1 Introduction and preparation
In this paper,
$\mathbb {D}$
denotes the open unit disk and
$\mathbb {T}$
be the unit circle. Let
$H(\mathbb {D})$
be the space of all analytic functions on
$\mathbb {D}$
. For
$0<p<\infty $
, the Hardy space
$H^p$
consists of analytic functions
$f\in \mathbb {D}$
such that
$$ \begin{align*}\|f\|^p_{H^p}=\sup_{0\leq r<1}\frac{1}{2\pi}\int^{2\pi}_{0}|f(re^{i\theta})|^p\,d\theta<\infty. \end{align*} $$
If
$p=\infty $
, then
$H^{\infty }$
is the space of bounded analytic functions f on
$H(\mathbb {D})$
with
$$ \begin{align*}\|f\|_{\infty}=\sup\{|f(z)|:z\in \mathbb{D}\}. \end{align*} $$
For
$0<p<\infty $
, the Bergman space
$A^p$
consists of all functions f analytic on
$\mathbb {D}$
such that
$$ \begin{align*}\|f\|^p_{A^p}=\int_{\mathbb{D}}|f(z)|^p \,dA(z)<\infty, \end{align*} $$
where
$dA(z)$
is the normalized Lebesgue area measure on
$\mathbb {D}$
. It is clear that
$H^p\subset A^p$
. Moreover,
$H^p\subset A^{2p}$
and
$\|f\|_{A^{2p}}\leq \|f\|_{H^p}$
for
$0<p<\infty $
. See [Reference Vukotić44] for example.
The Dirichlet-type space
$\mathcal {D}^p$
is the set of all functions
$f\in H(\mathbb {D})$
with
$$ \begin{align*}\|f\|^p_{\mathcal{D}^p}=|f(0)|^p+\int_{\mathbb{D}}|f'(z)|^p \,dA(z)<\infty. \end{align*} $$
The space of all conformal automorphisms of
$\mathbb {D}$
forms a group, called the Möbius group, and is denoted by
$\mathrm {Aut}\, (\mathbb {D})$
. It is well known that
$\varphi $
belongs to
$ \mathrm {Aut}\, (\mathbb {D})$
if and only if there exists a real number
$\theta $
and a point
$a\in \mathbb {D}$
such that
$$ \begin{align*} \varphi(z)=e^{i\theta}\sigma_{a}(z) ~\mbox{ and }~ \sigma_{a}(z)=\frac{a-z}{1-\overline{a}z}, \quad z\in\mathbb{D}. \end{align*} $$
Let X be a Banach space of analytic functions on
$\mathbb {D}$
. Then X is said to be Möbius-invariant whenever
$f\circ \varphi \in X$
for all
$f\in X$
and
$\varphi \in \mathrm {Aut}\, (\mathbb {D})$
and
$\|f\circ \varphi \|_{X}=\|f\|_{X}$
.
For
$1<p<\infty $
, the Besov space
$B_p$
consists of analytic functions f on
$\mathbb {D}$
such that
$$ \begin{align*}\int_{\mathbb{D}}(1-|z|^2)^{p-2}|f'(z)|^{p}\,dA(z)<\infty. \end{align*} $$
The norm of
$B_p$
is defined as
$$ \begin{align*}\|f\|_{B_p}=|f(0)|+\left(\int_{\mathbb{D}}(1-|z|^2)^{p-2}|f'(z)|^{p}\,dA(z)\right)^{\frac{1}{p}}. \end{align*} $$
When
$p=\infty $
,
$B_{\infty }=: \mathcal {B}$
is called the classical Bloch space. We define a norm on
$\mathcal {B}$
as
$$ \begin{align*}\|f\|_{\mathcal{B}}=|f(0)|+\sup_{z\in\mathbb{D}}(1-|z|^2)|f'(z)|<\infty. \end{align*} $$
When
$p=2$
,
$B_2=:\mathcal {D}$
is the classical Dirichlet space. When
$p=1$
, we get the analytic Besov space
$B_1$
, which is the minimal Möbius-invariant space consisting of all functions
$f\in H(\mathbb {D})$
with
$$ \begin{align*}f(z)=\sum^{\infty}_{k=1}c_k\sigma_{a_k}(z), \end{align*} $$
where the sequences
$\{c_k\}_{k\geq 1}\in \ell ^1$
and
$\{a_k\}_{k\geq 1}\in \mathbb {D}$
. An equivalent norm of
$B_1$
is defined as
$$ \begin{align*}\|f\|_{B_1}=|f(0)|+|f'(0)|+\int_{\mathbb{D}}|f''(z)|\,dA(z). \end{align*} $$
Arazy, Fisher, and Peetre [Reference Arazy, Fisher and Peetre5] first studied minimal Möbius-invariant space systematically. More results related to minimal Möbius-invariant space may be seen from [Reference Bao and Wulan6, Reference Blasco7, Reference Colonna and Li9, Reference Mitsis and Papadimitrakis32, Reference Ohno and Pavlović33, Reference Wulan and Xiong45].
Now, we define several operators on
$B_1$
. For
$g\in H(\mathbb {D})$
, the multiplication operator
$M_g$
on
$B_1$
is defined as
$$ \begin{align*}(M_{g})f(z)=f(z)g(z),~ f\in B_1,~ z\in\mathbb{D}. \end{align*} $$
The differentiation operator is given by
$Df=f'$
for each
$f\in H(\mathbb {D})$
. Given
$g\in H(\mathbb {D})$
, the Volterra-type operator
$T_g$
is defined as
$$ \begin{align*}(T_gf)(z)=\int^{z}_{0}f(w)g'(w)\,dw, ~\mbox{ for all } f\in B_1. \end{align*} $$
When
$g(z)=z$
, the operator
$T_zf(z)=\int ^{z}_{0}f(w)dw$
becomes the simplest Volterra operator. An integral operator related to
$T_g$
, denoted by
$I_g$
, is defined as
$$ \begin{align*}(I_gf)(z)=\int^{z}_{0}f'(w)g(w)\,dw, ~\mbox{ for all } f\in B_1. \end{align*} $$
The Volterra-type operator
$T_g$
was originally studied by Pommerenke [Reference Pommerenke38]. Later, a series of articles appeared on the study of Volterra-type integration operators on classical spaces of analytic functions, such as Hardy spaces, Bergman spaces, and Dirichlet-type spaces. For more details, please refer to [Reference Aleman and Cima1, Reference Aleman and Siskakis3, Reference Aleman and Siskakis4, Reference Galanopoulos, Girela and Peláez16, Reference Girela and Peláez18]. In [Reference Čučković and Paudyal13], Čučković and Paudyal describe the lattice of the closed invariant subspaces of Volterra-type operators. Lin, Liu, and Wu [Reference Lin, Liu and Wu22] generalized some of the works of [Reference Čučković and Paudyal13] to the general case when
$1\leq p<\infty $
, and obtained the boundedness of the Volterra-type operators
$T_g$
and
$I_g$
on the derivative Hardy space
$S_p(\mathbb {D})$
. And then, they also considered strict singularity of Volterra-type operators on Hardy spaces in [Reference Lin, Liu and Wu23]. Meanwhile, Lin [Reference Lin21] characterized the boundedness and compactness of the Volterra-type operators between Bloch-type spaces and weighted Banach spaces. In [Reference Miihkinen, Pau, Perälä and Wang31], Miihkinen et al. completely characterize the boundedness of the Volterra-type operators acting from the weighted Bergman spaces to the Hardy spaces of the unit ball.
In this paper, we mainly study the operator-theoretic properties in minimal Möbius-invariant space
$B_1$
. The structure of this article is as follows. In Section 2, we discuss the boundedness of the Volterra operator on
$B_1$
. In Section 3, it is shown whether the integral operator belongs to Deddens algebras. In Section 4, we will be concerned with the essential norms of integral operators on
$B_1$
. Section 5 is devoted to the study of the spectrum of integral operators on
$B_1$
.
Throughout this paper, we use the following convention. For two nonnegative functions F and G defined on some function space
$X,$
we write
$F \lesssim G$
if
$F(f) \leq C\cdot G(f)$
for all
$f \in X$
and for some positive constant C which is independent of F and
$G.$
Denote by
$F\approx G$
whenever
$F\lesssim G \lesssim F$
.
2 Volterra-type operators on
$B_1$
First, we need the following Hardy inequality.
Lemma 1 [Reference Duren15]
If
$f\in H^1$
and
$f(z)= \sum ^{\infty }_{n=0}a_nz^n$
, then
$$ \begin{align*}\sum^{\infty}_{n=0}\frac{|a_n|}{n+1}\leq\pi\|f\|_{H^1}. \end{align*} $$
The following lemma is a classic exercise in mathematical analysis, but it might be worth to give a brief details of the proof for completeness.
Lemma 2 Suppose that
$f(x)$
is a continuously increasing function on
$[a,b]$
. Then,
$$ \begin{align*}\int^{b}_{a}xf(x)dx\geq\frac{a+b}{2}\int^{b}_{a}f(x)\,dx. \end{align*} $$
Proof As
$f(x)$
is monotonically increasing on
$[a,b]$
, the “integral mean value theorem” shows that there exists a
$\xi \in [a,b]$
such that
$$ \begin{align*} \int^{b}_{a}\left(x-\frac{a+b}{2}\right)f(x)\,dx=&f(a)\int^{\xi}_{a}\left(x-\frac{a+b}{2}\right)dx+f(b)\int^{b}_{\xi}\left(x-\frac{a+b}{2}\right)dx\\ =&\frac{1}{2}(f(b)-f(a))(a\xi+b\xi-\xi^2-ab)\\ =&\frac{1}{2}(f(b)-f(a))(b-\xi)(\xi-a)\geq0. \end{align*} $$
This completes the proof of the lemma.
Lemma 3 If
$f\in \mathcal {D}^1$
, then
$\|f\|_{H^1}\leq \|f\|_{\mathcal {D}^1}.$
Proof Let
$f\in \mathcal {D}^1.$
Then, we can see that
$$ \begin{align*}|f(e^{i\theta})|-|f(0)|\leq|f(e^{i\theta})-f(0)|\leq\left|\int^1_0f'(re^{i\theta}) \,dr\right|\leq\int^1_0|f'(re^{i\theta})| \,dr. \end{align*} $$
Thus,
$$ \begin{align*} \|f\|_{H_1}=&\frac{1}{2\pi}\int^{2\pi}_{0}|f(e^{i\theta})| \,d\theta\\ \leq&\frac{1}{2\pi}\int^{2\pi}_{0}\left[|f(0)|+\int^{1}_{0}|f'(re^{i\theta})| \,dr\right]d\theta\\ \leq&|f(0)|+\frac{1}{2\pi}\int^{2\pi}_{0}\int^{1}_{0}|f'(re^{i\theta})| \,dr \,d\theta. \end{align*} $$
On the other hand,
$$ \begin{align*} \|f\|_{\mathcal{D}^1}=|f(0)|+\int_{\mathbb{D}}|f'(z)| \,dA(z) =&|f(0)|+\frac{1}{\pi}\int^{2\pi}_{0}\int^{1}_{0}r|f'(re^{i\theta})| \,dr \,d\theta. \end{align*} $$
By Hardy’s convexity theorem (see [Reference Mashreghi28, Reference Xiao and Zhu46]), we find that
$$ \begin{align*}F(r)=\frac{1}{2\pi}\int^{2\pi}_{0}|f'(re^{i\theta})|\,d\theta,\, ~0<r<1, \end{align*} $$
is a nondecreasing function of r. It follows from Lemma 2 that
$$ \begin{align*}\int^{1}_{0}F(r)dr\leq2\int^{1}_{0}rF(r) \,dr. \end{align*} $$
This shows that
$\|f\|_{H^1}\leq \|f\|_{\mathcal {D}^1}$
.
Remark 1 In Lemma 3, set
$f(z)=z$
. Then, we see that
$1=\|z\|_{H^1}\leq \|z\|_{\mathcal {D}^1}=1$
, showing that the norm estimate is sharp. This improves the previous conclusion, namely,
$\|f\|_{H^1}\leq 2\|f\|_{\mathcal {D}^1}$
from the works of Girela and Merchán [Reference Girela and Merchán17].
Lemma 4 If
$f\in B_1$
, then
$\|f\|_{\infty }\leq \pi \|f\|_{B_1}$
and
$B_1\subset H^{\infty }$
.
Proof Assume that
$f\in B_1$
and write
$f(z)=\sum ^{\infty }_{k=0}a_kz^k$
. By Lemmas 1 and 3, we have
$$ \begin{align*} |f(z)|=\left|\sum^{\infty}_{k=0}a_kz^k\right|\leq\sum^{\infty}_{k=0}|a_kz^k| \leq&\sum^{\infty}_{k=0}|a_k| \leq \pi\|f'\|_{H^1}+|f(0)| \leq \pi\|f'\|_{\mathcal{D}^1}+|f(0)|\leq\pi\|f\|_{B_1}, \end{align*} $$
for all
$z\in \mathbb {D}$
. Hence, we obtain that
$\|f\|_{\infty }\leq \pi \|f\|_{B_1}$
.
Remark 2 In [Reference Lin, Liu and Wu22, Theorem 1], they obtained that
$\|f\|_{\infty }\leq \pi \|f\|_{S^1}$
for each
$f\in S^1$
, where the space
$S^1$
is defined as
$S^1=\{f\in H^1:f'\in H^1\}$
. The norm on
$S^1$
is given by
$$ \begin{align*}\|f\|_{S^1}=|f(0)|+\|f'\|_{H^1}. \end{align*} $$
Moreover, we obtain the following norm estimate:
$$ \begin{align*} \|f\|_{\infty}\leq\pi\|f\|_{S^1}\leq\pi\|f\|_{B_1},\ \ \mbox{for all} f\in B_1. \end{align*} $$
In the following, we discuss the boundedness of
$T_g$
and
$I_g$
on minimal Möbius-invariant space.
Theorem 1 The operator
$T_g$
is bounded on
$B_1$
if and only if
$g\in B_1$
. Moreover,
$$ \begin{align*}\|g-g(0)\|\leq \|T_g\|\leq (1+\pi) \|g-g(0)\|_{B_1}. \end{align*} $$
Proof Let
$f\in B_{1}$
. By Hölder’s inequality and Lemma 3, we have
$$ \begin{align*} \int_{\mathbb{D}}|f'(z)g'(z)| \,dA(z)&\leq\left(\int_{\mathbb{D}}|f'(z)|^2 \, dA(z)\right)^{\frac{1}{2}}\left(\int_{\mathbb{D}}|g'(z)|^2 \,dA(z)\right)^{\frac{1}{2}} =\|f'\|_{A^2}\|g'\|_{A^2}\\ &\leq\|f'\|_{H^1}\|g'\|_{H^1} \\ &\leq\|f'\|_{\mathcal{D}^1}\|g'\|_{\mathcal{D}^1} \\ &\leq \|f\|_{B_1}\|g-g(0)\|_{B_1}. \end{align*} $$
Hence, we get
$$ \begin{align*} \|T_gf\|_{B_1}&=\|fg'\|_{\mathcal{D}^1(\mathbb{D})} = |f(0)g'(0)|+\!\int_{\mathbb{D}}|f'(z)g'(z)| \,dA(z)+\!\int_{\mathbb{D}}|f(z)g''(z)| \,dA(z)\\ &\leq \|f\|_{B_1}\|g-g(0)\|_{B_1}+\|f\|_{\infty}\|g-g(0)\|_{B_1}\\ &\leq (1+\pi) \|f\|_{B_1}\|g-g(0)\|_{B_1}, \end{align*} $$
showing that
$T_g$
is a bounded operator on
$B_1$
.
Conversely, assume that
$T_g$
is a bounded operator on
$B_1$
and let
$f=1$
. Then, we obtain
$$ \begin{align*}\|T_g\|\geq\|T_g1\|_{B_1}\geq \|g-g(0)\|_{B_1}, \end{align*} $$
which gives that
$g\in B_1$
. Thus,
$$ \begin{align*}\|g-g(0)\|\leq \|T_g\|\leq (1+\pi) \|g-g(0)\|_{B_1}.\\[-2.7pc] \end{align*} $$
For
$0<p<\infty $
,
$-2<q<\infty $
, and
$0\leq s<\infty $
, we define the general family of function spaces
$F(p,q,s)$
as the set of all analytic functions f on
$\mathbb {D}$
such that
$$ \begin{align*}\|f\|^{p}_{p,q,s}=|f(0)|+\sup_{a\in \mathbb{D}}\int_{\mathbb{D}}|f'(z)|^{p}(1-|z|^2)^{q}g^{s}(z,a) \,dA(z)<\infty, \end{align*} $$
where
$g(z,a)=\log \frac {1}{|\sigma _a(z)|}$
. These spaces were introduced by Zhao in [Reference Zhao47]. In 2003, Rättyä provided the following nth derivation characterization of functions in spaces
$F(p,q,s)$
.
Lemma 5 [Reference Rättyä39, Theorem 3.2]
Let f be an analytic function on
$\mathbb {D}$
, and let
$0<p<\infty , -2<q<\infty $
, and
$0\leq s<\infty $
. Let
$n\in \mathbb {N}$
and
$q+s>-1$
; or
$n=0$
and
$q+s-p>-1$
. Then,
$f\in F(p,q,s)$
if and only if
$$ \begin{align*}\sup_{a\in \mathbb{D}}\int_{\mathbb{D}}|f^{(n)}(z)|^p(1-|z|^2)^{np-p+q}(1-|\varphi_a(z)|^2)^s \,dA(z)<\infty. \end{align*} $$
For
$p>0$
, the space
$Z_p$
consists of all analytic functions f on
$\mathbb {D}$
such that
$$ \begin{align*}\|f\|_{Z_p}=|f(0)|+\sup_{a\in \mathbb{D}}\int_{\mathbb{D}}|(f\circ\sigma_a(z))'|(1-|z|^2)^{p-1} \,dA(z)<\infty. \end{align*} $$
It is clear that
$Z_1=F(1,-1,1)$
. For more results about
$Z_{p}$
space, see [Reference Liu, Lou and Zhu25, Reference Zhu48].
Theorem 2 The operator
$I_g$
is bounded on
$B_1$
if and only if
$g\in Z_{1}\cap H^{\infty }$
.
Proof Assume that
$g \in Z_{1}\cap H^{\infty }$
. Using Lemma
$6$
of [Reference Liu, Lou and Zhu25], we obtain that
$$ \begin{align*}\int_{\mathbb{D}}|f'(w)|\cdot|g'(w)| \,dA(w)\lesssim \|g\|_{Z_{1}} \left(|f'(0)|+\int_{\mathbb{D}}|f''(w)| \,dA(w)\right). \end{align*} $$
Thus, we have
$$ \begin{align*} &\|(I_gf)(z)\|_{B_1}=\|f'g\|_{\mathcal{D}^1}\\[3pt] & \quad \leq |f'(0)g(0)|+ \int_{\mathbb{D}}|f''(w)|\cdot|g(w)| \,dA(w)+\int_{\mathbb{D}}|f'(w)|\cdot|g'(w)| \,dA(w)\\[3pt] & \quad \lesssim |f'(0)g(0)|+ \|g\|_{\infty}\int_{\mathbb{D}}|f''(w)| \,dA(w)+\|g\|_{Z_{1}}\left(|f'(0)|+\int_{\mathbb{D}}|f''(w)| \,dA(w)\right)\\[3pt] & \quad \lesssim (\|g\|_{\infty}+\|g\|_{Z_{1}})\|f\|_{B_1} \end{align*} $$
for all
$f\in B_1$
. This implies that
$I_g$
is a bounded operator on
$B_1$
.
Conversely, suppose that
$I_g$
is a bounded operator on
$B_1$
. For each
$a\in \mathbb {D}$
, let
$f(z)=\sigma _{a}(z)$
. Then,
$\|f\|_{B_1}\lesssim 1$
and
$$ \begin{align*} \|I_g\| &\gtrsim \|I_gf\|_{B_1} =\|g(w)f'(w)\|_{\mathcal{D}^1}\\[3pt] &\geq \|g(w)f'(w)\|_{H^1} =\int^{2\pi}_{0}|g(e^{i\theta})|\cdot|\sigma'_{a}(e^{i\theta})| \,d\theta =\int^{2\pi}_{0}|g(\sigma_{a}(e^{i\theta}))| \,d\theta\\[3pt] &\geq |g(\sigma_{a}(0))|=|g(a)|, \end{align*} $$
which shows that
$\|I_g\|\gtrsim \|g\|_{\infty }$
. Hence,
$g\in H^{\infty }$
. Moreover,
$$ \begin{align*} \|I_g\| & \gtrsim \|I_g\sigma_a\|_{B_1}\\[3pt] &\geq\int_{\mathbb{D}}|\sigma'_{a}(w)|\cdot|g'(w)| \,dA(w)-\int_{\mathbb{D}}|\sigma"_{a}(w)|\cdot|g(w)| \,dA(w)\\[3pt] &\geq\int_{\mathbb{D}}|\sigma'_{a}(w)|\cdot|g'(w)| \,dA(w)-\|g\|_{\infty}\|\sigma_a\|_{B_1}, \end{align*} $$
from which it follows that
$$ \begin{align*}\|I_g\|+\|g\|_{\infty}\|\sigma_a\|_{B_1} \gtrsim \int_{\mathbb{D}}|\sigma'_{a}(w)|\cdot|g'(w)| \,dA(w). \end{align*} $$
This implies that
$g\in Z_1$
. Therefore,
$g\in Z_1\cap H^{\infty }$
. This completes the proof of the theorem.
Remark 3 Note that
$B_1\subset Z_1\cap H^{\infty }.$
By Lemma 4, we know that
$f\in H^{\infty }$
whenever
$f\in B_1$
. Using the second derivation characterization of functions on
$Z_1$
, we have
$$ \begin{align*} \|f\|_{Z_1}\approx |f(0)|+\sup_{a\in \mathbb{D}}\int_{\mathbb{D}}|f''(z)|(1-|\sigma_a(z)|^2) \,dA(z)\leq |f(0)|+\int_{\mathbb{D}}|f''(z)| \,dA(z)\leq\|f\|_{B_1}. \end{align*} $$
Then
$f\in Z_1$
, and thus
$B_1\subset Z_1\cap H^{\infty }$
.
Now, we define the space
$B^0_1$
as
$$ \begin{align*}B^0_1=\{f\in B_1:f(0)=0\}. \end{align*} $$
The following theorem gives the connection between the Volterra operator
$T_z$
on
$\mathcal {D}^1$
and the multiplication operator
$M_z$
on
$B^0_1$
.
Theorem 3 The Volterra-type operator
$T_{z}:\mathcal {D}^1\rightarrow B^0_1$
is bounded and invertible with
$T^{-1}_z= D .$
Proof First, we show that
$T_z(\mathcal {D}^1)=B^0_1.$
For
$f\in \mathcal {D}^1$
, we consider
$$ \begin{align*}F(z):=(T_zf)(z)=\int^{z}_{0}f(w)\,dw. \end{align*} $$
Clearly,
$F'=f\in \mathcal {D}^1$
so that
$F\in B^0_1$
and
$T_{z}(\mathcal {D}^1)\subseteq B^0_1.$
Conversely, for each
$F\in B^0_1$
, we have
$F' \in \mathcal {D}^1$
. Then,
$$ \begin{align*}(T_z(F'(z)))=\int^{z}_{0}F'(w)\,dw=F(z)-F(0)=F(z). \end{align*} $$
Then,
$B^0_1 \subseteq T_{z}(\mathcal {D}^1)$
. This implies that
$T_{z}(\mathcal {D}^1)=B^0_1$
.
Second, we show that
$T_z:\mathcal {D}^1 \rightarrow B^0_1$
is a bounded isomorphism, and its inverse
$T^{-1}_{z}= D.$
Recall from the above discussion that
$(T_z(F'(z)))=F(z)$
for
$F\in B^0_1$
. Then, for each
$f\in \mathcal {D}^1$
, we have that
$(D(T_zf))(z)=f(z)$
. This implies that
$T_z$
is a bijective operator from
$\mathcal {D}^1$
onto
$B^0_1$
, since
$T_z$
is linear and
$T_z$
is an isomorphism from
$\mathcal {D}^1$
onto
$B^0_1$
.
Finally, we need to prove that
$T_z$
is a bounded operator on
$\mathcal {D}^1$
. For
$f\in \mathcal {D}^1$
, we have
$$ \begin{align*}\|T_zf\|_{B_1}\lesssim ( |(T_zf)(0)|+\|f\|_{\mathcal{D}^1})\lesssim \|f\|_{\mathcal{D}^1}. \end{align*} $$
Therefore,
$T_z$
is a bounded isomorphism from
$\mathcal {D}^1$
onto
$B^0_1$
.
Let us introduce an addition operator P defined as
$$ \begin{align*}(Pf)(z)=(M_zf)(z)+(T_zf)(z)\ \mbox{ for } f\in H(\mathbb{D}) \mbox{ and } z\in\mathbb{D}. \end{align*} $$
Theorem 4 Let
$T_{z}:\mathcal {D}^1\rightarrow B^0_1$
and
$M_z:B^0_1\rightarrow \mathcal {D}^1$
be the Volterra-type operator and the multiplication operator, respectively. Then, P is an operator on
$\mathcal {D}^1$
with
$P=T^{-1}_{z}M_zT_z.$
Proof Let
$f\in \mathcal {D}^1$
and
$F=T_zf$
. Then, we get
$$ \begin{align*} (Pf)(z)=&(M_zf)(z)+(T_zf)(z) =zf(z)+F(z) =D(zF(z)) =(T_z^{-1}M_zT_zf)(z), \end{align*} $$
which shows that
$P=T^{-1}_{z}M_zT_z$
.
Theorem 5 If
$f,g\in B_1$
, then
$\|fg\|_{B_1}\leq (2\pi +2)\|f\|_{B_1}\|g\|_{B_1}$
.
Proof For
$f,g\in B_1$
, we get
$$ \begin{align*} \|f g\|_{B_1}&=|f(0)g(0)|+|(fg)'(0)|\\& \quad +\int_{\mathbb{D}}|f''(w)g(w)+2f'(w)g'(w)+f(w)g''(w)| \,dA(w)\\& \leq (|f(0)|+|f'(0)|)(|g(0)|+|g'(0)|)+\int_{\mathbb{D}}|f''(w)g(w)| \,dA(w)\\& \quad +2\int_{\mathbb{D}}|f'(w)g'(w)| \,dA(w)+\int_{\mathbb{D}}|f(w)g''(w)| \,dA(w)\\&\leq\|g\|_{B_1}(|f(0)|+|f'(0)|) +\|g\|_{\infty} (\|f\|_{B_1}-|f(0)|-|f'(0)|)\\& \quad +2\|f\|_{B_1}\|g\|_{B_1}+\|f\|_{\infty}\int_{\mathbb{D}}|g''(w)| \,dA(w)\\&\leq\|g\|_{B_1}(|f(0)|+|f'(0)|)+\pi\|g\|_{B_1}(\|f\|_{B_1}-|f(0)|-|f'(0)|)\\& \quad +2\|f\|_{B_1}\|g\|_{B_1}+\pi\|f\|_{B_1}\|g\|_{B_{1}}\\&\leq(2\pi+2)\|f\|_{B_1}\|g\|_{B_1}, \end{align*} $$
and the proof is complete.
Remark 4 In [Reference Arazy, Fisher and Peetre5, Theorem 10], Arazy et al. obtained
$\|fg\|\leq 7\|f\|\|g\|$
for
$f,g\in B_1$
, in which they defined the norm of
$f\in B_1$
as
$$ \begin{align*}\|f\|=\inf\left\{\sum^{\infty}_{k=1}|c_{k}|:f(z)=\sum^{\infty}_{k=1}c_k\sigma_{a_k}(z)\right\}. \end{align*} $$
Inspired by their work, we derived Theorem 5 and we are not sure whether the constant
$2\pi +2$
in Theorem 5 is optimal or not.
3 Deddens algebras
Let
$\mathcal {L}(X)$
denote the algebra of all bounded linear operators on a complex Banach space X. A nontrivial invariant subspace of an operator
$A\in \mathcal {L}(X)$
is, by definition, a closed subspace M of X such that
$M\neq \{0\}$
,
$M\neq X$
, and
$Ax\in M$
for every
$x\in M$
; or, briefly,
$A(M)\subset M$
.
Let
$A\in \mathcal {L}(X)$
. The operator T is said to belong to the Deddens algebra
$\mathcal {D}_A$
if there exists
$M=M(T)>0$
such that
$$ \begin{align*}\|A^nTf\|\leq M\|A^nf\| \end{align*} $$
for each
$n\in \mathbb {N}$
and
$f\in X$
.
The study of the Deddens algebra was originally introduced by Deddens [Reference Deddens14], where he assumed that A is an invertible operator and
$\sup _{n\in \mathbb {N}}\|A^nTA^{-n}\|<\infty $
. Later, it received the attention of many scholars (see [Reference Deddens14, Reference Karaev and Mustafayev19, Reference Lacruz20, Reference Petrovic34–Reference Petrovic and Sievewright37, Reference Sievewright42, Reference Todorov43]). Recently, Petrovic and Sievewright [Reference Petrovic and Sievewright37] studied the Deddens algebra associated with compact composition operators
$C_{\varphi }$
on Hardy spaces, where A is not necessarily invertible, and they have demonstrated that the operators
$M_g$
and
$T_z$
belong to the Deddens algebra
$\mathcal {D}_{C_{\varphi }}$
. It is worth to point out that compact operators on
$H^2$
are not invertible.
Let us begin to present the boundedness of composition operators and multiplication operators on
$B_1$
.
Lemma 6 [Reference Wulan and Xiong45]
Let
$\varphi $
be an analytic self-map of
$\mathbb {D}$
. Then, the composition operator
$C_{\varphi }$
is bounded on
$B_1$
if and only if
$$ \begin{align*}\sup_{a\in\mathbb{D}}\|C_{\varphi}\sigma_a\|_{B_1}<\infty. \end{align*} $$
In particular, we give another description of sufficiency condition for the boundedness of the composition operator in the following theorem.
Theorem 6 Let
$\varphi $
be an analytic self-map of
$\mathbb {D}$
such that
$\varphi (0)=0$
. Then, the composition operator
$C_{\varphi }$
is bounded on
$B_1$
whenever
$\varphi '\in Z_1\cap H^{\infty }$
.
Proof Suppose that
$\varphi '\in Z_1\cap H^{\infty }$
. Then, we get
$$ \begin{align*} \|C_{\varphi}f\|_{B_1}&=\|f(\varphi(z))\|_{B_1}\\&=|f(\varphi(0))|+|f'(\varphi(0))\varphi'(0)|+\int_{\mathbb{D}}|f''(\varphi(w))\cdot(\varphi'(w))^2\\ & \quad + f'(\varphi(w))\cdot\varphi''(w)| \,dA(w)\\&\leq |f(\varphi(0))|+|f'(\varphi(0))\varphi'(0)|+\int_{\mathbb{D}}|f''(\varphi(w))\cdot(\varphi'(w))^2| \,dA(w) \\ & \quad +\int_{\mathbb{D}}|f'(\varphi(w))\cdot\varphi''(w)| \,dA(w)\\ &\lesssim \|f\|_{\infty}+\|\varphi'\|_{\infty} \|f\|_{B_{1}}+\|\varphi'\|^{2}_{\infty} \|f\|_{B_{1}} +\|\varphi\|_{Z_{1}} \|\varphi'\|_{\infty} \|f\|_{B_{1}} \\ &\lesssim (\|\varphi'\|^{2}_{\infty}+\|\varphi\|_{Z_{1}} \|\varphi'\|_{\infty}+ \|\varphi'\|_{\infty} +1)\|f\|_{B_1}, \end{align*} $$
which implies that
$C_{\varphi }$
is bounded on
$B_1$
if
$\varphi '\in Z_1\cap H^{\infty }$
.
Theorem 7 Suppose that
$M_g$
is a multiplication operator on
$B_1$
. Then,
$M_g$
is bounded if and only if
$g\in B_1$
.
Proof For any
$f\in B_1$
, if
$g\in B_1$
, we have
$M_g$
is bounded on
$B_1,$
by Theorem 5.
Conversely, let
$M_g$
be a bounded operator on
$B_1.$
Then, with
$f=1$
, we get
$\|M_g\|\geq \|M_g1\|_{B_1}=\|g\|_{B_1}$
, which implies that
$g\in B_1$
.
In the following theorem, we will consider the algebra
$\mathcal {D}_{C_{\varphi }}$
, in which the operator
$C_{\varphi }$
is a bounded composition operator. For
$n\in \mathbb {N}$
, it clear that
$C^{n}_{\varphi }f=f\circ \varphi \circ \cdots \circ \varphi $
. For simplicity of the notation, we write
$\varphi _{n}$
instead of
$\varphi \circ \cdots \circ \varphi $
.
Theorem 8 Let
$g\in B_1$
, and let
$\varphi $
be an analytic self-map of
$\mathbb {D}$
with
$\varphi (0)=0$
such that
$C_{\varphi }$
is bounded on
$B_1$
. Then, the operators
$M_g$
,
$T_g$
, and
$I_g$
belong to the Deddens algebra
$\mathcal {D}_{C_{\varphi }}$
.
Proof For each
$n\in \mathbb {N}$
, we see that
$$ \begin{align*} C^n_{\varphi}M_gf=&C^{n}_{\varphi}(gf) =(g\circ\varphi_{n})(f\circ\varphi_{n}) =M_{g\circ\varphi_n}C^{n}_{\varphi}f. \end{align*} $$
Since
$\varphi _{n}(\mathbb {D})\subset \mathbb {D}$
, it follows that
$$ \begin{align*}\|M_{g\circ\varphi_n}f\|_{B_1}=\|(g\circ\varphi_n)f\|_{B_1}\lesssim\|g\|_{B_1}\|f\|_{B_1}, \end{align*} $$
and therefore
$$ \begin{align*}\|C^n_{\varphi}M_gf\|_{B_1}=\|M_{g\circ\varphi_n}C^{n}_{\varphi}f\|_{B_1}\lesssim\|g\|_{B_1}\|C^{n}_{\varphi}f\|_{B_1}, \end{align*} $$
where
$f\in B_{1}(\mathbb {D})$
. This implies that
$M_g\in \mathcal {D}_{C_{\varphi }}$
.
Next, we have
$$ \begin{align*}C^n_{\varphi}T_gf(z)=C^n_{\varphi}\left(\int^{z}_{0}f(w)g'(w)dw\right)=\int^{\varphi_{n}(z)}_{0}f(w)g'(w) \,dw. \end{align*} $$
Since
$\varphi $
is an analytic self-map of
$\mathbb {D}$
satisfying
$\varphi (0)=0$
, we have
$\varphi _{n}(0)=0$
, and therefore
$$ \begin{align*} T_{g\circ\varphi_n}C^{n}_{\varphi}f(z)&=T_{g\circ\varphi_n}\left(f(\varphi_n(z))\right)\\ &=\int^{z}_{0}f(\varphi_{n}(w))g'(\varphi_{n}(w))\varphi_{n}'(w) \,dw\\ &=\int^{\varphi_{n}(z)}_{0}f(w)g'(w) \,dw, \end{align*} $$
where
$f\in B_1$
. It shows that
$C^n_{\varphi }T_g=T_{g\circ \varphi _n}C^{n}_{\varphi }$
. By Theorem 1, we have
$$ \begin{align*} \|C^{n}_{\varphi}T_gf(z)\|_{B_1}=&\|T_{g\circ\varphi_n}C^{n}_{\varphi}f(z)\|_{B_1} \lesssim \|g\circ\varphi_n\|_{B_1}\|C^{n}_{\varphi}f(z)\|_{B_1} \lesssim \|g\|_{B_1}\|C^{n}_{\varphi}f(z)\|_{B_1}, \end{align*} $$
which gives
$T_g\in \mathcal {D}_{C_{\varphi }}$
.
Finally, we have
$$ \begin{align*}C^n_{\varphi}I_gf(z)=C^{n}_{\varphi}\left(\int^{z}_{0}f'(w)g(w) \,dw\right)=\int^{\varphi_{n}(z)}_{0}f'(w)g(w) \,dw \end{align*} $$
and
$$ \begin{align*} I_{g\circ\varphi_{n}}C^n_{\varphi}f(z)&=I_{g\circ\varphi_{n}}(f(\varphi_{n}(z)))\\ &=\int^{z}_{0}f'(\varphi_{n}(w))g(\varphi_{n}(w))\varphi_{n}'(w) \,dw\\ &=\int^{\varphi_n(z)}_{0}f'(w)g(w) \,dw, \end{align*} $$
where
$f\in B_1$
. Therefore,
$C^n_{\varphi }I_g=I_{g\circ \varphi _{n}}C^n_{\varphi }$
. By Theorem 2 and Remark 3, we find that
$$ \begin{align*}\|C^n_{\varphi}I_gf(z)\|_{B_1}=\|I_{g\circ\varphi_{n}}C^n_{\varphi}f(z)\|_{B_1} \lesssim\|g\circ\varphi_{n}\|_{B_1}\|C^n_{\varphi}f(z)\|_{B_1}\lesssim\|g\|_{B_1}\|C^n_{\varphi}f(z)\|_{B_1} \end{align*} $$
for all
$f\in B_1$
. We thus deduce that
$I_g\in \mathcal {D}_{C_{\varphi }}$
.
4 Essential norms of Volterra-type operators on
$B_{1}$
Suppose that X is a Banach space and T is a bounded linear operator on X. The essential norm of T is defined to be
$$ \begin{align*}\|T\|_{e}=\inf \{\|T-K\|:K ~\text{is a compact operator on } B_1\}. \end{align*} $$
Obviously, the essential norm of T is 0 if and only if T is compact. For more results, we invite the reader to refer to [Reference Liu, Liu, Xia and Yu26, Reference Shargorodsky41]. In this section, we characterize the essential norm of linear operator on
$B_1$
, which generalizes the conclusion of Liu et al. [Reference Liu, Lou and Xiong24]
Theorem 9 Every bounded operator
$T_g$
on
$B_1$
is compact.
Proof By definition, we know
$\|T_{g}\|_{e}\geq 0$
.
Next, we show that
$\|T_{g}\|_{e}\leq 0$
. To do this, we define the following operators:
$$ \begin{align*}T_{g_r}f:=\int^z_0rf(w)g_r'(w) \,dw, \end{align*} $$
where
$g_r(z)=g(rz)$
and
$r\in (0,1)$
. It is easy to see that
$T_{g_r}$
is a compact operator on
$B_1$
for
$g\in B_1$
. In fact, if
$T_{g}$
is a bounded operator on
$B_1$
, then
$g\in B_1$
. Suppose that
$\{ f_{n}\}_{n=1}^{\infty } \subset B_1$
with
$\|f_{n}\|_{B_1}\leq 1$
, and
$f_{n}\rightarrow 0$
uniformly on compact subsets of
$\mathbb {D}$
. Then,
$$ \begin{align*} \|T_{g_r}f_n\|_{B_1}&=\|f_{n}(rz)g'(rz)\|_{\mathcal{D}^1}\\ &\lesssim \int_{\mathbb{D}}|f'_{n}(rz)g'(rz)| \,dA(z)+\int_{\mathbb{D}}|f_{n}(rz)g''(rz)| \,dA(z)\\ &=\int_{\mathbb{D}_{\delta_1}}|f'_{n}(rz)g'(rz)| \,dA(z)+\int_{\mathbb{D}\backslash \mathbb{D}_{\delta_1}}|f'_{n}(rz)g'(rz)| \,dA(z)\\ & \quad +\int_{\mathbb{D}_{\delta_1}}|f_{n}(rz)g''(rz)| \,dA(z)+\int_{\mathbb{D}\backslash \mathbb{D}_{\delta_1}}|f_{n}(rz)g''(rz)| \,dA(z), \end{align*} $$
where
$\delta _1<1,$
and
$\mathbb {D}_{\delta _1}= \{z: |z|<\delta _1\}$
are compact subsets of
$\mathbb {D}$
.
Note that
$$ \begin{align*}\int_{\mathbb{D}}|f'_{n}(rz)g'(rz)| \,dA(z)\kern1.4pt{\lesssim}\kern1.4pt \|f_{n}\|_{B_1} \|g\|_{B_1} ~\text{ and }~\! \int_{\mathbb{D}}\kern-1pt|f_{n}(rz)g''(rz)| \,dA(z)\kern1.4pt{\lesssim}\kern1.4pt \|f_{n}\|_{B_1} \|g\|_{B_1}\kern-1pt. \end{align*} $$
Using the theorem of absolute continuity of Lebesgue measure, we conclude that
$$ \begin{align*}\int_{\mathbb{D}\backslash \mathbb{D}_{\delta_1}}|f'_{n}(rz)g'(rz)| \,dA(z) <\epsilon ~\text{ and }~ \int_{\mathbb{D}\backslash \mathbb{D}_{\delta_1}}|f_{n}(rz)g''(rz)| \,dA(z) <\epsilon. \end{align*} $$
On the other hand, using a basic result of complex analysis (see page 151 of [Reference Conway12]), we know that if
$f_{n}\rightarrow 0,$
then
$f_{n}' \rightarrow 0$
uniformly on compact subsets of
$\mathbb {D}$
. Consequently,
$$ \begin{align*}\int_{\mathbb{D}_{\delta_1}}|f'_{n}(rz)g'(rz)| \,dA(z) ~\text{ and }~ \int_{\mathbb{D}_{\delta_1}}|f_{n}(rz)g''(rz)| \,dA(z) \end{align*} $$
converge to
$0$
when
$n\rightarrow \infty $
. This implies that
$T_{g,r}$
is compact.
Meanwhile, we have
$$ \begin{align*}\|T_{g}\|_{e}\leq\|T_{g}-T_{g_r}\|=\|T_{g-g_r}\|\lesssim\|g-g_r\|_{B_1} \end{align*} $$
for
$r\in (0,1)$
. Similarly, with the above computation, we have
$$ \begin{align*}\lim_{r\rightarrow 1^{-}}\|g-g_r\|_{B_1}=0, \end{align*} $$
so that
$\|T_{g}\|_{e}\leq 0$
. Hence, we deduce that
$\|T_{g}\|_{e}=0$
, which completes the proof.
Theorem 10 If
$I_{g}$
is bounded operator on
$B_1$
, then
$\|g\|_{\infty }\leq \|I_{g}\|_{e} \lesssim ( \|g\|_{\infty }+\|g\|_{Z_{1}}).$
Proof From the proof of Theorem 2, we have
$$ \begin{align*}\|I_g\|_{e}=\inf\|I_g-K\|\leq\|I_g\|\leq C(\|g\|_{\infty}+\|g\|_{Z_{1}}), \end{align*} $$
where C is a positive constant.
We next show that
$\|I_g\|_{e}\geq \|g\|_{\infty }$
. Choose
$a_n\in \mathbb {D}$
such that
$|a_n|\rightarrow 1$
as
$n\rightarrow \infty $
. Let
$f_n(z)=\sigma _{a_n}(z)-a_n$
. It is obvious that
$\|f_n\|_{B_1}=1$
. Since
$\{f_n\}$
converges to zero uniformly on compact subsets of
$\mathbb {D}$
, for every compact operator K on
$B_1$
, we obtain
$\|Kf_n\|_{B_1}\rightarrow 0$
as
$n\rightarrow \infty $
. Therefore,
$$ \begin{align*} \|I_g-K\|&\geq\lim_{n\rightarrow\infty}\sup\|(I_g-K)f_n\|_{B_1}\\ &\geq\lim_{n\rightarrow\infty}\sup(\|I_gf_n\|_{B_1}-\|Kf_n\|_{B_1})\\ &=\lim_{n\rightarrow\infty}\sup\|I_gf_n\|_{B_1}. \end{align*} $$
Similarly to the proof of Theorem 2, we get
$$ \begin{align*} \|I_gf_n\|_{B_1}=\|g(z)f'_n(z)\|_{\mathcal{D}^{1}}\geq|g(\sigma_{a_n}(0))|=|g(a_{n})|. \end{align*} $$
As the choice of the sequence
$\{a_n\}\subset \mathbb {D}$
is arbitrary, we have
$\|I_g\|_{e}\geq \|g\|_{\infty }$
, which completes the proof.
Theorem 11 Every bounded operator
$M_g$
on
$B_1$
is compact.
Proof The proof is similar to Theorem 9, so we omit its details.
5 Spectrum of Volterra-type operators on
$B_{1}(\mathbb {D})$
The spectrum of integral operators on different spaces has attracted the attention of many scholars. The spectra of integral operators on weighted Bergman spaces are characterized by Aleman and Constantin [Reference Aleman and Constantin2]. Later, Constantin [Reference Constantin10] obtained the spectrum of Volterra-type operators on Fock spaces. Mengestie [Reference Mengestie29] studied the spectrum of Volterra-type operators on Fock–Sobolev spaces. Mengestie [Reference Mengestie30] also obtained the spectrum of
$T_g$
in terms of a closed disk of radius twice the coefficient of the highest degree term in a polynomial expansion of g. For more results, see [Reference Bonet8, Reference Malman27]. Recently, Lin et al. described the spectra of the multiplication operator and the Volterra-type operator
$I_g$
in [Reference Lin, Liu and Wu23], respectively. Inspired by the above results, it is natural to study the spectra of the multiplication operator and the Volterra-type operators on
$B_{1}$
.
Theorem 12 Suppose that
$M_g$
is bounded on
$B_1$
. Then, we have
$\sigma (M_g)=\overline {g(\mathbb {D})}.$
Proof Suppose that
$\lambda \notin \sigma (M_g)$
. Then,
$M_g-\lambda I$
is invertible. As
$1 \in B_1,$
there exists an
$f\in B_1$
such that
$(g(z)-\lambda )f(z)=1$
for all
$z\in \mathbb {D}$
, which implies that
$\lambda \notin g(\mathbb {D})$
. Thus,
$g(\mathbb {D})\subset \sigma (M_g)$
.
For the other way inclusion, we let
$\lambda \notin \overline {g(\mathbb {D})}$
. Then, we can choose a
$t>0$
such that
$|g(z)-\lambda |>t$
for all
$z\in \mathbb {D}$
. This shows that
$h=(g-\lambda )^{-1}$
is a bounded analytic function on
$\mathbb {D}$
. For all
$g\in B_1$
, we get
$$ \begin{align*} \|h\|_{B_1}&\leq t(|h(0)|+\|h'\|_{\mathcal{D}^{1}})\\&\leq \left(\frac{1}{|g(0)-\lambda|}+\|\frac{g'}{(g-\lambda)^2}\|_{\mathcal{D}^{1}}\right)\\[-14pt] \\&\leq \left(\frac{1}{|g(0)-\lambda|}+\left|\frac{g'(0)}{(g(0)-\lambda)^2}\right| +\int_{\mathbb{D}}\left|\frac{g''(z)}{(g(z)-\lambda)^2}-\frac{2(g')^2}{(g(z)-\lambda)^{3}}\right| \,dA(z)\right)\\&\lesssim \|g\|_{B_1}+ \|g\|^2_{B_1}. \end{align*} $$
Hence,
$h\in B_1$
. Then,
$M_h$
is bounded on
$B_1,$
by Theorem 7. Since
$M_h=M_{(g-\lambda )^{-1}}=M_{g-\lambda }^{-1}$
, we see that
$M_{g-\lambda }$
is invertible, and thus
$\lambda \notin \sigma (M_g)$
. Therefore,
$\sigma (M_g)\subset \overline {g(\mathbb {D})}$
. Since the spectrum set is closed, we conclude that
$\sigma (M_g)=\overline {g(\mathbb {D})}$
.
Lemma 7 [Reference Rudin40]
Let T be a bounded linear operator on a Banach space X, and let T be compact. If
$\dim X=\infty $
, then
$\sigma (T)=\{0\} {\cup} $
{eigen values of T}.
Theorem 13 Suppose that
$T_g$
is a bounded operator on
$B_1$
. Then,
$\sigma (T_g)=\{0\}.$
Proof Let
$T_g$
be a bounded operator on
$B_1.$
Then,
$T_g$
is compact, by Theorem 9. By Lemma 7, we obtain
$0\in \sigma (T)$
.
Next, we prove that
$T_g$
has no nonzero eigenvalue. Assume that
$T_g$
has an eigenvalue
$\lambda \neq 0$
with eigenvector
$f.$
Then,
$$ \begin{align} T_{g}f(z)=\int^{z}_{0}f(w)g'(w) \,dw=\lambda f(z). \end{align} $$
Differentiating equation (1) with respect to z, we get
$$ \begin{align*}f(z)g'(z)=\lambda f'(z). \end{align*} $$
All nonzero solutions of this equation are of the form
$f(z)=ce^{\frac {g(z)}{\lambda }}$
for some
$c\neq 0$
. Setting
$z=0$
in (1) shows that
$0=\lambda f(0)$
, which contradicts the last relation about
$f.$
Therefore, there is no nonzero eigenvalue for
$T_g$
. From this, we deduce that
$\sigma (T_g)=\{0\}$
.
Theorem 14 If
$I_g$
is a bounded operator on
$B_1$
, then
$$ \begin{align*}\sigma(I_g)=\{0\}\cup \overline{g(\mathbb{D})}. \end{align*} $$
Proof For any constant function a, we have
$$ \begin{align*}(I_ga)(z)=\int_{0}^{z}a'(w)g(w)\,dw=0, \end{align*} $$
which gives
$0\in \sigma (I_g)$
.
Suppose that
$\lambda \in \mathbb {C}\backslash \{0\}$
. Note that the equation
$$ \begin{align*}f-\frac{1}{\lambda}I_gf=h,~\hbox {for} ~h\in B_1, \end{align*} $$
has a unique analytic solution
$$ \begin{align*}f(z)=R_{\lambda,g}h(z)=\int^{z}_{0}\frac{h'(\xi)}{1-\frac{1}{\lambda}g(\xi)} \,d\xi+h(0)=I_{(1-\frac{1}{\lambda}g)^{-1}}h(z)+h(0) \end{align*} $$
(see [Reference Constantin and Persson11] for more details). Hence, the resolvent set
$\rho (I_g)$
of the bounded operator
$I_g$
consists precisely of all points
$\lambda \in \mathbb {C}$
for which
$R_{\lambda ,g}$
is a bounded operator on
$B_1$
.
If
$\lambda \in \mathbb {C}\backslash (\{0\}\cup \overline {g(D)})$
, then it is clear that
$1-\frac {1}{\lambda }g(z)$
is bounded away from
$0$
, which implies that
$\frac {1}{1-\frac {1}{\lambda }g(z)} \in H^{\infty }$
. If
$I_g$
is a bounded operator on
$B_1$
, then
$g\in H^{\infty }\bigcap Z_{1}$
by Theorem 2. Moreover, it is easy to show that
$ \frac {1}{1-\frac {1}{\lambda }g(z)} \in Z_{1}.$
This implies that the operator
$R_{\lambda ,g}$
is a bounded operator on
$B_1$
. It follows that
$\mathbb {C}\backslash (\{0\}\cup \overline {g(\mathbb {D})})\subset \rho (I_g)$
. Thus,
$\sigma (I_g)\subset (\{0\}\cup \overline {g(\mathbb {D})})$
.
On the other hand, if
$\lambda \in g(\mathbb {D})$
and
$\lambda \neq 0$
, then
$\frac {1}{1-\frac {1}{\lambda }g(\xi )}$
is not bounded, which shows that the operator
$R_{\lambda ,g}$
is not bounded on
$B_1$
. Therefore, we obtain that
$g(\mathbb {D})\backslash \{0\}\subset \sigma (I_g)$
. This together with the fact that
$0\in \sigma (I_g)$
shows that
$$ \begin{align*}g(\mathbb{D})\cup\{0\}\subset\sigma(I_g)\subset \overline{g(\mathbb{D})}\cup\{0\}. \end{align*} $$
Since the spectrum
$\sigma (I_g)$
is closed, we deduce that
$\sigma (I_g)=\overline {g(\mathbb {D})}\cup \{0\}$
.
Acknowledgement
We would like to thank the referee for his/her valuable comments.
