Fuglede–Putnam type theorems for $(p,k)$-quasihyponormal operators via hyponormal operators

Abstract

For Hilbert space operators S, X, and T, $(S,X,T)\in FP$ means Fuglede–Putnam theorem holds for triplet $(S,X,T)$, that is, $SX=XT$ ensures $S^{\ast }X=XT^{\ast }$. Similarly, $(S,T)\in FP$ means $(S,X,T)\in FP$ holds for each operator X. This paper is devoted to the study of Fuglede–Putnam type theorems for $(p,k)$-quasihyponormal operators via a class of operators based on hyponormal operators $FP(H):=\{S|(S,T)\in FP \mbox{ holds for each hyponormal operator } T^{\ast }\}$. Fuglede–Putnam type theorems involving $(p,k)$-quasihyponormal, dominant, and w-hyponormal operators, which are extensions of the results by Tanahashi, Patel, Uchiyama, et al., are obtained.

1 Introduction

Let $\mathcal{H}$ and $\mathcal{K}$ be complex Hilbert spaces, and let $\mathcal{B}(\mathcal{H})$ and $\mathcal{B}(\mathcal{H},\mathcal{K})$ be the set of all bounded linear operators on $\mathcal{H}$ and the set of all bounded linear operators from $\mathcal{H}$ to $\mathcal{K}$, respectively.

Theorem 1.1

(Fuglede–Putnam theorem, [3, 10])

Let $S\in \mathcal{B}(\mathcal{H})$, $T\in \mathcal{B}(\mathcal{K})$, and $X\in \mathcal{B}(\mathcal{H},\mathcal{K})$. If S, T are normal operators, then $SX=XT$ ensures $S^{\ast }X=XT^{\ast }$.

Theorem 1.2

([14])

Let $S\in \mathcal{B}(\mathcal{H})$, $T\in \mathcal{B}(\mathcal{K})$. The following assertions are equivalent.

• (1) If $SX=XT$, then $S^{\ast}X=XT^{\ast}$.

• (2) If $SX=XT$, then $[R(X)]$ reduces S, where $[R(X)]$ means the closure of range $R(X)$ of X, kerX reduces T, $S|_{[R(X)]}$ and $T|_{(\ker X)^{\bot }}$ are unitarily equivalent normal operators.

For $S\in \mathcal{B}(\mathcal{H})$, $T\in \mathcal{B}(\mathcal{K})$, and $X\in \mathcal{B}(\mathcal{H},\mathcal{K})$, $(S,X,T)\in FP$ means Fuglede–Putnam theorem holds for the triplet $(S,X,T)$, that is, $SX=XT$ ensures $S^{\ast }X=XT^{\ast }$. Similarly, $(S,T)\in FP$ means $(S,X,T)\in FP$ holds for all $X\in \mathcal{B}(\mathcal{H},\mathcal{K})$.

There are various extensions of Fuglede–Putnam theorem for non-normal operators including dominant operators (an operator T is called dominant if, for each complex number z, there exists $M_{z}>0$ such that $(T-z)^{\ast }(T-z)\geq M_{z}^{2}(T-z)(T-z)^{\ast }$), $(p,k)$-quasihyponormal operators (defined by $T^{\ast k}|T|^{2p}T ^{k}\geq T^{\ast k}|T^{\ast }|^{2p}T^{k}$, where $0< p\le 1$ and k is a nonnegative integer, a $(p,0)$-quasihyponormal operator means a p-hyponormal operator), w-hyponormal operators (defined by $(|T^{*}|^{\frac{1}{2}}|T||T^{*}|^{\frac{1}{2}})^{\frac{1}{2}}\ge |T ^{*}|$, the class of w-hyponormal operators coincides with class $A(\frac{1}{2},\frac{1}{2})$), and so on. See [1, 2, 11,12,13, 15, 18].

Among others, Tanahashi, Patel, and Uchiyama [15] proved three kinds of Fuglede–Putnam type theorems with kernel conditions as follows.

(I) Fuglede–Putnam type theorems with restrictions on kerS or $\ker T^{\ast }$.

Theorem 1.3

([15])

Let S be $(p,k)$-quasihyponormal and $T^{\ast }$ be $(p,k)$-quasihyponormal or dominant.

• (1) If $\ker S=\{0\}$ or $\ker T^{\ast }=\{0\}$, then $(S,T)\in FP$.

• (2) If $\ker S\subseteq \ker S^{\ast }$ and $\ker T^{\ast }\subseteq \ker T$, then $(S, T)\in FP$.

It is known that every dominant operator has a reducing kernel, so the condition $\ker T^{\ast }\subseteq \ker T$ in (2) of the above theorem in the case when $T^{\ast }$ is dominant holds.

(II) Fuglede–Putnam type theorems with restrictions on kerX or $\ker X^{\ast }$.

Theorem 1.4

([12, 15])

The following assertions hold.

• (1) Let S be $(p,k)$-quasihyponormal and T be normal. If X has a dense range, then $(S,X,T)\in FP$ and S is normal.

• (2) Let S be p-hyponormal and $T^{\ast }$ be $(p,k)$-quasihyponormal. If $\ker X=\{0\}$, then $(S,X,T)\in FP$ and T is normal.

(III) Fuglede–Putnam type theorems with restrictions on kerS, $\ker S^{\ast }$, and $\ker X^{\ast }$.

Theorem 1.5

([15])

Let S and $T^{\ast }$ be $(p,k)$-quasihyponormal. If $\ker S\subseteq \ker S^{\ast k}$ and $\ker S^{\ast k}\subseteq \ker X^{\ast }$, then $(S,X,T)\in FP$.

In this paper, we shall show extensions of Theorems 1.3–1.5 via the following classes of operators based on hyponormal operators.

$$\begin{aligned}& FP(N):= \bigl\{ S|(S,T)\in FP \mbox{ holds for each normal operator } T ^{\ast }\bigr\} . \\& FP(H):= \bigl\{ S|(S,T)\in FP \mbox{ holds for each hyponormal operator } T^{\ast }\bigr\} . \\& FP(p\mbox{-}H):= \bigl\{ S|(S,T)\in FP \mbox{ holds for each $p$-hyponormal operator } T^{\ast }\bigr\} . \end{aligned}$$

It is clear that $FP(N)\supseteq FP(H)\supseteq FP(p\mbox{-}H)$.

A part of an operator is its restriction to a closed invariant subspace. A class of operators is called hereditary if each part of an operator in the class also belongs to the class.

Remark 1.6

It is well known that the class $FP(p\mbox{-}H)$ includes many classes of operators, such as dominant operators [11,12,13, 18], $(p,k)$-quasihyponormal operators with reducing kernels [15, 17], and w-hyponormal operators with reducing kernels [1]. Moreover, it is known that the classes above also belong to the class of hereditary $FP(H)$ (denote this class by $HFP(H)$), that is, every restriction of an operator to its closed invariant subspace also belongs to the class. See [1, 7, 13, 16, 18].

In Sect. 2, some elementary properties of $FP(H)$ are considered. For example, the reducibility of invariant subspaces of $FP(N)$ operators; the relations between $HFP(H)$ and $HFP(p\mbox{-}H)$; the relations between Fuglede–Putnam type theorems with $\ker S=\{0\}$ or $\ker T^{\ast }=\{0\}$ and Fuglede–Putnam type theorems with reducing kernels. Sections 3–5 are devoted to generalizations of Theorems 1.3–1.5, respectively. Among others, it is proved that Theorem 1.3 holds if $T^{\ast }$ is a w-hyponormal operator, Theorem 1.4 holds if $T^{\ast }$ in Theorem 1.4(1) and S in Theorem 1.4(2) are replaced with a $(p,k)$-quasihyponormal operator, and Theorem 1.5 holds without the restriction $\ker S\subseteq \ker S^{\ast k}$. Lastly, an example is given which says that some kernel conditions in Fuglede–Putnam type theorems are inevitable.

2 Elementary properties of $FP(H)$

By observation, the definitions of $FP(N)$, $FP(H)$, and $FP(p \mbox{-}H)$ are equivalent to the following assertions.

$$\begin{aligned}& FP(N):= \bigl\{ T|\bigl(S,T^{\ast }\bigr)\in FP \mbox{ holds for each normal operator } S\bigr\} , \\& FP(H):= \bigl\{ T|\bigl(S,T^{\ast }\bigr)\in FP \mbox{ holds for each hyponormal operator } S\bigr\} , \\& FP(p\mbox{-}H):= \bigl\{ T|\bigl(S,T^{\ast }\bigr)\in FP \mbox{ holds for each $p$-hyponormal operator } S\bigr\} . \end{aligned}$$

In order to consider the reducibility of invariant subspaces of an operator, four properties are introduced in [20]. Let $\mathcal{M}$ be a nontrivial closed invariant subspace of T and $T|_{\mathcal{M}}$ be the restriction of T on $\mathcal{M}$.

$R_{1}$ :

If the restriction $T|_{\mathcal{M}}$ is normal, then $\mathcal{M}$ reduces T.

$R_{2}$ :

If there exists a positive integer k such that for each $\mathcal{M}\subseteq [R(T^{k})]$, the assertion that $T|_{\mathcal{M}}$ is normal ensures that $\mathcal{M}$ reduces T.

$R_{3}$ :

If $T|_{\mathcal{M}}$ is normal and injective, then $\mathcal{M}$ reduces T.

$R_{4}$ :

If $\lambda \neq 0$, then $\ker (T-\lambda )$ reduces T.

It is obvious that the property $R_{1}$ can be regarded as the case $k=0$ of $R_{2}$. An operator $T\in R_{i}$ means T has the property $R_{i}$, $i=1,2,3,4$. It is known that, for each $i\in \{1,2,3\}$, $T\in R_{i}$ implies $T\in R_{i+1}$ [20, Lemma 2.2]. There exists an operator T such that $T\in R_{3}$ and $T\notin R _{2}$ (Example 5.3(4)).

Lemma 2.1

The following assertions hold.

• (1) If $T\in FP(N)$, then $T\in R_{1}$.

• (2) If T is a $(p,k)$-quasihyponormal or w-hyponormal operator with reducing kernel, then $T\in R_{1}$.

• (3) If T is $(p,k)$-quasihyponormal, then $T\in R_{2}$.

Lemma 2.1 is a generalization of [15, Lemma 2.2].

Proof

(1) Let $\mathcal{M}$ be a nontrivial closed invariant subspace of T, $T=\left(\begin{array}{cc}{T}_{11}& T_{12}\\ 0& T_{22}\end{array}\right)$ on $\mathcal{M}\oplus \mathcal{M}^{\perp }$, $T_{11}=T|_{ \mathcal{M}}$ be normal, and $P=P_{\mathcal{M}}$ be a projection. Since $\left(\begin{array}{cc}{T}_{11}& 0\\ 0& 0\end{array}\right)$ is normal and $TP=\left(\begin{array}{cc}{T}_{11}& 0\\ 0& 0\end{array}\right)=P\left(\begin{array}{cc}{T}_{11}& 0\\ 0& 0\end{array}\right)$, $T\in FP(N)$ implies $S^{\ast }P=\left(\begin{array}{cc}{(T_{11})}^{\ast }& 0\\ {(T_{12})}^{\ast }& 0\end{array}\right)=P\left(\begin{array}{cc}{(T_{11})}^{\ast }& 0\\ 0& 0\end{array}\right)$. Then $T_{12}=0$ and $\mathcal{M}$ reduces T.

(2) The assertion follows by Remark 1.6.

(3) Let $\mathcal{M}\subseteq [R(T^{k})]$, $T=\left(\begin{array}{cc}{T}_{11}& T_{12}\\ 0& T_{22}\end{array}\right)$ on $\mathcal{M}\oplus \mathcal{M}^{\perp }$, $T_{11}=T|_{ \mathcal{M}}$ be normal and $P=P_{\mathcal{M}}$. Then

$$ \begin{aligned} &TT^{\ast }= \begin{pmatrix} T_{11}(T_{11})^{\ast }+T_{12}(T_{12})^{\ast }&T_{12}(T_{22})^{\ast } \\ T_{22}(T_{12})^{\ast }&T_{22}(T_{22})^{\ast } \end{pmatrix} , \\ &P|T|^{2p}P=PP_{[R(T^{k})]}|T|^{2p}P_{[R(T^{k})]}P\ge P \bigl\vert T^{\ast } \bigr\vert ^{2p}P. \end{aligned} $$

(1)

By Hansen’s inequality and Loewner–Heinz’ inequality [5], [4, p.127],

$$\begin{aligned} \begin{pmatrix} (T_{11})^{\ast }T_{11}&0 \\ 0&0 \end{pmatrix} ^{p} &= \bigl(P|T|^{2}P\bigr)^{p} \geq P|T|^{2p}P\geq P \bigl\vert T^{\ast } \bigr\vert ^{2p}P \\ &\geq P\bigl(TPT^{\ast }\bigr)^{p}P=\bigl(TPT^{\ast } \bigr)^{p} = \begin{pmatrix} T_{11}(T_{11})^{\ast })&0 \\ 0&0 \end{pmatrix} ^{p}. \end{aligned}$$

The normality of $T_{11}$ implies ${(T{T}^{\ast })}^p=\left(\begin{array}{cc}{|T_{11}|}^{2p}& A\\ A^{\ast }& B\end{array}\right)$, where A is an operator and B is a positive semidefinite operator.

Let ${(T{T}^{\ast })}^{\frac{p}{2}}=\left(\begin{array}{cc}X& Y\\ Y^{\ast }& Z\end{array}\right)$, again, by Hansen’s inequality and Loewner–Heinz’s inequality,

$$ \begin{pmatrix} |T_{11}|^{2p}&0 \\ 0&0 \end{pmatrix} ^{\frac{1}{2}} =\bigl(P \bigl(TT^{\ast }\bigr)^{p}P\bigr)^{\frac{1}{2}} \geq P \bigl(TT^{\ast }\bigr)^{ \frac{p}{2}}P \geq P\bigl(TPT^{\ast } \bigr)^{\frac{p}{2}}P = \begin{pmatrix} |T_{11}|^{2}&0 \\ 0&0 \end{pmatrix} ^{\frac{p}{2}}. $$

So $X=|T_{11}|^{p}$, ${(T{T}^{\ast })}^p={({(T{T}^{\ast })}^{\frac{p}{2}})}^2=\left(\begin{array}{cc}{|T_{11}|}^{2p}+Y{Y}^{\ast }& \ast \\ \ast \ast & \ast \ast \ast \end{array}\right)$, where ∗ means some elements of the matrix.

Thus $Y=0$, ${(T{T}^{\ast })}^{\frac{p}{2}}=\left(\begin{array}{cc}{|T_{11}|}^p& 0\\ 0& Z\end{array}\right)$, and $T{T}^{\ast }={(T{T}^{\ast })}^{\frac{p}{2}\cdot \frac{2}{p}}=\left(\begin{array}{cc}{|T_{11}|}^2& 0\\ 0& Z^{\frac{2}{p}}\end{array}\right)$. Then $T_{12}=0$ follows by (1). □

Aluthge introduced Aluthge transform $\widetilde{T}=|T|^{1/2}U|T|^{1/2}$ where the polar decomposition of T is $T=U|T|$. For each $s>0$ and $t>0$, $T(s,t)=|T|^{s}U|T|^{t}$ is called generalized Aluthge transform.

Lemma 2.2

([9])

Let $s>0$, $t>0$, $T\in A(s,t)$. If $T(s,t)$ is quasinormal (normal), then T is quasinormal (normal).

Lemma 2.3

([6, 19])

If T is p-hyponormal and $\alpha =\min \{p+s, p+t, s+t\}$, then

$$ \bigl(T(s,t)^{\ast }T(s,t)\bigr)^{\frac{\alpha }{s+t}} \ge \bigl(T(s,t)T(s,t)^{ \ast } \bigr)^{\frac{\alpha }{s+t}}. $$

Lemma 2.4

$HFP(H)=HFP(p\mbox{-}H)$.

Proof

It is sufficient to prove $HFP(H)\subseteq HFP(p\mbox{-}H)$.

Let $T^{\ast }\in HFP(H)$, S be p-hyponormal and $SX=XT$. Decompose S, T, X into

$$ \begin{aligned} &S= \begin{pmatrix} S_{11}&S_{12} \\ 0&S_{22} \end{pmatrix} \in \mathcal{B}\bigl(\bigl[R(X)\bigr]\oplus \ker X^{\ast }\bigr), \\ &T= \begin{pmatrix} T_{11}&0 \\ T_{21}&T_{22} \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X\bigr), \\ &X= \begin{pmatrix} X_{11}&0 \\ 0&0 \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X,\bigl[R(X)\bigr]\oplus \ker X^{ \ast }\bigr). \end{aligned} $$

(2)

Then

$$\begin{aligned}& SX=XT\quad \Longleftrightarrow\quad \begin{pmatrix} S_{11}X_{11}&0 \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ 0&0 \end{pmatrix} \quad \Longleftrightarrow\quad S_{11}X_{11}=X_{11}T_{11}, \end{aligned}$$

(3)

$$\begin{aligned}& \begin{aligned}[b] S^{\ast }X=XT^{\ast }& \quad \Longleftrightarrow\quad \begin{pmatrix} (S_{11})^{\ast }X_{11}&0 \\ (S_{12})^{\ast }X_{11}&0 \end{pmatrix} = \begin{pmatrix} X_{11}(T_{11})^{\ast }&X_{11}(T_{21})^{\ast } \\ 0&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad (S_{11})^{\ast }X_{11}=X_{11}(T_{11})^{\ast },\ (S _{12})^{\ast }X_{11}=0=X_{11}(T_{21})^{\ast }. \end{aligned} \end{aligned}$$

(4)

Since $S, T^{\ast }\in R_{1}$ by Lemma 2.1 and $X_{11}$ is quasiaffine, it is sufficient to prove $(S_{11},T_{11})\in FP$. By the assumption, $S_{11}$ is p-hyponormal and $T_{11}^{\ast }\in FP(H)$.

If $\frac{1}{2}\le p\le 1$, by Lemma 2.3, the Aluthge transform $S_{11}(\frac{1}{2},\frac{1}{2})$ of $S_{11}$ is hyponormal and $(S_{11}(\frac{1}{2},\frac{1}{2}),T_{11})\in FP$. So

$$\begin{aligned} &S_{11}X_{11}=X_{11}T_{11} \\ &\quad \Longrightarrow\quad S_{11}\biggl(\frac{1}{2}, \frac{1}{2} \biggr)|S_{11}|^{\frac{1}{2}}X_{11}=|S_{11}|^{\frac{1}{2}}X_{11}T _{11} \\ &\quad \Longrightarrow \quad S_{11}\biggl(\frac{1}{2}, \frac{1}{2}\biggr)=\biggl(S_{11}\biggl(\frac{1}{2}, \frac{1}{2}\biggr)\biggr)\bigg|_{[R(|S_{11}|^{\frac{1}{2}}X_{11})]}\oplus \biggl(S_{11} \biggl( \frac{1}{2},\frac{1}{2}\biggr)\biggr)\bigg|_{\ker (X_{11}^{\ast }|S_{11}|^{\frac{1}{2}})}, \end{aligned}$$

where $(S_{11}(\frac{1}{2},\frac{1}{2}))|_{[R(|S_{11}|^{\frac{1}{2}}X _{11})]}$ is normal. The assertion “$X_{11}$ is quasiaffine” implies that $[R(|S_{11}|^{\frac{1}{2}}X_{11})]=[R(|S_{11}|^{\frac{1}{2}})]$ and $\ker (X_{11}^{\ast }|S_{11}|^{\frac{1}{2}})=\ker (|S_{11}|)\subseteq \ker S_{11}(\frac{1}{2},\frac{1}{2})$. Then $S_{11}(\frac{1}{2}, \frac{1}{2})=(S_{11}(\frac{1}{2},\frac{1}{2}))|_{[R(|S_{11}|)]}\oplus 0$ is normal, $S_{11}$ is normal by Lemma 2.2, and $(S_{11},T_{11})\in FP$ for $T_{11}^{\ast }\in FP(H)$.

If $0< p\le \frac{1}{2}$, then $S_{11}(\frac{1}{2},\frac{1}{2})$ is $(p+\frac{1}{2})$-hyponormal and $(S_{11}(\frac{1}{2},\frac{1}{2}),T _{11})\in FP$ in the case $\frac{1}{2}\le p\le 1$. Similar to the proof of the case $\frac{1}{2}\le p\le 1$, $S_{11}(\frac{1}{2},\frac{1}{2})=(S _{11}(\frac{1}{2},\frac{1}{2}))|_{[R(|S_{11}|)]}\oplus 0$ is normal, $S_{11}$ is normal and $(S_{11},T_{11})\in FP$ for $T_{11}^{\ast } \in FP(H)$. □

Lemma 2.5

Let $C_{1}$, $C_{2}$ be two classes of operators with heredity. The following assertions (1)–(2) are equivalent to each other, (1) ensures (4) and (3) ensures (4).

• (1) If $S\in C_{1}$ with $\ker S=\{0\}$ and $T^{\ast }\in C_{2}$, then $(S,T)\in FP$.

• (2) If $S\in C_{2}$ and $T^{\ast }\in C_{1}$ with $\ker T^{\ast }=\{0\}$, then $(S,T)\in FP$.

• (3) If $S\in C_{1}$ and $T^{\ast }\in C_{2}$ with $\ker T^{\ast }=\{0\}$, then $(S,T)\in FP$.

• (4) If $S\in C_{1}$ with $\ker S\subseteq \ker S ^{\ast }$ and $T^{\ast }\in C_{2}$ with $\ker T^{\ast }\subseteq \ker T$, then $(S,T)\in FP$.

Proof

Since

$$ SX=XT\quad \Leftrightarrow \quad X^{\ast }S^{\ast }=T^{\ast }X^{\ast } \quad \mbox{and}\quad S^{\ast }X=XT^{\ast }\quad \Leftrightarrow\quad X^{\ast }S=TX^{\ast }, $$

we have

$$ (S,X,T)\in FP\quad \Leftrightarrow\quad \bigl(T^{\ast },X^{\ast },S^{\ast } \bigr)\in FP \quad \mbox{and}\quad (S,T)\in FP\quad \Leftrightarrow\quad \bigl(T^{\ast },S^{\ast }\bigr)\in FP. $$

(5)

By (5), it is sufficient to prove $(1)\Rightarrow (4)$ and $(3)\Rightarrow (4)$.

$(1)\Rightarrow (4)$ Let $\ker S\subseteq \ker S^{\ast }$ and $\ker T^{\ast }\subseteq \ker T$. Decompose S, T, X into

$$ \begin{aligned} &S= \begin{pmatrix} S_{11}&0 \\ 0&0 \end{pmatrix} \in \mathcal{B}\bigl(\bigl[R\bigl(S^{\ast } \bigr)\bigr]\oplus \ker S\bigr), \\ &T= \begin{pmatrix} T_{11}&0 \\ 0&0 \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R(T)\bigr]\oplus \ker T^{\ast }\bigr), \\ &X= \begin{pmatrix} X_{11}&X_{12} \\ X_{21}&X_{22} \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R(T)\bigr]\oplus \ker T^{\ast }, \bigl[R\bigl(S^{\ast } \bigr)\bigr]\oplus \ker S\bigr). \end{aligned} $$

(6)

Then $\ker S_{11}=\{0\}=\ker T_{11}^{\ast }$,

$$\begin{aligned}& \begin{aligned}[b] SX=XT&\quad \Longleftrightarrow \quad \begin{pmatrix} S_{11}X_{11}&S_{11}X_{12} \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ X_{21}T_{11}&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad S_{11}X_{11}=X_{11}T_{11},\ S_{11}X_{12}=0=X_{21}T _{11}, \end{aligned} \end{aligned}$$

(7)

$$\begin{aligned}& \begin{aligned}[b] S^{\ast }X=XT^{\ast }& \quad \Longleftrightarrow\quad \begin{pmatrix} (S_{11})^{\ast }X_{11}&(S_{11})^{\ast }X_{12} \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}(T_{11})^{\ast }&0 \\ X_{21}(T_{11})^{\ast }&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad (S_{11})^{\ast }X_{11}=X_{11}(T_{11})^{\ast },\ (S _{11})^{\ast }X_{12}=0=X_{21}(T_{11})^{\ast }. \end{aligned} \end{aligned}$$

(8)

By heredity and (1), $(S_{11},T_{11})\in FP$. Since $\ker S_{11}=\{0 \}=\ker T_{11}^{\ast }$, the assertion $S_{11}X_{12}=0=X_{21}T_{11}$ implies $X_{12}=0=X_{21}$. So that $(S,T)\in FP$.

$(3)\Rightarrow (4)$ The assertion holds in a similar manner to $(1)\Rightarrow (4)$. □

Lemma 2.6

Let C be a class of operators with heredity. The following assertion (1) ensures (2).

• (1) If $S\in C$ with $\ker S=\{0\}$ and $T^{\ast } \in FP(N)$, then $(S,T)\in FP$.

• (2) If $S\in C$ with $\ker S\subseteq \ker S^{ \ast }$ and $T^{\ast }\in FP(N)$, then $(S,T)\in FP$.

Proof

The proof is similar to the proof of [8, Theorem 7]. Let $\ker S\subseteq \ker S^{\ast }$. Decompose S, X into $S=S_{n} \oplus S_{p}$ on $\mathcal{H}=\mathcal{H}_{1}\oplus \mathcal{H}_{2}$ where $S_{n}$ and $S_{p}$ are normal part and pure part of S, respectively,

$$ X= \begin{pmatrix} X_{11} \\ X_{21} \end{pmatrix} : \mathcal{K} \longrightarrow \mathcal{H}_{1}\oplus \mathcal{H}_{2}. $$

Then $\ker S_{p}=\{0\}$,

$$\begin{aligned}& SX=XT\quad \Longleftrightarrow\quad \begin{pmatrix} S_{n}X_{11} \\ S_{p}X_{21} \end{pmatrix} = \begin{pmatrix} X_{11}T \\ X_{21}T \end{pmatrix} , \\& S^{\ast }X=XT^{\ast }\quad \Longleftrightarrow\quad \begin{pmatrix} S_{n}^{\ast }X_{11} \\ S_{p}^{\ast }X_{21} \end{pmatrix} = \begin{pmatrix} X_{11}T^{\ast } \\ X_{21}T^{\ast } \end{pmatrix} . \end{aligned}$$

Since $T^{\ast }\in FP(N)$, $(S_{n},T)\in FP$ follows. By $\ker S_{p}= \{0\}$ and (1), $(S_{p},T)\in FP$ and $(S,T)\in FP$. □

3 Extensions of Theorem 1.3

Theorem 3.1

Let S be $(p,k)$-quasihyponormal and $T^{\ast }$ be $(p,k)$-quasihyponormal, or dominant, or w-hyponormal.

• (1) If $\ker T^{\ast }=\{0\}$, then $(S,T)\in FP$.

• (2) If $T^{\ast }$ is $(p,k)$-quasihyponormal, $\ker S\subseteq \ker S^{\ast }$, and $\ker T^{\ast }\subseteq \ker T$, then $(S, T)\in FP$.

• (3) If $T^{\ast }$ is dominant and $\ker S\subseteq \ker S^{\ast }$, then $(S, T)\in FP$.

• (4) If $T^{\ast }$ is w-hyponormal, $\ker S\subseteq \ker S^{\ast }$, and $\ker T^{\ast }\subseteq \ker T$, then $(S, T)\in FP$.

Tanahashi et al. [15, Theorems 2.5, 2.7, 2.10–2.12] proved the case “$T^{\ast }$ is $(p,k)$-quasihyponormal or dominant” of Theorem 3.1. Here we prove Theorem 3.1 by using the class $HFP(H)$ (Remark 1.6). Theorem 3.1 means that Theorem 1.3 holds if $T^{\ast }$ is a w-hyponormal operator.

Lemma 3.2

([7, 16])

Let T be $(p,k)$-quasihyponormal.

• (1) If $T^{k}\mathcal{H}$ is not dense and $T=\left(\begin{array}{cc}{T}_{11}& T_{12}\\ 0& T_{22}\end{array}\right)$ on $[T^{k}\mathcal{H}]\oplus \ker T^{\ast k}$, then $T_{11}$ is p-hyponormal, $T_{22}^{k}=0$, and $\sigma (T)=\sigma (T_{11})\cup \{0\}$.

• (2) Each restriction $T|_{\mathcal{M}}$ of T to its invariant subspace $\mathcal{M}$ is also $(p,k)$-quasihyponormal.

Lemma 3.3

Let S be $(p,k)$-quasihyponormal and $T^{\ast }\in HFP(H)$.

• (1) If $\ker T^{\ast }=\{0\}$, then $(S,T)\in FP$.

• (2) If $\ker S\subseteq \ker S^{\ast }$, then $(S,T)\in FP$.

Proof

By Lemma 2.1(1), every $FP(N)$ operator has a reducing kernel. Thus, by Lemma 2.5, we only need to prove (1). Let $SX=XT$. As in the proof of Lemma 2.4, (2)–(4) hold. By $\ker T^{\ast }=\{0\}$ and Lemma 3.2(2),

$$ S_{11}X_{11}=X_{11}T_{11}\quad \Longrightarrow\quad \ker S_{11}^{\ast }=\{0\},\quad S _{11} \mbox{ is } (p,k)\mbox{-quasihyponormal}, $$

thus $S_{11}$ is p-hyponormal follows by Lemma 3.2(1). Hence $(S_{11},T_{11})\in FP$ by $T_{11}^{\ast }\in HFP(H)$ and Lemma 2.4. So $S_{11}$ is normal and injective. Lemma 2.1(3) ensures $S_{12}=0$. Since $X_{11}$ is quasiaffine, by Theorem 1.2 and Lemma 2.1(1), $T_{11}$ is normal and $T_{21}=0$ hold. So that the assertion holds by (4). □

Proof of Theorem 3.1

(1) If $\ker T^{\ast }=\{0\}$, then $T^{\ast }\in HFP(H)$ by Remark 1.6, and the assertion follows by Lemma 3.3(1).

(2) If $T^{\ast }$ is $(p,k)$-quasihyponormal and $\ker T^{\ast } \subseteq \ker T$, then $T^{\ast }\in HFP(H)$ (Remark 1.6). So $\ker S\subseteq \ker S^{\ast }$ and Lemma 3.3(2) ensure $(S, T)\in FP$.

(3)–(4) hold in a similar manner to (2). □

4 Extensions of Theorem 1.4

Theorem 4.1

The following assertions hold and they are equivalent to each other.

• (1) Let S be $(p,k)$-quasihyponormal, let $T^{\ast }$ be $(p,k)$-quasihyponormal with reducing kernel, or dominant, or w-hyponormal with reducing kernel. If $\ker X^{\ast }= \{0\}$, then $(S,X,T)\in FP$ and S is normal.

• (2) Let S be $(p,k)$-quasihyponormal with reducing kernel, or dominant, or w-hyponormal with reducing kernel, let $T^{\ast }$ be $(p,k)$-quasihyponormal. If $\ker X=\{0\}$, then $(S,X,T)\in FP$ and T is normal.

Theorem 4.1 implies that the normal operator $T^{\ast }$ in Theorem 1.4(1) can be replaced with a $(p,k)$-quasihyponormal operator with reducing kernel, or a dominant operator, or a w-hyponormal operator with reducing kernel; and the p-hyponormal operator S in Theorem 1.4(2) can be replaced with a $(p,k)$-quasihyponormal operator with reducing kernel, or a dominant operator, or a w-hyponormal operator with reducing kernel.

Lemma 4.2

The following assertions hold and (1) is equivalent to (2).

• (1) If S is $(p,k)$-quasihyponormal, $T^{\ast } \in HFP(H)$, and $\ker X^{\ast }=\{0\}$, then $(S,X,T)\in FP$ and S is normal.

• (2) If $S\in HFP(H)$, $T^{\ast }$ is $(p,k)$-quasihyponormal and X is injective, then $(S,X,T)\in FP$ and T is normal.

Proof

According to (5), it is sufficient to prove (1). Decompose S, T, X into

$$ \begin{aligned} &S= \begin{pmatrix} S_{11}&S_{12} \\ 0&S_{22} \end{pmatrix} \in \mathcal{B}\bigl([S\mathcal{H}]\oplus \ker S^{\ast }\bigr), \\ &T= \begin{pmatrix} T_{11}&0 \\ 0&0 \end{pmatrix} \in \mathcal{B} \bigl([T\mathcal{K}]\oplus \ker T^{\ast }\bigr), \\ &X= \begin{pmatrix} X_{11}&X_{12} \\ X_{21}&X_{22} \end{pmatrix} \in \mathcal{B} \bigl([T\mathcal{K}]\oplus \ker T^{\ast },[S\mathcal{H}] \oplus \ker S^{\ast }\bigr). \end{aligned} $$

(9)

Since X has a dense range,

$$\begin{aligned} SX=XT&\quad \Longrightarrow\quad [X T\mathcal{K}]=[SX \mathcal{K}]=[S\mathcal{H}] \\ &\quad \Longrightarrow\quad X_{21}=0,\ \ker X_{11}^{\ast }=\ker X_{22}^{\ast }=\{0 \}. \end{aligned}$$

(10)

Then

$$\begin{aligned}& \begin{aligned}[b] SX=XT&\quad \Longleftrightarrow \quad \begin{pmatrix} S_{11}X_{11}&S_{11}X_{12}+S_{12}X_{22} \\ 0&S_{22}X_{22} \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ 0&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad S_{11}X_{11}=X_{11}T_{11},\ S_{11}X_{12}+S_{12}X _{22}=S_{22}X_{22}=0, \end{aligned} \end{aligned}$$

(11)

$$\begin{aligned}& \begin{aligned}[b] S^{\ast }X=XT^{\ast } & \quad \Longleftrightarrow\quad \begin{pmatrix} (S_{11})^{\ast }X_{11}&(S_{11})^{\ast }X_{12} \\ (S_{12})^{\ast }X_{11}&(S_{12})^{\ast }X_{12}+(S_{22})^{\ast }X_{22} \end{pmatrix} = \begin{pmatrix} X_{11}(T_{11})^{\ast }&0 \\ 0&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad (S_{11})^{\ast }X_{11}=X_{11}(T_{11})^{\ast }, \\ &\hphantom{\quad \Longleftrightarrow\quad\ \ } (S _{11})^{\ast }X_{12}=(S_{12})^{\ast }X_{11}=(S_{12})^{\ast }X_{12}+(S _{22})^{\ast }X_{22}=0. \end{aligned} \end{aligned}$$

(12)

Since $S_{11}$ is $(p,k)$-quasihyponormal, $T_{11}^{\ast }\in HFP(H)$ and $\ker T_{11}^{\ast }=\{0\}$, $(S_{11},T_{11})\in FP$ by Lemma 3.3. So $S_{11}=S_{11}|_{[R(X_{11})]}$ is normal, $\ker S _{11}^{\ast }=\{0\}$ follows by $S_{11}X_{11}=X_{11}T_{11}$ and $\ker X_{11}^{\ast }=\ker T_{11}^{\ast }=\{0\}$.

Then $S_{12}=0$ holds by Lemma 2.1(3). Equation (11) and $\ker S_{11}=\ker X_{22}^{\ast }=\{0\}$ imply $X_{12}=S_{22}=0$. The assertion holds by (12). □

According to Remark 1.6, Theorem 4.1 follows by Lemma 4.2 directly.

5 Extensions of Theorem 1.5

Theorem 5.1

The following assertions hold and they are equivalent to each other.

• (1) Let S be $(p,k)$-quasihyponormal, let $T^{\ast }$ be $(p,k)$-quasihyponormal, or dominant, or w-hyponormal with reducing kernel. If $\ker S^{\ast k}\subseteq \ker X^{\ast }$, then $(S,X,T)\in FP$ and S is normal.

• (2) Let S be $(p,k)$-quasihyponormal with reducing kernel, or dominant, or w-hyponormal with reducing kernel, let $T^{\ast }$ be $(p,k)$-quasihyponormal. If $\ker T^{k}\subseteq \ker X$, then $(S,X,T)\in FP$ and T is normal.

Theorem 5.1(1) holds for every $(p,k)$-quasihyponormal operator $T^{\ast }$ and implies that the restriction $\ker S\subseteq \ker S ^{\ast k}$ in Theorem 1.5 is redundant.

Lemma 5.2

The following assertions hold and they are equivalent to each other.

• (1) If S is $(p,k)$-quasihyponormal, $T^{\ast } \in HFP(H)$ and $\ker S^{\ast k}\subseteq \ker X^{\ast }$, then $(S,X,T)\in FP$.

• (2) If $S\in HFP(H)$, $T^{\ast }$ is $(p,k)$-quasihyponormal and $\ker T^{k}\subseteq \ker X$, then $(S,X,T)\in FP$.

Proof

By (5), it is sufficient to prove (1). Decompose S, T, X into

$$ \begin{aligned} &S= \begin{pmatrix} S_{11}&S_{12} \\ 0&S_{22} \end{pmatrix} \in \mathcal{B}\bigl(\bigl[S^{k} \mathcal{H}\bigr]\oplus \ker S^{\ast k}\bigr), \\ &T= \begin{pmatrix} T_{11}&0 \\ T_{21}&T_{22} \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X\bigr), \\ &X= \begin{pmatrix} X_{11}&0 \\ X_{21}&0 \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X,\bigl[S^{k} \mathcal{H}\bigr]\oplus \ker S^{\ast k}\bigr). \end{aligned} $$

(13)

The condition $\ker S^{\ast k}\subseteq \ker X^{\ast }$ implies that $R(X)\subseteq [S^{k}\mathcal{H}]$, $X_{21}=0$ and $\ker X_{11}=\{0\}$. Thus

$$\begin{aligned}& SX=XT\quad \Longleftrightarrow\quad \begin{pmatrix} S_{11}X_{11}&0 \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ 0&0 \end{pmatrix} \quad \Longleftrightarrow \quad S_{11}X_{11}=X_{11}T_{11}, \end{aligned}$$

(14)

$$\begin{aligned}& S^{\ast }X=XT^{\ast }\quad \Longleftrightarrow\quad (S_{11})^{\ast }X_{11}=X_{11}(T _{11})^{\ast },\ (S_{12})^{\ast }X_{11}=X_{11}(T_{21})^{\ast }=0. \end{aligned}$$

(15)

The operator $S_{11}$ is p-hyponormal follows by Lemma 3.2. Since each p-hyponormal operator has a reducing kernel and $T_{11}^{\ast }\in HFP(H)$, $(S_{11},T_{11}) \in FP$ follows. Hence $S_{11}|_{[R(X_{11})]}$ ($=S|_{[R(X)]}$) and $T_{11}|_{[R(( X_{11})^{\ast })]}(=T_{11})$ are unitarily equivalent normal operators. So $T_{21}=0$ holds by Lemma 2.1(1), $S_{12}=0$ by $[R(X)]\subseteq [S^{k}\mathcal{H}]$, and (3) of Lemma 2.1. Therefore the assertion holds by (15). □

Proof of Theorem 5.1

It is sufficient to prove (1). If $T^{\ast }$ is dominant or w-hyponormal with reducing kernel, the assertion is a direct result of Lemma 5.2.

If $T^{\ast }$ is $(p,k)$-quasihyponormal, as in the proof of Lemma 5.2, (13)–(15) hold. Since $S_{11}$ is p-hyponormal and $\ker X_{11}=\{0\}$, $(S_{11},X_{11},T _{11})\in FP$ holds by Lemma 4.2(2). Then $S_{11}|_{[R(X _{11})]}(=S|_{[R(X)]})$ and $T_{11}$ are normal operators, and $S_{12}=0$ follows by $[R(X)]\subseteq [S^{k}\mathcal{H}]$ and Lemma 2.1(3).

Furthermore, let $P=P_{[R(S^{k})]}$ and $x\in \ker S_{11}$, then $P(S^{\ast }S)^{p}P\ge P(SS^{\ast })^{p}P$ and $S^{\ast }Sx=0=(S^{ \ast }S)^{p}x$. Hence $0=\langle (S^{\ast }S)^{p}Px,Px\rangle \ge \langle (SS^{\ast })^{p}Px,Px\rangle =\|(SS^{\ast })^{\frac{p}{2}}x\| ^{2}$, $x\in \ker (SS^{\ast })^{\frac{p}{2}}\cap [R(S^{k})]=\ker S ^{\ast }\cap [R(S^{k})]\subseteq \ker S^{\ast k}\cap [R(S^{k})]=\{0\}$. Therefore $\ker S_{11}=\{0\}$. Thus, by Lemma 2.1,

$$ S_{11}X_{11}=X_{11}T_{11}\quad \Longrightarrow \quad \ker T_{11}=\{0\}\quad \Longrightarrow\quad T_{21}=0. $$

So $(S,X,T)\in FP$ follows. □

At the end, we give an example which implies that some kernel conditions in Fuglede–Putnam type theorems above are crucial.

Example 5.3

Let $k\ge 2$ be a positive integer, S be an operator such that $S^{k-1}\neq 0$ and $S^{k}= 0$.

• (1) S and $S^{\ast }$ are $(p,k)$-quasihyponormal with $\ker S\neq 0$ and $\ker S^{\ast }\neq 0$, and $(S,S,S)\notin FP$.

• (2) Let $P=P_{[R(S^{k-1})]}$, then $\ker S\nsubseteq \ker S^{\ast }$ and $(S,P,1-P)\notin FP$.

• (3) Let $P=P_{[R(S^{k-1})]}$, then $\ker P^{\ast } \neq 0$ and $(S,P,1-P)\notin FP$.

• (4) If $k=2$, then S is a quasiclass A operator, $S\in R_{3}$ and $S\notin R_{2}$.

Example 5.3(1)–(2) says that, if $T^{\ast }$ is $(p,k)$-quasihyponormal, the kernel condition $\ker T^{\ast }=\{0\}$ in Theorem 3.1(1) is inevitable. Example 5.3(3) implies that the kernel condition $\ker X^{\ast }=\{0\}$ in Theorem 4.1(1) is crucial.

Lemma 5.4

([20])

If $\ker (T-\lambda )\subseteq \ker (T-\lambda )^{\ast }$ for a fixed number λ, then $\ker (T-\lambda )=\ker (T-\lambda )^{2}$ and $\ker (T-\lambda )\perp \ker (T-\mu )$ for each $\mu \neq \lambda$.

Lemma 5.5

([20])

Let k be a positive integer, $T\in k$-$QA(n)$, and $T=\left(\begin{array}{cc}{T}_{11}& T_{12}\\ 0& T_{22}\end{array}\right)$ on $\mathcal{M}\oplus \mathcal{M}^{\perp }$.

• (1) If $[R(T^{k})]\subseteq \mathcal{M}$, then $T_{22}^{k}=0$ and $\sigma (T)=\sigma (T_{11})\cup \{0\}$.

• (2) If $T\in k$-$QA(n)$ and $\mathcal{M}\subseteq [R(T ^{k})]$, then $T_{11}(=T|_{\mathcal{M}})\in A(n)$.

Proof of Example 5.3

(1) By $S^{k}= 0=S^{\ast k}$, S and $S^{\ast }$ are $(p,k)$-quasihyponormal. If $\ker S= 0$, then $\ker S^{k}=\ker S= 0$ and it contradicts the condition $\ker S^{k-1}\neq 0$. So $\ker S\neq 0$, and $\ker S^{\ast }\neq 0$ holds in a similar manner.

If $\ker S\subseteq \ker S^{\ast }$, then Lemma 5.4 implies $\ker S^{k}=\ker S$. It also contradicts the condition $S^{k-1}\neq 0$. Hence $\ker S\nsubseteq \ker S^{\ast }$, $S^{\ast }S\neq SS^{\ast }$, and $(S,S,S)\notin FP$.

(2) The assumption $S^{k}= 0$ implies $SP=0=P(1-P)$. By (1), kerS does not reduce S. So $S^{\ast }P\neq 0=P(1-P)$ and $(S,P,1-P)\notin FP$.

(3) If $\ker P^{\ast }=\ker S^{\ast (k-1)}= 0$, then $\ker S^{\ast }= \ker S^{\ast k}=0$. It contradicts the condition $S^{k}= 0$. Hence $\ker P^{\ast }\neq 0$ and $(S,P,1-P)\notin FP$.

(4) Since

$$ S^{2}= 0\quad \Longrightarrow \quad R(S)\subseteq \ker S\subseteq \ker S^{2}= \ker \bigl\vert S^{2} \bigr\vert \quad \Longrightarrow\quad S^{\ast } \bigl\vert S^{2} \bigr\vert S=0=S^{\ast }|S|^{2}S, $$

S is a quasiclass A operator. By Lemma 5.5 and $S^{2}= 0$, $S=\left(\begin{array}{cc}0& S_{12}\\ 0& 0\end{array}\right)$ on $\mathcal{H}=[S\mathcal{H}]\oplus \ker S^{\ast }$. The assumption $S\neq 0$ ensures $S_{12}\neq 0$, so $S|_{[S\mathcal{H}]}=0$ is normal and $[S\mathcal{H}]$ does not reduce S. Hence $S\in R_{3}$ [20, Theorem 2.4] and $S\notin R_{2}$. □