A Fuglede-Putnam property for N-class A(k) operators

1.   Introduction

For complex Hilbert spaces $\mathcal{H}$ and $\mathcal{K}, \mathcal{B}(\mathcal{H}), \mathcal{B}(\mathcal{K})$ and $\mathcal{B}(\mathcal{H}, \mathcal{K})$ denote the set of all bounded linear operator on $\mathcal{H}$, on $\mathcal{K}$ and from $\mathcal{H}$ to $\mathcal{K}$ respectively. A bounded operator $A\in \mathcal{B}(\mathcal{H})$ is called normal if $A^{*}A = AA^{*}$. An operator $A\in \mathcal{B}(\mathcal{H})$ is said to be a class $\mathcal{Y} _{\kappa}$ for $\kappa\leq 1$ if there exists a positive number $k_{\kappa}$ such that

It is known that $\mathcal{Y}_\kappa \subset \mathcal{Y}_\eta$ if $1\leq \kappa \leq \eta$. Let $\mathcal{Y} = \cup_{1\leq \kappa}\mathcal{Y}_\kappa$ ^[2].

The familiar Putnam-Fuglede's theorem asserts that if $A\in \mathcal{B}(\mathcal{H})$ and $B \in \mathcal{B}(\mathcal{K})$ are normal operators and $AX = XB$ for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $A^{*}X = XB^{*}$ (see ^[7]). Many authors have extended this theorem for several classes operators, recently S. Mecheri et al ^[6] proved that the Fuglede-Putnam theorem holds for $p$-hyponormal or class $\mathcal{Y}$, A. Bachir et al ^[1] proved that the theorem holds for $w$-hyponormal or class $\mathcal{Y}$ operators. We say that the pair $(A, B)$ satisfy Fuglede-Putnam theorem if $AX = XB$ implies $A^{*}X = XB^{*}$ for any $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$.

2.   Materials and method

Definition 1. An operator $A\in \mathcal{B}(\mathcal{H})$ is $N$-class $A(k)$ if

for a fixed integer $N$ and a positive number $k$.

Definition 2. We say that $A\in \mathcal{B}(\mathcal{H})$ has the single valued extension property at $\lambda$ (SVEP for short) if for every neighbourhood $U$ of $\lambda$, the only analytic function $f: U\rightarrow \mathcal{H}$ which satisfies the equation $(A-\lambda)f(\lambda) = 0$ for all $\lambda\in U$ is the function $f\equiv 0$. We say that $A\in \mathcal{B}(\mathcal{H})$ satisfies the SVEP property if $A$ has the single valued extension property at every $\lambda\in \mathbb{C}$.

We will prove and recall any known results which will be used in the sequel.

Lemma 3. Let $T\in \mathcal{B}(\mathcal{H})$ be a $N$-$A(k)$ class operator and $\mathcal{M} \subset \mathcal{H}$ an invariant subspace of $T$. Then $T|_ \mathcal{M}$ is $N$-$A(k)$ class operator as well.

Proof. Let $A = T|_ \mathcal{M}$ and $P$ be the orthogonal projection on $\mathcal{M}$.

Since $\mathcal{M}$ is an invariant for $T$, we get

Therefore

Since $T\in N$-$A(k)$ class, then

and so

From Hansen's inequality, we get

It follows from (2.1) and Hansen's inequality that $\vert A\vert^{2k}\geq P\vert T\vert^{2k}P$ and so

By Lowner-Heinz inequality, we get

Therefore, from (2.2) and (2.3), we get

which means that $A\in N$-$A(k)$ class.

We will need one more lemmas.

Lemma 4. ^[9] If $A$ is $N$-$A(k)$ class operator and $A = U\vert A\vert$, then the Aluthge transformation $\tilde{A} = \vert A\vert^{1/2} U\vert A\vert^{1/2}$ of $A$ is hyponormal.

Lemma 5. ^[1] If $A$ is hyponormal, then $A$ has SVEP.

Lemma 6. ^[13] Let $A$ be a class $(\mathcal{Y})$ and $\mathcal{M}\subset \mathcal{H}$ be an invariant subspace for $A$. If $A\mid _{\mathcal{M}}$ is normal, then $\mathcal{M}$ reduces $A$.

Lemma 7. ^[10] Let $A$ be a $N$-$A(k)$ class operator and $\mathcal{M}\subset \mathcal{H}$ be an invariant subspace for $A$. If $A\mid _{\mathcal{M}}$ is normal, then $\mathcal{M}$ reduces $A$.

Theorem 8. ^[12] Let $A_1\in \mathcal{B}(\mathcal{H}))$ and $A_2\in \mathcal{B}(\mathcal{K})$. Then the following assertions are equivalent

1. The pair $(A_1, A_2)$ satisfies Fuglede-Putnam theorem.

2. If $A_1X = XA_2$ for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $\overline{\text{ran} X}$ reduces $A_1$, $(\ker X)^{\perp}$ reduces $A_2$, and $A_1\mid_{\overline{\text{ran} X}}\;, A_2\mid_{(\ker X)^{\perp}}$ are normal operators.

3.   Results

Theorem 9. Let $A \in \mathcal{B}(\mathcal{H})$ be an injective $N$-$A(k)$ class operator and $B^{*}\in \mathcal{B}(\mathcal{K})$ be a class $(\mathcal{Y})$. If $AX = XB$ for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $A^*X = XB^{*}$.

Proof. Since $B^{*}$ is of class $(\mathcal{Y})$, there exist positive numbers $\kappa$ and $p_\kappa$ such that

Hence by ^[5], for all $x\in |BB^{*}-B^{*}B|^{\kappa/2}\mathcal{K}$ there exists a bounded function $g: \mathbb{C}\rightarrow \mathcal{K}$ such that

Let $A = U|A|$ be the polar decomposition of $A$ and defines its Aluthge transform by $\tilde{A} = |A|^{1/2}U|A|^{1/2}$. Then $\tilde{A}$ is hyponormal by lemma 4 and

We assert that $|A|^{1/2}Xx = 0$. Otherwise, if $|A|^{1/2}Xx\neq 0$, then from lemma 4 and by ^[11] the function $g: \mathbb{C}\rightarrow \mathcal{H}$ is bounded entire function and hence it is constant by Liouville theorem. Therefore, it follows from

that $g(\lambda) = 0$ and hence $|A|^{1/2}Xx = 0$. This is a contradiction.

Then

Since $\ker\, A = \ker |A| = \{0\}$, we get

Since $\overline{\text{ran} (X)}$ is invariant under $A$ and $(\ker X)^{\perp}$ is invariant under $B^{*}$, we can write

implying

Hence

Since $X_1$ is injective and has dense range,

This implies that the operator $B^{*}_1$ is hyponormal. Now, from the equality $AX = XB$, we get

where $A_1$ is $N$-$A(k)$ by Lemma 3 and $B^{*}_1$ is hyponormal. Let $A_1 = U\vert A_1\vert$ be the polar decomposition of $A_1$, and multiply in left both sides of (3.1) by $\vert A_1\vert^{1/2}$ to obtain

where $\tilde{A_1}$ and $B_1^{*}$ are hyponormal operators. By Fuglede-Putnam Theorem ^[8] it yields

Hence

And

Since $A_1$ is injective, then

Hence, $A_1$ and $B_1$ are normal by theorem 8 implying $A_2 = 0$ by lemma 7 and $B_2 = 0$ by lemma 6.

Consequently

Theorem 10. Let $A \in \mathcal{B}(\mathcal{H})$ be $N$-$A(k)$ class operator and $B^{*}\in \mathcal{B}(\mathcal{K})$ be a class $\mathcal{Y}$. If $AX = XB$, for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $A^*X = XB^{*}$.

Proof. Decompose $A$ into normal part $A_1$ and pure part $A_2$ as

and let

Since $A_2$ is an injective pure $N$-$A(k)$ class operator. $AX = XB$ implies

Hence

by applying theorem 9.

Theorem 11. Let $A\in \mathcal{B}(\mathcal{H})$ be class $\mathcal{Y}$ and $B^{*}\in \mathcal{B}(\mathcal{K})$ be $N$-$A(k)$ class operator. If $AX = XB$ for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $A^{*}X = XB^{*}$.

Proof. Case 1. If $B^{*}$ is injective. Suppose that $AX = XB$ for any $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$. Since $\overline{\text{ran} (X)}$ is invariant by $A$ and $(\ker X)^{\perp}$ is invariant by $B^{*}$, we consider the following decomposition:

Then it yields

and

From $AX = XB$, we get

Let $B_1^{*} = U^{*}\vert B_1^{*}\vert$ be the polar decomposition of $B_1^{*}$. Multiply both sides of (3.4) in the right by $\vert B_1^{*}\vert^{1/2}$, we obtain

Since $A_1$ is class $\mathcal{Y}$ and $(\tilde{B_1^{*}})^{*}$ is co-hyponormal, then $(A_1, (\tilde{B_1^{*}}))$ satisfies $(FP)$ property. Therefore $A\mid_{\overline{\text{ran} (X_1\vert B_1^{*}\vert^{1/2})}}$ and $\tilde{B_1^{*}}\mid_{(\ker X_1\vert B_1^{*}\vert^{1/2})^{\perp}}$ are normal operators by ^[12]. Since $X_1$ is injective with dense range and $\vert B_1^{*}\vert^{1/2}$ is injective, then

and

It follows that $\tilde{B_1^{*}}\mid_{(\ker X)^{\perp}}$ is normal and $(\ker X)^{\perp}$ reduces $B^{*}$, also $\tilde{\text{ran} (X)}$ reduces $A$. Thus $A_2 = B_2 = 0$. Since $A_1X_1 = X_1B_1$ with $A_1$ and $B_1$ being normal, then $A_1^{*}X_1 = X_1B_1^{*}$. Consequently, $A^{*}X = XB^{*}$.

Case 2. Decompose $B^{*}$ into normal part $B_1^{*}$ and pure part $B_2^{*}$ as $B^{*} = B_1^{*}\oplus B_2^{*}$ on $\mathcal{K} = \mathcal{K}_1\oplus \mathcal{K}_2$, where $B_2^{*}$ is an injective $N$-$A(k)$ class operator. Let

Since $B^{*}_1$ is an injective pure $N$-$A(k)$ class operator. $AX = XB$ implies

Hence

by Case 1.

Theorem 12. Let $A\in \mathcal{B}(\mathcal{H})$ be an injective $N_1$-$A(k_1)$ class operator and $B^{*}\in \mathcal{B}(\mathcal{K})$ be an injective $N_2$-$A(k_2)$ class operator. If $AX = XB$ for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $A^{*}X = XB^{*}$.

Proof. Since $\overline{\text{ran} (X)}$ is invariant by $A$ and $(\ker X)^{\perp}$ is invariant by $B^{*}$, if we consider the decomposition

then $A, B$ and $X$ can be written as

From $AX = XB$, we get

Let $A_1 = U_1\vert A_1\vert$ and $B_1^{*} = V^{*}\vert B_1^{*}\vert$ be the polar decomposition of $A_1$ and $B_1^{*}$ respectively. Multiply the both sides of (3.5) in left by $\vert A_1\vert^{1/2}$ and in the right by $\vert B_1^{*}\vert^{1/2}$ and uses the polar decomposition, we obtain

where $Y = \vert A_1\vert^{1/2}X\vert B_1^{*}\vert^{1/2}$. The last equality follows from the fact that $\tilde{T^{*}} = (\tilde{T})^{*}$. From the hyponormality of $\tilde{A_1}$ and $\tilde{B_1^{*}}$, we deduce that the pair $(\tilde{A_1}, \tilde{B_1^{*}})$ satisfies the Fuglede-Putnam, implying

Hence $\tilde{A_1}\mid_{\overline{\text{ran} (Y)}}$ and $\tilde{B_1}\mid_{(\ker Y)^{\perp}}$ are normal operators by ^[12].

Since $A_1, \, B_1^{*}$ and $X_1$ are injective, then $Y$ is injective i.e.,

It follows that $\tilde{B_1^{*}}$ is normal imlying ($B_1^{*}$ is normal), hence $(\ker X)^{\perp}$ reduces $B^{*}$. Therefore $B_2 = 0$. (We use the fact that if the Aluthge tranform of an operator $T$ is normal, then $T$ is normal). Also, since

By the same argument as before, we get $A_2 = 0$. Finally, we obtain $A^{*}_1X_1 = X_1B_{1}^{*}$, and therefore

This completes the proof.

Corollary 13. Let $A\in \mathcal{B}(\mathcal{H})$ be $N_1$-$A(k_1)$ class operator and $B^{*}\in \mathcal{B}(\mathcal{K})$ be $N_2$-$A(k_2)$ class operator. If $AX = XB$ for some $X\in \mathcal{B}(\mathcal{K}, \mathcal{H})$, then $A^{*}X = XB^{*}$.

Proof. Decompose $A$ (resp. $B^{*}$) into normal part $A_1$ (resp. $B^{*}_1$) and pure part $A_2$ (resp. $B^{*}_2$) as

and let

Here $A_1, B_1$ are normal, $A_2$ is an injective $N_1$-$A(k_1)$ class operator and $B_2^{*}$ is an injective $N_2$-$A(k_2)$ class operator. From $AX = XB$, we get

Hence

by applying theorem 12.

4.   Conclusions

The following Putnam-Fuglede theorem is very well known:

Theorem 14. (Putnam-Fuglede Theorem) ^[7]

Assume that $A, B\in B(\mathcal{H})$ are normal operators. If $AX = XB$ for some $X\in B(\mathcal{H})$, then $A^*X = XB^*$

These are many extensions of this theorem to several classes of operators. In 1978, S.K Berberian ^[4] showed that the Putnam-Fuglede theorem holds when $A$ and $B^*$ are hyponormal and $X$ a Hilbert-Schmidt operator. Radjapalipour ^[8] proved that the Putnam-Fuglede theorem remains valid for hyponormal operators. In 2002, Uchiyama and Tanahashi ^[14] proved that the theorem still holds for $p$-hyponormal and $\log$-hyponormal operators. Bachir and Lombarkia ^[1] gave an extension of Putnam-Fuglede theorem for $w$-hyponormal and class $\mathcal{Y}$. Recently, Bachir and Segres^[3] extended this theorem to class $(n, k)$-quasi-*-paranormal operators.

The novelty to this contribution is to extend the famous Putnam-Fuglede thorem to the $N$-$A(k)$ class operators which is a superclass containing the normal operators and in other hand, generalize the results obtained in ^[4,8].

Acknowledgments

The authors grateful to the referees for their time and effort in providing very help and valuable comments and suggestion which leads to improve the quality of the paper.

Conflict of interest

The authors declare that they have no conflicts of interest to report regarding the present study.