Journal of Fourier Analysis and Applications

, Volume 24, Issue 6, pp 1491–1517

# On the Characterization of Triebel–Lizorkin Type Spaces of Analytic Functions

Open Access
Article

## Abstract

We consider different characterizations of Triebel–Lizorkin type spaces of analytic functions on the unit disc. Even though our results appear in the folklore, detailed descriptions are hard to find, and in fact we are unable to discuss the full range of parameters. Without additional effort we work with vector-valued analytic functions, and also consider a generalized scale of function spaces, including for example so-called Q-spaces. The primary aim of this note is to generalize, and clarify, a remarkable result by Cohn and Verbitsky, on factorization of Triebel–Lizorkin spaces. Their result remains valid for functions taking values in an arbitrary Banach space, provided that the vector-valuedness “sits in the right factor”. On the other hand, if we impose vector-valuedness on the “wrong” factor, then the factorization theorem fails even for functions taking values in a separable Hilbert space.

## Keywords

Triebel–Lizorkin spaces Q-spaces Factorization Vector-valued analytic functions

## Mathematics Subject Classification

30H10 30H20 30H25 30H35 46E40

## 1 Introduction

### Definition 1.1

Let X, $$X_1$$ and $$X_2$$ be normed linear spaces of functions on $$\mathbb {D}$$. If for any $$f\in X$$ there exists $$f_1\in X_1$$ and $$f_2\in X_2$$ such that $$f=f_1f_2$$ and
\begin{aligned} \sup _{f\in X\setminus \left\{ 0\right\} }\inf _{f_1f_2=f}\frac{ \Vert f_1\Vert _{X_1} \Vert f_2\Vert _{X_2}}{ \Vert f\Vert _X}<\infty , \end{aligned}
then we say that $$X\subset X_1\cdot X_2$$. If for any $$f_1\in X_1$$ and $$f_2\in X_2$$ it holds that $$f_1f_2\in X$$ and
\begin{aligned} \sup _{f_1\in X_1\setminus \left\{ 0\right\} ,f_2\in X_2\setminus \left\{ 0\right\} }\frac{ \Vert f_1f_2\Vert _X}{ \Vert f_1\Vert _{X_1} \Vert f_2\Vert _{X_2}}<\infty , \end{aligned}
then we say that $$X_1\cdot X_2\subset X$$. If $$X\subset X_1\cdot X_2$$ and $$X_1\cdot X_2\subset X$$, then we say that $$X=X_1\cdot X_2$$.

Throughout this paper, we let $$\mathcal {X}$$ and $$\mathcal {H}$$ respectively denote a general Banach space and a separable Hilbert space, both complex. By $$\mathcal {A}\left( \mathcal {X}\right)$$ we denote the space of analytic $$\mathcal {X}$$-valued functions on the open unit disc $$\mathbb {D}$$. For short, we write $$\mathcal {A}=\mathcal {A}\left( \mathbb {C}\right)$$, i.e. we suppress $$\mathcal {X}=\mathbb {C}$$. The same principle will apply to all function spaces discussed below.

We let $$\mathbb {T}$$ denote be the unit circle in $$\mathbb {C}$$, and give it the parametrization $$x\mapsto \zeta _x$$, where $$\zeta _x=e^{2\pi i x}$$, $$x\in \mathbb {R}$$. For $$p\in \left( 0,\infty \right)$$, we denote by $$L^p\left( \mathbb {T},\mathcal {X}\right)$$ the class of strongly measurable functions $$f:\mathbb {T}\rightarrow \mathcal {X}$$ such that $$\Vert f\Vert _{L^p\left( \mathbb {T},\mathcal {X}\right) }^p=\int _{\zeta _x\in \mathbb {T}} \Vert f\left( \zeta _x\right) \Vert _\mathcal {X}^p\, dx <\infty$$, where we somewhat abusively write dx to indicate Lebesgue integration with respect to $$\zeta _x$$. We will often identify $$f\in L^1\left( \mathbb {T},\mathcal {X}\right)$$ with its Poisson extension $$\mathcal {P}\left[ f\right] :\mathbb {D}\rightarrow \mathcal {X}$$. Under this identification, the Fourier coefficients $$\hat{f}\left( n\right) = \int _{\mathbb {T}}f\left( \zeta _x\right) \overline{\zeta _x^n}\, dx$$ are the Taylor coefficients of $$\mathcal {P}\left[ f\right]$$. For short, we typically write f in place of $$\mathcal {P}\left[ f\right]$$. We denote the nth Taylor coefficient of a general function $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ by $$\hat{f}\left( n\right)$$, even though f is not necessarily the Poisson extension of an integrable function.

We define the Hardy space $$H^p\left( \mathcal {X}\right)$$ as the class of functions $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ such that $$\Vert f\Vert _{H^p\left( \mathcal {X}\right) }=\sup _{0<r<1} \Vert f_r\Vert _{L^p\left( \mathbb {T},\mathcal {X}\right) } <\infty$$, where $$f_r:w\mapsto f\left( rw\right)$$. In the case where $$\mathcal {X}=\mathcal {H}$$, we have the so-called square function characterization of $$H^p\left( \mathcal {H}\right)$$; in the language of Sect. 4, $$H^p\left( \mathcal {H}\right) =F_{p,2}^0\left( \mathcal {H}\right)$$. If $$p\ge 1$$, then $$H^p\left( \mathcal {H}\right)$$ also coincides with the space of $$f\in L^p\left( \mathbb {T},\mathcal {H}\right)$$ such that $$\hat{f}\left( n\right) =0$$ for $$n<0$$.

We define the pairing $$\left\langle f,\phi \right\rangle _{\mathcal {A}\left( \mathcal {H}\right) } = \sum \langle \hat{f}\left( n\right) ,\hat{\phi }\left( n\right) \rangle _\mathcal {H}$$, where $$f,\phi \in \mathcal {A}\left( \mathcal {H}\right)$$, and f is polynomial. If $$\phi \in H^1\left( \mathcal {H}\right)$$, then $$\left\langle f,\phi \right\rangle _{\mathcal {A}\left( \mathcal {H}\right) } = \int _{\mathbb {T}} \left\langle f, \phi \right\rangle _\mathcal {H}\, dx$$. With respect to this pairing, the dual of $$H^1\left( \mathcal {H}\right)$$ is given by $$BMOA\left( \mathcal {H}\right)$$, the space of $$\phi \in H^1\left( \mathcal {H}\right)$$ such that
\begin{aligned} \Vert \phi \Vert _{BMOA\left( \mathcal {H}\right) }= \Vert \phi \left( 0\right) \Vert _\mathcal {H}+\sup _{\text {arcs }I\subset \mathbb {T}}\frac{1}{|I|}\int _I \left\| \phi \left( \zeta _x\right) -\frac{1}{|I|}\int _I\phi \left( \zeta _y\right) dy \right\| _\mathcal {H}\, dx<\infty . \end{aligned}
The duality of $$H^1$$ and BMOA is a celebrated result by Fefferman . For a discussion on the vector-valued case, see for instance . A characterization of $$BMOA\left( \mathcal {H}\right)$$ relevant to this paper is, in the language of Sect. 4, that $$BMOA\left( \mathcal {H}\right) =F_{\infty ,2}^0\left( \mathcal {H}\right)$$. This is the so-called Carleson measure characterization of $$BMOA\left( \mathcal {H}\right)$$.
Given $$\alpha \in \mathbb {R}$$ and $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, we define the fractional derivative $$D^\alpha f$$ by
\begin{aligned} D^\alpha f\left( w\right) =\sum _{n=0}^\infty \left( 1+n\right) ^\alpha \hat{f}\left( n\right) w^n,\quad w\in \mathbb {D}. \end{aligned}
Consider the class $$D^\alpha H^p:=\left\{ f\in \mathcal {A};D^{-\alpha }f\in H^p\right\}$$ equipped with the norm $$\Vert f\Vert _{D^\alpha H^p} := \Vert D^{-\alpha }f\Vert _{H^p}$$. The following result is due to Cohn and Verbitsky [4, Theorem 2]:

### Theorem 1.2

Let $$\alpha >0$$, $$0<p,p_1,p_2<\infty$$, and $$p_1^{-1}+p_2^{-1}=p^{-1}$$. Then
\begin{aligned} D^\alpha H^p=H^{p_1}\cdot D^\alpha H^{p_2}. \end{aligned}
The present author’s interest in the above result arose while studying the following type of bilinear forms appearing naturally in control theory, e.g. [12, 18]: Given $$\phi \in \mathcal {A}$$ and $$\alpha >0$$, we define the bilinear Hankel type form
\begin{aligned} H_{\phi ,\alpha }:H^2\times H^2\ni \left( g,h\right) \mapsto \left\langle h,D^\alpha \left( \phi \overline{g}\right) \right\rangle _{\mathcal {A}}, \end{aligned}
(1)
on analytic polynomials. The next result on $$H^2$$-boundedness of $$H_{\phi ,\alpha }$$ has several proofs in the literature, e.g. [12, 18]. As an illustration, we prove it by applying Theorem 1.2:

### Proposition 1.3

$$H_{\phi ,\alpha }$$ is bounded if and only if $$D^\alpha \phi \in BMOA$$.

### Proof

Suppose that $$g,h\in H^2$$, and let $$f\in H^1$$ be a suitable function such that $$D^\alpha f= g\left( D^\alpha h\right)$$. Then
\begin{aligned} \left\langle h,D^\alpha \left( \phi \overline{g}\right) \right\rangle _\mathcal {A}= \left\langle D^\alpha h,\phi \overline{g}\right\rangle _\mathcal {A}= \left\langle g\left( D^\alpha h\right) ,\phi \right\rangle _\mathcal {A}= \left\langle D^\alpha f,\phi \right\rangle _\mathcal {A}= \left\langle f,D^\alpha \phi \right\rangle _\mathcal {A}. \end{aligned}
The statement now follows from the Fefferman $$H^1-BMOA$$ duality theorem. $$\square$$
The primary aim of this paper is to consider vector-valued generalizations of Theorem 1.2. Given $$\phi \in \mathcal {A}\left( \mathcal {H}\right)$$ and $$\alpha >0$$, there are two natural analogues of (1):
\begin{aligned} H_{\phi ,\alpha }:H^2\times H^2\left( \mathcal {H}\right) \ni \left( g,h\right) \mapsto \left\langle h,D^\alpha \left( \phi \overline{g}\right) \right\rangle _{\mathcal {A}\left( \mathcal {H}\right) }, \end{aligned}
(2)
and
\begin{aligned} H_{\phi ,\alpha }^*:H^2\left( \mathcal {H}\right) \times H^2\ni \left( g,h\right) \mapsto \left\langle h,D^\alpha \left( \left\langle \phi ,g\right\rangle _\mathcal {H}\right) \right\rangle _{\mathcal {A}}. \end{aligned}
(3)
The proof of Proposition 1.3 now leads us to the following questions:
\begin{aligned}&\hbox {Q1: Is }D^\alpha H^1\left( \mathcal {H}\right) = H^2 \cdot \left( D^\alpha H^2\left( \mathcal {H}\right) \right) ?\\&\hbox {Q2: Is }D^\alpha H^1\left( \mathcal {H}\right) =H^2\left( \mathcal {H}\right) \cdot \left( D^\alpha H^2\right) ? \end{aligned}
The first question will receive a positive answer. This yields that $$H_{\phi ,\alpha }$$ is bounded if and only if $$D^\alpha \phi \in H^1\left( \mathcal {H}\right) ^*=BMOA\left( \mathcal {H}\right)$$, a result also obtained in . The second question receives a negative answer. Indeed, if the answer was positive, then $$H_{\phi ,\alpha }$$ and $$H_{\phi ,\alpha }^*$$ would be simultaneously bounded. This would contradict the following result, essentially due to Davidson and Paulsen . See also [18, Sect. 4]:

### Proposition 1.4

Let $$\alpha >0$$, $$\phi \in \mathcal {A}\left( \mathcal {H}\right)$$, and define the forms $$H_{\phi ,\alpha }$$ and $$H_{\phi ,\alpha }^*$$ by (2) and (3) respectively. If $$H_{\phi ,\alpha }$$ is bounded, then $$H_{\phi ,\alpha }^*$$ is also bounded. The converse does not hold.

Before stating our main results, we define the concept of an outer analytic function:

### Definition 1.5

Let $$k\in L^1(\mathbb {T})$$ be a real-valued function. A function of the form $$e^{u+iv}$$, where u is the Poisson extension of k, and v is the harmonic conjugate of u, is called an outer analytic function.

For our purposes, the main feature of outer analytic functions is that they are never vanishing in $$\mathbb {D}$$.

As in , we state our main result in the language of Triebel–Lizorkin spaces $$F_{p,q}^s$$. Their definition is quite elaborate, and we refer to Sect. 4, where we discuss different characterizations. Remarkably, our results hold for analytic functions taking values in an arbitrary Banach space:

### Theorem 1.6

Let $$\mathcal {X}$$ be a complex Banach space, $$s<0$$, $$0<p<\infty$$ and $$1\le q<\infty$$. Then
\begin{aligned} F_{p,q}^s\left( \mathcal {X}\right) =H^p \cdot F_{\infty ,q}^s\left( \mathcal {X}\right) . \end{aligned}
Moreover, the $$H^p$$-factor can be constructed as an outer analytic function.

### Remark 1.7

The attentive reader will of course note that we are lacking a statement for $$0<q<1$$. This is perhaps the main shortcoming of this paper, and it relates to the fact that for this range of parameters we are not able to define the corresponding spaces.

Since $$H^p=H^{p_1}\cdot H^{p_2}$$ whenever $$p_1^{-1}+p_2^{-1}=p^{-1}$$, Theorem 1.6 implies that
\begin{aligned} F_{p,q}^s\left( \mathcal {X}\right) =H^p \cdot F_{\infty ,q}^s\left( \mathcal {X}\right) =H^{p_1} \cdot H^{p_2} \cdot F_{\infty ,q}^s\left( \mathcal {X}\right) =H^{p_1} \cdot F_{p_2,q}^s\left( \mathcal {X}\right) . \end{aligned}
Theorem 1.2 corresponds to the special case $$D^\alpha H^p=F_{p,2}^{-\alpha }$$. For emphasis, we state a corollary:

### Corollary 1.8

Let $$\mathcal {X}$$ be a complex Banach space, $$s<0$$, $$0<p,p_1,p_2<\infty$$ and $$1\le q<\infty$$. If $$p_1^{-1}+p_2^{-1}=p^{-1}$$, then
\begin{aligned} F_{p,q}^s\left( \mathcal {X}\right) =H^{p_1} \cdot F_{p_2,q}^s\left( \mathcal {X}\right) . \end{aligned}

We also obtain a non-factorization result:

### Theorem 1.9

Let $$\mathcal {X}$$ be a complex Banach space, $$s<0$$, $$0<p<\infty$$ and $$1\le q<\infty$$. Then
\begin{aligned} H^p\left( \mathcal {X}\right) \cdot F_{\infty ,q}^s\subset F_{p,q}^s\left( \mathcal {X}\right) . \end{aligned}
In general, this inclusion is strict. In particular, Proposition 1.4 shows that if $$\mathcal {H}$$ is an infinite-dimensional Hilbert space, then there exists $$f\in F_{1,2}^s\left( \mathcal {H}\right)$$, such that for any $$g\in H^1\left( \mathcal {H}\right)$$ and $$h\in F_{\infty ,2}^s$$, $$f\ne gh$$.

The first ingredient needed in order to generalize Theorem 1.2 is a factorization result for tent spaces, Theorem 3.1. We point out that the proofs from  go through also in the vector-valued case (replacing moduli with vector space norms). However, the scalar-valued result even implies the vector-valued one. We demonstrate how in Sect. 3.

We will also need some properties of Triebel–Lizorkin spaces of analytic functions on $$\mathbb {D}$$. These appear in the literature (e.g. ) but I have not been able to find any stringent justification. The vector-valued setting that we consider does not require any additional effort. Nevertheless, this setting does not appear to have been considered before. For these reasons we dedicate Sect. 4 to establishing some rudimentary theory. We develop our theory in the language of the more general class of distribution spaces $$F_{p,q}^{s,\tau }\left( \mathbb {R}^d\right)$$ introduced in by Yang and Yuan [23, 24]. Rather than increasing our efforts, $$F_{p,q}^{s,\tau }\left( \mathcal {X}\right)$$ unifies the spaces $$F_{p,q}^s\left( \mathcal {X}\right)$$ where $$p<\infty$$, and $$F_{\infty ,q}^s\left( \mathcal {X}\right)$$, perhaps even decreasing the amount of work needed. Another motivation to study the generalized scale is that it encompasses more spaces, for example the so-called Q-spaces introduced by Aulaskari et al. .

## 2 Preliminaries and Notation

We use the standard notation $$\mathbb {Z}$$, $$\mathbb {R}$$, and $$\mathbb {C}$$ for the respective rings of integers, real numbers, and complex numbers. In addition, $$\mathbb {D}=\left\{ w\in \mathbb {C};|w|<1\right\}$$ and $$\mathbb {T}=\left\{ \zeta \in \mathbb {C};|\zeta |=1\right\}$$. We will often identify $$\mathbb {T}$$ with $$\mathbb {R}/\mathbb {Z}$$, using the map $$x\mapsto e^{2\pi i x}$$. Subsets of $$\mathbb {R}/\mathbb {Z}$$ and $$\mathbb {T}$$ are identified in a similar way. In particular, we let the set of dyadic arcs $$\mathcal {D}\left( \mathbb {T}\right)$$ be the image of the set $$\mathcal {D}\left( \left[ 0,1\right) \right) =\left\{ \left[ 2^{-j}k,2^{-j}\left( k+1\right) \right) ;j\in \mathbb {N}_0,0\le k \le 2^j-1\right\}$$. We define the rank of $$I\in \mathcal {D}(\mathbb {T})$$ as $$\text {rk}(I)=-\log _2|I|$$. Note that in general, an arc $$I\subset \mathbb {T}$$ may correspond to the union of two intervals in $$\left[ 0,1\right)$$. We use the letters xyz to denote generic points on $$\mathbb {R}$$. By $$\zeta _x$$ we denote the point $$e^{2\pi i x}\in \mathbb {T}$$. The arc-wise distance between $$\zeta _x$$ and $$\zeta _y$$ is denoted by $$|\zeta _x-\zeta _y|$$. A Euclidean ball with radius r and center w is denoted $$B\left( w,r\right)$$. Given two parametrized sets of nonnegative numbers $$\left\{ A_i\right\} _{i\in I}$$ and $$\left\{ B_i\right\} _{i\in I}$$, we use the notation $$A_i\lesssim B_i$$, $$i\in I$$ to indicate the existence of a positive constant C such that $$A_i\le CB_i$$ whenever $$i\in I$$. Sometimes we allow ourselves to not mention the index set I and instead let it be implicit from the context. If $$A_i\lesssim B_i$$ and $$B_i\lesssim A_i$$, then we write $$A_i\approx B_i$$.

For a background on the Bochner–Lebesgue classes $$L^p\left( \mathbb {T},\mathcal {X}\right)$$ we refer to . Given a strongly measurable function $$f:\mathbb {T}\rightarrow \mathcal {X}$$, we define the corresponding Hardy–Littlewood maximal function by
\begin{aligned} Mf\left( \zeta _x\right) =\sup _{\begin{array}{c} \text {arcs }I\subset \mathbb {T};\\ I \text { centered at }\zeta _x \end{array}}\frac{1}{|I|}\int _{\zeta _y\in I} \Vert f\left( \zeta _y\right) \Vert _\mathcal {X}\, dy,\quad \zeta _x\in \mathbb {T}. \end{aligned}
The following periodic analogue of the vector-valued maximal theorem follows easily from [8, Theorem 1]:

### Theorem 2.1

Let $$\beta ,\gamma \in \left( 1,\infty \right)$$. Then there is a number $$K=K\left( \beta ,\gamma \right)$$ such that
\begin{aligned} \left\| \left( \sum _{n=1}^\infty |Mf_n|^\beta \right) ^{1/\beta }\right\| _{L^\gamma \left( \mathbb {T}\right) } \le K \left\| \left( \sum _{n=1}^\infty |f_n|^\beta \right) ^{1/\beta }\right\| _{L^\gamma \left( \mathbb {T}\right) } \end{aligned}
whenever $$\left( f_n\right) _{n=1}^\infty$$ is a sequence of measurable $$\mathbb {C}$$-valued functions on $$\mathbb {T}$$.
For $$f\in L^1\left( \mathbb {T},\mathcal {X}\right)$$, we define the Poisson extension
\begin{aligned} f_r\left( \zeta _x\right) =\mathcal {P}\left[ f\right] \left( w\right) =\int _\mathbb {T}f\left( \zeta _y\right) P_r\left( \zeta _{x-y}\right) \, dy,\quad w=r\zeta _x, \end{aligned}
where
\begin{aligned} P_r\left( \zeta _y\right) =\frac{1-r^2}{1-2r\cos \left( 2\pi y\right) +r^2}=\frac{1}{1-r\zeta _y}+\frac{r\overline{\zeta _y}}{1-r\overline{\zeta _y}},\quad y\in \mathbb {R}, \end{aligned}
is the Poisson kernel for $$\mathbb {D}$$. We also write $$\mathcal {P}[\mu ](w)=\int _\mathbb {T}P_r\left( \zeta _{x-y}\right) \, d\mu (y)$$, whenever $$w=r\zeta _x$$, and $$\mu$$ is a measure of finite total variation on $$\mathbb {T}$$.
By geometric summation, $$\hat{P}_r\left( n\right) =r^{|n|}$$. It is well-known that $$\Vert f_r\Vert _{L^p\left( \mathbb {T},\mathcal {X}\right) }\le \Vert f\Vert _{L^p\left( \mathbb {T},\mathcal {X}\right) }$$, and $$\Vert f_r\left( x\right) \Vert _\mathcal {X}\lesssim Mf\left( x\right)$$, e.g. [11, Sects. I.3 an I.4]. Note also that
\begin{aligned} P_r\left( \zeta _y\right) \approx \frac{1}{1-r}\frac{1}{1+\left( \frac{|\zeta _y-1|}{1-r}\right) ^2},\quad \zeta _y\in \mathbb {T}. \end{aligned}
(4)
A function $$v:\mathbb {D}\rightarrow \left[ -\infty ,\infty \right)$$ is called upper semi-continuous if for each $$w_0\in \mathbb {D}$$, $$\limsup _{w\rightarrow w_0}v\left( w\right) \le v\left( w_0\right)$$. If v is upper semi-continuous and if for each $$w_0\in \mathbb {D}$$ there exists $$r_0>0$$ such that for each $$0<r<r_0$$, $$B\left( w_0,r\right) \subset \mathbb {D}$$ and
\begin{aligned} v\left( w_0\right) \le \frac{1}{\pi r^2}\int _{B\left( w_0,r\right) }v\left( w\right) \, dA\left( w\right) , \end{aligned}
where dA denotes Lebesgue measure on $$\mathbb {C}$$, then we say that v is subharmonic. If v is subharmonic, then the function $$\left[ 0,1\right) \ni r\mapsto \int _\mathbb {T}v\left( r\zeta _x\right) \, dx$$ is increasing. If v is subharmonic and extends continuously to $$\mathbb {T}$$, then v is majorized by the Poisson extension of its boundary values, i.e.
\begin{aligned} v\left( w\right) \le \int _\mathbb {T}v\left( \zeta _y\right) P_r\left( \zeta _{x-y}\right) \, dy,\quad w=r\zeta _x, \end{aligned}
For proofs of these claims, we refer to . If $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, then for any $$0<p<\infty$$, the function $$\mathbb {D}\ni w\mapsto \Vert f\left( w\right) \Vert _\mathcal {X}^p$$ is subharmonic, e.g. [16, Chap. 4].
A Stolz angle $$\Gamma \left( \zeta _x\right)$$ is the convex hull of the set $$\left\{ \zeta _x\right\} \cup \frac{1}{2}\mathbb {T}$$. The non-tangential maximal function of $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ is given by
\begin{aligned} A_\infty f\left( \zeta _x\right) =\sup _{w\in \Gamma \left( \zeta _x\right) } \Vert f\left( w\right) \Vert _\mathcal {X},\quad \zeta _x\in \mathbb {T}. \end{aligned}
The functional $$A_\infty$$ is a classical tool in the characterization of $$H^p$$, and proves useful also in the $$\mathcal {X}$$-valued setting:

### Lemma 2.2

Let $$0<p\le \infty$$, and $$\mathcal {X}$$ be a complex Banach space. Then $$f\in H^p(\mathcal {X})$$ if and only if $$f\in \mathcal {A}(\mathcal {X})$$ and $$A_\infty f\in L^p$$.

### Proof

1The statement is obvious for $$p=\infty$$, as is the inclusion
\begin{aligned} \left\{ f\in \mathcal {A}\left( \mathcal {X}\right) ;\, A_\infty f\in L^p \right\} \subset H^p\left( \mathcal {X}\right) . \end{aligned}
For the reverse inclusion, suppose that $$f\in H^p(\mathcal {X})$$. We may also assume that f is not identically 0. By analyticity, $$\log \Vert f\Vert _\mathcal {X}$$ is subharmonic, as is $$\Vert f\Vert _\mathcal {X}^p$$, c.f. [16, Theorem 4.2.A]. A theorem by Littlewood  implies that the limit $$u\left( \zeta _x\right) =\lim _{r\rightarrow 1} \Vert f_r\left( \zeta _x\right) \Vert _\mathcal {X}$$ exists Lebesgue a.e. on $$\mathbb {T}$$.

Let $$\log ^+t=\max \{0,\log t\}$$, and note that $$\log ^+ t\le \frac{1}{p}t^p$$. Since $$f\in H^p(\mathcal {X})$$, it holds that $$\{\log ^+ \Vert f_r\Vert _\mathcal {X}\}_{0<r<1}$$ is a bounded (and increasing) family in $$L^1(\mathbb {T})$$. By [16, Theorem A.1.E], $$\log \Vert f\Vert _\mathcal {X}$$ has a least harmonic majorant $$h=\mathcal {P}[\mu ]$$, where the measure $$\mu$$ is the weak-$$*$$ limit point of $$\{\log \Vert f_r\Vert _\mathcal {X}\}_{0<r<1}$$ in $$L^1(\mathbb {T})$$ as $$r\rightarrow 1$$.

By Fatou’s theorem, $$h(\zeta _x)=\lim _{w\rightarrow \zeta _x}h(w)$$ exists as a nontangential limit for Lebesgue a.e. $$\zeta _x\in \mathbb {T}$$. Moreover, $$h\in L^1(\mathbb {T})$$, and $$d\mu = h(\zeta _x)\, dx +d\lambda _+-d\lambda _-$$, where $$\lambda _+,\lambda _-\ge 0$$ are singular measures with respect to dx.

Let $$\phi (t)=\exp (pt)$$, and note that $$\{\phi \left( \log ^+ \Vert f_r\Vert _\mathcal {X}\right) \}_{0<r<1}$$ is bounded in $$L^1(\mathbb {T})$$. By the Szegö–Solomentsev theorem [16, Appendix 2], $$\lambda _+=0$$ and $$h|_\mathbb {T}=\log u$$ Lebesgue a.e. It follows that $$h\le \mathcal {P}[\log u]$$.

Chose (some) $$H\in \mathcal {A}$$ with real part h, and define $$g=\exp (H)$$. Then $$\Vert f\Vert _\mathcal {X}\le |g|$$, and so $$\Vert f\Vert _{H^p\left( \mathcal {X}\right) }\le \Vert g\Vert _{H^p}$$. An application of Jensen’s inequality (with the probability measure $$P_r(\zeta _x)\, dx$$) shows that $$|g|^p=\phi (h)\le \phi (\mathcal {P}[\log u])\le \mathcal {P}\left[ \phi (\log u)\right] =\mathcal {P}\left[ u^p\right]$$. Thus $$\Vert g\Vert _{H^p}\le \Vert u\Vert _{L^p\left( \mathbb {T}\right) }\le \Vert f\Vert _{H^p\left( \mathcal {X}\right) }$$, where we have used, in turn, Minkowski’s inequality and Fatou’s lemma. The non-tangential maximal characterization of $$H^p\left( \mathcal {X}\right)$$ now follows from the scalar case, since
\begin{aligned} \Vert A_\infty f\Vert _{L^p}\le \Vert A_\infty g\Vert _{L^p}\lesssim \Vert g\Vert _{H^p} = \Vert f\Vert _{H^p\left( \mathcal {X}\right) }. \end{aligned}
$$\square$$
The square function of $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ is given by
\begin{aligned} S f\left( \zeta _x\right) =\left( \int _{w\in \Gamma \left( \zeta _x\right) } \Vert \left( Df\right) \left( w\right) \Vert _\mathcal {X}^2 \, dA\left( w\right) \right) ^{1/2},\quad \zeta _x\in \mathbb {T}. \end{aligned}
It is a famous result by Fefferman and Stein  that if $$f\in \mathcal {A}\left( \mathcal {H}\right)$$, then $$A_\infty f\in L^p\left( \mathbb {T}\right)$$ if and only if $$Sf\in L^p\left( \mathbb {T}\right)$$. In general, $$H^p\left( \mathcal {X}\right)$$ may fail to have a square function characterization, e.g. [5, Remark 4.11]. The duality between $$H^1\left( \mathcal {H}\right)$$ and $$BMOA\left( \mathcal {H}\right)$$ is a celebrated theorem by Fefferman , adapted to analytic functions on $$\mathbb {D}$$ (e.g. [11, Exercise VI.5]), with values in $$\mathcal {H}$$ (e.g. ).

If $$f\in H^p\left( \mathcal {H}\right)$$, then there exists $$bf\in L^p\left( \mathbb {T},\mathcal {H}\right)$$ such that $$bf\left( \zeta _x\right)$$ equals the non-tangential limit $$\lim _{w\rightarrow \zeta _x}f\left( w\right)$$ for almost every $$\zeta _x$$, and $$f_r\rightarrow bf$$ in $$L^p\left( \mathbb {T},\mathcal {H}\right)$$. Moreover, $$f=\mathcal {P}\left[ bf\right]$$, e.g. [16, Chap. 4]. For this reason we will typically not distinguish between a function $$f\in H^p\left( \mathcal {H}\right)$$, and its boundary values $$bf\in L^p\left( \mathcal {H}\right)$$.

The convolution of $$f\in \mathcal {A}$$ and $$g\in \mathcal {A}\left( \mathcal {X}\right)$$ is defined as $$f*g\in \mathcal {A}\left( \mathcal {X}\right)$$ with $$\left( f*g\right) ^{\hat{}}\left( n\right) =\hat{f} \left( n\right) \hat{g}\left( n\right)$$. If f and g are Poisson extensions of integrable functions, then so is $$f*g$$, and
\begin{aligned} \left( f*g\right) \left( \zeta _x\right) =\int _\mathbb {T}f\left( \zeta _{x-y}\right) g\left( \zeta _y\right) \, dy,\quad \zeta _x\in \mathbb {T}. \end{aligned}
Given a smooth function $$\varphi :\mathbb {R}\rightarrow \mathbb {C}$$, let $$\varphi ^{\left( k\right) }$$ denote its (classical) derivative of order $$k\in \mathbb {N}_0$$. The Schwartz space $$\mathcal {S}$$ is the class of functions $$\varphi :\mathbb {R}\rightarrow \mathbb {C}$$ for which all derivatives decay faster than any rational function, i.e. $$\varphi \in \mathcal {S}$$ if and only if for any $$k,N\in \mathbb {N}_0$$, $$\sup _{x\in \mathbb {R}}\left( 1+|x|\right) ^N|\varphi ^{\left( k\right) }\left( x\right) |<\infty$$. The Fourier transform of $$\varphi \in \mathcal {S}$$ is given by
\begin{aligned} \hat{\varphi }\left( \xi \right) =\int _{\mathbb {R}} \varphi \left( x\right) e^{-2\pi i x \xi }\, dx,\quad \xi \in \mathbb {R}. \end{aligned}
We will be interested in the 1-periodization of $$\varphi$$ given by
\begin{aligned} \Phi \left( \zeta _x\right) =\sum _{k\in \mathbb {Z}}\varphi \left( x-k\right) ,\quad x\in \mathbb {R}. \end{aligned}
It is easy to see that if $$N\ge 2$$, then
\begin{aligned} \sup _{\zeta _x\in \mathbb {T}}\left( 1+|\zeta _x-1|\right) ^N|\Phi \left( \zeta _x\right) |\lesssim \sup _{x\in \mathbb {R}}\left( 1+|x|\right) ^N|\varphi ^{\left( k\right) }\left( x\right) |. \end{aligned}
(5)
Moreover $$\hat{\Phi }\left( n\right) =\hat{\varphi }\left( n\right)$$, $$n\in \mathbb {Z}$$.

## 3 Tent Spaces

Given a subset $$E\subset \mathbb {T}$$, we define the “tent” over E as
\begin{aligned} T\left( E\right) =\mathbb {D}\setminus \left( \bigcup _{\zeta _x\in \mathbb {T}\setminus E}\Gamma \left( \zeta _x\right) \right) . \end{aligned}
Let $$f:\mathbb {D}\rightarrow \mathbb {C}$$ be a measurable function. For $$q\in \left( 0,\infty \right)$$, we define the functional $$A_q$$ by
\begin{aligned} A_qf\left( \zeta _x\right) =\left( \int _{\Gamma \left( x\right) }\frac{|f\left( z\right) |^q}{\left( 1-|z|\right) ^2}\, dA\left( z\right) \right) ^{1/q},\quad x\in \mathbb {T}. \end{aligned}
Note that the functional $$A_\infty$$ was defined in the previous section. For $$p,q\in \left( 0,\infty \right)$$, we say that $$f\in T_{p,q}$$ if $$\Vert f \Vert _{T_{p,q}}= \Vert A_{q}f \Vert _{L^p\left( \mathbb {T}\right) }<\infty$$. For $$q\in \left( 0,\infty \right)$$, we say that $$f\in T_{\infty ,q}$$ if
\begin{aligned} \Vert f\Vert _{T_{\infty ,q}}^q=\sup _{I\in \mathcal {D}}\frac{1}{|I|}\int _{T\left( I\right) }\frac{|f\left( z\right) |^q}{\left( 1-|z|\right) }\, dA\left( z\right) <\infty , \end{aligned}
i.e. if $$\frac{|f\left( z\right) |^q}{\left( 1-|z|\right) }\, dA\left( z\right)$$ is a Carleson measure.

For pq in the range discussed above, we define $$T_{p,q}\left( \mathcal {X}\right)$$ to be the set of functions $$f:\mathbb {D}\rightarrow \mathcal {X}$$ such that $$\Vert f\Vert _{\mathcal {X}}\in T_{p,q}$$. We equip this space with the obvious metric structure.

The main result of  can be explained in three steps. First, the authors obtain the inclusion $$T_{p,\infty }\cdot T_{\infty ,q} \subset T_{p,q}$$. This holds in particular when the first factor is analytic, but this is not a requirement. Second, the inclusion $$T_{p,q}\subset T_{p,\infty }\cdot T_{\infty ,q}$$ is proved. Here the first factor may be chosen to be outer analytic. Third, the authors use the fact that $$H^p=\mathcal {A}\cap T_{p,\infty }$$, i.e. the scalar version of Lemma 2.2. Put together, this reads that $$T_{p,q}= H^p\cdot T_{\infty ,q}$$, where the factor in $$H^p$$ may be constructed to be an outer analytic function. Using the first two steps together with Lemma 2.2, this result easily carries over to the $$\mathcal {X}$$-valued setting, provided that we take care which one of the factors is $$\mathcal {X}$$-valued:

### Theorem 3.1

Let $$\mathcal {X}$$ be a complex Banach space, and $$0<p,q<\infty$$. Then
\begin{aligned} T_{p,q}\left( \mathcal {X}\right) = H^p\cdot T_{\infty ,q}\left( \mathcal {X}\right) , \end{aligned}
and
\begin{aligned} H^p\left( \mathcal {X}\right) \cdot T_{\infty ,q} \subset T_{p,q}\left( \mathcal {X}\right) . \end{aligned}
If $$\mathcal {X}=\mathcal {H}$$ is an infinite-dimensional Hilbert space, then the inclusion is strict.

### Proof

If $$f\in T_{p,q}\left( \mathcal {X}\right)$$, then by the scalar-valued result $$\Vert f\Vert _\mathcal {X}=gH$$, where $$g\in H^p$$ is outer analytic, $$H\in T_{\infty ,q}$$, and $$\Vert g\Vert _{H^p} \Vert H\Vert _{T_{\infty ,q}}\lesssim \Vert f\Vert _{T_{p,q}\left( \mathcal {X}\right) }$$. Define $$h=\frac{f}{g}$$. Then $$\Vert h\Vert _\mathcal {X}= |H| \in T_{\infty ,q}$$. By definition $$h\in T_{\infty ,q} \left( \mathcal {X}\right)$$, and $$f=gh$$. This proves that $$T_{p,q}\left( \mathcal {X}\right) \subset H^p\cdot T_{\infty ,q}\left( \mathcal {X}\right)$$. The reverse inclusion $$H^p\cdot T_{\infty ,q}\left( \mathcal {X}\right) \subset T_{p,q}\left( \mathcal {X}\right)$$ also follows from the scalar-valued result: Let $$g\in H^p$$ and $$h\in T_{\infty ,q}\left( \mathcal {X}\right)$$. Then
\begin{aligned} \Vert gh \Vert _{ T_{p,q}\left( \mathcal {X}\right) } = \Vert g\Vert h \Vert _\mathcal {X}\Vert _{ T_{p,q}} \lesssim \Vert g \Vert _{H^p} \Vert \, \Vert h \Vert _\mathcal {X}\Vert _{ T_{\infty ,q}} = \Vert g \Vert _{H^p} \Vert h\Vert _{ T_{\infty ,q}\left( \mathcal {X}\right) }. \end{aligned}
The statement that $$H^p\left( \mathcal {X}\right) \cdot T_{\infty ,q} \subset T_{p,q}\left( \mathcal {X}\right)$$ follows similarly. For infinite-dimensional Hilbert spaces, this inclusion must be strict in order to not contradict Theorem 1.9. $$\square$$

## 4 Triebel–Lizorkin Type Spaces

The so called Triebel–Lizorkin spaces $$F^s_{p,q}\left( \mathbb {R}^d\right)$$, $$s\in \mathbb {R}$$, $$0<p,q\le \infty$$, are well-studied objects. An extensive treatment is given in Triebel’s monographs [19, 20, 21]. We also mention papers by Liang et al. , Peetre , Rychkov ,2 and Ullrich , all of which give a more direct introduction to many of the ideas to be used in this paper.

In this section we investigate some Triebel–Lizorkin type spaces of $$\mathcal {X}$$-valued analytic functions on $$\mathbb {D}$$. This setting is analogous to the corresponding theory of distributions on $$\mathbb {R}^d$$, but requires some new ideas. For instance, our spaces cannot be defined by mere finiteness of the appropriate norm, c.f. Definition 4.1 and Proposition 4.4. Moreover, the kernels for the local means, as defined for example in , poorly reflect the fact that the functions we deal with are analytic. A kernel that efficiently reflects the property of analyticity is the Poisson kernel. However, this kernel is not sufficiently regular for us to apply the typical techniques for analyzing $$F^s_{p,q}(\mathbb {R}^d)$$. Instead we need to use an intermediate substitute, c.f. the discussion following Theorem 4.3. We remark that the $$\mathcal {X}$$-valuedness does not present any major obstacles, since the proofs boil down to estimates of the scalar-valued integral kernels. Finally, for the convenience of the reader not yet familiar with the theory of $$F^s_{p,q}(\mathbb {R}^d)$$, or its generalization discussed below, we attempt to give reasonably complete proofs, the more involved parts of which are postponed to Sect. 4.1.

### Definition 4.1

Let $$0<p\le \infty$$, $$1\le q<\infty$$, $$s\in \mathbb {R}$$ and $$\alpha >s$$. We define the Triebel–Lizorkin space $$F_{p,q}^s\left( \mathbb {D},\mathcal {X}\right)$$ as the space of functions $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ such that
\begin{aligned} \Vert f\Vert _{F_{p,q}^s\left( \mathbb {D},\mathcal {X}\right) }:= \Vert \left( 1-|z|\right) ^{\alpha -s}D^\alpha f\Vert _{T_{p,q}\left( \mathcal {X}\right) }<\infty . \end{aligned}

The claim of this section is that the above definition is unambiguous, i.e. that it does not depend on the parameter $$\alpha$$. Moreover, the respective topologies defined for different choices of $$\alpha$$ are equivalent. This claim was made in , also for $$q<1$$, and obtaining a proof of this is one of the main motivations for this paper.

If we for the moment accept the unambiguity of Definition 4.1, then the proof of Theorem 1.6 is indeed short:

### Proof of Theorem 1.6

If $$s<0$$, then $$f\in F_{p,q}^s\left( \mathbb {D},\mathcal {X}\right)$$ if and only if $$\left( 1-|z|\right) ^{-s}f\in T_{p,q}\left( \mathcal {X}\right)$$. By Theorem 3.1, this happens if and only if $$f=gh$$, where $$g\in H^p$$ is an outer function and $$h=\left( 1-|z|\right) ^sH$$, with $$H\in T_{\infty ,q}\left( \mathcal {X}\right)$$. Since g is outer, $$h\in \mathcal {A}\left( \mathcal {X}\right)$$. Furthermore, $$h\in F_{\infty ,q}^s\left( \mathbb {D},\mathcal {X}\right)$$ by definition. $$\square$$

The remainder of this section is dedicated to the justification of Definition 4.1. In [23, 24], Yang and Yuan introduced the spaces $$F^{s,\tau }_{p,q}\left( \mathbb {R}^d\right)$$, $$s\in \mathbb {R}$$, $$\tau \ge 0$$, $$p\in \left( 0,\infty \right)$$, $$q\in \left( 0,\infty \right]$$. These include the standard Triebel–Lizorkin spaces: If $$0<p<\infty$$, then $$F^{s,0}_{p,q}\left( \mathbb {R}^d\right) =F^{s}_{p,q}\left( \mathbb {R}^d\right)$$, while $$F^{s,1/p}_{p,q}\left( \mathbb {R}^d\right) =F^{s}_{\infty ,q}\left( \mathbb {R}^d\right)$$. In contrast to $$F^{s}_{p,q}\left( \mathbb {R}^d\right)$$, the spaces $$F^{s,\tau }_{p,q}\left( \mathbb {R}^d\right)$$ are not always distinct for different choices of parameters. On the other hand, they include for example the spaces $$Q_\alpha \left( \mathbb {R}^d\right) = F^{\alpha ,1/2-\alpha /d}_{2,2}\left( \mathbb {R}^d\right)$$ introduced by Aulaskari et al. . This is in fact a motivation in . We chose to work with the more general scale of $$F^{s,\tau }_{p,q}$$-spaces, since this requires no additional effort.

A fundamental tool in the study of Triebel–Lizorkin spaces is the so-called Peetre maximal function: For $$a>0$$ and $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, we define
\begin{aligned} f_{r,a}^*\left( \zeta _x\right) =\sup _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_r\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+\frac{|\zeta _x-\zeta _y|}{1-r}\right) ^a},\quad r \in \left[ 0,1\right) ,\, \zeta _x\in \mathbb {T}. \end{aligned}
(6)

### Definition 4.2

Let $$\left( r_l\right) _{l\ge 0}$$ be a sequence such that $$0\le r_l<1$$ and $$2^l\left( 1-r_l\right) \approx 1$$. Furthermore, let $$0<p,q<\infty$$, $$s\in \mathbb {R}$$, $$\tau \ge 0$$ and $$a>\max \left\{ \frac{1}{p},\frac{1}{q}\right\}$$. Define then following (quasi-)norms for $$f\in \mathcal {A}\left( \mathcal {X}\right)$$:
\begin{aligned} \Vert f\Vert _1&{=} \sup _{I\in \mathcal {D}\left( \mathbb {T}\right) }\frac{1}{|I|^\tau }\left( \int _{\zeta _x\in I}\left[ \int _{\Gamma _I\left( \zeta _x\right) }\left( 1-|w|\right) ^{-2-sq} \Vert f\left( w\right) \Vert _\mathcal {X}^q\, dA\left( w\right) \right] ^{p/q} dx\right) ^{1/p}. \\ \Vert f\Vert _2&= \sup _{I\in \mathcal {D}\left( \mathbb {T}\right) }\frac{1}{|I|^\tau }\left( \int _{\zeta _x\in I}\left[ \int _{r=1-|I|}^1\left( 1-r\right) ^{-1-sq} \Vert f_r\left( \zeta _x\right) \Vert _\mathcal {X}^qdr\right] ^{p/q} dx\right) ^{1/p}. \\ \Vert f\Vert _3&= \sup _{I\in \mathcal {D}\left( \mathbb {T}\right) }\frac{1}{|I|^\tau }\left( \int _{\zeta _x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}^q\right] ^{p/q} dx\right) ^{1/p}. \\ \Vert f\Vert _4&= \sup _{I\in \mathcal {D}\left( \mathbb {T}\right) }\frac{1}{|I|^\tau }\left( \int _{\zeta _x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}f_{l,a}^*\left( \zeta _x\right) ^q\right] ^{p/q} dx\right) ^{1/p}. \\ \Vert f\Vert _5&= \sup _{I\in \mathcal {D}\left( \mathbb {T}\right) }\frac{1}{|I|^\tau }\left( \int _{\zeta _x\in I}\left[ \int _{r=1-|I|}^1\left( 1-r\right) ^{-1-sq}f_{r,a}^*\left( \zeta _x\right) ^q dr\right] ^{p/q} dx\right) ^{1/p}. \end{aligned}
If we wish to indicate the values of p, q, s and $$\tau$$, then we use the notation $$\Vert f|_{p,q}^{s,\tau }\Vert _k$$, $$1\le k\le 5$$. Similarly, $$\Vert f|\left( r_l\right) _{l\ge 0}\Vert _k$$, $$1\le k\le 5$$, indicates the choice of $$\left( r_l\right) _{l\ge 0}$$.

The following theorem is analogous to [13, Theorem 3.2]. Weaker theorems of the same form go back at least to Triebel .

### Theorem 4.3

The (quasi-)norms in Definition 4.2 are comparable for $$f\in \mathcal {A}\left( \mathcal {X}\right)$$.

Let $$\varphi \in \mathcal {S}$$ be a function such that for $$\xi \ge 0$$, $$\hat{\varphi }\left( \xi \right) =e^{-\xi }$$. With $$\left( r_l\right) _{l\ge 0}$$ as in Definition 4.2, let $$t_l= \log \left( \frac{1}{r_l}\right)$$, set $$\varphi _l\left( x\right) =\frac{1}{t_l}\varphi \left( \frac{x}{r_l}\right)$$, and let $$\Phi _l$$ denote the corresponding periodization. If $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, then $$\Phi _l*f=P_{r_l}*f$$, and so
\begin{aligned} \Vert f\Vert _3 = \sup _{I\in \mathcal {D}\left( \mathbb {T}\right) }\frac{1}{|I|^\tau }\left( \int _{\zeta _x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert \Phi _l*f\left( \zeta _x\right) \Vert _\mathcal {X}^q\right] ^{p/q}dx\right) ^{1/p}. \end{aligned}
This expression is a verbatim analogue of the defining (quasi-)norm for $$F_{p,q}^{s,\tau }\left( \mathbb {R}^d\right)$$. If $$s<0$$, then imposing finiteness of the above expression indeed gives us a space with the natural properties. However, for general $$s\in \mathbb {R}$$, such a definition would be severely flawed:

### Proposition 4.4

With notation as in Definition 4.2, and $$s\ge 0$$, if $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ and $$\Vert f\Vert _3<\infty$$, then $$f\equiv 0$$.

### Proof

We consider first the case where $$p=q$$. By interchanging orders of integration,
\begin{aligned} \Vert f\Vert _3 \ge \left( \sum _{l=0}^\infty 2^{slq}\int _{\zeta _x\in \mathbb {T}} \Vert f_{r_l}\left( \zeta _x\right) \Vert _\mathcal {X}^q\, dx\right) ^{1/q}. \end{aligned}
By subharmonicity, the function $$r\mapsto \int _{\zeta _x\in \mathbb {T}} \Vert f_{r}\left( \zeta _x\right) \Vert _\mathcal {X}^q\, dx$$ is increasing, and so the above right-hand side is infinite, unless $$f\equiv 0$$. The general case follows in the same way, with some simple modifications: If $$p>q$$, then we first apply Hölder’s inequality to the integral:
\begin{aligned} \int _{\zeta _x\in \mathbb {T}}\sum _{l=0}^\infty 2^{slq} \Vert f_{r_l}\left( \zeta _x\right) \Vert _\mathcal {X}^q\, dx \le \left( \int _{\zeta _x\in \mathbb {T}}\left[ \sum _{l=0}^\infty 2^{slq} \Vert f_{r_l}\left( \zeta _x\right) \Vert _\mathcal {X}^q\right] ^{p/q}\, dx\right) ^{q/p}. \end{aligned}
If $$p<q$$, then we instead use Hölder’s inequality on a partial sum:
\begin{aligned} \sum _{l=0}^N 2^{slp} \Vert f_{r_l}\left( \zeta _x\right) \Vert _\mathcal {X}^p\le \left( 1+N\right) ^{\frac{q-p}{q}}\left[ \sum _{l=0}^N 2^{slq} \Vert f_{r_l}\left( \zeta _x\right) \Vert _\mathcal {X}^q\right] ^{p/q}. \end{aligned}
Integrating the above inequality over $$\mathbb {T}$$, since $$s\ge 0$$, we may again argue by subharmonicity to conclude that
\begin{aligned} \left( 1+N\right) ^{p/q}\lesssim \int _{\zeta _x\in \mathbb {T}}\left[ \sum _{l=0}^N 2^{slq} \Vert f_{r_l}\left( \zeta _x\right) \Vert _\mathcal {X}^q\right] ^{p/q}\, dx. \end{aligned}
The result now follows by letting $$N\rightarrow \infty$$. $$\square$$
A related observation is that if $$s<0$$ and $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, then
\begin{aligned} \Vert f\left( w\right) \Vert _\mathcal {X}\lesssim \Vert f\Vert _1\left( 1-|w|\right) ^{s+\tau -1/p}, \end{aligned}
cf. Lemma 4.12. So if $$s+\tau -\frac{1}{p}>0$$ and $$\Vert f\Vert _1<\infty$$, then $$f\equiv 0$$ by the maximum principle for analytic functions. This motivates us to impose the restriction $$\tau \le \frac{1}{p}$$. The definition of $$F_{p,q}^{s,\tau }\left( \mathbb {D},\mathcal {X}\right)$$ is now inspired by the so-called lifting property $$D^\alpha :F_{p,q}^s\left( \mathbb {R}^d\right) \rightarrow F_{p,q}^{s-\alpha }\left( \mathbb {R}^d\right)$$, e.g. [19, Sect. 2.3.8].

### Definition 4.5

Let $$0<p<\infty$$, $$1\le q<\infty$$, $$s\in \mathbb {R}$$, $$0\le \tau \le \frac{1}{p}$$, and $$\alpha >s$$. We define the Triebel–Lizorkin type space $$F_{p,q}^{s,\tau }\left( \mathbb {D},\mathcal {X}\right)$$ as the space of functions $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ such that
\begin{aligned} \Vert f\Vert _{F_{p,q}^{s,\tau }\left( \mathbb {D},\mathcal {X}\right) }:= \Vert D^\alpha f|_{p,q}^{s-\alpha ,\tau }\Vert _1<\infty . \end{aligned}

Note that if $$0<p<\infty$$, then $$\Vert f|_{p,q}^{s,0}\Vert _1 = \Vert \left( 1-|z|\right) ^{-s} f\Vert _{T_{p,q}\left( \mathcal {X}\right) }$$ and $$\Vert f|_{q,q}^{s,1/q}\Vert _1\approx \Vert \left( 1-|z|\right) ^{-s} f\Vert _{T_{\infty ,q}\left( \mathcal {X}\right) }$$. Consequently $$F_{p,q}^{s,0}\left( \mathbb {D},\mathcal {X}\right) =F_{p,q}^{s}\left( \mathbb {D},\mathcal {X}\right)$$ and $$F_{q,q}^{s,1/q}\left( \mathbb {D},\mathcal {X}\right) =F_{\infty ,q}^{s}\left( \mathbb {D},\mathcal {X}\right)$$.

Definition 4.5 is justified by the following lemma:

### Lemma 4.6

Let $$0<p<\infty$$, $$1\le q<\infty$$, $$s\in \mathbb {R}$$, $$0\le \tau \le \frac{1}{p}$$, and $$\alpha >s$$. Then
\begin{aligned} \Vert D^\alpha f|_{p,q}^{s-\alpha ,\tau }\Vert _3\lesssim \Vert f|_{p,q}^{s,\tau }\Vert _4. \end{aligned}
(7)
By Proposition 4.4, the above lemma is trivial if $$s\ge 0$$. Assume therefore that $$s<0$$, and let $$\alpha >s$$. In particular (7) holds. Moreover, $$-\alpha >s-\alpha$$. By another application of Lemma 4.6, we obtain
\begin{aligned} \Vert f|_{p,q}^{s,\tau }\Vert _3= \Vert D^{-\alpha }\left( D^\alpha f\right) |_{p,q}^{s-\alpha -\left( -\alpha \right) ,\tau }\Vert _3\lesssim \Vert D^\alpha f|_{p,q}^{s-\alpha ,\tau }\Vert _4. \end{aligned}
Combined with Theorem 4.3, this yields that
\begin{aligned} \Vert f|_{p,q}^{s,\tau }\Vert _1\lesssim \Vert D^\alpha f|_{p,q}^{s-\alpha ,\tau }\Vert _1\lesssim \Vert f|_{p,q}^{s,\tau }\Vert _1,\quad f\in \mathcal {A}\left( \mathcal {X}\right) , \end{aligned}
whenever $$s<0$$ and $$s-\alpha < 0$$. This implies that if $$\alpha _1,\alpha _2 > s$$, then
\begin{aligned} \Vert D^{\alpha _1} f|_{p,q}^{s-\alpha _1,\tau }\Vert _1\approx \Vert D^{\alpha _2} f|_{p,q}^{s-\alpha _2,\tau }\Vert _1, \end{aligned}
and thus $$F_{p,q}^{s,\tau }\left( \mathbb {D},\mathcal {X}\right)$$ is well-defined, with topology independent of $$\alpha$$.

### 4.1 Proofs

In 4.1.1 we quantify the rigidity of analytic functions in a certain way (Lemma 4.7). We refer to this as “the first stability property”. This will imply that $$\Vert f\Vert _3$$ is essentially independent of $$\left( r_l\right) _{l\ge 0}$$, and also that $$\Vert f\Vert _1\lesssim \Vert f\Vert _2\lesssim \Vert f\Vert _3$$. The proof that $$\Vert f\Vert _3\lesssim \Vert f\Vert _1$$ is simpler. In 4.1.2, we deduce a similar stability property (the second) for the Peetre maximal function (Lemma 4.8). It follows that $$\Vert f\Vert _4\approx \Vert f\Vert _5$$. The estimate $$\Vert f\Vert _3\lesssim \Vert f\Vert _4$$ is trivial, since $$f_l\left( \zeta _x\right) \le f_{l,a}^*\left( \zeta _x\right)$$. We dedicate 4.1.3 to obtaining the reverse maximal control, the most involved part of this paper. In 4.1.4 we prove Lemma 4.6.

#### 4.1.1 The First Stability Property

Given $$I\in \mathcal {D}\left( \mathbb {T}\right)$$, we use the notation $$I_n=I+n|I|$$, for $$1-\frac{1}{2|I|}\le n\le \frac{1}{2|I|}$$. For other $$n\in \mathbb {Z}$$, we let $$I_n$$ be the empty set. Furthermore, we set $$\mathcal {I}_L=\cup _{|n|\le L}I_n$$.

### Lemma 4.7

Let $$\mathcal {X}$$ be a Banach space, $$\alpha \ge 0$$ and $$c_1,c_2,c_3,c_4>0$$. Then there exists constants $$K>0$$ and $$L\in \mathbb {N}$$ with the following property:

Let $$I\in \mathcal {D}\left( \mathbb {T}\right)$$. If $$r,r'\in \left[ 0,1\right)$$ satisfy
\begin{aligned} \left( 1-r\right) \le c_1\left( r'-r\right) \le c_2\left( 1-\frac{r}{r'}\right) \le c_3\left( 1-r\right) \le c_4 |I|, \end{aligned}
i.e.
\begin{aligned} \left( r'-r\right) \approx \left( 1-\frac{r}{r'}\right) \approx \left( 1-r\right) \lesssim |I|, \end{aligned}
and $$\zeta _x'\in \mathbb {T}$$ satisfies $$|\zeta _x-\zeta _x'|\le \alpha \left( 1-r\right)$$, where $$\zeta _x\in I$$, then for all $$\delta >0$$, for all $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, it holds that
\begin{aligned} \Vert f_r\left( \zeta _x'\right) \Vert _\mathcal {X}^\delta \le K \left( M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{r'}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) {+}\sum _{|n|> L}\frac{1}{\left( |n|{-}L\right) ^2|I|}\int _{I_n} \Vert f_{r'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy\right) . \end{aligned}
(8)

### Proof

By subharmonicity it holds that for any $$L\in \mathbb {N}$$
\begin{aligned} \Vert f_{r}\left( \zeta _x'\right) \Vert _\mathcal {X}^{\delta }\le {}&\int _\mathbb {T}P_{\frac{r}{r'}}\left( \zeta _{x'-y}\right) \Vert f_{r'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy \\ = {}&\int _{\mathcal {I}_L} P_{\frac{r}{r'}}\left( \zeta _{x'-y}\right) \Vert f_{r'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy\\&+ \sum _{|n|> L}\int _{I_n} P_{\frac{r}{r'}}\left( \zeta _{x'-y}\right) \Vert f_{r'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy. \end{aligned}
By the non-tangential maximal control of Poisson extensions,
\begin{aligned} \int _{\mathcal {I}_L} P_{\frac{r}{r'}}\left( \zeta _{x'-y}\right) \Vert f_{r'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy=\mathcal {P}\left[ \mathbb {1}_{\mathcal {I}_L} \Vert f_{r'}\Vert _\mathcal {X}^\delta \right] \left( \zeta _x'\right) \lesssim M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{r'}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) . \end{aligned}
Provided that L is sufficiently big, it follows from (4) that
\begin{aligned} P_{\frac{r_l}{r_l'}}\left( \zeta _{x'-y}\right) \lesssim \frac{1}{\left( |n|-L\right) ^2|I|},\quad \zeta _y\in I_n,|n|>L. \end{aligned}
$$\square$$
$$\Vert f\Vert _3$$is independent of$$\left( r_l\right) _{l\ge 0}$$: Let $$\left( r_l\right) _{l\ge 0}$$ and $$\left( r_l'\right) _{l\ge 0}$$ be sequences in $$\left[ 0,1\right)$$ such that
\begin{aligned} 2^{-l}c\le 1-r_l\le 2^{-l}C,\quad \text {and}\quad 2^{-l}c'\le 1-r_l'\le 2^{-l}C'. \end{aligned}
Chose $$M\in \mathbb {N}$$ such that $$2^{M-1}c\ge C'$$. For any $$l\in \mathbb {N}_0$$ it then holds that $$r_l< r_{l+M}'$$, and $$2^{-1-l}c\le r_{l+M}'-r_l\le 2^{-l}C$$. If we want to prove that $$\Vert f|\left( r_l\right) _{l\ge 0}\Vert _1\lesssim \Vert f|\left( r_l'\right) _{l\ge 0}\Vert _1$$, then by a shift of index we may assume that $$M=0$$. In this case our sequence has the asymptotic behavior
\begin{aligned} \left( r_l'-r_l\right) \approx \left( 1-\frac{r_l}{r_l'}\right) \approx \left( 1-r_l\right) \approx 2^{-l}. \end{aligned}
(9)
Let $$\delta <\min \left\{ p,q\right\}$$ and $$I\in \mathcal {D}\left( \mathbb {T}\right)$$. It follows from Lemma 4.7 (with $$\alpha =0$$) that
\begin{aligned} \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}^{\delta }\lesssim \underbrace{M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{l'}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) }_{=:A_l} +\sum _{|n|>L}\frac{1}{\left( |n|-L\right) ^2}\underbrace{\frac{1}{|I|}\int _{I_n} \Vert f_{l'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy}_{=:B_{l,n}}, \end{aligned}
for $$l\ge \text {rk}\left( I\right)$$ and $$x\in I$$. We now exploit the fact that $$p/q=\delta /q\cdot p/\delta$$. By Minkowski’s inequality,
\begin{aligned}&\left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}^{q/\delta }\right] ^{p/q} dx\right) ^{\delta /p} \lesssim {} \left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}A_l^{q/\delta }\right] ^{p/q} dx\right) ^{\delta /p} \nonumber \\&\quad + \sum _{|n|>L}\frac{1}{\left( |n|-L\right) ^2}\left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}B_{l,n}^{q/\delta }\right] ^{p/q} dx\right) ^{\delta /p}. \end{aligned}
(10)
By Theorem 2.1,
\begin{aligned} \left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}A_l^{q/\delta }\right] ^{p/q} dx\right) ^{\delta /p}&= \left\| \left( \sum _{l=\text {rk}\left( I\right) }^\infty \left( M\left( 2^{sl\delta }\mathbb {1}_{\mathcal {I}_L}\Vert f_{l'} \Vert _\mathcal {X}^\delta \right) \right) ^{q/\delta }\right) ^{\delta /q}\right\| _{L^{p/\delta }} \\&{\lesssim } \left\| \left( \sum _{l{=}\text {rk}\left( I\right) }^\infty 2^{slq}\mathbb {1}_{\mathcal {I}_L}\Vert f_{l'} \Vert _\mathcal {X}^q\right) ^{1/q}\right\| _{L^p}^\delta \\&\lesssim \left( \sum _{|n|\le L} \int _{I_n} \left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert f_{l'}\left( \zeta _y\right) \Vert _\mathcal {X}^q\right] ^{p/q} dy\right) ^{\delta /p} \\&\le |I|^{\delta \tau } \Vert f|\left( r_l'\right) _{l\ge 0}\Vert _{3}^{\delta }. \end{aligned}
By Minkowski’s and Jensen’s inequalities
\begin{aligned}&\left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}B_{l,n}^{q/\delta }\right] ^{p/q}dx\right) ^{\delta /p} \\&\quad = \left( \int _I\left( \left[ \sum _{l=\text {rk}\left( I\right) }^\infty \left( \frac{2^{sl\delta }}{|I|}\int _{I_n} \Vert f_{r_l'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy \right) ^{q/\delta }\right] ^{\delta /q}\right) ^{p/\delta }dx\right) ^{\delta /p} \\&\quad \le \left( \int _I\left( \frac{1}{|I|}\int _{I_n}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert f_{r_l'}\left( \zeta _y\right) \Vert _\mathcal {X}^q \right] ^{\delta /q}dy\right) ^{p/\delta }dx\right) ^{\delta /p} \\&\quad \le \left( \int _I\frac{1}{|I|}\int _{I_n}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert f_{r_l'}\left( \zeta _y\right) \Vert _\mathcal {X}^q \right] ^{p/q}dy\, dx\right) ^{\delta /p} \\&\quad \le |I|^{\delta \tau } \Vert f|\left( r_l'\right) _{l\ge 0}\Vert _3^\delta . \end{aligned}
Since $$\sum _{|n|>L}\frac{1}{\left( |n|-L\right) ^2}<\infty$$, this proves that $$\Vert f|\left( r_l\right) _{l\ge 0}\Vert _\lesssim \Vert f|\left( r_l'\right) _{l\ge 0}\Vert _3$$. $$\square$$
$$\Vert f\Vert _1\lesssim \Vert f\Vert _2$$: Let $$I\in \mathcal {D}\left( \mathbb {T}\right)$$. Obviously,
\begin{aligned}&\int _{\Gamma _I\left( x\right) }\left( 1-|w|\right) ^{-2-sq} \Vert f\left( w\right) \Vert _\mathcal {X}^q\, dA\left( w\right) \\&\quad \le \int _{r=1-|I|}^1\left( 1-r\right) ^{-2-sq}\int _{|\zeta _x-\zeta _y|\le \theta _r} \Vert f_r\left( \zeta _y\right) \Vert _\mathcal {X}^q\, dy\, dr, \end{aligned}
where $$\theta _r\approx 1-r$$. Let $$r'=\frac{1+r}{2}$$. By Lemma 4.7,
\begin{aligned} \Vert f_r\left( \zeta _x'\right) \Vert _\mathcal {X}^\delta \lesssim M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{r'}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) +\sum _{|n|> L}\frac{1}{\left( |n|-L\right) ^2}\frac{1}{|I|}\int _{I_n} \Vert f_{r'}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy, \end{aligned}
for $$|\zeta _x'-\zeta _x|\le \theta _r$$. It follows that
\begin{aligned} \int _{|\zeta _x-\zeta _y|\le \theta _r} \Vert f_r\left( \zeta _y\right) \Vert _\mathcal {X}^q\, dy \lesssim \left( 1-r\right) \left[ A_{r'}+\sum _{|n|> L}\frac{1}{\left( |n|-L\right) ^2}B_{r',n}\right] ^{q/\delta }, \end{aligned}
where
\begin{aligned} A_r:=M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{r}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) ,\quad \text {and}\quad B_{r,n}:=\frac{1}{|I|}\int _{I_n} \Vert f_{r}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy. \end{aligned}
By Minkowski’s inequality, and the change of variables $$r\mapsto r'$$,
\begin{aligned}&\left( \int _I\left[ \int _{\Gamma _I\left( \zeta _x\right) }\left( 1-|w|\right) ^{-2-sq} \Vert f\left( w\right) \Vert _\mathcal {X}^q\, dA\left( w\right) \right] ^{p/q} dx\right) ^{\delta /p} \\&\quad \lesssim {} \left( \int _I\left[ \int _{r=1-\frac{1}{2|I|}} \left( 1-r\right) ^{-1-sq} A_r^{q/\delta }dr\right] ^{p/q} dx\right) ^{\delta /p}\\&\qquad \qquad + \sum _{|n|>L}\frac{1}{\left( |n|-L\right) ^2}\left( \int _I\left[ \int _{r=1-\frac{|I|}{2}} \left( 1-r\right) ^{-1-sq} B_{r,n}^{q/\delta }dr\right] ^{p/q} dx\right) ^{\delta /p}. \end{aligned}
To complete the proof, we now proceed as from (10), with the obvious modifications. $$\square$$
$$\Vert f\Vert _2\lesssim \Vert f\Vert _3$$: It will prove convenient to work with $$r_l=1-2^{-l}$$. For any $$l\in N_0$$ and $$r\in \left[ r_{l},r_{l+1}\right]$$ it holds that
\begin{aligned} \left( r_{l+2}-r\right) \approx \left( 1-\frac{r}{r_{l+2}}\right) \approx \left( 1-r\right) \approx 2^{-l}. \end{aligned}
By Lemma 4.7, it follows that
\begin{aligned} \Vert f_r\left( \zeta _x\right) \Vert _\mathcal {X}^\delta \lesssim M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{r_{l+2}}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) + \sum _{|n|> L}\frac{1}{\left( |n|-L\right) ^2}\frac{1}{|I|}\int _{I_n} \Vert f_{r_{l+2}}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy, \end{aligned}
for $$I\in \mathcal {D}\left( \mathbb {T}\right)$$, $$x\in I$$, $$l\ge \text {rk}\left( I\right)$$ and $$r\in \left[ r_{l},r_{l+1}\right]$$. Since $$\int _{r_{l+1}}^{r_{l+2}} \left( 1-r\right) ^{-1-sq}\, dr\approx 2^{slq}$$, it follows that
\begin{aligned}&\left( \int _I\left[ \int _{r=1-|I|}\left( 1-r\right) ^{-1-sq} \Vert f_r\left( \zeta _x\right) \Vert _\mathcal {X}^q\, dr\right] ^{p/q} dx\right) ^{\delta /p}\\&\quad \lesssim {} \left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) +2} 2^{slq} A_l^{q/\delta }\right] ^{p/q} dx\right) ^{\delta /p}\\&\qquad \qquad + \sum _{|n|>L}\frac{1}{\left( |n|-L\right) ^2}\left( \int _I\left[ \sum _{l=\text {rk}\left( I\right) +2}2^{slq} B_{l,n}^{q/\delta }\right] ^{p/q} dx\right) ^{\delta /p}, \end{aligned}
where
\begin{aligned} A_l:=M\left( \mathbb {1}_{\mathcal {I}_L} \Vert f_{r_{l+2}}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) , \quad \text {and}\quad B_{l,n}:=\frac{1}{|I|}\int _{I_n} \Vert f_{r_{l+2}}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy. \end{aligned}
Once again, we proceed as from (10). $$\square$$
$$\Vert f\Vert _3\lesssim \Vert f\Vert _1$$: We work with the sequence given by $$r_l=1-2^{-1-l}$$. Given $$x\in \mathcal {I}\in \mathcal {D}\left( \mathbb {T}\right)$$, for each $$l\ge \text {rk}\left( I\right)$$ there exists a ball $$B_l=B\left( r_l\zeta _x,d_l\right) \subset \Gamma _I\left( x\right)$$. Moreover, these balls may be chosen so that they are disjoint and $$d_l\gtrsim 2^{-l}$$. By subharmonicity,
\begin{aligned} \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq} \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}^q\lesssim & {} \sum _{l=\text {rk}\left( I\right) }^\infty 2^{\left( 2+sq\right) l}\int _{B_l} \Vert f\left( w\right) \Vert _\mathcal {X}^q\, dA\left( w\right) \\\lesssim & {} \int _{\Gamma _I\left( \zeta _x\right) }\left( 1-|w|\right) ^{-2-sq} \Vert f\left( w\right) \Vert _\mathcal {X}^q\, dA\left( w\right) . \end{aligned}
The statement follows. $$\square$$

### Lemma 4.8

If $$x\in \mathbb {T}$$ and $$\left( r_l\right) _{l\ge 0},\left( r_l'\right) _{l\ge 0}\subset \left[ 0,1\right)$$ are sequences such that (9) holds, then
\begin{aligned} f^*_{l,a}\left( \zeta _x\right) \lesssim f^*_{l',a}\left( \zeta _x\right) . \end{aligned}

### Proof

The proof relies on an idea that will be useful several times in this section. Chose $$\varphi \in \mathcal {S}$$ such that $$\hat{\varphi }\left( \xi \right) =e^{-2\pi \xi }$$, $$\xi \ge -1$$. For $$t>0$$, let $$\varphi _t\left( x\right) =\frac{1}{t}\varphi \left( \frac{x}{t}\right)$$, and $$\Phi _r\left( \zeta _x\right) =\sum _{k\in \mathbb {Z}}\varphi _t\left( x-k\right)$$, where $$r=e^{-2\pi t}$$. For $$f\in \mathcal {A}\left( \mathcal {X}\right)$$ it holds that $$f_r=P_r*f=\Phi _r*f$$. Note that by (5), for any integer $$N\ge 2$$,
\begin{aligned} |\Phi _r\left( \zeta _x\right) |\lesssim \frac{1}{1-r}\frac{1}{\left( 1+\frac{|\zeta _x-1|}{1-r}\right) ^N},\quad r\in \left[ 0,1\right) ,\, \zeta _x\in \mathbb {T}. \end{aligned}
(11)
If we replaced $$\Phi _r$$ with the ordinary Poisson kernel, then the above inequality would hold only for $$N=2$$.
Now note that
\begin{aligned} f_{l}\left( \zeta _y\right) =\int _{\zeta _z\in \mathbb {T}}\Phi _\frac{r_l}{r_l'}\left( \zeta _{y-z}\right) f_{l'}\left( \zeta _z\right) \, dz. \end{aligned}
By the triangle inequality, together with (6) and the elementary inequality
\begin{aligned} 1+b|\zeta _x-\zeta _y|\le \left( 1+b|\zeta _x-\zeta _z|\right) \left( 1+b|\zeta _z-\zeta _y|\right) ,\quad x,y,z\in \mathbb {R},b>0, \end{aligned}
(12)
we obtain
\begin{aligned} \frac{ \Vert f_{l}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+\frac{|\zeta _x-\zeta _y|}{1-r_l}\right) ^a}\lesssim & {} \int _{\zeta _z\in \mathbb {T}}\frac{\left| \Phi _\frac{r_l}{r_l'}\left( y-z\right) \right| \Vert f_{l'}\left( z\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^a}\, dz \\\lesssim & {} f^*_{l',a}\left( \zeta _x\right) \int _{\zeta _z\in \mathbb {T}} \left| \Phi _\frac{r_l}{r_l'}\left( \zeta _{y-z}\right) \right| \frac{\left( 1+2^l|\zeta _x-\zeta _z|\right) ^a}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^a}\, dz \\\le & {} f^*_{l',a}\left( \zeta _x\right) \int _{\zeta _z\in \mathbb {T}}\left| \Phi _\frac{r_l}{r_l'}\left( \zeta _{y-z}\right) \right| \left( 1+2^l|\zeta _y-\zeta _z|\right) ^a\, dz. \end{aligned}
By (9) and (11), we see that if $$N>a+1$$, then
\begin{aligned} \int _{\zeta _z\in \mathbb {T}} \left| \Phi _\frac{r_l}{r_l'}\left( \zeta _{y-z}\right) \right| \left( 1+2^l|\zeta _y-\zeta _z|\right) ^a\, dz\lesssim 1. \end{aligned}
The statement follows. $$\square$$

$$\Vert f\Vert _4$$is independent of$$\left( r_l\right) _{l\ge 0}$$: Once again, we may assume that (9) holds. The statement follows immediately from Lemma 4.8. $$\square$$

$$\Vert f\Vert _4\approx \Vert f\Vert _5$$: We may chose $$\left( r_l\right) _{l\ge 0}$$ so that $$r_l=1-2^{-l}$$. Note that
\begin{aligned} \int _{r=1-|I|}^1\left( 1-r\right) ^{-1-sq}f^*_{r,a}\left( \zeta _x\right) ^q\, dr=\sum _{l=\text {rk}\left( I\right) }^\infty \int _{r=r_l}^{r_{l+1}}\left( 1-r\right) ^{-1-sq}f^*_{r,a}\left( \zeta _x\right) ^q\, dr. \end{aligned}
For an arbitrary sequence $$\left( r_l'\right) _{l\ge 0}$$ such that $$r_{l+2}'\in \left[ r_{l},r_{l+1}\right]$$, (9) holds. By an application of Lemma 4.8, we obtain
\begin{aligned} \int _{r=r_l}^{r_{l+1}}\left( 1-r\right) ^{-1-sq}f^*_{r,a}\left( \zeta _x\right) ^q\, dr&\lesssim \int _{r=r_l}^{r_{l+1}}\left( 1-r\right) ^{-1-sq}f^*_{l+2,a}\left( \zeta _x\right) ^q\, dr \\&\lesssim 2^{slq}f^*_{l+2,a}\left( \zeta _x\right) . \end{aligned}
It follows that $$\Vert f\Vert _5\lesssim \Vert f\Vert _4$$. The reverse estimate is similar. $$\square$$

#### 4.1.3 Reverse Maximal Control

It will be convenient to work with the sequence given by $$r_l=e^{-2\pi 2^{-l}}$$. Let $$\varphi _t$$ and $$\Phi _r$$ be as in the proof of Lemma 4.8, and set $$\varphi _m=\varphi _{2^{-m}}$$ and $$\Phi _m=\Phi _{r_m}$$. Then $$\Phi _m$$ is the 1-periodization of $$\varphi _m$$. Now choose $$\left\{ W_m\right\} _{m=1}^\infty \subset \mathcal {S}$$ such that $$\text {supp }\hat{W}_1\subset \left[ -\frac{1}{2},2\right]$$, $$\text {supp }\hat{W}_2\subset \left[ 1,4\right]$$, $$\hat{W}_m\left( \xi \right) =\hat{W}_{m-1}\left( \xi /2\right)$$ for $$m\ge 3$$, and $$\sum _{m=1}^\infty \hat{W}_m\left( \xi \right) =1$$ for $$\xi \ge 0$$. Define $$\left\{ \psi _m\right\} _{m=1}^\infty \subset \mathcal {S}$$ by
\begin{aligned} \hat{\psi }_m\left( \xi \right) =\frac{\hat{W}_m\left( \xi \right) }{\hat{\varphi }_m\left( \xi \right) },\quad m\in \mathbb {N},\xi \in \mathbb {R}. \end{aligned}
Then
\begin{aligned} \sum _{m=1}^\infty \hat{\psi }_m\left( \xi \right) \hat{\varphi }_m\left( \xi \right) =1\qquad \text {for }\xi \ge 0. \end{aligned}
Furthermore, for any $$l\in \mathbb {N}_0$$ we have that
\begin{aligned} \sum _{m=1}^\infty \hat{\psi }_m\left( 2^{-l}\xi \right) \hat{\varphi }_m\left( 2^{-l}\xi \right) = \sum _{m=1}^\infty \hat{\psi }_m\left( 2^{-l} \xi \right) \hat{\varphi }_{m+l}\left( \xi \right) =1,\quad \xi \ge 0. \end{aligned}
Define $$\lambda _{m,l}\in \mathcal {S}$$ by $$\hat{\lambda }_{m,l}\left( \xi \right) =\hat{\psi }_m\left( 2^{-l}\xi \right)$$, $$\xi \in \mathbb {R}$$, and let $$\Lambda _{m,l}$$ denote the corresponding 1-periodization. We thus obtain
\begin{aligned} \sum _{m=1}^\infty \hat{\Lambda }_{m,l}\left( n\right) \hat{\Phi }_{m+l}\left( n\right) = 1,\quad l,n\in \mathbb {N}_0. \end{aligned}
(13)
This is a so-called Calderon reproducing type formula for analytic functions.

We will need the following technical lemma:

### Lemma 4.9

If $$N\in \mathbb {N}$$, then there exists a $$K>0$$ such that
\begin{aligned} \sup _{\zeta _y\in \mathbb {T}} \left( 1+2^l|\zeta _y-1|\right) ^N |\Phi _{l}*\Lambda _{m,l}\left( \zeta _y\right) |\le K 2^l2^{-mN} \end{aligned}
for all $$m\in \mathbb {N}$$ and $$l\in \mathbb {N}_0$$.

### Proof

By (5), it suffices to show that
\begin{aligned} \sup _{y\in \mathbb {R}} |y^N\left( \varphi _{l}*\lambda _{m,l}\right) \left( y\right) |\le K 2^l2^{-\left( l+m\right) N} \end{aligned}
We consider the case $$m\ge 2$$. By elementary properties of the Fourier transform,
\begin{aligned} y^N\left( \varphi _{l}*\lambda _{m,l}\right) \left( y\right)&= \frac{1}{\left( 2\pi i\right) ^N}\int \left[ \left( \frac{d}{d\xi }\right) ^N\left( \hat{\varphi }_{l}\left( \xi \right) \hat{\lambda }_{m,l}\left( \xi \right) \right) \right] e^{2\pi i\xi y}\, d\xi \end{aligned}
Since $$\hat{\varphi }_{l}\left( \xi \right) \hat{\lambda }_{m,l}\left( \xi \right) =e^{-2\pi \left( 2^{-l}-2^{-l-m}\right) \xi } \hat{W}_2\left( 2^{2-l-m}\xi \right)$$, we may use Leibniz’s rule together with the support of $$\hat{W}_2$$ to obtain
\begin{aligned} |y^N\left( \varphi _{l}*\lambda _{m,l}\right) \left( y\right) |&\lesssim \sum _{k=0}^N2^{-l\left( N-1\right) }2^{-m\left( N-k-1\right) }e^{-2\pi \left( 2^m-1\right) }. \end{aligned}
This proves the statement for $$m\ge 2$$, since for any $$k\in \mathbb {N}_0$$,
\begin{aligned} 2^{-m\left( N-k-1\right) }e^{-2\pi \left( 2^m-1\right) }\lesssim 2^{-mN}. \end{aligned}
The case $$m=1$$ is similar. $$\square$$

The proof that $$\Vert \cdot \Vert _2\lesssim \Vert \cdot \Vert _1$$ is based on the following lemma:

### Lemma 4.10

Let $$N\in \mathbb {N}$$, $$0<a\le N$$, $$\delta >0$$ and $$r_l=e^{-2\pi 2^l}$$. Then there exists $$K=K\left( a,N,\delta \right) >0$$ such that
\begin{aligned} f_{l,a}^*\left( \zeta _x\right) ^\delta \le K \sum _{m=1}^\infty 2^{-mN\delta } 2^{m+l}\int _{y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{a\delta }}\, dy \end{aligned}
for $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, $$l\in \mathbb {N}_0$$ and $$x\in \mathbb {T}$$.

### Proof

By (13),
\begin{aligned} f_l\left( \zeta _x\right)&=\Phi _l*f\left( \zeta _x\right) \\&= \sum _{m=1}^\infty \Phi _l*\Lambda _{m,l}*\Phi _{l+m}*f\left( \zeta _x\right) \\&= \int _{\mathbb {T}} \left( \Phi _l*\Lambda _{m,l}\right) \left( \zeta _{x-y}\right) \left( \Phi _{l+m}*f\right) \left( \zeta _y\right) \, dy\\&= \int _{\mathbb {T}} \left( \Phi _l*\Lambda _{m,l}\right) \left( \zeta _{x-y}\right) \left( 1+2^l|\zeta _x-\zeta _y|\right) ^N\frac{\left( \Phi _{l+m}*f\right) \left( \zeta _y\right) }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^N}\, dy. \end{aligned}
By the triangle inequality and Lemma 4.9, we have that
\begin{aligned}&\Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}\nonumber \\&\quad \le \sum _{m=1}^\infty \sup _{\zeta _y\in \mathbb {T}} \left| \Phi _{r_l^\rho }*\Lambda _{m,l,\rho }\left( \zeta _y\right) \left( 1+2^l|\zeta _y-1|\right) ^N\right| \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^N}\, dy \nonumber \\&\quad \lesssim \sum _{m=1}^\infty 2^{-mN}2^l \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^N}\, dy. \end{aligned}
(14)
If $$\delta >1$$, then we proceed as follows: Since N is arbitrary, clearly
\begin{aligned} \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}\lesssim \sum _{m=1}^\infty 2^{-m\left( N+1\right) }2^l \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{N+1}}\, dy. \end{aligned}
Applying Hölder’s inequality twice we obtain
\begin{aligned} \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}^\delta&\lesssim \sum _{m=1}^\infty 2^{-mN\delta }2^{l\delta } \left( \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{N+1}}\, dy\right) ^\delta \\&\lesssim \sum _{m=1}^\infty 2^{-mN\delta }2^{l} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{N\delta }}\, dy. \end{aligned}
Now use that $$a\le N$$, divide by $$\left( 1+2^l|\zeta _z-\zeta _x|\right) ^{a\delta }$$ and use (12) with $$b=2^l$$ get that
\begin{aligned} \frac{ \Vert f_l\left( \zeta _x\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _z-\zeta _x|\right) ^{a\delta }} \lesssim \sum _{m=1}^\infty 2^{-mN\delta }2^{l} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _z-\zeta _y|\right) ^{a\delta }}\, dy. \end{aligned}
This completes the proof for $$\delta >1$$.
If $$\delta \le 1$$, then we instead do the following: By a shift of the index l in (14) we see that for each $$k\in \mathbb {N}_0$$,
\begin{aligned} 2^{-kN} \Vert f_{k+l}\left( \zeta _x\right) \Vert _\mathcal {X}&\lesssim \sum _{m=1}^\infty 2^{-\left( m+k\right) N}2^{l+k} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{k+l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^{l+k}|\zeta _x-\zeta _y|\right) ^N}\, dy \\&\le \sum _{m=1}^\infty 2^{-\left( m+k\right) N}2^{m+l+k} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{k+l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^{l}|\zeta _x-\zeta _y|\right) ^N}\, dy \\&= \sum _{m=1+k}^\infty 2^{-mN}2^{m+l} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^{l}|\zeta _x-\zeta _y|\right) ^N}\, dy \\&\le \sum _{m=1}^\infty 2^{-mN}2^{m+l} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^{l}|\zeta _x-\zeta _y|\right) ^N}\, dy. \end{aligned}
Using (12) again, we have that
\begin{aligned} \frac{2^{-kN} \Vert f_{l+k}\left( \zeta _x\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _z-\zeta _x|\right) ^a} \lesssim \sum _{m=1}^\infty 2^{-mN}2^{m+l} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^{l}|\zeta _z-\zeta _y|\right) ^a}\, dy. \end{aligned}
(15)
Introduce the maximal function
\begin{aligned} M_lf\left( \zeta _x\right) =\sup _{k\in \mathbb {N}_0}\sup _{\zeta _y\in \mathbb {T}}\frac{2^{-kN} \Vert f_{l+k}\left( \zeta _y\right) \Vert _\mathcal {X}}{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^a},\quad \zeta _x\in \mathbb {T}. \end{aligned}
By (15),
\begin{aligned} M_lf\left( \zeta _x\right) \lesssim \sum _{m=1}^\infty 2^{-mN\delta }2^{m+l}M_lf\left( \zeta _x\right) ^{1-\delta } \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^{l}|\zeta _z-\zeta _y|\right) ^{a\delta }}\, dy. \end{aligned}
By Lemma 4.11 below, this implies that
\begin{aligned} M_{l,\rho }\left( \zeta _x\right) ^\delta \lesssim \sum _{m=1}^\infty 2^{-mN\delta }2^{m+l} \int _{\zeta _y\in \mathbb {T}}\frac{ \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^{l}|\zeta _z-\zeta _y|\right) ^{a\delta }}\, dy, \end{aligned}
whenever the right-hand side i finite. This completes the proof. $$\square$$

### Lemma 4.11

If
\begin{aligned} \sum _{m=1}^\infty 2^{-mN\delta } 2^{m+l}\int _{y\in \mathbb {T}}\frac{ \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{a\delta }}\, dy <\infty , \end{aligned}
then $$M_lf\left( \zeta _x\right) <\infty$$.

### Proof

Since l is a fixed number and $$\mathbb {T}$$ is compact we may equivalently show that if
\begin{aligned} \sum _{m=1}^\infty 2^{m\left( 1-N\delta \right) }\int _{\zeta _y\in \mathbb {T}} \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy <\infty , \end{aligned}
(16)
then
\begin{aligned} \sup _{m\in \mathbb {N}_0}\sup _{\zeta _y\in \mathbb {T}} 2^{-mN\delta } \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta <\infty . \end{aligned}
By subharmonicity
\begin{aligned} \Vert f_{m+l}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta&\le \int \Vert f_{m+l+1}\Vert _\mathcal {X}^\delta P_{\frac{r_{l+m}}{r_{l+m+1}}}\left( \zeta _y\right) \, dy \\&\lesssim 2^{m+l} \int \Vert f_{m+l+1}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy \lesssim 2^l2^{mN\delta }, \end{aligned}
where the last inequality holds whenever (16) converges. $$\square$$
We now prove that $$\Vert \cdot \Vert _2\lesssim \Vert \cdot \Vert _1$$. Fix $$\delta \in \left( \frac{1}{a},\min \left\{ p,q\right\} \right)$$, and chose $$N\in \mathbb {N}$$ such that $$N>a$$ and $$1-N\delta -sq<0$$. Given $$I\in \mathcal {D}\left( \mathbb {T}\right)$$, define $$I_n=I+n|I|$$, where $$1-\frac{1}{2|I|}\le n\le \frac{1}{2|I|}$$, and $$3I=\cup _{|n|\le 1}I_n$$. By Lemma 4.10,
\begin{aligned} f_{l,a}^*\left( \zeta _x\right) ^\delta \lesssim {}&\sum _{m=1}^\infty 2^{-mN\delta } 2^{m+l}\int _{\mathbb {T}}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{a\delta }}\, dy. \end{aligned}
We treat the right-hand side as in the proof of Lemma 4.7.
\begin{aligned} f_{l,a}^*\left( \zeta _x\right) ^\delta \lesssim {}&\sum _{m=1}^\infty 2^{-mN\delta } 2^{m+l}\int _{3I}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{a\delta }}\, dy \\&\qquad + \sum _{|n|\ge 2}\sum _{m=1}^\infty 2^{-mN\delta } 2^{m+l}\int _{I_n}\frac{ \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta }{\left( 1+2^l|\zeta _x-\zeta _y|\right) ^{a\delta }}\, dy \\ \lesssim {}&\underbrace{\sum _{m=1}^\infty 2^{-mN\delta } 2^{m}M\left( \mathbb {1}_{3I} \Vert f_{m+l}\Vert _\mathcal {X}^\delta \right) \left( x\right) }_{=:A_l} \\&\qquad + \sum _{|n|\ge 2}\underbrace{\sum _{m=1}^\infty 2^{-mN\delta } 2^{m+l}\frac{1}{2^{la\delta }|n|^{a\delta }|I|^{a\delta }}\int _{I_n} \Vert f_{l+m}\left( \zeta _y\right) \Vert _\mathcal {X}^\delta \, dy}_{=:B_{l,n}}. \end{aligned}
By successive applications of Minkowski’s inequality,
\begin{aligned}&\left( \int _{\zeta _x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}f_{l,a}^*\left( \zeta _x\right) ^q\right] ^{p/q}dx\right) ^{\delta /p}\\&\quad \lesssim \left( \int _{x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}A_l^{q/\delta }\right] ^{p/q}dx\right) ^{\delta /p} \\&\qquad \qquad + \sum _{|n|\ge 2}\left( \int _{\zeta _x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}B_{l,n}^{q/\delta }\right] ^{p/q}dx\right) ^{\delta /p}. \end{aligned}
By Jensen’s inequality and rearrangement of terms,
\begin{aligned} \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}A_l^{q/\delta }&\lesssim \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}\sum _{m=1}^\infty 2^{m\left( 1-N\delta \right) } M\left( \mathbb {1}_{3I} \Vert f_{m+l}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) ^{q/\delta } \\&= \sum _{l=\text {rk}\left( I\right) +1}^\infty 2^{slq}\sum _{m=1}^{l-\text {rk}\left( I\right) } 2^{m\left( 1-N\delta -sq\right) } M\left( \mathbb {1}_{3I} \Vert f_{l}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) ^{q/\delta } \\&\lesssim \sum _{l=\text {rk}\left( I\right) +1}^\infty 2^{slq} M\left( \mathbb {1}_{3I} \Vert f_{l}\Vert _\mathcal {X}^\delta \right) \left( \zeta _x\right) ^{q/\delta }. \end{aligned}
It follows from Theorem 2.1 that
\begin{aligned} \left( \int _{x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}A_l^{q/\delta }\right] ^{p/q}dx\right) ^{\delta /p}\lesssim \Vert f\Vert _3|I|^{\tau \delta }. \end{aligned}
The corresponding estimate for $$B_{l,n}$$,
\begin{aligned} \sum _{|n|\ge 2}\left( \int _{\zeta _x\in I}\left[ \sum _{l=\text {rk}\left( I\right) }^\infty 2^{slq}B_{l,n}^{q/\delta }\right] ^{p/q}dx\right) ^{\delta /p}\lesssim \sum _{|n|\ge 2} \frac{ \Vert f\Vert _3 |I|^{\tau \delta }}{|n|^{a\delta }}\lesssim \Vert f\Vert _3 |I|^{\tau \delta }, \end{aligned}
is similar.

#### 4.1.4 Proof of Lemma 4.6

It suffices to consider the cases where $$\alpha >0$$ is large, and when $$\alpha =-1$$. The general case then follows from the diagram provided that $$N\in \mathbb {N}$$ is sufficiently big.
The case$$\alpha \ge a+2$$: We will prove that $$\Vert D^\alpha f|_{p,q}^{s-\alpha }\Vert _3\lesssim \Vert f|_{p,q}^{s}\Vert _4$$. Let $$r_l=e^{-2^{-l}}$$. Then $$r_l=r_{l+1}^2$$. Let $$\varphi _{l}^{\left( \alpha \right) }$$ be given by $$\hat{\varphi }_{l}^{\left( \alpha \right) }\left( \xi \right) =\left( 1+\xi \right) ^\alpha e^{-2^{-l}\xi }$$ for $$\xi \ge -1$$. Extend $$\hat{\varphi }_{l}^{\left( \alpha \right) }$$ by zero. If $$\Phi _l^{\left( \alpha \right) }$$ is the corresponding periodization, then $$D^\alpha f\left( r_l\zeta _x\right) =\Phi _{l+1}^{\left( \alpha \right) }*\Phi _{l+1}*f\left( \zeta _x\right)$$. This yields that
\begin{aligned} \Vert D^\alpha f\left( r_l\zeta _x\right) \Vert _\mathcal {X}\le f_{l+1,a}^*\left( \zeta _x\right) \int _\mathbb {T}|\Phi _{l+1}^{\left( \alpha \right) }\left( \zeta _y\right) |\left( 1+2^{l+1}|\zeta _y-1|\right) ^a\, dy. \end{aligned}
Similar to Lemma 4.9 we have that
\begin{aligned} |\Phi _{l}^{\left( \alpha \right) }\left( \zeta _y\right) |\lesssim \frac{2^{l\left( 1+\alpha \right) }}{\left( 1+2^l|\zeta _y-1|\right) ^N}, \end{aligned}
provided that $$\alpha >N$$. If $$\alpha > a +2$$, then we may choose N such that $$N>a+1$$. It follows that
\begin{aligned} \Vert D^\alpha f\left( r_l\zeta _x\right) \Vert _\mathcal {X}\lesssim 2^{l\alpha } f_{l+1,a}^*\left( \zeta _x\right) . \end{aligned}
The statement that $$D^\alpha :F_{p,q}^{s,\tau }\left( \mathbb {D},\mathcal {X}\right) \rightarrow F_{p,q}^{s-\alpha ,\tau }\left( \mathbb {D},\mathcal {X}\right)$$ now follows from Theorem 4.3. $$\square$$

The following estimate follows by the usual tricks, e.g. the proof of [11, Eq. I.3.9]:

### Lemma 4.12

Let $$0<p<\infty$$. There exists $$K=K\left( p\right) >0$$ such that,
\begin{aligned} \Vert f\left( w\right) \Vert _\mathcal {X}\le K \left( 1-|w|\right) ^{s+\tau -\frac{1}{p}} \Vert f|_{p,q}^{s,\tau }\Vert _1 \end{aligned}
for $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, $$0<q<\infty$$, $$s,\tau \in \mathbb {R}$$. In particular, if $$s+\tau -\frac{1}{p}>0$$ and $$f\in \mathcal {A}\left( \mathcal {X}\right)$$, then $$\Vert f|_{p,q}^{s,\tau }\Vert _1<\infty$$ implies that $$f\equiv 0$$.

The following straightforward, yet clever, adaptation of Hardy’s inequality is from [10, p. 758]:

### Lemma 4.13

Let $$q\ge 1$$ and $$\mu <0$$. There exists $$K=K\left( q,\mu \right)$$ such that
\begin{aligned} \int _0^1\left( 1-r\right) ^{-1-\mu q}\left\{ \int _0^r h\left( \rho \right) \, d\rho \right\} ^qdr\le K \int _0^1\left( 1-r\right) ^{-1-\left( \mu -1\right) q}h\left( \rho \right) ^q\, dr, \end{aligned}
whenever $$h:\left[ 0,1\right) \rightarrow \left[ 0,\infty \right]$$ is measurable.
The case$$\alpha =-1$$: The operator $$D^{-1}:\mathcal {A}\left( \mathcal {X}\right) \rightarrow \mathcal {A}\left( \mathcal {X}\right)$$ has the integral representation
\begin{aligned} \left( D^{-1}f\right) \left( r\zeta _x\right) =\frac{1}{r}\int _0^r f\left( \rho \zeta _x\right) \, d\rho , \end{aligned}
as is verified by term-wise integration of the Taylor series. Let $$r_0=1-|I|$$, and assume for simplicity that $$I\ne \mathbb {T}$$, which yields $$\frac{1}{r_0}\le 2$$. We then have
\begin{aligned} \int _{r=r_0}^1 \left( 1-r\right) ^{-1-\left( s+1\right) q} \Vert \left( D^{-1}f\right) _r\left( \zeta _x\right) \Vert _\mathcal {X}^q \, dr \lesssim A + B, \end{aligned}
where
\begin{aligned} A = \int _{r=r_0}^1 \left( 1-r\right) ^{-1-\left( s+1\right) q}\left\{ \int _{\rho = 0}^{r_0} \Vert f_{\rho }\left( \zeta _x\right) \Vert _\mathcal {X}\, d\rho \right\} ^q \, dr, \end{aligned}
and
\begin{aligned} B = \int _{r=r_0}^1 \left( 1-r\right) ^{-1-\left( s+1\right) q}\left\{ \int _{\rho = r_0}^{1} \Vert f_{\rho }\left( \zeta _x\right) \Vert _\mathcal {X}\, d\rho \right\} ^q \, dr. \end{aligned}
By Lemma 4.12, we trivially obtain the estimate
\begin{aligned} A \lesssim \Vert f|_{p,q}^{s,\tau }\Vert _3^q|I|^{\tau q+q}. \end{aligned}
Applying Lemma 4.13, with $$h\left( \rho \right) =\mathbb {1}_{\left[ r_0,1\right) }\left( \rho \right) \Vert f_\rho \left( \zeta _x\right) \Vert _\mathcal {X}$$ and $$\mu = s+1$$, we obtain
\begin{aligned} B\lesssim \int _{r=r_0}^1\left( 1-r\right) ^{-1-sq} \Vert f_r\left( \zeta _x\right) \Vert _\mathcal {X}^q\, dr. \end{aligned}
These estimates together show that $$D^{-1}:F_{p,q}^{s,\tau }\left( \mathbb {D},\mathcal {X}\right) \rightarrow F_{p,q}^{s+1,\tau }\left( \mathbb {D},\mathcal {X}\right)$$ is bounded under the conditions in Theorem 4.3. This concludes the proof. $$\square$$

## Footnotes

1. 1.

This is a known result, e.g. [2, Lemma 1.1], but we include a proof for the convenience of the reader.

2. 2.

This paper contains a mistake, which is corrected in .

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