---
abstract: 'Let $\mathbb B_n$ be the open unit ball in $\mathbb C^n$. We characterize the spectra of pointwise multipliers $M_u$ acting on Banach spaces of analytic functions on $\mathbb B_n$ satisfying some general conditions.  These spaces include Bergman-Sobolev spaces $A^p_{\alpha,\beta}$, Bloch-type spaces $\mathcal B_\alpha$, weighted Hardy spaces $H^p_w$ with Muckenhoupt weights and Hardy-Sobolev Hilbert spaces $H^2_\beta$. Moreover, we describe the essential spectra of multipliers in most of the aforementioned spaces, in particular, in those spaces for which the set of multipliers is a subset of the ball algebra.' 
address:
- 'Mikael Lindstr[ö]{}m.  Department of Mathematics, Åbo Akademi University.  FI-20500 Åbo, Finland. * e*.mail: mikael.lindstrom@abo.fi'
- 'Santeri Miihkinen.  Department of Mathematics, Abo Akademi University.  FI-20500 Åbo, Finland. * e*.mail: santeri.miihkinen@abo.fi'
- 'David Norrbo.  Department of Mathematics, Åbo Akademi University.  FI-20500 Åbo, Finland. * e*.mail: dnorrbo@abo.fi '
author:
- Mikael Lindström
- Santeri Miihkinen
- David Norrbo
title: Unified approach to spectral properties of multipliers
---

Introduction and preliminaries
==============================

In a very recent article [@CHZ], Cao, He, and Zhu considered the multiplication operator $M_u$ acting on the Hardy-Sobolev Hilbert space and characterized the spectrum and essential spectrum of $M_u.$ In the present work, we extend and generalize the results obtained there from Hardy-Sobolev Hilbert space to the Bergman-Sobolev and Bloch-type spaces of the open unit ball $\mathbb{B}_n$ of ${{\mathbb{C}}}^n$ and weighted Hardy spaces of the open unit disk ${{\mathbb{D}}}$ with Muckenhoupt weights.  In particular, our main focus is to allow the multiplier space $M(X(\mathbb B_n))$ to be contained in the ball algebra, which holds for example for certain Bergman-Sobolev spaces and Bloch-type spaces.  We formulate our results on spectral properties of $M_u$ acting on a Banach space $X(\mathbb{B}_n)$ of analytic functions in $\mathbb{B}_n$, where $X(\mathbb{B}_n)$ satisfies very general and natural properties regarding its multiplier space and the norm topology.  Consequently, we approach the spectral properties of multipliers in a unified manner and key examples of such spaces include the aforementioned spaces.  Aside from obtaining a description of the spectrum for all spaces satisfying the mentioned properties, we also have to develop some new techniques to determine the essential spectrum of $M_u$ regarding the non-Hilbert space case.  Other previous work regarding spectral and related properties of multiplication operators on analytic function spaces includes [@A], [@BB1], [@BDL], [@CH], [@FX], [@OF] and [@V]. 

The article is organised as follows.  In section 2, we introduce general Banach spaces $X(\mathbb B_n)$ of analytic functions on $\mathbb B_n$ and give central concrete examples of them.  Section 3 focuses on the spectrum of $M_u$ by first establishing a characterization of invertibility of $M_u$ and then obtaining the spectrum of $M_u$ and giving admissible examples of spaces on which $M_u$ can be defined.  In section 4, we begin with a characterization of the essential spectrum in the high-dimensional case $n > 1$. Then we consider the case $n=1$ by first establishing a characterization of the Fredholmness of $M_u$ when conditions **(I)**, **(IV)** and $M(X({{\mathbb{D}}}))= H^\infty({{\mathbb{D}}})$ hold.  Examples of spaces satisfying the previous conditions are also given.  Next, we consider the difficult case when $M(X({{\mathbb{D}}})) \subset A({{\mathbb{D}}})$ (or $u \in M(X({{\mathbb{D}}})) \cap A({{\mathbb{D}}})$) and starting off with the space $X({{\mathbb{D}}})=\mathcal{B}_\alpha ({{\mathbb{D}}})$ for $0 < \alpha \le 1$ and showing that the condition, earlier observed to be sufficient for the Fredholmness of $M_u$, is also necessary.  Finally, we show the necessity of the condition in the case of those Bergman-Sobolev spaces $A^p_{\alpha,\beta}({{\mathbb{D}}})$ for which $M(A^p_{\alpha,\beta}({{\mathbb{D}}})) \subset A({{\mathbb{D}}}).$ From these two cases we obtain the essential spectrum of $M_u$ for several scales of spaces $\mathcal{B}_\alpha({{\mathbb{D}}})$ and $A^p_{\alpha,\beta}({{\mathbb{D}}})$ as the main result of Section 4. 

To conclude, our main result regarding the spectra of multiplication operators acting on $X(\mathbb B_n)$ is Theorem \[MainThmSpec\].  The essential spectra of operators $M_u$ acting on certain spaces $X({{\mathbb{D}}})$ having their multiplier spaces $M(X({{\mathbb{D}}}))$ contained in the disk algebra are described in Theorem \[MainResEssSpec\].  In the case of general $X({{\mathbb{D}}})$ with $M(X({{\mathbb{D}}})) = H^\infty({{\mathbb{D}}})$, the essential spectra of operators $M_u$ are characterized in Theorem \[EssSpecMlutiHInfty\].  In Theorem \[EssSpecForMultiDim\], we present the high-dimensional case $n>1$ concerning the essential spectra of operators $M_u$ acting on general spaces $X(\mathbb B_n)$.

Now we introduce some definitions and notations.  Throughout this article, let $\mathbb Z_{\geq a} = \{n\in\mathbb Z : n\geq a\}$ and $\mathbb Z_{>a}=\{n\in\mathbb Z : n> a\}$, where $a\in\mathbb R$. Furthermore, let $\mathbb B_n = \{z\in\mathbb{C}^n : {\left\vertz\right\vert}<1 \}$, $n\in \mathbb Z_{\geq 1}$, be the open unit ball in $\mathbb C^n$ and ${{\mathbb{D}}}= \mathbb B_1$. Moreover, let ${\mathcal{H}}(\mathbb B_n)$ be the space of all analytic functions $f\colon \mathbb B_n\to \mathbb{C}$ and ${\mathcal P}(\mathbb B_n)$ be the set of all analytic polynomials $p\colon\mathbb B_n\to \mathbb C$ such that $p(z)=\sum_{k \in J} c_k z^k$, where $J \subset {{\mathbb{Z}}}^n_{\geq 0}$ is a finite set, $k = (k_1,\ldots, k_n) \in {{\mathbb{Z}}}^n_{\geq 0}, \, |k| = k_1+\ldots+k_n$, $z^k = z_1^{k_1}\cdot \ldots \cdot z_n^{k_n}$ and $c_k\in \mathbb C$ for $ k \in J$.

We also recall that a bounded linear operator $T$ acting on a Banach space is Fredholm if it has closed range and both kernel and cokernel of $T$ are finite dimensional.  The essential spectrum $\sigma_e(T)$ of an operator $T$ is defined as $\sigma_e(T)=\{\lambda \in {{\mathbb{C}}}: T-\lambda I \textup{ is not Fredholm}\},$ where $I$ is the identity operator, and the reader may observe that $\sigma_e(T)$ is a subset of the spectrum $\sigma(T)$. See [@AA] for more details on Fredholm properties of bounded operators. 

For any $f\in {\mathcal{H}}(\mathbb B_n),$ the gradient of $f$ is given by $$\nabla f(z) = \left( \frac{\partial f}{\partial {z_1}},...,  \frac{\partial f}{\partial {z_n}}\right)$$ and will be denoted $Df(z)$ in the case $n=1$.

Let $\beta\in\mathbb{R}$ and $f\in {\mathcal{H}}(\mathbb B_n)$. The fractional radial derivative $R^\beta$ is given by $$R^\beta f (z) = \sum_{k=1}^\infty k^\beta f_k(z),$$ where $f(z)=\sum_{k=0}^\infty f_k(z)$ is the homogeneous expansion of $f\in {\mathcal{H}}(\mathbb B_n)$. Let $I\colon {\mathcal{H}}(\mathbb B_n) \to  {\mathcal{H}}(\mathbb B_n) $ be the identity operator.  The operator $(I+R)^\beta$ will also be used and is naturally defined by $$(I+R)^\beta f (z) = \sum_{k=0}^\infty (1+k)^\beta f_k(z).$$

For expressing asymptotic behaviour, the notation $a_k\sim b_k$, as $k\to\infty$, means $\lim_{k\to\infty} \frac{a_k}{b_k}=1$. Moreover, by $a(x)\gtrsim b(x)$ (or $a(x)\lesssim b(x)$) we indicate the existence of a constant $C>0$ independent of $x$ such that $a(x)\geq C b(x) $ (or $a(x)\leq C b(x)$) for all $x$ in some implicit set.  If both $a(x)\gtrsim b(x)$ and $a(x)\lesssim b(x)$ hold, we write $a(x)\asymp b(x)$. When two Banach spaces $X_1$ and $X_2$ are isomorphic, we use the notation $X_1 \simeq  X_2$.

Conditions and Examples
=======================

We deal with a vector space $X(\mathbb B_n)$ of analytic functions on $\mathbb B_n$ and a norm $\|\cdot\|_X$ on it, that renders $X(\mathbb B_n)$ a Banach space.  As usual, for each $z\in \mathbb B_n,$ the evaluation functional $\delta_z$ is defined by $\delta_z(f) = f(z)$ for all $f \in X(\mathbb B_n)$. We assume that $X(\mathbb B_n)$ contains the constant functions, so then all $\delta_z$ are non-zero.  Furthermore, we associate to $X(\mathbb B_n)$ another Banach space $Y(\mathbb B_n)\subset {\mathcal{H}}(\mathbb B_n)$ containing the constant functions and equipped with the norm ${\left\Vert\cdot\right\Vert}_Y$ as will be explained below. 

The Banach spaces $X(\mathbb B_n)$ and $Y(\mathbb B_n)$ are often assumed to satisfy the first three conditions below:

1.   \[finerThanCOTop\] *The topologies induced by ${\left\Vert\cdot\right\Vert}_X$ and ${\left\Vert\cdot\right\Vert}_Y$ are both finer than the compact-open topology $\tau_0.$*

In particular, for every $z\in \mathbb B_n$, $\delta_z$ is a bounded linear functional on both $X(\mathbb B_n)$ and $Y(\mathbb B_n)$.

Let $$M(X(\mathbb B_n))=\left\{    u\in {\mathcal{H}}(\mathbb B_n) :  uf \in X(\mathbb B_n)   \  \text{for all}     \  f\in X(\mathbb B_n)  \right\}.$$ Using condition \[finerThanCOTop\] and the closed graph theorem, it follows that every $u\in M(X(\mathbb B_n))$ induces a bounded linear operator $M_u: X(\mathbb B_n)\to X(\mathbb B_n).$

1.   \[RelationXtoY\] *For some $N\in \mathbb Z_{\geq 1}$ it holds that ${\left\Vertf\right\Vert}_X \asymp {\left\vertf(0)\right\vert}+{\left\VertR^N f\right\Vert}_Y$ for all $f\in {\mathcal{H}}(\mathbb B_n)$.*

Condition \[RelationXtoY\] describes a relationship between the Banach spaces $X(\mathbb B_n)$ and $Y(\mathbb B_n)$ such that Lemma \[MuInvert\] holds.  Since the lemma is trivial for spaces $X(\mathbb B_n)$ with $ M(X(\mathbb B_n))=H^\infty (\mathbb B_n)$, this condition may be omitted when such spaces are considered.  For these spaces we have $Y(\mathbb B_n)=X(\mathbb B_n)$.

1.   \[HinftySubsetMultiplY\] *$H^\infty(\mathbb B_n)\subset M(Y(\mathbb B_n))$.*

By condition \[finerThanCOTop\] it is well-known that $\sup_{z\in\mathbb B_n}|u(z)|\le ||M_u||$ for all $u\in M(X(\mathbb B_n))$, so $M(X(\mathbb B_n))\subset H^\infty(\mathbb B_n) \ \ \text{ and } \ \ M(Y(\mathbb B_n)) = H^\infty(\mathbb B_n) ,$ where also condition \[HinftySubsetMultiplY\] is used in the second statement.  Since $u\mapsto M_u $ is bounded according to the bounded inverse theorem, it follows from the boundedness of $M_u$, that there exists a constant $C>0$ such that $\|M_ug\|_Y\le  C\|u\|_\infty  ||g||_Y$ for all $g\in Y(\mathbb B_n)$ and $u\in M(Y(\mathbb B_n))$.

When considering the case $n=1$, we will need the following condition to determine the essential spectra of the multiplication operator generated by $u\in M(X({{\mathbb{D}}}))$.

1.   \[divideByPolynom\] *If $f\in X(\mathbb D)$ has a zero at $z_0\in \mathbb D$, then $\frac{f(z)}{z- z_0}\in X(\mathbb D).$*

\[dividingWithPolynomial\] Let $f\in \mathcal H (\mathbb D )$ and $v \colon \mathbb D \to [0,\infty)$ be a bounded function such that $v(z)=v(|z|)$ for all $z\in\mathbb D$. Moreover, let $N\in\mathbb Z_{\geq 0}$ be such that $$\sup_{z\in\mathbb{D}} v(z) |D^N f(z)| <\infty.$$ If $z_0\in \mathbb D$ is a zero of $f$, then $$\sup_{z\in\mathbb{D}}v(z) \left|D^N\frac{f(z)}{z-z_0}\right|<\infty.$$

Let $g(z)=\frac{f(z)}{z-z_0}$. Since $f$ is analytic with a zero at $z_0$ we have $D^N g \in \mathcal H (\mathbb D)$. Thus, $h(z) = v(z) |D^N g(z)|$ is bounded on $\mathbb D$ if and only if $h$ is bounded near the boundary.  For $z\in T=\left\{ z\in\mathbb D:|z|>\frac{1+|z_0|}{2}\right\}$, we have the following estimate

$$\label{GBlochUpperApprox}
|D^N g (z)| = \left|\sum_{j=0}^N \binom{N}{j} D^j f (z) D^{N-j}(z-z_0)^{-1}\right| \leq \sum_{j=0}^N \frac{ N!|D^j f(z)| }{  (|z|-|z_0|)^{N-j+1}   }.$$

Furthermore, for $k\geq 0$ we have

$$\begin{aligned}
|D^{k} f (z)| &\leq \left| \int_{C_z} D^{k+1} f(w) dw\right| + |D^k f(0)| \\
&\leq |z| \sup_{|w|=|z|}| D^{k+1} f(w)| + |D^k f(0) |\\
&\leq \sup_{|w|=|z|}| D^{k+1} f(w)| + |D^k f(0) |,\end{aligned}$$

where $C_z$ is the line from $0$ to $z$ in $\mathbb D$.

By induction, it can be shown that $$|D^{k} f (z)| \leq \sup_{|y|=|z|}  | D^{N} f(y)| + \sum_{j=0}^{N-k-1}|D^{k+j}f(0) |$$ for $0\leq k \leq N$. Moreover, from the fact that $\sup_{z\in\mathbb D}\sup_{|w|=|z|}$ is interchangable with $\sup_{w\in\mathbb D}$ and $v(z) = v(|z|)$ we now obtain

$$\begin{aligned}
 v(z) |D^{k} f (z)| &\leq \sup_{|y|=|z|} v(z) | D^{N} f(y)| + v(z)\sum_{j=0}^{N-k-1}|D^{k+j}f(0) |\\
&= \sup_{|y|=|z|} v(y) | D^{N} f(y)| + v(z)\sum_{j=0}^{N-k-1}|D^{k+j}f(0) |\\
&\leq \sup_{z\in \mathbb D}  v(z)| D^{N} f(z)| + \sup_{z\in \mathbb D}  v(z) \sum_{j=0}^{N-1}|D^{j}f(0) |= M_{f,N,v}<\infty\end{aligned}$$

for all $z\in\mathbb D$. Especially for $z\in T$, using (\[GBlochUpperApprox\]), we have

$$\begin{aligned}
 v(z)|D^N g (z)|&\leq \sum_{k=0}^N \frac{N!   v(z) |D^k f(z)| }{   (|z|-|z_0|)^{N-k+1}   }\leq  M_{f,N,v}\sum_{k=0}^N \frac{N!2^{N-k+1}   }{   (1-|z_0|)^{N-k+1}   }<\infty,\end{aligned}$$

which proves the lemma. 

Next, we list a number of spaces satisfying the above conditions \[finerThanCOTop\]-\[divideByPolynom\].  However, in part **(c)** we consider spaces $X(\mathbb B_n)$ for which $M(X(\mathbb B_n))=H^\infty(\mathbb B_n)$, implying that condition \[RelationXtoY\] is irrelevant. 

[**Examples.**]{}  $\textbf{(a)}$ For $\alpha>0$ the Bloch-type space $X(\mathbb B_n)=\mathcal B _\alpha(\mathbb B_n)$ is the space of all $f\in {\mathcal{H}}(\mathbb B_n)$ satisfying ${\left\Vertf\right\Vert}_{\mathcal B_ \alpha}=|f(0)| + \sup_{z\in\mathbb B_n}(1-{\left\vertz\right\vert}^2)^\alpha {\left\vert\nabla f(z)\right\vert} < \infty$, see [@Z].  To these spaces correspond $$Y(\mathbb B_n)= H_{\alpha}^\infty(\mathbb B_n)=\{f\in\mathcal H(\mathbb B_n) : {\left\Vertf\right\Vert}_{H^\infty_\alpha}=\sup_{z\in\mathbb B_n} (1-{\left\vertz\right\vert}^2)^\alpha {\left\vertf(z)\right\vert}<\infty\},$$ see [@BS].  The little Bloch-type space $\mathcal B _{0,\alpha}(\mathbb B_n)$ is the subspace of $\mathcal B _\alpha(\mathbb B_n)$ satisfying $\lim_{|z|\to1} (1-{\left\vertz\right\vert}^2)^\alpha {\left\vert\nabla f(z)\right\vert}=0$. It is well-known that these spaces obey \[finerThanCOTop\].  Let ${\left\Vertf\right\Vert}_{{\mathcal{B}}R_\alpha}=|f(0)| + \sup_{z\in\mathbb B_n}(1-{\left\vertz\right\vert}^2)^\alpha {\left\vertRf(z)\right\vert}$. According to Theorem 7.1 in [@Z], it holds that $$\{f\in \mathcal H (\mathbb B_n):{\left\Vertf\right\Vert}_{{\mathcal{B}}_\alpha}<\infty\}=\{f\in \mathcal H (\mathbb B_n):{\left\Vertf\right\Vert}_{{\mathcal{B}}R_\alpha}<\infty\}.$$ Therefore it follows from the bounded inverse theorem that ${\left\Vert\cdot\right\Vert}_{\mathcal B _\alpha}\asymp {\left\Vert\cdot\right\Vert}_{{\mathcal{B}}R_\alpha}$ and, hence, both of these spaces satisfy condition \[RelationXtoY\].  Condition \[HinftySubsetMultiplY\] holds by definition.  We will consider the space $({\mathcal{B}}_\alpha(\mathbb D),{\left\Vert\cdot\right\Vert}_{\mathcal B _\alpha})$ in the one-dimensional case. 

By Theorem [2.1 (i)]{} in [@OSZ], for $0 < \alpha < 1$, $u \in M({\mathcal{B}}_\alpha({{\mathbb{D}}}))  \text{  if and only if} \  u \in {\mathcal{B}}_\alpha({{\mathbb{D}}}) \cap H^\infty({{\mathbb{D}}})=  {\mathcal{B}}_\alpha({{\mathbb{D}}})\subset A({{\mathbb{D}}})$, where the inclusion is found in Theorem 7.9 in [@Z].  When $\alpha = 1$, we get from Theorem [2.1 (ii)]{} in [@OSZ] that $u \in M({\mathcal{B}}_\alpha({{\mathbb{D}}}))$ if and only if $$\sup_{z\in{{\mathbb{D}}}} {\left\vertu'(z)\right\vert}(1 - {\left\vertz\right\vert}^2)\log\left(\frac{e}{1 - {\left\vertz\right\vert}^2}\right) < \infty 
    \quad \text{and} \quad  u\in  H^\infty({{\mathbb{D}}}).$$ Therefore $u\in {\mathcal{B}}_{0,1}({{\mathbb{D}}}).$ Finally by Theorem [2.1 (iii)]{} in [@OSZ] we have for $\alpha > 1$ that $u \in M({\mathcal{B}}_\alpha({{\mathbb{D}}}))  \text{ if and only if}  \   u \in {\mathcal{B}}_1({{\mathbb{D}}}) \cap H^\infty({{\mathbb{D}}})=  H^\infty({{\mathbb{D}}}).$ According to Lemma \[dividingWithPolynomial\] a function belonging to ${\mathcal{B}}_\alpha({{\mathbb{D}}})$ will remain in ${\mathcal{B}}_\alpha({{\mathbb{D}}})$ after removing a finite number of zeros $z_0$ through division by $z-z_0$, which proves \[divideByPolynom\].  The notations ${\mathcal{B}}({{\mathbb{D}}})$ and ${\mathcal{B}}_0({{\mathbb{D}}})$ stand for ${\mathcal{B}}_1({{\mathbb{D}}})$ and ${\mathcal{B}}_{0,1}({{\mathbb{D}}})$ respectively.\

$\textbf{(b)}$ Let $\beta\ge 0$, $\alpha\ge -1$ and $1\leq p <\infty$. The holomorphic Sobolev space $A_{\alpha,\beta}^p(\mathbb B_n)$ is defined by $$A_{\alpha,\beta}^p(\mathbb B_n) = \{  f\in{\mathcal{H}}(\mathbb B_n) : {\left\Vertf\right\Vert}_{A_{\alpha,\beta}^p}  <\infty \},$$ where the norm is defined by $${\left\Vertf\right\Vert}_{A_{\alpha,\beta}^p} ={\left\Vert(I+R)^\beta f\right\Vert}_{A_\alpha^p} =\left( \int_{\mathbb B_n} |(I+R)^\beta f(z)|^p  dA_\alpha(z)  \right)^\frac{1}{p}$$ for $\alpha>-1$ and $${\left\Vertf\right\Vert}_{A_{-1,\beta}^p} ={\left\Vert(I+R)^\beta f\right\Vert}_{H^p} =\left( \int_{\partial\mathbb B_n} |(I+R)^\beta f(z)|^p  dS(z)  \right)^\frac{1}{p}.$$ Furthermore, $dA_\alpha(z)=  \frac{\Gamma(n+\alpha+1)}{n!  \Gamma(\alpha+1)} (1-{\left\vertz\right\vert}^2)^\alpha dA(z)$, where $dA(z)$ is the $2n$-dimensional Lebesgue measure normalized so that $\int_{\mathbb B_n} dA(z) = 1$, and hence, $\int_{\mathbb B_n} dA_\alpha(z) = 1$ for every $\alpha>-1$. The notation $dS(z)$ stands for the surface measure satisfying $\int_{\partial \mathbb B _n} dS(z)=1$. The holomorphic Sobolev spaces can be partitioned into the Bergman-Sobolev spaces, $\alpha>-1$, and the Hardy-Sobolev spaces, $H^p_\beta(\mathbb B_n) =A^p_{-1,\beta}(\mathbb B_n) .$ In case of $\beta=0$, these spaces are called the weighted Bergman spaces $A^p_\alpha(\mathbb B_n) = A^p_{\alpha,0}(\mathbb B_n)$ with $\alpha > -1$ and the Hardy spaces $H^p (\mathbb B_n) = A^p_{-1,0}(\mathbb B_n)$.

For $p\ge 1$, $\alpha_j > -1$, $\beta_j \ge 0$ $(j=1,2)$, with $\alpha_1- \alpha_2 = p(\beta_1 -\beta_2)$, the following equivalence holds by Theorem 5.12 in [@BB] (see also [@CKS]): $$\label{formimp}
A_{\alpha_1,\beta_1}^p(\mathbb B_n)   \simeq A_{\alpha_2,\beta_2}^p(\mathbb B_n),$$ where the isomorphism is given by the identity operator, and hence, the spaces have equivalent norms.  By the same theorem, one also obtains the statement (\[formimp\]) for $\alpha_1=-1$ and $p=2$. From this it follows that for $\beta_1 < \frac{1+\alpha_1}{p}$, where equality may be used in the case of $p=2$, we have $A_{\alpha_1,\beta_1}^p(\mathbb B_n)   \simeq A_{\alpha_1-\beta_1 p,0}^p(\mathbb B_n)$. The right-hand side is a weighted Bergman space or $H^2(\mathbb B_n)$, hence, $M(A_{\alpha,\beta}^p(\mathbb B_n)) = H^\infty (\mathbb{B}_n)$ for $\beta<\frac{1+\alpha}{p}$, where equality may be used in the case of $p=2$. Regarding the case $n=1$, if $\beta > \frac{2+\alpha}{p}$, then $A_{\alpha,\beta}^p({{\mathbb{D}}})$ is an algebra and $M(A_{\alpha,\beta}^p({{\mathbb{D}}}))= A_{\alpha,\beta}^p({{\mathbb{D}}})$, see [@BB1].  In this setting, there is a $b<\beta$ satisfying $0 < b-\frac{2+\alpha}{p} <1$, so that $$A_{\alpha,\beta}^p({{\mathbb{D}}}) \subset A_{\alpha,b}^p({{\mathbb{D}}}) \subset \Lambda_{ b-\frac{2+\alpha}{p}}({{\mathbb{D}}}) \subset A({{\mathbb{D}}}),$$ where $\Lambda_{ b-\frac{2+\alpha}{p}}$ is a Lipschitz space, see [@Z].  The first inclusion follows from (\[formimp\]), the second inclusion can be found in Theorem 5.5 in [@BB] and the last one is given by Theorem 7.9 in [@Z].  Furthermore, by Proposition 2.2 in [@CKS], we have for $p\ge 1$, $\alpha\ge -1,$ and every positive integer $N$ that $$\label{good}
{\left\Vertf\right\Vert}_{A_{\alpha,N}^p} \asymp \sum_{j=0}^{N-1} |D^jf(0)| + {\left\VertD^Nf\right\Vert}_{A_\alpha ^p}$$ for $f\in{\mathcal{H}}(\mathbb D)$. Next, we check the conditions \[finerThanCOTop\] - \[divideByPolynom\]. 

The topology generated by ${\left\Vert\cdot\right\Vert}_{A^p_{\alpha,\beta}}$ is finer than the compact-open topology $\tau_0$, so condition \[finerThanCOTop\] holds.  Indeed, the statement follows from Lemma 5.6 in [@BB] with the use of supremum over an arbitrary compact subset of $\mathbb B_n$. Hereafter, we will assume that $\beta\geq \frac{1+\alpha}{p}$. For smaller $\beta$ it was mentioned that the multiplier space is $H^\infty(\mathbb B_n)$ which is considered in **Examples** **(c)**, where the the space $A_{\alpha,\beta}^p(\mathbb B_n)$ can be viewed as a weighted Bergman space. 

In the case $N>\beta -\frac{\alpha+1}{p}\geq 0$, an application of (\[formimp\]) gives that $$f\in A_{\alpha,\beta}^p(\mathbb B_n)  \ \text{if and only if} \   f\in A_{(N-\beta)p+\alpha,N}^p(\mathbb B_n) \ \text{if and only if} \  R^Nf\in  A_{(N-\beta)p+\alpha}^p(\mathbb B_n).$$ Therefore, let $X(\mathbb B_n) =A_{\alpha,\beta}^p(\mathbb B_n)$ and $Y(\mathbb B_n) =A_{(N-\beta)p+\alpha}^p(\mathbb B_n)$, where $N=\inf\{\hat{N}\in\mathbb Z_{\geq 1}:\hat{N}>\beta -\frac{\alpha+\frac{1}{2}}{p}\}$. Moreover, for $f\in {\mathcal{H}}(\mathbb B_n)$ we have $$\label{BergSobNormEquivRel}
{\left\Vert(I+R)^\beta f\right\Vert}_{A_\alpha^p} \asymp {\left\vertf(0)\right\vert}+{\left\VertR^\beta f\right\Vert}_{A_\alpha^p},$$ according to Lemma \[BergSobEquivNorm\].  Condition \[RelationXtoY\] follows by first using the equivalence of the norms ${\left\Vert\cdot\right\Vert}_{A_{\alpha,\beta}^p}$ and ${\left\Vert\cdot\right\Vert}_{A_{(N-\beta)p+\alpha,N}^p}$ by (\[formimp\]), and then applying (\[BergSobNormEquivRel\]) to the latter norm.  Furthermore, it holds that $M(A^p_\alpha(\mathbb B_n)) = H^\infty(\mathbb B_n)$, which shows that condition \[HinftySubsetMultiplY\] is satisfied. 

Let us check the condition \[divideByPolynom\] for $A_{\alpha,\beta}^p({{\mathbb{D}}})$. We assume that $f \in A_{\alpha,\beta}^p({{\mathbb{D}}})$ has a zero at $z=z_0$. Let us show that $\frac{f}{z-z_0} \in  A_{\alpha,\beta}^p({{\mathbb{D}}})$ by establishing that $R^N\left(\frac{f}{z-z_0}\right)  \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}})$. Let us take $|z_0| < r < 1.$ We may assume that $|z| \ge r,$ since $R^N\left(\frac{f}{z-z_0}\right) \in H({{\mathbb{D}}})$ is bounded on $r{{\mathbb{D}}}.$ We will utilize the following formula given in Proposition 6 in [@CHZ]: $$\begin{aligned}
R^N\left(\frac{f(z)}{z-z_0}\right) = \frac{(-1)^N}{(z-z_0)^{N+1}} \sum_{k=0}^N (-1)^k \binom{N+1}{k} (z-z_0)^k R^N((z-z_0)^{N-k}f),\end{aligned}$$ where $r \le |z| < 1$. It suffices to show that $R^N((z-z_0)f) \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}})$, which implies that $R^N((z-z_0)^{N-k}f) \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}})$ for $k=0,1,\ldots,N$. Using the general Leibniz rule we obtain $$\begin{aligned}
R^N((z-z_0)f) = \sum_{k=0}^N \binom{N}{k} R^{N-k}(z-z_0)R^k(f) =  \sum_{k=0}^{N} \binom{N}{k} z R^k(f).\end{aligned}$$ We observe that $\|zR^kf\|_{A^p_{(N-\beta)p + \alpha}} \le \|R^kf\|_{A^p_{(N-\beta)p + \alpha}}$ and $R^kf \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}})$ if and only if $$(1-|z|^2)^{N-k}R^{(N-k)}R^kf = (1-|z|^2)^{N-k}R^{N}f \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}}),$$ see [@Z p. 75].  The last statement holds, since $${\left\Vert(1-|z|^2)^{N-k}R^{N}f \right\Vert}_{A^p_{(N-\beta)p + \alpha}} \le {\left\VertR^{N}f\right\Vert}_{A^p_{(N-\beta)p + \alpha}} < \infty,$$ where we used the fact $R^Nf  \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}}).$ So we have that $R^kf \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}})$ for $k = 0,1,\ldots,N$ and consequently $R^N((z-z_0)f) \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}})$. Therefore $R^N\left(\frac{f}{z-z_0}\right)  \in A^p_{(N-\beta)p + \alpha}({{\mathbb{D}}}).$

It should also be mentioned that the spaces $A^p_{\alpha,\beta}(\mathbb B_n)$ are reflexive for $p>1$, see Proposition 5.7 (iv) in [@BB]. 

$\textbf{(c)}$ We consider all spaces $X(\mathbb B_n)$ that satisfies \[finerThanCOTop\] and $M(X(\mathbb B_n))=H^\infty(\mathbb B_n)$. Furthermore, condition \[divideByPolynom\] is also assumed to hold if $n=1$. Letting $Y(\mathbb B_n)=X(\mathbb B_n)$ condition \[HinftySubsetMultiplY\] is also satisfied and condition \[RelationXtoY\] is irrelevant, see the remark after condition \[RelationXtoY\].  These spaces include growth spaces $H^\infty_\alpha({{\mathbb{D}}}), \, \alpha  >0,$ and weighted Hardy spaces $H_w^p(\mathbb D),\ p >1$, where $w\in (A^p)$, that is, $w$ satisfies the Muckenhoupt $(A^p)$-condition, see details in [@BGZ].  Considering the weighted Hardy spaces, condition \[finerThanCOTop\] follows from the proof of Lemma 2.1 in [@BGZ].  Notice that if $w\in (A^p), \ p>1$, then the critical exponent $q_w<p$. For $f\in\mathcal H (\mathbb D)$ we have ${\left\Vertf\right\Vert}_{H_w^p}<\infty$ if and only if $$\lim_{r\to1^-} \int_{-\pi}^\pi \left| f(re^{i\theta}) \right|^p w(\theta) d\theta   <\infty.$$ Since, for every $z_0\in\mathbb D$, there exists $r<1$ such that $\frac{1}{z-z_0}$ is bounded on $\mathbb D\setminus r\mathbb D$, condition \[divideByPolynom\] follows.  Condition \[divideByPolynom\] is proved by similar arguments for many spaces, for example, weighted Bergman spaces, growth spaces and Hardy spaces. 

\[BergSobEquivNorm\] Let $\beta\geq 0$. If either $\alpha> -1$ and $p\geq 1$, or $\alpha= -1$ and $p=2$, it holds that

$${\left\Vertf\right\Vert}_{A^p_{\alpha,\beta}} \asymp {\left\vertf(0)\right\vert}+{\left\VertR^\beta f\right\Vert}_{A^p_\alpha} , \ f\in\mathcal H (\mathbb B_n).$$ Moreover, the space $A^p_{\alpha,\beta}(\mathbb B_n)$ endowed with the norm defined as ${\left\Vertf\right\Vert}_{p,\alpha,\beta}={\left\vertf(0)\right\vert}+{\left\VertR^\beta f\right\Vert}_{A^p_\alpha}$ is a Banach space. 

Let $N$ be the smallest integer in the set $\mathbb Z_{> \beta+1}$ and $\gamma=p(N-\beta)+\alpha>-1$. By (\[formimp\]) and Lemma 1 in [@CHZ] we have ${\left\Vertf\right\Vert}_{A^p_{\alpha,\beta}} \asymp{\left\Vertf\right\Vert}_{A^p_{\gamma,N}} $ and ${\left\Vertf\right\Vert}_{p,\alpha, \beta} \asymp {\left\Vertf\right\Vert}_{p, \gamma, N}$ respectively.  The norm equivalences are also well known to experts in the case $\alpha=-1$ and $p = 2$. Therefore $$\left(A^p_{\alpha,\beta}(\mathbb B_n),   {\left\Vert\cdot\right\Vert}_{p,\alpha,\beta}   \right)$$ is a Banach space, since this is true for $$\left(A^p_{\alpha,\beta}(\mathbb B_n), {\left\Vert\cdot\right\Vert}_{p,\gamma',N}\right)$$ for all $\gamma'>-1$, see [@ZZ].  It now suffices to show that ${\left\Vertf\right\Vert}_{A^p_{\gamma,N}}  \asymp  {\left\Vertf\right\Vert}_{p, \gamma, N}$. Using Jensen’s inequality we have $$\begin{aligned}
\left|(I+R)^N f(z)\right| &=\left|\sum_{j=0}^N \binom{N}{j} R^j f(z)\right|\\
& \leq2^N\sum_{j=0}^N \frac{1}{2^N}\binom{N}{j} \left|R^j f(z)\right|\\
& \leq 2^N \left(\sum_{j=0}^N \frac{1}{2^N}\binom{N}{j} \left|R^j f(z)\right|^p\right)^\frac{1}{p}.\end{aligned}$$ Furthermore, [ $$\begin{aligned}
{\left\Vertf\right\Vert}_{A^p_{\gamma, N}}&=\left(\int_{\mathbb B _n}\left|(I+R)^N f(z)\right|^p dA_\gamma (z)\right)^\frac{1}{p}\\
&\leq \left( 2^{N(p-1)} \sum_{j=0}^N \binom{N}{j}\int_{\mathbb B _n} \left|R^j f(z)\right|^p dA_\gamma(z)\right)^\frac{1}{p}\\
&\asymp   \left(\sum_{j=0}^N\int_{\mathbb B _n} \left|R^N f(z)\right|^p dA_{p(N-j)+\gamma}(z)\right)^\frac{1}{p}\\
&\lesssim \left(  \int_{\mathbb B _n} \left|R^{N} f(z)\right|^p   dA_\gamma(z)\right)^\frac{1}{p} +{\left\vertf(0)\right\vert} =  {\left\Vertf\right\Vert}_{p, \gamma, N}.  \end{aligned}$$ ]{}Our approach to prove the converse is very similar.  It holds that [ $$\begin{aligned}
\int_{\mathbb B _n}\left|R^N f(z)\right|^p dA_\gamma (z)&=\int_{\mathbb B _n}\left|(I+R-I)^N f(z)\right|^p dA_\gamma (z)\\
&\leq 2^{N(p-1)} \sum_{j=0}^N \binom{N}{j}\int_{\mathbb B_n} \left|(I+R)^j f(z)\right|^p dA_\gamma(z)\\
&\asymp  \sum_{j=0}^N \int_{\mathbb B _n} \left|(I+R)^{N} f(z)\right|^p   dA_{p(N-j)+\gamma}(z)\\
&\lesssim \int_{\mathbb B _n} \left|(I+R)^N f(z)\right|^p   dA_\gamma(z).\end{aligned}$$ ]{}From this and Lemma 5.6 in [@BB], it follows that there exists a constant $M=M(n,N,p,\gamma)>0$ such that $$\begin{aligned}
{\left\VertR^Nf\right\Vert}_{A^p_\gamma} + {\left\vertf(0)\right\vert} &\leq M{\left\Vert(I+R)^Nf\right\Vert}_{A^p_\gamma}+{\left\vertf(0)\right\vert}\leq 2M{\left\Vert(I+R)^Nf\right\Vert}_{A^p_\gamma},\end{aligned}$$ which finishes the proof. 

The spectrum of $M_u$
=====================

Next, we will characterize the spectra of multiplication operators acting on $X(\mathbb B_n)$ in the case that there exists a space $Y(\mathbb B_n)$ such that conditions \[finerThanCOTop\] - \[HinftySubsetMultiplY\] are satisfied.  Condition \[RelationXtoY\] is crucial in the following lemma.  The corresponding results for the Hardy-Sobolev Hilbert spaces were obtained in [@CHZ]. 

\[MuInvert\] Assume that *\[finerThanCOTop\]*, *\[RelationXtoY\]* and *\[HinftySubsetMultiplY\]* are satisfied and let $u\in M(X(\mathbb B_n))$. The following statements are equivalent:

1.   $\frac{1}{u}\in M(X(\mathbb B_n))$,\[reciprocal\_uMX\]\

2.   $\frac{1}{u}\in H^\infty \mathbb (\mathbb B_n)$,\[reciprocal\_uHinfty\]\

3.   $M_u$ is invertible.\[muInvert\]

Assuming $\frac{1}{u}\in M(X(\mathbb B_n))$ we obtain immediately, by the remark after condition \[HinftySubsetMultiplY\], that $\frac{1}{u}\in H^\infty (\mathbb B_n)$. To prove the converse implication we will use the formula $$\label{RadialDerivOfQuotient}
R^N \left( \frac{f}{u} \right) = \frac{  (-1)^N   }{   u^{N+1}  } \sum_{k=0}^N (-1)^k \binom{N+1}{k} u^k  R^N \left( u^{N-k} f  \right), \ f\in {\mathcal{H}}(\mathbb B_n),$$ which can be found in Corollary 5 in [@CHZ].  The proof of the formula uses the derivative $D$, but the formula remains valid for all linear operators $S$ that admit the law $S(fg) = fSg+gSf, \ f,g\in\mathcal H (\mathbb D),$ and for which the formula is valid for $N=1$. Moreover, the dimension $n$ is irrelevant for the proof, and therefore we may replace $D$ with $R$ and also consider the formula in higher dimensions.  Notice that (\[RadialDerivOfQuotient\]) is invalid for $N=0$.

If $\frac{1}{u}\in H^\infty(\mathbb B_n)$, then $u$ is uniformly bounded from below, that is, there exists a $0<c<1$ such that $\inf_{z\in\mathbb  B_n} |u(z)|\geq c$ and hence formula (\[RadialDerivOfQuotient\]) is applicable.  By condition \[RelationXtoY\] we have that $f\in X(\mathbb B_n)$ if and only if $R^N f  \in Y(\mathbb B_n)$.

One should also notice that $u^k\in M(X(\mathbb B_n))$ and $(\frac{1}{u})^k\in H^\infty(\mathbb B_n)$ for all $k\in\mathbb Z_{\geq 0}$. For $f\in X(\mathbb B_n)$ we obtain that $u^kf\in X(\mathbb B_n)$ for all $k\in\mathbb Z_{\geq 0}$, and therefore,

[ $$\begin{aligned}
{\left\Vert   \frac{f}{u}  \right\Vert}_X&\asymp {\left\Vert   R^N \left( \frac{f}{u} \right)  \right\Vert}_Y +{\left\vert\frac{f(0)}{u(0)}\right\vert}\\
&\leq  \sum_{k=0}^N  \binom{N+1}{k}  {\left\Vert  \left(   \frac{1}{u}   \right)^{N+1-k}  R^N \left( u^{N-k} f  \right) \right\Vert}_Y+{\left\vert\frac{f(0)}{u(0)}\right\vert}\\
&\lesssim   \sum_{k=0}^N   {\left\Vert  \left(   \frac{1}{u}   \right)^{N+1-k} \right\Vert}_\infty {\left\Vert R^N \left( u^{N-k} f  \right) \right\Vert}_Y+{\left\vert\frac{f(0)}{u(0)}\right\vert}\\
&\leq  \left(   \frac{1}{c}   \right)^{N+1}  \sum_{k=0}^N {\left\Vert R^N \left( u^{N-k} f  \right) \right\Vert}_Y+{\left\vert\frac{f(0)}{u(0)}\right\vert}\\
&\lesssim  \sum_{k=0}^N   {\left\Vert  u^{N-k} f  \right\Vert}_X+{\left\vert\frac{f(0)}{u(0)}\right\vert} < \infty,\end{aligned}$$ ]{}where the remark after condition \[HinftySubsetMultiplY\] gives the second inequality.  Hence, we have shown that the two statements \[reciprocal\_uMX\] and \[reciprocal\_uHinfty\] are equivalent.  If $\frac{1}{u}\in  M(X(\mathbb B_n))$, then clearly $f\mapsto \frac{f}{u}$ is the inverse of $M_u$. Conversely, if $M_u$ is invertible, then $M_{\frac{1}{u}}$ must be the unique bounded inverse, so $\frac{1}{u}\in M(X(\mathbb B_n)).$

\[MainThmSpec\] Assume that *\[finerThanCOTop\]*, *\[RelationXtoY\]* and *\[HinftySubsetMultiplY\]* are satisfied and let $M_u:X(\mathbb B_n) \to X(\mathbb B_n)$ be a multiplication operator generated by $u\in M(X(\mathbb B_n))$. The spectrum of $M_u$ is given by $\sigma(M_u) =
 {\overline{u(\mathbb B_n)}}$.

Let $\lambda\in\mathbb C$.  Clearly $u - \lambda\in  M(X(\mathbb B_n)).$ If $\lambda\in {\overline{u(\mathbb B_n)}}$, then $|u(z)-\lambda|$ is not bounded from below so $M_u-\lambda I = M_{u-\lambda}$ is not invertible by Lemma \[MuInvert\].  Using again Lemma \[MuInvert\], it follows that for any $\lambda\in\mathbb{C} \setminus{\overline{u(\mathbb B_n)}}$ the operator $M_u-\lambda I $ is invertible since $|u(z)-\lambda|$, in this case, is bounded from below.  Hence, the spectrum is given by $\sigma(M_u) = {\overline{u(\mathbb B_n)}}$.

The above result implies that $r(M_u) = ||u||_\infty\le ||M_u||$. Moreover, since the spectrum $\sigma(M_u) =
 {\overline{u(\mathbb B_n)}}$ is connected, when $u$ is continuous, any nonzero spectral radius would imply an uncountable number of points in the spectrum, from which it follows that the operator is not compact.  Consequently, $M_u$ is never compact if $u\ne 0.$

\[examplesOfSpectr\] Let $X(\mathbb B _n)$ be any of the following spaces

1.   $A^p_{\alpha,\beta}(\mathbb B _n),\ p\geq 1, \ \beta\geq 0, \text{ and }\alpha>-1$;\

2.   $\mathcal B_\alpha (\mathbb B _n),\ \alpha >0$;\

3.   $H_\beta^2(\mathbb B _n),\ \beta\geq 0$;\

4.   $H_w^p(\mathbb D),\ p >1, \ w \in (A^p)$.

Then the spectrum of a multiplication operator $M_u:X(\mathbb B _n)\to X(\mathbb B _n)$ is given by $\sigma(M_u) = {\overline{u(\mathbb B_n)}}$.

The essential spectrum of $M_u$ {#sect4}
===============================

Examining the essential spectrum of a multiplication operator when the domain is $\mathbb B_n, \ n>1$, the result concerning $H_\beta^2$, obtained by Cao, He and Zhu, can be made quite general, see Theorem \[EssSpecForMultiDim\].  In the case $n=1$, we have obtained a sufficient condition for Fredholmness in Lemma \[ThenFredholm\_ForX\], where all four conditions \[finerThanCOTop\]-\[divideByPolynom\] were assumed.  For the spaces mentioned in our main result, namely Theorem \[MainResEssSpec\], this condition is also necessary for Fredholmness, see lemmas \[IfFredholm\_BlochType\] and \[IfFredholm\_BergmanSobolev\], but for this to be proved, space-specific properties were used.  An asymptotic approximation for the behaviour of the norm of the peak functions is necessary for the result concerning Bergman-Sobolev spaces.  The estimate given in Lemma 11 in [@CHZ] is insufficient for our purposes, not only because it only considers $p=2$, but also because it is not a sharp lower bound.  The necessity of an asymptotic approximation instead of a non-sharp lower bound of the behaviour is clear when an arbitrary $p\in(1,\infty)$ is considered in Theorem \[IfFredholm\_BergmanSobolev\]. 

\[EssSpecForMultiDim\] Assume that condition *\[finerThanCOTop\]* is satisfied and $n >1$. Furthermore, let $u\in M(X(\mathbb B_n))$ and $P_j\colon \mathbb B_n \to {{\mathbb{C}}}, \, P_j(z) = z_j$ for every $j = 1, \ldots , n.$ Suppose that $P_j \in  M(X(\mathbb B_n))$ for every $j$.

Then $  \sigma_e(M_u)  = \bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}} = {\overline{u(\mathbb B_n)}}  = \sigma(M_u) $. 

Let $\lambda\in u(\mathbb B_n).$ Since $n >1$, the function $u(z)  - \lambda$ has infinitely many distinct zeros, and therefore, there must exist an infinite subset $\{\alpha_k\}_{k=1}^\infty, \ \alpha_k=(\alpha_{k,1},...,\alpha_{k,n}),$ of these zeros such that for some $j = 1,\ldots,n$ we have $\alpha_{k,j} \ne \alpha_{l,j}$ whenever $k \ne l$. We first show, by induction, that $(\delta_{\alpha_k})_{k=1}^\infty$ are linearly independent in $Ker \ M^*_{u-\lambda}$. Clearly all $\delta_{\alpha_k}\in Ker \ M^*_{u-\lambda}.$ Suppose that $$\sum_{k=1}^m c_k \delta_{\alpha_k}=0$$ for some $m\in\mathbb Z_{\geq 1}$. If $m=1$, it follows that $c_1=0$. Assume that $m\geq 2$. For arbitrary $f\in X(\mathbb B_n)$ we have by assumption that $P_j f\in X(\mathbb B_n)$, so $$\sum_{k=1}^m c_k \alpha_{k,j} \delta_{\alpha_k}(f)=0  \ \text{and}\ \sum_{k=1}^m c_k \delta_{\alpha_k}(f)=0.$$ Hence $$\sum_{k=2}^m c_k (\alpha_{k,j} - \alpha_{1,j})\delta_{\alpha_k}(f)= \sum_{k=1}^m c_k (\alpha_{k,j} - \alpha_{1,j})\delta_{\alpha_k}(f)=0 \ \text{for all} \  f \in  X(\mathbb B_n),$$ and therefore, by the induction hypothesis, $c_k(\alpha_{k,j} - \alpha_{1,j})=0$ for all $k=2,...,m.$ This implies that $c_k=0$ for $k=2,...,m,$ and consequently $c_1=0.$ Then $Ker \ M^*_{u-\lambda}$ is infinite dimensional so that $M^*_{u-\lambda}$, and equivalently $M_{u-\lambda}$, is not Fredholm.  It follows that $u(\mathbb B_n)\subset \sigma_e(M_u)$ and, moreover, that $$\bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}}\subset {\overline{u(\mathbb B_n)}} \subset  \sigma_e(M_u)\subset \sigma(M_u).$$ For the converse conclusion, let $\lambda \notin \bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}}.$ Hence, there are $r\in (0,1)$ and $\delta>0$ such that $|\lambda - u(z)| \ge \delta$ for all $r < |z| < 1.$ Then $v(z) = (u(z) - \lambda)^{-1}$ is holomorphic and bounded on $\mathbb B_n\setminus r\overline{\mathbb B}_n.$ As in [@CHZ], using Hartogs’ extension theorem and the identity theorem, we can extend $v$ to a function $\tilde v\in{\mathcal{H}}(\mathbb B_n)$ such that $\tilde v(z) = (u(z) - \lambda)^{-1}$ for all $z\in \mathbb B_n,$ and therefore $\tilde v\in H^\infty(\mathbb B_n).$ Now $M_{u-\lambda}$ is invertible by Lemma \[MuInvert\], so $\lambda\notin\sigma(M_u).$

\[nOneIndependentPointEval\] Following the proof of Theorem \[EssSpecForMultiDim\] it is clear that $(\delta_{\alpha_k})_{k=1}^\infty$ are linearly independent when $n=1$.

Now we proceed to the case $n=1$.

The following result is based on ideas due to Axler [@A] that was carried on in [@BDL].  It holds for all spaces $X(\mathbb D)$ such that $M(X(\mathbb D)) = H^\infty(\mathbb D).$

\[IfFredholm\_ForX\] Assume that condition *\[finerThanCOTop\]* is satisfied and let $u\in M(X(\mathbb D))=H^\infty(\mathbb D)$. If $M_u \colon X(\mathbb D)\to X(\mathbb D)$ is Fredholm, then there are $r\in (0,1)$ and $\delta>0$ such that $|u(z)| \ge \delta$ for all $r\le |z| < 1.$

Assume we can find a sequence $(z_n)_{n=1}^\infty \subset { \mathbb D}$ with $|z_n|\to 1$ and $|u(z_n)|\to 0$ when $n\to\infty.$ Then we can assume that $(z_n)_n$ is an interpolating sequence in $H^\infty(\mathbb D)$ by going to a subsequence if necessary.  Therefore, (see e.g. [@An], Ch.  7.3) there is a constant $M> 0$ such that for each $N\in\mathbb N$ there is a function $u_N\in H^\infty(\mathbb D)$ with $$u_N(z_n) = \left\{
\begin{array}{@{}ll}
u(z_n), &  n\ge N,\\
0, &  n < N
\end{array} \right.$$ and ${\left\Vertu_N\right\Vert}_\infty \le M \sup_{n\ge N} {\left\vertu(z_n)\right\vert}$. Let $$Z_N = \{ f \in X(\mathbb D): \delta_{z_n}(f) = 0 \ \text{for all} \ n\ge N\},$$ which is a closed subspace of $X(\mathbb D)$. From Remark \[nOneIndependentPointEval\] we know that the $\delta_{z_n}\in X(\mathbb D)^*$ are linearly independent, which implies that $Z_N ^\perp$ is infinite-dimensional.  Since $\delta_{z_n}(u- u_N) =0$ for all $n\ge N$, we get $M_{u- u_N}(X(\mathbb D))\subset Z_N.$ Now $(X(\mathbb D)/Z_N)^* = Z_N^\perp$, so $X(\mathbb D)/Z_N$ is infinite-dimensional.  Hence $X(\mathbb D)/ M_{u- u_N}(X(\mathbb D))$ is also infinite-dimensional, and $M_{u -u_N}\colon X(\mathbb D) \to X(\mathbb D)$ is not Fredholm.  As $M(X(\mathbb D)) = H^\infty(\mathbb D)$ and the set of non-Fredholm operators is closed, it follows from $${\left\VertM_{u - u_N} - M_u\right\Vert} = {\left\VertM_{u_N}\right\Vert}\le C {\left\Vertu_N\right\Vert}_\infty\le C M \sup_{n\ge N} {\left\vertu(z_n)\right\vert}\to 0 \ \text{as} \ N\to\infty,$$ that $M_u$ is not Fredholm. 

\[ThenFredholm\_ForX\] Assume that *\[finerThanCOTop\]*, *\[RelationXtoY\]*, *\[HinftySubsetMultiplY\]* and *\[divideByPolynom\]* are satisfied and let $u\in M(X(\mathbb D))$. If there are $r\in (0,1)$ and $\delta>0$ such that $|u(z)| \ge \delta$ for all $r\le |z| < 1$, then $M_u \colon X(\mathbb D)\to X(\mathbb D)$ is Fredholm. 

By assumption we have that $u$ can have only finitely many zeros $\alpha_1,...\alpha_n$ inside $\mathbb D$ with multiplicities $m_1,...,m_n$ respectively.  Then for all $z\in\mathbb D$, $$u(z) = v(z) (z - \alpha_1)^{m_1}... (z - \alpha_n)^{m_n}= v(z) p(z),$$ where $v\in{\mathcal{H}}(\mathbb D)$ and $\frac{1}{v}\in H^\infty(\mathbb D)$.

Let us now define the point evaluation maps for derivatives by $\delta^{(k)}_z(f) = f^{(k)}(z)$ for all $z\in \mathbb D$ and all $k \in\mathbb Z_{\geq 0}$. By assumption \[finerThanCOTop\], it holds that $\delta^{(k)} _z  \in X(\mathbb D)^*$ for all $k$ and $z$. Clearly, $$M_u(X(\mathbb D)) \subset \bigcap_{i=1}^n\bigcap _{k=0}^{m_i-1} {\text  Ker} \ \delta_{\alpha_i}^{(k)}.$$ Let $f \in  \bigcap_{i=1}^n\bigcap _{k=0}^{m_i-1} {\text  Ker} \ \delta_{\alpha_i}^{(k)},$ so $ f^{(k)}(\alpha_i)=0$ for all $i=1,...,n$ and all $k=0,...,{m_i}-1.$ Then $\frac{f}{u} \in
 {\mathcal{H}}(\mathbb D).$ Now assumption \[divideByPolynom\] implies that $v\in M(X(\mathbb D)).$ Indeed, if $g\in X(\mathbb D)$, then $ug\in X(\mathbb D)$ and by assumption \[divideByPolynom\] it follows that $v g = \frac{ug}{p} \in X(\mathbb D).$ Therefore, $\frac{1}{v} \in M(X(\mathbb D))$ by Lemma \[MuInvert\], so that $\frac{f}{u} = \frac{f/p}{v}\in X(\mathbb D)$ by assumption \[divideByPolynom\].  As a result, $f=u\frac{f}{u}\in M_u(X(\mathbb D))$, and thus $$M_u(X(\mathbb D)) = \bigcap_{i=1}^n\bigcap _{k=0}^{m_i-1} \text{ Ker} \ \delta_{\alpha_i}^{(k)}.$$ Consequently, $M_u$ has closed range, and since $M_u\colon  X(\mathbb D)\to X(\mathbb D)$ is always injective, the dimension of the kernel of $M_u$ is finite.  Since $^\perp(\operatorname{span}\{\delta_{\alpha_i}^{(k)}\}) = {\text Ker} \  \delta_{\alpha_i}^{(k)}$, it follows that the $w^*$-closed one-dimensional space $\operatorname{span}\{\delta_{\alpha_i}^{(k)}\}  = ( {\text  Ker} \ \delta_{\alpha_i}^{(k)})^\perp$, see [@M Theorem 11 on p. 341].  Therefore, by [@M Theorem 13 on p. 342], we have $$M_u(X(\mathbb D))^\perp = \sum_{i=1}^n\sum _{k=0}^{m_i-1} (\text{ Ker} \ \delta_{\alpha_i}^{(k)})^\perp =\sum_{i=1}^n\sum _{k=0}^{m_i-1} \operatorname{span}\{ \delta_{\alpha_i}^{(k)} \},$$ and hence, the dimension of the co-kernel of $M_u$ is finite, and $M_u$ is Fredholm. 

\[EssSpecMlutiHInfty\] Assume that *\[finerThanCOTop\]*, *\[HinftySubsetMultiplY\]* and *\[divideByPolynom\]* are satisfied and $M(X(\mathbb D)) = H^\infty(\mathbb D).$ Let $M_u\colon X(\mathbb D) \to X(\mathbb D)$ be a multiplication operator generated by $u\in M(X(\mathbb D))$. The essential spectrum of $M_u$ is given by $\sigma_e(M_u) = \bigcap_{0 < r < 1} {\overline{u(\mathbb D \setminus r\mathbb D)}}$.

We have that $\lambda\in{\overline{u(\mathbb D \setminus r\mathbb D)}}$ for all $r\in(0,1)$ if and only if for all $r\in(0,1)$ there is a sequence $(z_n)_{n=1}^\infty\subset\mathbb D$ such that $|z_n| \ge r$ for all $n\in\mathbb N$ and $|u(z_n) -\lambda| \to 0$ when $n \to\infty.$ Since $M_u-\lambda I = M_{u-\lambda}$, we can now apply lemmas \[IfFredholm\_ForX\] and \[ThenFredholm\_ForX\] to conclude that the last statement equivalently means that $M_u-\lambda I$ is not Fredholm, that is $\lambda\in\sigma_e(M_u).  $ The use of Lemma \[ThenFredholm\_ForX\] is justified by the remark after condition [\[RelationXtoY\]]{}. 

In [**Examples**]{} it was stated that the multiplier spaces for $A_{\alpha,\beta}^p({{\mathbb{D}}})$ with $p \geq 1, \ \alpha>-1 , \ \beta < \frac{1+\alpha}{p} $; $H_w^p({{\mathbb{D}}})$ with $p>1, \ w\in (A^p)$ and ${\mathcal{B}}_\alpha({{\mathbb{D}}})$ with $\alpha>1$ are $H^\infty({{\mathbb{D}}})$. Thus, we obtain the following results. 

\[exampleOfEssSpectr\] In each of the following three cases:

1.   $p\geq 1,\ \alpha>-1$ and $\beta < \frac{1+\alpha}{p}$ with $u\in M(A^p_{\alpha,\beta}({{\mathbb{D}}}))$;\

2.   $\alpha>1$ with $u\in M({\mathcal{B}}_\alpha({{\mathbb{D}}})  )$;\

3.   $p >1, \ w \in (A^p)$ with $u\in M(H_w^p({{\mathbb{D}}}))$,

the essential spectrum of $M_u$ is given by $$\sigma_e(M_u) = \bigcap_{0 < r < 1} {\overline{u(\mathbb D \setminus r\mathbb D)}}.$$

It was shown in Theorem \[EssSpecForMultiDim\] that in higher dimensions, $n>1$, the essential spectra of multiplication operators coincide with their spectra for many spaces.  This is seldom true for $n=1$. In corollaries \[examplesOfSpectr\] and \[exampleOfEssSpectr\] and Theorem \[MainResEssSpec\], we list some spaces, on which multiplier operators have the spectrum given by $ {\overline{u(\mathbb B_n)}}$ and the essential spectrum given by $\bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}} $.  Although the sets may differ, their spectral and essential spectral radii coincide according to the following remark. 

\(a) Let $n\in\mathbb Z_{\geq 1}$. Since the decreasing sequence $\left({\overline{u(\mathbb B_n \setminus \left(1-\frac{1}{k})\mathbb B_n\right)}}\right)_{k=2}^\infty$ consists of compact and connected sets, the intersection $\bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}}$ is compact and connected.\
\
(b) For $n\in \mathbb Z_{\geq 1}$ and $u\in H^\infty (\mathbb B_n)$ we have $$\sup \left\{  {\left\vert\lambda\right\vert} : \lambda \in \bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}} \right\}=  {\left\Vertu\right\Vert}_\infty =\sup_{\lambda \in {\overline{u(\mathbb B_n)}}}    {\left\vert\lambda\right\vert}.$$ Moreover, both suprema are attained.  Clearly $${\left\Vertu\right\Vert}_\infty = \sup_{z\in\mathbb B_n} {\left\vertu(z)\right\vert} =  \sup_{\lambda\in u(\mathbb B_n)} {\left\vert\lambda\right\vert} =  \sup_{\lambda\in {\overline{u(\mathbb B_n)}}} {\left\vert\lambda\right\vert}\geq \sup \left\{  {\left\vert\lambda\right\vert} : \lambda \in \bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}} \right\}.$$ Furthermore, since $u\in H^\infty$ there is a sequence $(z_j)_{j=1}^\infty$ such that $z_j\in \mathbb B_n\setminus r_j\mathbb B_n$ and $\lim_{j\to\infty} {\left\vertu(z_j)\right\vert}= {\left\Vertu\right\Vert}_{\infty}$, where $r_j=1-j^{-1}$. The sequence $(u(z_j))_{j=1}^\infty$ is bounded, and therefore, by Bolzano - Weierstrass theorem, there is a convergent subsequence $(\lambda_k)_{k=1}^\infty$, where $\lambda_k = u(z_{j_k}) \in {\overline{u(\mathbb B_n\setminus r_{j_k}\mathbb B_n)}}$. Since the sets $U_k = {\overline{u(\mathbb B_n\setminus r_{j_k}\mathbb B_n)}}$ are compact and $U_{k+1}\subset U_k, \ k=1,2,...$, it holds that $\lim_{k\to\infty}\lambda_k = \lambda\in U_j$ for every $j$, and hence, we have $\lambda \in \bigcap_{0 < r < 1} {\overline{u(\mathbb B_n \setminus r\mathbb B_n)}}$ and ${\left\vert\lambda\right\vert} = {\left\Vertu\right\Vert}_\infty$.

For $\xi\in\partial \mathbb D$ and $k\in\mathbb Z_{\geq 1}$, let $f_{\xi,k}:\mathbb D\to\mathbb D$ be a peak function defined by $$f_{\xi,k}(z) = \left(\frac{1+{\overline{\xi}}z}{2}\right)^k.$$

For $\alpha > 0 $ it is well-known that ${\mathcal{B}}_{0,\alpha}({{\mathbb{D}}})^{*} \simeq { A^1_0}(\mathbb D)$ and ${A^1_0}(\mathbb D)^{*} \simeq {\mathcal{B}}_\alpha({{\mathbb{D}}})$ via an integral pairing, see [@Z]. 

\[usefulBloch\] Let $0< \alpha \le 1$, $\xi\in{{\mathbb{D}}}$, and $g_{\xi,k}(z)= \left( \frac{1+{\overline{\xi}}z}{2} \right)^k{\left\Vert\left( \frac{1+{\overline{\xi}}z}{2} \right)^k\right\Vert}_{{\mathcal{B}}_\alpha}^{-1}$ be the normalized peak function.  Then we have $g^{(m)}_{\xi,k}\to 0, \ m\in\mathbb Z_{\geq 0}$ uniformly on every set $A_\delta= \{z\in{{\mathbb{D}}}: |z -\xi| \ge \delta\}$, $\delta>0,$ and $g_{\xi, k} \to  0$ weakly in ${\mathcal{B}}_\alpha({{\mathbb{D}}})$ as $k\to \infty$.

For the Bloch-type spaces ${\mathcal{B}}_\alpha({{\mathbb{D}}})$, it can be shown that $${\left\Vert\left( \frac{1+{\overline{\xi}}z}{2} \right)^k\right\Vert}_{{\mathcal{B}}_\alpha} \asymp k^{1-\alpha}.$$ The property $g^{(m)}_{\xi,k}\to 0, \ m\in\mathbb Z_{\geq 0},$ uniformly on the sets $A_\delta$ as $k\to \infty$ is a consequence of the definition of $g_{\xi,k}$. Moreover, the sequence $(g_{\xi,k})_{k=1}^\infty$ is a $weak^*$ null sequence by using Lemma 3.1 in [@CPPR].  Since $(g_{\xi,k})_k \subset \mathcal P(\mathbb D)\subset {\mathcal{B}}_{0,\alpha}(\mathbb D)$, we conclude that $g_{\xi, k}\to 0$ weakly when $k\to\infty$.

\[IfFredholm\_BlochType\] Let us assume that either $u\in M({\mathcal{B}}({{\mathbb{D}}}))\cap  A(\mathbb D)$ or $u\in M({\mathcal{B}}_\alpha({{\mathbb{D}}}))=\mathcal B_\alpha({{\mathbb{D}}})$ with $0 < \alpha <1.$ If $M_u\colon {\mathcal{B}}_\alpha({{\mathbb{D}}}) \to {\mathcal{B}}_\alpha({{\mathbb{D}}})$ is Fredholm, then there are $r\in (0,1)$ and $\delta>0$ such that $|u(z)| \ge \delta$ for all $r\le |z| < 1.$

Suppose there is a sequence $(z_k)_{k=1}^\infty \subset\mathbb D$ such that $|z_k|\to 1$ and $|u(z_k)| \to 0$ when $k\to\infty.$ Then, by going to a subsequence if necessary, we can assume that $z_k\to\xi\in\partial\mathbb D$ when $k\to\infty.$ Since $u$ is continuous up to the boundary in both cases, $u(\xi)=0.$ Now by Lemma \[usefulBloch\] it holds that $g_{\xi,k}\to 0,$ $g'_{\xi,k}\to 0$ uniformly on every set $A_\delta= \{z\in{{\mathbb{D}}}: |z -\xi| \ge \delta\}$, $\delta>0,$ and $g_{\xi, k} \to  0$ weakly as $k\to \infty$. We consider the two cases: $(i)$ when $\alpha =1$ and $(ii)$ when $0 < \alpha <1.$

$(i)$ It holds that $\sup_{k\in \mathbb Z_{\geq 1}}||g_{\xi,k}||_\infty<\infty.$ Since $u\in  M({\mathcal{B}}({{\mathbb{D}}}))$, we know that $u\in {\mathcal{B}}_0({{\mathbb{D}}})\cap H^\infty({{\mathbb{D}}}).$ Let $B_\delta= \{z\in{{\mathbb{D}}}: |z -\xi| < \delta\},$ so ${{\mathbb{D}}}= A_\delta \cup B_\delta.$ Let $\varepsilon> 0$ be given, and choose $\delta >0$ such that $|u(z)| < \varepsilon$ and $|u'(z)| (1 -|z|^2) < \varepsilon$ for $z\in B_\delta.$ The following estimates hold,

$$\begin{aligned}
{\left\VertM_u(g_{\xi,k})\right\Vert}_{\mathcal{B}}& \le   I_{k,{A_\delta}} + { II}_{k,{B_\delta}}+{\left\vertg_{\xi,k}(0)u(0)\right\vert},\end{aligned}$$

where $$\begin{aligned}
 I_{k,{A_\delta}}&=\sup_{z\in A_\delta}|u(z)| |g'_{\xi,k}(z)| (1 - |z|^2)+\sup_{z\in A_\delta}|u'(z)| |g_{\xi,k}(z)| (1 - |z|^2)  \text{ and }\\
{ II}_{k,{B_\delta}} &=\sup_{z\in B_\delta}|u(z)| |g'_{\xi,k}(z)| (1 - |z|^2) +  \sup_{z\in B_\delta}|u'(z)| |g_{\xi,k}(z)| (1 - |z|^2).\end{aligned}$$ Consequently, we get that $\lim_{k\to\infty}  I_{k,{A_\delta}} =0$ and $\lim_{k\to\infty} { II}_{k,{B_\delta}} \le  2\varepsilon.$ We also have ${\left\vertg_{\xi,k}(0)\right\vert} \asymp 2^{-k}$. Thus ${\left\Vert{M_u}(g_{\xi,k})\right\Vert}_{\mathcal{B}}\to 0 \ \text{when} \ k\to\infty,$ which means by Lemma 4.3.15 in [@D] that $0\in\sigma_e(M_u)$. Therefore $M_u$ is not Fredholm. 

$(ii)$ The result follows similarily from showing that ${\left\Vertu  g_{\xi, k}\right\Vert}_{{\mathcal{B}}_\alpha} \to  0$ as $k\to \infty$. Take $\varepsilon>0$ and choose $\delta>0$ such that ${\left\vertu(z)\right\vert}<\varepsilon$ on $B_\delta$. It is clear that $I_{k,{A_\delta}}\to 0$ as $k\to \infty$. From the definition of $g_{\xi,k}$ we have ${\left\Vertg_{\xi,k}\right\Vert}_\infty \asymp k^{\alpha-1} $, so ${\left\Vertg_{\xi,k}\right\Vert}_\infty\to 0$ as $k\to \infty$, hence, ${ II}_{k,{B_\delta}}<2\varepsilon$ for $k$ large enough. 

Let us now consider the space $X({{\mathbb{D}}}) = A_{\alpha,\beta}^p({{\mathbb{D}}})$ with $1 < p <\infty$. The following lemma will be used to obtain an estimate for the Bergman-Sobolev norm of the peak function. 

\[someAsympBehav\] Let $L,M\geq 0$. Then $$\frac{\Gamma(K+L)}{\Gamma(K)}  \sim K^L\text{ and } \frac{\Gamma(2K+L)}{\Gamma(K+L)\Gamma(K+M)}\sim \frac{2^{2K+L-1}}{\sqrt{\pi} }K^{\frac{1}{2}-M},$$ as $K\to \infty$.

According to Stirling’s approximation, $\Gamma(x)\sim \sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^x$ as $x\to\infty$, we have $$\begin{aligned}
\frac{\Gamma(K+L)}{K^L\Gamma(K)} &\sim  \sqrt{\frac{K}{K+L}} e^{K-(K+L)}\frac{(K+L)^{K+L}}{K^L K^K} \\
&=\left( 1+\frac{L}{K} \right)^{-\frac{1}{2}}  e^{-L} \left(1+\frac{L}{K}\right)^K\left(1+\frac{L}{K}\right)^L \rightarrow 1\end{aligned}$$ as $K\to\infty$. Moreover, $$\begin{aligned}
\frac{\sqrt{\pi }K^{M-\frac{1}{2}}\Gamma(2K+L)}{2^{L+2K-1}\Gamma(K+L)\Gamma(K+M)} &\sim  \frac{   \sqrt{\pi}K^{M-\frac{1}{2}}     }{    2^{L+2K-1}     } \sqrt{\frac{(K+L)(K+M)}{2\pi(2K+L)}} \frac{e^M(2K+L)^{2K+L}}{(K+L)^{K+L} (K+M)^{K+M}} \\
&=e^M \sqrt{\frac{\left(1+\frac{L}{K}\right)\left(1+\frac{M}{K}\right)}{\left(1+\frac{L}{2K}\right)}}  \frac{\left(1+\frac{L}{2K}\right)^{2K+L}
}{\left(1+\frac{L}{K}\right)^{K+L}\left(1+\frac{M}{K}\right)^{K+M}}\to 1\end{aligned}$$ as $K\to\infty$.

In the following important lemma a fairly good approximation of the behaviour of the Bergman-Sobolev norm of the peak functions is obtained.  The proof also gives an exact asymptotic formula for ${\left\VertD^j f_{\xi,k}\right\Vert}_{A^p_\alpha}$ as $k\to \infty$ in the case of $p\in\mathbb Z_{\geq 1}$, namely, $${\left\VertD^j f_{\xi,k}\right\Vert}_{A^p_\alpha}^p\sim \frac{ \Gamma(\alpha+2) 2^{2\alpha+\frac{5}{2}-jp}}{   \sqrt{ \pi }   p^{\alpha+\frac{3}{2}}         }(k+1)^{jp-(\alpha+\frac{3}{2})}.$$ Furthermore, some properties for the normalized peak function are given in order to prove Lemma \[IfFredholm\_BergmanSobolev\], from which a part of the main result follows.  Observe that, as already mentioned in the beginning of section \[sect4\], the following lemma is a necessary refinement of Lemma 11 in [@CHZ] and, as a sharp estimate, it is also of independent interest. 

\[bra\] Let $p \geq 1$, $\alpha > -1$ or $p = 2$, $\alpha =-1$. If $\beta\geq 0$, then $${\left\Vertf_{\xi,k}\right\Vert}_{A_{\alpha,\beta}^p}^p \asymp ( k+1)^{-\alpha + \beta p  -\frac{3}{2}}$$ for $k\in\mathbb Z$ large enough.  Consequently, if $\beta > \frac{2 +\alpha}{p}$ and $\xi\in \partial \mathbb D$, then the functions $g_{\xi, k} = f_{\xi,k} / {\left\Vertf_{\xi,k}\right\Vert}_{A_{\alpha,\beta}^p} \in{\mathcal P}({{\mathbb{D}}})$ have the properties that ${\left\Vertg_{\xi, k}\right\Vert}_{A_{\alpha,\beta}^p}=1$; $g_{\xi,k}\to 0$; $R^mg_{\xi,k}\to 0$, $m\in\mathbb Z_{\geq 1}$, uniformly on every set $A_\delta= \{z\in{{\mathbb{D}}}: |z -\xi| \ge \delta\}$, $\delta>0,$ and for $p>1$ it also holds that $g_{\xi, k} \to  0$ weakly in $A_{\alpha,\beta}^p$ as $k\to \infty$.

Let $N$ be a positive integer satisfying $N > \beta- \frac{\frac{1}{2} +\alpha}{p}$. By (\[formimp\]) and (\[good\]) we have that $${\left\Vertf_{\xi,k}\right\Vert}_{A_{\alpha,\beta}^p}\asymp{\left\Vertf_{\xi,k}\right\Vert}_{A_{(N-\beta)p+\alpha,N}^p}\asymp \sum_{l=0}^{N-1} |D^l f_{\xi,k}(0)| + {\left\VertD^Nf_{\xi,k}\right\Vert}_{A_{(N-\beta)p+\alpha}^p} \asymp {\left\VertD^Nf_{\xi,k}\right\Vert}_{A_{(N-\beta)p+\alpha}^p}.$$ The last equivalence follows from $$0\leq {\left\vert D^l f_{\xi,k}(0) \right\vert} \leq {\left\vert D^N f_{\xi,k}(0) \right\vert} \leq  {\left\VertD^N f_{\xi,k}\right\Vert}_{A_{(N-\beta)p+\alpha}^p}$$ for $l\leq N$. To finish the proof, it will be shown that for $\gamma>-1$ and $j\in\mathbb Z_{\geq 0}$ we have $${\left\VertD^j f_{\xi,k}\right\Vert}_{A_\gamma^ p}^p \asymp (k+1)^{jp-(\gamma+\frac{3}{2})} ,$$ from which the lemma follows by letting $\gamma = (N-\beta)p + \alpha$ and $j=N$.

Let $q$ be the smallest integer greater than or equal to $p$ and $k\geq j$. We have [ $$\begin{aligned}
\frac{{\left\VertD^j f_{\xi,k}\right\Vert}_{A_\gamma^ p}^p}{\gamma+1}&=\int_{\mathbb D} |D^j f_{\xi,k} (z)|^p (1-|z|^2)^\gamma  dA(z) \\
&= \left(\frac{k!}{(k-j)!}\right)^p\int_{\mathbb D} \left(\frac{    \left|1+{\overline{\xi}} z\right|^{(k-j)}    }{2^{k}}\right)^p (1-|z|^2)^\gamma  dA(z) \\
&\stackrel{(*)}{\geq} \left(\frac{k!}{(k-j)!}\right)^p\int_{\mathbb D} \left(\frac{    \left|1+{\overline{\xi}} z\right|^{(k-j)}    }{2^{k}}\right)^q (1-|z|^2)^\gamma  dA(z)  \\
&\stackrel{(**)}{\geq} \left(\frac{k!}{(k-j)!}\right)^p\int_{\mathbb D} \frac{1}{     2^{jq}  }    \frac{     \left|1+{\overline{\xi}} z\right|^{2K}          }{    2^{2K}           } (1-|z|^2)^\gamma  dA(z)=U_{k,\xi,j,\gamma}.  \end{aligned}$$ ]{}The $(*)$ indicates that choosing $q$ to be the greatest integer smaller than $p$ we similarily obtain the opposite strict inequality.  The function $K\colon\mathbb Z_{\geq j}\to \mathbb Z_{\geq 0}$ is defined as $K=K_j(k)=\frac{(k-j)q}{2}$ if $k-j$ is even.  In this case $\stackrel{(**)}{\geq}$ is an equality.  If $k-j$ is odd, then $K$ is defined by $K=\frac{(k+1-j)q}{2}$ or $K=\frac{(k-1-j)q}{2}$ depending on which inequality we want to obtain.  In the latter case $\stackrel{(**)}{\geq}$ is replaced by $\leq$.

We continue the proof by evaluating the integral with respect to the angle.  It is enough to examine the expression for $\xi = 1$. Now consider the functions $g_r\in L^2([0,2\pi)), \ g_r(t)=(1+re^{it})^K=\sum_{n=0}^K\binom{K}{n} r^n e^{itn}$ for $r\geq 0$. From Parseval’s equality we obtain $$\begin{aligned}
\int_0^{2\pi} |1+re^{it}|^{2K} dt = 2\pi \sum_{n=0}^K \binom{K}{n}^2 r^{2n},\end{aligned}$$ for every $0\leq r<1$ and thus, [ $$\begin{aligned}
U_{k,\xi,j,\gamma} &=\left(\frac{k!}{(k-j)!}\right)^p \frac{2}{2^{(jq+2K)}} \int_0^1    \sum_{n=0}^K \binom{K}{n}^2 r^{2n} (1-r^2)^\gamma   rdr\\
&=\left(\frac{k!}{(k-j)!}\right)^p \frac{1}{2^{(jq+2K)}}   \sum_{n=0}^K \binom{K}{n}^2 \int_0^1 r^{2n} (1-r^2)^\gamma   2rdr\\
&=\left(\frac{k!}{(k-j)!}\right)^p \frac{1}{2^{(jq+2K)}}   \sum_{n=0}^K \binom{K}{n}^2 \int_0^1 r^n (1-r)^\gamma   dr.\end{aligned}$$ ]{}Moreover, [ $$\begin{aligned}
U_{k,\xi,j,\gamma} &= \left(\frac{k!}{(k-j)!}\right)^p\frac{1}{2^{(jq+2K)}}   \sum_{n=0}^K \binom{K}{n}^2 \beta(n+1,\gamma+1)\\
&=\left(\frac{k!}{(k-j)!}\right)^p \frac{1}{2^{(jq+2K)}}   \sum_{n=0}^K \binom{K}{n}^2\frac{\Gamma(\gamma+1)\Gamma(n+1)}{\Gamma(n+\gamma+2)}\\
&=\left(\frac{\Gamma(k-j+1+j)}{\Gamma(k-j+1)}\right)^p  \frac{1}{2^{(jq+2K)}}  \frac{\Gamma(\gamma+1)}{\Gamma(K+\gamma+2)}    \frac{\Gamma(2K+\gamma+2)}{\Gamma(K+\gamma+2)}\\
&\sim k^{jp} \frac{ \Gamma(\gamma+1)}{2^{(jq+2K)}}  K^{-\gamma-\frac{3}{2}} \frac{2^{\gamma+1+2K}}{\sqrt{\pi }}\\
&\sim \Gamma(\gamma+1) k^{jp} (kq)^{-\gamma-\frac{3}{2}} \frac{2^{2\gamma+\frac{5}{2}-jq}}{\sqrt{\pi }} \\
&=\frac{ \Gamma(\gamma+1) 2^{2\gamma+\frac{5}{2}-jq}}{   \sqrt{ \pi }   q^{\gamma+\frac{3}{2}}         }k^{jp-(\gamma+\frac{3}{2})},\end{aligned}$$ ]{}as $k\to\infty$, where the first asymptotic approximation is given by Lemma \[someAsympBehav\] and $(k-c)^a\sim k^a$ as $k\to\infty$ for every $c\in\mathbb R$. The third equality follows from the Chu-Vandermonde identity, see [@K p. 32] with the parameters $n=K,b=-K$ and $c=\gamma+2$.

To prove that $(g_{\xi,k})_{k=1}^\infty$ is a weak null sequence, let $B_{A^p_{\alpha,\beta}(\mathbb D)}$ denote the closed unit ball of the Bergman-Sobolev space $A^p_{\alpha,\beta}(\mathbb D),\ p>1$. Let $\tau_p$ denote the topology of pointwise convergence.  Notice that $(B_{A^p_{\alpha,\beta}(\mathbb D)},\tau_p)$ is a Hausdorff space and that $B_{A^p_{\alpha,\beta}(\mathbb D)}$ is weakly compact, since the space is reflexive.  Since $\delta_z\in A^p_{\alpha,\beta}(\mathbb D)^*$ by condition \[finerThanCOTop\], the identity map $$\operatorname{id}\colon(B_{A^p_{\alpha,\beta}(\mathbb D)},w)\to (B_{A^p_{\alpha,\beta}(\mathbb D)},\tau_p)$$ is continuous, and hence, it represents a homeomorphism between the spaces $(B_{A^p_{\alpha,\beta}(\mathbb D)},w)$ and $(B_{A^p_{\alpha,\beta}(\mathbb D)},\tau_p)$. Since $\operatorname{id}^{-1}\colon (B_{A^p_{\alpha,\beta}(\mathbb D)},\tau_p) \to (B_{A^p_{\alpha,\beta}(\mathbb D)},w)$ is continuous, we conclude that $g_{\xi, k}\to 0$ weakly, when $k\to\infty$.

\[IfFredholm\_BergmanSobolev\] Let $\alpha>-1$, $p>1$ or $\alpha=-1$, $p=2$ and assume $\beta > \frac{2+\alpha}{p}$. If $M_u\colon A^p_{\alpha,\beta} \to A^p_{\alpha,\beta}$ is Fredholm, then there exist $\delta >0$ and $r \in (0,1)$ such that $|u(z)| \ge \delta$ for all $r \le |z| < 1$.

The proof will be carried out by contraposition.  Since $u$ belongs to the disk algebra it is continuous up to the boundary of $\mathbb D.$ Assume there is a point $\xi\in \partial \mathbb D$ such that $u(\xi)=0$. This assumption is equivalent to $u$ not being bounded from below arbitrarily close to the boundary, since $u$ is continuous.  It will be shown that $${\left\Vertug_{\xi,k}\right\Vert}_{A^p_{p(N-\beta)+\alpha,N}}\to 0  \ \text{ as} \ k\rightarrow \infty,$$ which by (\[formimp\]) implies that $$\label{ugNullSeq}
{\left\Vertug_{\xi,k}\right\Vert}_{A^p_{\alpha,\beta}}\to 0  \ \text{ as} \ k\rightarrow \infty,$$ where $N$ is the positive integer satisfying $$0<N-\beta + \frac{\frac{1}{2}+\alpha}{p}\leq 1,$$ and $g_{\xi,k}$ is the function defined in Lemma \[bra\].  The lemma follows from Lemma \[bra\], (\[ugNullSeq\]) and Lemma 4.3.15 in [@D]. 

To prove the null sequence statement, we will make use of (\[BergSobNormEquivRel\]).  First, notice that by Lemma \[bra\] we obtain $$\begin{aligned}
{\left\vertu(0)g_{\xi,k}(0)\right\vert}\lesssim \frac{   {\left\vertu(0)f_{\xi,k}(0)\right\vert}   }{(k+1)^{-\frac{\alpha}{p}+\beta-\frac{3}{2p}}   } \to 0\end{aligned}$$ as $k\to \infty$. Using the general Leibniz formula we have $$R^N(u g_{\xi,k}) = \sum_{j=0}^N  \binom{N}{j} R^j u R^{N-j}  g_{\xi,k},$$ from which it follows that $$\label{eq: leibniz}
{\left\VertR^N(u g_{\xi,k}) \right\Vert}_{A^p_{p(N-\beta)+\alpha}} \le \sum_{j=0}^N  \binom{N}{j}{\left\Vert R^j u R^{N-j}  g_{\xi,k} \right\Vert}_{A^p_{p(N-\beta)+\alpha}}.$$ Therefore, it suffices to show that $$I_{k,j} = \int_\mathbb D |R^j u R^{N-j} g_{\xi,k}|^p dA_{p(N-\beta)+\alpha}$$ approaches zero for $j=0,1,...,N$ as $k$ tends to infinity.  To prove the assertion for the case $j=0$, we take $\varepsilon>0$ and choose $\delta>0$ such that $|u(z)|^p<\varepsilon$ for all $$z\in B_\delta= \{z\in{{\mathbb{D}}}: |z -\xi| < \delta\}.$$ We can now choose a $K>0$ such that $$\int_{A_\delta} |R^N g_{\xi,k}(z) | ^p dA_{p(N-\beta)+\alpha} (z)  < \varepsilon,$$ which implies $$\int_{A_\delta} |u(z)R^N g_{\xi,k}(z) | ^p dA_{p(N-\beta)+\alpha} (z)  < {\left\Vertu\right\Vert}_\infty^p \varepsilon$$ for $k>K$, where Lemma \[bra\] has been used and $A_\delta= \{z\in{{\mathbb{D}}}: |z -\xi| \geq \delta\}$. Thus, for $k > K$ $$I_{k,0} <( {\left\Vertu\right\Vert}^p_{\infty} + {\left\VertR^N g_{\xi,k}\right\Vert}_{A^p_{p(N-\beta)+\alpha} }^p)\varepsilon \leq ( {\left\Vertu\right\Vert}^p_{\infty} + M{\left\Vertg_{\xi,k}\right\Vert}_{A^p_{\alpha,\beta} }^p)\varepsilon$$ where (\[formimp\]) gives the second inequality for some $M>0$. Since $u\in A_{\alpha,\beta}^p(\mathbb D)\subset H^\infty(\mathbb D)$ and $\|g_{\xi,k}\|_{A^p_{\alpha, \beta}}=1$ for every $k$, the result follows.  To assure the result in the case $j\geq 1$, we will use the following approximation:

$$\begin{aligned}
I_{k,j}&\leq \frac{(k+1)^{p(N-j)}}{{\left\Vertf_{\xi,k}\right\Vert}^p_{A_{\alpha,\beta}^p}} \int_\mathbb D {\left\vertR^j u(z)\right\vert}^p \left|\frac{1+{\overline{\xi}}z}{2}\right|^{(k-(N-j))p}  dA_{p(N-\beta)+\alpha}(z).\end{aligned}$$

From Lemma \[bra\] it follows that $$\begin{aligned}
\begin{split}
\label{IntegralIApproxMAIN}
I_{k,j}&\lesssim \frac{    (k+1)^{p(N-j)}    }{      ( k+1)^{   -\alpha + \beta p  -\frac{3}{2} }       } \int_\mathbb D |R^j u(z)|^p \left|\frac{1+{\overline{\xi}}z}{2}\right|^{(k-(N-j))p} (1-|z|^2)^{p(N-\beta)+\alpha} dA(z)\\
&=  (k+1)^{p(N-\beta+\frac{\alpha +\frac{3}{2}   }{ p}-j)}       \int_\mathbb D |R^j u(z)|^p \left|\frac{1+{\overline{\xi}}z}{2}\right|^{(k-(N-j))p} (1-|z|^2)^{p(N-\beta)+\alpha} dA(z). 
\end{split}\end{aligned}$$

For integers $j\in[2,N]$ the result $I_{k,j}\to 0$ as $k\to\infty$ is obtained from the following three facts:

$$\begin{aligned}
&u\in A^p_{\alpha,\beta} \simeq A^p_{p(N-\beta)+\alpha,N}\subset A^p_{p(N-\beta)+\alpha,j}; \\
&{\left\Vertf_{\xi,k} \right\Vert}_\infty \leq 1 \ \forall k\in\mathbb Z_{\geq 1}; \\
&p(N-\beta+\frac{ \frac{3}{2} + \alpha}{p}-j)\leq 1+p-pj<0.\end{aligned}$$

For $j=1$ we make an additional partition.  We will, once at a time, assume that $N-\beta+\frac{\frac{3}{2} +  \alpha }{p}-1$ is stricly less than zero, equal to zero or strictly larger than zero.  In the first case we can apply the procedure used for $j\geq 2$. In the second case we may utilize the Lebesgue dominated convergence theorem to functions $$\left|\frac{1+{\overline{\xi}}z}{2}\right|^{(k-(N-j))p} \le 1$$ for all $z \in {{\mathbb{D}}}$ and $k \in \mathbb Z_{\geq N}$ to obtain the result. 

The only thing that remains to show is that $I_{k,1}\to 0$ as $k\to \infty$ when $N-\beta+\frac{\frac{3}{2} + \alpha }{p}-1>0$. This condition implies that $$N >\beta - \frac{\frac{3}{2} +  \alpha  }{p}+1 >1,$$ so that $N\geq 2$.

To prove that $(I_{k,1})_{k=1}^\infty$ is a null sequence we will use Lemma 5.4 in [@BB] and Lemma \[bra\].  Lemma 5.4 in [@BB] gives us three different approximations for the behaviour of $|Du(z)|$, depending on values of some parameters.  Hence, it suffices to prove the null convergence for all of these approximations, one at a time.  Notice that $q=\alpha+1$ when comparing notations with [@BB].  First, assume $\beta<\frac{2+\alpha}{p}+ 1$. Then we have [ $$\begin{aligned}
I_{k,1} &\lesssim {\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p\int_{{{\mathbb{D}}}} |R^{N-1} g_{\xi,k}|^p (1-|z|^2)^{-p(\frac{2+\alpha}{p}+1-\beta)}dA_{p(N-\beta)+\alpha}(z)\\
&\lesssim\frac{{\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p}{{\left\Vertf_{\xi,k}\right\Vert}_{A^p_{\alpha,\beta}}^p}\int_{{{\mathbb{D}}}} |R^{N-1} f_{\xi,k}|^p dA_{p(N-1)-2}(z)\\
&=\frac{{\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p}{{\left\Vertf_{\xi,k}\right\Vert}_{A^p_{\alpha,\beta}}^p}{\left\Vertf_{\xi,k}\right\Vert}^p_ {A^p_{p(N-1)-2,N-1}}\\
&\asymp{\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p(k+1)^{2  +\alpha-\beta p   },\end{aligned}$$ ]{}therefore $(I_{k,1})_k$ is a null sequence in this case.  If $\beta\geq \frac{2+\alpha}{p}+ 1$, then a worse upper bound than the one stated in Lemma 5.4 is given by $C   {\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p \frac{        1       }{(1-|z|^2)^r}$ for some positive constant $C$ and any $r>0$. In this case we have, for $0<r<\frac{1}{2}$, that $$\begin{aligned}
I_{k,1} &\lesssim {\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p\int_{{{\mathbb{D}}}} |R^{N-1} g_{\xi,k}|^p (1-|z|^2)^{-r}dA_{p(N-\beta)+\alpha}(z)\\
&\lesssim\frac{{\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p}{{\left\Vertf_{\xi,k}\right\Vert}_{A^p_{\alpha,\beta}}^p}\int_{{{\mathbb{D}}}} |R^{N-1} f_{\xi,k}|^p dA_{p(N-\beta)+\alpha-r}(z)\\
&=\frac{{\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p}{{\left\Vertf_{\xi,k}\right\Vert}_{A^p_{\alpha,\beta}}^p}{\left\Vertf_{\xi,k}\right\Vert}^p_ {A^p_{p(N-\beta)+\alpha-r,N-1}}\\
&\asymp{\left\Vertu\right\Vert}_{A^p_{\alpha,\beta}}^p(k+1)^{r-p},\end{aligned}$$ which completes the proof. 

We are now ready to present the main result. 

\[MainResEssSpec\] Let $X(\mathbb D)$ be any of the following spaces:

1.   ${\mathcal{B}}_\alpha(\mathbb D), \, 0 <\alpha  <  1,$ with $u\in M({\mathcal{B}}_\alpha({{\mathbb{D}}})) ={\mathcal{B}}_\alpha({{\mathbb{D}}}) \subset  A(\mathbb D)$;\

2.   ${\mathcal{B}}(\mathbb D)$ with $u\in M({\mathcal{B}}({{\mathbb{D}}}))\cap  A(\mathbb D)$;\

3.   $A^p_{\alpha,\beta}(\mathbb D)$ with $u\in M(A^p_{\alpha,\beta}(\mathbb D)) =A^p_{\alpha,\beta}(\mathbb D)\subset  A(\mathbb D)$, where $p>1, \, \alpha>-1\text{ and } \beta > \frac{2+\alpha}{p}$;\

4.   $H^2_\beta(\mathbb D)$ with $u\in M(H^2_\beta(\mathbb D)) =H^2_\beta(\mathbb D)\subset  A(\mathbb D),$ where $\beta > \frac{1}{2}$.

Then the essential spectrum of $M_u:X(\mathbb D) \to X(\mathbb D)$ is given by $$\sigma_e(M_u) = \bigcap_{0 < r < 1} {\overline{u(\mathbb D \setminus r\mathbb D)}}=u(\partial \mathbb D).$$

As in the proof of Theorem \[EssSpecMlutiHInfty\], now using lemmas \[ThenFredholm\_ForX\], \[IfFredholm\_BlochType\] and \[IfFredholm\_BergmanSobolev\], we obtain $\sigma_e(M_u) = \bigcap_{0 < r < 1} {\overline{u(\mathbb D \setminus r\mathbb D)}}$ whenever $u\in M(X(\mathbb D))\cap A(\mathbb D)$ and $X(\mathbb D)$ is any of the spaces listed above.  To prove the last equality, we utilize the continuity of $u$ on $\overline{{{\mathbb{D}}}}$, which implies the first equality below $$\begin{aligned}
\bigcap_{0 < r < 1} {\overline{u(\mathbb D \setminus r\mathbb D)}}&=\bigcap_{0 < r < 1} u({\overline{\mathbb D \setminus r\mathbb D}})\\
&=\bigcap_{0 < r < 1} u({\overline{\mathbb D}} \setminus r\mathbb D)\\
&\supset u(\partial \mathbb D).\end{aligned}$$ To show the opposite inclusion, take $z\in\bigcap_{0 < r < 1} u({\overline{\mathbb D}} \setminus r\mathbb D)$. Now there is a sequence $(y_n)_{n=1}^\infty, \ 1-\frac{1}{n}\leq|y_n|\leq 1$ such that $u(y_n) = z$. Since $(y_n)_{n=1}^\infty$ is bounded there is a convergent subsequence $(y_{n_k})_{k=1}^\infty$ such that $y_{n_k}\to y\in \partial \mathbb D$ as $k\to \infty$. Since $u$ is continuous on ${\overline{\mathbb D}}$ we have $$z=\lim_{k\to\infty} u(y_{n_k}) = u(y),$$ so $z\in u(\partial \mathbb D)$, which proves the theorem. 

Acknowledgements {#acknowledgements .unnumbered}
----------------

The first two authors were supported in part by the Academy of Finland project 296718.  The third author acknowledges support from the Magnus Ehrnrooth Foundation. 

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