\documentclass[CRMATH,Unicode,biblatex,XML]{cedram} \TopicFR{Analyse fonctionnelle, Théorie des opérateurs} \TopicEN{Functional analysis, Operator theory} \addbibresource{CRMATH_Bhuia_20240349.bib} %\DeclareMathAlphabet{\oldmathcal}{OMS}{cmsy}{m}{n} \newcommand{\clb}{{\mathcal{B}}} \newcommand{\cle}{{\mathcal{E}}} \newcommand{\clf}{{\mathcal{F}}} \newcommand{\clh}{{\mathcal{H}}} \newcommand{\clk}{{\mathcal{K}}} \newcommand{\clm}{{\mathcal{M}}} \newcommand{\cln}{{\mathcal{N}}} \newcommand{\clr}{{\mathcal{R}}} \newcommand{\N}{\mathbb{N}} \newcommand\restr[2]{\ensuremath{\left.#1\right|_{#2}}} \newcommand{\D}{\mathbb{D}} \newcommand{\T}{\mathbb{T}} \newcommand{\dd}{\mathrm{d}} \makeatletter \DeclareFontFamily{U}{mathx}{\hyphenchar\font45} \DeclareFontShape{U}{mathx}{m}{n}{ <5> <6> <7> <8> <9> <10> <10.95> <12> <14.4> <17.28> <20.74> <24.88> mathx10 }{} \DeclareSymbolFont{mathx}{U}{mathx}{m}{n} \DeclareFontSubstitution{U}{mathx}{m}{n} \DeclareMathAccent{\widecheck} {0}{mathx}{"71} \makeatother \numberwithin{equation}{section} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \makeatletter \let\save@mathaccent\mathaccent \newcommand*\if@single[3]{% \setbox0\hbox{${\mathaccent"0362{#1}}^H$}% \setbox2\hbox{${\mathaccent"0362{\kern0pt#1}}^H$}% \ifdim\ht0=\ht2 #3\else #2\fi } %The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath: \newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna} %If there's a superscript following the bar, then no negative kern may follow the bar; %an additional {} makes sure that the superscript is high enough in this case: \newcommand*\widebar[1]{\@ifnextchar^{{\wide@bar{#1}{0}}}{\wide@bar{#1}{1}}} %Use a separate algorithm for single symbols: \newcommand*\wide@bar[2]{\if@single{#1}{\wide@bar@{#1}{#2}{1}}{\wide@bar@{#1}{#2}{2}}} \newcommand*\wide@bar@[3]{% \begingroup \def\mathaccent##1##2{% %Enable nesting of accents: \let\mathaccent\save@mathaccent %If there's more than a single symbol, use the first character instead (see below): \if#32 \let\macc@nucleus\first@char \fi %Determine the italic correction: \setbox\z@\hbox{$\macc@style{\macc@nucleus}_{}$}% \setbox\tw@\hbox{$\macc@style{\macc@nucleus}{}_{}$}% \dimen@\wd\tw@ \advance\dimen@-\wd\z@ %Now \dimen@ is the italic correction of the symbol. \divide\dimen@ 3 \@tempdima\wd\tw@ \advance\@tempdima-\scriptspace %Now \@tempdima is the width of the symbol. \divide\@tempdima 10 \advance\dimen@-\@tempdima %Now \dimen@ = (italic correction / 3) - (Breite / 10) \ifdim\dimen@>\z@ \dimen@0pt\fi %The bar will be shortened in the case \dimen@<0 ! 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The subspaces $\clr (T)=\{Tx:x\in \cle\}$ and $\cln (T)=\{x\in \cle:Tx=0\}$ denotes the range space and the null space, respectively. A linear operator $T\in \clb(\cle)$ is said to be bounded below if there exists a constant $M\geq 0$ such that $\left\|Tx\right\|\geq M\left\|x\right\|$ for every $x\in \cle$, $T\in \clb(\cle)$ is said to normal if $T^*T=TT^*$, hyponormal if $TT^*\leq T^*T$, and isometry if $T^*T=I$. Let $\D$ be the open unit disc in $\mathbb{C}$, and let $\cle$ be a Hilbert space. The $\cle$-valued \emph{Hardy space} $H^2_\cle(\D)$ ($H^2(\D)$ if $\cle = \mathbb{C}$ ) over $\D$ is the Hilbert space of all $\cle$-valued analytic functions $f(z) = \sum_{n=0}^{\infty} a_n z^n$, $a_n \in \cle$, $z\in \D$ such that \[ \|f\| = \left(\sum_{n=0}^{\infty} \|a_n\|^2\right)^{\frac{1}{2}} < \infty. \] \looseness-1 Let $\cle_*$ be another Hilbert space. Then denote by $H^\infty_{\clb(\cle, \cle_*)}(\D)$ ($H^\infty_{\clb(\cle)}(\D)$ if $\cle_* = \cle$) the set of $\clb( \cle,\cle_*)$-valued bounded analytic functions on $\D$ which is a Banach space with the norm defined by \[\left\|\Theta\right\|_\infty=\sup\{\left\|\Theta(z)\right\|:|z|<1\}.\] An operator-valued analytic function $\Theta \in H^\infty_{\clb(\cle, \cle_*)}(\D)$ is said to be inner if $\Theta(z)^*\Theta(z)=I_\cle$ a.e. $z\in \T$. \subsection{Toeplitz and Hankel Operators} In this section we give definitions of Toeplitz and Hankel operators and their basic properties. Let $\cle$ be any Hilbert space. Let $L^2_\cle(\T)$, where $\T$ denotes the unit circle in the complex plane, denote the Hilbert space of all square $\cle$-valued Lebesgue integrable functions on $\T$, that is \[ L^2_\cle(\T) = \left \{ f:\T \to \cle \text{ measurable}: \|f\|^2= \int_{\T} \|f(z)\|^2_\cle\, \dd m(z) <\infty \right \}, \] where $m$ represent the normalized Lebesgue measure on $\T$. The Hardy space $H^2_\cle(\D)$ can also be identified (via radial limits) with a subspace of $\cle$-valued functions in $L^2_\cle(\T)$, which we will also denote by $H^2_\cle(\D)$. This subspace consists of functions $f$ for which $\hat{f}(n) = 0$ for all $n < 0$, where $\hat{f}(n)$ denotes the $n$-th Fourier coefficient of $f$. Given another Hilbert space $\cle_*$, we denote by $L^\infty_{\clb(\cle,\cle_*)}(\T)$ ( $L^\infty_{\clb(\cle)}(\T)$ if $\cle_* = \cle$) the set of $\clb(\cle,\cle_*)$-valued bounded functions on $\T$. Let $ \cle_*$ and $\cle$ be Hilbert spaces. For $\Phi \in L^\infty_{\clb(\cle,\cle_*)}(\T)$, the \emph{Laurent operator} $L_\Phi : L^2_{\cle}(\T) \to L^2_{\cle_*} (\T)$ is defined by $(L_\Phi f)(z) = \Phi(z) f(z)$, $z \in \T$. In this case, $L_\Phi$ is bounded and $\|L_\Phi\| = \|\Phi\|_{\infty}$. The \emph{Toeplitz operator} $T_\Phi : H^2_{\cle} (\D) \to H^2_{\cle_*} (\D)$ with (operator-valued) symbol $\Phi$ is defined by \[ T_\Phi = P_{H^2_{\cle_*} (\D)}L_\Phi|_{H^2_{\cle} (\D)}, \] where $P_{H^2_{\cle_*} (\D)}$ (in short $P$) is the orthogonal projection of $L^2_{\cle_*}(\T)$ onto $H^2_{\cle_*} (\D)$. It is well known that $\|T_\Phi\|=\|\Phi\|_\infty$ (cf.~\cite[Theorem 1.7, p.~112]{Hari}). The Toeplitz operator $T_{\Phi}=A$ is characterized by the operator equation \[T_{z}^{\ast}AT_{z}=A.\] Let $\cle$ be a Hilbert space, and let $\Phi \in H^\infty_{\clb(\cle)}(\D)$. The \emph{analytic Toeplitz operator} $T_\Phi:H^2_\cle(\D) \to H^2_\cle(\D)$ with symbol $\Phi$ is defined by \[ (T_\Phi f)(z) = \Phi(z) f(z) \qquad (f\in H^2_\cle(\D), z\in \D). \] It is known that $\|T_\Phi\| = \|\Phi\|_\infty$, and $T_\Phi$ is an isometry if and only if $\Phi$ is inner~\cite[Proposition 2.2]{NF}. Let $A\in \clb(H^2_\cle(\D))$. Then $AT_z=T_z A$ if and only if $A$ is an \emph{analytic Toeplitz operator} that is, $A=T_{\Phi}$ for some $\Phi \in H^\infty_{\clb(\cle)}(\D)$. Let $J$ be defined on $L_{\cle}^{2}(\T)$ by \[ Jf(z)=\bar{z}f(\bar{z}),f \in L_{\cle}^{2}(\T). \] The $J$ maps $\bar{zH_{\cle}^{2}(\D)}$ onto $H_{\cle}^{2}(\D)$, and $J$ maps $H_{\cle}^{2}(\D)$ onto $\bar{zH_{\cle}^{2}(\D)}.$ This $J$ is a unitary operator with the following properties: \[ \postdisplaypenalty1000000 J^{\ast}=J,J^{2}=I,\text{ }JM_{z}^{\ast}=M_{z}J,\text{ }JQ=PJ,\text{ and }JP=QJ, \] where $Q:=I-P$ the projection from $L_{\cle}^{2}(\T)$ to $\bar{zH_{\cle}^{2}(\D)}:=L_{\cle}^{2}(\T)\ominus H_{\cle}^{2}(\D).$ The Hankel operator $H_{\Phi}$ from $H_{\cle}^{2}(\D)$ into $H_{\cle_*}^{2}(\D)$ is defined by \[ H_{\Phi}h=JQ\left(\Phi h\right) =PJ\left( \Phi h\right) , \ \ h \in H_{\cle}^{2}(\D). \] The Hankel operator $H_{\Phi}=A$ is characterized by the operator equation \begin{equation}\label{Hankel operator equation} AT_{z}=T^*_{z}A. \end{equation} It is easy to verify that $H_{\Phi}^{\ast}% =H_{\Phi(\bar{z})^{\ast}}.$ The Toeplitz and Hankel operators are related by the following equation \begin{equation*} T_{\Phi\Psi}-T_\Phi T_\Psi=H^*_{\Phi^*}H_\Psi\quad\text{equivalently,} \quad T_{\Check{\Phi}\Tilde{\Psi}}-T_{\Check{\Phi}} T_{\Tilde{\Psi}}=H_{\Phi}H^*_\Psi, \end{equation*} where $\Check{\Phi}(z)=\Phi(\bar{z})$ and $\Tilde{\Phi}(z)=\Phi(\bar{z})^*$. \begin{theorem}[{\cite[p.~69]{NF}}]\label{CLT} Let $T\in \clb(\clh)$ and $T^{\prime}\in \clb(\clh^{\prime})$ be two contractions. \begin{enumerate}[label=(\arabic*)] \item\label{CLT1} Let $U\in \clb(\clk)$ and $U^{\prime}\in \clb(\clk^{\prime})$ be the minimal isometric dilation of $T$ and $T^{\prime},$ respectively. Then for any operator $W\in \clb(\clh,\clh^{\prime})$ satisfying $WT=T^{\prime}W,$ there exists $W_{1}\in \clb(\clk,\clk^{\prime})$ such that $W_{1}U=U^{\prime}W_{1},$ $\left\Vert W_{1}\right\Vert =\left\Vert W\right\Vert ,$ $W=P_{\clh^{\prime}}\restr{W_{1}}{\clh}$ and $W(\clk\ominus \clh)\subset \clk^{\prime }\ominus \clk^{\prime}.$ \item\label{CLT2} Let $U\in \clb(\clk)$ and $U^{\prime}\in \clb(\clk^{\prime})$ be the co-isometric extension of $T$ and $T^{\prime},$ respectively. Then for any operator $W\in \clb(\clh,\clh^{\prime})$ satisfying $WT=T^{\prime}W,$ there exists $W_{1}\in \clb(\clk,\clk^{\prime})$ such that $W_{1}U=U^{\prime}W_{1},$ $\left\Vert W_{1}\right\Vert =\left\Vert W\right\Vert ,$ $W=\restr{W_{1}}{\clh}$. \end{enumerate} \end{theorem} \begin{theorem}[{Douglas Theorem~\cite[Theorem 1]{Douglas_Lemma}}]\label{DL} If $A, B \in \clb(\clh)$, then the following statements are equivalent: \begin{enumerate}[label=(\arabic*)] \item\label{theo2_1} $\clr(A) \subseteq \clr (B)$; \item\label{theo2_2} $AA^* \leq \lambda^2 BB^*$ for some $\lambda\geq 0$; \item\label{theo2_3} $A=BC$ for some $C \in \clb(\clh)$. \end{enumerate} Moreover, if~\ref{theo2_1},~\ref{theo2_2}, and~\ref{theo2_3} are valid, then there exists a unique operator $C$ so that \begin{enumerate}[label=(\alph*)] \item $\|C\|^2=\inf\{\mu|AA^*\leq \mu BB^*\}$; \item $\cln(A) = \cln(C)$; \item $\clr(C) \subseteq \bar {\clr(B^{*})}$. \end{enumerate} Here $C=D^*$, where $D:\bar{\clr(B^*) }\to\bar{\clr(A^*) }$ defined by $D(B^*f)=A^*f$ and $D=0$ on the orthogonal complement of $\bar{\clr(B^*) }.$ \end{theorem} \begin{theorem}[{Beurling-Lax-Halmos theorem~\cite[Chapter V, Theorem 3.3]{NF}}]\label{BLH} Any $T_z$-invariant subspace of $\clm$ of $H^2_\cle (\D)$ is of the form \begin{equation*} \clm=\Theta H^2_{\clf}(\D), \end{equation*} where $\clf$ is a Hilbert space and $\Theta:\D\to\clb(\clf,\cle)$ is an inner function. \end{theorem} The main aim of this paper is to give the factorization of Hankel operators using Douglas theorem on range inclusion, factorization and majorization (cf. Theorem~\ref{DL}). We also studied the case when the range of the Toeplitz operator is included in the range of the Hankel operator (cf. Theorem~\ref{main theorem}). We have extended the results of~\cite{Lotto} in the vector-valued Hardy space setup. We conclude by discussing the hyponormality of Hankel operators. \section{Main results} We begin this section with some observations. \begin{proposition}\label{prelim results} Let $\Psi\in H^\infty_{\clb(\cle)}(\D)$ and $\Phi\in L^\infty_{\clb(\cle)}(\T)$. Then \begin{enumerate}[label=(\arabic*)] \item\label{prop4_1} $H_\Phi T_\Psi=H_{\Phi\Psi}$. \item\label{prop4_2}\label{prelim2} $T^*_\Psi H_\Phi=H_\Phi T_{\tilde{\Psi}}$. \end{enumerate} \end{proposition} \begin{proof} The proof of~\ref{prop4_1} follows directly from the definition. Let us prove~\ref{prop4_2}. \[ T^*_\Psi H_\Phi=(H^*_\Phi T_\Psi)^*=(H_{\tilde{\Phi}} T_\Psi)^*=H^*_{\tilde{\Phi}\Psi}=H_{\Phi\tilde{\Psi}} =H_\Phi T_{\tilde{\Psi}}.\qedhere \] \end{proof} \begin{remarks}\ \\[-1.5em] \begin{enumerate}[label=(\arabic*)] \item\label{rema5_1} For analytic function $\Phi \in H^\infty_{\clb(\cle)}(\D)$, the subspace $\cln (T_\Phi)$ is $T_z$-invariant, equivalently, $\bar{\clr (T^*_\Phi)}$ is $T^*_z$-invariant. By Theorem~\ref{BLH}, there exists a Hilbert space $\clf$ and an inner function $\Theta\in H^\infty_{\clb(\clf,\cle)}(\D)$ such that $\cln (T_\Phi)=T_\Theta H^2_\clf(\D)$. \item\label{rema5_2} For any $\Phi \in L^\infty_{\clb(\cle)}(\T)$, $\bar{\clr (H^*_\Phi)}$ is $T^*_z$-invariant. By Theorem~\ref{BLH}, there exists a Hilbert space $\clf$ and an inner function $\Theta\in H^\infty_{\clb(\clf,\cle)}(\D)$ such that $\cln (H_\Phi)=T_\Theta H^2_\clf(\D)$. If $\clr (H_{\Phi_1})\subseteq \clr (H_{\Phi_2})$. Then, by Theorem~\ref{DL}, there exists $C\in\clb(H^2_\cle(\D))$ such that $H_{\Phi_1}=H_{\Phi_2}C$. Therefore, for any $\Delta\in H^\infty_{\clb(\cle)}(\D)$, using Proposition~\ref{prelim results}, we have \begin{equation*} \begin{split} &C^*T^*_{\Delta}H^*_{\Phi_2}=C^*H^*_{\Phi_2}T_{\Tilde{\Delta}}=H^*_{\Phi_1}T_{\Tilde{\Delta}}=T^*_{\Delta}H^*_{\Phi_1}=T^*_{\Delta}C^* H^*_{\Phi_2}\\ &\left (C^*T^*_{\Delta}- T^*_{\Delta}C^*\right)H^*_{\Phi_2}=0. \end{split} \end{equation*} Therefore, $C^*T^*_{\Delta}- T^*_{\Delta}C^*=0$ on $\bar{\clr (H^*_{\Phi_2})}$ and $\clr (CT_{\Delta}- T_{\Delta}C)\subseteq \cln (H_{\Phi_2})=T_\Theta H^2_\clf(\D)$. \end{enumerate} \end{remarks} Let $\Theta_1,\Theta_2\in H^\infty_{\clb(\cle)}(\D)$ with $\Theta_2$ be inner such that $\clr (T_{\Theta_1})\subseteq \clr (T_{\Theta_2})$. Then, by Theorem~\ref{DL}, there exists $C\in \clb(H^2_\cle(\D) )$ such that $T_{\Theta_1}=T_{\Theta_2}C$. Next, observe that \[T_{\Theta_2}T_z C=T_z T_{\Theta_2}C=T_z T_{\Theta_1}=T_{\Theta_1} T_z=T_{\Theta_2}C T_z.\] Since $T_{\Theta_2}$ is isometry, we have $CT_z=T_z C$ which implies $C=T_{\Theta_3}$ for some $\Theta_3\in H^\infty_{\clb(\cle)}(\D)$. Thus $T_{\Theta_1}=T_{\Theta_2} T_{\Theta_3}$. Again if $\Theta_1,\Theta_2\in H^\infty_{\clb(\cle)}(\D)$ with $\Theta_1$ is isometry and $\left\|\Theta_2\right\|_\infty\leq 1$ such that $T_{\Theta_1}T^*_{\Theta_1}\leq T_{\Theta_2}T^*_{\Theta_2}$, then $T_{\Theta_1}=T_{\Theta_2} T_{\Theta_3}$ for some $\Theta_3\in H^\infty_{\clb(\cle)}(\D)$ which is inner. This fact is an easy consequence of~\cite[Proposition V.5.3]{NF}. By using the result in~\cite{Leech:IEOT}, we observe that if $\Theta_1,\Theta_2\in H^\infty_{\clb(\cle)}(\D)$ with $\left\|\Theta_i\right\|_\infty\leq 1$ for $i=1,2$ such that $T_{\Theta_1}T^*_{\Theta_1}\leq T_{\Theta_2}T^*_{\Theta_2}$, then $T_{\Theta_1}=T_{\Theta_2} T_{\Theta_3}$ for some $\Theta_3\in H^\infty_{\clb(\cle)}(\D)$ with $\left\|\Theta_3\right\|_\infty\leq 1$. In contrast to the aforementioned observations, we prove the following theorem, which also generalizes the factorization of contractive analytic functions~\cite[p.~205]{NF}. Although this result may be known to experts, we provide a different proof using Douglas theorem (cf. Theorem~\ref{DL}). \begin{theorem} Let $\Theta_1,\Theta_2\in H^\infty_{\clb(\cle)}(\D)$. Then $\clr(T_{\Theta_1})\subseteq \clr(T_{\Theta_2})$ if and only if $\Theta_1=\Theta_2\Theta_3$ for some $\Theta_3\in H^\infty_{\clb(\cle)}(\D)$. \end{theorem} \begin{proof} Suppose that $\clr(T_{\Theta_1})\subseteq \clr(T_{\Theta_2})$. Then, by Theorem~\ref{DL}, there exists an operator $W:\bar{\clr(T^*_{\Theta_2})}\to \bar{\clr(T^*_{\Theta_1})}$ such that $T^*_{\Theta_1}=WT^*_{\Theta_2}$. Since $T_{\Theta_i}T_z=T_z T_{\Theta_i}$, and $\bar{\clr (T^*_{\Theta_{i}})}$ is $T^*_z$-invariant for $i=1,2$, we have \begin{equation*} \begin{aligned} WT^*_z T^*_{\Theta_2}=W T^*_{\Theta_2} T^*_z&=T^*_{\Theta_1} T^*_z=T^*_zT^*_{\Theta_1} =T^*_z WT^*_{\Theta_2} \\ W\left( \restr{T^*_z}{\bar{\clr (T^*_{\Theta_{2}})}} \right)&=\left( \restr{T^*_z}{\bar{\clr (T^*_{\Theta_{1}})}} \right)W. \end{aligned} \end{equation*} Therefore, by using~\ref{CLT2} of Theorem~\ref{CLT}, there exists $W_1$ such that $W_1 T^*_z=T^*_z W_1$ and $W=\restr{W_1}{\bar{\clr (T^*_{\Theta_{2}})}}$. Therefore, we get $W^*_1$ is an analytic Toeplitz operator $T_{\Theta_3}$. Thus $T_{\Theta_1}=T_{\Theta_2}T_{\Theta_3}$ which implies $\Theta_1=\Theta_2\Theta_3$. The converse trivially follows from the Theorem~\ref{DL}. \end{proof} \begin{theorem}\label{main theorem} Let $\Phi,\Phi_1,\Phi_2,\Psi \in L^\infty_{\clb(\cle)}(\T)$. Then the following are true: \begin{enumerate}[label=(\arabic*)] \item\label{Range1} $\clr (T_\Phi)\subseteq \clr (H_\Psi)$ if and only if $\Phi=0$. \item\label{Range2}Suppose that $\Phi$ is not identically zero, then $\clr (H^*_\Phi)\subseteq \clr (T^*_\Phi)$ if and only if $P$ is bounded below on $\overline{ \Phi H^2_\cle(\D)}$. \item\label{Hankel Factorization} $\clr (H_{\Phi_1})\subseteq \clr (H_{\Phi_2})$ if and only if there exists $\Omega\in H^\infty_{\clb(\cle)}$ such that $H_{\Phi_1}=H_{\Phi_2}T_\Omega=H_{\Phi_2\Omega}$. \end{enumerate} \end{theorem} \goodbreak \begin{proof}\ \\[-2em] \begin{proof}[\ref{Range1}] Suppose that $\clr (T_\Phi)\subseteq \clr (H_\Psi)$. Then, for some $\lambda\geq 0$ \begin{equation*} \begin{split} T_\Phi T^*_\Phi &\leq \lambda^2 H_\Psi H^*_\Psi\\ T_\Phi T^*_\Phi &\leq \lambda^2 (T_{\Check{\Psi}\Tilde{\Psi}}-T_{\Check{\Psi}} T_{\Tilde{\Psi}}). \end{split} \end{equation*} Thus for every $f\in H^2_\cle(\D)$, we have \begin{equation*} \begin{split} \left\|T_{\Phi^*}f\right\|^2+\lambda^2\left\|T_{\Tilde{\Psi}}f\right\|^2&\leq \lambda^2\left\|\Tilde{\Psi}f\right\|^2\\ \left\|P(\Phi^*f)\right\|^2+\lambda^2\left\|P(\Tilde{\Psi}f)\right\|^2&\leq \lambda^2\left\|\Tilde{\Psi}f\right\|^2. \end{split} \end{equation*} Since $\{z^m\eta:m\in \N,\, \eta\in\cle \}$ is total in $H^2_\cle(\D)$, we have \begin{equation*} \left\|P(\Phi^* z^m\eta)\right\|^2+\lambda^2\left\|P(\Tilde{\Psi} z^m\eta)\right\|^2\leq \lambda^2\left\|\Tilde{\Psi} (z^m\eta)\right\|^2. \end{equation*} As $m\to\infty$, from above expression, we conclude that \begin{equation*} \begin{split} \left\|\Phi^*\right\|^2+\lambda^2\left\|\Tilde{\Psi} \right\|^2&\leq \lambda^2\left\|\Tilde{\Psi} \right\|^2, \end{split} \end{equation*} which implies $\Phi=0$. The converse is evident. \let\qed\relax \end{proof} \begin{proof}[\ref{Range2}] Suppose that $\clr (H^*_\Phi)\subseteq \clr (T^*_\Phi)$. Then, by Theorem~\ref{DL}, there exists $\lambda\geq 0$ such that $ H^*_\Phi H_\Phi\leq \lambda^2 T^*_\Phi T_\Phi $, which implies \begin{equation*} T_{\Phi^* \Phi}-T_{\Phi^*} T_{\Phi}\leq \lambda^2 T^*_\Phi T_\Phi. \end{equation*} Therefore, for every $f\in H^2_{\cle}(\D)$, we have \begin{equation*} \begin{split} \left\|\Phi f\right\|^2-\left\|T_\Phi f\right\|^2&\leq \lambda^2 \left\|T_\Phi f\right\|^2\\ \left\|P(\Phi f)\right\|^2&\geq (1+\lambda^2)^{-1}\left\|\Phi f\right\|^2. \end{split} \end{equation*} Which shows that the orthogonal projection operator $P$ is bounded below on $\overline{ \Phi H^2_\cle(\D)}$. Conversely, if $P$ is bounded below on $\overline{ \Phi H^2_\cle (\D)}$, then there exists a constant $M\geq 0$ such that $\left\|P(\Phi f)\right\|^2 \geq M\left\|\Phi f\right\|^2$ for every $f\in H^2_\cle (\D)$. Therefore, \[ \left\|P(\Phi (\eta z^n))\right\|^2 \geq M\left\|\Phi (\eta z^n)\right\|^2. \] As $n\to\infty$, we have $M\leq 1$. Then choose $M=(1+\mu^2)^{-1}$ for some $\mu\geq 0$, and apply the above argument in the necessity part to obtain the desired conclusion. \let\qed\relax \end{proof} \begin{proof}[\ref{Hankel Factorization}] First, we assume that $\clr (H_{\Phi_1})\subseteq \clr (H_{\Phi_2})$. Then, by Theorem~\ref{DL}, we have \begin{equation*} H_{\Phi_1}H^*_{\Phi_1}\leq \lambda^2 H_{\Phi_2}H^*_{\Phi_2},\quad H_{\Phi_1}=H_{\Phi_2}C \end{equation*} such that $\cln (C)=\cln (H_{\Phi_1})$, $\|C\|\leq \lambda^2$, and $\clr (C)\subseteq \bar{\clr (H^*_{\Phi_2})}$. Note that $C=W^*$, where $W:\bar{\clr (H^*_{\Phi_2})}\to \bar{\clr (H^*_{\Phi_1})}$ is defined by $W(H^*_{\Phi_2}f)=H^*_{\Phi_1}f$ and $W=0$ on $\clr (H^*_{\Phi_2})^\perp$. Since $T_z^*H_{\Phi_2}=H_{\Phi_2}T_z$, we have $\cln (H_{\Phi_2})$ is $T_z$-invariant, equivalently $\bar{\clr (H^*_{\Phi_2})}$ is $T^*_z$-invariant. Therefore, by using~\eqref{Hankel operator equation}, we get \begin{equation*} WT^*_zH^*_{\Phi_2}=WH^*_{\Phi_2}T_z=H^*_{\Phi_1}T_z=T^*_zH^*_{\Phi_1}=T^*_zWH^*_{\Phi_2} \end{equation*} which implies \begin{equation*} W\left(\restr{T^*_z}{\bar{\clr (H^*_{\Phi_2})}} \right)=\left(\restr{T^*_z}{\bar{\clr (H^*_{\Phi_1})}} \right)W. \end{equation*} Using~\ref{CLT2} of Theorem~\ref{CLT}, there exists $W_1:H^2_\cle(\D)\to H^2_\cle(\D)$ such that $W_1T^*_z=T^*_z W_1$ and $W=\restr{W_1}{\bar{\clr (H^*_{\Phi_2})}}$. Thus $W^*_1$ is an analytic Toeplitz operator $T_\Omega$ and $H_{\Phi_1}=H_{\Phi_2}W^*=H_{\Phi_2}T_\Omega$. The converse is obvious. \end{proof} \let\qed\relax \end{proof} \begin{corollary}\label{hyponormal condition} Let $\Phi\in L^\infty_{\clb(\cle)}(\T)$. Then $H_\Phi$ is hyponormal if and only if $H_\Phi=H^*_\Phi T_\Omega$ for some $\Omega\in H^\infty_{\clb(\cle)}(\D)$ with $\left\|\Omega\right\|_\infty\leq 1$. \end{corollary} \begin{proof} Suppose that $H_\Phi$ is hyponormal that is, $H_\Phi H^*_\Phi\leq H^*_\Phi H_\Phi$ equivalently, $H_\Phi H^*_\Phi\leq H_{\tilde{\Phi}}H^*_{\tilde{\Phi}}$. Now set $\Phi_1=\Phi$ and $\Phi_2=\tilde{\Phi}$. By applying~\ref{Hankel Factorization} of Theorem~\ref{main theorem}, we have $H_\Phi=H_{\tilde{\Phi}} T_\Omega =H^*_\Phi T_\Omega$ for some $\Omega\in H^\infty_{\clb(\cle)}(\D)$ with $\left\|\Omega\right\|_\infty\leq 1$. Conversely, suppose $H_\Phi=H^*_\Phi T_\Omega$ for some $\Omega\in H^\infty_{\clb(\cle)}(\D)$ with $\left\|\Omega\right\|_\infty\leq 1$. Then $\left\|T_\Omega\right\|=\left\|\Omega\right\|_\infty\leq 1$. Therefore, \[ H^*_\Phi H_\Phi-H_\Phi H^*_\Phi= H^*_\Phi(I-T_\Omega T^*_\Omega)H_\Phi\geq 0. \] Hence $H_\Phi$ is hyponormal. \end{proof} \begin{corollary} Let $\Phi\in L^\infty_{\clb(\cle)}(\T)$. If $H_\Phi$ is hyponormal, then it is normal. \end{corollary} \begin{proof} Suppose $H_\Phi$ is hyponormal. By Corollary~\ref{hyponormal condition} and~\ref{prelim2} of Proposition~\ref{prelim results}, we observe that \[ H_{\tilde{\Phi}}=H^*_{\Phi}=T^*_\Omega H_{\Phi}=H_\Phi T_{\tilde{\Omega}}=(H^*_\Phi)^* T_{\tilde{\Omega}}=H^*_{\tilde{\Phi}}T_{\tilde{\Omega}}. \] \looseness1 Which shows that if $H_\Phi$ is hyponormal, then $H^*_\Phi$ is also hyponormal, and consequently,\break normal. \end{proof} \section*{Declaration of interests} The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations. \printbibliography \end{document}