# Asymmetric Putnam-Fuglede Theorem for (n,k)-Quasi-∗-Paranormal Operators

^{1}

^{2}

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## Abstract

**:**

## 1. Introduction

- (i)
- A and ${B}^{\ast}$ are n-∗-paranormal operators
- (ii)
- A is a $(n,k)$-quasi-∗-paranormal operator with reduced kernel and ${B}^{\ast}$ are n-∗-paranormal operator;
- (iii)
- A is a n-∗-paranormal operator and ${B}^{\ast}$ are $(n,k)$-quasi-∗-paranormal operator with reduced kernel (an operator T with reduced kernel means that its kernel is invariant under ${T}^{\ast}$).

- if T is a $(n,k)$-quasi-∗-paranormal operator with reduced kernel (resp. n-∗-paranormal operator or a n-quasi-∗-class A with reduced kernel), then T has a part in the class ${C}_{00}$ on a stable subspace ${\mathcal{H}}_{0}$ and a compression quasi-affine transform to an isometry on the orthogonal complement of ${\mathcal{H}}_{0}$.
- Next, we prove that if T is completely non-normal $(n,k)$-quasi-∗-paranormal operator; for $k=0,1$ and verifying the defect operator ${D}_{T}$ is a Hilbert-Schmidt class, then $T\in {C}_{10}$.

**Definition**

**1.**

- $glim$ is positive, i.e., if ${x}_{n}\ge 0$ for all $n\in \mathbb{N}$ then $glim\left({x}_{n}\right)\ge 0$;
- $glim\left(1\right)=1$, where $\left(1\right)=(1,1,1,\dots .)$;
- $glim$ is shift-invariant, i.e., $glim\left({x}_{n}\right)=glim\left({x}_{n+1}\right)$.

- (i)
- A power-bounded operator T is said to be of class ${C}_{1.}$ if the sequence $\{\parallel {T}^{n}x\parallel :n\in N\}$ does not converge to 0 for any non-zero vector x i.e., ${\mathcal{H}}_{0}\left(T\right)=\left\{0\right\}$.
- (ii)
- T is said to be strongly stable if ${\mathcal{H}}_{0}\left(T\right)=\mathcal{H}$ and we write $T\in {C}_{0.}$;
- (iii)
- T is of class ${C}_{.j}:j=0,1$ if ${T}^{\ast}$ is of class ${C}_{j.};j=0,1$;
- (iv)
- T is of class ${C}_{ij}:i,j=0,1$ if $T\in {C}_{i.}\cap {C}_{.j}$.

**Definition**

**2.**

- (i)
- the joint point spectrum, denoted by ${\sigma}_{jp}\left(T\right)$ is the set$${\sigma}_{jp}\left(T\right)=\{\lambda \in \mathbb{C}:Tx=\lambda x\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}and\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{T}^{\ast}x=\overline{\lambda}x\}.$$
- (ii)
- the joint approximate point spectrum, denoted by ${\sigma}_{ja}\left(T\right)$ is the set of scalars λ for which there exists a normalized sequence $\left\{{x}_{n}\right\}\subset \mathcal{H}$ verifying$$(T-\lambda ){x}_{n}\to 0and{(T-\lambda )}^{\ast}{x}_{n}\to 0.$$

**Definition**

**3.**

## 2. Properties of $(\mathit{n},\mathit{k})$-Quasi-∗-Hyponormal Operators

**Lemma**

**1.**

**Lemma**

**2.**

**Proof.**

**Lemma**

**3.**

**Proof.**

**Proposition**

**1.**

- 1.
- If T is a power-bounded n-∗-paranormal operator and there is an invariant subspace $\mathcal{M}$ for which the restriction ${T|}_{\mathcal{M}}=N$ of T on $\mathcal{M}$ is a normal operator, then $\mathcal{M}$ reduces T and $N=U\oplus 0$ where U is unitary.
- 2.
- If T is a power-bounded $(n,k)$-quasi-∗-paranormal operator and there is an invariant subspace $\mathcal{M}$ for which the restriction ${T|}_{\mathcal{M}}=N$ of T on $\mathcal{M}$ is an injective normal operator then $\mathcal{M}$ reduces T and N is a unitary operator. In particular, if $\mathcal{M}=\overline{Ran{T}^{k}}$ and ${T}_{1}$ as in the previous Lemma, is normal operator, then $\overline{Ran{T}^{k}}$ reduces T.

**Proof.**

- Let T be a power-bounded n-∗-paranormal operator and let us consider an invariant subspace $\mathcal{M}\subset \mathcal{H}$ for T such that ${T|}_{\mathcal{M}}=N$ is normal. The operator T has the following matrix decomposition$$T=\left[\begin{array}{cc}N& R\\ 0& \ast \end{array}\right]$$On the one hand, we have that N is a power-bounded normal operator, since N is normaloid it follows that N is a contraction. It is well known that $N=U\oplus {N}_{0}$ where U is unitary and ${N}_{0}$ is of class ${C}_{00}$ for possible ${N}_{0}=0$. On the other hand, Since the operator T is n-∗-paranormal, it follows that$$\parallel {R}^{\ast}{x\parallel}^{2}+{\parallel Nx\parallel}^{2}=\parallel {T}^{\ast}{x\parallel}^{2}\le \parallel {T}^{n}{x\parallel}^{\frac{2}{n}}={\parallel {N}^{n}x\parallel}^{\frac{2}{n}},$$Since the kernel of N reduces N, hence $N={N}_{1}\oplus 0$.If $x\in kerN$ then from (7) we get $x\in ker{R}^{\ast}$. Thus$${R}^{\ast}=0\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}on\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{4pt}{0ex}}kerN$$For each unit vector $x\in \mathcal{M}\ominus kerN$ we have, $(\parallel {R}^{\ast}{x\parallel}^{2}{+1)}^{n}\le {\displaystyle \frac{1}{{\parallel Nx\parallel}^{2n}}}{\parallel {N}^{n}x\parallel}^{2}$From Lemma 1, we get for all $k\ge 1$, that$$(\parallel {R}^{\ast}{x\parallel}^{2}{+1)}^{nk}\le {\displaystyle \frac{1}{{\parallel Nx\parallel}^{2nk}}}\parallel {N}^{n}{x\parallel}^{2k}\le {\displaystyle \frac{1}{{\parallel Nx\parallel}^{2nk}}}{\parallel {N}^{nk}x\parallel}^{2}$$If ${N}_{0}\ne 0$ then$$(\parallel {R}^{\ast}{x\parallel}^{2}{+1)}^{nk}\le {\displaystyle \frac{1}{{(\parallel Ux\parallel}^{2}+\parallel {N}_{0}x{{\parallel}^{2})}^{nk}}}(\parallel {U}^{nk}{x\parallel}^{2}+\parallel {N}_{0}^{nk}x{\parallel}^{2})$$Since U is unitary and $\parallel x\parallel =1$ then$$(\parallel {R}^{\ast}{x\parallel}^{2}{+1)}^{nk}\le {\displaystyle \frac{1}{(1+\parallel {N}_{0}x{{\parallel}^{2})}^{nk}}}(1+\parallel {N}_{0}^{nk}x{\parallel}^{2})$$Since ${N}_{0}\in {C}_{00}$ and $1+\parallel {N}_{0}{x\parallel}^{2}>1$ then for $k\to \infty $ we get $(\parallel {R}^{\ast}x{{\parallel}^{2}+1)}^{nk}\to 0$. However, $\parallel {R}^{\ast}{x\parallel}^{2}+1\ge 1$ which is a contradiction. Hence, ${N}_{0}=0$ and then $N=U$ on $\mathcal{M}\ominus kerN$.
- As in the case $\left(1\right)$, let us consider an invariant subspace $\mathcal{M}\subset \mathcal{H}$ for T such that ${T|}_{\mathcal{M}}=N$ is normal. The operator T has the following matrix$$T=\left[\begin{array}{cc}N& R\\ 0& \ast \end{array}\right],$$Since T is $(n,k)$-quasi-∗-paranormal operator, then$$\parallel {T}^{\ast}\left({T}^{k}x\right){\parallel}^{2}\le \parallel {T}^{(n+1)}{T}^{k}{x\parallel}^{\frac{2}{(n+1)}}{\parallel {T}^{k}x\parallel}^{\frac{2n}{(n+1)}}$$Put $y={T}^{k}x={N}^{k}x\in \mathcal{M}=\overline{ran{N}^{k}}$ for all $x\in \mathcal{M}$, we get,$$(\parallel {R}^{\ast}{y\parallel}^{2}+{\parallel Ny\parallel}^{2}{)}^{n+1}\le \parallel {N}^{n+1}{y\parallel}^{2}{\parallel y\parallel}^{2n},$$Since N is a contraction and $y={N}^{k}x$ we get ${\parallel y\parallel}^{2n}\le 1$. Hence,$$(\parallel {R}^{\ast}{y\parallel}^{2}+{\parallel Ny\parallel}^{2}{)}^{n+1}\le {\parallel {N}^{n+1}y\parallel}^{2},$$

**Lemma**

**4.**

**Proof.**

**Corollary**

**1.**

- 1.
- If $k=0$ (i.e., T is n-∗-paranormal) and $\lambda \in {\sigma}_{p}\left(T\right)$, then $ker(T-\lambda )$ reduces T. Also, if $k>0$ and $\phantom{\rule{0.166667em}{0ex}}\lambda \in {\sigma}_{p}\left(T\right)-\left\{0\right\}$, then $ker(T-\lambda )$ reduces T.
- 2.
- If T is n-∗-paranormal and $Tx=\lambda x$ such that $x\ne 0$, then ${T}^{\ast}x=\overline{\lambda}x$ and ${\sigma}_{p}\left(T\right)={\sigma}_{jp}\left(T\right)$. The same result holds in case $k>0$ and $\lambda \ne 0$ and ${\sigma}_{p}\left(T\right)-\left\{0\right\}={\sigma}_{jp}\left(T\right)-\left\{0\right\}$.
- 3.
- If $\lambda \ne \mu $, then $ker\phantom{\rule{0.166667em}{0ex}}(T-\lambda )\perp ker\phantom{\rule{0.166667em}{0ex}}(T-\mu )$.
- 4.
- $T=N\oplus A$ on the decomposition $\mathcal{H}=\mathcal{M}\oplus {\mathcal{M}}^{\perp}$, where $\mathcal{M}$ is the subspace spanned by the eigenspaces of T, N is a normal operator and A is a power-bounded $(n,k)$-quasi-∗-paranormal operator with ${\sigma}_{r}\left({A}^{\ast}\right)=\varnothing $. Moreover,$$\sigma \left({A}^{\ast}\right)\subseteq {\sigma}_{p}\left({A}^{\ast}\right)\cup {\sigma}_{c}\left({A}^{\ast}\right)\subseteq {\sigma}_{a}\left({A}^{\ast}\right).$$
- 5.
- If T is n-∗-paranormal, then ${\sigma}_{a}\left(T\right)={\sigma}_{ja}\left(T\right)$; also, and if $k>0$, then ${\sigma}_{a}\left(T\right)-\left\{0\right\}={\sigma}_{ja}\left(T\right)-\left\{0\right\}$.

**Proof.**

- The result follows immediately from Proposition 1 by taking $\mathcal{M}=ker(T-\lambda )$ and $N=\lambda I$ which is normal.
- It follows from item (1).
- If $Tx=\lambda x$ and $Ty=\mu y$ with $\lambda \ne \mu $, then$$\lambda \langle x,y\rangle =\langle Tx,y\rangle =\langle x,{T}^{\ast}y\rangle =\langle x,\overline{\mu}y\rangle =\mu \langle x,y\rangle $$
- From items (1), (2), (3) and according to the decomposition $\mathcal{H}=\mathcal{M}\oplus {\left(\mathcal{M}\right)}^{\perp}$, where $\mathcal{M}$ is the subspace spanned by the eigenspaces, the operator T can be written$$T=\left[\begin{array}{cc}N& 0\\ 0& A\end{array}\right],$$$$ker{(A-\lambda )}^{\perp}={\left\{0\right\}}^{\perp}\mathrm{i}.\mathrm{e}.,\overline{\mathrm{ran}({A}^{\ast}-\overline{\lambda})}=\mathcal{K},$$Therefore, the residual spectrum of A is empty. From the decomposition of the spectrum, we get$$\sigma \left({A}^{\ast}\right)={\sigma}_{p}\left({A}^{\ast}\right)\cup {\sigma}_{c}\left({A}^{\ast}\right)\subseteq {\sigma}_{a}\left({A}^{\ast}\right).$$
- the last statement follows from Lemma 1 and the assertion (1).

**Lemma**

**5.**

**Proof.**

## 3. Main Theorems

**Definition**

**4.**

**Lemma**

**6.**

- 1.
- $A,B$ satisfy Fuglede-Putnam theorem;
- 2.
- if $AX=XB$ for any operator $X\in \mathcal{B}(\mathcal{H},\mathcal{K})$, then $\overline{\mathrm{ran}\left(X\right)}$ reduces A, ${(kerX)}^{\perp}$ reduces B and ${A|}_{\overline{\mathrm{ran}\left(X\right)}}$, ${B|}_{{(kerX)}^{\perp}}$ are unitarily equivalent normal operators.

**Proposition**

**2.**

**Proof.**

**Theorem**

**1.**

**Proof.**

**Theorem**

**2.**

**Proof.**

**Remark**

**1.**

**Definition**

**5.**

- (i)
- k-quasi-∗-class A if ${T}^{\ast k}|{T}^{2}|{T}^{k}\ge {T}^{\ast k}{\left|{T}^{\ast}\right|}^{2}{T}^{k}$ for non-negative integer k;
- (ii)
- $(n,k)$-quasi-paranormal operator if$$\parallel T\left({T}^{k}x\right){\parallel}^{(1+n})\le \parallel {T}^{(1+n)}\left({T}^{k}x\right)\parallel \parallel {T}^{k}{x\parallel}^{n}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}$$

**Lemma**

**7.**

- (i)
- (k-quasi-∗-class A) ⊂ (k-quasi-∗-paranormal);
- (ii)
- the class $(n,k)$-quasi-∗-paranormal operator is normaloid, for $k=0,1$.

**Proof.**

**Corollary**

**2.**

- 1.
- $\mathcal{A}$ is k-quasi-∗-class A operator with reduced kernel and $\mathcal{B}$ is n-∗-paranormal operator;
- 2.
- $\mathcal{A}$ is n-∗-paranormal operator and $\mathcal{B}$ is k-quasi-∗-class A operator with reduced kernel;
- 3.
- $\mathcal{A},\mathcal{B}\in \mathcal{B}\left(\mathcal{H}\right)$ are k-quasi-∗-class A operators with 0 not in their approximate spectrum.

**Corollary**

**3.**

- (i)
- n-∗-paranormal operator;
- (ii)
- $(n,k)$-quasi-∗-paranormal operator with reduced kernel;
- (iii)
- k-quasi-∗-class A operator with reduced kernel.

## 4. Application

**Definition**

**6.**

**Proposition**

**4.**

**Proof.**

**Lemma**

**8.**

**Remark**

**2.**

**Proposition**

**5.**

- (i)
- ${T}_{0}\in {C}_{00}$;
- (ii)
- ${T}_{1}\in {C}_{10}$;
- (iii)
- ${T}_{1}$ is quasi-affine transform of an isometry.

**Proof.**

**Proposition**

**6.**

- (i)
- $T\in {C}_{0.}$;
- (ii)
- $T\in {C}_{0}$;
- (iii)
- $ind\left(T\right)=0$.

**Proof.**

**Remark**

**3.**

**Proposition**

**7.**

**Proof.**

**Theorem**

**3.**

**Proof.**

## 5. Discussion and Further Studies

**Theorem**

**4.**

- Characterization of $(n,k)$-quasi-∗-paranormal operators with reduced kernel.
- Characterization of completely non-normal $(n,k)$-quasi-∗-paranormal operators. These generalizes the results given by

## Author Contributions

## Acknowledgments

## Conflicts of Interest

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**MDPI and ACS Style**

Bachir, A.; Segres, A.
Asymmetric Putnam-Fuglede Theorem for (*n*,*k*)-Quasi-∗-Paranormal Operators. *Symmetry* **2019**, *11*, 64.
https://doi.org/10.3390/sym11010064

**AMA Style**

Bachir A, Segres A.
Asymmetric Putnam-Fuglede Theorem for (*n*,*k*)-Quasi-∗-Paranormal Operators. *Symmetry*. 2019; 11(1):64.
https://doi.org/10.3390/sym11010064

**Chicago/Turabian Style**

Bachir, Ahmed, and Abdelkader Segres.
2019. "Asymmetric Putnam-Fuglede Theorem for (*n*,*k*)-Quasi-∗-Paranormal Operators" *Symmetry* 11, no. 1: 64.
https://doi.org/10.3390/sym11010064