# The Brown measure of a sum of two free random variables, one of which is
triangular elliptic

Serban Belinschi, Zhi Yin, and Ping Zhong  Serban Belinschi, CNRS, France.
<EMAIL_ADDRESS>Zhi Yin, School of Mathematics and
Statistics, Central South University, Changsha, Hunan 410083, China.
<EMAIL_ADDRESS>Ping Zhong, Department of Mathematics and Statistics,
University of Wyoming, Laramie, WY 82070, USA<EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

The triangular elliptic operators are natural extensions of the elliptic
deformation of circular operators. We obtain a Brown measure formula for the
sum of a triangular elliptic operator $g_{{}_{\alpha\beta,\gamma}}$ with an
arbitrary random variable $x_{0}$, which is $*$-free from
$g_{{}_{\alpha\beta,\gamma}}$ with amalgamation over certain unital
subalgebra. Let $c_{t}$ be a circular operator. We prove that the Brown
measure of $x_{0}+g_{{}_{\alpha\beta,\gamma}}$ is the push-forward measure of
the Brown measure of $x_{0}+c_{t}$ by an explicitly defined map on
$\mathbb{C}$ for some suitable $t$. We show that the Brown measure of
$x_{0}+c_{t}$ is absolutely continuous with respect to Lebesgue measure on
$\mathbb{C}$ and its density is bounded by $1/\pi{t}$. This work generalizes
earlier results on the addition with a circular operator, semicircular
operator, or elliptic operator to a larger class of operators. We extend
operator-valued subordination functions, due to Biane and Voiculescu, to
certain unbounded operators. This allows us to extend our results to unbounded
operators.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Extension of subordination functions to unbounded operators
  4. 4 The sum with a triangular elliptic operator
  5. 5 The Brown measure formulas
  6. 6 Some regularity results on the pushforward map
  7. 7 The sum of an elliptic operator and an unbounded operator

##  1\. Introduction

###  1.1. Brown measure and its regularization

The limit objects of random matrix models can often be identified as operators
in free probability theory. In [11], Brown introduced the Brown measure of
operators in the context of semi-finite von Neumann algebras, which is a
natural generalization of the eigenvalue counting measure of square matrices
with finite dimension. In his breakthrough paper [28], Voiculescu discovered
that independent Gaussian random matrices are asymptotically free as the
dimension of matrices goes to infinity. The Brown measure of a free random
variable is regarded as a candidate for the limit empirical spectral
distribution of suitable random matrix models.

Let $\mathcal{A}$ be a von Neumann algebra with a faithful, normal, tracial
state $\phi$. Recall that the Fuglede-Kadison determinant of $x\in\mathcal{A}$
is defined by

$\Delta(x)=\exp\left(\int_{0}^{\infty}\log td\mu_{|x|}(t)\right),$

where $\mu_{|x|}$ is the spectral measure of $|x|$ with respect to $\phi$. It
is known [11] that the function $\lambda\mapsto\log\Delta(x-\lambda{\bf 1})$
is a subharmonic function whose Riesz measure is the unique probability
measure $\mu_{x}$ on $\mathbb{C}$ with the property that

$\log\Delta(x-\lambda)=\int_{\mathbb{C}}\log|z-\lambda|d\mu_{x}(z),\qquad\lambda\in\mathbb{C}.$

The measure $\mu_{x}$ is called the _Brown measure_ of $x,$ and it can be
recovered as the distributional Laplacian of the function
$\log\Delta(x-\lambda{\bf 1})$ as follows

(1.1)
$\mu_{x}=\frac{1}{2\pi}\nabla^{2}\log\Delta(x-\lambda)=\frac{2}{\pi}\frac{\partial}{\partial\lambda}\frac{\partial}{\partial\bar{\lambda}}\log\Delta(x-\lambda).$

When $\mathcal{A}=M_{n}({\mathbb{C}})$ and $\phi$ is the normalized trace on
$M_{n}(\mathbb{C})$, for $x\in M_{n}({\mathbb{C}})$, we have

$\mu_{x}=\frac{1}{n}\left(\delta_{\lambda_{1}}+\cdots+\delta_{\lambda_{n}}\right)$

where $\lambda_{1},\cdots,\lambda_{n}$ are the eigenvalues of $x$.

Let $\widetilde{\mathcal{A}}$ be the set of operators which are affiliated
with $\mathcal{A}$. Denote by $\log^{+}(\mathcal{A})$ the set of operators
$T\in\widetilde{\mathcal{A}}$ satisfying

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

It is known [17] that the set $\log^{+}(\mathcal{A})$ is a subspace of
$\widetilde{\mathcal{A}}$. In particular, for $T\in\log^{+}(\mathcal{A})$ and
$\lambda\in\mathbb{C}$, $T-\lambda\in\log^{+}(\mathcal{A})$. Moreover, the
function $\lambda\mapsto\log\Delta(T-\lambda)$ is again subharmonic in
$\mathbb{C}$. In the noncommutative probability space $(\mathcal{A},\phi)$,
one can extend the definition of the Brown measure to $\log^{+}(\mathcal{A})$
[11, 17] using the same formula (1.1).

It is useful to consider the regularized function

(1.2)
$S(x,\lambda,\varepsilon)=\phi\bigg{(}\log\big{(}(x-\lambda)^{*}(x-\lambda)+\varepsilon^{2}\big{)}\bigg{)}.$

Then, $S(x,\lambda,0)=\lim_{\varepsilon\rightarrow
0}S(x,\lambda,\varepsilon)$. The regularized Brown measure is defined as

$\mu_{x}^{(\varepsilon)}=\frac{1}{4\pi}\nabla^{2}S(x,\lambda,\varepsilon).$

It is known that $\mu_{x}^{(\varepsilon)}\rightarrow\mu_{x}$ weakly as
$\varepsilon\rightarrow 0$. It turns out that the regularized Brown measure
approximates the Brown measure in a much stronger sense for the operators we
are interested in.

###  1.2. Main results

Given two $*$-free random variables $x,y$ in a $W^{*}$-probability space
$(\mathcal{A},\phi)$, it desirable to find a general method to calculate the
Brown measure of $x+y$. In the special case when $y$ is an elliptic operator,
the third author obtained Brown measure formulas for an arbitrary $x$ in [33].
This generalized previous results of Hall–Ho [20] and Ho–Zhong [22], and
introduced a new approach based on the Hermitian reduction method and
subordination functions. The purposes of this paper are two-folds. First, we
adapt techniques developed in [5, 33] to study the Brown measure of the sum of
a larger family of operators $g_{{}_{\alpha\beta,\gamma}}$, called triangular
elliptic operators, and another random variable $x_{0}$ that is $*$-free from
$g_{{}_{\alpha\beta,\gamma}}$ in the sense of operator-valued free
probability. Second, we extend our main results to unbounded operator $x_{0}$.
Our results for the case $\alpha=\beta$ extend main results obtained by the
third author [33], and generalize main results in [21] obtained by Ho for the
special case $x_{0}$ being self-adjoint, and another result by
Bordenave–Caputo–Chafaï [9] for unbounded normal operators.

It turns out the family of Brown measures $x_{0}+g_{{}_{\alpha\beta,\gamma}}$
can be retrieved from the Brown measure $x_{0}+c_{t}$ by a family of
explicitly constructed self-maps of $\mathbb{C}$, where $c_{t}$ is a circular
operator with variance $t$. Hence, it is crucial to understand the Brown
measure $\mu_{x_{0}+c_{t}}$. We introduce a new approach to study regularity
properties for the Brown measure of $x_{0}+c_{t}$ via regularized Brown
measure, and obtain a complete description of the Brown measure
$\mu_{x_{0}+c_{t}}$. We prove that the density of the regularized Brown
measure $\mu_{x_{0}+c_{t}}^{(\varepsilon)}$ is bounded by $1/\pi{t}$.
Consequently, we show that $\mu_{x_{0}+c_{t}}$ has no singular part and this
settles an open question in [33].

Our method is completely different from the PDE method used by Ho [21], where
he studied the sum of an unbounded selfadjoint operator and an elliptic
operator (corresponding to $\alpha=\beta$). Although our approach is an
extension of the method used in [33], the details are substantially more
technical and some new ideas are developed to study the distinguished measure
$\mu_{x_{0}+c_{t}}$. In addition, we extend the subordination functions in
operator-valued free probability, due to Biane and Voiculescu, to unbounded
operators under some assumptions. The subordination result is of interest by
its own and it allows us to extend our results to unbounded operators.

Consider a random matrix $A_{N}=(a_{ij})_{1\leq i,j\leq N}$ such that the
joint distribution of random variables $(\Re a_{ij},\Im a_{ij})_{1\leq i,j\leq
N}$ is centered Gaussian with covariance given by

$\mathbb{E}a_{ij}\overline{a_{kl}}=\begin{cases}\frac{\alpha}{N}\delta_{ik}\delta_{jl}\quad&\text{if}\,i<j,\\\
\frac{\alpha+\beta}{2N}\delta_{ik}\delta_{jl}\quad&\text{if}\,i=j,\\\
\frac{\beta}{N}\delta_{ik}\delta_{jl}\quad&\text{if}\,i>j,\end{cases}\quad\text{and}\quad\mathbb{E}a_{ij}a_{kl}=\frac{\gamma}{N}\delta_{il}\delta_{jk},$

where $\alpha,\beta>0$ and $\gamma\in\mathbb{C}$ such that
$|\gamma|\leq\sqrt{\alpha\beta}$. One can show that $A_{N}$ converges in
$*$-moments to some operator $g_{\alpha\beta,\gamma}$, called a triangular
elliptic operator, in $\mathcal{A}$.

When $\alpha=\beta=t$ and $\gamma=0$, then $A_{N}$ converges in $*$-moments to
the circular operator $c_{t}$ with variance $t$, which has the same
distribution as $\sqrt{t/2}(s_{1}+is_{2})$, where $\\{s_{1},s_{2}\\}$ is a
semicircular system. Let $\\{g_{t_{1}},g_{t_{2}}\\}$ be freely independent
semicircular operators with mean zero and variances $t_{1},t_{2}$
respectively. Choose $t_{1}=(t+|\gamma|)/2,t_{2}=(t-|\gamma|)/2$ and
$\theta\in[0,2\pi]$ such that $e^{i2\theta}=\gamma/|\gamma|$. When
$\alpha=\beta=t$ and $|\gamma|\leq t$, then $A_{N}$ converges in $*$-moments
to twisted elliptic operator $g_{t,\gamma}$, which has the same distribution
as $e^{i\theta}(g_{t_{1}}+ig_{t_{2}})$.

Consider the open set

(1.3)
$\Xi_{t}=\left\\{\lambda\in\mathbb{C}:\phi\left[\big{(}(x_{0}-\lambda)^{*}(x_{0}-\lambda)\big{)}^{-1}\right]>\frac{1}{t}\right\\}.$

Fix $\lambda\in\Xi_{t}$ and let $w=w(0;\lambda,t)$ be a function of $\lambda$
taking positive values such that

$\phi\left[((x_{0}-\lambda)^{*}(x_{0}-\lambda)+w^{2})^{-1}\right]=\frac{1}{t},$

and let $w(0;\lambda,t)=0$ for $\lambda\in\mathbb{C}\backslash\Xi_{t}$. We
denote

$\Phi_{t,\gamma}(\lambda)=\lambda+\gamma\cdot p_{\lambda}^{(0)}(w),$

where

$p_{\lambda}^{(0)}(w)=-\phi\bigg{[}(x_{0}-\lambda)^{*}\big{(}(x_{0}-\lambda)(x_{0}-\lambda)^{*}+w(0;\lambda,t)^{2}\big{)}^{-1}\bigg{]}.$

######  Theorem 1.1.

[33, Theorem 4.2 and 5.4] The Brown measure of $x_{0}+c_{t}$ has no atom and
it is supported in the closure of $\Xi_{t}$. It is absolutely continuous with
respect to the Lebesgue measure in $\Xi_{t}$. Moreover, the density in
$\Xi_{t}$ is strictly positive and can be expressed explicitly in terms of
$w(0;\lambda,t)$.

For any $\gamma\in\mathbb{C}$ such that $|\gamma|\leq t$, the Brown measure of
$x_{0}+g_{t,\gamma}$ is the push-forward measure of the Brown measure of
$x_{0}+c_{t}$ under the map $\Phi_{t,\gamma}$.

The push-forward property of a similar type was first observed for some
special cases when $g_{t,\gamma}$ is a semicircular element, and self-adjoint
operator $x_{0}$ in [22] by PDE methods. In a followup work [20], it was
proved for the sum of an imaginary multiple of semicircular elements and a
self-adjoint operator $x_{0}$.

When $\alpha\neq\beta$, one can show [13, 5] that the limit operator
$y=g_{{}_{\alpha\beta,\gamma}}$ of $A_{N}$ lives in some operator-valued
$W^{*}$-probability space $(\mathcal{A},\mathbb{E},\mathcal{B})$ where the
unital subalgebra $\mathcal{B}$ can be identified as
$\mathcal{B}=L^{\infty}[0,1]$ and
$\mathbb{E}:\mathcal{A}\rightarrow\mathcal{B}$ is the conditional expectation
(see Section 4.1). Given $x\in\mathcal{A}$, by identifying $\mathbb{E}(x)=f$
for some $f\in L^{\infty}[0,1]$, the tracial state
$\phi:\mathcal{A}\rightarrow\mathbb{C}$ is determined by

$\phi(x)=\phi(\mathbb{E}(x))=\int_{0}^{1}f(s)ds.$

Let $x_{0}\in\log^{+}(\mathcal{A})$ be an operator that is $*$-free from the
family of operators
$\\{g_{{}_{\alpha\beta,\gamma}}\\}_{|\gamma|\leq\sqrt{\alpha\beta}}$ with
amalgamation over $\mathcal{B}$ such that $x_{0}$ is $*$-free from
$\mathcal{B}$ with respect to the tracial state $\phi$. Our main interest is
the Brown measure of $x_{0}+g_{\alpha\beta,\gamma}$.

We now describe our main results. Given $\alpha,\beta>0$, we denote

${\Xi_{\alpha,\beta}}=\left\\{\lambda:\phi(|x_{0}-\lambda|^{-2})>\frac{\log\alpha-\log\beta}{\alpha-\beta}\right\\}.$

For $\lambda\in{\Xi_{\alpha,\beta}}$, let $s(\lambda)$ be the positive number
$s$ determined by

$\frac{\log\alpha-\log\beta}{\alpha-\beta}=\phi\left\\{\left[(\lambda-
x_{0})^{*}(\lambda-x_{0})+s^{2}\right]^{-1}\right\\},$

and for $\lambda\in\mathbb{C}\backslash{\Xi_{\alpha,\beta}}$, we put
$s(\lambda)=0$. For $\gamma\in\mathbb{C}$ such that
$|\gamma|\leq\sqrt{\alpha\beta}$, we define the map
$\Phi_{\alpha\beta,\gamma}:\mathbb{C}\rightarrow\mathbb{C}$ by

$\Phi_{\alpha\beta,\gamma}(\lambda)=\lambda+\gamma\phi\left\\{(\lambda-
x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+s(\lambda)^{2}\right]^{-1}\right\\}.$

######  Theorem 1.2 (See Theorem 5.4 and Theorem 5.8).

Let $x_{0}\in\log^{+}(\mathcal{A})$ be an operator that is $*$-free from the
family of operators
$\\{g_{{}_{\alpha\beta,\gamma}}\\}_{|\gamma|\leq\sqrt{\alpha\beta}}$ with
amalgamation over $\mathcal{B}$ such that $x_{0}$ is $*$-free from
$\mathcal{B}$ with respect to the tracial state $\phi$. Given $\alpha,\beta>0$
such that $\alpha\neq\beta$, the support of the Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ of $x_{0}+g_{{}_{\alpha\beta,0}}$ is the
closure of the set ${\Xi_{\alpha,\beta}}$. The Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ is absolutely continuous and the density
is strictly positive in ${\Xi_{\alpha,\beta}}$. Moreover, the density formula
can be expressed in terms of $s(\lambda)$.

Given $\gamma\in\mathbb{C}$ such that $|\gamma|\leq\sqrt{\alpha\beta}$, the
Brown measure $\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}$ is the push-forward
measure of $\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ under the map
$\Phi_{\alpha\beta,\gamma}$. If, in addition, the map
$\Phi_{\alpha\beta,\gamma}$ is non-singular at any
$\lambda\in{\Xi_{\alpha,\beta}}$, then it is also one-to-one in
${\Xi_{\alpha,\beta}}$.

The above Theorem 1.2 recovers the Brown measure formula obtained in [2,
Section 4] and [5, Section 6] as a special case for $x_{0}=0$. See Example
5.9. We note that the map $\Phi_{{}_{\alpha\beta,\gamma}}$ could be singular
in general. This restriction does not allow us to calculate the Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}$ directly. We adapt an approach
introduced in [33] as follows. We show that the regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}^{(\varepsilon)}$ is the pushforward
measure of the Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}^{(\varepsilon)}$ by some regularized map
$\Phi_{{}_{\alpha\beta,\gamma}}^{(\varepsilon)}$. We then show that the
pushforward connection is preserved by passing to the limit. Hence, the
following diagram commutes.

${\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}^{(\varepsilon)}}$${\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}^{(\varepsilon)}}$${\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}}$${\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}}$$\varepsilon\rightarrow
0$$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$$\Phi_{\alpha\beta,\gamma}$$\varepsilon\rightarrow
0$

In [33], it was left open if the Brown measure of $\mu_{x_{0}+c_{t}}$ has any
singular continuous part. We show that it is absolutely continuous with
respect to the Lebesgue measure on $\mathbb{C}$ and the statement of Theorem
1.1 still holds for unbounded operators.

######  Theorem 1.3 (See Theorem 7.12).

The Brown measure $\mu_{x_{0}+c_{t}}$ is absolutely continuous with respect to
the Lebesgue measure on $\mathbb{C}$ and the result in Theorem 1.1 holds for
any $x_{0}\in\log^{+}(\mathcal{A})$ that is $*$-free from
$\\{c_{t},g_{t,\gamma}\\}$. Moreover, the density function $\rho_{t,x_{0}}$ of
the Brown measure has an upper bound

$\rho_{t,x_{0}}(\lambda)\leq\frac{1}{\pi t}.$

In [9], Bordenave, Caputo and Chafaï investigated the spectrum of the Markov
generator of random walk on a randomly weighted oriented graph. They
identified the limit distribution as the Brown measure of $a+c$, where $a$ is
a Gaussian distributed normal operator and $c$ is the standard circular
operator, freely independent from $a$. Hence, Theorem 1.3 implies that
$\mu_{a+c}$ is absolutely continuous and its density is bounded by $1/\pi$.
Our main results also provide potential applications to unify the deformed
i.i.d. random matrix model, the deformed Wigner random matrix model, and the
deformed elliptic random matrix model. We hope to study this question
somewhere else.

By comparing precise formulas in Theorem 1.1 with Theoorem 1.2, we can view
the main results in [33] as the special case of our results when
$\alpha=\beta=t$, where $\frac{\log\alpha-\log\beta}{\alpha-\beta}$ is
regarded as $\frac{1}{t}$.

######  Corollary 1.4.

Under the same assumption as Theorem 1.2, given $\alpha,\beta>0$, set
$t=\frac{\alpha-\beta}{\log\alpha-\log\beta}$. Suppose $x_{0}$ is $*$-free
from $\\{c_{t},g_{t,\gamma}\\}_{|\gamma|\leq t}$ in $(\mathcal{A},\phi)$.
Then,

  1. (1)

the Brown measure of $x_{0}+g_{{}_{\alpha\beta,0}}$ is the same as the Brown
measure of $x_{0}+c_{t}$. In particular, it is absolutely continuous in
$\mathbb{C}$ and its density is bounded by $1/\pi{t}$;

  2. (2)

the push-forward map $\Phi_{\alpha\beta,\gamma}$ between Brown measures
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ and
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}$ is the same as the push-forward map
$\Phi_{t,\gamma}$ between Brown measures $\mu_{x_{0}+c_{t}}$ and
$\mu_{x_{0}+g_{t,\gamma}}$.

The paper has six more sections. We review some results from free probability
theory in Section 2. We study the subordination functions in Section 3 and
regularized Brown measure. We calculate the Brown measure formulas and prove
the push-forward property in Section 4. In Section 5, we study some further
properties of the push-forward map. In Section 6, we study the Brown measure
of an addition with an elliptic operator.

##  2\. Preliminaries

###  2.1. Free probability and subordination functions

An _operator-valued $W^{*}$-probability space_
$({\mathcal{A}},\mathbb{E},{\mathcal{B}})$ consists of a von Neumann algebra
${\mathcal{A}}$, a unital $*$-subalgebra ${\mathcal{B}}\subset{\mathcal{A}}$,
and a _conditional expectation_
$\mathbb{E}:{\mathcal{A}}\rightarrow\mathcal{B}$. Thus, $\mathbb{E}$ is a
unital linear positive map satisfying: (1) $\mathbb{E}(b)=b$ for all
$b\in{\mathcal{B}}$, and (2) $\mathbb{E}(b_{1}xb_{2})=b_{1}\mathbb{E}(x)b_{2}$
for all $x\in{\mathcal{A}}$, $b_{1},b_{2}\in{\mathcal{B}}$. Let
$({\mathcal{A}}_{i})_{i\in I}$ be a family of subalgebras
${\mathcal{B}}\subset{\mathcal{A}}_{i}\subset{\mathcal{A}}$. We say that
$({\mathcal{A}}_{i})_{i\in I}$ are _free with amalgamation_ over
${\mathcal{B}}$ with respect to the conditional expectation $\mathbb{E}$ _(_
or free with amalgamation in $({\mathcal{A}},\mathbb{E},{\mathcal{B}}))$ if
for every $n\in\mathbb{N}$,

$\mathbb{E}(x_{1}x_{2}\cdots x_{n})=0$

whenever $x_{j}\in{\mathcal{A}}_{i_{j}}(j=1,2,\cdots,n)$ for some indices
$i_{1},i_{2},\cdots,i_{n}\in I$ such that $i_{1}\neq i_{2},i_{2}\neq
i_{3},\cdots,i_{n-1}\neq i_{n}$, and $\mathbb{E}(x_{j})=0$ for all
$j=1,2\cdots,n$.

Let $({\mathcal{A}},\mathbb{E},{\mathcal{B}})$ be an operator-valued
$W^{*}$-probability space. The elements in ${\mathcal{A}}$ are called
(noncommutative) random variables. We call

$\mathbb{H}^{+}({\mathcal{B}})=\\{b\in{\mathcal{B}}:\exists\varepsilon>0,\Im(b)\geq\varepsilon{\bf
1}\\}$

the Siegel upper half-plane of ${\mathcal{B}}$, where we use the notation
$\Im(b)=\frac{1}{2i}(b-b^{*})$. We set
$\mathbb{H}^{-}({\mathcal{B}})=\\{-b:b\in\mathbb{H}^{+}({\mathcal{B}})\\}$.
The $\mathcal{B}$-valued Cauchy transform $G_{X}$ of any self-adjoint operator
$X\in{\mathcal{A}}$ is defined by

$G_{X}(b)=\mathbb{E}[(b-X)^{-1}],\quad b\in\mathbb{H}^{+}({\mathcal{B}}).$

Note that the $\mathcal{B}$-valued Cauchy transform $G_{X}$ is a map from
$\mathbb{H}^{+}({\mathcal{B}})$ to $\mathbb{H}^{-}({\mathcal{B}})$, and it is
one-to-one in $\\{b\in\mathbb{H}^{+}({\mathcal{B}}):||b^{-1}||<\varepsilon\\}$
for sufficiently small $\varepsilon$. Voiculescu’s amalgamated $R$-transform
is now defined for $X\in{\mathcal{A}}$ by

$R_{X}(b)=G_{X}^{\langle-1\rangle}(b)-b^{-1}$

for $b$ being invertible element of ${\mathcal{B}}$ suitably close to zero.

Let $X,Y$ be two self-adjoint bounded operators that are free with
amalgamation in $(\mathcal{A},\mathbb{E},\mathcal{B})$. The $R$-transform
linearizes the free convolution in the sense that if $X,Y$ are free with
amalgamation in $({\mathcal{A}},\mathbb{E},{\mathcal{B}})$, then

$R_{X+Y}(b)=R_{X}(b)+R_{Y}(b)$

for $b$ in some suitable domain. There exist two analytic self-maps
$\Omega_{1},\Omega_{2}$ of the upper half-plane $\mathbb{H}^{+}(\mathcal{B})$
so that

(2.1)
$(\Omega_{1}(b)+\Omega_{2}(b)-b)^{-1}=G_{X}(\Omega_{1}(b))=G_{Y}(\Omega_{2}(b))=G_{X+Y}(b),$

for all $b\in\mathbb{H}^{+}(\mathcal{B})$. We refer the reader to [27, 29] for
basic operator-valued free probability theory and [4, 8, 30] for operator-
valued subordination functions.

###  2.2. Hermitian reduction method and subordination functions

Our approach is based on a Hermitian reduction method and (operator-valued)
subordination functions. The Hermitian reduction method was used for the
calculation of the Brown measure of quasi-nilpotent DT operators in
Aagaardd–Haagerup’s work [2]. The idea was refined in recent work [5]. The
usage of such idea in Brown measure calculation also appears earlier in some
physics papers (see [15, 23] for example). The idea of the Hermitian reduction
method has a connection to the work of Girko [16] on circular law, though it
is written in a different form.

Let $({\mathcal{A}},\mathbb{E},{\mathcal{B}})$ be an operator-valued
$W^{*}$-probability space and let $\phi:\mathcal{A}\rightarrow\mathbb{C}$ be a
tracial state on ${\mathcal{A}}$ such that $\phi(x)=\phi(\mathbb{E}(x))$ for
any $x\in{\mathcal{A}}$. We equip the algebra $M_{2}({\mathcal{A}})$, the
$2\times 2$ matrices with entries from ${\mathcal{A}}$, with the conditional
expectation $M_{2}(\mathbb{E}):M_{2}({\mathcal{A}})\rightarrow
M_{2}({\mathcal{B}})$ given by

(2.2) $M_{2}(\mathbb{E})\begin{bmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}\mathbb{E}(a_{11})&\mathbb{E}(a_{12})\\\
\mathbb{E}(a_{21})&\mathbb{E}(a_{22})\end{bmatrix},$

and $M_{2}(\phi):M_{2}({\mathcal{A}})\rightarrow M_{2}(\mathbb{C})$ by

(2.3) $M_{2}(\phi)\begin{bmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}\phi(a_{11})&\phi(a_{12})\\\
\phi(a_{21})&\phi(a_{22})\end{bmatrix}.$

Then the triple $(M_{2}(\mathcal{A}),M_{2}(\mathbb{E}),M_{2}(\mathcal{B}))$ is
a $W^{*}$-probability space such that $M_{2}(\phi)\circ
M_{2}(\mathbb{E})=M_{2}(\phi)$. Given $x\in{\mathcal{A}}$, let

(2.4) $X=\begin{bmatrix}0&x\\\ x^{\ast}&0\end{bmatrix}\in
M_{2}({\mathcal{A}}),$

which is a selfadjoint element in $M_{2}({\mathcal{A}}).$ The
$M_{2}(\mathcal{B})$-valued Cauchy transform is defined as

(2.5) $G_{X}(b)=M_{2}(\mathbb{E})\left[(b-X)^{-1}\right],\qquad
b\in\mathbb{H}^{+}(M_{2}(\mathcal{B}))$

which is an analytic function on $\mathbb{H}^{+}(M_{2}(\mathcal{B}))$. In
particular, for $\varepsilon>0$ the element

(2.6) $\Theta(\lambda,\varepsilon)=\begin{bmatrix}i\varepsilon&\lambda\\\
\bar{\lambda}&i\varepsilon\end{bmatrix}\in M_{2}({\mathbb{C}})\subset
M_{2}(\mathcal{B})$

belongs to the domain of $G_{X}$, and

$\displaystyle(\Theta(\lambda,\varepsilon)-X)^{-1}$ $\displaystyle=$
$\displaystyle\begin{bmatrix}-i\varepsilon\big{(}(\lambda-x)(\lambda-x)^{\ast}+\varepsilon^{2}\big{)}^{-1}&(\lambda-x)\big{(}(\lambda-x)^{\ast}(\lambda-x)+\varepsilon^{2}\big{)}^{-1}\\\
(\lambda-x)^{\ast}\big{(}(\lambda-x)(\lambda-x)^{\ast}+\varepsilon^{2}\big{)}^{-1}&-i\varepsilon\big{(}(\lambda-x)^{\ast}(\lambda-x)+\varepsilon^{2}\big{)}^{-1}\end{bmatrix}.$

Hence, by taking entry-wise trace on the $M_{2}(\mathcal{B})$-valued Cauchy
transform, we have

(2.7) $\displaystyle
M_{2}(\phi)\big{(}G_{X}(\Theta(\lambda,\varepsilon))\big{)}=M_{2}(\phi)\big{(}(\Theta(\lambda,\varepsilon)-X)^{-1}\big{)}$
$\displaystyle=\begin{bmatrix}g_{X,11}(\lambda,\varepsilon)&g_{X,12}(\lambda,\varepsilon)\\\
g_{X,21}(\lambda,\varepsilon)&g_{X,22}(\lambda,\varepsilon)\end{bmatrix},$

where

(2.8) $\displaystyle g_{X,11}(\lambda,\varepsilon)$
$\displaystyle=-i\varepsilon\phi\left(\big{(}(\lambda-x)(\lambda-x)^{*}+\varepsilon^{2}\big{)}^{-1}\right),$
$\displaystyle g_{X,12}(\lambda,\varepsilon)$
$\displaystyle=\phi\left((\lambda-x)\big{(}(\lambda-x)^{*}(\lambda-x)+\varepsilon^{2}\big{)}^{-1}\right),$
$\displaystyle g_{X,21}(\lambda,\varepsilon)$
$\displaystyle=\phi\left((\lambda-x)^{*}\big{(}(\lambda-x)(\lambda-x)^{*}+\varepsilon^{2}\big{)}^{-1}\right),$
$\displaystyle g_{X,22}(\lambda,\varepsilon)$
$\displaystyle=-i\varepsilon\phi\left(\big{(}(\lambda-x)^{*}(\lambda-x)+\varepsilon^{2}\big{)}^{-1}\right).$

Recall that $S(x,\lambda,\varepsilon)$ is defined in (1.2). We see immediately
that

$ig_{X,11}(\lambda,\varepsilon)=\frac{1}{2}\frac{\partial
S(x,\lambda,\varepsilon)}{\partial\varepsilon},\qquad
g_{X,21}(\lambda,\varepsilon)=\frac{\partial
S(x,\lambda,\varepsilon)}{\partial\lambda}.$

Hence, the entries of operator-valued Cauchy transform carry important
information about the Brown measure.

Consider now two operators $x,y\in\mathcal{A}$ that are $*$-free with
amalgamation over $\mathcal{B}$. We have to understand the
$M_{2}(\mathcal{B})$-valued distribution of

$\begin{bmatrix}0&x+y\\\ (x+y)^{*}&0\end{bmatrix}=X+Y=\begin{bmatrix}0&x\\\
x^{*}&0\end{bmatrix}+\begin{bmatrix}0&y\\\ y^{*}&0\end{bmatrix}$

in terms of the $M_{2}(\mathcal{B})$-valued distributions of $X$ and of $Y$.
Note that $X$ and $Y$ are free in
$(M_{2}(\mathcal{A}),M_{2}(\mathbb{E}),M_{2}(\mathcal{B}))$ with amalgamation
over $M_{2}(\mathcal{B})$. The subordination functions in this context are two
analytic self-maps $\Omega_{1},\Omega_{2}$ of the upper half-plane
$\mathbb{H}^{+}(M_{2}(\mathcal{B}))$ so that

(2.9)
$(\Omega_{1}(b)+\Omega_{2}(b)-b)^{-1}=G_{X}(\Omega_{1}(b))=G_{Y}(\Omega_{2}(b))=G_{X+Y}(b),$

for every $b\in\mathbb{H}^{+}(M_{2}(\mathcal{B}))$. We shall be concerned with
a special type of $b$, namely $b=\Theta(\lambda,\varepsilon)$. More precisely,
we want to understand the entries of the $M_{2}(\mathbb{C})$-valued Cauchy-
transform

$M_{2}(\phi)(G_{X+Y}(\Theta(\lambda,\varepsilon)))=M_{2}(\phi)\left(\begin{bmatrix}i\varepsilon&\lambda-
x-y\\\ (\lambda-x-y)^{*}&i\varepsilon\end{bmatrix}^{-1}\right)$

The idea of calculating the Brown measure of $x+y$ is to separate the
information of $X$ and $Y$ in some tractable way using subordination functions
(2.9).

##  3\. Extension of subordination functions to unbounded operators

Let $(\mathcal{A},\mathbb{E},\mathcal{B})$ be an operator-valued
$W^{*}$-probability space. Consider arbitrary $x,y\in\tilde{\mathcal{A}}$
which are $*$-free over $\mathcal{B}$ with respect to $\mathbb{E}$, meaning
that $W^{*}(\mathcal{B},x)$ and $W^{*}(\mathcal{B},y)$ are free with respect
to $\mathbb{E}$ (when $x$ is unbounded, $W^{*}(\mathcal{B},x)$ denotes the von
Neumann algebra generated by $\mathcal{B}$, $u$, and the spectral projections
of $\sqrt{x^{*}x}$ from the polar decomposition $x=u\sqrt{x^{*}x}$).

We denote $X=\begin{bmatrix}0&x\\\ x^{*}&0\end{bmatrix}$ and
$Y=\begin{bmatrix}0&y\\\ y^{*}&0\end{bmatrix}$. By the similar result for
bounded operators, then, $X,Y$ are free with amalgamation over
$M_{2}(\mathcal{B})$ with respect to
$M_{2}(\mathbb{E})\begin{bmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{bmatrix}=\begin{bmatrix}\mathbb{E}(a_{11})&\mathbb{E}(a_{12})\\\
\mathbb{E}(a_{21})&\mathbb{E}(a_{22})\end{bmatrix}$. The following result is
an extension of Voiculescu’s subordination functions [30] in the operator-
valued framework (see also [7]). The method is adapted from the approach used
in [4]. It is essentially covered by [3, Corollary 6.6], and by Williams [32],
but we provide here a more elementary argument. The reader is referred to [4,
Section 2] for some details.

######  Theorem 3.1.

If either $\mathcal{B}$ is finite dimensional or $y$ is bounded and
$\Im\mathbb{E}\left[(b_{0}-X)^{-1}\right]<0$ for some $b_{0}\in
M_{2}(\mathcal{B}),\Im b_{0}>0$, then there exist analytic maps

$\Omega_{1},\Omega_{2}\colon\mathbb{H}^{+}(M_{2}(\mathcal{B}))\to\mathbb{H}^{+}(M_{2}(\mathcal{B}))$

such that

(3.1) $\displaystyle\Omega_{1}(b)+\Omega_{2}(b)-b$
$\displaystyle=\mathbb{E}\left[(\Omega_{1}(b)-X)^{-1}\right]^{-1}\\!\\!$
$\displaystyle=\mathbb{E}\left[(\Omega_{2}(b)-Y)^{-1}\right]^{-1}\\!\\!=\mathbb{E}\left[(b-\\!X-\\!Y)^{-1}\right]^{-1}\\!.$

###### Proof.

For simplicity, denote
$F_{X+Y}(b)=\mathbb{E}\\!\left[(b-\\!X\\!-\\!Y)^{-1}\right]^{-1}$, and
$h_{X+Y}(b)=F_{X+Y}(b)-b$.

Case one: assume that both $x$ and $y$ are unbounded and
dim$(\mathcal{B})<\infty$. We shall obtain the $\Omega_{1},\Omega_{2}$ as
fixed points of an analytic map. We will use a slightly different trick, due
to Hari Bercovici. Observe that, if one knows the two subordination functions
to exist, then one has
$h_{X}(\Omega_{1}(b))+b=\Omega_{2}(b),h_{Y}(\Omega_{2}(b))+b=\Omega_{1}(b)$.
Hence, $\Omega_{1}$ satisfies

$\Omega_{1}(b)=b+h_{Y}(b+h_{X}(\Omega_{1}(b)))$

and similarly, $\Omega_{2}$ satisfies

$\Omega_{2}(b)=b+h_{X}(b+h_{Y}(\Omega_{2}(b))).$

This turns out to be equivalent to the map
$\boldsymbol{\Omega}:=(\Omega_{1},\Omega_{2})$ being the inverse to the right
of another map, namely
$\boldsymbol{\Phi}(w_{1},w_{2})=(w_{1}-h_{Y}(w_{2}),w_{2}-h_{X}(w_{1}))$. More
precisely, let us define the following extensions:

$\Omega_{1}(b_{1},b_{2})=b_{1}+h_{Y}(b_{2}+h_{X}(\Omega_{1}(b_{1},b_{2}))),$

and

$\Omega_{2}(b_{1},b_{2})=b_{2}+h_{X}(b_{1}+h_{Y}(\Omega_{2}(b_{1},b_{2}))).$

For $N>0$, observe that $\chi_{[-N,N]}(Y)$ is well-defined and belongs to
$M_{2}(W^{*}(\mathcal{B},y))$. Indeed, $-N\leq Y\leq N\iff Y^{2}\leq
N^{2}\iff\begin{bmatrix}yy^{*}&0\\\ 0&y^{*}y\end{bmatrix}\leq N^{2}\iff
yy^{*}\leq N^{2}.$ Now, $y=\mathfrak{u}\sqrt{y^{*}y}$,
$y^{*}=\mathfrak{v}\sqrt{yy^{*}}$ for some partial isometries
$\mathfrak{u,v}\in W^{*}(\mathcal{B},y)$, so that $\begin{bmatrix}0&y\\\
y^{*}&0\end{bmatrix}=Y=\mathfrak{U}\sqrt{Y^{*}Y}=\begin{bmatrix}0&\mathfrak{u}\\\
\mathfrak{v}&0\end{bmatrix}\begin{bmatrix}\sqrt{yy^{*}}&0\\\
0&\sqrt{y^{*}y}\end{bmatrix}$. Thus, $Y_{N}:=\chi_{[-N,N]}(Y)Y$ is free from
$X$ over $M_{2}(\mathcal{B})$ with respect to $M_{2}(\mathbb{E})$ as well.

The derivative of $\boldsymbol{\Phi}$ is

$D\boldsymbol{\Phi}(w_{1},w_{2})=\begin{bmatrix}{\rm
id}_{M_{2}(\mathcal{B})}&-h_{Y}^{\prime}(w_{2})\\\ -h_{X}^{\prime}(w_{1})&{\rm
id}_{M_{2}(\mathcal{B})}\end{bmatrix},$

with a similar expression for $D\boldsymbol{\Phi}_{N}$, except that $h_{X}$ is
replaced by $h_{X_{N}}$ and $h_{Y}$ is replaced by $h_{Y_{N}}$. We intend to
show that $D\boldsymbol{\Phi}(i\varepsilon,i\varepsilon)$ is close (in
operator norm) to the identity on $M_{2}(\mathcal{B})\times
M_{2}(\mathcal{B})$ for $\varepsilon\in(0,+\infty)$ large. Let us perform a
more detailed analysis of the behavior of $h_{Y}$. We recall that
$h_{Y}(w)=F_{Y}(w)-w=\mathbb{E}\left[(w\\!-\\!Y)^{-1}\right]^{-1}-w$. The
expression of its derivative is

$h^{\prime}_{Y}(w)(\alpha)=\mathbb{E}\left[(w\\!-\\!Y)^{-1}\right]^{-1}\mathbb{E}\left[(w\\!-\\!Y)^{-1}\alpha(w\\!-\\!Y)^{-1}\right]^{-1}\mathbb{E}\left[(w\\!-\\!Y)^{-1}\right]^{-1}-\alpha.$

For a fixed $w=\Re w+i\Im w\in\mathbb{H}^{+}(M_{2}(\mathcal{B}))$, let us look
at $\mathbb{C}^{+}\ni z\mapsto(\Re w+z\Im w\\!-\\!Y)^{-1}$. For any positive
continuous linear functional $\varphi$ on $M_{2}(\mathcal{A})$, the function
$\mathbb{C}^{+}\ni z\mapsto\varphi\left((\Re w\\!+\\!z\Im
w\\!-\\!Y)^{-1}\right)$ sends $\mathbb{C}^{+}$ to $\mathbb{C}^{-}$ and
satisfies

$\displaystyle\lim_{\varepsilon\to+\infty}\\!i\varepsilon\varphi\\!\left((i\varepsilon\Im
w\\!-\\!(Y\\!-\\!\Re w))^{-1}\right)\allowbreak=\varphi\left((\Im
w)^{-1}\right).$

Since $\varphi$ is arbitrary, it follows that

$\lim_{\varepsilon\to+\infty}i\varepsilon(i\varepsilon\Im w\\!-\\!(Y-\Re
w))^{-1}=(\Im w)^{-1}$

in the weak sense. However, this tells us something more: it guarantees that
$\mathbb{C}^{+}\ni z\mapsto\varphi\left((\Re w\\!+\\!z\Im
w\\!-\\!Y)^{-1}\right)$ is the Cauchy transform of a positive Borel measure on
$\mathbb{R}$ of total mass $\varphi\left((\Im w)^{-1}\right)\leq\|(\Im
w)^{-1}\|$, independently of the state $\varphi$. For given state $\psi$ on
$M_{2}(\mathcal{B})$, recall that $\mathbb{E}$ is weakly continuous, we then
apply the above to $\varphi=\psi\circ\mathbb{E}$ to obtain the weak
convergence

$\lim_{\varepsilon\to+\infty}i\varepsilon\mathbb{E}\left[\left(i\varepsilon\Im
w-(Y-\Re w)\right)^{-1}\right]=\left(\Im w\right)^{-1}$

bounded by $\|\left(\Im w\right)^{-1}\|$. Since $\mathcal{B}$ is finite-
dimensional, the weak and norm topologies coincide. Thus, one has

$\lim_{\varepsilon\to+\infty}i\varepsilon\mathbb{E}\left[\left(i\varepsilon\Im
w-(Y-\Re w)\right)^{-1}\right]=\left(\Im w\right)^{-1}\text{ in norm.}$

As a consequence,
$\lim_{\varepsilon\to+\infty}i\varepsilon\mathbb{E}\left[\left(i\varepsilon\Im
w-(Y-\Re w)\right)^{-1}\right]^{-1}=\Im w$, again in norm. Thus, for any
$w\in\mathbb{H}^{+}(M_{2}(\mathcal{B}))$, there is an $\varepsilon_{w}>0$ such
that $\|h^{\prime}_{Y}(\Re w+i\varepsilon\Im w)\|<1$ for all
$\varepsilon\geq\varepsilon_{w}$. Using the formula of
$D\boldsymbol{\Phi}(w_{1},w_{2})$ and elementary arithmetic operations, it
follows that there exists an open set of elements
$(w_{1},w_{2})\in\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))$
on which $D\boldsymbol{\Phi}(w_{1},w_{2})$ is invertible as a linear operator
from $M_{2}(\mathcal{B})\times M_{2}(\mathcal{B})$ to itself. The implicit
function theorem guarantees the existence of a right inverse in a neighborhood
of each point in this set. We argue next that these inverses extend to a
unique $\boldsymbol{\Omega}$ defined on all of
$\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))$.

We let
$\boldsymbol{\Phi}_{N}(w_{1},w_{2})=(w_{1}-h_{Y_{N}}(w_{2}),w_{2}-h_{X_{N}}(w_{1}))$.
Direct computation shows that $h_{Y_{N}}(w_{2})\to h_{Y}(w_{2})$ pointwise in
the weak topology, and hence, since $\mathcal{B}$ is finite dimensional,
uniformly on compact subsets of $\mathbb{H}^{+}(M_{2}(\mathcal{B}))$ in the
norm topology of $M_{2}(\mathcal{B})$. Thus,
$\boldsymbol{\Phi}_{N}\to\boldsymbol{\Phi}$ in the same topology, but this
time on
$\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))$.
As $X_{N},Y_{N}$ are bounded, there exist functions

$\boldsymbol{\Omega}_{N}\colon\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))\allowbreak\to\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))$

such that $\boldsymbol{\Phi}_{N}\circ\boldsymbol{\Omega}_{N}={\rm
id}_{\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))}$.
We let $N\to\infty$, so that $\boldsymbol{\Phi}_{N}\to\boldsymbol{\Phi},$ with
$\boldsymbol{\Phi}(w_{1},w_{2})=(w_{1}-h_{Y}(w_{2}),w_{2}-h_{X}(w_{1}))$ as
above. Clearly $\\{\boldsymbol{\Omega}_{N}\\}_{N\in\mathbb{N}}$ is a normal
family, so it has a convergent subsequence. By the identity principle and the
existence of a local inverse for $\boldsymbol{\Phi}$, any limit point must
still satisfy $\boldsymbol{\Phi}\circ\boldsymbol{\Omega}={\rm
id}_{\mathbb{H}^{+}(M_{2}(\mathcal{B}))\times\mathbb{H}^{+}(M_{2}(\mathcal{B}))}$.
This completes the argument.

We have established that analytic subordination functions exist for any
$\mathcal{B}$-free $x,y\in\tilde{\mathcal{A}}$ if dim$(\mathcal{B})<\infty$.

Case two: assume that $y$ is bounded and $x$ is unbounded, but such that the
Cauchy transform $\mathbb{E}\left[(b-X)^{-1}\right]$ takes values in the
operator lower half-plane whenever $b\in\mathbb{H}^{+}(\mathcal{B})$; that
immediately implies $h_{X}$ is a well-defined analytic map, with values in
$\mathbb{H}^{+}(\mathcal{B})$. Observe also that it is enough to show that
$\Im\mathbb{E}\left[(b-X)^{-1}\right]<0$ for a single
$b\in\mathbb{H}^{+}(\mathcal{B})$ in order to conclude that the inequality is
satisfied for all $b\in\mathbb{H}^{+}(\mathcal{B})$. Pick
$b_{1},b_{2}\in\mathbb{H}^{+}(\mathcal{B})$ (since there are no supplementary
conceptual difficulties involved in proving the full analogue of the result
from Case one, we choose to prove it, even though we will only apply it with
$b_{1}=b_{2}=b$). Define

$f_{b_{1}b_{2}}(w)=b_{1}+h_{Y}(b_{2}+h_{X}(w)),\quad
w\in\mathbb{H}^{+}(\mathcal{B}).$

Since $\Im\mathbb{E}\left[(w-X)^{-1}\right]<0$ for all $w$ in the operator
upper half-plane, it follows that $h_{X}$ is an analytic map defined on
$\mathbb{H}^{+}(\mathcal{B})$ and taking values in its closure. Thus,
$b_{2}+h_{X}(\mathbb{H}^{+}(\mathcal{B}))\subsetneq\mathbb{H}^{+}(\mathcal{B})+{b_{2}}/{2}$.
By [4, Lemma 2.12] it follows that
$f_{b_{1},b_{2}}(\mathbb{H}^{+}(\mathcal{B}))$ is a bounded set which is
bounded away in norm from the complement of $\mathbb{H}^{+}(\mathcal{B})$.
Specifically, the lemma shows that if $\varepsilon\in(0,+\infty)$ is fixed,
then for all $w\in\mathbb{H}^{+}(\mathcal{B})+i\varepsilon 1$, one has

$h_{Y}(\mathbb{H}^{+}(\mathcal{B})+i\varepsilon
1)\subset\\{v\in\mathbb{H}^{+}(\mathcal{B})\colon\|v\|\leq
4\|Y\|(1+2\varepsilon^{-1}\|Y\|)\\}.$

Thus, if we choose $\varepsilon>0$ such that $\Im b_{j}>\varepsilon 1$,
$j=1,2$, then

$\displaystyle f_{b_{1},b_{2}}(\mathbb{H}^{+}(\mathcal{B}))$
$\displaystyle\subset\\!b_{1}+h_{Y}(\mathbb{H}^{+}(\mathcal{B})+i\varepsilon
1)$ $\displaystyle\subset\\!\\{v\in\mathbb{H}^{+}(\mathcal{B})+i\varepsilon
1\colon\|v\|\leq 4\|Y\|(1+2\varepsilon^{-1}\|Y\|)\\}.$

Thus, one may apply the Earle-Hamilton Theorem (See [4, Section 2.1]) to the
function $f_{b_{1},b_{2}}$ defined on the set
$\\{v\in\mathbb{H}^{+}(\mathcal{B})+i\frac{\varepsilon}{2}1\colon\|v\|\leq
4\|Y\|(1+2\varepsilon^{-1}\|Y\|)+2\|b_{1}\|+2\|b_{2}\|\\}$ in order to
conclude that $\\{f_{b_{1},b_{2}}^{\circ n}\\}_{n\in\mathbb{N}}$ converges in
norm to a fixed point $\omega_{1}(b_{1},b_{2})$ which depends analytically on
$b_{1}$ and $b_{2}$. One simply defines
$\omega_{2}(b_{1},b_{2})=b_{2}+h_{X}(\omega_{1}(b_{1},b_{2}))$. Choosing
$b_{1}=b_{2}=b$ proves our claim. ∎

######  Remark 3.2.

The results in Theorem 3.1 have a fully matricial extension following [30,
31]. For any $n\in\mathbb{N}$, we define the map
$G_{X,n}(b)=(\mathbb{E}\otimes Id_{\mathbb{C}^{n\times n}})[(b-X\otimes
I_{n})^{-1}]$ for $b\in\mathcal{B}\otimes M_{n}(\mathbb{C})$ such that
$\Im(b)>\varepsilon$ for some $\varepsilon>0$. The map is a noncommutative
(nc) function and a remarkable observation due to Voiculescu is that the
family $\\{G_{X,n}\\}$ can retrieve the distribution of $X$. In this
framework, by applying the same proof to Theorem 3.1, we deduce that there are
nc functions $\Omega_{j,n}(j=1,2)$ such that

$G_{X+Y,n}(b)=G_{X,n}(\Omega_{1,n}(b))=G_{Y,n}(\Omega_{2,n}(b)).$

The interested reader is referred to [4, 30, 31] for some details.

##  4\. The sum with a triangular elliptic operator

###  4.1. A review on triangular elliptic operator

By enlarging the operator-valued $W^{*}$-probability space
$(\mathcal{A},\mathbb{E},\mathcal{B})$ if necessary, we assume that
$\mathcal{M},\mathcal{N}$ are unital von Neumann subalgebras of $\mathcal{A}$
such that:

  * •

$\mathcal{M},\mathcal{N}$ are free with amalgamation over $\mathcal{B}$ with
respect to $\mathbb{E}$;

  * •

the tracial state $\phi$ on $\mathcal{A}$ satisfies
$\phi(x)=\phi(\mathbb{E}(x))$ for any $x\in\mathcal{A}$;

  * •

$\mathbb{E}(x)=\phi(x)$ for any $x\in\mathcal{N}$.

The space we work on is the special case when $\mathcal{B}=L^{\infty}[0,1]$.
The tracial state $\phi:\mathcal{A}\rightarrow\mathbb{C}$ satisfy
$\phi(y)=\phi({\mathbb{E}}(y))$ for $y\in\mathcal{A}$, and

$\phi(f)=\int_{0}^{1}f(s)ds$

for $f\in\mathcal{B}$. The existence of such subalgebra $\mathcal{N}$ can be
verified as follows: let $\mathcal{B}\subset\mathcal{M}$ and let
${\mathcal{N}}$ be a unital subalgebra that is $*$-free from $\mathcal{B}$
with respect to the tracial state $\phi$ and let ${\mathcal{N}}_{1}$ be the
unital subalgebra generated by $\mathcal{N}\cup\mathcal{B}$. We then choose
$\mathcal{A}=\mathcal{M}*_{\mathcal{B}}{\mathcal{N}}_{1}$ and identify
$\mathcal{N}$ as some subalgebra in ${\mathcal{N}}_{1}\subset\mathcal{A}$. We
note that $\mathcal{N}$ is $*$-free from $\mathcal{B}$ in
$(\mathcal{A},\phi)$.

Let us review some basic properties of DT-operators introduced by
Dykema–Haagerup [13, 14]. Assume that
$(X_{i})_{i=1}^{\infty}\subset\mathcal{M}$ is a standard semicircular family
of random variables that are $*$-free from $\mathcal{B}$ in the
$W^{*}$-probability space $(\mathcal{A},\phi)$. Then for $i\in\mathbb{N}$, one
can construct a quasi-nilpotent DT operator $T_{i}$ as the norm limit of

$T^{(i)}_{n}=\sum_{j=1}^{n}1_{\left[\frac{j-1}{n},\frac{j}{n}\right]}X_{i}1_{\left[\frac{j}{n},1\right]}$

as $n\rightarrow\infty$, where $1_{[a,b]}$ is viewed as an element in
$\mathcal{B}$. It follows that $T_{i}\in W^{*}(\mathcal{B}\cup\\{X_{i}\\})$
for all $i\in\mathbb{N}$ and $(T_{i})_{i=1}^{\infty}$ are free with
amalgamation over $\mathcal{B}$ in $(\mathcal{A},\mathbb{E},\mathcal{B})$. It
is shown that $(T_{i},T_{i}^{*})_{i=1}^{\infty}$ is a centered
$\mathcal{B}$-Gaussian family in [14, Appendix A].

Recall that $\alpha,\beta>0$, and $\gamma\in\mathbb{C}$ such that
$|\gamma|\leq\sqrt{\alpha\beta}$. The _triangular elliptic element_
$y=g_{{}_{\alpha\beta,\gamma}}\in\mathcal{M}$ is an operator, introduced in
[5, Section 6], whose only nonzero free cumulants are given by

(4.1) $\displaystyle\kappa(y,fy)(t)$ $\displaystyle=\gamma\int_{0}^{1}f(s)ds,$
$\displaystyle\kappa(y^{*},fy^{*})(t)$
$\displaystyle=\bar{\gamma}\int_{0}^{1}f(s)ds,$
$\displaystyle\kappa(y,fy^{*})(t)$
$\displaystyle=\alpha\int_{t}^{1}f(s)ds+\beta\int_{0}^{t}f(s)ds,$
$\displaystyle\kappa(y^{*},fy)(t)$
$\displaystyle=\alpha\int_{0}^{t}f(s)ds+\beta\int_{t}^{1}f(s)ds$

for every $f\in\mathcal{B}$. We say that such triangular elliptic operator $y$
has parameter $(\alpha,\beta,\gamma)$. One can construct the triangular
elliptic operator ($\mathcal{B}$-circular element as studied by Dykema [12])
as follows. Let $T$ be a quasi-nilpotent DT operator. By [14, Appendix A], the
distribution of the pair $T,T^{*}$ is a $\mathcal{B}$-Gaussian distribution
determined by the free cumulants given by

$\kappa(T,fT^{*})(t)=\int_{t}^{1}f(s)ds,\qquad\kappa(T^{*},fT)(t)=\int_{0}^{t}f(s)ds,$

and $\kappa(T,fT)=\kappa(T^{*},fT^{*})=0$. Write $\gamma=e^{2\theta
i}|\gamma|$. Then the triangular elliptic operator $y=g_{\alpha\beta,\gamma}$
has the same $*$-moments as $e^{i\theta}(\sqrt{\alpha}T+\sqrt{\beta}T^{*})$.
In particular, a quasi-nilpotent DT operator is a triangular elliptic element
with parameter $\alpha=1,\beta=\gamma=0$. When $\alpha=\beta=t$, the operator
$g_{\alpha\alpha,0}$ is just the Voiculescu’s circular operator $c_{t}$ of
variance $t$, and the operator $g_{\alpha\alpha,\gamma}$ is the so-called
_twisted elliptic operator_ $g_{t,\gamma}$ as studied in [33, Section 2.4].

Let $x_{0}\in\log^{+}(\mathcal{N})$ that is $*$-freely independent from
$g_{\alpha\beta,\gamma}$ with amalgamation over $\mathcal{B}$. The main object
of this paper is to calculate the Brown measure of
$x_{0}+g_{\alpha\beta,\gamma}$. In the $M_{2}(\mathcal{B})$-valued
$C^{*}$-probability
space$(M_{2}(\mathcal{A}),M_{2}(\mathbb{E}),M_{2}(\mathcal{B}))$, we denote

$X=\begin{bmatrix}0&x_{0}\\\
x_{0}^{*}&0\end{bmatrix},\;\;Y=\begin{bmatrix}0&y\\\ y^{*}&0\end{bmatrix}.$

Then $X,Y$ are free over $M_{2}(\mathcal{B})$. This can be deduced from the
connection between $\mathcal{B}$-valued free cumulants and
$M_{2}(\mathcal{B})$-valued free cumulants similar to [24, Section 9.3,
Proposition 13].

The following result adapts the identification used in [1, Lemma 5.6], where
the case $x_{0}=0$ was considered.

######  Proposition 4.1.

Given $\alpha\geq\beta=t>0$, $|\gamma|\leq t$, and a bounded operator
$x_{0}\in\mathcal{N}$, let $g_{t,\gamma}$ be a twisted elliptic operator
$*$-free from $\mathcal{M}\cup\mathcal{N}$ with respect to $\phi$. Then,

$x_{0}+g_{\alpha\beta,\gamma}\mathop{\sim}\limits^{\mathrm{*-dist.}}x_{0}+g_{(\alpha-\beta)0,0}+g_{t,\gamma},$

where $g_{(\alpha-\beta)0,0}$ denotes a triangular elliptic operator in
$\mathcal{M}$ with parameters $(\alpha-\beta,0,0)$.

In particular, when $\alpha=\beta$, the operator $g_{\alpha\beta,\gamma}$ is
$*$-free from $\mathcal{N}$ in the scalar-valued $W^{*}$-probability space
$(\mathcal{A},\phi)$.

###### Proof.

It is clear that $g_{\alpha\beta,\gamma}$ has the same$*$-distribution as
$g_{t,\gamma}+g_{(\alpha-\beta)0,0}$, where $g_{t,\gamma}$ and
$g_{(\alpha-\beta)0,0}$ are free with amalgamation over $\mathcal{B}$. We
express $g_{t,\gamma}$ as a linear combination $e^{2\theta
i}\sqrt{t}(T_{i}+T_{i}^{*})$ for some quasi-nilpotent DT operators $T_{i}$.
The operator-valued free cumulants of $g_{t,\gamma}$ in
$(\mathcal{A},\mathbb{E},\mathcal{B})$ are complex numbers. By applying [25,
Theorem 3.6], it follows that $g_{t,\gamma}$ is $\mathbb{C}$-Gaussian and is
$*$-free from $\mathcal{B}$ with respect to $\phi$. Since $T_{j}\in
W^{*}(\mathcal{B}\cup\\{X_{j}\\})$ and
$(X_{i})_{i=1}^{\infty}\subset\mathcal{M}$ is a standard semicircular family
of random variables that are $*$-free from $\mathcal{B}$ in the
$W^{*}$-probability space $(\mathcal{A},\phi)$, it follows that
$g_{t,\gamma}=e^{2\theta i}\sqrt{t}(T_{i}+T_{i}^{*})$ is $*$-free with respect
to $\phi$ from other quasi-nilpotent DT operator $T_{j}$ in
$(\mathcal{A},\phi)$ for $j\neq i$. Hence, $g_{t,\gamma}$ is also $*$-free
from $x_{0}+g_{(\alpha-\beta)0,0}$ with respect to $\phi$. ∎

Hence, if $|\gamma|\leq\min\\{\alpha,\beta\\}$, the Brown measure of
$x_{0}+g_{\alpha\beta,\gamma}$ is the same as the Brown measure of
$y_{0}+g_{t,\gamma}$ for $y_{0}=x_{0}+g_{|\alpha-\beta|0,0}$ where
$g_{|\alpha-\beta|0,0}$ denotes a triangular elliptic operator with parameter
$(|\alpha-\beta|,0,0)$ and has the same distribution as a scalar multiple of a
quasi-nilpotent DT operator. By [33], we can express the Brown measure formula
in terms of certain subordination functions determined by $y_{0}$ if
$|\gamma|\leq\min\\{\alpha,\beta\\}$ and $x_{0}$ is bounded. We will extend
the method in [33] such that we are able to express the Brown measure formula
for $x_{0}+g_{\alpha\beta,\gamma}$ in terms of some functions determined
directly by $x_{0}$ for any $\gamma$ such that
$|\gamma|\leq\sqrt{\alpha\beta}$. In addition, we show that the main results
hold for any $x_{0}\in\log^{+}(\mathcal{N})$.

###  4.2. The operator-valued subordination function

For parameter ${\bf t}=(\alpha,\beta,\gamma)$ where $\alpha,\beta>0$ and
$\gamma\in\mathbb{C}$ such that $|\gamma|\leq\sqrt{\alpha\beta}$, we denote
$y=g_{{}_{\alpha\beta,\gamma}}$, and we take the convention that $y=0$ if
$\alpha=\beta=\gamma=0$. For $\lambda\in\mathbb{C}$, we have

(4.2)
$S(x_{0}+y,\lambda,\varepsilon)=\log\triangle\left[(x_{0}+y-\lambda)^{*}(x_{0}+y-\lambda)+\varepsilon^{2}\right],\;\;\varepsilon>0.$

We introduce the following notations

(4.3) $\begin{split}p_{\lambda}^{{\bf t}}(\varepsilon)&=\phi\left\\{(\lambda-
x_{0}-y)^{*}\left[(\lambda-x_{0}-y)(\lambda-
x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\},\\\ p_{\bar{\lambda}}^{{\bf
t}}(\varepsilon)&=\phi\left\\{(\lambda-x_{0}-y)\left[(\lambda-
x_{0}-y)^{*}(\lambda-x_{0}-y)+\varepsilon^{2}\right]^{-1}\right\\},\\\
q_{\varepsilon}^{{\bf t}}(\lambda)&=\varepsilon\phi\left\\{\left[(\lambda-
x_{0}-y)^{*}(\lambda-
x_{0}-y)+\varepsilon^{2}\right]^{-1}\right\\}.\end{split}$

and

(4.4) $\begin{split}P_{\lambda}^{{\bf
t}}(\varepsilon)&={\mathbb{E}}\left\\{(\lambda-x_{0}-y)^{*}\left[(\lambda-
x_{0}-y)(\lambda-x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\},\\\
P_{\bar{\lambda}}^{{\bf t}}(\varepsilon)&={\mathbb{E}}\left\\{(\lambda-
x_{0}-y)\left[(\lambda-x_{0}-y)^{*}(\lambda-
x_{0}-y)+\varepsilon^{2}\right]^{-1}\right\\},\\\ Q_{\varepsilon}^{{\bf
t}}(\lambda)&=\varepsilon{\mathbb{E}}\left\\{\left[(\lambda-x_{0}-y)(\lambda-
x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\},\\\
\widetilde{Q}_{\varepsilon}^{{\bf
t}}(\lambda)&=\varepsilon{\mathbb{E}}\left\\{\left[(\lambda-
x_{0}-y)^{*}(\lambda-
x_{0}-y)+\varepsilon^{2}\right]^{-1}\right\\}.\end{split}$

Note that $p_{\lambda}^{{\bf t}}(\varepsilon)$, $p_{\bar{\lambda}}^{{\bf
t}}(\varepsilon)$ and $q_{\varepsilon}^{{\bf t}}(\lambda)$ are derivatives of
$S(x_{0}+y,\lambda,\varepsilon)$ with respect to $\lambda,\overline{\lambda}$
and $\varepsilon$ (up to some constant). They are also related to entries of
the Cauchy transform (2.8) of the Hermitian reduction for $x_{0}+y$. It is
clear that, for $\lambda\in\mathbb{C}$ and $\varepsilon>0$, we have

(4.5) $\phi(P_{\lambda}^{{\bf t}}(\varepsilon))=p_{\lambda}^{{\bf
t}}(\varepsilon),\;\phi(P_{\bar{\lambda}}^{{\bf
t}}(\varepsilon))=p_{\bar{\lambda}}^{{\bf t}}(\varepsilon),$

and, by the tracial property, we have

(4.6) $\phi(Q_{\varepsilon}^{{\bf
t}}(\lambda))=\phi(\widetilde{Q}_{\varepsilon}^{{\bf
t}}(\lambda))=q_{\varepsilon}^{{\bf t}}(\lambda).$

Therefore, we have

$G_{X+Y}\left(\begin{bmatrix}i\varepsilon&\lambda\\\
\overline{\lambda}&i\varepsilon\end{bmatrix}\right)=M_{2}(\mathbb{E})\left(\bigg{(}X+Y-\begin{bmatrix}i\varepsilon&\lambda\\\
\overline{\lambda}&i\varepsilon\end{bmatrix}\bigg{)}^{-1}\right)=\begin{bmatrix}-iQ_{\varepsilon}^{\bf
t}(\lambda)&P_{\bar{\lambda}}^{{\bf t}}(\varepsilon)\\\ P_{\lambda}^{{\bf
t}}(\varepsilon)&-i\widetilde{Q}_{\varepsilon}^{{\bf
t}}(\lambda)\end{bmatrix}.$

For any $\varepsilon_{1},\varepsilon_{2}\in\mathcal{B}$, set

(4.7)
$\begin{split}{g}_{11}(\lambda,\varepsilon_{1},\varepsilon_{2})&=-i\varepsilon_{2}{\mathbb{E}}\left\\{\left[(\lambda-
x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\},\\\
{g}_{12}(\lambda,\varepsilon_{1},\varepsilon_{2})&={\mathbb{E}}\left\\{(\lambda-
x_{0})\left[(\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\},\\\
{g}_{21}(\lambda,\varepsilon_{1},\varepsilon_{2})&={\mathbb{E}}\left\\{(\lambda-
x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\},\\\
{g}_{22}(\lambda,\varepsilon_{1},\varepsilon_{2})&=-i\varepsilon_{1}{\mathbb{E}}\left\\{\left[(\lambda-
x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\}.\\\ \end{split}$

For convenience, we also denote

$g_{ij}=g_{ij}(\lambda,\varepsilon_{1},\varepsilon_{2}),\qquad
i,j\in\\{1,2\\}.$

Then, we have

$G_{X}\left(\begin{bmatrix}i\varepsilon_{1}&\lambda\\\
\overline{\lambda}&i\varepsilon_{2}\end{bmatrix}\right)=\begin{bmatrix}{g}_{11}&{g}_{12}\\\
{g}_{21}&{g}_{22}\end{bmatrix}.$

######  Proposition 4.2.

The operator $Y$ is an operator-valued semicircular element in
$(M_{2}(\mathcal{A}),M_{2}(\mathbb{E}),M_{2}(\mathcal{B}))$. For any
$b=\begin{bmatrix}a_{11}&a_{12}\\\ a_{21}&a_{22}\end{bmatrix}\in
M_{2}(\mathcal{B})$, the $R$-transform of $Y$ is given by

$R_{Y}(b)=M_{2}(\mathbb{E})(YbY)=\begin{bmatrix}\kappa(y,a_{22}y^{*})&\kappa(y,a_{21}y)\\\
\kappa(y^{*},a_{12}y^{*})&\kappa(y^{*},a_{11}y)\end{bmatrix}.$

###### Proof.

It follows from a result connecting matrix-valued free cumulants with free
cumulants as in [24, Section 9.3, Proposition 13], but we have to modify it
appropriately by replacing scalar-valued free cumulants by operator-valued
free cumulants in $(\mathcal{A},\mathbb{E},\mathcal{B})$. ∎

######  Lemma 4.3.

Let $y=g_{{}_{\alpha\beta,\gamma}}\in{\mathcal{M}}$ and
$x_{0}\in\widetilde{\mathcal{N}}$ be a random variable that is $*$-free from
$y$ with amalgamation over $\mathcal{B}$. For any $\varepsilon>0$ and
$z\in\mathbb{C}$, set

(4.8) $\lambda=z-\kappa(y,P_{z}^{\bf t}(\varepsilon)y),$

where $P_{z}^{\bf t}(\varepsilon)$ is defined in (4.4). For
$\Theta(z,\varepsilon)=\begin{bmatrix}i\varepsilon&z\\\
\bar{z}&i\varepsilon\end{bmatrix}$ we have subordination relation

$G_{X+Y}(\Theta(z,\varepsilon))=G_{X}(\Omega_{1}(\Theta(z,\varepsilon))),$

where

(4.9)
$\Omega_{1}(\Theta(z,\varepsilon))=\begin{bmatrix}i\varepsilon_{1}&\lambda\\\
\bar{\lambda}&i\varepsilon_{2}\end{bmatrix},$

and

(4.10) $\varepsilon_{1}=\varepsilon+\kappa(y,\widetilde{Q}_{\varepsilon}^{\bf
t}(z)y^{*}),\quad\varepsilon_{2}=\varepsilon+\kappa(y^{*},Q_{\varepsilon}^{{\bf
t}}(z)y).$

In other words, the subordination relation $G_{X+Y}(b)=G_{X}(\Omega_{1}(b))$
for $b=\Theta(z,\varepsilon)$ can be written as

(4.11) $\mathbb{E}\left(\begin{bmatrix}i\varepsilon&z-(x_{0}+y)\\\
\overline{z}-(x_{0}+y)^{*}&i\varepsilon\end{bmatrix}^{-1}\right)=\mathbb{E}\left(\begin{bmatrix}i\varepsilon_{1}&\lambda-
x_{0}\\\
\overline{\lambda}-x_{0}^{*}&i\varepsilon_{2}\end{bmatrix}^{-1}\right).$

Moreover, the above subordination relation is equivalent to

(4.12) $\begin{split}{g}_{11}=-iQ_{\varepsilon}^{\bf
t}(z),\;{g}_{22}=-i\widetilde{Q}_{\varepsilon}^{\bf
t}(z),\;{g}_{12}=P_{\bar{z}}^{\bf t}(\varepsilon),\;{g}_{21}=P_{z}^{\bf
t}(\varepsilon),\end{split}$

where ${\bf t}=(\alpha,\beta,\gamma)$ and $g_{ij}$ was defined in (4.7).

###### Proof.

We first verify that Theorem 3.1 applies to our context. Recall that we have
$(\mathcal{B},\phi)=(L^{\infty}[0,1],ds)$ and
$\mathbb{E}|_{W^{*}(x_{0})}=\phi$, where ${W^{*}(x_{0})}$ denotes the von
Neumann algebra generated by $x_{0}$. It follows that

$\displaystyle-\mathbb{E}[(i-X)^{-1}]$
$\displaystyle=-\mathbb{E}\left(\begin{bmatrix}i&-x_{0}\\\
-x_{0}^{*}&i\end{bmatrix}^{-1}\right)$
$\displaystyle=\begin{bmatrix}i\phi((1+x_{0}x_{0}^{*})^{-1})&\phi((1+x_{0}x_{0}^{*})^{-1}x_{0})\\\
\phi(x_{0}^{*}(1+x_{0}x_{0}^{*})^{-1})&i\phi((1+x_{0}^{*}x_{0})^{-1})\end{bmatrix}.$

The imaginary part of the above matrix is

$-\Im\mathbb{E}[(i-X)^{-1}]=\begin{bmatrix}\phi((1+x_{0}x_{0}^{*})^{-1})&0\\\
0&\phi((1+x_{0}^{*}x_{0})^{-1})\end{bmatrix},$

which is strictly positive by the faithfulness of the trace. Hence, the
condition for Case two in Theorem 3.1 is satisfied.

For $b=\Theta(z,\varepsilon)$, by (3.1) in Theorem 3.1, the subordination
functions satisfy

(4.13) $\Omega_{1}(b)+\Omega_{2}(b)=b+(G_{X+Y}(b))^{-1}.$

The subordination relation $G_{Y}(\Omega_{2}(b))=G_{X+Y}(b)$ yields

$\Omega_{2}(b)=R_{Y}\big{(}G_{X+Y}(b)\big{)}+(G_{X+Y}(b))^{-1},$

provided that $\lVert b^{-1}\rVert$ is small enough. It follows that

(4.14) $\Omega_{1}(b)=b-R_{Y}(G_{X+Y}(b)),$

and this is true for any $b\in\mathbb{H}^{+}(M_{2}(\mathbb{B}))$ because
$R_{Y}\circ G_{X+Y}$ is defined for any
$b\in\mathbb{H}^{+}(M_{2}(\mathbb{B}))$ thanks to Proposition 4.2.
Consequently, (4.14) holds for any $b=\Theta(z,\varepsilon)$ with
$\varepsilon>0$.

By definitions (4.4), the Cauchy transform can be expressed as

(4.15) $\begin{split}G_{X+Y}(b)&=M_{2}(\mathbb{E})\left[(b-X-Y)^{-1}\right]\\\
&=\begin{bmatrix}-iQ_{\varepsilon}^{\bf t}(z)&P_{\bar{z}}^{\bf
t}(\varepsilon)\\\ P_{z}^{\bf
t}(\varepsilon)&-i\widetilde{Q}_{\varepsilon}^{\bf
t}(z)\end{bmatrix}.\end{split}$

The formula for $R$-transform of $Y$ (Proposition 4.2) implies

$\begin{split}R_{Y}(G_{X+Y}(b))=\begin{bmatrix}-i\kappa(y,\widetilde{Q}_{\varepsilon}^{\bf
t}(z)y^{*})&\kappa(y,P_{z}^{\bf t}(\varepsilon)y)\\\
\kappa(y^{*},P_{\bar{z}}^{\bf
t}(\varepsilon)y^{*})&-i\kappa(y^{*},Q_{\varepsilon}^{{\bf
t}}(z)y)\end{bmatrix}.\end{split}$

Hence

(4.16) $\begin{split}\Omega_{1}(b)&=b-R_{Y}(G_{X+Y}(b))\\\
&=\begin{bmatrix}i\varepsilon+i\kappa(y,\widetilde{Q}_{\varepsilon}^{\bf
t}(z)y^{*})&z-\kappa(y,P_{z}^{\bf t}(\varepsilon)y)\\\
\bar{z}-\kappa(y^{*},P_{\bar{z}}^{\bf
t}(\varepsilon)y^{*})&i\varepsilon+i\kappa(y^{*},Q_{\varepsilon}^{{\bf
t}}(z)y)\end{bmatrix}\\\ &=\begin{bmatrix}i\varepsilon_{1}&\lambda\\\
\bar{\lambda}&i\varepsilon_{2}\end{bmatrix},\end{split}$

where $\lambda=z-\kappa(y,P_{z}^{\bf t}(\varepsilon)y)\in\mathbb{C}$ (note
that $\kappa(y,P_{z}^{\bf
t}(\varepsilon)y)=\overline{\kappa(y^{*},P_{\bar{z}}^{\bf
t}(\varepsilon)y^{*})}$). Therefore, we have

(4.17)
$\begin{split}G_{X}(\Omega_{1}(b))&=M_{2}(\mathbb{E})\left[(\Omega_{1}(b)-X)^{-1}\right]\\\
&=\begin{bmatrix}{g}_{11}&{g}_{12}\\\ {g}_{21}&{g}_{22}\end{bmatrix}\in
M_{2}(\mathcal{B}).\end{split}$

This establishes (4.9) and (4.11). By comparing (4.15) with (4.17), we have

$\begin{split}{g}_{11}=-iQ_{\varepsilon}^{\bf
t}(z),\;{g}_{22}=-i\widetilde{Q}_{\varepsilon}^{\bf
t}(z),\;{g}_{12}=P_{\bar{z}}^{\bf t}(\varepsilon),\;{g}_{21}=P_{z}^{\bf
t}(\varepsilon).\end{split}$

This finishes the proof. ∎

######  Lemma 4.4.

Fix $z\in\mathbb{C}$ and $\varepsilon>0$. Using notations in Lemma 4.3, let

$D={\mathbb{E}}\left[((\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2})^{-1}\right],$

and

$\widetilde{D}={\mathbb{E}}\left[((\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{1}\varepsilon_{2})^{-1}\right].$

If $\mathcal{N}$ is $*$-free from $\mathcal{B}$ in $(\mathcal{A},\phi)$, then
$\varepsilon_{1}\varepsilon_{2}$ and $D$ are constant functions in
$\mathcal{B}=L^{\infty}[0,1].$ Consequently,

(4.18) $\begin{split}D&=\widetilde{D}=\phi\left[((\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{1}\varepsilon_{2})^{-1}\right].\end{split}$

###### Proof.

We note that ${Q}_{\varepsilon}^{\bf t}(z),\widetilde{Q}_{\varepsilon}^{\bf
t}(z)$ are strictly positive functions in $\mathcal{B}=L^{\infty}[0,1]$ by the
definition (4.4). Hence, $\kappa(y,\widetilde{Q}_{\varepsilon}^{\bf
t}(z)y^{*})>0$ and $\kappa(y^{*},Q_{\varepsilon}^{{\bf t}}(z)y)>0$.
Consequently, by the defining identity (4.10)

$\varepsilon_{1}=\varepsilon+\kappa(y,\widetilde{Q}_{\varepsilon}^{\bf
t}(z)y^{*})>\varepsilon.$

Similarly, $\varepsilon_{2}>\varepsilon$. Since for any $x\in\mathcal{N}$, we
have

(4.19) $\mathbb{E}[p(x,x^{*})]=\phi[p(x,x^{*})]$

where $p$ is an arbitrary polynomial of two indeterminates. Given any
$x\in\mathcal{N}$, $b\in\mathcal{B}$ and $n\in\mathbb{N}$, since $\mathcal{N}$
is $*$-free from $\mathcal{B}$ in $(\mathcal{A},\phi)$, we hence have

$\mathbb{E}((x^{*}xb)^{n})=\mathbb{E}((xx^{*}b)^{n}),$

where we used the tracial property of $\phi$. If $b\in\mathcal{B}$ is
invertible and $\|b\|$ is large enough, we can write

$(xx^{*}+b)^{-1}=\sum_{n=0}^{\infty}(-1)^{n}b^{-1}(xx^{*}b^{-1})^{n}.$

It follows that, if $\|b\|$ is large enough, we have

(4.20)
$\mathbb{E}\left[(xx^{*}+b)^{-1}\right]=\mathbb{E}\left[(x^{*}x+b)^{-1}\right].$

The function $b\mapsto{\mathbb{E}}\left[(xx^{*}+b)^{-1}\right]$ is a
holomorphic function at $b$ that is strictly positive in the sense that
$b\geq\delta>0$ for some $\delta\in\mathbb{R}$ and
$b\in\mathbb{H}^{+}(\mathcal{B})$. Hence, the identify (4.20) holds for any
$b\geq\delta>0$ by the uniqueness of holomorphic functions.

For any $x\in\widetilde{\mathcal{N}},$ let $x=u|{x}||$ be the polar
decomposition of $x$. Consider the truncated operator
$x_{N}:=u|x|\cdot\chi_{[0,N]}(|{x}|).$ It is known that $x_{N}\in\mathcal{N}$
and $x_{N}$ converges to $x$ respect to the strong operator topology. We may
assume that $\mathcal{A}$ is a subalgebra of the set of all bounded operators
$B(\mathrm{H})$ acting on some Hilbert space $\mathrm{H}$. By the identity
$A^{-1}-B^{-1}=B^{-1}(B-A)A^{-1}$, we have

$\begin{split}&\left\|\left[(xx^{*}+b)^{-1}-(x_{N}x_{N}^{*}+b)^{-1}\right]h\right\|\\\
&=\left\|(x_{N}x_{N}^{*}+b)^{-1}(x_{N}x_{N}^{*}-xx^{*})(xx^{*}+b)^{-1}h\right\|\\\
&\leq\left\|(x_{N}x_{N}^{*}+b)^{-1}\right\|\cdot\left\|(x_{N}x_{N}^{*}-xx^{*})(xx^{*}+b)^{-1}h\right\|\end{split}$

for any vector $h\in\mathrm{H}$. Note that
$\left\|(x_{N}x_{N}^{*}+b)^{-1}\right\|\leq\|{b^{-1}}\|$ and $x_{N}x_{N}^{*}$
converges to $xx^{*}$ in strong operator topology. Hence,
$(x_{N}x_{N}^{*}+b)^{-1}$ converges to $(xx^{*}+b)^{-1}$ in the strong
operator topology. Similarly, we have $(x_{N}^{*}x_{N}+b)^{-1}$ converges to
$(x^{*}x+b)^{-1}.$ In particular, by letting $x=\lambda-x_{0}$ and combining
(4.20), we have

(4.21) ${\mathbb{E}}\left[((\lambda-x_{0})(\lambda-
x_{0})^{*}+b)^{-1}\right]={\mathbb{E}}\left[((\lambda-x_{0})^{*}(\lambda-
x_{0})+b)^{-1}\right].$

Since $\varepsilon_{1}\varepsilon_{2}\in\mathcal{B}$ and
$\varepsilon_{1}\varepsilon_{2}\geq\varepsilon^{2}$, this implies that (4.21)
holds for $b=\varepsilon_{1}\varepsilon_{2}$. Hence, $D=\widetilde{D}$.

We next show that $\varepsilon_{1}\varepsilon_{2}$ is a constant. By
differentiating respect to $t$, we have

$\varepsilon_{1}^{\prime}=(\varepsilon+k(y,\widetilde{Q}_{\varepsilon}^{\bf
t}(z)y^{*}))^{\prime}=-(\alpha-\beta)\widetilde{Q}_{\varepsilon}^{\bf t}(z)$

and

$\varepsilon_{2}^{\prime}=(\varepsilon+\kappa(y^{*},Q_{\varepsilon}^{{\bf
t}}(z)y))^{\prime}=(\alpha-\beta)Q_{\varepsilon}^{{\bf t}}(z).$

Lemma 4.3 implies that

$\widetilde{Q}_{\varepsilon}^{\bf
t}(z)=ig_{22}=\varepsilon_{1}\widetilde{D}=\varepsilon_{1}D$

and similarly ${Q}_{\varepsilon}^{\bf t}(z)=ig_{11}=\varepsilon_{2}D$. Hence

$(\varepsilon_{1}\varepsilon_{2})^{\prime}=(\alpha-\beta)\varepsilon_{1}\varepsilon_{2}(D-D)=0.$

Hence $\varepsilon_{1}\varepsilon_{2}$ is a constant function in
$L^{\infty}[0,1]$. Then (4.19) implies that $D$ is given by

$D={\mathbb{E}}\left[((\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2})^{-1}\right]=\phi\left[((\lambda-
x_{0})(\lambda-x_{0})^{*}+\varepsilon_{1}\varepsilon_{2})^{-1}\right].$

This finishes the proof. ∎

######  Lemma 4.5.

Let $g_{ij}$ be entries of the $2\times 2$ matrix-valued Cauchy transform in
(4.17) as defined in (4.7) and assume $\alpha\neq\beta$. Then ${g}_{12}$ and
${g}_{21}$ are constant functions given by

${g}_{12}=\phi\left\\{(\lambda-x_{0})\left[(\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\},$

and

${g}_{21}=\phi\left\\{(\lambda-x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\}.$

Moreover,

(4.22) $\begin{split}{g}_{11}(t)=\frac{i\varepsilon D(\alpha-\beta)}{\beta
e^{(\alpha-\beta)D}-\alpha}e^{(\alpha-\beta)Dt},\;\;{g}_{22}(t)=\frac{i\varepsilon
D(\beta-\alpha)}{\alpha
e^{(\beta-\alpha)D}-\beta}e^{(\beta-\alpha)Dt}.\end{split}$

###### Proof.

By Lemma 4.4, the product $\varepsilon_{1}\varepsilon_{2}$ is a constant.
Recall that $\mathbb{E}(x)=\phi(x)$ for any $x\in\mathcal{N}$. It follows that
$g_{12},g_{21}$ are constants given by the formulas above.

We next observe that

$g_{11}=-i\varepsilon_{2}D,\qquad g_{22}=-i\varepsilon_{1}D.$

Taking the derivative as in the proof of Lemma 4.4, we have

$g_{11}^{\prime}=-iD(\varepsilon_{2})^{\prime}=-iD(\alpha-\beta)(Q_{\varepsilon}^{\bf
t}(z))=(\alpha-\beta)Dg_{11}$

where we used $g_{11}=-iQ_{\varepsilon}^{\bf t}(z)$ from Lemma 4.3. Hence, by
solving the ODE, we see that $g_{11}(t)=Ce^{(\alpha-\beta)Dt}$ for some
constant $C$. We rewrite the definition of $\varepsilon_{2}$ using the
subordination relation in Lemma 4.3 as

$\varepsilon_{2}=\varepsilon+\kappa(y^{*},Q_{\varepsilon}^{{\bf
t}}(z)y)=\varepsilon+i\kappa(y^{*},g_{11}y).$

Hence, $g_{11}(t)=Ce^{(\alpha-\beta)Dt}$ and

$g_{11}(t)=-i\varepsilon_{2}D=-(i\varepsilon-\kappa(y^{*},g_{11}y))D.$

The cumulant formula (4.1) of $y$ reads

$\kappa(y^{*},g_{11}y)=\alpha\int_{0}^{t}g_{11}(s)ds+\beta\int_{t}^{1}g_{11}(s)ds,$

which determines the value $C$ by solving the integral equation. A direct
verification shows

$C=\frac{i\varepsilon D(\alpha-\beta)}{\beta e^{(\alpha-\beta)D}-\alpha}.$

Similarly, one can obtain the formula for $g_{22}$. ∎

######  Corollary 4.6.

For $\alpha\neq\beta$, the functions $\varepsilon_{1}$ and $\varepsilon_{2}$
in $\mathcal{B}=L^{\infty}[0,1]$ are given by

(4.23) $\varepsilon_{1}(t)=\frac{\varepsilon(\alpha-\beta)}{(\alpha-\beta
e^{(\alpha-\beta)D})}e^{(1-t)(\alpha-\beta)D},$

and

$\varepsilon_{2}(t)=\frac{\varepsilon(\alpha-\beta)}{(\alpha-\beta
e^{(\alpha-\beta)D})}e^{t(\alpha-\beta)D}$

Consequently,

(4.24)
$\varepsilon_{1}\varepsilon_{2}=\varepsilon^{2}\left(\frac{\alpha-\beta}{\alpha-\beta
e^{(\alpha-\beta)D}}\right)^{2}e^{(\alpha-\beta)D}$

###### Proof.

It follows directly from $g_{11}=-i\varepsilon_{2}D$,
$g_{22}=-i\varepsilon_{1}D$ and the formulas for $g_{11}$ and $g_{22}$ as in
(4.22). ∎

######  Remark 4.7.

For any $\varepsilon>0$, we have $D<\frac{\log\alpha-\log\beta}{\alpha-\beta}$
where $D$ is defined in Lemma 4.4. Indeed, as we shown in the proof of Lemma
4.4, we have $\varepsilon_{1}(t)>\varepsilon>0$, hence by the formula (4.23)
for $\varepsilon_{1}$, we see that $\frac{\alpha-\beta}{\alpha-\beta
e^{(\alpha-\beta)D}}>0$ which yields that
$D<\frac{\log\alpha-\log\beta}{\alpha-\beta}$ whenever $\alpha\neq\beta$.

It follows from Corollary 4.6 that

(4.25)
$\phi(\varepsilon_{1})=\phi(\varepsilon_{2})=\frac{\varepsilon(e^{(\alpha-\beta)D}-1)}{D(\alpha-\beta
e^{(\alpha-\beta)D})}.$

Then, since $D$ is constant and $g_{11}=-i\varepsilon_{2}D$, we have

(4.26)
$i\phi(g_{11})=\phi(\varepsilon_{2})D=\phi(\varepsilon_{2})\phi\left\\{\left[(\lambda-
x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\}.$

Since $\frac{\alpha-\beta}{\alpha-\beta e^{(\alpha-\beta)D}}>0$, we then set

(4.27)
$\varepsilon_{0}=\sqrt{\varepsilon_{1}\varepsilon_{2}}=\frac{\varepsilon(\alpha-\beta)}{\alpha-\beta
e^{(\alpha-\beta)D}}e^{(\alpha-\beta)D/2}.$

We can then rewrite $\phi(\varepsilon_{i})$ as

$\phi(\varepsilon_{1})=\phi(\varepsilon_{2})=\varepsilon_{0}\frac{(e^{(\alpha-\beta)D}-1)}{D(\alpha-\beta)}e^{-(\alpha-\beta)D/2}.$

Our approach is to fix $\varepsilon>0$ and $\lambda\in\mathbb{C}$, and we will
show that $z,\varepsilon_{1},\varepsilon_{2}$ can be regarded as functions of
$\lambda$ and $\varepsilon$.

In light of (4.24), the constant $D$ satisfies the following implicit formula

(4.28) $D=\phi\left\\{\left[(\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon^{2}\left(\frac{\alpha-\beta}{\alpha-\beta
e^{(\alpha-\beta)D}}\right)^{2}e^{(\alpha-\beta)D}\right]^{-1}\right\\}.$

We shall show that (4.28) determines a unique solution for
$D<\frac{\log\alpha-\log\beta}{\alpha-\beta}$. Put $\sigma=(\alpha-\beta)D$,
and denote

$F(\sigma,\lambda,\varepsilon)=\frac{1}{\sigma}\cdot\phi\left\\{\left[(\lambda-
x_{0})^{*}(\lambda-
x_{0})+\varepsilon^{2}\frac{(\alpha-\beta)^{2}}{(\alpha-\beta
e^{\sigma})^{2}e^{-\sigma}}\right]^{-1}\right\\}.$

Then (4.28) is rewritten as

(4.29) $F(\sigma,\lambda,\varepsilon)=\frac{1}{\alpha-\beta}.$

######  Lemma 4.8.

Assume that $\alpha>\beta>0$, let $0<\sigma<\log(\alpha/\beta)$. For any fixed
$\varepsilon>0$ and $\lambda\in\mathbb{C}$, the function $\sigma\mapsto
F(\sigma,\lambda,\varepsilon)$ is a decreasing function on $(0,\infty)$, and

$\lim_{\sigma\rightarrow
0^{+}}F(\sigma,\lambda,\varepsilon)=\infty,\qquad\lim_{\sigma\rightarrow\log(\alpha/\beta)}F(\sigma,\lambda,\varepsilon)=0.$

###### Proof.

It is easy to check that $\sigma\mapsto(\alpha-\beta
e^{\sigma})^{2}e^{-\sigma}$ is increasing if $\sigma>\log(\alpha/\beta)$ and
is decreasing if $\sigma<\log(\alpha/\beta)$. Hence, $\frac{\partial
F(\sigma,\lambda,\varepsilon)}{\partial\sigma}<0$ when
$\sigma<\log(\alpha/\beta))$. ∎

######  Proposition 4.9.

Fix $\lambda\in\mathbb{C}$ and $\varepsilon>0$, the equation (4.28) determines
$D$ uniquely, and the function $D=D(\lambda,\varepsilon)$ is a
$C^{\infty}$-function of $(\lambda,\varepsilon)$ over
$\mathbb{C}\times(0,\infty)$ determined by (4.28). Consequently, the function
$\varepsilon_{0}$ defined by (4.27) is a $C^{\infty}$-function of
$(\lambda,\varepsilon)$ over $\mathbb{C}\times(0,\infty)$, which we denoted by
$\varepsilon_{0}=\varepsilon_{0}(\lambda,\varepsilon)$.

###### Proof.

Without losing generality, we may assume that $\alpha>\beta$ (the proof for
the case $\alpha<\beta$ is similar by symmetric consideration). It is known
from Remark 4.7 that $D<\frac{\log\alpha-\log\beta}{\alpha-\beta}$ which means
$\sigma=(\alpha-\beta)D<\log(\alpha/\beta)$. The uniqueness then follows from
Lemma 4.8. The function $(\sigma,\lambda,\varepsilon)\mapsto
F(\sigma,\lambda,\varepsilon)$ is real analytic in $(\Re\lambda,\Im\lambda)$
and complex analytic in $(\sigma,\varepsilon)$. Hence $D(\lambda,\varepsilon)$
is a $C^{\infty}$-function by implicit function theorem. ∎

For any $\varepsilon>0$ and $\lambda\in\mathbb{C},$ the following equations

(4.30) $\begin{cases}D&=\phi\left\\{\left[(\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{0}^{2}\right]^{-1}\right\\}\\\
\varepsilon_{0}^{2}&=\frac{\varepsilon^{2}(\alpha-\beta)^{2}}{(\alpha-\beta
e^{(\alpha-\beta)D})^{2}}\cdot e^{(\alpha-\beta)D}\end{cases}$

determine uniquely $\varepsilon_{0}=\varepsilon_{0}(\lambda,\varepsilon)>0$
and $D=D(\lambda,\varepsilon)<\frac{\log\alpha-\log\beta}{\alpha-\beta}$. We
now define
$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}:\mathbb{C}\rightarrow\mathbb{C}$ as
follows

(4.31)
$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)=\lambda+\gamma\phi\left\\{(\lambda-
x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{0}(\lambda,\varepsilon)^{2}\right]^{-1}\right\\}.$

We summarize the relation between $\lambda$ and $z$:

  * •

fix $\varepsilon>0$, for any $z\in\mathbb{C}$, let
$\lambda=z-\kappa(y,P_{z}^{\bf t}(\varepsilon)y)$ as in Lemma 4.3, then by the
cumulant formula, we have

$\displaystyle\lambda$ $\displaystyle=z-\int_{0}^{1}\big{[}P_{z}^{\bf
t}(\varepsilon)\big{]}(t)dt=z-\phi(P_{z}^{\bf t}(\varepsilon))$ (4.32)
$\displaystyle=z-\gamma\phi\left\\{({z}-x_{0}-y)^{*}\left[({z}-x_{0}-y)({z}-x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\},$

where $y=g_{\alpha\beta,\gamma}$.

  * •

fix $\varepsilon>0$, for any $\lambda\in\mathbb{C}$, we set
$z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)$. We would like to see
if $\lambda$ can be retrieved by the previous construction (4.32).

The following result shows that there is a natural one-to-one correspondence
between $\lambda$ and $z$ via the subordination relation (4.32) and
$z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)$.

######  Lemma 4.10.

The map $\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$ is a homeomorphism of the
complex plane for any $\varepsilon>0$. Its inverse map is

$J^{(\varepsilon)}_{\alpha\beta,\gamma}(z)=z-\gamma\phi\left\\{({z}-x_{0}-y)^{*}\left[({z}-x_{0}-y)({z}-x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\},$

where $y=g_{\alpha\beta,\gamma}$.

###### Proof.

Using notation in the proof of Lemma 4.3, for
$b\in\mathbb{H}^{+}(\mathcal{B})$, we have (see 4.14)

$\Omega_{1}(b)=b-R_{Y}(G_{X+Y}(b))=b-R_{Y}(G_{X}(\Omega_{1}(b))).$

Set $H_{1}(b)=b+R_{Y}(G_{X}(b))$, then $H_{1}(\Omega_{1}(b))=b$ for any
$b\in\mathbb{H}^{+}(\mathcal{B})$. Hence, for any
$p\in\Omega_{1}(\mathbb{H}^{+}(\mathcal{B}))$, we have
$\Omega_{1}(H_{1}(p))=p$.

Observe that

$\|({z}-x_{0}-y)^{*}\big{(}({z}-x_{0}-y)({z}-x_{0}-y)^{*}+\varepsilon^{2}\big{)}^{-1}\|\leq\frac{1}{2\varepsilon}.$

Hence $J^{(\varepsilon)}_{\alpha\beta,\gamma}(z)\approx z$ for large $z$.
Since $J^{(\varepsilon)}_{\alpha\beta,\gamma}$ is a $C^{\infty}$ map of
$\mathbb{C}$ to itself, we then deduce that
$J^{(\varepsilon)}_{\alpha\beta,\gamma}$ is surjective. By Theorem 7.3, we
deduce that
$\Omega_{1}\big{(}\\{\Theta(z,\varepsilon):z\in\mathbb{C}\\}\big{)}$ contains
any element of the form

$p(\lambda)=\begin{bmatrix}i\varepsilon_{1}&\lambda\\\
\overline{\lambda}&i\varepsilon_{2}\end{bmatrix},\qquad\lambda\in\mathbb{C},$

where $\varepsilon_{1},\varepsilon_{2}$ are defined as in Lemma 4.3.

Suppose $z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)$ for
$\lambda\in\mathbb{C}$. We want to show that
$\lambda=J_{\alpha\beta,\gamma}^{(\varepsilon)}(z)$. Suppose
$p(\lambda)=\Omega_{1}(b)$, where $b=\Theta(z^{\prime},\varepsilon)$ for some
$z^{\prime}\in\mathbb{C}$. Then $b=H_{1}(\Omega_{1}(b))=H_{1}(p(\lambda))$.
One can check that

$H_{1}(p(\lambda))=\Theta(z,\varepsilon)=\begin{bmatrix}i\varepsilon&z\\\
\overline{z}&i\varepsilon\end{bmatrix},$

which yields that $b=\Theta(z^{\prime},\varepsilon)=\Theta(z,\varepsilon)$.
Hence, $z=z^{\prime}$ and $p(\lambda)=\Omega_{1}(\Theta(z,\varepsilon))$.
Therefore,

$\lambda=z-\kappa(y,P_{z}^{\bf
t}(\varepsilon)y)=J_{\alpha\beta,\gamma}^{(\varepsilon)}(z).$

This finishes the proof. ∎

The relation (4.33) between $\lambda$ and $z$ in Lemma 4.3 plays a fundamental
role in our work. We will show that the regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}^{(\varepsilon)}$ is the push-forward
measure of the regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}^{(\varepsilon)}$ under the map
$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$.

######  Theorem 4.11.

For $\alpha\neq\beta$, let $y=g_{{}_{\alpha\beta,\gamma}}\in\mathcal{M}$ and
$x_{0}\in\widetilde{\mathcal{N}}$ be a random variable that is $*$-free from
$y$ with amalgamation over $\mathcal{B}$ in
$(\mathcal{A},\mathbb{E},\mathcal{B})$, and $\mathcal{N}$ is $*$-free from
$\mathcal{B}$ in $(\mathcal{A},\phi)$. For any $\varepsilon>0$ and
$\lambda\in\mathbb{C},$ let $\varepsilon_{0}>0$ and
$D<\frac{\log\alpha-\log\beta}{\alpha-\beta}$ be the solution for the system
of equations (4.30), and put

(4.33) $z=\lambda+\gamma\cdot\phi\left\\{(\lambda-x_{0})^{*}\left[(\lambda-
x_{0})(\lambda-x_{0})^{*}+\varepsilon_{0}^{2}\right]^{-1}\right\\}.$

We have the subordination equation

(4.34) $M_{2}(\phi)[G_{X+Y}(b)]=M_{2}(\phi)[G_{X}(\Omega_{1}(b))]$

where

(4.35) $\Omega_{1}\left(\begin{bmatrix}i\varepsilon&z\\\
\bar{z}&i\varepsilon\end{bmatrix}\right)=\begin{bmatrix}i\varepsilon_{1}&\lambda\\\
\bar{\lambda}&i\varepsilon_{2}\end{bmatrix},$

and $\varepsilon_{1},\varepsilon_{2}$ are given by Corollary 4.6. Moreover,
the subordination equation (4.34) for $b=\begin{bmatrix}i\varepsilon&z\\\
\bar{z}&i\varepsilon\end{bmatrix}$ gives

(4.36) $p_{\lambda}^{\bf t_{0}}(\varepsilon_{0})=p_{z}^{\bf
t}(\varepsilon),\;\;p_{\bar{\lambda}}^{\bf
t_{0}}(\varepsilon_{0})=p_{\bar{z}}^{\bf t}(\varepsilon),$

where ${\bf t}=(\alpha,\beta,\gamma)$ and ${\bf t_{0}}=(0,0,0)$ following
notation in (4.3), and

(4.37) $q_{\varepsilon}^{\bf
t}(z)=\frac{\varepsilon(e^{(\alpha-\beta)D}-1)}{\alpha-\beta
e^{(\alpha-\beta)D}}.$

###### Proof.

By Lemma 4.10, fix $\varepsilon>0$, $\lambda$ and $z$ are determined by each
other by two maps $\Phi_{{}_{\alpha\beta,\gamma}}^{(\varepsilon)}$ and
$J_{{\alpha\beta,\gamma}}^{(\varepsilon)}$. We note that, by Lemma 4.5 and the
subordination relation, we have

$z=\lambda+\kappa(y,P_{z}^{\bf
t}(\varepsilon)y)=\lambda+\kappa(y,g_{21}(\lambda,\varepsilon_{1},\varepsilon_{2})y)=\lambda+\gamma\cdot
p_{\lambda}^{\bf t_{0}}(\varepsilon_{0}),$

where we used the fact that $g_{21}(\lambda,\varepsilon_{1},\varepsilon_{2})$
is constant and $\varepsilon_{0}^{2}=\varepsilon_{1}\varepsilon_{2}$ (see
Lemma 4.5 and (4.5)) to derive the second identity. The formula for
subordination function $\Omega_{1}$ is the same as Lemma 4.3. The (4.36)
follows by taking the trace for (4.12). Note that, by the first subordination
relation in (4.12), we have

$\displaystyle i\phi(g_{11})=\phi(Q_{\varepsilon}^{\bf
t}(z))=q_{\varepsilon}^{\bf t}(z),$

where $Q_{\varepsilon}^{\bf t}(z)$ and $q_{\varepsilon}^{\bf t}(z)$ are
defined in (4.3). We also have $i\phi(g_{11})=\phi(\varepsilon_{2})D$. By
(4.25), we then have

$q_{\varepsilon}^{\bf
t}(z)=\phi(\varepsilon_{2})D=\frac{\varepsilon(e^{(\alpha-\beta)D}-1)}{\alpha-\beta
e^{(\alpha-\beta)D}}.$

Finishing the proof. ∎

###  4.3. The regularized Brown measures and regularized map

The following result is a reformulation of the argument from [33, Theorem
5.2]. For completeness, we provide a proof.

######  Lemma 4.12.

Let $S_{1},S_{2}$ be two $C^{\infty}$ subharmonic function on $\mathbb{C}$.
Given $\gamma\in\mathbb{C}$, denote the map $\Phi$ by

$\Phi(\lambda)=\lambda+\gamma\frac{\partial S_{1}(\lambda)}{\partial\lambda}.$

Assume that:

  1. (1)

the map $\Phi$ is a homeomorphism of the complex plane, and

  2. (2)

for any $\lambda\in\mathbb{C}$, we have

$\frac{\partial S_{1}(\lambda)}{\partial\lambda}=\frac{\partial
S_{2}(z)}{\partial z},\quad\frac{\partial
S_{1}(\lambda)}{\partial\overline{\lambda}}=\frac{\partial
S_{2}(z)}{\partial\overline{z}},$

where $z=\Phi(\lambda)$.

Then, the Riesz measure $\mu_{2}$ of $S_{2}$ is the push-forward measure of
the Riesz measure $\mu_{1}$ of $S_{1}$ under the map $\Phi$. In other words,
for any Borel measurable set $E$, we have

$\mu_{2}(E)=\mu_{1}(\Phi^{-1}(E)).$

###### Proof.

Let $\gamma=\gamma_{1}+i\gamma_{2}$. For $\lambda=\lambda_{1}+i\lambda_{2}$
and $z=z_{1}+iz_{2}$, we denote the vector fields

$P^{(1)}(\lambda_{1},\lambda_{2})=\frac{1}{2}\frac{\partial
S_{1}(\lambda)}{\partial\lambda_{1}},\quad
Q^{(1)}(\lambda_{1},\lambda_{2})=\frac{1}{2}\frac{\partial
S_{1}(\lambda)}{\partial\lambda_{2}}$

and

$P^{(2)}(z_{1},z_{2})=\frac{1}{2}\frac{\partial S_{2}(\lambda)}{\partial
z_{1}},\quad Q^{(2)}(z_{1},z_{2})=\frac{1}{2}\frac{\partial
S_{2}(\lambda)}{\partial z_{2}}.$

Then, $z=\Phi(\lambda)$ can be expressed as

$\displaystyle z_{1}$
$\displaystyle=\lambda_{1}+\left(\gamma_{1}P^{(1)}(\lambda_{1},\lambda_{2})+\gamma_{2}Q^{(1)}(\lambda_{1},\lambda_{2})\right),$
$\displaystyle z_{2}$
$\displaystyle=\lambda_{2}+\left(\gamma_{2}P^{(1)}(\lambda_{1},\lambda_{2})-\gamma_{1}Q^{(1)}(\lambda_{1},\lambda_{2})\right).$

Denote the differential $1-$form

$d_{2}=-Q^{(2)}(z_{1},z_{2})dz_{1}+P^{(1)}(z_{1},z_{2})dz_{2}.$

Let $d_{1}$ be the pulled-back $1-$form of $d_{2}$ under the map $\Phi$. In
this case, it means that we change the variable from $(z_{1},z_{2})$ to
$(\lambda_{1},\lambda_{2})$ because $\Phi$ is one-to-one. We have

$\displaystyle d_{1}$
$\displaystyle=-Q^{(2)}(z_{1},z_{2})d\big{(}\lambda_{1}+\gamma_{1}P^{(1)}+\gamma_{2}Q^{(1)}\big{)}$
$\displaystyle\qquad\qquad+P^{(2)}(z_{1},z_{2})d\big{(}\lambda_{2}+\gamma_{2}P^{(1)}-\gamma_{1}Q^{(1)}\big{)}$
$\displaystyle=-Q^{(1)}d\big{(}\lambda_{1}+\gamma_{1}P^{(1)}+\gamma_{2}Q^{(1)}\big{)}$
$\displaystyle\qquad\qquad+P^{(1)}d\big{(}\lambda_{2}+\gamma_{2}P^{(1)}-\gamma_{1}Q^{(1)}\big{)}.$

We can rewrite it as

$d_{1}=-Q^{(1)}d\lambda_{1}+P^{(1)}d\lambda_{2}+d\left[-\gamma_{1}P^{(1)}Q^{(1)}+\frac{1}{2}\big{(}(P^{(1)})^{2}-(Q^{(1)})^{2}\big{)}\right].$

Hence, for any simply connected domain ${D}$ with piecewise smooth boundary,
we have

$\int_{\partial\Phi(D)}d_{2}=\int_{\partial D}d_{1}.$

By Green’s formula and the definitions of $1-$form $d_{1}$ and $1-$form
$d_{2}$, we have

$\mu_{2}(\Phi(D))=\frac{1}{2\pi}\int_{\Phi(D)}\nabla^{2}S_{2}dz_{1}dz_{2}=\frac{1}{2\pi}\int_{\partial\Phi(D)}d_{2}$

and similarly,

$\mu_{1}(D)=\frac{1}{2\pi}\int_{\partial D}d_{1}.$

This finishes the proof. ∎

######  Theorem 4.13.

For $\alpha\neq\beta$, the regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}^{(\varepsilon)}$ is the push-forward
measure of the regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}^{(\varepsilon)}$ under the map
$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$.

###### Proof.

Recall that
$S(x,\lambda,\varepsilon)=\phi\bigg{(}\log\big{(}(x-\lambda)^{*}(x-\lambda)+\varepsilon^{2}\big{)}\bigg{)}$.
We set $S_{1}(\lambda)=S(x_{0}+g_{\alpha\beta,0},\lambda,\varepsilon)$ and
$S_{2}(z)=S(x_{0}+g_{\alpha\beta,\gamma},z,\varepsilon)$, and
$S_{0}(\lambda)=S(x_{0},\lambda,\varepsilon_{0})$, where
$\varepsilon_{0}=\varepsilon_{0}(\lambda,\varepsilon)$ and
$z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)$ defined as (4.31). Then
the map $\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}$ can be rewritten as

$\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}(\lambda)=\lambda+\gamma\frac{\partial
S_{0}}{\partial\lambda}.$

The map $\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}$ is a homeomorphism by
Lemma 4.10. By Definition (4.3), we have

$p_{\lambda}^{\bf t_{0}}(\varepsilon_{0})=\frac{\partial
S_{0}(\lambda)}{\partial\lambda},\qquad p_{z}^{\bf
t}(\varepsilon)=\frac{\partial S_{2}(z)}{\partial z},$

where ${\bf t_{0}}=(0,0,0)$ and ${\bf t}=(\alpha,\beta,\gamma)$. In addition,
since $\Phi_{{}_{\alpha\beta,0}}^{(\varepsilon)}(\lambda)=\lambda$, we also
have

$p_{\lambda}^{\bf t(0)}(\varepsilon)=\frac{\partial
S_{1}(\lambda)}{\partial\lambda},$

where ${\bf t(0)}=(\alpha,\beta,0)$. Hence, by choosing $\gamma=0$ and an
arbitrary eligible $\gamma$, the subordination relation $p_{\lambda}^{\bf
t_{0}}(\varepsilon_{0})=p_{z}^{\bf t}(\varepsilon)$ from (4.36) shows

$\frac{\partial S_{0}(\lambda)}{\partial\lambda}=\frac{\partial
S_{1}(\lambda)}{\partial\lambda}=\frac{\partial S_{2}(z)}{\partial z}$

and

$\frac{\partial S_{0}(\lambda)}{\partial\overline{\lambda}}=\frac{\partial
S_{1}(\lambda)}{\partial\overline{\lambda}}=\frac{\partial
S_{2}(z)}{\partial\overline{z}}.$

Hence,
$\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}(\lambda)=\lambda+\gamma\frac{\partial
S_{1}(\lambda)}{\partial\lambda}$ and conditions in Lemma 4.12 are satisfied,
which yields the desired result. ∎

##  5\. The Brown measure formulas

In this section, we calculate the Brown measure of
$x_{0}+g_{{}_{\alpha\beta,0}}$ where
$x_{0}\in\log^{+}(\mathcal{N})\subset\widetilde{\mathcal{N}}$ is a random
variable that is $*$-free from $g_{{}_{\alpha\beta,0}}$ with amalgamation over
$\mathcal{B}$ in the operator-valued $W^{*}$-probability space
$(\mathcal{A},\mathbb{E},\mathcal{B})$. Recall that $\mathcal{N}$ is $*$-free
from $\mathcal{B}$ in $(\mathcal{A},\phi)$.

###  5.1. The limit of subordination functions

Throughout this subsection, we choose $\alpha\neq\beta$. Recall that the
subordination relation $G_{X+Y}(b)=G_{X}(\Omega_{1}(b))$ gives

$\begin{split}{g}_{11}=-iQ_{\varepsilon}^{\bf
t}(z),\;{g}_{22}=-i\widetilde{Q}_{\varepsilon}^{\bf
t}(z),\;{g}_{12}=P_{\bar{z}}^{\bf t}(\varepsilon),\;{g}_{21}=P_{z}^{\bf
t}(\varepsilon),\end{split}$

where ${\bf t}=(\alpha,\beta,\gamma)$ and $g_{ij}$ was defined in (4.7).

We now turn to study the limit of $\varepsilon_{0}$ as $\varepsilon$ goes to
zero. Set

$H(\lambda)=(\phi(|x_{0}-\lambda|^{-2}))^{-1}=\left(\int_{0}^{\infty}\frac{1}{u^{2}}d\mu_{|x_{0}-\lambda|}(u)\right)^{-1}.$

If $\alpha>\beta>0$, then $\sigma=(\alpha-\beta)D>0$. We have

$\frac{1}{\alpha-\beta}=F(\sigma,\lambda,\varepsilon)=\frac{1}{\sigma}\cdot\phi\left\\{\left[(\lambda-
x_{0})^{*}(\lambda-
x_{0})+\varepsilon^{2}\frac{(\alpha-\beta)^{2}}{(\alpha-\beta
e^{\sigma})^{2}e^{-\sigma}}\right]^{-1}\right\\}\leq\frac{1}{\sigma H},$

which implies that

$\sigma\leq\frac{\alpha-\beta}{H}.$

Similarly, if $0\leq\alpha<\beta$, then

$\sigma\geq\frac{\alpha-\beta}{H}.$

We now set

(5.1)
${\Xi_{\alpha,\beta}}=\left\\{\lambda:H<\frac{\alpha-\beta}{\log\alpha-\log\beta}\right\\},$

and denote
$\sigma_{0}(\lambda)=\min\\{\log\left(\frac{\alpha}{\beta}\right),\frac{\alpha-\beta}{H}\\}$
if $\alpha>\beta>0$.

For any $\varepsilon\geq 0$, denote the function $f_{\varepsilon}$ of
$\lambda$ by

$f_{\varepsilon}(\lambda)=\phi\big{(}(|x_{0}-\lambda|^{2}+\varepsilon)^{-1}\big{)}.$

For any $\varepsilon>0$, the function $f_{\varepsilon}$ is a continuous
function of $\lambda$. As the increasing limit of the
$(f_{\varepsilon})_{\varepsilon>0}$, we deduce that $f_{0}$ is lower semi-
continuous. Hence, the set ${\Xi_{\alpha,\beta}}$ is open.

######  Lemma 5.1.

If $\lambda\in{\Xi_{\alpha,\beta}}$, then $\lim_{\varepsilon\rightarrow
0^{+}}\varepsilon_{0}(\lambda,\varepsilon)$ exists and the limit is a finite
positive number $s$ determined by

(5.2) $\frac{\log\alpha-\log\beta}{\alpha-\beta}=\phi\left\\{\left[(\lambda-
x_{0})^{*}(\lambda-x_{0})+s^{2}\right]^{-1}\right\\}.$

If $\lambda\notin{\Xi_{\alpha,\beta}}$, then $\lim_{\varepsilon\rightarrow
0^{+}}\varepsilon_{0}(\lambda,\varepsilon)=0$. Denote the limit by
$s=s(\lambda)$ for $\lambda\in\mathbb{C}$, then $s(\lambda)$ is a $C^{\infty}$
function of $\lambda\in{\Xi_{\alpha,\beta}}$.

###### Proof.

For $\varepsilon>0$, recall that $\sigma=(\alpha-\beta)D$ is the unique
solution of $F(\sigma)=1/(\alpha-\beta)$ such that
$\sigma<\log(\alpha/\beta)$. By symmetric consideration, we may assume that
$\alpha>\beta>0$. Then, $\lambda\in\Xi_{\alpha,\beta}$ if and only if

$\log\left(\frac{\alpha}{\beta}\right)<\frac{\alpha-\beta}{H},$

which implies $\sigma_{0}(\lambda)=\log(\alpha/\beta)$. By Proposition 4.9,
$\sigma=\sigma(\lambda,\varepsilon)$ is a $C^{\infty}$-function of
$(\lambda,\varepsilon)$ over $\mathbb{C}\times(0,\infty)$. We first claim that
$\lim_{\varepsilon\rightarrow
0^{+}}\sigma(\lambda,\varepsilon)=\log(\alpha/\beta)$. Assume that there
exists some $x_{k}\rightarrow 0^{+}$ such that
$\sigma=\sigma(\lambda,x_{k})=(\alpha-\beta)D(\lambda,x_{k})$ converges to
some $\theta<\log(\alpha/\beta)$. Then, we have

$\displaystyle\lim_{x_{k}\rightarrow
0^{+}}\varepsilon_{0}(\lambda,\varepsilon)=\lim_{x_{k}\rightarrow
0^{+}}\frac{\varepsilon(\alpha-\beta)}{\alpha-\beta
e^{(\alpha-\beta)D}}e^{(\alpha-\beta)D/2}=0.$

Consequently,

$\lim_{x_{k}\rightarrow 0^{+}}\sigma F(\sigma)=H^{-1}.$

On the other hand,

$\lim_{x_{k}\rightarrow 0^{+}}\sigma
F(\sigma)=\frac{\theta}{\alpha-\beta}<\frac{\log\alpha-\log\beta}{\alpha-\beta}.$

This contradicts to $\lambda\in{\Xi_{\alpha,\beta}}$. Hence,
$\lim_{\varepsilon\rightarrow
0^{+}}\sigma=\log(\alpha/\beta)=\sigma_{0}(\lambda)$ if
$\lambda\in{\Xi_{\alpha,\beta}}$. By using the identity $D=\sigma
F(\sigma)=\frac{\sigma}{\alpha-\beta}$, we have

$\frac{\log\alpha-\log\beta}{\alpha-\beta}=\lim_{\varepsilon\rightarrow
0^{+}}\sigma F(\sigma)=\lim_{\varepsilon\rightarrow
0^{+}}\phi\big{\\{}[|\lambda-x_{0}|^{2}+\varepsilon_{0}^{2}]^{-1}\big{\\}}$

which implies that $\lim_{\varepsilon\rightarrow 0^{+}}\varepsilon_{0}$ exists
and is determined by (5.2).

For any $s>0$ and $\lambda\in\mathbb{C}$, the map
$s\mapsto\phi\left\\{\left[|\lambda-x_{0}|^{2}+s^{2}\right]^{-1}\right\\}$ is
$C^{\infty}$ on $[x,\infty)$, and for any $s>x$, the map
$\lambda\mapsto\phi\left\\{\left[|\lambda-
x_{0}|^{2}+s^{2}\right]^{-1}\right\\}$ is $C^{\infty}$ on $\mathbb{C}$. Then,
the implicit function theorem implies that as unique solution of (5.2), the
function $s(\lambda)$ is $C^{\infty}$ on ${\Xi_{\alpha,\beta}}$.

Similarly, if $\lambda\notin{\Xi_{\alpha,\beta}}$, by assuming
$\lim_{\varepsilon\rightarrow 0^{+}}\varepsilon_{0}(\lambda,\varepsilon)\neq
0$, we must have $\lim_{\varepsilon\rightarrow
0^{+}}\sigma=\log(\alpha/\beta)$. This yields that
$H<\frac{\alpha-\beta}{\log\alpha-\log\beta}$. This contradiction implies that
if $\lambda\notin{\Xi_{\alpha,\beta}}$, then $\lim_{\varepsilon\rightarrow
0^{+}}\varepsilon_{0}(\lambda,\varepsilon)=0$. ∎

######  Remark 5.2.

The function $s(\lambda)$ can be viewed as the boundary value of scalar-valued
subordination function parameterized by $\lambda\in\mathbb{C}$. See Definition
7.2 and [33, Definition 3.3 and Proposition 3.5] for details.

######  Lemma 5.3.

The convergence $\lim_{\varepsilon\rightarrow
0}\varepsilon_{0}(\lambda,\varepsilon)=s(\lambda)$ is uniform over any compact
subset in $\lambda\in\mathbb{C}$. If $x_{0}$ is bounded, the convergence is
uniform over $\mathbb{C}$.

###### Proof.

We first show that $s(\lambda)$ is a continuous function in $\mathbb{C}$.
Since $s(\lambda)$ is $C^{\infty}$ in the open set $\Xi_{\alpha,\beta}$, to
show $s(\lambda)$ is continuous, it remains to show that for any
$\lambda_{0}\in\mathbb{C}\backslash\Xi_{\alpha,\beta}$ and a sequence
$\\{\lambda_{n}\\}\subset\Xi_{\alpha,\beta}$ converging to $\lambda_{0}$, we
have

$\lim_{n\rightarrow\infty}s(\lambda_{n})=s(\lambda_{0})=0.$

Assume this is not true, by dropping to a subsequence if necessary, we may
assume that there exists $\delta>0$ such for all $n$, $s(\lambda_{n})>\delta$.
In this case, note that

$\phi\left\\{\left[(\lambda_{n}-x_{0})^{*}(\lambda_{n}-x_{0})+s(\lambda_{n})^{2}\right]^{-1}\right\\}<\phi\left\\{\left[(\lambda_{n}-x_{0})^{*}(\lambda_{n}-x_{0})+\delta^{2}\right]^{-1}\right\\}.$

By using 5.2 and passing to the limit, we have

$\frac{\log\alpha-\log\beta}{\alpha-\beta}\leq\phi\left\\{\left[(\lambda_{0}-x_{0})^{*}(\lambda_{0}-x_{0})+\delta^{2}\right]^{-1}\right\\}$

which yields that $s(\lambda_{0})\geq\delta>0$. It contradicts to our choice
$\lambda_{0}\in\mathbb{C}\backslash\Xi_{\alpha,\beta}$. Hence
$\lim_{n\rightarrow\infty}s(\lambda_{n})\rightarrow s(\lambda_{0})=0$ and the
function $s(\lambda)$ is continuous in $\mathbb{C}$.

Without losing generality, we now assume that $\alpha>\beta$. For any
$\lambda\in\mathbb{C}$, for $0<\varepsilon_{1}<\varepsilon_{2}$, note that
$F(\sigma,\lambda,\varepsilon)$ decreases as $\varepsilon$ increases, hence we
have

$\frac{1}{\alpha-\beta}=F(\sigma_{1},\lambda,\varepsilon_{1})=F(\sigma_{2},\lambda,\varepsilon_{2})<F(\sigma_{2},\lambda,\varepsilon_{1})$

where $\sigma_{i}=\sigma(\lambda,\varepsilon_{i})\,(i=1,2)$. By Lemma 4.8, we
deduce that $\sigma_{1}>\sigma_{2}$. Recall that
$\sigma(\lambda,\varepsilon_{i})=(\alpha-\beta)D(\lambda,\varepsilon_{i})$ and

$D(\lambda,\varepsilon_{i})=\phi\left\\{\left[(\lambda-x_{0})^{*}(\lambda-
x_{0})+\varepsilon_{0}(\lambda,\varepsilon_{i})^{2}\right]^{-1}\right\\}.$

Hence,
$\varepsilon_{0}(\lambda,\varepsilon_{1})<\varepsilon_{0}(\lambda,\varepsilon_{2})$.
We then conclude that $\varepsilon_{0}(\lambda,\varepsilon)$ converges to
$s(\lambda)$ uniformly in any compact subset of $\mathbb{C}$ as $\varepsilon$
tends to zero by Dini’s theorem.

Assume now that $x_{0}$ is a bounded operator. Since $\Xi_{\alpha,\beta}$ is a
bounded set, to prove the convergence is uniform over $\mathbb{C}$, it
suffices to show that $\varepsilon_{0}(\lambda,\varepsilon)<2\varepsilon$ if
$|\lambda|$ is sufficiently large. Recall that

$\varepsilon_{0}(\lambda,\varepsilon)=\frac{\varepsilon(\alpha-\beta)}{\alpha-\beta
e^{\sigma}}e^{\sigma/2},$

where $\sigma=\sigma(\lambda,\varepsilon)$. Hence,
$\varepsilon_{0}(\lambda,\varepsilon)\geq 2\varepsilon$ if and only if

(5.3) $\alpha-\beta>2\left(\alpha e^{-\sigma/2}-\beta e^{\sigma/2}\right).$

But $\sigma=(\alpha-\beta)D\leq(\alpha-\beta)/(M^{2}+4\varepsilon^{2})$ if
$|\lambda|>M+||x_{0}||$ and $\varepsilon_{0}(\lambda,\varepsilon)\geq
2\varepsilon$ by the formula for $D$ in (4.30). It follows that $\sigma$ could
be arbitrarily small by choosing $M$ large enough. Observe that (5.3) can not
hold when $\sigma$ is sufficiently small. Hence, we may choose $M$ large
enough, so that $\varepsilon_{0}(\lambda,\varepsilon)<2\varepsilon$ for any
$|\lambda|>M+\|x_{0}\|$. ∎

###  5.2. The Brown measure formula

We adapt the proof of [33, Theorem 4.6] to calculate the density formula
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ in ${\Xi_{\alpha,\beta}}$.

######  Theorem 5.4.

The support of the Brown measure $\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ of
$x_{0}+g_{{}_{\alpha\beta,0}}$ is the closure of the set
${\Xi_{\alpha,\beta}}$ defined by (5.1), which can be rewritten as

${\Xi_{\alpha,\beta}}=\left\\{\lambda:\phi(|x_{0}-\lambda|^{-2})>\frac{\log\alpha-\log\beta}{\alpha-\beta}\right\\}.$

Let $s(\lambda)$ be the function determined by

(5.4) $\frac{\log\alpha-\log\beta}{\alpha-\beta}=\phi\left\\{\left[(\lambda-
x_{0})^{*}(\lambda-x_{0})+s(\lambda)^{2}\right]^{-1}\right\\}$

as in Lemma 5.1, and set

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

and

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

The probability measure $\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ is absolutely
continuous in the open set ${\Xi_{\alpha,\beta}}$ with density given by

(5.5)
$\frac{1}{\pi}s^{2}(\lambda)\phi(h^{-1}k^{-1})+\frac{1}{\pi}\frac{|\phi((\lambda-
x_{0})(h^{-1})^{2})|^{2}}{\phi((h^{-1})^{2})}.$

We need the following result. It should be well-known, but we include a proof
for the sake of completeness.

######  Lemma 5.5.

  1. (1)

Let $x\in\widetilde{\mathcal{N}},$ for any $a\in(0,\infty)$ we have

(5.6) $x(x^{*}x+a)^{-1}=(xx^{*}+a)^{-1}x.$

  2. (2)

For $h(\lambda,s)$ and $k(\lambda,s)$ given in Theorem 5.4, we have

(5.7)
$\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}\right)=-\phi\left[h^{-1}\left((\lambda-
x_{0})+2s\frac{\partial s}{\partial\overline{\lambda}}\right)h^{-1}\right],$

and

(5.8)
$\begin{split}\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}(\lambda-
x_{0})^{*}\right)=\phi\left[h^{-1}-h^{-1}\left((\lambda-
x_{0})+2s\frac{\partial
s}{\partial\overline{\lambda}}\right)h^{-1}\right].\end{split}$

###### Proof.

For (1), it is equivalent to show that $(xx^{*}+a)x(x^{*}x+a)^{-1}=x$. Since
the affiliated algebra $\widetilde{\mathcal{N}}$ is an algebra, this follows
from the following calculation

$xx^{*}x(x^{*}x+a)^{-1}=x(x^{*}x+a-a)(x^{*}x+a)^{-1}=x-ax(x^{*}x+a)^{-1}.$

For (2), if $x_{0}$ is bounded, we can obtain the result by applying [19,
Lemma 3.2]. We denote

$\displaystyle h=h(\lambda,\overline{\lambda},s)$ $\displaystyle=(\lambda-
x_{0})^{*}(\lambda-x_{0})+s^{2},$ $\displaystyle
k=k(\lambda,\overline{\lambda},s)$ $\displaystyle=(\lambda-x_{0})(\lambda-
x_{0})^{*}+s^{2},$

where $s=s(\lambda,\overline{\lambda})$ determined by (5.4). By the identity
$a^{-1}-b^{-1}=b^{-1}(b-a)a^{-1}$ we have

$\displaystyle
h(\lambda,\overline{\lambda^{\prime}},s)^{-1}-h(\lambda,\overline{\lambda},s)^{-1}$
$\displaystyle=-h(\lambda,\overline{\lambda^{\prime}},s)^{-1}((\overline{\lambda^{\prime}}-\overline{\lambda})(\lambda-
x_{0})+s(\lambda,\overline{\lambda^{\prime}})^{2}-s(\lambda,\overline{\lambda})^{2})h(\lambda,\overline{\lambda},s)^{-1}$

Since $s$ is a $C^{\infty}$ function of $\lambda$ and $s>0$, it follows that

$\lim_{\overline{\lambda^{\prime}}\rightarrow\overline{\lambda}}h(\lambda,\overline{\lambda^{\prime}},s(\lambda,\overline{\lambda^{\prime}}))^{-1}=h(\lambda,\overline{\lambda},s(\lambda,\overline{\lambda}))^{-1}.$

Therefore, by using the notation of complex derivative, the partially
derivation with respect to $\overline{\lambda}$ can be calculated as

$\displaystyle\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}\right)$
$\displaystyle=\lim_{\overline{\lambda^{\prime}}\rightarrow\lambda}\frac{\phi(h(\lambda,\overline{\lambda^{\prime}},s)^{-1})-\phi(h(\lambda,\overline{\lambda},s)^{-1}))}{\overline{\lambda^{\prime}}-\overline{\lambda}}$
$\displaystyle=-\lim_{\overline{\lambda^{\prime}}\rightarrow\lambda}\phi\big{[}h(\lambda,\overline{\lambda^{\prime}},s)^{-1}(\lambda-
x_{0})h(\lambda,\overline{\lambda},s)^{-1}\big{]}$
$\displaystyle\qquad-\lim_{\overline{\lambda^{\prime}}\rightarrow\lambda}\left[\frac{s(\lambda,\overline{\lambda^{\prime}})^{2}-s(\lambda,\overline{\lambda})^{2})}{\overline{\lambda^{\prime}}-\overline{\lambda}}\cdot\phi\big{[}h(\lambda,\overline{\lambda^{\prime}},s)^{-1}h(\lambda,\overline{\lambda},s)^{-1}\big{]}\right]$
$\displaystyle=-\phi\left[h^{-1}\left((\lambda-x_{0})+2s\frac{\partial
s}{\partial\overline{\lambda}}\right)h^{-1}\right].$

The proof for (5.8) is similar. ∎

###### Proof of Theorem 5.4.

Set ${\bf t}=(\alpha,\beta,0)$, ${\bf t_{0}}=(0,0,0)$ and
$b=\begin{bmatrix}i\varepsilon&\lambda\\\
\overline{\lambda}&i\varepsilon\end{bmatrix}$, we have

$M_{2}(\phi)[G_{X+Y}(b)]=\begin{bmatrix}-iq_{\varepsilon}^{\bf
t}(\lambda)&p_{\bar{\lambda}}^{\bf t}(\varepsilon)\\\ p_{\lambda}^{\bf
t}(\varepsilon)&-i{q}_{\varepsilon}^{\bf t}(\lambda)\end{bmatrix}$

Applying (4.36) for $\gamma=0$, we have

$p_{\lambda}^{\bf t}(\varepsilon)=p_{\lambda}^{\bf
t_{0}}(\varepsilon_{0})=\phi\left\\{(\lambda-x_{0})^{*}\left[(\lambda-
x_{0})(\lambda-x_{0})^{*}+\varepsilon_{0}^{2}\right]^{-1}\right\\}.$

By Lemma 5.1, we have

(5.9) $\displaystyle\lim_{\varepsilon\rightarrow 0^{+}}\phi\left\\{(\lambda-
x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{0}^{2}\right]^{-1}\right\\}$ $\displaystyle=$
$\displaystyle\phi\left\\{(\lambda-x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+s(\lambda)^{2}\right]^{-1}\right\\}$ $\displaystyle=$
$\displaystyle\phi\left\\{\left[(\lambda-x_{0})^{*}(\lambda-
x_{0})+s(\lambda)^{2}\right]^{-1}(\lambda-x_{0})^{*}\right\\}$
$\displaystyle=$ $\displaystyle\phi\left(h^{-1}(\lambda-x_{0})^{*}\right).$

By [10, Lemma 4.19] (up to a sign change), the Brown measure of
$x_{0}+g_{{}_{\alpha\beta,0}}$ is equal to

(5.10) $\displaystyle\mu_{x_{0}+g_{\alpha\beta,0}}$
$\displaystyle=\frac{1}{\pi}\lim_{\varepsilon\rightarrow
0^{+}}\frac{\partial}{\partial\overline{\lambda}}p_{\lambda}^{\bf
t}(\varepsilon)=\frac{1}{\pi}\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}(\lambda-
x_{0})^{*}\right)$

in the distribution sense. Applying Lemma 5.5, we have

(5.11)
$\displaystyle\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}(\lambda-
x_{0})^{*}\right)=\phi\left[h^{-1}\bigg{(}{\bf 1}-(\lambda-
x_{0})h^{-1}(\lambda-x_{0})^{*}\bigg{)}\right]$
$\displaystyle\qquad\qquad-2s(\lambda)\frac{\partial
s(\lambda)}{\partial\overline{\lambda}}\phi\left[(\lambda-
x_{0})^{*}(h^{-1})^{2}\right].$

Applying the identity
$x(x^{*}x+\varepsilon)^{-1}x^{*}=(xx^{*}+\varepsilon)^{-1}xx^{*}$, we obtain

(5.12) $\displaystyle\phi\left[h^{-1}\bigg{(}{\bf 1}-(\lambda-
x_{0})h^{-1}(\lambda-
x_{0})^{*}\bigg{)}\right]=s(\lambda)^{2}\phi(h^{-1}k^{-1}).$

The function $s(\lambda)$ is the implicit function determined by (5.4) when
$\lambda\in\Xi_{\alpha,\beta}$, which can be rewritten as (by tracial
property)

$\phi(h^{-1})=\frac{\log\alpha-\log\beta}{\alpha-\beta}.$

Applying implicit differentiation
$\frac{\partial}{\partial\overline{\lambda}}$, we have

$\phi\left[h^{-1}\left((\lambda-x_{0})+2s(\lambda)\frac{\partial
s(\lambda)}{\partial\overline{\lambda}}\right)h^{-1}\right]=0$

which yields

(5.13) $-2s(\lambda)\frac{\partial
s(\lambda)}{\partial\overline{\lambda}}=\frac{\phi((\lambda-
x_{0})(h^{-1})^{2})}{\phi((h^{-1})^{2})}.$

Since $\overline{\phi((\lambda-x_{0})(h^{-1})^{2})}=\phi((\lambda-
x_{0})^{*}(h^{-1})^{2})$, we hence get the density formula 5.5 by plugging
(5.15) and (5.13) to (5.11).

Note that if $\lambda\notin\overline{{\Xi_{\alpha,\beta}}}$, then
$s(\lambda)=0$ in some neighborhood of $\lambda$, and

(5.14) $\phi(|\lambda-
x_{0}|^{-2})\leq\frac{\log\alpha-\log\beta}{\alpha-\beta},$

which implies that the above limit (5.9) is finite by an estimation using
Cauchy-Schwartz inequality. It follows that we can take the limit

(5.15) $\lim_{\varepsilon\rightarrow 0^{+}}p_{\lambda}^{\bf
t}(\varepsilon)=\phi\left\\{(\lambda-x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}\right]^{-1}\right\\}=\lim_{\varepsilon\rightarrow
0}p_{\lambda}^{\bf t_{0}}(\varepsilon).$

Hence, $\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ and $\mu_{x_{0}}$ coincide in the
open set $\mathbb{C}\backslash\overline{\Xi_{\alpha,\beta}}$. By the
inequality (5.14) and [33, Theorem 4.5], for any
$\lambda\in\mathbb{C}\backslash\overline{{\Xi_{\alpha,\beta}}}$, then
$\lambda$ is not in the support of $\mu_{x_{0}}$. Consequently the density of
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ is zero on
$\mathbb{C}\backslash\overline{{\Xi_{\alpha,\beta}}}$ by (5.10) and (5.15). ∎

######  Remark 5.6.

Let $t=\frac{\alpha-\beta}{\log\alpha-\log\beta}$. The proof for Theorem 5.4
shows that the partial derivatives of $\log
S(x_{0}+g_{\alpha\beta,0},\lambda,0)$ are the same as the partial derivatives
of $\log S(x_{0}+c_{t},\lambda,0)$ by comparing the proof of Theorem 7.10 (see
also [33, Theorem 4.2] for bounded case). Hence, the Brown measure of
$x_{0}+g_{\alpha\beta,0}$ is the same as the Brown measure of $x_{0}+c_{t}$.
In particular, by [33, Theorem 4.6], the measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ is absolutely continuous with respect to
the Lebesgue measure on $\mathbb{C}$. We refer to Corollary 7.8 and Theorem
7.10 for $x_{0}$ to be unbounded operator.

###  5.3. The push-forward map

In this section, we show that the Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}$ is the push-forward measure of
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ under the map $\Phi_{\alpha\beta,\gamma}$
defined by

$\Phi_{\alpha\beta,\gamma}(\lambda)=\lambda+\gamma\phi\left\\{(\lambda-
x_{0})^{*}\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+s(\lambda)^{2}\right]^{-1}\right\\},$

where $s(\lambda)=\lim_{\varepsilon\rightarrow
0}\varepsilon_{0}(\lambda,\varepsilon)$ as in Lemma 5.1. However, this map
might be singular. We explain how the strategy in [33, Section 5] applies to
our case.

######  Lemma 5.7.

The function $\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}$ converges uniformly
to $\Phi_{\alpha\beta,\gamma}$ in any compact subset of $\mathbb{C}$ as
$\varepsilon$ tends to zero. The convergence is uniform over $\mathbb{C}$ if
$x_{0}$ is bounded.

###### Proof.

For $\varepsilon_{2}>\varepsilon_{1}>0$, apply the same estimation as in [33,
Lemma 5.3], we have

$|p_{\lambda}^{\bf
t_{0}}(\varepsilon_{0}(\lambda,\varepsilon_{2}))-p_{\lambda}^{\bf
t_{0}}(\varepsilon_{0}(\lambda,\varepsilon_{1}))|\leq\frac{\log\alpha-\log\beta}{\alpha-\beta}|\varepsilon_{0}(\lambda,\varepsilon_{2})-\varepsilon_{0}(\lambda,\varepsilon_{1})|.$

Since $\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}$ can be written as

$\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}(\lambda)=\lambda+\gamma
p_{\lambda}^{\bf t_{0}}(\varepsilon_{0}(\lambda,\varepsilon))$

where ${\bf t_{0}}=(0,0,0)$. The result then follows from convergence result
of $\varepsilon_{0}(\lambda,\varepsilon)$ to $s(\lambda)$ showed in Lemma 5.3.
∎

######  Theorem 5.8.

The Brown measure $\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}$ is the push-
forward measure of $\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}$ under the map
$\Phi_{\alpha\beta,\gamma}$. Hence, we have the following commutative diagram.

${\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}^{(\varepsilon)}}$${\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}^{(\varepsilon)}}$${\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}}$${\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}}$$\varepsilon\rightarrow
0$$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$$\Phi_{\alpha\beta,\gamma}$$\varepsilon\rightarrow
0$

###### Proof.

The regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,\gamma}}}^{(\varepsilon)}$ is the push-forward
measure of the regularized Brown measure
$\mu_{x_{0}+g_{{}_{\alpha\beta,0}}}^{(\varepsilon)}$ under the map
$\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$. For any
$x\in\log^{+}(\mathcal{A})$, the regularized Brown measure
$\mu_{x}^{(\varepsilon)}$ converges to the Brown measure $\mu_{x}$ weakly as
$\varepsilon$ tends to zero. Since $\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}$
converges to $\Phi_{\alpha\beta,\gamma}$ uniformly in any compact subset of
$\mathbb{C}$, we then conclude the desired result. ∎

######  Example 5.9 (The Brown measure of triangular elliptic operator).

Given $\alpha,\beta>0$, set $t=\frac{\alpha-\beta}{\log\alpha-\log\beta}$.
When $x_{0}=0$, the set ${\Xi_{\alpha,\beta}}$ is the circle centered at the
origin with radius $\sqrt{t}$,

${\Xi_{\alpha,\beta}}=\\{\lambda\in\mathbb{C}:|\lambda|<t\\}$

and $s(\lambda)^{2}+|\lambda|^{2}=t$ for $\lambda\in{\Xi_{\alpha,\beta}}$.
Using notation in Theorem 5.4,

$h(\lambda,s)=k(\lambda,s)=|\lambda|^{2}+s(\lambda)^{2}=t.$

Hence, the density of the Brown measure of $g_{{}_{\alpha\beta,0}}$ is

$d\mu_{g_{\alpha\beta,0}}=\frac{1}{\pi}\frac{s^{2}}{t^{2}}+\frac{1}{\pi}\frac{|\lambda|^{2}}{t^{2}}=\frac{1}{\pi
t}dxdy.$

In other words, it has the same Brown measure as $c_{t}$, the circular
operator with variance $t$. If $T$ is a quasi-nilpotent DT operator, and
$c_{\varepsilon}$ is a circular operator with variance $\varepsilon$, let
$\alpha={1+\varepsilon},\beta={\varepsilon}$, then $T+c_{\varepsilon}$ has the
same $*$-moments as $g_{{}_{\alpha\beta,0}}$. Hence, the Brown measure
$T+c_{\varepsilon}$ is the uniform measure on the circle
$\Big{\\{}\lambda:|\lambda|\leq\frac{1}{\sqrt{\log(1+\varepsilon^{-1})}}\Big{\\}}$.
This recovers a result of Aagaard-Haagerup [2, Theorem 4.3].

For $\gamma\in\mathbb{C}$ such that $|\gamma|\leq\sqrt{\alpha\beta}$, we now
have

$\Phi_{\alpha\beta,\gamma}(\lambda)=\lambda+\gamma\cdot\frac{\overline{\lambda}}{|\lambda|^{2}+s^{2}},$

for $|\lambda|<t$. Hence, the Brown measure of $g_{{}_{\alpha\beta,\gamma}}$
is supported in the ellipse with parametrization

$\sqrt{t}e^{i\theta}+\frac{|\gamma|}{\sqrt{t}}e^{i(\psi-\theta)}$

where $\psi=\arg(\gamma)$. For $0\leq r<\sqrt{t}$, we have

$\Phi_{\alpha\beta,\gamma}(re^{i\theta})=re^{i\theta}+\frac{|\gamma|r}{t}e^{i(\psi-\theta)}.$

Hence, for $\gamma=\gamma_{1}+i\gamma_{2}$, the Jacobian matrix of $\Phi_{\bf
t}$ at $\lambda=\lambda_{1}+i\lambda_{2}$ is given by

$\text{Jacobian}(\Phi_{\alpha\beta,\gamma})=\begin{bmatrix}1+\frac{\gamma_{1}}{t}&\frac{\gamma_{2}}{t}\\\
\frac{\gamma_{2}}{t}&1-\frac{\gamma_{1}}{t}\end{bmatrix}$

whose determinant is equal to $1-\frac{|\gamma|^{2}}{t}$. Hence, the Brown
measure of $g_{{}_{\alpha\beta,\gamma}}$ is the uniform measure in the ellipse
by Theorem 5.8. This recovers the result obtained in [5, Section 6]

The interested reader is referred to [20, 33] for more examples of the push-
forward maps and explicit Brown measure formulas.

##  6\. Some regularity results on the pushforward map

The push-forward map $\Phi_{\alpha\beta,\gamma}$ is the limit of the family of
homeomorphisms $\Phi^{(\varepsilon)}_{\alpha\beta,\gamma}$, which could be
singular in general. In this section, we study its properties under some
regularity assumptions. Recall that $\mathcal{N}$ is $*$-free from
$\mathcal{M}$ with amalgamation over $\mathcal{B}$ in the operator-valued
$W^{*}$-probability space $(\mathcal{A},\mathbb{E},\mathcal{B})$ constructed
in Section (4.1). The unital subalgebra $\mathcal{N}$ is also $*$-free from
$\mathcal{B}$ in $(\mathcal{A},\phi)$.

######  Lemma 6.1.

Let $y=g_{{}_{\alpha\beta,0}}\in\mathcal{M}$ and
$x_{0}\in\log^{+}(\mathcal{N})$. The function $\lambda\mapsto
S(x_{0}+y,\lambda,0)=\log\Delta(|x_{0}+y-\lambda|^{2})$ is a
$C^{\infty}$-function of $\lambda$ in the open set ${\Xi_{\alpha,\beta}}$.

###### Proof.

By the definition (4.2) of the function $S$, we observe that

$\frac{1}{2}\frac{d}{d\varepsilon}S(x_{0}+y,\lambda,\varepsilon)=q_{\varepsilon}^{\bf
t}(\lambda),$

where ${\bf t}=(\alpha,\beta,0)$, and by (4.3)

$q_{\varepsilon}^{\bf t}(\lambda)=\varepsilon\phi\left\\{\left[(\lambda-
x_{0}-y)^{*}(\lambda-x_{0}-y)+\varepsilon^{2}\right]^{-1}\right\\}.$

We now apply Lemma 4.3 to the case $\gamma=0$. In this case, the free cumulant
$\kappa(y,fy)=0$ for any $f\in\mathcal{B}$, and hence
$z=\Phi_{\alpha\beta,0}^{(\varepsilon)}(\lambda)=\lambda$. The subordination
relation (4.12) yields

$g_{11}(\lambda,\varepsilon_{1},\varepsilon_{2})=-iQ^{\bf
t}_{\varepsilon}(\lambda),$

and by (4.4) and (4.7)

${g}_{11}(\lambda,\varepsilon_{1},\varepsilon_{2})=-i\varepsilon_{2}{\mathbb{E}}\left\\{\left[(\lambda-
x_{0})(\lambda-x_{0})^{*}+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\}$

and

$Q^{\bf
t}_{\varepsilon}(\lambda)=\varepsilon{\mathbb{E}}\left\\{\left[(\lambda-
x_{0}-y)(\lambda-x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\}.$

By taking trace, we showed in Section 5.1 (see (4.25) and (4.26)) that

$i\phi(g_{11})=\phi(\varepsilon_{2})D=\phi(\varepsilon_{2})\phi\left\\{\left[(\lambda-
x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{1}\varepsilon_{2}\right]^{-1}\right\\},$

and $\varepsilon_{0}^{2}=\varepsilon_{1}\varepsilon_{2}$, and

$\phi(\varepsilon_{2})=\varepsilon_{0}\frac{(e^{\sigma}-1)}{\sigma}e^{-\sigma/2},$

where $\sigma=\sigma(\lambda,\varepsilon)=(\alpha-\beta)D$. We now put

$f(\sigma)=\frac{e^{\sigma/2}+e^{-\sigma/2}}{\sigma}.$

Then we can rewrite $i\phi(g_{11})$ as

$i\phi(g_{11})=\left(\varepsilon_{0}\phi\left\\{\left[(\lambda-x_{0})(\lambda-
x_{0})^{*}+\varepsilon_{0}^{2}\right]^{-1}\right\\}\right)\cdot
f(\sigma(\lambda,\varepsilon)).$

Recall that $\phi(Q_{\varepsilon}^{\bf t}(\lambda))=q_{\varepsilon}^{\bf
t}(\lambda)$. Denote ${\bf t_{0}=(0,0,0)}$. Following definition (4.3), the
identity $i\phi(g_{11})=\phi(Q_{\varepsilon}^{\bf t}(\lambda))$ can be
rewritten as

$q_{\varepsilon_{0}(\lambda,\varepsilon)}^{\bf t_{0}}(\lambda)\cdot
f(\sigma(\lambda,\varepsilon))=q_{\varepsilon}^{\bf t}(\lambda).$

Therefore, by taking the integration of the above identity over some interval
$[\varepsilon,\delta]$, we obtain

(6.1)
$S(x_{0}+y,\lambda,\delta)-S(x_{0}+y,\lambda,\varepsilon)=2\int^{\delta}_{\varepsilon}f(\sigma(\lambda,u))q_{\varepsilon_{0}(\lambda,u)}^{\bf
t_{0}}(\lambda)\,du.$

By Proposition 4.9, we know that $\varepsilon_{0}(\lambda,\varepsilon)$ and
$D(\lambda,\varepsilon)$ are $C^{\infty}$-functions of
$(\lambda,\varepsilon)$. Hence, $(\lambda,\varepsilon)\rightarrow
f(\sigma(\lambda,\varepsilon))$ is a $C^{\infty}$-function of
$(\lambda,\varepsilon)$ over ${\Xi_{\alpha,\beta}}\times[0,\infty)$. Moreover,
by Lemma 5.1, recall that, for $\lambda\in{\Xi_{\alpha,\beta}}$, we have
$\lim_{\varepsilon\rightarrow 0^{+}}\varepsilon_{0}=s(\lambda)>0$ and
$s(\lambda)$ is a $C^{\infty}$-function of $\lambda.$ Therefore,

$\displaystyle S(x_{0}+y,\lambda,0)$
$\displaystyle=S(x_{0}+y,\lambda,\delta)-\lim_{\varepsilon\rightarrow
0^{+}}2\int_{\varepsilon}^{\delta}f(\sigma(\lambda,u))q_{\varepsilon_{0}(\lambda,u)}^{\bf
t_{0}}(\lambda)\,du$
$\displaystyle=S(x_{0}+y,\lambda,\delta)-2\int_{0}^{\delta}f(\sigma(\lambda,u))q_{\varepsilon_{0}(\lambda,u)}^{\bf
t_{0}}(\lambda)\,du.$

By the above discussion, the right hand side of the above equation is a
$C^{\infty}$-function of $\lambda\in{\Xi_{\alpha,\beta}}$. We remark that it
is clear that $S(x_{0}+y,\lambda,\delta)$ is a $C^{\infty}$-function of
$\lambda$ due to the regularity of $S(x_{0}+y,\lambda,\varepsilon)$ for
$\varepsilon>0$. Hence, $S(x_{0}+y,\lambda,0)$ is a $C^{\infty}$-function of
$\lambda$ in ${\Xi_{\alpha,\beta}}$. ∎

######  Proposition 6.2.

Given ${\bf t(\gamma)}=(\alpha,\beta,\gamma)$ and set ${\bf t}={\bf
t(0)}=(\alpha,\beta,0)$ and ${\bf t_{0}}=(0,0,0)$, let
$x_{0}\in\log^{+}(\mathcal{N})$ be an operator that is $*$-free from
$\\{g_{{}_{\alpha\beta,\gamma}},g_{{}_{\alpha\beta,0}}\\}$ with amalgamation
over $\mathcal{B}$. If the map $\Phi_{\alpha\beta,\gamma}$ is non-singular at
some $\lambda\in{\Xi_{\alpha,\beta}}$, then the map $(z,\varepsilon)\mapsto
S(x_{0}+y,z,\varepsilon)$ has a $C^{\infty}$-extension in some neighborhood of
$(\Phi_{\alpha\beta,\gamma}(\lambda),0)$. Hence,

(6.2) $\lim_{\varepsilon\rightarrow
0^{+}}p_{\Phi_{\alpha\beta,\gamma}(\lambda)}^{\bf
t(\gamma)}(\varepsilon)=\lim_{\varepsilon\rightarrow 0^{+}}p_{\lambda}^{\bf
t}(\varepsilon)=\lim_{\varepsilon\rightarrow 0^{+}}p_{\lambda}^{{\bf
t_{0}}}(\varepsilon_{0}(\lambda,\varepsilon)).$

Morever, if the map $\Phi_{\alpha\beta,\gamma}$ is non-singular at any
$\lambda\in{\Xi_{\alpha,\beta}}$, then it is also one-to-one.

###### Proof.

Again, by the definition (4.2) of the function $S(x_{0}+y,z,\varepsilon)$, we
observe that

(6.3)
$\frac{1}{2}\frac{d}{d\varepsilon}S(x_{0}+g_{{}_{\alpha\beta,\gamma}},z,\varepsilon)=q_{\varepsilon}^{\bf
t(\gamma)}(z).$

Recall that the subordination relation in (4.12) reads

$g_{11}(\lambda,\varepsilon_{1},\varepsilon_{2})=-iQ^{\bf
t(\gamma)}_{\varepsilon}(z)$

where

$z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)=\lambda+\gamma
p_{\lambda}^{\bf t_{0}}(\varepsilon_{0}).$

If $\gamma=0$, then $\Phi_{\alpha\beta,0}^{(\varepsilon)}(\lambda)=\lambda$.
By choosing $\gamma=0$ and an arbitrary eligible $\gamma$ respectively, we
have

$Q^{\bf t(\gamma)}_{\varepsilon}(z)=Q^{\bf
t}_{\varepsilon}(\lambda)=ig_{11}(\lambda,\varepsilon_{1},\varepsilon_{2}),$

By taking the trace, we have $q_{\varepsilon}^{\bf t(\gamma)}(z)=q^{\bf
t}_{\varepsilon}(\lambda)$. Hence, for
$z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)$, by integrating (6.3),
we obtain

(6.4)
$\begin{split}S(x_{0}+g_{{}_{\alpha\beta,\gamma}},z,\varepsilon)&-S(x_{0}+g_{{}_{\alpha\beta,\gamma}},z,1)\\\
&=S(x_{0}+g_{{}_{\alpha\beta,0}},\lambda,\varepsilon)-S(x_{0}+g_{{}_{\alpha\beta,0}},\lambda,1).\end{split}$

Lemma 6.1 shows that the right hand side of the above equation is a
$C^{\infty}$-function of $\lambda$ when $\varepsilon$ goes to zero. Recall
that, for $\lambda\in{\Xi_{\alpha,\beta}}$, $\lim_{\varepsilon\rightarrow
0^{+}}\varepsilon_{0}(\lambda,\varepsilon)=s(\lambda)$. We then have

$\lim_{\varepsilon\rightarrow
0^{+}}\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)=\lim_{\varepsilon\rightarrow
0^{+}}\lambda+\gamma\cdot p_{\lambda}^{\bf
t_{0}}(\varepsilon_{0})=\lambda+\gamma\cdot p_{\lambda}^{\bf
t_{0}}(s(\lambda))=\Phi_{\alpha\beta,\gamma}(\lambda).$

Thus, the assumption that $\Phi_{\alpha\beta,\gamma}$ is non-singular at
$\lambda$ implies that the map $(\lambda,\varepsilon)\mapsto(z,\varepsilon)$
is non-singular at $(\lambda,0)$. This yields that the analyticity of right
hand side of (6.4) implies that the left hand side of (6.4) has a
$C^{\infty}$-extension in some neighborhood of
$(\Phi_{\alpha\beta,\gamma}(\lambda),0)$.

Recall that $g_{21}=g_{21}(\lambda,\varepsilon_{1},\varepsilon_{2})$ is a
constant by Lemma 4.5. Again by choosing $\gamma=0$ and an arbitrary $\gamma$
respectively in the subordination relation $g_{21}=P_{z}^{\bf
t(\gamma)}(\varepsilon)=p_{z}^{\bf t(\gamma)}(\varepsilon)$, we have

$p_{z}^{\bf t(\gamma)}(\varepsilon)=p_{\lambda}^{\bf
t}(\varepsilon)=p_{\lambda}^{\bf t_{0}}(\varepsilon_{0}),$

where $z=\Phi_{\alpha\beta,\gamma}^{(\varepsilon)}(\lambda)$ and
$\varepsilon_{0}=\varepsilon_{0}(\lambda,\varepsilon)$ is given by (4.30).
Recall that

$p_{z}^{\bf
t(\gamma)}(\varepsilon)=\frac{d}{d\varepsilon}S(x_{0}+g_{{}_{\alpha\beta,\gamma}},z,\varepsilon).$

Recall that, by definitions (4.3) and (4.2), for $y=g_{\alpha\beta,\gamma}$,
we have

$\displaystyle p_{z}^{{\bf t(\gamma)}}(\varepsilon)$
$\displaystyle=\phi\left\\{(z-x_{0}-y)^{*}\left[(z-x_{0}-y)(z-x_{0}-y)^{*}+\varepsilon^{2}\right]^{-1}\right\\}$
$\displaystyle=\frac{\partial S(x_{0}+y,z,\varepsilon)}{\partial{z}}.$

Hence, the regularity of $S(x_{0}+g_{{}_{\alpha\beta,\gamma}},z,\varepsilon)$
in the neighborhood of $(\Phi_{\alpha\beta,\gamma}(\lambda),0)$ implies that

(6.5) $\lim_{\varepsilon\rightarrow
0^{+}}p_{\Phi_{\alpha\beta,\gamma}(\lambda)}^{\bf
t(\gamma)}(\varepsilon)=\lim_{\varepsilon\rightarrow 0^{+}}p_{\lambda}^{\bf
t}(\varepsilon)=\lim_{\varepsilon\rightarrow 0^{+}}p_{\lambda}^{{\bf
t_{0}}}(\varepsilon_{0}(\lambda,\varepsilon)).$

Assume now that the map $\Phi_{\alpha\beta,\gamma}$ is non-singular at any
$\lambda\in{\Xi_{\alpha,\beta}}$, then (6.2) holds at any
$\lambda\in{\Xi_{\alpha,\beta}}$. Hence, if
$\lambda_{1},\lambda_{2}\in{\Xi_{\alpha,\beta}}$ and
$\Phi_{{}_{\alpha\beta,\gamma}}(\lambda_{1})=\Phi_{{}_{\alpha\beta,\gamma}}(\lambda_{2})=z$,
we have

$z=\lambda_{1}+\gamma p_{\lambda_{1}}^{\bf
t_{0}}(s(\lambda_{1}))=\lambda_{2}+\gamma p_{\lambda_{2}}^{\bf
t_{0}}(s(\lambda_{2})).$

Then (6.5) shows

$\lim_{\varepsilon\rightarrow 0^{+}}p_{\lambda_{1}}^{\bf
t_{0}}(\varepsilon_{0}(\lambda_{1},\varepsilon))=\lim_{\varepsilon\rightarrow
0^{+}}p_{\lambda_{2}}^{\bf
t_{0}}(\varepsilon_{0}(\lambda_{2},\varepsilon))=p_{z}^{\bf t(\gamma)}(0).$

That is $p_{\lambda_{1}}^{\bf t_{0}}(s(\lambda_{1}))=p_{\lambda_{2}}^{\bf
t_{0}}(s(\lambda_{2}))$. Hence, we deduce that $\lambda_{1}=\lambda_{2}$ and
the map $\Phi_{{}_{\alpha\beta,\gamma}}$ is one-to-one in
${\Xi_{\alpha,\beta}}$. ∎

##  7\. The sum of an elliptic operator and an unbounded operator

We proceed to study the case when $\alpha=\beta$, in which case
$g_{\alpha\beta,\gamma}$ is a twisted elliptic operator. Consider now an
operator $x_{0}\in\log^{+}({\mathcal{A}})$ and a twisted elliptic operator
$g_{t,\gamma}$ that is $*$-free from $x_{0}$. By using operator-valued
subordination functions for unbounded operators from Lemma 4.3, we are able to
study the Brown measure of $x_{0}+g_{t,\gamma}$. Since $x_{0}$ could be
unbounded, we extend main results in [33]. We denote

(7.1) $X=\begin{bmatrix}0&x_{0}\\\ x_{0}^{*}&0\end{bmatrix},\qquad
Y=\begin{bmatrix}0&g_{t,\gamma}\\\ g_{t,\gamma}^{*}&0\end{bmatrix}.$

Then $X$ is affiliated with ${M}_{2}(\mathcal{A})$, and $\\{X,Y\\}$ are free
with amalgamation over $M_{2}(\mathbb{C})$ in the noncommutative probability
space $(M_{2}(\mathcal{A}),M_{2}(\phi),M_{2}(\mathbb{C}))$. By Theorem 3.1,
there exist two analytic self-maps $\Omega_{1},\Omega_{2}$ of upper half-plane
$\mathbb{H}^{+}({M}_{2}(\mathbb{C}))$ of $M_{2}(\mathbb{C})$ such that

(7.2)
$(\Omega_{1}(b)+\Omega_{2}(b)-b)^{-1}=G_{X}(\Omega_{1}(b))=G_{Y}(\Omega_{2}(b))=G_{X+Y}(b),$

for all $b\in M_{2}(\mathbb{C})$ with $\Im b>0$. In other words, subordination
relation (2.1) for $X+Y$ is still available, where $X$ could be unbounded.

We introduce a new approach via regularized Brown measure to study the
distinguished measure $\mu_{x_{0}+c_{t}}$. This allows us to prove several
regularity results for this Brown measure. In particular, we show that
$\mu_{x_{0}+c_{t}}$ is absolutely continuous with respect to the Lebesgue
measure.

###  7.1. The subordination functions

We first calculate a formula for $\Omega_{1}(\Theta(\lambda,\varepsilon))$ and
study its entries following the strategy in [33, Section 3]. The next result
is [33, Proposition 3.5] and we provide a direct proof for convenience.

######  Lemma 7.1.

Let $\mu_{1}$ be a symmetric probability measure, not necessarily compactly
supported, and let $\mu_{2}$ be the semicircular distribution with variance
$t$. Denote $\mu=\mu_{1}\boxplus\mu_{2}$. Let $\omega_{1},\omega_{2}$ be
subordination function such that

$G_{\mu}(z)=G_{\mu_{1}}(\omega_{1}(z))=G_{\mu_{2}}(\omega_{2}(z)).$

Set $w(\varepsilon)=\Im\omega_{1}(i\varepsilon)$. Then, $w=w(\varepsilon)$ is
the unique solution in $(0,\infty)$ of the following equation

(7.3) $w=\varepsilon+tw\int_{\mathbb{R}}\frac{1}{u^{2}+w^{2}}d\mu_{1}(u).$

###### Proof.

Note that the $R$-transform of $\mu_{2}$ is $R_{\mu_{2}}(z)=tz$. Since
$R_{\mu}=R_{\mu_{1}}+R_{\mu_{2}}$, we have

$G_{\mu}^{\langle-1\rangle}(z)=G^{\langle-1\rangle}_{\mu_{1}}(z)+R_{\mu_{2}}(z)=G^{-1}_{\mu_{1}}(z)+tz.$

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

(7.4) $\omega_{1}(i\varepsilon)=i\varepsilon-
tG_{\mu}(i\varepsilon)=i\varepsilon-tG_{\mu_{1}}(\omega_{1}(i\varepsilon)).$

It is clear that $\omega_{1}(i\varepsilon)$ is a pure imaginary number and
hence $\omega_{1}(i\varepsilon)=iw(\varepsilon)$. Note that,

$G_{\mu_{1}}(iw)=\int_{\mathbb{R}}\frac{1}{iw-u}d\mu_{1}(u)=-iw\int_{\mathbb{R}}\frac{1}{u^{2}+w^{2}}d\mu_{1}(u)$

by the symmetry of $\mu_{1}$. Taking the imaginary part of (7.4), we see that
$w(\varepsilon)$ satisfy (7.3).

We next show that (7.3) has a unique solution in $(0,\infty)$. Denote

(7.5)
$k(w,\varepsilon)=\frac{w}{w-\varepsilon}\left(\int_{\mathbb{R}}\frac{1}{u^{2}+w^{2}}d_{\mu_{1}}(u)\right).$

Then (7.3) is equivalent to $k(w,\varepsilon)=1/t$. Observe that $w\mapsto
k(w,\varepsilon)$ is a strictly decreasing function in $(0,\infty)$. Hence,
(7.3) has a unique solution in $(0,\infty)$ ∎

######  Definition 7.2.

For any probability measure $\mu$ on $\mathbb{R}$, we denote by
$\widetilde{\mu}$ the symmetrization of $\mu$ defined by

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

for any Borel measurable set $B$ on $\mathbb{R}$. Given
$\lambda\in\mathbb{C}$, let $\mu_{1}=\widetilde{\mu}_{|\lambda-x_{0}|}$ and
let $\mu_{2}$ be the semicircular distribution with variance $t$. Let
$\omega_{1}$ be the subordination function (depending on $\lambda$) such that

$G_{\mu_{1}\boxplus\mu_{2}}(z)=G_{\mu_{1}}(\omega_{1}(z)).$

We denote $w(\varepsilon;\lambda,t)=\Im\omega_{1}(i\varepsilon)$.

We choose

$b=\Theta(z,\varepsilon)=\begin{bmatrix}i\varepsilon&z\\\
\overline{z}&i\varepsilon\end{bmatrix}$

where $\varepsilon>0$ and $z\in\mathbb{C}$. The following result extends [33,
Theorem 3.8] to unbounded operator $x_{0}$.

######  Theorem 7.3.

Let $x_{0}\in\log^{+}(\mathcal{A})$ be an operator that is $*$-free from the
elliptic operator $g_{t,\gamma}$. For any $\varepsilon>0$ and
$z\in\mathbb{C}$, we set

(7.6)
$\lambda=z-\gamma\cdot\phi\bigg{(}(z-x_{0}-g_{t,\gamma})^{*}\big{(}(z-x_{0}-g_{t,\gamma})(z-x_{0}-g_{t,\gamma})^{*}+\varepsilon^{2}\big{)}^{-1}\bigg{)}.$

Then
$\Omega_{1}(\Theta(z,\varepsilon))=\Theta(\lambda,w(\varepsilon;\lambda,t))$.
That is,

(7.7) $\displaystyle\Omega_{1}\left(\begin{bmatrix}i\varepsilon&z\\\
\overline{z}&i\varepsilon\end{bmatrix}\right)=\begin{bmatrix}iw(\varepsilon;\lambda,t)&\lambda\\\
\overline{\lambda}&iw(\varepsilon;\lambda,t)\end{bmatrix}.$

The subordination relation
$G_{X+Y}(\Theta(z,\varepsilon))=G_{X}(\Omega_{1}(\Theta(z,\varepsilon)))$ is
expressed as

(7.8)
$\displaystyle\mathbb{E}\left(\begin{bmatrix}i\varepsilon&z-(x_{0}+g_{t,\gamma})\\\
\overline{z}-(x_{0}+g_{t,\gamma})^{*}&i\varepsilon\end{bmatrix}^{-1}\right)=\mathbb{E}\left(\begin{bmatrix}iw(\varepsilon;\lambda,t)&\lambda-
x_{0}\\\
\overline{\lambda}-x_{0}^{*}&iw(\varepsilon;\lambda,t)\end{bmatrix}^{-1}\right),$

which is also equivalent to

(7.9)
$\displaystyle\varepsilon\phi\bigg{(}\big{(}({z}-x_{0}-g_{t,\gamma})({z}-x_{0}-g_{t,\gamma})^{*}+\varepsilon^{2}\big{)}^{-1}\bigg{)}$
$\displaystyle\qquad\qquad=w(\varepsilon;\lambda,t)\phi\bigg{(}\big{(}(\lambda-
x_{0})(\lambda-x_{0})^{*}+w(\varepsilon;\lambda,t)^{2}\big{)}^{-1}\bigg{)},$
$\displaystyle\phi\bigg{(}({z}-x_{0}-g_{t,\gamma})^{*}\big{(}({z}-x_{0}-g_{t,\gamma})({z}-x_{0}-g_{t,\gamma})^{*}+\varepsilon^{2}\big{)}^{-1}\bigg{)}$
$\displaystyle\qquad\qquad=\phi\bigg{(}(\lambda-x_{0})^{*}\big{(}(\lambda-
x_{0})(\lambda-x_{0})^{*}+w(\varepsilon;\lambda,t)^{2}\big{)}^{-1}\bigg{)}.$

###### Proof.

Apply Lemma 4.3 to the case $\alpha=\beta=t$. In this case,

$Q_{\varepsilon}^{\bf t}(z)=\widetilde{Q}^{\bf t}_{\varepsilon}(z)$

and hence $\varepsilon_{1}=\varepsilon_{2}\in(\varepsilon,\infty)$. Moreover,
by (4.12) and the definition (4.7), then $\varepsilon_{1}$ can be expressed as

$\varepsilon_{1}=\varepsilon+\varepsilon_{1}\phi\bigg{(}\big{(}(\lambda-
x_{0})(\lambda-x_{0})^{*}+\varepsilon_{1}^{2}\big{)}^{-1}\bigg{)}.$

This shows that $\varepsilon_{1}=\varepsilon_{2}=w(\varepsilon;\lambda,t)$ by
(7.3) and Definition 7.2. ∎

Recall that the open set $\Xi_{t}$ is defined in (1.3) as

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

######  Definition 7.4.

For $\lambda\in\Xi_{t}$, let $w(0;\lambda,t)$ be the unique solution
$w\in(0,\infty)$ to the following equation

(7.10)
$\phi\left[\big{(}(x_{0}-\lambda)^{*}(x_{0}-\lambda)+w(0;\lambda,t)^{2}\big{)}^{-1}\right]=\frac{1}{t}.$

For $\lambda\in\mathbb{C}\backslash\Xi_{t}$, set $w(0;\lambda,t)=0$.

We define the function $\Phi^{(\varepsilon)}_{t,\gamma}$ on $\mathbb{C}$ by

(7.11) $\Phi^{(\varepsilon)}_{t,\gamma}(\lambda)=\lambda+\gamma\cdot
p_{\lambda}^{(0)}(w(\varepsilon;\lambda,t)),\qquad\lambda\in\mathbb{C}$

where

$p_{\lambda}^{(0)}(w(\varepsilon;\lambda,t))=\phi\bigg{[}(\lambda-
x_{0})^{*}\big{(}(\lambda-x_{0})(\lambda-
x_{0})^{*}+w(\varepsilon;\lambda,t)^{2}\big{)}^{-1}\bigg{]}.$

This can also be rewritten as

$\Phi^{(\varepsilon)}_{t,\gamma}(\lambda)=\lambda+\gamma\cdot\frac{\partial
S}{\partial\lambda}(x_{0},\lambda,w(\varepsilon;\lambda,t)).$

The map $\Phi_{t,\gamma}$ is defined as

$\Phi_{t,\gamma}(\lambda)=\lambda+\gamma\cdot
p_{\lambda}^{(0)}(w(0;\lambda,t)),\qquad\lambda\in\mathbb{C}.$

The following result is a special case of Lemma 4.10 by letting
$\alpha=\beta=t$.

######  Corollary 7.5.

The map $\Phi^{(\varepsilon)}_{t,\gamma}$ is a homeomorphism of the complex
plane for any $\varepsilon>0.$ Its inverse map is

(7.12) $J^{(\varepsilon)}_{t,\gamma}(z)=z-\gamma\cdot
p_{z}^{(t,\gamma)}(\varepsilon),$

where

$p_{z}^{(t,\gamma)}(\varepsilon)=\phi\bigg{(}({z}-x_{0}-g_{t,\gamma})^{*}\big{(}({z}-x_{0}-g_{t,\gamma})({z}-x_{0}-g_{t,\gamma})^{*}+\varepsilon^{2}\big{)}^{-1}\bigg{)}.$

######  Lemma 7.6.

For any $\varepsilon>0$, the function $\lambda\mapsto
w(\varepsilon;\lambda,t)$ is a $C^{\infty}$ function. The function
$\lambda\mapsto w(0;\lambda,t)$ is a continuous function of $\lambda$ on
$\mathbb{C}$ and is a $C^{\infty}$ function in the open set $\Xi_{t}$.

Moreover, the function $w(\varepsilon;\lambda,t)$ converges uniformly to
$w(0;\lambda,t)$ in any compact subset of $\mathbb{C}$ as $\varepsilon$ tends
to zero.

###### Proof.

By Definition 7.2, for any $\varepsilon>0$, the function
$w(\varepsilon;\lambda,t)$ is the imaginary part of the subordination function
parameterized by $\lambda$. In particular,
$w(\varepsilon;\lambda,t)=\omega_{1}(i\varepsilon)>\varepsilon$, where
$\omega_{1}$ is the subordination as in Definition 7.2. By Lemma 7.1, the
function $w(\varepsilon;\lambda,t)$ is the unique $w$ such that

(7.13) $\displaystyle w$
$\displaystyle=\varepsilon+tw\int_{\mathbb{R}}\frac{1}{u^{2}+w^{2}}d\,\widetilde{\mu}_{|\lambda-
x_{0}|}(u)$ $\displaystyle=\varepsilon+tw\phi[((\lambda-x_{0})^{*}(\lambda-
x_{0})+w^{2})^{-1}].$

Following (7.5), we rewrite it as

(7.14) $\displaystyle k(w,\varepsilon)$
$\displaystyle=\frac{w}{w-\varepsilon}\left(\int_{\mathbb{R}}\frac{1}{u^{2}+w^{2}}d\,\widetilde{\mu}_{|\lambda-
x_{0}|}(u)\right)$ $\displaystyle=\frac{w}{w-\varepsilon}\phi[((\lambda-
x_{0})^{*}(\lambda-x_{0})+w^{2})^{-1}]=\frac{1}{t}.$

Note that $\frac{\partial}{\partial w}k(w,\varepsilon)<0$ for $w>\varepsilon$
and $\lambda\mapsto k(w,\varepsilon)$ is a smooth function. By implicit
function theorem, $\lambda\mapsto w(\varepsilon;\lambda,t)$ is a $C^{\infty}$
function. Similarly, for $\lambda\in\Xi_{t}$, the function $w(0;\lambda,t)$ is
determined by (7.10), which yields that $\lambda\mapsto w(0;\lambda,t)$ is a
$C^{\infty}$ function in the open set $\Xi_{t}$ by applying the implicit
function theorem.

Recall that $w(\varepsilon;\lambda,t)$ is the unique solution of
$k(w,\varepsilon)=1/t$ as in (7.14). We observe that

$k(w(\varepsilon_{1};\lambda,t),\varepsilon_{1})<k(w(\varepsilon_{1};\lambda,t),\varepsilon_{2})$

if $0<\varepsilon_{1}<\varepsilon_{2}<w(\varepsilon_{1};\lambda,t)$. Hence,
$w(\varepsilon_{1};\lambda,t)<w(\varepsilon_{2};\lambda,t)$ since
$\frac{\partial}{\partial w}k(w,\varepsilon)<0$. By Dini’s theorem, the
convergence of $w(\varepsilon;\lambda,t)$ to $w(0;\lambda,t)$ is uniform in
any compact subset of $\mathbb{C}$. ∎

###  7.2. The Fuglede-Kadison determinant formula

By using Theorem 7.3, we can apply exactly the same methods as in [33, Lemma
3.11, Theorem 3.12] to obtain a formula for
$\Delta\big{(}(x_{0}+g_{t,\gamma})^{*}(x_{0}+g_{t,\gamma})+\varepsilon^{2}\big{)}$.
We only state a result for circular operator below (See [6, Section 5.2] for
an alternative proof).

######  Theorem 7.7.

[33, Theorem 3.12] For $\lambda\in\mathbb{C}$, we have the following Fuglede-
Kadison determinant formulas.

  1. (1)

If $\lambda\in\Xi_{t}$, then

(7.15) $\displaystyle\Delta\big{(}x_{0}+c_{t}-\lambda\big{)}^{2}$
$\displaystyle={\Delta\big{(}(x_{0}-\lambda)^{*}(x_{0}-\lambda)+w(0;\lambda,t)^{2}\big{)}}$
$\displaystyle\qquad\qquad\times{\exp\bigg{(}-\frac{(w(0;\lambda,t))^{2}}{t}\bigg{)}}.$

  2. (2)

If $\lambda\notin\Xi_{t}$, then

(7.16) $\Delta(x_{0}+c_{t}-\lambda)=\Delta(x_{0}-\lambda).$

The following proof is adapted from [33, Theorem 4.6]. We include a proof for
the reader’s convenience.

######  Corollary 7.8.

For any $\lambda\in\mathbb{C}$, we have

$\log\Delta(x_{0}+c_{t}-\lambda)>-\infty.$

The Brown measure $\mu_{x_{0}+c_{t}}$ has no atom and the support of
$\mu_{x_{0}+c_{t}}$ is contained in the closure $\overline{\Xi_{t}}$.

###### Proof.

If $\lambda\in\Xi_{t}$, since $w(0;\lambda,t)>0$, it follows from (7.15) that
$\log\Delta(x_{0}+c_{t}-\lambda)>-\infty$. If $\lambda\notin\Xi_{t}$, then
$\phi(|x_{0}-\lambda|^{-2})\leq 1/t$, and hence

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

Therefore, $\lambda\notin\Xi_{t}$,

$\log\Delta(x_{0}+c_{t}-\lambda)=\log\Delta(x_{0}-\lambda)>-\infty$

By [17, Proposition 2.14, 2.16], we deduce that
$\mu_{x_{0}+c_{t}}(\\{\lambda\\})=\mu_{x_{0}+c_{t}-\lambda}(\\{0\\})=0$ for
any $\lambda\in\mathbb{C}$.

The Brown measure of $\mu_{(x_{0}-\lambda)^{-1}}$ is the pushforward measure
of $\mu_{x_{0}}$ under the map $z\mapsto(z-\lambda)^{-1}$. By Weil’s
inequality for operators in $L^{p}(\mathcal{A})$ (see [17, Theorem 2.19]), we
have

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

Consequently, if $\lambda\notin\overline{\Xi_{t}}$, then

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

in some neighborhood of $\lambda$. This implies that $\lambda$ is not in the