---
abstract: 'We investigate some bounded linear operators $T$ on a Hilbert space which satisfy the condition $|T|\le |{\rm Re} T|$. We describe the maximum invariant subspace for a contraction $T$ on which $T$ is a partial isometry to obtain that, in certain cases, the above condition ensures that $T$ is self-adjoint.  In other words we show that the Fong-Tsui conjecture holds for partial isometries, contractive quasi-isometries, or $2$-quasi-isometries, and Brownian isometries of positive covariance, or even for a more general class of operators.' 
address:
- ' UFR de Mathématiques, Laboratoire Paul Painlevé, UMR CNRS 8524, Université Lille 1, 59655 Villeneuve d’Ascq Cedex, France'
- 'Department of Mathematics, “Lucian Blaga” University of Sibiu, Dr. Ion Raţiu 5-7, Sibiu, 550012, Romania'
author:
- Mostafa Mbekhta
- Laurian Suciu
title: |
    Partial isometries and the conjecture\
    of C. K. Fong and S. K. Tsui
---

Introduction and terminology
============================

For two complex Hilbert spaces ${\mathcal{H}}$ and ${\mathcal{K}}$ we denote by $\mathcal{B}({\mathcal{H}},{\mathcal{K}})$ the Banach space of all bounded linear operators from ${\mathcal{H}}$ into ${\mathcal{K}}$, and ${\mathcal{B(H)}}=\mathcal{B}({\mathcal{H}},{\mathcal{H}})$ considered as a Banach algebra with $I=I_{{\mathcal{H}}}$ the identity operator on ${\mathcal{H}}$. For $T\in \mathcal{B}({\mathcal{H}},{\mathcal{K}})$, ${\mathcal{R}}(T)$ and ${\mathcal{N}}(T)$ stand for the range and the null-space of $T$, respectively.  For a subspace $\mathcal{G}$ of ${\mathcal{H}}$ its closure is denoted by $\overline{\mathcal{G}}$. As usually, a closed subspace $\mathcal{G}$ of ${\mathcal{H}}$ is invariant (reducing) for $T$ if $T\mathcal{G} \subset \mathcal{G}$ (and $T^* \mathcal{G} \subset \mathcal{G})$. Also, $T^* \in \mathcal{B}({\mathcal{K}},{\mathcal{H}})$ stands for the adjoint operator of $T$, while the orthogonal projection associated to a closed subspace $\mathcal{G}$ of ${\mathcal{H}}$ is denoted by $P_{\mathcal{G}}$, that is $P_{\mathcal{G}} \in \mathcal{B}({\mathcal{H}})$ with $P_{\mathcal{G}}^2=P_{\mathcal{G}}=P_{\mathcal{G}}^*$.

An operator $T\in \mathcal{B}({\mathcal{H}},{\mathcal{K}})$ is a [*contraction*]{} if $\|T\|\le 1$, and $T$ is a [*partial isometry*]{} when $T^*T$ is an orthogonal projection.  In particular $T$ is an [*isometry*]{} if $T^*T=I_{{\mathcal{H}}}$, and [*unitary*]{} if $T$ is a surjective isometry.  A unitary operator $U \in {\mathcal{B(H)}}$ with $U^*=U$ is called a [*symmetry*]{}.  A contraction $T$ is called [*pure*]{} if $\|Tx\|<\|x\|$ for any $x\in {\mathcal{H}}$, $x\neq 0$.

Also, we say that an operator $T \in {\mathcal{B(H)}}$ is a [*$m$-quasi-isometry*]{} for some integer $m\ge 1$, if $T|_{{\mathcal{R}}(T^m)}$ is an isometry.  The $1$-quasi-isometries are shortly called [*quasi-isometries*]{}, such operators being firstly studied \[18, 19\] and latterly in \[20, 21\], and other articles.  The generalization to $m$-quasi-isometries for $m\ge 2$ appear in \[8, 15, 16\].  It was proved in \[16\] that a quasi-isometric contraction $T$ is subnormal, that is it has a normal extension, hence $T$ is hyponormal that is $TT^* \le T^*T$.

As usually, for $T\in {\mathcal{B(H)}}$ we denote the module of $T$ by $|T|=(T^*T)^{1/2}$, and the real part of $T$ by ${\rm Re}T=\frac{1}{2}(T+T^*)$.

An interesting conjecture formulated by C. K. Fong and S. K. Tsui in \[11\] says that if $T$ satisfies the condition $$\begin{aligned}
\label{ec11}
|T|\le |{\rm Re}T|\end{aligned}$$ then $T$ is self-adjoint. 

This conjecture was partially proved in \[11\] in finite dimensional Hilbert spaces, in finite von Neumann algebras, and for compact operators in any Hilbert space. 

Originally, C. K. Fong and V. Istrăţescu proved in \[10\] that $T$ is self-adjoint if and only if $|T|^2\le ({\rm Re}T)^2$. In particular, by Lemma 1.5 \[10\] it follows that the Fong-Tsui conjecture holds for hyponormal operators. 

Recently, M. H. Mortad shows in \[17\] that this conjecture is also true when $T$ commutes with the partial isometry $U$ which appears in the polar decomposition of ${\rm Re}T$ ($U$ being a symmetry on $\overline{{\mathcal{R}}({\rm Re}T)}$). 

A difficulty for the solution of this conjecture is the fact that $|T|$ and $|{\rm Re}T|$ cannot be easily expressed in terms of $T$ (and $T^*$), by contrast to $|T|^2$ and $({\rm Re}T)^2$. An idea is to transfer the conjecture on some reducing or just invariant parts of $T$ which may be easily expressed in $T, T^*$, and to investigate the condition on each such part. 

The purpose of this paper is to show that the Fong-Tsui conjecture is also true for some operators which are related to partial isometries, as well as those before. 

In Section 2 we describe the maximum invariant subspace for a contraction $T$ on which $T$ is a partial isometry (Theorem \[te21\]).  We use some block matrix forms for $T$, and we refer to the case when this subspace is just ${\mathcal{N}}(T^*T-(T^*T)^2)$, and also to a special case when this latter subspace is reducing for $T$.

In Section 3 we give the main result (Theorem \[te31\]) which asserts that for a contraction $T$ which satisfies the condition the fixed points of $T^*T$, $TT^*$ and $|{\rm Re}T|$ coincide, and their subspace reduces $T$ to a symmetry.  So, in this case the invariant partial isometric part of $T$ in ${\mathcal{H}}$ is ${\mathcal{N}}(I-T^*T)\oplus {\mathcal{N}}(T)$, which does not reduce $T$, in general. 

As consequences, we derive that the Fong-Tsui conjecture holds for partial isometries, quasi-isometric or $2$-quasi-isometric contractions.  In the case of an $m$-quasi-isometric contraction $T$ with $m\ge 3$ we obtain that $T$ is a symmetry on $\overline{{\mathcal{R}}(T^m)}$, this subspace being even the unitary part of $T$ in ${\mathcal{H}}$. In this case $T=S\oplus Q$ with $S$ a symmetry on $\overline{{\mathcal{R}}(T^m)}$ and $Q^m=0$. So $T=T^*$ if and only if $Q=0$.

Another special class of non-contractive operators for which the Fong-Tsui conjecture can be shown true is given by the Brownian isometries.  This class containing the Brownian unitaries was extensively studied by J. Agler and M. Stankus in \[1-3\].  Such operators arise naturally in the context of $2$-isometries, that is of operators $T$ on ${\mathcal{H}}$ satisfying the identity $T^{*2}T^2-2T^*T+I=0$. According to \[2, Proposition 5.37\], a $2$-isometry $T$ on ${\mathcal{H}}$ is a Brownian isometry of covariance $\sigma >0$ if $\sigma ^2 =\|T^*T-I\|$ and, with respect to a decomposition ${\mathcal{H}}={\mathcal{H}}_0\oplus {\mathcal{H}}_1$, $T$ has a block matrix form $$\begin{aligned}
\label{ec12}
T=
\begin{pmatrix}
V & \sigma E\\
0 & U
\end{pmatrix},\end{aligned}$$ where $V$ is an isometry on ${\mathcal{H}}_0$, $E\in \mathcal{B}({\mathcal{H}}_1,{\mathcal{H}}_0)$ is an injective contraction with ${\mathcal{R}}(E)\subset {\mathcal{N}}(V^*)$, while $U$ is unitary on ${\mathcal{H}}_1$ such that $UE^*E =E^*EU$.

Finally, we remark that, under the condition for a contraction $T$, $T^*$ has a similar block matrix form like that of $T$ (given by Theorem \[te31\]), without imposing the condition for $T^*$. But these matrix representations of $T$ and $T^*$ cannot lead to a symmetric condition (in $T$ and $T^*$) as $T=T^*$, from a non-symmetric one as .  The Fong-Tsui conjecture remains an interesting open problem, in particular for pure contractions. 

The invariant partial isometric part of a contraction
=====================================================

The main result of this section is the following

\[te21\] For every contraction $T$ on ${\mathcal{H}}$ there exists the maximum subspace ${\mathcal{M}}$ which is invariant for $T$, on which $T$ is a partial isometry.  More precisely, one has ${\mathcal{N}}(T) \subset {\mathcal{M}}\subset {\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)$, and $T$ has the block matrix form $$\begin{aligned}
\label{ec21}
T=
\begin{pmatrix}
W & R\\
0 & Q
\end{pmatrix}\end{aligned}$$ on ${\mathcal{H}}={\mathcal{M}}\oplus {\mathcal{M}}^{\perp} $, where $W$ is a partial isometry on ${\mathcal{M}}$, $R\in \mathcal{B}({\mathcal{M}}^{\perp},{\mathcal{M}})$ is a contraction with $W^*R=0$, and $Q$ is a contraction on ${\mathcal{M}}^{\perp}$ such that ${\mathcal{N}}(I-Q^*Q)\subset {\mathcal{N}}(I-T^*T)$.

Moreover, we have $$\begin{aligned}
\label{ec22}
{\mathcal{N}}(T)\oplus {\mathcal{N}}(I-T^*T)={\mathcal{M}}\oplus {\mathcal{N}}(I-Q^*Q)\end{aligned}$$ if and only if ${\mathcal{N}}(I-T^*T) \cap {\mathcal{M}}^{\perp} \subset {\mathcal{N}}(R)$.

The required subspace ${\mathcal{M}}$ need to satisfy the inclusions $${\mathcal{N}}(T)\subset {\mathcal{M}}\subset {\mathcal{N}}(T^*T-(T^*T)^2)={\mathcal{N}}(T)\oplus {\mathcal{N}}(I-T^*T).$$ But with respect to the decomposition $${\mathcal{H}}={\mathcal{N}}(T)\oplus {\mathcal{N}}(I-T^*T)\oplus {\mathcal{H}}'$$ where ${\mathcal{H}}'=\overline{{\mathcal{R}}(T^*T-(T^*T)^2)}$, $T$ has a block matrix form $$\begin{aligned}
\label{ec23}
T=
\begin{pmatrix}
0 & T_0 & T_0'\\
0 & T_1 & T_1'\\
0 & T_2 & T_2'
\end{pmatrix},\end{aligned}$$ with some appropriate contractions $T_j$ and $T_j'$. In particular one has $T_2=P_{{\mathcal{H}}'}T|_{{\mathcal{N}}(I-T^*T)}$, while the subspace ${\mathcal{H}}_1:={\mathcal{N}}(T_2)$ is invariant for $T_2$. Therefore ${\mathcal{N}}(T)\oplus {\mathcal{H}}_1$ is invariant for $T$, and from the above representation of $T$ we can infer another block matrix form for $T$ on the decomposition ${\mathcal{H}}={\mathcal{H}}_0\oplus {\mathcal{H}}_1 \oplus {\mathcal{H}}_2$, where ${\mathcal{H}}_0={\mathcal{N}}(T)$ and ${\mathcal{H}}_2:= [{\mathcal{N}}(I-T^*T)\ominus {\mathcal{H}}_1] \oplus {\mathcal{H}}'$, as follows $$\begin{aligned}
\label{ec24}
T=
\begin{pmatrix}
0 & W_0 & R_0\\
0 & W_1 & R_1\\
0 & 0 & Q
\end{pmatrix}.\end{aligned}$$ Here all operators are contractions between the corresponding subspaces of the decomposition of ${\mathcal{H}}$ and, in particular, for the operator $W\in \mathcal{B}({\mathcal{H}}_0,{\mathcal{H}}_1)$ with the block matrix form $$W=
\begin{pmatrix}
0 & W_0\\
0 & W_1
\end{pmatrix},$$ we have $W_0 =T_0|_{{\mathcal{H}}_1}=P_{{\mathcal{H}}_0}T|_{{\mathcal{H}}_1}$, $W_1=T_1|_{{\mathcal{H}}_1}=P_{{\mathcal{H}}_1}T|_{{\mathcal{H}}_1}$. Also, by considering the operator $R= \begin{pmatrix} R_0\\ R_1 \end{pmatrix} : {\mathcal{H}}_2 \to {\mathcal{H}}_0 \oplus {\mathcal{H}}_1 =: {\mathcal{M}}$, we can write $T$ in the form $$T=
\begin{pmatrix}
W & R\\
0 & Q
\end{pmatrix}$$ on ${\mathcal{H}}={\mathcal{M}}\oplus {\mathcal{M}}^{\perp}$. This representation gives $$T^*T=
\begin{pmatrix}
W^*W & W^*R\\
R^*W & R^*R+Q^*Q
\end{pmatrix},$$ and since ${\mathcal{H}}_1 \subset {\mathcal{N}}(I-T^*T)$ one has $W^*W|_{{\mathcal{H}}_1}=I_{{\mathcal{H}}_1}$. This means $$W^*W=
\begin{pmatrix}
0 & 0\\
0 & W_0^*W_0+W_1^*W_1
\end{pmatrix}
=0\oplus I_{{\mathcal{H}}_1},$$ that is $W$ is a partial isometry on ${\mathcal{M}}$. On the other hand, we have $$TT^*=
\begin{pmatrix}
WW^*+RR^* & RQ^*\\
QR^* & QQ^*
\end{pmatrix}
\le I,$$ whence it follows $RR^* \le I_{{\mathcal{M}}}-P_{{\mathcal{R}}(W)}=P_{{\mathcal{N}}(W^*)}$, which means $W^*R=0$.

To show that ${\mathcal{M}}$ is the maximum invariant subspace for $T$ in ${\mathcal{H}}$ on which $T$ is a partial isometry, let us consider another such subspace ${\mathcal{M}}'\subset {\mathcal{H}}$. So, $T$ has a similar block matrix form on ${\mathcal{H}}={\mathcal{M}}'\oplus ({\mathcal{M}}')^{\perp}$, namely $$T=
\begin{pmatrix}
W' & R'\\
0 & Q'
\end{pmatrix}$$ with $W'$ a partial isometry on ${\mathcal{M}}'$ satisfying (as above) $W^{'*}R'=0$. Then $$T^*T=
\begin{pmatrix}
W'^*W' & 0\\
0 & R'^*R'+Q'^*Q'
\end{pmatrix}$$ and this gives $${\mathcal{N}}(W')\subset {\mathcal{N}}(T) \subset {\mathcal{M}}.$$ As $W'$ is a partial isometry we have the inclusion $${\mathcal{R}}(W'^*)={\mathcal{N}}(I-W'^*W')\subset {\mathcal{N}}(I-T^*T) \cap {\mathcal{M}}'$$ and, in fact, the equality holds here.  Indeed, if $x=T^*Tx\in {\mathcal{M}}'$ and $x\perp {\mathcal{R}}(W'^*)$ then $0=W'x=TT^*x$ so $Tx=0$ that is $x=0$. Hence we get $${\mathcal{M}}'={\mathcal{N}}(W') \oplus {\mathcal{R}}(W'^*) \subset {\mathcal{N}}(T) \oplus {\mathcal{R}}(W'^*),$$ which implies $$T{\mathcal{R}}(W'^*) =W'{\mathcal{R}}(W'^*) \subset {\mathcal{M}}'\subset {\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)\cap {\mathcal{M}}'.$$ We infer (by using the operator $T_2$ in ) that $$T_2{\mathcal{R}}(W'^*)=P_{{\mathcal{H}}'}T|_{{\mathcal{N}}(I-T^*T)\cap {\mathcal{M}}'}=0,$$ which means ${\mathcal{R}}(W'^*)\subset {\mathcal{N}}(T_2)={\mathcal{H}}_1$. Conclude (by an above inclusion) that $${\mathcal{M}}'\subset {\mathcal{N}}(T) \oplus {\mathcal{H}}_1 ={\mathcal{M}},$$ that is ${\mathcal{M}}$ has the required maximality property. 

Now we prove the other properties concerning the operator $Q$. Firstly, if $x\in {\mathcal{N}}(I-Q^*Q)$ then $$\|x\|=\|Qx\|=\|P_{{\mathcal{M}}^{\perp}}Tx\| \le \|Tx\|\le \|x\|,$$ so $x\in {\mathcal{N}}(I-T^*T)$. This gives ${\mathcal{N}}(I-Q^*Q) \subset {\mathcal{N}}(I-T^*T).$

Clearly, the equality is equivalent to the following : $${\mathcal{N}}(I-T^*T) ={\mathcal{H}}_1 \oplus {\mathcal{N}}(I-Q^*Q).$$ Let $x\in {\mathcal{N}}(I-T^*T)\cap {\mathcal{M}}^{\perp}$, so $x=T^*Tx$. Then $$Qx=P_{{\mathcal{M}}^{\perp}}TT^*Tx=P_{{\mathcal{M}}^{\perp}}Tx$$ and we have $$\begin{aligned}
\|x\|^2 =\|Tx\|^2&=&\|P_{{\mathcal{M}}}Tx\|^2 + \|P_{{\mathcal{M}}^{\perp}}Tx\|^2\\
&=& \|Rx\|^2 + \|Qx\|^2.\end{aligned}$$ Hence $x\in {\mathcal{N}}(I-Q^*Q)$ if and only if $Rx=0$. This proves the last assertion of theorem, having in view the inclusion of ${\mathcal{N}}(I-Q^*Q)$ into ${\mathcal{N}}(I-T^*T)$ already quoted. 

\[co22\] Let $T$ be a contraction on ${\mathcal{H}}$. Then the subspace ${\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)$ is invariant for $T$ if and only if $Q$ (in ) is a pure contraction on ${\mathcal{M}}^{\perp}$.

If ${\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)$ is invariant for $T$ then it is reduced to ${\mathcal{M}}$, hence ${\mathcal{N}}(I-Q^*Q)=\{0\}$ by Theorem \[te21\].  This means that the contraction $Q$ is pure.  Conversely, assuming $Q$ pure, we have also $Q^* =T^*|_{{\mathcal{M}}^{\perp}}$ pure.  So, ${\mathcal{M}}^{\perp}= \overline{{\mathcal{R}}(I-QQ^*)} \subset \overline{{\mathcal{R}}(I-TT^*)}$ which implies $$T{\mathcal{N}}(I-T^*T) ={\mathcal{N}}(I-TT^*) \subset {\mathcal{M}}\subset {\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T).$$ Hence ${\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)$ is invariant for $T$.

\[re23\] Recall \[14\] that the maximum subspace invariant for a contraction $T$ on which $T$ is an isometry is ${\mathcal{N}}(I-S_T)$, where $S_T$ is the asymptotic limit of $T$, defined as the strong limit of the powers $T^{*n}T^n$, $n\ge 1$. Therefore ${\mathcal{N}}(I-S_T) \subset {\mathcal{M}}\cap {\mathcal{N}}(I-T^*T)$ and, in general, the inclusion is strict, because the powers $T^n$ are not always partial isometries on ${\mathcal{M}}$.

In the case that ${\mathcal{M}}={\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)$ and ${\mathcal{N}}(T^*)$ is invariant for $T$ (that is ${\mathcal{N}}(T^*) \subset {\mathcal{N}}(T)$), and if ${\mathcal{M}}^* \subset {\mathcal{H}}$ is the corresponding subspace for $T^*$ given by Theorem \[te21\], then ${\mathcal{M}}^* \subset {\mathcal{N}}(T^*)\oplus {\mathcal{N}}(I-TT^*)={\mathcal{N}}(T^*) \oplus T{\mathcal{N}}(I-T^*T) \subset {\mathcal{M}}$. In addition, if ${\mathcal{N}}(I-TT^*)$ is invariant for $T$ then $$T{\mathcal{M}}^* ={\mathcal{N}}(T^*) \oplus T{\mathcal{N}}(I-TT^*) \subset {\mathcal{N}}(T^*) \oplus {\mathcal{N}}(I-TT^*)={\mathcal{M}}^*,$$ hence ${\mathcal{M}}^* $ reduces $T$ to a partial isometry. 

Clearly, the maximum subspace which reduces $T$ to a partial isometry exists always, but it is different of ${\mathcal{M}}\cap {\mathcal{M}}^*$, in general.  Its structure is more complicated, and will not be given here. 

In the following section we see that the subspace ${\mathcal{M}}$ has the form from Corollary \[co22\] with ${\mathcal{N}}(I-T^*T)$ reducing for $T$, under the condition , but ${\mathcal{M}}$ does not reduce $T$, in general. 

On the Fong-Tsui conjecture
===========================

Remark firstly that in the Fong-Tsui conjecture one can suppose that $T$ is a contraction, because the condition works simultaneously for $T$ and $T|_{\|T\|}$.

Concerning the structure of such a contraction we have the following main result. 

\[te31\] Let $T$ be a contraction on ${\mathcal{H}}$ satisfying the condition .  Then $$\begin{aligned}
\label{ec31}
{\mathcal{N}}(I-T^*T)={\mathcal{N}}(I-TT^*)={\mathcal{N}}(I-|{\rm Re}T|)\end{aligned}$$ and this subspace reduces $T$ to a symmetry.  Also, we have $$\begin{aligned}
\label{ec32}
{\mathcal{N}}({\rm Re}T)={\mathcal{N}}(T)\cap {\mathcal{N}}(T^*), \quad {\mathcal{N}}(T^*)={\mathcal{N}}({\rm Re}T)\oplus {\mathcal{N}}(T^*|_{\overline{{\mathcal{R}}(T^*)}}).\end{aligned}$$

Moreover, one has ${\mathcal{N}}(T)={\mathcal{N}}({\rm Re}T)$ if and only if $T=U\oplus Z$ with respect to a decomposition ${\mathcal{H}}=\mathcal{G} \oplus \mathcal{G}^{\perp}$, where $U$ is a symmetry, and $Z$ is a pure contraction satisfying the condition . 

Consider the block matrix of $T$ that is $$T=
\begin{pmatrix}
0 & W_0 & R_0\\
0 & W_1 & R_1\\
0 & 0 & Q
\end{pmatrix}$$ on ${\mathcal{H}}={\mathcal{H}}_0 \oplus {\mathcal{H}}_1 \oplus {\mathcal{H}}_2$, where ${\mathcal{H}}_0:= {\mathcal{N}}(T)$, ${\mathcal{H}}_1:= {\mathcal{N}}(P'T|_{{\mathcal{N}}(I-T^*T)})$, while $P'$ is the orthogonal projection onto ${\mathcal{H}}':= \overline{{\mathcal{R}}(T^*T-(T^*T)^2)}$, and ${\mathcal{H}}_2= [{\mathcal{N}}(I-T^*T) \ominus {\mathcal{H}}_1] \oplus {\mathcal{H}}'$. In addition, from the proof of Theorem \[te21\] we have the relations $$\begin{aligned}
\label{ec33}
W_0^*W_0 + W_1^*W_1=I, \quad W_0^*R_0 + W_1^*R_1=0,\end{aligned}$$ and $$\begin{aligned}
\label{ec34}
W_0W_0^* +R_0R_0^*\le I, \quad W_1W_1^*+R_1R_1^*\le I.\end{aligned}$$

By a simple computation we get the representations : $$T^*T=
\begin{pmatrix}
0 & 0 & 0\\
0 & I & 0\\
0 & 0 & R_0^*R_0+R_1^*R_1+ Q^*Q
\end{pmatrix},
\quad {\rm Re}T=\frac{1}{2}
\begin{pmatrix}
0 & W_0 & R_0\\
W_0^* & 2 {\rm Re}W_1 & R_1\\
R_0^* & R_1^* & 2 {\rm Re}Q
\end{pmatrix}$$ and respectively (by using the second relation of ) $$({\rm Re}T)^2=\frac{1}{4}
\begin{pmatrix}
W_0W_0^* +R_0R_0^* & 2W_0 {\rm Re}W_1 +R_0R_1^* & W_0R_1 +2 R_0 {\rm Re}Q\\
2 ({\rm Re}W_1)W_0^*+R_1R_0^* & W_0^*W_0 + 4 ({\rm Re}W_1)^2 + R_1R_1^* & W_1R_1 + 2 R_1 {\rm Re}Q\\
R_1^* W_0^*+ 2 ({\rm Re}Q)R_0^* & R_1^* W_1^* + 2 ({\rm Re}Q)R_1^* & R_0^*R_0 +R_1^*R_1 + 4 ({\rm Re}Q)^2
\end{pmatrix}.$$ Since the subspace ${\mathcal{H}}_1$ reduces $|T|$ to $I_{{\mathcal{H}}_1}$ and as ${\rm Re}T$ is a contraction, the condition implies that ${\mathcal{H}}_1$ also reduces $|{\rm Re}T|$ and $|{\rm Re}T||_{{\mathcal{H}}_1}=I_{{\mathcal{H}}_1}$. As $|{\rm Re}T|^2=({\rm Re}T)^2$ we get from the above matrix representation that $$\begin{aligned}
\label{ec35}
\frac{1}{4} (W_0^*W_0+4({\rm Re}W_1)^2 + R_1R_1^*)=I_{{\mathcal{H}}_1}.\end{aligned}$$ By using the former relation in and the second relation in we obtain $$I_{{\mathcal{H}}_1}\le {\rm Re} W_1^2,$$ which means by \[11, Corollary 3\] that $W_1^2=I$. So, the relation becomes $W_1W_1^*+R_1R_1^*=I_{{\mathcal{H}}_1}$ which leads to $I_{{\mathcal{H}}_1} +W_1R_1R_1^*W_1^*=W_1W_1^*$. But this gives $R_1=0$ and $W_1W_1^*=I_{{\mathcal{H}}_1}$, hence $W_1=W_1^*$, while by this yields $W_0=0$. So, the block matrices of $T$, $T^*T$, ${\rm Re}T$ and $({\rm Re}T)^2$ have simpler forms. 

Now, we have by Theorem \[te21\] that ${\mathcal{N}}(I-Q^*Q)\subset {\mathcal{N}}(I-T^*T)\cap {\mathcal{H}}_2$, and we show next that ${\mathcal{N}}(I-T^*T) \cap {\mathcal{H}}_2 \subset {\mathcal{N}}(I-QQ^*)$. Indeed, let $x=T^*Tx \in {\mathcal{H}}_2$. To use the condition we consider $|{\rm Re}T|$ to have the following block matrix form on ${\mathcal{H}}={\mathcal{H}}_0\oplus {\mathcal{H}}_1\oplus {\mathcal{H}}_2$ (having in view that $|{\rm Re}T||_{{\mathcal{H}}_1}=I_{{\mathcal{H}}_1}$) : $$|{\rm Re}T|=
\begin{pmatrix}
A & 0 & B\\
0 & I & 0\\
B^* & 0 & C
\end{pmatrix},$$ with some appropriate contractions $A, B$ and $C$ with $A,C\ge 0$. As $|{\rm Re}T|^2=({\rm Re}T)^2$ we obtain that $B^*B+C^2=\frac{1}{4}[R_0^*R_0 + (Q + Q^*)^2]$, so for $x$ as above we get (by ) $$\|x\|^2 =\langle |T|x,x \rangle \le \langle |{\rm Re}T|x,x\rangle = \langle Cx,x \rangle \le \|x\|^2$$ which means $Cx=x$ (because $C\ge 0$).  Then the above equality together with the fact that $(R_0^*R_0+Q^*Q)x=T^*Tx=x$, lead to the relation $$\frac{3}{4} \|x\|^2 + \|Bx\|^2 =\frac{1}{4} \|Q^*x\|^2 + \frac{1}{2} \langle {\rm Re}Q^2x,x \rangle.$$ Since $Q^2$ is a contraction it follows that $$\frac{1}{4} (\|x\|^2 - \|Q^*x\|^2) + \|Bx\|^2 \le 0,$$ hence $Bx=0$ and $\|Q^*x\|=\|x\|$. So $x\in {\mathcal{N}}(I-QQ^*)$ and the inclusion ${\mathcal{N}}(I-T^*T) \cap {\mathcal{H}}_2 \subset {\mathcal{N}}(I-QQ^*)$ is proved. 

Next, we consider the block matrix form of $TT^*$ on ${\mathcal{H}}={\mathcal{H}}_0 \oplus {\mathcal{H}}_1 \oplus {\mathcal{H}}_2$, namely $$TT^*=
\begin{pmatrix}
R_0R_0^* & 0 & R_0Q^*\\
0 & I & 0 \\
QR_0^* & 0 & QQ^*
\end{pmatrix}.$$ Since $T^*|_{{\mathcal{H}}_2}=Q^*$ and $T$ is a contraction we have ${\mathcal{N}}(I-QQ^*) \subset {\mathcal{N}}(I-TT^*)$, and from this representation of $TT^*$ one has also ${\mathcal{H}}_1 \subset {\mathcal{N}}(I-TT^*)$. So we infer that $$\begin{aligned}
\label{ec36}
{\mathcal{N}}(I-T^*T) ={\mathcal{H}}_1 \oplus {\mathcal{N}}(I-T^*T) \cap {\mathcal{H}}_2 \subset {\mathcal{H}}_1 \oplus {\mathcal{N}}(I-QQ^*) \subset {\mathcal{N}}(I-TT^*).\end{aligned}$$ This means that ${\mathcal{N}}(I-TT^*)$ is invariant for $T^*$, hence ${\mathcal{N}}(I-T^*T)$ is also invariant for $T^*$.

To see that ${\mathcal{N}}(I-T^*T)$ just reduces $T$ we firstly remark from that $${\mathcal{N}}(I-T^*T) ={\mathcal{N}}(I- |T|) \subset {\mathcal{N}}(I- |{\rm Re}T|)={\mathcal{N}}(I- ({\rm Re}T)^2).$$ In fact, these subspaces coincide, they containing the subspace ${\mathcal{H}}_1$. To see this equality, let us consider the polar decomposition $${\rm Re}T=\widetilde{U}|{\rm Re}T|,$$ where $\widetilde{U}$ is a symmetry on $\overline{{\mathcal{R}}({\rm Re}T)}=\overline{{\mathcal{R}}(|{\rm Re}T|)}$. Clearly, one has $${\mathcal{N}}(I-|{\rm Re}T|)={\mathcal{N}}(\widetilde{U}-{\rm Re}T)$$ and this subspace reduces the operators ${\rm Re}T$ and $\widetilde{U}$.

Let $x\in {\mathcal{N}}(\widetilde{U}-{\rm Re}T) \cap {\mathcal{H}}_2$ such that $x$ is orthogonal on ${\mathcal{N}}(I-T^*T)$, hence $\|Tx\|<\|x\|$. As ${\mathcal{H}}_2 \subset \overline{{\mathcal{R}}({\rm Re}T)}$ and $\widetilde{U}$ is unitary on this range, we get $$\|x\|=\|Ux\|=\|({\rm Re}T)x\|\le \frac{1}{2} (\|Tx\|+\|T^*x\|)< \|x\|$$ which forces to have $x=0$. Therefore ${\mathcal{N}}(I-|{\rm Re}T|) \cap {\mathcal{H}}_2 \subset {\mathcal{N}}(I-T^*T)$, and since ${\mathcal{H}}_1 \subset {\mathcal{N}}(I-|{\rm Re}T|)$ we conclude that ${\mathcal{N}}(I-T^*T) ={\mathcal{N}}(I-|{\rm Re}T|)$. But this subspace reduces ${\rm Re}T$ and it is invariant for $T^*$ (as we have seen before).  Hence ${\mathcal{N}}(I-T^*T)$ reduces $T$, which also gives the inclusion ${\mathcal{N}}(I-TT^*) \subset {\mathcal{N}}(I-T^*T)$. Finally, we conclude that $${\mathcal{N}}(I-T^*T) ={\mathcal{N}}(I-TT^*)={\mathcal{N}}(I- |{\rm Re}T|)={\mathcal{H}}_1,$$ where for the last equality we have in view the maximality of the subspace ${\mathcal{N}}(T) \oplus {\mathcal{H}}_1$ relative to $T$. The identities are proved. 

In addition, from the inclusions we infer that ${\mathcal{N}}(I-QQ^*)=\{0\}$, so $Q^*$ like $Q$ are pure contractions on ${\mathcal{H}}_2$.

Now, the condition yields ${\mathcal{N}}({\rm Re}T)={\mathcal{N}}(T)\cap {\mathcal{N}}(T^*)$, and this subspace reduces $T$. Therefore we have ${\mathcal{N}}({\rm Re}T)={\mathcal{N}}(T)$ if and only if ${\mathcal{N}}(T)$ reduces $T$, that is $R_0=0$. Equivalently, this means that $T$ has the diagonal representation $T=W_1\oplus 0\oplus Q$ on ${\mathcal{H}}={\mathcal{H}}_1\oplus {\mathcal{H}}_0 \oplus {\mathcal{H}}_2$, with $W_1, Q$ as above, that is $T=W_1\oplus Z$ where $Z=0\oplus Q$ is a pure contraction which satisfies the condition . 

Finally, for the second relation in we have from the above block matrix form of $T$ (with $W_0=0$, $R_1=0$) $${\mathcal{N}}(T^*)=\{y\oplus z \in {\mathcal{N}}(T) \oplus {\mathcal{H}}_2 : R_0^*y+Q^* z=0\}.$$ But ${\mathcal{N}}(T^*) \cap {\mathcal{N}}(T)={\mathcal{N}}(R_0^*)$ and ${\mathcal{N}}(T^*|_{{\mathcal{H}}_2})={\mathcal{N}}(Q^*)={\mathcal{N}}(T^*)\cap {\mathcal{H}}_2= {\mathcal{N}}(T^*|_{\overline{{\mathcal{R}}(T^*)}})$. So we obtain $${\mathcal{N}}(T^*)={\mathcal{N}}(R_0^*)\oplus {\mathcal{N}}(Q^*) ={\mathcal{N}}({\rm Re}T)\oplus {\mathcal{N}}(T^*|_{\overline{{\mathcal{R}}(T^*)}}),$$ and this finishes the proof. 

\[co32\] A real scalar multiple of a partial isometry which satisfies the condition is self-adjoint. 

If $T$ is a partial isometry satisfying then ${\mathcal{H}}={\mathcal{N}}(I-T^*T) \oplus {\mathcal{N}}(T)$, so $T=U\oplus 0=T^*$ by Theorem \[te31\].  More general, if $T=\alpha T_0$ with $\alpha \in {\mathbb{R}}$ and $T_0$ a partial isometry, then $\frac{1}{\alpha}T=\frac{1}{\alpha}T^*$ by the previous remark, so $T=T^*$.

A more general result than the previous corollary is the following

\[pr33\] Let $0 \neq T \in {\mathcal{B(H)}}$ having with respect to a decomposition ${\mathcal{H}}=\mathcal{G} \oplus \mathcal{G}^{\perp}$ the block matrix form $$\begin{aligned}
\label{ec37}
T=
\begin{pmatrix}
S & R\\
0 & Q
\end{pmatrix},\end{aligned}$$ where $\frac{1}{\|T\|}S$ is an isometry on $\mathcal{G}$ and, in addition, either $R\in \mathcal{B}(\mathcal{G}^{\perp}, \mathcal{G})$ is injective and $Q\in \mathcal{B}(\mathcal{G}^{\perp})$ is arbitrary, or $Q^2=0$. If $T$ satisfies the condition then $T$ is self-adjoint. 

Let $T \neq 0$ as in , and let $T_0 =\frac{1}{\alpha}T$, $S_0=\frac{1}{\alpha}S$, $R_0=\frac{1}{\alpha}R$, $Q_0=\frac{1}{\alpha}Q$, where $\alpha =\|T\|$. Then the contraction $T_0$ satisfies , and by Theorem \[te31\] we infer ${\mathcal{N}}(I-T_0^*T_0)={\mathcal{N}}(I-T_0T_0^*)$. To use this fact we have in view the representations $$I-T_0^*T_0=
\begin{pmatrix}
0 & 0\\
0 & I-R_0^*R_0-Q_0^*Q_0
\end{pmatrix},
\quad I-T_0T_0^*=
\begin{pmatrix}
I- S_0S_0^*-R_0R_0^* & -R_0Q_0^*\\
-Q_0R_0^* & I-Q_0Q_0^*
\end{pmatrix},$$ where for $I-T_0^*T_0$ we used that $S_0$ is an isometry on $\mathcal{G}$. These representations together with the above kernels give that $\mathcal{G} \subset {\mathcal{N}}(I-T_0T_0^*)$, that is the relations $$\begin{aligned}
\label{ec38}
I-S_0S_0^*-R_0R_0^*=0, \quad Q_0R_0^*=0.\end{aligned}$$

From the first relation we have $R_0R_0^*=P_{{\mathcal{N}}(S_0^*)}$, so $R_0$ is a partial isometry and ${\mathcal{R}}(R_0)={\mathcal{N}}(S_0^*)$.

Now, if $R$ is injective, then from the second relation in one obtains $Q_0=0$. In this case $R_0$ is an isometry on $\mathcal{G}^{\perp}$, and since by the above matrix representations we get $$\mathcal{G}={\mathcal{N}}(I-T_0T_0^*)={\mathcal{N}}(I-T_0^*T_0)=\mathcal{G}\oplus {\mathcal{N}}(I-R_0^*R_0)=\mathcal{G}\oplus \mathcal{G}^{\perp},$$ it follows $\mathcal{G}^{\perp}=\{0\}$. Hence $T_0=S_0$ is a symmetry on ${\mathcal{H}}$, and consequently $T=T^*$.

Assume next the other condition from hypothesis, namely for $Q^2=0$. Since ${\mathcal{N}}(I-T_0^*T_0)=\mathcal{G}\oplus {\mathcal{N}}(I-Q_0Q_0^*)$ is invariant for $T_0^*$, ${\mathcal{N}}(I-Q_0Q_0^*)$ will be invariant for $Q_0^*=T_0^*|_{\mathcal{G}^{\perp}}$. So, if $b\in {\mathcal{N}}(I-Q_0Q_0^*)$ we have $Q_0^*b=Q_0Q_0^{*2}b=0$, which means $b=Q_0Q_0^*b\in {\mathcal{N}}(Q_0^*)$ that is $b=0$. Hence ${\mathcal{N}}(I-Q_0Q_0^*)=\{0\}$ which gives ${\mathcal{N}}(I-T_0T_0^*)=\mathcal{G}$ and this subspace reduces $T_0$. Thus $R_0=0$, while $S_0$ and $Q_0$ satisfy the condition .  Therefore $S_0$ will be a symmetry on $\mathcal{G}$, and as $Q_0^2=0$ it is easy to see that $Q=0$. Indeed, since $Q^2=0$, $Q$ will have on $\mathcal{G}^{\perp}=\overline{{\mathcal{R}}(Q)}\oplus {\mathcal{N}}(Q^*)$ the block matrix form $$Q=
\begin{pmatrix}
0 & Q_1\\
0 & 0
\end{pmatrix}.$$

By a simple computation we get $$|Q|=0\oplus |Q_1|, \quad |{\rm Re}Q|=\frac{1}{2} (|Q_1^*|\oplus |Q_1|)$$ on the above decomposition of $\mathcal{G}^{\perp}$. So, the condition for $Q$ implies $Q_1=0$ that is $Q=0$.

We conclude that $T=S \oplus 0=\alpha S_0\oplus 0=T^*$, $S_0$ being a symmetry.  This ends the proof. 

To apply this proposition for $2$-quasi-isometries, we recall (see \[8, Remark 3.10\]; or \[15, Remark 2.7\]) that such an operator has a block matrix form as in on ${\mathcal{H}}=\overline{{\mathcal{R}}(T^2)}\oplus {\mathcal{N}}(T^{*2})$, with $S$ an isometry and $Q^2=0$. So the previous proposition gives the following

\[co34\] A contractive $2$-quasi-isometry which satisfies the condition is self-adjoint. 

It is still unknown if the second assumption in the hypothesis of Proposition \[pr33\], namely $Q^2=0$, can be replaced by the weaker condition $Q^m=0$ for some $m\ge 3$, in order to preserve the conclusion; that is to prove that $Q=0$ under the condition . 

However, for $m$-quasi-isometries we have the following

\[co35\] If $T\in {\mathcal{B(H)}}$ is a contractive $m$-quasi-isometry for an integer $m\ge 3$ which satisfies the condition then ${\mathcal{N}}(I-T^*T)={\mathcal{N}}(I-TT^*)=\overline{{\mathcal{R}}(T^m)}$ reduces $T$ to a symmetry.  In addition, $T$ is self-adjoint if and only if $T=S\oplus 0$.

If $T$ is a $m$-quasi-isometry then $T|_{{\mathcal{R}}(T^m)}$ is an isometry.  Such an operator has the form on ${\mathcal{H}}= \overline{{\mathcal{R}}(T^m)} \oplus {\mathcal{N}}(T^{*m})$ with $S$ an isometry and $Q^m=0$. In the case when $\|T\|=1$ we also have (as in the proof of Proposition \[pr33\]) that $S^*R=0$ and $${\mathcal{N}}(I-T^*T)={\mathcal{N}}(I-TT^*)= \overline{{\mathcal{R}}(T^m)} \oplus {\mathcal{N}}(I-QQ^*),$$ this subspace reducing $T$ to a symmetry.  So, if $b\in {\mathcal{N}}(I-QQ^*)$ and $m\ge 3$ then $Q^*b \in {\mathcal{N}}(I-QQ^*)$ which leads (by recurrence) to $b=QQ^*b=Q^mQ^{*m}b=0$. Hence ${\mathcal{N}}(I-QQ^*)=\{0\}$, ${\mathcal{N}}(I-T^*T)=\overline{{\mathcal{R}}(T^m)}$, and $S$ is a symmetry on this subspace.  Then the condition $S^*R=0$ yields $R=0$, consequently $T=S\oplus Q$.

Assume now $T=T^*$ that is $Q=Q^*$. Since $Q^m=0$ one has ${\mathcal{R}}(Q^{m-1})\subset {\mathcal{N}}(Q)={\mathcal{N}}(Q^*) \subset {\mathcal{N}}(Q^{*(m-1)})$, hence $Q^{m-1}=0$. By recurrence one infers $Q=0$, so $T=S \oplus 0$. The converse implication for the second assertion of corollary being trivial, the proof is finished. 

We remarked before that a non-null nilpotent operator of order $2$ cannot satisfy the condition .  We can also use this fact to obtain the following result

\[pr36\] Let $T$ be a contraction on ${\mathcal{H}}$ having with respect to the decomposition ${\mathcal{H}}=\mathcal{G} \oplus \mathcal{G}^{\perp}$ the block matrix form $$\begin{aligned}
\label{ec39}
T=
\begin{pmatrix}
W & R\\
0 & W'
\end{pmatrix}\end{aligned}$$ where $W$ and $W'$ are partial isometries on $\mathcal{G}$ and $\mathcal{G}^{\perp}$, respectively, while $R\in \mathcal{B}(\mathcal{G}^{\perp}, \mathcal{G})$. If $T$ satisfies the condition then $T$ is self-adjoint. 

Consider firstly that $W'=0$ in .  Then $\overline{{\mathcal{R}}(T)} \subset \mathcal{G}$ and, assuming , by Theorem \[te21\] and Theorem \[te31\] one has $\mathcal{G} \subset {\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T)=:{\mathcal{M}}$. Therefore ${\mathcal{M}}^{\perp}\subset \mathcal{G}^{\perp}\subset {\mathcal{N}}(T^*)$ that is $T^*|_{{\mathcal{M}}^{\perp}}=0$, and using the $3 \times 3$ block matrix of $T$ on ${\mathcal{H}}={\mathcal{N}}(T) \oplus {\mathcal{N}}(I-T^*T) \oplus {\mathcal{M}}^{\perp}$ given by the proof of Theorem \[te31\] (with $W_0=0$, $R_1=0$, $Q=0$ by the previous remark, $W_1$ a symmetry on ${\mathcal{N}}(I-T^*T)$, and $R_0 \in \mathcal{B}({\mathcal{M}}^{\perp}, {\mathcal{N}}(T))$) we get the representation $T=W_1 \oplus \widetilde{R}$ on ${\mathcal{H}}={\mathcal{N}}(I-T^*T) \oplus [{\mathcal{N}}(T)\oplus {\mathcal{M}}^{\perp}]$, where $\widetilde{R} \in \mathcal{B}({\mathcal{N}}(T) \oplus {\mathcal{M}}^{\perp})$ has the block matrix form $$\widetilde{R}=
\begin{pmatrix}
0 & R_0\\
0 & 0
\end{pmatrix}.$$ Since $\widetilde{R}^2=0$ and $\widetilde{R}$ satisfies one has $\widetilde{R}=0$, which ensures $T=T^*$.

In the general case, as $W'$ is a partial isometry on $\mathcal{G}^{\perp}$ and ${\mathcal{N}}(I-T^*T)={\mathcal{N}}(I-TT^*)$ by , we have $\mathcal{G}^{\perp}\subset {\mathcal{N}}(I-T^*T)\oplus {\mathcal{N}}(T^*)$, and as above $\mathcal{G} \subset {\mathcal{M}}$. So, ${\mathcal{M}}^{\perp}\subset \mathcal{G}^{\perp}$ which by the previous inclusion of $\mathcal{G}^{\perp}$ gives ${\mathcal{M}}^{\perp} \subset {\mathcal{N}}(T^*)$. This shows that $T$ has on ${\mathcal{H}}={\mathcal{M}}\oplus {\mathcal{M}}^{\perp}$ a block matrix of the form $$T=
\begin{pmatrix}
\widetilde{W} & \widetilde{R}\\
0 & 0
\end{pmatrix},$$ with $\widetilde{W}=0\oplus W_1$ a partial isometry and then it follows that $T=T^*$ by the previous conclusion. 

The contractions mentioned in this proposition are not hyponormal, in general.  But among these one gets the quasi-isometries (the case when $W$ is an isometry), which are subnormal (as we already quoted in the introduction). 

\[re37\] An operator $T\in {\mathcal{B(H)}}$ having the block matrix form as in Proposition \[pr33\] is not a contraction, in general.  But in the case when $T$ is a contraction with $S$ an isometry and $R$, $Q$ arbitrary contractions (in ), then the condition $S^*R=0$ is also true and $R$ will be a partial isometry (as in the proof of Proposition \[pr33\]), if $T$ satisfies .  In general $R\neq 0$, but if $Q$ is pure then $R=0$ because ${\mathcal{N}}(I-T^*T)$ reduces $T$ to a symmetry and we have $${\mathcal{N}}(I-T^*T)={\mathcal{N}}(I-TT^*)=\mathcal{G} \oplus {\mathcal{N}}(I-R^*R-Q^*Q)=\mathcal{G} \oplus {\mathcal{N}}(I-QQ^*).$$ This latter case occurs, for instance, when $\mathcal{G}={\mathcal{N}}(I-S_T)$ in the representation of $T$, and then the Fong-Tsui conjecture for $T$ one reduces to its pure part $Q$.

In fact, every operator $T\neq 0$ on ${\mathcal{H}}$ has the form with $R,Q$ arbitrary operators, on the decomposition ${\mathcal{H}}={\mathcal{N}}(I-S_{T_0})\oplus \overline{{\mathcal{R}}(I-S_{T_0})}$, where $T_0=\frac{1}{\|T\|}T$, $S_{T_0}$ is the asymptotic limit of $T_0$, and so $\frac{1}{\|T\|}S$ is an isometry (in ).  Thus, the assumptions on the operators $R$ or $Q$ in Proposition \[pr33\] and the condition force such operator $T$ to be self-adjoint. 

In turn to Proposition \[pr36\], it is clear that in one can consider multiples of partial isometries with the scalar $\alpha = \|T\|$ instead of $W$ and $W'$ respectively, in order to preserve the conclusion. 

Another special class of non-contractive operators which contains some $2$-isometries as well as the Brownian isometries, and for which the Fong-Tsui conjecture holds, is mentioned by the following

\[pr38\] Let $T\in {\mathcal{B(H)}}$ such that $T^*T\ge I$ and having the block matrix form on ${\mathcal{H}}=\mathcal{G} \oplus \mathcal{G}^{\perp}$, where $S$ is an isometry on $\mathcal{G}$, $R\in \mathcal{B}(\mathcal{G}^{\perp}, \mathcal{G})$ with $S^*R=0$, while $Q$ is a contraction on $\mathcal{G}^{\perp}$. If $T$ satisfies the condition then $T$ is self-adjoint, in fact a symmetry. 

By using with $S,R,Q$ as above one obtains $$T^*T=
\begin{pmatrix}
I & 0\\
0 & R^*R+Q^*Q
\end{pmatrix},
\quad ({\rm Re}T)^2= \frac{1}{4}
\begin{pmatrix}
(S+S^*)^2+RR^* & SR+R(Q+Q^*)\\
R^*S^* + (Q+Q^*)R^* & R^*R+ (Q+Q^*)^2
\end{pmatrix}.$$

Since $T^*T \ge I$ the condition $|T|\le |{\rm Re}T|$ implies $|{\rm Re}T|\ge I$. Therefore we have $|T|\le |{\rm Re}T|\le ({\rm Re}T)^2$ which gives for any $x\in \mathcal{G}$ (as $R^*S=0$), $$\begin{aligned}
\|x\|^2&=& \|Sx\|^2= \langle |T|Sx,Sx \rangle \le \langle ({\rm Re}T)^2 Sx,Sx\rangle \\
&=& \frac{1}{4} \langle ((S+S^*)^2 +RR^*)Sx,Sx \rangle =\frac{1}{4} \langle (S^3+S^* +2S)x, Sx\rangle \\
& =& \frac{1}{2} (\|x\|^2 + \langle ({\rm Re}S^2)x,x \rangle ).\end{aligned}$$ This means that $I_{\mathcal{G}} \le {\rm Re}S^2$, which by \[10, Corollary 3\] yields $S^2=I_{\mathcal{G}}$. As $S$ is an isometry it will be just a symmetry, and as $S^*R=0$ it follows $R=0$.

On the other hand, since $I\le ({\rm Re}T)^2$ we have for each $y \in \mathcal{G}^{\perp}$ ($Q$ being a contraction) $$\|y\|^2 \le \frac{1}{4}\langle (Q+Q^*)^2y,y \rangle \le \frac{1}{4} \langle (Q^2+Q^{*2} +2I)y,y \rangle,$$ whence $I_{\mathcal{G}^{\perp}}\le {\rm Re} Q^2$. Then as above $Q$ is a symmetry, hence $T=S \oplus Q=T^*$. This ends the proof. 

In concordance with the representations and for Brownian isometries, from the previous proposition we derive the following

\[co39\] A Brownian isometry of positive covariance which satisfies the condition is a symmetry. 

\[re310\] Proposition \[pr38\] one refers to a larger class of operators than that of Brownian isometries, but not to all $2$-isometries, because we need to impose that $Q$ is a contraction. 

Recall (see \[1, Theorem 1.26\]) that the block matrix of a $2$-isometry on ${\mathcal{H}}={\mathcal{N}}(T^*T-I) \oplus \overline{{\mathcal{R}}(T^*T-I)}$ has the form , with $S$ an isometry, $S^*R=0$, $R^*R+Q^*Q-I$ injective and $$Q^*(R^*R+Q^*Q-I)Q=R^*R+Q^*Q-I.$$ But if such $T$ satisfies the condition , then as in the previous proof $S$ will be a symmetry, hence $R=0$. In this case, the last relation before means $Q^*(Q^*Q-I)Q=Q^*Q-I$, that is $Q$ is a $2$-isometry and $Q^*Q\ge I$. So, the Fong-Tsui conjecture for $T$ one reduces to its $2$-isometric pure part $Q$.

Final Remarks
=============

In turn to Theorem \[te31\] which plays an essential role for our considerations concerning the Fong-Tsui conjecture, we make some comments. 

\[re41\] The operator $T^*$ has a similar form like $T$, under the condition , and in the corresponding decomposition of ${\mathcal{H}}$. Indeed, we have from the proof of Theorem \[te31\] the representation $$\begin{aligned}
\label{ec41}
T=
\begin{pmatrix}
W_1 & 0 & 0\\
0 & 0 & R_0\\
0 & 0 & Q
\end{pmatrix}\end{aligned}$$ on the decomposition ${\mathcal{H}}={\mathcal{H}}_1 \oplus {\mathcal{H}}_0 \oplus {\mathcal{H}}_2$ where ${\mathcal{H}}_1={\mathcal{N}}(I-T^*T)$ and ${\mathcal{H}}_0={\mathcal{N}}(T)$. To see here ${\mathcal{N}}(T^*)$ given by , we refine by considering ${\mathcal{N}}(T)={\mathcal{N}}({\rm Re}T) \oplus {\mathcal{H}}_0'$ and ${\mathcal{H}}_2={\mathcal{N}}(Q^*)\oplus \overline{{\mathcal{R}}(Q)}$. So, on the decomposition ${\mathcal{H}}={\mathcal{H}}_1 \oplus {\mathcal{N}}({\rm Re}T)\oplus {\mathcal{N}}(Q^*)\oplus {\mathcal{H}}_0'\oplus \overline{{\mathcal{R}}(Q)}$ we have the representations $$T=
\begin{pmatrix}
W_1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & R_{00} & 0 & R_{01}\\
0 & 0 & Q_0 & 0 & Q_1
\end{pmatrix},
\quad T^*=
\begin{pmatrix}
W_1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & R_{00}^* & Q_0^*\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & R_{01}^* & Q_1 ^*
\end{pmatrix},$$ where $R_{00}=P_{{\mathcal{H}}_0'}R_0|_{{\mathcal{N}}(Q^*)}$, $R_{01}=P_{{\mathcal{H}}_0'}R_0|_{\overline{{\mathcal{R}}(Q)}}$, and $Q_0=Q|_{{\mathcal{N}}(Q^*)}$, $Q_1=Q|_{\overline{{\mathcal{R}}(Q)}}$. Here we used the fact that $P_{{\mathcal{N}}({\rm Re}T)}R_0=0$ and $P_{{\mathcal{N}}(Q^*)}Q=0$. We get that $T^*$ has on ${\mathcal{H}}={\mathcal{H}}_1 \oplus {\mathcal{N}}(T^*) \oplus {\mathcal{H}}_3$ with ${\mathcal{H}}_3={\mathcal{H}}_0'\oplus {\mathcal{H}}_3$ the block matrix form $$T^*=
\begin{pmatrix}
W_1 & 0 & 0\\
0 & 0 & R_*\\
0 & 0 & Q_*
\end{pmatrix},$$ where $R_*:= \begin{pmatrix} 0 & 0 \\ R_{00}^* & Q_0^* \end{pmatrix}$ and $Q_*:= \begin{pmatrix} 0 & 0 \\ R_{01}^* & Q_1^* \end{pmatrix}.$

Hence $T^*$ has the same form as $T$ in on the corresponding decomposition ${\mathcal{H}}={\mathcal{H}}_1 \oplus {\mathcal{N}}(T^*) \oplus {\mathcal{H}}_3$, which was obtained under the condition , even if $T^*$ does not satisfy this condition. 

These representations of $T$ and $T^*$ also provide that the invariant partial isometric parts for $T$ and $T^*$ can be different under the condition , that is ${\mathcal{N}}(T) \neq {\mathcal{N}}(T^*)$, in general.  But the condition ${\mathcal{N}}(T)={\mathcal{N}}(T^*)$ is necessary for $T$ to be self-adjoint.  In this latter case we have $R_0=0$ and $Q_0=0$, while $Q=Q_1$ like $Q^*$ are injective and pure contractions, hence $$\overline{{\mathcal{R}}(Q)}=\overline{{\mathcal{R}}(Q^*)}=\overline{{\mathcal{R}}(I-Q^*Q)}=\overline{{\mathcal{R}}(I-QQ^*)}.$$ But only these information on $Q$ are not sufficient to obtain that $Q=Q^*$ that is $T=T^*$.

The major difficulty to use the condition in order to obtain the self-adjointness of $T$ consists in the fact that $|{\rm Re}T|=[({\rm Re}T)^2]^{1/2}$ cannot be easily expressed in terms of $T$ (that is using the block matrix form ). 

Clearly, on can have in view the polar decomposition ${\rm Re}T=U|{\rm Re}T|$, where $U$ is a symmetry on $\overline{{\mathcal{R}}({\rm Re}T)}$. Since ${\mathcal{N}}({\rm Re}T)={\mathcal{N}}(U)$ reduces $U$ and $T$, one can just only consider the condition on $\overline{{\mathcal{R}}({\rm Re}T)}$ which reduces $T$ and $U$.

But even in the case when ${\mathcal{N}}(T)$ reduces $T$ ($R_0=0$ in ), hence when $U$ is a symmetry on the subspace ${\mathcal{H}}_2$, we cannot use this to obtain $Q=Q^*$.

But, as we have seen before, this is possible for some classes of operators related to partial isometries, or for the operators $T$ which commute with $U$, as it was recently shown in \[17\].  Let us remark that this last class of operators and that of partial isometries are not contained one into the other.  For example, one can consider the partial isometry $T\in \mathcal{B}({\mathcal{H}}\oplus {\mathcal{H}})$ and the corresponding partial isometry for ${\rm Re}T$ with the block matrices $$T=
\begin{pmatrix}
0 & I\\
0 & 0
\end{pmatrix},
\quad U=
\begin{pmatrix}
0 & I\\
I & 0
\end{pmatrix},$$ and it is clear that $TU=I \oplus 0$ and $UT=0 \oplus I$.

\[re42\] The condition means that there exists a contraction $A$ on ${\mathcal{H}}$ satisfying the relation $$\begin{aligned}
\label{ec42}
A|{\rm Re}T|^{1/2}=|T|^{1/2},\end{aligned}$$ or equivalently $$\begin{aligned}
\label{ec43}
|{\rm Re}T|^{1/2}A^*=|T|^{1/2}.\end{aligned}$$ From these relations we have ${\mathcal{N}}(A^*) \subset {\mathcal{N}}(T)$ and also $$A\overline{{\mathcal{R}}({\rm Re}T)}\subset \overline{{\mathcal{R}}(T^*)} \subset \overline{{\mathcal{R}}({\rm Re}T)}$$ that is $\overline{{\mathcal{R}}({\rm Re}T)}$ is invariant for $A$.  Denoting $$A_0=A|_{\overline{{\mathcal{R}}({\rm Re}T)}},$$ we infer that $$\overline{{\mathcal{R}}(T^*)} =\overline{A|{\rm Re}T|^{1/2}{\mathcal{H}}} =\overline{A_0{\mathcal{R}}({\rm Re}T)}= \overline{{\mathcal{R}}(A_0)}\subset \overline{{\mathcal{R}}(A)}.$$

In addition, if ${\mathcal{N}}(T)$ reduces $T$ then $$\overline{{\mathcal{R}}(A)}=\overline{{\mathcal{R}}(A_0)}=\overline{{\mathcal{R}}(T^*)}=\overline{{\mathcal{R}}({\rm Re}T)},$$ and some relationship between the closeness of the ranges of $T$, ${\rm Re}T$ and $A$ can be obtained, but unessential for the Fong-Tsui conjecture.  However, from we infer $|T|^2 \le A |{\rm Re}T|^2A^*$. So, if $A^*$ is a $|{\rm Re} T|^2$-contraction that is $A|{\rm Re}T|^2A^*\le |{\rm Re}T|^2$, then $|T|^2 \le |{\rm Re}T|^2$ and by \[10, Theorem 1.3\] it follows that $T$ is self-adjoint.  In particular, if $A$ satisfies some conditions of self-adjointness, for instance those quoted in \[13\] then by $A$ and $|{\rm Re}T|$ commute, therefore by the previous remark one has $|T|^2 \le |{\rm Re}T|^2$. But, in general the condition does not impose other restriction on the contraction $A$ in .  In conclusion we doubt that for $T$ implies always $T=T^*$.

Notice finally that an interesting context where it is possible to show that Fong-Tsui conjecture is true is that of $(A,m)$-expansive operators which were recently studied in \[4\] and \[12\], but we do not refer to them here. 

Another interesting context for investigations on this conjecture is that of $A$-contractions (for some positive operator $A$ on ${\mathcal{H}}$), or even for $A$-bounded operators, which were extensively studied in the last years.  Here is natural to use the concept of $A$-adjoint operator and $A$-projection, and an important role have the $A$-partial isometries recently investigated in \[5, 6\], \[7\], \[9\], or just the quasi-isometries in the context of $A$-contractions which were studied in \[16\]. 

References {#references .unnumbered}
==========

\[1\] J. Agler and M. Stankus, [*$m$-Isometric Transformations of Hilbert spaces*]{}, Integr.  Equat.  Oper.  Theory, 21, 4 (1995), 383-429. 

\[2\] J. Agler and M. Stankus, [*$m$-Isometric Transformations of Hilbert spaces II*]{}, Integr.  Equat.  Oper.  Theory, 23, 10 (1995), 1-48. 

\[3\] J. Agler and M. Stankus, [*$m$-Isometric Transformations of Hilbert spaces III*]{}, Integr.  Equat.  Oper.  Theory, 24 (1996), 379-421. 

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