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
Anzeige
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 \(X\in \mathcal{B}(\mathcal{H},\mathcal{K})\).
Anzeige
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, 11‐13, 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
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 [11‐13, 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.
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 \(\mathcal{M}\).
$$\begin{aligned}& FP(N):= \bigl\{ T|\bigl(S,T^{\ast }\bigr)\in FP \mbox{ holds for each normal operator } S\bigr\} , \\& FP(H):= \bigl\{ T|\bigl(S,T^{\ast }\bigr)\in FP \mbox{ holds for each hyponormal operator } S\bigr\} , \\& FP(p\mbox{-}H):= \bigl\{ T|\bigl(S,T^{\ast }\bigr)\in FP \mbox{ holds for each $p$-hyponormal operator } S\bigr\} . \end{aligned}$$
\(R_{1}\)
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
\(T\in R_{2}\).
Proof
(1) Let \(\mathcal{M}\) be a nontrivial closed invariant subspace of T, on \(\mathcal{M}\oplus \mathcal{M}^{\perp }\), \(T_{11}=T|_{ \mathcal{M}}\) be normal, and \(P=P_{\mathcal{M}}\) be a projection. Since is normal and , \(T\in FP(N)\) implies . Then \(T_{12}=0\) and \(\mathcal{M}\) reduces T.
(2) The assertion follows by Remark 1.6.
(3) Let \(\mathcal{M}\subseteq [R(T^{k})]\), on \(\mathcal{M}\oplus \mathcal{M}^{\perp }\), \(T_{11}=T|_{ \mathcal{M}}\) be normal and \(P=P_{\mathcal{M}}\). Then
By Hansen’s inequality and Loewner–Heinz’ inequality [5], [4, p.127],
The normality of \(T_{11}\) implies , where A is an operator and B is a positive semidefinite operator.
$$ \begin{aligned} &TT^{\ast }= \begin{pmatrix} T_{11}(T_{11})^{\ast }+T_{12}(T_{12})^{\ast }&T_{12}(T_{22})^{\ast } \\ T_{22}(T_{12})^{\ast }&T_{22}(T_{22})^{\ast } \end{pmatrix} , \\ &P|T|^{2p}P=PP_{[R(T^{k})]}|T|^{2p}P_{[R(T^{k})]}P\ge P \bigl\vert T^{\ast } \bigr\vert ^{2p}P. \end{aligned} $$
(1)
$$\begin{aligned} \begin{pmatrix} (T_{11})^{\ast }T_{11}&0 \\ 0&0 \end{pmatrix} ^{p} &= \bigl(P|T|^{2}P\bigr)^{p} \geq P|T|^{2p}P\geq P \bigl\vert T^{\ast } \bigr\vert ^{2p}P \\ &\geq P\bigl(TPT^{\ast }\bigr)^{p}P=\bigl(TPT^{\ast } \bigr)^{p} = \begin{pmatrix} T_{11}(T_{11})^{\ast })&0 \\ 0&0 \end{pmatrix} ^{p}. \end{aligned}$$
Let , again, by Hansen’s inequality and Loewner–Heinz’s inequality,
So \(X=|T_{11}|^{p}\), , where ∗ means some elements of the matrix.
$$ \begin{pmatrix} |T_{11}|^{2p}&0 \\ 0&0 \end{pmatrix} ^{\frac{1}{2}} =\bigl(P \bigl(TT^{\ast }\bigr)^{p}P\bigr)^{\frac{1}{2}} \geq P \bigl(TT^{\ast }\bigr)^{ \frac{p}{2}}P \geq P\bigl(TPT^{\ast } \bigr)^{\frac{p}{2}}P = \begin{pmatrix} |T_{11}|^{2}&0 \\ 0&0 \end{pmatrix} ^{\frac{p}{2}}. $$
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
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
Then
$$ \begin{aligned} &S= \begin{pmatrix} S_{11}&S_{12} \\ 0&S_{22} \end{pmatrix} \in \mathcal{B}\bigl(\bigl[R(X)\bigr]\oplus \ker X^{\ast }\bigr), \\ &T= \begin{pmatrix} T_{11}&0 \\ T_{21}&T_{22} \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X\bigr), \\ &X= \begin{pmatrix} X_{11}&0 \\ 0&0 \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X,\bigl[R(X)\bigr]\oplus \ker X^{ \ast }\bigr). \end{aligned} $$
(2)
$$\begin{aligned}& SX=XT\quad \Longleftrightarrow\quad \begin{pmatrix} S_{11}X_{11}&0 \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ 0&0 \end{pmatrix} \quad \Longleftrightarrow\quad S_{11}X_{11}=X_{11}T_{11}, \end{aligned}$$
(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
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 \(T_{11}^{\ast }\in FP(H)\).
$$\begin{aligned} &S_{11}X_{11}=X_{11}T_{11} \\ &\quad \Longrightarrow\quad S_{11}\biggl(\frac{1}{2}, \frac{1}{2} \biggr)|S_{11}|^{\frac{1}{2}}X_{11}=|S_{11}|^{\frac{1}{2}}X_{11}T _{11} \\ &\quad \Longrightarrow \quad S_{11}\biggl(\frac{1}{2}, \frac{1}{2}\biggr)=\biggl(S_{11}\biggl(\frac{1}{2}, \frac{1}{2}\biggr)\biggr)\bigg|_{[R(|S_{11}|^{\frac{1}{2}}X_{11})]}\oplus \biggl(S_{11} \biggl( \frac{1}{2},\frac{1}{2}\biggr)\biggr)\bigg|_{\ker (X_{11}^{\ast }|S_{11}|^{\frac{1}{2}})}, \end{aligned}$$
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
we have
$$ SX=XT\quad \Leftrightarrow \quad X^{\ast }S^{\ast }=T^{\ast }X^{\ast } \quad \mbox{and}\quad S^{\ast }X=XT^{\ast }\quad \Leftrightarrow\quad X^{\ast }S=TX^{\ast }, $$
$$ (S,X,T)\in FP\quad \Leftrightarrow\quad \bigl(T^{\ast },X^{\ast },S^{\ast } \bigr)\in FP \quad \mbox{and}\quad (S,T)\in FP\quad \Leftrightarrow\quad \bigl(T^{\ast },S^{\ast }\bigr)\in FP. $$
(5)
\((1)\Rightarrow (4)\) 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 \(\ker S_{11}=\{0\}=\ker T_{11}^{\ast }\),
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 \((S,T)\in FP\).
$$\begin{aligned}& \begin{aligned}[b] SX=XT&\quad \Longleftrightarrow \quad \begin{pmatrix} S_{11}X_{11}&S_{11}X_{12} \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ X_{21}T_{11}&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad S_{11}X_{11}=X_{11}T_{11},\ S_{11}X_{12}=0=X_{21}T _{11}, \end{aligned} \end{aligned}$$
(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)
\((3)\Rightarrow (4)\) 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 \(\ker S_{p}=\{0\}\),
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 \((S,T)\in FP\). □
$$\begin{aligned}& SX=XT\quad \Longleftrightarrow\quad \begin{pmatrix} S_{n}X_{11} \\ S_{p}X_{21} \end{pmatrix} = \begin{pmatrix} X_{11}T \\ X_{21}T \end{pmatrix} , \\& S^{\ast }X=XT^{\ast }\quad \Longleftrightarrow\quad \begin{pmatrix} S_{n}^{\ast }X_{11} \\ S_{p}^{\ast }X_{21} \end{pmatrix} = \begin{pmatrix} X_{11}T^{\ast } \\ X_{21}T^{\ast } \end{pmatrix} . \end{aligned}$$
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
Let
T
be
\((p,k)\)-quasihyponormal.
(1)
If
\(T^{k}\mathcal{H}\)
is not dense and
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),
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). □
$$ S_{11}X_{11}=X_{11}T_{11}\quad \Longrightarrow\quad \ker S_{11}^{\ast }=\{0\},\quad S _{11} \mbox{ is } (p,k)\mbox{-quasihyponormal}, $$
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,
Then
$$\begin{aligned} SX=XT&\quad \Longrightarrow\quad [X T\mathcal{K}]=[SX \mathcal{K}]=[S\mathcal{H}] \\ &\quad \Longrightarrow\quad X_{21}=0,\ \ker X_{11}^{\ast }=\ker X_{22}^{\ast }=\{0 \}. \end{aligned}$$
(10)
$$\begin{aligned}& \begin{aligned}[b] SX=XT&\quad \Longleftrightarrow \quad \begin{pmatrix} S_{11}X_{11}&S_{11}X_{12}+S_{12}X_{22} \\ 0&S_{22}X_{22} \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ 0&0 \end{pmatrix} \\ &\quad \Longleftrightarrow\quad S_{11}X_{11}=X_{11}T_{11},\ S_{11}X_{12}+S_{12}X _{22}=S_{22}X_{22}=0, \end{aligned} \end{aligned}$$
(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 \(\ker X_{11}^{\ast }=\ker T_{11}^{\ast }=\{0\}\).
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
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
$$ \begin{aligned} &S= \begin{pmatrix} S_{11}&S_{12} \\ 0&S_{22} \end{pmatrix} \in \mathcal{B}\bigl(\bigl[S^{k} \mathcal{H}\bigr]\oplus \ker S^{\ast k}\bigr), \\ &T= \begin{pmatrix} T_{11}&0 \\ T_{21}&T_{22} \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X\bigr), \\ &X= \begin{pmatrix} X_{11}&0 \\ X_{21}&0 \end{pmatrix} \in \mathcal{B} \bigl(\bigl[R\bigl(X^{\ast }\bigr)\bigr]\oplus \ker X,\bigl[S^{k} \mathcal{H}\bigr]\oplus \ker S^{\ast k}\bigr). \end{aligned} $$
(13)
$$\begin{aligned}& SX=XT\quad \Longleftrightarrow\quad \begin{pmatrix} S_{11}X_{11}&0 \\ 0&0 \end{pmatrix} = \begin{pmatrix} X_{11}T_{11}&0 \\ 0&0 \end{pmatrix} \quad \Longleftrightarrow \quad S_{11}X_{11}=X_{11}T_{11}, \end{aligned}$$
(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,
So \((S,X,T)\in FP\) follows. □
$$ S_{11}X_{11}=X_{11}T_{11}\quad \Longrightarrow \quad \ker T_{11}=\{0\}\quad \Longrightarrow\quad T_{21}=0. $$
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
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 is a quasiclass A operator. By Lemma 5.5 and \(S^{2}= 0\), 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 \(S\notin R_{2}\). □
$$ S^{2}= 0\quad \Longrightarrow \quad R(S)\subseteq \ker S\subseteq \ker S^{2}= \ker \bigl\vert S^{2} \bigr\vert \quad \Longrightarrow\quad S^{\ast } \bigl\vert S^{2} \bigr\vert S=0=S^{\ast }|S|^{2}S, $$
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