On absolutely norm attaining operators

Abstract

We give the necessary and sufficient conditions for a bounded operator defined between complex Hilbert spaces to be absolutely norm attaining. We discuss the structure of such operators in the case of self-adjoint and normal operators separately. Finally, we discuss several properties of absolutely norm attaining operators.

1 Introduction and preliminaries

The class of absolutely norm attaining operators (or shortly, ${\mathcal {AN}}$-operators) between complex Hilbert spaces was introduced and several important class of examples and properties of these operators were discussed by Carvajal and Neves in [3]. Later, a structure of these operators on separable Hilbert spaces was proposed in [11]. But, an example of ${\mathcal {AN}}$-operator which does not fit into the characterization of [11] was given in [10] and the authors discussed the structure of positive ${\mathcal {AN}}$-operators between arbitrary Hilbert spaces. In this article, first, we give necessary and sufficient conditions for an operator to be positive and ${\mathcal {AN}}$. In fact, we show that a bounded operator T defined on an infinite dimensional Hilbert space is positive and ${\mathcal {AN}}$ if and only if there exists a unique triple $(K,F,\alpha )$, where K is a positive compact operator, F is a positive finite rank operator, $\alpha$ is a positive real number such that $T=K-F+\alpha I$ and $KF=FK=0,\; F\le \alpha I$ (see Theorem 2.5). In fact, here $\alpha =m_e(T)$, the essential minimum modulus of T. This is an improvement of [10, Theorem 5.1]. Using this result, we give explicit structure of self-adjoint and ${\mathcal {AN}}$-operators as well as normal and ${\mathcal {AN}}$-operators. Finally, we also obtain structure of general ${\mathcal {AN}}$-operators. In the process, we also prove several important properties of ${\mathcal {AN}}$-operators. All these results are new.

We organize the article as follows: In the remaining part of this section, we explain the basic terminology, notations and necessary results that will be needed for proving main theorems. In section 2, we give a characterization of positive $\mathcal { AN}$-operators and prove several important properties. In section 3, we discuss the structure of self-adjoint and normal ${\mathcal {AN}}$-operators and in section 4, we discuss about the general ${\mathcal {AN}}$-operators.

Throughout the article we consider complex Hilbert spaces which will be denoted by $H,H_1,H_2$, etc. The inner product and the induced norm are denoted by $\langle , \rangle$ and $\Vert \cdot \Vert$ respectively. The unit sphere of a closed subspace M of H is denoted by $S_M:={\{x\in M: \Vert x\Vert =1}\}$ and $P_M$ denotes the orthogonal projection $P_M:H\rightarrow H$ with range M. The identity operator on M is denoted by $I_M$.

A linear operator $T: H_1\rightarrow H_2$ is said to be bounded if there exists a number $k>0$ such that $\Vert Tx\Vert \le k \Vert x\Vert$ for all $x\in H_1$. If T is bounded, the quantity $\Vert T\Vert =\sup {\{\Vert Tx\Vert : x\in S_{H_1}}\}$ is finite and is called the norm of T. We denote the space of all bounded linear operators between $H_1$ and $H_2$ by ${\mathcal {B}}(H_1,H_2)$. In case if $H_1=H_2=H$, then $\mathcal B(H_1,H_2)$ is denoted by ${\mathcal {B}}(H)$. For $T\in \mathcal B(H_1,H_2)$, there exists a unique operator denoted by $T^*:H_2\rightarrow H_1$ satisfying

$$\begin{aligned} \langle Tx,y\rangle =\langle x, T^*y\rangle \; \text {for all}\; x\in H_1 \; \text {and} \; \text {for all}\; y\in H_2. \end{aligned}$$

This operator $T^*$ is called the adjoint of T. The null space and the range spaces of T are denoted by N(T) and R(T) respectively.

Let $T\in {\mathcal {B}}(H)$. Then T is said to be normal if $T^*T=TT^*$, self-adjoint if $T=T^*$. If $\langle Tx,x\rangle \ge 0$ for all $x\in H$, then T is called positive. It is well known that for a positive operator T, there exists a unique positive operator $S\in {\mathcal {B}}(H)$ such that $S^2=T$. We write $S=T^{\frac{1}{2}}$ and it is called as the positive square root of T.

If $S,T\in {\mathcal {B}}(H)$ are self-adjoint and $S-T\ge 0$, then we write this by $S\ge T$.

If $P\in {\mathcal {B}}(H)$ is such that $P^2=P$, then P is called a projection. If N(P) and R(P) are orthogonal to each other, then P is called an orthogonal projection. It is a well known fact that a projection P is orthogonal if and only if it is self-adjoint if and only if it is normal.

We call an operator $V\in {\mathcal {B}}(H_1,H_2)$ to be an isometry if $\Vert Vx\Vert =\Vert x\Vert$ for each $x\in H_1$. An operator $V\in {\mathcal {B}}(H_1,H_2)$ is said to be a partial isometry if $V|_{N(V)^{\bot }}$ is an isometry. That is $\Vert Vx\Vert =\Vert x\Vert$ for all $x\in N(V)^{\bot }$. If $V\in {\mathcal {B}}(H)$ is isometry and onto, then V is said to be a unitary operator.

In general, if $T\in {\mathcal {B}}(H_1,H_2)$, then $T^*T\in \mathcal B(H_1)$ is positive and $|T|:=(T^*T)^{\frac{1}{2}}$ is called the modulus of T. In fact, there exists a unique partial isometry $V\in {\mathcal {B}}(H_1,H_2)$ such that $T=V|T|$ and $N(V)=N(T)$. This factorization is called the polar decomposition of T.

If $T\in {\mathcal {B}}(H)$, then $T=\frac{T+T^*}{2}+i(\frac{T-T^*}{2i})$. The operators $\mathrm{Re}(T):=\frac{T+T^*}{2}$ and $\mathrm{Im}(T):=\frac{T-T^*}{2i}$ are self-adjoint and are called the real and the imaginary parts of T respectively.

A closed subspace M of H is said to be invariant under $T\in {\mathcal {B}}(H)$ if $TM\subseteq M$ and reducing if both M and $M^\bot$ are invariant under T.

For $T\in {\mathcal {B}}(H)$, the set

$$\begin{aligned} \rho (T):={\{\lambda \in \mathbb C: T-\lambda I \text { is invertible and} \;(T-\lambda I)^{-1}\in {\mathcal {B}}(H) }\} \end{aligned}$$

is called the resolvent set and the complement $\sigma (T)={\mathbb {C}}\setminus \rho (T)$ is called the spectrum of T. It is well known that $\sigma (T)$ is a non empty compact subset of ${\mathbb {C}}$. The point spectrum of T is defined by

$$\begin{aligned} \sigma _p(T)={\{\lambda \in {\mathbb {C}}: T-\lambda I\ \text {is not one-to-one}}\}. \end{aligned}$$

Note that $\sigma _{p}(T)\subseteq \sigma (T)$.

A self-adjoint operator $T\in {\mathcal {B}}(H)$ is positive if and only if $\sigma (T)\subseteq [0,\infty )$.

A bounded linear operator $T:H_1\rightarrow H_2$ is called finite rank if R(T) is finite dimensional. The space of all finite rank operators between $H_1$ and $H_2$ is denoted by ${\mathcal {F}}(H_1,H_2)$ and we write ${\mathcal {F}}(H,H)={\mathcal {F}}(H)$. If $T\in {\mathcal {B}}(H_1,H_2)$, then T is said to be compact if for every bounded set S of $H_1$, the set T(S) is pre-compact in $H_2$. Equivalently, for every bounded sequence $(x_n)$ of $H_1$, $(Tx_n)$ has a convergent subsequence in $H_2$. We denote the set of all compact operators between $H_1$ and $H_2$ by ${\mathcal {K}}(H_1,H_2)$. In case if $H_1=H_2=H$, then ${\mathcal {K}}(H_1,H_2)$ is denoted by ${\mathcal {K}}(H)$.

All the above mentioned basics of operator theory can be found in [4, 8, 12, 13].

An operator $T\in {\mathcal {B}}(H_1,H_2)$ is said to be norm attaining if there exist $x\in S_{H_1}$ such that $\Vert Tx\Vert =\Vert T\Vert$. We denote the class of norm attaining operators by $\mathcal N(H_1,H_2)$. It is known that ${\mathcal {N}}(H_1,H_2)$ is dense in ${\mathcal {B}}(H_1,H_2)$ with respect to the operator norm of $\mathcal B(H_1,H_2)$. We refer [5] for more details on this topic.

We say $T\in {\mathcal {B}}(H_1,H_2)$ to be absolutely norm attaining or ${{\mathcal {AN}}}$-operator (shortly), if $T|_M$, the restriction of T to M, is norm attaining for every non zero closed subspace M of $H_1$. That is, $T|_M\in {\mathcal {N}}(M, H_2)$ for every non zero closed subspace M of $H_1$ [3]. This class contains ${\mathcal {K}}(H_1,H_2)$, and the class of partial isometries with finite dimensional null space or finite dimensional range space.

We have the following characterization of norm attaining operators.

PROPOSITION 1.1

[3, Proposition 2.4]

Let $T\in {\mathcal {B}}(H)$ be self-adjoint. Then

• (1) $T\in {\mathcal {N}}(H)$ if and only if either $\Vert T\Vert \in \sigma _p(T)$ or $-\Vert T\Vert \in \sigma _p(T),$

• (2) if $T\ge 0$, then $T\in {\mathcal {N}}(H)$ if and only if $\Vert T\Vert \in \sigma _p(T)$.

For $T\in {{\mathcal {B}}}(H_1,H_2)$, the quantity

$$\begin{aligned} m(T):=\inf {\{\Vert Tx\Vert : x\in S_{H_1}}\} \end{aligned}$$

is called the minimum modulus of T. If $H_1=H_2=H$ and $T^{-1}\in {\mathcal {B}}(H)$, then $m(T)=\dfrac{1}{\Vert T^{-1}\Vert }$ (see [1, Theorem 1] for details).

The following definition is available in [9] for densely defined closed operators (not necessarily bounded) on a Hilbert space, and this holds true automatically for bounded operators.

DEFINITION 1.2

[9, Definition 8.3 p. 178]

Let $T=T^*\in \mathcal B(H)$. Then the discrete spectrum $\sigma _d(T)$ of T is defined as the set of all eigenvalues of T with finite multiplicities which are isolated points of the spectrum $\sigma (T)$ of T. The complement set $\sigma _{\mathrm{ess}}(T)=\sigma (T)\setminus \sigma _d(T)$ is called the essential spectrum of T.

By the Weyl’s theorem we can assert that if $T=T^*$ and $K=K^*\in {\mathcal {K}}(H)$, then $\sigma _{\mathrm{ess}}(T+K)=\sigma _{\mathrm{ess}}(T)$ (see [9, Corollary 8.16, p. 182] for details). If H is a separable Hilbert space, the essential minimum modulus of T is defined to be $m_e(T):=\inf {\{\lambda : \lambda \in \sigma _{\mathrm{ess}}(|T|)}\}$ (see [1] for details). The same result in the general case is dealt in [2].

Let $H=H_1\oplus H_2$ and $T\in {\mathcal {B}}(H)$. Let $P_j:H\rightarrow H$ be an orthogonal projection onto $H_j$ for $j=1,2$. Then $T=\left( \begin{array}{cc} T_{11} &{}T_{12} \\ T_{21} &{}T_{22} \\ \end{array} \right)$, where $T_{ij}:H_j\rightarrow H_i$ is the operator given by $T_{ij}=P_iTP_j|_{H_j}$. In particular, $T(H_1)\subseteq H_1$ if and only if $T_{12}=0$. Also, $H_1$ reduces T if and only if $T_{12}=0=T_{21}$ (for details, see [4, 13]).

2 Positive ${\mathcal {AN}}$-operators

In this section, we describe the structure of operators which are positive and satisfy the ${\mathcal {AN}}$-property. First, we recall the results that are necessary for proving our results.

Theorem 2.1

[10, Theorem 5.1]. Let H be a complex Hilbert space of arbitrary dimension and let P be a positive operator on H. Then P is an ${\mathcal {AN}}$-operator iff P is of the form $P = \alpha I + K + F$, where $\alpha \ge 0$, K is a positive compact operator and F is a self-adjoint finite rank operator.

Theorem 2.2

[10, Theorem 3.8]. Let $T\in B(H)$ be positive and $T\in {\mathcal {AN}}(H)$. Then

$$\begin{aligned} T=\displaystyle \sum _{\alpha \in \Lambda } \beta _{\alpha }v_{\alpha }\otimes v_{\alpha }, \end{aligned}$$

(2.1)

where ${\{v_{\alpha }:\alpha \in \Lambda }\}$ is an orthonormal basis consisting of entirely eigenvectors of T and for every $\alpha \in \Lambda$, $Tv_{\alpha }=\beta _{\alpha }v_{\alpha }$ with $\beta _{\alpha }\ge 0$ such that

• (1) for every non empty set $\Gamma$ of $\Lambda$, we have

  $$\begin{aligned} \sup {\{\beta _{\alpha }:\alpha \in \Gamma }\}=\max {\{\beta _{\alpha }:\alpha \in \Gamma }\}\mathrm{,} \end{aligned}$$

• (2) the spectrum $\sigma (T)=\overline{{\{\beta _{\alpha }:\alpha \in \Lambda }\}}$ has at most one limit point. Moreover, this unique limit point (if exists) can only be the limit of an increasing sequence in the spectrum,

• (3) the set ${\{\beta _{\alpha }:\alpha \in \Lambda }\}$ of eigenvalues of T, without counting multiplicities, is countable and has atmost one eigenvalue with infinite multiplicity,

• (4) if $\sigma (T)$ has both, i.e., a limit point and an eigenvalue with infinite multiplicity, then they must be the same.

Here $(v_{\alpha }\otimes v_{\alpha })(x)=\langle x,v_{\alpha }\rangle v_{\alpha }$ for each $\alpha \in \Lambda$ and for each $x\in H$.

Remark 2.3

In Theorem 2.1, the standing hypothesis is that the operator T is positive. In particular, if K is a positive compact operator, F is a self-adjoint finite rank operator and $\alpha \ge 0$, it is interesting to know when $T:=K+F+\alpha I$ is positive? We will answer this question later in this article and also discuss about the uniqueness of the representation of T given in Theorem 2.1.

Lemma 2.4

Let $S,T\in {\mathcal {B}}(H)$ be positive such that $S\le T$. Then $N(T)\subseteq N(S)$.

Proof

If $x\in H$, then $\Vert S^{\frac{1}{2}}x\Vert ^2=\langle Sx,x\rangle \le \langle Tx,x\rangle =\Vert T^{\frac{1}{2}}x\Vert ^2$. By observing the fact that for any $A\in {\mathcal {B}}(H)$ with $A\ge 0$, $N(A^{\frac{1}{2}})=N(A)$, the conclusion follows. $\square$

Theorem 2.5

Let H be an infinite dimensional Hilbert space and $T\in \mathcal B(H)$. Then the following statements are equivalent:

• (1) $T\in {{\mathcal {AN}}}(H)$ and positive,

• (2) there exists a unique triple $(K,F,\alpha )$, where

  • (a) $K\in {\mathcal {K}}(H)$ is positive, $\alpha \ge 0$,

  • (b) $F\in {\mathcal {F}}(H)$ and $0\le F\le \alpha I$,

  • (c) $KF=0$,

  such that $T=K-F+\alpha I$.

Proof

Proof of (1) $\Rightarrow$ (2). By Theorem 2.1, $T=K'-F'+\alpha I$, where $K'\in \mathcal K(H)$ is positive, $F'=F'^{*}\in {\mathcal {F}}(H)$ and $\alpha \ge 0$. As $K'-F'$ is compact, self-adjoint, there exists an orthonormal set ${\{\phi _n:n\in {\mathbb {N}}}\}$ of eigenvectors corresponding to the eigenvalues ${\{\lambda _n:n\in {\mathbb {N}}}\}$. That is, $(K'-F')(\phi _n)=\lambda _n\phi _n$ for each $n\in {\mathbb {N}}$. By [10, Lemma 4.8], $K'-F'$ can have at most finitely many negative eigenvalues. Without loss of generality, assume that $\lambda _1,\lambda _2,\ldots , \lambda _k$ are those negative eigenvalues. Define $F(x)= \sum _{n=1}^{k}(-\lambda _n)\langle x,\phi _{n}\rangle \phi _n$ and $K(x)= \sum _{n=k+1}^{\infty }\lambda _n\langle x,\phi _{n}\rangle \phi _n$ for all $x\in H$. Then clearly, $FK=KF=0$ and F is a positive finite rank operator and K is a positive compact operator. Also, $K'-F'=K-F$ and hence $T=\alpha I+K-F$.

Next, $TF=(\alpha I-F)F=FT$. Since T and F are positive, it follows that TF is positive. Let $\lambda \in \sigma (F)$. Then $\lambda \ge 0$ and since $FT\ge 0$, by the spectral mapping theorem, we have that $\lambda (\alpha -\lambda )\ge 0$. From this, we can conclude that $\alpha -\lambda \ge 0$ for each $\lambda \in \sigma (F)$. As $\alpha I-F$ is self-adjoint and $\sigma (\alpha I-F)\subseteq [0,\infty )$, $\alpha I-F$ must be positive. This concludes that $F\le \Vert F\Vert I\le \alpha I$.

Next, we show that the triple satisfying the given conditions is unique. Suppose there exists two triples $(K_1,F_1,\alpha _1), (K_2,F_2,\alpha _2 )$ satisfying the stated conditions. We prove this by considering all possible cases.

Case 1. $\alpha _1=0$. In this case, $F_1=0$. Hence $K_1=T=K_2-F_2+\alpha _2I$. This shows that $\alpha _2I=K_1-K_2+F_2$, a compact operator. Since H is infinite dimensional, it follows that $\alpha _2=0$. Thus $F_2=0$. Hence we can conclude that $K_1=K_2$.

Case 2. $F_1=0$, $\alpha _1>0$. In this case,

$$\begin{aligned} K_1+\alpha _1I=K_2-F_2+\alpha _2I. \end{aligned}$$

(2.2)

Then $(\alpha _2-\alpha _1)I=(K_1-K_2)+F_2$ , a compact operator. Hence $\alpha _1=\alpha _2$.

Now equation (2.2) can be written as $K_2=F_2+K_1\ge F_2$. By Lemma  2.4, we have that $N(K_2)\subseteq N(F_2)$. But, by the condition $K_2F_2=0$, we have $R(F_2)\subseteq N(K_2)$, and hence $R(F_2)\subseteq N(F_2)$. Thus, $F_2=0$. From this, we can conclude that $K_1=K_2$.

Case 3. $K_1=0$, $F_1\ne 0$, $\alpha _1>0$. We have $F_1+\alpha _1 I=K_2-F_2+\alpha _2 I$. Using the same argument as in the above cases, we can conclude that $\alpha _1=\alpha _2$. Thus we have $F_2=K_2+F_1\ge K_2$. Now, by Lemma 2.4, $N(F_2)\subseteq N(K_2)$. But by the property $K_2F_2=0$, it follows that $R(F_2)\subseteq N(K_2)$. Hence $H=N(F_2)\oplus R(F_2)\subseteq N(K_2)$. This shows that $K_2=0$. Finally, using this we can get $F_1=F_2$.

Case 4. $K_1\ne 0$, $F_1\ne 0$, $\alpha _1>0$. We can prove $\alpha _1=\alpha _2$ by arguing as in the earlier cases. With this, we have

$$\begin{aligned} K_1-F_1=K_2-F_2. \end{aligned}$$

(2.3)

As $F_1$ commute with $K_1$ and $F_1$, it commute with $K_2-F_2$. So $F_1$ must commute with $(K_2-F_2)^2=K_2^2+F_2^2=(K_2+F_2)^2$. Thus, it commute with $K_2+F_2$. Hence we can conclude that $F_1$ commute with both $K_2$ and $F_2$. Since $N(F_1)$ is invariant under $K_1$ and $F_1$, by equation (2.3), $N(F_1)$ is invariant under $K_2-F_2$.

Now if $x\in N(F_1)$, then by equation (2.3), we have $(K_2-K_1)x=F_2x$. Using the fact that $F_2\ge 0$, we can conclude that $K_2\ge K_1$ on $N(F_1)$. We also show that this will happen on $R(F_1)$.

For $x\in H$, we have $F_1x\in R(F_1)$. Now,

$$\begin{aligned} \langle (F_2-F_1)(F_1x),F_1x\rangle =\langle (K_2-K_1)(F_1x),F_1x\rangle =\langle K_2(F_1x),F_1x\rangle \ge 0. \end{aligned}$$

This shows that $K_2-K_1=F_2-F_1\ge 0$ on $R(F_1)$. Combining with the earlier argument, we can conclude that $K_1\le K_2$. Now, interchanging the roles of $K_1$ and $K_2$, we can conclude that $K_2\le K_1$ and hence $K_1=K_2$. By equation (2.3), we can conclude that $F_1=F_2$.

Proof of (2) $\Rightarrow$ (1). If $T=K-F+\alpha I$, where $K\in {\mathcal {K}}(H)$ is positive, $F\in {\mathcal {F}}(H)$ is positive, $\alpha \ge 0$ and $KF=0$. Then by Theorem 2.1, $T\in {{\mathcal {AN}}}(H)$. Since $K\ge 0$ and $-F+\alpha I\ge 0$, T must be positive. $\square$

Remark 2.6

Let T be as in Theorem 2.5. Then we have the following:

• (1) if $\alpha =0$, then $F=0$ and hence $T=K$. In this case $\sigma _{\mathrm{ess}}(T)={\{\alpha }\}$,

• (2) if $\alpha >0$ and $F=0$, then $T=K+\alpha I$. In this case, $\sigma _{\mathrm{ess}}(T)={\{\alpha }\}$ and $m_{e}(T)=\alpha =m(T)$,

• (3) if $\alpha >0,\; K=0$ and $F\ne 0$, then $T=\alpha I-F$. In this case also, $\sigma _{\mathrm{ess}}(T)={\{\alpha }\}$ and $m_{e}(T)=\alpha$,

• (4) if $\alpha >0,\; F\ne 0$ and $K\ne 0$, then by the Weyl’s theorem, $\sigma _{\mathrm{ess}}(T)={\{\alpha }\}$ and $m_{e}(T)=\alpha$,

• (5) if $\alpha =0$ and $K=0$, then $T=0$,

• (6) if N(T) is infinite dimensional, then 0 is an eigenvalue with infinite multiplicity and hence $\alpha =0$, by Theorem 2.2. In this case, $F=0$ and hence $T=K$.

Remark 2.7

If we take $F=0$ in Theorem 2.5, then we get the structure obtained in [11].

Here we prove some important properties of ${\mathcal {AN}}$-operators.

PROPOSITION 2.8

Let $T=K-F+\alpha I$, where $K\in {\mathcal {K}}(H)$ is positive, $F\in {\mathcal {F}}(H)$ is positive with $KF=0$ and $F\le \alpha I$. If $\alpha >0$, then the following statements hold:

• (1) R(T) is closed,

• (2) N(T) is finite dimensional,

• (3) $N(T)\subseteq N(K)$,

• (4) $Fx=\alpha x$ for all $x\in N(T)$. Hence $N(T)\subseteq R(F)$. In this case, $\Vert F\Vert =\alpha$,

• (5) T is one-to-one if and only if $\Vert F\Vert <\alpha$,

• (6) T is Fredholm and $m_e(T)=\alpha$.

Proof

Proof of (1). Since $K-F$ is a compact operator, R(T) is closed. Here we have used the fact that for any $A\in {\mathcal {K}}(H)$ and $\lambda \in {\mathbb {C}}\setminus {\{0}\}$, $R(A+\lambda I)$ is closed.

Proof of (2). Let $x\in N(T)$. Then

$$\begin{aligned} (K-F)x=-\alpha x, \end{aligned}$$

(2.4)

that is, $\alpha I_{N(T)}$ is compact. This concludes that N(T) is finite dimensional.

Proof of (3). Let $x\in N(T)$. Multiplying Equation (2.4) by K and using the fact that $KF=FK=0$, we have $K^2x=-\alpha Kx$. If $Kx\ne 0$, then $-\alpha \in \sigma _p(K)$ contradicts the positivity of K. Hence $Kx=0$.

Proof of (4). Clearly, if $Tx=0$, then by (3), we have $Fx=\alpha x$. This also concludes that $N(T)\subseteq R(F)$ and $\Vert F\Vert =\alpha$.

Proof of (5). If T is not one-to-one, then $Fx=\alpha x$ for $x\in N(T)$ by (3). Suppose T is one-to-one and $\Vert F\Vert =\alpha$. Since F is norm attaining by Proposition 1.1, there exists $x\in S_H$ such that $Fx=\alpha x$. Then $Tx=Kx-Fx+\alpha x=Kx$. But $KF=0$ implies that $x\in N(K)$. So, $Tx=Kx$=0. By the injectivity of T, we have that $x=0$. This contradicts the fact that $x\in S_H$. Hence $\Vert F\Vert <\alpha$.

Proof of (6). Note that $\sigma _\mathrm{ess}(T)={\{\alpha }\}$ by the Weyl’s theorem on essential spectrum. Hence $m_e(T)=\alpha =m_e(T^*)$. Now T is Fredholm operator by [1, Theorem 2] with index zero. $\square$

Theorem 2.9

Let $T\in {\mathcal {B}}(H)$ and positive. Then $T\in {{\mathcal {AN}}}(H)$ if and only if $T^2\in {{\mathcal {AN}}}(H)$.

Proof

First we assume that $T\in {{\mathcal {AN}}}(H).$ Then there exists a triple $(K, F,\alpha )$ as in Theorem 2.5(2). Then $T^2=K_1-F_1+\beta I$, where $K_1=K^2+2\alpha K$ is a positive compact operator, $F_1=2\alpha F-F^2=(2\alpha I-F)F$ and $\beta =\alpha ^2$. Clearly, $F_1\ge 0$ as it is the product of two commuting positive operators. Also, $F_1\in {\mathcal {F}}(H)$. Next, we show that $F_1\le \alpha ^2I$. Clearly, $\alpha ^2I-F_1$ is self-adjoint and $\alpha ^2I-F_1=(\alpha I-F)^2\ge 0$. It can be easily verified that $K_1F_1=0$. So, $T^2$ is also in the same form. Hence by Theorem 2.5, $T^2\in {{\mathcal {AN}}}(H)$.

Now, let $T^2\in {{\mathcal {AN}}}(H)$. Then by Theorem 2.5, $T^2= K-F + \alpha I$, where $K\in {{\mathcal {K}}}(H)$ is positive, $F\in {\mathcal {F}}(H)$ is positive with $FK=KF=0$ and $F\le \alpha I$. If $\alpha > 0,$ then $(T-\sqrt{\alpha } I)(T+\sqrt{\alpha } I)= K-F$. Since T is positive $T+\sqrt{\alpha } I$ is a positive invertible operator. Hence $T-\sqrt{\alpha } I= (K-F)(T+\sqrt{\alpha } I)^{-1}$. Hence there is a positive compact operator, namely $K_1 = K (T+\sqrt{\alpha })^{-1}$ and a finite rank positive operator, namely $F_1 = F (T+\sqrt{\alpha } I)^{-1},$ such that $T-\sqrt{\alpha } I= K_1 - F_1.$ Hence $T= K_1-F_1 + \sqrt{\alpha }I$. Also, note that since F and K commute with $T^2$, hence commutes with T. Thus, we can conclude that $F_1K_1=0$. Finally,

$$\begin{aligned} \Vert F_1\Vert \le \Vert F\Vert \, \Vert (T+\sqrt{\alpha } I)^{-1}\Vert&\le \alpha \, \frac{1}{m(T+\sqrt{\alpha } I)}\\&=\frac{\alpha }{\sqrt{\alpha }+m(T)} \\&\le \frac{\alpha }{\sqrt{\alpha }}=\sqrt{\alpha }. \end{aligned}$$

In the third step of the above inequalities, we used the fact that $m(T+\sqrt{\alpha } I)=\sqrt{\alpha }+m(T)$, which follows by [11, Proposition 2.1].

If $\alpha =0$, then clearly $F=0$ and hence $T^2=K$. So, $T=K^{\frac{1}{2}}$, a compact operator which is clearly an ${\mathcal {AN}}$-operator. $\square$

COROLLARY 2.10

Let $T\in {\mathcal {B}}(H)$ and positive. Then $T\in {{\mathcal {AN}}}(H)$ if and only if $T^{\frac{1}{2}}\in {{\mathcal {AN}}}(H)$.

Proof

Let $S=T^{\frac{1}{2}}$. Then $S\ge 0$. The conclusion follows by Theorem 2.9. $\square$

COROLLARY 2.11

Let $T\in {\mathcal {B}}(H_1,H_2)$. Then $T\in {{\mathcal {AN}}}(H_1,H_2)$ if and only $T^*T\in {{\mathcal {AN}}}(H_1)$.

Proof

Proof follows from the following arguments: $T^*T\in {{\mathcal {AN}}}(H_1)\Leftrightarrow |T|^2\in {{\mathcal {AN}}}(H_1)\Leftrightarrow |T|\in {{\mathcal {AN}}}(H_1) \Leftrightarrow T \in {{\mathcal {AN}}}(H_1, H_2).$ $\square$

We have the following consequence.

Theorem 2.12

Let $T\in {{\mathcal {AN}}}(H)$ be self-adjoint and $\lambda$ be a purely imaginary number. Then $T \pm \lambda I \in {\mathcal {AN}}(H)$.

Proof

Let $S=T \pm \lambda I$. Then $S^*S=T^2+|\lambda |^2I =K-F+(\alpha +|\lambda |^2)I$, where the triple $(K,F,\alpha )$ satisfy conditions Theorem 2.5(2) applied to $T^2$. Hence by Corollary 2.11, $S\in {\mathcal {AN}}(H)$. $\square$

The following result is well known.

Lemma 2.13

Let $S,T\in {\mathcal {B}}(H)$ be such that $S^{-1},T^{-1}\in \mathcal B(H)$. Then $S^{-1}-T^{-1}=T^{-1}(T-S)S^{-1}$.

Theorem 2.14

Let $T=K-F+\alpha I$, where $(K, F, \alpha )$ satisfy conditions of Theorem 2.5(2). Then

• (1) R(F) reduces T,

• (2) $T=\left( \begin{array}{cc} K_0+\alpha I_{N(F)}&{} 0 \\ 0 &{} \alpha I_{R(F)}-F_0 \\ \end{array} \right)$, where $K_0=K|_{N(F)}$ and $F_0=F|_{R(F)}$,

• (3) if T is one-to-one and $\alpha >0$, then $T^{-1}\in {\mathcal {B}}(H)$ and

  $$\begin{aligned} T^{-1}=\left( \begin{array}{cc} \alpha ^{-1}I_{N(F)}-\alpha ^{-1}K_0(K_0+\alpha I_{N(F)})^{-1} &{}0 \\ 0 &{} \alpha ^{-1}I_{R(F)}+\alpha ^{-1}F_0(\alpha I_{R(F)}-F_0)^{-1} \\ \end{array} \right) . \end{aligned}$$

Proof

Proof of (1). First note that $T\ge 0$ and $T\in {\mathcal {AN}}(H)$. Let $y=Fx$ for some $x\in H$. Then $Ty=TFx=(K-F+\alpha I)Fx=(\alpha I-F)(Fx)=F(\alpha I-F)x\in R(F)$. This shows that R(F) is invariant under T. As T is positive, it follows that R(F) is a reducing subspace for T.

Proof of (2). First, we show that $K_0$ is a map on N(F). For this we show that N(F) is invariant under K. If $x\in N(F)$, then $FKx=0$ since $FK=0$. This proves that N(F) is invariant under K. Thus $K_0\in {\mathcal {K}}(N(F))$. Also, clearly, R(F) is invariant under F. Thus $F_0: R(F)\rightarrow R(F)$ is a finite dimensional operator. With respect to the pair of subspaces (N(F), R(F)), K has the decomposition

$$\begin{aligned} \left( \begin{array}{ll} K_0 &{} \quad 0 \\ 0 &{}\quad 0 \\ \end{array} \right) . \end{aligned}$$

Similarly the operators F and $\alpha I$ has the following block matrix forms:

$$\begin{aligned} \left( \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad F_0 \\ \end{array} \right) \; \; \text {and}\;\; \left( \begin{array}{ll} \alpha I_{N(F)} &{}\quad 0 \\ 0 &{}\quad \alpha I_{R(F)} \\ \end{array} \right) . \end{aligned}$$

With these representations of K, F and $\alpha I$, by definition, T can be represented as in (2).

Proof of (3). By Proposition 2.8(1), R(T) is closed. As T is one-to-one, T is bounded below. Since T is positive, $T^{-1}\in {\mathcal {B}}(H)$. In this case, $\Vert F_0\Vert =\Vert F\Vert <\alpha$, by (5) of Proposition 2.8. Hence we have

$$\begin{aligned} T^{-1}=\left( \begin{array}{cc} (K_0+\alpha I_{N(F)})^{-1} &{} 0 \\ 0 &{} (\alpha I_{R(F)}-F_0)^{-1} \\ \end{array} \right) . \end{aligned}$$

(2.5)

By Lemma 2.13, we have

$$\begin{aligned} (K_0+\alpha I_{N(F)})^{-1}-\alpha ^{-1} I_{N(F)}=-\alpha ^{-1}K_0(K_0+\alpha I_{N(F)})^{-1}, \end{aligned}$$

and hence

$$\begin{aligned} (K_0+\alpha I_{N(F)})^{-1}=\alpha ^{-1} I_{N(F)}-\alpha ^{-1}K_0(K_0+\alpha I_{N(F)})^{-1}. \end{aligned}$$

With similar arguments, we can obtain $(\alpha I_{R(F)}-F_0)^{-1}=\alpha ^{-1}I_{R(F)}+\alpha ^{-1}F_0(\alpha I_{R(F)}-F_0)^{-1}$. Substituting these quantities in equation (2.5), we obtain the representation of $T^{-1}$ as in (3). $\square$

Remark 2.15

Let

$$\begin{aligned} \beta&=\alpha ^{-1},\\ K_1&= \left( \begin{array}{cc} \alpha ^{-1}K_0(K_0+\alpha I_{N(F)})^{-1} &{} 0 \\ 0 &{} 0 \\ \end{array}\right) \end{aligned}$$

and

$$\begin{aligned} F_1&=\left( \begin{array}{cc} 0&{} 0 \\ 0 &{} \alpha ^{-1}F_0(\alpha I_{R(F)}-F_0)^{-1} \\ \end{array} \right) . \end{aligned}$$

Then $T^{-1}=\beta I-K_1+F_1$. Note that $\Vert K_1\Vert \le \beta$, since $\Vert K_0(\alpha I_{N(F)}+K_0)^{-1}\Vert \le 1$. Clearly, by definition, $K_1F_1=0$. This is exactly the structure of absolutely minimum attaining operators (shortly $\mathcal {AM}$-operators) in the case when T is positive and one-to-one. We refer [6] for more details of the structure of these operators. We recall that $A\in {\mathcal {B}}(H_1,H_2)$ is said to be minimum attaining if there exists $x_0\in S_{H_1}$ such that $\Vert Ax_0\Vert =m(A)$ and absolutely minimum attaining if $A|_{M}$ is minimum attaining for each non zero closed subspace M of $H_1$.

PROPOSITION 2.16

Let $T\in {\mathcal {B}}(H)$ satisfy condition (2) of Theorem 2.5. Then, with respect the pair of subspace $(N(K), N(K)^{\bot })$, T has the following decomposition:

$$\begin{aligned} T=\left( \begin{array}{cc} \alpha I_{N(K)}-F_0 &{} 0 \\ 0 &{} K_0+\alpha I_{N(K)^{\bot }} \\ \end{array} \right) , \end{aligned}$$

where $F_0=F|_{N(K)}$ and $K_0=K|_{N(K)^{\bot }}$.

Proof

First, we show that N(K) is a reducing subspace for T. Since $T\ge 0$, it suffices to show that N(K) is invariant under T. For this, let $x\in N(K)$. Then $Tx=(\alpha I-F)(x)$ and $K(Tx)=(\alpha I-F)(Kx)=0$. This proves the claim. Next, if $x\in N(K)$, then $Tx=(\alpha I-F)(x)$. That is, $T|_{N(K)}=\alpha I_{N(K)}-F|_{N(K)}$.

If $y\in N(K)^{\bot }=\overline{R(K)}$, then there exists a sequence $(x_n)\subset H$ such that $y= \lim _{n\rightarrow \infty }Kx_n$. So $Fy=\lim _{n\rightarrow \infty }FKx_n=0$. Thus we have $Ty=Ky+\alpha y$. So $T|_{N(K)^{\bot }}=K_{N(K)^{\bot }}+\alpha I_{N(K)^{\bot }}$. $\square$

3 Self-adjoint and normal ${\mathcal {AN}}$-operators

In this section, we first discuss the structure of self-adjoint ${\mathcal {AN}}$-operators. Later, we extend this to the case of normal operators.

Theorem 3.1

Let $T=T^*\in {{\mathcal {AN}}}(H)$. Then there exists an orthonormal basis consisting of eigenvectors of T.

Proof

The proof follows along the similar lines of [10, Theorem 3.1]. For the sake of completeness, we provide the details here. Let $\mathcal B=\{x_\alpha : \alpha \in I\}$ be the maximal set of orthonormal eigenvectors of T. This set is non empty as $T=T^*\in {\mathcal {AN}}(H)$. Let $M=\overline{\text{ span }}\{x_\alpha : \alpha \in I\}$. Then we claim that $M=H$. If not, $M^{\bot }$ is a proper non-zero closed subspace of H and it is invariant under T. Since $T=T^* \in {\mathcal {AN}}(H)$, we have either $||T|M^{\bot }||$ or $-||T|M^{\bot }||$ is an eigenvalue for $T|M^{\bot }$. Hence there is a non-zero vector, say $x_0$ in $M^{\bot }$, such that $Tx_0=\pm ||T|M^{\bot }|| x_0.$ Hence $x_{0}\in M$. Since $M\cap M^{\bot }={\{0}\},$ a contradiction. $\square$

PROPOSITION 3.2

Let $T=T^*\in {{\mathcal {AN}}}(H)$. Then the following holds:

• (1) T can have atmost two eigenvalues with infinite multiplicity. Moreover, if $\alpha$ and $\beta$ are such eigenvalues, then $\alpha =\pm \beta$,

• (2) if T has an eigenvalue $\alpha$ with infinite multiplicity and $\beta$ is a limit point of $\sigma (T)$, then $\alpha =\pm \beta$,

• (3) $\sigma (T)$ can have atmost two limit points. If $\alpha$ and $\beta$ are such points, then $\alpha =\pm \beta$.

Proof

Proof of (1). Let $\alpha _j\in \sigma _p(T)$ be such that $N(T-\alpha _jI)$ is infinite dimensional for each $j=1,2,3$. Then $\alpha _j^2\in \sigma _p(T^2)$ and we have $N(T-\alpha _jI)\subseteq N(T^2-\alpha _j^2I)$ for each $j=1,2,3$. Since $T^2\in {\mathcal {AN}}(H)$ and positive, by (3) of Theorem 2.2, it follows that $\alpha _1^2=\alpha _2^2=\alpha _3^2$. Thus $\alpha _1=\pm \alpha _2=\pm \alpha _3$.

Proof of (2). Let $\alpha \in \sigma _p(T)$ with infinite multiplicity and $\beta \in \sigma (T)$, which is a limit point. Since $\sigma (T^2)={\{\lambda ^2:\lambda \in \sigma (T)}\}$, it follows that $\alpha ^2$ is an eigenvalue of $T^2$ with infinite multiplicity as $N(T-\alpha I)\subseteq N(T^2-\alpha ^2I)$ and $\beta ^2$ is a limit point $\sigma (T^2)$. Since $T^2\in {\mathcal {AN}}(H)$ is positive, by (4) of Theorem (2.2), $\alpha ^2=\beta ^2$. Thus $\alpha =\pm \beta$.

Proof of (3). Let $\alpha ,\beta \in \sigma (T)$ be limit points of $\sigma (T)$. Then $\alpha ^2,\beta ^2\in \sigma (T^2)$ are limit points of $\sigma (T^2)$ and since $T^2\in {\mathcal {AN}}(H)$ and positive, by Theorem 2.2(2), $\alpha ^2=\beta ^2$, concluding that $\alpha =\pm \beta$. By arguing as in Proof of (1), we can show that there are at most two limit points for the spectrum. $\square$

The following decomposition of a self-adjoint operator is used in the sequel:

Let $T=T^*\in {\mathcal {B}}(H)$ and have the polar decomposition $T=V|T|$. Let $H_0=N(T),\; H_{+}=N(I-V)$ and $H_{-}=N(I+V)$. Then $H=H_0\oplus H_{+}\oplus H_{-}$. All these subspaces are invariant under T. Let $T_0=T|_{N(T)},\; T_{+}=T|_{H_{+}}$ and $T_ {-}=T|_{H_{-}}$. Then $T=T_0\oplus T_{+}\oplus T_{-}$. Further more, $T_{+}$ is strictly positive, $T_{-}$ is strictly negative and $T_0=0$ if $N(T)\ne {\{0}\}$. Let $P_0=P_{N(T)},\; P_{\pm }=P_{H_{\pm }}$. Then $P_0=I-V^2$ and $P_{\pm }=\frac{1}{2}(V^2\pm V)$. Thus $V=P_{+}-P_{-}$. Moreover, $|T|=T_{+}\oplus (-T_{-})\oplus T_0$. For details, see [9, Example 7.1, p. 139]. Note that the operators $T_{+}$ and $T_{-}$ are different than those used in Theorem 2.5.

Theorem 3.3

Let $T\in {{\mathcal {AN}}}(H)$ be self-adjoint with the polar decomposition $T=V|T|$. Then

• (1) the operator T has the representation

  $$\begin{aligned} T=K-F+\alpha V, \end{aligned}$$

  where $K\in {\mathcal {K}}(H),\; F\in {\mathcal {F}}(H)$ are self-adjoint with $KF=0=FK,\; \alpha \ge 0$ and $F^2\le \alpha ^2I$,

• (2) if T is not a compact operator, then $V\in {\mathcal {AN}}(H)$,

• (3) $K^2+2\alpha \text {Re}(VK)\ge 0$.

Proof

Proof of (1). Since T is self-adjoint, by considering $H_0,H_{\pm }$ and $T_0,T_{\pm }$ as in the earlier discussion, we can write $H=H_{0}\oplus H_{+}\oplus H_{-}$ and $T=T_0\oplus T_{+}\oplus T_{-}$. Note that $T_0=0$ if $H_{0}\ne {\{0}\}$. Since $H_{\pm }$ reduces T, we have $T_{\pm }\in \mathcal B(H_{\pm })$. As $T\in {{\mathcal {AN}}}(H)$, we have that $T_{\pm }\in {\mathcal {AN}}(H_{\pm })$. Hence by Theorem 2.5, we have $T_{+}=K_{+}-F_{+}+\alpha I_{H_{+}}$ such that $K_{+}$ is a positive compact operator, $F_{+}$ is a finite rank positive operator with the property that $K_{+}F_{+}=0$ and $F_{+}\le \alpha I_{H_{+}}$. As $T_{+}$ is strictly positive, $\alpha >0$, in fact, $\alpha =m_e(T_{+})$.

Similarly, $T_{-}\in {{\mathcal {AN}}}(H_{-})$ and strictly negative. Hence there exists a triple $(K_{-},F_{-},\beta )$ such that $-T_{-}=K_{-}-F_{-}+\beta I_{H_{-}}$, where $K_{-}\in \mathcal K(H_{-})$ is positive, $F_{-}\in {\mathcal {F}}(H_{-})$ is positive with $K_{-}F_{-}=0,\; F_{-}\le \beta I_{H_{-}}$ and $\beta >0$, in fact, $\beta =m_e(T_{-})$. Hence we can write $T_{-}=-K_{-}+F_{-}-\beta I_{H_{-}}$. So, we have

$$\begin{aligned} T&=\left( \begin{array}{ccc} T_{+} &{} 0 &{} 0 \\ 0 &{} T_{-} &{} 0 \\ 0 &{} 0 &{} T_0 \\ \end{array} \right) \\&=\left( \begin{array}{ccc} K_{+}-F_{+}+\alpha I_{H_{+}}&{} 0 &{} 0 \\ 0 &{} -K_{-}+F_{-}-\beta I_{H_{-}} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) \\&=\left( \begin{array}{ccc} K_{+} &{}0 &{}0 \\ 0 &{} -K_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) -\left( \begin{array}{ccc} F_{+} &{} 0 &{} 0 \\ 0 &{} -F_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) +\left( \begin{array}{ccc} \alpha I_{H_{-}} &{}0 &{} 0 \\ 0 &{} -\beta I_{H_{-}} &{}0 \\ 0 &{}0 &{}0 \\ \end{array} \right) . \end{aligned}$$

Note that $|T|=T_{+}\oplus (-T_{-})\oplus T_0$ (see [9, Example 7.1, p. 139] for details). That is,

$$\begin{aligned} |T|&=\left( \begin{array}{ccc} K_{+} &{}0 &{}0 \\ 0 &{} K_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) -\left( \begin{array}{ccc} F_{+} &{} 0 &{} 0 \\ 0 &{}F_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) +\left( \begin{array}{ccc} \alpha I_{H_{-}} &{}0 &{} 0 \\ 0 &{} -\beta I_{H_{-}} &{}0 \\ 0 &{}0 &{}0 \\ \end{array} \right) \\&=\left( \begin{array}{ccc} K_{+} &{}0 &{}0 \\ 0 &{} K_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) -\left( \begin{array}{ccc} F_{+} &{} 0 &{} 0 \\ 0 &{}F_{-} &{} 0 \\ 0 &{} 0 &{} \alpha I_{H_0} \\ \end{array} \right) +\left( \begin{array}{ccc} \alpha I_{H_{-}} &{}0 &{} 0 \\ 0 &{} -\beta I_{H_{-}} &{}0 \\ 0 &{}0 &{}\alpha I_{H_0}\\ \end{array} \right) . \end{aligned}$$

Let $K=\left( \begin{array}{ccc} K_{+} &{}0 &{}0 \\ 0 &{} -K_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right)$ and $F=\left( \begin{array}{ccc} F_{+} &{} 0 &{} 0 \\ 0 &{}-F_{-} &{} 0 \\ 0 &{} 0 &{} \alpha I_{H_0} \\ \end{array} \right)$.

Since $|T|\in {\mathcal {AN}}(H)$ and positive, by the uniqueness of the representation, we get that $\alpha =\beta$.

Now let, $V=\left( \begin{array}{ccc} I_{H_{+}} &{}0 &{} 0 \\ 0 &{} - I_{H_{-}} &{}0 \\ 0 &{}0 &{}0 \\ \end{array} \right)$. Then $T=K-F+\alpha V$ and K and F satisfy the stated properties.

If T is one-to-one, then $N(T)={\{0}\}$. In this case, $H=H_{+}\oplus H_{-}$ and $T=T_{+}\oplus T_{-}$. Arguing as above, we can obtain the representation for $T=K-F+\alpha V$, where

$$\begin{aligned} K=\left( \begin{array}{cc} K_{+} &{} 0 \\ 0 &{} -K_{-} \\ \end{array} \right) ,\; F=\left( \begin{array}{cc} F_{+} &{} 0 \\ 0 &{} -F_{-} \\ \end{array} \right) ,\; V=\left( \begin{array}{cc} \alpha I_{H_+} &{} 0 \\ 0 &{} -\alpha I_{H_{-}} \\ \end{array} \right) . \end{aligned}$$

Proof of (2). Note that if $\alpha =0$, then T is compact. If $\alpha >0$ and V is a finite rank operator, then also T is compact. Since we assumed that T is not compact, it must be the case that $\alpha >0$ and R(V) is infinite dimensional. But by Proposition 2.8, $N(T)=N(V)$ is finite dimensional. So the conclusion follows by [3, Proposition 3.14].

Proof of (3). First note that since $T=T^*$, it follows that $V=V^*$. As $VK=KV$, KV is self-adjoint. Hence $K^2+2\,\mathrm{Re}(V^*K)=K^2+2VK$. Thus

$$\begin{aligned} K^2+2VK=\left( \begin{array}{ccc} K_{+}^2+2K_{+} &{}0 &{}0 \\ 0 &{} K_{-}^2-2K_{-} &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right) . \end{aligned}$$

Since the (1, 1) entry of the above matrix is positive, to get the conclusion, it suffices to prove that the (2, 2) entry is positive. Clearly, $K_{-}^2-2K_{-}$ is self-adjoint. Next, we show that $\sigma (K_{-}^2-2K_{-})$ is positive. Let $\lambda \in \sigma (K_{-})$. Then $\lambda \le 0$ and $\lambda ^2-2\lambda \in \sigma (K_{-}^2-2K_{-})$. But $\lambda ^2-2\lambda =\lambda (\lambda -2)\ge 0$. Hence $K_{-}^2-2K_{-}$ is positive. $\square$

COROLLARY 3.4

Let $T=T^*\in {\mathcal {AN}}(H)$. Then $\sigma (T)$ is countable.

Proof

Since $T=T_{+}\oplus T_{-}\oplus T_0$, all these operators $T_{+},T_{-}$ and $T_0$ are ${\mathcal {AN}}$ operators. We know that $\sigma (T_{+}),\sigma (T_0)$ are countable, as they are positive. Also, $-T_{-}$ is a positive ${{\mathcal {AN}}}$-operator and hence $\sigma (T_{-})$ is countable. Hence, we can conclude that $\sigma (T)=\sigma (T_{+})\cup \sigma (T_{-})\cup \sigma (T_0)$ is countable. $\square$

Next, we can get the structure of normal ${\mathcal {AN}}$-operators. Here we use a different approach to the one used in Theorem 3.3.

PROPOSITION 3.5

Let $T\in {\mathcal {AN}}(H)$ be normal with the polar decomposition $T=V|T|$. Then there exists a compact normal operator K, a finite rank normal operator $F\in {\mathcal {B}}(H)$ and $\alpha \ge 0$ such that

• (1) T has the representation

  $$\begin{aligned} T=K-F+\alpha V \end{aligned}$$

  (3.1)

  with $KF=0$ and $F^*F\le \alpha ^2I$,

• (2) $K^*K+2\alpha \text {Re}(V^*K)\ge 0$,

• (3) V, K, F commutes mutually,

• (4) if $\alpha >0$, then $V\in {{\mathcal {AN}}}(H)$.

Proof

Proof of (1). Since T is normal, $|T|=|T^*|$. Hence $|T^*|^2=TT^*=V|T|^2V^*$. That is $V^*|T^*|^2=V^*V|T|^2V^*=|T|^2V^*$ or $V|T|^2=|T|^2V$. By the square root property, it follows that $V|T|=|T|V$.

Since $|T|\in {\mathcal {AN}}(H)$, we have $|T|=K_1-F_1+\alpha I$, where $K_1\in {\mathcal {K}}(H)$ is positive, $F_1\in {\mathcal {F}}(H)$ is positive and $F_1\le \alpha I$.

Next, we show that V is normal. We have $N(T^*)=N(T)=N(V)$. Hence

$$\begin{aligned} V^*V=P_{N(V)^{\bot }}=P_{N(T)^{\bot }}=P_{N(T^*)^{\bot }}=P_{\overline{R(T)}}=P_{R(V)}=VV^*. \end{aligned}$$

So $T=K-F+\alpha V$, where $K=VK_1$ and $F=VF_1$. Next, we show that K and F are normal. As T is normal, V commutes with |T|, we have

$$\begin{aligned} V(K_1-F_1)=(K_1-F_1)V. \end{aligned}$$

(3.2)

Since V commutes with $K_1-F_1$, it also commutes with $(K_1-F_1)^2$. But, $(K_1-F_1)^2=K_1^2+F_1^2=(K_1+F_1)^2$. With this, we can conclude that $V(K_1+F_1)^2=(K_1+F_1)^2V$. Hence,

$$\begin{aligned} V(K_1+F_1)=(K_1+F_1)V. \end{aligned}$$

(3.3)

Thus by equations (3.2) and (3.3), we can conclude that $VK_1=K_1V$ and $VF_1=F_1V$. By the Fuglede’s theorem we can conclude that $V^*K_1=K_1V^*$ and $V^*F_1=F_1V^*$. Next,

$$\begin{aligned} K^*K=K_1V^*VK_1=K_1VV^*K_1 =VK_1V^*K_1=VK_1K_1V^*=KK^*. \end{aligned}$$

With similar arguments we can show that F is normal.

Next, we show that $KF=0$. Since V commutes with $K_1$ and $F_1$, we have $KF=VK_1VF_1=V^2K_1F_1=0$.

Finally, $F^*F=F_1V^*VF_1\le \Vert V\Vert ^2F_1^2\le \alpha ^2I$.

Proof of (2). Using the relations $VK_1=K_1V$ and $V^*K_1=K_1V^*$, we get

$$\begin{aligned} K^*K+\alpha (V^*K+K^*V)&=K_1V^*VK_1+\alpha (V^*VK_1+K_1V^*V)\\&=V^*V(K_1^2+2\alpha K_1) \\&=P_{N(V)^{\bot }} (K_1^2+2\alpha K_1)\\&=K_1^2+2\alpha K_1\\&\ge 0. \end{aligned}$$

In the fourth step of the above equations, we have used the fact that $P_{N(V)^{\bot }}K_1=P_{R(V)}K_1=P_{R(|T|)}K_1=K_1$.

Proof of (3). We have $VK=VVK_1=VK_1V=KV$ and $VF=VVF_1=VF_1V=FV$. Also, $KF=0=FK$.

Proof of (4). Note that by applying Proposition 2.8(2) to |T|, we can conclude that $N(|T|)=N(T)=N(V)$ is finite dimensional. Now the conclusion follows by [3, Proposition 3.14]. $\square$

COROLLARY 3.6

Let $T\in {\mathcal {B}}(H)$ be normal. Then $T\in {\mathcal {AN}}(H)$ if and only if $T^*\in {\mathcal {AN}}(H)$.

Proof

We know that $T\in {\mathcal {AN}}(H)$ if and only if $T^*T\in {\mathcal {AN}}(H)$, by Corollary 2.11. Since $T^*T=TT^*$, by Corollary 2.11 again, it follows that $TT^*\in {\mathcal {AN}}(H)$ if and only if $T^*\in {\mathcal {AN}}(H)$. $\square$

4 General case

In this section, we prove the structure of absolutely norm attaining operators defined between two different Hilbert spaces.

Theorem 4.1

Let $T\in {{\mathcal {AN}}}(H_1,H_2)$ with the polar decomposition $T=V|T|$. Then

$$\begin{aligned} T=K-F+\alpha V, \end{aligned}$$

where $K\in {\mathcal {K}}(H_1,H_2), \; F\in {\mathcal {F}}(H_1,H_2)$ such that $K^*F=0=KF^*$ and $\alpha ^2I\ge F^*F$.

Proof

Since $|T|\in {\mathcal {AN}}(H_1)$ and positive, we have by Theorem 2.5, $|T|=K_1-F_1+\alpha I$, where the triple $(K_1,F_1,\alpha )$ satisfy conditions in Theorem 2.5(2). Now, $T=K-F+\alpha V$, where $K=VK_1,\; F=VF_1$. Clearly,

$$\begin{aligned} K^*F=K_1V^*VF_1=K_1P_{N(V)^{\bot }}F_1&=K_1(I-P_{N(V)})F_1\\&=K_1F_1-K_1P_{N(V)}F_1\\&=0 \;(\text {since }\; N(V)=N(|T|)\subseteq N(K_1)) . \end{aligned}$$

Also, clearly, $KF^*=VK_1F_1V^*=0$.

Finally, $F^*F=F_1V^*VF_1\le \Vert V^*V\Vert F_1^2\le F_1^2 \le \alpha ^2 I$. $\square$

PROPOSITION 4.2

Let $T\in {\mathcal {B}}(H)$ and $U\in {\mathcal {B}}(H)$ be unitary such that $T^*=U^*TU$. Then $T\in {{\mathcal {AN}}}(H)$ if and only if $T^*\in {{\mathcal {AN}}}(H)$.

Proof

This follows by [3, Theorem 3.5]. $\square$

Next, we discuss a possible converse in the general case.

Theorem 4.3

Let $K\in {\mathcal {K}}(H_1,H_2),\; F\in {\mathcal {F}}(H_1,H_2)$, $\alpha \ge 0$ and $V\in {\mathcal {B}}(H_1,H_2)$ be a partial isometry. Further, assume that

• (1) $V\in {\mathcal {AN}}(H_1,H_2)$,

• (2) $K^*K+\alpha (V^*K+K^*V)\ge 0$.

Then $T:=K-F+\alpha V \in {{\mathcal {AN}}}(H_1,H_2)$.

Proof

If $\alpha =0$, then $T\in {\mathcal {K}}(H_1,H_2)$. Hence $T\in {{\mathcal {AN}}}(H_1,H_2)$. Next assume that $\alpha >0$. We prove this case by showing $T^*T\in {\mathcal {AN}}(H_1)$. By a simple calculation, we can get $T^*T={\mathcal {K}}-{\mathcal {F}} +\alpha ^2 P_{N(V)^{\bot }}$, where

$$\begin{aligned} {\mathcal {K}}= K^*K+\alpha (V^*K+K^*V),\quad {\mathcal {F}}= F^*F-F^*K-K^*F-\alpha (V^*F+F^*V). \end{aligned}$$

Since $V\in {\mathcal {AN}}(H_1,H_2)$, either N(V) or $N(V)^{\bot }$ is finite dimensional by [3, Proposition 3.14]. If $N(V)^{\bot }$ is finite dimensional, then $T^*T\in {\mathcal {K}}(H_1)$. Hence $T\in {\mathcal {K}}(H_1,H_2)$.

If N(V) is finite dimensional, then $T^*T={\mathcal {K}}-(\mathcal F-\alpha ^2 P_{N(V)})+\alpha ^2I$. Note that the operator $\mathcal F-\alpha ^2 P_{N(V)}$ is a finite rank self-adjoint operator. Hence $T^*T\in {\mathcal {AN}}(H_1)$, by Theorem 2.1. Now the conclusion follows by Corollary 2.11. $\square$

COROLLARY 4.4

Suppose that $K\in {\mathcal {K}}(H)$, $F\in {\mathcal {F}}(H)$ are normal and $V\in {\mathcal {B}}(H)$ is a normal partial isometry such that V, F, K commute mutually. Let $\alpha \ge 0$. Then

• (1) $T:=K-F+\alpha V$ is normal,

• (2) if $K^*K+2\alpha V^*K \ge 0$ and $V\in {{\mathcal {AN}}}(H)$, then $T\in {{\mathcal {AN}}}(H)$.

Proof

Proof of (1). We observe that if A and B are commuting normal operators, then $A+B$ is normal (see [12, p. 342, Exercise 12] for details). By this observation, it follows that T is normal.

Proof of (2). Since $VK=KV$, by Fuglede’s theorem [12, p. 315], $V^*K=KV^*$. With this observation and Theorem 4.3, the conclusion follows. $\square$

COROLLARY 4.5

Suppose that $K\in {\mathcal {K}}(H)$, $F\in {\mathcal {F}}(H)$ are self-adjoint and $V\in {\mathcal {B}}(H)$ is a self-adjoint, partial isometry and $\alpha \ge 0$ such that

• (a) $V\in {\mathcal {AN}}(H)$,

• (b) $K^2+2\alpha (VK)\ge 0$.

Then $T:=K-F+\alpha V$ is self-adjoint ${\mathcal {AN}}$-operator.

Proof

The proof directly follows by Theorem 4.3. $\square$

DEFINITION 4.6

[4, p. 349]

Let $T\in {\mathcal {B}}(H_1,H_2)$. Then T is called left-semi-Fredholm if there exists $B\in \mathcal B(H_2,H_1)$ and $K\in {\mathcal {K}}(H_1)$ such that $BT=K+I$, and right-semi-Fredholm if there exists $A\in \mathcal B(H_2,H_1)$ and $K'\in {\mathcal {K}}(H_2)$ such that $TA=K'+I$.

If T is both left-semi-Fredholm and right-semi-Fredholm, then T is called Fredholm.

Remark 4.7

Note that T is left semi-Fredholm if and only if $T^*$ is right semi-Fredholm (see [4, section 2, p. 349] for details).

COROLLARY 4.8

Let $T\in {\mathcal {AN}}(H_1,H_2)$ but not compact. Then T is left-semi-Fredholm.

Proof

Let $T=V|T|$ be the polar decomposition of T. Then $|T|=V^*T$. As $|T|\in {\mathcal {AN}}(H_1)$, by Theorem 2.5, there exists a triple $(K,F,\alpha )$ satisfying conditions in Theorem 2.5, such that $V^*T=K-F+\alpha I$. Let $K^{'}=K-F$. Then $V^*T=K^{'}+\alpha I$. By Definition 4.6, it follows that T is left-semi-Fredholm. $\square$