#  $\mu$-norm of an operator

D.Treschev
Steklov Mathematical Institute of Russian Academy of Sciences

###### Abstract

Let $({\cal X},\mu)$ be a measure space. For any measurable set $Y\subset{\cal
X}$ let ${\bf 1}_{Y}:{\cal X}\to\mathbb{R}$ be the indicator of $Y$ and let
$\pi_{Y}$ be the orthogonal projector $L^{2}({\cal X})\ni
f\mapsto\pi_{Y}f={\bf 1}_{Y}f$. For any bounded operator $W$ on $L^{2}({\cal
X},\mu)$ we define its $\mu$-norm
$\|W\|_{\mu}=\inf_{\chi}\sqrt{\sum\mu(Y_{j})\|W\pi_{Y}\|^{2}}$, where the
infinum is taken over all measurable partitions
$\chi=\\{Y_{1},\ldots,Y_{J}\\}$ of ${\cal X}$. We present some properties of
the $\mu$-norm and some computations. Our main motivation is the problem of
the construction of a quantum entropy.

##  1 Introduction and motivation

Let ${\cal X}$ be a nonempty set and let ${\cal B}$ be a $\sigma$-algebra of
subsets $X\subset{\cal X}$. Consider the measure space $({\cal X},{\cal
B},\mu)$, where $\mu$ is a probability measure: $\mu({\cal X})=1$.

Consider the Hilbert space ${\cal H}=L^{2}({\cal X},\mu)$ with the scalar
product and the norm

$\langle f,g\rangle=\int_{\cal X}f\overline{g}\,d\mu,\quad\|f\|=\sqrt{\langle
f,f\rangle}.$

For any bounded operator $W$ on ${\cal H}$ let $\|W\|$ be its norm defined by

$\|W\|=\sup_{\|f\|=1}\|Wf\|.$

For any $X\in{\cal B}$ consider the orthogonal projector

$\pi_{X}:{\cal H}\to{\cal H},\qquad{\cal H}\ni f\mapsto\pi_{X}f={\bf
1}_{X}\cdot f,$ (1.1)

where ${\bf 1}_{X}$ is the indicator of $X$ (the function equal to 1 on $X$
and vanishing at other points). Then $\mu(X)>0$ implies $\|\pi_{X}\|=1$ and
for any $X^{\prime},X^{\prime\prime}\in{\cal B}$

$\pi_{X^{\prime}}+\pi_{X^{\prime\prime}}=\pi_{X^{\prime}\cup
X^{\prime\prime}}+\pi_{X^{\prime}\cap
X^{\prime\prime}},\quad\pi_{X^{\prime}}\pi_{X^{\prime\prime}}=\pi_{X^{\prime}\cap
X^{\prime\prime}}.$

We say that $\chi=\\{Y_{1},\ldots,Y_{J}\\}$ is a (finite, measurable)
partition (of ${\cal X}$) if

$Y_{j}\in{\cal B},\quad\mu\big{(}{\cal X}\setminus\cup_{1\leq j\leq
J}Y_{j}\big{)}=0,\quad\mu(Y_{j}\cap Y_{k})=0\quad\mbox{ for any
$j,k\in\\{1,\ldots,J\\}$, $k\neq j$}.$

For any two partitions $\chi=\\{Y_{1},\ldots,Y_{J}\\}$ and
$\kappa=\\{X_{1},\ldots,X_{K}\\}$ we denote

$\chi\vee\kappa=\\{Y_{j}\cap X_{k}\\}_{j=1,\ldots,J,\,k=1,\ldots,K}.$

Obviously, $\chi\vee\kappa$ is also a partition.

Let $W$ be a bounded operator on ${\cal H}$. For any partition
$\chi=\\{Y_{1},\ldots,Y_{J}\\}$ of ${\cal X}$ we define

${\cal M}_{\chi}(W)=\sum_{j=1}^{J}\mu(Y_{j})\|W\pi_{Y_{j}}\|^{2}.$ (1.2)

We define the $\mu$-norm111in fact, a seminorm of $W$ by

$\|W\|_{\mu}=\inf_{\chi}\sqrt{{\cal M}_{\chi}(W)}.$ (1.3)

Recall that the operator $U$ is said to be an isometry if

$\langle f,g\rangle=\langle Uf,Ug\rangle,\qquad f,g\in{\cal H}.$

If the isometry $U$ is invertible then $U$ is called a unitary operator.

For any bounded $W$, any $Y\in{\cal B}$, and any isometry $U$

$\|W\pi_{Y}\|\leq\|W\|,\quad\|UW\|=\|W\|.$

This implies the following obvious properties of the $\mu$-norm:

$\displaystyle\|\operatorname{id}\|_{\mu}$ $\displaystyle=$ $\displaystyle
1,\qquad\|W\|_{\mu}\;\leq\;\|W\|,$ (1.4) $\displaystyle\|W_{1}W_{2}\|_{\mu}$
$\displaystyle\leq$ $\displaystyle\|W_{1}\|\|W_{2}\|_{\mu},$ (1.5)
$\displaystyle\|\lambda W\|_{\mu}$ $\displaystyle=$
$\displaystyle|\lambda|\,\|W\|_{\mu}\quad\mbox{ for any
$\lambda\in\mathbb{C}$},$ (1.6) $\displaystyle\|W\|_{\mu}$ $\displaystyle=$
$\displaystyle\|UW\|_{\mu}\quad\mbox{ for any unitary $U$}.$ (1.7)

The first question which probably comes to the reader’s mind is ‘‘why?’’. Why
such a construction may be useful or interesting? Now we are going to explain
our motivations.

Let $F:{\cal X}\to{\cal X}$ be an endomorphism of the measure space ${\cal
X},{\cal B},\mu$. This means that for any $X\in{\cal B}$ the set $F^{-1}(X)$
(the complete preimage) also lies in ${\cal B}$ and $\mu(X)=\mu(F^{-1}(X))$.
Invertible endomorphisms are called automorphisms. Let $End({\cal X})$ denote
the semigroup of all endomorphisms of $({\cal X},{\cal B},\mu)$. There are two
standard constructions associated with any $F\in End({\cal X},\mu)$.

(1). Any such $F$ generates the isometry (a unitary operator if $F$ is an
automotphism) $U_{F}$ on $L^{2}({\cal X},\mu)$ (the Koopman operator):

$L^{2}({\cal X},\mu)\ni f\mapsto U_{F}f=f\circ F,\qquad
U_{F}=\operatorname{Koop}(F).$

(2) For any $F\in End({\cal X},\mu)$ it is possible to compute the measure
entropy (another name is the Kolmogorov-Sinai entropy) $h(F)$.

Our question is as follows. Is it possible to determine in some ‘‘natural
way’’ a real nonnegative function on the semigroup $Iso(L^{2}({\cal X},\mu))$
so that the diagram

$\begin{array}[]{rcl}&End({\cal X},\mu)&\\\
h\swarrow&&\searrow\operatorname{Koop}\\\
\mathbb{R}_{+}&\stackrel{{\scriptstyle{\mathfrak{h}}}}{{\longleftarrow}}&Iso(L^{2}({\cal
X},\mu))\end{array}$

is commutative.

Recall the construction of the measure entropy of an endomorphism. Let $J_{N}$
be the set of multiindices $j=(j_{0},\ldots,j_{N})$, where any component
$j_{n}$ takes values in the set $\\{0,\ldots,K\\}$. For any partition
$\chi=\\{X_{0},\ldots,X_{K}\\}$ and $j\in J_{N}$ we define

${\bf X}_{j}=F^{-N}(X_{j_{N}})\cap\ldots\cap F^{-1}(X_{j_{1}})\cap X_{j_{0}}.$

We define $h_{F}(\chi,N+1)$ by

$h_{F}(\chi,N+1)=-\sum_{j\in J_{N}}\mu({\bf X}_{j})\log\mu({\bf X}_{j}).$

The function $h_{F}$, as a function of the second argument, is subadditive:
$h_{F}(\chi,n+m)\leq h_{F}(\chi,n)+h_{F}(\chi,m)$. This implies existence of
the limit

$h_{F}(\chi)=\lim_{n\to\infty}\frac{1}{n}h_{F}(\chi,n).$

Finally, the measure entropy is defined by

$h(F)=\sup_{\chi}h_{F}(\chi).$

Our idea is to construct the entropy of a unitary operator $U$ analogously
with the following difference. Instead of ${\bf X}_{j}$ we take

${\mathfrak{X}}_{j}=\pi_{X_{j_{N}}}U\pi_{X_{j_{N-1}}}U\ldots
U\pi_{X_{j_{1}}}U\pi_{X_{j_{0}}}.$

We define

${\mathfrak{h}}_{U}(\chi,N+1)=-\sum_{j\in
J_{N}}\|{\mathfrak{X}}_{j}\|_{\mu}^{2}\log\|{\mathfrak{X}}_{j}\|_{\mu}^{2}.$

Other details are the same:

${\mathfrak{h}}_{U}(\chi)=\lim_{n\to\infty}\frac{1}{n}{\mathfrak{h}}(\chi,n),\qquad{\mathfrak{h}}(U)=\sup_{\chi}{\mathfrak{h}}_{U}(\chi).$
(1.8)

Corollary 2.5 (below) implies that for any automorphism $F$

${\mathfrak{h}}(U_{F})=h(F).$ (1.9)

It is interesting to see what this construction gives in finite-dimensional
case. Suppose ${\cal X}$ is finite, $\\#{\cal X}=J$. Then the space ${\cal H}$
is isomorphic to $\mathbb{C}^{J}$. Operators on ${\cal H}$ are identified with
$J\times J$ matrices. By using results of Section 3.3, it is possible to
compute for any unitary $U=(U_{jk})$

${\mathfrak{h}}(U)=-\frac{1}{J}\sum_{\sigma_{0},\sigma_{1}=1}^{J}|U_{\sigma_{0}\sigma_{1}}|^{2}\,\log|U_{\sigma_{0}\sigma_{1}}|^{2}.$
(1.10)

Note that the numbers $P_{jk}=|U_{jk}|^{2}$ can be regarded as elements of a
bi-stochastic matrix $P$. Then the vector $\nu=(1/J,\ldots,1/J)$ is a
stationary distribution for the corresponding Markov chain. It is known [7]
that if a Markov chain is irreducible, aperiodic and positive recurrent then
its stationary distribution is unique and its entropy rate is determined by
$-\sum_{j,k}\mu_{j}P_{jk}\log P_{jk}$. This quantity coincides with
${\mathfrak{h}}(U)$.

In the literature there exist several attempts to extend the concept of the
measure entropy to quantum systems, see [6, 17, 18, 2, 3, 15, 13] and many
others. In [1] several mutual relations between these approaches are given. In
some definitions of quantum entropy (see, for example, [2, 21, 19, 4, 14])
equations similar to (1.10) appear. In [8, 9] a construction for measure
entropy is proposed for doubly stochastic operators on various spaces of
functions on a measure space. The question, which approach to quantum
generalization of the measure entropy is ‘‘more physical’’ remains unclear.

Concerning our definition of ${\mathfrak{h}}(U)$, many technical questions
appear, including the question on subadditivity222This subadditivity is
important for the existence of the limit (1.8). of
${\mathfrak{h}}_{U}(\chi,n)$, on the possibility to include into the
construction isometries $U$ together with unitary operators, on methods of
computation (or at least, on effective lower and upper estimates) of the
quantities ${\mathfrak{h}}(U)$, and many others. Answers to these questions
depend on the detailed analysis of the operation $\|\cdot\|_{\mu}$. We start
the analysis in this paper.

The main results we present in this paper are as follows (below $W$ is an
arbitrary bounded operator).

  * •

$\|\pi_{X}\|_{\mu}^{2}=\mu(X)$ for any $X\in{\cal B}$ (Lemma 2.1).

  * •

If $\chi^{\prime}$ is a subpartition of $\chi$ then ${\cal
M}_{\chi^{\prime}}(W)\leq{\cal M}_{\chi}(W)$ (Lemma 2.2 and Corollary 2.1).
Hence the quantities ${\cal M}_{\chi}(W)$ approach the infinum (1.3) on fine
partitions.

  * •

For any two bounded operators $W_{1}$ and $W_{2}$ we prove that

$\|W_{1}+W_{2}\|_{\mu}\leq\|W_{1}\|_{\mu}+\|W_{2}\|_{\mu}$

(Lemma 2.3). This triangle inequality combined with (1.6) imply that
$\|\cdot\|_{\mu}$ is a seminorm on the space of bounded operators on ${\cal
H}$.

  * •

Let $F$ be an automorphism of $({\cal X},{\cal B},\mu)$ and let $U_{F}$ be the
corresponding unitary operator on ${\cal H}$, defined by $f\mapsto
U_{F}=f\circ F$. Then

$\|WU_{F}\|_{\mu}=\|W\|_{\mu}$

(Lemma 2.4). This implies $\|U_{F}^{-1}WU_{F}\|_{\mu}=\|W\|_{\mu}$. Informally
speaking, this means that measure preserving coordinate changes on ${\cal X}$
preserve the $\mu$-norm.

  * •

$\|\cdot\|_{\mu}$ is a continuous function in the $L^{2}({\cal X})$ operator
topology (Corollary 2.2).

  * •

If the measure $\mu$ has no atoms then $\|W+W_{0}\|_{\mu}=\|W\|_{\mu}$ for any
bounded $W$ and compact $W_{0}$ (Corollary 3.2).

  * •

Given $g\in L^{\infty}({\cal X})$ let $\widehat{g}$ be the operator defined by
$f\mapsto\widehat{g}f=gf$. Then $\|\widehat{g}\|_{\mu}=\|g\|$, (Lemma 3.2).

  * •

Suppose ${\cal X}=\\{1,\ldots,J\\}$ is finite and the measure of any element
equals $1/J$. Let $W=(W_{jk})_{j,k=1,\ldots,J}$ be an operator on ${\cal
H}\cong\mathbb{C}^{J}$. Then by Lemma 3.3

$\|W\|_{\mu}^{2}=\frac{1}{J}\sum_{j,k=1}^{J}|W_{jk}|^{2}.$

  * •

Let $\pi_{H}:{\cal H}\to H$ be the orthogonal projector to the closed subspace
$H\subset{\cal H}$. We define the $\mu$-dimension of $H$ by
$\dim_{\mu}(H):=\|\pi_{H}\|_{\mu}^{2}$. According to Lemma 3.4 the following
statements hold.

1\. If $\mu$ has no atoms and $H$ is finite-dimensional then
$\dim_{\mu}(H)=0$.

2\. If $H=\pi_{X}({\cal H})$ for some $X\in{\cal B}$ then
$\dim_{\mu}(H)=\mu(X)$.

3\. If ${\cal X}$ is finite and $\mu$ is uniformly distributed among the
points then ${\cal H}=\mathbb{C}^{J}$, $J=\\#{\cal X}$ and for any subspace
$H\subset{\cal H}\;$ $\dim_{\mu}(H)=\dim(H)/J$, where $\dim$ is the ordinary
dimension.

For any $\mu$-preserving almost free action of the cyclic group
$\mathbb{Z}_{q}$ on ${\cal X}$ we consider the space $H_{0}\subset{\cal H}$ of
$\mathbb{Z}_{q}$-invariant functions. It is natural to call these functions
periodic with period $1/q$. We prove that $\dim_{\mu}(H_{0})=1/q$ (Lemma 3.5).

  * •

For any partition $\\{X_{1},\ldots,X_{K}\\}$

$\|W\|_{\mu}^{2}=\sum_{k=1}^{K}\|W\pi_{X_{k}}\|_{\mu}^{2}$

(the right additivity of $\|\cdot\|_{\mu}^{2}$, Lemma 4.1).

  * •

Instead of the left additivity we have the weaker property (Lemma 4.2):

$\|W\|_{\mu}^{2}\leq\sum_{k=1}^{K}\|\pi_{X_{k}}W\|_{\mu}^{2}.$

  * •

In Section 5 we consider the case when ${\cal X}$ is a compact metric space
and $\mu$ is a Borel measure w.r.t. the corresponding topology. Let
$B_{r}(x)\subset{\cal X}$ denote the open ball with center at $x$ and radius
$r$. We prove (Proposition 5.1) that for any $x$ the limit

$\vartheta(x)=\lim_{\varepsilon\searrow 0}\|W\pi_{B_{\varepsilon}(x)}\|^{2}$

exists, the function $\vartheta:{\cal X}\to\mathbb{R}$ is measurable and
$\|W\|_{\mu}^{2}\leq\int_{\cal X}\vartheta\,d\mu$. In general this inequality
is strict (Example 5.1).

However we prove that $\|W\|_{\mu}^{2}=\int_{\cal X}\vartheta\,d\mu$ provided
two additional conditions C1 and C2 hold (Proposition 5.2).

  * •

In Section 6 we compute the $\mu$-norm of a convolution operator on ${\cal
H}=L^{2}(\mathbb{T})$, where $\mathbb{T}=\mathbb{R}/(2\pi\mathbb{Z})$ is a
circle. For any bounded sequence $\\{\lambda_{k}\\}_{k\in\mathbb{Z}}$ we
consider the distribution $\lambda(x)=\sum\lambda_{k}e^{ikx}$. Then the
operator $f\mapsto\operatorname{Conv}_{\lambda}f=\lambda*f$ is bounded:
$\|\operatorname{Conv}_{\lambda}\|=\sup_{k\in\mathbb{Z}}|\lambda_{k}|$. We
prove (Proposition 6.1) that

$\|\operatorname{Conv}_{\lambda}\|_{\mu}^{2}=\rho(\lambda),\qquad\rho(\lambda)=\limsup_{\\#I\to\infty}\frac{1}{\\#I}\sum_{k\in
I}|\lambda_{k}|^{2},$

where $I\subset\mathbb{Z}$ are integer intervals.

  * •

In Section 7 we compute $\mu$-norm for a wide class of bounded operators on
$L^{2}(\mathbb{T})$, the so-called operators of diagonal type. Consider the
operator $W=(W_{j,k})_{j,k\in\mathbb{Z}}$

$f=\sum f_{k}e^{ikx}\mapsto Wf=\sum_{j,k\in\mathbb{Z}}W_{j,k}f_{k}e^{ijx}.$

We say that $W$ is of diagonal type ($W\in{\cal DT}$) if

$|W_{j,k}|\leq c_{j-k}\quad\mbox{for some sequence
$\\{c_{s}\\}_{s\in\mathbb{Z}}$ with $\sum c_{s}={\bf c}<\infty$}.$ (1.11)

We put $\|W\|_{\cal DT}=\inf{\bf c}$ where infimum is taken over all sequences
$c_{s}$, satisfying (1.11).

As simple examples we have the following operators of diagonal type.

(a) Convolution operators from Section 6.

(b) Operators of multiplication by functions with absolutely converging
Fourier series.

(c) Linear combinations and products of operators of diagonal type.

(d) If $W\in{\cal DT}$ then the conjugated operator $W^{*}$ also lies in
${\cal DT}$.

We prove (Corollary 7.1) that the space $({\cal DT},\|\cdot\|_{\cal DT})$ is a
$C^{*}$-algebra.

  * •

We have the following inequality between norms ((1.4) and Lemma 7.3):

$\|W\|_{\mu}\leq\|W\|\leq\|W\|_{\cal DT}.$

  * •

We associate with $W\in{\cal DT}$ and any point $a\in\mathbb{T}$ the
distribution $L_{a}$,

$L_{a}=\sum_{j\in\mathbb{Z}}w_{j}(a)e^{ijx},\qquad
w_{j}(a)=\sum_{k\in\mathbb{Z}}W_{j,k}e^{i(j-k)a}.$

We prove that

(1) the function $a\mapsto\rho(L_{a})$ is continuous (Lemma 7.4),

(2)
$\displaystyle\|W\|_{\mu}^{2}=\frac{1}{2\pi}\int_{\mathbb{T}}\rho(L_{a})\,da$
(Proposition 7.1).

  * •

For any operator $W\in{\cal DT}$ we introduce ‘‘average trace’’ of $W^{*}W$ by

${\bf T}(W)=\limsup_{\\#I\to\infty}\frac{1}{\\#I}\sum_{j\in\mathbb{Z},\,l\in
I}|W_{l,j}|^{2},$

where $I\subset\mathbb{Z}$ are intervals. Then (Proposition 8.1)
$\|W\|_{\mu}^{2}\geq{\bf T}(W)$.

We also prove (Proposition 8.2) that if $U$ is a unitary operator of diagonal
type then ${\bf T}(W)={\bf T}(U^{-1}WU)$.

##  2 General properties of the $\mu$-norm

###  2.1 $\mu$-norm of $\pi_{X}$

###### Lemma 2.1

For any $X\in{\cal B}$

$\|\pi_{X}\|_{\mu}^{2}=\mu(X).$ (2.1)

Proof. For any partition $\chi=\\{Y_{1},\ldots,Y_{J}\\}$

${\cal
M}_{\chi}(\pi_{X})=\sum_{j=1}^{J}\mu(Y_{j})\|\pi_{X}\pi_{Y_{j}}\|^{2}=\sum_{\mu(Y_{j}\cap
X)>0}\mu(Y_{j})\geq\mu(X).$

This implies ${\cal M}_{\chi}(\pi_{X})\geq\mu(X)$.

On the other hand, for $\chi=\\{X,Y\\}$, $Y={\cal X}\setminus X$ we have:

${\cal
M}_{\chi}(\pi_{X})=\mu(X)\|\pi_{X}\|^{2}+\mu(Y)\|\pi_{X}\pi_{Y}\|^{2}=\mu(X).$

Hence $\inf_{\chi}{\cal M}_{\chi}(\pi_{X})=\mu(X)$.

###  2.2 $\mu$-norm and subpartitions

###### Lemma 2.2

Let $W$ be a bounded operator and $\chi=\\{Y_{1},\ldots,Y_{J}\\}$ a partition
of the set $Y\in{\cal B}$. Then

$\mu(Y)\|W\pi_{Y}\|^{2}\geq\sum_{j=1}^{J}\mu(Y_{j})\|W\pi_{Y_{j}}\|^{2}.$
(2.2)

Proof. The following two simple remarks

$\|W\pi_{Y_{j}}\|\leq\|W\pi_{Y}\|\quad\mbox{and}\quad\sum_{j=1}^{J}\mu(Y_{j})=\mu(Y)$

imply (2.2).

###### Corollary 2.1

Lemma 2.2 implies that ${\cal M}_{\chi^{\prime}}(W)\leq{\cal M}_{\chi}(W)$ if
$\chi^{\prime}$ is a subpartition of $\chi$.

###  2.3 Triangle inequality

###### Lemma 2.3

For any two bounded operators $W_{1}$ and $W_{2}$

$\|W_{1}+W_{2}\|_{\mu}\leq\|W_{1}\|_{\mu}+\|W_{2}\|_{\mu}.$ (2.3)

Proof. First we show that for any partition $\chi=\\{X_{1},\ldots,X_{J}\\}$

$\sqrt{{\cal M}_{\chi}(W_{1}+W_{2})}\leq\sqrt{{\cal
M}_{\chi}(W_{1})}+\sqrt{{\cal M}_{\chi}(W_{2})}.$ (2.4)

We put

$a_{j}=\|W_{1}\pi_{X_{j}}\|,\quad b_{j}=\|W_{2}\pi_{X_{j}}\|,\quad
c_{j}=\|(W_{1}+W_{2})\pi_{X_{j}}\|,\quad\mu_{j}=\mu(X_{j}).$

Then

$0\leq c_{j}\leq a_{j}+b_{j}$ (2.5)

and inequality (2.4) is equivalent to

$\sqrt{\sum\mu_{j}c_{j}^{2}}\leq\sqrt{\sum\mu_{j}a_{j}^{2}}+\sqrt{\sum\mu_{j}b_{j}^{2}}.$

By (2.5) to prove this inequality, it is sufficient to check that

$\sqrt{\sum\mu_{j}(a_{j}+b_{j})^{2}}\leq\sqrt{\sum\mu_{j}a_{j}^{2}}+\sqrt{\sum\mu_{j}b_{j}^{2}}.$

This inequality is equivalent to

$\Big{(}\sum\mu_{j}a_{j}b_{j}\Big{)}^{2}\leq\sum\mu_{j}a_{j}^{2}\cdot\sum\mu_{j}b_{j}^{2}$

which follows from Cauchy-Bunyakovsky-Schwartz.

To derive (2.3) from (2.4), we take a partition $\chi$ such that

$\|W_{1}\|_{\mu}>\sqrt{{\cal
M}_{\chi}(W_{1})}-\varepsilon,\quad\mbox{and}\quad\|W_{2}\|_{\mu}>\sqrt{{\cal
M}_{\chi}(W_{2})}-\varepsilon.$

Then by (2.4)

$\displaystyle\|W_{1}+W_{2}\|_{\mu}$ $\displaystyle\leq$
$\displaystyle\sqrt{{\cal M}_{\chi}(W_{1}+W_{2})}\leq\sqrt{{\cal
M}_{\chi}(W_{1})}+\sqrt{{\cal M}_{\chi}(W_{2})}$ $\displaystyle\leq$
$\displaystyle\|W_{1}\|_{\mu}+\|W_{2}\|_{\mu}+2\varepsilon.$

Since $\varepsilon>0$ is arbitrary, (2.3) follows.

An equivalent form of inequality (2.3) is

$\big{|}\|W_{2}\|_{\mu}-\|W_{1}\|_{\mu}\big{|}\leq\|W_{2}-W_{1}\|_{\mu}.$

###### Corollary 2.2

By (1.5) we have:

$\big{|}\|W_{2}\|_{\mu}-\|W_{1}\|_{\mu}\big{|}\leq\|W_{2}-W_{1}\|.$

Therefore, the $\mu$-norm is a continuous function of $W$ in the $L^{2}$
operator topology.

###  2.4 $U_{auto}({\cal H})$-invariance

Let $F:{\cal X}\hookleftarrow$ be a measure preserving map: for any $X\in{\cal
B}$

$F^{-1}(X)\in{\cal B},\quad\mbox{and}\quad\mu(X)=\mu(F^{-1}(X)).$

Such maps are called endomorphisms of the measure space $({\cal X},{\cal
B},\mu)$. If an endomorphism $F$ is invertible and the inverse map is
measurable, then $F$ is said to be an automorphism.

Any endomorphism $F:{\cal X}\hookleftarrow$ generates an operator $U=U_{F}$ on
${\cal H}$, where for any $f\in{\cal H}$

$U_{F}f=f\circ F.$ (2.6)

The operator (2.6) preserves the scalar product and satisfies the following
identities:

$\displaystyle U_{F}(fg)$ $\displaystyle=$ $\displaystyle U_{F}f\cdot
U_{F}g\qquad\mbox{for any }f,g\in{\cal H},$ (2.7) $\displaystyle
U_{F}\pi_{\xi}$ $\displaystyle=$
$\displaystyle\pi_{F^{-1}(\xi)}U_{F}\qquad\mbox{ for any }\xi\in{\cal B}.$
(2.8)

Operators $U_{F}$ corresponding to automorphisms are invertible and therefore,
unitary. They form a subgroup $U_{\operatorname{auto}}({\cal H})$ in the group
$U({\cal H})$ of unitary operators on ${\cal H}$.

Equation (2.8) implies the following

###### Corollary 2.3

For any endomorphism $F$, any bounded $W$ and any $\xi\in{\cal B}$

$\|WU_{F}\pi_{\xi}\|\leq\|W\pi_{F^{-1}(\xi)}\|.$

Indeed, for any $f\in L^{2}({\cal X})$ (2.8) implies

$\|WU_{F}\pi_{\xi}f\|=\|W\pi_{F^{-1}(\xi)}U_{F}f\|,\quad\|f\|=\|U_{F}f\|.$

###### Lemma 2.4

For any bounded $W$ and any automorphism $F$

$\|WU_{F}\|_{\mu}=\|W\|_{\mu}.$ (2.9)

###### Corollary 2.4

If $F$ is an automorphism then by (2.9) and (1.7)
$\|U_{F}^{-1}WU_{F}\|_{\mu}=\|W\|_{\mu}$. Hence the $\mu$-norm is invariant
w.r.t. conjugations by elements of $U_{\operatorname{auto}}({\cal H})$.

Proof of Lemma 2.4. Let $\chi=\\{Y_{1},\ldots,Y_{J}\\}$ be such that
$\|W\|_{\mu}^{2}\geq{\cal M}_{\chi}(W)-\varepsilon$. Putting
$\kappa=\\{F(Y_{1}),\ldots,F(Y_{J})\\}$, we have:

$\|WU_{F}\|_{\mu}^{2}\leq{\cal
M}_{\kappa}(WU_{F})=\sum_{j=1}^{J}\mu(F(Y_{j}))\|WU_{F}\pi_{F(Y_{j})}\|^{2}.$

By Corollary 2.3 and the equations $F^{-1}(F(Y_{j}))=Y_{j}$ this sum does not
exceed

$\sum_{j=1}^{J}\mu(Y_{j})\|W\pi_{Y_{j}}\|^{2}={\cal
M}_{\chi}(W)\leq\|W\|_{\mu}^{2}+\varepsilon.$

Since $\varepsilon>0$ is arbitrary, $\|WU_{F}\|_{\mu}\leq\|W\|_{\mu}$. Since
$F$ is invertible, we have analogously $\|W\|_{\mu}\leq\|WU_{F}\|_{\mu}$.

###### Corollary 2.5

Equation (2.8) and Lemma 2.1 imply that for any collection of sets
$Y_{0},\ldots,Y_{K}\in{\cal B}$ and any automorphism $F$

$\|\pi_{Y_{K}}U_{F}\pi_{Y_{K-1}}U_{F}\ldots
U_{F}\pi_{Y_{1}}U_{F}\pi_{Y_{0}}\|_{\mu}^{2}=\mu\big{(}Y_{K}\cap\ldots\cap
F^{-K+1}(Y_{1})\cap F^{-K}(Y_{0})\big{)}.$

Indeed, by (2.8), (2.9)

$\displaystyle\|\pi_{Y_{K}}U_{F}\pi_{Y_{K-1}}U_{F}\ldots\pi_{Y_{1}}U_{F}\pi_{Y_{0}}\|_{\mu}^{2}$
$\displaystyle=$
$\displaystyle\|\pi_{Y_{K}}\pi_{F^{-1}(Y_{K-1})}U_{F}^{2}\pi_{Y_{K-2}}\ldots\pi_{Y_{1}}U_{F}\pi_{Y_{0}}\|_{\mu}^{2}$
$\displaystyle=$
$\displaystyle\ldots\;=\;\|\pi_{Y_{K}}\pi_{F^{-1}(Y_{K-1})}\ldots\pi_{F^{-K}(Y_{0})}U_{F}^{K}\|_{\mu}^{2}$
$\displaystyle=$ $\displaystyle\|\pi_{Y_{K}\cap\ldots\cap
F^{-K}(Y_{0})}\|_{\mu}^{2}.$

By (2.1) the latter quantity equals $\mu\big{(}Y_{K}\cap\ldots\cap
F^{-k}(Y_{0})\big{)}$.

##  3 Examples

###  3.1 $\mu$-norm of a compact operator

The measure $\mu$ is said to have no atoms if for any $X\in{\cal B}$,
$\mu(X)>0$ there exists a set $Y\in{\cal B}$ such that $0<\mu(Y)<\mu(X)$.

###### Lemma 3.1

Suppose $\mu$ has no atoms and $W$ is a finite rank operator. Then
$\|W\|_{\mu}=0$.

Proof. Any finite rank operator in ${\cal H}$ has the form, (see, for example,
[16])

$f\mapsto Wf,\qquad Wf=\sum_{l=1}^{L}\lambda_{l}\langle f,v_{l}\rangle u_{l}$

for two orthonormal systems $\\{v_{l}\\}$ and $\\{u_{l}\\}$. Given
$\varepsilon>0$ we can choose a partition $\chi=\\{Y_{1},\ldots,Y_{J}\\}$ with
elements so small that

$\|\pi_{Y_{j}}v_{l}\|\leq\varepsilon\quad\mbox{ for any $1\leq j,l\leq J$}.$

Then for any $f\in{\cal H}$, $\|f\|=1$ and any $j\in\\{1,\ldots,J\\}$

$\|W\pi_{Y_{j}}f\|^{2}=\Big{\|}\sum_{l=1}^{L}\lambda_{l}\langle\pi_{Y_{j}}f,v_{l}\rangle
u_{l}\Big{\|}^{2}\leq\Lambda\varepsilon^{2},\qquad\Lambda=\sum_{1}^{L}\lambda^{2}_{l}.$

Therefore ${\cal
M}_{\chi}(W)\leq\sum\mu(Y_{j})\Lambda\varepsilon^{2}=\Lambda\varepsilon^{2}$.

###### Corollary 3.1

Suppose $\mu$ has no atoms and $W$ is a compact operator. Then
$\|W\|_{\mu}=0$.

Indeed, any compact operator in ${\cal H}$ may be approximated by finite rank
operators. It remains to use Corollary 2.2.

###### Corollary 3.2

Suppose $\mu$ has no atoms. Then for any bounded $W$ and compact $W_{0}$ we
have: $\|W+W_{0}\|_{\mu}=\|W\|_{\mu}$.

This statement is a combination of (2.3) and Corollary 3.1.

###  3.2 Multiplication operator

For any $g\in L^{\infty}({\cal X})$ consider the operator $\widehat{g}$ of
multiplication by $g$:

$L^{2}({\cal X})\ni f\mapsto\widehat{g}f=gf.$ (3.1)

###### Lemma 3.2

$\|\widehat{g}\|_{\mu}=\|g\|$.

Proof. First, suppose that $g$ is simple i.e., for some partition
$\kappa=\\{X_{1},\ldots,X_{K}\\}$

$g=\sum_{k=1}^{K}c_{k}{\bf 1}_{X_{k}},\qquad c_{k}\in\mathbb{C}.$

Then $\|\widehat{g}\pi_{Y}\|^{2}=\max|c_{k}|^{2}$, where maximum is taken for
$k$ such that $\mu(Y\cap X_{k})>0$.

By Corollary 2.1 it is sufficient to consider partitions
$\chi=\\{Y_{1},\ldots,Y_{J}\\}$ which are subpartitions of $\kappa$. Then

${\cal M}_{\chi}(\widehat{g})=\sum_{j=1}^{J}\mu(Y_{j})\max_{\mu(Y_{j}\cap
X_{k})>0}|c_{k}|^{2}=\sum_{k=1}^{K}\mu(X_{k})|c_{k}|^{2}=\|g\|^{2}.$

If $g$ is not simple then neglecting a set of a small measure, we can
approximate $g$ in the $L^{\infty}$-norm by simple functions.

###  3.3 Finite-dimensional case

Let ${\cal X}=\\{1,\ldots,J\\}$ be a finite set. Then ${\cal B}$ consists of
all subsets $X\subset{\cal X}$. We assume that $\mu(X)=\\#X/J$ i.e., the
measure $\mu$ is uniformly distributed between points of ${\cal X}$. Then
${\cal H}=\mathbb{C}^{J}$ with the product
$\langle\,,\rangle=\frac{1}{J}(\,,)$, where $(\,,)$ is the standard Hermitian
product.

Linear operators on ${\cal H}$ are identified with $J\times J$-matrices. For
any $X\subset{\cal X}$

$\pi_{X}=\operatorname{diag}(\delta_{1X},\ldots,\delta_{JX}),\qquad\delta_{jX}=\left\\{\begin{array}[]{cc}1\quad\mbox{if
}j\in X,\\\ 0\quad\mbox{if }j\not\in X.\end{array}\right.$

###### Lemma 3.3

Suppose $\\#{\cal X}=J<\infty$. Let $W=(W_{jk})$ be an operator on ${\cal H}$.
Then333The quantity in the right-hand side of (3.2) equals the trace of
$W^{*}W$ divided by $J$. In particular, if $W$ is unitary then
$\|W\|_{\mu}=1$. Below such and analogous quantities will be said to be the
average trace of $W^{*}W$.

$\displaystyle\|W\|_{\mu}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{J}\sum_{j,k=1}^{J}|W_{kj}|^{2},$ (3.2)
$\displaystyle\|W\|_{\mu}$ $\displaystyle=$
$\displaystyle\|WU\|_{\mu}\quad\mbox{for any unitary operator $U$ on ${\cal
H}$}.$ (3.3)

Proof. By Lemma 2.2 to compute $\|W\|_{\mu}$, we may use the finest partition

$\chi=\big{\\{}\\{1\\},\ldots,\\{J\\}\big{\\}}.$ (3.4)

Then for any $j$ the unit vector $e_{j}=\pi_{\\{j\\}}e_{j}$ has the
coordinates $e_{jk}=\sqrt{J}\delta_{jk}$. Therefore

$\|W\|_{\mu}^{2}=\frac{1}{J}\sum_{j=1}^{J}\frac{\|W\pi_{\\{j\\}}\|^{2}}{J}=\frac{1}{J^{2}}\sum_{j=1}^{J}\|We_{j}\|^{2}=\frac{1}{J}\sum_{j,k=1}^{J}|W_{kj}|^{2}.$

To prove (3.3) we may use (3.2). Then

$J\|WU\|_{\mu}^{2}=\sum_{j,k,l,s}W_{kl}U_{lj}\overline{W}_{ks}\overline{U}_{sj}=\sum_{j,k,l,s}W_{kl}\overline{W}_{ks}U_{lj}U^{*}_{js}=\sum_{k,l,s}W_{kl}\overline{W}_{ks}\delta_{ls}=\sum_{k,l}|W_{kl}|^{2}.$

The latter quantity equals $J\|W\|_{\mu}^{2}$.

###  3.4 Dimension of a subspace

Let $H\subset{\cal H}$ be a closed subspace. We define its $\mu$-dimension by

$\dim_{\mu}(H):=\|\pi_{H}\|_{\mu}^{2},\quad\mbox{where $\pi_{H}:{\cal H}\to H$
is the orthogonal projector}.$

###### Lemma 3.4

1\. If $\mu$ has no atoms and $H$ is finite-dimensional then
$\dim_{\mu}(H)=0$.

2\. If $H=\pi_{X}({\cal H})$ for some $X\in{\cal B}$ then
$\dim_{\mu}(H)=\mu(X)$.

3\. Suppose ${\cal X}$ is finite, $\\#{\cal X}=J$, and for any $X\subset{\cal
X}$ $\mu(X)=\\#X/J$. Then for any subspace $H\subset{\cal H}=\mathbb{C}^{J}$
its $\mu$-dimension equals its relative dimension: $\dim_{\mu}(H)=\dim(H)/J$,
where $\dim(\cdot)$ is the conventional dimension of a vector space.

Proof. The first statement follows from Lemma 3.1. The second statement
follows from Lemma 2.1. The third statement follows from Lemma 3.3.

Suppose a cyclic group $\mathbb{Z}_{q}$ acts on $({\cal X},{\cal B},\mu)$ by
automorphisms i.e., for any $s\in\mathbb{Z}_{q}$ there is an automorphism
$F_{s}$ such that the following identities hold:

$F_{s^{\prime}}\circ
F_{s^{\prime\prime}}=F_{s^{\prime}+s^{\prime\prime}}\quad\mbox{and}\quad
F_{-s}=F^{-1}_{s}.$ (3.5)

We assume that the action is almost free. This means that there is a set
$X\in{\cal B}$ such that the sets $X_{s}=F_{s}(X)$ do not intersect pairwise
and $\mu({\cal X}\setminus\cup X_{s})=0$. Let $U_{s}=U_{F_{s}}$ be the unitary
operators, determined by (2.6). Then by (3.5)

$U_{s^{\prime}}U_{s^{\prime\prime}}=U_{s^{\prime}+s^{\prime\prime}}\quad\mbox{and}\quad
U_{-s}=U^{-1}_{s}.$

Putting $r=e^{2\pi i/q}$, we consider the spaces

$H_{n}=\\{f\in{\cal H}:U_{s}f=r^{ns}f\;\mbox{ for any }s\in\mathbb{Z}_{q}\\}.$

Functions $f\in H_{0}$ are invariant w.r.t. the $\mathbb{Z}_{q}$-action.

###### Lemma 3.5

The spaces $H_{n}$ are pairwise orthogonal, ${\cal H}=\oplus H_{n}$, and
$\dim_{\mu}(H_{n})=1/q$.

Proof. For any functions $f_{j}\in H_{j}$ and $f_{k}\in H_{k}$ and any
$s\in\mathbb{Z}_{q}$ we have:

$\langle f_{j},f_{k}\rangle=\langle
U_{s}f_{j},U_{s}f_{k}\rangle=r^{(j-k)s}\langle f_{j},f_{k}\rangle.$

This proves the first statement of the lemma. The second statement follows
from the equations ($f\in{\cal H}$ is arbitrary)

$f=\sum_{n\in\mathbb{Z}_{q}}\pi_{H_{n}}f,\qquad\pi_{H_{n}}f=\frac{1}{q}\sum_{k\in\mathbb{Z}_{q}}r^{-nk}U_{k}f\in
H_{n}.$ (3.6)

To compute the $\mu$-dimension of $H_{n}$, consider a partition
$\chi=\\{Y_{1},\ldots,Y_{J}\\}$. We may assume that each element of the
partition has nonempty intersection only with one set $X_{s}$. Then for any
function $f=\pi_{Y_{j}}f$ the supports of the terms $r^{-nk}U_{n}f$ do not
intersect pairwise. Therefore

$\|\pi_{H_{n}}f\|^{2}=\frac{1}{q^{2}}\sum_{k\in\mathbb{Z}_{q}}\|r^{-nk}U_{n}f\|^{2}=\frac{\|f\|^{2}}{q}.$

Hence ${\cal M}_{\chi}(\pi_{H_{n}})=\|f\|^{2}/q$. This finishes the proof.

##  4 Additivity

###### Lemma 4.1

For any partition $\kappa=\\{X_{1},\ldots,X_{K}\\}$

$\|W\|_{\mu}^{2}=\sum_{k=1}^{K}\|W\pi_{X_{k}}\|_{\mu}^{2}.$ (4.1)

Proof. Equation (4.1) follows from the simple observation that for any
subpartition $\chi=\\{Y_{1},\ldots,Y_{J}\\}$ of the partition $\kappa$

$\sum_{k=1}^{K}{\cal M}_{\chi}(W\pi_{X_{k}})={\cal M}_{\chi\vee\kappa}(W).$

It is natural to call the simple property of the $\mu$-norm presented in Lemma
4.1 the right additivity. The question about the left additivity turns out to
be nontrivial. The following lemma presents a partial result in this
direction.

###### Lemma 4.2

For any partition $\kappa=\\{X_{1},\ldots,X_{K}\\}$

$\|W\|_{\mu}^{2}\leq\sum_{k=1}^{K}\|\pi_{X_{k}}W\|_{\mu}^{2}.$ (4.2)

Proof. For any $f\in{\cal H}$ and $Y\in{\cal B}$

$\|W\pi_{Y}f\|^{2}=\sum_{k=1}^{K}\|\pi_{X_{k}}W\pi_{Y}f\|^{2},$ (4.3)

Therefore for any partition $\chi$

${\cal M}_{\chi}(W)\leq\sum_{k=1}^{K}{\cal M}_{\chi}(\pi_{X_{k}}W).$ (4.4)

###### Remark 4.1

Unfortunately, we do not have in (4.4) an equation instead of the inequality.
Indeed, for different terms in (4.3) $\sup_{\|f\|=1}$ may be taken (almost
taken) on different functions $f$.

Let $\chi_{1},\ldots,\chi_{K}$ be partitions such that

${\cal
M}_{\chi_{k}}(\pi_{X_{k}}W)\leq\|\pi_{X_{k}}W\|_{\mu}^{2}+\varepsilon,\qquad
k=1,\ldots,K.$

We put $\chi=\vee_{k=1}^{K}\chi_{k}$. Then by (4.4) and Corollary 2.1

$\|W\|_{\mu}^{2}\leq{\cal M}_{\chi}(W)\leq\sum_{k=1}^{K}{\cal
M}_{\chi}(\pi_{X_{k}}W)\leq\sum_{k=1}^{K}{\cal
M}_{\chi_{k}}(\pi_{X_{k}}W)\leq\sum_{k=1}^{K}\|\pi_{X_{k}}W\|_{\mu}^{2}+K\varepsilon.$

Since $\varepsilon>0$ is arbitrary, we obtain (4.2).

Let $\widehat{g}$ be the operator of multiplication by the function $g\in
L^{\infty}({\cal X})$.

###### Lemma 4.3

For any collection of functions $g_{1},\ldots,g_{K},g\in L^{\infty}({\cal X})$
such that

$\sum|g_{k}|^{2}=|g|^{2}$ (4.5)

and any bounded operator $W$

$\sum_{k=1}^{K}\|W\widehat{g}_{k}\|_{\mu}^{2}=\|W\widehat{g}\|_{\mu}^{2}.$
(4.6)

Proof. First, suppose that the functions $g_{k}$ are simple:

$g_{k}=\sum_{j=1}^{J_{k}}\widetilde{g}_{kj}{\bf 1}_{X_{kj}},\qquad
k=1,\ldots,K,$ (4.7)

where $\widetilde{g}_{kj}\in\mathbb{C}$ and
$\chi_{k}=\\{X_{k1},\ldots,X_{kJ_{k}}\\}$ are partitions. Note that instead of
the partitions $\chi_{k}$ we can use their common refinement
$\chi=\chi_{1}\vee\ldots\vee\chi_{K}=\\{X_{1},\ldots,X_{J}\\}$. Hence, we can
replace (4.7) by

$g_{k}=\sum_{j=1}^{J}g_{kj}{\bf 1}_{X_{j}},\qquad g_{kj}\in\mathbb{C}.$

The function $g$ in this case is also simple: $g=\sum g_{j}{\bf 1}_{X_{j}}$.
Equation (4.5) implies

$\sum_{k=1}^{K}|g_{kj}|^{2}=|g_{j}|^{2}\quad\mbox{for any }j=1,\ldots,J.$
(4.8)

Equation (4.1) implies

$\displaystyle\|W\widehat{g}\|_{\mu}^{2}$ $\displaystyle=$
$\displaystyle\sum_{j}\|W\widehat{g}\pi_{X_{j}}\|_{\mu}^{2}=\sum_{j}\|Wg_{j}\pi_{X_{j}}\|_{\mu}^{2}=\sum_{j}|g_{j}|^{2}\|W\pi_{X_{j}}\|_{\mu}^{2},$
$\displaystyle\|W\widehat{g}_{k}\|_{\mu}^{2}$ $\displaystyle=$
$\displaystyle\sum_{j}\|W\widehat{g}_{k}\pi_{X_{j}}\|_{\mu}^{2}=\sum_{j}\|Wg_{kj}\pi_{X_{j}}\|_{\mu}^{2}=\sum_{j}|g_{kj}|^{2}\|W\pi_{X_{j}}\|_{\mu}^{2}.$

Now (4.6) follows from (4.8).

If the functions $g_{k}$ are not simple, we approximate them in
$L^{\infty}({\cal X})$-norm by simple functions and use Corollary 2.2.

##  5 An additional structure

Let the function $\operatorname{dist}(\cdot,\cdot):{\cal X}\times{\cal
X}\to\mathbb{R}_{+}$ determine on ${\cal X}$ the structure of a compact metric
space such that the balls

$B_{r}(x)=\\{y\in{\cal X}:\operatorname{dist}(y,x)<r\\}$

are measurable. Moreover, we assume that ${\cal B}$ is the corresponding Borel
$\sigma$-algebra while $\mu$ is a Borel measure.

###### Proposition 5.1

For any $x\in{\cal X}$ there exists the limit

$\vartheta(x)=\lim_{\varepsilon\searrow
0}\vartheta_{\varepsilon}(x),\qquad\vartheta_{\varepsilon}(x)=\|W\pi_{B_{\varepsilon}(x)}\|^{2},$
(5.1)

the function $\vartheta:{\cal X}\to[0,\|W\|^{2}]$ is measurable and

$\|W\|_{\mu}^{2}\leq\int_{\cal X}\vartheta\,d\mu.$ (5.2)

Proof of Proposition 5.1. For any $x\in{\cal X}$ the function
$\varepsilon\mapsto\vartheta_{\varepsilon}(x)$, $\varepsilon>0$ is non-
decreasing and non-negative. This implies the existence of limits (5.1).

Consider a sequence $\\{\varepsilon_{j}\\}_{j\in\mathbb{N}}$,
$\varepsilon_{j}\to 0$ as $j\to\infty$. For any $j\in\mathbb{N}$ let ${\cal
N}_{j}=\\{y_{1},\ldots,y_{K}\\}\subset{\cal X}$ be a finite
$\varepsilon_{j}$-net444The number $K\in\mathbb{N}$ as well as the points
$y_{k}$ depend on $j$, but for brevity we do not indicate this in the
notation.:

$\mbox{for any $x\in{\cal X}$ there exists $y_{k}\in{\cal N}_{j}$ such that
$\operatorname{dist}(y_{k},x)<\varepsilon_{j}$}.$

We define

$y(x,j)\in\\{y_{k}\in{\cal N}_{j}:\operatorname{dist}(y_{k},x)=\min_{y\in{\cal
N}_{j}}\operatorname{dist}(y,x)\\}$

as the element with minimal possible index. Then
$\operatorname{dist}(y(x,j),x)<\varepsilon_{j}$.

For any $x\in{\cal X}$ and $j\in\mathbb{N}$

$B_{\varepsilon_{j}}(x)\subset B_{2\varepsilon_{j}}(y(x,j))\subset
B_{3\varepsilon_{j}}(x).$ (5.3)

We put
$\widetilde{\vartheta}_{j}(x)=\|W\pi_{B_{2\varepsilon_{j}}(y(x,j))}\|^{2}$.
Then $\widetilde{\vartheta}_{j}$ takes a finite number of values and by
construction of $y(x,j)$ the preimage of each value is measurable. Therefore
$\widetilde{\vartheta}_{j}$ is measurable.

Inclusions (5.3) imply

$\vartheta_{\varepsilon_{j}}(x)\leq\widetilde{\vartheta}_{j}(x)\leq\vartheta_{3\varepsilon_{j}}(x).$
(5.4)

Inequalities (5.4) imply

$\lim_{j\to\infty}\widetilde{\vartheta}_{j}(x)=\lim_{\varepsilon\searrow
0}\vartheta_{\varepsilon}(x)=\vartheta(x).$

Hence $\vartheta$ is measurable as a pointwise limit of measurable functions,
[11].

Now we turn to the proof of inequality (5.2). We may assume that
$\varepsilon_{j+1}\leq\varepsilon_{j}/3$. Then by (5.3)
$B_{2\varepsilon_{j+1}}(y(x,j+1))\subset B_{2\varepsilon_{j}}(y(x,j))$.
Therefore $\widetilde{\vartheta}_{j+1}(x)\leq\widetilde{\vartheta}_{j}(x)$ for
any $x\in{\cal X}$. The sequences
$\\{\widetilde{\vartheta}_{j}(x)\\}_{j\in\mathbb{N}}$ monotonically tend to
$\vartheta(x)$ as $j\to\infty$. The sets

$S_{j,\sigma}=\\{x\in{\cal
X}:\widetilde{\vartheta}_{j}-\vartheta(x)>\sigma\\}$

are measurable and satisfy for any $\sigma>0$

$S_{j+1,\sigma}\subset
S_{j,\sigma},\quad\lim_{j\to\infty}\mu(S_{j,\sigma})=0.$ (5.5)

We put $\varepsilon^{\prime}=\varepsilon_{j^{\prime}}/2$, where
$\mu(S_{j^{\prime},\sigma})<\sigma_{1}$. By (generalization of) Lusin’s
theorem [12] there exists $X_{\sigma_{1}}\in{\cal B}$ such that
$\mu(X_{\sigma_{1}})\leq\sigma_{1}$, ${\cal X}_{\sigma_{1}}={\cal X}\setminus
X_{\sigma_{1}}$ is compact, and $\vartheta$ is continuous on ${\cal
X}_{\sigma_{1}}$.

The function $\vartheta|_{{\cal X}_{\sigma_{1}}}$ is uniformly continuous.
Hence there exists $\varepsilon^{\prime\prime}>0$ such that for any
$x,y\in{\cal X}_{\sigma_{1}}$

$|x-y|<\varepsilon^{\prime\prime}\quad\mbox{implies}\quad|\vartheta(x)-\vartheta(y)|<\sigma_{2}.$
(5.6)

For $0<\varepsilon<\min\\{\varepsilon^{\prime},\varepsilon^{\prime\prime}\\}$
let ${\cal N}_{\varepsilon}=\\{y_{1},\ldots,y_{K}\\}\subset{\cal
X}_{\sigma_{1}}$ be a finite $\varepsilon$-net on ${\cal X}_{\sigma_{1}}$. The
balls $B_{\varepsilon}(y_{1}),\ldots,B_{\varepsilon}(y_{K})$ cover ${\cal
X}_{\sigma_{1}}$. Consider the partition $\chi=\\{S,X_{1},\ldots,X_{K}\\}$,
where

$S=X_{\sigma_{1}}\cup S_{j^{\prime},\sigma},\quad
X_{1}=B_{\varepsilon}(y_{1})\setminus S,\quad
X_{j}=B_{\varepsilon}(y_{j})\setminus(S\cup
B_{\varepsilon}(y_{1})\cup\ldots,\cup B_{\varepsilon}(y_{j-1}))$

for $j\geq 2$. Then

$\Theta(x)=\left\\{\begin{array}[]{ccc}\vartheta(y_{j})&\mbox{if}&x\in
X_{j},\\\ 0&\mbox{if}&x\in S\end{array}\right.$

is a simple function on ${\cal X}$ which approximates $\vartheta$ on the set
${\cal X}\setminus S$ with precision $\sigma_{2}$:

$\big{|}(\Theta-\vartheta)|_{{\cal X}\setminus S}\big{|}\leq\sigma_{2}.$

Moreover,

$\Big{|}\int_{\cal X}\Theta\,d\mu-\int_{\cal
X}\vartheta\,d\mu\Big{|}\leq\sigma_{2}+\int_{S}\vartheta\,d\mu\leq\sigma_{2}+2\|W\|^{2}\sigma_{1}.$
(5.7)

Consider the quantity $\displaystyle{\cal
M}_{\chi}(W)=\mu(S)\|W\pi_{S}\|^{2}+\sum_{k=1}^{K}\mu(X_{k})\|W\pi_{X_{k}}\|^{2}$.
Then

${\cal M}_{\chi}(W)-\int_{\cal X}\Theta\,d\mu\leq
2\sigma_{1}\|W\|^{2}+\sum_{k=1}^{K}\mu(X_{k})(\|W\pi_{X_{k}}\|^{2}-\vartheta(y_{k})).$
(5.8)

By (5.3) $X_{k}\subset B_{2\varepsilon_{j^{\prime}}}(y(y_{k},j^{\prime}))$.
Therefore by (5.6)

$\|W\pi_{X_{k}}\|^{2}\leq\vartheta(y_{k})+\sigma_{2}.$

This implies $\sum\mu(X_{k})\|W\pi_{X_{k}}\|^{2}\leq\int_{\cal
X}\Theta\,d\mu+\sigma_{2}$. Combining this estimate with (5.7) and (5.8), we
obtain:

${\cal M}_{\chi}(W)\leq\int_{\cal
X}\vartheta\,d\mu+2\sigma_{2}+2\|W\|^{2}\sigma_{1}.$

Since $\sigma_{1}$ and $\sigma_{2}$ may be chosen arbitrary small, we obtain
(5.2).

###### Example 5.1

Inequality in (5.2) may be strict. Indeed, let ${\cal X}$ be the circle
$\mathbb{T}=\mathbb{R}/(2\pi\mathbb{Z})$ with the Lebesgue measure
$d\mu=dx/(2\pi)$. Let $X\subset\mathbb{T}$ be the complement to a Cantorian
set of positive measure. Then $X$ is open, dense, and $\mu(X)<1$. Take
$W=\pi_{X}$. Then $\vartheta\equiv 1$. However

$\|W\|_{\mu}^{2}=\mu(X)<1=\frac{1}{2\pi}\int_{\mathbb{T}}\vartheta(x)\,dx.$

To replace in (5.2) the inequality by the equation, we need additional
assumptions.

C1. The function $\vartheta$ is continuous.

C2. There exists $c>0$ such that for any open ${\cal O}\subset{\cal X}$,
$\operatorname{diam}({\cal O})\leq\varepsilon$ and any $x\in{\cal O}$ there
exists $f=\pi_{\cal O}f$ satisfying

$\Big{|}\|Wf\|^{2}-\vartheta(x)\|f\|^{2}\Big{|}\leq\gamma(\varepsilon)\|f\|^{2},\quad
c<\big{|}f|_{\cal O}\big{|}<c^{-1},$

where $\gamma(\varepsilon)\to 0$ as $\varepsilon\to 0$.

###### Proposition 5.2

Suppose conditions C1–C2 hold. Then

$\|W\|_{\mu}^{2}=\int_{\cal X}\vartheta\,d\mu.$

Proof. According to Proposition 5.1 it is sufficient to prove the inequality

$\|W\|_{\mu}^{2}\geq\int_{\cal X}\vartheta\,d\mu.$

By C1 $\vartheta$ is continuous on ${\cal X}$. Therefore it is uniformly
continuous. Given a small $\sigma>0$ we choose $\varepsilon>0$ such that for
any ball $B\subset{\cal X}$ of radius $\varepsilon$

$x,y\in B\quad\mbox{implies}\quad|\vartheta(x)-\vartheta(y)|\leq\sigma.$ (5.9)

Suppose

$\|W\|_{\mu}^{2}<\int_{\cal X}\vartheta\,d\mu-\alpha\quad\mbox{for some
$\alpha>0$}.$

Then for some partition $\chi=\\{Y_{1},\ldots,Y_{K}\\}$ we have:

${\cal M}_{\chi}(W)<\int_{\cal X}\vartheta\,d\mu-\alpha.$

We are going to show that this inequality holds only for
$\alpha=\alpha(\varepsilon)\to 0$ as $\varepsilon\to 0$.

By Lemma 2.2 we may assume that $\operatorname{diam}(Y_{k})<\varepsilon$,
$k=1,\ldots,K$. We take arbitrary $Y\in\chi$ such that

$\mu(Y)\|W\pi_{Y}\|^{2}<\int_{Y}\vartheta\,d\mu-\alpha.$ (5.10)

By [11] the measure $\mu$ is regular as any Borel probability measure on a
compact metric space. Hence for any $\gamma_{1}>0$ there exists a ball
$B_{\varepsilon}(x)$ and an open set $A$ such that

$Y\subset A\subset B_{\varepsilon}(x),\quad\mu(Y)\geq\mu(A)(1-\gamma_{1}).$
(5.11)

Putting $D=A\setminus Y$, we have by (5.11):

$\mu(D)\leq\gamma_{1}\mu(A).$ (5.12)

Let $f=\pi_{A}f$ be a function which by C2 satisfies

$\Big{|}\|Wf\|^{2}-\vartheta(x)\|f\|^{2}\Big{|}\leq\gamma(\varepsilon)\|f\|^{2},\quad
c<\big{|}f|_{A}\big{|}<c^{-1},\quad x^{\prime}\in A.$ (5.13)

Since $f=\pi_{A}f=\pi_{A\cap Y}f+\pi_{D}f$, the triangle inequality implies

$\|W\pi_{Y}f\|\geq\|Wf\|-\|W\pi_{D}f\|$

We estimate $\|Wf\|$ by (5.13) and $\|W\pi_{D}f\|\leq\|W\|\,\|\pi_{D}f\|$ by
Lemma 9.1:

$\|Wf\|\geq(\vartheta(x)-\gamma(\varepsilon))\|f\|^{2},\quad\|\pi_{D}f\|\leq\frac{\sqrt{\mu(D)}}{c^{2}\sqrt{\mu(A)}}\|f\|.$

Therefore for any $x^{\prime}\in A\cap Y$

$\displaystyle\|W\pi_{Y}f\|^{2}$ $\displaystyle\geq$
$\displaystyle\bigg{(}\sqrt{\vartheta(x^{\prime})-\gamma(\varepsilon)}\|f\|-\|W\|\frac{\sqrt{\mu(D)}}{c^{2}\sqrt{\mu(A)}}\|f\|\bigg{)}^{2}$
$\displaystyle\geq$
$\displaystyle\bigg{(}\vartheta(x^{\prime})-\gamma(\varepsilon)-2\frac{\sqrt{\vartheta(x)}\|W\|}{c^{2}}\frac{\sqrt{\mu(D)}}{\sqrt{\mu(A)}}\bigg{)}\|f\|^{2}.$

By using (5.9), (5.12), and the estimate $\vartheta\leq\|W\|^{2}$, we obtain

$\frac{\|W\pi_{Y}f\|^{2}}{\|f\|^{2}}\geq\vartheta(x)-\sigma-\gamma(\varepsilon)-2\sqrt{\gamma_{1}}\frac{\|W\|^{2}}{c^{2}}.$

This estimate implies that

$\|W\pi_{Y}\|^{2}-\frac{1}{\mu(Y)}\int_{Y}\vartheta\,d\mu\geq\frac{\|W\pi_{Y}f\|^{2}}{\|f\|^{2}}-\frac{1}{\mu(Y)}\int_{Y}\vartheta\,d\mu\geq-\alpha,$

where

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\Big{|}\vartheta(x)-\frac{1}{\mu(Y)}\int_{Y}\vartheta\,d\mu\Big{|}+\gamma_{1}\vartheta(x)+\sigma+\gamma(\varepsilon)+2\sqrt{\gamma_{1}}\frac{\|W\|^{2}}{c^{2}}$
$\displaystyle=$ $\displaystyle
2\sigma+\gamma_{1}\vartheta(x)+\gamma(\varepsilon)+2\sqrt{\gamma_{1}}\frac{\|W\|^{2}}{c^{2}}.$

Hence $\alpha$ in (5.10) has to be arbitrarily small if $\varepsilon$ is small
enough.

##  6 Convolutions on $L^{2}(\mathbb{T})$

###  6.1 Preliminary constructions

a. Let $\mathbb{T}=\mathbb{R}/(2\pi\mathbb{Z})$ be a circle. The measure $\mu$
and the Hermitian product have the form

$d\mu=\frac{1}{2\pi}\,dx,\quad\langle
f,g\rangle=\frac{1}{2\pi}\int_{\mathbb{T}}f\overline{g}\,dx.$

For any bounded sequence $\\{\lambda_{k}\\}_{k\in\mathbb{Z}}$ consider the
distribution

$\lambda(x)=\sum_{k\in\mathbb{Z}}\lambda_{k}e^{ikx}.$ (6.1)

In particular, if all $\lambda_{k}$ equal $1$ then $\lambda=2\pi\delta$, where
$\delta=\delta(x)$ is the $\delta$-function on $\mathbb{T}$.

b. The operator of convolution

$L^{2}(\mathbb{T})\ni
f\mapsto\operatorname{Conv}_{\lambda}f=\lambda*f=\frac{1}{2\pi}\int_{\mathbb{T}}\lambda(x-y)f(y)\,dy$

is bounded and

$\|\operatorname{Conv}_{\lambda}\|=c_{\lambda},\qquad
c_{\lambda}=\sup_{k\in\mathbb{Z}}|\lambda_{k}|.$ (6.2)

###### Remark 6.1

Below to reduce the number of brackets, we assume that the convolution has a
higher priority in comparison with the arithmetic operations. Hence, $fg*h$
means $f\cdot(g*h)$ for any functions $f,g,h$ on $\mathbb{T}$.

If the sequence $\\{\lambda_{k}\\}_{k\in\mathbb{Z}}$ lies in the space $l^{2}$
then $\lambda\in L^{2}(\mathbb{T})$ and

$\|\lambda\|^{2}=\|\\{\lambda_{k}\\}_{k\in\mathbb{Z}}\|^{2}_{l^{2}}.$ (6.3)

c. For any $n\in\mathbb{Z}$, any $a\in L^{\infty}(\mathbb{T})$ and any $g\in
L^{2}(\mathbb{T})$

$\lambda*(e^{inx}g)=e^{inx}((e^{-inx}\lambda)*g),\quad\|ag\|\leq\|a\|_{\infty}\|g\|,$
(6.4)

where $\|\cdot\|_{\infty}$ is the norm in $L^{\infty}$.

d. For any $\Lambda\subset\mathbb{Z}$ consider the sequence
$\\{\lambda_{k}\\}_{k\in\mathbb{Z}}$, where $\lambda_{k}=1$ if $k\in\Lambda$
and $\lambda_{k}=0$ if $k\not\in\Lambda$. Let $d_{\Lambda}$ denote the
corresponding distribution (6.1). Then $\operatorname{Conv}_{d_{\Lambda}}$ is
the orthogonal projector to the subspace of $L^{2}(\mathbb{T})$ spanned by the
vectors $e^{ikx}$, $k\in\Lambda$.

Given $a\in\mathbb{T}$ consider the distribution $d_{a,\Lambda}$, defined by

$d_{a,\Lambda}(x)=d_{\Lambda}(x-a)\quad\mbox{for any }x\in\mathbb{T}.$

If the sets $A,B\subset\mathbb{Z}$ have empty intersection then $d_{a,A\cup
B}=d_{a,A}+d_{a,B}$, and for any $a\in\mathbb{T}$ and any $g\in
L^{2}(\mathbb{T})$

$\|d_{a,A\cup B}*g\|^{2}=\|d_{a,A}*g\|^{2}+\|d_{a,B}*g\|^{2}.$ (6.5)

For any $k\in\mathbb{Z}$ let $\\{k\\}$ denote the corresponding one-point
subset of $\mathbb{Z}$. Then for any $f\in L^{2}(\mathbb{T})$

$d_{a,\\{k\\}}*f=f_{k}d_{a,\\{k\\}}.$ (6.6)

where $f_{k}$ is the Fourier coefficient of $f$ with the number $k$.

e. For any interval $J=\\{n,n+1,\ldots,m\\}\subset\mathbb{Z}$ let $\\#J=m-n+1$
be the number of points on $J$.

###### Lemma 6.1

Let $Y=[a-\varepsilon,a+\varepsilon]$ and $\displaystyle
f=\pi_{Y}f=\sum_{k\in\mathbb{Z}}f_{k}e^{ikx}\in L^{2}(\mathbb{T})$. Then for
any integer interval $J$ and any $m\in\mathbb{Z}$

$\displaystyle\|f-e^{im(x-a)}f\|$ $\displaystyle\leq$
$\displaystyle|m|\varepsilon\|f\|,$ (6.7)
$\displaystyle|f_{m}-e^{ila}f_{m+l}|$ $\displaystyle\leq$
$\displaystyle\frac{\varepsilon^{3/2}}{\sqrt{\pi}}|l|\|f\|,$ (6.8)
$\displaystyle\Big{|}\sum_{k\in\mathbb{Z}}e^{-ima}f_{k}\overline{f}_{k+m}-\|f\|^{2}\Big{|}$
$\displaystyle\leq$ $\displaystyle|m|\varepsilon\|f\|^{2}.$ (6.9)

Proof. We put $g=f-e^{im(x-a)}f$. By using inequality (6.4), we have:

$\|g\|^{2}=\frac{1}{2\pi}\int_{a-\varepsilon}^{a+\varepsilon}|1-e^{im(x-a)}|^{2}|f|^{2}\,dx\leq\|\pi_{Y}(1-e^{im(x-a)})\|_{\infty}^{2}\|f\|^{2}\leq
m^{2}\varepsilon^{2}\|f\|^{2}.$

This implies (6.7).

Now we turn to inequality (6.8). We have:

$\displaystyle|f_{m}-f_{m+l}e^{ila}|$ $\displaystyle=$
$\displaystyle\bigg{|}\frac{1}{2\pi}\int_{a-\varepsilon}^{a+\varepsilon}\Big{(}f(x)e^{-ilx}-f(x)e^{-ilx-
il(x-a)}\Big{)}\,dx\bigg{|}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2\pi}\int_{a-\varepsilon}^{a+\varepsilon}\Big{|}f(x)(1-e^{-il(x-a)})\Big{|}\,dx$
$\displaystyle\leq$
$\displaystyle\|f\|\,\big{\|}\pi_{Y}(1-e^{-il(x-a)})\big{\|}\;\leq\;\frac{\varepsilon^{3/2}}{\sqrt{\pi}}|l|\,\|f\|.$

To prove (6.9), we consider the Hermitian product

$\langle
g,f\rangle=\|f\|^{2}-\sum_{k\in\mathbb{Z}}e^{-ima}f_{k}\overline{f}_{k+m}.$

By (6.7) $|\langle g,f\rangle|\leq\|g\|\cdot\|f\|\leq|m|\varepsilon\|f\|^{2}$.
Hence, (6.9) follows.

f. For any distribution (6.1) we put

$\rho(\lambda)=\limsup_{\\#I\to\infty}\rho_{I}(\lambda),\qquad\rho_{I}(\lambda)=\frac{1}{\\#I}\sum_{k\in
I}|\lambda_{k}|^{2},$ (6.10)

where $I$ denotes integer intervals.

###### Proposition 6.1

$\|\operatorname{Conv}_{\lambda}\|_{\mu}^{2}=\rho(\lambda)$.

###### Remark 6.2

Corollary 3.1 suggests the question: does the equation $\|W\|_{\mu}=0$ imply
compactness of the operator $W$? Proposition 6.1 gives an example of a
noncompact operator $W$ with $\|W\|_{\mu}=0$. We may take
$W=\operatorname{Conv}_{d_{\Lambda}}$, where
$\Lambda=\\{k^{2}:k\in\mathbb{N}\\}$.

We prove Proposition 6.1 in Sections 6.2 and 6.3

###  6.2 Computation of
$\|\operatorname{Conv}_{\lambda}\pi_{(-\varepsilon,\varepsilon)}\|$

We put $Y=(a-\varepsilon,a+\varepsilon)\subset\mathbb{T}$. It is possible to
change the variable $x\mapsto x-a$ on $\mathbb{T}$. This will change
$\lambda(x)$ by $\lambda(x-a)$, but $\rho$ will remain the same. Hence we can
assume that $a=0$ and $Y=(-\varepsilon,\varepsilon)$.

###### Lemma 6.2

For any $\gamma>0$ and any $f=\pi_{Y}f\in L^{2}(\mathbb{T})$

$\|\lambda*f\|^{2}\leq(\rho+\gamma)\|f\|^{2}\quad\mbox{if $\varepsilon$ is
sufficiently small.}$ (6.11)

For any $\gamma>0$ there exists $f=\pi_{Y}f$, such that $|f|=\mbox{\rm const}$
on $Y$ and

$\|\lambda*f\|^{2}\geq(\rho-\gamma)\|f\|^{2}\quad\mbox{if $\varepsilon$ is
sufficiently small.}$ (6.12)

###### Corollary 6.1

Estimates (6.11)–(6.12) imply that for any sufficiently small interval $Y$

$\Big{|}\|\operatorname{Conv}_{\lambda}\pi_{Y}\|-\rho\Big{|}=o(\mu(Y)).$
(6.13)

Proof of Lemma 6.2. Let $M^{\prime}(\sigma)$ be such that

$\rho_{I}(\lambda)\leq\rho+\sigma\quad\mbox{for any interval
$I\subset\mathbb{Z}$, \quad$\\#I\geq M^{\prime}(\sigma)$}.$ (6.14)

Let $I=-I\subset\mathbb{Z}$ be an interval with
$\\#I=2B+1>M^{\prime}(\sigma)$. We put

$F=\frac{1}{\\#I}d_{I}f=\frac{1}{2B+1}\sum_{k\in\mathbb{Z},\,|l|\leq
B}f_{k+l}e^{ikx}.$ (6.15)

By Lemma 6.1

$\|f-F\|\leq\frac{1}{\\#I}\sum_{l\in
I}\|f-e^{ilx}f\|\leq\frac{1}{2B+1}\sum_{|l|\leq B}|l|\varepsilon\|f\|.$

Hence

$\|f-F\|\leq(B+1)\varepsilon\|f\|,\quad\|\lambda*f-\lambda*F\|\leq
c_{\lambda}(B+1)\varepsilon\|f\|$ (6.16)

We have:

$\displaystyle\|\lambda*F\|^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{(2B+1)^{2}}\sum_{k\in\mathbb{Z}}|\lambda_{k}|^{2}\sum_{|l|,|n|\leq
B}f_{k+l}\overline{f}_{k+n}$ (6.17) $\displaystyle=$
$\displaystyle\frac{1}{(2B+1)^{2}}\sum_{k\in\mathbb{Z}}|\lambda_{k-l}|^{2}\sum_{|l|,|m+l|\leq
B}f_{k}\overline{f}_{k+m}$ $\displaystyle=$ $\displaystyle\sum_{|m|\leq
2B}\sum_{k\in\mathbb{Z}}b_{m}\rho_{I_{k,m}}f_{k}\overline{f}_{k+m},$

where

$b_{m}=\frac{2B+1-|m|}{(2B+1)^{2}},\quad
I_{k,m}=\left\\{\begin{array}[]{cc}[k-B,k+B-m]\cap\mathbb{Z}&\mbox{ if }m\geq
0,\\\ {}[k-B-m,k+B]\cap\mathbb{Z}&\mbox{ if }m<0.\end{array}\right.$ (6.18)

Note that

$\sum_{|m|\leq 2B}b_{m}=1,\quad\\#I_{k,m}=2B+1-|m|,\quad\rho_{I_{k,m}}\leq
c_{\lambda},$ (6.19)

and by (6.14) $\rho_{I_{k,m}}\leq\rho+\sigma$ if $\\#I_{k,m}\geq
M^{\prime}=M^{\prime}(\sigma)$.

By using (6.9) with $a=0$, we have the estimate

$\displaystyle\|\lambda*F\|^{2}$ $\displaystyle\leq$
$\displaystyle\sum_{2B+1-|m|<M^{\prime},\,k\in\mathbb{Z}}b_{m}c_{\lambda}f_{k}\overline{f}_{k+m}+\sum_{2B+1-|m|\geq
M^{\prime},\,k\in\mathbb{Z}}b_{m}(\rho-\sigma)f_{k}\overline{f}_{k+m}$
$\displaystyle\leq$
$\displaystyle\bigg{(}\frac{M^{\prime}(1+M^{\prime})}{(2B+1)^{2}}c_{\lambda}+\frac{(2B+1)^{2}-M^{\prime}(1+M^{\prime})}{(2B+1)^{2}}(\rho-\sigma)\bigg{)}(1-B\varepsilon)\|f\|^{2}.$

This implies (6.11) if $\sigma$, $M^{\prime}/B$, and $B\varepsilon$ are small.

Now we check estimate (6.12). We fix small positive $\sigma$ and
$\varepsilon$. By Lemma 9.2 we choose an interval $J\subset\mathbb{Z}$ such
that

$\displaystyle\\#J\geq 20M^{\prime}(\sigma)\mbox{ is
odd},\quad\rho_{J}(\lambda)>\rho-\sigma,$ (6.20) $\displaystyle
2(\sigma_{1}\varepsilon)^{-1}+1\leq\\#J\leq
4(\sigma_{1}\varepsilon)^{-1}+3,\quad\sigma_{1}=\sigma^{1/3}.$ (6.21)

Consider the function

$\varphi=\pi_{Y}\varphi=\frac{{\bf
1}_{Y}}{\sqrt{2\varepsilon}}=\frac{1}{\sqrt{2\varepsilon}}\sum_{k\in\mathbb{Z}}\frac{\sin
k\varepsilon}{\pi k}e^{ikx},\quad\|\varphi\|=1.$

For any interval $I=-I\subset\mathbb{Z}$ such that $\\#I=2b+1$ we have the
estimate

$\|\varphi-
d_{I}*\varphi\|^{2}\leq\frac{1}{2\varepsilon}\sum_{|k|>b}\frac{1}{\pi^{2}k^{2}}<\frac{1}{b\varepsilon}.$
(6.22)

By (6.20) $\\#J$ is odd. Hence there exists $q\in\mathbb{Z}$ such that
$(J-q)=-(J-q)$. We define $f=e^{iqx}\varphi$ and show that this $f$ satisfies
(6.12).

Obviously $f=\pi_{Y}f$ and $|f|=\mbox{const}$ on $Y$. We put

$I=[-B,B]\cap\mathbb{Z},\quad
I^{\prime}=[-M^{\prime},M^{\prime}]\cap\mathbb{Z},\quad
M^{\prime}=M^{\prime}(\sigma),\quad B>M^{\prime}.$

and consider $\widetilde{J}=-\widetilde{J}\subset\mathbb{Z}$ such that
$\widetilde{J}+4I+I^{\prime}=J-q$. We will assume

$(\sigma_{1}\varepsilon)^{-1}>8B+2M^{\prime}.$ (6.23)

Then by (6.21)

$\frac{1}{\varepsilon\sigma_{1}}<\\#\widetilde{J}=\\#J-8B-2M^{\prime}<\frac{4}{\varepsilon\sigma_{1}}$

and by (6.22) with $b=(\sigma_{1}\varepsilon)^{-1}$

$\displaystyle\|f-d_{\widetilde{J}+q}*f\|^{2}$ $\displaystyle\leq$
$\displaystyle\sigma_{1}\|f\|^{2},$ (6.24)
$\displaystyle\|\lambda*f-\lambda*d_{\widetilde{J}+q}*f\|$ $\displaystyle\leq$
$\displaystyle c_{\lambda}\|f-d_{\widetilde{J}+q}*f\|\leq
c_{\lambda}\sqrt{\sigma_{1}}\|f\|.$ (6.25)

We determine $F$ by (6.15) and put

$\Phi=\sum_{|m|\leq
2B}\sum_{k\in\widetilde{J}+2I+q}b_{m}\rho_{I_{k,m}}f_{k}\overline{f}_{m+k}.$
(6.26)

Since $\rho_{I}\leq c_{\lambda}$, for any interval $I\subset\mathbb{Z}$, we
have by (6.17):

$\displaystyle\Big{|}\Phi-\|\lambda*F\|^{2}\Big{|}$ $\displaystyle=$
$\displaystyle\sum_{|m|\leq
2B}\sum_{l\not\in\widetilde{J}+2I}b_{m}\rho_{I_{l+q,m}}\varphi_{l}\overline{\varphi}_{m+l}$
$\displaystyle\leq$ $\displaystyle\sum_{|m|\leq
2B}\sum_{l\not\in\widetilde{J}+2I}b_{m}c_{\lambda}|\varphi_{l}\varphi_{m+l}|$
$\displaystyle\leq$ $\displaystyle\sum_{|m|\leq
2B}\sum_{l\not\in\widetilde{J}}c_{\lambda}b_{m}|\varphi_{l}|^{2}\;\leq\;2c_{\lambda}\sigma_{1}\|f\|^{2}.$

Note that the conditions $|m|\leq 2B$, $k\in\widetilde{J}+2I+q$ in (6.26)
imply $I_{k,m}\subset J$, where $I_{k,m}$ are defined in (6.18). Therefore by
Lemma 9.3

$\rho_{I_{k,m}}=\rho-s_{m},\qquad s_{m}\leq\frac{2\sigma\\#J}{2B+1-|m|}.$

This implies that $\Phi$ is estimated as follows:

$\Phi\geq\sum_{|m|\leq
2B}\sum_{k\in\widetilde{J}+2I+q}\Big{(}b_{m}\rho-\frac{2\sigma\\#J}{(2B+1)^{2}}\Big{)}f_{k}\overline{f}_{m+k}$

We perform summation in $k$ with the help of (6.9) and (6.24):

$\displaystyle\sum_{k\in\widetilde{J}+2I+q}f_{k}\overline{f}_{m+k}$
$\displaystyle=$
$\displaystyle\bigg{(}\sum_{k\in\mathbb{Z}}\;\;-\\!\\!\sum_{k\not\in\widetilde{J}+2I+q}\bigg{)}f_{k}\overline{f}_{m+k}$
$\displaystyle\geq$
$\displaystyle(1-|m|\varepsilon)\|f\|^{2}-\sum_{k\not\in\widetilde{J}+q}|f_{k}|^{2}\;\geq\;(1-|m|\varepsilon-\sigma_{1})\|f\|^{2}.$

Hence

$\displaystyle\Phi$ $\displaystyle\geq$ $\displaystyle\sum_{|m|\leq
2B}(1-2\varepsilon
B-\sigma_{1})\Big{(}b_{m}\rho-\frac{2\sigma\\#J}{(2B+1)^{2}}\Big{)}\|f\|^{2}$
$\displaystyle\geq$ $\displaystyle(1-2\varepsilon
B-\sigma_{1})\Big{(}\rho-\frac{2\sigma\\#J}{2B+1}\Big{)}\|f\|^{2}.$

This estimate implies that

$\Phi\geq(\rho-\gamma)\|f\|^{2},$

where $\gamma$ is small provided $\sigma_{1}$, $\sigma\\#J/B$, and
$\varepsilon B$ are small. These quantities should be small, and conditions
(6.21) and (6.23) should hold. This can be made if we choose
$\sigma_{1}=\sigma^{1/3}$, $\\#J/11<B<\\#J/10$.

###  6.3 Application of Proposition 5.2

Corollary 6.1 implies that the function $\vartheta$, corresponding by (5.1) to
the operator $W=\operatorname{Conv}_{\lambda}$, equals to $\rho$ identically
on $\mathbb{T}$. Hence $\vartheta$ is continuous and moreover, by Lemma 6.2
Condition C2 from Section 5 holds. Therefore Proposition 5.2 implies

$\|\operatorname{Conv}_{\lambda}\|_{\mu}^{2}=\frac{1}{2\pi}\int_{\mathbb{T}}\rho\,dx=\rho.$

##  7 Operators of diagonal type

###  7.1 Definition and properties

In this section we continue the study of bounded operators on
$L^{2}(\mathbb{T})$. Consider the operator $W=(W_{j,k})_{j,k\in\mathbb{Z}}$

$f\mapsto F=Wf,\quad F_{j}=\sum_{k\in\mathbb{Z}}W_{j,k}f_{k},\qquad
f=\sum_{k\in\mathbb{Z}}f_{k}e^{ikx},\quad
F=\sum_{k\in\mathbb{Z}}F_{k}e^{ikx}.$

Let $\Lambda_{k}$, $k\in\mathbb{Z}$ be the distributions

$\Lambda_{k}(x)=\sum_{j\in\mathbb{Z}}W_{k+j,j}e^{ijx}.$ (7.1)

Then by (6.2)

$\|\operatorname{Conv}_{\Lambda_{k}}\|=c_{k},\qquad
c_{k}=\sup_{j\in\mathbb{Z}}|W_{k+j,j}|.$ (7.2)

By using the distributions $\Lambda_{k}$, we obtain another form of $W$ (see
Remark 6.1):

$Wf=\sum_{k\in\mathbb{Z}}e^{ikx}\Lambda_{k}*f.$

###### Definition 7.1

The operator $W$ is said to be of diagonal type if
$\sum_{k\in\mathbb{Z}}c_{k}={{\bf c}}<\infty$. Below ${\cal DT}$ denotes the
space of such operators. The sequence $\\{c_{k}\\}_{k\in\mathbb{Z}}$ will be
said to be the majorating sequence for $W$. We also put

$\|W\|_{{\cal DT}}={\bf c}.$

If $W\in{\cal DT}$ then $|W_{j,k}|\leq c_{j-k}$. Hence, elements $W_{j,k}$
which are not too close to zero are concentrated near the main diagonal of the
matrix $W$. This motivates the terminology.

Examples. 1. For any distribution (6.1) with finite $c_{\lambda}$ the operator
$\operatorname{Conv}_{\lambda}\in{\cal DT}$ because all $c_{k}$ vanish except
$c_{0}=c_{\lambda}$.

2\. Let $g:\mathbb{T}\to\mathbb{C}$ be a function with absolutely converging
Fourier series. Then the multiplication operator $f\mapsto\widehat{g}f=gf$ is
of diagonal type. Indeed, $\widehat{g}_{j,k}=g_{j-k}$, where $g=\sum
g_{k}e^{ikx}$. Hence, $c_{k}=|g_{k}|$.

We have the obvious estimate

$\|g\|_{\infty}\leq\|\widehat{g}\|_{{\cal DT}}.$

3\. If $W\in{\cal DT}$ then the conjugated operator $W^{*}$ is also of
diagonal type with the majorating sequence
$\\{c_{k}^{*}\\}_{k\in\mathbb{Z}}=\\{c_{-k}\\}_{k\in\mathbb{Z}}$. This follows
from the equation $\sup_{j}|W^{*}_{k+j,j}|=\sup_{j}|W_{-k+j,j}|$. Moreover, we
have the equation $\|W^{*}\|_{{\cal DT}}=\|W\|_{{\cal DT}}$.

4\. Linear combination of operators of diagonal type is an operator of
diagonal type and

$\|\lambda_{1}W_{1}+\lambda_{2}W_{2}\|_{{\cal
DT}}\leq|\lambda_{1}|\|W_{1}\|_{{\cal DT}}+|\lambda_{2}|\|W_{2}\|_{{\cal
DT}}.$

###### Lemma 7.1

Product of two operators of diagonal type is also an operator of diagonal type
and $\|W^{\prime}W^{\prime\prime}\|_{{\cal DT}}\leq\|W^{\prime\prime}\|_{{\cal
DT}}\|W^{\prime}\|_{{\cal DT}}$.

Proof. Suppose $W^{\prime}$ and $W^{\prime\prime}$ are of diagonal type. Let
$c^{\prime}_{k}$ and $c^{\prime\prime}_{k}$ be the majorating sequences. Then

$|(W^{\prime}W^{\prime\prime})_{jl}|=\Big{|}\sum_{k}W^{\prime}_{jk}W^{\prime\prime}_{kl}\Big{|}\leq\tilde{c}_{j-l},\qquad\tilde{c}_{j-l}=\sum_{k}c^{\prime}_{j-k}c^{\prime\prime}_{k-l}.$

In particular, the majorating sequence $c_{k}$ of the operator
$W^{\prime}W^{\prime\prime}$ satisfies the inequality $c_{k}\leq\tilde{c}_{k}$
for any $k\in\mathbb{Z}$.

The obvious computation
$\sum\tilde{c}_{s}=\sum_{s,k}c^{\prime}_{s-k}c^{\prime\prime}_{k}={\bf
c}^{\prime}{\bf c}^{\prime\prime}$ finishes the proof.

###### Lemma 7.2

The space ${\cal DT}$ is closed w.r.t. the norm $\|\cdot\|_{{\cal DT}}$.

Proof. Suppose $\\{W_{n}\\}_{n\in\mathbb{N}}$ is a Cauchy sequence: for any
$\varepsilon>0$ there exists $N\in\mathbb{N}$ such that

$m,n>N\quad\mbox{implies}\quad\|W_{m}-W_{n}\|_{{\cal DT}}<\varepsilon.$ (7.3)

Then for any $j,k\in\mathbb{Z}$ the matrix elements
$\\{(W_{n})_{jk}\\}_{n\in\mathbb{N}}$ also form a Cauchy sequence. Hence,
there exists a matrix $W=\lim_{n\to\infty}W_{n}$, where the limit is taken
element-wise:

$W_{jk}=\lim_{n\to\infty}(W_{n})_{jk},\qquad j,k\in\mathbb{Z}.$

We put

$d_{k}=\sup_{j\in\mathbb{Z}}\big{|}(W_{n})_{k+j,j}-W_{k+j,j}\big{|},\quad
d=\|W_{n}-W\|_{{\cal DT}}.$

Then $d=\sum_{k\in\mathbb{Z}}d_{k}$. There exists $K>0$ such that

$\Big{|}d-\sum_{k=-K}^{K}d_{k}\Big{|}<\varepsilon.$ (7.4)

There exists an integer sequence $\\{j_{k}\\}$ such that putting
$\delta_{k}=\big{|}(W_{n})_{k+j_{k},j_{k}}-W_{k+j_{k},j_{k}}\big{|}$, we have:

$|d_{k}-\delta_{k}|<\frac{\varepsilon}{2K+1}\qquad\mbox{for any
$k\in[-K,K]$},$

Then

$\sum_{k=-K}^{K}|d_{k}-\delta_{k}|<\varepsilon.$ (7.5)

Putting
$\widetilde{\delta}_{k}=\big{|}(W_{m})_{k+j_{k},j_{k}}-(W_{n})_{k+j_{k},j_{k}}\big{|}$,
we take $m>n$ such that

$\widetilde{\delta}_{k}<\frac{\varepsilon}{2K+1},\qquad-K\leq k\leq K.$

Hence,

$\sum_{k=-K}^{K}\widetilde{\delta}_{k}<\varepsilon.$ (7.6)

By (7.3)

$\sum_{k=-K}^{K}\big{|}(W_{n})_{k+j_{k},j_{k}}-(W_{m})_{k+j_{k},j_{k}}\big{|}\leq\|W_{n}-W_{m}\|_{{\cal
DT}}<\varepsilon.$ (7.7)

Combining (7.4)–(7.7), we obtain $d\leq 4\varepsilon$.

###### Corollary 7.1

The space ${\cal DT}$ endowed with the norm $\|\cdot\|_{{\cal DT}}$ is a
$\mathbb{C}^{*}$-algebra.

###### Lemma 7.3

Suppose $W\in{\cal DT}$. Then $\|W\|\leq\|W\|_{{\cal DT}}$.

Proof. For any $f\in L^{2}(\mathbb{T})$ we have:

$\displaystyle\|Wf\|^{2}$ $\displaystyle=$
$\displaystyle\sum_{k}\Big{|}\sum_{j}W_{kj}f_{j}\Big{|}^{2}=\sum_{k,j,l}W_{kj}\overline{W}_{kl}f_{j}\overline{f}_{l}\leq\sum_{k,j,l}|W_{kj}W_{kl}f_{j}f_{l}|$
(7.8) $\displaystyle\leq$
$\displaystyle\sum_{k,j,l}c_{k-j}c_{k-l}|f_{j}f_{l}|=\sum_{j,l}A_{j-l}|f_{j}f_{l}|,$

where $A_{j-l}=\sum_{k\in\mathbb{Z}}c_{k-j}c_{k-l}$. Note that
$\sum_{s\in\mathbb{Z}}A_{s}=\sum_{s,k}c_{k}c_{s+k}={\bf c}^{2}.$ Continuing
computation (7.8), we obtain:

$\|Wf\|^{2}\leq\sum_{s}A_{s}\sum_{l}|f_{l+s}|\,|f_{l}|\leq\|f\|^{2}\sum_{s}A_{s}\leq{\bf
c}^{2}\|f\|^{2}.$

The middle inequality follows from Cauchy-Bunyakovsky-Schwarz. This implies
the lemma.

For any $a\in\mathbb{T}$ consider the distribution $L_{a}$,

$L_{a}(x)=\sum_{k\in\mathbb{Z}}e^{ik(x+a)}\Lambda_{k}(x).$ (7.9)

Then

$\|\operatorname{Conv}_{L_{a}}\|\leq\sum\|\operatorname{Conv}_{\Lambda_{k}}\|={\bf
c}.$ (7.10)

By (7.1)

$L_{a}(x)=\sum_{j,k\in\mathbb{Z}}e^{ik(x+a)}W_{k+j,j}e^{ijx}=\sum_{l\in\mathbb{Z}}w_{l}(a)e^{ilx},\qquad
w_{l}(a)=\sum_{j\in\mathbb{Z}}W_{l,j}e^{i(l-j)a}.$ (7.11)

For any $l\in\mathbb{Z}$ and $a\in\mathbb{T}$

$|w_{l}(a)|\leq\sum_{j\in\mathbb{Z}}|W_{l,j}|\leq{\bf c}.$ (7.12)

###  7.2 Computation of the $\mu$-norm

Consider the function $a\mapsto\rho(L_{a})$ determined by (6.10):

$\rho(L_{a})=\limsup_{\\#I\to\infty}\rho_{I}(L_{a}),\qquad\rho_{I}(L_{a})=\frac{1}{\\#I}\sum_{l\in
I}|w_{l}(a)|^{2},$ (7.13)

where $I\subset\mathbb{Z}$ are intervals.

###### Lemma 7.4

If $W\in{\cal DT}$ then $\rho(L_{a})$ is a continuous function of
$a\in\mathbb{T}$.

We prove Lemma 7.4 in Section 9.3.

###### Proposition 7.1

Suppose $W\in{\cal DT}$. Then

$\|W\|_{\mu}^{2}=\frac{1}{2\pi}\int_{\mathbb{T}}\rho(L_{a})\,da.$ (7.14)

###### Definition 7.2

We say that the matrix $(W_{k,j})$ is $\tau$-periodic, $\tau\in\mathbb{N}$, if

$W_{k+\tau,j+\tau}=W_{k,j}\quad\mbox{for any }k,j\in\mathbb{Z}.$ (7.15)

###### Lemma 7.5

Let $W^{\prime}$ and $W^{\prime\prime}$ be two bounded operators with
$\tau$-periodic matrices. Then matrices of the operators

$W^{\prime*},\quad W^{\prime}W^{\prime\prime},\quad\mbox{and
}a^{\prime}W^{\prime}+a^{\prime\prime}W^{\prime\prime},\qquad
a^{\prime},a^{\prime\prime}\in\mathbb{C}$

are also $\tau$-periodic.

We skip an obvious proof.

###### Corollary 7.2

The space of ${\cal DT}$ operators with $\tau$-periodiс matrices form a
$\mathbb{C}^{*}$ subalgebra in ${\cal DT}$.

###### Example 7.1

Suppose the matrix $(W_{k,j})$ is $\tau$-periodic, $\tau\in\mathbb{N}$. Then
$w_{l+\tau}(a)=w_{l}(a)$ and therefore,
$\rho(L_{a})=\frac{1}{\tau}\sum_{l=0}^{\tau-1}|w_{l}(a)|^{2}$.

In particular, if $\tau=1$ then $w_{l}=w_{0}$ for any $l\in\mathbb{Z}$, and
$W$ is the operator of multiplication by $w_{0}(x)$. In this case by (7.14)

$\|W\|_{\mu}^{2}=\frac{1}{2\pi}\int_{\mathbb{T}}|w_{0}(a)|^{2}\,da=\|w_{0}\|^{2}.$

This result is a particular case of Lemma 3.2 under additional assumptions
that $({\cal X},\mu)=(\mathbb{T},\frac{1}{2\pi}dx)$ and the Fourier series of
the function $g$ absolutely converges.

Proof of Proposition 7.1.

###### Lemma 7.6

Suppose $W\in{\cal DT}$, $Y=[a-\varepsilon,a+\varepsilon]\subset\mathbb{T}$
and $f=\pi_{Y}f\in L^{2}(\mathbb{T})$. Given $\sigma>0$ for any
$a\in\mathbb{T}$ and any $0<\varepsilon\leq\varepsilon_{0}(\sigma,W)$

$\|Wf-L_{a}*f\|\leq 3\sigma\|f\|.$ (7.16)

Proof of Lemma 7.6. We define the cut-off $W_{K}$,

$W_{K}f=\sum_{|k|\leq K}e^{ikx}\Lambda_{k}*f.$

Then if $K=K(\sigma)$ is large enough then

$\Big{\|}Wf-
W_{K}f\Big{\|}=\Big{\|}\sum_{|k|>K}e^{ikx}\Lambda_{k}*f\Big{\|}\leq\sum_{|k|>K}\|\Lambda_{k}*f\|\leq\sum_{|k|>K}c_{k}\|f\|\leq\sigma\|f\|.$
(7.17)

By (6.4)

$\displaystyle W_{K}f$ $\displaystyle=$ $\displaystyle\sum_{|k|\leq
K}(e^{ik(x+a)}\Lambda_{k})*(e^{ik(x-a)}f)=L_{a,K}*f+\Delta,$ $\displaystyle
L_{a,K}(x)$ $\displaystyle=$ $\displaystyle\sum_{|k|\leq
K}e^{ik(x+a)}\Lambda_{k}(x),\quad\Delta\;=\;\sum_{|k|\leq
K}(e^{ik(x+a)}\Lambda_{k})*\big{(}(e^{ik(x-a)}-1)f\big{)}.$

By (6.4) and (7.2)

$\|\Delta\|\leq\sum_{|k|\leq K}c_{k}\|(e^{ik(x-a)}-1)f\|\leq\sum_{|k|\leq
K}c_{k}\varepsilon K\|f\|<\varepsilon K{\bf c}\|f\|.$

Therefore

$\|W_{K}f-L_{a,K}*f\|\leq\varepsilon K{\bf c}\|f\|.$ (7.18)

We also have:

$\|L_{a,K}*f-L_{a}*f\|\leq\|\operatorname{Conv}_{L_{a,K}-L_{a}}\|\cdot\|f\|\leq\sum_{|k|>K}c_{k}\|f\|\leq\sigma\|f\|.$
(7.19)

Combining estimates (7.17), (7.18), and (7.19), we obtain (7.16) if
$\varepsilon_{0}=\sigma/(K{\bf c})$.

Now we return to the proof of Proposition 7.1. We associate with $W$ the
function $\vartheta$ by (5.1). According to Lemmas 6.2 and 7.6 for any
$a\in\mathbb{T}$ we have: $\vartheta(a)=\rho(L_{a})$. The function $\vartheta$
is continuous (Lemma 7.4) and Condition C2 from Section 5 holds because by
Lemma 6.2 it holds for convolutions. Hence (7.14) follows from Proposition
5.2.

##  8 $\mu$-norm and average trace

###  8.1 Definition of ${\bf T}(W)$

By Lemma 3.3 in the finite-dimensional case the $\mu$-norm coincides with the
average trace of $W^{*}W$. The following lemma compares $\mu$-norm with the
average trace for operators of diagonal type.

For any operator $W\in{\cal DT}$ we define average trace of $W^{*}W$ by

${\bf T}(W)=\limsup_{\\#I\to\infty}{\bf T}(I,W),\qquad{\bf
T}(I,W)=\frac{1}{\\#I}\sum_{j\in\mathbb{Z},\,l\in I}|W_{l,j}|^{2}.$ (8.1)

###### Proposition 8.1

Suppose $W\in{\cal DT}$. Then

$\|W\|_{\mu}^{2}\geq{\bf T}(W).$ (8.2)

###### Corollary 8.1

Combining (5.2) and (8.2), we have:

${\bf T}(W)\leq\|W\|_{\mu}^{2}\leq\int_{\mathbb{T}}\vartheta\,d\mu.$

Proof of Proposition 8.1. By (7.11)

$|w_{l}(a)|^{2}=\sum_{j,k\in\mathbb{Z}}W_{l,j}\overline{W}_{l,k}e^{i(k-j)a}.$
(8.3)

Then by (7.13), (7.14)

$\|W\|_{\mu}^{2}=\frac{1}{2\pi}\int_{\mathbb{T}}\limsup_{\\#I\to\infty}\frac{1}{\\#I}\sum_{j,k\in\mathbb{Z},\,l\in
I}W_{l,j}\overline{W}_{l,k}e^{i(k-j)a}\,da$ (8.4)

Given an arbitrary big $M>0$ and arbitrarily small $\sigma>0$ let
$J\subset\mathbb{Z}$ be an interval such that $\\#J>M$ and ${\bf T}(J,W)>{\bf
T}(W)-\sigma$. Then by (8.4)

$\|W\|_{\mu}^{2}\geq\frac{1}{2\pi}\int_{\mathbb{T}}\frac{1}{\\#J}\sum_{j,k\in\mathbb{Z},\,l\in
J}W_{l,j}\overline{W}_{l,k}e^{i(k-j)a}\,da={\bf T}(J,W).$

Hence $\|W\|_{\mu}^{2}\geq{\bf T}(W)-\sigma$. This implies (8.2).

###  8.2 Multiplication by a unitary operator

###### Proposition 8.2

Suppose $W,U\in{\cal DT}$ and $U$ is unitary. Then

${\bf T}(WU)={\bf T}(W)={\bf T}(UW).$ (8.5)

Proof. First, note that $W^{\prime}=WU$ and $W^{\prime\prime}=UW$ are
operators of diagonal type and

$W^{\prime}_{l,j}=\sum_{m\in\mathbb{Z}}W_{l,m}U_{m,j},\quad
W^{\prime\prime}_{l,j}=\sum_{m\in\mathbb{Z}}U_{l,m}W_{m,j}.$

Then the equation $U^{-1}=U^{*}$ implies

$\sum_{j\in\mathbb{Z},\,l\in
I}|W^{\prime}_{l,j}|^{2}=\sum_{m,n,j\in\mathbb{Z},\,l\in
I}W_{l,m}U_{m,j}\overline{W}_{l,n}\overline{U}_{n,j}=\sum_{m\in\mathbb{Z},\,l\in
I}|W_{l,m}|^{2}.$

Hence, the first equation (8.5) follows.

The second equation (8.5) requires a larger effort. Let
$U_{jl}^{-1}=\overline{U}_{lj}$ denote elements of the matrix $U^{-1}$. Then

$\sum_{j\in\mathbb{Z}}|W^{\prime\prime}_{lj}|^{2}=\sum_{m,n,j\in\mathbb{Z}}U_{l,m}W_{m,j}\overline{U}_{l,n}\overline{W}_{n,j}=\sum_{m,n,j\in\mathbb{Z}}U_{l,m}U_{n,l}^{-1}W_{m,j}\overline{W}_{n,j}.$

Let $I\subset\mathbb{Z}$ be an interval. Since

${\bf T}(I,W)=\frac{1}{\\#I}\sum_{l\in
I,\,m,n,j\in\mathbb{Z}}\delta_{n,l}\delta_{l,m}W_{m,jn}\overline{W}_{n,j},$

we have:

$\displaystyle|{\bf T}(I,W^{\prime\prime})-{\bf T}(I,W)|$ $\displaystyle=$
$\displaystyle\frac{1}{\\#I}\bigg{|}\sum_{l\in
I,\,m,n,j\in\mathbb{Z}}(U_{n,l}^{-1}U_{lm}-\delta_{n,l}\delta_{lm})W_{m,j}\overline{W}_{n,j}\bigg{|}$
$\displaystyle\leq$
$\displaystyle\sum_{m,n,j\in\mathbb{Z}}\bigg{|}\frac{1}{\\#I}\sum_{l\in
I}(U_{n,l}^{-1}U_{l,m}-\delta_{n,l}\delta_{l,m})\bigg{|}c_{m-l}c_{n-l}\;=\;\Delta,$
$\displaystyle\Delta=\sum_{m,n\in\mathbb{Z}}\Gamma_{m,n}\widetilde{c}_{m-n},\\!\\!\\!\\!\\!\\!\\!\\!$
$\displaystyle\quad\Gamma_{m,n}=\bigg{|}\frac{1}{\\#I}\sum_{l\in
I}(U_{n,l}^{-1}U_{l,m}-\delta_{n,l}\delta_{l,m})\bigg{|},$

where $\widetilde{c}_{m-n}=\sum_{j\in\mathbb{Z}}c_{m-l}c_{n-l}$,
$\sum_{m\in\mathbb{Z}}\widetilde{c}_{m}={\bf c}^{2}$.

Since $U,U^{-1}\in{\cal DT}$, we have: $|U_{kl}|\leq d_{l-k}$ ,
$\sum_{j\in\mathbb{Z}}d_{j}={\bf d}<\infty$. Given $\sigma>0$ there exists
$M>0$ such that

$\sum_{|j|>M}d_{j}<\sigma.$

We put $\displaystyle
e_{n-l}=\sum_{m\in\mathbb{Z}}d_{m-l}\widetilde{c}_{m-n}$,

$I_{M}^{+}=\\{k:[k-M,k+M]\cap I\neq\emptyset\\},\quad
I_{M}^{-}=\\{k:[k-M,k+M]\subset I\\}.$

Then $\sum_{j\in\mathbb{Z}}e_{j}={\bf d}{\bf c}^{2}$.

We have: $\Delta=\Delta_{1}+\Delta_{2}+\Delta_{3}$:

$\displaystyle\Delta_{1}$ $\displaystyle=$
$\displaystyle\sum_{m\in\mathbb{Z},\,n\not\in
I^{+}_{M}}\Gamma_{m,n}\widetilde{c}_{m-n},$ $\displaystyle\Delta_{2}$
$\displaystyle=$ $\displaystyle\sum_{m\in\mathbb{Z},\,n\in
I^{-}_{M}}\Gamma_{m,n}\widetilde{c}_{m-n},$ $\displaystyle\Delta_{3}$
$\displaystyle=$ $\displaystyle\sum_{m\in\mathbb{Z},\,n\in I^{+}_{M}\setminus
I^{-}_{M}}\Gamma_{m,n}\widetilde{c}_{m-n}.$

First, we estimate $\Delta_{1}$:

$\displaystyle\Delta_{1}$ $\displaystyle\leq$
$\displaystyle\frac{1}{\\#I}\sum_{m\in\mathbb{Z},\,n\not\in
I^{+}_{M}}\sum_{l\in I}d_{n-l}d_{m-l}\widetilde{c}_{m-n}$ $\displaystyle\leq$
$\displaystyle\frac{1}{\\#I}\sum_{n\not\in I^{+}_{M},\,l\in
I}d_{n-l}e_{n-l}\;\leq\;{\bf d}{\bf c}^{2}\sigma.$

Now we turn to $\Delta_{2}$. Since
$\sum_{l\in\mathbb{Z}}U^{-1}_{n,l}U_{l,m}=\sum_{l\in\mathbb{Z}}\delta_{n,l}\delta_{m,l}=\delta_{n,m}$,
we have:

$\displaystyle\Delta_{2}$ $\displaystyle=$
$\displaystyle\sum_{m\in\mathbb{Z},\,n\in
I^{-}_{M}}\bigg{|}\frac{1}{\\#I}\sum_{l\in
I}(U^{-1}_{n,l}U_{l,m}-\delta_{n,l}\delta_{m,l})\bigg{|}\,\widetilde{c}_{m-n},$
$\displaystyle=$ $\displaystyle\frac{1}{\\#I}\sum_{m\in\mathbb{Z},\,n\in
I^{-}_{M}}\bigg{|}\sum_{l\in\mathbb{Z}\setminus
I}U^{-1}_{n,l}U_{l,m}\bigg{|}\,\widetilde{c}_{m-n},$ $\displaystyle\leq$
$\displaystyle\frac{1}{\\#I}\sum_{m\in\mathbb{Z},\,n\in
I^{-}_{M},\,l\in\mathbb{Z}\setminus I}d_{n-l}d_{m-l}\widetilde{c}_{m-n},$
$\displaystyle=$ $\displaystyle\frac{1}{\\#I}\sum_{n\in
I^{-}_{M},\,l\in\mathbb{Z}\setminus
I}d_{n-l}e_{n-l}\;\leq\;\sum_{|m|>M}d_{m}e_{m}\;\leq\;{\bf d}{\bf
c}^{2}\sigma.$

Finally, we estimate $\Delta_{3}$:

$\displaystyle\Delta_{3}$ $\displaystyle\leq$
$\displaystyle\frac{1}{\\#I}\sum_{m\in\mathbb{Z},\,n\in I^{+}_{M}\setminus
I^{-}_{M},\,l\in
I}(d_{n-l}d_{m-l}+\delta_{n,l}\delta_{l,m})\widetilde{c}_{m-n}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\\#I}\sum_{n\in I^{+}_{M}\setminus
I^{-}_{M},\,l\in I}(d_{n-l}e_{n-l}+\delta_{n,l}\widetilde{c}_{l-n})$
$\displaystyle=$ $\displaystyle\frac{1}{\\#I}\sum_{n\in I^{+}_{M}\setminus
I^{-}_{M}}({\bf d}^{2}{\bf c}^{2}+{\bf c}^{2})\;=\;\frac{2(M+1)}{\\#I}({\bf
d}^{2}+1){\bf c}^{2}.$

We see that $\Delta$ can be made arbitrarily small if we choose sufficiently
small $\sigma$ and $M/\\#I$. This implies

$\limsup_{\\#I\to\infty}{\bf T}(I,W)=\limsup_{\\#I\to\infty}{\bf
T}(I,W^{\prime\prime}).$

##  9 Technical statements

###  9.1 Norm of a function on a subset

###### Lemma 9.1

Let $D\subset A\subset{\cal X}$ be two measurable sets and let $f=\pi_{A}f\in
L^{2}({\cal X})$ satisfy $c<\big{|}f|_{A}\big{|}<c^{-1}$. Then

$\|\pi_{D}f\|\leq\frac{\sqrt{\mu(D)}}{c^{2}\sqrt{\mu(A)}}\|f\|.$ (9.1)

Proof. We have: $\|f\|^{2}\geq c^{2}\mu(A)$ and $\|\pi_{D}f\|^{2}\leq
c^{-2}\mu(D)$. Therefore

$\|\pi_{D}f\|^{2}\leq c^{-2}\mu(D)\frac{\|f\|^{2}}{c^{2}\mu(A)}.$

This implies (9.1).

###  9.2 The function $\rho(\lambda)$

In this section we present two lemmas on the function $\rho=\rho(\lambda)$ for
a distribution $\lambda$, determined by (6.1).

###### Lemma 9.2

For any $\beta>0$ and any $C>1$ there exists an interval $J\subset\mathbb{Z}$
such that

$\rho_{J}(\lambda)>\rho-\beta\quad\mbox{and}\quad C\leq\\#J\leq 2C+1.$

Proof. There exists $J_{0}$ such that

$\rho_{J_{0}}(\lambda)>\rho-\beta\quad\mbox{and}\quad\\#J_{0}>2C+1.$

We break $J_{0}$ into two nonintersecting intervals $J^{\prime}_{0}$ and
$J^{\prime\prime}_{0}$ such that
$|\\#J^{\prime}_{0}-\\#J^{\prime\prime}_{0}|\leq 1$. One of these intervals
(we redenote it by $J_{1}$) satisfies $\rho_{J_{1}}(\lambda)>\rho-\beta$. If
$\\#J_{1}>2C+1$, we repeat the argument.

###### Lemma 9.3

Let $M^{\prime}=M^{\prime}(\sigma)$ satisfy (6.14). Let $\widetilde{J}$ be an
integer interval such that

$\rho_{\widetilde{J}}(\lambda)\geq\rho-\widetilde{\sigma},\quad\\#\widetilde{J}>2M^{\prime}.$
(9.2)

Let $J\subset\widetilde{J}$ be the interval which is obtained from
$\widetilde{J}$ by removing two intervals: extreme left $J_{l}$ and extreme
right $J_{r}$, $\\#J_{l}=\\#J_{r}=M^{\prime}$.

Then for any integer interval $J_{0}\subset J$

$\rho_{J_{0}}(\lambda)=\rho-\sigma_{0},\quad\sigma_{0}\leq(\widetilde{\sigma}+\sigma)\frac{\\#\widetilde{J}}{\\#J_{0}}-\sigma.$
(9.3)

Proof. We have: $\widetilde{J}=\widetilde{J}_{l}\cup
J_{0}\cup\widetilde{J}_{r}$, where $\widetilde{J}_{l}\supset J_{l}$ and
$\widetilde{J}_{r}\supset J_{r}$ are integer intervals and
$\\#\widetilde{J}_{l}\geq M^{\prime}$, $\\#\widetilde{J}_{r}\geq M^{\prime}$.
Hence, by (9.2)

$\\#\widetilde{J}(\rho-\widetilde{\sigma})\leq\sum_{j\in\widetilde{J}}|\lambda_{j}|^{2}\leq(\rho-\sigma_{0})\\#J_{0}+(\rho+\sigma)(\\#\widetilde{J}_{l}+\\#\widetilde{J}_{r}).$

Therefore
$\\#J_{0}\sigma_{0}\leq\\#\widetilde{J}\widetilde{\sigma}+(\\#\widetilde{J}-\\#J_{0})\sigma$.
This implies (9.3).

###  9.3 The function $\rho(L_{a})$

In this section we study the distribution $L_{a}$, determined by (7.9).

###### Lemma 9.4

Suppose $W\in{\cal DT}$. Then the functions $a\mapsto\rho_{I}(L_{a})$ are
equicontinuous i.e., for any $\sigma>0$ there exists $\gamma(\sigma)$ such
that

$|\rho_{I}(L_{a})-\rho_{I}(L_{b})|<\sigma\quad\mbox{whenever
$|a-b|<\gamma(\sigma)$}.$

It is important that $\gamma$ does not depend on $I$.

Proof of Lemma 9.4. By (7.11) and (7.13) we have the estimate

$\displaystyle|\rho_{I}(L_{a})-\rho_{I}(L_{b})|$ $\displaystyle=$
$\displaystyle\frac{1}{\\#I}\Big{|}\sum_{l\in
I}\sum_{j,n\in\mathbb{Z}}W_{l,j}\overline{W}_{l,n}(e^{i(n-j)a}-e^{i(n-j)b})\Big{|}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\\#I}\sum_{l\in
I,\,j,n\in\mathbb{Z}}c_{l-j}c_{l-n}|(e^{i(n-j)a}-e^{i(n-j)b}|$
$\displaystyle=$
$\displaystyle\sum_{j,n\in\mathbb{Z}}c_{n}c_{j}|(e^{i(n-j)a}-e^{i(n-j)b}|\;\leq\;\Sigma_{1}+\Sigma_{2},$

where

$\Sigma_{1}=\sum_{|j-n|\leq
K}c_{n}c_{j}|(e^{i(n-j)a}-e^{i(n-j)b}|,\quad\Sigma_{2}=\sum_{|j-n|>K}c_{n}c_{j}|(e^{i(n-j)a}-e^{i(n-j)b}|.$

We have:

$\displaystyle\Sigma_{1}$ $\displaystyle\leq$ $\displaystyle
K|a-b|\sum_{|j-n|\leq K}c_{n}c_{j}\leq{\bf c}^{2}K|a-b|,$
$\displaystyle\Sigma_{2}$ $\displaystyle\leq$ $\displaystyle
2\sum_{|j-n|>K}c_{n}c_{j}\leq
2\sum_{j\in\mathbb{Z},|n|>K/2}c_{n}c_{j}+2\sum_{n\in\mathbb{Z},|j|>K/2}c_{n}c_{j}\leq
4{\bf c}\sigma_{0}(K),$ $\displaystyle\sigma_{0}(K)$ $\displaystyle=$
$\displaystyle\sum_{|n|>K/2}c_{n}.$

Hence, for $|a-b|<\gamma$

$|\rho_{I}(L_{a})-\rho_{I}(L_{b})|\leq{\bf c}^{2}K\gamma+4{\bf
c}\sigma_{0}(K).$

To make the right-hand side of this inequality arbitrarily small, we first,
choose sufficiently large $K$ and then take $\gamma=\gamma(K)>0$ sufficiently
small.

Now Lemma 7.4 turns out to be a simple corollary of Lemma 9.4. For
completeness we present the argument.

Given $\sigma>0$ take an interval $I\subset\mathbb{Z}$ with arbitrarily large
$\\#I$ and such that $\rho_{I}(L_{a})>\rho(L_{a})-\sigma$. Then for any
$b\in\mathbb{T}$ such that $|b-a|<\gamma(\sigma)$, where $\gamma(\sigma)$ is
defined in Lemma 9.4, we have:

$\rho_{I}(L_{b})\geq\rho_{I}(L_{a})-\sigma>\rho(L_{a})-2\sigma.$

Analogously if $N(\sigma)$ is sufficiently large then for any
$I\subset\mathbb{Z}$ with $\\#I>N(\sigma)$ we have:
$\rho_{I}(L_{a})<\rho(L_{a})+\sigma$. Then for any $b\in\mathbb{T}$ satisfying
$|b-a|<\gamma(\sigma)$

$\rho_{I}(L_{b})\leq\rho_{I}(L_{a})+\sigma<\rho(L_{a})+2\sigma.$

Hence $|\rho(L_{a})-\rho(L_{b})|<2\sigma$ whenever $|a-b|<\gamma(\sigma)$.

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