The study of spectral properties of Toeplitz and Hankel operators acting on the Bergman space is a difficult topic. In recent years, some progress has been made in understanding spectral properties of Toeplitz operators. We highlight the result by Sundberg and Zheng [Reference Sundberg and ZhengSZ10] who have proved that the spectra and essential spectra of Toeplitz operators on the unit disc
$\mathbb {D}$
need not be connected. Their result shows a sharp difference with the spectra of Toeplitz operators on the Hardy space. As is well-known, Widom [Reference WidomWid64] showed that for any
$L^{\infty }$
symbol
$\psi $
on the unit circle, the spectrum of the Toeplitz operator
$T_{\psi }$
is a connected subset of
$\mathbb {C}$
. Similarly Douglas [Reference DouglasDou98, Theorem 7.45] proved that the essential spectrum of
$T_{\psi }$
is also connected. In this context, we also mention papers [Reference Zhao and ZhengZZ16, Reference Guo, Zhao and ZhengGZZ23] which study spectra of certain classes of the Bergman Toeplitz operators on the unit disk. On the other hand, we are not aware of much work done about the spectral properties of Hankel operators.
Since the Hankel operator
$H_{\psi }$
does not map the Bergman space into itself, we will consider the Hermitian square
$H_{\psi }^* H_{\psi }$
and we will obtain some initial results about the spectrum. In this paper, we will only be concerned with symbols that are continuous up to the boundary of the domain. In the case of the unit disc,
$H_{\psi }$
is compact for
$\psi \in C(\overline {\mathbb {D}})$
(see, for instance, [Reference Axler, Conway and McDonaldACM82, Proposition 8]); hence the spectrum of
$H_{\psi }^* H_{\psi }$
is discrete. Like the situation on the unit disc, on bounded strongly pseudoconvex domains in
$\mathbb {C}^n$
,
$H_{\psi }$
is compact [Reference Čučković and ŞahutoğluČŞ09]. Then on such domains the spectrum of
$H_{\psi }^* H_{\psi }$
is discrete as well. In this paper, we focus on the polydisc and find some sufficient conditions in terms of behavior
$\psi $
on the boundary so that the spectrum of
$H_{\psi }^* H_{\psi }$
contains intervals. One of the reasons for this departure from the one-dimensional and strongly pseudoconvex case is the fact that Hankel operators with symbols continuous on
$\overline {\mathbb {D}^2}$
may not be compact (see [Reference Clos and ŞahutoğluCŞ18a, Reference LeLe10]). We note that, in case the symbol is smooth on the closure the same result was proven in [Reference Čučković and ŞahutoğluČŞ09]. With regard to Sundberg–Zheng result, from the same papers mentioned in this paragraph, we know that if the symbol
$\psi $
is holomorphic along any disc in the boundary of a convex domain, then
$H_{\psi }$
is compact [Reference Čučković and ŞahutoğluČŞ09, Reference Çelik, Şahutoğlu and StraubeÇŞS23, Reference ZimmerZim23]; hence the spectrum of
$H_{\psi }^* H_{\psi }$
is disconnected.
At this point we would like to mention the well known formula connecting Hankel and Toeplitz operators
$$\begin{align*}H_{\psi}^* H_{\psi} = T_{|\psi|^2} - T_{\overline{\psi}} T_{\psi}.\end{align*}$$
Thus our results could shed a new light on the spectra of semicommutators of Toeplitz operators. For more information about Hankel and Toeplitz operators on the unit disc, we refer the reader to a standard reference [Reference ZhuZhu07].
In the rest of the paper, we prove two sufficient conditions for the spectrum of
$H^*_{\psi } H_{\psi }$
to contain intervals. We also compute the spectrum of
$H^*_{\psi } H_{\psi }$
in the case
$\psi $
is monomial.
1 Main results
Let
$\Omega $
be a bounded domain in
$\mathbb {C}^n$
and
$A^2(\Omega )$
denote the Bergman space, the set of square integrable holomorphic functions on
$\Omega $
. We denote the Bergman projection by
$P^{\Omega }$
. Then for a bounded measurable function
$\psi $
on
$\Omega $
, Hankel operator
$H^{\Omega }_\psi $
on
$A^2(\Omega )$
with symbol
$\psi $
is defined as
$$\begin{align*}H^{\Omega}_{\psi} f=(I-P^{\Omega})(\psi f),\end{align*}$$
for
$f\in A^2(\Omega ).$
Here I denotes the identity operator. For simplicity, we will simply write
$H_\psi $
when there is no confusion about the domain. We note that the Toeplitz operator
$T_{\psi }$
is defined as
$$\begin{align*}T_{\psi}f=P(\psi f),\end{align*}$$
for
$f\in A^2(\Omega )$
.
The spectrum for
$H^*_{\psi }H_{\psi }$
, for symbols continuous up to the boundary, is a discrete set for a large class of domains on which the operator is compact [Reference Čučković and ŞahutoğluČŞ09]. So it would be interesting to know the sufficient conditions for the spectrum to contain intervals.
Let
$\sigma (T)$
denote the spectrum of a linear map T. The set of eigenvalues is called the point spectrum
$\sigma _p(T)$
[Reference ConwayCon90, VII Definition 6.2]. The discrete spectrum,
$\sigma _d(T)$
is composed of eigenvalues with finite (algebraic) multiplicity that are isolated points of
$\sigma (T)$
. Finally, the essential spectrum
$\sigma _e(T)$
is defined as
$\sigma _e(T)=\sigma (T)\setminus \sigma _d(T)$
. A characterization of the essential spectrum in our set-up, called Weyl’s Criterion, is presented in Theorem A below. We note that for self-adjoint operators
$T_1$
and
$T_2$
we have
$\sigma _e(T_1)=\sigma _e(T_2)$
if
$T_1-T_2$
is compact (see, for instance, [Reference Hislop and SigalHS96, Theorems 5.10 and 7.2] or [Reference ConwayCon90, XI Proposition 4.2]).
Our first result is about the essential spectrum of
$H^*_{\psi }H_{\psi }$
on
$A^2(\mathbb {D}^n)$
when
$\psi $
is a product of two functions that depend on different variables. We note that,
$H^{\mathbb {D}^{n-1*}}_{\varphi }$
in the theorem below denotes the adjoint of the operator
$H^{\mathbb {D}^{n-1}}_{\varphi }$
.
Theorem 1.1 Let
$\varphi \in C(\overline {\mathbb {D}^{n-1}}),\chi \in C(\overline {\mathbb {D}})$
and
$\psi (z',z_n)=\varphi (z')\chi (z_n)$
for
$z'\in \overline {\mathbb {D}^{n-1}},$
and
$z_n\in \overline {\mathbb {D}}$
. Then the essential spectrum of
$H^*_{\psi }H_{\psi }$
contains the set
$$\begin{align*}\left\{|\chi(q)|^2\mu:q\in b\mathbb{D}, \mu\in \sigma(H^{\mathbb{D}^{n-1*}}_{\varphi}H^{\mathbb{D}^{n-1}}_{\varphi})\right\}.\end{align*}$$
In case the symbol is separable we have the following corollary, proof of which will be presented in Section 2.
Corollary 1.2 Let
$\chi _j\in C(\overline {\mathbb {D}})$
for
$j=1,\ldots , n$
and
$\psi (z_1,\ldots ,z_n)=\chi _1(z_1)\chi _2(z_2)\cdots \chi _n(z_n)$
. Then the essential spectrum of
$H^*_{\psi }H_{\psi }$
contains the set
$$\begin{align*}\bigcup_{j=1}^n\left\{\mu\prod_{k\neq j}|\chi_k(q_k)|^2:q_k\in b\mathbb{D}, \mu\in \sigma(H^{\mathbb{D}^*}_{\chi_j}H^{\mathbb{D}}_{\chi_j})\right\}.\end{align*}$$
In the next theorem, we give a sufficient condition for the spectrum to contain an interval for more general symbols than the ones in Theorem 1.1.
Theorem 1.3 Let
$\psi \in C(\overline {\mathbb {D}^n}), 1\leq k\leq n-1,$
and
$\psi _q(z')=\psi (z',q)$
for
$z'\in \mathbb {D}^{n-1}$
and
$q\in b\mathbb {D}$
. Assume that
$q\to \| H^{\mathbb {D}^{n-1}}_{\psi _q}\|$
is non-constant. Then the essential spectrum of
$H^*_{\psi }H_{\psi }$
contains an open interval.
The following corollary can be proved using Theorems 1.1 or 1.3. We will mention both in Section 2.
Corollary 1.4 Let
$\varphi \in C(\overline {\mathbb {D}^{n-1}})$
and
$\chi \in C(\overline {\mathbb {D}})$
such that
$|\chi |$
is not constant on the unit circle and
$\varphi $
is not holomorphic. Then the essential spectrum of
$H^*_{\psi }H_{\psi }$
contains an open interval where
$\psi (z',z_n)=\varphi (z')\chi (z_n)$
for
$z'\in \overline {\mathbb {D}^{n-1}}$
and
$z_n\in \overline {\mathbb {D}}$
.
Example 1.5 Let
$\psi (z_1,z_2)=\overline {z}_1(\overline {z}_2+1)$
. Since
$\overline {z}_1$
is not holomorphic and
$|\overline {z}_2+1|$
is not constant on the unit circle, Corollary 1.4 implies that
$\sigma _e(H^*_{\psi }H_{\psi })$
contains an interval. It is worth noting that (see Theorem 1.6 below) the spectrum of the Hermitian square of a Hankel operator with a monomial symbol is a discrete set; yet for the a quadratic polynomial
$\overline {z}_1(\overline {z}_2+1)$
the spectrum contains an interval.
In Theorem 1.6 below, we give a complete characterization of the spectrum on the polydisc for monomial symbols.
We note that
$\mathbb {N}_0$
denotes the set of non-negative integers,
$\{0,1,2,\ldots \}$
. For
$\mathbf {m},\mathbf {n},\boldsymbol {\alpha }\in \mathbb {N}^n_0$
and
$\emptyset \neq B\subset B_n=\{1,2,3,\ldots , n\}$
we define
$$ \begin{align*} \lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}= \prod_{k\in B}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1}, \end{align*} $$
whenever
$ \alpha _k< m_k-n_k$
for some
$k\in B$
and
$$ \begin{align*} \lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}= \prod_{k\in B}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1} -\prod_{k\in B}\frac{(\alpha_k+1)(\alpha_k+n_k-m_k+1)}{(\alpha_k+n_k+1)^2}, \end{align*} $$
whenever
$\alpha _k\geq m_k-n_k$
for all
$k\in B$
.
Theorem 1.6 Let
$\psi (z)=z^{\mathbf {n}}\overline {z}^{\mathbf {m}}$
for some
$\mathbf {m},\mathbf {n}\in \mathbb {N}^n_0$
. Then
$H^*_{\psi }H_{\psi }$
on
$A^2(\mathbb {D}^n)$
has the following spectrum
$$\begin{align*}\sigma(H^*_{\psi}H_{\psi})=\{0\}\cup \{\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}: \boldsymbol{\alpha}\in \mathbb{N}_0^n, \emptyset\neq B\subset B_n\}.\end{align*}$$
Furthermore, if
$n_k+m_k\geq 1$
for all
$k\in B_n$
and
$m_k\geq 1$
for some
$k\in B_n$
then all of the eigenvalues have finite multiplicities. On the other hand, if
$n_k+m_k=0$
for some
$k\in B_n$
then all of the eigenvalues have infinite multiplicities.
Since the spectrum of
$H^*_{\psi }H_{\psi }$
on
$A^2(\mathbb {D}^n)$
for
$\psi (z)=z^{\mathbf {n}}\overline {z}^{\mathbf {m}}$
contains countably many points it has empty interior.
Corollary 1.7 Let
$\psi (z)=z^{\mathbf {n}}\overline {z}^{\mathbf {m}}$
for some
$\mathbf {m},\mathbf {n}\in \mathbb {N}^n_0$
. Then the spectrum of
$H^*_{\psi }H_{\psi }$
on
$A^2(\mathbb {D}^n)$
has empty interior.
In case the symbol is a pure conjugate holomorphic monomial in
$\mathbb {C}^2$
, we have the following corollary.
Corollary 1.8 Let
$\psi (z_1,z_2)=\overline {z}_1^n\overline {z}_2^m$
on
$\mathbb {D}^2$
for some positive integers
$n,m$
. Then
$$ \begin{align*} \sigma(H^*_{\psi}H_{\psi})=\{0\}\cup&\bigcup_{\alpha_1=0}^{n-1} \bigcup_{\alpha_2=0}^{m-1}\left\{\frac{\alpha_1+1}{\alpha_1+n+1}, \frac{\alpha_2+1}{\alpha_2+m+1}\right\} \\ &\bigcup_{\alpha_1=n}^{\infty}\bigcup_{\alpha_2=m}^{\infty} \left\{\frac{n^2}{(\alpha_1+n+1)(\alpha_1+1)},\frac{m^2}{(\alpha_2+m+1)(\alpha_2+1)}\right\}\\ &\bigcup_{\alpha_1=n}^{\infty}\bigcup_{\alpha_2=m}^{\infty} \left\{\frac{n^2(\alpha_2+1)^2+m^2(\alpha_1+1)^2-n^2m^2}{(\alpha_1+n+1)(\alpha_2+m+1) (\alpha_1+1)(\alpha_2+1)}\right\}\\ &\bigcup_{\alpha_1=0}^{n-1}\bigcup_{\alpha_2=0}^{\infty} \left\{\frac{(\alpha_1+1)(\alpha_2+1)}{(\alpha_1+n+1)(\alpha_2+m+1)}\right\} \\ & \bigcup_{\alpha_1=0}^{\infty}\bigcup_{\alpha_2=0}^{m-1} \left\{\frac{(\alpha_1+1)(\alpha_2+1)}{(\alpha_1+n+1)(\alpha_2+m+1)}\right\}. \end{align*} $$
Furthermore, all of the eigenvalues are of finite multiplicity.
A precursor to the next corollary has appeared in [Reference Čučković and ŞahutoğluČŞ18b, Remark 4.2].
Corollary 1.9 Let
$\psi (z_1,z_2)=\overline {z}_1^n$
on
$\mathbb {D}^2$
for some positive integer n. Then
$$ \begin{align*} \sigma(H^*_{\psi}H_{\psi})=\{0\}\cup\bigcup_{\alpha_1=0}^{n-1} \left\{\frac{\alpha_1+1}{\alpha_1+n+1}\right\} \bigcup_{\alpha_1=n}^{\infty} \left\{\frac{n^2}{(\alpha_1+n+1)(\alpha_1+1)}\right\}. \end{align*} $$
Furthermore, all of the eigenvalues are of infinite multiplicity.
2 Proof of Theorems 1.1 and 1.3
In the proofs, below
$K^{\Omega }$
and
$k^{\Omega }_z=K^{\Omega }_z/\|K^{\Omega }_z\|$
denote the Bergman kernel and the normalized Bergman kernel of
$\Omega $
, respectively. We will drop the superscript
$\Omega $
when the domain is clear.
To prove Theorems 1.1 and 1.3, we will need the following result. We refer the reader to [Reference Hislop and SigalHS96] for a proof.
Theorem A (Weyl’s Criterion)
Let A be a bounded self-adjoint linear operator on a Hilbert space H. Then
-
i.
$\lambda $
is in the spectrum of A if and only if there exists a sequence
$\{u_n\}\subset H$
such that
$\|u_n\|=1$
for all n and
$\|(A-\lambda )u_n\|\to 0$
as
$n\to \infty $
. -
ii.
$\lambda $
is in the essential spectrum of A if and only if there exists a sequence
$\{u_n\}\subset H$
such that
$\|u_n\|=1$
for all n,
$u_n\to 0$
weakly, and
$\|(A-\lambda )u_n\|\to 0$
as
$n\to \infty $
.
The following lemma is probably well known. We include it here for completeness. We would like to thank Tomas Miguel P. Rodriguez for the proof.
Lemma 2.1 Let
$\Omega $
be a domain in
$\mathbb {C}^n$
and
$\{f_j\}$
be a sequence in
$A^2(\Omega )$
. Then
$f_j\to 0$
weakly as
$j\to \infty $
if and only if
$\{f_j\}$
is bounded in
$A^2(\Omega )$
and
$f_j\to 0$
as
$j\to \infty $
uniformly on compact subsets in
$\Omega $
.
Proof. Let us assume that
$\{f_j\}$
is bounded in
$A^2(\Omega )$
and
$f_j\to 0$
as
$j\to \infty $
uniformly on compact subsets in
$\Omega $
. Let
$f\in A^2(\Omega )$
and
$\varepsilon>0$
. Then there exists a compact set
$K\subset \Omega $
such that
$\|f\|_{L^2(\Omega \setminus K)}<\varepsilon $
. Then for large
$j,$
we have
$\sup \{|f_j(z)|:z\in K\}<\varepsilon $
and
$$ \begin{align*} |\langle f, f_j\rangle| \leq \, & \|f\|_{L^2(K)}\| f_j\|_{L^2(K)}+\|f\|_{L^2(\Omega\setminus K)} \|f_j\|_{L^2(\Omega)} \\ \leq \,& \varepsilon\left(\|f\|_{L^2(K)}\sqrt{V(K)}+\sup\{\|f_j\|_{L^2(\Omega)}:j\in \mathbb{N}\}\right), \end{align*} $$
where
$V(K)$
denotes the Lebesgue volume of K. Hence
$\langle f, f_j\rangle \to 0$
. That is,
$f_j\to 0$
weakly as
$j\to \infty $
.
For the converse, we assume that
$f_j\to 0$
weakly as
$j\to \infty $
. We define
$S_{f_j}(f)=\langle f,f_j\rangle $
for
$f\in A^2(\Omega )$
. Then
$S_{f_j}(f)\to 0$
as
$j\to \infty $
. That is
$\sup \{|S_{f_j}(f)|:j\in \mathbb {N}\}<\infty $
for all
$f\in A^2(\Omega )$
. Then by the uniform boundedness principle
$\sup \{\|S_{f_j}\|:j\in \mathbb {N}\}=\sup \{ \|f_j\|:j\in \mathbb {N}\}<\infty $
. That is
$\{f_j\}$
is bounded in
$A^2(\Omega )$
. Furthermore,
$f_j(z)=\langle f_j, K_z\rangle \to 0$
as
$j\to \infty $
for all
$z\in \Omega $
. We will use this fact to conclude that
$f_j\to 0$
uniformly on compact subsets as follows. Let K be a compact subset of
$\Omega $
and
$\{f_{j_k}\}$
be a subsequence of
$\{f_j\}$
. Then Montel’s theorem implies that there is a further subsequence
$\{f_{j_{k_l}}\}$
that is convergent to a holomorphic function f uniformly on K. However,
$f_j(z)=\langle f_j, K_z\rangle \to 0$
as
$j\to \infty $
. Then
$f=0$
. Therefore, every subsequence
$\{f_{j_k}\}$
has a further subsequence
$\{f_{j_{k_l}}\}$
converging to zero uniformly on K. Then
$f_j\to 0$
uniformly on K as
$j\to \infty $
. Since K is arbitrary we conclude that
$f_j\to 0$
uniformly on compact subsets.
Proof of Theorem 1.1
Let
$ q\in b\mathbb {D}$
. For
$p\in \mathbb {D}$
and
$j\in \mathbb {N}$
we define
$$\begin{align*}f_{j,p}(z',z_n)=g_j(z')k^{\mathbb{D}}_p(z_n),\end{align*}$$
where
$z'=(z_1,\ldots ,z_{n-1}), k^{\mathbb {D}}_p$
is the normalized kernel for
$A^2(\mathbb {D})$
centered at p and
$\{g_j\}\subset A^2(\mathbb {D}^{n-1})$
, to be determined later, such that
$\|g_j\|_{L^2(\mathbb {D}^{n-1})}=1$
. We note that
$\{g_j\}$
is uniformly bounded on compact subsets of
$\mathbb {D}^{n-1}$
. Then
$\sup \{|f_{j,p}(z)|:j\in \mathbb {N},z\in K\} \to 0$
as
$p\to q$
for any compact set
$K\subset \mathbb {D}^n$
. Hence, by Lemma 2.1, for any
$p_j\to q$
we conclude that
$f_{j,p_j}\to 0$
weakly
$j\to \infty $
.
Let us fix j and denote
$\psi _q(z',z_n)=\psi (z',q)=\chi (q)\varphi (z')$
. Since
$\psi _q$
is independent of
$z_n$
and
$$\begin{align*}P(fg)(z)=(P^{\mathbb{D}^{n-1}}f)(z')\cdot (P^{\mathbb{D}}g)(z_n),\end{align*}$$
whenever f is a function of
$z'$
and g is a function of
$z_n$
, we have
$$ \begin{align} \nonumber H_{\psi_q}f_{j,p} = \, & \psi_q g_jk^{\mathbb{D}}_p-P(\psi_qg_jk^{\mathbb{D}}_p) \\ = \,& \chi(q)\varphi g_j k^{\mathbb{D}}_p - \chi(q)\left(P^{\mathbb{D}^{n-1}}(\varphi g_j)\right)k^{\mathbb{D}}_p \\ \nonumber =\, &\chi(q)(H^{\mathbb{D}^{n-1}}_{\varphi}g_j)k^{\mathbb{D}}_p. \end{align} $$
Then
$\|H_{\psi _q}f_{j,p}\| =|\chi (q)|\|H^{\mathbb {D}^{n-1}}_{\varphi }g_j\|_{L^2(\mathbb {D}^{n-1})}.$
Now we write
$\psi =\psi -\psi _q+\psi _q$
and
$$\begin{align*}H^*_{\psi}H_{\psi}f_{j,p}=H^*_{\psi_q}H_{\psi_q}f_{j,p} +H^*_{\psi}H_{\psi-\psi_q}f_{j,p} +H^*_{\psi-\psi_q}H_{\psi_q}f_{j,p}.\end{align*}$$
Hence for
$\lambda \in \mathbb {R}$
we have
$$ \begin{align} \nonumber \left\|H^*_{\psi_q}H_{\psi_q}f_{j,p}-\lambda f_{j,p}\right\| -&\left\|H^*_{\psi}H_{\psi-\psi_q}f_{j,p} +H^*_{\psi-\psi_q}H_{\psi_q}f_{j,p} \right\| \\ \leq\,& \left\|H^*_{\psi}H_{\psi}f_{j,p}-\lambda f_{j,p}\right\| \\ \nonumber \leq\, & \left\|H^*_{\psi_q}H_{\psi_q}f_{j,p}-\lambda f_{j,p}\right\|\\ \nonumber &+\left\|H^*_{\psi}H_{\psi-\psi_q}f_{j,p} +H^*_{\psi-\psi_q}H_{\psi_q}f_{j,p} \right\|. \end{align} $$
Let
$\{h_j\}$
be a bounded sequence in
$L^2(\mathbb {D}^{n-1})$
. Then
$\|(\psi -\psi _q)h_jk^{\mathbb {D}}_p\|\to 0$
as
$p\to q$
for any fixed j. This can be seen as follows. By continuity of
$\psi $
, for
$\varepsilon>0$
there exists
$\delta>0$
such that
$$\begin{align*}|\psi(z',z_n)-\psi_q(z',z_n)|<\varepsilon \text{ for } |z_n-q|<\delta.\end{align*}$$
Then we have
$$ \begin{align*} &\left\|(\psi-\psi_q) h_jk^{\mathbb{D}}_p\right\|^2 \\ &\quad= \left\|(\psi-\psi_q)h_jk^{\mathbb{D}}_p\right\|^2_{L^2(\{(z',z_n)\in \mathbb{D}^n: |z_n-q|< \delta\})} +\left\|(\psi-\psi_q)h_jk^{\mathbb{D}}_p\right\|^2_{L^2(\{(z',z_n)\in \mathbb{D}^n: |z_n-q|\geq \delta\})}\\ &\quad\leq\, \varepsilon^2 \|h_j\|^2_{L^2(\mathbb{D}^{n-1})}\|k^{\mathbb{D}}_p\|^2_{L^2(\mathbb{D})} \\ &\qquad+ \pi \|h_j\|^2_{L^2(\mathbb{D}^{n-1})} \sup\left\{\left|(\psi(z',z_n)-\psi_q(z',z_n))k^{\mathbb{D}}_p(z_n)\right|{}^2: z'\in \mathbb{D}^{n-1}, |z_n-q|\geq \delta\right\}. \end{align*} $$
However,
$$\begin{align*}\sup\left\{\left|k^{\mathbb{D}}_p(z_n)\right|:|z_n-q|\geq \delta\right\} \to 0 \text{ as } p\to q.\end{align*}$$
Then, for any j we have
$\limsup _{p\to q}\|(\psi -\psi _q) h_jk^{\mathbb {D}}_p\|\leq \varepsilon \|h_j\|$
for all
$\varepsilon>0$
. Therefore,
$$ \begin{align} \lim_{p\to q}\left\|(\psi-\psi_q) h_jk^{\mathbb{D}}_p\right\| =0 \text{ for any } j. \end{align} $$
This fact together with
$h_j=1$
imply that for any j we have
$$\begin{align*}\left\|H^*_{\psi}H_{\psi-\psi_q}f_{j,p}\right\| = \left\|H^*_{\psi}(I-P)\Big((\psi-\psi_q)f_{j,p}\Big)\right\|\to 0 \text{ as }p \to q.\end{align*}$$
Next we will use the following fact (see [Reference Čučković and ŞahutoğluČŞ14, Lemma 1])
$$ \begin{align} H^*_u H_v = PM_{\overline{u}}(I-P)M_v \text{ for } u,v\in L^{\infty}(\Omega), \end{align} $$
where
$M_v$
denotes the multiplication operator by v.
Since
$\mathbb {D}^n$
is a product domain and
$\psi _q$
is independent of
$z_n$
, the fact above, (2.1) and (2.3) with
$h_j=H^{\mathbb {D}^{n-1}}_{\varphi }g_j$
imply that for any j we have
$$\begin{align*}\left\|H^*_{\psi-\psi_q}H_{\psi_q}f_{j,p}\right\| =|\chi(q)|\left\|P\left((H^{\mathbb{D}^{n-1}}_{\varphi}g_j)(\overline{\psi} -\overline{\psi_q})k^{\mathbb{D}}_p\right)\right\|\to 0 \text{ as } p\to q\end{align*}$$
and
$$\begin{align*}H^*_{\psi_q}H_{\psi_q}f_{j,p} = |\chi(q)|^2(H^{\mathbb{D}^{n-1*}}_{\varphi}H^{\mathbb{D}^{n-1}}_{\varphi}g_j)k^{\mathbb{D}}_p.\end{align*}$$
Then using (2.2) for any j we have
$$ \begin{align} \left\|H^*_{\psi}H_{\psi}f_{j,p}-\lambda f_{j,p}\right\| \to \left\||\chi(q)|^2H^{\mathbb{D}^{n-1*}}_{\varphi}H^{\mathbb{D}^{n-1}}_{\varphi}g_j -\lambda g_j\right\|_{L^2(\mathbb{D}^{n-1})} \text{ as } p\to q. \end{align} $$
Let
$\mu \in \sigma (H^{\mathbb {D}^{n-1*}}_{\varphi }H^{\mathbb {D}^{n-1}}_{\varphi })$
. Then, by i. in Theorem A, we choose a sequence
$\{g_j\}\subset A^2(\mathbb {D}^{n-1})$
such that
$\|g_j\|=1$
for all j and
$$\begin{align*}\left\|H^{\mathbb{D}^{n-1*}}_{\varphi}H^{\mathbb{D}^{n-1}}_{\varphi}g_j -\mu g_j\right\|_{L^2(\mathbb{D}^{n-1})}\to 0 \text{ as } j\to \infty. \end{align*}$$
Then for
$\lambda =|\chi (q)|^2\mu ,$
we have
$$\begin{align*}\left\||\chi(q)|^2H^{\mathbb{D}^{n-1*}}_{\varphi}H^{\mathbb{D}^{n-1}}_{\varphi}g_j -\lambda g_j\right\|_{L^2(\mathbb{D}^{n-1})}\to 0 \text { as } j\to\infty.\end{align*}$$
Then using (2.5) and the limit above we choose
$p_j\in \mathbb {D}$
such that
$p_j\to q$
as
$j\to \infty $
and
$$\begin{align*}\left\|H^*_{\psi}H_{\psi}f_{j,p_j}-\lambda f_{j,p_j}\right\| < \left\||\chi(q)|^2H^{\mathbb{D}^{n-1*}}_{\varphi}H^{\mathbb{D}^{n-1}}_{\varphi}g_j -\lambda g_j\right\|_{L^2(\mathbb{D}^{n-1})}+\frac{1}{j}\end{align*}$$
for all j. Hence
$$\begin{align*}\left\|H^*_{\psi}H_{\psi}f_{j,p_j}-\lambda f_{j,p_j}\right\| \to 0 \text{ as } j\to \infty.\end{align*}$$
Finally, we use ii. in Theorem A to conclude that
$\lambda =|\chi (q)|^2\mu \in \sigma _e(H^*_{\psi }H_{\psi })$
for any
$q\in b\mathbb {D}$
and
$\mu \in \sigma (H^{\mathbb {D}^{n-1*}}_{\varphi }H^{\mathbb {D}^{n-1}}_{\varphi })$
.
Proof of Corollary 1.2
Without loss of generality let
$\mu \in \sigma (H^{\mathbb {D}^*}_{\chi _1}H^{\mathbb {D}}_{\chi _1})$
. Then, by Theorem 1.1 for
$n=2$
, we have
$\mu |\chi _2(q_2)|^2 \in \sigma (H^{\mathbb {D}^{2*}}_{\chi _1\chi _2}H^{\mathbb {D}^2}_{\chi _1\chi _2})$
. Hence, inductively, we get
$$\begin{align*}\mu |\chi_2(q_2)\cdots\chi_n(q_n)|^2\in \sigma_e(H^*_{\psi}H_{\psi})\end{align*}$$
completing the proof of the corollary.
The proof of Theorem 1.3 is similar to the proof of Theorem 1.1. So we will not provide all the details but will highlight the differences and similarities below.
Proof of Theorem 1.3
Let
$q\in b\mathbb {D}$
and
$$\begin{align*}\lambda_q =\left\|H^{\mathbb{D}^{n-1*}}_{\psi_q}H^{\mathbb{D}^{n-1}}_{\psi_q}\right\|_{L^2(\mathbb{D}^{n-1})} =\left\|H^{\mathbb{D}^{n-1}}_{\psi_q}\right\|^2_{L^2(\mathbb{D}^{n-1})}.\end{align*}$$
Then
$\lambda _q\in \sigma (H^{\mathbb {D}^{n-1*}}_{\psi _q}H^{\mathbb {D}^{n-1}}_{\psi _q})$
(see, for instance, [Reference Hislop and SigalHS96, Theorem 5.14]). By i. in Theorem A there exists
$\{g_{j,q}\}\subset A^2(\mathbb {D}^{n-1})$
such that
$\|g_{j,q}\|=1$
and
$$ \begin{align} \left\|H^{\mathbb{D}^{n-1*}}_{\psi_q}H^{\mathbb{D}^{n-1}}_{\psi_q} g_{j,q} -\lambda_qg_{j,q}\right\|_{L^2(\mathbb{D}^{n-1})}\to 0 \text{ as } j\to\infty. \end{align} $$
Let
$$\begin{align*}f_{j,q,p}(z',z_n)=g_{j,q}(z')k^{\mathbb{D}}_p(z_n)\end{align*}$$
(again here
$k^{\mathbb {D}}_p$
is the normalized kernel for
$\mathbb {D}$
centered at p). As in the proof of Theorem 1.1 we have
$f_{j,q,p_j}\to 0$
weakly as
$j\to \infty $
for any
$p_j\to q$
. Similar to (2.1), one can show that
$$\begin{align*}H_{\psi_q}f_{j,q,p}= (H^{\mathbb{D}^{n-1}}_{\psi_q}g_{j,q})k^{\mathbb{D}}_p.\end{align*}$$
Then
$\|H_{\psi _q}f_{j,q,p}\| =\|H^{\mathbb {D}^{n-1}}_{\psi _q}g_{j,q}\|_{L^2(\mathbb {D}^{n-1})}$
and as in (2.2) we have
$$ \begin{align*} \left\|H^*_{\psi_q}H_{\psi_q}f_{j,q,p}-\lambda_q f_{j,q,p}\right\| -&\left\|H^*_{\psi}H_{\psi-\psi_q}f_{j,q,p} +H^*_{\psi-\psi_q}H_{\psi_q}f_{j,q,p} \right\| \\ \leq\,& \left\|H^*_{\psi}H_{\psi}f_{j,q,p}-\lambda_qf_{j,q,p}\right\| \\ \leq\, & \left\|H^*_{\psi_q}H_{\psi_q}f_{j,q,p} -\lambda_q f_{j,q,p}\right\| \\ &+\left\|H^*_{\psi}H_{\psi-\psi_q}f_{j,q,p} +H^*_{\psi-\psi_q}H_{\psi_q}f_{j,q,p} \right\|. \end{align*} $$
As in the proof of Theorem 1.1 one can show that
$\|(\psi -\psi _q)f_{j,q,p}\|\to 0$
as
$p\to q$
. This fact implies that
$$\begin{align*}\left\|H^*_{\psi}H_{\psi-\psi_q}f_{j,q,p}\right\| =\left\|H^*_{\psi}(I-P)\Big((\psi-\psi_q)f_{j,q,p}\Big)\right\|\to 0 \text{ as } p\to q.\end{align*}$$
Furthermore, since
$\mathbb {D}^n$
is a product domain and
$\psi _q$
is independent of
$z_n$
using (2.3) with
$$\begin{align*}h_{j,q}=H^{\mathbb{D}^{n-1}}_{\psi_q}g_{j,q,}\end{align*}$$
we have
$$\begin{align*}\left\|H^*_{\psi-\psi_q}H_{\psi_q}f_{j,q,p}\right\| =\left\|P\left((H^{\mathbb{D}^{n-1}}_{\psi_q}g_{j,q})(\overline{\psi}-\overline{\psi_q}) k^{\mathbb{D}}_{p}\right)\right\| \to 0 \text{ as } p\to q.\end{align*}$$
Furthermore, by (2.1) we have
$$\begin{align*}H^*_{\psi_q}H_{\psi_q}f_{j,q,p} = (H^{\mathbb{D}^{n-1*}}_{\psi_q}H^{\mathbb{D}^{n-1}}_{\psi_q}g_{j,q})k^{\mathbb{D}}_p.\end{align*}$$
Then for any fixed
$j,$
we have
$$ \begin{align*} \left\|H^*_{\psi}H_{\psi}f_{j,q,p} -\lambda_q f_{j,q,p}\right\| \to \left\|H^{\mathbb{D}^{n-1*}}_{\psi_q}H^{\mathbb{D}^{n-1}}_{\psi_q}g_{j,q} -\lambda_q g_{j,q}\right\|_{L^2(\mathbb{D}^{n-1})} \text { as } p\to q. \end{align*} $$
However, by (2.6) we can pass to subsequence of
$\{g_{j,q}\}$
, if necessary, and get
$$\begin{align*}\left\|H^{\mathbb{D}^{n-1*}}_{\psi_q}H^{\mathbb{D}^{n-1}}_{\psi_q} g_{j,q} -\lambda_qg_{j,q}\right\|_{L^2(\mathbb{D}^{n-1})}<\frac{1}{j}\end{align*}$$
for all j. Next we choose
$\{p_j\}\subset \mathbb {D}$
such that
$p_j\to q$
as
$j\to \infty $
and
$$\begin{align*}\left\|H^*_{\psi}H_{\psi}f_{j,q,p_j} -\lambda_q f_{j,q,p_j}\right\|<\frac{2}{j}\end{align*}$$
for all j. Therefore, we have
$$\begin{align*}\left\|H^*_{\psi}H_{\psi}f_{j,q,p_j} -\lambda_q f_{j,q,p_j}\right\| \to 0 \text { as } j\to \infty.\end{align*}$$
Then ii. in Theorem A implies that
$\lambda _q\in \sigma _e(H^*_{\psi }H_{\psi })$
.
Next we note that
$\|H^{\mathbb {D}^{n-1*}}_{\psi _q}H^{\mathbb {D}^{n-1}}_{\psi _q}\|$
depends on q continuously because
$\psi _q$
depends on q continuously and
$$ \begin{align*} \left\|H^{\mathbb{D}^{n-1*}}_{\psi_{q_1}}H^{\mathbb{D}^{n-1}}_{\psi_{q_1}} -H^{\mathbb{D}^{n-1*}}_{\psi_{q_2}}H^{\mathbb{D}^{n-1}}_{\psi_{q_2}}\right\| = & \left\|H^{\mathbb{D}^{n-1*}}_{\psi_{q_1}-\psi_{q_2}}H^{\mathbb{D}^{n-1}}_{\psi_{q_1}} +H^{\mathbb{D}^{n-1*}}_{\psi_{q_2}}H^{\mathbb{D}^{n-1}}_{\psi_{q_1}-\psi_{q_2}} \right\| \\ \leq \,& \left\|H^{\mathbb{D}^{n-1}}_{\psi_{q_1}-\psi_{q_2}}\right\| \left(\left\|H^{\mathbb{D}^{n-1}}_{\psi_{q_1}}\right\| +\left\|H^{\mathbb{D}^{n-1}}_{\psi_{q_2}}\right\|\right)\\ \leq \, &\left\| \psi_{q_1}-\psi_{q_2}\right\|_{L^{\infty}} \left(\left\|H^{\mathbb{D}^{n-1}}_{\psi_{q_1}}\right\| +\left\|H^{\mathbb{D}^{n-1}}_{\psi_{q_2}}\right\|\right). \end{align*} $$
Then we conclude that
$\{\lambda _q:q\in b\mathbb {D}\}$
is connected. Therefore, since we assume that the mapping
$q\to \| H^{\mathbb {D}^{n-1}}_{\psi _q}\|$
is non-constant, we conclude that
$\{\lambda _q:q\in b\mathbb {D}\}$
contains an open interval in
$(0, \infty )$
. That is,
$\sigma _e(H^*_{\psi }H_{\psi })$
contains an open interval in
$(0, \infty )$
.
The proof of Theorem 1.3 above implies the following corollary.
Corollary 2.2 Let
$\psi \in C(\overline {\mathbb {D}^2})$
and
$\psi _{1,\theta }(\xi )=\psi (e^{i\theta },\xi ), \psi _{2,\theta }(\xi )=\psi (\xi ,e^{i\theta })$
for
$\xi \in \mathbb {D}$
. Then
$$\begin{align*}\left\{\left\| H^{\mathbb{D}}_{\psi_{1,\theta}}\right\|^2_{L^2(\mathbb{D})}:\theta\in [0,2\pi]\right\} \cup \left\{\left\| H^{\mathbb{D}}_{\psi_{2,\theta}}\right\|^2_{L^2(\mathbb{D})}:\theta\in [0,2\pi]\right\} \subset \sigma_e(H^*_{\psi}H_{\psi}).\end{align*}$$
Remark 2.3 We note that Theorem 1.6 shows that the inclusion in Corollary 2.2 is not an equality in general. Indeed, for
$\psi (z_1,z_2)=\overline {z}_1^n\overline {z}_2^m$
the quantities
$\| H^{\mathbb {D}}_{\psi _{1,\theta }}\|$
and
$\| H^{\mathbb {D}}_{\psi _{2,\theta }}\|$
are constant as functions of
$\theta $
. Therefore, the left-hand side of the inclusion in Corollary 2.2 contains at most two numbers whereas the right hand side, by Theorem 1.6, contains infinitely many numbers.
Proof of Corollary 1.4
If
$\varphi $
is not holomorphic then
$H^{\mathbb {D}^{n-1}}_{\varphi }$
is non-zero operator as
$H^{\mathbb {D}^{n-1}}_{\varphi }1\neq 0$
. Hence
$\|H^{\mathbb {D}^{n-1}}_{\varphi }\|>0$
. Furthermore, since
$|\chi |$
is non-constant on the unit circle, the image of the mapping
$$\begin{align*}q\to \left\| H^{\mathbb{D}^{n-1}}_{\psi_q}\right\|_{L^2(\mathbb{D}^{n-1})} =|\chi(q)|\left\|H^{\mathbb{D}^{n-1}}_{\varphi}\right\|_{L^2(\mathbb{D}^{n-1})}\end{align*}$$
contains an open interval. Therefore, by Theorem 1.3, the essential spectrum of
$H^*_{\psi }H_{\psi }$
contains an open interval.
Remark 2.4 One can also prove of Corollary 1.4 using Theorem 1.1 as follows. Since
$H^{\mathbb {D}^{n-1}}_{\varphi }$
is non-zero operator, there exists a positive number
$\mu \in \sigma (H^{\mathbb {D}^{n-1*}}_{\varphi }H^{\mathbb {D}^{n-1}}_{\varphi })$
. Then the image of the mapping
$q\to |\chi (q)|^2\mu $
contains an open interval as
$|\chi |$
is non-constant on the unit circle. Therefore, by Theorem 1.1, the essential spectrum of
$H^*_{\psi }H_{\psi }$
contains an open interval.
3 Proof of Theorem 1.6
Before we present the proof of Theorem 1.6, we make some elementary computations. Let
$\Omega $
be a complete Reinhardt domain in
$\mathbb C^n$
and again
$\mathbb {N}_0=\{0,1,2,3,\ldots \}$
. The Bergman kernel function K has the expression
$$\begin{align*}K(z,w)=\sum_{\boldsymbol{\alpha}\in \mathbb{N}_0^n}c_{\boldsymbol{\alpha}} z^{\boldsymbol{\alpha}}\overline{w}^{\boldsymbol{\alpha}},\end{align*}$$
where
$c_{\boldsymbol {\alpha }}= 0$
when
$ \|z^{\boldsymbol {\alpha }}\|_{L^2(\Omega )}=\infty $
and
$c_{\boldsymbol {\alpha }}= \|z^{\boldsymbol {\alpha }}\|_{L^2(\Omega )}^{-2}$
otherwise.
A function
$\psi $
is called quasi-homogeneous if there exists
$f:[0,\infty )^n\to \mathbb {C}$
,
$(k_1,\ldots ,k_n)\in \mathbb {Z}^n$
such that
$$ \begin{align} \psi(r_1e^{i\theta_1},\ldots,r_ne^{i\theta_n}) =f(r_1, \ldots,r_n)e^{i(k_1\theta_1+\cdots +k_n\theta_n)}. \end{align} $$
For
$z=(z_1,\dots ,z_n)\in \mathbb C^n$
and
$\boldsymbol {\alpha }=(\alpha _1,\dots ,\alpha _n)\in \mathbb Z^n$
, we will use the notation
$|z|$
for
$(|z_1|,\dots ,|z_n|)$
and write
$|z|^{\boldsymbol {\alpha }}$
for the product
$\prod _{j=1}^{n}|z_j|^{\alpha _j}$
. Then (3.1) can also be expressed as:
$$\begin{align*}\psi(z) =f(|z|)e^{i\mathbf{k}\cdot\boldsymbol{\theta}},\end{align*}$$
where
$\mathbf {k}=(k_1,\dots ,k_n)$
,
$\boldsymbol {\theta }=(\theta _1,\dots ,\theta _n)$
, and
$z_j=|z_j|e^{i\theta _j}$
. Now we consider the spectrum of
$H^*_{\psi }H_{\psi }$
when
$\psi $
is bounded and quasi-homogeneous on the complete Reinhardt domain
$\Omega $
.
By (2.4) we have
$$ \begin{align*} H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} &=PM_{\overline{\psi}}(I-P)M_\psi(z^{\boldsymbol{\alpha}})\\ &=PM_{\overline{\psi}}(I-P)f(|z|)|z|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha})\cdot \boldsymbol{\theta}}\\ &=PM_{\overline{\psi}}(z^{\boldsymbol{\alpha}}\psi-P(|z|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha} )\cdot \boldsymbol{\theta}}f(|z|))), \end{align*} $$
for
$\boldsymbol {\alpha }\in \mathbb N_0^n$
. If the multi-index
$\mathbf {k}+\boldsymbol {\alpha }\notin \mathbb {N}_0^n$
, then
$P(|z|^{\boldsymbol {\alpha }} e^{i(\mathbf {k}+\boldsymbol {\alpha } )\cdot \boldsymbol {\theta }}f(|z|))=0$
which yields that
$$ \begin{align} H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =P(z^{\boldsymbol{\alpha}}|\psi|^2) =\frac{\|z^{\boldsymbol{\alpha}}\psi\|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}}z^{\boldsymbol{\alpha}}. \end{align} $$
Otherwise,
$$ \begin{align*} P(|z|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha} )\cdot \boldsymbol{\theta}}f(|z|)) &=\int_{\Omega}K(z,w)|w|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha} )\cdot \boldsymbol{\theta}}f(|w|)dV(w)\\ &=\sum_{\boldsymbol{\beta}\in \mathbb{N}_0^n}c_{\boldsymbol{\beta}}z^{\boldsymbol{\beta}} \int_{\Omega} \overline{w}^{\boldsymbol{\beta}} |w|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha} )\cdot \boldsymbol{\theta}}f(|w|)dV(w)\\ &=c_{\boldsymbol{\alpha}+\mathbf{k}} z^{\boldsymbol{\alpha}+\mathbf{k}}\int_{\Omega}\overline{w}^{\boldsymbol{\alpha}+\mathbf{k}} |w|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha} )\cdot \boldsymbol{\theta}}f(|w|)dV(w)\\ &=\frac{\int_{\Omega}|w|^{2\boldsymbol{\alpha}+k} f(|w|)dV(w)}{\|z^{\boldsymbol{\alpha}+\mathbf{k}}\|^2_{L^2}} z^{\mathbf{k}+\boldsymbol{\alpha}}, \end{align*} $$
which implies that
$$ \begin{align} \nonumber H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =\,&PM_{\overline{\psi}}(z^{\boldsymbol{\alpha}}\psi-P(|z|^{\boldsymbol{\alpha}} e^{i(\mathbf{k}+\boldsymbol{\alpha} )\cdot \boldsymbol{\theta}}f(|z|)))\\ =\,&PM_{\overline{\psi}}\left(z^{\boldsymbol{\alpha}}\psi-\frac{\int_{\Omega}|w|^{2\boldsymbol{\alpha}+k} f(|w|)dV(w)}{\|z^{\boldsymbol{\alpha} +\mathbf{k}}\|^2_{L^2}}z^{\mathbf{k}+\boldsymbol{\alpha}}\right)\nonumber\\ =\,&P\left(z^{\boldsymbol{\alpha}} |\psi|^2-\frac{\int_{\Omega}|w|^{2\boldsymbol{\alpha}+k} f(|w|)dV(w)}{\|z^{\boldsymbol{\alpha} +\mathbf{k}}\|^2_{L^2}}|z|^{\mathbf{k}}z^{\boldsymbol{\alpha}}\overline{f(|z|)}\right)\nonumber\\ =\,&z^{\boldsymbol{\alpha}}\left(\frac{\|z^{\boldsymbol{\alpha}} \psi\|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}} -\frac{\left|\int_{\Omega}|w|^{2\boldsymbol{\alpha}+k} f(|w|)dV(w)\right|{}^2}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}\| z^{\boldsymbol{\alpha}+\mathbf{k}}\|^2_{L^2}}\right). \end{align} $$
Hence the spectrum
$\sigma (H^*_{\psi }H_{\psi })$
contains the eigenvalues
$$\begin{align*}\frac{\|z^{\boldsymbol{\alpha}} \psi\|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}} -\frac{\left|\int_{\Omega}|w|^{2\boldsymbol{\alpha}+k} f(|w|)dV(w)\right|{}^2}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}\| z^{\boldsymbol{\alpha}+\mathbf{k}}\|^2_{L^2}}\end{align*}$$
corresponding to the eigenfunction
$z^{\boldsymbol {\alpha }}$
for
$\mathbf {k}+\boldsymbol {\alpha }\in \mathbb {N}_0^n$
and
$$\begin{align*}\frac{\|z^{\boldsymbol{\alpha}} \psi\|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}}\end{align*}$$
for
$\mathbf {k}+\boldsymbol {\alpha }\not \in \mathbb {N}_0^n$
. Furthermore, since the eigenfunctions form an orthogonal basis for
$A^2(\Omega )$
, the closure of this set is the whole spectrum
$\sigma (H^*_{\psi }H_{\psi })$
(see by Lemma 3.4 below).
Example 3.1 When
$f(|z|)=|z|^{\mathbf {k}}$
for
$\mathbf {k}\in \mathbb {N}_0^n$
, Equation (3.3) becomes
$$\begin{align*}H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =z^{\boldsymbol{\alpha}}\left(\frac{\|z^{\boldsymbol{\alpha}+\mathbf{k}} \|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}}- \frac{\|z^{2\boldsymbol{\alpha}+2\mathbf{k}}\|^2_{L^1}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2} \|z^{\boldsymbol{\alpha}+\mathbf{k}}\|^2_{L^2}}\right)=0.\end{align*}$$
This is not surprising since in this case the symbol
$\psi (z)=z^{\mathbf {k}}$
is holomorphic.
Example 3.2 When
$\psi (z)=e^{i\mathbf {k}\cdot \boldsymbol {\theta }}$
for
$\mathbf {k}\in \mathbb {N}_0^n$
, Equation (3.3) becomes
$$\begin{align*}H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =z^{\boldsymbol{\alpha}}\left(1 -\frac{\|z^{2\boldsymbol{\alpha}+\mathbf{k}}\|^2_{L^1}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2} \|z^{\boldsymbol{\alpha}+\mathbf{k}}\|^2_{L^2}}\right).\end{align*}$$
Example 3.3 When
$\mathbf {k}=\mathbf {0}$
,
$\psi (z)$
is radial and (3.3) becomes
$$\begin{align*}H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =z^{\boldsymbol{\alpha}} \left(\frac{\|z^{\boldsymbol{\alpha}}\psi \|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}} -\frac{\|z^{2\boldsymbol{\alpha}}\psi\|^2_{L^1}}{\|z^{\boldsymbol{\alpha}}\|^4_{L^2}}\right).\end{align*}$$
Lemma 3.4 Let
$T:H\to H$
be a bounded linear map on a separable Hilbert space H. Assume that
$\lambda _j$
is an eigenvalue with a corresponding eigenvector
$u_j$
for
$j\in \mathbb {N}$
and
$\{u_j:j\in \mathbb {N}\}$
forms an orthogonal basis for H. Then
$\sigma (T)=\overline {\{\lambda _j:j\in \mathbb {N}\}}$
. Furthermore,
$\lambda \in \sigma _e(T)$
if and only if it is the limit of a subsequence of
$\{\lambda _j\}$
.
Proof. Since
$\sigma (T)$
is a compact set containing
$\{\lambda _j:j\in \mathbb {N}\},$
we have
$\overline {\{\lambda _j:j\in \mathbb {N}\}}\subset \sigma (T)$
. To prove the converse, assume that
$\lambda \not \in \overline {\{\lambda _j:j\in \mathbb {N}\}}$
. Then we define
$Su_j=\frac {1}{\lambda _j-\lambda }u_j$
for all j. Then S is a bounded operator, as
$\{(\lambda _j-\lambda )^{-1}\}$
is a bounded sequence, and it is the inverse of
$T-\lambda I$
, as
$$\begin{align*}(T-\lambda I)Su_j=S(T-\lambda I)u_j=u_j,\end{align*}$$
for all j. Then,
$\lambda \not \in \sigma (T)$
. That is,
$\sigma (T)\subset \overline {\{\lambda _j:j\in \mathbb {N}\}}$
.
We recall that
$\sigma _e(T)=\sigma (T)\setminus \sigma _d(T)$
where
$\sigma _d(T)$
is composed of isolated eigenvalues with finite multiplicity. To prove the last statement, let
$\lambda \in \sigma _e(T)$
. Then
$\lambda \in \sigma (T)$
is not an isolated eigenvalue with finite multiplicity. That is, either
$\lambda $
is not isolated in
$\sigma (T)$
or it is an eigenvalue with infinite multiplicity. Either way, there exists a subsequence of
$\{\lambda _j\}$
converging to
$\lambda $
. Conversely, if
$\lambda \in \sigma (T)$
is the limit of a subsequence
$\{\lambda _{j_k}\}$
of
$\{\lambda _j\}$
, then either
$\lambda =\lambda _{j_k}$
for infinitely many ks (hence, it is an eigenvalue with infinite multiplicity) or
$\lambda $
is not isolated in
$\sigma (T)$
. Again, either way
$\lambda \in \sigma _e(T)$
.
Finally, we present the proof of Theorem 1.6.
Proof of Theorem 1.6
Note that
$\psi (z) =z^{\mathbf {n}}\overline {z}^{\mathbf {m}} =|z|^{\mathbf {n}+\mathbf {m}}e^{i(\mathbf {n}-\mathbf {m})\cdot \boldsymbol {\theta }}$
. By (3.2) and (3.3), it follows that
$$ \begin{align} H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =\begin{cases} z^{\boldsymbol{\alpha}} \frac{\|z^{\boldsymbol{\alpha}+\mathbf{n}}\overline{z}^{\mathbf{m}}\|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}} & \boldsymbol{\alpha}+\mathbf{n}-\mathbf{m}\notin \mathbb{N}_0^n\\ z^{\boldsymbol{\alpha}}\left(\frac{\|z^{\boldsymbol{\alpha}+\mathbf{n}}\overline{z}^{\mathbf{m}}\|^2_{L^2}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2}} -\frac{\|z^{2\boldsymbol{\alpha}+2\mathbf{n}-\mathbf{m}}\overline{z}^{\mathbf{m}}\|^2_{L^1}}{\|z^{\boldsymbol{\alpha}}\|^2_{L^2} \|z^{\boldsymbol{\alpha}+\mathbf{n}-\mathbf{m}}\|^2_{L^2}}\right) & \boldsymbol{\alpha}+\mathbf{n}-\mathbf{m}\in \mathbb N_0^n. \end{cases} \end{align} $$
We note that for multi-index
$\boldsymbol {\beta }\in \mathbb N_0^n$
we have,
$$\begin{align*}\|z^{\boldsymbol{\beta}}\|^2_{L^2}=\|z^{2\boldsymbol{\beta}}\|_{L^1}=\frac{\pi^n}{\prod_{k=1}^{n}(\beta_k+1)}.\end{align*}$$
Substituting this into (3.4) yields that
$$ \begin{align*} H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}} =\begin{cases} z^{\boldsymbol{\alpha}}\prod_{k=1}^{n}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1} & \boldsymbol{\alpha}+\mathbf{n}-\mathbf{m}\notin \mathbb{N}_0^n\\ z^{\boldsymbol{\alpha}}\left(\prod_{k=1}^{n}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1} - \prod_{k=1}^{n}\frac{(\alpha_k+1)(\alpha_k+n_k-m_k+1)}{(\alpha_k+n_k+1)^2}\right) & \boldsymbol{\alpha}+\mathbf{n}-\mathbf{m}\in \mathbb{N}_0^n. \end{cases} \end{align*} $$
Hence
$$\begin{align*}H^*_{\psi} H_{\psi} z^{\boldsymbol{\alpha}}=\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B_n}z^{\boldsymbol{\alpha}},\end{align*}$$
for
$\boldsymbol {\alpha }\in \mathbb {N}_0^n$
where
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B_n}$
is defined before Theorem 1.6 and
$B_n=\{1,2,3,\ldots , n\}.$
That is,
$z^{\boldsymbol {\alpha }}$
is an eigenvector corresponding to the eigenvalue
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B_n}$
. Furthermore, since
$\{z^{\boldsymbol {\alpha }}:\boldsymbol {\alpha }\in \mathbb {N}_0^n\}$
forms an orthogonal basis for
$A^2(\mathbb {D}^n)$
, Lemma 3.4 implies that
$$\begin{align*}\sigma(H^*_{\psi} H_{\psi}) =\overline{\left\{\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B_n}: \boldsymbol{\alpha}\in \mathbb{N}_0^n\right\}}.\end{align*}$$
We note that in case
$\mathbf {m}=0$
the operator
$H^*_{\psi } H_{\psi } z^{\boldsymbol {\alpha }}$
is the zero operator and hence 0 is the only eigenvalue with infinite multiplicity. For the rest of the proof, we will assume that
$\mathbf {m}\neq 0$
.
Lemma 3.4 implies that each element in the spectrum is either an eigenvalue or a limit of a sequence of eigenvalues. To describe the spectrum outside of the point spectrum, we assume that
$\lambda \in \sigma (H^*_{\psi } H_{\psi })$
such that
$$\begin{align*}\lambda=\lim_{j\to\infty}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n,} \end{align*}$$
for a sequence
$\{\boldsymbol {\alpha }(j)\}\subset \mathbb {N}_0^n$
. Next we will show that either
$\lambda =0$
or
$\lambda =\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B}$
for some
$\emptyset \neq B\subset B_n$
and
$\boldsymbol {\alpha }\in \mathbb {N}_0^n$
. We assume that
$\lambda \neq 0$
.
We note that
$z^{\boldsymbol {\alpha }(j)}$
is an eigenvector corresponding to the eigenvalue
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}$
. Let us assume that
$$\begin{align*}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n} =\prod_{k=1}^{n} \frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1}.\end{align*}$$
For each j there exists
$k\in B_n$
such that
$\alpha _k(j)<m_k-n_k$
. Then there exists k such that
$\alpha _k(j)<m_k-n_k$
for infinitely many js. That is, the kth sequence
$\{\alpha _k(j)\}_{j=1}^{\infty }$
has a subsequence bounded by
$m_k-n_k$
. We pass to that subsequence of
$\{\boldsymbol {\alpha }(j)\}$
and still call it
$\{\boldsymbol {\alpha }(j)\}$
.
Next we construct B as follows. Let
$k_1\in B_n$
be the smallest integer so that
$\{\alpha _{k_1}(j)\}_{j=1}^{\infty }$
has a bounded subsequence. Then we pass onto a subsequence, still calling it
$\{\boldsymbol {\alpha }(j)\}$
, so that
$\{\alpha _{k_1}(j)\}_{j=1}^{\infty }$
is a constant sequence. If
$\alpha _{k}(j)\to \infty $
as
$j\to \infty $
for all
$k\geq k_1+1$
then we stop here and
$B=\{k_1\}$
. Otherwise, we choose
$k_2\geq k_1+1$
to be the smallest integer so that
$\{\alpha _{k_2}(j)\}_{j=1}^{\infty }$
has a bounded subsequence. Now we pass onto a subsequence, again calling it
$\{\boldsymbol {\alpha }(j)\}$
, so that
$\{\alpha _{k_2}(j)\}_{j=1}^{\infty }$
is a constant sequence. After finitely many steps we obtain
$B=\{k_1,\ldots ,k_p\}$
and a subsequence
$\{\boldsymbol {\alpha }(j)\}$
so that
$\{\alpha _{k}(j)\}$
is a constant sequence for every
$k\in B$
and
$\alpha _{k}(j)\to \infty $
as
$j\to \infty $
for
$k\not \in B$
.
Hence for this subsequence of
$\{\boldsymbol {\alpha }(j)\}$
, taking limit as
$j\to \infty $
, we have
$$ \begin{align*} \lambda=\lim_{j\to\infty}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n} =&\, \lim_{j\to\infty}\prod_{k=1}^{n} \frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} \\ =&\, \prod_{k\in B}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1} =\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B,} \end{align*} $$
where
$\boldsymbol {\alpha }\in \mathbb {N}_0^n$
such that
$\alpha _k$
is arbitrary for
$k\not \in B$
and
$\alpha _k=\alpha _k(j)$
for
$k\in B$
. We note that
$\alpha _k<m_k-n_k$
for some
$k\in B$
.
Similarly, if
$\lim _{j\to \infty }\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n} =\lambda $
where
$$\begin{align*}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n} =\left(\prod_{k=1}^{n}\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} - \prod_{k=1}^{n}\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2}\right)\end{align*}$$
then
$$\begin{align*}\lambda=\lim_{j\to\infty}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n} =\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B},\end{align*}$$
and
$\alpha _k\geq m_k-n_k$
for all
$k\in B$
. Hence we have shown that
$$ \begin{align*} \sigma(H^*_{\psi}H_{\psi})\subset \{0\}\cup \{\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}: \boldsymbol{\alpha}\in \mathbb{N}_0^n, \emptyset\neq B\subset B_n\}. \end{align*} $$
Next we prove the converse of the inclusion above. First, we choose
$\alpha _k(j)=j$
for all k. Then it is easy to see that
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}\to 0$
as
$j\to \infty $
. Hence
$0\in \sigma (H^*_{\psi }H_{\psi })$
. Secondly, let us assume that
$\emptyset \neq B \subset B_n$
and
$\alpha _k<m_k-n_k$
for some
$k\in B$
. We choose
$\alpha _k(j)=j$
for
$k\not \in B$
and
$\alpha _k(j)=\alpha _k$
for
$k\in B$
. Then
$$\begin{align*}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n} \to \lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}\in \sigma(H^*_{\psi}H_{\psi}) \text{ as } j\to \infty.\end{align*}$$
Similarly, if
$\alpha _k\geq m_k-n_k$
for all
$k\in B$
we choose
$\{\boldsymbol {\alpha }(j)\}$
as above again and conclude that
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B}\in \sigma (H^*_{\psi }H_{\psi })$
. Therefore,
$$ \begin{align*} \sigma(H^*_{\psi}H_{\psi})=\{0\}\cup \{\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}: \boldsymbol{\alpha}\in \mathbb{N}_0^n, \emptyset\neq B\subset B_n\}. \end{align*} $$
We finish the proof by proving the claims about multiplicities. Let us assume that
$n_k+m_k\geq 1$
for all
$k\in B_n$
and
$\lambda $
is a non-zero eigenvalue of infinite multiplicity. Then there exists a sequence of multi-indices
$\{\boldsymbol {\alpha }(j)\}$
such that
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}=\lambda $
for all j. After passing to a subsequence, we may assume that
$\lim _{j\to \infty } \alpha _k(j)$
exists allowing infinity as the limit for all k.
If
$B=\theta $
then
$\lambda =0$
. For the rest of the proof, we assume that
$B\neq \theta $
and, by passing to a subsequence if necessary,
$\alpha _k(j)=\alpha _k$
for
$k\in B$
and
$\alpha _k(j)\to \infty $
for
$k\not \in B$
. In case
$\alpha _k(j)< m_k-n_k$
for some
$k\in B$
, it follows that
$$\begin{align*}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n} <\prod_{k\in B}\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} =\prod_{k\in B}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1}=\lambda,\end{align*}$$
which is a contradiction with
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n} =\lambda $
for all j.
On the other hand, if
$\alpha _k(j)\geq m_k-n_k$
for all
$k=1,2,\dots ,n$
, the equality
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}=\lambda $
for all j implies that
$$ \begin{align}\nonumber \lambda=&\prod_{k\in B}\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} -\prod_{k\in B}\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2}\\ =& \prod_{k=1}^n\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} -\prod_{k= 1}^n\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2}. \end{align} $$
Next we will prove that this is impossible for all j. Let us set
$$ \begin{align*} a&=\prod_{k\in B}\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1}, \\b&=\prod_{k\in B}\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2}. \end{align*} $$
Then (3.5) becomes
$$ \begin{align} a-b-&a\prod_{k\not\in B}\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1}\\ &+b\prod_{k\not\in B}\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2} =0. \nonumber \end{align} $$
Rearranging this equation, we obtain a polynomial equation
$$ \begin{align*} (a-b)&\prod_{k\not\in B}(\alpha_k(j)+n_k+1)^2(\alpha_k(j)+n_k+m_k+1)\\ &-a \prod_{k\not\in B}(\alpha_k(j)+n_k+1)^2(\alpha_k(j)+1)\\ \nonumber &+b\prod_{k\not\in B}(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)(\alpha_k(j)+n_k+m_k+1) =0. \end{align*} $$
Rewriting the equality above, we get
$$ \begin{align} \nonumber (a-b)&\prod_{k\not\in B}(\alpha_k(j)+n_k+1)^2 \left(\prod_{k\not\in B}(\alpha_k(j)+n_k+m_k+1)-\prod_{k\not\in B}(\alpha_k(j)+1)\right)\\ &-b\prod_{k\not\in B}(\alpha_k(j)+1)m^2_k =0. \end{align} $$
If
$m_k\geq 1$
for some
$k\in B$
then
$a-b>0$
and
$$\begin{align*}\prod_{k\not\in B}(\alpha_k(j)+n_k+m_k+1)-\prod_{k\not\in B}(\alpha_k(j)+1)\geq 1.\end{align*}$$
Furthermore,
$\prod _{k\not \in B}(\alpha _k(j)+n_k+1)^2$
dominates
$\prod _{k\not \in B}(\alpha _k(j)+1)m^2_k$
as
$j\to \infty $
. Hence, the left-hand side of (3.7) converges to
$\infty $
as
$j\to \infty $
, reaching a contradiction.
On the other hand, if
$m_k=0$
for all
$k\in B$
and since
$\mathbf {m}\neq 0$
, we have
$m_k\geq 1$
for some
$k\not \in B$
. Then
$a-b=0$
and (3.6) implies
$$\begin{align*}-\prod_{k\not\in B}{\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1}} +\prod_{k\not\in B}{\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2}}=0. \end{align*}$$
However, the left-hand side of the equation above is equal to a positive multiple of the following expression
$$\begin{align*}-\prod_{k\not\in B} (\alpha_k(j)+n_k+1)^2 +\prod_{k\not\in B} \left((\alpha_k(j)+n_k+1)^2 -m_k^2\right) \end{align*}$$
which is negative, reaching a contradiction again. Therefore, we showed that, if
$m_k+n_k\geq 1$
for all
$k\in B_n$
then each eigenvalue is of finite multiplicity.
Finally, let us define
$B_0=\{k\in B_n:m_k=n_k=0\}$
. Now we will show that all of the eigenvalues have infinite multiplicities when
$B_0\neq \theta $
. We note that since
$\mathbf {m}\neq 0$
we know that
$B_0\subsetneqq B_n$
.
Assume that
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B_n}$
is an eigenvalue. Then either
$\alpha _k<m_k-n_k$
for some k or
$\alpha _k\geq m_k-n_k$
for all k. In the first case, we have
$\alpha _k<m_k-n_k$
for some
$k\not \in B_0$
. Then
$$\begin{align*}\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n}= \prod_{k=1 }^{n}\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} = \prod_{k\not\in B_0}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1} =\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B_n,}\end{align*}$$
for
$\alpha _k(j)=j$
for
$k\in B_0$
and
$\alpha _k(j)=\alpha _k$
for
$k\not \in B_0$
. Similarly, if
$\alpha _k\geq m_k-n_k$
for all k we make the same choice
$\alpha _k(j)=j$
for
$k\in B_0$
and
$\alpha _k(j)=\alpha _k$
for
$k\not \in B_0$
and one can see that
$$ \begin{align*} \lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha}(j),B_n}=& \, \prod_{k=1}^n\frac{\alpha_k(j)+1}{\alpha_k(j)+n_k+m_k+1} -\prod_{k= 1}^n\frac{(\alpha_k(j)+1)(\alpha_k(j)+n_k-m_k+1)}{(\alpha_k(j)+n_k+1)^2}\\ =\,&\prod_{k\not\in B_0}\frac{\alpha_k+1}{\alpha_k+n_k+m_k+1} -\prod_{k\not\in B_0}\frac{(\alpha_k+1)(\alpha_k+n_k-m_k+1)}{(\alpha_k+n_k+1)^2}\\ =\,&\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B_n}. \end{align*} $$
Therefore, in either case the eigenvalues have infinite multiplicities.
By the proof of Theorem 1.6, we can also characterize the essential spectrum of Hankel operators with monomial symbols.
Corollary 3.5 Let
$\psi (z)=z^{\mathbf {n}}\overline {z}^{\mathbf {m}}$
for some
$\mathbf {m},\mathbf {n}\in \mathbb {N}^n_0$
. If
$m_k+n_k=0$
for some
$k\in B_n$
, then the essential spectrum of
$H^*_{\psi }H_{\psi }$
on
$A^2(\mathbb {D}^n)$
$$\begin{align*}\sigma_e(H^*_{\psi}H_{\psi})=\sigma(H^*_{\psi}H_{\psi}) =\{0\}\cup \left\{\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}:\boldsymbol{\alpha}\in \mathbb{N}_0^n, \theta\neq B\subset B_n\right\}.\end{align*}$$
On the other hand, if
$n_k+m_k\geq 1$
for all
$k\in B_n$
and
$m_k\geq 1$
for some
$k\in B_n$
, then the essential spectrum of
$H^*_{\psi }H_{\psi }$
on
$A^2(\mathbb {D}^n)$
$$\begin{align*}\sigma_e(H^*_{\psi}H_{\psi})=\{0\}\cup \left\{\lambda_{\mathbf{n},\mathbf{m},\boldsymbol{\alpha},B}: \boldsymbol{\alpha}\in \mathbb{N}_0^n, \theta\neq B\subset B_n \text{ and } B\neq B_n\right\}.\end{align*}$$
Proof. First let us assume that
$B_0=\{k\in B_n:m_k+n_k=0\}\neq \theta $
. Then all of the eigenvalues have infinite multiplicity. So the discrete spectrum is empty and the essential spectrum is identical to the spectrum.
Next we assume that
$B_0=\theta $
and
$m_k\geq 1$
for some
$k\in B_n$
. Then 0 is in the essential spectrum because
$0=\lim _{j\to \infty }\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}$
for
$\alpha _k(j)=j$
for all
$k\in B_n$
. Next one can see that for any
$\boldsymbol {\alpha }\in \mathbb N^n_0$
, the value
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B}$
is in the essential spectrum for
$\theta \neq B\subset B_n$
and
$B\neq B_n$
as follows. Let
$\{\boldsymbol {\alpha }(j)\}$
be a sequence in
$\mathbb N^n_0$
with
$\alpha _k(j)=\alpha _k$
for
$k\in B$
and
$\alpha _k(j)=j$
for
$k\notin B$
. Then as shown in the proof of Theorem 1.6, the constant
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}$
is an eigenvalue and
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B} =\lim _{j\to \infty }\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha }(j),B_n}$
.
On the other hand, Lemma 3.4 implies that every
$\lambda $
in the essential spectrum is the limit of a sequence of eigenvalues. By passing to subsequences argument as in Theorem 1.6 and the fact that
$B_0$
is empty, we can see that such a
$\lambda $
will be of form
$\lambda _{\mathbf {n},\mathbf {m},\boldsymbol {\alpha },B}$
for some
$\boldsymbol {\alpha }\in \mathbb N^n_0$
and some proper subset B of
$B_n$
.
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