# Higher Rank Numerical Ranges of Normal Operators and unitary dilations

Pankaj Dey and Mithun Mukherjee  Pankaj Dey, School of Mathematics, Indian
Institute of Science Education and Research, Thiruvananthapuram, Maruthamala
PO, Vithura, Thiruvananthapuram-695551, Kerala, India.
<EMAIL_ADDRESS>Mithun Mukherjee, School of Mathematical &
Computational Sciences, Indian Association for the Cultivation of Science, 2A
& 2B Raja S C Mullick Road, Jadavpur, Kolkata-700032, INDIA.
<EMAIL_ADDRESS>
###### Abstract.

We describe here the higher rank numerical range, as defined by Choi, Kribs
and Życzkowski, of a normal operator on an infinite dimensional Hilbert space
in terms of its spectral measure. This generalizes a result of Avendaño for
self-adjoint operators. An analogous description of the numerical range of a
normal operator by Durszt is derived for the higher rank numerical range as an
immediate consequence. It has several interesting applications. We show using
Durszt’s example that there exists a normal contraction $T$ for which the
intersection of the higher rank numerical ranges of all unitary dilations of
$T$ contains the higher rank numerical range of $T$ as a proper subset.
Finally, we strengthen and generalize a result of Wu by providing a necessary
and sufficient condition for the higher rank numerical range of a normal
contraction being equal to the intersection of the higher rank numerical
ranges of all possible unitary dilations of it.

††Keywords: Higher rank numerical range, Normal operator, Spectral measure,
Unitary dilation
MSC(2020): 47A12, 47A20, 47B15, 15A60, 81P68.

##  1\. Introduction

The notion of the quadratic forms associated with matrix was extended for
operators on Hilbert space which is known as the numerical range. Let
$\mathscr{H}$ be a Hilbert space and $\mathscr{B(\mathscr{H})}$ be the algebra
of bounded linear maps on $\mathscr{H}$. Suppose
$T\in\mathscr{B\mathscr{(H)}}$. The numerical range of $T$ denoted by $W(T)$
is defined as

$W(T):=\left\\{\langle Tf,f\rangle:f\in\mathscr{H},\|f\|=1\right\\}.$

It was studied extensively. A celebrated result in the theory of the numerical
range is that $W(T)$ is a convex set which is known as Toeplitz Hausdorff
Theorem (cf. [10]). Durszt [5] showed that the numerical range of a normal
operator $T$ is the intersection of all convex Borel set $s$ in $\mathbb{C}$
such that $E(s)=I$ where $E$ is the unique spectral measure associated with
$T$ and used it to settle a conjecture of Halmos [12] in negative.

Choi, Kribs and Życzkowski have first defined the higher rank numerical range
in the context of “quantum error correction” [1] which is defined in the
following way. Let $T\in\mathscr{B\mathscr{(H)}}$ and
$k\in\mathbb{N}_{\infty}:=\mathbb{N}\cup\\{\infty\\}$. The $k$-rank numerical
range of $T$ denoted by $\Lambda_{k}(T)$ is defined as

$\Lambda_{k}(T):=\left\\{\lambda\in\mathbb{C}:PTP=\lambda P,\text{ for some
projection }P\text{ of rank }k\right\\}$

or, equivalently $\lambda\in\Lambda_{k}(T)$ if and only if there is an
orthonormal set $\\{f_{j}\\}_{j=1}^{k}$ such that $\langle
Tf_{j},f_{k}\rangle=\lambda\delta_{j,k}$ for $j,k\in\\{1,2,\cdots,k\\}$.
Clearly,

$W(T)=\Lambda_{1}(T)\supseteq\Lambda_{2}(T)\supseteq\cdots\supseteq\Lambda_{k}(T)\supseteq\cdots.$

Choi, Kribs and Życzkowski have given a description of the higher rank
numerical range of self-adjoint matrices in terms of its eigenvalues [1]. Let
$M_{n}$ be the algebra of $n$-by-$n$ matrices with complex entries. Suppose
$T\in M_{n}$ be a self-adjoint matrix and $1\leq k\leq n$. CKZ showed

$\displaystyle\Lambda_{k}(T)=\left[\lambda_{n-k+1},\lambda_{k}\right]$ (1.1)

where $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n}$ are eigenvalues of
$T$. In the same article, they put forward a conjecture (CKZ conjecture) on
the higher rank numerical range of normal matrices. Let $T\in M_{n}$ be a
normal matrix and $1\leq k\leq n$. CKZ conjecture says

$\displaystyle\Lambda_{k}(T)=\bigcap\limits_{1\leq j_{1}<\cdots<j_{n-k+1}\leq
n}\text{ conv }\left\\{\lambda_{j_{1}},\ldots,\lambda_{j_{n-k+1}}\right\\}$
(1.2)

where $\lambda_{1},\ldots\lambda_{n}$ are eigenvalues of $T$. Condition for
the higher rank numerical range of a matrix being non-empty was also studied
in [1] and [13].

Li, Poon and Sze have described the higher rank numerical range of a matrix as
the intersection of the closed half plane determined by eigenvalue which
settled CKZ conjecture in affirmative [14]. Let $T\in M_{n}$ and $1\leq k\leq
n$. They showed

$\displaystyle\Lambda_{k}(T)=\bigcap\limits_{\xi\in[0,2\pi)}\left\\{\mu\in\mathbb{C}:\Re({e^{i\xi}\mu})\leq\lambda_{k}(\Re({e^{i\xi}T}))\right\\}.$
(1.3)

It also proved that the higher rank numerical range of a matrix is always
convex. However, the convexity of the higher rank numerical range of any
operator was first independently shown by Woerdeman [20] (see also [2]).

There are some work on extending the finite dimensional results on the higher
rank numerical range to infinite dimension. Avendaño has described the higher
rank numerical range of self-adjoint operators [16] on infinite dimensional
Hilbert space in terms of the spectral measure associated with it which
extends (1.1). Let $T\in\mathscr{B(\mathscr{H})}$ be self-adjoint operator and
$E$ be the unique spectral measure associated with $T$. Suppose
$k\in\mathbb{N}_{\infty}$. Let

$\displaystyle A_{k}:=\left\\{a\in\mathbb{R}:\text{ dim
ran}E(-\infty,a]<k\right\\},$ $\displaystyle
B_{k}:=\left\\{b\in\mathbb{R}:\text{ dim ran}E[b,\infty)<k\right\\},$
$\displaystyle\widetilde{\Omega}_{k}:=A_{k}^{c}\cap B_{k}^{c}.$

Avendaño proved the following.

######  Theorem 1.1 (Theorem 3.4, [16]).

Let $T\in\mathscr{B(\mathscr{H})}$ be self-adjoint operator and
$k\in\mathbb{N}_{\infty}$. Then

$\displaystyle\Lambda_{k}(T)=\bigcap\limits_{V\in\mathscr{V}_{k}}W(V^{*}TV)=\widetilde{\Omega}_{k}$

where $\mathscr{V}_{k}$ be the set of all isometries
$V:\mathscr{H}\rightarrow\mathscr{H}$ such that codimension of ran $V$ is less
than $k$.

In other words, the above theorem states that $\lambda\in\Lambda_{k}(T)$ if
and only if $\text{dim ran}E(-\infty,\lambda]\geq k$ and $\text{dim
ran}E[\lambda,\infty)\geq k$ for $k\in\mathbb{N}_{\infty}$ and self-adjoint
$T$. We also mention here the fact that the $k$-rank numerical range of an
operator on infinite dimensional Hilbert space is always non-empty for
$k\in\mathbb{N},$ obtained as a corollary of Theorem 1.1 (Corollary 3.5,
[16]). There is also an independent proof of the above fact in [13]. However,
$\infty$-rank numerical range of an operator could be empty (cf. [15], [16]).

Li, Poon and Sze extended (1.3) for any operator. Suppose
$T\in\mathscr{B\mathscr{(H)}}$ and $k\in\mathbb{N}$. Let

$V_{k}(T):=\bigcap\limits_{\xi\in[0,2\pi)}\left\\{\mu\in\mathbb{C}:\Re({e^{i\xi}\mu})\leq\lambda_{k}(\Re({e^{i\xi}T}))\right\\}$

where

$\lambda_{k}(H):=\sup\left\\{\lambda_{k}(V^{*}HV):V:\mathbb{C}^{k}\rightarrow\mathscr{H}\text{
such that }V^{*}V=I\right\\}$

for some selfadjoint operator $H\in\mathscr{B(\mathscr{H})}$. Let
$\text{{Int}}(S)$ denote relative interior of $S$ and $\overline{S}$ denote
the closure of $S$ where $S\subseteq\mathbb{C}$. Li and etc. showed the
following.

######  Theorem 1.2 (Theorem 2.1, [15]).

Let $T\in\mathscr{B(\mathscr{H})}$ and $k\in\mathbb{N}$. Then

$\text{{Int}}\left(V_{k}(T)\right)\subseteq\Lambda_{k}(T)\subseteq
V_{k}(T)=\overline{\Lambda_{k}(T)}.$

In the same article, Li and etc. have also obtained that
$\Lambda_{\infty}(T)=\bigcap\limits_{k\geq 1}\Lambda_{k}(T)$ (Theorem 5.1,
[15]) for $T\in\mathscr{B(\mathscr{H})}$. For more details on the higher rank
numerical range of operators on infinite dimensional Hilbert space, the reader
may look at [9], [15], [16].

In this paper, we describe the higher rank numerical range of a normal
operator on an infinite dimensional Hilbert space in terms of the spectral
measure associated with it (Theorem 3.2) which extends (1.2). An analogous
description of the numerical range of normal operators by Durszt [5] is
derived for the higher rank numerical ranges as an immediate corollary
(Corollary 3.3). It has several other applications. (Corollary 5.4, Corollary
5.5). An elementary alternative proof of Theorem 4.4 [15] is also given
(Corollary 5.7). The description of the higher rank numerical range of the
bilateral shift is immediate using the main theorem of this paper (Example
5.1).

As an application of the main theorem of this paper, we show using Durszt’s
[5] example that there exists a normal contraction $T$ for which the higher
rank numerical range of all unitary dilations of $T$ contains the higher rank
numerical range of $T$ as a proper subset (Theorem 6.4). We strengthen and
generalize a result of Wu [19] by providing a necessary and sufficient
condition on a normal contraction $T$ that its higher rank numerical range
equals the intersection of the higher rank numerical ranges of unitary
dilations of $T$ (Theorem 6.9). Finally, we propose a necessary and sufficient
condition on a contraction $T$ acting on an infinite dimensional Hilbert space
that its higher rank numerical range equals the intersection of the higher
rank numerical ranges of unitary dilations of $T$ (Conjecture 6.10).

##  2\. Preliminaries

We, throughout this paper, denote by $\mathscr{H}$, an infinite dimensional
separable Hilbert space and $\mathscr{B(\mathscr{H})}$, the algebra of all
bounded linear maps on $\mathscr{H}.$ Set
$\mathbb{N}_{\infty}:=\mathbb{N}\cup\\{\infty\\}$. For
$T\in\mathscr{B(\mathscr{H})}$, we denote by $\sigma_{e}(T)$, the eigen-
spectrum of $T$ and $\sigma(T)$, the spectrum of $T$. Let us denote
$\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$ and
$\mathbb{T}:=\\{z\in\mathbb{C}:|z|=1\\}$. We will also continue to use
$S^{\circ}$, the topological interior of $S$ and $\delta S$, the boundary of
$S$ for $S\subseteq\mathbb{C}$.

Let $k\in\mathbb{N}_{\infty}$. If the dimension of a subspace is less than $k$
then we mean that the dimension of that subspace is at most $k-1$ whenever $k$
is finite and same is finite whenever $k$ is infinite.

Let us start with few basic lemmas.

######  Lemma 2.1.

Let $X$ be a topological space and $S\subseteq X$. Then
$\overline{S}=S^{\circ}\cup\delta S.$

###### Proof.

Let $x\in\overline{S}$ such that $x\notin S^{\circ}$. Suppose $U$ be an open
set containing $x$. As $x\in\overline{S}$, we have $S\cap U\neq\emptyset$.
Now, we claim that $U\cap S^{c}\neq\emptyset$. If possible let $U\cap
S^{c}=\emptyset$. Then $U\subseteq S$ which contradicts that $x\notin
S^{\circ}$. So, $U\cap S^{c}\neq\emptyset$. Hence $x\in\delta S$. The other
containment is immediate. ∎

######  Lemma 2.2.

Let $S$ be a convex set in $\mathbb{C}$ and $\lambda\notin S$. Then there
exists a closed half plane $H$ with the line $\delta H$ passing through
$\lambda$ such that $H$ contains $S$.

###### Proof.

It is immediate using separation theorem of convex set by half plane. ∎

######  Lemma 2.3.

Let $S$ be a convex set in $\mathbb{C}$. Then
$(\overline{S})^{\circ}=S^{\circ}$.

###### Proof.

Let $\lambda\in(\overline{S})^{\circ}$. Then there exists an open ball $B$
containing $\lambda$ such that $B\subseteq\overline{S}$. Let $\lambda\notin
S^{\circ}$. Then by Lemma 2.1, $\lambda\in\delta S$. So, $B\cap
S^{c}\neq\emptyset$. Let $\mu\in B\cap S^{c}$. As $\mu\notin S$ and $S$ is a
convex set, there exists a closed half plane $H$ with the line $\delta H$
passing through $\mu$ such that $S\subseteq H$ by Lemma 2.2. Again
$\mu\in(\overline{S})^{\circ}$. Then there exists an open ball $C$ containing
$\mu$ such that $C\subseteq\overline{S}$. Take $D=C\cap H^{c}$. Note that $D$
is non empty (as $\mu$ lies on $\delta H$), open set and contains boundary
point(s) of $S$ as $D=C\cap H^{c}\subseteq\overline{S}\cap H^{c}=\delta S\cap
H^{c}$. So, $D\cap S\neq\emptyset$ which is a contradiction. Hence $\lambda\in
S^{\circ}$. The other containment is trivial. ∎

######  Lemma 2.4 (Lemma, [5]).

Let $S$ be a convex set in the complex plane and $\mu$ be a non-negative Borel
measure defined on $S$ such that $\mu(S)=1$. Then $\int\limits_{S}zd\mu\in S$.

######  Lemma 2.5.

Let $k\in\mathbb{N}$ and $\\{P_{n}\\}_{n=1}^{\infty}$ be a sequence of
projection with $\text{dim ran}P_{n}<k$ for all $n$ such that
$P_{n}\xrightarrow{\text{SOT}}P$ where $P$ is a projection. Then $\text{dim
ran}P<k$.

###### Proof.

Let $\text{dim ran}P\geq k$. Suppose $\\{e_{r}:1\leq r\leq k\\}$ be an
orthonormal set in $\text{ran}P$. Let $A_{n}=\left(\langle
P_{n}e_{j},e_{i}\rangle\right)_{i,j=1}^{k}\in M_{k}$. As
$P_{n}\xrightarrow{\text{SOT}}P$, we have
$\lim\limits_{n\to\infty}P_{n}e_{r}=Pe_{r}=e_{r}$ for all
$r\in\\{1,\cdots,k\\}$. So, $A_{n}\rightarrow I_{k}$ in norm. Since the set of
all invertible matrices is open, there exists $n_{\circ}\in\mathbb{N}$ such
that $A_{n_{\circ}}$ is invertible which implies $\\{P_{n_{\circ}}e_{r}:1\leq
r\leq k\\}$ is linearly independent. It contradicts the fact that $\text{dim
ran}P_{n}<k$ for all $n$. ∎

######  Remark 2.6.

Let $\mathscr{H}$ be an infinite dimensional Hilbert space with orthonormal
basis $\\{e_{r}:r\geq 1\\}$. Suppose $P_{n}$ be projection on
$\text{span}\\{e_{1},e_{1},\cdots,e_{n}\\}$. Then
$P_{n}\xrightarrow{\text{SOT}}I$ where $I$ is the identity operator on
$\mathscr{H}$. Note that $\text{dim ran}P_{n}$ is finite for all $n$ but
$\text{ran}E(I)=\mathscr{H}$ is infinite dimensional. It shows that the above
lemma does not hold for $k=\infty$.

There are various versions of the spectral theorem. We will use the following
one.

######  Theorem 2.7.

(Spectral Theorem) Let $T\in\mathscr{B\mathscr{(H)}}$ be normal. Then there
exists a unique spectral measure $E$ defined on the Borel subsets of
$\mathbb{C}$ and supported on $\sigma(T)$ such that

$T=\int_{\mathbb{C}}x\;dE(x)$

Furthermore, every projection $E(\omega)$ commutes with every
$S\in\mathscr{B{\mathscr{(}H)}}$ which commutes with $T$ where $\omega$ is a
Borel set in $\mathbb{C}$.

A good reference for the spectral measure (resolution of identity) and the
spectral theorem is [18]. Finally, we will end this section by recalling the
following proposition.

######  Proposition 2.8 (Proposition $2.3$, [16]).

Let $T\in\mathscr{B\mathscr{(H)}}$ and $k\in\mathbb{N}_{\infty}$. Then

$\displaystyle\Lambda_{k}(T)\subseteq\bigcap\limits_{V\in\mathscr{V}_{k}}W(V^{*}TV)$

where $\mathscr{V}_{k}$ is the set of all isometries
$V:\mathscr{H}\rightarrow\mathscr{H}$ such that codimension of ran $V$ is less
than $k$.

##  3\. Higher rank numerical range of a normal operator

Let us start with the following definition.

######  Definition 3.1.

A subset $H\subseteq\mathbb{C}$ is said to be a half closed-half plane in
$\mathbb{C}$ if $H$ can be written as the union of an open half plane $\omega$
and a ray $l$ on the line $\delta\omega$. Moreover, if $\lambda$ be the
initial point of $l$ then we sometimes call $H$ as a half closed-half plane at
$\lambda$ generated by the line $\delta\omega$.

For example, $\\{z\in\mathbb{C}:\Im{(z)}>0\\}\cup[0,\infty)$ is a half closed-
half plane at $0$ generated by the real axis.

Let $T\in\mathscr{B(\mathscr{H})}$ be a normal operator and $E$ be the unique
spectral measure associated with $T$ defined on the Borel $\sigma$-algebra in
$\mathbb{C}$ supported on $\sigma(T)$. Suppose

$\displaystyle\mathcal{S}_{k}:=\left\\{H:H\text{ is a half closed-half plane
in }\mathbb{C}\text{ and dim ran }E(H)<k\right\\}.$ (3.1)

Define

$\displaystyle\Omega_{k}(T):=\bigcap\limits_{H\in\mathcal{S}_{k}}H^{c}.$ (3.2)

As arbitrary intersection of convex set is always convex, $\Omega_{k}(T)$ is a
convex set in $\mathbb{C}$. It can be checked that $\Omega_{k}(T)$ is non-
empty for $k\in\mathbb{N}$ (cf. Proposition 4.4). Later, we will also see an
example that $\Omega_{\infty}(T)$ could be empty (Example 5.2).

We are now ready to state the main theorem of this paper. A finite dimensional
version of our result is found at [14] (see also [3],[7]). It also extends the
description of the higher rank numerical range of self-adjoint operator acting
on an infinite dimensional Hilbert space [16].

######  Theorem 3.2.

Let $T\in\mathscr{B(\mathscr{H})}$ be a normal operator and
$k\in\mathbb{N}_{\infty}$. Then

$\displaystyle\Lambda_{k}(T)=\bigcap\limits_{V\in\mathscr{V}_{k}}W(V^{*}TV)=\Omega_{k}(T)$

where $\mathscr{V}_{k}$ is the set of all isometries
$V:\mathscr{H}\rightarrow\mathscr{H}$ such that codimension of ran $V$ is less
than $k$.

The proof of Theorem 3.2 will be given in the next section. It has many
applications. In this section, we, motivated by the description of the
numerical range of normal operators by Durszt [5], derive an analogous
description for the higher rank numerical range of normal operators as an
immediate consequence of Theorem 3.2.

######  Corollary 3.3.

Let $T\in\mathscr{B}\mathscr{(H)}$ be normal and $k\in\mathbb{N}_{\infty}$.
Suppose $\mathcal{R}_{k}$ is the collection of all convex Borel set $s$ with
$\text{dim ran}E(s^{c})<k$. Then
$\Lambda_{k}(T)=\bigcap\limits_{s\in\mathcal{R}_{k}}s$.

###### Proof.

Let $V_{k}(T)=\bigcap\limits_{s\in\mathcal{R}_{k}}s$. Let $H$ be a half
closed-half plane with $\text{dim ran}E(H)<k$. As $H^{c}$ is a convex Borel
set with $\text{dim ran}E(H)<k$, we have $H^{c}\in\mathcal{R}_{k}$. So,
$V_{k}(T)\subseteq\Omega_{k}(T)$. Hence by Theorem 3.2, we obtain
$V_{k}(T)\subseteq\Lambda_{k}(T)$.

Let $\lambda\notin V_{k}(T)$. Then there exists a convex Borel set $s$ with
$\text{dim ran}E(s^{c})<k$ such that $\lambda\notin s$. As $s$ is convex and
$\lambda\notin s$, there exists a closed half plane $H$ with the line $\delta
H$ passing through $\lambda$ such that $s\subseteq H$ by Lemma 2.2. Let
$l_{1},l_{2}$ be two opposite directed rays with initial point $\lambda$ on
the line $\delta H$. Now, $s$ can not have non empty intersection with both
$l_{1}$ and $l_{2}$ as otherwise $\lambda\in s$ because of convexity of $s$.
Without loss of generality, let $s\cap l_{1}=\emptyset$. Take $K=H^{c}\cup
l_{1}\subseteq s^{c}$. As $\text{dim ran}E(s^{c})<k$, we have $\text{dim
ran}E(K)<k$. So, $K$ is a half closed-half plane at $\lambda$ with $\text{dim
ran}E(K)<k$. Now, by Theorem 3.2 and the definition of $\Omega_{k}(T)$, we
have $\Lambda_{k}(T)=\Omega_{k}(T)\subseteq K^{c}$. As $\lambda\notin K^{c}$,
we have $\lambda\notin\Lambda_{k}(T)$. This completes the proof. ∎

##  4\. Proof of the Main Theorem

We begin with few lemmas.

######  Lemma 4.1.

Let $T\in\mathscr{B(\mathscr{H})}$ be a normal operator and
$k\in\mathbb{N}_{\infty}$. Then $\lambda\in\Omega_{k}(T)$ if and only if for
any half closed-half plane $H$ at $\lambda$, $\text{dim ran }E(H)\geq k$.

###### Proof.

‘$\Rightarrow$’. Let there exists a half closed-half plane $H$ at $\lambda$
such that $\text{dim ran }E(H)<k$. Then by the definition of $\Omega_{k}(T)$,
we have $\Omega_{k}(T)\subseteq H^{c}$. As $\lambda\notin H^{c}$, we have
$\lambda\notin\Omega_{k}(T)$.
‘$\Leftarrow$’. Let $\lambda\notin\Omega_{k}(T)$. Then by the definition of
$\Omega_{k}(T)$, there exists a half closed-half plane $H$ containing
$\lambda$ such that $\text{dim ran}E(H)<k$. Let $L$ be the line passing
through $\lambda$ and parallel to the line $\delta H$. Consider a half closed-
half plane $H^{\prime}$ at $\lambda$ generated by $L$ such that
$H^{\prime}\subseteq H$. Then $\text{dim ran}E(H^{\prime})<k$. ∎

Let $T$ be normal and $A,B\subseteq\mathbb{C}$. We denote $A\subseteq B$ a.e.
to mean $E(A\setminus B)=0$.

######  Lemma 4.2.

Let $T\in\mathscr{B(\mathscr{H})}$ be a normal operator and $k\in\mathbb{N}$.
Suppose $\lambda\in\Omega_{k}(T)$. Then $\lambda\in\delta\Omega_{k}(T)$ if and
only if there exists a half closed-half plane $H$ at $\lambda$ such that
$\text{dim ran}E(H^{\circ})<k$.

###### Proof.

‘$\Leftarrow$’. Let $\lambda\notin\delta\Omega_{k}(T)$. Then
$\lambda\in\Omega_{k}(T)^{\circ}$ as
$\Omega_{k}(T)=\Omega_{k}(T)^{\circ}\cup\left(\delta\Omega_{k}(T)\cap\Omega_{k}(T)\right)$
by Lemma 2.1. Let $H$ be a half closed-half plane at $\lambda$. As
$\lambda\in\Omega_{k}(T)^{\circ}$, it is possible to choose a point
$\mu\in\Omega_{k}(T)$ and a half closed-half plane $H^{\prime}$ at $\mu$ such
that $H^{\prime}\subseteq H^{\circ}$. Now, by Lemma 4.1, we have $\text{dim
ran}E(H^{\prime})\geq k$ and hence $\text{dim ran}E(H^{\circ})\geq k$ (note
that we do not need $k$ to be finite to prove this direction).
‘$\Rightarrow$’. Let $\lambda\in\delta\Omega_{k}(T)$. Then there exists a
sequence $\\{\lambda_{n}\\}_{n=1}^{\infty}$ not in $\Omega_{k}(T)$ and lying
on a ray with initial point $\lambda$ (which is possible as $\Omega_{k}(T)$ is
convex) such that $\lim\limits_{n\to\infty}\lambda_{n}=\lambda$. Since
$\lambda_{n}\notin\Omega_{k}(T)$, by Lemma 4.1, there exists a half closed-
half plane $H_{n}$ at $\lambda_{n}$ such that $\text{dim ran}E(H_{n})<k$ for
each $n$. Let $L_{n}=\delta H_{n}$. Suppose $L_{n}^{\prime}$ be the line
passing through $\lambda$ and parallel to $L_{n}$ and $H_{n}^{\prime}$ be a
half closed-half plane generated by $L_{n}^{\prime}$ lying on the side of
$H_{n}$. Let $\theta_{n}$ be the slope of $L_{n}^{\prime}$. As
$\theta_{n}\in[0,\pi]$ for all $n$ and $[0,\pi]$ is compact,
$\\{\theta_{n}\\}_{n=1}^{\infty}$ has a convergent subsequence which we call
$\\{\theta_{n}\\}_{n=1}^{\infty}$ itself and let
$\lim\limits_{n\to\infty}\theta_{n}=\theta$. Let $L$ be the line with slope
$\theta$ and passing through $\lambda$. Suppose $H$ is a half closed-half
plane generated by $L$ such that $H^{\circ}$ lies on the side of $H_{n}$ for
all but finitely many $n$. Now, we will show $H$ is the desired half closed-
half plane at $\lambda$.

Let $\text{dim ran}E(H^{\circ})\geq k$. We claim that there exists a closed
half plane $\omega$ generated by a line $L^{\prime}$ parallel to $L$ such that
$\omega\subseteq H^{\circ}$ and $\text{dim ran}E(\omega)\geq k$. Indeed, if it
be not true then $H^{\circ}=\bigcup\limits_{n=1}^{\infty}\omega_{n}$ where
$\\{\omega_{n}\\}_{n=1}^{\infty}$ is an increasing sequence of closed half
plane with $\text{dim ran}E(\omega_{n})<k$ for all $n$. Then
$E(\omega_{n})\xrightarrow{\text{SOT}}E(H^{\circ})$. As $\text{dim
ran}E(\omega_{n})<k$ for all $n$, by Lemma 2.5, we obtain $\text{dim
ran}E(H^{\circ})<k$ which is a contradiction. Hence the claim follows.

Let $Q$ be a closed and bounded rectangle such that $\sigma(T)\subseteq Q$.
So, $E(Q^{c})=0$. Let $S$ be a strip generated by $L^{\prime}$ and the mirror
image of $L^{\prime}$ with respect to $L$. As
$\lim\limits_{n\to\infty}\theta_{n}=\theta$ and $L_{n}$ being parallel to
$L_{n}^{\prime},\lim\limits_{n\to\infty}\lambda_{n}=\lambda,E(Q^{c})=0$, there
exists $n_{\circ}\in\mathbb{N}$ such that $H_{n_{\circ}}^{\prime}\setminus
H_{n_{\circ}}\subseteq S$ a.e. which implies $\omega\subseteq H_{n_{\circ}}$
a.e. As $\text{dim ran}E(H_{n_{\circ}})<k$, we have $\text{dim
ran}E(\omega)<k$ which is a contradiction. Hence $\text{dim
ran}E(H^{\circ})<k$. ∎

######  Lemma 4.3.

Let $T\in\mathscr{B(\mathscr{H})}$ be a normal operator. Then
$\Omega_{\infty}(T)=\bigcap\limits_{k\geq 1}\Omega_{k}(T)$.

###### Proof.

It is clear that $\Omega_{\infty}(T)\subseteq\bigcap\limits_{k\geq
1}\Omega_{k}(T)$ by Lemma 4.1. Let $\lambda\in\bigcap\limits_{k\geq
1}\Omega_{k}(T)$. Then $\lambda\in\Omega_{k}(T)$ for all $k\geq 1$. Let $H$ be
a half closed-half plane at $\lambda$. Then by Lemma 4.1, $\text{dim ran
}E(H)\geq k$ for all $k\geq 1$. So, $\text{dim ran }E(H)\geq\infty$. Again, by
Lemma 4.1, we get $\lambda\in\Omega_{\infty}(T)$. ∎

Let us now observe the following property before going to the proof of the
main theorem.

######  Proposition 4.4.

Let $T\in\mathscr{B(\mathscr{H})}$ be normal and $k\in\mathbb{N}$. Then
$\Omega_{k}(T)$ is non-empty.

###### Proof.

Let $V_{k}(T)=\bigcap\limits_{H\in\mathcal{S}_{k}}\overline{H^{c}}$. Then note
that $\lambda\notin V_{k}(T)$ if and only if there exists a half closed half
plane $H$ with $\text{dim ran}E(H)<k$ such that $\lambda\in H^{\circ}$.
Clearly $\Omega_{k}(T)\subseteq V_{k}(T)$. Let us first show that
$V_{k}(T)\neq\emptyset$. Suppose $V_{k}(T)=\emptyset$. Then
$\mathbb{C}=\bigcup\limits_{H\in\mathcal{S}_{k}}H^{\circ}$. Let $Q$ be a
closed and bounded rectangle in $\mathbb{C}$ such that $\sigma(T)\subseteq Q$.
Then $\text{dim ran}E(Q^{c})=0$. Now,
$Q=\bigcup\limits_{H\in\mathcal{S}_{k}}(H^{\circ}\cap Q)$. As $Q$ is compact,
there exist $H_{1},\cdots,H_{n}\in\mathcal{S}_{k}$ such that
$Q=\bigcup\limits_{i=1}^{n}({H_{i}}^{\circ}\cap Q)$. Since
${H_{i}}^{\circ}\cap Q\subseteq H_{i}$ and $H_{i}\in\mathcal{S}_{k}$ for all
$i\in\\{1,2,\cdots,n\\}$, we have $\text{dim ran}E(Q)$ is finite which is a
contradiction. So, $V_{k}(T)$ is a non-empty, closed and convex set. We will
now consider the following cases and show that $\Omega_{k}(T)\neq\emptyset$ in
each case .

Case I: Let $V_{k}(T)$ be a singleton set. Without loss of generality, we may
take $V_{k}(T)=\\{0\\}$. It is sufficient to show $\text{dim
ran}E(\mathbb{C}\setminus\\{0\\})$ is finite as then $0$ is an eigenvalue of
$T$ with infinite multiplicity and hence $0\in\Omega_{k}(T)$. Let
$(p_{n})_{n=1}^{\infty}$ be a sequence of positive real numbers converging to
$0$. As $p_{n}\notin V_{k}(T)$, there exists a half closed-half plane $H_{n}$
at $p_{n}$ such that $\text{dim ran}E(H_{n})<k$. Let $L_{n}$ be the line
passing through $0$ and parallel to the line $\delta H_{n}$. Suppose
$\theta_{n}$ be the slope of $L_{n}$. As $\theta_{n}\in[0,\pi]$ for all $n$
and $[0,\pi]$ is compact, $(\theta_{n})_{n=1}^{\infty}$ has a convergent
subsequence say $(\theta_{n})_{n=1}^{\infty}$ itself and
$\lim\limits_{n\to\infty}\theta_{n}=\theta$. As $k\in\mathbb{N}$, by the same
reason argued in Lemma 4.2, there exists a half closed-half plane $H$ at $0$
with $\theta$ being the slope of the line $\delta H$ such that $\text{dim
ran}E(H^{\circ})<k$. Clearly $H$ contains $p_{n}$ for all $n$.

Now, by applying the same argument to a sequence of negative real numbers
converging to $0$, there exists a half closed-half plane $K_{1}$ at $0$ such
that $\text{dim ran}E(K_{1}^{\circ})<k$. If $\delta K_{1}=\delta H$ then
$\text{dim ran}E(\mathbb{C}\setminus\\{0\\})<4k$ as $\text{dim ran}E((\delta
K_{1})^{c})<2k$ and by applying the same argument to a sequence converging to
$0$ lying on each $L_{i}\setminus\\{0\\}$, we have $\text{dim
ran}E(L_{i}\setminus\\{0\\})<k$ for all $i\in\\{1,2\\}$ where $L_{1},L_{2}$ be
are two opposite directed rays with initial point $0$ on the line $\delta
K_{1}$. If $\delta K_{1}\neq\delta H$ then there exists a triangle
$\Delta_{1}$ with $0$ as one of the vertices such that $\text{dim
ran}E(\Delta_{1}^{c})<2k$. Now, choose a sequence $(q_{n})_{n=1}^{\infty}$
converging to $0$ lying on $l\setminus\\{0\\}$ where $l$ is the line segment
joining $0$ and the midpoint of the side of $\Delta_{1}$ opposite to $0$.
Again applying the same argument to $(q_{n})_{n=1}^{\infty}$, there exists a
half closed-half plane $K_{2}$ at $0$ such that $\text{dim
ran}E(K_{2}^{\circ})<k$. If the line $\delta K_{2}$ does not intersect
$\Delta_{1}$ other than $0$ then we get $\text{dim
ran}E(\mathbb{C}\setminus\\{0\\})<3k$ as $\text{dim ran}E(\Delta_{1}^{c})<2k$
and $\text{dim ran}E(\Delta_{1}\setminus\\{0\\})\leq\text{dim
ran}E(K_{2}^{\circ})<k$ (since $\Delta_{1}\setminus\\{0\\}\subseteq
K_{2}^{\circ}$). Now, if the line $\delta K_{2}$ passes through any side of
$\Delta_{1}$ adjacent to $0$, then we have $\text{dim
ran}E(\mathbb{C}\setminus\\{0\\})<4k$ as $\text{dim ran}E((\delta
K_{2})^{c})<2k$ and $\text{dim ran}E(M_{i}\setminus\\{0\\})<k$ for all
$i\in\\{1,2\\}$ where $M_{1},M_{2}$ be are two opposite directed rays with
initial point $0$ on the line $\delta K_{2}$. Finally, if the line $\delta
K_{2}$ intersects $\Delta_{1}^{\circ}$ then there exists another triangle
$\Delta_{2}$ with $0$ as one of the vertices such that
$\Delta_{2}\subsetneq\Delta_{1}$ and $\text{dim ran}E(\Delta_{2}^{c})<2k$. If
the final case continues to occur repeatedly then we will end up with
$\text{dim ran}E(\mathbb{C}\setminus\\{0\\})<4k$. Therefore, in any case, we
have $\text{dim ran}E(\mathbb{C}\setminus\\{0\\})<4k$. Hence we are done in
this case.

Case II: Let $V_{k}(T)$ be a nondegenerate line segment. Without loss of
generality, we may assume $V_{k}(T)=\left[-p,p\right]\subseteq\mathbb{R}$
where $p>0$. Let $\omega_{1}:=\\{z\in\mathbb{C}:\Im(z)>0\\}$. Let
$(z_{n})_{n=1}^{\infty}$ be a sequence lying on the positive imaginary axis
converging to $0$. Claim: $\text{dim ran}E(\omega_{1})<k$. If not then as
$k\in\mathbb{N}$, there exists a closed half plane $\omega$ generated by a
line parallel to the real axis such that $\omega\subseteq\omega_{1}$ and
$\text{dim ran}E(\omega)\geq k$ (by the similar reason argued in the proof of
Lemma 4.2). As $z_{n}\notin V_{k}(T)$, there exists a half closed-half plane
$H_{n}$ with $\text{dim ran}E(H_{n})<k$ such that $z_{n}\in H_{n}^{\circ}$.
Observe that the line $\delta H_{n}$ can not intersect $(-p,p)$. Indeed, if so
at $q\in(-p,p)$ then either $\frac{q+p}{2}\notin V_{k}(T)$ or
$\frac{-p+q}{2}\notin V_{k}(T)$ as $V_{k}(T)\subseteq\overline{H_{n}^{c}}$.
This is a contradiction. Now, since $\delta H_{n}$ does not intersect $(-p,p)$
for every $n\in\mathbb{N},\lim\limits_{n\to\infty}z_{n}=0$ and $E(Q^{c})=0$,
it follows that there exists $n_{\circ}\in\mathbb{N}$ such that
$\omega\subseteq H_{n_{\circ}}$ a.e. which implies $\text{dim
ran}E(\omega)<k$. This is a contradiction. So, $\text{dim
ran}E(\omega_{1})<k$. Similarly, we have $\text{dim ran}E(\omega_{2})<k$ where
$\omega_{2}:=\\{z\in\mathbb{C}:\Im(z)<0\\},\text{ dim ran}E(p,\infty)<k$ and
$\text{ dim ran}E(-\infty,-p)<k$. Therefore, $\text{dim ran}E([-p,p]^{c})<4k$
which implies $\text{dim ran}E([-p,p])$ is infinite. Now, as $k\in\mathbb{N}$,
by Lemma 3.2, [16] and Lemma 4.1, we have
$\emptyset\neq\Omega_{k}(P_{\text{ran}E([-p,p])}T|_{\text{ran}E([-p,p])})\subseteq\Omega_{k}(T)$.
Hence $\Omega_{k}(T)\neq\emptyset$ for $k\in\mathbb{N}$.

Case III: Let $V_{k}(T)$ contains atleast three non-collinear points say $p,q$
and $r$. As $V_{k}(T)$ is a closed and convex set, $V_{k}(T)$ contains the
closed triangle $\Delta$ with vertices $p,q$ and $r$. Let $s$ be an interior
point of $\Delta$. If possible let $s\notin\Omega_{k}(T)$. By Lemma 4.1, there
exist a half closed-half plane $H$ at $s$ such that $\text{dim ran}E(H)<k$.
Then $V_{k}(T)\subseteq\overline{H^{c}}$, which contradicts that $V_{k}(T)$
contains $\Delta$. So, $s\in\Omega_{k}(T)$. Hence we are done in this case
also. ∎

######  Remark 4.5.

Let $k\in\mathbb{N}$. Observe that if $V_{k}(T)=\\{0\\}$ then $0$ is an
eigenvalue of $T$ with infinite multiplicity. However, this is not true for
$k=\infty$ as discussed later (Remark 5.3).

Let $\lambda_{1},\lambda_{2}\in\mathbb{C}$. We denote by
$(\lambda_{1},\lambda_{2})=\\{t\lambda_{1}+(1-t)\lambda_{2}:0<t<1\\}$, the
open line segment joining $\lambda_{1}$ and $\lambda_{2}$ and
$[\lambda_{1},\lambda_{2}]=\\{t\lambda_{1}+(1-t)\lambda_{2}:0\leq t\leq 1\\}$,
the closed line segment joining $\lambda_{1}$ and $\lambda_{2}$. We are now
ready to prove the main theorem of this paper.

######  Proof of Theorem 3.2.

We will first prove

$\displaystyle\Lambda_{k}(T)\subseteq\bigcap\limits_{V\in\mathscr{V}_{k}}W(V^{*}TV)\subseteq\Omega_{k}(T)$

for $k\in\mathbb{N}_{\infty}$. Let $k\in\mathbb{N}_{\infty}$ and $H$ be a half
closed-half plane with $\text{dim ran}E(H)<k$. Since codimension of
$\text{ran}E(H^{c})$ is finite and $\mathscr{H}$ is infinite dimensional,
there exists an isometry $V_{H}$ with range $\text{ran}E(H^{c})$. Clearly
$V_{H}\in\mathscr{V}_{k}$ and $V_{H}V_{H}^{*}=E(H^{c})$. Denote
$\mathscr{K}=\text{ran}E(H^{c})$. Observe that
$W(V_{H}^{*}TV_{H})=W(P_{\mathscr{K}}T|_{\mathscr{K}})$. Let $\lambda\in
W(P_{\mathscr{K}}T|_{\mathscr{K}})$. Then there exists $f\in\mathscr{K}$ with
$\|f\|=1$ such that

$\displaystyle\lambda=\langle
P_{\mathscr{K}}T|_{\mathscr{K}}f,f\rangle=\langle
Tf,f\rangle=\int_{\mathbb{C}}x\;d\langle
E(\centerdot)f,f\rangle=\int_{H^{c}}x\;d\|E(\centerdot)f\|^{2}$

Also $E_{f,f}(H^{c})=\langle E(H^{c})f,f\rangle=\|f\|^{2}=1$. So, by Lemma
2.4, we have $\lambda\in H^{c}$. Hence we obtain $W(V_{H}^{*}TV_{H})\subseteq
H^{c}$. Now, using Proposition 2.8, we have

$\displaystyle\Lambda_{k}(T)\subseteq\bigcap\limits_{V\in\mathscr{V}_{k}}W(V^{*}TV)\subseteq\bigcap\limits_{H\in\mathcal{S}_{k}}W(V_{H}^{*}TV_{H})\subseteq\bigcap\limits_{H\in\mathcal{S}_{k}}H^{c}=\Omega_{k}(T).$

We will now prove

$\displaystyle\Omega_{k}(T)\subseteq\Lambda_{k}(T)$ (4.1)

for $k\in\mathbb{N}$. We will show it by induction on $k$. Let
$\lambda\in\Omega_{1}(T)$. If possible let $\lambda\notin W(T)$. As $W(T)$ is
convex, by Lemma 2.2, there exists a closed half plane $H^{\prime}$ with the
line $\delta H^{\prime}$ passing through $\lambda$ such that $W(T)\subseteq
H^{\prime}$. Let $L_{1}^{\prime},L_{2}^{\prime}$ be two opposite directed rays
with initial point $\lambda$ on the line $\delta H^{\prime}$. Now,
$L_{1}^{\prime}$ and $L_{2}^{\prime}$ both can not have non empty intersection
with $W(T)$ as otherwise $\lambda\in W(T)$ because of convexity of $W(T)$. So,
without loss of generality, let $L_{1}^{\prime}$ has empty intersection with
$W(T)$. Let $\mathscr{K}_{1}=\text{ran}E(L_{1}^{\prime})$. We claim that
$\text{dim}\mathscr{K}_{1}=0$. If $\text{dim }\mathscr{K}_{1}$ be positive and
finite then $P_{\mathscr{K}_{1}}T|_{\mathscr{K}_{1}}$ has eigenvalue on
$L_{1}^{\prime}$ which implies $T$ has eigenvalue on $L_{1}^{\prime}$ (since
$\mathscr{K}_{1}$ is reducible subspace with respect to $T$) which contradicts
that $L_{1}^{\prime}$ has empty intersection with $W(T)$. If $\text{dim
}\mathscr{K}_{1}$ be infinite then $\emptyset\neq
W\left(P_{\mathscr{K}_{1}}T|_{\mathscr{K}_{1}}\right)\subseteq W(T)\cap
L_{1}^{\prime}$ which again contradicts that $L_{1}^{\prime}$ has empty
intersection with $W(T)$. So, $\text{dim }\mathscr{K}_{1}=0$. Then
$\widetilde{H}={H^{\prime}}^{c}\cup L_{1}^{\prime}$ is a half closed-half
plane at $\lambda$ with $\text{dim ran}E(\widetilde{H})=0$. So, by Lemma 4.1,
$\lambda\notin\Omega_{1}(T)$ which contradicts our hypothesis. Hence
$\lambda\in W(T)$. It shows (4.1) is true for $k=1$.

Let us now assume that (4.1) is true for any $m<k$. Let us first show
$\Omega_{k}(T)\subseteq\overline{\Lambda_{k}(T)}$. Since for any
$a,b\in\mathbb{C},\;\overline{\Lambda_{k}(aT+bI)}=a\overline{\Lambda_{k}(T)}+b,$
$\Omega_{k}(aT+bI)=a\Omega_{k}(T)+b$ and $aT+bI$ is normal, without loss of
generality we show that whenever $0\in\Omega_{k}(T),$ it would imply
$0\in\overline{\Lambda_{k}(T)}$. Let $0\in\Omega_{k}(T)$. Let
$\theta\in[0,2\pi)$. Suppose $L_{\theta}$ be a straight line passing through
$0$ making angle $\theta$ with positive imaginary axis. Let
$\omega_{1}^{\theta},\omega_{2}^{\theta}$ are two open half planes induced by
$L_{\theta}$ and $L_{1}^{\theta},L_{2}^{\theta}$ are two opposite directed
rays with initial point $0$ on $L_{\theta}$. Then
$H_{1}^{\theta}=\omega_{1}^{\theta}\cup L_{1}^{\theta}$ and
$H_{2}^{\theta}=\omega_{2}^{\theta}\cup L_{2}^{\theta}$ are two half closed-
half planes at $0$. As $0\in\Omega_{k}(T)$, we have $\text{dim
ran}E(H_{j}^{\theta})\geq k$ for all $j\in\\{1,2\\}$ by Lemma 4.1. Let
$S_{\theta}=\Re{(e^{i\theta}T)}$ and $E_{S}^{\theta}$ be the spectral measure
associated with $S_{\theta}$. Then we have $\text{dim
ran}E_{S}^{\theta}\left(\Re(e^{i\theta}H_{j}^{\theta})\right)\geq k$ for all
$j\in\\{1,2\\}$ where we denote $\Re(C):=\\{\Re(z):z\in C\\}$ for some
$C\subseteq\mathbb{C}$. Let

$\displaystyle A_{k}^{\theta}:=\\{a_{\theta}\in\mathbb{R}:\text{dim
ran}E_{S}^{\theta}(-\infty,a_{\theta}]<k\\},$ $\displaystyle
B_{k}^{\theta}:=\left\\{b_{\theta}\in\mathbb{R}:\text{dim
ran}E_{S}^{\theta}[b_{\theta},\infty)<k\right\\}.$

As $\text{dim
ran}E_{S}^{\theta}\left(\Re(e^{i\theta}H_{j}^{\theta})\right)\geq k$ for all
$j\in\\{1,2\\}$, we obtain $0\notin A_{k}^{\theta}$ and $0\notin
B_{k}^{\theta}$. So,
$0\in{A_{k}^{\theta}}^{c}\cap{B_{k}^{\theta}}^{c}=\Lambda_{k}(\Re({e^{i\theta}T)})$
by Theorem 1.1. Then there exists an isometry
$V_{\theta}:\mathbb{C}^{k}\rightarrow\mathscr{H}$ such that
${V_{\theta}}^{*}\Re({e^{i\theta}T)}V_{\theta}=0I_{k}$. So,
$0=\lambda_{k}({V_{\theta}}^{*}\Re({e^{i\theta}T)}V_{\theta})\leq\lambda_{k}(\Re({e^{i\theta}T)})$
for all $\theta\in[0,2\pi)$. By Theorem 1.2, we get
$0\in\overline{\Lambda_{k}(T)}$. Hence we obtain
$\Omega_{k}(T)\subseteq\overline{\Lambda_{k}(T)}$. As $\Lambda_{k}(T)$ is a
convex set in $\mathbb{C}$, we have
$\overline{\Lambda_{k}(T)}^{\circ}=\Lambda_{k}(T)^{\circ}$ by Lemma 2.3. So,

$\displaystyle\Omega_{k}(T)^{\circ}\subseteq\Lambda_{k}(T)^{\circ}$ (4.2)

Next, we will show
$\delta\Omega_{k}(T)\cap\Omega_{k}(T)\subseteq\Lambda_{k}(T)$. Let
$\lambda\in\delta\Omega_{k}(T)\cap\Omega_{k}(T)$. By Lemma 4.2, there exist a
half closed-half plane $H$ at $\lambda$ with $\text{dim ran }E(H^{\circ})<k$.
Let $L_{1},L_{2}$ be two opposite directed rays with initial point $\lambda$
on the line $\delta H$. Then $H_{1}=H^{\circ}\cup L_{1}$ and
$H_{2}=H^{\circ}\cup L_{2}$ are two half closed-half planes at $\lambda$. As
$\lambda\in\Omega_{k}(T)$, we have $\text{dim ran}E(H_{1})\geq k$ and
$\text{dim ran}E(H_{2})\geq k$ by Lemma 4.1. As $\text{dim
ran}E(H^{\circ})<k$, we have $\text{dim ran}E(L_{1})\geq 1$ and $\text{dim
ran}E(L_{2})\geq 1$. Let us now consider the following cases.

Case I: Let $\lambda$ be an eigenvalue of $T$ with eigenvector $f$. Let
$T=[\lambda]\oplus T_{0}$ with respect to the decomposing
$\mathscr{H}=\text{span}\\{f\\}\oplus\text{span}\\{f\\}^{\bot}$. Clearly
$T_{0}$ is normal. As $\lambda\in\Omega_{k}(T)$, we have
$\lambda\in\Omega_{k-1}(T_{0})$ by Lemma 4.1. So, by induction hypothesis, we
have $\lambda\in\Lambda_{k-1}(T_{0})$. Therefore, there exists a projection
$P_{0}$ of rank $k-1$ such that $P_{0}T_{0}P_{0}=\lambda P_{0}$. Take
$P=[1]\oplus P_{0}$. Then $P$ is a projection of rank $k$ such that
$PTP=\lambda P$. Hence $\lambda\in\Lambda_{k}(T)$.

Case II: Let $\lambda$ be not an eigenvalue of $T$ but $T$ has eigenvalues
both on $L_{1}$ and $L_{2}$ say $\lambda_{1}$ and $\lambda_{2}$ respectively.
Let $f_{1},f_{2}$ be two orthonormal eigenvectors of $T$ corresponding to
eigenvalues on $\lambda_{1},\lambda_{2}$ respectively. Let $T=T_{1}\oplus
T_{2}$ with respect to the decomposition of
$\mathscr{H}=\text{span}\\{f_{1},f_{2}\\}\oplus\text{span}\\{f_{1},f_{2}\\}^{\bot}$.
Clearly, $T_{1},T_{2}$ are both normal and $T_{1}$ is unitarily similar to
$\text{diag}(\lambda_{1},\lambda_{2})$. As $\lambda\in\Omega_{k}(T)$, we have
$\lambda\in\Omega_{1}(T_{1})$ and $\lambda\in\Omega_{k-1}(T_{2})$ by Lemma
4.1. Then by induction hypothesis, we have $\lambda\in\Lambda_{1}(T_{1})$ and
$\lambda\in\Lambda_{k-1}(T_{2})$. So, there exists an unit vector
$\\{g_{1}\\}$ in $\text{span}\\{f_{1},f_{2}\\}$ such that $\langle
T_{1}g_{1},g_{1}\rangle=\lambda$ and an orthonormal set
$\\{g_{j}\\}_{j=2}^{k}$ in $\text{ span}\\{f_{1},f_{2}\\}^{\bot}$ such that
$\langle T_{2}g_{j},g_{r}\rangle=\lambda\delta_{j,r}$ for
$j,r\in\\{2,\cdots,k\\}$. Then $\\{g_{j}\\}_{j=1}^{k}$ is an orthonormal set
and $\langle Tg_{j},g_{r}\rangle=\lambda\delta_{j,r}$ for all
$j,r\in\\{1,2,\cdots,k\\}$. Hence $\lambda\in\Lambda_{k}(T)$.

Case III: Suppose $T$ does not have eigenvalues on $L_{1}$ and $L_{2}$ both.
If both $\text{dim ran}E(L_{1})$ and $\text{dim ran}E(L_{2})$ are atleast $k,$
then by Theorem 1.1, we have

$\displaystyle\lambda\in\Lambda_{k}\left(P_{\text{ran}E(L)}T|_{\text{ran
}E(L)}\right)\subseteq\Lambda_{k}(T).$

Let one of $\text{dim ran}E(L_{1})$ and $\text{dim ran}E(L_{2})$ be less than
$k$. Without loss of generality, let $\text{dim ran}E(L_{2})=r<k$. Then
$E(L_{2})TE(L_{2})$ has $r$ eigenvalues on $L_{2}$ say
$\lambda_{1},\lambda_{2},\cdots,\lambda_{r}$ with
$0<|\lambda-\lambda_{1}|\leq\cdots\leq|\lambda-\lambda_{r}|$. As $T$ does not
have eigenvalues on $L_{1}$ and $L_{2}$ both, $\text{ dim ran}E(L_{1})$ is
infinite. Then there exists $\mu\in L_{1}$ such that $\text{dim
ran}E([\lambda,\mu])$ and $\text{dim ran}E\left([\lambda,\mu]^{c}\cap
L_{1}\right)$ are both infinite. Let $T=T_{1}\oplus T_{2}$ with respect to the
decomposing
$\mathscr{H}=\text{ran}E([\lambda_{1},\mu])\oplus\text{ran}E([\lambda_{1},\mu])^{\bot}$.
Clearly $T_{1},T_{2}$ are both normal. As $\lambda\in\Omega_{k}(T)$, we have
$\lambda\in\Omega_{1}(T_{1})$ and $\lambda\in\Omega_{k-1}(T_{2})$ by Lemma
4.1. So, by induction hypothesis, we have $\lambda\in\Lambda_{1}(T_{1})$ and
$\lambda\in\Lambda_{k-1}(T_{2})$. Then there exists an unit vector
$\\{g_{1}\\}$ in $\text{ran}E([\lambda_{1},\mu])$ such that $\langle
T_{1}g_{1},g_{1}\rangle=\lambda$ and an orthonormal set
$\\{g_{j}\\}_{j=2}^{k}$ in $\text{ran}E([\lambda_{1},\mu])^{\bot}$ such that
$\langle T_{2}g_{j},g_{r}\rangle=\lambda\delta_{j,r}$ for
$j,r\in\\{2,\cdots,k\\}$. Then $\\{g_{j}\\}_{j=1}^{k}$ is an orthonormal set
and $\langle Tg_{j},g_{r}\rangle=\lambda\delta_{j,r}$ for all
$j,r\in\\{1,2,\cdots,k\\}$. Hence $\lambda\in\Lambda_{k}(T)$.

So, combining all three cases, we conclude

$\displaystyle\delta\Omega_{k}(T)\cap\Omega_{k}(T)\subseteq\Lambda_{k}(T)$
(4.3)

Now, by (4.2), (4.3) and Lemma 2.1, we obtain
$\Omega_{k}(T)\subseteq\Lambda_{k}(T)$. Hence by principle of mathematical
induction, we have $\Omega_{k}(T)\subseteq\Lambda_{k}(T)$ for
$k\in\mathbb{N}$.

Finally, we will prove $\Omega_{\infty}(T)\subseteq\Lambda_{\infty}(T)$. Let
$\lambda\in\Omega_{\infty}(T)$. Then by Lemma 4.3, $\lambda\in\Omega_{k}(T)$
for all $k\geq 1$. By (4.1), we have $\lambda\in\Lambda_{k}(T)$ for all $k\geq
1$. So, $\lambda\in\bigcap\limits_{k\geq
1}\Lambda_{k}(T)=\Lambda_{\infty}(T)$. ∎

##  5\. Examples and Consequences

We compute here the higher rank numerical ranges of some standard normal
operators using Theorem 3.2. Our first example is the bilateral shift.

######  Example 5.1.

Let $S:l^{2}(\mathbb{Z})\rightarrow l^{2}(\mathbb{Z})$ such that

$\displaystyle S\left((\ldots,x_{-1},\boxed{x_{0}},x_{1},\ldots)\right)$
$\displaystyle=(\ldots,x_{-2},\boxed{x_{-1}},x_{0},\ldots),$
$\displaystyle\qquad\text{ where
}(\ldots,x_{-1},\boxed{x_{0}},x_{1},\ldots)\in l^{2}(\mathbb{Z}).$

Note that $S$ is unitary with $\sigma_{e}(S)=\emptyset$ and
$\sigma(S)=\mathbb{T}$. Let $S_{n}$ be $n$-dimensional backward shift. As
$S_{n}$ is a compression of $S$, we have
$W(S_{n})=\\{z\in\mathbb{C}:|z|\leq\cos\frac{\pi}{n+1}\\}\subseteq W(S)$.
Taking $n\to\infty$, we obtain $\mathbb{D}\subseteq W(S)$. Again,
$W(S)\subseteq\overline{W(S)}=\text{conv }\sigma(S)=\overline{\mathbb{D}}$.
Let $\lambda\in\mathbb{T}$ such that $\lambda\in W(S)$. Then there exists
$x\in l^{2}(\mathbb{Z})$ with $\|x\|=1$ such that $\lambda=\langle
Sx,x\rangle$. Now,

$\displaystyle 1=|\lambda|=|\langle Sx,x\rangle|\leq\|Sx\|\|x\|=1$

As Cauchy-Schwartz inequality is being achieved, we get that $\lambda$ is an
eigenvalue of $S$ which contradicts $\sigma_{e}(S)=\emptyset$. So,
$W(S)\subseteq\mathbb{D}$. Hence $W(S)=\mathbb{D}$.

Let $k\in\mathbb{N}_{\infty}$. Note that
$\Omega_{k}(S)\subseteq\Omega_{1}(S)=W(S)=\mathbb{D}$. Let $\mu\in\mathbb{D}$.
Suppose $H$ is a half closed-half plane at $\mu$. We claim that $\text{dim
ran}E(H)\geq k$. If $0<\text{dim ran}E(H)<k$ then $S$ has eigenvalue which is
a contradiction. If $\text{dim ran}E(H)=0$ then $\Omega_{1}(S)\subseteq H^{c}$
which contradicts that $\Omega_{1}(S)=\mathbb{D}$. So, $\text{dim ran}E(H)\geq
k$. By Lemma 4.1, we obtain $\mu\in\Omega_{k}(S)$. So,
$\Omega_{k}(S)=\mathbb{D}$. Hence by Theorem 3.2, $\Lambda_{k}(S)=\mathbb{D}$
for $k\in\mathbb{N}_{\infty}$.

However, the reader may look at Section $4$, [16] (Corollary 4.7, [16]) for a
different proof of the description of the higher rank numerical range of the
bilateral shift. The following example shows that $\infty$-rank numerical
range could be empty.

######  Example 5.2.

Consider $T=\bigoplus\limits_{n\geq 2}\begin{pmatrix}-\frac{1}{n}&0\\\
0&\frac{e^{\frac{i\pi}{n}}}{n}\end{pmatrix}$. Then

$\displaystyle\sigma_{e}(T)=\\{-\frac{1}{n}:n\geq
2\\}\cup\\{\frac{e^{\frac{i\pi}{n}}}{n}:n\geq 2\\},$
$\displaystyle\sigma(T)=\sigma_{e}(T)\cup\\{0\\}.$

Let $\omega=\\{z\in\mathbb{C}:\Im(z)<0\\}$. Then $H_{0}=\omega\cup[0,\infty)$
is a half closed-half plane at $0$ with $\text{dim ran}E(H_{0})=0$. So,
$0\notin\Omega_{1}(T)$ and hence $0\notin\Omega_{\infty}(T)$ by Lemma 4.3. Let
$\lambda\neq 0$ be any point in $\mathbb{C}$. Then there exists a half closed-
half plane $H_{\lambda}$ at $\lambda$ such that $\text{dim
ran}E(H_{\lambda})<r$ for some $r\in\mathbb{N}$. So, by Lemma 4.1,
$\lambda\notin\Omega_{r}(T)$ and hence $\lambda\notin\Omega_{\infty}(T)$ by
Lemma 4.3. Therefore, $\Omega_{\infty}(T)=\emptyset$. Hence by Theorem 3.2,
$\Lambda_{\infty}(T)=\Omega_{\infty}(T)=\emptyset$.

######  Remark 5.3.

In Example 5.2, $V_{\infty}(T)=\\{0\\}$ (recalling $V_{k}(T)$ from Proposition
4.4) since $0\in V_{\infty}(T)$ and for any $\lambda\neq 0$, there exists a
half closed-half plane $H$ with $\lambda\in H^{\circ}$ such that $\text{dim
ran}E(H)<r$ for some $r\in\mathbb{N}$ which implies $\lambda\notin V_{r}(T)$
and hence $\lambda\notin V_{\infty}(T)=\bigcap\limits_{k\geq 1}V_{k}(T)$.

We now list down some consequences of Theorem 3.2.

######  Corollary 5.4.

Let $T\in\mathscr{B(\mathscr{H})}$ be a normal operator with no eigenvalue.
Then

$\Lambda_{k}(T)=W(T),\quad\text{ for all }k\in\mathbb{N}_{\infty}.$

###### Proof.

Let $H\in\mathcal{S}_{k}$. Then $H$ is a half closed-half plane in
$\mathbb{C}$ with $\text{dim ran}E(H)<k$. We want to show
$H\in\mathcal{S}_{1}$. If possible let $\text{ dim ran }E(H)\geq 1$. As
$E(H)T=TE(H),\text{ ran }E(H)$ is a reducible subspace and
$P_{\text{ran}E(H)}T|_{\text{ran}E(H)}$ is a normal matrix which implies $T$
has eigenvalue which contradicts our hypothesis. So, $\text{dim ran }E(H)<1$.
Therefore, $H\in S_{1}$ and hence $\mathcal{S}_{k}\subseteq\mathcal{S}_{1}$.
Clearly $\mathcal{S}_{1}\subseteq\mathcal{S}_{k}$ also. So,
$\mathcal{S}_{1}=\mathcal{S}_{k}$. Now, by Theorem 3.2,

$\displaystyle W(T)=\Lambda_{1}(T)=\Omega_{1}(T)=\bigcap\limits_{H\in
S_{1}}H^{c}=\bigcap\limits_{H\in S_{k}}H^{c}=\Omega_{k}(T)=\Lambda_{k}(T).$

This completes the proof. ∎

######  Corollary 5.5.

Let $T\in\mathscr{B(\mathscr{H})}$ be normal with
$\Lambda_{\infty}(T)=\emptyset$. Then $T$ is diagonalizable.

###### Proof.

Let $T$ be not diagonalizable. Suppose $T=T_{1}\oplus T_{2}$ with respect to
the decomposition $\mathscr{H}=\text{ran }E(\sigma_{e}(T))\oplus\text{ran
}E(\sigma_{e}(T)^{c})$. Clearly $T_{1},T_{2}$ are both normal and $T_{2}$ has
no eigenvalue. As $\Lambda_{\infty}(T)=\emptyset$, we have
$\Lambda_{\infty}(T_{2})=\emptyset$. Since $T_{2}$ is normal with no
eigenvalue, by Corollary 5.4, we obtain $W(T_{2})=\Lambda_{\infty}(T_{2})$
which contradicts the fact that $W(T_{2})$ is always non-empty. Hence $T$ is
diagonalizable. ∎

######  Remark 5.6.

Let $I$ be the identity operator acting on $\mathscr{H}$. Then
$\Lambda_{k}(I)=W(I)=\\{1\\}$ for all $k\in\mathbb{N}_{\infty}$. Note that $I$
is a diagonalizable operator with the eigenvalue $1$. So, the converses of
Corollary 5.4 and Corollary 5.5 are not necessarily true.

Let us now give an alternative proof of Theorem 4.4, [15].

######  Corollary 5.7 (Theorem 4.4, [15]).

Suppose $T\in\mathscr{B(\mathscr{H})}$ is normal and $1\leq k<\infty$. Then
$\lambda\notin\Lambda_{k}(T)$ if and only if $T$ can be decomposed into
$T_{1}\oplus T_{2}$ such that $T_{1}$ has dimension at most $k-1$,
$W(T_{1})\subseteq\lambda+S$ and
$W(T_{2})\subseteq\mathbb{C}\setminus(\lambda+S)$ where
$S=e^{it}(\mathcal{P}\cup L)$ with $\mathcal{P}=\\{z\in\mathbb{C}:\Im(z)>0\\}$
and $L=(-\infty,0]$ or $[0,\infty)$ for some $t\in\mathbb{R}$.

###### Proof.

‘$\Rightarrow$’. Let $\lambda\notin\Lambda_{k}(T)$. Then by Theorem 3.2,
$\lambda\notin\Omega_{k}(T)$. By Lemma 4.1, there exists a half closed-half
plane $H$ at $\lambda$ such that $\text{dim ran}E(H)<k$. Take
$T_{1}=P_{\text{ran}E(H)}T|_{\text{ran}E(H)}$ and
$T_{2}=P_{\text{ran}E(H^{c})}T|_{\text{ran}E(H^{c})}$. Then $T=T_{1}\oplus
T_{2}$ such that dimension of $T_{1}$ is at most $k-1$ and $W(T_{1})\subseteq
H$ and $W(T_{2})\subseteq H^{c}$.
‘$\Leftarrow$’. Let $T$ can be decomposed into $T_{1}\oplus T_{2}$ such that
$T_{1}$ has dimension at most $k-1$, $W(T_{1})\subseteq\lambda+S$ and
$W(T_{2})\subseteq\mathbb{C}\setminus(\lambda+S)$ where
$S=e^{it}(\mathcal{P}\cup L)$ with $\mathcal{P}=\\{z\in\mathbb{C}:\Im(z)>0\\}$
and $L=(-\infty,0]$ or $[0,\infty)$ for some $t\in\mathbb{R}$. Suppose
$H=\lambda+S$. Then $H$ is a half closed-half plane at $\lambda$ with
$\text{dim ran}E(H)<k$. So, by Lemma 4.1, $\lambda\notin\Omega_{k}(T)$. Hence
by Theorem 3.2, $\lambda\notin\Lambda_{k}(T)$. ∎

We end this section by the following observation.

######  Corollary 5.8.

Let $T\in\mathscr{B(\mathscr{H})}$ be normal. Then
$\lambda\notin\Lambda_{\infty}(T)$ if and only if $T$ can be decomposed into
$T_{1}\oplus T_{2}$ such that $T_{1}$ has dimension at most finite,
$W(T_{1})\subseteq\lambda+S$ and
$W(T_{2})\subseteq\mathbb{C}\setminus(\lambda+S)$ where
$S=e^{it}(\mathcal{P}\cup L)$ with $\mathcal{P}=\\{z\in\mathbb{C}:\Im(z)>0\\}$
and $L=(-\infty,0]$ or $[0,\infty)$ for some $t\in\mathbb{R}$.

###### Proof.

It follows similarly as in the proof of Corollary 5.7 using Theorem 3.2. ∎

##  6\. Unitary dilations of a normal contraction

In this section, we will see some interesting applications of the main theorem
in this paper to the dilation theory. Let us first recall the definition of a
dilation of an operator. Let $\mathscr{H}$ be a Hilbert space and
$T\in\mathscr{B(\mathscr{H})}$. Suppose $\mathscr{K}$ is a Hilbert space
containing $\mathscr{H}$. An operator $S\in\mathscr{B(\mathscr{K})}$ is said
to be a dilation of $T$ (or $T$ is said to be a compression of $S$) if there
exists a projection $P\in\mathscr{B(\mathscr{K})}$ on $\mathscr{H}$ such that
$T=P_{\mathscr{H}}S|_{\mathscr{H}}$ or equivalently $S$ is unitarily similar
to $2\times 2$ operator matrix $\begin{pmatrix}T&*\\\ *&*\\\ \end{pmatrix}$.
Moreover, if $S$ being a dilation of $T$ be unitary then $S$ is said to be a
unitary dilation of $T$ (cf. [17] for general background on the dilation
theory).

Halmos [11] showed that every contraction $T\in\mathscr{B(\mathscr{H})}$ has a
unitary dilation $U\in\mathscr{B(\mathscr{H}\oplus\mathscr{H})}$ of the
following form

$U=\begin{pmatrix}T&-\sqrt{I-TT^{*}}\\\ \sqrt{I-T^{*}T}&T^{*}\\\
\end{pmatrix}.$

It generated a lot of research including far reaching Sz.-Nagy dilation
theorem regarding power unitary dilation of a contraction. Let
$T\in\mathscr{B(\mathscr{H})}$ be a contraction. Halmos [12] conjectured in
1962 that

$\displaystyle W(T)=\bigcap\left\\{W(U):U\text{ is a unitary dilation of
T}\right\\}.$ (6.1)

Durszt [5] gave an example of a normal contraction to settle (6.1) in negative
(1964). Later, Choi and Li have proved the following (2001).

######  Theorem 6.1 (Theorem 2.4, [4]).

Let $T$ be a contraction acting on a separable Hilbert space $\mathscr{H}$.
Then

$\displaystyle\overline{W(T)}=\bigcap\left\\{\overline{W(U)}:U\in\mathscr{B}\mathcal{(H\oplus
H)}\text{ is a unitary dilation of A}\right\\}.$

Gau, Li and Wu generalised Theorem 6.1 for the higher rank numerical range of
a contraction acting on a finite dimensional Hilbert space.

######  Theorem 6.2.

(Theorem 1.2, [8]) Let $T\in\mathbb{M}_{n}$ be a contraction and $1\leq k\leq
n$. Set $d_{T}=\mbox{dim~{}ran}(I-T^{*}T)^{\frac{1}{2}}.$ Then

$\displaystyle\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\in
M_{n+d_{T}}\text{ is a unitary dilation of T}\right\\}.$

We have further generalized Theorem 6.2 for the higher rank numerical range of
a contraction acting on an infinite dimensional separable Hilbert space [6].

######  Theorem 6.3 ([6]).

Let $T\in\mathscr{B(\mathscr{H})}$ be a contraction and
$k\in\mathbb{N}_{\infty}$. Then

$\displaystyle\overline{\Lambda_{k}(T)}=\bigcap\left\\{\overline{\Lambda_{k}(U)}:U\in\mathscr{B}\mathcal{(H\oplus
H)}\text{ is a unitary dilation of A}\right\\}.$

The following theorem generalizes Theorem 2, [5] by Durszt. It says that the
closure sign on the higher rank numerical range in Theorem 6.3 can not be
omitted.

######  Theorem 6.4.

Let $k\in\mathbb{N}_{\infty}$. Then there exists a normal contraction $T$ for
which the $k$-rank numerical range of all unitary dilations of $T$ contains
$\Lambda_{k}(T)$ as a proper subset.

###### Proof.

Let $T$ be a normal operator with $0$ being the only eigenvalue of $T$ with
multiplicity $k$ and $\sigma(T)=\Gamma\cup\\{0\\}$ where
$\Gamma=\\{z\in\mathbb{C}:\Im{(z)}\geq 0,\;|z|=1\\}$. Clearly $T$ is a
contraction. We can construct such normal contraction by considering
$T=T_{1}\oplus 0I_{k}$ where $T_{1}$ is the multiplication operator by the
variable $z$ on the Hilbert space $L^{2}(\mu)$ where $\mu$ is the Lebesgue
measure supported on $\Gamma$. Let
$R=\\{z\in\mathbb{C}:\Im{(z)}>0,\;|z|<1\\}$. As $T$ does not have any
eigenvalue on $\Gamma$ and $E_{T}$ has positive measure on any subarc of
$\Gamma$, we get $R\cup\\{0\\}\subseteq\Lambda_{k}(T)$ by Corollary 5.4. Now,
by Lemma 4.1 and Theorem 3.2, we obtain $\Lambda_{k}(T)\subseteq
R\cup\\{0\\}$. Therefore, $\Lambda_{k}(T)=R\cup\\{0\\}$.

Let $U$ be a unitary dilation of $T$. Then
$R\cup\\{0\\}\subseteq\Lambda_{k}(U)$. We claim that
$(-1,1)\subseteq\Lambda_{k}(U)$. If possible let there exists $r\in(-1,1)$
such that $r\notin\Lambda_{k}(U)$. Without loss of generality, let $r>0$.
Again by Theorem 3.2 and Lemma 4.1, there exists a half closed-half plane
$H_{r}$ at $r$ such that $\text{dim ran}E_{U}(H_{r})<k$. As
$R\subseteq\Lambda_{k}(U)$, $\delta H_{r}$ has to be the real axis. Again as
$0\in\Lambda_{k}(U)$, the only possible choice of
$H_{r}=\\{z\in\mathbb{C}:\Im{(z)}<0\\}\cup[r,\infty)$. Let
$H_{0}=\\{z\in\mathbb{C}:\Im{(z)}<0\\}\cup[0,\infty)$. As $\text{dim
ran}E_{U}(H_{r})<k$, we have $\text{dim ran}E_{U}(H_{0})<k$. So, by Lemma 4.1,
$0\notin\Omega_{k}(U)$ and hence $0\notin\Lambda_{k}(U)$ by Theorem 3.2 which
is a contradiction. Therefore, $(-1,1)\subseteq\Lambda_{k}(U)$. This completes
the proof. ∎

In fact, more generally, we obtain the following.

######  Proposition 6.5.

Let $k\in\mathbb{N}_{\infty}$ and $T$ be a normal strict contraction with

$\displaystyle\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\text{ is a
unitary dilation of T}\right\\}$

Then whenever $\Lambda_{k}(T)$ contains a point
$\lambda\in\delta\Lambda_{k}(T)$ it also contains every open line segment
$(\lambda,\lambda^{\prime})$ in $\delta\Lambda_{k}(T)$.

###### Proof.

Let $\Lambda_{k}(T)$ contains a point $\lambda\in\delta\Lambda_{k}(T)$.
Suppose $(\lambda,\lambda^{\prime})$ is any open line segment in
$\delta\Lambda_{k}(T)$. Let $L$ be the line extending
$\lambda,\lambda^{\prime}$ and $H$ be the open half plane generated by $L$
such that $H\cap\Lambda_{k}(T)=\emptyset$. Suppose $U$ is a unitary dilation
of $T$. Then $\Lambda_{k}(T)\subseteq\Lambda_{k}(U)$. We claim that
$(\lambda,\lambda^{\prime})\subseteq\Lambda_{k}(U)$. Let there exists
$\lambda_{\circ}\in(\lambda,\lambda^{\prime})$ such that
$\lambda_{\circ}\notin\Lambda_{k}(U)$. By Theorem 3.2 and Lemma 4.1, there
exists a half closed-half plane $H_{\lambda_{\circ}}$ at $\lambda_{\circ}$
such that $\text{dim ran}E_{U}(H_{\lambda_{\circ}})<k$. We now argue that
$\delta H_{\lambda_{\circ}}=L$. If $\delta H_{\lambda_{\circ}}\neq L$ then
there exists a ball $B$ with centre at
$\frac{\lambda_{\circ}+\lambda^{\prime}}{2}$ (or
$\frac{\lambda_{\circ}+\lambda}{2}$) such that $B\subseteq
H_{\lambda_{\circ}}$. As $\Lambda_{k}(T)\subseteq\Lambda_{k}(U)\subseteq
H_{\lambda_{\circ}}^{c}$, we get $B\cap\Lambda_{k}(T)=\emptyset$ which
contradicts the fact that $\frac{\lambda_{\circ}+\lambda^{\prime}}{2}$ (or
$\frac{\lambda_{\circ}+\lambda}{2}$) is a boundary point of $\Lambda_{k}(T)$.
So, we conclude that $\delta H_{\lambda_{\circ}}=L$. Again as
$\lambda\in\Lambda_{k}(U)$, the only possible choice of
$H_{\lambda_{\circ}}=H\cup l_{\lambda_{\circ}}$ where $l_{\lambda_{\circ}}$ is
a ray with initial point $\lambda_{\circ}$ passing through $\lambda^{\prime}$.
Let $H_{\lambda}=H\cup l_{\lambda}$ where $l_{\lambda}$ is a ray with initial
point $\lambda$ passing through $\lambda_{\circ}$. Then $H_{\lambda}$ is a
half closed-half plane at $\lambda$ with $\text{dim ran}E_{U}(H_{\lambda})<k$
as $\text{dim ran}E_{U}(H_{\lambda_{\circ}})<k$ and $|\lambda|<1$. So, by
Lemma 4.1 and Theorem 3.2, we get $\lambda\notin\Lambda_{k}(U)$ which is a
contradiction. Hence $(\lambda,\lambda^{\prime})\subseteq\Lambda_{k}(U)$. As
$U$ is arbitrary unitary dilation of $T$, we have
$(\lambda,\lambda^{\prime})\subseteq\Lambda_{k}(T)$ by hypothesis. ∎

The converse of Proposition 6.5 is not true for $k\geq 2$ as shown in the
following example.

######  Example 6.6.

Let
$R=\left\\{z\in\mathbb{C}:-\frac{1}{2}\leq\Re{(z)},\Im{(z)}\leq\frac{1}{2}\right\\}$
and $\mu$ be the planer Lebesgue measure on $R$. Suppose $k\geq 2$. Let
$T=T_{1}\oplus T_{2}$ where $T_{1}$ is the multiplication operator by the
variable $z$ on the Hilbert space $L^{2}(\mu)$ and
$T_{2}=(\frac{1}{2}+\frac{i}{4})I_{k-1}\oplus\left[\frac{1}{2}-\frac{i}{4}\right]\in
M_{k}$. Clearly $T$ is a normal strict contraction with
$\sigma_{e}(T)=\left\\{\frac{1}{2}+\frac{i}{4},\frac{1}{2}-\frac{i}{4}\right\\}$
and $\sigma(T)=R$. Now, by Lemma 4.1 and Theorem 3.2, we obtain

$\displaystyle\Lambda_{k}(T)=\left\\{z\in\mathbb{C}:-\frac{1}{2}<\Re{(z)},\Im{(z)}<\frac{1}{2}\right\\}.$

Note that $\frac{1}{2}\notin\Lambda_{k}(T)$ and the condition whenever
$\Lambda_{k}(T)$ contains a point $\lambda\in\delta\Lambda_{k}(T)$ it also
contains every open line segment $(\lambda,\lambda^{\prime})$ in
$\delta\Lambda_{k}(T)$ is being satisfied here vacuously.

Let $U$ be a unitary dilation of $T$. Suppose
$\frac{1}{2}\notin\Lambda_{k}(U)$. Then by Theorem 3.2 and Lemma 4.1, there
exists a half closed-half plane $H$ at $\frac{1}{2}$ such that $\text{dim
ran}E_{U}(H)<k$. As $\Lambda_{k}(T)\subseteq\Lambda_{k}(U)$, we can see $H$ is
the union of $\\{z\in\mathbb{C}:\Re{(z)}>\frac{1}{2}\\}$ with either $L_{1}$
or $L_{2}$ where $L_{1},L_{2}$ are two opposite directed rays with initial
point $\frac{1}{2}$ on the line $\\{z\in\mathbb{C}:\Re{(z)}=\frac{1}{2}\\}$.
Clearly $U=U_{1}\oplus U_{2}$ where $U_{1}$ is a unitary dilation of $T_{1}$
and $U_{2}$ is the same for $T_{2}$. As $\text{dim ran}E_{U}(H)<k$, we have
$\text{dim ran}E_{U_{2}}(H)<k$. Denote $\Gamma_{1}=\mathbb{T}\cap H$ and
$\Gamma_{2}=\mathbb{T}\cap H^{c}$. Let $U_{2}=U_{2}^{\prime}\oplus
U_{2}^{\prime\prime}$ with respect to the decomposition
$\text{ran}E_{U_{2}}(\Gamma_{1})\oplus\text{ran}E_{U_{2}}(\Gamma_{2})$. As
codimension of $\text{ran}E_{U_{2}}(\Gamma_{2})$ is less than $k$ and
$T_{2}\in M_{k}$, there exists
$\xi\in\text{ran}E_{U_{2}}(\Gamma_{2})\cap\mathbb{C}^{k}$ with $\|\xi\|=1$
such that $\langle T_{2}\xi,\xi\rangle=\langle
U_{2}^{\prime\prime}\xi,\xi\rangle\in\text{conv}\Gamma_{2}$ which is a
contradiction as
$W(T_{2})=\left[\frac{1}{2}-\frac{i}{4},\frac{1}{2}+\frac{i}{4}\right]$. So,
$\frac{1}{2}\in\Lambda_{k}(U)$. Therefore, the higher rank numerical range of
every unitary dilations of $T$ contains $\Lambda_{k}(T)$ as a proper subset.

######  Example 6.7.

Let $\Gamma=\\{z\in\mathbb{C}:\Re{(z)}\leq 0,\;|z|=1\\}$ and $\mu$ is a
Lebesgue measure supported on $\Gamma$. Suppose $k\in\mathbb{N}_{\infty}$. Let
$T=T_{1}\oplus iI_{k}$ where $T_{1}$ is the multiplication operator by the
variable $z$ on the Hilbert space $L^{2}(\mu)$. Note that
$\sigma_{e}(T)=\\{i\\}$ and $\sigma(T)=\Gamma$. By Lemma 4.1, Theorem 3.2 and
Corollary 5.4, we obtain

$\displaystyle\Lambda_{k}(T)=\\{z\in\mathbb{C}:\Re{(z)}<0,\;|z|<1\\}\cup\\{i\\}.$

We have here $\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\text{ is a
unitary dilation of T}\right\\}$ (as $T$ is unitary) but $\Lambda_{k}(T)$ does
not contain $(i,-i)\subseteq\delta\Lambda_{k}(T)$ despite containing
$i\in\delta\Lambda_{k}(T)$. Note here that $T$ is not a strict contraction.

The next theorem gives a necessary and sufficient condition on a normal
contraction $T$ acting on an infinite dimensional separable Hilbert space that
its higher rank numerical range equals the intersection of the higher rank
numerical ranges of unitary dilations of $T$. This theorem strengthens and
generalises Theorem 3.2, [19] by Wu. The condition and the proof given there
are erroneous as can be seen in the following example.

######  Example 6.8.

Consider $T=\bigoplus\limits_{n\geq 2}\begin{pmatrix}-\frac{1}{n}&0\\\
0&\frac{e^{\frac{i\pi}{n}}}{n}\end{pmatrix}$. As $T$ is diagonalizable, it can
be computed directly (or otherwise) that
$W(T)=\text{conv}[\\{-\frac{1}{n}:n\geq
2\\}\cup\\{\frac{e^{\frac{i\pi}{n}}}{n}:n\geq 2\\}]$. Note that $0\notin W(T)$
and whenever $W(T)$ contains a point $\lambda\in\delta W(T)$ it also contains
every open line segment $(\lambda,\lambda^{\prime})$ in $\delta W(T)$.

Let $U$ be a unitary dilation of $T$. Then $W(T)\subseteq W(U)$. Let $0\not\in
W(U)$. Then by Theorem 3.2 and Lemma 4.1, there exists a half closed-half
plane $H_{0}$ at $0$ such that $E_{U}(H_{0})=0$. As $W(T)\subseteq W(U)$,
$\delta H_{0}$ has to be the real axis. Again as $\\{-\frac{1}{n}:n\geq
2\\}\subseteq W(U)$, the only possible choice of
$H_{0}=\\{z\in\mathbb{C}:\Im{(z)}<0\\}\cup[0,\infty)$. Then
$H_{-\frac{1}{2}}=\\{z\in\mathbb{C}:\Im{(z)}<0\\}\cup[-\frac{1}{2},\infty)$ is
a half closed-half plane at $-\frac{1}{2}$ with $E_{U}(H_{-\frac{1}{2}})=0$ as
$E_{U}(H_{0})=0$. Again by Lemma 4.1 and Theorem 3.2, we have
$-\frac{1}{2}\notin W(U)$ which is a contradiction. Hence, $0\in W(U)$.
Therefore, the numerical range of every unitary dilations of $T$ contains
$W(T)$ as a proper subset.

The following is the main theorem of this section.

######  Theorem 6.9.

Let $T$ be a normal strict contraction and $k\in\mathbb{N}_{\infty}$. Then

$\displaystyle\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\text{ is a
unitary dilation of T}\right\\}$

holds if and only if for any
$\lambda\in\delta\Lambda_{k}(T)\setminus\Lambda_{k}(T)$, there exists a closed
half plane $H$ with the line $\delta H$ passing through $\lambda$ such that
$\text{dim ran}E(H)<k$.

###### Proof.

“$\Rightarrow$”. Suppose $\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\text{
is a unitary dilation of T}\right\\}$ holds. Let
$\lambda\in\delta\Lambda_{k}(T)\setminus\Lambda_{k}(T)$. Then by hypothesis,
there exists a unitary dilation $U$ of $T$ such that
$\lambda\in\delta\Lambda_{k}(U)\setminus\Lambda_{k}(U)$. So, by Theorem 3.2
and Lemma 4.1, there exists a half closed-half plane $H_{\lambda}$ at
$\lambda$ such that $\text{dim ran}E_{U}(H_{\lambda})<k$. Denote
$\Gamma_{1}=\mathbb{T}\cap H_{\lambda}$ and $\Gamma_{2}=\mathbb{T}\cap
H_{\lambda}^{c}$. Let $U=U_{1}\oplus U_{2}$ with respect to the decomposition
$\text{ran}E_{U}(\Gamma_{1})\oplus\text{ran}E_{U}(\Gamma_{2})$. Suppose
$H=\overline{H_{\lambda}}$. Then $H$ is a closed half plane with the line
$\delta H$ passing through $\lambda$. Let $T=T_{1}\oplus T_{2}$ with respect
to the decomposition $\text{ran}E(H)\oplus\text{ran}E(H^{c})$. Let $\text{dim
ran}E(H)\geq k$. As codimension of $\text{ran}E_{U}(\Gamma_{2})$ is less than
$k$, there exists $\xi\in\text{ran}E_{U}(\Gamma_{2})\cap\text{ran}E(H)$ with
$\|\xi\|=1$ such that $\langle T_{1}\xi,\xi\rangle=\langle
U_{2}\xi,\xi\rangle$. As $T_{1}$ is a strict contraction, we have $\langle
T_{1}\xi,\xi\rangle=\langle U_{2}\xi,\xi\rangle\in H^{c}$ which contradicts
that $W(T_{1})\subseteq H$. So, $\text{dim ran}E(H)<k$.
“$\Leftarrow$”. It is clear that

$\displaystyle\Lambda_{k}(T)\subseteq\bigcap\left\\{\Lambda_{k}(U):U\text{ is
a unitary dilation of T}\right\\}.$

Let $\lambda\notin\Lambda_{k}(T)$. We need to show that there exists a unitary
dilation $U$ of $T$ such that $\lambda\notin\Lambda_{k}(U)$. Suppose
$\lambda\notin\overline{\Lambda_{k}(T)}$. Then by Theorem 1.2 and Theorem 5.2
[15], there exists $\alpha\in[0,2\pi)$ such that
$\Re{(e^{i\alpha}\lambda)}>\lambda_{k}(\Re{(e^{i\alpha}T)})$. We choose

$\displaystyle U=\begin{pmatrix}T&-e^{-i\alpha}\sqrt{I-TT^{*}}\\\
e^{-i\alpha}\sqrt{I-T^{*}T}&e^{-2i\alpha}T^{*}\\\ \end{pmatrix}.$

Then $U$ is a unitary dilation of $T$ with
$\Re{(e^{i\alpha}\lambda)}>\lambda_{k}(\Re{(e^{i\alpha}T)})=\lambda_{k}(\Re{(e^{i\alpha}U)})$.
So, again by Theorem 1.2 and Theorem 5.2, [15], we obtain
$\lambda\notin\overline{\Lambda_{k}(U)}$ and hence
$\lambda\notin\Lambda_{k}(U)$.

Next, let $\lambda\in\delta\Lambda_{k}(T)\setminus\Lambda_{k}(T)$. Then by
hypothesis, there exists a closed half plane $H$ with the line $\delta H$
passing through $\lambda$ such that $\text{dim ran}E(H)<k$. Let $T=T_{1}\oplus
T_{2}$ with respect to the decomposition
$\text{ran}E(H)\oplus\text{ran}E(H^{c})$. Clearly $T_{1},T_{2}$ are both
normal. Without loss of generality, we can take
$T_{1}=\text{diag}(d_{1},d_{2},\cdots,d_{r})$ for some $r<k$. Denote
$\Gamma_{1}=\mathbb{T}\cap H$ and $\Gamma_{2}=\mathbb{T}\cap H^{c}$. Let
$j\in\\{1,2,\cdots,r\\}$. Choose $\xi_{j}\in\Gamma_{1}$ and
$\eta_{j}\in\Gamma_{2}$ such that $d_{j}$ lies on the line segment joining
$\xi_{j}$ and $\eta_{j}$. Then $V_{j}=\text{diag}(\xi_{j},\eta_{j})$ is a
unitary dilation of $(d_{j})$. So, $U_{1}=\sum^{\oplus}_{j}V_{j}$ is a unitary
dilation of $T_{1}$ with $\text{dim ran}E_{U_{1}}(H)=r<k$. Note that
$W(T_{2})\subseteq H^{c}$. Then there exists $\beta\in[0,2\pi)$ and
$\mu\in\mathbb{R}$ such that
$W(e^{i\beta}T_{2})\subseteq\\{z\in\mathbb{C}:\Re{(z)}<\mu\\}$. We again
choose

$\displaystyle
U_{2}=\begin{pmatrix}T_{2}&-e^{-i\beta}\sqrt{I-T_{2}T_{2}^{*}}\\\
e^{-i\beta}\sqrt{I-T_{2}^{*}T_{2}}&e^{-2i\beta}T_{2}^{*}\\\ \end{pmatrix}.$

Then $U_{2}$ is a unitary dilation of $T_{2}$ with
$W(e^{i\beta}U_{2})\subseteq\\{z\in\mathbb{C}:\Re{(z)}<\mu\\}$ which implies
$W(U_{2})\subseteq H^{c}$. So, $\text{dim ran}E_{U_{2}}(H)=0$. Then
$U=U_{1}\oplus U_{2}$ is a unitary dilation of $T$ with $\text{dim
ran}E_{U}(H)=r<k$. Hence by Lemma 4.1 and Theorem 3.2, we obtain
$\lambda\notin\Lambda_{k}(U)$. ∎

In the following conjecture, we propose a necessary and sufficient condition
on a contraction $T$ acting on an infinite dimensional separable Hilbert space
that its higher rank numerical range equals the intersection of the higher
rank numerical ranges of unitary dilations of $T$ which, in particular, gives
a necessary and sufficient condition for the Halmos conjecture (6.1) to be
true.

######  Conjecture 6.10.

Let $T$ be a strict contraction and $k\in\mathbb{N}_{\infty}$. Then

$\displaystyle\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\text{ is a
unitary dilation of T}\right\\}$

holds if and only if for any
$\lambda\in\delta\Lambda_{k}(T)\setminus\Lambda_{k}(T)$, there exists
$\theta\in[0,2\pi)$ such that

$\displaystyle\text{dim ran}E_{S}([0,\infty))<k\text{ where
}S=\Re{(e^{i\theta}T-\lambda)}.$

We will see that the condition given in the Conjecture 6.10 is necessary as
discussed below. However, we strongly believe that the given condition is also
sufficient which we wish to investigate in future.

###### Proof of “$\Rightarrow$” in Conjecture 6.10.

Suppose

$\displaystyle\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(U):U\text{ is a
unitary dilation of T}\right\\}$

holds. Let $\lambda\in\delta\Lambda_{k}(T)\setminus\Lambda_{k}(T)$. Then by
hypothesis, there exists a unitary dilation $U$ of $T$ such that
$\lambda\in\delta\Lambda_{k}(U)\setminus\Lambda_{k}(U)$. Now, by Theorem 3.2
and Lemma 4.1, there exists a half closed-half plane $H_{\lambda}$ at
$\lambda$ such that $\text{dim ran}E_{U}(H_{\lambda})<k$. Set
$\Gamma_{1}=\mathbb{T}\cap H_{\lambda}$ and $\Gamma_{2}=\mathbb{T}\cap
H_{\lambda}^{c}$. Let $U=U_{1}\oplus U_{2}$ with respect to the decomposition
$\text{ran}E_{U}(\Gamma_{1})\oplus\text{ran}E_{U}(\Gamma_{2})$. Suppose
$H=\overline{H_{\lambda}}$. Then $H$ is a closed half plane with the line
$\delta H$ passing through $\lambda$. Then there exists $\theta\in[0,2\pi)$
such that $H^{\prime}=e^{i\theta}H-\lambda=\\{z\in\mathbb{C}:\Re{(z)\geq
0}\\}$. Denote
$\Gamma_{1}^{\prime}=e^{i\theta}\Gamma_{1}-\lambda,\Gamma_{2}^{\prime}=e^{i\theta}\Gamma_{2}-\lambda,T^{\prime}=e^{i\theta}T-\lambda,U^{\prime}=e^{i\theta}U-\lambda$
and $U^{\prime}=U_{1}^{\prime}\oplus U_{2}^{\prime}$ with respect to the
decomposition
$\text{ran}E_{U^{\prime}}(\Gamma_{1}^{\prime})\oplus\text{ran}E_{U^{\prime}}(\Gamma_{2}^{\prime})$.
Let $\text{dim ran}E_{S}([0,\infty))\geq k$. As codimension of
$\text{ran}E_{U^{\prime}}(\Gamma_{2}^{\prime})$ is less than $k$, there exists
$\xi\in\text{ran}E_{U^{\prime}}(\Gamma_{2}^{\prime})\cap\text{ran}E_{S}([0,\infty))$
with $\|\xi\|=1$ such that $\langle T^{\prime}\xi,\xi\rangle=\langle
U_{2}^{\prime}\xi,\xi\rangle$. So, $\langle T\xi,\xi\rangle=\langle
U_{2}\xi,\xi\rangle$. As $T$ is a strict contraction, we have $\langle
T\xi,\xi\rangle=\langle U_{2}\xi,\xi\rangle\in H^{c}$ which implies $\langle
S\xi,\xi\rangle=\langle\Re{(U_{2}^{\prime})}\xi,\xi\rangle<0$. As
$\xi\in\text{ran}E_{S}([0,\infty))$, we also have

$\displaystyle\langle S\xi,\xi\rangle=\int_{\mathbb{R}}x\;d\langle
E_{S}(.)\xi,\xi\rangle=\int_{0}^{\infty}x\;d\langle
E_{S}(.)\xi,\xi\rangle=\int_{0}^{\infty}x\;d\|E_{S}(.)\xi\|^{2}\geq 0$

which leads to a contradiction. Hence, $\text{dim ran}E_{S}([0,\infty))<k$. ∎

We end this section by posing the following question which we could not able
to answer at this stage. In fact, the answer is not known even for $k=1$.

######  Question 6.11.

Let $T\in\mathscr{B(\mathscr{H})}$ and $k\in\mathbb{N}_{\infty}$. Is

$\displaystyle\Lambda_{k}(T)=\bigcap\left\\{\Lambda_{k}(N):N\text{ is a normal
dilation of T}\right\\}?$

## Acknowledgements

The research of the first author is supported by the Integrated Ph.D
fellowship of Indian Institute of Science Education and Research
Thiruvananthapuram.

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