Proof. That the conditions (i)–(iii) are equivalent is immediate from the following identities (see Proposition 2.1):

(3.1)\begin{equation} T'^*T' = (T^*T)^{{-}1}\quad \text{and}\quad T^{\prime}T^*= P_{{\mathcal{R}}(T)}. \end{equation}

To see that the conditions (i) and (iv) are equivalent, note that $T'|_{{\mathcal {R}}(T')}$ is hyponormal if and only if

\[ \|(T'|_{{\mathcal{R}}(T')})^*T'x\| \leqslant \|T'^2x\|,\quad x \in \mathcal{H}. \]

Using (3.1) and the equalities $(T'|_{{\mathcal {R}}(T')})^* = P_{{\mathcal {R}}(T')}T'^*|_{{\mathcal {R}}(T')}$ and ${\mathcal {R}}(T')={\mathcal {R}}(T)$, we conclude that the operator $T'|_{{\mathcal {R}}(T')}$ is hyponormal if and only if

\[ \|T'T^*x\| \leqslant \|T'Tx\|, \quad x \in \mathcal{H}. \]

The above inequality coincides with (1.3). This shows that the conditions (i)–(iv) are equivalent.

To show the remaining part, recall that any concave operator satisfies (1.1) and (1.2). Since the Cauchy dual of a concave operator is hyponormal (see [Reference Shimorin36, p. 294]) and the restriction of a hyponormal operator to its closed invariant subspace is hyponormal (see [Reference Conway18, Proposition II.4.4]), an application of the implication (iv)$\Rightarrow$(i) completes the proof.

Proof. (i) Assume that $T$ is invertible. Since $\partial \sigma (T) \subseteq \sigma _{ap}(T)$, we infer from (1.1) that $\sigma (T) \subseteq \partial \mathbb {D}$. Hence

(3.2)\begin{equation} \sigma(T^*) = \sigma(T)^* \subseteq\partial \mathbb{D}. \end{equation}

Noting that the invertibility of $T$ implies the invertibility of $T'$, we deduce that ${\mathcal {H}_u'=\mathcal {H}}$. From the assumption that $T'|_{\mathcal {H}_u'}$ is hyponormal, we see that $T'$ is hyponormal. However, by invertibility of $T$, $T'=T^{*-1},$ which by [Reference Conway18, Proposition II.4.4] implies that $T^*$ is hyponormal. Hence, by (3.2) and the second corollary on [Reference Stampfli40, Pg 473], $T$ is unitary.

(ii) Assume now that $T$ is analytic. Suppose to the contrary that $T$ has a non-zero normal direct summand, say $N$. Then by (1.1), we have

\[ \sigma(N) = \sigma_{ap}(N) \subseteq \sigma_{ap}(T) \subseteq\partial \mathbb{D}. \]

Hence, $N$ is unitary. This contradicts the fact that every analytic operator is completely non-unitary.

Proof. First, observe that if $T\in \mathcal {B}(\mathcal {H}),$ then for any positive integer $m$, $T$ is left-invertible if and only if $T^m$ is left-invertible. Hence, $W^k$ is left-invertible if and only if $\inf _{n \in \mathbb {Z}_+} w_n > 0$. This means that there is no loss of generality in assuming that all powers of $W$ are left-invertible. Noting that

\begin{align*} W^k e_n & = \left(\prod_{j=0}^{k-1} w_{n+j}\right) e_{k+n}, \quad n \in \mathbb{Z}_+, \\ W^{*k} e_{k+n} & = \left(\prod_{j=0}^{k-1} w_{n+j}\right) e_{n}, \quad n \in \mathbb{Z}_+, \end{align*}

we deduce that

(4.1)\begin{equation} W^{*k}(W^{*k}W^k)^{{-}1}W^k e_n =\prod_{j=0}^{k-1} \frac{w_{n+j}^2}{w_{k+n+j}^2} e_{n}, \quad n \in \mathbb{Z}_+. \end{equation}

Clearly

(4.2)\begin{equation} P_{{\mathcal{R}}(W^k)} e_n =\begin{cases} 0 & \text{if } n=0, \ldots, k-1, \\ e_n & \text{if } n \geqslant k. \end{cases} \end{equation}

Combining (4.1) and (4.2) with the equivalence (i) $\Leftrightarrow$ (iii) of Proposition 3.1 completes the proof.

Proof. By [Reference Abdullah and Le1, Theorem 2.1] (see also [Reference Jabłoński, Jung and Stochel25, Proposition A.2] and [Reference Bermúdez, Martinón and Negrín8, Theorem 3.4]), $W$ is an $m$-isometry if and only if there exists a polynomial $p\in \mathbb {R}[x]$ of degree at most $m-1$ such that (4.3) is valid. Therefore, there is no loss of generality in assuming that $W$ is an $m$-isometry with $p\in \mathbb {R}[x]$ of degree at most $m-1$ that satisfies (4.3). By [Reference Agler and Stankus3, Lemma 1.21], $W$ satisfies (1.1). Hence $0\notin \sigma _{ap}(W)$, or equivalently $W$ is left-invertible. Moreover, $W$ is an expansion if and only if $w_n \geqslant 1$ for every $n \in \mathbb {Z}_+$, or equivalently if and only if (4.4) holds. Hence, applying (4.3) and Proposition 4.1 with $k=1$, we see that the statements (i) and (ii) are equivalent (the condition (4.5) corresponds to (1.3)).

Finally, if $p$ is as in Proposition 4.2, then $\{1/p(n)\}_{n=0}^{\infty }$ is a Hausdorff moment sequence which as such is monotonically decreasing and logarithmically convex [the former is equivalent to (4.4), while the latter implies (4.5)].

Proof. It follows from [Reference Richter31, Lemma 1(b)] that

(4.6)\begin{equation} \|T^n h\|^2 \leqslant \|h\|^2 + n (\|Th\|^2 - \|h\|^2), \quad h \in \mathcal{H}, \, n\in \mathbb{Z}_+. \end{equation}

By [Reference Agler and Stankus3, p. 389] (see also [Reference Jabłoński, Jung and Stochel25, Theorem 3.3]), for every $h \in \mathcal {H}$ the expression $\|T^n h\|^2$ is a polynomial inFootnote ¹ $n$ of degree at most $m-1$. Combined with (4.6), this implies that $m \leqslant 2$, which completes the proof.

Proof. We begin by noting that

(5.5)\begin{equation} L-\lambda I = L (I-\lambda T) ={-}\lambda L(T-\lambda^{{-}1}I), \quad \lambda \in \mathbb{C}_{{\bullet}}. \end{equation}

Second, there is no loss of generality in assuming that $0\in \sigma (T)$ (because otherwise $L=T^{-1}$, and so we can use (5.1) to get $\phi (\sigma (L)) =\sigma (T)$). Thus, by (5.2), $L$ is not invertible. Since $T$ is left-invertible, $T$ is not right-invertible. In turn, because $L$ is right-invertible, $L$ is not left-invertible. This yields

(5.6)\begin{equation} 0 \in \sigma_r(T) {\setminus} \sigma_l(T)\quad\text{and}\quad 0 \in \sigma_l(L) {\setminus} \sigma_r(L). \end{equation}

We claim that

(5.7)\begin{equation} \phi(\mathbb{C}_{{\bullet}} {\setminus}\sigma_{l}(L)) \subseteq \sigma_{r}(T){\setminus} \sigma_{l}(T). \end{equation}

For, assume that $\lambda \in \mathbb {C}_{\bullet } {\setminus} \sigma _{l}(L)$. Then there exists $L_1\in \mathcal {B}(\mathcal {H})$ such that

(5.8)\begin{equation} L_1(L-\lambda I)=I. \end{equation}

Hence, by (5.5), $T-\lambda ^{-1}I$ is left-invertible, meaning that $\lambda ^{-1}\in \mathbb {C} {\setminus} \sigma _{l}(T)$. Thus, to get (5.7), it suffices to show that $\lambda ^{-1} \in \sigma _{r}(T).$ Suppose, to the contrary, that $\lambda ^{-1} \in \mathbb {C}{\setminus} \sigma _{r}(T)$. As a consequence, $T-\lambda ^{-1}I$ is invertible. Right-multiplying (5.5) by the inverse of $T-\lambda ^{-1}I$ yields

(5.9)\begin{equation} (L-\lambda I) (T-\lambda^{{-}1}I)^{{-}1} ={-} \lambda L. \end{equation}

Now left-multiplying (5.9) by $L_1$ leads to

(5.10)\begin{equation} (T - \lambda^{{-}1} I)^{{-}1}\mathop =\limits^{(5.8)} L_1(L-\lambda I)(T - \lambda^{{-}1}I)^{{-}1} ={-} \lambda L_1 L. \end{equation}

Finally, left-multiplying (5.10) by $T - \lambda ^{-1} I$ gives

\[ I = \left(-\lambda (T - \lambda^{{-}1} I) L_1\right) L. \]

This means that $L$ is left-invertible, showing that $0 \in \mathbb {C} \backslash \sigma _{l}(L)$, which contradicts (5.6). This justifies (5.7). Combined with (5.6), this implies the first inclusion in (5.3).

To prove (5.4), we first check that $0 \notin \partial \sigma (L)$. Note that

(5.11)\begin{equation} 0 \in \partial \sigma(L) \Longleftrightarrow 0 \in \partial \sigma(L^*). \end{equation}

Next, since $L$ is right-invertible and

\[ \sigma_{ap}(L^*)=\sigma_{l}(L^*)=\sigma_{r}(L)^{*}, \]

we see that $0 \notin \sigma _{ap}(L^*)$. Combined with (5.11) and $\partial \sigma (L^*) \subseteq \sigma _{ap}(L^*),$ this implies that $0 \notin \partial \sigma (L)$. Since, by (5.7), $\phi (\mathbb {C}_{\bullet } {\setminus} \sigma (L)) \subseteq \sigma _{r}(T) {\setminus} \sigma _{l}(T),$ and the right spectrum of any bounded linear operator is closed, the inclusion $\phi (\partial \sigma (L)) \subseteq \sigma _r(T)$ now follows from the continuity of $\phi.$ This completes the proof.

Proof. By Theorem 5.1, $0 \notin \partial \sigma (L)$ and so $r(L) > 0$. According to the definition of the spectral radius, $\mathbb {D}_{1/r(L)} \cap \phi (\sigma (L)) = \emptyset$, and thus the desired dichotomy is immediate from Theorem 5.1. The inequality $r(T)r(L) \geqslant 1$ follows from (5.12).

Proof. (i) Suppose that $\sigma _r(T') \subseteq \overline {\mathbb {D}}$. Set $L=T^{\prime *}$. Then $L$ is a left-inverse of $T$ and

\[ \sigma_l(L) = \sigma_l(T^{\prime *}) =\sigma_r(T^{\prime})^* \subseteq \overline{\mathbb{D}}. \]

This combined with Theorem 5.1 yields

\[ \mathbb{D}=\mathbb{C} {\setminus}\phi(\overline{\mathbb{D}}) \subseteq \mathbb{C} {\setminus} \phi(\sigma_l(L)) \subseteq\sigma_r(T) {\setminus} \sigma_l(T), \]

which justifies (i).

(ii) Assume now that $\|T\| \leqslant 1$ and $\sigma _r(T') \subseteq \overline {\mathbb {D}}$. Then using (i) and the well-known fact that the spectral radius of a bounded linear operator is at most its operator norm, we get

\[ {\mathbb{D}} \subseteq \sigma_r(T) {\setminus}\sigma_l(T) \subseteq \sigma_r(T) \subseteq \sigma(T) \subseteq \overline{{\mathbb{D}}}. \]

Since $\sigma _r(T)$ and $\sigma (T)$ are closed subsets of $\mathbb {C}$, the proof is complete.

Proof. Recall the fact that for Hilbert space operators, left spectrum and approximate point spectrum coincide. Note that $T'$ is a contraction because $T$ is an expansion. Further, since $T$ is not invertible, $T'$ is not invertible. Also, because the boundary of the spectrum is contained in the approximate point spectrum and $0 \in \sigma (T),$ we infer from $\sigma _{ap}(T) \subseteq \partial \mathbb {D}$ that $\sigma (T) = \overline {\mathbb {D}}$ and $\sigma _{ap}(T)=\partial \mathbb {D}.$ By applying Corollary 5.5(ii) to $T'$, we obtain $\sigma (T')=\overline {\mathbb {D}}$ and $\partial \mathbb {D} \subseteq \sigma _{ap}(T')$ (see Remark 5.6). An application of Corollary 5.5(i) to $T'$ yields

\[ \mathbb{D} \subseteq \sigma(T') {\setminus}\sigma_l(T') = \overline{\mathbb{D}} {\setminus} \sigma_{ap}(T') \subseteq \mathbb{D}, \]

and hence $\sigma _{ap}(T')=\partial \mathbb {D}$. This completes the proof.

Proof of Theorem 1.3 In view of Corollary 3.3(i), we may assume that $0 \in \sigma (T).$ Since the subspace $\mathcal {H}'_u$ [see (1.4)] is invariant for $T'$, we can consider the operator $R:=T'|_{\mathcal {H}'_u} \in \mathcal {B} (\mathcal {H}'_u).$ Note that $T'$ is bounded from below and $R$ maps $\mathcal {H}'_u$ onto $\mathcal {H}'_u$, hence $R$ is invertible. By Corollary 5.7(ii), we have

(6.1)\begin{equation} \sigma_{ap}(R) \subseteq \sigma_{ap}(T') =\partial \mathbb{D}. \end{equation}

Since $0 \notin \sigma (R)$ and $\partial \sigma (R) \subseteq \sigma _{ap}(R)$, we deduce from (6.1) that $\sigma (R) \subseteq \partial \mathbb {D}.$ Further, by Proposition 3.1, $R$ being a restriction of the hyponormal operator $T'|_{{\mathcal {R}}(T')}$ to $\mathcal {H}'_u$ is hyponormal. Hence, by [Reference Stampfli40, second corollary, p. 473], $R$ is unitary. Recall that if $\mathcal {M}$ is a closed invariant subspace for a contraction $A \in \mathcal {B}(\mathcal {H})$ such that $A|_{\mathcal {M}}$ is unitary, then $\mathcal {M}$ reduces $A$ (see e.g., [Reference Chavan13, Lemma 2.5]). Applying this fact to $A=T'$ ($T'$ is a contraction because $T$ is an expansion) and $\mathcal {M}=\mathcal {H}'_u,$ we deduce that $\mathcal {H}'_u$ reduces $T'$. By [Reference Anand, Chavan, Jabłoński and Stochel6, Proposition 2.2] applied to $T^{\prime }$, $\mathcal {H}'_u$ reduces $T$ and $T|_{\mathcal {H}'_u} = R^{\prime }$. Consequently, $T|_{\mathcal {H}'_u}$ is unitary and so $T(\mathcal {H}'_u)=\mathcal {H}'_u.$ This in turn implies that $\mathcal {H}'_u \subseteq \mathcal {H}_u$ (see (1.4)). On the other hand, by [Reference Shimorin35, Proposition 2.7(i)] we have

\[ (\mathcal{H}'_u)^{{\perp}}=\bigvee_{n \in \mathbb{Z}_+} T^n(\ker T^*). \]

Since $T|_{\mathcal {H}'_u}$ is unitary, we conclude that $T|_{\mathcal {H} \ominus \mathcal {H}'_u}$ has the wandering subspace property. This completes the proof.

Proof. It follows from [Reference Shimorin35, Proposition 3.4] that $\mathcal {H}_u$ reduces $T$ to a unitary operator, so by (2.1) and Theorem 1.3 we conclude that $\mathcal {H}_u = \mathcal {H}'_u$, which completes the proof.

Proof. Suppose that $\lambda \in \mathbb {C} {\setminus} \sigma _{ap}(T)$. Then $T-\lambda I$ is left-invertible, so the space ${\mathcal {R}}(T-\lambda I)$ is closed and $\dim \ker (T-\lambda I)=0$. According to (6.2), $\dim \ker (T^*-\bar \lambda I)< \infty$. This means that the operator $T-\lambda I$ is Fredholm, so $\lambda \in \mathbb {C}{\setminus} \sigma _{e}(T).$ This proves (6.3). Moreover, we have

\begin{align*} \mathrm{ind}_{T}(\lambda)& = \dim \ker (T-\lambda I) - \dim \ker (T^*-\bar \lambda I) \\ & ={-} \dim \ker (T^*-\bar \lambda I). \end{align*}

This completes the proof.

Proof of Theorem 1.4 By (6.2), $\ker T^*$ is finite dimensional. Combined with [Reference Shimorin35, Proposition 2.7(ii)] and the hypothesis that $T$ is analytic, this implies that $T'$ is finitely cyclic with the cyclic space $\ker T^*$. Moreover, since the range of $T'$ is closed, we have

(6.5)\begin{equation} \bigvee_{n \in \mathbb{Z}_+} (T'|_{{\mathcal{R}}(T')})^n\big(T'(\ker T^*)\big) = T' \Big(\bigvee_{n \in \mathbb{Z}_+} T'^n(\ker T^*)\Big) ={\mathcal{R}}(T'). \end{equation}

It follows that $T'|_{{\mathcal {R}}(T')}$ is finitely cyclic with the cyclic space $T'(\ker T^*)$. By Proposition 3.1, $T'|_{{\mathcal {R}}(T')}$ is hyponormal. Hence, by the Berger–Shaw Theorem (see [Reference Conway18, Theorem IV.2.1]), $T'|_{{\mathcal {R}}(T')}$ has a trace class self-commutator. Since $T'$ is a finite rank perturbation of $T'|_{{\mathcal {R}}(T')} \oplus 0$, where $0$ is the zero operator on $\ker {T^{\prime *}} = \ker {T^*}$, and the set of finite rank operators is a $*$-ideal, one can see that $T'$ has a trace class self-commutator. Observing that for any left-invertible operator $A \in \mathcal {B}(\mathcal {H})$ the following operator identities hold (cf. the proof of [Reference Chavan12, Proposition 2.21]):

(6.6)\begin{equation} \begin{aligned}\ [A^*,A]A &\mathop =\limits^{(3.1)} -A^*A[A'^*,A']AA^*A, \\ [A^*, A] &\mathop =\limits^{(3.1)} ([A^*,A]A)A'^* + [A^*,A]P, \end{aligned} \end{equation}

where $P$ denotes the orthogonal projection of $\mathcal {H}$ onto $\ker A^*,$ and using the fact that the set of trace class operators is an ideal (see [Reference Simon39, Theorem 3.6.6]), we deduce that $T$ has a trace class self-commutator.

To prove the ‘moreover’ part we proceed as follows. First, we show that

(6.7)\begin{equation} \sigma(S) = \bar{\mathbb{D}}\ {\rm and}\ \sigma_{e}(S) = \sigma_{ap}(S) =\partial \mathbb{D}\ {\rm for}\ S \in \{T,T'\}. \end{equation}

Since $T$ is analytic, we see that $0\in \sigma (T)$. In view of the above discussion, $T$ and $T'$ are finitely cyclic. Let $S$ be any of the operators $T$ or $T'.$ It follows from Corollary 5.7 and Lemma 6.3 that $\sigma (S)=\overline {\mathbb {D}}$ and $\sigma _{e}(S) \subseteq \sigma _{ap}(S)=\partial {\mathbb {D}}$. However, if $\lambda \in \partial {\mathbb {D}}=\partial \sigma (S)$, then by [Reference Conway17, Theorem XI.6.8] either $\lambda$ is an isolated point of $\sigma (S)$ (which is not the case) or $\lambda \in \sigma _{e}(S)$, meaning that $\partial {\mathbb {D}} \subseteq \sigma _{e}(S)$. Putting all this together, we conclude that $\sigma _{e}(S) = \sigma _{ap}(S) =\partial \mathbb {D}$, which completes the proof of (6.7).

Next, applying (6.7) and the fact that $\mathrm {ind}_S(\cdot )$ is constant on connected components of $\mathbb {C} \backslash \sigma _{e}(S)$ (use [Reference Conway17, Corollary XI.3.13]), we deduce that

(6.8)\begin{align} \mathrm{ind}_{T}(\lambda) & =\mathrm{ind}_{T}(0)\nonumber\\ & \mathop =\limits^{(6.4)} - \dim\ker T^*\nonumber\\ & ={-} \dim \ker T^{\prime *} \nonumber\\ & \mathop =\limits^{(6.4)}\mathrm{ind}_{T'}(0) =\mathrm{ind}_{T'}(\lambda), \quad \lambda\in\mathbb{D}. \end{align}

It follows from [Reference Carey and Pincus11, Theorem 8.1] that for any $\lambda \in \mathbb {C} \backslash \sigma _{e}(S)$, the Pincus principal function $\mathcal {P}_S(\cdot )$ of a completely non-normal operator $S$ with a trace class self-commutator is equal to $\mathrm {ind}_{S}(\lambda )$ a.e. with respect to the planar Lebesgue measure in a neighbourhood of $\lambda$. By Corollary 3.3(ii) and [Reference Anand, Chavan, Jabłoński and Stochel6, Proposition 2.2], $T$ and $T'$ are completely non-normal, and thus by (6.7) and (6.8), we deduce that

(6.9)\begin{equation} \mathcal{P}_T(\lambda)= \mathcal{P}_{T'}(\lambda)={-} \dim \ker T^* \cdot \chi_{\mathbb{D}}(\lambda)\quad \text{for a.e.}\ \lambda\in \mathbb{C}, \end{equation}

where $\chi _{\mathbb {D}}$ stands for the characteristic function of $\mathbb {D}$. Since the Carey–Pincus trace formula for polynomials in $S$ and $S^*$ is valid for completely non-normal operators $S$ with a trace class self-commutator (see [Reference Carey and Pincus11, Theorem 5.1]; see also [Reference Helton and Howe21, Theorem, p. 148]), the proof is complete.

Proof. A careful look at [Reference Conway17, Theorem XI.6.8] and the proofs of (6.7) and (6.8) yields (6.10) and (6.11). That $\dim \ker (T^*)$ is a positive integer can be inferred from (6.10), (6.11) and the fact that the Fredholm index $\mathrm {ind}_S(\cdot )$ of any $S\in \mathcal {B}(\mathcal {H})$ is constant on connected components of $\mathbb {C}\backslash \sigma _{e}(S).$

Proof. According to Proposition 3.1, the case $k=1$ is precisely Theorem 1.4 (by (7.1), this implies the case $k=0$). Assume now that $k \geqslant 2.$ Since $T$ is finitely cyclic, we infer from (6.2) that $\dim \ker T^* < \infty$. By the analyticity of $T$, $\mathcal {H}_u=\{0\}$, so by [Reference Shimorin35, Proposition 2.7(ii)] $T'$ is finitely cyclic with the cyclic space $\ker T^*$. Since the range of $T^{\prime k}$ is closed, we deduce that $R_k:=T'|_{{\mathcal {R}}(T^{\prime k})}$ is finitely cyclic with the cyclic space $T^{\prime k}(\ker T^*)$ (cf. (6.5)). By (${\rm iii}_{k}$), $R_k$ is hyponormal, so by [Reference Conway18, Theorem IV.2.1], $[R_k^*,R_k]$ is of trace class. It follows from [Reference Shimorin35, Lemma 2.1] that

\[ {\mathcal{R}}(T^{\prime k})^{{\perp}}=\ker{T^{\prime * k}}=\bigvee_{j=0}^{k-1} T^j(\ker T^*), \]

which yields $\dim {\mathcal {R}}(T^{\prime k})^{\perp } <\infty$. This implies that $T'$ is a finite rank perturbation of the operator $R_k \oplus 0$ which has a trace class self-commutator. Thus, the self-commutator of $T'$ and consequently that of $T$ are of trace class (see (6.6)). It is also easy to see that (6.7) holds (with the same proof). Next, arguing as in last paragraph of the proof of Theorem 1.4 [with Proposition 3.2(ii) in place of Corollary 3.3(ii)], we conclude that (6.9) holds. Finally, applying the Carey–Pincus trace formula completes the proof.

Proof. By Corollary 6.4, $n$ is a positive integer. Next, note that $U_+^n$ is a concave operator which is analytic and finitely cyclic with $\dim \ker (U_+^{n})^{*}=n$. Applying Corollary 6.4 to the operators $T$ and $U_+^n$, we conclude that

(7.2)\begin{align} \sigma_{e}(T)& =\sigma_{e}(U_+^n)=\partial \mathbb{D}, \end{align}

(7.3)\begin{align} \mathrm{ind}_{T}(\lambda) & =\mathrm{ind}_{U_+^n}(\lambda), \quad \lambda\in \mathbb{C} {\setminus}\partial \mathbb{D}. \end{align}

By Theorem 7.4, the operators $T$ and $U_+^n$ have compact self-commutators. This together with (7.2), (7.3) and [Reference Douglas19, Theorem 2] completes the proof.

Question 7.7 Is an analytic finitely cyclic operator $T$ of class $\mathcal {A}_{\infty }$ unitarily equivalent to a compact perturbation of $U^n_+$, where $n=\dim \ker (T^*)$?

Proof. Using [Reference Shimorin35, Lemma 2.1], the fact that the range of $T^{\prime k}$ is closed and finally Proposition 2.1(iii), we get

\begin{align*} {\mathcal{R}}(T^{\prime k}) & = \left(\bigvee_{j=0}^{k-1} T^j(\ker T^*)\right)^{{\perp}} \\ & = \bigcap_{j=0}^{k-1} \left(T^j(\ker T^*)\right)^{{\perp}} \\ & = \bigcap_{j=0}^{k-1}\left({\mathcal{R}}(T^j(I-P_{{\mathcal{R}}(T)}))\right)^{{\perp}} \\ & = \bigcap_{j=0}^{k-1} \ker((I - T'T^*)T^{*j}). \end{align*}

This completes the proof.

Proof. Since $T^{\prime k}$ is left-invertible, $C_k$ is invertible and, by Proposition 2.1(iii), we have

(8.1)\begin{equation} P_{{\mathcal{R}}(T^{\prime k})} = T^{\prime k}(T^{\prime *k}T^{\prime k})^{{-}1}T^{\prime *k} = T^{\prime k}C_k^{{-}1}T^{\prime *k}. \end{equation}

It is easily seen that $T'|_{{\mathcal {R}}(T^{\prime k})}$ is hyponormal if and only if

\[ T^{\prime *k} T^{\prime} P_{{\mathcal{R}}(T^{\prime k})}T^{\prime *} T^{\prime k} \leqslant T^{\prime *(k+1)} T^{\prime (k+1)}, \]

or equivalently, by (8.1), if and only if

\[ C_k T' C_k^{{-}1} T^{\prime *} C_k \leqslant T^{\prime *} C_k T^{\prime}, \]

which in turn is equivalent to

\[ C_k^{1/2} T' C_k^{{-}1} T^{\prime *} C_k^{1/2} \leqslant C_k^{{-}1/2} T^{\prime *} C_k T^{\prime} C_k^{{-}1/2}. \]

As a consequence, $T'|_{{\mathcal {R}}(T^{\prime k})}$ is hyponormal if and only if

\[ \|(C_k^{1/2} T^{\prime} C_k^{{-}1/2})^* h\|^2 \leqslant \|(C_k^{1/2} T^{\prime} C_k^{{-}1/2}) h\|^2, \quad h \in \mathcal{H}, \]

which completes the proof.

Proof. It follows from Proposition 2.1(i) that

\[ C_1^{1/2} T' C_1^{{-}1/2} = |T|^{{-}1} T |T|^{{-}1} = |T|^{{-}1} U. \]

Combined with Proposition 8.2, this completes the proof.

Proof. By Proposition 7.2, the operator $S_k$ is hyponormal. It follows that

\[ I - 2S_k^*S_k + S_k^{*2}S_k^2 \geqslant I - 2S_k^*S_k + S_k^{*}S_kS_k^*S_k = (I-S_k^*S_k)^2 \geqslant 0, \]

which gives (8.2) for $m=2$ (this is true for all hyponormal operators). It is now easy to see that (8.2) for $m=2$ is equivalent to

\[ C_k - 2T^{\prime *}C_kT^{\prime} + T^{\prime *2}C_kT^{\prime 2} \geqslant 0, \]

which in turn is equivalent to (8.3).

Since $T^*T \geqslant I,$ we deduce that

\[ C_1=T^{\prime *} T^{\prime} = (T^*T)^{{-}1} \leqslant I. \]

Consequently, we have

\[ C_{k+1} = T^{\prime * k} C_1 T^{\prime k} \leqslant T^{\prime * k} T^{\prime k} = C_k, \quad k \in\mathbb{Z}_+, \]

or equivalently that

\[ C_k^{{-}1/2}C_{k+1} C_k^{{-}1/2} \leqslant I, \quad k\in\mathbb{Z}_+. \]

This gives

\[ S_k^*S_k = C^{{-}1/2}_k T^{\prime *} C_k T' C^{{-}1/2}_k = C^{{-}1/2}_k C_{k+1} C^{{-}1/2}_k \leqslant I, \quad k\in \mathbb{Z}_+, \]

which shows that (8.2) holds for $m=1.$ This completes the proof.

Proof. Set $P_k=P_{{\mathcal {R}}(T^{\prime k})}$ for $k \in \mathbb {Z}_+$ and $P_{\infty } = P_{\mathcal {H}_{u}'}$. Since ${\mathcal {R}}(T^{\prime k}) \searrow \mathcal {H}_{u}'$ as $k\nearrow \infty$, we deduce that $P_k \searrow P_{\infty }$ as $k\nearrow \infty$ in SOT. By the sequential SOT-continuity of multiplication in $\mathcal {B}(\mathcal {H})$, we deduce that

(8.4)\begin{align} \left. \begin{array}{c}the\ sequences\ \{P_k T^{\prime *} P_k\}_{k=0}^{\infty}\ and\ \{(P_k T^{\prime *} P_k)^*P_k T^{\prime *} P_k\}_{k=0}^{\infty}\\ converge\ in\ SOT\ to\ P_{\infty} T^{\prime *} P_{\infty}\ and\ P_{\infty} T^{\prime} P_{\infty} T^{\prime *} P_{\infty}, respectively,\end{array}\right\} \end{align}

(8.5)\begin{align} \left. \begin{array}{c}the\ sequences\ \{T' P_k\}_{k=0}^{\infty}\ and\ \{(T'P_k)^*T'P_k\}_{k=0}^{\infty}\ converge\\ in\ SOT\ to\ T'P_{\infty}\ and\ P_{\infty} T^{\prime *} T^{\prime} P_{\infty} = P_{\infty} (T^*T)^{{-}1} P_{\infty}, respectively.\end{array}\right\} \end{align}

Observe now that,

\[ P_k T^{\prime *} P_k \mathop =\limits^{(8.1)} T^{\prime k}C_k^{{-}1}T^{\prime *k} T^{\prime *} T^{\prime k}C_k^{{-}1}T^{\prime *k} = T^{\prime k}C_k^{{-}1}T^{\prime *(k+1)}, \quad k\in \mathbb{Z}_+, \]

which yields

(8.6)\begin{align} (P_k T^{\prime *} P_k)^*P_k T^{\prime *} P_k & = T^{\prime (k+1)}C_k^{{-}1} T^{\prime * k} T^{\prime k}C_k^{{-}1}T^{\prime *(k+1)}\nonumber\\ & =T^{\prime (k+1)}C_k^{{-}1} T^{\prime *(k+1)}, \quad k\in\mathbb{Z}_+. \end{align}

Combined with (8.4), this implies that the sequence $\{T^{\prime (k+1)}C_k^{-1} T^{\prime *(k+1)}\}_{n=0}^{\infty }$ converges in SOT to $A=P_{\infty } T^{\prime } P_{\infty } T^{\prime *} P_{\infty }$, and

(8.7)\begin{equation} \langle{Ah},h \rangle= \lim_{k\to \infty} \|(C_k^{1/2}T'C_k^{{-}1/2})^* C_k^{{-}1/2} T^{\prime * k}h\|^2, \quad h\in \mathcal{H}. \end{equation}

A similar argument shows that

(8.8)\begin{equation} (T'P_k)^*T'P_k = T^{\prime k}C_k^{{-}1} C_{k+1}C_k^{{-}1}T^{\prime *k}, \quad k\in \mathbb{Z}_+, \end{equation}

so by (8.5) the sequence $\{T^{\prime k}C_k^{-1} C_{k+1} C_k^{-1} T^{\prime * k}\}_{n=0}^{\infty }$ converges in SOT to $B=P_{\infty } (T^*T)^{-1} P_{\infty }$, and moreover:

(8.9)\begin{equation} \langle{Bh},h \rangle= \lim_{k\to \infty} \|(C_k^{1/2}T'C_k^{{-}1/2}) C_k^{{-}1/2} T^{\prime * k}h\|^2,\quad h\in \mathcal{H}. \end{equation}

(i)$\Leftrightarrow$(ii) It is easy to see that $T'|_{\mathcal {H}_{u}'}$ is hyponormal if and only if

\[ \|P_{\infty} T^{\prime *} P_{\infty} h\|^2 \leqslant \|T' P_{\infty} h\|^2, \quad h \in \mathcal{H}, \]

or equivalently, by (8.4) and (8.5), if and only if

\[ \lim_{k\to \infty} \|P_k T^{\prime *} P_k h\|^2 \leqslant \lim_{k\to \infty} \|T'P_k h\|^2, \quad h\in \mathcal{H}. \]

Combined with (8.6) and (8.8), this shows that (i) and (ii) are equivalent.

(ii)$\Leftrightarrow$(iii) This follows from (8.7) and (8.9).

(ii)$\Leftrightarrow$(iv) This is a consequence of the identities $A=P_{\infty } T^{\prime } P_{\infty } T^{\prime *} P_{\infty }$ and $B=P_{\infty } (T^*T)^{-1} P_{\infty }$.

Proof. (i) By definition, $p_k=p_k^*$ and

(9.3)\begin{equation} p_k^2 = t^{\prime k}c_k^{{-}1}(t^{\prime *k} t^{\prime k})c_k^{{-}1}t^{\prime *k} =p_k, \quad k\in \mathbb{Z}_+, \end{equation}

so $p_k=p_k^*p_k$ for all $k\in \mathbb {Z}_+.$ Since

(9.4)\begin{equation} p_kp_{k+1} = t^{\prime k}c_k^{{-}1}(t^{\prime *k} t^{\prime k})t' c_{k+1}^{{-}1}t^{\prime *(k+1)}= p_{k+1}, \quad k\in \mathbb{Z}_+, \end{equation}

and thus $p_kp_{k+1}=p_{k+1}p_k$ for all $k\in \mathbb {Z}_+,$ we deduce that

\[ (p_k-p_{k+1})^2 \mathop =\limits^{(9.3)} p_k - 2 p_k p_{k+1} + p_{k+1} \mathop =\limits^{(9.4)} p_k-p_{k+1}, \quad k \in \mathbb{Z}_+. \]

Hence, $p_{k+1} \leqslant p_k$ for all $k\in \mathbb {Z}_+$. Since $e-p_k$ is a self-adjoint idempotent, we see that $p_k \leqslant e$ for all $k\in \mathbb {Z}_+.$ This proves (i).

(ii) We can argue as in the second paragraph of the proof of Corollary 8.5 (the arguments given there are purely $C^*$-algebraic).

(iii) The identity $u_k=t^{\prime } p_k t^{\prime *}$ is obvious. It follows from (ii) that $c_k^{-1} \geqslant e$ and $c_k^{-1} \leqslant c_{k+1}^{-1}$ for all $k\in \mathbb {Z}_+$, so

\begin{align*} 0 \leqslant t^{\prime (k+1)} t^{\prime * (k+1)} & \leqslant\mathop{\overbrace{t^{\prime (k+1)} c_k^{{-}1} t^{\prime * (k+1)}}}\limits^{u_k} \\ & \leqslant t^{\prime (k+1)} c_{k+1}^{{-}1} t^{\prime * (k+1)} = p_{k+1} \mathop{\leqslant}\limits^{\mathrm{(i)}} e,\quad k\in \mathbb{Z}_+, \end{align*}

which gives the second part of (iii). The third part of (iii) follows from

\[ u_{k+1} = t^{\prime} p_{k+1} t^{\prime *}\mathop{\leqslant}\limits^{\mathrm{(i)}} t^{\prime} p_k t^{\prime *} = u_k, \quad k \in \mathbb{Z}_+. \]

(iv) It is easy to see that the identity $v_k = p_k t^{\prime *}t^{\prime } p_k$ is valid. Hence, by (i) and (ii), we have

\[ 0 \leqslant v_k = p_k c_1 p_k \leqslant p_k^2=p_k \leqslant e,\quad k \in \mathbb{Z}_+, \]

which proves the third part of (iv). Finally, we come to the conclusion that

\[ v_{k+1} = p_{k+1} (t^{\prime *} t^{\prime}) p_{k+1} \mathop =\limits^{\mathrm{(i)}} p_{k+1} (p_k(t^{\prime *} t^{\prime})p_k) p_{k+1} = p_{k+1} v_k p_{k+1}, \quad k\in \mathbb{Z}_+, \]

which gives the first identity in (iv).

The proof of the ‘moreover’ part is as follows. That $t$ is an expansive element of $C^*(t)$ and $t^{\prime }, c_k, c_k^{-1} \in C^*(t)$ for every $k\in \mathbb {Z}_+$ follows from (9.1) and the fact that $\sigma _{C^*(t)}(s)=\sigma _{\mathcal {C}}(s)$ for every $s \in C^*(t)$ (see [Reference Conway17, Proposition VIII.1.14]; see also [Reference Conway17, Theorem VIII.3.6]). As a consequence, each of the elements $p_k$, $u_k$ and $v_k$ belongs to $C^*(t).$ This completes the proof.

Proof. (i)&(ii) By Proposition 9.3(i), $\{p_k\}_{k=0}^{\infty }$ is a uniformly bounded monotonically decreasing sequence of projections of $\mathcal {W}.$ It can be deduced from [Reference Sakai34, Lemma 1.7.4, Theorem 1.16.7 and Corollary 1.15.6] and [Reference Mlak29, Theorem 4.3.5] that the sequence $\{p_k\}_{k=0}^{\infty }$ is $\sigma (\mathcal {W},\mathcal {W}_{*})$-convergent to a projection of $\mathcal {W}$, call it $p_{\infty }$, which is equal to the infimum of $\{p_k\}_{k=0}^{\infty }$ computed in the set of all self-adjoint elements of $\mathcal {W}$. Consequently, $p_{\infty }$ coincides with the infimum of $\{p_k\}_{k=0}^{\infty }$ computed in the set of all projections of $\mathcal {W}$.

(iii) It follows from [Reference Sakai34, Theorem 1.16.7] that there exists a faithful $W^*$-representation $\pi \colon \mathcal {W} \to \mathcal {B}(\mathcal {H})$ of $\mathcal {W}.$ By [Reference Sakai34, Proposition 1.16.2], $\pi (\mathcal {W})$ is $\sigma (\mathcal {B}(\mathcal {H}),\mathcal {B}(\mathcal {H})_{*})$-closed. In turn, according to [Reference Sakai34, Proposition 1.15.1], $\pi (\mathcal {W})$ is SOT-closed. Since $\pi$ is an isometric $W^*$-homomorphism, we deduce from [Reference Shimorin36, Corollary 1.15.6] that for every $r\in (0,\infty )$, the mapping

\[ \pi_r:=\pi|_{\mathcal{W}_r} \colon \mathcal{W}_r\to \pi(\mathcal{W})_r \]

is continuous if $\mathcal {W}_r$ and $\pi (\mathcal {W})_r$ are equipped with the topologies $\sigma (\mathcal {W}, \mathcal {W}_{*})$ and $\sigma$-WOT (the $\sigma$-weak operator topology), respectively. Because $\pi _r$ is a bijection, we infer from the Banach–Alaoglu theorem that $\pi _r$ is a homeomorphism. By setting $T=\pi (t)$, it can be verified that

(9.5)\begin{equation} \pi(v_k) = T^{\prime k}C_k^{{-}1} C_{k+1} C_k^{{-}1} T^{\prime * k}, \quad k\in \mathbb{Z}_+. \end{equation}

It follows from Proposition 8.7 that $\{\pi (v_k)\}_{k=0}^{\infty }$ converges in SOT, and consequently in $\sigma$-WOT to a positive operator $B\in \mathcal {B}(\mathcal {H})$. Since $\pi (\mathcal {W})$ is SOT-closed, there exists a (unique) positive element $b\in \mathcal {W}$ such that $B=\pi (b).$ By the uniform boundedness principle, there exists $r\in (0,\infty )$ such that $\pi (b), \pi (v_k) \in \pi (\mathcal {W})_r$ for all $k\in \mathbb {Z}_+$. Since $\pi _r$ is a homeomorphism, the sequence $\{v_k\}_{k=0}^{\infty }$ converges in $\sigma (\mathcal {W}, \mathcal {W}_{*})$ to $b$. The same reasoning can be applied to the sequence $\{u_k\}_{k=0}^{\infty }$.

The ‘moreover’ part is a direct consequence of the statements (i) and (iii), the ‘moreover’ part of Proposition 9.3 and [Reference Sakai34, Corollary 1.7.9]. This completes the proof.

Proof. Without loss of generality, we can assume that $t$ is an expansive element of $\mathcal {W}.$ Let $\pi \colon \mathcal {W} \to \mathcal {B}(\mathcal {H})$ be a faithful $W^*$-representation of $\mathcal {W}$ (see [Reference Sakai34, Theorem 1.16.7]). Set $T=\pi (t)$ and $\mathcal {H}_u'=\bigcap _{n=0}^{\infty }{\mathcal {R}}(T^{\prime n})$. By Proposition 9.5 and [Reference Sakai34, Corollary 1.15.6], $\{\pi (p_k)\}_{k=0}^{\infty }$ is $\sigma$-WOT-convergent to $\pi (p_{\infty }).$ Since by (8.1), $\pi (p_k) = P_{{\mathcal {R}}(T^{\prime k})}$ for every $k \in \mathbb {Z}_+,$ we see that $\pi (p_k) \searrow P_{\mathcal {H}_{u}'}$ as $k\nearrow \infty$ in SOT (see the proof of Proposition 8.7). As a consequence,

(9.6)\begin{equation} \pi(p_{\infty}) = P_{\mathcal{H}_{u}'}. \end{equation}

According to the proof of Proposition 9.5(iii), $\{\pi (v_k)\}_{k=0}^{\infty }$ is SOT-convergent to $\pi (b).$ By (9.5), the ‘moreover’ part of Proposition 8.7 and (9.6), we have

(9.7)\begin{equation} \pi(b) = B = P_{\mathcal{H}_{u}'} (T^*T)^{{-}1} P_{\mathcal{H}_{u}'} = \pi(p_{\infty} (t^*t)^{{-}1} p_{\infty}). \end{equation}

The same reasoning applied to the sequence $\{u_k\}_{k=0}^{\infty }$ leads to the identity

(9.8)\begin{equation} \pi(a) = A=P_{\mathcal{H}_{u}'} T^{\prime} P_{\mathcal{H}_{u}'} T^{\prime *} P_{\mathcal{H}_{u}'} = \pi(p_{\infty} t^{\prime} p_{\infty} t^{\prime *} p_{\infty}). \end{equation}

Clearly, the condition (iii) of Definition 9.7 is equivalent to $\pi (b-a) \geqslant 0,$ which by (9.7) and (9.8) is equivalent to the condition (iii) of Definition 9.6. This completes the proof.

Proof. It follows from Proposition 9.5 and [Reference Conway17, Proposition VIII.1.14] that the infimum of $\{p_k\}_{k=0}^{\infty }$ computed in the set of all projections of $\mathcal {W}$ coincides with the infimum of $\{p_k\}_{k=0}^{\infty }$ computed in the set of all projections of $W^*(t).$ Hence, by Theorem 9.8, we can apply Definition 9.6 and [Reference Conway17, Proposition VIII.1.14] again.