1 Introduction
Let
$\mathbb {D}$
and
$\partial \mathbb {D}=\{z:|z|=1\}$
denote, respectively, the open unit disc and the unit circle in the complex plane
$\mathbb {C}.$
We denote as
$\mathrm{Mer}(\mathbb {D})$
the set of meromorphic functions defined in the unit disc
$\mathbb {D}$
. Let
$\mathrm{Hol}(\mathbb {D})$
be the space of all analytic functions on
$\mathbb {D}$
.
Given
$g\in \mathrm{Hol}(\mathbb {D})$
, the Volterra-type operator
$T_{g}$
is defined as
$$ \begin{align*}T_{g}f(z)=\int_{0}^{z}g'(\omega)f(\omega)\,d\omega,\quad\forall f\in \mathrm{Hol}(\mathbb{D}).\end{align*} $$
If
$g=\log \frac {1}{1-z}$
, the Volterra-type operator
$T_{g}$
is known as the
$Ces\grave {a}ro$
operator. Another related integration operator is defined as
$$ \begin{align*}S_{g}f(z)=\int_{0}^{z}f'(\omega)g(\omega)\,d\omega,\quad\forall f\in \mathrm{Hol}(\mathbb{D}).\end{align*} $$
Denote the multiplication operator by
$M_{g}f=fg$
, which satisfies the fundamental identity
$$ \begin{align*}M_{g}f(z)=T_{g}f(z)+S_{g}f(z)+f(0)g(0).\end{align*} $$
On
$H^2$
, the operator
$T_g$
is bounded if and only if
$g\in \mathrm {BMOA}$
[Reference Pommerenke11]; this extends to all
$H^p$
with
$1\le p<\infty $
[Reference Aleman and Siskakis2], and there is also a complete description of the boundedness of
$T_g:H^p\to H^q$
[Reference Aleman and Cima1]. For
$g(z)=-\log (1-z),$
one recovers the shifted Cesàro operator, which is bounded on every
$H^p$
[Reference Miao10, Reference Siskakis12], yet there exist functions
$f\notin H^p$
with
$C(f)\in H^p$
. These facts motivated Curbera–Ricker to introduce the optimal domain viewpoint for C and, more generally, for
$T_g$
[Reference Curbera and Ricker5, Reference Curbera and Ricker6]. Conceptually, the Hardy theory shows that
$T_g$
measures the boundary oscillation of g through
$g'$
, while the target space fixes the integrability scale. The optimal-domain perspective decouples “what the operator can accept” from any preassigned source space, which is precisely the lever one needs when the symbol has sparse regularity or large oscillation.
Definition 1.1 The space of the bounded analytic functions
$H^{\infty }(\mathbb {D})$
consists of all functions f in
$\mathrm{Hol}(\mathbb {D})$
such that
$$ \begin{align*}\|f\|_{\infty}=\sup\{|f(z)|:z\in \mathbb{D}\}. \end{align*} $$
Definition 1.2 If
$0<\alpha \leq 1$
, the Lipschitz space
$\Lambda _\alpha $
is defined as
$$ \begin{align*}\Lambda_\alpha=\left\lbrace f \in \mathrm{Hol}(\mathbb{D}):\; \|f\|_{\Lambda_\alpha}=\sup\nolimits_{z \in \mathbb{D}}(1-|z|^2)^{1-\alpha} |f^{\prime}(z)|<\infty\right\rbrace. \end{align*} $$
For an in-depth description of the spaces
$\Lambda _\alpha $
, we refer to [Reference Zhu14].
Definition 1.3 For
$0\leq \alpha <1$
, we also consider the Korenblum space
$K_\alpha $
, defined as
$$ \begin{align*}K_\alpha=\left\lbrace f \in \mathrm{Hol}(\mathbb D) :\; \|f\|_{K_\alpha}=\sup_{z \in \mathbb D}(1-|z|^2)^\alpha |f(z)|<\infty\right\rbrace.\end{align*} $$
We recall here some of the properties of
$K_\alpha $
that we will use. First of all, it is clear that
$f \in \Lambda _\alpha $
if and only if
$f' \in K_{1-\alpha }$
. Moreover, if
$f \in K_{\alpha }$
, then
$f \in K_\beta $
, for every
$\beta>\alpha $
. For an introduction to Korenblum spaces, we refer the reader to [Reference Korenblum8].
Definition 1.4 Let
$\alpha>0$
, the Bloch-type space
$\mathcal {B}_{\alpha }$
is defined to be the space of
${f\in \mathrm{Hol}(\mathbb {D})}$
such that
$$ \begin{align*}\|f\|_{\mathcal{B}_{\alpha}}:=|f(0)|+\sup\{(1-|z|^2)^{\alpha}|f'(z)|:z\in\mathbb{D}\}<\infty.\end{align*} $$
It is clear that
$f \in \Lambda _\alpha $
if and only if
$f \in \mathcal {B}_{1-\alpha }$
. When
$\alpha =1,$
it is the classical Bloch space
$\mathcal {B}.$
For further properties of Bloch-type spaces, we refer the reader to [Reference Hedenmalm, Korenblum and Zhu7, Reference Zhu15] and the references therein.
Definition 1.5 Given
$0<\lambda < 2$
and
$p>0$
, a function f in
$\mathrm{Hol}(\mathbb {D})$
is said to belong to Bergman–Morrey space
$A^{p,\lambda }$
if
$$ \begin{align*}\|f\|_{A^{p,\lambda}}:=\left(|f(0)|^p+ \sup\limits_{a\in\mathbb{D}} (1-|a|^{2})^{2-\lambda}\int_{\mathbb{D}}|f'(z)|^{p}(1-|z|^{2})^{p}|\sigma^{\prime}_{a}(z)|^{2}dA(z)\right)^{\frac{1}{p}}<\infty,\end{align*} $$
where
$\sigma _{a}(z)=\frac {a-z}{1-\overline {a}z}$
.
The Bergman–Morrey scale
$A^{p,\lambda }$
may be viewed as a Morrey-type refinement of classical Bergman theory, aligned with the tent/Carleson-box geometry built into its definition. In particular, Yang–Liu developed a precise Carleson embedding for
$A^{p,\lambda }$
and established a mapping theory on this scale for the two Volterra-type integrals
$T_g$
and
$I_g$
– including norm and essential-norm formulations (e.g.,
$\|I_g\|\simeq \|g\|_{\infty }$
and
$\|T_g\|\simeq \|g\|_{\mathcal B}$
), as well as compactness criteria (see [Reference Yang and Liu13]). Against this background, the optimal-domain program becomes natural: for a fixed target
$A^{p,\lambda }$
, we seek the largest (holo/mero)morphic input classes that
$T_g$
can admit while still landing in
$A^{p,\lambda }$
. Allowing meromorphic inputs with the constraint
$fg'\in \mathrm{Hol}(\mathbb {D})$
neutralizes the zeros of
$g'$
without losing analytic control of the primitive.
In short, the space
$A^{p,\lambda }$
supplies the right Carleson geometry and growth scale to quantify how much of
$g'$
the Volterra primitive can absorb, and the (meromorphic) optimal domains record exactly that amount. In particular, when
$\lambda =0$
,
$A^{p,0}=A^{p}$
. If
$0<\lambda \leq 2+p<\infty $
and
$\alpha =\frac {p+2-\lambda }{p},$
then
$A^{p,\lambda }$
is contained in the Bloch-type spaces
$\mathcal {B}_{\alpha }$
.
For an arc
$I\subset \partial \mathbb D$
, write
$|I|$
for its normalized arc length, so that
$|\partial \mathbb D|=1$
. The Carleson box (or Carleson square) over I is defined by
$$\begin{align*}S(I):=\bigl\{\, r e^{i\theta}\in\mathbb D:\ e^{i\theta}\in I,\ 1-|I|\le r<1 \,\bigr\}. \end{align*}$$
With this convention, if
$I=\partial \mathbb D,$
then
$S(I)=\mathbb D$
(since
$|\partial \mathbb D|=1$
and hence
$0\le r<1$
).
Definition 1.6 [Reference Li, Liu and Lou9]
For
$0<p<\infty ,$
a nonnegative measure
$\mu $
on
$\mathbb {D}$
is called a p-Carleson measure if
$$ \begin{align*}\sup _{I \subset \partial \mathbb{D}} \frac{\mu(S(I))}{|I|^{p}}<\infty.\end{align*} $$
When
$p=1, ~1$
-Carleson measure is the classical Carleson measure.
In [Reference Curbera and Ricker5], Curbera and Ricker showed that the optimal domain of the classical
$Ces\grave {a}ro$
operator acting on Hardy spaces
$H^{p}(\mathbb {D})$
is strictly larger than
$H^{p}(\mathbb {D})$
. Then Bellavita et al. [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis4] considered the optimal domain of
$T_{g}$
acting on
$H^{p}(\mathbb {D})$
and described its structural properties. The optimal domain of the Volterra-type operator
$T_{g}$
acting on
$A^{p,\lambda }$
is defined as follows:
$$ \begin{align*}[T_{g},A^{p,\lambda}]:=\{f\in \mathrm{Hol}(\mathbb{D})\,:\,T_{g}f\in A^{p,\lambda}\}.\end{align*} $$
Clearly,
$[T_{g},A^{p,\lambda }]$
is a linear subspace of
$\mathrm{Hol}(\mathbb {D})$
. If g is constant, then
$[T_{g},A^{p,\lambda }]=\mathrm{Hol}(\mathbb {D})$
. If g is non-constant, this space is equipped with the norm
$$ \begin{align*}\|f\|_{[T_{g},A^{p,\lambda}]}:=\|T_{g}f\|_{A^{p,\lambda}}.\end{align*} $$
The notation for the optimal domain follows that used in [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis4].
Definition 1.7 Let
$g\in \mathcal {B}$
and
$1\leq p<\infty $
. The meromorphic optimal domain
$(T_g,A^{p,\lambda } )$
consists of all the functions
$f\in \mathrm{Mer}(\mathbb {D})$
, such that
$f g'\in \mathrm{Hol}(\mathbb {D})$
and
$T_g(f)\in A^{p,\lambda }.$
Zeros of
$g'$
prevent a purely holomorphic domain from capturing all inputs f that are mapped by
$T_g$
into the target space. Allowing f to be meromorphic, under the constraint
$fg'\in \mathrm{Hol}(\mathbb {D})$
, restores a robust framework that mirrors the Hardy spaces setting and guarantees
$T_g(f)$
is well defined. Also, the same condition implies that the poles of f belong to the zero set of
$g'$
with at most the same multiplicity. In practice, this is the minimal enlargement that is stable under the algebra dictated by the identity
$(T_g f)'=g' f$
: it retains analytic control where it matters (on
$fg'$
) while accommodating the unavoidable singularities forced by the zeros of
$g'$
.
In order to study the properties of the meromorphic optimal domains, we remark that
$(T_g,A^{p,\lambda })$
has the following description:
$$ \begin{align} (T_g,A^{p,\lambda})=\biggl\{ f\in \mathrm{Mer}(\mathbb{D})\colon f(z)=\frac{h'(z)}{g'(z)} \text{ for some } h\in A^{p,\lambda}\biggr\} \end{align} $$
for non-constant
$g \in \mathcal {B}.$
Viewed this way,
$g'$
acts as a “change of density” between f and h: the zeros of
$g'$
are exactly the geometric barrier for bounded evaluations, and multiplier rigidity reflects that only bounded coefficients can be pushed through the Volterra primitive without losing control in the target space.
Characterizing all the functions
$h \in \mathcal {B}$
such that
$(T_h,A^{p,\lambda })= (T_g,A^{p,\lambda })$
is an interesting problem. To address it, we define and study the following space.
Definition 1.8 Let
$1\leq p<\infty $
and
$g\in \mathcal {B}$
. We consider the spaces
$$ \begin{align*} W_g^p&:=\left\lbrace h \in \mathcal{B} \colon (T_g,A^{p,\lambda})\subseteq (T_h,A^{p,\lambda})\right\rbrace,\\ V_g^p&:=\left\lbrace h \in \mathcal{B} \colon [T_g,A^{p,\lambda}]\subseteq [T_h,A^{p,\lambda}]\right\rbrace. \end{align*} $$
Our questions are symbol-theoretic: for which symbols do (mero)holomorphic optimal domains coincide, and how do these domains embed across symbols? The class
$$\begin{align*}W_g:=\{\,h\in\mathcal B:\ (T_g,A^{p,\lambda})\subseteq (T_h,A^{p,\lambda})\,\} \end{align*}$$
is designed to capture meromorphic inclusion at the level of derivatives. Indeed,
$(T_hF)'=h'F$
and
$(T_gF)'=g'F$
show that inclusions are driven by factorizations
$h'=k\,g'$
. Our multiplier results identify the admissible coefficients as bounded analytic functions, which forces the structural formula
$$\begin{align*}W_g=T_g(H^\infty)+\mathbb C, \end{align*}$$
and explains why
$W_g$
does not depend on p. In turn, this yields practical equivalence criteria for meromorphic optimal domains and keeps the bookkeeping entirely at the symbolic level, exactly where the Volterra mechanism lives.
Operationally, this says that the symbol is determined up to adding a Volterra primitive of a bounded function and a constant – an equivalence that packages the effect of g on optimal domains while stripping away inessential primitives. This p-independence is a strong hint that the phenomenon is structural rather than an artifact of a particular scale.
We emphasize that the general meromorphic optimal-domain framework for generalized Volterra operators, including the definition that allows meromorphic inputs f under the natural holomorphy condition
$f g' \in \mathrm{Hol}(\mathbb{D})$
, as well as the associated symbol classes
$W_g$
and
$V_g$
and their core structural properties, was introduced and developed by Bellavita, Belli, Nikolaidis and Stylogiannis in [Reference Bellavita, Belli, Nikolaidis and Stylogiannis3]. Our contribution is to adapt and extend this
$W_g/V_g$
-based machinery to the Bergman–Morrey spaces, thereby obtaining sharp boundedness and (holo/mero)morphic optimal-domain descriptions for Volterra-type operators between Bergman–Morrey spaces.
The article is organized as follows.
Section 2 collects preliminaries on Bergman–Morrey spaces, including equivalent norm characterizations and a sharp derivative growth estimate. Section 3 establishes sharp boundedness criteria for the Volterra-type maps between scales:
$T_g:A^{p,\lambda _1}\!\to A^{p,\lambda _2}$
is bounded iff
$g\in \mathcal B_{1+\frac {\lambda _1-\lambda _2}{p}}$
(for
$0<\lambda _1<\lambda _2<2$
,
$p>\lambda _2-\lambda _1$
), and
$S_g:A^{p,\lambda _2}\!\to A^{p,\lambda _1}$
is bounded iff
$g\in H^\infty $
. Section 4 develops the holomorphic optimal domain
$[T_g,A^{p,\lambda }]$
: we prove completeness (Banach structure), bounded point evaluations on compact sets, identify its multiplier algebra with
$H^\infty $
, and show strict inclusions as the parameters vary. Section 5 treats the meromorphic optimal domain
$(T_g,A^{p,\lambda })$
: it coincides with
$\mathrm{Mer}(\mathbb D)$
exactly when g is constant; otherwise, it is a Banach space of meromorphic functions with bounded point evaluations off
$\mathcal Z(g')$
, multiplier algebra
$H^\infty $
, and monotonicity in p. Section 6 introduces the symbol classes
$W_g$
and
$V_g^p$
and analyzes symbol equivalence: we show
$W_g=T_g(H^\infty )+\mathbb C$
(independent of p), obtain criteria for equality of optimal domains via bounded analytic multipliers, derive a comparison principle
$g'=Bu'\Rightarrow V_g^p\subseteq V_u^p$
, and give a locally univalent corollary.
We use the symbol
$\mathcal {Z}(h)$
to denote the zero set of
$h\in \mathrm{Hol}(\mathbb {D})$
, where the elements of
$\mathcal {Z}(h)$
are taken according to their multiplicities. For two functions F and
$G,$
we write
$F \lesssim G$
if there is a constant
$C>0$
, dependent only on parameters, such as
$p,\lambda ,\ldots $
, such that
$F \le CG.$
Furthermore, we denote
$F \approx G$
(i.e., F is comparable to G) if
$F \lesssim G \lesssim F$
holds.
2 Preliminaries
Lemma 2.1 [Reference Yang and Liu13, Remark 2.2]
Let
$0<\lambda <2$
and
$p>0,$
then
$f \in \mathrm{Hol}(\mathbb {D})$
has the following equivalent norms in
$A^{p, \lambda } :$
$$ \begin{align*} \|f\|_{A^{p, \lambda}} & \approx|f(0)|+\sup _{a \in \mathbb{D}}\left(\left(1-|a|^{2}\right)^{2-\lambda} \int_{\mathbb{D}}\left|f^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p}\left|\sigma_{a}^{\prime}(z)\right|{}^{2} d A(z)\right)^{\frac{1}{p}} \\ & \approx|f(0)|+\sup _{I \subset \partial \mathbb{D}}\left(\frac{1}{|I|^{\lambda}} \int_{S(I)}\left|f^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z)\right)^{\frac{1}{p}} \\ & \approx|f(0)|+\sup _{a \in \mathbb{D}}\left(\int_{\mathbb{D}}\left|f^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p-\lambda}\left(1-\left|\sigma_{a}(z)\right|{}^{2}\right)^{\lambda} d A(z)\right)^{\frac{1}{p}}. \end{align*} $$
Estimating the growth rate of functions in
$A^{p,\lambda }$
will be crucial for the subsequent proof of some later results.
Lemma 2.2 Let
$0<\lambda <2$
and
$p \ge 1$
. If
$f\in A^{p,\lambda }$
, then
$$ \begin{align*}|f'(z)|\lesssim\frac{\|f\|_{A^{p,\lambda}}}{(1-|z|^2)^{1+\frac{2-\lambda}{p}}},\quad \quad \forall z\in \mathbb{D}.\end{align*} $$
Proof Fix
$b\in \mathbb {D}$
. Choose
$a=b$
in the above equivalent norm characterization and perform the change of variables
$z=\sigma _b(w)$
. By [Reference Zhu15, Lemma 4.12], we obtain
$$ \begin{align*} \|f\|_{A^{p, \lambda}} & \approx|f(0)|+\sup _{a \in \mathbb{D}}\left[\left(1-|a|^{2}\right)^{2-\lambda} \int_{\mathbb{D}}\left|f^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p}\left|\sigma_{a}^{\prime}(z)\right|{}^{2} d A(z)\right]^{\frac{1}{p}} \\ & \geq\left[\left(1-|b|^{2}\right)^{2-\lambda} \int_{\mathbb{D}}\left|f^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p}\left|\sigma_{b}^{\prime}(z)\right|{}^{2} d A(z)\right]^{\frac{1}{p}} \\ & =\left[\left(1-|b|^{2}\right)^{2-\lambda} \int_{\mathbb{D}}\left|f^{\prime}\left(\sigma_{b}(w)\right)\right|{}^{p}\left(1-\left|\sigma_{b}(w)\right|{}^{2}\right)^{p} d A(w)\right]^{\frac{1}{p}} \\ & \gtrsim\left(1-|b|^{2}\right)^{1+\frac{2-\lambda}{p}}\left|f^{\prime}(b)\right|. \end{align*} $$
Since b is arbitrary, we conclude that
$$ \begin{align*}\left | f'(z) \right | \lesssim \frac{\left \| f \right \|_{A^{p,\lambda} } }{(1-\left | z \right |^{2} ) ^{1+\frac{2-\lambda}{p}}},~~~~\forall z \in \mathbb{D}.\end{align*} $$
This completes the proof.
The next result leverages the derivative growth bound together with Carleson-box estimates to read off the exact Bloch-type regularity of the symbol.
3 Boundedness of Volterra operators
Having fixed the basic growth and box characterizations on
$A^{p,\lambda },$
we now turn to the mapping theory of the Volterra kernels. The next section isolates the precise symbol regularity that governs boundedness between Bergman–Morrey scales.
Theorem 3.1 For
$0< \lambda _{1}<\lambda _{2}<2$
and
$p>\lambda _{2}-\lambda _{1} $
, the operator
$T_{g}:A^{p,\lambda _{1}} \to A^{p,\lambda _{2}}$
is bounded if and only if
$g \in \mathcal {B}_{\alpha }$
, where
$\alpha =1+\frac {\lambda _{1}-\lambda _{2}}{p}.$
Proof Suppose
$g \in \mathcal {B}_{\alpha }$
,
$\alpha =1+\frac {\lambda _{1}-\lambda _{2}}{p}.$
For any
$f \in A^{p,\lambda _{1}}$
and any arc
$I\subset \partial \mathbb {D},$
let
$\zeta \in \partial \mathbb {D}$
be the center of the arc I, and set
$b=(1-\left | I \right | )\zeta \in \mathbb {D}.$
Then we have
$$ \begin{align*} \frac{1}{|I|^{\lambda_{2}}} & \int_{S(I)}\left|\left(T_{g} f\right)^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z) \\& =\frac{1}{|I|^{\lambda_{2}}} \int_{S(I)}|f(z)|^{p}\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z) \\& \lesssim \frac{1}{|I|^{\lambda_{2}}} \int_{S(I)}|f(b)|^{p}\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z)\\&\quad + \frac{1}{|I|^{\lambda_{2}}}\int_{S(I)}|f(z)-f(b)|^{p}\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z) \\& \triangleq I_{1}+I_{2}. \end{align*} $$
For
$I_{1}$
, by [Reference Yang and Liu13, Lemma 2.3], we obtain
$$ \begin{align*}|f(b)| \lesssim \frac{\|f\|_{A^{p,\lambda_{1}}}}{\left(1-|b|^{2}\right)^{\frac{2-\lambda_{1}}{p}}} \lesssim \frac{\|f\|_{A^{p,\lambda_{1}}}}{|I|^{\frac{2-\lambda_{1}}{p}}}.\end{align*} $$
We could estimate
$$ \begin{align*} I_{1}&=\frac{1}{|I|^{\lambda_{2}}} \int_{S(I)}|f(b)|^{p}\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z)\\ &\lesssim\left \| f \right \|^{p} _{A^{p,\lambda_{1}}} \frac{1}{\left | I \right |^{2+\lambda_{2}-\lambda_{1}}} \int_{S(I)}\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z)\\ &\lesssim\|g\|_{\mathcal{B}_{\alpha}}^{p}\left \| f \right \|^{p} _{A^{p,\lambda_{1}}}. \end{align*} $$
Now, we estimate the term
$I_{2}$
. Let
$d\mu _{g}(z)=|{g}' (z)|^{p}(1-|z|^{2})^{p}dA(z).$
For any
$z\in S(I)$
, we know that
$$ \begin{align*}1-\left|\sigma_{b}(z)\right|{}^{2} \gtrsim \frac{1-|z|^{2}}{|I|}.\end{align*} $$
For any
$z,b\in \mathbb {D}$
, it is easy to see
$(1-|z|)\leq |1-\overline {b}z|.$
Since
$g\in \mathcal {B} _{\alpha },\alpha =1+\frac {\lambda _{1}-\lambda _{2}}{p},$
then
$d\mu _{g}(z)$
is a
$(2+\lambda _{2}-\lambda _{1})$
-Carleson measure, combining the above conditions yields
$$ \begin{align*} I_{2} & =\frac{1}{|I|^{\lambda_{2}}} \int_{S(I)}|f(z)-f(b)|^{p}\left|g^{\prime}(z)\right|{}^{p}(1-|z|^{2})^{p}dA(z) \\ & \approx (1-|b|^{2})^{2-\lambda_{2}} \int_{S(I)}\frac{|f(z)-f(b)|^{p}(1-|b|^{2})^{2}}{|1-\bar{b}z|^{4}}\left|g^{\prime}(z)\right|{}^{p}\left(1-\left|z\right|{}^{2}\right)^{p} dA(z) \\ & \approx (1-|b|^{2})^{2} \int_{S(I)}\frac{|f(z)-f(b)|^{p}(1-|b|^{2})^{2}}{|1-\bar{b}z|^{4+\lambda_{2}}}\left|g^{\prime}(z)\right|{}^{p}\left(1-\left|z\right|{}^{2}\right)^{p} d A(z) \\ &\approx (1-|b|^{2})^{2-\lambda_{1}} \int_{S(I)}\frac{|f(z)-f(b)|^{p}(1-|b|^{2})^{2}}{|1-\bar{b}z|^{4+\lambda_{2}-\lambda_{1} }}\left|g^{\prime}(z)\right|{}^{p}\left(1-\left|z\right|{}^{2}\right)^{p} d A(z) \\ & \lesssim\left \| g \right \| _{\mathcal{B_{\alpha}}}^{p} (1-|b|^{2})^{2-\lambda_{1}} \int_{\mathbb{D} }\frac{|f(z)-f(b)|^{p}(1-|b|^{2})^{2}}{|1-\bar{b}z|^{4+\lambda_{2}-\lambda_{1}}}\left(1-\left|z\right|{}^{2}\right)^{\lambda_{2}-\lambda_{1}} d A(z) \\ &\lesssim\left \| g \right \| _{\mathcal{B_{\alpha}}}^{p} (1-|b|^{2})^{2-\lambda_{1}} \int_{\mathbb{D} }\frac{|f(z)-f(b)|^{p}(1-|b|^{2})^{2}}{|1-\bar{b}z|^{4}}d A(z) \\ & \lesssim\left \| g \right \| _{\mathcal{B_{\alpha}}}^{p}\left \| f \circ \sigma_{b}(z)-f(b) \right \|_{A^{p}}^{p}\lesssim \left \| g \right \| _{\mathcal{B_{\alpha}}}^{p}\left \| f \right \|_{A^{p, \lambda_{1}}}^{p}.\\ \end{align*} $$
Consequently,
$$ \begin{align*}\left\|T_{g} f\right\|_{A^{p, \lambda_{2}}}^{p} \lesssim I_{1}+I_{2} \lesssim\|g\|_{\mathcal{B}_{\alpha}}^{p}\|f\|_{A^{p, \lambda_{1}}}^{p},\end{align*} $$
which implies
$\left \|T_{g}\right \| \lesssim \|g\|_{\mathcal {B_{\alpha }}}.$
On the other hand, assume
$T_{g}:A^{p,\lambda _{1}} \to A^{p,\lambda _{2}}$
is bounded. For any
$I \subset \partial \mathbb {D},$
let
$b=(1-|I|) \zeta \in \mathbb {D},$
where
$\zeta $
is the center of
$I.$
Then
$$ \begin{align*}\left(1-|b|^{2}\right) \approx|1-\bar{b} z| \approx|I|, \quad z \in S(I)\end{align*} $$
(see [Reference Zhu15]). Let us consider the test function
$$ \begin{align*}F_{b}(z)=(1-\left | b \right |^{2} )(|1-\bar{b} z|)^{\frac{\lambda_{1}-2}{p}-1}, ~ \text{ where}~ z\in \mathbb{D}. \end{align*} $$
It follows from [Reference Yang and Liu13, Lemma 4.2] that
$F_{b}\in A^{p,\lambda _{1}} $
and
$\left \| F_{b} (z)\right \| _{A^{p,\lambda _{1}}} \lesssim 1$
. Accordingly,
$$ \begin{align*} \frac{1}{|I|^{2+\lambda_{2}-\lambda_{1}}} &\int_{S(I)}\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z) \\ &\lesssim \frac{1}{|I|^{\lambda_{2}}} \int_{S(I)} \frac{(1-\left | b \right |^{2} )^{p}}{\left | 1-\overline{b} z \right |^{2+p-\lambda_{1}} }\left|g^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z) \\ &=\frac{1}{|I|^{\lambda_{2}}} \int_{S(I)}\left|F_{b}(z)\right|{}^{p}\left|g'(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z)\\ &=\frac{1}{|I|^{\lambda_{2}}} \int_{S(I)}\left|\left(T_{g} F_{b}\right)^{\prime}(z)\right|{}^{p}\left(1-|z|^{2}\right)^{p} d A(z)\\ & \lesssim\left\|T_{g} F_{b}\right\|_{A^{p,\lambda_{2}}}^{p} \lesssim\left\|F_{b}\right\|_{A^{p,\lambda_{1}}}^{p}. \end{align*} $$
Since I is arbitrary, we conclude that
$g \in \mathcal {B}_{1+\frac {\lambda _{1}-\lambda _{2}}{p}}$
, which completes the proof.
Complementing the
$T_g$
, the companion operator
$S_g$
exhibits a sharper rigidity: boundedness collapses to boundedness of the symbol itself.
Theorem 3.2 Let
$0< \lambda _{1}<\lambda _{2}<2$
. Then
$S_{g}:A^{p,\lambda _{2}} \to A^{p,\lambda _{1}}$
is bounded if and only if
$g \in H^{\infty }(\mathbb {D})$
.
Proof Let
$g \in H^{\infty }(\mathbb {D}).$
For any
$f \in A^{p,\lambda _{2}}(\mathbb {D}),$
we have
$$ \begin{align*} \left\|S_{g} f\right\|^{p}_{A^{p,\lambda_{1}}} &=\sup_{b \in \mathbb{D}}(1-|b|^{2})^{2-\lambda_{1}}\int_{\mathbb{D}}|{f}'(z) |^{p}|g(z)|^{p}(1-|z|^{2})^{p}|{\sigma_{b}}'(z)|^{2}dA(z)\\ &\lesssim \left \| g \right \|^{p}_{\infty} \sup_{b\in \mathbb{D}}(1-|b|^{2})^{\lambda_{2}-\lambda_1}(1-|b|^{2})^{2-\lambda_{2}}\int_{\mathbb{D} }|{f}'(z) |^{p}|{\sigma_{b}}'(z)|^{2}(1-|z|^{2})^{p}dA(z)\\ &\lesssim\left \| g \right \|^{p}_{\infty} \left \| f \right \|^{p}_{A^{p,\lambda_{2}}}\sup_{b\in \mathbb{D}}(1-|b|^{2})^{\lambda_{2}-\lambda_{1}}\\ &\lesssim \left \| g \right \|^{2}_{\infty}\left \| f \right \|^{2}_{A^{p,\lambda_{2}}}. \end{align*} $$
This leads to the boundedness of
$S_{g}$
from
$A^{p,\lambda _{2}}$
into
$A^{p,\lambda _{1}}$
.
Conversely, if
$S_{g}$
is bounded from
$A^{p,\lambda _{2}}(\mathbb {D})$
to
$A^{p,\lambda _{1}}(\mathbb {D}).$
For any
$b \in \mathbb {D},$
set
$$ \begin{align*}f_{b}(z)=(1-|b|^{2})^{\frac{\lambda_{2}-2}{p}} (\sigma _{b}(z)-b),\end{align*} $$
then
$f_{b}(z) \in A^{p,\lambda _{2}}$
and
$\|f_{b}\|_{A^{p,\lambda _{2}}}\lesssim 1$
(e.g., [Reference Yang and Liu13, Lemma 4.2]). Thus, we get
$$ \begin{align*} \|f_{b}\|^{p}_{A^{p,\lambda_{2}}} & \gtrsim\left\|S_{g} f_{b}\right\|^{p}_{A^{p,\lambda_{1}}} \\ &\approx \sup _{a \in \mathbb{D}}\left(1-|a|^{2}\right)^{2-\lambda_{1}} \int_{\mathbb{D}}\left|f_{b}^{\prime}(z)\right|{}^{p}|g(z)|^{p}(1-|z|^{2})^{p}\left|\sigma_{a}^{\prime}(z)\right|{}^{2} d A(z)\\ & \geqslant\int_{\mathbb{D}}\left|\sigma_{b}^{\prime}(z)\right|{}^{p}|g(z)|^{p}(1-|z|^{2})^{p}\left|\sigma_{b}^{\prime}(z)\right|{}^{2} d A(z)\\ & =\int_{\mathbb{D}}\left|g\left(\sigma_{b}(w)\right)\right|{}^{p}\left(1-|w|^{2}\right)^{p} d A(w)\\ & \gtrsim|g(b)|^{p}, \end{align*} $$
where we used [Reference Zhu15, Lemma 4.12] again in the last inequality. Since b is arbitrary, we have
$g\in H^{\infty }(\mathbb {D})$
.
The boundedness criteria determine what
$T_g$
can do once a source space is fixed. To decouple the operator from any preassigned source, we pass to the optimal-domain viewpoint and endow
$[T_{g},A^{p,\lambda }]$
with its natural norm.
4 The structure of the optimal domain
In this section, we identify the optimal domain of the Volterra operator
$T_g$
on the Bergman–Morrey space
$A^{p,\lambda },$
namely,
$[T_{g},A^{p,\lambda }],$
and develop its fundamental properties as a Banach space.
Firstly, for
$1\leq p<\infty $
, it is easy to verify
$([T_{g},A^{p,\lambda }],\|\cdot \|_{[T_{g},A^{p,\lambda }]})$
be a normed linear space. Non-negativity and absolute homogeneity can be directly verified. For triangle inequality, by the Minkowski inequality, we get
$$ \begin{align*} \|f+h\|_{[T_{g},A^{p,\lambda}]}&=\sup\limits_{a\in\mathbb{D}} (1-|a|^{2})^{2-\lambda}\left[\int_{\mathbb{D}}|f(z)+h(z)|^{p}|g'(z)|^{p}(1-|z|^{2})^{p}|\sigma^{\prime}_{a}(z)|^{2}dA(z)\right]^{\frac{1}{p}}\\&\leq\sup\limits_{a\in\mathbb{D}} (1-|a|^{2})^{2-\lambda}\left[ \int_{\mathbb{D}}|f(z)|^{p}|g'(z)|^{p}(1-|z|^{2})^{p}|\sigma^{\prime}_{a}(z)|^{2}dA(z)\right]^{\frac{1}{p}}\\& \quad +\sup\limits_{a\in\mathbb{D}} (1-|a|^{2})^{2-\lambda}\left[\int_{\mathbb{D}}|h(z)|^{p}|g'(z)|^{p}(1-|z|^{2})^{p}|\sigma^{\prime}_{a}(z)|^{2}dA(z)\right]^{\frac{1}{p}}\\&=\|f\|_{[T_{g},A^{p,\lambda}]}+\|h\|_{[T_{g},A^{p,\lambda}]}. \end{align*} $$
Now, we give the boundedness of the point evaluation.
Lemma 4.1 Let K be a compact subset of
$\mathbb {D}$
, for any
$z_{0} \in K\subseteq \mathbb {D}$
, then
$\delta _{z_{0}}(f)=f(z_{0})$
is bounded linear functional on
$\left [ T_{g},A^{p,\lambda } \right ].$
Proof Let
$\delta _{z_{0}}(f)=f(z_{0})$
be a point evaluation of
$\left [ T_{g},A^{p,\lambda } \right ]$
. In every compact subset K of
$\mathbb {D},$
for any
$z_{0} \in \mathbb {D},$
we need to prove
$\delta _{z_{0}}(f):A^{p,\lambda } \longrightarrow \mathbb {C}$
is bounded, it means to demonstrate
$|f(z_{0})|\le C\left \| f \right \|_{\left [ T_{g},A^{p,\lambda } \right ]}.$
For any
$f \in \left [ T_{g},A^{p,\lambda } \right ]$
, namely,
$T_{g}f \in A^{p,\lambda },$
then by Lemma 2.2,
$$ \begin{align*} |f(z_0)\cdot g'(z_0)|&=\,|(T_gf)'(z_0)|\\ &\lesssim \, \dfrac{\|T_g(f)\|_{A^{p,\lambda}}}{(1-|z_0|^2)^{1+\frac{2-\lambda}{p}}} =\dfrac{\|f\|_{[T_{g},A^{p,\lambda}]}}{(1-|z_0|^2)^{1+\frac{2-\lambda}{p}}}.\end{align*} $$
If
$g'(z_{0} )\ne 0$
, then
$$\begin{align*}|f(z_0)|\lesssim \frac{\|f\|_{[T_g,A^{p,\lambda}]}}{|g'(z_0)|(1-|z_0|^2)^{1+\frac{2-\lambda}{p}}}. \end{align*}$$
If
$g'(z_{0} ) = 0,$
there exists an annulus
$A_{z_{0}} \subseteq K \subseteq \mathbb {D}$
centered at
$z_{0}$
such that
$$ \begin{align*}\min_{\zeta\in A_{z_0}}|g'(\zeta)|>\varepsilon>0. \end{align*} $$
Combining [Reference Zhu15, Lemma 4.12] and
$\left |{A_{z_0}}\right |$
the area of the annulus
$A_{z_{0}}$
shows
$$ \begin{align*} |f(z_0)|&\leq \frac{1}{|A_{z_0}|}\int_{A_{z_0}}|f(w)|\,dA(w)\\ &\lesssim \frac{1}{|A_{z_0}|}\int_{A_{z_0}}\frac{\|f\|_{[T_g,A^{p,\lambda}]}}{(1-|w|^2)^{1+\frac{2-\lambda}{p}}\min_{\zeta\in A_{z_0}}|g'(\zeta)|}\,dA(w)\\ & =\frac{\|f\|_{[T_g,A^{p,\lambda}]}}{\varepsilon|A_{z_0}|}\int_{A_{z_0}}\frac{1}{(1-|w|^2)^{1+\frac{2-\lambda}{p}}}\,dA(w)= C\,\|f\|_{[T_g,A^{p,\lambda}]}, \end{align*} $$
where
$C>0$
depends only on
$z_0.$
In summary,
$~~\delta _{z_{0}} $
is bounded linear functional on
$\left [ T_{g},A^{p,\lambda } \right ].$
With point evaluations under control on compacta, completeness follows by a standard Cauchy-limit argument transported through
$T_g.$
Theorem 4.2 Let
$0<\lambda <2, p \ge 1,$
and
$g\in \mathcal {B}$
. Then
$[T_g,A^{p,\lambda } ]$
equipped with the norm
$\|\cdot \|_{[T_g,A^{p,\lambda }] }$
is a Banach space.
Proof Let
$\{f_n\}$
be a Cauchy sequence in
$[T_g,A^{p,\lambda }]$
, that is,
$$ \begin{align*} \|f_n-f_m\|_{[T_g,A^{p,\lambda}]}=\|T_g(f_n-f_m)\|_{A^{p,\lambda}}\rightarrow 0 \quad \text{as} \quad m,n\to \infty.\ \end{align*} $$
So
$\left \{ T_{g}(f_{n}) \right \}$
is a Cauchy sequence in
$A^{p,\lambda }(\mathbb {D})$
. Since [Reference Yang and Liu13] already proved
$A^{p,\lambda }(\mathbb {D})$
is a Banach space under
$0<\lambda <2, p \ge 1$
, then there exists
$k \in A^{p,\lambda }$
such that
$\|T_{g}(f_{n})-k\|_{A^{p,\lambda }} \rightarrow 0$
as
$n\rightarrow \infty .$
For any
$f \in A^{p,\lambda },$
by Lemma 2.2,
$$ \begin{align*}\left | f'(z) \right | \lesssim \frac{\|f\|_{A^{p,\lambda}}}{(1-|z|^2)^{1+\frac{2-\lambda}{p}}},\quad \quad \forall z\in \mathbb{D}. \end{align*} $$
Then for any
$z \in \mathbb {D},$
$$ \begin{align*}\left | (T_{g}f_{n}-k)^{'} (z) \right | \lesssim \frac{\|(T_{g}f_{n}-k)\|_{A^{p,\lambda}}}{(1-|z|^2)^{1+\frac{2-\lambda}{p}}} \rightarrow 0, \end{align*} $$
which means that
$T_g'(f_n)(z)\rightarrow k'(z)$
as
$n\rightarrow \infty .$
By Lemma 4.1, we have
$$ \begin{align*}\left | \delta _{z_{0} }(f_{n}-f_{m} ) \right | \lesssim \left \| f_{n}-f_{m} \right \| _{[T_g,A^{p,\lambda}]},\quad \forall z_{0} \in K\subseteq \mathbb{D}. \end{align*} $$
Thus,
$\left | f_{n}(z_{0})-f_{m}(z_{0}) \right | \rightarrow 0$
as
$n,m \rightarrow \infty .$
Since we also know
$f_{n}(z_{0}) \rightarrow f(z_{0}) $
as
$n \rightarrow \infty .$
Taking the primitives, then
$k(z)=\int _{0}^{z} f(w)g^{'}(w)dw+C.$
And because
$\left \| T_{g}(f_{n})-k \right \| _{A^{p,\lambda }}\rightarrow 0$
as
$n \rightarrow \infty , $
we have
$\left | T_{g}(f_{n})(0)-k(0) \right |\rightarrow 0 $
as
$n \rightarrow \infty $
. Finally, we get
$0=T_{g}(f_{n})(0)\rightarrow k(0)=C,$
namely,
$C=0.$
So
$k=T_{g}(f),$
which completes the proof.
Theorem 4.3 For
$0<\lambda <2$
and
$p\ge 1$
,
$$\begin{align*}A^{p,\lambda}\ \subset\ \bigcap_{g\in\mathcal B}[T_g,A^{p,\lambda}]. \end{align*}$$
Proof If
$g\in \mathcal B$
, then by [Reference Yang and Liu13, Theorem 4.4], we know
$T_g:A^{p,\lambda }\to A^{p,\lambda }$
is bounded. Hence, for any
$f\in A^{p,\lambda }$
, we have
$T_g(f)\in A^{p,\lambda }$
, that is,
$f\in [T_g,A^{p,\lambda }]$
. Intersect over all
$g\in \mathcal B$
.
Remark 4.4 Let
$0<\lambda <2$
,
$p\ge 1$
, and
$g\in \mathcal B.$
Then
$$\begin{align*}A^{p,\lambda}\ \subsetneq\ [T_g,A^{p,\lambda}]. \end{align*}$$
Theorem 4.5 Let
$0<\lambda <2, p \ge 1,$
and
$g\in \mathcal {B}$
. The multiplication operator
$$ \begin{align*}M_h:\,[T_g,A^{p,\lambda}]\to [T_g,A^{p,\lambda}] \end{align*} $$
is bounded if and only if
$h \in H^{\infty }$
. Moreover, the norm of operator satisfies
$\left \| M_{h} \right \| \approx \left \| h \right \| _{\infty }.$
Proof Let
$f \in [T_g,A^{p,\lambda }]$
and
$h \in H^{\infty }.$
It is straightforward to show that
$$ \begin{align*} T_{g}(hf)=\int_{0}^{z} h(w) f(w)g^{'}(w)dw=S_{h}(T_{g}(f)). \end{align*} $$
By [Reference Yang and Liu13, Theorem 4.3],
$S_{h}$
is bounded on
$A^{p,\lambda }$
if and only if
$h \in H^{\infty }$
. Therefore, we obtain
$$ \begin{align*} \|M_h(f)\|_{[T_g,A^{p,\lambda}]}&=\|T_g(hf)\|_{ A^{p,\lambda}}=\|S_h(T_g(f))\|_{ A^{p,\lambda}}\\ &\leq C\,\|T_g(f)\|_{ A^{p,\lambda}}\,=\, C\,\|f\|_{[T_g,A^{p,\lambda}]}\,<\,\infty, \end{align*} $$
where
$C\,=\,\| S_h\|_{A^{p,\lambda } \to A^{p,\lambda } }\,\approx \|h\|_{\infty }$
.
Conversely, for any
$h\in \mathrm{Hol}(\mathbb {D})$
, assume that
$ M_h\colon [T_g,A^{p,\lambda }]\rightarrow [T_g,A^{p,\lambda }]$
is bounded. Let
$z\in \mathbb {D}$
and denote by
$\delta _{z}$
the point evaluation functional in
$[T_g,A^{p,\lambda }]$
. By Lemma 4.1,
$\delta _{z}$
is a bounded linear functional. Since
$[T_g,A^{p,\lambda }]$
contains the constant functions, then
$\|\delta _z\|\neq 0$
and
$$ \begin{align*}\|\delta_z\|:=\|\delta_z\|_{[T_g,A^{p,\lambda}]}\neq 0,\qquad \forall z\in\mathbb{D}.\end{align*} $$
Let
$f\in [T_g,A^{p,\lambda }]$
with
$\|f\|_{[T_g,A^{p,\lambda }]}=1$
, then we get
$$ \begin{align*} |h(z)f(z)|&=|\delta_z(M_h(f))|\leq \|\delta_z\|\|M_h\|. \end{align*} $$
By taking the supremum on
$f\in [T_g,A^{p,\lambda }]$
with
$\|f\|_{[T_g,A^{p,\lambda }]}=1$
, we get
$$ \begin{align*}|h(z)|\|\delta_z\|\leq \sup_{\|f\|_{[T_g,A^{p,\lambda}]}=1}|h(z)f(z)| \leq \|M_h\|\|\delta_z\|. \end{align*} $$
Thus,
$h\in H^{\infty }$
follows from
$\|\delta _z\|\neq 0$
. Moreover,
$$ \begin{align*} \left \| M_{h} \right \| &=\sup\limits_{\|f\|_{[T_g,A^{p,\lambda}]}=1} \frac{\left \| M_{h}(f) \right \|_{[T_g,A^{p,\lambda}]} }{\left \| f \right \|_{[T_g,A^{p,\lambda}]} } = \sup\limits_{f\in [T_g,A^{p,\lambda}]}\left \| M_{h}(f) \right \|_{[T_g,A^{p,\lambda}]}\\ &\le \left \| S_{h} \right \| \left \| f \right \| _{[T_g,A^{p,\lambda}]}=\left \| S_{h} \right\| \approx \left \| h \right \| _{\infty }. \end{align*} $$
This completes the proof.
Since multipliers are exactly
$H^{\infty },$
domain inclusions reflect only bounded analytic coefficients, a theme that reappears in the comparison principles below.
To complete the proof of our next theorem, we need to give two lemmas.
Lemma 4.6 [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis4, Proposition 5]
Let
$0\leq \gamma <\delta $
. Then
$M_g:\;K_\gamma \to K_\delta $
is bounded if and only if
$\,g \in K_{\delta -\gamma }$
.
Lemma 4.7 [Reference Bellavita, Daskalogiannis, Nikolaidis and Stylogiannis4, Lemma 2]
Let
$g \in K_{\gamma }$
with
$0<\gamma <1$
and let
$0<\varepsilon <\min \{\gamma , 1-\gamma \}.$
Then, there exists a function
$\phi \in \mathrm{Hol}(\mathbb {D})$
such that
$\phi \cdot g \in K_{\gamma +\varepsilon } \backslash K_{\gamma }.$
Theorem 4.8 Let
$g \in \mathcal {B}$
, for
$0< \lambda _{1}<\lambda _{2}<2 $
and
$p \ge 1$
, then
$$ \begin{align*}[T_g,A^{p,\lambda_{2}}]\subsetneq [T_g,A^{p,\lambda_{1}}].\end{align*} $$
Proof It is easy to prove
$A^{p,\lambda _{2}}\subsetneq A^{p,\lambda _{1}},$
and
$$ \begin{align*}\left \| f \right \| _{[T_g,A^{p,\lambda_{1}}]}=\left \| T_{g}(f) \right \|_{A^{p,\lambda_{1}}} \le \left \| T_{g}(f) \right \|_{A^{p,\lambda_{2}}}=\left \| f \right \| _{[T_g,A^{p,\lambda_{2}}]}\end{align*} $$
yields that
$[T_g,A^{p,\lambda _{2}}]\subset [T_g,A^{p,\lambda _{1}}].$
Now, we will prove that this inclusion is strict for all
$g \in \mathcal {B}.$
To obtain a contradiction, suppose that
$[T_g,A^{p,\lambda _{2}}]=[T_g,A^{p,\lambda _{1}}]$
. Let
$Z(g')$
denote the discrete sequence of the zeros of
$g'$
, repeated according to their multiplicity. Set X be a subset of
$\mathcal {B}$
, defined by
$$ \begin{align*}X=\{G\in \mathcal{B}:\; Z(g')\subseteq Z(G')\}. \end{align*} $$
For any function
$G\in X \subseteq \mathcal {B}$
, then
$\frac {G'}{g'} \in \mathrm{Hol}(\mathbb D)$
. Let
$h\in A^{p,\lambda _{1}}$
, then
$\varphi =\frac {G'}{g'}\,h \in H{(\mathcal {D})}$
and
$$ \begin{align*}T_g(\varphi )(z)=\int_0^z g'(w)\,\frac{G'(w)}{g'(w)}\,h(w)\,dw\,=\,T_G(h)(z). \end{align*} $$
Since
$G\in \mathcal {B}$
, it follows that
$\,T_G(h)(z)\in A^{p,\lambda _{1}}$
, and hence
$T_g(\varphi )(z)\in A^{p,\lambda _{1}}$
. This implies that
$\varphi \in [T_g,A^{p,\lambda _{1}}]$
. By assumption, we obtain that
$\varphi \in [T_g,A^{p,\lambda _{2}}]$
. This means that for every
$h\in A^{p,\lambda _{1}}$
,
$T_G(h) \in A^{p,\lambda _{2}}$
. Therefore,
$T_G:\;A^{p,\lambda _{1}}\to A^{p,\lambda _{2}}$
is bounded. It follows from Theorem 3.1 that G must be in
$\Lambda _{\frac {\lambda _{2}-\lambda _{1}}{p}}$
.
Note that
$G \in \Lambda _{\frac {\lambda _{2}-\lambda _{1}}{p}}$
is equivalent to
$G' \in K_{1+\frac {\lambda _{1}-\lambda _{2}}{p}}$
. By Lemma 4.7, we can find a function
$\phi $
such that
$\phi \cdot G' \in K_{1+\frac {\lambda _{1}-\lambda _{2}}{p}+ \varepsilon } \setminus K_{1+\frac {\lambda _{1}-\lambda _{2}}{p}}$
. Considering the function
$$\begin{align*}H_1(z)=\int_0^z \phi(w)\cdot G'(w)\,dw, \end{align*}$$
we see that
$ Z(G')\subseteq Z(H_{1}')$
and for
$G' \in K_{1-\alpha },$
where
$\alpha =\frac {\lambda _{2}-\lambda _{1}}{p}$
. Accordingly,
$$ \begin{align*} \frac{1}{\left | I \right |^{2} } \int_{S(I)}&\left | H_{1}'(z) \right | ^{2}(1-\left | z \right |^{2} )dA(z)\\ &=\frac{1}{\left | I \right |^{2} } \int_{S(I)}\left | \phi(w)G'(z) \right | ^{2}(1-\left | z \right |^{2} )dA(z)\\ &\lesssim \frac{\left \| \phi G' \right \|^{2} _{K_{1+\frac{\lambda_{1}-\lambda_{2}}{p}+ \varepsilon}}}{\left | I \right |^{2} } \int_{S(I)}\frac{1}{(1-\left | z \right |^{2} )^{\frac{2}{p}(\lambda_{1}-\lambda_{2})+2\varepsilon}} dA(z)\\ &\lesssim \left \| \phi G' \right \|^{2} _{K_{1+\frac{\lambda_{1}-\lambda_{2}}{p}+ \varepsilon}}. \end{align*} $$
This shows that
$H_{1}\in \mathcal {B}$
, and so
$H_1 \in X$
. But
$H^{\prime }_1 \notin K_{1-\alpha }$
or, equivalently,
$H_1 \notin \Lambda _\alpha $
, which leads to a contradiction.
5 The structure of the meromorphic optimal domain
Zeros of
$g'$
prompt a minimal enlargement of the holomorphic domain. This motivates the meromorphic optimal domain
$(T_g,A^{p,\lambda }),$
where poles are constrained to the zero set of
$g'$
and the norm is transported through
$T_g.$
We begin by specifying
$(T_g,A^{p,\lambda })$
in the constant-symbol case.
Lemma 5.1 Let
$g \in \mathcal {B}$
and
$1\leq p<\infty , 0<\lambda <2.$
Then
$(T_g,A^{p,\lambda })=\mathrm{Mer}(\mathbb {D})$
if and only if g is constant.
Proof Let f be meromorphic. If g is constant, then
$g'\equiv 0$
; hence,
$f\,g'\equiv 0$
is analytic and
$T_g(f)\equiv 0$
. Therefore, the optimal meromorphic domain equals
$\mathrm{Mer}(\mathbb {D})$
.
For the converse, assume
$(T_g,A^{p,\lambda })=\mathrm{Mer}(\mathbb {D})$
. Fix
$z_0\in \mathbb {D}$
and consider
$$\begin{align*}f(z)=\frac{1}{z-z_0},\qquad z\in\mathbb{D}, \end{align*}$$
which is meromorphic with a simple pole at
$z_0$
. By the hypothesis,
$f\in (T_g,A^{p,\lambda })$
; consequently,
$$\begin{align*}\frac{g'(z)}{z-z_0},\qquad z\in\mathbb{D}, \end{align*}$$
is analytic on
$\mathbb {D}$
. Hence,
$g'$
vanishes at
$z_0$
(with multiplicity at least one). Since
$z_0$
was arbitrary, we obtain
$g'\equiv 0$
on
$\mathbb {D}$
, and thus g is constant.
For
$g\in \mathcal {B,}$
we have
$A^{p,\lambda }\subset (T_g,A^{p,\lambda })$
. Lemma 5.1 shows that
$(T_g,A^{p,\lambda })$
coincides with
$\mathrm{Mer}(\mathbb {D})$
only in the constant-symbol case. When g is not constant, we equip
$(T_g,A^{p,\lambda })$
with the norm
$$ \begin{align} \|f\|_{(T_g,A^{p,\lambda})}:= \|T_g(f)\|_{A^{p,\lambda}}. \end{align} $$
Having isolated the trivial constant-symbol case, we show that non-constant symbols make the meromorphic norm genuine.
Since
$T_g$
is linear,
$(T_g,A^{p,\lambda })$
is a vector space of meromorphic functions on the unit disc. The functional
$\|\cdot \|_{(T_g,A^{p,\lambda })}$
defined in (5.1) is a seminorm: it is subadditive and absolutely homogeneous.
Lemma 5.2 Let
$g\in \mathcal {B}$
be non-constant. Then
$\|f\|_{(T_g,A^{p,\lambda })}=0$
if and only if
$f=0$
. Hence,
$\|\cdot \|_{(T_g,A^{p,\lambda })}$
is a norm in
$(T_g,A^{p,\lambda })$
.
Proof Clearly,
$T_g(0)=0$
. Now, take
$f\in \mathrm{Mer}(\mathbb {D})$
with
$fg'\in \mathrm{Hol}(\mathbb {D})$
and
$\|T_g(f)\|_{A^{p,\lambda }}=0$
. Then
$T_g(f)(z)=0$
for all
$z\in \mathbb {D}$
. Differentiating gives
$$ \begin{align*}g'(z)f(z)=0\,\, \text{ for all } z\in\mathbb{D}. \end{align*} $$
Because
$g'\not \equiv 0$
, it follows that
$f(z)=0$
for every
$z \in \mathbb {D}\setminus \mathcal {Z}(g')$
. Since
$\mathcal {Z}(g')$
is at most countable, we have
$\displaystyle {\lim _{z\to w}f(z)=0}$
for each
$w\in \mathcal {Z}(g')$
. Hence, f has no poles and vanishes almost everywhere on
$\mathbb {D}$
; therefore,
$f=0$
.
Lemma 5.3 Let
$g\in \mathcal {B}$
be non-constant. Let
$z_0\in \mathbb {D}$
and
$\delta _{z_0}$
denote the point evaluation functional at
$z_0.$
The functional
$\delta _{z_0}$
is bounded in
$(T_g,A^{p,\lambda })$
if and only if
$z_0 \in \mathbb D\setminus \mathcal {Z}(g')$
.
Proof Assume first that
$z_0\in \mathbb D\setminus \mathcal {Z}(g')$
and let
$f\in (T_g,A^{p,\lambda })$
. By the standard estimate for point evaluations of derivatives of
$A^{p,\lambda }$
functions, we obtain
$$ \begin{align*} |f(z_0)\cdot g'(z_0)|&=\,|(T_gf)'(z_0)|\\ &\lesssim \, \dfrac{\|T_g(f)\|_{A^{p,\lambda}}}{(1-|z_0|^2)^{1+\frac{2-\lambda}{p}}} =\dfrac{\|f\|_{(T_{g},A^{p,\lambda})}}{(1-|z_0|^2)^{1+\frac{2-\lambda}{p}}}.\\ \end{align*} $$
Consequently,
$$\begin{align*}|f(z_0)|\lesssim \frac{\|f\|_{(T_g,A^{p,\lambda})}}{|g'(z_0)|(1-|z_0|^2)^{1+\frac{2-\lambda}{p}}}. \end{align*}$$
Thus,
$\delta _{z_0}$
is bounded. Conversely, let
$z_0\in \mathcal {Z}(g')$
. Since
$\frac {1}{g'}\in (T_g,A^{p,\lambda })$
and
$\lim _{z\to z_0}\frac {1}{g'(z)}=\infty $
, it follows that
$\delta _{z_0}$
is not bounded.
Theorem 5.4 Assume
$g\in \mathcal {B}$
is non-constant and
$1\leq p<\infty $
. The pair
$ ( (T_g,A^{p,\lambda }), \|\cdot \|_{(T_g,A^{p,\lambda })})$
constitutes a Banach space of meromorphic functions, and the point evaluation functional is bounded at every
$z_0 \in \mathbb D\setminus \mathcal {Z}(g')$
.
Proof Let
$\{f_n\}$
be Cauchy in
$(T_g,A^{p,\lambda })$
, namely,
$$ \begin{align*} \|f_n-f_m\|_{(T_g,A^{p,\lambda})}=\|T_g(f_n-f_m)\|_{A^{p,\lambda}}\rightarrow 0 \quad \text{as} \quad m,n\to \infty. \end{align*} $$
Hence,
$\left \{ T_{g}(f_{n}) \right \}$
is a Cauchy sequence in
$A^{p,\lambda }(\mathbb {D})$
. Since
$A^{p,\lambda }(\mathbb {D})$
is complete for
$0<\lambda <2, p \ge 1$
, there exists
$k \in A^{p,\lambda }$
with
$\|T_{g}(f_{n})-k\|_{A^{p,\lambda }} \rightarrow 0$
as
$n\rightarrow \infty .$
For any
$f \in A^{p,\lambda }$
, Lemma 2.2 yields
$$ \begin{align*}\left | f'(z) \right | \lesssim \frac{\|f\|_{A^{p,\lambda}}}{(1-|z|^2)^{1+\frac{2-\lambda}{p}}},\quad \quad \forall z\in \mathbb{D}. \end{align*} $$
Therefore, for all
$z \in \mathbb {D}$
,
$$ \begin{align*}\left | (T_{g}f_{n}-k)^{'} (z) \right | \lesssim \frac{\|(T_{g}f_{n}-k)\|_{A^{p,\lambda}}}{(1-|z|^2)^{1+\frac{2-\lambda}{p}}} \rightarrow 0, \end{align*} $$
so
$T_g'(f_n)(z)\rightarrow k'(z)$
as
$n\rightarrow \infty .$
Invoking Lemma 5.3, we have
$$ \begin{align*}\left | \delta _{z_{0} }(f_{n}-f_{m} ) \right | \lesssim \left \| f_{n}-f_{m} \right \| _{(T_g,A^{p,\lambda})},\quad \forall z_{0} \in K\subseteq \mathbb{D}. \end{align*} $$
Thus,
$\left | f_{n}(z_{0})-f_{m}(z_{0}) \right | \rightarrow 0$
when
$n,m \rightarrow \infty .$
Consequently,
$f_{n}(z_{0}) \rightarrow f(z_{0}) $
as
$n \rightarrow \infty .$
Taking primitives gives
$k(z)=\int _{0}^{z} f(w)g^{'}(w)dw+C.$
Since
$\left \| T_{g}(f_{n})-k \right \| _{A^{p,\lambda }}\rightarrow 0$
as
$n \rightarrow \infty , $
it follows that
$\left | T_{g}(f_{n})(0)-k(0) \right |\rightarrow 0 $
as
$n \rightarrow \infty $
. Hence,
$0=T_{g}(f_{n})(0)\rightarrow k(0)=C,$
which forces
$C=0.$
Therefore,
$k=T_{g}(f),$
completing the proof.
Theorem 5.5 For
$1\leq p_1 \leq p_2<\infty $
and
$0<\lambda <2,$
we have
$$ \begin{align*}(T_g,A^{p_2,\lambda})\subsetneq(T_g,A^{p_1,\lambda}).\end{align*} $$
Proof The containment
$(T_g,A^{p_2,\lambda })\subseteq (T_g,A^{p_1,\lambda })$
is immediate since
$A^{p_2,\lambda }\subset A^{p_1,\lambda }$
. To show that the inclusion is proper, consider
$$ \begin{align*}\phi(z)=\frac{1}{(1-z)^{\frac{2-\lambda}{p_2}+1+\epsilon}}\frac{1}{g'(z)} \qquad \text{for}\quad 0<\epsilon<(2-\lambda)\left(\frac{1}{p_1}-\frac{1}{p_2}\right). \end{align*} $$
Because every zero of
$g'$
in
$\mathbb {D}$
has finite multiplicity, we have
$\phi \in \mathrm{Mer}(\mathbb {D})$
. In addition,
$\phi \,g'\in \mathrm{Hol}(\mathbb {D})$
and
$$\begin{align*}T_g(\phi)(z)= \frac{p_2}{2-\lambda+\epsilon p_2}[(1-z)^{\frac{\lambda-2}{p_2}-\epsilon }-1] \end{align*}$$
A routine verification yields
$T_g(\phi )\in A^{p_1,\lambda }\setminus A^{p_2,\lambda }$
, which establishes the claim.
Theorem 5.6 Let
$\mathcal {M}(T_g,A^{p,\lambda })$
denote the multiplier space of
$(T_g,A^{p,\lambda })$
. Then
$\mathcal {M}(T_g,A^{p,\lambda })=H^{\infty }$
.
Proof Recall that for
$g,F\in \mathrm{Hol}(\mathbb {D}),$
we set
$$ \begin{align*}S_g(F)(z)=\int_{0}^{z}g(\zeta)F'(\zeta)\,d\zeta, \end{align*} $$
which we call the
$T_g$
-companion operator.
Fix
$f\in (T_g,A^{p,\lambda })$
and choose
$k\in H^\infty $
. We claim that
$kf\in (T_g,A^{p,\lambda })$
. By (1.1), pick
$h\in A^{p,\lambda }$
with
$f=\frac {h'}{g'}.$
Then
$$ \begin{align*}k(z)f(z)g'(z)=k(z)h'(z)\qquad \forall z\in\mathbb{D}. \end{align*} $$
Since
$fg'\in \mathrm{Hol}(\mathbb {D})$
, the product
$kf$
is meromorphic and
$kfg'\in \mathrm{Hol}(\mathbb {D})$
. Moreover,
$$ \begin{align*}T_g(kf)(z)=S_{k}(h)(z)\qquad z\in \mathbb{D}. \end{align*} $$
Appealing to [Reference Yang and Liu13, Theorem 4.3], we obtain
$S_k(h)\in A^{p,\lambda }$
, hence
$T_g(kf)\in A^{p,\lambda }$
.
Conversely, let
$\phi \in \mathrm{Mer}(\mathbb {D})$
be such that
$\phi f\in (T_g,A^{p,\lambda })$
for all
$f\in (T_g,A^{p,\lambda })$
. Because
$\frac {1}{g'}\in (T_g,A^{p,\lambda })$
, the very definition of
$(T_g,A^{p,\lambda })$
forces
$\phi \in \mathrm{Hol}(\mathbb {D})$
. By Lemma 5.3, if
$z_0\in \mathbb {D}\setminus \mathcal {Z}(g'),$
then
$\delta _{z_0}$
is bounded; and since
$(T_g,A^{p,\lambda })$
contains constants, we have
$\|\delta _{z_0}\|>0$
. For any
$f \in (T_g,A^{p,\lambda })$
with
$\|f\|_{(T_g,A^{p,\lambda })}=1$
,
$$ \begin{align*}|\phi(z_0)f(z_0)| = |\delta_{z_0}(M_\phi(f))| \le \|\delta_{z_0}\|\|M_\phi\|. \end{align*} $$
Taking the supremum over all such f yields
$$ \begin{align*}|\phi(z_0)|\le \|M_\phi\|. \end{align*} $$
Since
$z_0$
was arbitrary, this estimate holds on
$\mathbb {D}\setminus \mathcal {Z}(g')$
. As
$\phi \in \mathrm{Hol}(\mathbb {D})$
and
$\mathcal {Z}(g')$
is countable, continuity extends the bound to the points of
$\mathcal {Z}(g')$
; thus, for every
$z\in \mathbb {D}$
,
$$ \begin{align*}|\phi(z)|\le \|M_\phi\|. \end{align*} $$
Therefore,
$\phi \in H^\infty $
, completing the proof.
6 The spaces
$W^{p}_g$
and
$V^{p}_g$
Once the domains are set, the remaining questions are symbolic: when do different symbols induce the same (mero)optimal domain, and how do these domains embed across symbols? The classes
$W^{p}_g$
and
$V^{p}_g$
package precisely these inclusion relations. Our goal is to determine for which
$g_1,g_2\in \mathcal {B}$
the associated meromorphic optimal domains coincide, namely,
$(T_{g_1},A^{p,\lambda })=(T_{g_2},A^{p,\lambda })$
. By the very definition of
$W_{g}^p$
, the identity
$(T_{g_1},A^{p,\lambda })=(T_{g_2},A^{p,\lambda })$
holds exactly when
$g_1\in W^p_{g_2}$
and
$g_2\in W^p_{g_1}$
. Thus, an explicit characterization of
$W_g^p$
is central.
Theorem 6.1 Let
$g\in \mathcal {B}$
and
$1\leq p<\infty $
. Then
$h\in W_g^p$
if and only if there exists
$f\in T_g(H^{\infty })$
such that
$h=f+h(0)$
. Equivalently,
$$ \begin{align*}W_g^p=T_g(H^{\infty})+\mathbb{C}.\end{align*} $$
Proof By Lemma 5.1, constant functions are contained in
$W^p_g$
. Suppose
$f-f(0)=T_g(k)$
for some
$k\in H^{\infty }$
. Then, for every
$z \in \mathbb {D}$
,
$$ \begin{align*}f'(z)=k(z)g'(z). \end{align*} $$
We claim
$f\in W_g^p$
, that is,
$(T_g,A^{p,\lambda })\subseteq (T_f,A^{p,\lambda })$
. Let
$G\in (T_g,A^{p,\lambda })$
. By Theorem 5.6,
$kG\in (T_g,A^{p,\lambda })$
, and hence
$$ \begin{align*}T_f(G)=T_g(kG)\in A^{p,\lambda}, \end{align*} $$
which proves the desired inclusion.
Conversely, assume
$f \in W^p_g$
. Since
$\frac {1}{g'}\in (T_g,A^{p,\lambda })$
and
$(T_g,A^{p,\lambda })\subseteq (T_f,A^{p,\lambda })$
, it follows that
$\frac {1}{g'}\in (T_f,A^{p,\lambda })$
. By the definition of the meromorphic optimal domain,
$$ \begin{align*}\frac{f'}{g'} \in \mathrm{Hol}(\mathbb{D}). \end{align*} $$
Now, take any
$G\in (T_g,A^{p,\lambda })\subseteq (T_f,A^{p,\lambda })$
. Then
$T_f(G)\in A^{p,\lambda }$
and
$$ \begin{align*}T_g\!\left(\frac{f'}{g'}\,G\right)=T_f(G)\in A^{p,\lambda}. \end{align*} $$
Hence,
$\frac {f'}{g'}$
acts as a multiplier of
$(T_g,A^{p,\lambda })$
and, by Theorem 5.6, we conclude
$\frac {f'}{g'}\in H^{\infty }$
. Therefore,
$f-f(0)\in T_g(H^{\infty })$
, completing the proof.
Lemma 5.1 shows that every constant belongs to
$W_g^p$
for all
$g \in \mathcal {B}$
. Theorem 6.1 further yields
$T_g(H^{\infty })\subset \mathcal {B}$
whenever
$g\in \mathcal {B}$
. In addition, by Theorem 6.1, the spaces
$W^p_g$
do not depend on p for
$1\leq p<\infty $
; we will therefore drop the superscript and write simply
$W_g$
. The structural formula for
$W_g$
immediately yields a symbolic criterion for equality of meromorphic optimal domains.
Theorem 6.2 Assume
$1\leq p<\infty $
and
$g_1,g_2\in \mathcal {B}$
are non-constant. The statements below are equivalent:
-
a)
$(T_{g_1},A^{p,\lambda })=(T_{g_2},A^{p,\lambda })$
. -
b) There are
$k_1,k_2\in H^{\infty }$
with
$$ \begin{align*}g_1=T_{g_2}(k_1)+g_1(0) \text{ and } g_2=T_{g_1}(k_2)+g_2(0). \end{align*} $$
In addition, one has
$k_2=\frac {1}{k_1}$
.
Proof Observe first that
$(T_{g_1},A^{p,\lambda })=(T_{g_2},A^{p,\lambda })$
holds exactly when
$W_{g_1}=W_{g_2}$
. Assume
$(T_{g_1},A^{p,\lambda })=(T_{g_2},A^{p,\lambda })$
. For any
$f \in W_{g_1}$
, we have
$$ \begin{align*}(T_{g_1},A^{p,\lambda})\subseteq (T_f,A^{p,\lambda}).\end{align*} $$
Since
$(T_{g_2},A^{p,\lambda })=(T_{g_1},A^{p,\lambda })$
, it follows that
$$ \begin{align*}(T_{g_2},A^{p,\lambda})\subseteq (T_f,A^{p,\lambda}),\end{align*} $$
which implies
$f \in W_{g_2}$
. Hence,
$ W_{g_1}\subseteq W_{g_2}$
; the reverse inclusion is obtained by symmetry.
Conversely, suppose
$ W_{g_1}= W_{g_2}$
. Then
$g_1 \in W_{g_1}=W_{g_2}$
, and thus
$$ \begin{align*}(T_{g_2},A^{p,\lambda})\subseteq (T_{g_1},A^{p,\lambda}).\end{align*} $$
Likewise,
$g_2 \in W_{g_1}$
, giving
$(T_{g_1},A^{p,\lambda })\subseteq (T_{g_2},A^{p,\lambda })$
. Therefore,
$(T_{g_1},A^{p,\lambda })= (T_{g_2},A^{p,\lambda })$
.
Assertion (b) follows from Theorem 6.1, using that for each
$g\in \mathcal {B,}$
one has
$W_{g}=T_g(H^{\infty })+\mathbb {C}$
. The functions
$k_1,k_2\in H^{\infty }$
satisfy the relations
$$ \begin{align*}\begin{cases} g_1'=k_1 g_2'\\ g_2'=k_2g_1'. \end{cases}\end{align*} $$
Hence,
$k_2=\frac {1}{{k_1}}.$
We next examine structural features of
$W_g$
. To that end, endow
$W_g$
with the norm: for
$h \in W_g$
and
$h-h(0)=T_g(k)$
with
$k \in H^\infty $
,
$$ \begin{align*}\|h\|_{W_g}=\|T_g(k)\|_{\mathcal{B}}+|h(0)|+\|k\|_{H^\infty}.\end{align*} $$
Theorem 6.3 Let
$g\in \mathcal {B}$
be non-constant and
$1<p<\infty $
. The pair
$\left ( W_g, \|\cdot \|_{W_g}\right )$
is a linear subspace of
$\mathcal {B}$
and, moreover, a Banach space.
Proof Since
$W_g=T_g(H^\infty )$
and
$T_g:H^\infty \to \mathcal {B}$
is bounded, the graph of
$T_g$
– which coincides with
$W_g$
– is a closed subspace of
$H^\infty \times \mathcal {B}$
. Consequently, endowed with the graph norm
$\|\cdot \|_{W_g}$
, the space
$W_g$
is complete and thus Banach.
In addition, for each fixed
$z\in \mathbb {D}$
, the point-evaluation functionals are bounded on both
$H^\infty $
and
$\mathcal {B}$
. Hence, there exists a constant
$\delta _z>0$
, depending only on z, such that
$$ \begin{align*}|f(z)|\leq \delta_z \|f\|_{W_g}. \end{align*} $$
Therefore, point evaluations define bounded linear functionals on
$W_g$
.
Theorem 6.4 Let
$g\in \mathcal {B}$
be non-constant and
$1< p<\infty $
. We have
$$ \begin{align*}\bigcap_{h \in W_g\colon (T_g,A^{p,\lambda})\subsetneq(T_h,A^{p,\lambda}) } (T_h,A^{p,\lambda})=(T_g,A^{p,\lambda}).\end{align*} $$
Proof The inclusion
$\supseteq $
is immediate. For the converse, take
$$ \begin{align} f \in \bigcap_{h \in W_g\colon (T_g,A^{p,\lambda})\subsetneq(T_h,A^{p,\lambda}) } (T_h,A^{p,\lambda}). \end{align} $$
We first verify that
$fg'\in \mathrm{Hol}(\mathbb {D})$
. Assume to the contrary that there is
$z_0\in \mathbb {D}$
which is a pole of
$fg'$
of order
$\lambda _1>0$
. Since (6.1) holds, choose
$h_1\in W_g$
with
$(T_g,A^{p,\lambda })\subsetneq (T_{h_1},A^{p,\lambda })$
. By Theorem 6.1, there exists
$k_{h_1}\in H^{\infty }$
such that
$h^{\prime }_1=g' k_{h_1}$
. Because
$fh^{\prime }_1\in \mathrm{Hol}(\mathbb {D})$
, the function
$k_{h_1}$
must vanish at
$z_0$
with order
$\lambda _2\geq \lambda _1$
. Define
$$ \begin{align*}k_{h_2}(z)=\frac{k_{h_1}(z)}{(z-z_0)^{\lambda_2}}(1-z)^{1+\frac{2-\lambda}{p}},\qquad z\in\mathbb{D}, \end{align*} $$
and let
$h_2= T_g(k_{h_2})$
. Then
$k_{h_2}\in H^\infty $
,
$k_{h_2}(z_0)\neq 0$
, and, by Theorem 6.1,
$h_2\in W_g$
. Moreover,
$(T_g,A^{p,\lambda })\subsetneq (T_{h_2},A^{p,\lambda })$
since
$$ \begin{align*}F(z)=\frac{1}{g'(z)}\frac{1}{(1-z)^{2+\frac{2-\lambda}{p}}} \in (T_{h_2},A^{p,\lambda})\setminus (T_g,A^{p,\lambda}). \end{align*} $$
Indeed,
$$ \begin{align*}T_g(F)(z)=\int_0^z \frac{1}{(1-\xi)^{2+\frac{2-\lambda}{p}}}d\xi\notin A^{p,\lambda} \end{align*} $$
and
$$ \begin{align*}T_{h_2}(F)(z)=\int_{0}^z\frac{k_{h_2}(\xi)}{(1-\xi)^{2+\frac{2-\lambda}{p}}}d\xi=\int_{0}^z \frac{k_{h_1}(\xi)}{(\xi-z_0)^{\lambda_2}}\frac{1}{1-\xi}d\xi \in A^{p,\lambda}, \end{align*} $$
since
$\frac {k_{h_1}(z)}{(z-z_0)^\lambda }\in H^\infty $
and
$-\log (1-z) \in A^{p,\lambda }$
.
As f satisfies (6.1), the function
$fh_2'\in \mathrm{Hol}(\mathbb {D})$
, which is impossible, since
$ fh^{\prime }_2=fg'k_{h_2} $
has a pole at
$z_0$
.
Now, we prove that
$T_g(f)\in A^{p,\lambda }$
. We chose
$k_{1}(z)=(1+z)$
and
$k_{2}(z)=(1-z)$
. Because of Theorem 6.1,
$T_g(k_i) \in W_g$
and, as shown in the first part of the proposition,
$(T_g,A^{p,\lambda })\subsetneq (T_{T_g(k_i)},A^{p,\lambda })$
.
Let
$\Sigma _+=\{z\in \mathbb {D}:\Re z\ge 0\}$
and
$\Sigma _-=\{z\in \mathbb {D}:\Re z\le 0\}$
. Then for all
$z\in \Sigma _+$
, we have
$|k_1(z)|=|1+z|\ge 1,$
and for
$z\in \Sigma _-,$
we have
$\ |k_2(z)|=|1-z|\ge 1,$
while always
$|k_i(z)|\le 2$
in
$\mathbb {D}$
. For each arc
$I,$
write
$E_\pm :=S(I)\cap \Sigma _\pm $
. Then
$$\begin{align*}\frac{1}{|I|^\lambda}\!\!\int_{S(I)}\!\!|f|^p|g'|^p(1-|z|^2)^p\,dA(z) =\sum_{\sigma\in\{+,-\}}\frac{1}{|I|^\lambda}\!\!\int_{E_\sigma}\!\!|f|^p|g'|^p(1-|z|^2)^p\,dA(z). \end{align*}$$
On
$E_+$
, we use
$|k_1|\ge 1$
to get
$|f|\le |k_1 f|$
; similarly on
$E_-$
,
$|f|\le |k_2 f|$
. Hence,
$$\begin{align*}\frac{1}{|I|^\lambda}\!\!\int_{S(I)}\!\!|f|^p|g'|^p(1-|z|^2)^p\,dA(z) &\le\ \frac{1}{|I|^\lambda}\!\!\int_{S(I)}\!\!|k_1f|^p|g'|^p(1-|z|^2)^p\,dA(z)\\ &\quad +\frac{1}{|I|^\lambda}\!\!\int_{S(I)}\!\!|k_2f|^p|g'|^p(1-|z|^2)^p\,dA(z). \end{align*}$$
Taking the supremum over all
$I\subset \partial \mathbb {D}$
and using the box characterization for
$T_g(k_i f)$
yields
$$\begin{align*}\|T_g f\|_{A^{p,\lambda}}^p\ \lesssim\ \|T_g(k_1f)\|_{A^{p,\lambda}}^p+\|T_g(k_2f)\|_{A^{p,\lambda}}^p=\|T_{T_{g}(k_1)}f\|_{A^{p,\lambda}}^p+\|T_{T_{g}(k_2)}f\|_{A^{p,\lambda}}^p <\ \infty. \end{align*}$$
Thus,
$T_g f\in A^{p,\lambda }$
.
Theorem 6.5 Let
$g\in \mathcal {B}$
be non-constant and
$1<p<\infty $
. Denote by
$\mathcal {M}(W_g)$
the set of multiplier of
$W_g$
. Then
$\phi \in \mathcal {M}(W_g)$
if and only if
$\phi \in H^{\infty }$
and the integral operator
$T_\phi $
acts boundedly on
$W_g$
.
Proof Assume first that
$\phi \in H^{\infty }$
and that
$T_\phi $
is bounded on
$W_g$
. We show that
$\phi f\in W_g$
for every
$f\in W_g$
. By Theorem 6.1, it suffices to verify that
$$\begin{align*}\frac{(\phi f)'}{g'}\in H^{\infty}. \end{align*}$$
Indeed,
$$ \begin{align} \frac{(\phi(z) f(z))'}{g'(z)} &=\frac{\phi'(z) f(z)}{g'(z)}+\frac{\phi(z)f'(z)}{g'(z)} =\frac{T_\phi(f)'(z)}{g'(z)}+\phi(z)\frac{f'(z)}{g'(z)}. \end{align} $$
Since
$\frac {f'}{g'}$
and
$\frac {T_\phi (f)'}{g'}$
lie in
$H^{\infty }$
, it follows that
$$\begin{align*}\frac{(\phi f)'}{g'}\in H^{\infty}. \end{align*}$$
Conversely, let
$\phi $
be analytic, which is a bounded multiplier on
$W_g$
. Fix
$z\in \mathbb {D}$
and let
$\delta _z$
denote point evaluation on
$W_g$
. Because
$W_g$
is a closed subspace of
$\mathcal {B}$
containing constants, we have
$0\neq \|\delta _z\|<\infty $
. As in the proof of Theorem 5.6,
$$\begin{align*}|\phi(z)|\le \|M_\phi\|\frac{\|\delta_z\|}{\|\delta_z\|}=\|M_\phi\|, \end{align*}$$
hence
$\phi \in H^{\infty }$
. It remains to show that
$T_\phi $
is bounded on
$W_g$
. For
$f\in W_g$
,
$$\begin{align*}\frac{\left(T_{\phi}(f)\right)'}{g'}=\frac{\phi' f}{g'}=\frac{(\phi f)'}{g'}-\frac{\phi f'}{g'}. \end{align*}$$
The first term belongs to
$H^\infty $
because
$\phi $
is a bounded multiplier of
$W_g$
, and the second term lies in
$H^\infty $
as a product of two
$H^\infty $
functions. Thus,
$T_\phi (f)\in W_g$
for all
$f\in W_g$
, and the boundedness of
$T_\phi :W_g\to W_g$
follows by a standard application of the closed graph theorem.
Theorem 6.6 Let
$1\leq p<\infty $
and
$g\in \mathcal {B}$
. Then we have
$$ \begin{align*}T_g(H^\infty)+\mathbb{C}\subseteq V_g^p. \end{align*} $$
Moreover, if there exist
$u\in \mathcal {B}$
and
$B \in H^\infty $
with
$g'=Bu'$
, then
$$ \begin{align*}V_g^p \subseteq V^p_u.\end{align*} $$
Proof Write
$h=T_{g}(q)+h(0)$
for some
$q\in H^\infty $
. For the first inclusion, it suffices to check that
$T_h(f)\in A^{p,\lambda }$
for each
$f\in [T_g, A^{p,\lambda }]$
. Invoking Theorem 4.5, we know
$fq\in [T_g,A^{p,\lambda }]$
and hence
$$ \begin{align*} T_{h}(f)=T_{g}(qf)\in A^{p,\lambda}. \end{align*} $$
For the second inclusion, assume
$g'=Bu'$
and show
$V_g^p\subseteq V_u^p$
. Indeed, for every
$f \in [T_u,A^{p,\lambda }]$
,
$$ \begin{align*}T_g(f)=T_u(Bf) \in A^{p,\lambda}\, \end{align*} $$
by Theorem 4.5. Thus,
$[T_u,A^{p,\lambda }]\subseteq [T_g,A^{p,\lambda }]$
, which yields the claim.
Theorem 6.7 Let
$1\leq p<\infty $
and
$g_1,g_2\in \mathcal {B}$
with
$h=\frac {g^{\prime }_2}{g^{\prime }_1}\in \mathrm{Hol}(\mathbb D)$
. Then
$[T_{g_1},A^{p,\lambda }]=[T_{g_2},A^{p,\lambda }]$
if and only if
$h \in H^\infty (\mathbb D)$
and the multiplier
$M_h$
on
$[T_{g_1},A^{p,\lambda }]$
is bounded from below.
Proof Assume first
$[T_{g_1},A^{p,\lambda }]=[T_{g_2},A^{p,\lambda }]$
. By the Open Mapping Theorem, for all
$f \in [T_{g_1},A^{p,\lambda }]$
,
$$ \begin{align*}\|f\|_{[T_{g_1},A^{p,\lambda}]}\sim \|f\|_{[T_{g_2},A^{p,\lambda}]}. \end{align*} $$
Moreover,
$$ \begin{align*}\|hf\|_{[T_{g_1},A^{p,\lambda}]}=\|T_{g_1}(hf)\|_{A^{p,\lambda}}=\|T_{g_2}(f)\|_{A^{p,\lambda}}=\|f\|_{[T_{g_2},A^{p,\lambda}]}\lesssim \|f\|_{[T_{g_1},A^{p,\lambda}]}. \end{align*} $$
Hence,
$h \in \mathcal {M}([T_{g_1},A^{p,\lambda }])$
, and Theorem 4.5 gives
$h \in H^\infty (\mathbb D)$
.
Conversely, for every
$f \in [T_{g_1},A^{p,\lambda }],$
we have
$$ \begin{align*} \|hf\|_{[T_{g_1},A^{p,\lambda}]} &=\|T_{g_2}(f)\|_{A^{p,\lambda}} =\|f\|_{[T_{g_2},A^{p,\lambda}]} \gtrsim \|f\|_{[T_{g_1},A^{p,\lambda}]}. \end{align*} $$
The reverse implication follows from the same estimates.
This criterion alone does not settle the question, since multipliers of
$[T_{g},A^{p,\lambda }]$
that are bounded below have not been characterized. Thus, we next compare holomorphic optimal domains under additional natural assumptions on the symbols
$g_i$
.
Locally univalent: When g is locally univalent,
$g'$
never vanishes and the criteria simplify: equality of holomorphic optimal domains is equivalent to the existence of a bounded analytic change of density between
$g'$
and
$h'$
. If g is locally univalent, the space
$V_g^p$
admits an explicit description. As in Theorem 6.2, under this hypothesis, we can decide when two holomorphic optimal domains coincide.
Corollary 6.8 For
$1\leq p<\infty $
, let
$g_1,g_2\in \mathcal {B}$
be non-constant. The following are equivalent:
-
a)
$[T_{g_1},A^{p,\lambda }]=[T_{g_2},A^{p,\lambda }]$
; -
b) there exist
$k_1,k_2\in H^{\infty }$
such that
$$ \begin{align*}g_1=T_{g_2}(k_1)-g_1(0) \text{ and } g_2=T_{g_1}(k_2)-g_2(0). \end{align*} $$
Moreover,
$k_2=\frac {1}{k_1}$
.
Data availability statement
Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.
Competing interests
The authors declare that they have no competing interests.
