Frames, their relatives and reproducing kernel Hilbert spaces

TL;DR

This paper explores the relationship between reproducing kernel Hilbert spaces (RKHS) and frame theory, revealing structural insights and limitations in extending Riesz bases to general measure spaces.

Contribution

It introduces a new RKHS-based approach to analyze frames, proves finite redundancy implies atomic measure spaces, and characterizes the RKHS structure of analysis operator ranges.

Findings

Finite redundancy of continuous frames implies atomic measure spaces.

All attempts to extend Riesz bases to general measure spaces are limited to discrete cases.

Range of analysis operators of reproducing pairs can be endowed with an RKHS structure.

Abstract

This paper considers different facets of the interplay between reproducing kernel Hilbert spaces (RKHS) and stable analysis/synthesis processes: First, we analyze the structure of the reproducing kernel of a RKHS using frames and reproducing pairs. Second, we present a new approach to prove the result that finite redundancy of a continuous frame implies atomic structure of the underlying measure space. Our proof uses the RKHS structure of the range of the analysis operator. This in turn implies that all the attempts to extend the notion of Riesz basis to general measure spaces are fruitless since every such family can be identified with a discrete Riesz basis. Finally, we show how the range of the analysis operators of a reproducing pair can be equipped with a RKHS structure.

Equations

$$m\Vert f{\Vert }^2\le {\int }_X|⟨f,{\Psi }_x⟩{|}^{2}d\mu (x)\le M\Vert f{\Vert }^2,\forall f\in \mathcal{H}.$$

$$S_{\Psi ,\Phi }f={\int }_X⟨f,{\Psi }_x⟩{\Phi }_{x}d\mu (x)\in GL(\mathcal{H}),$$

$$R(\Psi ):=\dim (\mathsf{Ran}{C}_{\Psi }{}^{\perp }).$$

$$\inf\big{\{}\mu(A):\ A\mbox{ measurable and }\mu(A)>0\big{\}}=C>0,$$

$$\mu ={\mu }_a+{\mu }_c.$$

$$m{\Vert f\Vert }^2\le {\int }_X{\left|⟨f,{\Psi }_x⟩\right|}^{2}d\mu (x)\le M{\Vert f\Vert }^2,\forall f\in \mathcal{H}.$$

$$0<{\int }_X{\left|⟨f,{\Psi }_x⟩\right|}^{2}d\mu (x).$$

$$D_{\Psi }:L^2(X,\mu )\to \mathcal{H},D_{\Psi }F:={\int }_{X}F(x){\Psi }_{x}d\mu (x),$$

$$S_{\Psi }:\mathcal{H}\to \mathcal{H},S_{\Psi }f:=D_{\Psi }{C}_{\Psi }f={\int }_X⟨f,{\Psi }_x⟩{\Psi }_{x}d\mu (x).$$

$$f:=D_{\Psi }{C}_{\Phi }f=D_{\Phi }{C}_{\Psi }f,\forall f\in \mathcal{H},$$

$$0<{\int }_X{\left|⟨f,{\Psi }_x⟩\right|}^{2}d\mu (x)\le M\Vert f{\Vert }^2,\forall f\in \mathcal{H}\backslash \{0\},$$

$$m\Vert f{\Vert }^2\le {\int }_X{\left|⟨f,{\Psi }_x⟩\right|}^{2}d\mu (x),$$

$$⟨S_{\Psi ,\Phi }f,g⟩:={\int }_X⟨f,{\Psi }_x⟩⟨{\Phi }_x,g⟩d\mu (x),$$

$$K(z,z)=\langle k_{z},k_{z}\rangle=\|k_{z}\|^{2}=\sup\big{\{}|f(z)|^{2}:\ f\in\mathcal{H}_{K},\ \|f\|=1\big{\}}.$$

$$K(z,w)=\sum _{i\in \mathcal{I}}{\varphi }_i(z)\overline{{\varphi }_i(w)},$$

$$0<\sum _{i\in \mathcal{I}}|{\varphi }_i(z){|}^2=K(z,z)<\infty ,\forall z\in \Omega .$$

$$K(z,w)=\sum _{i\in \mathcal{I}}(A{\varphi }_i)(z)\overline{{\psi }_i(w)}=\sum _{i\in \mathcal{I}}(A^{\ast }{\psi }_i)(z)\overline{{\varphi }_i(w)},\forall z,w\in \Omega ,$$

$$K(z,w)=⟨k_w,k_z⟩=\sum _{i\in \mathcal{I}}⟨k_w,{\psi }_i⟩⟨S_{\Psi ,\Phi }^{-1}{\varphi }_i,k_z⟩=\sum _{i\in \mathcal{I}}\overline{{\psi }_i(w)}(S_{\Psi ,\Phi }^{-1}{\varphi }_i)(z).$$

$$⟨f,g⟩$$

$$=\sum _{n,m=1}^N{\alpha }_n\overline{{\beta }_m}\sum _{i\in \mathcal{I}}(A{\varphi }_i)(w_m)\overline{{\psi }_i(z_n)}$$

$$=\sum _{n,m=1}^N{\alpha }_n\overline{{\beta }_m}\sum _{i\in \mathcal{I}}⟨k_{z_n},{\psi }_i⟩⟨A{\varphi }_i,k_{w_m}⟩$$

$$\displaystyle=\sum_{i\in\mathcal{I}}\Big{\langle}\sum_{n=1}^{N}\alpha_{n}k_{z_{n}},\psi_{i}\Big{\rangle}\Big{\langle}A\phi_{i},\sum_{m=1}^{N}\beta_{m}k_{w_{m}}\Big{\rangle}$$

$$=\sum _{i\in \mathcal{I}}⟨f,{\psi }_i⟩⟨A{\varphi }_i,g⟩=⟨A{S}_{\Psi ,\Phi }f,g⟩.$$

$$\frac{{\sum }_{i\in \mathcal{I}}|{\psi }_i(z){|}^2}{K(z,z)}\le M,\forall z\in \Omega .$$

$$0<m\le \frac{{\sum }_{i\in \mathcal{I}}|{\psi }_i(z){|}^2}{K(z,z)},\forall z\in \Omega .$$

$$\sum _{i\in \mathcal{I}}|{\psi }_i(z){|}^2=\sum _{i\in \mathcal{I}}|⟨k_z,{\psi }_i⟩{|}^2\le M\Vert k_z{\Vert }^2=MK(z,z).$$

$$\Vert D_{\Psi }a{\Vert }_{{\ell }^2(\mathbb{N})}^2={\Vert \left(1,\frac{1}{2},\frac{1}{2},\frac{1}{3},\frac{1}{3},\frac{1}{3},\dots \right)\Vert }_{{\ell }^2(\mathbb{N})}^2=\sum _{k\in \mathbb{N}}\frac{1}{k}=\infty .$$

$${\mathcal{H}}_K:=\{V_{\phi }f:f\in L^2(\mathbb{R})\}\subset L^2({\mathbb{R}}^2),$$

$$\sum _{\lambda \in \mathbb{Z}}|⟨F,K(\cdot ,\lambda )⟩{|}^2=\sum _{\lambda \in \mathbb{Z}}|⟨f,\pi (\lambda )\phi ⟩{|}^2,$$

$$z\mapsto \sum _{\lambda \in {\mathbb{Z}}^2}|{\psi }_{\lambda }(z){|}^2=\sum _{\lambda \in {\mathbb{Z}}^2}{e}^{-\pi |z-\lambda {|}^2}>e^{-\pi /2},$$

$$\Big{\|}\sum_{i\in\mathcal{I}}\psi_{i}(z)\psi_{i}\Big{\|}\leq\sqrt{M\cdot K(z,z)}<\infty,\qquad\forall\ z\in\Omega.$$

$$0<\sum _{i\in \mathcal{I}}|{\psi }_i(z){|}^2=C_z<\infty ,\forall z\in \Omega ,$$

Full text

MSC2010: 42C15, 46E22 (primary), 47B32 (secondary)

Keywords: continuous frames, reproducing pairs, reproducing kernel Hil- bert spaces, redundancy, atomic measures

1 Introduction

Reproducing kernel Hilbert spaces (RKHS) were introduced by Zaremba [34] and Mercer [25] and systematically studied by Aronszajn [8] in 1950. These spaces play an important role in many diverse branches of mathematics such as complex analysis [19], and learning theory [29]. Another field with manifold connections to RKHS, which is the main focus of this paper, is frame theory and related concepts.

A mapping $x\mapsto\Psi_{x}\in\mathcal{H}$ is called a continuous frame if there exist constants $m,M>0$ such that

[TABLE]

Frames have proven to be a viable tool in many different fields such as signal processing [12] acoustics [13] or mathematical physics [2, 11]. It is however not always possible to satisfy both inequalities, which is why new concepts like semi-frames [3, 4] and reproducing pairs [5, 32, 33] were introduced recently. An upper (resp. lower) semi-frame is a complete system that only satisfies the upper (resp. lower) frame inequality in (1). A reproducing pair is a pair of mappings $(\Psi,\Phi)$ that generates a bounded and boundedly invertible analysis/synthesis process, i.e.,

[TABLE]

without assuming any frame inequality for neither $\Psi$ nor $\Phi$.

This paper is divided into three main parts, portraying different facets of the interplay between frames, reproducing pairs, and RKHS.

In Section 3, we study systems taking values in RKHS. In particular, we give an explicit expression for the reproducing kernel in terms of a reproducing pair in Theorem 4 that extends the results from [26, 28] and introduce a necessary condition for a family of vectors to form a frame, see Proposition 6.

In Section 4 we study the dependence of the redundancy of (semi-)frames on the structure of the measure space. In the discrete case, the redundancy of a frame measures, loosely speaking, how much a Hilbert space is oversampled by that frame, see for example [15, 16]. It is however impossible to directly translate this concept to continuous (semi-)frames. In [9] the authors therefore used a property of Riesz bases to define redundancy. A Riesz basis is a discrete, non-redundant frame, i.e., its analysis operator $C_{\Psi}:\mathcal{H}\rightarrow\ell^{2}$ is surjective. We therefore define redundancy of a (semi-)frame $\Psi$ by

[TABLE]

It has been observed in several articles [9, 23, 24] that $R(\Psi)$ depends on the underlying measure space $(X,\mu)$. In particular, if a (lower semi-)frame has finite redundancy, then it follows that $(X,\mu)$ is atomic. The proofs in the aforementioned papers all rely in one way or the other on the following argument: If the redundancy of a frame is zero (finite), then

[TABLE]

which implies that $(X,\mu)$ is atomic. We present a new proof in Section 4.1 using that ${\sf Ran}\,C_{\Psi}$ is a RKHS, which, in our opinion, better explains the underlying structure of the problem. This result also shows that all efforts to generalize the notion of Riesz bases ($R(\Psi)=0$), see [7, 20], to the setting of measure spaces that are not atomic, are essentially an attempt to square the circle. The measure space of a continuous Riesz basis is always atomic and every such system can therefore be written as a discrete family of vectors, see Corollary 19.

For upper semi-frames on the other hand, there is no connection between the structure of the measure space and the redundancy. In particular, we give a sufficient condition for the existence of upper semi-frames indexed by a non-atomic measure space with redundancy zero, see Proposition 20.

Finally, Section 5 is concerned with characterizing the ranges of the analysis operators of a reproducing pair. The omission of the frame inequalities causes the inconvenience that ${\sf Ran}\,C_{\Psi}$ and ${\sf Ran}\,C_{\Phi}$ need no longer be contained in $L^{2}(X,\mu)$. Therefore, in [5] a pair of Hilbert spaces, intrinsically generated by a reproducing pair, was introduced to study this problem. We demonstrate that these spaces are actually RKHS and calculate the reproducing kernels.

2 Preliminaries

2.1 Atomic and non-atomic measures

Throughout this paper we assume that $(X,\mu)$ is a measure space, where $\mu$ is a non-trivial, $\sigma$-finite, and positive measure. A measurable set $A\subset X$ is called an atom if $\mu(A)>0$, and for any measurable subset $B\subset A$, with $\mu(B)<\mu(A)$, it holds $\mu(B)=0$. A measure space is called atomic if there exists a partition $\{A_{n}\}_{n\in\mathbb{N}}$ of $X$ consisting of atoms and null sets. The space $(X,\mu)$ is called non-atomic if there are no atoms in $(X,\mu)$. To our knowledge there is no term to denote a measure space that is not atomic. In order to avoid any confusion with non-atomic spaces, we therefore call a measure space an-atomic if it is not atomic.

A well-known result by Sierpiński states that non-atomic measures take a continuity of values.

Theorem 1** (Sierpiński [30])**

Let $(X,\mu)$ be non-atomic and let $A\subset X$ be measurable with $\mu(A)>0$. For every $0\leq b\leq\mu(A)$, there exists $B\subset A$ such that $\mu(B)=b$.

Since we could not find any reference for the second part of the following lemma, we provide a proof in the appendix.

Lemma 2

Let $(X,\mu)$ be a $\sigma$-finite measure space.

1. (i)

There exists $\mu_{a}$ atomic and $\mu_{c}$ non-atomic such that

[TABLE] 2. (ii)

If $(X,\mu)$ is an-atomic, then there exists $A\subset X$ with $\mu(A)>0$ and $(A,\mu)$ non-atomic.

2.2 Continuous frames, semi-frames and reproducing pairs

Let $\mathcal{H}$ be a separable Hilbert space. We denote by $GL(\mathcal{H})$ the space of bounded linear operators on $\mathcal{H}$ with bounded inverse.

Definition 1

A mapping $\Psi:X\rightarrow\mathcal{H}$ is called a continuous frame if

1. (i)

$\Psi$* is weakly measurable, that is, $x\mapsto\langle f,\Psi_{x}\rangle$ is a measurable function for every $f\in\mathcal{H}$,* 2. (ii)

there exist positive constants $m,M>0$ such that

[TABLE]

The mapping $\Psi$ is called complete, if for every $f\in\mathcal{H}\backslash\{0\}$

[TABLE]

The constants $m,M$ are called the frame bounds and $\Psi$ is called Bessel if the second inequality in (3) is satisfied. If $m=M=1$, then $\Psi$ is called a Parseval frame. If $(X,\mu)$ is a countable set equipped with the counting measure, then one recovers the classical definition of a discrete frame, see for example [17]. For a self-contained introduction to continuous frames, we refer the reader to [27].

The fundamental operators in frame theory are given by the analysis operator $C_{\Psi}:\mathcal{H}\rightarrow L^{2}(X,\mu)$, $C_{\Psi}f(x):=\langle f,\Psi_{x}\rangle,$ and the synthesis operator

[TABLE]

where the integral is defined weakly. The frame operator $S_{\Psi}\in GL(\mathcal{H})$ is defined as the composition of $C_{\Psi}$ and $D_{\Psi}$

[TABLE]

Every frame $\Phi$ satisfying

[TABLE]

is called a dual frame for $\Psi$.

There exists a variety of interesting complete systems that do not meet both frame conditions. Several concepts to generalize the frame property were thus introduced and studied. An upper semi-frame is a complete Bessel system, i.e.,

[TABLE]

whereas a lower semi-frame satisfies

[TABLE]

see [3, 4].

Another generalization is the concept of reproducing pairs, defined in [32] and further investigated in [5, 6, 33]. Here, one considers a pair of mappings instead of a single one without any assumption on frame inequalities.

Definition 2

Let $\Psi,\Phi:X\rightarrow\mathcal{H}$ be weakly measurable. The pair $(\Psi,\Phi)$ is called a reproducing pair for $\mathcal{H}$ if the operator $S_{\Psi,\Phi}:\mathcal{H}\rightarrow\mathcal{H}$, defined by

[TABLE]

is an element of $GL(\mathcal{H})$.

2.3 Reproducing kernel Hilbert spaces (RKHS)

Let $\mathcal{F}(\Omega,\mathbb{C})$ denote the vector space of all functions $f:\Omega\rightarrow\mathbb{C}$.

Definition 3

A Hilbert space $\mathcal{H}_{K}\subset\mathcal{F}(\Omega,\mathbb{C})$ is called a reproducing kernel Hilbert space (RKHS), if the point evaluation functional $\delta_{z}:\mathcal{H}_{K}\rightarrow\mathbb{C}$, $\delta_{z}(f):=f(z)$ is bounded for every $z\in\Omega$, that is, if there exists $C_{z}>0$ such that $|\delta_{z}(f)|\leq C_{z}\|f\|$, for every $f\in\mathcal{H}_{K}$.

As $\delta_{z}$ is bounded, there exists a unique vector $k_{z}\in\mathcal{H}_{K}$ such that $f(z)=\langle f,k_{z}\rangle,$ for every $f\in\mathcal{H}_{K}$. The function $K(z,w):=k_{w}(z)=\langle k_{w},k_{z}\rangle$ is called the reproducing kernel for $\mathcal{H}_{K}$. The reproducing kernel is unique, $K(z,w)=\overline{K(w,z)}$ and its diagonal reads

[TABLE]

The following result can be found in [1, Theorem 3.1 and 3.2].

Theorem 3

If $\mathcal{H}_{K}$ is a RKHS and $\{\phi_{i}\}_{i\in\mathcal{I}}\subset\mathcal{H}_{K}$ an orthonormal basis, then

[TABLE]

with pointwise convergence of the series. In particular,

[TABLE]

Conversely, if there exists an orthonormal basis for a Hilbert space $\mathcal{H}_{K}\subset\mathcal{F}(\Omega,\mathbb{C})$ that satisfies (6), then $\mathcal{H}_{K}$ can be identified with a RKHS consisting of functions $f:\Omega\rightarrow\mathbb{C}$.

For a thorough introduction to RKHS we refer the reader to [8, 26].

3 Frames and reproducing pairs taking values in a RKHS

In this section, we investigate the pointwise behavior of frames in RKHS, characterize the reproducing kernel and introduce sufficient conditions on a frame that ensures the existence of a reproducing kernel.

The following result adapts the arguments of [26, Theorem 3.12] to reproducing pairs.

Theorem 4

Let $\mathcal{H}_{K}$ be a RKHS and $\Psi=\{\phi_{i}\}_{i\in\mathcal{I}},\ \Phi=\{\psi_{i}\}_{i\in\mathcal{I}}\subset\mathcal{H}_{K}$. The pair $(\Psi,\Phi)$ is a reproducing pair for $\mathcal{H}_{K}$ if and only if there exists $A\in GL(\mathcal{H}_{K})$ such that

[TABLE]

where the series converges pointwise. In particular, $A$ is uniquely given by $S_{\Psi,\Phi}^{-1}$.

**Proof: **If $(\Psi,\Phi)$ is a reproducing pair, then

[TABLE]

Conversely, assume that $K$ is given by (7). If $f,g\in\mbox{span}\{k_{z}:z\in\Omega\}$, that is, there exist $\alpha_{n},\beta_{m}\in\mathbb{C}$, and $z_{n},w_{m}\in\Omega$, such that $f=\sum_{n}^{N}\alpha_{n}k_{z_{n}}$ and $g=\sum_{m}^{N}\beta_{m}k_{w_{m}}$, then

[TABLE]

In [26, Proposition 3.1] it is shown that $\mbox{span}\{k_{z}:z\in\Omega\}$ is dense in $\mathcal{H}_{K}$. Therefore, it follows that $AS_{\Psi,\Phi}=I$. As $A\in GL(\mathcal{H}_{K})$ we may conclude that $S_{\Psi,\Phi}\in GL(\mathcal{H}_{K})$, that is, $(\Psi,\Phi)$ is a reproducing pair. $\Box$

Remark 5

For certain special cases, Theorem 4 is already known. In particular, the result can be found in [28, Theorem 7] if $\Psi$ and $\Phi$ are dual frames, and in [26, Theorem 3.12] if $\Psi=\Phi$ is a Parseval frame.

Proposition 6

Let $\mathcal{H}_{K}$ be a RKHS and $\{\psi_{i}\}_{i\in\mathcal{I}}\subset\mathcal{H}_{K}$.

1. (i)

If the family $\{\psi_{i}\}_{i\in\mathcal{I}}$ is Bessel, then

[TABLE] 2. (ii)

If $\{\psi_{i}\}_{i\in\mathcal{I}}$ satisfies the lower frame inequality, then

[TABLE]

**Proof: **If $\{\psi_{i}\}_{i\in\mathcal{I}}$ is Bessel, then, for every $z\in\Omega$, it holds

[TABLE]

The same argument shows the lower bound in (9) if $\{\psi_{i}\}_{i\in\mathcal{I}}$ is a lower semi-frame. $\Box$

Remark 7

(i) The converse statements of Proposition 6 are in general false. First, consider the sequence $\psi_{1}=(1,0,\ldots),\ \psi_{2}=(0,1,1,0,\ldots),\psi_{3}=(0,0,0,1,1,1,0\ldots),\ldots\in\mathcal{H}_{K}=\ell(\mathbb{N})$, i.e. $\psi_{n}(l)=1,$ if $l\in n(n-1)/2+1,\ldots,n(n+1)/2$, and zero otherwise. This system satisfies (8) with $M=1$ but $\{\psi_{n}\}_{n\in\mathbb{N}}$ is not Bessel. To see this, recall that a sequence is Bessel if and only if its synthesis operator $D_{\Psi}$ is bounded. Taking $a=\{1/n\}_{n\in\mathbb{N}}$ gives

[TABLE]

To see that the reverse of (9) is false, consider $\mathcal{H}_{K}=\mathbb{C}^{2}$, $\Omega=\{1,2\}$ and $\psi=(1,1)$. Then $K(z,z)=1$ and $|\psi(1)|^{2}=|\psi(2)|^{2}=1$, but $\psi$ is not spanning.

The reverse statement remains false if one additionally assumes that $\{\psi_{i}\}_{i\in\mathcal{I}}$ is complete. Let us construct a counterexample from the system of integer time-frequency shifts of the Gaussian window $\varphi(t)=2^{1/4}e^{-\pi t^{2}}$, which is a complete Bessel sequence (i.e. an upper semi-frame) but not a frame in $L^{2}(\mathbb{R})$, see [21, 33]. Let $z=(x,\omega)\in\mathbb{R}^{2}$, and $\pi(z)f(t):=f(t-x)e^{2\pi i\omega t}$. The short-time Fourier transform with window $\varphi$ is defined as $V_{\varphi}f(z)=\langle f,\pi(z)\varphi\rangle,$ and the space

[TABLE]

forms a RKHS with kernel $K(z,w)=\langle\pi(w)\varphi,\pi(z)\varphi\rangle$. If $F=V_{\varphi}f\in\mathcal{H}_{K}$ for some $f\in L^{2}(\mathbb{R})$, then

[TABLE]

and the isometry property $\|F\|=\|f\|$ implies that $\{\psi_{\lambda}\}_{\lambda\in\mathbb{Z}^{2}}=\{K(\hskip 1.42271pt\cdot\hskip 1.42271pt,\lambda)\}_{\lambda\in\mathbb{Z}^{2}}$ is a complete Bessel sequence for $\mathcal{H}_{K}$, but not a frame. Moreover, we have $|K(z,w)|=e^{-\pi|z-w|^{2}/2}$ by [21, Lemma 1.5.2], which gives that

[TABLE]

is a strictly positive, continuous, and $\mathbb{Z}^{2}$-periodic function. Therefore, as $|K(z,z)|=1$, (9) is satisfied with $m\geq e^{-\pi/2}$.

(ii) Another consequence of Proposition 6 is that, if $\{\psi_{i}\}_{i\in\mathcal{I}}\subset\mathcal{H}_{K}$ is a Bessel sequence, then

[TABLE]

Let us now consider the converse part of Theorem 3, which gives a sufficient condition on orthonormal bases that ensures that the space is a RKHS. We generalize the result to a condition on discrete frames.

Theorem 8

Let $\mathcal{H}_{K}\subset\mathcal{F}(X,\mathbb{C})$ be a Hilbert space. If there exists a discrete frame $\Psi$ for $\mathcal{H}_{K}$ such that

[TABLE]

then $\mathcal{H}_{K}$ is a RKHS.

**Proof: **If $\Psi$ is a frame such that (11) holds, then

[TABLE]

is well-defined in $\mathcal{H}_{K}$ for every $z\in\Omega$. Moreover, for $f\in\mathcal{H}_{K}$, we have

[TABLE]

Hence, point evaluation is continuous. $\Box$

Note that all results of this section can be reformulated for continuous frames and reproducing pairs, if one replaces the index set $\mathcal{I}$ by $(X,\mu)$.

4 Continuous (semi-)frames and their redundancy

Redundancy of a discrete frame measures, roughly speaking, how much the frame is oversampling the Hilbert space $\mathcal{H}$. A non-redundant discrete frame is in fact a Riesz basis, that is, a frame with the additional property that ${\sf Ran}\,C_{\Psi}=({\sf Ker}\,D_{\Psi})^{\bot}=\ell^{2}(\mathcal{I})$. This justifies to define redundancy for general families $\Psi$ by

[TABLE]

Note that $C_{\Psi}$ is a Fredholm operator if $\Psi$ is Bessel and has finite redundancy. In that case, $R(\Psi)=-\mbox{ind}(C_{\Psi})=\mbox{ind}(D_{\Psi})$, where $\mbox{ind}(A)$ denotes the Fredholm index of a bounded operator $A$.

Example 1

Let $\{e_{n}\}_{n\in\mathbb{N}}$ be an orthonormal basis. If we define $\Psi=\{e_{1}\}\cup\{e_{n}\}_{n\in\mathbb{N}}$, and $\Phi=\{e_{n}\}_{n\in\mathbb{N}}\cup\{e_{n}\}_{n\in\mathbb{N}}$, then $R(\Psi)=1$ and $R(\Phi)=\infty$.

If $\Psi=\{f_{n}\}_{n=1}^{N}$ is a finite frame for $\mathbb{C}^{d}$, then $R(\Psi)=N-d$, whereas the classical definition of redundancy for finite frames gives $N/d$.

4.1 Frames and lower semi-frames

The main goal of this section is to give a new proof for Theorem 9, which connects finite redundancy with the structure of the measure space, a result that has already been stated in [23, Theorem 2], [9, Theorem 2.2] and [24, Proposition 3.3]. We thereby use the property that ${\sf Ran}\,C_{\Psi}$ forms a RKHS. In our opinion, this proof does a better job explaining the inherent structure of continuous frames.

Theorem 9

If a (lower semi-)frame $\Psi$ has finite redundancy $R(\Psi)<\infty$, then the measure space $(X,\mu)$ is atomic.

The converse is obviously not true, take for example the discrete family $\Phi$ in Example 1.

Corollary 10

If $(X,\mu)$ is an-atomic and $\Psi$ a frame, then $R(\Psi)=\infty$ and there exist infinitely many dual frames for $\Psi$.

**Proof: **Take an arbitrary pointwise defined function $\theta\in{\sf Ran}\,C_{\Psi}^{\bot}$, an arbitrary vector $g\in\mathcal{H}$ and define $\phi(x)=\theta(x)g$. Then $S_{\Psi}^{-1}\Psi+\Phi$ is a dual frame for $\Psi$. $\Box$

We need to collect some auxiliary results in order to prove Theorem 9.

Proposition 11** ([20], Corollary 2.9)**

Let $\mathcal{H}_{K}$ be a subspace of $L^{2}(X,\mu)$. If $\Psi$ satisfies the lower frame inequality, then $({\sf Ran}\,C_{\Psi},\left\|\cdot\right\|_{2})$ is a RKHS. Moreover, the following are equivalent:

1. (i)

There exists a continuous frame $\Psi$ such that ${\sf Ran}\,(C_{\Psi})=\mathcal{H}_{K}$, 2. (ii)

$\mathcal{H}_{K}$* is a RKHS.*

In the following, we prove that $L^{2}(X,\mu)$ is not a RKHS if $(X,\mu)$ is an-atomic. This statement needs some elaboration to be precise. By definition, $L^{2}(X,\mu)$ is a space of equivalence classes of functions and does therefore not allow for pointwise evaluation. However, if we take an orthonormal basis $\{\phi_{i}\}_{i\in\mathcal{I}}$ for $L^{2}(X,\mu)$ and fix one particular representative for every basis element, then every vector $F\in L^{2}(X,\mu)$ can formally be written as

[TABLE]

where the series converges for almost every $x\in X$.

Proposition 12

If $(X,\mu)$ is a an-atomic measure space, then $L^{2}(X,\mu)$ is not a reproducing kernel Hilbert space.

**Proof: **Every an-atomic measure space contains a non-atomic subspace by Lemma 2 $(ii)$. Without loss of generality we may therefore assume that $(X,\mu)$ is non-atomic. Take $A\subset X$ with $\mu(A)>0$. We may assume without loss of generality that $\mu(A)=1$ as $(X,\mu)$ is $\sigma$-finite. Let $\{A_{m}\}_{m=1}^{n}$ be a partition of $A$ satisfying $\mu(A_{m})=1/n,$ for every $m=1,...,n$. Such a partition exists by Theorem 1. Define $B^{n}\subset A$ by

[TABLE]

Clearly, $\mu(B^{n})=\mu(A)$ for every $n\in\mathbb{N}$. If we assume that $L^{2}(X,\mu)$ is a RKHS, then, for every $x\in B^{n}$, and some $m\in\{1,...,n\},$ one has

[TABLE]

In particular, $\|k_{x}\|^{2}\geq n$ for every $x\in B^{n}$. Setting $B:=\bigcap_{n\in\mathbb{N}}B^{n}$, one gets that $\mu(B)=\mu(A)$. Consequently, $\|k_{x}\|^{2}=K(x,x)=\infty$ for almost every $x\in A$, a contradiction to (6). $\Box$

Corollary 13

Let $(X,\mu)$ be an-atomic. There is no orthonormal basis $\{\phi_{i}\}_{i\in\mathcal{I}}\subset L^{2}(X,\mu)$, such that

[TABLE]

In particular, for every orthonormal basis $\{\phi_{i}\}_{i\in\mathcal{I}}\subset L^{2}(X,\mu)$, there exists a set $A\subset X$ of positive measure, such that

[TABLE]

Proposition 14

If $(X,\mu)$ is an-atomic and $\mathcal{H}_{K}\subset L^{2}(X,\mu)$ is a RKHS, then $\dim(\mathcal{H}_{K}{}^{\bot})=\infty$.

**Proof: **If we assume that $\dim(\mathcal{H}_{K}{}^{\bot})=N<\infty$, then any orthonormal basis $\{\phi_{i}\}_{i\in\mathcal{I}}$ of $\mathcal{H}_{K}$ can be complemented to a orthonormal basis of $L^{2}(X,\mu)$ using $N$ vectors $\{u_{n}\}_{n=1}^{N}$. In particular, we have by Theorem 3 that

[TABLE]

a contradiction to Corollary 13. $\Box$

Proof of Theorem 9: The range of $C_{\Psi}$ is a RKHS by Propositions 11. By Proposition 14 it thus follows that either $R(\Psi)=\infty$ or $(X,\mu)$ is atomic. $\Box$

Remark 15

The results of this section lead to interesting consequences in the context of quantum mechanics. Let us assume that $\Psi:X\rightarrow\mathcal{H}$ is a system of coherent states, see [2]. The probabilistic interpretation of quantum mechanics then states that the probability distribution of finding a system $f$ in the state $\Psi_{x}$ is given by $|\langle{f},{\Psi_{x}}\rangle|^{2}$. Hence, in light of Theorem 9, it follows that there is always a infinite dimensional subspace of probability distributions that does not correspond to any physically feasible system.

4.2 Strictly continuous mappings

In the following, we show that the discrete components of a continuous frame can be separated and we call the remaining remaining mapping a strictly continuous mapping.

Definition 4

A mapping $\Psi:X\rightarrow\mathcal{H}$ is called strictly continuous if $(X,\mu)$ is non-atomic and there exists no set $A\subset X$, $\mu(A)>0$, such that $C_{\Psi}f|_{A}$ is constant for every $f\in\mathcal{H}$.

Example 2

Let $G$ be a locally compact group, $\pi:G\rightarrow\mathcal{U}(\mathcal{H})$ a square-integrable group representations for $G$, and $\psi\in\mathcal{H}$ an admissible vector, see e.g. [22] for more information. If the left Haar measure of $G$ is non-atomic, then $\Psi=\{\pi(g)\psi\}_{g\in G}$ is a strictly continuous frame, as $\langle f,\pi(g_{1})\psi\rangle=\langle f,\pi(g_{2})\psi\rangle$ for every $\forall f\in\mathcal{H}$, implies $g_{1}=g_{2}$. Short-time Fourier systems or continuous wavelet systems, see [21], are just two instances from this large class of strictly continuous mappings.

Throughout the rest of this section we show that every continuous frame can be decomposed into a discrete and a strictly continuous Bessel system.

Lemma 16** ([31], Theorem 3.8.1)**

If $A\subset X$ is an atom, and $F:X\rightarrow\mathbb{C}$ is measurable, then $F$ is constant almost everywhere on $A$.

Lemma 17

If $\Psi$ is Bessel, $A\subset X$ such that $\mu(A)>0$, and $\langle{f},{\Psi(\cdot)}\rangle$ is constant on $A$ for every $f\in\mathcal{H}$, then there exists a unique $\psi\in\mathcal{H}$ such that

[TABLE]

In particular, $\psi$ is weakly given by

[TABLE]

**Proof: **First, observe that $\psi$ defined by (15) is unique for every $n\in\mathbb{N}$ by Riesz representation theorem

[TABLE]

where $M$ is the upper frame bound of $\Psi$. Moreover,

[TABLE]

where we used (15) and the fact that $\langle{f},{\Psi(\cdot)}\rangle$ is almost everywhere constant on $A$. $\Box$

Theorem 18

Every frame $\Psi$ can be written as $\Psi=\Psi_{d}\cup\Psi_{c}$, where $\Psi_{d}$ is a discrete Bessel system and $\Psi_{c}:(X_{c},\mu_{c})\rightarrow\mathcal{H}$ is a strictly continuous Bessel mapping with $X_{c}\subset X$. In particular, if $(X,\mu)$ is atomic, then $\Psi$ can be written as a discrete frame.

**Proof: **By Lemma 2 $(i)$, every measure $\mu$ can be written as $\mu=\mu_{a}+\mu_{c}$, where $\mu_{a}$ is atomic and $\mu_{c}$ is non-atomic. By Lemma 16 and 17 we deduce that $\Psi$ defined on $(X,\mu_{a})$ can be identified with a discrete Bessel system $\Psi_{a}$. Let $X_{d}=\bigcup X_{i}\subset X$ be the disjoint union of all sets $X_{i}\subset X$, such that $\mu_{c}(X_{i})>0$, and $C_{\Psi}f|_{X_{i}}$ is constant for all $f\in\mathcal{H}$. Setting $A=X_{i}$ in (15), then defines a family of vectors $\{\psi_{i}\}_{i\in\mathcal{I}}$. By definition $\Psi_{c}:=\Psi|_{X\backslash X_{d}}$ is a strictly continuous Bessel mapping. It therefore remains to show that $\mathcal{I}$ is countable. This, however, is a direct consequence from the fact that $\sigma$-finite measure spaces can only be partitioned into countably many sets of positive measure. Hence, $\Psi_{d}:=\Psi_{a}\cup\{\psi_{i}\}_{i\in\mathcal{I}}$ is a discrete Bessel sequence the result follows. $\Box$

In an attempt to generalize the concept of Riesz bases, continuous Riesz bases [7] and Riesz-type mappings [20] were introduced. It turns out that the these notions are equivalent and characterized as frames with redundancy zero [7, Proposition 2.5 & Theorem 2.6].

Corollary 19

Every continuous Riesz basis (Riesz-type mapping) can be written as a discrete Riesz basis.

**Proof: **If $\Psi$ is a continuous Riesz basis, then $R(\Psi)=0$ by definition. By Theorem 9, $(X,\mu)$ is atomic. Consequently, $\Psi$ corresponds to a discrete Riesz basis by Theorem 18. $\Box$

4.3 Upper semi-frames

Let us now illustrate how upper semi-frames behave fundamentally different than (lower semi-)frames. In particular, the closure of the range of the analysis operator is not necessarily a reproducing kernel Hilbert space and there exist upper semi-frames on non-atomic measure spaces with redundancy zero (compare to Proposition 11 and Theorem 9).

Example 3

In [3, 5] the following upper semi-frame has been studied. Take $\mathcal{H}_{n}:=L^{2}(\mathbb{R}^{+},r^{n-1}dr)$, where $n\in\mathbb{N}$, and $(X,\mu)=(\mathbb{R},dx)$. We use the following convention to denote the Fourier transform

[TABLE]

For $\psi\in\mathcal{H}_{n}$, we define the affine coherent state $\Psi_{x}$ by

[TABLE]

The mapping $\Psi$ forms an upper semi-frame if $\operatorname{ess\,sup}_{r\in{\mathbb{R}}^{+}}r^{n-1}|\psi(r)|^{2}<\infty,$ and $|\psi(r)|\neq 0$ for a.e. $r\in\mathbb{R}^{+}$. The frame operator is then given by a multiplication operator on $\mathcal{H}_{n}$

[TABLE]

It is thus easy to see that $\Psi$ cannot form a frame since for every $\psi\in\mathcal{H}_{n}$, $\operatorname{ess\,inf}_{r\in{\mathbb{R}}^{+}}r^{n-1}|\psi(r)|^{2}=0$. In [5, Section 5.2] it is shown that ${\sf Ker}\,D_{\Psi}=\mathcal{F}_{+}$, where

[TABLE]

Clearly, $\overline{{\sf Ran}\,C_{\Psi}}=({\sf Ker}\,D_{\Psi})^{\bot}=\mathcal{F}_{+}{}^{\bot}=\mathcal{F}_{-}$, where

[TABLE]

Therefore, $\Psi$ has infinite redundancy and a short argument shows that $\mathcal{F}_{-}$ is not a RKHS:

The dilation operator $D_{a}$, defined by $D_{a}f(x):=a^{-1/2}f(x/a)$, $a\in\mathbb{R}^{+}$, acts isometrically on $\mathcal{F}_{-}$. Take $f\in\mathcal{F}_{-}$ with $\|f\|=1$ and $f(0)\neq 0$, then $|D_{a}f(0)|=|a^{-1/2}f(0)|\rightarrow\infty$, as $a\rightarrow 0$. Consequently, point evaluation cannot be continuous and $\overline{{\sf Ran}\,C_{\Psi}}=\mathcal{F}_{-}$ is not a RKHS.

The mapping $\Psi$ possesses several other interesting properties, see [5]. For instance, it forms a total Bessel system with no dual, i.e., there is no mapping $\Phi$ such that $(\Psi,\Phi)$ generates a reproducing pair.

Proposition 20

If $(X,\mu)$ is a measure space, such that there exists an orthonormal basis $\{\psi_{n}\}_{n\in\mathbb{N}}$ of $L^{2}(X,\mu)$ satisfying

[TABLE]

then there exists an upper semi-frame $\Psi$ for $\mathcal{H}$ such that $\overline{{\sf Ran}\,C_{\Psi}}=L^{2}(X,\mu)$. In particular, $R(\Psi)=0$.

**Proof: **Take an arbitrary orthonormal basis $\{e_{n}\}_{n\in\mathbb{N}}$ of $\mathcal{H}$, and define

[TABLE]

The series converges absolutely in every point, and $\Psi$ is an upper semi-frame with the desired properties. To see this, we first observe that $\Psi:X\rightarrow\mathcal{H}$ is well-defined as, for $x\in X$ fixed,

[TABLE]

[TABLE]

where we used (16) and Cauchy-Schwarz inequality. Moreover,

[TABLE]

[TABLE]

Since $\{\psi_{n}\}_{n\in\mathbb{N}}$ is an orthonormal basis of $L^{2}(X,\mu)$, it follows that $\Psi$ is total in $\mathcal{H}$, as

[TABLE]

[TABLE]

for every $f\neq 0$. Finally, the range of the analysis operator of the system $\{n^{-1}e_{n}\}_{n\in\mathbb{N}}$ is dense in $l^{2}(\mathbb{N})$, which implies that ${\sf Ran}\,C_{\Psi}$ is dense in $L^{2}(X,\mu)$. $\Box$

Example 4

Take the non-atomic measure space $(X,\mu)=(\mathbb{T},dx)$, where $\mathbb{T}$ denotes the torus, and the orthonormal Fourier basis $\psi_{n}(x)=e^{2\pi ixn},$ $n\in\mathbb{Z}$, then

[TABLE]

Hence, there exists an upper semi-frame $\Psi$ with the closure of ${\sf Ran}\,C_{\Psi}$ being $L^{2}(\mathbb{T},dx)$, i.e., $R(\Psi)=0$.

4.4 Existence of duals for lower semi-frames

In this section, we fill a gap in the arguments of the proof of [3, Proposition 2.6], which states that for every lower semi-frame $\Psi$ there exists a dual Bessel mapping $\Phi$ such that $S_{\Psi,\Phi}=I$ on ${\sf Dom}\,C_{\Psi}$. While the result itself is correct, the construction of the dual system $\Phi$ in [3] is in general not well-defined. In particular, $\Phi$ is defined there by

[TABLE]

where $V:L^{2}(X,\mu)\rightarrow\mathcal{H}$ is a bounded operator depending on $\Psi$ only and $\{\phi_{n}\}_{n\in\mathbb{N}}$ is an orthonormal basis for $L^{2}(X,\mu)$. However, if $(X,\mu)$ is an-atomic, then by Corollary 13 there exists a set of positive measure $A$ such that $\sum|\phi_{n}(x)|^{2}=\infty,$ for all $x\in A$. Thus, $\Phi$ may be not well-defined on a set of positive measure.

Proposition 21** ([3], Proposition 2.6)**

If $\Psi$ is a lower semi-frame in $\mathcal{H}$, then there exists an upper semi-frame $\Phi$ such that

[TABLE]

Moreover, if ${\sf Dom}\,C_{\Psi}$ is dense in $\mathcal{H}$, then

[TABLE]

**Proof: **If $\Psi$ is a lower semi-frame, then ${\sf Ran}\,C_{\Psi}$ is a RKHS in $L^{2}(X,\mu)$ by Proposition 11. Moreover, let $P$ denote the orthogonal projection from $L^{2}(X,\mu)$ onto ${\sf Ran}\,C_{\Psi}$, and $\{e_{n}\}_{n\in\mathbb{N}}$ be an orthonormal basis for $\mathcal{H}$. Define the linear operator $V:L^{2}(X,\mu)\rightarrow\mathcal{H}$ by $V:=C_{\Psi}^{-1}$ on ${\sf Ran}\,C_{\Psi}$ and $V:=0$ on $({\sf Ran}\,C_{\Psi})^{\bot}$. Then $V$ is bounded and for all $f\in{\sf Dom}\,C_{\Psi},\ g\in\mathcal{H}$, it holds

[TABLE]

where $\Phi_{x}:=\sum_{n\in\mathbb{N}}\overline{(PV^{\ast}e_{n})}(x)e_{n}$. It remains to show that $\Phi_{x}$ is well-defined for every $x\in X$. Since $\{e_{n}\}_{n\in\mathbb{N}}$ is an orthonormal basis, if follows that $\Phi_{x}$ is well-defined if and only if

[TABLE]

By Proposition 6, it is sufficient to show that $\{PV^{\ast}e_{n}\}_{n\in\mathbb{N}}$ is a Bessel sequence on ${\sf Ran}\,C_{\Psi}$. If $F\in{\sf Ran}\,C_{\Psi}$, then

[TABLE]

as $PF=F$ and $V$ is bounded. Finally, it remains to show that $\Phi$ is Bessel. If $f\in\mathcal{H}$, then

[TABLE]

[TABLE]

as $\{PV^{\ast}e_{n}\}_{n\in\mathbb{N}}$ is Bessel. $\Box$

Remark 22

There is no analogue result of Proposition 21 if $\Psi$ is an upper semi-frame. In [5] it is shown that the affine coherent state system presented in Section 4.3 is a complete Bessel mapping with no dual.

5 Reproducing pairs and RKHSs

The absence of frame bounds causes trouble analyzing ${\sf Ran}\,C_{\Psi}$ and ${\sf Ran}\,C_{\Phi}$ of a reproducing pair $(\Psi,\Phi)$. Without an upper frame bound it is no longer guaranteed that ${\sf Ran}\,C_{\Psi}$ is a subspace of $L^{2}(X,\mu)$. The lower frame inequality, on the other hand, ensures that ${\sf Ran}\,C_{\Psi}$ is a RKHS. In [5], a construction of two mutually dual Hilbert spaces that are intrinsically generated by the pair $(\Psi,\Phi)$ is given. Let us first recall some of the results before explaining how RKHS enter the picture.

Let ${\mathcal{V}}_{\Phi}(X,\mu)$ be the space of all measurable functions $F:X\to\mathbb{C}$ such that

[TABLE]

Note that in general neither ${\mathcal{V}}_{\Phi}(X,\mu)\subset L^{2}(X,\mu)$ nor $L^{2}(X,\mu)\subset{\mathcal{V}}_{\Phi}(X,\mu)$. The linear map $T_{\Phi}:{\mathcal{V}}_{\Phi}(X,\mu)\rightarrow\mathcal{H}$ given weakly by

[TABLE]

is thus well-defined by Riesz representation theorem, and can be seen as the natural extension of the synthesis operator $D_{\Phi}$ (defined on ${\sf Dom}\,D_{\Phi}\subseteq L^{2}(X,\mu)$) to ${\mathcal{V}}_{\Phi}(X,\mu)$.

Let $(\Psi,\Phi)$ be a reproducing pair. According to [5], it then holds that

[TABLE]

This observation, together with the fact that $T_{\Phi}$ is general not one-to-one, motivates to define the redundancy for arbitrary complete mappings via

[TABLE]

We expect that similar results as in Section 4.1 hold for this definition.

Conjecture 1

If $R(\Phi)<\infty$, then $(X,\mu)$ is atomic.

The main difficulty is that there is no characterization of ${\mathcal{V}}_{\Phi}(X,\mu)$ that would allow to treat the problem in a similar manner than in Section 4.1 using (18). It is in particular not even clear if ${\mathcal{V}}_{\Phi}(X,\mu)$ is normable.

Let us introduce the following vector space

[TABLE]

equipped with the inner product

[TABLE]

This is indeed an inner product as $\langle{F},{F}\rangle_{\Phi}=0$ if and only if $F\in{\sf Ker}\,T_{\Phi}$. Hence, $V_{\Phi}(X,\mu)$ forms a pre-Hilbert space and $T_{\Phi}:V_{\Phi}(X,\mu)\rightarrow\mathcal{H}$ is an isometry. By (17) $\langle{\cdot},{\cdot}\rangle_{\Phi}$ can be written explicitly as

[TABLE]

The following result answers the question if, given a mapping $\Phi$, there exist another mapping $\Psi$ such that $(\Psi,\Phi)$ forms a reproducing pair.

Theorem 23** ([5], Theorem 4.1)**

Let $\Phi:X\rightarrow\mathcal{H}$ be a weakly measurable mapping and $\{e_{i}\}_{i\in\mathcal{I}}$ an orthonormal basis of $\mathcal{H}$. There exists another family $\Psi$, such that $(\Psi,\Phi)$ is a reproducing pair if and only if

1. (i)

${\sf Ran}\,T_{\phi}=\mathcal{H}$, 2. (ii)

there exists $\{\mathcal{E}_{i}\}_{i\in\mathcal{I}}\subset{\mathcal{V}}_{\Phi}(X,\mu)$ satisfying $T_{\Phi}\mathcal{E}_{i}=e_{i},\ \forall\ i\in\mathcal{I},$ and

[TABLE]

A reproducing partner $\Psi$ is then given by

[TABLE]

Theorem 23 is a powerful tool for the study of complete systems. It has for example been used to construct a reproducing partner for the Gabor system of integer time-frequency shifts of the Gaussian window [33] and to prove the non-existence of a dual for the system of affine coherent states in Example 3.

Let us briefly discuss the conditions $(i)$ and $(ii)$ and some conceptual interpretations. For a complete system one can show that (under mild conditions [3, Lemma 2.2]) $\overline{{\sf Ran}\,D_{\Phi}}=\mathcal{H}$. It might therefore seem that $(i)$ is trivially satisfied for complete systems since $T_{\Phi}$ extends $D_{\Phi}$ to its domain ${\mathcal{V}}_{\Phi}(X,\mu)$. The complete upper semi-frame from Example 3 however does not satisfy $(i)$, see [5, Section 6.2.3].

Coefficient map interpretation: Property $(i)$ ensures the existence of a linear coefficient map $A:\mathcal{H}\rightarrow{\mathcal{V}}_{\Phi}(X,\mu)$ satisfying $f=T_{\Phi}A(f)$ for every $f\in\mathcal{H}$. Property $(ii)$ then guarantees that $A(f)$ can be calculated taking inner products of $f$ with a second mapping $\Psi:X\rightarrow\mathcal{H}$.

RKHS interpretation: If $(i)$ and $(ii)$ are satisfied, then it follows that $\{\mathcal{E}_{i}\}_{i\in\mathcal{I}}$ forms an orthonormal family with respect to $\langle{\cdot},{\cdot}\rangle_{\Phi}$, since $(ii)$ implies

[TABLE]

Hence, $\{\mathcal{E}_{i}\}_{i\in\mathcal{I}}$ forms an orthonormal basis for

[TABLE]

Theorem 3 together with (21) thus ensure that $\mathcal{H}_{K}^{\Phi}$ is a RKHS. Moreover, the definition of the reproducing partner $\Psi$ in (22) yields that

[TABLE]

To put it another way, $(i)$ and $(ii)$ guarantee the existence of a RKHS $\mathcal{H}_{K}^{\Phi}\subset{\mathcal{V}}_{\Phi}(X,\mu)$ reproducing $\mathcal{H}$ in the sense that $T_{\Phi}(\mathcal{H}_{K}^{\Phi})=\mathcal{H}$.

Let us assume that $(\Psi,\Phi)$ is a reproducing pair. There is a natural way to generate frames on $\mathcal{H}$ and $\mathcal{H}_{K}^{\Phi}$ via the analysis and synthesis operators.

Proposition 24

Let $(\Psi,\Phi)$ be a reproducing pair for $\mathcal{H}$, $\{g_{i}\}_{i\in\mathcal{I}}$ a frame for $\mathcal{H}$, and $\{G_{i}\}_{i\in\mathcal{I}}$ a frame for $\mathcal{H}_{K}^{\Phi}$. If $H_{i}(x):=\langle{g_{i}},{\Psi_{x}}\rangle$ and $h_{i}:=T_{\Phi}G_{i}$, then $\{H_{i}\}_{i\in\mathcal{I}}$ is a frame for $\mathcal{H}_{K}^{\Phi}$ and $\{h_{i}\}_{i\in\mathcal{I}}$ is a frame for $\mathcal{H}$.

**Proof: **If $F\in\mathcal{H}_{K}^{\Phi}$, then

[TABLE]

[TABLE]

[TABLE]

The lower bound follows from the same argument as $(S_{\Psi,\Phi})^{\ast}$ is boundedly invertible. Hence, $\{H_{i}\}_{i\in\mathcal{I}}$ is a frame for $\mathcal{H}_{K}^{\Phi}$.

For $f\in\mathcal{H}$, we have

[TABLE]

which, together with

[TABLE]

yields that $\{h_{i}\}_{i\in\mathcal{I}}$ is a frame for $\mathcal{H}$. $\Box$

The rest of this section is concerned with the explicit calculation of the reproducing kernel for $\mathcal{H}_{K}^{\Phi}$. For a reproducing pair $(\Psi,\Phi)$, there exists a similar characterization of the range of the analysis operators as for frames. In particular, if $R_{\Psi,\Phi}(x,y):=\langle{S_{\Psi,\Phi}^{-1}\Phi_{y}},{\Psi_{x}}\rangle$ defines the integral operator

[TABLE]

then by [32, Proposition 2] $\mathcal{R}_{\Psi,\Phi}(F)(x)=F(x)$ if and only if there exists $f\in\mathcal{H}$ such that $F(x)=\langle{f},{\Psi_{x}}\rangle$, for all $x\in X$. However, $R_{\Psi,\Phi}$ is not the reproducing kernel for $\mathcal{H}_{K}^{\Phi}$ since the reproducing formula is based on the inner product of $L^{2}(X,\mu)$ and not on $\langle\cdot,\cdot\rangle_{\Phi}$.

If $F\in{\sf Ran}\,C_{\Psi}$, then (20) and the identity $f=T_{\Phi}C_{\Psi}S_{\Psi,\Phi}^{-1}f$ yield

[TABLE]

Hence, using $(S_{\Psi,\Phi}^{-1})^{\ast}=S_{\Phi,\Psi}^{-1}$, we finally obtain that

[TABLE]

is the reproducing kernel for $\mathcal{H}_{K}^{\Phi}$.

6 Conclusion

With the results of Section 4 in mind, we suggest to use the term continuous frame only in the case of a strictly continuous frame, and semi-continuous or semi-discrete frame if it can be decomposed into nontrivial strictly continuous and discrete parts. Moreover, the notion of a continuous Riesz basis (Riesz type mapping) should not be used any further as every such system can be written as a discrete Riesz basis.

An interesting topic for future research is to find and study alternative notions of redundancy for continuous frames. A promising approach that may be adapted can be found in [10]. Exploring the dependence of any notion of redundancy on the underlying measure space should remain a key task.

Appendix

Proof of Lemma 2: Ad $(i)$: See [18].

Ad $(ii)$: Let $(X,\mu)$ be an-atomic. Let us assume to the contrary that for every measurable set $A\subset X$ with $\mu(A)>0$ there exists an atom $B\subset A$, and take $\{A_{n}\}_{n\in\mathbb{N}}\subset X$ to be a countable partition of $X$ by sets of finite measure. We show that each $A_{n}$ can be partitioned into atoms and null sets, a contradiction. Assume without loss of generality that $\mu(A_{1})>0$. By assumption, there exists an atom $B_{1}\subset A_{1}$. If $\mu(B_{1})=\mu(A_{1})$, then $A_{1}$ is an atom. If $0<\mu(B_{1})<\mu(A_{1})$, then $\mu(A_{1}\backslash B_{1})>0$. Hence, there exists an atom $B_{2}\subset A_{1}\backslash B_{1}$ and the preceding argument can be repeated. If one has \mu\big{(}A_{1}\backslash\big{(}\bigcup_{k=1}^{K}B_{k}\big{)}\big{)}>0 for all iteration steps $K$, then \mu_{K}:=\mu\big{(}\bigcup_{k=1}^{K}B_{k}\big{)} defines a strictly increasing sequence, bounded by $\mu(A_{1})$. Hence, $\mu_{K}$ is convergent to some $\mu^{\ast}$ and the limit equals $\mu(A_{1})$. Indeed, if $\mu^{\ast}<\mu(A_{1})$ then, by assumption, there exists an atom $B^{\ast}\subset A_{1}\backslash\bigcup_{k\in\mathbb{N}}B_{k}$ and

[TABLE]

a contradiction. Consequently, $A_{1}=\bigcup_{k\in\mathbb{N}}B_{k}\cup N,$ where $N=A_{1}\backslash\bigcup_{k\in\mathbb{N}}B_{k}$ is of measure zero. In particular, we constructed a partition of $A_{1}$ consisting of atoms and null sets. Repeating this argument for every $A_{n}$, $n\in\mathcal{I}$, with $\mu(A_{n})>0$ shows that $(X,\mu)$ is atomic, a contradiction. $\Box$

Acknowledgement

This work was funded by the Austrian Science Fund (FWF) START-project FLAME (’Frames and Linear Operators for Acoustical Modeling and Parameter Estimation’; Y 551-N13).

The authors would like to thank Jean-Pierre Antoine for fruitful discussions on the physical interpretation of the result on the redundancy of continuous frames. The second authors thanks Nora Simovich for help with typing.