# Frame-related sequences in chains and scales of Hilbert spaces

Peter Balazs  ,  Giorgia Bellomonte  and  Hessam Hosseinnezhad  Acoustics
Research Institute, Austrian Academy of Sciences, Wohllebengasse 12-14, A-1040
Vienna, Austria<EMAIL_ADDRESS>Dipartimento di Matematica e
Informatica, Università degli Studi di Palermo, I-90123 Palermo, Italy;
<EMAIL_ADDRESS>Department of Basic Sciences, Shahid Sattari
Aeronautical University of Science and Technology, Tehran, Iran;
<EMAIL_ADDRESS>
###### Abstract.

Frames for Hilbert spaces are interesting for mathematicians but also
important for applications e.g. in signal analysis and in physics. Both in
mathematics and physics it is natural to consider a full scale of spaces, and
not only a single one. In this paper, we study how certain frame-related
properties, as completeness or the property of being a (semi-)frame, of a
certain sequence in one of the spaces propagate to other spaces in a scale. We
link that to the properties of the respective frame-related operators, like
analysis or synthesis. We start with a detailed survey of the theory of
Hilbert chains. Using a canonical isomorphism the properties of frame
sequences are naturally preserved between different spaces. We also show that
some results can be transferred if the original sequence is considered, in
particular that the upper semi-frame property is kept in larger spaces, while
the lower one to smaller ones. This leads to a negative result: a sequence can
never be a frame for two Hilbert spaces of the scale if the scale is non-
trivial, i.e. spaces are not equal.

###### Key words and phrases:

Frames; Scales of Hilbert Spaces; Hilbert chains; Bessel sequences; Semi-
frames.

###### 2010 Mathematics Subject Classification:

42C15, 46C99, 47A70

##  1\. Introduction

Frames have been used as a powerful alternative to Hilbert space bases, and
they allow a deep theory (for an overview see [21, 23, 37]). They are also
very important for applications, e.g. in physics [1, 26], signal processing
[7, 17, 19], numerical treatment of operator equations [27, 43] and acoustics
[11, 39].

There have been numerous generalizations of the concept of frames, see e.g.
[1, 13, 14, 22, 24, 33], among others.

The basic idea in the work by Duffin and Schaeffer [28] was to have a sequence
of elements in a Hilbert space, that allow redundant and stable
representations. It is often more natural to not only consider a single
Hilbert space, but a whole chain or scale of spaces [2, 34, 45, 9]. Therefore,
aiming at an extension of the concept of frames to such a setting is very
natural. Several approaches have already been done, see Gelfand frames [27],
or Riesz-like bases in rigged Hilbert spaces [15]. Those concepts “only” deal
with a triplet of spaces, while here we will work on the concept of frames and
related objects on a full scale. Applications of frames for scales of Hilbert
spaces can be found in the discretization of operators (as in [6, 8, 10, 36,
43]).

For localized frames [8, 32] a natural scale of (Banach) spaces is associated
to a frame in a central Hilbert space. For this concept the frame-related
properties get naturally shared on all spaces. In this manuscript we
investigate how frame-related properties are transferred in a general scale of
Hilbert space, motivated by the applications using e.g. Sobolev spaces [29].

Let the increasing chain of Hilbert spaces be given [2]:

$...\subseteq\mathcal{H}_{2}\subseteq\mathcal{H}_{1}\subseteq\mathcal{H}_{0}\subseteq\mathcal{H}_{-1}\subseteq\mathcal{H}_{-2}\subseteq...,$

with dense inclusions and $\mathcal{H}_{-n}=\mathcal{H}_{n}^{\times}$, where
$\mathcal{H}_{n}^{\times}$ is the antidual/conjugate dual of $\mathcal{H}_{n}$
with respect to the inner product of $\mathcal{H}_{0}$ [4] i.e. the space of
continuous conjugate linear functionals on the Hilbert space $\mathcal{H}_{n}$
[41].

If such a chain is generated by the domain of an operator [5], we call this a
scale of Hilbert spaces. Note that, a chain of three Hilbert spaces generates
always a scale of Hilbert spaces, see Section 3.3.2. For a similar definition,
see the one of nested Hilbert space [35], where only a partial order is
assumed. See also the similar concept of Gelfand chain [40]. In this paper, we
will look at frame-like sequences for such a scale, and how properties of
those sequence are transferred between spaces.

This paper is organized as follows: In Section 2 we collect results from the
literature and fix the notation. In Section 3 we give a survey for chains of
Hilbert spaces, following literature to some extent but describing it from a
new point of view, focusing on the so-called Berezanskii isomorphisms between
two arbitrary spaces of the scale. In Section 4 we study how certain frame-
related properties of a certain sequence in a space of a scale of Hilbert
spaces and of operators directly linked to it propagate in the whole scale.

##  2\. Known facts, definitions and notation

Before going forth, let us introduce some notations and recall the main
definitions in literature we are going to use.

Let $\mathcal{H},\mathcal{K}$ be two separable infinite dimensional Hilbert
spaces, with inner products
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{\mathcal{H}}$ resp.
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{\mathcal{K}}$, chosen
to be linear in the first entry, and the induced norms
$\|\cdot\|_{\mathcal{H}}$ resp. $\|\cdot\|_{\mathcal{K}}$. A bounded operator
$A\in\mathcal{B}(\mathcal{H},\mathcal{K})$ is said to be unitary if it is
isometric, i.e. $\|Af\|_{\mathcal{K}}=\|f\|_{\mathcal{H}}$ for every
$f\in\mathcal{H}$ and is onto, i.e. its range ${\sf Ran}(A)=\mathcal{K}$. If
we refer to a full sequence we will denote it by a letter without an index,
i.e. $c=\left(c_{k}\right)$.

###  2.1. Frames in Hilbert Spaces

Let now $\mathcal{H}$ be a separable Hilbert space with inner product
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle$ and norm $\|\cdot\|$. A
sequence $(\psi_{k})\subset\mathcal{H}$ is said to be

  * •

complete (or total) if ${\sf span}(\psi_{k})$, the linear span of
$(\psi_{k})$, is dense in $\mathcal{H}$;

  * •

a frame for $\mathcal{H}$ if there exist $A>0$ and $B<\infty$ such that

(1)
$A\|f\|^{2}\leq\sum\limits_{k\in\mathbb{N}}|\left\langle{f}\left|{\psi_{k}}\right.\right\rangle|^{2}\leq
B\|f\|^{2},\qquad\forall f\in\mathcal{H};$

  * •

a Bessel sequence in $\mathcal{H}$ if there exists $B>0$ such that the upper
inequality in (1) holds true. It is called an upper semi-frame [3], if the
Bessel sequence is also complete (this is equivalent to
$0<\sum\limits_{k\in\mathbb{N}}|\left\langle{f}\left|{\psi_{k}}\right.\right\rangle|^{2}\leq
B||f||^{2},\,\,\forall f\not=0$);

  * •

a lower semi-frame for $\mathcal{H}$ if it satisfies the lower frame
inequality in (1);

  * •

a Riesz basis for $\mathcal{H}$ if there exist an orthonormal basis $(e_{k})$
for $\mathcal{H}$ and a bounded bijective operator
$T:\mathcal{H}\to\mathcal{H}$ such that $\psi_{k}=Te_{k}$ for all
$k\in\mathbb{N}$.

There are several operators canonically associated to a sequence
$\psi=(\psi_{k})$ of elements of a Hilbert space $\mathcal{H}$. The analysis
operator $C_{\psi}:{\sf Dom}(C_{\psi})\subseteq\mathcal{H}\to\ell^{2}$ of
$(\psi_{k})$ is defined by

$C_{\psi}f=(\left\langle{f}\left|{\psi_{k}}\right.\right\rangle),\qquad\forall
f\in{\sf Dom}(C_{\psi}),\text{ where }$ ${\sf
Dom}(C_{\psi})=\left\\{f\in\mathcal{H}:\sum_{k\in\mathbb{N}}|\left\langle{f}\left|{\psi_{k}}\right.\right\rangle|^{2}<\infty\right\\}.$

The synthesis operator $D_{\psi}:{\sf
Dom}(D_{\psi})\subseteq\ell^{2}\to\mathcal{H}$ of $(\psi_{k})$ is defined on
the dense domain

${\sf
Dom}(D_{\psi}):=\left\\{{c}\in\ell^{2}:\sum_{k\in\mathbb{N}}c_{k}\psi_{k}\text{
is convergent in }\mathcal{H}\right\\}$

by

$D_{\psi}(c_{k})=\sum_{k\in\mathbb{N}}c_{k}\psi_{k},\qquad\forall(c_{k})\in{\sf
Dom}(D_{\psi}).$

It is known [12] that we have for any sequence $C_{\psi}=D_{\psi}^{*}$ where,
as usual, $D_{\psi}^{*}$ indicates the adjoint of the operator $D_{\psi}$.
Other two operators associated to a sequence
$\psi=(\psi_{k})\subset\mathcal{H}$ are the frame operator111Note that for the
definition of the frame operator as a potentially unbounded operator the
sequence does not have to be a frame. $S_{\psi}:{\sf
Dom}(S_{\psi})\subseteq\mathcal{H}\to\mathcal{H}$ of $\psi$

$S_{\psi}f:=\sum_{k\in\mathbb{N}}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle\psi_{k}$

where

${\sf
Dom}(S_{\psi})=\\{f\in\mathcal{H}:\sum_{k\in\mathbb{N}}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle\psi_{k}\mbox{
converges in }\mathcal{H}\\};$

and the Gram operator $G_{\psi}:{\sf
Dom}(G_{\psi})\subseteq\ell^{2}\to\ell^{2},$ with

${\left(G_{\psi}c\right)_{k}:=\sum_{l\in\mathbb{N}}(G_{\psi})_{k,l}\cdot
c_{l}},$

where

${\sf
Dom}(G_{\psi})=\\{{c}\in\ell^{2}:\sum_{l\in\mathbb{N}}(G_{\psi})_{k,l}c_{l}\mbox{
converges }\forall k\in\mathbb{N}\mbox{ and is in }\ell^{2}\\}$

and the Gram matrix $((G_{\psi})_{k,l})_{k,l}$ is defined by
$(G_{\psi})_{k,l}=\left\langle{\psi_{l}}\left|{\psi_{k}}\right.\right\rangle$,
${k,l}\in\mathbb{N}$.

If we combine those operators we end up at the following definition: let
$\psi=(\psi_{k})$ and $\phi=(\phi_{k})$ be two sequences in $\mathcal{H}$. The
pair $\left(\psi,\phi\right)$ is called

  * •

a reproducing pair if the cross-frame operator defined by

$\left<S_{\psi,\phi}f,g\right>=\sum\limits_{k}\left<f,\psi_{k}\right>\left<\phi_{k},g\right>$

is an invertible, bounded operator [42].

  * •

a weakly dual pair [31] if it is a reproducing pair and
$S_{\psi,\phi}=\mathsf{id}$.

Clearly, for every reproducing pair, the pair
$\left(\psi,S_{\psi,\phi}^{-1}\phi\right)$ is weakly dual.

###  2.2. Rigged Hilbert spaces

In a formulation using a topology viewpoint, where not all involved spaces
have to be normed, see e.g. [15, 44], a rigged Hilbert space (RHS) consists of
a triplet $(\mathcal{D},\mathcal{H},\mathcal{D}^{\times})$ where $\mathcal{D}$
is a dense subspace of $\mathcal{H}$ endowed with a locally convex topology
$t$, finer than that induced by the Hilbert norm of $\mathcal{H}$, and
$\mathcal{D}^{\times}$ is the conjugate dual of $\mathcal{D}[{t}]$, endowed
with the strong topology
${t}^{\times}:=\beta(\mathcal{D}^{\times},\mathcal{D})$. We have

$\mathcal{D}[t]\hookrightarrow\mathcal{H}\hookrightarrow\mathcal{D}^{\times}[t^{\times}],$

where $\hookrightarrow$ denotes a continuous embedding and since
$\mathcal{D}^{\times}$ contains a subspace that can be identified with
$\mathcal{H}$, we will read (2.2) as a chain of topological inclusions:
$\mathcal{D}[t]\subset\mathcal{H}\subset\mathcal{D}^{\times}[t^{\times}]$.
These identifications imply that the sesquilinear form $B(\cdot,\cdot)$ that
puts $\mathcal{D}$ and $\mathcal{D}^{\times}$ in duality is an extension of
the inner product of $\mathcal{D}$:
$B(f,g)=\left\langle{f}\left|{g}\right.\right\rangle$, for every
$f,g\in\mathcal{D}$ (as usual, to simplify notations we adopt the symbol
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle$ for both of them).

Let now
$\mathcal{D}[t]\subset\mathcal{H}\subset\mathcal{D}^{\times}[t^{\times}]$ be a
rigged Hilbert space, and let
${\mathfrak{L}}(\mathcal{D},\mathcal{D}^{\times})$ denote the vector space of
all continuous linear maps from $\mathcal{D}[t]$ into
$\mathcal{D}^{\times}[t^{\times}]$. If $\mathcal{D}[t]$ is barreled (e.g.
reflexive), an involution $X\mapsto X^{\mbox{\scriptsize${\dagger}$}}$ can be
introduced in ${\mathfrak{L}}(\mathcal{D},\mathcal{D}^{\times})$ by the
equality

(2)
$\left\langle{X^{\mbox{\scriptsize${\dagger}$}}\eta}\left|{\xi}\right.\right\rangle=\overline{\left\langle{X\xi}\left|{\eta}\right.\right\rangle},\quad\forall\xi,\eta\in\mathcal{D}$

and ${\mathfrak{L}}(\mathcal{D},\mathcal{D}^{\times})$ results to be a
†-invariant vector space.
We will show in Section 3.3.1 how this can be expressed and extended in a
Hilbert space setting.

###  2.3. Scales of Hilbert Spaces

Let $\mathcal{H}_{0}$ be a Hilbert space with inner product
$\left<.,.\right>_{0}$. Let $A$ be a self-adjoint, strictly positive,
unbounded operator with domain ${\sf Dom}(A)\subset\mathcal{H}_{0}$ and range
${\sf Ran}(A)\subset\mathcal{H}_{0}$, without loss of generality let
$A=A^{*}\geq 1$. By assumption, this operator has dense domain, dense range
and is closed. It has closed range, and is therefore onto. Its inverse
$A^{-1}:\mathcal{H}_{0}\rightarrow{\sf Dom}(A)$ is bounded and self-adjoint.

The same argument holds for any $A^{k}$ and $A^{-k}$, $k\in\mathbb{N}$.
Define, for each $k\geq 0$, the Hilbert space $\mathcal{H}_{+k}=\left\\{{\sf
Dom}(A^{k/2}),\|\cdot\|_{+k}\right\\}$ i.e. the domain of $A^{k/2}$ equipped
with the norm $\|\cdot\|_{+k}$ induced by the inner product
$\left\langle{x}\left|{y}\right.\right\rangle_{+k}:=\left\langle{A^{k/2}x}\left|{A^{k/2}y}\right.\right\rangle_{0},x,y\in{\sf
Dom}(A^{k/2})$. Therefore $A^{k}:\mathcal{H}_{+2k}\rightarrow\mathcal{H}_{0}$
is bounded, even unitary.

Define $\mathcal{H}_{-k}$ as the completion of $\mathcal{H}_{0}$ with respect
to the norm $\|\cdot\|_{-k}$ induced by the inner product
$\left\langle{f}\left|{g}\right.\right\rangle_{-k}:=\left\langle{A^{-{k/2}}f}\left|{A^{-{k/2}}g}\right.\right\rangle_{0},f,g\in\mathcal{H}_{0}$.
Clearly, $\mathcal{H}_{0}\subset(\mathcal{H}_{+k})^{\prime}$ densely (where we
apply the Riesz isomorphism for
$\mathcal{H}_{0}\cong(\mathcal{H}_{0})^{\prime}$). We have for
$f\in\mathcal{H}_{0}$

$\|f\|_{\left(\mathcal{H}_{+k}\right)^{\prime}}=\sup\limits_{\|g\|_{+k}=1}|\left<f,g\right>_{0}|=\sup\limits_{\|A^{k/2}g\|_{0}=1}\left|\left<f,g\right>_{0}\right|=$
$=\sup\limits_{\|h\|_{0}=1}|\left<f,A^{-k/2}h\right>_{0}|=\sup\limits_{\|h\|_{0}=1}|\left<A^{-k/2}f,h\right>_{0}|=\|A^{-k/2}f\|_{0}.$

and so the norms are equivalent. Therefore,
$(\mathcal{H}_{+k})^{\prime}=\mathcal{H}_{-k}$. For each fixed $k>0$ the
triplet

$\mathcal{H}_{+k}\subset\mathcal{H}_{0}\subset\mathcal{H}_{-k}$

is a rigged Hilbert space with dense inclusions. The chain of Hilbert spaces
infinite in both sides

$...\subset\mathcal{H}_{+k}\subset...\subset\mathcal{H}_{0}\subset...\subset\mathcal{H}_{-k}\subset...$

is called $A$-scale of Hilbert spaces [38].

An important property of an $A$-scale is the invariance of the structure of
rigged triple
$\mathcal{H}_{+k}\subseteq\mathcal{H}_{0}\subseteq\mathcal{H}_{-k}$ under
shift along the $A$-scale, i.e. shift of the index $k$, this property has been
called the _first invariance principle_ of the $A$-scale [38]: for any fixed
$k>0$ and an arbitrary $n$, the triple of spaces
$\mathcal{H}_{n+k}\subseteq\mathcal{H}_{n}\subseteq\mathcal{H}_{n-k}$ is a new
rigged Hilbert space, where in particular the Hilbert space
$\mathcal{H}_{n-k}$ is the dual of $\mathcal{H}_{n+k}$, which becomes
apparent, if we do the above construction taking $\mathcal{H}_{n}$ as the
“central space”.

##  3\. Hilbert chains

In this section we include a detailed introduction to Hilbert chains, to some
extent following [18], albeit extended and reformulated. We have done this to
make the manuscript self-contained, stress results not well known in the frame
theory community and adapt the respective point-of-view. We have added details
and covered new side-aspects.

###  3.1. Hilbert Triplets

This section deals with canonical Hilbert triplets and how one can define
operators between them.

Let $\mathcal{H}_{0}$ be a separable, infinite dimensional Hilbert space with
scalar product $\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{0}$ and
norm $\|\cdot\|_{0}$. Fix any $i\in\mathbb{N}\setminus\\{0\\}$, and let
$\mathcal{H}_{+i}$ be a dense subspace of $\mathcal{H}_{0}$,
$\mathcal{H}_{+i}\subseteq\mathcal{H}_{0}$, complete w.r.to the norm
$\|\cdot\|_{+i}$ induced by the inner product
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{+i}$ with

(3) $\|x\|_{0}\leq\|x\|_{+i},\quad\forall x\in\mathcal{H}_{+i}.$

Let $\iota_{+i,0}$ be the inclusion of $\mathcal{H}_{+i}$ into
$\mathcal{H}_{0}$, which by (3) is bounded. We have

(4)
$\left\langle{f}\left|{x}\right.\right\rangle_{0}=\left\langle{f}\left|{\iota_{+i,0}x}\right.\right\rangle_{0}=\left\langle{\iota_{+i,0}^{*}f}\left|{x}\right.\right\rangle_{+i},\quad\forall
f\in\mathcal{H}_{0},x\in\mathcal{H}_{+i}.$

Define

(5)
$\iota_{0,+i}:=\iota_{+i,0}^{*}:\mathcal{H}_{0}\rightarrow\mathcal{H}_{+i},$

where we consider the Hilbert space adjoints, i.e. apply the Riesz
isomorphisms for both Hilbert spaces.

We have $\|\iota_{+i,0}\|\leq 1$ and $\iota_{+i,0}$ is injective (one-to-one)
and has dense range, but cannot be onto. Therefore
$\iota_{0,+i}:=\iota_{+i,0}^{*}$ has the same properties.

On the other hand, let us note that, naturally,

$\left\langle{x}\left|{y}\right.\right\rangle_{+i}\not=\left\langle{\iota_{+i,0}x}\left|{\iota_{+i,0}y}\right.\right\rangle_{0},\quad
x,y\in\mathcal{H}_{+i}$

because else (3) would be an equality and the two spaces would collapse to
only one space.

Let us define now the scalar product

(6)
$\left\langle{f}\left|{g}\right.\right\rangle_{-i}:=\left\langle{\iota_{0,+i}f}\left|{g}\right.\right\rangle_{0}=\left\langle{\iota_{0,+i}f}\left|{\iota_{0,+i}g}\right.\right\rangle_{+i},\quad
f,g\in\mathcal{H}_{0}$

and consider the completion $\mathcal{H}_{-i}$ of $\mathcal{H}_{0}$ with
respect to the norm $\|\cdot\|_{-i}$ induced by the inner product defined in
(6). Then, from $\|\iota_{0,+i}\|\leq 1$ it follows that
$\|f\|_{-i}\leq\|f\|_{0}$ $\forall f\in\mathcal{H}_{0}$, and we have:

$\mathcal{H}_{+i}\subset\mathcal{H}_{0}\subset\mathcal{H}_{-i},$

with dense inclusions, and

$\|x\|_{-i}\leq\|x\|_{0}\leq\|x\|_{+i},\quad\forall x\in\mathcal{H}_{+i}.$

Since $\iota_{0,+i}$ maps a dense subspace of $\mathcal{H}_{-i}$ (i.e.
$\mathcal{H}_{0}$) into $\mathcal{H}_{+i}$ we consider the (unique!) bounded
extension
$I_{-i,+i}=\overline{\iota_{0,+i}}\in\mathcal{B}(\mathcal{H}_{-i},\mathcal{H}_{+i})$.

As $\iota_{0,+i}$ has dense range, $I_{-i,+i}$ is onto (surjective). By (6),
$I_{-i,+i}$ is an isometry,

$\left\langle{\alpha}\left|{\beta}\right.\right\rangle_{-i}=\left\langle{I_{-i,+i}\alpha}\left|{I_{-i,+i}\beta}\right.\right\rangle_{+i},\quad\alpha,\beta\in\mathcal{H}_{-i}.$

Therefore, $I_{-i,+i}$ is a unitary operator, hence

$I_{+i,-i}:=I_{-i,+i}^{*}=I_{-i,+i}^{-1}\in\mathcal{B}(\mathcal{H}_{+i},\mathcal{H}_{-i})$

is a unitary operator too. The operators
$I_{-i,+i}\in\mathcal{B}(\mathcal{H}_{-i},\mathcal{H}_{+i})$ and
$I_{+i,-i}\in\mathcal{B}(\mathcal{H}_{+i},\mathcal{H}_{-i})$ are called
Berezanskii canonical isomorphisms (see e.g. [38]). This is a particular
instance of the Riesz isomorphism, if a particular duality (with the pivot
space $\mathcal{H}_{0}$) is chosen.

###  3.2. Duality by Pivot Spaces

Now we show that $\mathcal{H}_{0}$ can be considered a pivot space of
$\mathcal{H}_{-i}$ and $\mathcal{H}_{+i}$ in the sense that the scalar product
$\left\langle{\alpha}\left|{x}\right.\right\rangle_{0}$ defines a duality
relation. Consider the bilinear form
$b(f,x)=\left\langle{f}\left|{x}\right.\right\rangle_{0}$ defined on
$\mathcal{H}_{0}\times\mathcal{H}_{+i}$. Extend it by continuity to the
bilinear form

$B_{-i,+i}:(\alpha,x)\in\mathcal{H}_{-i}\times\mathcal{H}_{+i}\to\mathbb{C}.$

For $f\in\mathcal{H}_{0}$ and $x\in\mathcal{H}_{+i}$ we have

$\displaystyle|\left\langle{f}\left|{x}\right.\right\rangle_{0}|$
$\displaystyle=$
$\displaystyle|\left\langle{\iota_{0,+i}f}\left|{x}\right.\right\rangle_{+i}|=|\left\langle{I_{-i,+i}f}\left|{x}\right.\right\rangle_{+i}|$
$\displaystyle\leq$
$\displaystyle\|I_{-i,+i}f\|_{+i}\|x\|_{+i}=\|f\|_{-i}\|x\|_{+i}$

By a limit argument we obtain

(7) $|B_{-i,+i}(\alpha,x)|\leq\|\alpha\|_{-i}\|x\|_{+i},$

for $\alpha\in\mathcal{H}_{-i},x\in\mathcal{H}_{+i}$. Therefore $B_{-i,+i}$ is
continuous.

######  Remark 3.1.

The form $B_{-i,+i}(\cdot,\cdot)$ that puts $\mathcal{H}_{+i}$ and
$\mathcal{H}_{-i}$ in duality is an extension of the inner product
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{0}$ and we will use the
latter symbol for both of them.

By a limit argument (and the conjugate symmetry of any inner product), we
obtain for $\alpha,\beta\in\mathcal{H}_{-i},x,y\in\mathcal{H}_{+i}$:

(8)
$\left\langle{\alpha}\left|{x}\right.\right\rangle_{0}=\left\langle{I_{-i,+i}\alpha}\left|{x}\right.\right\rangle_{+i}\text{
and
}\left\langle{x}\left|{y}\right.\right\rangle_{+i}=\left\langle{I_{+i,-i}x}\left|{y}\right.\right\rangle_{0}$

and

$\left\langle{\alpha}\left|{\beta}\right.\right\rangle_{-i}=\left\langle{I_{-i,+i}\alpha}\left|{\beta}\right.\right\rangle_{0}=\left\langle{\alpha}\left|{I_{-i,+i}\beta}\right.\right\rangle_{0}=\left\langle{I_{-i,+i}\alpha}\left|{I_{-i,+i}\beta}\right.\right\rangle_{+i}.$

By Remark 3.1 we see that $\alpha\in\mathcal{H}_{-i}$ is in
$(\mathcal{H}_{+i})^{\prime}$ with the same norm. On the other hand, any
functional $L$ on $\mathcal{H}_{+i}$ can be represented by a
$y\in\mathcal{H}_{+i}$, i.e.

$L(x)=\left\langle{y}\left|{x}\right.\right\rangle_{+i}=\left\langle{I_{+i,-i}\,y}\left|{x}\right.\right\rangle_{0}.$

As $I_{+i,-i}$ is an isometry we have shown:

######  Remark 3.2.

This construction corresponds to considering the dual pair
$(\mathcal{H}_{+i},\mathcal{H}_{0})$ using
$\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{0}$ and choosing
$\mathcal{H}_{-i}=\overline{\mathcal{H}_{0}}^{\sigma(\mathcal{H}_{+i},\mathcal{H}_{0})}$
using the weak dual topology [25]. $\mathcal{H}_{-i}$ is therefore a
representation of the dual space of $\mathcal{H}_{+i}$.

Note that this is not a pure abstract baublery, but important for concrete
spaces. While it is true that “everything is isomorphic anyway”, this
isomorphisms cannot be thought of as equality if concrete choices for
$\mathcal{H}_{+i}$ and $\mathcal{H}_{-i}$ are worked with, for example in the
setting of Sobolev spaces [36]. See [8] for an application of this topology to
the more general setting of coorbit spaces of localized frames.

######  Remark 3.3.

One might expect that there is a clear link of $I_{+i,-i}$ to the inclusion
$\iota_{+i,-i}$. But clearly, the relation cannot be trivial, e.g. an
extension, as the former operator is bijective, the latter is injective and
has dense range and both are bounded on all of $\mathcal{H}_{+i}$.

At the same time, it could be interesting to look at the role of the inclusion
of $\mathcal{H}_{0}$ into $\mathcal{H}_{-i}$, i.e. $\iota_{0,-i}$ in this
setting. But again, it cannot be trivially linked to $I_{-i,+i}$, as e.g.
$\iota_{0,-i}$ cannot be $\overline{{I_{-i,+i}}_{|_{\mathcal{H}_{0}}}}$
because else it would be onto. For more on that, see Remark 3.7.

###  3.3. Hilbert Chains

Now let us add a step more. Consider a dense subspace $\mathcal{H}_{+j}$ of
the Hilbert space $\mathcal{H}_{+i}$, with $j>i$ ($i,j\in\mathbb{N}$),
complete w.r. to the norm $\|\cdot\|_{+j}$, induced on $\mathcal{H}_{+j}$ by
the inner product $\left\langle{\cdot}\left|{\cdot}\right.\right\rangle_{+j}$,
such that $\|x\|_{+i}\leq\|x\|_{+j}$ with $x\in\mathcal{H}_{+j}$. Then
$\mathcal{H}_{+j}$ is a dense subspace of $\mathcal{H}_{0}$, too, and
$\|x\|_{0}\leq\|x\|_{+j}$ with $x\in\mathcal{H}_{+j}$. If we consider the
completion $\mathcal{H}_{-j}$ of $\mathcal{H}_{0}$ with respect to the norm
$\|\cdot\|_{-j}$ defined like in (6) (with $i=j$ and
$\iota_{0,+j}:\mathcal{H}_{0}\to\mathcal{H}_{+j}\subset\mathcal{H}_{+i}$,
defined like in (5) as the adjoint of the inclusion), we have for every
$x\in\mathcal{H}_{+j}$, and every $j>i$, $i\in\mathbb{N}$:

$\mathcal{H}_{+j}\subset\mathcal{H}_{+i}\subset\mathcal{H}_{0}\subset\mathcal{H}_{-i}\subset\mathcal{H}_{-j},$

where every inclusion is dense. Indeed, let us prove that
$\mathcal{H}_{-i}\subset\mathcal{H}_{-j}$ with $j>i$. For
$f\in\mathcal{H}_{0}$ we have from (6) and (7)

$\|f\|_{-j}^{2}=\left\langle{\iota_{0,+j}f}\left|{f}\right.\right\rangle_{0}\leq\|\iota_{0,+j}f\|_{+i}\|f\|_{-i}=\|I_{-j,+j}f\|_{+i}\|f\|_{-i}\leq$
$\leq\|I_{-j,+j}f\|_{+j}\|f\|_{-i}=\|f\|_{-j}\|f\|_{-i},$

for every $f\in\mathcal{H}_{0}$.

Because $\mathcal{H}_{-i}\subset\mathcal{H}_{-j}$ are the closures of
$\mathcal{H}_{0}$ with respect to those norms we have the dense inclusions and

$\|x\|_{-j}\leq\|x\|_{-i}\leq\|x\|_{0}\leq\|x\|_{+i}\leq\|x\|_{+j},\quad\forall
x\in\mathcal{H}_{+j}.$

Since we have $\iota_{+j,0}=\iota_{+i,0}\iota_{+j,+i}$ we also have
$\iota_{0,+j}=\iota_{+i,+j}\iota_{0,+i}$. Again note that (in the non-trivial
case) we have that $\iota_{-i,-j}I_{+i,-i}\iota_{+j,+i}\not=I_{+j,-j}$. But,
naturally, we have $\iota_{k,i}={\iota_{j,i}}_{{}_{|_{\mathcal{H}_{k}}}}$ for
$k\geq j\geq i$.

Note that we can change the role of the central pivot space: for every $r<p$,
let

(9) $\iota_{r,p}:\mathcal{H}_{r}\to\mathcal{H}_{p}$

be the adjoint of the inclusion of the Hilbert space $\mathcal{H}_{p}$ into
the space $\mathcal{H}_{r}$. As it is known

(10) $\|x\|_{r}\leq\|x\|_{p},\quad\forall x\in\mathcal{H}_{p}.$

Consider the bilinear form

$B_{r,p}:(f,x)\in\mathcal{H}_{r}\times\mathcal{H}_{p}\to\left\langle{f}\left|{x}\right.\right\rangle_{r}\in\mathbb{C},$

then by (10) it is continuous for $f\in\mathcal{H}_{r}$ and for
$x\in\mathcal{H}_{p}$, hence it can be represented as a scalar product both in
$\mathcal{H}_{r}$ and in $\mathcal{H}_{p}$:

$\left\langle{f}\left|{x}\right.\right\rangle_{r}=B_{r,p}(f,x)=\left\langle{\iota_{r,p}f}\left|{x}\right.\right\rangle_{p},\quad
f\in\mathcal{H}_{r},x\in\mathcal{H}_{p}$

where $\iota_{r,p}:\mathcal{H}_{r}\to\mathcal{H}_{p}$ is a bounded operator in
$\mathcal{B}(\mathcal{H}_{r},\mathcal{H}_{p})$.

######  Remark 3.4.

Let $\\{\mathcal{H}_{r};{r}\in\mathbb{Z}\\}$ be the family of the Hilbert
spaces of a Hilbert chain. For every $r,p\in\mathbb{Z}$, with $p\geq{r}$ the
maps $\iota_{r,p}:\mathcal{H}_{r}\to\mathcal{H}_{p}$ are injective, such that
$\|\iota_{r,p}x\|_{p}\leq\|x\|_{r}$, $\forall x\in\mathcal{H}_{r}$,
$\iota_{r,r}$ is the identity of $\mathcal{H}_{r}$ and
$\iota_{r,n}=\iota_{p,n}\iota_{r,p}$, ${r}\leq p\leq n$. Hence, the family
$\\{\mathcal{H}_{r},\iota_{r,p},r,p\in\mathbb{Z},p\geq r\\}$ is a directed
contractive system of Hilbert spaces, see [16]. It produces two spaces
$\mathcal{D}$ and $\mathcal{D}^{\times}$ in conjugate duality: the latter is
obtained as the inductive limit of the system, whereas the former is proved to
be the projective limit of the spaces $\mathcal{H}_{p}$’s, with respect to the
adjoint maps of the $\iota_{r,p}$’s.

We will use this in the $A$-scale framework and extend the notion of
Berezanskii canonical isomorphism for each ordered pair of spaces
$\mathcal{H}_{k}$ and $\mathcal{H}_{l}$, both $k,l\neq 0$ by defining unitary
operators as $I_{k,l}:\mathcal{H}_{k}\rightarrow\mathcal{H}_{l}$ by means of
fractional powers of the operator $A$, its closure and their inverses.

####  3.3.1. Different Adjoints

Before going forth, we give some thoughts about the adjoints, as we consider
different spaces here, and “the adjoint” depends on which spaces are
considered.

Consider a chain of Hilbert spaces as before. Fix any
$j,i\in\mathbb{N}\setminus\\{0\\}$. For
$A:\mathcal{H}_{+i}\to\mathcal{H}_{+j}$ we have already indicated by $A^{*}$
the Hilbertian adjoint $A^{*}:\mathcal{H}_{+j}\to\mathcal{H}_{+i}$, i.e.

$\left\langle{Ax}\left|{y}\right.\right\rangle_{+j}=\left\langle{x}\left|{A^{*}y}\right.\right\rangle_{+i},\quad\forall
x\in\mathcal{H}_{+i},\forall y\in\mathcal{H}_{+j}.$

This adjoint is not always useful, as two Riesz isomorphisms are applied,
which is not compatible with a structure where we consider Hilbert spaces
included in each other, i.e.
$\mathcal{H}_{+j}\subseteq\mathcal{H}_{+i}\subseteq\mathcal{H}_{0}\subseteq\mathcal{H}_{-i}\subseteq\mathcal{H}_{-j}$
but distinguished from each other. (Note that even though all Hilbert spaces
are isomorphic to each other, still it makes a lot of sense to treat them
differently, e.g. $L^{2}(\mathbb{R})$ and $\ell^{2}(\mathbb{N})$). By
identifying the dual of $\mathcal{H}_{-j}$ with $\mathcal{H}_{+j}$ the whole
scale would collapse. Therefore, taking $i,j\geq 0$, we now consider another
adjoint $A^{\bigstar}\in\mathcal{B}(\mathcal{H}_{-j},\mathcal{H}_{-i})$ for
$A\in\mathcal{B}(\mathcal{H}_{+i},\mathcal{H}_{+j})$, the “pivot adjoint”
defined as follows

(11)
$\left\langle{\alpha}\left|{Ax}\right.\right\rangle_{0}=\left\langle{A^{\bigstar}\alpha}\left|{x}\right.\right\rangle_{0}\quad\forall\alpha\in\mathcal{H}_{-j},x\in\mathcal{H}_{+i}.$

Here the number in the circle indicates the subscript of the pivot Hilbert
space $\mathcal{H}_{0}$ w.r. to which the dual is taken. We can, very
naturally, define the adjoint for any pivot space
$\mathcal{H}_{p}\not=\mathcal{H}_{0}$.

The same construction is possible for
$A\in\mathcal{B}(\mathcal{H}_{-j},\mathcal{H}_{-i})$,
$\mathcal{B}(\mathcal{H}_{-j},\mathcal{H}_{+i})$ or
$\mathcal{B}(\mathcal{H}_{+i},\mathcal{H}_{-j})$.

We highlight that the notion of pivot adjoint is a generalization of the
involuted $A^{\dagger}$ recalled in (2).

Moreover, similar to (4), if
$A\in\mathcal{B}(\mathcal{H}_{+i},\mathcal{H}_{+j})$, we have for
$\alpha\in\mathcal{H}_{-j},x\in\mathcal{H}_{+i}$

$\displaystyle\left\langle{\alpha}\left|{Ax}\right.\right\rangle_{0}$
$\displaystyle=$
$\displaystyle\left\langle{I_{-j,+j}\alpha}\left|{Ax}\right.\right\rangle_{+j}=\left\langle{A^{*}I_{-j,+j}\alpha}\left|{x}\right.\right\rangle_{+i}$
$\displaystyle=$
$\displaystyle=\left\langle{I_{+i,-i}A^{*}I_{-j,+j}\alpha}\left|{x}\right.\right\rangle_{0}=$
$\displaystyle=$
$\displaystyle\left\langle{A^{\bigstar}\alpha}\left|{x}\right.\right\rangle_{0}$

therefore we deduce that

(12) $A^{\bigstar}=(I_{-i,+i})^{-1}A^{*}I_{-j,+j}=I_{+i,-i}A^{*}I_{-j,+j}$

and

$\|A^{\bigstar}\|=\|A^{*}\|=\|A\|.$

######  Remark 3.5.

In general, if $p,q\in\mathbb{Z}$ and
$A\in\mathcal{B}(\mathcal{H}_{p},\mathcal{H}_{q})$, then
$A^{*}\in\mathcal{B}(\mathcal{H}_{q},\mathcal{H}_{p})$ and

(13)
$A^{\bigstar}=I_{p,-p}A^{*}I_{-q,q}\in\mathcal{B}(\mathcal{H}_{-q},\mathcal{H}_{-p}).$

######  Remark 3.6.

Because $A^{\bigstar}$ is an adjoint, two properties follow immediately:

  1. (1)

The double pivot adjoint of an operator
$A\in\mathcal{B}(\mathcal{H}_{p},\mathcal{H}_{q})$, $p,q\in\mathbb{Z}$, is

$A^{\bigstar\bigstar}=(A^{\bigstar})^{\bigstar}=A\quad\mbox{ and
}\quad\|A^{\bigstar\bigstar}\|=\|A\|.$

  2. (2)

Let $A\in\mathcal{B}(\mathcal{H}_{0},\mathcal{H}_{p})$ and
$B\in\mathcal{B}(\mathcal{H}_{m},\mathcal{H}_{0})$, $p,m\in\mathbb{N}$, then
$(AB)^{\bigstar}=B^{\bigstar}A^{\bigstar}.$ Indeed, if $x\in\mathcal{H}_{m}$
and $y\in\mathcal{H}_{p}$, by (11)

$\left\langle{ABx}\left|{y}\right.\right\rangle_{0}=\left\langle{Bx}\left|{A^{\bigstar}y}\right.\right\rangle_{0}=\left\langle{x}\left|{B^{\bigstar}A^{\bigstar}y}\right.\right\rangle_{0}.$

As a side remark to (12) note that $I_{+i,-i}$ is a unitary operator with
respect to the pair $(\mathcal{H}_{+i},\mathcal{H}_{-i})$, but is selfadjoint
with respect to $\mathcal{H}_{0}$.

Note that $A^{\bigstar}$ also corresponds to the Banach space adjoint of
$A:\mathcal{H}_{+i}\rightarrow\mathcal{H}_{+j}$ where the Riesz isomorphism at
the pivot space level is considered to be an equality, i.e.
$\mathcal{H}_{0}=\mathcal{H}_{0}^{\times}$. It maps $\mathcal{H}_{-j}$ into
$\mathcal{H}_{-i}$.

######  Remark 3.7.

Define

$\iota_{0,-i}:=\iota_{+i,0}^{\bigstar}:\mathcal{H}_{0}\rightarrow\mathcal{H}_{-i},$

we have $\|\iota_{0,-i}\|\leq 1$. As we have seen in (13)

$\iota_{0,-i}=(I_{-i,+i})^{-1}\iota_{+i,0}^{*}I_{0,0}=I_{+i,-i}\iota_{0,+i}.$

The operator $\iota_{0,-i}$ is effectively the inclusion of $\mathcal{H}_{0}$
in $\mathcal{H}_{-i}$ indeed, by (8),

$\left\langle{f}\left|{x}\right.\right\rangle_{0}=\left\langle{\iota_{0,+i}f}\left|{x}\right.\right\rangle_{+i}=\left\langle{I_{+i,-i}\iota_{0,+i}f}\left|{x}\right.\right\rangle_{0},\quad
f\in\mathcal{H}_{0},x\in\mathcal{H}_{+i}.$

####  3.3.2. Putting all together

Let us go back to the operator $\iota_{0,+1}$; it acts continuously from
$\mathcal{H}_{0}$ to $\mathcal{H}_{+1}$. Since
$\mathcal{H}_{+1}\subseteq\mathcal{H}_{0}$, this operator may be considered as
acting in $\mathcal{H}_{0}$; denote it as the operator
$\hat{I}:\mathcal{H}_{0}\rightarrow\mathcal{H}_{0}$. We have

(14) $\hat{I}=\iota_{+1,0}\iota_{0,+1}.$

The operator $\hat{I}$ is continuous with bound less or equal to $1$, positive
and invertible onto ${\sf Ran}(\hat{I})$, with ${\sf Ran}(\hat{I})={\sf
Dom}(\hat{I}^{-1})$ dense in $\mathcal{H}_{0}$.

It is easy to see that $\hat{I}^{-1}$ is also self-adjoint and positive in
$\mathcal{H}_{0}$, (later it will be clear that ${\sf Ran}(\hat{I})={\sf
Dom}(\hat{I}^{-1})=\mathcal{H}_{+2}$, and
$\hat{I}^{-1}=I_{+2,0}:\mathcal{H}_{+2}\to\mathcal{H}_{0}$, see (15)).

The operator $\hat{I}^{-1}$ is densely defined as an operator in
$\mathcal{H}_{0}$, has closed range and is one-to-one. Because it has a
bounded inverse, it is closed. It is positive and self-adjoint and we have
${\left(\hat{I}^{-1}\right)}^{1/2}={\left({\hat{I}}^{1/2}\right)}^{-1}=:\hat{I}^{-1/2}$.
The properties of this operator can be summarized by

######  Theorem 3.8.

[18, Theorem I.1.1] Consider the operator $F=(\hat{I}^{-1})^{1/2}$ in the
space $\mathcal{H}_{0}$. It is a positive self-adjoint operator for which
${\sf Dom}(F)=\mathcal{H}_{+1}$ and ${\sf Ran}(F)=\mathcal{H}_{0}$. This
operator acts isometrically from $\mathcal{H}_{+1}$ onto $\mathcal{H}_{0}$:

$\left\langle{x}\left|{y}\right.\right\rangle_{+1}=\left\langle{Fx}\left|{Fy}\right.\right\rangle_{0}~{}~{}~{}(x,y\in\mathcal{H}_{+1}).$

Consider $F$ as an operator acting from $\mathcal{H}_{0}$ to
$\mathcal{H}_{-1}$ and form the closure by continuity, denote this operator by
$\mathbf{F}$. $\mathbf{F}$ acts isometrically from the whole $\mathcal{H}_{0}$
onto $\mathcal{H}_{-1}$:

$\left\langle{f}\left|{g}\right.\right\rangle_{0}=\left\langle{\mathbf{F}f}\left|{\mathbf{F}g}\right.\right\rangle_{-1}~{}~{}~{}(f,g\in\mathcal{H}_{0}),$

and moreover $I_{+1,-1}=I_{-1,+1}^{-1}=\mathbf{F}F$. The relation

$\left\langle{f}\left|{Fx}\right.\right\rangle_{0}=\left\langle{\mathbf{F}f}\left|{x}\right.\right\rangle_{0}~{}~{}~{}(f\in\mathcal{H}_{0},x\in\mathcal{H}_{+1}),$

holds, therefore $\mathbf{F}=F^{\bigstar}$.

Let us write $B=F^{-1}$ and $\mathbf{B}=\mathbf{F}^{-1}$.

From the factorization of $I_{-1,+1}$ it follows immediately that

$I_{+1,-1}=I_{-1,+1}^{-1}=\mathbf{B}^{-1}B^{-1}=\mathbf{F}F=F^{\bigstar}F.$

thus obtaining a factorization of $I_{-1,+1}$ in terms of isometric operators.

Using the Hilbert adjoint we can write

$B=F^{*}\text{ and }\mathbf{B}=\mathbf{F}^{*}.$

We have the following scheme

$\mathcal{H}_{0}$$\mathcal{H}_{+1}$$\mathcal{H}_{-1}$$B$$F$$I_{+1,-1}=F^{\bigstar}F$$I_{-1,+1}=BB^{\bigstar}$$F^{\bigstar}$$B^{\bigstar}$

There is a well-known connection between infinite chains of Hilbert spaces and
positive self-adjoint operators in $\mathcal{H}_{0}$, if $\mathcal{H}_{0}$ is
the central space, see Section 2.3.

######  Remark 3.9.

Not every infinite chain of Hilbert spaces is a scale, see [5, pag 161].

However, see e.g. [18, Theorem I.1.1], given a chain of Hilbert spaces there
exists a positive self-adjoint operator $A=A^{*}$ on $\mathcal{H}_{0}$ such
that the triple
$\mathcal{H}_{+i}\subset\mathcal{H}_{0}\subset\mathcal{H}_{-i}$ of the chain
coincides with the “central” part of the $A$-scale of Hilbert spaces generated
by $A$.

We now proceed as in Section 2.3.

Consider the $A$-scale of Hilbert spaces where $A=F^{2}=\hat{I}^{-1}$. Since
$F$ is self-adjoint, then all its powers $F^{n}$, $n>0$ exist. For every
$n>0$, the domain ${\sf Dom}(F^{n})$ can be made into a Hilbert space by
setting

(15)
$\left\langle{x}\left|{y}\right.\right\rangle_{+n}=\left\langle{F^{n}x}\left|{F^{n}y}\right.\right\rangle_{0},\qquad
x,y\in{\sf Dom}(F^{n}).$

$F$ is injective, hence the following norm is induced:

$\|x\|_{+n}=(\left\langle{x}\left|{x}\right.\right\rangle_{+n})^{1/2}=(\left\langle{F^{n}x}\left|{F^{n}x}\right.\right\rangle_{0})^{1/2}=\|F^{n}x\|_{0},\quad
x\in{\sf Dom}(F^{n}).$

Put $\mathcal{H}_{+n}={\sf Dom}(F^{n})$, $n\geq 0$. As usual, for every $n>0$
we can consider the Hilbert spaces $\mathcal{H}_{-n}$ the conjugate dual of
$\mathcal{H}_{+n}$ with respect to the inner product of $\mathcal{H}_{0}$. We
get like e.g. in [4, Example 10.1.1] an $A$-scale of Hilbert spaces.

We have also that for $B=F^{-1}$

(16)
$\left\langle{f}\left|{g}\right.\right\rangle_{0}=\left\langle{B^{n}f}\left|{B^{n}g}\right.\right\rangle_{+n},\qquad
f,g\in\mathcal{H}_{0}.$

By

$\left\langle{f}\left|{g}\right.\right\rangle_{0}=\left\langle{F^{\bigstar}f}\left|{F^{\bigstar}g}\right.\right\rangle_{-1},~{}~{}~{}f,g\in\mathcal{H}_{0}$

we deduce

$\left\langle{\alpha}\left|{\beta}\right.\right\rangle_{-1}=\left\langle{B^{\bigstar}\alpha}\left|{B^{\bigstar}\beta}\right.\right\rangle_{0},~{}~{}~{}\alpha,\beta\in\mathcal{H}_{-1},$

$\left\langle{x}\left|{y}\right.\right\rangle_{+1}=\left\langle{Fx}\left|{Fy}\right.\right\rangle_{0}=\left\langle{F^{\bigstar}Fx}\left|{F^{\bigstar}Fy}\right.\right\rangle_{-1},~{}~{}~{}\forall
x,y\in\mathcal{H}_{+1},$

and

$\left\langle{\alpha}\left|{\beta}\right.\right\rangle_{-n}=\left\langle{(B^{\bigstar})^{n}\alpha}\left|{(B^{\bigstar})^{n}\beta}\right.\right\rangle_{0},~{}~{}~{}\alpha,\beta\in\mathcal{H}_{-n},\quad
n>0.$

######  Remark 3.10.

We already know that $I_{+1,-1}=F^{\bigstar}F=(B^{\bigstar})^{-1}F$. For
$0\leq r\leq p$, put $I_{p,r}=B^{r}F^{p}$. Let $p,r<0$, put
$F^{p}:=(F^{\bigstar})^{p}=(B^{\bigstar})^{-p}$ and
$B^{r}:=(B^{\bigstar})^{r}=(F^{\bigstar})^{-r}$. With this convention,
whatever $p,r$ are in $\mathbb{Z}$, we can decompose

$I_{p,r}=B^{r}F^{p}:\mathcal{H}_{p}\to\mathcal{H}_{r}.$

All these operators are unitary because they are products of unitary
operators. We highlight however that they are not Berezanskii isomorphisms
unless $(p-r)/2\in\mathbb{Z}$, in fact, only in this case $\mathcal{H}_{p}$
and $\mathcal{H}_{r}$ are extreme spaces of a rigged Hilbert space.
Furthermore, being $(B^{r}F^{p})^{*}=(F^{p})^{*}(B^{r})^{*}=B^{p}F^{r}$, for
$0\leq r\leq p$ by (15) and (16) we have

$\left\langle{f}\left|{g}\right.\right\rangle_{p}=\left\langle{F^{p}f}\left|{F^{p}g}\right.\right\rangle_{0}=\left\langle{B^{r}F^{p}f}\left|{B^{r}F^{p}g}\right.\right\rangle_{r}=\left\langle{I_{p,r}f}\left|{I_{p,r}g}\right.\right\rangle_{r},\,\mbox{
with }f,g\in\mathcal{H}_{p}$

similarly,

$\left\langle{f}\left|{g}\right.\right\rangle_{r}=\left\langle{B^{p}F^{r}f}\left|{B^{p}F^{r}g}\right.\right\rangle_{p}=\left\langle{I_{r,p}f}\left|{I_{r,p}g}\right.\right\rangle_{p},\quad\mbox{
with }f,g\in\mathcal{H}_{r}.$

If now $f\in\mathcal{H}_{p},g\in\mathcal{H}_{r}$

$\begin{split}\left\langle{I_{p,r}f}\left|{g}\right.\right\rangle_{r}&=\left\langle{B^{r}F^{p}f}\left|{g}\right.\right\rangle_{r}=\left\langle{F^{p}f}\left|{F^{r}g}\right.\right\rangle_{0}=\left\langle{B^{p}F^{p}f}\left|{B^{p}F^{r}g}\right.\right\rangle_{p}\\\
&=\left\langle{f}\left|{B^{p}F^{r}g}\right.\right\rangle_{p}=\left\langle{f}\left|{I_{r,p}g}\right.\right\rangle_{p}\end{split}$

hence, predictably, $I_{r,p}=I_{p,r}^{*}$.
With the due changes, the same results are obtained for any
$p,r\in\mathbb{Z}$.

###  3.4. Generator of a scale and shifting of the central space

Can we shift the central space? I.e. how does the dual of
$\mathcal{H}_{n_{0}+n}$ in respect to $\mathcal{H}_{n_{0}}$ look?

######  Corollary 3.11.

Let $r<p$ and let ${\mathcal{H}_{p}^{\times}}^{(r)}$ be the dual of
$\mathcal{H}_{p}$ with respect to the topology of $\mathcal{H}_{r}$ (i.e.
consider the triplet
$\mathcal{H}_{p}\subseteq\mathcal{H}_{r}\subseteq{\mathcal{H}_{p}^{\times}}^{(r)}$).

Then ${\mathcal{H}_{p}^{\times}}^{(r)}=\mathcal{H}_{2r-p}$.

###### Proof.

Let us consider an $A$-scale Hilbert spaces with $A=(\hat{I}^{-1})$, with
$\hat{I}$ defined as in (14), so by the first invariant principle of
$A$-scales, see the Section 3.1, for $k=p-r$ the triple

$\mathcal{H}_{r+k}=\mathcal{H}_{p}\subseteq\mathcal{H}_{r}\subseteq\mathcal{H}_{2r-p}=\mathcal{H}_{r-k},$

forms a chain of Hilbert scale. ∎

In other words: the dual of $\mathcal{H}_{n_{0}+n}$ in respect to
$\mathcal{H}_{n_{0}}$ is $\mathcal{H}_{n_{0}-n}$.

######  Remark 3.12.

Clearly, if $r<p$ then the space $\mathcal{H}_{r}$ can be considered the dual
space of $\mathcal{H}_{p}$ with respect to some space $\mathcal{H}_{n_{0}}$
with $\mathcal{H}_{p}\subseteq\mathcal{H}_{n_{0}}\subseteq\mathcal{H}_{r}$ if
and only if $p+r$ is an even number. If this is the case,
$n_{0}=\frac{p+r}{2}$.

Now, for a fixed $A$-scale, we wonder how the operator $A$ changes if we fix
another Hilbert space of the scale as the “central” space, for example
$\mathcal{H}_{k}$, $k\in\mathbb{Z}$. Following [38] the answer to this
question is given by using what has been called the second invariance
principle of the A-scale. Indeed, it results that $A$ is unitarily equivalent
to its image under any “shift” along the $A$-scale.

Just to fix some ideas let $k\geq 2$, consider the $A$-scale

$...\subseteq\mathcal{H}_{2+k}\subseteq...\subseteq\mathcal{H}_{+k}\subseteq...\subseteq\mathcal{H}_{2}\subseteq...\subseteq\mathcal{H}_{2-k}\subseteq...\subseteq\mathcal{H}_{-k}\subseteq...,$

and the operators

$A_{+k}:=I_{2+k,k}:\mathcal{H}_{2+k}\to\mathcal{H}_{+k}\quad\mbox{ and }\quad
A_{-k}:=I_{2-k,-k}:\mathcal{H}_{2-k}\to\mathcal{H}_{-k},$

with $I_{p,r}$ defined as before (in particular $A=A_{0}=I_{+2,0}$). Then it
is easy to see that [38] $A_{+k}=A_{{}_{|_{\mathcal{H}_{2+k}}}}$ i.e. it is
the restriction of $A$ to $\mathcal{H}_{2+k}$, hence $A_{+k}$ is self-adjoint
in $\mathcal{H}_{2+k}$. The operator $A_{-k}$ is self-adjoint on
$\mathcal{H}_{2-k}$ and it results $A_{-k}=\overline{A}^{\|\cdot\|_{-k}}$ the
closure of $A$ in $\mathcal{H}_{-k}$. Both operators $A_{\pm k}$,
$k\in\mathbb{N}$ are unitary images of the original operator $A$ on
$\mathcal{H}_{0}$, in fact

$A_{+k}=I_{0,k}AI_{2+k,2}=B^{k}F^{0}B^{0}F^{2}B^{2}F^{2+k}=B^{k}F^{2+k}$

and

$A_{-k}=I_{0,-k}AI_{2-k,2}=B^{-k}F^{0}B^{0}F^{2}B^{2}F^{2-k}=B^{-k}F^{2-k},$

i.e. for the sake of brevity, for $p\in\mathbb{Z}$

$A_{p}=B^{p}F^{2+p}=I_{2+p,p}.$

We remark that the operator $A$ is essentially self-adjoint in each space
$\mathcal{H}_{-k}$, $k\geq 1$.

##  4\. Frame-related properties on Hilbert scales

We will look at a scale of Hilbert space, and let’s fix:

$\mathcal{H}_{m}\subseteq\mathcal{H}_{p}\subseteq\mathcal{H}_{r},$

i.e. $r\leq p\leq m$.

We look at a sequence $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ which has a
property in some of the other spaces and investigate on how these properties
“spread” to the other spaces: if it is complete, forms a Bessel sequence, a
frame, a basis, has a Riesz property, etc.. for a $\mathcal{H}_{r}$ (or
$\mathcal{H}_{p}$), what can we say about this sequence in the other spaces?

If we allow the sequences to be changed the results are trivial consequences
of the following straightforward generalization of [23, Cor. 5.3.4]:

######  Corollary 4.1.

Given two Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$, if $\psi=(\psi_{k})$
is a frame for $\mathcal{H}$ with frame bounds $A,B$ and
$U:\mathcal{H}\rightarrow\mathcal{K}$ is a unitary operator, then
$(U\psi_{k})$ is a frame for $\mathcal{K}$ with the same frame bounds.

######  Remark 4.2.

If we have a unitary operator between two Hilbert spaces, all the frame
properties naturally transfer from a Hilbert space to the other.

In the sequel we will use the unitary operators $I_{r,p}$ defined like in
Remark 3.10, and get easy results for $\left(I_{r,p}\psi_{k}\right)$. See
Corollary 4.8.

We need some preparation before that.

###  4.1. Completeness

######  Lemma 4.3.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$. Then the following statements
hold.

  * i)

If $(\psi_{k})$ is complete in $\mathcal{H}_{p}$, then it is also complete in
$\mathcal{H}_{r}$ for $r\leq p\leq m$.

  * ii)

If $(\psi_{k})$ is complete in $\mathcal{H}_{r}$ $r\leq m$, then
$(I_{r,p}\psi_{k})$ is a complete sequence in $\mathcal{H}_{p}$ for any $p$.

###### Proof.

i) Since $\mathcal{H}_{p}\subseteq\mathcal{H}_{r}$ densely, then for each
$f\in\mathcal{H}_{r}$ and $\epsilon>0$, there is an element
$x\in\mathcal{H}_{p}$ with $\|f-x\|_{r}<\epsilon/2$. Now, let
$(\psi_{k})\subseteq\mathcal{H}_{m}$ be a complete sequence in
$\mathcal{H}_{p}$. Then there exists $y\in{{\sf span}(\psi_{k})}$ such that
$\|x-y\|_{p}<\epsilon/2$. So

$\|f-y\|_{r}\leq\|f-x\|_{r}+\|x-y\|_{r}\leq\|f-x\|_{r}+\|x-y\|_{p}<\epsilon.$

It follows that $\psi=(\psi_{k})$ is a complete sequence in $\mathcal{H}_{r}$
for $r\leq p\leq m$.
$ii)$ Trivial by Remark 4.2. ∎

Later, in Lemma 4.12, we will see that the converse of Lemma 4.3 $(i)$ is also
true.

###  4.2. Unbounded Frame-related operators on Hilbert chains

Let us consider an arbitrary sequence
$\psi=(\psi_{k})\subseteq\mathcal{H}_{m}{\subseteq\mathcal{H}_{p}}$, and as in
[12] define the analysis operator $C_{\psi}^{p}:{\sf
Dom}(C_{\psi}^{p})\subseteq\mathcal{H}_{p}\to\ell^{2}$ of $(\psi_{k})$ by

${\sf
Dom}(C_{\psi}^{p})=\left\\{f\in\mathcal{H}_{p}:\sum_{k}|\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}|^{2}<\infty\right\\}$
$C_{\psi}^{p}f:=(\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}),\qquad\forall
f\in{\sf Dom}(C_{\psi}^{p}).$

In an analogous way we can define the synthesis operator

$D_{\psi}^{p}:{\sf
Dom}(D_{\psi}^{p})\subseteq\ell^{2}\rightarrow\mathcal{H}_{p}$

associated with the sequence $\psi$ by

${\sf
Dom}(D_{\psi}^{p})=\left\\{c=(c_{k})\in\ell^{2}:\sum_{k}c_{k}\psi_{k}\mbox{
converges in }\mathcal{H}_{p}\right\\}$
$D_{\psi}^{p}c=\sum_{k}c_{k}\psi_{k},\qquad\forall c\in{\sf
Dom}(D_{\psi}^{p}).$

As it is known ${\sf Dom}(D_{\psi}^{p})$ is dense in $\ell^{2}$, since it
contains the finite sequences which form a dense subset of $\ell^{2}$ and
$C_{\psi}^{p}=(D_{\psi}^{p})^{*}$, hence $C_{\psi}^{p}$ is closed.

Clearly, we have for $r\leq p$ that

(17) ${\sf Dom}{D_{\psi}^{p}}\subseteq{\sf Dom}{D_{\psi}^{r}}\mbox{,
\quad}{\sf Ran}{D_{\psi}^{p}}\subseteq{\sf Ran}{D_{\psi}^{r}}$
$\ker{D_{\psi}^{p}}\subseteq\ker{D_{\psi}^{r}}\text{ all inclusion are dense.
}$

Clearly $D_{\psi}^{p}\subset D_{\psi}^{r}$, since their domain are such that
${\sf Dom}{D_{\psi}^{p}}\subseteq{\sf Dom}{D_{\psi}^{r}}$ and
$D_{\psi}^{p}c=D_{\psi}^{r}c$ for every $c\in{\sf Dom}{D_{\psi}^{p}}$. Then,
if we look at $D_{\psi}^{p}$ as an operator into a subspace of
$\mathcal{H}_{r}$ we can say that $C_{\psi}^{r}\subset(D_{\psi}^{p})^{*}_{r}$
where $(D_{\psi}^{p})^{*}_{r}$ is the adjoint of $D_{\psi}^{p}$ as on operator
from $\ell^{2}$ into $\mathcal{H}_{r}$.

Furthermore, let us consider $D_{\psi}^{00}$ defined as operator on $c_{00}$
the space of finite sequences. By the above definition we have that
$D_{\psi}^{p}$ is the closure of $D_{\psi}^{00}$ for any $p\geq m$. So,
$D_{\psi}^{p}=\overline{D_{\psi}^{00}}^{\mathcal{H}_{p}}$. Let us now use that
$r\leq p$ and so $\|\cdot\|_{r}\leq\|\cdot\|_{p}$ and therefore
$D_{\psi}^{r}=\overline{D_{\psi}^{00}}^{\mathcal{H}_{r}}=\overline{\left(\overline{D_{\psi}^{00}}^{\mathcal{H}_{p}}\right)}^{\mathcal{H}_{r}}=\overline{D_{\psi}^{p}}^{\mathcal{H}_{r}}$.
So, in summary

$D_{\psi}^{r}=\overline{D_{\psi}^{p}}^{\mathcal{H}_{r}},\quad\mbox{ for }r\leq
p.$

Let us also introduce the combination of those operators. Consider the “frame
operator” $S^{p}_{\psi}:{\sf
Dom}(S^{p}_{\psi})\subseteq\mathcal{H}_{p}\to\mathcal{H}_{p}$ of $\psi$

$S^{p}_{\psi}f:=\sum_{k\in\mathbb{N}}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}\psi_{k}$

where

${\sf
Dom}(S^{p}_{\psi})=\\{f\in\mathcal{H}_{p}:\sum_{k\in\mathbb{N}}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}\psi_{k}\mbox{
converges in }\mathcal{H}_{p}\\};$

and the Gram operator $G^{p}_{\psi}:{\sf
Dom}(G^{p}_{\psi})\subseteq\ell^{2}\to\ell^{2},$ with

$G^{p}_{\psi}(c_{k})_{k}:=\left(\sum_{l\in\mathbb{N}}(G^{p}_{\psi})_{k,l}c_{l}\right)_{k},$

where

$\displaystyle{\sf Dom}(G^{p}_{\psi})=\\{(c_{k})_{k}\in\ell^{2}:$
$\displaystyle\sum_{l\in\mathbb{N}}(G^{p}_{\psi})_{k,l}c_{l}\mbox{ converges
}\forall k\in\mathbb{N}\mbox{ and }$
$\displaystyle\left(\sum_{l\in\mathbb{N}}(G^{p}_{\psi})_{k,l}c_{l}\right)_{k}\in\ell^{2}\\}$

and the Gram matrix $((G^{p}_{\psi})_{k,l})_{k,l}$ is defined by
$(G^{p}_{\psi})_{k,l}=\left\langle{\psi_{l}}\left|{\psi_{k}}\right.\right\rangle_{p}$,
${k,l}\in\mathbb{N}$.

###  4.3. Frame Properties of $I_{r,p}\psi$

By using the unitary operator introduced in Remark 3.10 we naturally get

######  Lemma 4.4.

For a given $p\in\mathbb{Z}$ let $\psi=(\psi_{k})\subseteq\mathcal{H}_{p}$ be
an arbitrary sequence. Then for every $r\in\mathbb{Z}$,
$C^{r}_{I_{p,r}(\psi)}=C^{p}_{\psi}I_{r,p}$ and
$C^{p}_{\psi}=C^{r}_{I_{p,r}(\psi)}I_{p,r}$.

###### Proof.

We have

$\begin{split}{\sf Dom}(C^{p}_{\psi}I_{r,p})&=\\{f\in{\sf
Dom}(I_{r,p}):I_{r,p}f\in{\sf Dom}C^{p}_{\psi}\\}\\\
&=\\{f\in\mathcal{H}_{r}:(\langle
I_{r,p}f,\psi_{k}\rangle_{p})\in\ell^{2}\\}\\\
&=\\{f\in\mathcal{H}_{r}:(\langle
f,I_{p,r}\psi_{k}\rangle_{r})\in\ell^{2}\\}\\\ &={\sf
Dom}(C^{r}_{I_{p,r}(\psi)}).\end{split}$

And by the same argument the operators also are the same.

The other result is obtained by multiplying the operator $I_{p,r}$, inverse of
$I_{r,p}$, on the right in the equality
$C^{r}_{I_{p,r}(\psi)}=C^{p}_{\psi}I_{r,p}$. ∎

An analogue result is true for the other frame-related operators. We start
with a result for the synthesis operator:

######  Lemma 4.5.

For a given $p\in\mathbb{Z}$ let $\psi=(\psi_{k})\subseteq\mathcal{H}_{p}$ be
an arbitrary sequence. Then for every $r\in\mathbb{Z}$,
$D^{r}_{I_{p,r}(\psi)}=I_{p,r}D^{p}_{\psi}$ and
$D^{p}_{\psi}=I_{r,p}D^{r}_{I_{p,r}(\psi)}$.

###### Proof.

If $c\in{\sf Dom}(D_{I_{p,r}(\psi)}^{r})\subseteq\ell^{2}$ then there exists
$f\in\mathcal{H}_{r}$ such that $D_{I_{p,r}(\psi)}^{r}c=f$, i.e. for every
$\varepsilon>0$ there exists $n_{\varepsilon}\in\mathbb{N}$ such that for
every $n\geq n_{\varepsilon}$

$\displaystyle\left\|\sum_{k=1}^{n}c_{k}I_{p,r}\psi_{k}-f\right\|_{r}$
$\displaystyle=$
$\displaystyle\left\|I_{r,p}\sum_{k=1}^{n}c_{k}I_{p,r}\psi_{k}-I_{r,p}f\right\|_{p}$
$\displaystyle=$
$\displaystyle\left\|\sum_{k=1}^{n}c_{k}\psi_{k}-I_{r,p}f\right\|_{p}<\varepsilon.$

Hence $D_{\psi}^{p}c=I_{r,p}f$, ${\sf Dom}(D_{I_{p,r}(\psi)}^{r})={\sf
Dom}(I_{p,r}D_{\psi}^{p})$ and $D_{I_{p,r}(\psi)}^{r}=I_{p,r}D_{\psi}^{p}$.
The other result is obtained by multiplying the operator $I_{r,p}$, inverse of
$I_{p,r}$, on the left in the equality
$D_{I_{p,r}(\psi)}^{r}=I_{p,r}D_{\psi}^{p}$.∎

######  Lemma 4.6.

For a given $p\in\mathbb{Z}$ let $\psi=(\psi_{k})\subseteq\mathcal{H}_{p}$ be
an arbitrary sequence. Then for every $r\in\mathbb{Z}$,
$S^{p}_{I_{r,p}(\psi)}=I_{r,p}S^{r}_{\psi}I_{p,r}$.

###### Proof.

It is a consequence of the previous Lemma 4.4 and Lemma 4.5 and of $(iii)$ in
[12, Proposition 3.3]. ∎

######  Lemma 4.7.

For a given $p\in\mathbb{Z}$ let $\psi=(\psi_{k})\subseteq\mathcal{H}_{p}$ be
an arbitrary sequence. Then $C^{p}_{\psi}D^{p}_{\psi}\subseteq G^{p}_{\psi}$
and $G^{p}_{\psi}=G^{r}_{I_{p,r}(\psi)}$ for every $r\in\mathbb{Z}$.

###### Proof.

By $(iv)$ in [12, Proposition 3.3] we have that
$C^{p}_{\psi}D^{p}_{\psi}\subseteq G^{p}_{\psi}$, for every $p\leq n$. Now,
recall that
$\left\langle{\psi_{l}}\left|{\psi_{k}}\right.\right\rangle_{p}=\left\langle{I_{p,r}\psi_{l}}\left|{I_{p,r}\psi_{k}}\right.\right\rangle_{r}$,
we have

$\displaystyle{\sf Dom}(G^{p}_{\psi})$ $\displaystyle=$
$\displaystyle\left\\{c\in\ell^{2}:\sum_{l\in\mathbb{N}}\left\langle{\psi_{l}}\left|{\psi_{k}}\right.\right\rangle_{p}c_{l}\mbox{
converges }\forall k\in\mathbb{N}\mbox{ and }\right.$
$\displaystyle\left.\sum_{k}\left|\sum_{l\in\mathbb{N}}\left\langle{\psi_{l}}\left|{\psi_{k}}\right.\right\rangle_{p}c_{l}\right|^{2}<\infty\right\\}$
$\displaystyle=$
$\displaystyle\left\\{c\in\ell^{2}:\sum_{l\in\mathbb{N}}\left\langle{I_{p,r}\psi_{l}}\left|{I_{p,r}\psi_{k}}\right.\right\rangle_{r}c_{l}\mbox{
converges }\forall k\in\mathbb{N}\mbox{ and }\right.$
$\displaystyle\left.\sum_{k}\left|\sum_{l\in\mathbb{N}}\left\langle{I_{p,r}\psi_{l}}\left|{I_{p,r}\psi_{k}}\right.\right\rangle_{r}c_{l}\right|^{2}<\infty\right\\}={\sf
Dom}(G^{r}_{I_{p,r}(\psi)}).$

∎

Note that all the above results lead to statements saying that if $\psi$ is a
frame (Bessel sequence, Riesz basis,…) for some Hilbert space
$\mathcal{H}_{r}$, $I_{r,p}\psi$ is one for $\mathcal{H}_{p}$. They all are
trivial consequences of Remark 4.2.

######  Corollary 4.8.

For a given $p\in\mathbb{Z}$ let $\psi=(\psi_{k})\subseteq\mathcal{H}_{p}$ be
an arbitrary sequence. Then for any $r\in\mathbb{Z}$ the following is true.

  1. (1)

If $(\psi_{k})$ is a Bessel sequence in $\mathcal{H}_{p}$, then
$(I_{p,r}\psi_{k})$ is a Bessel sequence in $\mathcal{H}_{r}$.

  2. (2)

If $(\psi_{k})$ is a semi-frame in $\mathcal{H}_{p}$, then $(I_{p,r}\psi_{k})$
is a semi-frame in $\mathcal{H}_{r}$ with the same bounds.

  3. (3)

If $(\psi_{k})$ is a frame in $\mathcal{H}_{p}$, then $(I_{p,r}\psi_{k})$ is a
frame in $\mathcal{H}_{r}$ with the same bounds.

  4. (4)

If $(\psi_{k})$ and $(\phi_{k})\subset\mathcal{H}_{p}$ are a reproducing pair,
then $(I_{p,r}\phi_{k})$ and $(I_{p,r}\psi_{k})$ are a reproducing pair in
$\mathcal{H}_{r}$ with the same bounds.

  5. (5)

If $(\phi_{k})\subset\mathcal{H}_{p}$ is a dual sequence of $(\psi_{k})$ in
$\mathcal{H}_{p}$, then $(I_{p,r}\phi_{k})$ is a dual sequence of
$(I_{p,r}\psi_{k})$ in $\mathcal{H}_{r}$.

  6. (6)

If $(\psi_{k})$ is an orthonormal basis of $\mathcal{H}_{p}$, then
$(I_{p,r}\psi_{k})$ is an orthonormal basis of $\mathcal{H}_{r}$.

  7. (7)

If $(\psi_{k})$ is a Riesz basis of $\mathcal{H}_{p}$ and
$T\in\mathcal{B}(\mathcal{H}_{p})$ is the bijective operator such that
$Te_{k}=\psi_{k}$ for every $k$ with $\\{e_{k}\\}$ an orthonormal basis of
$\mathcal{H}_{p}$. Then $(I_{p,r}T^{-1}\psi_{k})$ is an orthonormal basis of
$\mathcal{H}_{r}$.

  8. (8)

If $(\psi_{k})$ is a Riesz basis of $\mathcal{H}_{p}$ then $(I_{p,r}\psi_{k})$
is a Riesz basis of $\mathcal{H}_{r}$ with the same bounds.

On the other hand, if $p<r$ then $\psi$ is also a sequence in
$\mathcal{H}_{p}$. So, in addition to looking at
$\left<f,I_{r,p}\psi_{k}\right>_{p}$ we can also look at
$\left<f,\psi_{k}\right>_{p}$. This is done in the next section.

###  4.4. Frame-related Operators for the Original Sequence $\psi$

If we do not apply the operator we can still show

######  Lemma 4.9.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ be an arbitrary sequence. Then
for every $r\leq p\leq m$,

(18) $C^{r}_{\psi}=C^{p}_{\psi}\iota_{r,p}.$

where $\iota_{r,p}$ is defined as in (9).

###### Proof.

We have

$\left<f,\psi_{k}\right>_{r}=\left<f,\iota_{p,r}\psi_{k}\right>_{r}=\left<\iota_{r,p}f,\psi_{k}\right>_{p}$

and

$\begin{split}{\sf Dom}(C^{p}_{\psi}\iota_{r,p})&=\\{f\in{\sf
Dom}(\iota_{r,p});\iota_{r,p}f\in{\sf Dom}C^{p}_{\psi}\\}\\\
&=\\{f\in\mathcal{H}_{r};(\langle\iota_{r,p}f,\psi_{k}\rangle_{p})\in\ell^{2}\\}\\\
&=\\{f\in\mathcal{H}_{r};(\langle f,\psi_{k}\rangle_{r})\in\ell^{2}\\}\\\
&={\sf Dom}(C^{r}_{\psi}).\end{split}$

∎

In consequence, we have

######  Lemma 4.10.

If $\psi\subset\mathcal{H}_{m}$ and $\psi$ is a Bessel sequence for
$\mathcal{H}_{p}$ then, for $r\leq p\leq m$, $\psi$ is a Bessel sequence for
$\mathcal{H}_{r}$ with the same bound.

###### Proof.

By [23, Cor. 3.2.4 and Theor. 3.2.3], $\psi$ is a Bessel sequence in
$\mathcal{H}_{p}$ if and only if ${\sf Dom}{D_{\psi}^{p}}=\ell^{2}$. By (17)
${\sf Dom}{D_{\psi}^{p}}\subseteq{\sf Dom}{D_{\psi}^{r}}$, this is true if and
only if $\psi$ is also a Bessel sequence for $\mathcal{H}_{r}$.

Furthermore, we have that $\|C_{\psi}^{r}\|\leq\|C_{\psi}^{p}\|$ by (18). ∎

By [3, Lemma 3.2] $\psi\subset\mathcal{H}_{m}$ is an upper semi-frame for
$\mathcal{H}_{p}$, $p<m$ if and only if it is a total Bessel sequence for
$\mathcal{H}_{p}$. Then putting together Lemma 4.10 and Lemma 4.3 $(i)$ we
obtain

######  Lemma 4.11.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ and $r\leq p\leq m$. If $\psi$
is an upper semi-frame for $\mathcal{H}_{p}$, then $\psi$ is an upper semi-
frame for $\mathcal{H}_{r}$ with the same bound.

Let us now note that $\iota_{r,p}^{-1}$ is a densely defined, bijective
operator, which is not bounded. It is “never” bounded, in the sense that if it
were bounded, the involved norms would be equivalent and the whole scale of
Hilbert spaces would collapse, however, we can use it to show

######  Lemma 4.12.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ be an arbitrary sequence. Then
for every $r\leq p\leq m$,

(19) $C^{p}_{\psi}=C^{r}_{\psi}\iota_{r,p}^{-1}\quad\text{ on }{\sf
Dom}(C_{\psi}^{p})$

Therefore, ${\iota_{r,p}}_{{}_{|_{{\sf Dom}(C^{r}_{\psi})}}}$ is a bijective
operator from ${\sf Dom}(C_{\psi}^{r})$ onto ${\sf Dom}(C_{\psi}^{p})$.

Furthermore, if $\psi\subset\mathcal{H}_{m}$ and $\psi$ is complete for
$\mathcal{H}_{r}$, then if $r\leq p\leq m$, then $\psi$ is complete for
$\mathcal{H}_{p}$.

###### Proof.

The first statement is a direct consequence of (18).

For the converse of Lemma 4.3 (i) we use [12, Prop.4.1(g)]: $\psi$ is complete
in $\mathcal{H}_{r}$ if and only if $C_{\psi}^{r}$ is injective and since
$\iota_{r,p}^{-1}$ is injective, this implies that $C_{\psi}^{p}$ is injective
too, which is equivalent to $\psi$ being complete in $\mathcal{H}_{p}$. ∎

By Lemma 4.10 we know that a Bessel sequence in $\mathcal{H}_{p}$ is also one
in $\mathcal{H}_{r}$ for $r\leq p$. We can show an opposite direction for
lower semi-frame.

######  Lemma 4.13.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ be a lower semi-frame for
$\mathcal{H}_{r}$ then it is also a lower semi-frame for $\mathcal{H}_{p}$ for
every $r\leq p\leq m$ with the same lower bound.

###### Proof.

A sequence is a lower semi-frame if and only if the analysis operator is
boundedly invertible. By (19) we have that
${C^{p}_{\psi}}^{-1}=\iota_{r,p}{C^{r}_{\psi}}^{-1}$.

Also $\|{C^{p}_{\psi}}^{-1}\|\leq\|{C^{r}_{\psi}}^{-1}\|$. ∎

####  4.4.1. Frames

Combining those results we get

######  Corollary 4.14.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ be a frame for $\mathcal{H}_{r}$
then it is an upper semi-frame for $\mathcal{H}_{p}$ for every $r\leq p\leq m$
and a lower semi-frame for $\mathcal{H}_{q}$, for every $q\leq r\leq m$.

####  4.4.2. Duality

The sequence $\psi=(\psi_{k})\subset\mathcal{H}_{p}$ is a lower semi-frame for
$\mathcal{H}_{p}$ if and only if [20] there exists a Bessel sequence
$\phi=(\phi_{k})\subset\mathcal{H}_{p}$ such that

(20)
$f=\sum_{k}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}\phi_{k}=D^{p}_{\phi}C^{p}_{\psi}(f),\quad\forall
f\in{\sf Dom}(C^{p}_{\psi}).$

Let $\psi\subseteq\mathcal{H}_{m}$ be a lower semi-frame for
$\mathcal{H}_{r}$, $r\leq m$; then by Lemma 4.13 it is one in
$\mathcal{H}_{m}$. Hence there exists a dual sequence
$\phi\subset\mathcal{H}_{m}$, which is Bessel in $\mathcal{H}_{m}$ and
therefore also in all spaces $\mathcal{H}_{r}$, $r\leq m$. By assumption (20)
is valid on ${\sf Dom}(C^{m}_{\psi})$.

Now let $f\in{\sf Dom}(C^{p}_{\psi})$, then by Lemma 4.13 (20) converges for
all $p$, with $r\leq p\leq m$. Consider $g=\iota_{p,m}f$, then $g\in{\sf
Dom}(C^{m}_{\psi})$.

$g=\sum_{k}\left\langle{g}\left|{\psi_{k}}\right.\right\rangle_{m}\phi_{k}=\sum_{k}\left\langle{\iota_{p,m}\iota_{p,m}^{-1}g}\left|{\psi_{k}}\right.\right\rangle_{m}\phi_{k}=\sum_{k}\left\langle{\iota_{p,m}^{-1}g}\left|{\iota_{m,p}\psi_{k}}\right.\right\rangle_{p}\phi_{k}=$
$=\sum_{k}\left\langle{\iota_{p,m}^{-1}g}\left|{\psi_{k}}\right.\right\rangle_{p}\phi_{k}=\sum_{k}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}\phi_{k}.$

Furthermore

$f=\iota_{p,m}^{-1}g=\iota_{p,m}^{-1}\sum_{k}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}\phi_{k}.$

So, in summary, this shows

######  Proposition 4.15.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ be a lower semi-frame for
$\mathcal{H}_{r}$ then there exists a Bessel sequence
$\phi\subseteq\mathcal{H}_{m}$ such that

$f=\iota_{p,m}^{-1}\sum_{k}\left\langle{f}\left|{\psi_{k}}\right.\right\rangle_{p}\phi_{k},\quad\forall
f\in{\sf Dom}(C^{p}_{\psi})\subseteq\mathcal{H}_{p}.$

for all $r\leq p\leq m$.

####  4.4.3. A negative result

######  Proposition 4.16.

Let $\psi=(\psi_{k})\subseteq\mathcal{H}_{m}$ be a frame for $\mathcal{H}_{p}$
and one for $\mathcal{H}_{q}$ , for every $q\leq p\leq m$. Then the norms are
equivalent and so $\mathcal{H}_{q}=\mathcal{H}_{r}=\mathcal{H}_{p}$ for $q\leq
r\leq p$.

###### Proof.

By (19) $C^{p}_{\psi}=C^{q}_{\psi}\iota_{q,p}^{-1}$. As $\psi$ is a frame for
$\mathcal{H}_{q}$ we have that
$\iota_{q,p}^{-1}=({S^{q}_{\psi}})^{-1}D^{q}_{\psi}C^{p}_{\psi}$ and it is
therefore bounded. ∎

######  Remark 4.17.

In particular, this means that if two Hilbert spaces, one contained into the
other one, do not coincide, a sequence can never be a frame for both of them.

######  Remark 4.18.

For this statement it is important we have considered Hilbert spaces and
(standard) frames with the sequence space $\ell^{2}$. If we consider Banach
spaces, associated to a weighted $\ell^{p}$ space [30] or Hilbert spaces with
weighted $\ell^{2}$ sequence spaces [27, 29], this result does not hold. Quite
the opposite, e.g. for localized frames [8, Theorem 1] one can show that a
frame on the pivot space $\mathcal{H}_{0}$ is also a frame for the other
(Banach) spaces $\mathcal{H}_{q}$.

### Acknowledgements

G.B. acknowledges that this work has been partially supported by the Gruppo
Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni
(GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). The work of
P.B. was supported by the Innovation project of the Austrian Academy of
SciencesIF_2019_24_Fun (Frames and Unbounded Operators) and the project P
34624 (“Localized, Fusion and Tensors of Frames” - LoFT) of the Austrian
Science Fund (FWF).

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# Aggregated Customer Engagement Model

Priya Gupta<EMAIL_ADDRESS>Amazon and  Cuize Han<EMAIL_ADDRESS>Amazon

###### Abstract.

E-commerce websites use machine learned ranking models to serve shopping
results to customers. Typically, the websites log the customer search events,
which include the query entered and the resulting engagement with the shopping
results, such as clicks and purchases. Each customer search event serves as
input training data for the models, and the individual customer engagement
serves as a signal for customer preference. So a purchased shopping result,
for example, is perceived to be more important than one that is not. However,
new or under-impressed products do not have enough customer engagement signals
and end up at a disadvantage when being ranked alongside popular products. In
this paper, we propose a novel method for data curation that aggregates all
customer engagements within a day for the same query to use as input training
data. This aggregated customer engagement gives the models a complete picture
of the relative importance of shopping results. Training models on this
aggregated data leads to less reliance on behavioral features. This helps
mitigate the cold start problem and boosted relevant new products to top
search results. In this paper, we present the offline and online analysis and
results comparing the individual and aggregated customer engagement models
trained on e-commerce data.

data aggregation, product search, learning to rank

##  1\. Introduction

Retail e-commerce accounts for billions of US dollars in sales worldwide every
year. Optimizing product search is critical for customer satisfaction. This
means finding the right products to match the customer’s intent, and ranking
them in the order that is most important to the customer. Learning to Rank
(LTR) (Liu, 2011) is a common approach to rank search results. It is a
supervised learning algorithm that uses past customer behavior or manual human
labels as the signal for product preference.

Product search, unlike web search, poses unique challenges in collecting
labeled data. While human annotation of web search results can be useful (John
Le, 2010), it can lead to misleading labels in product search. Different
product facets might have different importance to customers. For example,
price might be a primary factor for some while brand name might be more
important to others. The preferences also change based on product type. In
studies where human judges were used to rate the products, there was
significant disagreement in the ratings and true customer engagement signals
(Shubhra Kanti Karmaker Santu, 2017; Omar Alonso, 2009).

Using past customer behavior as labels in LTR has similar challenges. For the
same query, individual customer engagement (ICE) such as view, click, add-to-
cart, add-to-wishlist, purchase etc. might be different. Hence, this ICE based
data leads to different labels for the same query-product pairs. With the new
model, we propose using aggregated customer engagements (ACE) within a day for
the same query across different customer search sessions as labels. The labels
now encode the total number of times all customers engaged with a product for
the same query within a day. This aggregation gives an estimate of how
important a product is in relation to other products since customer engagement
is our proxy for relevance. This relative importance is the fundamental idea
behind pairwise ranking of LambdaRank algorithm (Burges, [n.d.]).

Customer implicit feedback is known to suffer from position and selection bias
(Aman Agarwal, 2019). Position bias occurs because people are more likely to
examine the products that are ranked high. Since customers can only interact
with the products that are presented to them, the implicit feedback is biased
towards products that are selected by the current ranking model. This bias
leads to the “rich get richer and poor get poorer” phenomenon. This is
aggravated by the ranking model’s heavy dependence on behavioral features.
These are features that directly capture the historic customer behavior data
and memorize their preference. While behavioral features help with variance
reduction when fitting the ranking model to the customer feedback data, over-
reliance on these features leads to less generalizability. For new products or
rarely purchased products that do not have enough customer engagement, a model
that predominantly ranks based on behavioral features might rank them poorly.
In this paper, we show empirically and theoretically, that the ACE model
relies less on behavioral features and is better at ranking new or under-
impressed products. Long term this can mitigate the impact of position bias.

The rest of the paper is organized as follows: In section 2 we describe the
methodology behind ACE model. Then we share results comparing the ICE and ACE
models in section 3. In section 4, we dive deep into the theoretical
explanation of the efficacy of the ACE model. The final section 5 details our
conclusions and future work.

##  2\. ACE Model

For each customer search event with a query $q_{i}$, a list of $N_{i}$
products $\\{A_{ij},j=1,.,N_{i}\\}$ are returned in search results. Each of
these products $A_{ij}$ has a feature vector
$x_{ij}=(x^{1}_{ij},..x^{d}_{ij})$ associated with it that the LTR model uses
for ranking. Customers may engage with some products in the search results.
This engagement is used as labels $l_{ij}$ for the training data. In case of
binary labeling, if a customer engaged with a product, its label $l_{ij}=1$,
otherwise it is zero. In case of the ICE model, each query $q_{i}$, the
corresponding products in search results $\\{A_{ij},j=1,.,N_{i}\\}$ and
associated feature vectors $\\{x_{ij},j=1,.,N_{i}\\}$ and labels
$\\{l_{i1}.,.l_{iN_{i}}\\}$ form an instance of ICE training data. We use
millions of instances to train and test the ICE model.

In the ACE model, we aggregate the labels for the same query-product pairs
$(q,A)$ across different customer search events within the same day:

$\tilde{l}_{q,A}=\sum_{(q_{i},A_{j})=(q,A)}l_{ij}$

The motivation behind selecting one day as the aggregation window was to have
a large enough window such that the aggregation includes several instances of
query-products pairs even for infrequent queries, but small enough window to
avoid day-over-day trends in customer preferences in the aggregation. The
aggregation of labels can lead to unbounded values for $\tilde{l}_{q,A}$,
particularly for popular query-product pairs that occur tens, hundreds or even
thousand times in the data. So we bucket these aggregated labels
$\tilde{l}_{q,A}$ using quantiles and map onto a finite set of integer labels.
For ACE model, each unique daily query along with all the corresponding
distinct search results and their associated features and the aggregated
labels form an instance of ACE training data. Like the ICE model, we use
millions of ACE instances to train and test the ACE model.

The capping of labels in ACE data has the consequence of limiting the
influence of highly popular query-product pairs on the ranking model. In other
words, down-sampling the signals from these frequently impressed query-product
pairs leads to a ranking model that relies less on behavioral features. As a
result, products that have little engagement, either because they are new or
preferred by a minority of customers, have a chance to move to top search
results. This is how the ACE model helps alleviate the cold start problem that
is common in product search.

##  3\. Results

We trained ICE and ACE models on three data sets from a large e-commerce
website. Each set had more than 3M data points. We tested the models offline
and ran A/B tests online. With each experimental result, we iterated over the
ACE method to improve its efficacy at mitigating the cold start problem
without impacting other customer engagement related metrics.

Aggregating customer engagement and collapsing several searches with the same
query into one data point leads to a larger percentage of distinct queries and
products in the data. For example, in one of our data sets 86% of the queries
in the ACE data were distinct while only 49% in the ICE data were unique.
Similarly, 89% of the query-product pairs in the ACE data were distinct while
62% in ICE data were unique. Greater diversity in the data leads to smaller
errors and more generalized models (Trevor Hastie, 2009).

During model assessment, we measure the variance reduction in the labels due
to each feature selected by the model. This variance reduction is correlated
with feature importance. We compared the total variance reduction due to all
the behavioral features in the ACE and ICE models trained on the three data
sets. The larger the variance reduction due to a feature, the more the model
relies on it for ranking. As is clear in Table. 1, the ACE models rely less on
behavioral features when compared to ICE model. Although the difference is
small, it is enough to allow more textual and product newness (days since
launch date) related features to be picked up by the ACE model. This helps
surface products that are better exact matches with the query even if they are
new and have not accumulated enough customer engagement related behavioral
features.

Table 1. Total variance reduction on model prediction due to all behavioral features picked up by feature selection during model development Dataset | ICE Model | ACE Model
---|---|---
1 | 85.34% | 80.54%
2 | 76.8% | 76.38%
3 | 83.23% | 80.81%

In the offline analysis of the models, we measured how effective the ACE model
was at surfacing newer products, that were less than seven days old, in top 16
search results. When compared to the ICE model, on one data set the ACE model
served 30% more new product impressions while on another data set it served
110% more. The big jump from 30% to 110% came due to an improvement in feature
selection strategy that was applied to the latter data set.

After incorporating improvements and learnings from offline analysis, we ran
an A/B test on the e-commerce website in the US. We measured the difference
between new product impressions, clicks and purchases between the ICE and ACE
models. For this online experiment, ’new product’ was defined as any product
launched on the e-commerce website in the past three days. The experiment ran
for two weeks during which the ACE model served 12.4% more new product
impressions than the ICE model when measured over millions of customer search
sessions. Because of the increase in new product impressions, the ACE model
led to 10.6% more clicks and 17.54% more number of purchases of these new
products. These results validated our hypothesis that the ACE model can help
with the cold start problem.

##  4\. Theoretical Explanation

The ACE model down-samples the signals from highly impressed products whose
behavioral features have enough power to predict the customer action well.
Thus, models trained on ACE data focus more on non-behavioral features which
leads to better generalizability.

Given a query $q$, denote $A$ to be a random selected product that presented
to the customer. Let $Y$ to be the $0/1$ random variable that represents the
customer action for that pair. The feature vector of the pair can be
transformed and decomposed into behavioral features ($X_{bhv}$) and non-
behavioral features ($X_{nbhv}$) that are random elements in $\mathcal{X}$.
For $(x,x^{\prime})\in\mathcal{X}$, denote the conditional expectations as

$\displaystyle f_{bhv}(x)=$ $\displaystyle\mathbb{E}\left[Y|X_{bhv}=x\right]$
$\displaystyle f_{nbhv}(x^{\prime})=$
$\displaystyle\mathbb{E}\left[Y|X_{nbhv}=x^{\prime}\right]$

Then $f_{bhv}(X_{bhv})$ and $f_{nbhv}(X_{nbhv})$ are theoretically the best
prediction of the customer action given the knowledge of $X_{bhv}$ or
$X_{nbhv}$. We also assume the effects have been orthogonalized and the
residues are uncorrelated:

(1)
$\mathbb{E}\left(Y-\mathbb{E}\left[Y|X_{bhv}\right]\right)\left(Y-\mathbb{E}\left[Y|X_{nbhv}\right]\right)=0$

Suppose the scoring function behind the ranking takes the form of
$f_{w}(x,x^{\prime})=wf_{bhv}(x)+(1-w)f_{nbhv}(x^{\prime})$ with a weight
parameter $w\in[0,1]$, then the optimal weight that minimizes the expected
mean square error can be represented as :

$\displaystyle\mathbb{E}\left[\left(Y-f_{w}(X_{bhv},X_{nbhv})\right)^{2}\right]$
$\displaystyle=w^{2}\mathbb{E}\epsilon_{bhv}(X_{bhv})+$ (2)
$\displaystyle(1-w)^{2}\mathbb{E}\epsilon_{nbhv}(X_{nbhv})$

where $\mathbb{E}\epsilon_{bhv}(X_{bhv})$ and
$\mathbb{E}\epsilon_{nbhv}(X_{nbhv})$ represent the variances that cannot be
explained by only $X_{bhv}$ or $X_{nbhv}$ respectively and

$\displaystyle\epsilon_{bhv}(x)=$
$\displaystyle\mathbb{E}\left[(Y-f_{bhv}(X_{bhv}))^{2}|X_{bhv}=x\right]$
$\displaystyle\epsilon_{nbhv}(x^{\prime})=$
$\displaystyle\mathbb{E}\left[(Y-f_{nbhv}(X_{nbhv}))^{2}|X_{nbhv}=x^{\prime}\right].$

(2) also uses the assumption (1). Then solving the quadratic function of $w$
in (2), we have the optimal weight $w^{*}$ equals

(3)
$w^{*}=\dfrac{\mathbb{E}\epsilon_{nbhv}(X_{nbhv})}{\mathbb{E}\epsilon_{bhv}(X_{bhv})+\mathbb{E}\epsilon_{nbhv}(X_{nbhv})}$

So the model will put less weight on behavioral features if non-behavioral
features get better ability to reduce the variance (corresponding to the
decrease of $\mathbb{E}\epsilon_{nbhv}(X_{nbhv})$ ) or/and behavioral features
reduce less variance (corresponding to the increase of
$\mathbb{E}\epsilon_{bhv}(X_{bhv})$).

We claim that for the current data distribution, ACE will decrease
$\mathbb{E}\epsilon_{nbhv}(X_{nbhv})$ and increase
$\mathbb{E}\epsilon_{bhv}(X_{bhv})$. Consequently, the model will put more
weight on non-behavioral features.

Suppose there are $m$ customer search sessions with the query $q$ in the
training data. Given $m$ labels $Y_{1},..,Y_{m}$, ACE model makes a new label
$\tilde{Y}$ out of $Y_{1},..,Y_{m}$ for the product $A$ and features
$(X_{bhv},X_{nbhv})$. For simplicity, we consider the extreme cut-off
situation where we define

(4) $\tilde{Y}=\begin{cases}1&\text{at least one of $Y_{1},..,Y_{m}$ is
$1$}\\\ 0&\text{otherwise}\end{cases}$

Then for a feature $X$ if we denote $p(x)=\mathbb{P}(Y=1|X=x),$ then for the
new label

$\displaystyle\tilde{p}(x)$ $\displaystyle=\mathbb{P}(\tilde{Y}=1|X=x)$
$\displaystyle=1-\mathbb{P}(\tilde{Y}=0|X=x)$
$\displaystyle=1-\left(\mathbb{P}(Y=0|X=x)\right)^{m}$ (5)
$\displaystyle=1-(1-p(x))^{m}$

Now the variance that cannot be explained by $X$ for label $Y$ is
$\mathbb{E}\epsilon(X)$ where

$\displaystyle\epsilon(x)$
$\displaystyle=\mathbb{E}\left[\left(Y-\mathbb{E}(Y|X)\right)^{2}|X=x\right]$
$\displaystyle=\mathbb{E}[Y^{2}|X=x]-(\mathbb{E}(Y|X=x))^{2}$
$\displaystyle=\mathbb{E}[Y|X=x]-(\mathbb{E}(Y|X=x))^{2}$ (6)
$\displaystyle=p(x)(1-p(x))$

The derivation of (6) uses the fact that $Y^{2}\equiv Y$ and
$\mathbb{E}[Y|X=x]\equiv\mathbb{P}(Y=1|X=x)$. Then the corresponding quantity
for $\tilde{Y}$ is

$\displaystyle\tilde{\epsilon}(x)$
$\displaystyle=\mathbb{E}\left[\left(\tilde{Y}-\mathbb{E}(\tilde{Y}|X)\right)^{2}|X=x\right]$
$\displaystyle=\tilde{p}(x)(1-\tilde{p}(x))$ (7)
$\displaystyle=(1-(1-p(x))^{m})(1-p(x))^{m}$

Figure 1. Plot of unexplained variance in ACE model $\tilde{\epsilon}(x)$
against probability of customer action $p(x)$ when aggregating different
number of customer sessions m. This figure shows that as number of sessions
aggregated goes up, the unexplained variance quickly drops to zero as
probability of customer action increases.

The error $\epsilon(x)$ will be small if $p(x)$ is close to $0$ or $1$, that
is, given $X=x$, we are able to tell with more certainty whether the customer
will take action or not. On the other hand, it is maximized when $p(x)=1/2$,
corresponding to given $X=x$, we are still randomly guessing the result. So
the plot between $\epsilon(x)$ and $p(x)$ looks like a bell curve and with the
effect of ACE, the curve skews to the left with the increase of aggregated
session number $m$ as shown in Figure 1.

We observe that currently majority of the variance of the label is explained
by the behavioral features. That is $\mathbb{E}\epsilon_{bhv}(X_{bhv})$, the
unexplained variance of behavioral features is much smaller than
$\mathbb{E}\epsilon_{nbhv}(X_{nbhv})$. In order for
$\mathbb{E}\epsilon_{bhv}(X_{bhv})$ to be small, the majority of the mass of
the distribution of $X_{bhv}$ needs to accumulate at the two ends of $p(x)$.
Since the label $Y$ itself skews to the left (most products won’t receive
action), we claim that for behavioral features, majority of the mass of the
distribution accumulates to the left end of $p(x)$. From Figure 1, we see that
the error $\tilde{\epsilon}(x)$ increases at the left end of $p(x)$ with the
increase of $m$. Thus, the unexplained variance for behavioral features in ACE
becomes larger:
$\mathbb{E}\tilde{\epsilon}_{bhv}(X_{bhv})>\mathbb{E}\epsilon_{bhv}(X_{bhv})$.
On the other hand, relatively most of the mass of $X_{nbhv}$ concentrates in
the middle part of $p(x)$ as the unexplained variance of $X_{nbhv}$ is large.
From the figure we see at the middle part of $p(x)$, the error
$\tilde{\epsilon}(x)$ quickly drop to near zero as $m$ increases. Thus, for
non-behavioral features, the unexplained variance becomes smaller:
$\mathbb{E}\tilde{\epsilon}_{nbhv}(X_{nbhv})<\mathbb{E}\epsilon_{nbhv}(X_{nbhv})$.
Hence, the optimal weight $\tilde{w^{*}}$ on the behavioral features in ACE
reduces as

$\dfrac{\mathbb{E}\tilde{\epsilon}_{nbhv}(X_{nbhv})}{\mathbb{E}\tilde{\epsilon}_{bhv}(X_{bhv})+\mathbb{E}\tilde{\epsilon}_{nbhv}(X_{nbhv})}<\dfrac{\mathbb{E}\epsilon_{nbhv}(X_{nbhv})}{\mathbb{E}\epsilon_{bhv}(X_{bhv})+\mathbb{E}\epsilon_{nbhv}(X_{nbhv})}$

##  5\. Conclusion

In this paper, we described a new way of processing training data for
developing ranking models for e-commerce. We showed that aggregating customer
engagement across different search sessions for the same query leads to better
ranking of new and under-impressed products. This is because the ACE model
relies less on behavioral features which otherwise tend to dominate the
ranking. Products that are popular have the benefit of position and selection
bias leading to strong behavioral features. With a model that under-samples
these customer engagement signals, we can mitigate the cold start problem.
Products that are under-impressed, perhaps because they are preferred by a
minority of customers, also get a chance to show in top search results. We
plan to conduct more online experiments with the ACE model to further validate
the methodology using empirical results.

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