{2}}d\mu_{x}(u)>\frac{1}{\varepsilon}.$

Note the map $b\mapsto\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+b^{2}}d\mu_{x}(u)$
is monotonically decreasing. If $\lambda=a+bi$ satisfies (5.20), we let
$v_{\varepsilon}(a)$ be the unique value so that

(5.21)
$\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+v_{\varepsilon}(a)^{2}}d\mu_{x}(u)=\frac{1}{\varepsilon}.$

Then, by comparing with (5.16), we have

$v_{\varepsilon}(a)^{2}=b^{2}+s_{1}(0)^{2}.$

This formulation appeared in Biane work [7]. See [17, Section 2.3 and Theorem
3.10].

######  Theorem 5.12.

Let $x_{0}\in{\mathcal{M}}^{\Delta}$ be a self-adjoint random variable freely
independent from the circular operator $c_{\varepsilon}$ with variance
$\varepsilon>0$. Let $\mu_{x_{0}}$ be the spectral measure of $x_{0}$. The
Brown measure of $x_{0}+c_{\varepsilon}$ is supported in the closure of the
set

(5.22)
$\Lambda_{\varepsilon}=\left\\{\lambda=a+bi\in\mathbb{C}:\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+b^{2}}d\mu_{x_{0}}(u)>\frac{1}{\varepsilon}\right\\}.$

The Brown measure in this set is given by

(5.23)
$d\mu_{x_{0}+c_{\varepsilon}}(\lambda)=\frac{1}{\pi\varepsilon}\left(1-\frac{\varepsilon}{2}\frac{d}{da}\int_{\mathbb{R}}\frac{u}{(u-a)^{2}+v_{\varepsilon}(a)^{2}}d\mu_{x_{0}}(u)\right).$

Moreover, the density is strictly positive and analytic within the interior of
the support.

###### Proof.

The positivity of the Brown measure density follows from Theorem 5.10. By
Proposition 4.12, we have

$\displaystyle\frac{\partial}{\partial\lambda}\log\Delta(|x_{0}+c_{\varepsilon}-\lambda{\bf
1}|^{2})$ $\displaystyle=\phi((x_{0}-\lambda{\bf 1})^{*}(|x_{0}-\lambda{\bf
1}|^{2}+s_{1}(0)^{2})^{-1})$
$\displaystyle=\int_{\mathbb{R}}\frac{u-a+ib}{(u-a)^{2}+b^{2}+s_{1}(0)^{2}}d\mu_{x_{0}}(u)$
$\displaystyle=\int_{\mathbb{R}}\frac{u-a}{(u-a)^{2}+v_{\varepsilon}(a)^{2}}d\mu_{x_{0}}(u)+\frac{ib}{\varepsilon}$

where we used (5.21). Note that the real part of the above formula depends
only on $a$. Hence,

$d\mu_{x_{0}+c_{\varepsilon}}(\lambda)=\frac{1}{\pi}\frac{\partial}{\partial\overline{\lambda}}\left(\int_{\mathbb{R}}\frac{u-a}{(u-a)^{2}+v_{\varepsilon}(a)^{2}}d\mu_{x_{0}}(u)+\frac{ib}{\varepsilon}\right)$

which yields the density formula. Using the same proof as [17, Lemma 3.11], we
also see that $\mu_{x_{0}+c_{\varepsilon}}(\Lambda_{\varepsilon})=1$. ∎

The following generalizes [24, Theorem 6.10] where the support and density of
the Brown measure of $g_{t}+u$ was studied.

######  Corollary 5.13.

Let $T\in{\mathcal{M}}^{\Delta}$ be an $R$-diagonal operator and $g_{t}$ be a
semicircular operator. The support of the Brown measure of $T+g_{t}$ is
contained in a ring where the inner boundary is a circle with radius
$\sqrt{(\lambda_{1}(T)^{2}-t)_{+}}$ and the outer boundary is an ellipse with
semi-axes
$\big{(}\frac{2t+\lambda_{2}(T)^{2}}{t+\lambda_{2}(T)^{2}},\frac{\lambda_{2}(T)^{2}}{t+\lambda_{2}(T)^{2}}\big{)}$.

###### Proof.

We apply Theorem 4.1 for $x_{0}=g_{t}$. We have

$||g_{t}-\lambda{\bf 1}||_{2}=\phi\big{(}(g_{t}-\lambda{\bf
1})^{*}(g_{t}-\lambda{\bf 1})\big{)}=t+|\lambda|^{2}.$

Hence, condition $||g_{t}-\lambda{\bf 1}||_{2}||T^{-1}||_{2}>1$ is written as

(5.24) $|\lambda|^{2}>1/||T^{-1}||_{2}-t=\lambda_{1}(T)^{2}-t.$

For other condition $||(g_{t}-\lambda{\bf 1})^{-1}||_{2}||T||_{2}>1$. We
replace $T$ by a circular operator $c_{s}$ with
$s=||T||_{2}^{2}=\lambda_{2}(T)^{2}$. This condition can then be rewritten as

(5.25)
$\int_{\mathbb{R}}\frac{1}{(u-a)^{2}+b^{2}}d\sigma(u)>\frac{1}{\lambda_{2}(T)^{2}}=\frac{1}{s}$

where $\sigma$ is the semicircular distribution with variance $t$. We
recognize this is the same equation (see (5.22)) when we study the Brown
measure of the elliptic operator $g_{t}+c_{s}$ whose support is known to be
the ellipse with semi-axes $(\frac{2t+s}{\sqrt{t+s}},\frac{2s}{\sqrt{s+t}})$.
That means the solution of (5.25) is the region enclosed by the ellipse with
semi-axes $(\frac{2t+s}{\sqrt{t+s}},\frac{2s}{\sqrt{s+t}})$ where
$s=\lambda_{2}(T)^{2}$. ∎

###  5.3. The sum of a circular Cauchy operator and a free random variable

In this section, we study $R$-diagonal operator $z$ with $\tilde{\mu}_{z}$ as
the Cauchy distribution. We give a short proof of various results of
Haaagerup-Schultz [15] on this operator. Those results played a key role in
their seminal paper [15] on the invariant subspace problem in type II1
factors.

######  Definition 5.14.

The operator $z\in{\mathcal{M}}^{\Delta}$ is called free circular Cauchy
operator if $z$ has the same distribution as $uh$ where $u$ and $h$ are
$*$-free, $u$ is a Haar unitary and $h=h^{*}$ is Cauchy distributed,

$d\mu_{h}=\frac{1}{\pi}\frac{1}{x^{2}+1}dx.$

In particular, $z$ is $R$-diagonal. A simple computation shows that its Cauchy
transform is

$G_{\mu_{h}}(z)=\frac{1}{z+i}$

and its $R$-transform is

$\mathcal{R}_{\mu_{h}}(z)=-i.$

Let $\\{x,y\\}$ be a circular system and $\\{s_{1},s_{2},s_{3},s_{4}\\}$ be a
semicircular system and $U$ be a Haar unitary operator free from
$\\{s_{1},s_{2},s_{3},s_{4}\\}$ in the finite von Neumann algebra
${\mathcal{M}}$ equipped with faithful normal trace state $\phi$ . Then
$x\underset{\ast{\mathcal{D}}}{\sim}\frac{s_{1}+is_{2}}{\sqrt{2}}\underset{\ast{\mathcal{D}}}{\sim}Us_{1}$.
The probability measure $\tilde{\mu}_{|x|}$ of the symmetrization of
$\mu_{|x|}$ has the semicircle distribution

$d\tilde{\mu}_{|x|}=\frac{1}{2\pi}\sqrt{4-t^{2}}1_{[-2,2]}(t)dt.$

The following result was mentioned inexplicitly in [14, Section 5].

######  Proposition 5.15.

The free circular Cauchy operator $z$ has the same distribution as $xy^{-1}$
and $s_{1}s_{2}^{-1}$. The $S$-transform of $|z|^{2}$ is given by

$S_{|z|^{2}}(t)=-\frac{t}{1+t},\quad t\in(-1,0),$

and the Cauchy transform

(5.26)
$G_{\mu_{|z|^{2}}}(\lambda)=\frac{1}{\lambda-\sqrt{-\lambda}},\quad\lambda<0.$

###### Proof.

Let $T=xy^{-1}$. The operators $x$ and $y^{-1}$ are $*$-free and hence $T$ is
$R$-diagonal. The distribution of $|T|$ is given by [14, Theorem 5.2 (b)] as

$d\mu_{|T|}(t)=\frac{2}{\pi}\frac{1}{1+t^{2}}1_{(0,\infty)}(t)dt.$

Hence the symmetric measure $\tilde{\mu}_{|T|}$ is Cauchy distributed. Hence,

$z\underset{\ast{\mathcal{D}}}{\sim}U|T|\underset{\ast{\mathcal{D}}}{\sim}xy^{-1}.$

The semicircular operators $s_{1}$ and $s_{2}$ are even operators, hence the
product $s_{1}s_{2}^{-1}$ is $R$-diagonal. Observe that $|s_{1}s_{2}^{-1}|$
has the same distribution as $|xy^{-1}|$. Hence, $z$, $xy^{-1}$ and
$s_{1}s_{2}^{-1}$ have the same distribution. The $S$-transform and Cauchy
transform of $|z|^{2}$ was calculated in the proof of [14, Theorem 5.2]. ∎

From the distribution of $|z|^{2}$, one can check that (see [14, Theorem
5.4]), for any $n\in\mathbb{N}$,

(5.27)
$h_{\mu_{|z|^{n}}}(s)=\int_{0}^{\infty}\frac{s}{s^{2}+u^{2}}d\mu_{|z|^{n}}(u)=\left(s+s^{\frac{n-1}{n+1}}\right)^{-1}.$

In particular,

(5.28) $h_{\mu_{|z|}}(s)=\frac{1}{s+1}.$

We then study the Brown measure of $x_{0}+z$ where
$x_{0}\in{\mathcal{M}}^{\Delta}$ is $*$-free from $z$ with an arbitrary
distribution. We now set

$\mu_{1}=\tilde{\mu}_{|x_{0}-\lambda{\bf
1}|},\quad\text{and}\quad\mu_{2}=\tilde{\mu}_{|z|}.$

and $\mu=\mu_{1}\boxplus\mu_{2}$. We hence have

$\tilde{\mu}_{|x_{0}+z-\lambda{\bf 1}|}=\mu_{1}\boxplus\mu_{2}.$

For $\lambda\in\mathbb{C}$ fixed, following notations in Section 3, we set in
this subsection

$T_{1}=V(x_{0}-\lambda{\bf 1}),\qquad T_{2}=z.$

where $V$ is a Haar unitary that is $*$-free from $\\{x_{0},z\\}$. We recall
that for any probability measure $\mu$ on $\mathbb{R}^{+}$, we have
$h_{\mu}(s)=h_{\tilde{\mu}}(s)$, where

$h_{\mu}(s)=\int_{\mathbb{R}}\frac{s}{s^{2}+u^{2}}d\mu(u),\quad s>0.$

######  Lemma 5.16.

We have $s_{1}(t)=t+1$ and

(5.29) $\Delta(|T_{2}|^{2}+s_{2}(t)^{2})h_{\mu}(t)^{2}=1.$

###### Proof.

The function $s_{1}(t)$ satisfies the following fixed point equation from
(3.9)

$s_{1}(t)=t+f_{\mu_{2}}(t+f_{\mu_{1}}(s_{1}(t))),\quad t>0$

where $f_{\mu_{i}}(i=1,2)$ was defined in (3.2). In particular, due to (5.28),
we have

$f_{\mu_{2}}(s)=1.$

Consequently,

(5.30) $s_{1}(t)=t+1$

and

$s_{2}(t)=t+\frac{1}{h_{\mu}(t)}-s_{1}(t)=\frac{1}{h_{\mu}(t)}-1.$

Taking the derivative with respect to $t$, we obtain

$s_{2}^{\prime}(t)=-\frac{h_{\mu}^{\prime}(t)}{h_{\mu}(t)^{2}}$

and therefore (by using $h_{\mu}(t)=h_{\mu_{2}}(s_{2}(t))$)

(5.31)
$h_{\mu_{2}}(s_{2}(t))s_{2}^{\prime}(t)+\frac{h_{\mu}^{\prime}(t)}{h_{\mu}(t)}=0.$

The above equation is exactly the one half of the logarithmic derivative of
(5.29). Finally, observe that

$\lim_{t\rightarrow\infty}\Delta(|T_{2}|^{2}+s_{2}(t)^{2})h_{\mu}(t)^{2}=1.$

The result then follows. ∎

######  Lemma 5.17.

Let $T\in{\mathcal{M}}^{\Delta}$ be $*$-free from the circular Cauchy operator
$z\in{\mathcal{M}}^{\Delta}$. We then have

(5.32) $\Delta\large((T+z)^{*}(T+z)+t^{2}{\bf
1}\large)=\Delta\large(T^{*}T+(t+1){\bf 1}\large)$

###### Proof.

It follows from Theorem 4.5 and Lemma 5.16. ∎

We are now be able to extend [15, Proposition 4.5] to any unbounded free
random variable in ${\mathcal{M}}^{\Delta}$.

######  Theorem 5.18.

Let $T\in{\mathcal{M}}^{\Delta}$ and $z=xy^{-1}$, where $\\{x,y\\}$ is a
circular system. Then for any $a>0$, we have

(5.33) $\Delta(T+az)=\Delta(T^{*}+a^{2}{\bf 1})^{\frac{1}{2}},\quad a>0.$

###### Proof.

By rescaling with a factor of $1/a$, it suffices to prove (5.33) when $a=1$.
We have

$\lim_{t\rightarrow 0}\Delta\large((T+z)^{*}(T+z)+t^{2}{\bf
1}\large)=\large(\Delta(T+z)\large)^{2}$

and

$\lim_{t\rightarrow 0}\Delta(T^{*}T+(t+1){\bf 1})=\Delta(T^{*}T+{\bf 1}).$

Then (5.33) follows from Lemma 5.17. ∎

The following result generalizes [15, Corollary 4.6] to unbounded operators.

######  Corollary 5.19.

Let $T\in{\mathcal{M}}^{\Delta}$ be $*$-free from the circular Cauchy operator
$z\in{\mathcal{M}}^{\Delta}$. The Brown measure of $T+az$ is given by

(5.34)
$d\mu_{T+az}(\lambda)=\frac{a^{2}}{\pi}\phi\large((T(\lambda)^{*}T(\lambda)+a^{2}{\bf
1})^{-1}(T(\lambda)T(\lambda)^{*}+a^{2}{\bf
1})^{-1}\large)d\lambda_{1}d\lambda_{2},$

where $\lambda_{1}=\Re\lambda,\lambda_{2}=\Im\lambda$, and
$T(\lambda)=T-\lambda{\bf 1}$. In particular, the Brown measure of $z$ is
given by

$d\mu_{z}(\lambda)=\frac{1}{\pi(1+|\lambda|^{2})^{2}}d\lambda_{1}d\lambda_{2}.$

###### Proof.

The proof is identical to the proof of [15, Corollary 4.6], thanks to Theorem
5.18. ∎

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