Aluthge transform of operators on the Bergman space

Abstract

The aim of this paper is to explore some sufficient conditions for Aluthge transform of Toeplitz operators on the Bergman space to be unitary, average of unitaries and normal.

1 Introduction

Let H, K be two separable, infinite dimensional, complex Hilbert spaces and let ${\mathcal {L}}(H)$ denote the algebra of all bounded linear operators on H. An arbitrary operator T in ${\mathcal {L}}(H)$ has a polar decomposition $T = U|T|$, where $|T| = (T^* T)^{\frac{1}{2}}$ and U is the appropriate partial isometry (with $\mathrm{ker}(U) = \mathrm{ker}(T)$ and $\mathrm{ker}(U^{*})=\mathrm{ker}(T^{*})$). The Aluthge transformation $\Delta (T)=|T|^{\frac{1}{2}}U|T|^{\frac{1}{2}}$ was first introduced by Aluthge [2]. An operator T is said to be normal if $T^{*}T = TT^{*}$. The operator T is called quasinormal if $(TT^{*})T = (T^{*}T)T$. An operator T is called hyponormal if $T^{*}T\ge TT^{*}$. An operator T is paranormal if $\Vert Tx\Vert ^{2}\le \Vert T^{2}x\Vert \Vert x\Vert$. Here, we denote $\Delta ^*(T)= (\Delta (T))^*$ for $T \in {\mathcal {L}}(H)$. Let sp(T) denotes the spectrum of T. Define $m(T)= \mathrm{inf}\{\Vert Tx\Vert :\Vert x\Vert =1\}$. An operator T is called w-hyponormal if $|\Delta (T)|\ge |T|\ge |\Delta ^{*}(T)|$. An operator T is said to be binormal if $[| T |, | T^{*} |] = 0$, where $[T,S]= TS-ST.$ The tensor product of $x\in H$ and $y \in K$ is a conjugate bilinear functional $x \otimes y:H \times K\longrightarrow {\mathbb {C}}$ defined by $(x \otimes y) (u, v) = \langle x, u \rangle \langle y , v \rangle$ for every $(u, v) \in H \times K$. The collection of all (finite) sums of tensors $x_i \otimes y_i$ with $x_i \in H$ and $y_i \in K$, denoted by $H\otimes K$, is a complex linear space equipped with an inner product $\langle . _{,} . \rangle :(H\otimes K)\times (H\otimes K) \longrightarrow {\mathbb {C}}$ defined for arbitrary $\sum \nolimits ^N_{i=1} x_i \otimes y_i$, and $\sum \nolimits ^M_{j=1}w_j \otimes z_j$ in $H\otimes K$, by $\left\langle \sum \nolimits ^N_{i=1} x_i \otimes y_i, \sum \nolimits ^M_{j=1}w_j \otimes z_j \right\rangle = \sum \nolimits ^N_{i=1} \sum \nolimits ^M_{j=1}\left\langle x_i,w_j \right\rangle \left\langle y_i,z_j \right\rangle$   (the same notation for the inner products on H, K and $H \otimes K$). The tensor product on $H \otimes K$ of two operators T in ${\mathcal {L}}(H)$ and S in ${\mathcal {L}}(K)$ is the operator $T \otimes S : H \otimes K \longrightarrow H \otimes K$ defined by $(T \otimes S) \sum \nolimits ^N_{i=1} x_i \otimes y_i = \sum \nolimits ^N_{i=1} T x_i \otimes S y_i$ for every $\sum \nolimits ^N_{i=1} x_i \otimes y_i \in H \otimes K,$ which lies in ${\mathcal {L}}(H\otimes K).$ The complete inner product space $H\otimes K$ is denoted by $H {\hat{\otimes }} K$, which is the tensor product space of H and K . The extension of $T\otimes S$ over the Hilbert space $H {\hat{\otimes }} K$ denoted by $T {\hat{\otimes }} S$ is the tensor product of T and S on the tensor product space, which lies in ${\mathcal {L}}(H {\hat{\otimes }} K)$ (see [6, 12] and [17]).

Definition 1.1

If the sequence

$$\begin{aligned} \{\cdots , T^3 (T^3)^*, T^2 (T^2)^*, T T^*,T^* T, (T^2)^* T^2, (T^3)^* T^3, (T^4)^* T^4, \cdots \} \end{aligned}$$

is commutative, the operator $T \in {\mathcal {L}}(H)$ is called centered operator (see [10]).

Definition 1.2

A bounded linear operator T is said to be Fredholm operator if and only if its range is closed, dim $\ker T$ and dim $\ker T^*$ are finite.

The index of an operator $T \in {\mathcal {L}}(H)$ is a function from ${\mathcal {F}}(H)$ to ${\mathbb {Z}}$ which is denoted by i(T) and defined by $i(T)=$ dim $\ker T$ - dim $\ker T^*,$ where ${\mathcal {F}}(H)$ is a collection of fredholm operators on H (see [14]). An operator T is normaloid if $\Vert T\Vert = sup\{|\langle Tx , x\rangle |; \Vert x\Vert =1\}.$

The Bergman space $A^2({\mathbb {D}})$ is the Hilbert space consisting of all analytic functions f on the open unit disk ${\mathbb {D}}=\{z\in {\mathbb {C}}: |z|<1\}$ for which

$$\begin{aligned} \Vert f\Vert _{A^2({\mathbb {D}})}= \left( \int _{{\mathbb {D}}} |f(z)|^2 dA(z)\right) ^{\frac{1}{2}} < \infty , \end{aligned}$$

where dA(z) is the Lebesgue area measure on the open unit disk ${\mathbb {D}}$, normalized so that the measure of the disk ${\mathbb {D}}$ equals 1. If $h(z)=\sum \nolimits _{n=0}^\infty a_n z^n$ and $k(z)=\sum \nolimits _{n=0}^\infty b_n z^n$ are two functions in $A^2({\mathbb {D}}),$ then the inner product of h and k is given by

$$\begin{aligned} \langle h,k \rangle = \displaystyle \int _{{\mathbb {D}}}h(z)\overline{k(z)} dA(z)= \displaystyle \sum _{n=0}^\infty \frac{a_n \overline{b_n}}{n+1}. \end{aligned}$$

For $z \in {{\mathbb {D}}}$, the Bergman reproducing kernel is the function $K_z \in {A^2({\mathbb {D}})}$ such that f(z) = $\langle f,K_z\rangle$ for every $f \in {A^2({\mathbb {D}})}$. The normalized reproducing kernel $k_z$ is the function $\frac{K_z}{\Vert K_z\Vert _2}$. For any $n\ge 0, n \in {\mathbb {Z}}$, let $e_n(z)=\sqrt{n+1}z^n .$ Then $\{e_n\}$ forms an orthonormal basis for $A^2({\mathbb {D}})$. For $\phi \in L^\infty ({\mathbb {D}})$, the Toeplitz operator $T_\phi$ with symbol $\phi$ is the operator on ${A^2({\mathbb {D}})}$ defined by $T_\phi f = P(\phi f)$; here, P is the orthogonal projection from $L^2({\mathbb {D}},dA)$ onto ${A^2({\mathbb {D}})}$.

Let $\overline{A^2({\mathbb {D}})}$ be the space of conjugate analytic functions in $L^2({\mathbb {D}},dA)$. Then, $\overline{A^2({\mathbb {D}})} = \{{\overline{g}}: g \in A^2({\mathbb {D}}) \}$ is closed in $L^2({\mathbb {D}},dA)$. Let $\phi \in L^\infty ({\mathbb {D}}),$ the little Hankel operator $h_{\phi } :A^2({\mathbb {D}})\rightarrow \overline{A^2({\mathbb {D}})}$ be defined by $h_{\phi }f = {\overline{P}}(\phi f), f\in A^2({\mathbb {D}})$ where ${\overline{P}}$ is the orthogonal projection from $L^2({\mathbb {D}},dA)$ onto $\overline{A^2({\mathbb {D}})}$. There are also numerous equivalent ways of defining little Hankel operators on ${A^2({\mathbb {D}})}$. For illustration, define the map $S_{\phi }:A^2({\mathbb {D}})\rightarrow A^2({\mathbb {D}})$ by $S_{\phi }f = PJ(\phi f)$, where J is self-adjoint, unitary mapping from $L^2({\mathbb {D}},dA)$ into itself given by $Jh(z) = h({\overline{z}})$. Observe that $JS_{\phi } = h_{\phi }$. So $S_{\phi }$ is unitarily equivalent to $h_{\phi }$.

Let $Aut({\mathbb {D}})$ be the Lie group of all automorphisms (biholomorphic mappings) of ${\mathbb {D}}$. We can define for each $a \in {\mathbb {D}}$, an automorphism $\phi _a$ in $Aut({\mathbb {D}})$ such that

• (i) $(\phi _a~ o ~ \phi _a)(z)\equiv z$;

• (ii) $\phi _a(0)= a , \phi _a(a)=0$;

• (iii) $\phi _a$ has a unique fixed point in ${\mathbb {D}}$.

In fact, $\phi _a(z)=\frac{a-z}{1-{\overline{a}}z}$ for all a and z in ${\mathbb {D}}$. It is easy to verify that the derivative of $\phi _a$ at z is equal to $-k_a(z)$. It implies the real Jacobian determinant of $\phi _a$ at z is $J_{\phi _a(z)}=\left| k_a(z)\right| ^2 =\frac{\left( 1-|a|^2 \right) ^2}{\left| 1-{\overline{a}}z\right| ^4}$. Given $a \in {\mathbb {D}}$ and f (any measurable function on ${\mathbb {D}}$), let us define a function $U_a f$ on ${\mathbb {D}}$ by $U_a f(w)=k_a(w)f(\phi _a(w))$. Notice that $U_a$ is a bounded linear operator on $L^2({\mathbb {D}},dA)$ and $A^2({\mathbb {D}})$ for all $a \in {\mathbb {D}}$. Further, it can be verified that $U^2_a =I$, the identity operator, $U^*_a = U_a, U_a(A^2({\mathbb {D}})) \subset A^2({\mathbb {D}})$ and $U_a((A^2({\mathbb {D}}))^\perp )\subset (A^2({\mathbb {D}}))^\perp$ for all $a \in {\mathbb {D}}$. Thus, $U_a P = P U_a$ for all $a \in {\mathbb {D}}$ (see [20]). Let $H^\infty ({\mathbb {D}})$ denote the space of bounded analytic functions on ${\mathbb {D}}.$

In this article, we establish some sufficient conditions for Aluthge transform of Toeplitz operators and elementary operators on Bergman space to be unitary and average of unitaries. Furthermore, we explore some sufficient conditions for Aluthge transform of tensor product of Toeplitz operators to be average of unitaries on the tensor product of Bergman spaces. Again, we explain some sufficient conditions on $T_\phi$ and $S_\psi$ which imply $\Delta (T_\phi {{\hat{\otimes }}} S_\psi )$ is normal.

2 Unitary operator

This section contains a few sufficient conditions by which Aluthge transform of Toeplitz operators on $A^2({\mathbb {D}})$ express as an average of unitaries.

Theorem 2.1

Let $\phi \ge 0$ with $\Vert \phi \Vert _\infty \le 1$ for $\phi \in L^\infty ({\mathbb {D}})$. If $\Vert T_{1+\phi }\Vert < 1$. Then, the Aluthge transform $\Delta (T_\phi )$ of $T_\phi$ on $A^2({\mathbb {D}})$ can be represented as an average of two unitary operators.

Proof

Since $\phi$ is positive, $T_\phi \ge 0$ on $A^2({\mathbb {D}}).$ Then, by [1], we obtain $\Vert V-T_\phi \Vert \le \Vert I+T_\phi \Vert =\Vert T_{1+\phi }\Vert <1,$ where V is any unitary operator on $A^2({\mathbb {D}}).$ Since $\Vert V-T_\phi \Vert <1,$ so it implies $\Vert I-V^*T_\phi \Vert <1.$ Hence, $V^*T_\phi$ and $T_\phi$ are invertible. Then, by ([13], Lemma 1.1), $\Delta (T_\phi )$ is invertible. Let $\Delta (T_\phi )= U |\Delta (T_\phi )|$ be the polar decomposition of $\Delta (T_\phi ).$ As $\Delta (T_\phi )$ is invertible, the partial isometry U becomes a unitary operator. But from ([13] , Proposition 1.6), we obtain $\Vert \Delta (T_\phi )\Vert \le \Vert T_\phi \Vert \le 1.$ This implies $\Vert |\Delta (T_\phi )|\Vert \le 1.$ Therefore, $I- |\Delta (T_\phi )|^2 \ge 0$ and $\Vert I-|\Delta (T_\phi )|^2\Vert \le 1.$ Now, we define two operators as $W_1= |\Delta (T_\phi )| + i (I- |\Delta (T_\phi )|^2)^\frac{1}{2}$ and $W_2= |\Delta (T_\phi )| - i (I- |\Delta (T_\phi )|^2)^\frac{1}{2}$ such that $W^*_1=W_2.$ Therefore, it is clear that $W_1W_1^*=I$ and $W_1^*W_1=I.$ Thus, $W_1W_1^*=W_1^*W_1=I$ and $W_2W_2^*=W_2^*W_2=I.$ This implies that $W_1$ and $W_2$ are two unitary operators on $A^2({\mathbb {D}}).$ Hence, $\Delta (T_\phi )= U |\Delta (T_\phi )|= U (\frac{W_1+W_2}{2})= \frac{1}{2}(UW_1+ UW_2)= \frac{U_1+U_2}{2}$, where $U_1= UW_1$ and $U_2=UW_2$ are the two unitary operators on the Bergman space $A^2({\mathbb {D}}).$ The proof is thus completed. $\square$

Corollary 2.2

Let $\Vert \phi \Vert _\infty \le 1,$ where $\phi \ge 0$ for $\phi \in L^\infty ({\mathbb {D}})$. If $\Vert U_a - T_\phi \Vert < 1, ~~ \forall a \in {\mathbb {D}}$. Then, the Aluthge transform $\Delta (T_{\phi \circ \phi _a})$ of $T_{\phi \circ \phi _a}$ can be written as an average of the two unitaries.

Proof

Since $\Vert U_a - T_\phi \Vert < 1,$ by ([8] , Corollary-1), $T_\phi$ is invertible. By ([4], Proposition 3.3), we have $\Delta (T_{\phi \circ \phi _a})= \Delta (U_a T_{\phi } U_a)= U_a \Delta (T_\phi )U_a.$ Hence, the result holds from Theorem 2.1. $\square$

Corollary 2.3

Let $\phi$ be an essentially bounded Lebesgue measurable function on ${\mathbb {D}}$ such that $\phi \ge 0$ and $\Vert \phi \Vert _\infty \le 1.$ If $\Vert U_a - T_\phi \Vert < 1$, the Aluthge transform $\Delta (T_{\phi })$ of $T_{\phi }$ can be represented as $\Delta (T_\phi )= \sum \nolimits _{j=1}^4 \frac{(-1)^{j+1}}{4}V_j$, where $V_j$ is a unitary operator.

Proof

Let $\Delta (T_\phi )= U |\Delta (T_\phi )|$ be the polar decomposition of $\Delta (T_\phi ).$ As $\Vert \phi \Vert _\infty \le 1,$ so $\Vert T_\phi \Vert \le 1.$ Given that $\Vert U_a - T_\phi \Vert < 1$, from ([8], Corollary-1), $T_\phi$ is invertible. Therefore, $\Delta (T_\phi )$ is also invertible. So U is a unitary operator. It is known that $\Vert \Delta (T_\phi )\Vert \le \Vert T_\phi \Vert \le 1.$ This implies $\Vert ~|\Delta (T_\phi )|~\Vert \le 1.$ Hence, $I- |\Delta (T_\phi )|^2$ is a positive operator and $\Vert I-|\Delta (T_\phi )|^2\Vert \le 1.$ So we can define the four operators as $W_1= |\Delta (T_\phi )| + i (I- |\Delta (T_\phi )|^2)^\frac{1}{2}, W_3= |\Delta (T_\phi )| - i (I- |\Delta (T_\phi )|^2)^\frac{1}{2}, W_2= -|\Delta (T_\phi )| + i (I- |\Delta (T_\phi )|^2)^\frac{1}{2}$ and $W_4= - |\Delta (T_\phi )| - i (I- |\Delta (T_\phi )|^2)^\frac{1}{2}.$ Hence, $W_1, W_2, W_3$ and $W_4$ are unitary operators on the Bergman space $A^2({\mathbb {D}}).$ Thus, $\Delta (T_\phi )=U|\Delta (T_\phi )|=U(\frac{W_1-W_2+W_3-W_4}{4})=\frac{1}{4}(UW_1- UW_2 + UW_3- UW_4)= \frac{V_1-V_2+ V_3 - V_4}{4}$, where $V_1= UW_1, V_2= UW_2, V_3= UW_3$ and $V_4=UW_4$ are four unitary operators on $A^2({\mathbb {D}}).$ $\square$

Corollary 2.4

Let $\phi _1,\phi _2 \in L^\infty ({\mathbb {D}})$ with $\Vert \phi _1\Vert _\infty$, $\Vert \phi _2\Vert _\infty \le 1$ and let $T_{\phi _1} = Q_1 |T_{\phi _1}|$ and $T_{\phi _2} = Q_2 |T_{\phi _2}|$ be the polar decompositions of $T_{\phi _1}$ and $T_{\phi _2}$, respectively. If $T_{\phi _1}$ doubly commutes with $T_{\phi _2}$ (that is, $[T_{\phi _1} ,T_{\phi _2}]=0$ and $[T_{\phi _1}, T^*_{\phi _2}]=0$) and $\dim \ker (T_{\phi _1} T_{\phi _2})= \dim \ker (T_{\overline{\phi _2}} T_{\overline{\phi _1}}).$ Then, $\Delta (T_{\phi _1} T_{\phi _2})$ can be represented as $\Delta (T_{\phi _1} T_{\phi _2})=\sum \nolimits _{j=1}^4 \frac{(-1)^{j+1}}{4}V_j$, where $V_j$ is a unitary operator.

Proof

Let $T_{\phi _1} = Q_1 |T_{\phi _1}|$ and $T_{\phi _2} = Q_2 |T_{\phi _2}|$ be the polar decompositions of $T_{\phi _1}$ and $T_{\phi _2}$, respectively, and $T_{\phi _1}$ doubly commutes with $T_{\phi _2}$ then by [7], $T_{\phi _1} T_{\phi _2}= Q_1 Q_2 |T_{\phi _1} T_{\phi _2}|$ is also the polar decomposition of $T_{\phi _1} T_{\phi _2}.$ Further, since $\dim \ker (T_{\phi _1} T_{\phi _2})= \dim \ker (T_{\overline{\phi _2}} T_{\overline{\phi _1}})$, $Q_1Q_2$ is unitary. Thus, the result is obvious by Corollary 2.3. $\square$

Corollary 2.5

Let $T_\phi = U |T_\phi |$ be the polar decomposition of $T_\phi$ with the symbol $\phi \in L^\infty ({\mathbb {D}})$ and $\Vert \phi \Vert _\infty \le 1$. If $\dim \ker (T^2_\phi )= \dim \ker (T^2_{{\overline{\phi }}})$, then $\Delta (T^2_\phi )$ can be represented as $\frac{1}{4}$ of the alternating finite series of four unitary operators.

Proof

Since $T_\phi ^2 = U^2 |T^2_\phi |$ is the polar decomposition of $T_\phi ^2$ and $\Delta (T^2_\phi )= U^2 |\Delta (T^2_\phi )|$ is also the polar decomposition of $\Delta (T^2_\phi )$ , the result follows from Corollary 2.4. $\square$

Corollary 2.6

Let $\Vert \phi \Vert _\infty \le 1$ for $\phi \in L^\infty ({\mathbb {D}})$ and $T_\phi = U |T_\phi |$ be the polar decomposition of $T_\phi .$ If $T_\phi$, centered with index of $T^n_\phi$ is equal to zero. Then, $\Delta (T^n_\phi )$ can be represented as $\frac{1}{4}$ times the alternating finite series of four unitary operators.

Proof

If $T_\phi$ is centered then by [10], we obtain $T^n_\phi = U^n |T^n_\phi |$ is the polar decomposition for all $n \in {\mathbb {N}}.$ Again $\Delta (T^n_\phi )=U^n |\Delta (T^n_\phi )|$ is also the polar decomposition of $\Delta (T^n_\phi )$. Thus, the proof is evident from Corollary 2.4. $\square$

Here, we introduce the tensor product of Toeplitz operators $T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n}$ on tensor product of Bergman spaces $A^2({\mathbb {D}}) ~{\hat{\otimes }} \cdots {\hat{\otimes }}A^2({\mathbb {D}}).$

Let $T_{\phi _j}= Q_j |T_{\phi _j}|$ be the polar decompositions of $T_{\phi _j}$ for each $j=1,2,\cdots ,n.$ Then, $T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n} = (Q_1 {\hat{\otimes }} Q_2 {\hat{\otimes }} \cdots {\hat{\otimes }} Q_n)|T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n}|$ is the polar decomposition of $T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n}.$ For more details, see [17].

Corollary 2.7

Let $\phi _j \in L^\infty ({\mathbb {D}})$ for $j=1,2,\cdots ,n$ such that $\phi _j \ge 0$ and $\Vert \phi _j\Vert _\infty \le 1.$ If $\Vert T_{1+\phi _j}\Vert < 1$. Then, the Aluthge transform $\Delta (T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n} )$ of $T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n}$ can be represented as the average of $2^n$ unitary operators on $A^2({\mathbb {D}}) ~{\hat{\otimes }} \cdots {\hat{\otimes }}A^2({\mathbb {D}}).$

Proof

Let there be $\phi _j \in L^\infty ({\mathbb {D}}),$ $\phi _j \ge 0$ with $\Vert \phi _j\Vert _\infty \le 1,$ $\Vert T_{1+\phi _j}\Vert < 1$ for $j=1,2,\cdots ,n$, then by Theorem - 2.1, we find that $\Delta (T_{\phi _j})$ can be represented as an average of two unitary operators for each j. It is known from the properties of Aluthge transform [17] of operators that $\Delta (T_{\phi _1} {\hat{\otimes }} T_{\phi _2} {\hat{\otimes }} \cdots {\hat{\otimes }} T_{\phi _n})= \Delta (T_{\phi _1}) {\hat{\otimes }} \Delta (T_{\phi _2}) {\hat{\otimes }} \cdots {\hat{\otimes }} \Delta (T_{\phi _n}).$ Hence, by induction the result is obvious from Theorem -2.1. $\square$

For single operator $A \in {\mathcal {L}}(H),$ we define two elementary operators $L_A$ and $R_A$ on ${\mathcal {L}}(H)$ by $L_A(X)=AX$ and $R_A(X)=XA$ for every $X \in {\mathcal {L}}(H).$ For more details, see [15]. Here, our immediate corollary is about the Aluthge transform of tensor product of two elementary operators.

Corollary 2.8

Let $T_{\phi _1} = Q_1 |T_{\phi _1}|, ~T_{\phi _2} = Q_2 |T_{\phi _2}|, ~T_{\psi _1} = R_1 |T_{\psi _1}|$ and $T_{\psi _2}= R_2 |T_{\psi _2}|$ be the polar decompositions of $T_{\phi _1},T_{\phi _2},T_{\psi _1}$ and $T_{\psi _2}$, respectively, and let $\Vert \phi _j\Vert _\infty \le 1 ,$ $\Vert \psi _j\Vert _\infty \le 1$ for $j=1,2$ and $T_{\phi _1}$ doubly commutes with $T_{\phi _2}$ and $T_{\psi _1}$ doubly commutes with $T_{\psi _2}.$ If $\dim \ker (T_{\phi _1} T_{\phi _2})= \dim \ker (T_{\overline{\phi _2}} T_{\overline{\phi _1}})$ and $\dim \ker (T_{\psi _1} T_{\psi _2})= \dim \ker (T_{\overline{\psi _2}} T_{\overline{\psi _1}})$. Then, the Aluthge transform

$$\begin{aligned} \Delta (L_{T_{\phi _1}}(T_{\phi _2}) {\hat{\otimes }} R_{T_{\psi _2}}(T_{\psi _1}))~~ of~~L_{T_{\phi _1}}(T_{\phi _2}) {\hat{\otimes }} R_{T_{\psi _2}}(T_{\psi _1}) \end{aligned}$$

is the average of four unitary operators.

Proof

Since $T_{\phi _1}$ doubly commutes with $T_{\phi _2}$ and $T_{\psi _1}$ doubly commutes with $T_{\psi _2}$ then by [7], $T_{\phi _1} T_{\phi _2}= Q_1 Q_2 |T_{\phi _1}T_{\phi _2}|$ and $T_{\psi _1}T_{\psi _2}=R_1 R_2 |T_{\psi _1}T_{\psi _2}|$ are the polar decompositions of $T_{\phi _1}T_{\phi _2}$ and $T_{\psi _1}T_{\psi _2}$, respectively. Further, as $\Delta (T_{\phi _1}T_{\phi _2})= Q_1 Q_2 |\Delta (T_{\phi _1}T_{\phi _2})|$ and $\Delta (T_{\psi _1}T_{\psi _2})= R_1 R_2 |\Delta (T_{\psi _1}T_{\psi _2})|$ are the polar decompositions of $T_{\phi _1}T_{\phi _2}$ and $T_{\psi _1}T_{\psi _2}$, respectively. Thus, $T_{\phi _1}T_{\phi _2} {\hat{\otimes }}T_{\psi _1}T_{\psi _2}= (Q_1 Q_2 {\hat{\otimes }} R_1 R_2) |T_{\phi _1}T_{\phi _2} {\hat{\otimes }}T_{\psi _1}T_{\psi _2}|$ is also the polar decomposition of $T_{\phi _1}T_{\phi _2} {\hat{\otimes }}T_{\psi _1}T_{\psi _2}.$ It is clear from the properties of Aluthge transform of an operator that $\Delta (T_{\phi _1}T_{\phi _2} {\hat{\otimes }}T_{\psi _1}T_{\psi _2})= \Delta (T_{\phi _1}T_{\phi _2}) {\hat{\otimes }}\Delta (T_{\psi _1}T_{\psi _2}).$ Given that $\dim \ker (T_{\phi _1} T_{\phi _2})= \dim \ker (T_{\overline{\phi _2}} T_{\overline{\phi _1}})$ and $\dim \ker (T_{\psi _1} T_{\psi _2})= \dim \ker (T_{\overline{\psi _2}} T_{\overline{\psi _1}})$, this implies $Q_1 Q_2$ and $R_1 R_2$ are unitary operators. Now,

$$\begin{aligned} \Vert \Delta (T_{\phi _1}T_{\phi _2} {\hat{\otimes }}T_{\psi _1}T_{\psi _2})\Vert&= \Vert \Delta (T_{\phi _1}T_{\phi _2}) {\hat{\otimes }}\Delta (T_{\psi _1}T_{\psi _2})\Vert \\ {}&\le \Vert T_{\phi _1}T_{\phi _2}\Vert \Vert T_{\psi _1}T_{\psi _2}\Vert \\ {}&\le 1. \end{aligned}$$

Hence, the result is evident from Theorem 2.1. $\square$

3 Normal operator

In this section, we state some sufficient conditions for Aluthge transform of tensor product of Toeplitz and little Hankel operators to be normal.

Theorem 3.1

Let $\phi \ge 0$, $\psi \ge 0$ $\in L^\infty ({\mathbb {D}})$ be such that $\Vert 1+ \phi \Vert _\infty < 1$ and $\Vert 1+ \psi \Vert _\infty < 1$. Suppose that $T_\phi =U|T_\phi |$ and $S_\psi =V|S_\psi |$ be the polar decompositions of $T_\phi$ and $S_\psi$ on $A^2({\mathbb {D}})$, respectively, with sp(U) and sp(V) are contained in some open semi-circles. Then, $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal iff $T_\phi$, $S_\psi$ are normal.

Proof

Since $\phi , \psi \ge 0$ then $T_\phi$ and $S_\psi$ are positive on $A^2({\mathbb {D}})$. Thus by ([1], theorem-3.1), for every unitary operators $W_1$ and $W_2$ on $A^2({\mathbb {D}})$ we have $\Vert W_1 - T_\phi \Vert \le \Vert I + T_\phi \Vert = \Vert T_{1+\phi }\Vert \le \Vert 1+ \phi \Vert _\infty < 1$ and $\Vert W_2 - S_\psi \Vert \le \Vert I + S_\psi \Vert = \Vert S_{1+\psi }\Vert \le \Vert 1+ \psi \Vert _\infty < 1$. Hence, $T_\phi$ and $S_\psi$ are invertible. Since sp(U) contained in some open semi-circle then by [11], $\Delta (T_\phi )$ is normal iff $T_\phi$ is normal. Similarly using the same argument for $S_\psi$, We get $\Delta (S_\psi )$ is normal iff $S_\psi$ is normal. As $\Delta (T_\phi {\hat{\otimes }}S_\psi )= \Delta (T_\phi ){\hat{\otimes }}\Delta (S_\psi )$ for $T_\phi$, $S_\psi$ $\in \mathcal {L^\infty }({\mathbb {D}})$. Therefore, by ([12], Proposition-5), $\Delta (T_\phi {\hat{\otimes }}S_\psi )$ is normal iff $T_\phi$ and $S_\psi$ are normal. $\square$

Corollary 3.2

Let $\phi$, $\psi$ $\in L^\infty ({\mathbb {D}})$ be such that $\phi , \psi \ge 0$ and $\Vert 1+ \phi \Vert _\infty < 1$ and $\Vert 1+ \psi \Vert _\infty < 1$. If $T_\phi$ and $S_\psi$ are hyponormal operators with spectrum of $T_\phi$ and $S_\psi$ are arcs, respectively, then $\Delta (T_\phi {\hat{\otimes }}S_\psi )$ is normal.

Proof

As $T_\phi$ and $S_\psi$ are hyponormal with their spectrums as arcs, $T_\phi$ and $S_\psi$ are normal. Since $T_\phi$ is invertible, so by [3], $T_\phi$ and $\Delta (T_\phi )$ are similar operators. Similarly, $S_\psi$ and $\Delta (S_\psi )$ are similar operators. Again by [3], $|T_\phi |$ and $|S_\psi |$ are invertible as $T_\phi$ and $S_\psi$ are invertible. Therefore, $T_\phi$ and $S_\psi$ can be expressed as $|T_\phi |^{-\frac{1}{2}}\Delta (T_\phi )|T_\phi |^{\frac{1}{2}}$ and $|S_\psi |^{-\frac{1}{2}}\Delta (S_\psi )|S_\psi |^{\frac{1}{2}}$, respectively. Hence

$$\begin{aligned} T_\phi ^{*}T_\phi = |T_\phi |^{-\frac{1}{2}}\Delta ^{*}(T_\phi )\Delta (T_\phi )|T_\phi |^{\frac{1}{2}}. \end{aligned}$$

(3.1)

Similarly

$$\begin{aligned} T_\phi T_\phi ^{*} = |T_\phi |^{-\frac{1}{2}}\Delta (T_\phi )\Delta ^{*}(T_\phi )|T_\phi |^{\frac{1}{2}}. \end{aligned}$$

(3.2)

As $T_\phi$ is normal, it is clear from equations (3.1) and (3.2) that $\Delta (T_\phi )$ is normal. Similarly, $\Delta (S_\psi )$ is normal. Thus, by Theorem 3.1, $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal. $\square$

Corollary 3.3

Let $T_\phi =U|T_\phi |$ and $S_\psi =V|S_\psi |$ be the polar decompositions of $T_\phi$ and $S_\psi$, respectively. If U and V are hyponormal with sp(U) and sp(V) as the set of real numbers, then $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal.

Proof

Let $T_\phi =U|T_\phi |$ and $S_\psi =V|S_\psi |$ be the polar decomposition of $T_\phi$ and $S_\psi$, respectively. Now

$$\begin{aligned} \Delta ^{*}(T_\phi {\hat{\otimes }} S_\psi )&= |T_\phi {\hat{\otimes }} S_\psi |^{\frac{1}{2}}(U^{*}{\hat{\otimes }}V^{*})|T_\phi {\hat{\otimes }} S_\psi |^{\frac{1}{2}}\\&=(|T_\phi |^{\frac{1}{2}}U^{*}|T_\phi |^{\frac{1}{2}}){\hat{\otimes }}(|S_\psi |^{\frac{1}{2}}V^{*}|S_\psi |^{\frac{1}{2}}) \end{aligned}$$

As U and V are hyponormal and sp(U), sp(V) are the set of real numbers. So by [16], U and V are self-adjoint. Therefore

$$\begin{aligned} \Delta ^{*}(T_\phi {\hat{\otimes }} S_\psi )&=(|T_\phi |^{\frac{1}{2}}U|T_\phi |^{\frac{1}{2}}){\hat{\otimes }}(|S_\psi |^{\frac{1}{2}}V|H_\psi |^{\frac{1}{2}})\\ {}&= \Delta (T_\phi {\hat{\otimes }}S_\psi ) \end{aligned}$$

So $\Delta (T_\phi {\hat{\otimes }}S_\psi )$ is self-adjoint. Since self-adjoint operators are normal, so $\Delta (T_\phi {\hat{\otimes }}S_\psi )$ is normal. $\square$

Theorem 3.4

Let $\phi$, $\psi$ $\in L^{\infty }({\mathbb {D}})$ and $X \in {\mathcal {L}}(A^2({\mathbb {D}}))$. If $\phi$ and $\psi$ $\ge 0$ with $T_\phi X = I$ and $S_\psi X = I$ then $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal.

Proof

As $\phi$ and $\psi$ are positive, $T_\phi$ and $S_\psi$ are positive on $A^{2}({\mathbb {D}})$. Since $T_\phi X = I$, $T_\phi$ is right invertible. Now $X^{*}T_\phi = (T_\phi X)^{*}= I^{*} = I$. This implies that $T_\phi$ is left invertible. Hence, $T_\phi$ is invertible. Similarly, $S_\psi$ is invertible. Then by [5], $\Delta (T_\phi )$, $\Delta (S_\psi )$ are positive. Since positive operators are normal, then $\Delta (T_\phi )$, $\Delta (S_\psi )$ are normal. Then by [12], $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal. $\square$

Theorem 3.5

Let $T_\phi , S_\psi \in {\mathcal {L}}(A^2({\mathbb {D}}))$ with $m(T_\phi ), m(T_\phi ^*)>0$ and $m(S_\psi ), m(S_\psi ^*)>0$. If $T_\phi$ and $S_\psi$ are partial isometry satisfying ${T_\phi ^*}^{2}T_\phi ^{2}+2kT_\phi ^*T_\phi +k^{2}\ge 0$ and ${S_\psi ^*}^{2}S_\psi ^{2}+2kS_\psi ^*S_\psi +k^{2}\ge 0$ for all real k then $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal.

Proof

Let $m(T_\phi )= inf\{\Vert T_\phi x\Vert :\Vert x\Vert =1\}$ and $m(T_\phi ^*)=inf\{\Vert T_\phi ^{*}x\Vert :\Vert x\Vert =1\}$. Since $m(T_\phi ) , m(T_\phi ^*)> 0$ then by [18], $T_\phi$ is invertible. Similarly, $S_\psi$ is invertible. Again since ${T_\phi ^*}^{2}T_\phi ^{2}+2kT_\phi ^*T_\phi +k^{2}\ge 0$ and ${S_\psi ^*}^{2}S_\psi ^{2}+2kS_\psi ^*S_\psi +k^{2}\ge 0$ for any real number k , so $T_\phi$ and $S_\psi$ are paranormal. As partial isometry and paranormal operators are quasinormal. So $T_\phi$ and $S_\psi$ are quasinormal. Therefore by [13], $\Delta (T_\phi )=T_\phi$ and $\Delta (S_\psi )=S_\psi$. Thus,

$$\begin{aligned} \Delta (T_\phi {\hat{\otimes }} S_\psi )&= \Delta (T_\phi ) {\hat{\otimes }} \Delta (S_\psi )\\&= T_\phi {\hat{\otimes }} S_\psi . \end{aligned}$$

Since $T_\phi$ and $S_\psi$ are quasinormal and invertible operators, this implies that $T_\phi$ and $S_\psi$ are normal. Hence by [12], $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal. $\square$

Corollary 3.6

Let $T_\phi = T_{\phi _1}T_{\phi _2}$ , $S_\psi = S_{\psi _1}S_{\psi _2}$. If $T_{\phi _1},T_{\phi _2},S_{\psi _1}$ and $S_{\psi _2}$ are invertible quasinormal operators with $[T_{\phi _1},T_{\phi _2}]=[T_{\phi _1},T_{\phi _2}^{*}] =0$ and $[S_{\psi _1},S_{\psi _2}]=[S_{\psi _1},S_{\psi _2}^{*}] =0$ then $\Delta (T_\phi {\hat{\otimes }}S_\psi )$ is normal.

Proof

Since $T_{\phi _1}$ and $T_{\phi _2}$ are invertible quasinormal operators and also $[T_{\phi _1},T_{\phi _2}]=[T_{\phi _1},T_{\phi _2}^{*}] =0$ then by [9], $T_{\phi _1}T_{\phi _2}$ is normal. Similarly, $S_{\psi _1}S_{\psi _2}$ is normal. Now

$$\begin{aligned} \Delta ((T_{\phi _1}T_{\phi _2}){\hat{\otimes }}(S_{\psi _1}S_{\psi _2})) =\Delta (T_{\phi _1}T_{\phi _2}){\hat{\otimes }}\Delta (S_{\psi _1}S_{\psi _2}). \end{aligned}$$

(3.3)

As $T_{\phi _1}$ and $T_{\phi _2}$ are quasinormal with $T_{\phi _1}T_{\phi _2}=T_{\phi _2}T_{\phi _1}$ and $T_{\phi _1}T_{\phi _2}^*=T_{\phi _2}^*T_{\phi _1}$ then $T_{\phi _1}T_{\phi _2}$ is quasinormal. Similarly, $S_{\psi _1}S_{\psi _2}$ is quasinormal. Hence by [13], $\Delta (T_{\phi _1}T_{\phi _2})= T_{\phi _1}T_{\phi _2}$ and $\Delta (S_{\psi _1}S_{\psi _2})= S_{\psi _1}S_{\psi _2}.$ Therefore, from Eq. (3.3), we obtained

$$\begin{aligned} \Delta ((T_{\phi _1}T_{\phi _2}){\hat{\otimes }}(S_{\psi _1}S_{\psi _2}))&=\Delta (T_{\phi _1}T_{\phi _2}){\hat{\otimes }}\Delta (S_{\psi _1}S_{\psi _2})\\ {}&= T_{\phi _1}T_{\phi _2} {\hat{\otimes }}S_{\psi _1}S_{\psi _2}. \end{aligned}$$

Thus, the proof is evident from [12]. $\square$

Theorem 3.7

Let $\phi = \phi _1 + \phi _2 \in L^\infty ({\mathbb {D}})$ and $\psi = \psi _1 + \psi _2 \in L^\infty ({\mathbb {D}})$ are positive measurable functions. Suppose $T_{\phi }, S_{\psi }$ are invertible operators with $[\Delta (T_{\phi _1}),\Delta (T_{\phi _2}^{*})] =0$ and $[\Delta (S_{\psi _1}),\Delta (S_{\psi _2}^{*})] =0$ then $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal.

Proof

Since $\phi$ and $\psi$ $\ge 0$ then $T_{\phi }$, $S_{\psi }$ are positive operators. Therefore

$$\begin{aligned} T_\phi = T_{\phi _1} + T_{\phi _2} \quad ~\text {and}~\quad S_\psi = S_{\psi _1} + S_{\psi _2} \end{aligned}$$

(3.4)

are positive. So $T_{\phi _1}$, $T_{\phi _2}$ are positive and as $T_{\phi _1}, T_{\phi _2}$ are invertible operators, By [5], $\Delta (T_{\phi _1})$ and $\Delta (T_{\phi _2})$ are positive. Since positive operators are normal. So $\Delta (T_{\phi _1})$ and $\Delta (T_{\phi _2})$ are normal. Similarly, we can easily seen that $\Delta (S_{\psi _1})$ and $\Delta (S_{\psi _2})$ are normal. Therefore

$$\begin{aligned} \Delta (T_\phi {\hat{\otimes }} S_\psi )&= \Delta (T_\phi ) {\hat{\otimes }}\Delta (S_\psi )\\&= \Delta (T_{\phi _1}+T_{\phi _2}){\hat{\otimes }}\Delta (S_{\psi _1}+S_{\psi _2})\\&= (\Delta (T_{\phi _1})+ \Delta (T_{\phi _2})){\hat{\otimes }}(\Delta (S_{\psi _1})+ \Delta (S_{\psi _2})). \end{aligned}$$

Further, since $[\Delta (T_{\phi _1}),\Delta (T_{\phi _2}^{*})] =0$ and $[\Delta (S_{\psi _1}),\Delta (S_{\psi _2}^{*})] =0$, $\Delta (T_{\phi _1})+ \Delta (T_{\phi _2})$ and $\Delta (S_{\psi _1})+ \Delta (S_{\psi _2})$ are normal. Hence, $\Delta (T_\phi {\hat{\otimes }}S_\psi )$ is normal. $\square$

Theorem 3.8

Let $T_\phi = T_{\phi _1}+iT_{\phi _2}$ and $S_\psi = S_{\psi _1}+iS_{\psi _2}$ be the cartesian decompositions of $T_\phi$ and $S_\psi$, respectively. Suppose $T_\phi$ and $S_\psi$ are invertible operators. If $T_\phi$, $S_\psi$ are hyponormal operators and $T_{\phi _2}$, $S_{\psi _2}$ are compact, then $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal.

Proof

Let $T_\phi = T_{\phi _1}+iT_{\phi _2}$ and $S_\psi = S_{\psi _1}+iS_{\psi _2}$ be the cartesian decompositions of $T_\phi$ and $S_\psi$, respectively. Since $T_\phi$, $S_\psi$ are hyponormal operators whose imaginary parts are compact, so by [19], $T_\phi$, $S_\psi$ are normal. Again, since $T_\phi$ and $S_\psi$ are invertible, therefore by [3], $T_\phi$ and $\Delta (T_\phi )$ are similar operators. Similarly, $S_\psi$ and $\Delta (S_\psi )$ are similar operators and also $|T_\phi |$ and $|S_\psi |$ are invertible as $T_\phi$ and $S_\psi$ are invertible. Therefore, we can express $T_\phi$ and $S_\psi$ as $T_\phi = |T_\phi |^{\frac{-1}{2}}\Delta (T_\phi )|T_\phi |^{\frac{1}{2}}$ and $S_\psi = |S_\psi |^{\frac{-1}{2}}\Delta (S_\psi )|S_\psi |^{\frac{1}{2}}$, respectively. So, as $T_\phi$ and $S_\psi$ are normal it can be easily shown that $\Delta (T_\phi )$ and $\Delta (S_\psi )$ are normal. Hence, $\Delta (T_\phi {\hat{\otimes }} S_\psi )$ is normal. $\square$