# Operators between Hilbert Spaces Viewed as Only Linear Topological – Towards
a Classification

Eliahu Levy  Department of Mathematics, Technion – Israel Institute of
Technology, Haifa 32000, Israel<EMAIL_ADDRESS>
###### Abstract.

In topological equivalence, a bounded linear operator between Banach spaces –
we focus on the case of Hilbert spaces – is viewed as only acting linearly and
continuously between them qua different spaces with the structure of linear
topological space. For instance, invertible operators in Banach spaces (that
is, isomorphisms among them) will make up one equivalence class for each class
of isomorphic spaces. On the other hand, compact and non-compact operators, or
operators with or without a kernel, clearly will not. We make some crucial
steps towards describing invariants that will characterize these topological
equivalence classes

###### Key words and phrases:

Topological equivalence of operators, Hilbert space, Spectral decomposition,
Cardinal numbers-valued measure, Sequences of eigenvalues, Uniformly log-
bounded

###### Contents

  1. 1 Introduction, Topological Equivalence of Operators
  2. 2 $I$-Spaces for Operators, Reduction to Positive Semi-Definite
  3. 3 Isomorphism Property of Maximal $[0,\mu]$-Spaces and their Quotients
  4. 4 The Cardinal Number-Valued ‘Dimension Spectral Measure’ as an Invariant
  5. 5 Focusing on Invariants to Characterize Topological Equivalence
  6. 6 Towards a Classification, ‘Eigenvalue Sequences with Infinity Intermissions’
  7. A Dimension in Hilbert Spaces
  8. B In Hilbert Space, the Subspace of Vectors Where the Norm is Attained

##  1\. Introduction, Topological Equivalence of Operators

In topological equivalence of operators between Banach spaces – we focus on
the case of Hilbert spaces – we wish to view an operator as only acting
linearly and continuously between the spaces, qua different spaces with the
structure of linear topological space. Thus

######  Definition 1.

Two operators $T_{1}:X_{1}\to Y_{1}$ and $T_{2}:X_{2}\to Y_{2}$ between Banach
spaces will be called Topologically equivalent if there exist isomorphisms
onto $U:X_{1}\to X_{2}$ and $V:Y_{1}\to Y_{2}$ so that $T_{2}=V^{-1}T_{1}U$.

Note that even if $X_{1}=X_{2}=Y_{1}=Y_{2}$, $U$ and $V$ are allowed to be
different isomorphisms.

Of course, for $T_{1}$ and $T_{2}$ to be topologically equivalent, $X_{1}$ and
$X_{2}$, and likewise $Y_{1}$ and $Y_{2}$, must be isomorphic.

We make some crucial steps towards describing invariants that will
characterize the topological equivalence classes.

First, any operator $T:\mathcal{H}\to\mathcal{H}^{\prime}$ between Hilbert is
topologically (even ‘isometrically’) equivalent to $A:=\sqrt{T^{\star}T}$ – a
positive semi-definite operator in $\mathcal{H}$. Indeed
$\|Tx\|=\|Ax\|,\,x\in\mathcal{H}$ ,

One focuses on $I$-subspaces, where $I$ is the interval in the positive half-
line, consisting of the numbers $<$, $\leq$, $>$, or $\geq$ from some fixed
$\mu$. An $I$-subspace $Y$ for $T$ (here the same for $A:=\sqrt{T^{\star}T}$
mentioned above) is then defined by requiring that $\|Tx\|/\|x\|\in I$ for all
$0\neq x\in Y$.

And we prove that for maximal $[0,\mu]$-subspaces, and for $\mu>\mu^{\prime}$
the quotient of a maximal $[0,\mu]$-subspaces over some maximal
$[0,\mu^{\prime}]$-subspace contained in it, are all isomorphic, a special
instant featuring in the spectral decomposition of the a positive semi-
definite $A$.

That allows us to relate features of that spectral decomposition to notions
more unaffected when a Hilbert norm is replaced by an equivalent Hilbert norm,
thus to formulate some necessary condition for topological equivalence,
Theorem 8.

To obtain sufficient conditions, one notes that an operator is characterized
up to topological equivalence (indeed isometric) by the dimensions of the
parts in the latter spectral decomposition – these forming a cardinal number-
valued ($\sigma$-additive) ‘measure’ on the Borel subsets of
$\mathbb{R}^{+}=[0,\infty[$.

And one sees what manipulations in the latter preserve topological
equivalence, toward attaching to an operator what we call an eigenvalue
sequence with infinity intermissions with some criterions for topological
equivalence, Theorem 12.

##  2\. $I$-Spaces for Operators, Reduction to Positive Semi-Definite

Let $T:X_{1}\to X_{2}$ be a continuous linear operator between Banach spaces.

######  Definition 2.

Let $I$ be an interval of the form $[0,\mu]$ or $[0,\mu[$, $\mu\geq 0$, in
$\mathbb{R}^{+}=[0,+\infty[$, or of the ‘complement’ form $]\mu,\infty[$ or
$[\mu,\infty[$. We say that a linear subspace $Y\subset X_{1}$ (here, not
necessarily closed) is an $I$-subspace (with respect to $T$) if for every
$0\neq x\in Y$ $\|Tx\|/\|x\|\in I$.

In particular, one may consider maximal (by inclusion) $I$-subspaces in $X$
with respect to $T$ .

Clearly, if $I$ is closed, (i.e. is of the form $[0,\mu]$ or $[\mu,\infty[$)
then the closure of an $I$-subspace is an $I$-subspace. Therefore if $I$ is
closed a maximal $I$-subspace is closed.

By a straightforward application of Zorn’s lemma, every $I$-subspace is
contained in a maximal $I$-subspace.

Let now $\mathcal{H}$ and $\mathcal{H}^{\prime}$ be Hilbert spaces (assumed
complex, or if real complexify them, since we shall need the spectral
decomposition) and let $T:\mathcal{H}\to\mathcal{H}^{\prime}$ be a bounded
linear operator.

We have for any $x\in\mathcal{H}$

$\|Tx\|^{2}=\langle Tx,Tx\rangle=\langle T^{\star}Tx,x\rangle,$

where $T^{\star}T$ is a positive semi-definite operator from $\mathcal{H}$ to
itself.

$T^{\star}T$ has a unique positive semi-definite square root
$A=\sqrt{T^{\star}T}$, and we have, for any $x\in\mathcal{H}$

$\|Tx\|^{2}=\langle T^{\star}Tx,x\rangle=\langle A^{2}x,x\rangle=\langle
Ax,Ax\rangle=\|Ax\|^{2}.$

(As a consequence $\mathrm{Ker}T=\mathrm{Ker}A$).

We shall thus have an isometry $U$ between the image of $A$ in $\mathcal{H}$
and the image of $T$ in $\mathcal{H}^{\prime}$, by mapping $Ax\mapsto Tx$.
This isometry can, of course, be uniquely extended to an isometry $U$ from the
closure $\mathrm{cl}(A\,\mathcal{H})$ in $\mathcal{H}$ to the closure
$\mathrm{cl}(T\,\mathcal{H})$ in $\mathcal{H}^{\prime}$. Of course $T=UA$.

Thus, $T$, as operator from $\mathcal{H}\to\mathrm{cl}(T\,\mathcal{H})$, is
topologically equivalent to $A=\sqrt{T^{\star}T}$ from
$\mathcal{H}\to\mathrm{cl}(A\mathcal{H})$.

Hence, when the linear topological spaces can be normed as Hilbert, we, up to
topological equivalence, may restrict ourself to positive semi-definite
operators in a Hilbert space $\mathcal{H}$ as operators from $\mathcal{H}$ to
the closure of their image.

A particular case is when $T$, consequently $A^{2}$ and $A$, have no kernel.
Since $A$ is Hermitian, for it kernel $=$ orthogonal complement of the image,
so in that case it has a dense image in $\mathcal{H}$. Thus $U$ is an isometry
from $\mathcal{H}$ onto $\mathcal{H}^{\prime}$. Consequently, in this case
($\mathcal{H}\simeq\mathcal{H}^{\prime}$) and $T$ is topologically equivalent
to $A$.

Returning to the general case, $A$, a (semi-definite) Hermitian operator, has
a spectral decomposition [5], given by a ‘decomposition of unity’, i.e. a
spectral measure on $\mathcal{H}$, $d\mathcal{E}(\lambda)$, defined on
$\mathbb{R}^{+}=[0,+\infty[$,111That is, $\mathcal{E}(\tau)$ is an (operator
on $\mathcal{H}$)-valued measure on the Borel subsets $\tau$ of
$\mathbb{R}^{+}$, strongly $\sigma$-additive, whose value on sets is always an
orthogonal projection, and such that the measure of the empty set is the
operator $0$, the measure of the whole $\mathbb{R}^{+}$ is the operator
$\mathbf{1}$, and the measure of the intersection of two Borel sets is the
product of the measures of the sets, which thus commute. In particular, the
measures of disjoint sets are orthogonal projections on orthogonal subspaces.
such that $A=\int_{[0,+\infty[}\lambda\,d\mathcal{E}(\lambda)$, i.e.,
$\forall\,x,x^{\prime}\in\mathcal{H}$

$\langle Ax,x^{\prime}\rangle=\int_{[0,+\infty[}\lambda\langle
d\mathcal{E}(\lambda)x,x^{\prime}\rangle,$

which implies, by the rules of manipulating spectral measures, that
$A^{2}=\int_{[0,+\infty[}\lambda^{2}\,d\mathcal{E}(\lambda)$, i.e.,

$\langle T^{\star}Tx,x^{\prime}\rangle=\langle
A^{2}x,x^{\prime}\rangle=\int_{[0,+\infty[}\lambda^{2}\langle
d\mathcal{E}(\lambda)x,x^{\prime}\rangle.$

##  3\. Isomorphism Property of Maximal $[0,\mu]$-Spaces and their Quotients

######  Proposition 3.

Referring to the above spectral decomposition of $A=\sqrt{T^{\star}T}$, let
$I$ be of the form $[0,\mu]$, $[0,\mu[$, $]\mu,\infty[$ or $[\mu,\infty[$,
$\mu\geq 0$.

Then $\mathcal{E}(I)\mathcal{H}$ is a maximal $I$-subspace with respect to
$T$, equivalently with respect to $A$.

######  Proof .

Let $x\in\mathcal{E}(I)\mathcal{H}$. Then $x$ is orthogonal to
$\mathcal{E}(I^{c})\mathcal{H}$ and we have (here $I^{2}$ denotes the set of
squares of members of $I$),

$\displaystyle\|Tx\|^{2}=\langle
A^{2}x,x\rangle=\int_{[0,+\infty[}\lambda^{2}\langle
d\mathcal{E}(\lambda)x,x\rangle=\int_{I}\lambda^{2}\langle
d\mathcal{E}(\lambda)x,x\rangle\in$ $\displaystyle\in
I^{2}\cdot\int_{I}\langle
d\mathcal{E}(\lambda)x,x\rangle=I^{2}\cdot\langle\mathcal{E}(I)x,x\rangle=I^{2}\cdot\|x\|^{2}.$

Consequently $\|Tx\|\in I\cdot\|x\|$, making $\mathcal{E}(I)\mathcal{H}$ an
$I$-subspace.

Now denote by $I^{c}$ the complement of $I$ in $\mathbb{R}^{+}=[0,\infty[$.
Since $I^{c}$ is also an interval of our form, $\mathcal{E}(I^{c})\mathcal{H}$
is an $I^{c}$-subspace with respect to $A=\sqrt{T^{\star}T}$, thus to $T$.

And we have the orthogonal decomposition
$\mathcal{H}=\mathcal{E}(I)\mathcal{H}\oplus\mathcal{E}(I^{c})\mathcal{H}$.

Now suppose $\mathcal{E}(I)\mathcal{H}$ were not maximal, i.e. contained in a
larger $I$-subspace $Y$. Then $Y$ must intersect
$\mathcal{E}(I^{c})\mathcal{H}$ nontrivially, i.e. there would be a nonzero
$a\in Y\cap\mathcal{E}(I^{c})\mathcal{H}$. Then $\|Ta\|/\|a\|$ must be both in
$I$ and $I^{c}$ – impossible. ∎

######  Remark 4.

For our record here, note trivially that for $\mu=0$, $I=[0,0]=\\{0\\}$, and
we have: for $Y$ to be an $I$-subspace means that $T$ vanishes on $Y$; for $Y$
to be a maximal $I$-subspace means that $Y=\mathrm{Ker}\,T$; and the latter is
$\mathcal{E}(\\{0\\})\mathcal{H}$ in the above spectral decomposition.

######  Proposition 5.

Suppose $I$ is an interval of the form $[0,\mu]$, $\mu\geq 0$.

($I$ being closed, clearly the closure $\mathrm{cl}\,Y$ of an $I$-subspace $Y$
(with respect to $T$, equivalently to $A=\sqrt{T^{\star}T}$) is also an
$I$-subspace. Therefore a maximal $I$-subspece must be closed.)

(i) Let $Y$ be an $I$-subspace.

Then the restriction to $Y$ of the orthogonal projection on
$\mathcal{E}(I)\mathcal{H}$ is injective on $Y$.

(ii) Let $Y$ be a maximal $I$-subspace (hence closed).

Then the restriction to $Y$ of the orthogonal projection on $\mathcal{E}(I)$
is also onto, hence is an isomorphism between $Y$ and
$\mathcal{E}(I)\mathcal{H}$.

######  Proof .

(i) Recall that we denote by $I^{c}$ the complement of $I$ in
$\mathbb{R}^{+}=[0,\infty[$, here $I^{c}=]\mu,\infty[$.

We have the orthogonal decomposition
$\mathcal{H}=\mathcal{E}(I)\mathcal{H}\oplus\mathcal{E}(I^{c})\mathcal{H}$,
where, by Proposition 3, $\mathcal{E}(I)\mathcal{H}$ is an $I$-subspace and
$\mathcal{E}(I^{c})\mathcal{H}$ is an $I^{c}$-subspace with respect to $T$,
equivalently with respect to $A=\sqrt{T^{\star}T}$. Therefore, if $Y$ and
$\mathcal{E}(I^{c})\mathcal{H}$ had a nontrivial intersection, i.e. there was
an $a\neq 0$ contained in both, then $\|Ta\|/\|a\|$ would have been both in
$I$ and $I^{c}$ – impossible.

Consequently $Y\cap\mathcal{E}(I^{c})=0$, making the restriction to $Y$ of the
orthogonal projection on $\mathcal{E}(I)$ in the above orthogonal
decomposition injective.

(ii) Suppose now that $Y$ is a maximal $I$-subspace, and, toward a
contradiction, that the projection is not onto, i.e. there is an
$a\in\mathcal{E}(I)$ not in the image of the above projection.

Let $y$ be any member of $Y$.

Denote by $y_{I}$ and $y_{I^{c}}$ the projections of $y$ on
$\mathcal{E}(I)\mathcal{H}$ and $\mathcal{E}(I^{c})\mathcal{H}$, resp.

Now, $a+y_{I}$ is in $\mathcal{E}(I)$ and $y_{I^{c}}$ is in
$\mathcal{E}(I^{c})$, which are orthogonal and $A$-invariant. Therefore the
above projection sends $a+y$ to $a+y_{I}$, hence different from all
projections of members of $Y$, implying $a\notin Y$.

Thus, the linear space $Y^{\prime}$ generated by $Y$ and $a$ is strictly
bigger than $Y$. Therefore, as $y$ is a maximal $I$-subspace, $Y^{\prime}$
must contain some $a+y$ with $\|A(a+y)\|\notin I\|a+y\|$ (while $\|A(a)\|\in
I\|a\|$ since $a\in\mathcal{E}(I)\mathcal{H}$, and $\|A(y)\|\in I\|y\|$ since
$y\in Y$).

So (recall that $I=[0,\mu]$) we have

$\|A(a+y)\|^{2}>\mu^{2}\|a+y\|^{2},$

i.e.,

(1)
$\mu^{2}\|a\|^{2}-\|Aa\|^{2}+\mu^{2}\|y\|^{2}-\|Ay\|^{2}+2\mathrm{Re}\Big{(}\mu^{2}\langle
a,y\rangle-\langle Aa,Ay\rangle\Big{)}<0.$

On the other hand, consider the segment $\\{ta+(1-t)y\,|\,0\leq t\leq 1\\}$.
The quadratic expression in $t$

$\|A(ta+(1-t)y)\|^{2}-\mu^{2}\|ta+(1-t)y\|^{2}$

is (recall that $I=[0,\mu]$), $\leq 0$ for $t=0$ (since
$a\in\mathcal{E}(I)\mathcal{H}$) and for $t=1$ (since $y\in Y$), but positive
for $t=1/2$. Therefore its derivative at $t=0$ must be positive.

The derivative at $t=0$ being positive means

$\mathrm{Re}\langle Ay,A(a-y)\rangle-\mu^{2}\mathrm{Re}\langle
y,a-y\rangle>0,$

or

$\mathrm{Re}\Big{(}\mu^{2}\langle y,a\rangle-\langle
Aa,Ay\rangle\Big{)}<\mu^{2}\|y\|^{2}-\|Ay\|^{2},$

And one may exchange $a$ and $y$ here to find

$\mathrm{Re}\Big{(}\mu^{2}\langle y,a\rangle-\langle
Aa,Ay\rangle\Big{)}<\mu^{2}\|a\|^{2}-\|Aa\|^{2}.$

Replacing $y$ by $-y$, one gets similar inequalities for absolute values

$\displaystyle\Big{|}\mathrm{Re}\Big{(}\mu^{2}\langle y,a\rangle-\langle
Aa,Ay\rangle\Big{)}\Big{|}<\mu^{2}\|y\|^{2}-\|Ay\|^{2},$
$\displaystyle\Big{|}\mathrm{Re}\Big{(}\mu^{2}\langle y,a\rangle-\langle
Aa,Ay\rangle\Big{)}\Big{|}<\mu^{2}\|a\|^{2}-\|Aa\|^{2}.$

Which makes the LHS expression in (1) positive, a contradiction. ∎

Let now $\mu>\mu^{\prime}\geq 0$, $I=[0,\mu]$,
$I^{\prime}=[0,\mu^{\prime}]\subset I$. Of course, every $I^{\prime}$-subspace
is ipso facto an $I$-subspace, so is contained in some maximal $I$-subspace.

######  Proposition 6.

Let $\mu>\mu^{\prime}\geq 0$, $I^{\prime}=[0,\mu^{\prime}]$, $I=[0,\mu]$.

Consider a maximal $I^{\prime}$-subspace $Y^{\prime}$ contained in a maximal
$I$-subspace $Y$. Then the quotient space $Y/Y^{\prime}$ is isomorphic to
$\mathcal{E}(I\setminus I^{\prime})\mathcal{H}$.

######  Proof .

$\mathcal{E}(I)\mathcal{H}$ has the orthogonal decomposition
$\mathcal{E}(I^{\prime})\mathcal{H}\oplus\mathcal{E}(I\setminus
I^{\prime})\mathcal{H}$. The orthogonal projection here on
$\mathcal{E}(I^{\prime})\mathcal{H}$ will be the restriction to
$\mathcal{E}(I^{\prime})\mathcal{H}$ of the orthogonal projection from the
whole space $\mathcal{H}$ on $\mathcal{E}(I^{\prime})\mathcal{H}$.

Of course, $\mathcal{E}$ being a spectral decomposition, we have

$\mathcal{E}(I^{\prime})\mathcal{E}(I)=\mathcal{E}(I)\mathcal{E}(I^{\prime})=\mathcal{E}(I^{\prime}).$

We know that the restriction to $Y$ of the orthogonal projection
$\mathcal{E}(I)$ from the whole space $\mathcal{H}$ on
$\mathcal{E}(I)\mathcal{H}$ is an isomorphism with its image
$\mathcal{E}(I)\mathcal{H}$. It maps the subspace $Y^{\prime}$ to the subspace
$\mathcal{E}(I)Y^{\prime}\subset\mathcal{E}(I)\mathcal{H}$ there which the
‘further’ orthogonal projection $\mathcal{E}(I^{\prime})$ on
$\mathcal{E}(I^{\prime})\mathcal{H}$ will map isomorphically onto the latter.

Hence, if we refer to $\mathcal{E}(I)\mathcal{H}$ as the product of
$\mathcal{E}(I^{\prime})\mathcal{H}$ and $\mathcal{E}(I\setminus
I^{\prime})\mathcal{H}$, then $\mathcal{E}(I)Y^{\prime}$ is the ‘graph’ of
some bounded linear operator
$B:\mathcal{E}(I^{\prime})\mathcal{H}\to\mathcal{E}(I\setminus
I^{\prime})\mathcal{H}$.

This makes $Y/Y^{\prime}$ isomorphic to the quotient space of
$\mathcal{E}(I)\mathcal{H}$, identified with the product of
$\mathcal{E}(I^{\prime})\mathcal{H}$ and $\mathcal{E}(I\setminus
I^{\prime})\mathcal{H}$, over that ‘graph’ $\\{y,By\\}$, and the latter is
isomorphic to $\mathcal{E}(I\setminus I^{\prime})\mathcal{H}$ via the map
$(y,z)\mapsto z-By$. ∎

######  Corollary 7.

All maximal $[0,\mu]$-subspaces (with respect to $T$, equivalently to the
positive semi-definite $A=\sqrt{T^{\star}T}$) are isomorphic, in particular
are isomorphic to $\mathcal{E}([0,\mu])\mathcal{H}$ (hence have the same
(Hilbert) dimension $\dim\mathcal{E}(I)$). And if $\mu\geq\mu^{\prime}$ then
all quotients of a maximal $[0,\mu]$-subspace over an included maximal
$[0,\mu^{\prime}]$-subspace are isomorphic to
$\mathcal{E}(]\mu^{\prime},\mu])\mathcal{H}$.

##  4\. The Cardinal Number-Valued ‘Dimension Spectral Measure’ as an
Invariant

As usual, by the (Hilbert) dimension $\mathrm{dim}\,\mathcal{H}$ of a Hilbert
space $\mathcal{H}$ we mean the cardinality of an orthonormal basis to
$\mathcal{H}$ – see §A (as is well known and easily proved, all such bases
have the same cardinality).

So, returning to our operator $T$ in $\mathcal{H}$ and the spectral
decomposition $\mathcal{E}$ of $A=\sqrt{T^{\star}T}$, it defines a cardinal-
number-valued ($\sigma$-additive) measure on Borel subsets
$S\subset\mathbb{R}^{+}=[0,\infty[$, namely
$\mathrm{dim}\,\mathcal{E}(S)\mathcal{H}$, which we call the dimension
spectral measure of $A$ and $T$.

Clearly the dimension spectral measure determines a positive semi-definite
operator $A$ up to replacing $A$ by $UAU^{-1}$, $U$ unitary.

Thus Corollary 7 implies that for any $\mu\geq 0$, all the dimensions of
maximal $[0,\mu]$-subspaces are given by that dimension (cardinal number-
valued) spectral measure $\mathrm{dim}\,\mathcal{E}(S)\mathcal{H}$ of
$S=[0,\mu]$; and for any $\mu>\mu^{\prime}\geq 0$ all the dimensions of
quotients of a maximal $[0,\mu]$-subspace over an included maximal
$[0,\mu^{\prime}]$-subspace are given by the dimension spectral measure
$\mathrm{dim}\,\mathcal{E}(S)\mathcal{H}$ of $S=]\mu^{\prime},\mu]$.

But suppose we replace some bounded operator $T$ between Hilbert spaces by a
topologically equivalent $T^{\prime}$, i.e. $T^{\prime}=V^{-1}TU$ where $U$
and $V$ are isomorphisms onto between Hilbert spaces. We may think about that
as $T$ a continuous operator between locally convex linear topological spaces
with different equivalent Hilbert norms defining the topology taken in them.

Then there is some constant $K_{1}>0$ so that a maximal $[0,\mu]$-subspace $Y$
of $T$ with respect to one of the norms is a $[0,K_{1}\mu]$-subspace, hence
included in a maximal $[0,K_{1}\mu]$-subspace, for some other. And since by
Corollary 7 the dimensions of quotients of such characterize the ‘dimension
spectral measure’ of the operators, we conclude that

######  Theorem 8.

For operators between Hilbert spaces, to be topologically equivalent their
dimension (cardinal number-valued) spectral measures must be equivalent in the
following sense:

######  Definition 9.

Two cardinal-number-valued ($\sigma$-additive) measures are called equivalent
if there exists a constant $K\geq 1$ such that:

The weight given to an interval $[0,\mu]$ (resp. $]\mu^{\prime},\mu]$) by one
of the measures is always $\leq$ than the weight given to $[0,K\mu]$ (resp.
$]\mu^{\prime}/K,K\mu]$) by the other.

And, of course, since linear operators, to be continuous must be bounded, the
spectral measure, hence the dimension (cardinal number-valued) spectral
measure of an operator (between Hilbert spaces) is supported on some bounded
interval $[0,M]$.

##  5\. Focusing on Invariants to Characterize Topological Equivalence

  1. (i)

Let $\mathcal{M}$ be some cardinal-number-valued ($\sigma$-additive) measure.

  2. (ii)

If a compact $\mathcal{K}\subset[0,\infty[$ has $\mathcal{M}(\mathcal{K})$
finite, while all its members $x\in\mathcal{K}$ are with
$\mathcal{M}(\\{x\\})=0$ then $\mathcal{M}(\mathcal{K})=0$; By
$\sigma$-additivity ‘compact’ can be replaced here by ‘$\sigma$-compact’, i.e.
a countable union of compacta.

######  Proof .

We prove that every $x\in\mathcal{K}$ has an open neighborhood
$]x-\varepsilon,x+\varepsilon[$ with
$\mathrm{M}(\mathcal{K}\cap]x-\varepsilon,x+\varepsilon[)=0$.

Indeed, $\mathrm{M}(\mathcal{K}\cap]x-\varepsilon,x+\varepsilon[)$ is a
decreasing function of $\varepsilon$, $\varepsilon>0$ with finite values,
because $\mathrm{M}(\mathcal{K}\subset[0,\infty[)$ is finite. So it has a
minimum value attained for some $\varepsilon\leq\varepsilon_{0}$. But that
implies that the weight of differences
$\mathrm{M}(]x-\varepsilon_{0}/(n-1),x-\varepsilon_{0}/n]\cup[x+\varepsilon_{0}/n,x+\varepsilon_{0}/(n-1)[=0$,
and by $\sigma$-additivity
$\mathrm{M}(]x-\varepsilon,x[\cup]x,x+\varepsilon[=0$; Since by assumption
$\mathrm{M}(\\{x\\})=0$, we have $\mathrm{M}(]x-\varepsilon,x+\varepsilon[=0$.

Now, these neighborhoods for all $x\in\mathcal{K}$ are an open cover of
$\mathcal{K}$; it has a finite subcover, whose union will have weight $0$;
therefore $\mathcal{K}$ has weight $0$. ∎

  3. (iii)

If a $\sigma$-compact $\mathcal{K}\subset[0,\infty[$ has
$\mathrm{M}(\mathcal{K})$ finite, then there is a finite set
$F\subset\mathcal{K}$ such that $\mathrm{M}(\mathcal{K}\setminus F)=0$.

######  Proof .

Let $F$ be the set of all $x\in\mathcal{K}$ with $\mathrm{M}(\\{x\\})>0$, the
latter thus a positive integer. Clearly $F$ has no more than
$\mathrm{M}(F)\leq\mathrm{M}(\mathcal{K})$ members, so it is finite. And then
by the previous item $\mathcal{K}\setminus F$ must have weight $0$, being a
$\sigma$-compact with every singleton of weight $0$. ∎

  4. (iv)

If in a compact $\mathcal{K}\subset[0,\infty[$ every point has a neighborhood
$U$ with $\mathrm{M}(\mathcal{K}\cup U)$ finite then $\mathcal{K}$ has finite
weight.

######  Proof .

This is standard compactness argument: the neighborhoods with finite weight
form an open cover of $\mathcal{K}$; it has a finite subcover; therefore the
union of the latter has finite weight; hence so has $\mathcal{K}$. ∎

  5. (v)

Denote by $\mathbf{FIN}_{\mathcal{M}}$ the set of all $x\in[0,\infty[)$ with
some neighborhood $]x-\varepsilon,x+\varepsilon[$ of finite weight, and by
$\mathbf{INF}_{\mathcal{M}}$ its complement
$\mathbf{INF}_{\mathcal{M}}:=[0,\infty[\setminus\mathbf{FIN}_{\mathcal{M}}$.

This makes $\mathbf{FIN}_{\mathcal{M}}$ the union of all open intervals of
finite weight. Therefore it is an open subset of $[0,\infty[$, consequently a
finite or countable union of open intervals there – its connected components,
and $\mathbf{INF}_{\mathcal{M}}$ is closed.

By the above, any compact interval $[a,b]\subset\mathrm{FIN}_{\mathcal{M}}$
must have finite weight. Thus it has a finite subset where all the mass is
concentrated. In any connected component (thus an open interval) $C$ of
$\mathbf{FIN}_{\mathcal{M}}$, the mass will thus be concentrated in some
discrete finite or infinite subset $\Sigma$, in which every point $x$ not the
last (resp. first) will have an immediate successor (resp. predecessor) – the
biggest element of $\Sigma$ strictly less than $x$ (resp. the smallest element
of $\Sigma$ strictly greater than $x$).

Therefore $\Sigma$ (of some component – open interval – of
$\mathbf{FIN}_{\mathcal{M}}$) can be written as an increasing sequence, finite
or infinite forward or backward or both, where if infinite it will converge to
an extremity of $C$ (necessarily a member of $\mathbf{INF}_{\mathcal{M}}$).

  6. (vi)

Returning to $\mathcal{D}$, the dimension spectral measure of an operator, the
above $\Sigma$ will consist of eigenvalues of $A$, each with a finite-
dimension space of eigenvectors, these spaces for different eigenvalues
orthogonal. And in such space consisting of eigenvector of some eigenvalue
$\lambda$, say of dimension $n$, take an orthonormal basis of $n$
eigenvectors. Then $\Sigma$ (counting such $\lambda$ $n$ times) will turn into
a non-decreasing sequence of eigenvalues with a corresponding orthonormal set
of eigenvectors.

  7. (vii)

And whenever we have a sequence $(\lambda_{k})$ of eigenvalues of $A$ and
corresponding orthonormal eigenvectors $(v_{k})$, $Av_{k}=\lambda_{k}v_{k}$,
take some sequence of scalars $\gamma_{k}$ ‘uniformly log-bounded’ (put
otherwise, ‘bounded and bounded away from $0$’), i.e. there exists a $K>1$
such that for all $k$,

$K^{-1}\leq\gamma_{k}\leq K.$

Define an operator $B$ in the Hilbert space by $Bv_{k}:=\gamma_{k}v_{k}$ and
$B$ the identity on the orthogonal complement of the subspace spanned by the
$v_{k}$.

Then $B$ is invertible – its inverse obtains as the analog replacing each
$\gamma_{k}$ by $\gamma_{k}^{-1}$. $B$ also commutes with $A$, and their
product (thus, topologically equivalent with $A$) is an operator with each
$v_{k}$ now an eigenvector of $\gamma_{k}\lambda_{k}$ – in the dimension
spectral measure the $1$-mass was ‘transferred’ from $\lambda_{k}$ to
$\gamma_{k}\lambda_{k}$. Hence

######  Corollary 10.

Suppose $\gamma_{k}$ is ‘uniformly log-bounded’ as above, then two operators
$T$ and $T^{\prime}$ with their dimension (cardinal-number-valued) spectral
measures obtained one from the other by ‘transferring $1$-masss from
$\lambda_{k}$ to $\gamma_{k}\lambda_{k}$’, must necessarily be topologically
equivalent.

We will sometimes refer to such transfer of masses, in a cardinal-number-
valued ($\sigma$-additive) measure, to uniformly log-bounded-ly distant places
as a uniformly log-bounded shift in the measure.

  8. (viii)

Now let us be given some countable partition of $]0,\infty[$ into Borel
subsets:

$]0,\infty[=\cup_{k}S_{k},\quad\mathrm{the}\,S_{k}\,\mathrm{disjoint}\,\mathrm{Borel}\,\mathrm{sets},$

and some corresponding sequence of points $a_{k}\in]0,\infty[$, such that the
log-diameters of $S_{k}\cup\\{a_{k}\\}$ are uniformly bounded: there is a
$K>1$ such that for all $k$,

$y,z\in S_{k}\cap\\{x_{k}\\}\Rightarrow K^{-1}\leq y/z\leq K.$

As an example, let $\beta>1$ and take
$S_{k}=]b\beta^{k},b\beta^{k-1}],\,\,k\in\mathbb{Z}$ (or
$S_{k}=[b\beta^{k},b\beta^{k-1}[,\,\,k\in\mathbb{Z}$) and
$a_{k}=a\beta^{k},\,\,k\in\mathbb{Z}$ for some fixed $a,b>0$.

Returning to our operator $T$ between Hilbert spaces and (the positive semi-
definite) $A:=\sqrt{T^{\star}T}$, with spectral decomposition $\mathcal{E}$.
Recall that

$A=\int_{]0,+\infty[}\lambda\,d\mathcal{E}(\lambda).$

Let the scalar function $\varphi$ on $]0,\infty[$ be defined by
$\varphi(\lambda):=a_{k}/\lambda$ when $\lambda\in S_{k}$. Then by our
assumptions we will have $K^{-1}\leq\varphi(\lambda)\leq K$ for
$\lambda\in]0,\infty[$.

Therefore if we define an operator from $\mathcal{H}$ to itself:

$B:=\mathcal{E}(\\{0\\})+\int_{]0,+\infty[}\varphi(\lambda)\,d\mathcal{E}(\lambda),$

then $B$ is invertible – its inverse obtains as the analogous integral
replacing the integrand by its inverse. $B$ also commutes with $A$, and their
product (thus, topologically equivalent with $A$) is

$BA:=\int_{]0,+\infty[}\lambda\varphi(\lambda)\,d\mathcal{E}(\lambda)=\sum_{k}\int_{S_{k}}a_{k}\,d\mathcal{E}(\lambda)=\sum_{k}a_{k}\mathcal{E}(S_{k}).$

In particular, we found that the operators $T$ and $A$ are topologically
equivalent to a (positive semi-definite) operator whose spectral decomposition
is supported on $0$ and the $a_{k}$, and moreover depends only on, in fact is
a combination of, the $a_{k}$’s and the value $\mathcal{E}$ gives to
$(\\{0\\})$ and to the $S_{k}$’s.

Therefore

######  Corollary 11.

With $S_{k}$ and $a_{k}$ as above, two operators $T$ and $T^{\prime}$ with
their dimension (cardinal-number-valued) spectral measure giving the same
respective weights to $(\\{0\\})$ and to the $S_{k}$’s must necessarily be
topologically equivalent.

Or, to put it another way,

For an operator between Hilbert spaces, up to replacing the operator by a
topologically equivalent one, we may, in the dimension spectral measure,
‘transfer’ all the mass of $S_{k}$ into $a_{k}$ for each $k$.

Which is another case of a uniformly log-bounded shift – ‘transfer of mass (in
a cardinal-number-valued ($\sigma$-additive) measure) to uniformly log-
bounded-ly distant places’.

##  6\. Towards a Classification, ‘Eigenvalue Sequences with Infinity
Intermissions’

We are given our operator $T$ between Hilbert spaces, topologically equivalent
to the positive semi-definite $A:=T^{\star}T$, with its dimension spectral
measure $\mathcal{D}$ – the latter characterizing it up to isometry, i.e.
conjugation by a unitary, hence a fortiori up to topological equivalence.

For $\mathcal{D}$ we will have the open (subset of $]0,\infty[$)
$\mathbf{FIN}_{\mathcal{D}}$, thus an at most countable union of open
intervals – its connected components; and its complement – the closed (in
$]0,\infty[[$
$\mathbf{INF}_{\mathcal{D}}=]0.\infty[\setminus\mathbf{FIN}_{\mathcal{D}}$.

Then, if necessary making uniformly log-bounded shifts in the dimension
spectral measure $\mathcal{D}$, thus replacing an operator by a topologically
equivalent, one may

  * •

Fixing a $\beta>1$, ‘transfer’ all masses to, say, $0$ and
$\beta^{k},\,k\in\mathbb{Z}$.

  * •

Recall that the mass is necessarily supported in some interval $[0,a]$ since
the operator is bounded. And we can make $a$ an arbitrarily smaller
$a^{\prime}<a$ by ‘transferring’ the mass of $[a,a^{\prime}[$ into
$a^{\prime}$.

  * •

Recall the decomposition of $]0,\infty[$ into $\mathbf{FIN}_{\mathcal{D}}$ and
$\mathbf{INF}_{\mathcal{D}}$. In $\mathbf{INF}_{\mathcal{D}}$ no point $x$ has
a neighborhood with finite weight. Thus either $]x-\varepsilon,x]$ has
infinite weight for every $\varepsilon>0$, or $[x,x+\varepsilon[$ are so (or
both). Then, applying a uniformly log-bounded shift, we may assume $\\{x\\}$
itself has infinite weight.

  * •

And if in a component $C$ of $\mathbf{FIN}_{\mathcal{D}}$, the set $\Sigma$ of
eigenvalues turns to be infinite forward or backward, thus tending to an
extremity $a$ of $C$, necessarily $a\in\mathbf{INF}_{\mathcal{D}}$, the
infinite part from any fixed place can be ‘transferred’ to $a$, thus
‘truncating’ the sequence to leave only a finite sequence of eigenvalues, and
surely making $\mathcal{D}(\\{a\\}$ infinite.

  * •

More generally, a point $a$ with infinite weight may serve to ‘erase’, by
uniformly log-bounded shift (‘transferring them to $a$), any discrete points
with finite weight uniformly log-bounded distanced from $a$.

So, after some uniformly log-bounded shift, one may assume the mass
concentrated in $0,\lambda_{k}$, the $\lambda_{k}$ a (given in advance)
strictly decreasing sequence, tending to $0$ and with log-distances uniformly
bounded below. i.e. $\lambda_{k}/\lambda_{k+1}\geq K$ for some fixed $K>1$ –
as a particular case we may choose $\lambda_{k}:=\beta^{k_{0}-k}$, $\beta>1$
given.

Some points $0,\lambda_{k}$ will have an infinite-cardinal weight.

The other, with finite weight, may be arranges as a non-increasing sequence of
eigenvalues, counting $n$ times an eigenvalue $\lambda$ with multiplicity $n$,
i.e. with weight $n$ in the dimension spectral measure (and of course
$\lambda$’s with weight $0$ not appearing). Recall that this non-increasing
sequence is such that quotients of a term by its successor, if not $1$, are
uniformly bounded away from $1$.

We call that an eigenvalue sequence with infinity intermissions.

To repeat, up to topological equivalence, making a uniformly log-bounded shift
if necessary, every operator is characterized by such eigenvalue sequence with
infinity intermissions.

Also, by uniformly log-bounded shift, one can ‘erase’ some or all eigenvalues
in the eigenvalue sequence with uniformly bounded log-distance from points
$0,\lambda_{k}$ with infinite weight.

Recall that Theorem 8 with Definition 9 gave a necessary condition for
operators between Hilbert spaces, with corresponding dimension spectral
measures $\mathcal{D}$ and $\mathcal{D}^{\prime}$, to be topologically
equivalent: namely, that there exists a constant $K\geq 1$ such that:

The weight given to an interval $[0,\mu]$ (resp. $]\mu^{\prime},\mu]$) by one
of the measures is always $\leq$ than the weight given to $[0,K\mu]$ (resp.
$]\mu^{\prime}/K,K\mu]$) by the other.

What does that say in our scenario?

If $\lambda>\lambda^{\prime}$ are terms in the eigenvalue sequence with no
infinity intermissions between them and with $\lambda^{\prime}$ $n$ steps
after $\lambda$, then clearly for any $\varepsilon>0$,

The weight of $]\lambda^{\prime}-\varepsilon,\lambda]\geq n\geq$ the weight of
$]\lambda^{\prime},\lambda-\varepsilon]$.

Therefore the necessary condition translates here to:

In the eigenvalue sequences of the operators, starting from some $\lambda$,
even allowing a uniformly log-bounded shift in it – and then going $n$ steps
($n$ finite) in the sequence for either of the operators without encountering
infinity intermissions – will give for one operator results at uniformly log-
bounded distance (that is, quotient) from the other operator.

And also, that points with infinite cardinal weight for one of the operators
must have, at uniformly log-bounded distance, points with no less weight for
the other operator.

Where, recall, as long as we are in uniformly log-bounded distance from points
with infinite weight, eigenvalues in the sequence may be erased (by uniformly
log-bounded shift).

Thus in starting, as above, from some $\lambda$, even allowing a uniformly
log-bounded shift in it, then going $n$ steps ($n$ finite) in the sequence –
thing matter only as long as we are not at uniform log-bounded distance from
points with infinite weight, for one operator consequently for the other – our
necessary condition then says that going $n$ steps for one operator or the
other will keep us at uniformly log-bounded distance.

And clearly if all that is satisfied, the eigenvalue sequence with infinity
intermissions for one operator obtains from that for the other by some
uniformly log-bounded shift. Therefore

######  Theorem 12.

For two operators between Hilbert spaces, a necessary and sufficient condition
to be topologically equivalent is that their ‘eigenvalue sequence with
infinity intermissions’ can be obtained one from the other by uniformly log-
bounded shift.

discussion

  * •

Of course, $\mathcal{D}(\\{0\\})$ – the (Hilbert) dimension of the kernel, is
an invariant for topological equivalence.

  * •

So are $\mathcal{D}([0,\infty[)$ – the (Hilbert) dimension of the space and
$\mathcal{D}(]0,\infty[)$ – the (Hilbert) dimension of the image, in
particular when they are finite.

  * •

When the number of eigenvalues between the infinity intermissions is uniformly
bounded we may, by a uniform log-bounded shift assume that all points with
non-zero weight have infinite weight.

  * •

The case where there are no infinity intermissions is clearly the case of a
compact operator – obviously a topological equivalence invariant.

  * •

What may yield non-topologically-equivalent operators is, for instance,
unbounded ‘chunks’ in the eigenvalue sequence between ‘infinities’– recall
these are non-increasing, with quotients of a term by its successor, if not
$1$, uniformly bounded away from $1$. Hence what can differentiate here are
thing like unbounded lengths of rows of equal terms, i.e. eigenvalues with
multiplicity; and kog-unbounded ‘jumps’ from an eigenvalue term to the next.

  * •

Also, especially when the Hilbert space is non-separable, which infinite
cardinalities appear and their distribution in the infinity intermissions –
again inasmuch as not ‘uniformly bounded’,

##  Appendix A Dimension in Hilbert Spaces

Recall that, as usual, by the dimension $\mathrm{dim}\,\mathcal{H}$ of a
Hilbert space $\mathcal{H}$ we mean the cardinality of an orthogonal basis to
$\mathcal{H}$ (as is well known and easily proved, all such bases have the
same cardinality).

######  Proposition 13.

If $\mathcal{H}$ and $\mathcal{H}^{\prime}$ are Hilbert spaces and
$L:\mathcal{H}\to\mathcal{H}^{\prime}$ is an injective bounded linear
operator, then $\dim\mathcal{H}\leq\dim\mathcal{H}^{\prime}$.

######  Proof .

Let $\delta=\dim\mathcal{H}$ (resp.
$\delta^{\prime}=\dim\mathcal{H}^{\prime}$) and let $(e_{i})_{i\in\delta}$ be
an orthonormal basis to $\mathcal{H}$.

If $\delta$ is finite then $L\mathcal{H}\subset\mathcal{H}^{\prime}$ is also a
finite-dimensional space of dimension $\delta$ hence
$\delta\leq\dim\mathcal{H}^{\prime}=\delta^{\prime}$.

Assume now that $\delta$, hence $\delta^{\prime}$, is infinite.

$L$ is strongly, hence weakly, continuous, and is injective. Therefore it
defines a homeomorphism, with respect to the weak topologies, from the unit
ball of $\mathcal{H}$, which is weakly compact, to its image, consequently
also weakly compact.

The set $\\{e_{i}\\}$ has weakly the discrete topology in the unit ball of
$\mathcal{H}$, therefore also the set $\\{Le_{i}\\}$ has weakly the discrete
topology in $\mathcal{H}^{\prime}$.

Now, $\mathcal{H}^{\prime}$ has an open basis to the weak topology of
cardinality $\max(\delta^{\prime},\aleph_{0})=\delta^{\prime}$ (consisting of
weakly open sets defined by inequalities on the inner product with finite
rational combinations of the orthonormal basis). Since $\\{Le_{i}\\}$ is
weakly discrete, every $Le_{i}$ belongs to some member $U_{i}$ of that basis
containing it and no other $Le_{j}$, $j\neq i$. Of course, all the $U_{i}$’s
must be different. Thus $\delta$, the cardinality of $\\{Le_{i}\\}$, is less
than $\delta^{\prime}$. ∎

######  Corollary 14.

Two isomorphic Hilbert spaces (i.e. such that there is a 1-1 and onto, i.e.
invertible, operator between them), have the same Hilbert dimension, hence
are, in fact, isometric.

##  Appendix B In Hilbert Space, the Subspace of Vectors Where the Norm is
Attained

We are in the scenario of §2 and §3.

For a vector $x\in\mathcal{H}$, we have

(2) $\|Tx\|^{2}=\langle
A^{2}x,x\rangle=\left\langle\int_{[0,+\infty[}\lambda^{2}\,d\mathcal{E}(\lambda)x\,\,,\,\,x\right\rangle=\int_{[0,+\infty[}\lambda^{2}\,\left\langle\,d\mathcal{E}(\lambda)x,x\right\rangle.$

This means that $\|T\|$ is ‘the upper end of the support of the spectral
measure $\mathcal{E}$’. Indeed, $\mathcal{E}(]\|T\|,\infty[)=0$, since
otherwise the subspace $\mathcal{E}(]\|T\|,\infty[)\mathcal{H}$ would be a
non-zero subspace, thus including some $x\neq 0$ for which by (2) $\|Tx\|$
would be $>\|T\|\|x\|$; while on the other hand for any $\varepsilon>0$
$\mathcal{E}(]\|T\|-\varepsilon,\infty[)\neq 0$, otherwise by (2) for every
$x\in\mathcal{H}$ $\|Tx\|\leq(\|T\|-\varepsilon)\|x\|$, contrary to the
definition of $\|T\|$.

Moreover, one finds that for a vector $x\in\mathcal{H}$ to satisfy
$\|Tx\|=\|T\|\|x\|$, one must have, by (2)

$\displaystyle\int_{[0,\|T\|]}\lambda^{2}\,\left\langle\,d\mathcal{E}(\lambda)x,x\right\rangle=\|T\|^{2}\int_{[0,+\infty[}\left\langle\,d\mathcal{E}(\lambda)x,x\right\rangle,$
$\displaystyle\int_{[0,\|T\|[}(\|T\|^{2}-\lambda^{2})\,\left\langle\,d\,\mathcal{E}(\lambda)x,x\right\rangle=0,$

which, since $\|T\|^{2}-\lambda^{2}$ is strictly positive in $[0,+\|T\|[$, can
happen only if
$\|\mathcal{E}([0,+\|T\|[)x\|^{2}=\int_{[0,+\|T\|[}\left\langle\,d\mathcal{E}(\lambda)x,x\right\rangle=0$,
meaning that $x\in\mathcal{E}(\\{\|T\|\\})\mathcal{H}$. Thus we reach

######  Corollary 15.

For any operator $T$ on a Hilbert space $\mathcal{H}$, and $\mathcal{E}$ the
spectral measure of $A:=\sqrt{T^{\star}T}$ as above, the set of vectors
$x\in\mathcal{H}$ satisfying $\|Tx\|=\|T\|\|x\|$ is the (closed) subspace
$\mathcal{E}(\\{\|T\|\\})\mathcal{H}$.

## References

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#
$2$-Coupon Coloring of Cubic Graphs Containing $3$-Cycle or $4$-Cycle

S<EMAIL_ADDRESS>M<EMAIL_ADDRESS>A. Fazli
<EMAIL_ADDRESS>B<EMAIL_ADDRESS>E.
<EMAIL_ADDRESS>
###### Abstract

Let $G$ be a graph. A total dominating set in a graph $G$ is a set $S$ of
vertices of $G$ such that every vertex in $G$ is adjacent to a vertex in $S$.
Recently, the following question was proposed:
“Is it true that every connected cubic graph containing a $3$-cycle has two
vertex disjoint total dominating sets?”
In this paper, we give a negative answer to this question. Moreover, we prove
that if we replace $3$-cycle with $4$-cycle the answer is affirmative. This
implies every connected cubic graph containing a diamond (the complete graph
of order $4$ minus one edge) as a subgraph can be partitioned into two total
dominating sets, a result that was proved in $2017$.

##  1 Introduction

Throughout this paper, all graphs are simple that is with no loop and multiple
edges. Let $G$ be a graph. We denote the vertex set and the edge set of $G$ by
$V(G)$ and $E(G)$, respectively. For $v\in V(G)$, $d(v)$ denotes the degree of
$v$. A vertex $v$ is called an $i$-vertex if $d(v)=i$. A total dominating set
in a graph $G$ is a set $S\subseteq V(G)$ such that every vertex in $G$ is
adjacent to a vertex in $S$. The study of cubic graphs whose vertex set can be
partitioned into two total dominating sets is an attractive topic and many
authors have investigated this problem. A $k$-coloring of a graph $G$ is an
assignment of colors from the set $[k]=\left\\{0,\ldots,k-1\right\\}$ to the
vertices of $G$. A $k$-coupon coloring of a graph $G$ without isolated
vertices is a $k$-coloring of $G$ such that the neighborhood of every vertex
of $G$ contains vertices of all colors from $[k]$. The maximum $k$ for which a
$k$-coupon coloring exists is called the coupon coloring number of $G$. We say
that a vertex has coupon property if all $k$ colors are appeared in the
neighbors of that vertex. The concept of $k$-coupon coloring of graphs has
been studied by several authors, for instance, see [References], [References]
and [References].
The open neighborhood of a vertex $v$ in $G$ denoted by $N_{G}(v)$ is the set
of all vertices adjacent to $v$. In this paper for any $S\subseteq V(G)$,
$G[S]$ denotes the subgraph of $G$ induced by $S$. A
$\left\\{1,2\right\\}$-factor is a spanning subgraph of $G$ in which each
component is either $1$-regular or $2$-regular. We use $C_{n}$ and $P_{n}$,
for the cycle and the path of order $n$, respectively. Also, $C_{n}$ is called
an $n$-cycle. A vertex $v$ is called a good vertex or a bad vertex, if it has
or has not the coupon property, respectively. It is not hard to see that a
$2$-coupon coloring of a graph is equivalent to the vertex partitioning of a
graph into two total dominating sets. In [References] the following
interesting question was proposed:
“Is it true that every connected cubic graph containing a $3$-cycle has two
vertex disjoint total dominating sets?”
Here, we construct a connected cubic graph of order $60$ containing a
$3$-cycle whose vertex set cannot be partitioned into two total dominating
sets. The complete graph of order $4$ minus one edge is called a diamond. In
[References] it is shown that the vertex set of a cubic graph containing a
diamond as a subgraph can be partitioned into two total dominating sets. In
this paper, we strengthen this result by showing that the vertex set of every
cubic graph containing a $4$-cycle can be partitioned into two total
dominating sets.

For a graph $G$, we use the symbol $G^{*}$ for a bipartite graph that is
constructed from $G$ in the following way:
Let $V(G)=\left\\{v_{1},\ldots,v_{n}\right\\}$. Form a bipartite cubic graph
$G^{*}=(U,W)$ of order $2n$, where $U=\left\\{u_{1},\ldots,u_{n}\right\\}$ and
$W=\left\\{w_{1},\ldots,w_{n}\right\\}$. Join $u_{i}$ to $w_{j}$ and $u_{j}$
to $w_{i}$ if and only if $v_{i}v_{j}\in E(G)$.

##  2 Main Results

In this paper, we denote the Heawood graph by $\mathcal{H}$.

The Heawood graph

Note that $\mathcal{H}=(X,Y)$ is a vertex transitive bipartite cubic graph.

We start this section with the following lemma.

######  Lemma 1.

In any $2$-coloring of $\mathcal{H}$, there are at least two bad vertices.

###### Proof.

By contradiction and using [References, Observation $7$], there exists a
$2$-coloring of $\mathcal{H}=(X,Y)$ with exactly one bad vertex. With no loss
of generality assume that $X$ has no bad vertex. Since $\mathcal{H}$ is vertex
transitive, one can color the part $Y$ such as $X$ to obtain a $2$-coupon
coloring for $\mathcal{H}$, a contradiction. The proof is complete. ∎

######  Lemma 2.

Let $e=uv\in E(\mathcal{H})$. If there is a $2$-coloring for
$\mathcal{H}\backslash e$ such that all vertices in
$V(\mathcal{H})\backslash\left\\{u,v\right\\}$ are good, then $u$ and
$N_{\mathcal{H}\backslash e}(v)$ have the same color.

###### Proof.

By contradiction assume that $w\in N_{\mathcal{H}\backslash e}(v)$ and the
color of $u$ and $w$ are different. It is straightforward to see that it is a
$2$-coloring of $\mathcal{H}$ in which every vertex in
$V(\mathcal{H})\backslash{u}$ is good which contradicts Lemma 1. The proof is
complete. ∎

######  Theorem 1.

There exists a connected cubic graph containing a $3$-cycle with no $2$-coupon
coloring.

###### Proof.

Consider $4$ vertex disjoint copies of $\mathcal{H}$ and denote them by
$\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}_{3},\mathcal{H}_{4}$. Let
$e_{i}=u_{i}v_{i},$ for $i=1,\ldots,4$. Now, we would like to form a cubic
graph containing a triangle. Let $K$ be the following graph:

$\mathit{x}$$\mathit{z}$$\mathit{y}$$\mathit{t}$

Consider $(\bigcup_{i=1}^{4}\mathcal{H}_{i}\backslash e_{i})\cup K$. Join $t$,
$y$, and $z$ to $\left\\{v_{1},v_{2}\right\\}$, $u_{3}$, and $u_{4}$,
respectively. Also join $u_{1}$ and $v_{4}$ to $u_{2}$ and $v_{3}$,
respectively and call the resultant cubic graph by $G$.

$\mathit{u_{1}}$$\mathit{u_{2}}$$\mathit{v_{1}}$$\mathit{v_{2}}$$\mathit{u_{4}}$$\mathit{v_{4}}$$\mathit{u_{3}}$$\mathit{v_{3}}$$\mathit{x}$$\mathit{z}$$\mathit{y}$$\mathcal{H}_{1}\backslash
e_{1}$$\mathcal{H}_{4}\backslash e_{4}$$\mathcal{H}_{2}\backslash
e_{2}$$\mathcal{H}_{3}\backslash e_{3}$$\mathit{t}$

We claim that $G$ has no $2$-coupon coloring.
By contradiction assume that $c$ is a $2$-coupon coloring for $G$. With no
loss of generality suppose that $c(v_{1})=0$, where $c(v_{1})$ denotes the
color of $v_{1}$. By Lemma 2, every vertex in $N_{\mathcal{H}_{1}\backslash
e_{1}}(u_{1})$ has color $0$. Since $u_{1}$ is a good vertex, $c(u_{2})=1$.
Thus by Lemma 2, every vertex in $N_{\mathcal{H}_{2}\backslash e_{2}}(v_{2})$
has color $1$. Since $v_{2}$ is a good vertex, one can see that $c(t)=0$. By
Lemma 2, two vertices of $N_{\mathcal{H}_{1}\backslash e_{1}}(v_{1})$ have the
same color, and since $v_{1}$ is a good vertex, each vertex of
$N_{\mathcal{H}_{1}}(v_{1})$ has color $1$. On the other hand, by Lemma 2,
since $u_{2}$ is a good vertex, every vertex of $N_{\mathcal{H}_{2}}(u_{2})$
has color $0$. Now, since $t$ is good, we have $c(x)=1$. Since $x$ is good, we
find that $c(y)=1$ or $c(z)=1$. With no loss of generality assume that
$c(y)=1$. Since $u_{3}$ is good by Lemma 2, every vertex of
$N_{\mathcal{H}_{3}}(u_{3})$ has color $0$. Since $v_{4}$ is good, by Lemma 2,
we find that $c(u_{4})=1$ which contradicts $z$ is good. The proof is
complete. ∎

Now, we have the following corollary.

######  Corollary 2.

There exists a connected cubic graph containing a triangle that has no two
vertex disjoint total dominating sets.

###### Proof.

Consider the graph $G$ stated in Theorem 1. If $G$ has two vertex disjoint
total dominating sets $S_{0}$ and $S_{1}$, then color all vertices of $S_{0}$
and $S_{1}$ by $0$ and $1$, respectively, and color $V(G)\backslash(S_{0}\cup
S_{1})$ by $1$. Obviously, this is a $2$-coupon coloring of $G$ which
contradicts Theorem 1. ∎

######  Theorem 3.

Let $G$ be a connected cubic graph and $S\subsetneq V(G)$. If $G[S]$ has a
$2$-coupon coloring, and $G\backslash S$ has a $\left\\{1,2\right\\}$-factor,
then $G$ has a $2$-coupon coloring.

###### Proof.

First assume that $G=(X,Y)$ is a connected bipartite cubic graph. By
contradiction assume that $S$ is the largest set of vertices of $G$ satisfying
the assumption and $G\backslash S$ has a $\left\\{1,2\right\\}$-factor $F$,
and $G$ has no $2$-coupon coloring. With no loss of generality, one may assume
that $F$ contains no even cycle because every even cycle has a perfect
matching.
Since $G$ is connected, then there exists an edge $uv$ such that $u\in S$ and
$v\in V(G\backslash S)$. Let $A$ be the set of all edges of $P_{2}$-components
of $F$. By a good walk in $G$, we mean a walk starting at $u$ whose edges are
alternatively contained in $E(G)\backslash A$ and $A$ such that its first edge
is $uv$ and each edge in $E(G)\backslash A$ appears at most once.
Let $W$ be the longest good walk in $G$. Suppose that
$W:ue_{1}x_{1}e_{2}y_{1}e_{3}x_{2}e_{4}y_{2}\ldots e_{k}w$, where $e_{i}$ is
in $E(G)\backslash A$ and $A$ if $i$ is odd and even, respectively. Note that
since $G$ is bipartite, we have
$\left\\{x_{1},x_{2},\ldots\right\\}\cap\left\\{y_{1},y_{2},\ldots\right\\}=\varnothing$.
Since $W$ has the longest length among all good walks and for every
$x_{i}y_{i}\in A$, $d(x_{i})=d(y_{i})=3$, it implies that the number of edges
in $E(G)\backslash A$ incident with $x_{i}$ is the same as the number of edges
in $E(G)\backslash A$ incident with $y_{i}$. Thus $W$ terminates with an edge
in $E(G)\backslash A$. So $w\in S$ because if $w\in V(G\backslash S)$, then
$w$ has an edge in $A$ and this contradicts $W$ has the longest length. Color
$y_{1}$ by $1-c(u)$, where $c(u)$ is the color of $u$. If $y_{i}$ is not
colored before, color $y_{i}$ by $1-c(y_{i-1})$, for
$i=1,\ldots,\frac{k-1}{2}$. Let $i$ be an arbitrary index, $v=x_{i}$ and
$e=vz\in A$. Suppose that we meet $e$ for the first time in $W$. Obviously,
$z$ is still not colored and after we color $z$, $v$ becomes a good vertex. So
every $x_{i}$ has coupon property. Define $c(x_{\frac{k-1}{2}})=1-c(w)$. Also
if $x_{j}$ is not colored, define $c(x_{j})=1-c(x_{j+1})$, for
$j=\frac{k-1}{2}-1,\ldots,1$ and then with the previous argument, every
$y_{i}$ becomes a good vertex. Note that every vertex in $S\cup V(W)$ has the
coupon property and $G\backslash(S\cup V(W))$ has a
$\left\\{1,2\right\\}$-factor which contradicts the assumption.
Next, assume that $G$ is a connected non-bipartite cubic graph. Let
$V(G)=\left\\{v_{1},\ldots,v_{n}\right\\}$. Consider the bipartite cubic graph
$G^{*}$. Since $G$ contains an odd cycle, it is easy to see that $G^{*}$ is a
connected bipartite cubic graph. With no loss of generality, suppose that
$S=\left\\{v_{1},\ldots,v_{r}\right\\}$. In the graph $G^{*}$, define
$S’=\left\\{u_{1},\ldots,u_{r}\right\\}\cup\left\\{w_{1},\ldots,w_{r}\right\\}$.
Assume that $c$ is a $2$-coupon coloring for $G[S]$. If we define
$c(u_{i})=c(w_{i})=c(v_{i})$, then clearly we obtain a $2$-coupon coloring for
$G^{*}[S’]$. By assumption, $G\backslash V(S)$ has a
$\left\\{1,2\right\\}$-factor $F$. If $v_{i_{1}},v_{i_{2}},\ldots,v_{i_{2t}}$
is an even cycle of $F$, then $u_{i_{1}}w_{i_{2}}u_{i_{3}}w_{i_{4}}\ldots
w_{i_{2t}}$ and $w_{i_{1}}u_{i_{2}}w_{i_{3}}u_{i_{4}}\ldots u_{i_{2t}}$ are
two vertex disjoint even cycles in $G^{*}$. If $v_{i_{1}}v_{i_{2}}\ldots
v_{i_{2t+1}}$ is an odd cycle in $G$, then
$u_{i_{1}}w_{i_{2}}u_{i_{3}}w_{i_{4}}\ldots u_{i_{2t+1}}w_{i_{1}}$ is an even
cycle of order $4t+2$ in $G^{*}$. Moreover, if $v_{i}v_{j}$ is a
$P_{2}$-component of $F$, then $u_{i}w_{j}$ and $u_{j}w_{i}$ form a two
matching edges in $G^{*}$. These imply that $H\backslash V(S’)$ has a
$\left\\{1,2\right\\}$-factor.
Now, by the previous case, $G^{*}$ has a $2$-coupon coloring $c$. If we define
$c(v_{i})=c(u_{i})$, for $i=1,\ldots,n$, then one can see that $c$ is a
$2$-coupon coloring for $G$. ∎

Now, we have the following corollary.

######  Corollary 4.

If $G$ is a connected cubic graph containing a $4$-cycle, then $G$ has a
$2$-coupon coloring.

###### Proof.

First assume that $G=(X,Y)$ is bipartite. If $C$ is a $4$-cycle of $G$, we
claim that $G\backslash V(C)$ has a perfect matching. Let
$H=(X’,Y’)=G\backslash V(C)$ and $S\subseteq X’$. By Marriage Theorem
[References] it suffices to show that $|N_{H}(S)|\geq|S|$. Clearly, the number
of edges between $S$ and $Y’$ is at least $3|S|-2$. If $|N_{H}(S)|\leq|S|-1$,
then there exists a vertex in $Y’$ whose degree is at least
$\frac{3|S|-2}{|S|-1}>3$, a contradiction. So $H$ has a perfect matching and
the claim is proved. Since $C$ has a $2$-coupon coloring, Theorem 3 yields
that $G$ has a $2$-coupon coloring.
Next, suppose that $G$ is non-bipartite. Consider the connected bipartite
cubic graph $G^{*}$. Since $G$ contains a $4$-cycle, $G^{*}$ contains two
vertex disjoint $4$-cycles say $C_{1}$ and $C_{2}$. By the same method as we
did before, one can see that $G^{*}\backslash V(C_{1})$ has a
$\left\\{1,2\right\\}$-factor. Since $C_{1}$ has a $2$-coupon coloring, by
Theorem 3, $G^{*}$ and so $G$ has a $2$-coupon coloring, as desired. ∎

######  Remark 1.

For every positive integer $r\,(r\geq 3)$, not divisible by $4$, there exists
a connected cubic graph that contains an induced $r$-cycle and has no
$2$-coupon coloring.

###### Proof.

We define $H_{1},H_{2}$ and $H_{3}$ as follows:

$\mathcal{H}\backslash e$ |  $\mathcal{H}\backslash e$$\mathcal{H}\backslash e$ |  $\mathcal{H}\backslash e$$\mathcal{H}\backslash e$ $H_{3}$
---|---|---
$H_{1}$ $H_{2}$

Now, three cases can be considered.
Case 1. $r=1\,(mod\,4)$. Consider an $r$-cycle $C$ with the vertex set
$\left\\{v_{1},\ldots,v_{r}\right\\}$. We join $v_{1}$ to the $2$-vertex of
$H_{1}$. Now, consider $\frac{r-1}{2}$ disjoint copies of $H_{2}$. Join
$v_{2k}$ to one $2$-vertex and $v_{2k+1}$ to another $2$-vertex of $k$-th copy
of $H_{2}$, for $k=1,2,\ldots,\frac{r-1}{2}$.

$\mathit{\ldots}$$\mathit{C}$$\mathcal{H}\backslash e$$\mathcal{H}\backslash
e$$\mathcal{H}\backslash e$$\mathcal{H}\backslash e$$\mathcal{H}\backslash e$
$r=1\,(mod\,4)$

Case 2. $r=1\,(mod\,4)$. Consider an $r$-cycle $C$ with the vertex set
$\left\\{v_{1},\ldots,v_{r}\right\\}$. We join $v_{1}$ to the $2$-vertex of
$H_{3}$. Now, consider $\frac{r-1}{2}$ disjoint copies of $H_{2}$. Join
$v_{2k}$ to one $2$-vertex and $v_{2k+1}$ to another $2$-vertex of $k$-th copy
of $H_{2}$, for $k=1,2,\ldots,\frac{r-1}{2}$.

$\mathit{\ldots}$$\mathit{C}$$\mathcal{H}\backslash e$$\mathcal{H}\backslash
e$$\mathcal{H}\backslash e$$\mathcal{H}\backslash e$$\mathcal{H}\backslash
e$$\mathcal{H}\backslash e$ $r=3\,(mod\,4)$

Case 3. $r=2\,(mod\,4)$. Since $\frac{r}{2}$ is odd, by Cases $1$ and $2$,
there exists a connected cubic graph $G$ with an induced $\frac{r}{2}$-cycle
which has no $2$-coupon coloring. Now, $G^{*}$ satisfies the condition.
By a similar argument as we did in the proof of Theorem 1, one can see that
the connected cubic graphs introduced in Cases $1,2$ and $3$ have no
$2$-coupon coloring. ∎

## References

  * [1] B. Chen, J. H. Kim, M. Tait, and J. Verstraete, On coupon colorings of graphs, Discrete Applied Math, 193(2015), 94-101 .
  * [2] W.J. Desormeaux , T.W. Haynes, M.A. Henning, Partitioning the vertices of a cubic graph into two total dominating sets, Discrete Applied Mathematics 223 (2017) 52-63.
  * [3] P. Hall, On Representatives of Subsets, J. London Math. Soc. 10(1935), no. 1,26-30.
  * [4] Z. L. Nagy, Coupon-coloring and total domination in Hamiltonian planar triangulations, Graphs Combin. 34(2018), no. 6, 1385-1394.
  * [5] Y. Shi, M. Wei, J. Yun, Y. Zhao, Coupon coloring of some special graphs, J. Comb. Optim. 33(2017), no. 1, 156-164.


& \times \cdots \times \iint_{\dom^2} \ud z_{n-1} \, \ud z_{n-1}'\, \widetilde{G}_D(s_{n-2}-s_{n-1}, z_{n-2}, z_{n-2}', z_{n-1}, z_{n-1}') f(z_{n-1}-z_{n-1}') \\
& \times \iint_{\dom^2} \ud z_{n} \, \ud z_{n}'\,
\widetilde{G}_D(s_{n-1}-s_{n}, z_{n-1}, z_{n-1}', z_{n}, z_{n}') f(z_{n}-z_{n}')
\widetilde{G}_D(s_{n}, z_{n}, z_{n}', y, y').
\end{align*}

Then, we derive a lower bound by integrating on the smaller intervals

\begin{align}\label{interval-si}
\begin{split}
s_1 & \in \left[ \left(1-\frac{1}{4n}\right)\frac t 2, \frac{t}{2}\right],     \\
s_2 & \in \left[\left(1-\frac{1}{2n}\right)\frac t 2, s_1-\frac{t}{8n}\right], \\
s_3 & \in \left[\left(1-\frac{2}{2n}\right)\frac t 2, s_2 - (s_1-s_2)\right],  \\
& \vdotswithin{\in}                                                        \\
s_n & \in \left[\left(1-\frac{n-1}{2n}\right)\frac t 2, s_{n-1}-(s_{n-2}-s_{n-1})\right].
\end{split}
\end{align}

$0 < \eps_0 < 1$ be given by (<ref>). For each $\eps \in (0,
\eps_0)$ and $x \in \dom_\eps$, recall the subset $V(x) \subset \dom$ defined
by (<ref>). Consider

z_1, z_1'     ∈A_1           V(x) ∩B(x, √((t-s_1)/(5n)) ∧(_0/2)),

z_2           ∈A_2           V(x) ∩B(x, √(s_1-s_2) ∧(_0/2)) ∩B(z_1 , √(s_1-s_2) ∧(_0/2)),

z_2'          ∈A_2'          V(x) ∩B(x, √(s_1-s_2) ∧(_0/2)) ∩B(z_1', √(s_1-s_2) ∧(_0/2)),

z_n           ∈A_n           V(x) ∩B(x, √(s_n-1-s_n) ∧(_0/2)) ∩B(z_n-1 , √(s_n-1-s_n) ∧(_0/2)),

z_n'          ∈A_n'          V(x) ∩B(x, √(s_n-1-s_n) ∧(_0/2)) ∩B(z_n-1', √(s_n-1-s_n) ∧(_0/2)).

Note that by (<ref>), we have $(t-s_1)/(5n) \le t/(8n) \le s_1-s_2
\le s_2-s_3 \le \dots \le s_{n-1}-s_n$, which ensures that for $2 \le i \le
n$, both $z_{i-1}$ and $z_{i-1}'$ lie in $V(x) \cap B(x,
\sqrt{s_{i-1}-s_i}\wedge (\eps_0/2))$. Then, by Lemma <ref>, for $2
\le i \le n$,

\begin{align*}
\mathrm{Vol}(A_i) \wedge \mathrm{Vol}(A_i') \ge C (\sqrt{s_{i-1}-s_i} \wedge (\eps_0/2))^d
\ge C (\eps_0/2)^d ((s_{i-1}-s_i) \wedge 1)^{d/2}.
\end{align*}

By Lemma <ref> and (<ref>), we have

\begin{gather*}
\mathrm{Vol}(A_1) \ge C \left(\sqrt{\frac{t-s_1}{5n}}\wedge (\eps_0/2)\right)^d \ge C (\eps_0/2)^d (5n)^{-d/2} ((t-s_1) \wedge 1)^{d/2}, \\
1\wedge \Phi_1(z) \ge 1 \wedge (c_0^{-1} \eps^{a_1}) \quad \text{for all $z \in V(x)$}.
\end{gather*}

Also, on $A_1 \times A_1$, we have $f(z_1 - z_1') \ge C(((t-s_1)/n) \wedge 1)^{-\beta/2}$,
and on $A_i \times A_i'$, where  $2 \le i \le n$, we have $f(z_i - z_i') \ge C((s_{i-1}-s_i) \wedge 1)^{-\beta/2}$.
Then, by the lower bound in (<ref>) and the inequalities

\begin{align*}
| z_n-z_{n-1} | ^2 + | z_n' - z_{n-1}' |^2 & \le 2(s_{n-1}-s_n), \\
|z_n - y|^2 + |z_n' - y'|^2                & \le 5(s_{n-1}-s_n) + 2|x-y|^2 + 3|x'-y'|^2 + 3|x-x'|^2,
\end{align*}

we have

\begin{gather*}
\iint_{A_n\times A_n'} \ud z_{n} \, \ud z_{n}'\,
\widetilde{G}_D(s_{n-1}-s_{n}, z_{n-1}, z_{n-1}', z_{n}, z_{n}') f(z_{n}-z_{n}')
\widetilde{G}_D(s_{n}, z_{n}, z_{n}', y, y') \\
\ge C_\eps^4 e^{-2\mu_1 s_{n-1}}(1\wedge (s_{n-1}-s_n))^{-\frac \beta 2}
\frac{1}{1 \wedge t^d}
e^{-c_2\frac{3|x-x'|^2 + 3|x-y|^2 + 3|x'-y'|^2 + 5s_{n-1}}{s_n}}.
\end{gather*}

For $i = 2, \dots, n-1$, by $|z_i - z_{i-1}|^2 + |z_i' - z_{i-1}'|^2 \le
2(s_{i-1}-s_i)$, we have

\begin{align*}
\iint_{A_i\times A_i'} \ud z_{i} \, \ud z_{i}'\, \widetilde{G}_D(s_{i-1}-s_{i}, z_{i-1}, z_{i-1}', z_{i}, z_{i}') f(z_{i}-z_{i}')
\ge C_\eps^2 e^{-2\mu_1 (s_{i-1}-s_i)} (1\wedge (s_{i-1}-s_i))^{-\frac \beta 2}.
\end{align*}

For $i = 1$, by the inequality

\begin{align*}
| z_1 - x | ^2 + | z_1' - x' | ^2 &\le 3(t-s_1) + 2|x-x'|^2,
\end{align*}

we have

\begin{align*}
\iint_{A_1\times A_1} \ud z_{1} \, \ud z_{1}'\, \widetilde{G}_D(t-s_{1}, x, x', z_1, z_1') f(z_1-z_1')
\ge \frac{C_\eps^2}{n^{d/2}} e^{-2\mu_1 (t-s_1)} e^{-2c_2\frac{|x-x'|^2}{t-s_1}} \left(1\wedge \frac{t-s_1}{n} \right)^{-\frac \beta 2}.
\end{align*}

Combining these estimates and using (<ref>), we get that

\begin{align*}
\MoveEqLeft \widetilde{G}_D^{\triangleright (n+1)}(t, x, x', y, y')                                                                               \\
& \ge C_\eps^{2(n+1)} \frac{e^{-2\mu_1 t}}{1 \wedge t^d} \int_{(1-\frac{1}{4n})\frac{t}{2}}^{\frac{t}{2}} \ud s_1 \, \left(1 + \left(\frac{t-s_1}{n}\right)^{-\frac \beta 2}\right) e^{-2c_2 \frac{|x-x'|^2}{t-s_1}}       \\
& \quad \times \int_{(1-\frac{1}{2n})\frac{t}{2}}^{s_1-\frac{t}{8n}} \ud s_2\, \left(1+ (s_1-s_2)^{-\frac \beta 2}\right)\times
\int_{(1-\frac{2}{2n})\frac{t}{2}}^{s_2-(s_1-s_2)} \ud s_3\, \left(1+ (s_2-s_3)^{-\frac \beta 2}\right)\times \cdots \\
& \quad \cdots \times \int_{(1-\frac{n-2}{2n})\frac{t}{2}}^{s_{n-2}-(s_{n-3}-s_{n-2})} \ud s_{n-1} \, \left(1 + (s_{n-2}-s_{n-1})^{-\frac \beta 2}\right) \\
& \quad \times \int_{(1-\frac{n-1}{2n})\frac{t}{2}}^{s_{n-1}-(s_{n-2}-s_{n-1})} \ud s_n\, \left(1 + (s_{n-1}-s_n)^{-\frac\beta 2}\right) e^{-c_2\frac{3|x-x'|^2 + 3|x-y|^2 + 3|x'-y'|^2 + 5t}{t/4}}.
\end{align*}

Let $I$ denote the above multiple integral for $s_1, \dots, s_n$. By
(<ref>), we have $s_{i-1}-s_i \ge \frac{t}{8n}$ and
$s_{i-1}-(1-\frac{i}{2n})\frac{t}{2} \ge (1-\frac{i-1}{2n})\frac{t}{2} -
(1-\frac{i}{2n})\frac{t}{2} = \frac{t}{4n}$ for $2 \le i \le n$. Fixing $s_1
\in \left[\left(1-\frac{1}{4n} \right)\frac{t}{2}, \frac{t}{2}\right]$, by the
change of variables $s_n \mapsto s_{n-1}-s_n$, $s_{n-1} \mapsto
s_{n-2}-s_{n-1}$, $\dots$, $s_2 \mapsto s_1-s_2$, we have

\begin{align*}
s_n & \in \left[s_{n-2}-s_{n-1}, s_{n-1}-\left(1-\frac{n-1}{2n}\right)\frac t 2\right] \supset \left[\frac{t}{8n}, \frac{t}{4n}\right], \\
& \vdotswithin{\in}                                                                                \\
s_3 & \in \left[s_1-s_2, s_2 - \left(1-\frac{2}{2n}\right)\frac t 2\right] \supset \left[\frac{t}{8n}, \frac{t}{4n}\right],             \\
s_2 & \in \left[\frac{t}{8n}, s_1 - \left(1-\frac{1}{2n}\right)\frac t 2\right] \supset \left[\frac{t}{8n}, \frac{t}{4n}\right].
\end{align*}

It follows that

\begin{align*}
I \ge \:
& e^{-10c_2} e^{-16c_2\frac{|x-x'|^2}{t}} e^{-12c_2 \frac{|x-y|^2+|x'-y'|^2}{t}} \\
& \times \int_{(1- \frac{1}{4n})\frac t 2}^{\frac{t}{2}} \ud s_1 \, \left(1 + \left(\frac{t-s_1}{n}\right)^{-\frac \beta 2}\right)
\times \int_{\frac{t}{8n}}^{\frac{t}{4n}} \ud s_2 \, \left(1+s_2^{-\frac \beta 2}\right)\times \cdots \\
& \times \int_{\frac{t}{8n}}^{\frac{t}{4n}} \ud s_{n-1} \, \left(1+s_{n-1}^{-\frac \beta 2}\right)
\times \int_{\frac{t}{8n}}^{\frac{t}{4n}} \ud s_n \, \left(1+s_n^{-\frac \beta 2}\right)\\
& \eqqcolon e^{-10c_2} e^{-16c_2\frac{|x-x'|^2}{t}} e^{-12c_2 \frac{|x-y|^2+|x'-y'|^2}{t}} \prod_{i=1}^{n} I_i.
\end{align*}

The above $\ud s_2\cdots\ud s_n$ integrals can be evaluated explicitly, which
is equal to

\begin{align*}
\prod_{i=2}^{n} I_i
= \left(\frac{t}{8n} + \frac{\left(\frac{t}{4n}\right)^{1-\frac \beta 2}-\left(\frac{t}{8n}\right)^{1-\frac \beta 2}}{1-\frac \beta 2}\right)^{n-1}
\ge c^{n-1} \left(\frac{t}{n}\right)^{n-1} \left(1+\left(\frac{t}{n}\right)^{-\frac \beta 2}\right)^{n-1}.
\end{align*}

As for the $\ud s_1$ integral, we have that

\begin{align*}
I_1 \ge \int_{(1- \frac{1}{4n})\frac t 2}^{\frac{t}{2}} \ud s_1 \, \left(1 + \left(\frac{t- t/2}{n}\right)^{-\frac \beta 2}\right)
\ge c \: \frac{t}{n} \left(1+\left(\frac{t}{n}\right)^{-\frac \beta 2}\right).
\end{align*}

Therefore, we see that

\begin{align*}
I & \ge c^n n^{-\beta/2} e^{-10c_2} e^{-16c_2\frac{|x-x'|^2}{t}} e^{-12c_2 \frac{|x-y|^2+|x'-y'|^2}{t}} \left(\frac tn\right)^n \left(1 + \left(\frac{t}{n}\right)^{-\beta/2} \right)^n \\
& = c^n n^{-\beta/2} e^{-10c_2} e^{-16c_2\frac{|x-x'|^2}{t}} e^{-12c_2 \frac{|x-y|^2+|x'-y'|^2}{t}} \sum_{k = 0}^n \frac{n!}{k!(n-k)!} \left(\frac tn\right)^{n-k\beta/2}.
\end{align*}

By Stirling's formula, $\frac{n^{k\beta/2}}{n^n} \ge C^n
\frac{(k!)^{\beta/2}}{n!}$ for all $k = 0, 1, \dots, n$ and $n \ge 2$. It
follows that

\begin{align*}
\widetilde{G}_D^{\triangleright (n+1)}(t, x, x', y, y')
\ge C_\eps^{2(n+1)} c^n \frac{e^{-2\mu_1 t}}{1 \wedge t^d} e^{-16c_2\frac{|x-x'|^2}{t}}e^{-12c_2 \frac{|x-y|^2 + |x'-y'|^2}{t}} \sum_{k = 0}^n \frac{t^{n-k\beta/2}}{(n-k)! (k!)^{1-\beta/2}}.
\end{align*}

Finally, recalling the definition of $\mathcal{K}^\lambda_D$, interchanging
the order of summation, and using (<ref>), we get that

\begin{align*}
\MoveEqLeft \mathcal{K}^\lambda_D(t, x, x', y, y')
= \lambda^2 \sum_{n = 0}^\infty \lambda^{2n} \widetilde{G}_D^{\triangleright (n+1)}(t, x, x', y, y')\\
& \ge C_\eps^2 \lambda^2 \frac{e^{-2\mu_1 t}}{1 \wedge t^d}e^{-16c_2\frac{|x-x'|^2}{t}} e^{-12c_2\frac{|x-y|^2 + |x'-y'|^2}{t}}
\sum_{n = 0}^\infty {(C_\eps^2 c\lambda^2 t)}^n
\sum_{k = 0}^n \frac{t^{-k\beta/2}}{(n-k)! (k!)^{1-\beta/2}}\\
& \ge C_\eps^2 C \lambda^2 \frac{e^{-2\mu_1 t}}{1 \wedge t^d} e^{-16c_2\frac{|x-x'|^2}{t}}e^{-12c_2\frac{|x-y|^2 + |x'-y'|^2}{t}}
e^{t\Big(K \lambda^2 + K' \lambda^{\frac{4}{2-\beta}} \Big)}.
\end{align*}

The proof of Proposition <ref> is complete.

Let $\dom$ be a bounded Lipschitz domain. Then:

(i) There exist positive finite constants $C$, $c$, $c'$ such that
for all $t > 0$, for all $x, x', y, y' \in \dom$,

\begin{equation}\label{K-n-ub}
\mathcal{K}_N^\lambda(t, x, x', y, y')
\le \frac{C\lambda^2}{1\wedge t^d} e^{-c_3\left(\frac{|x-y|^2}{t} + \frac{|x'-y'|^2}{t}\right)}
e^{t\Big(c\lambda^2 + c' \lambda^{\frac{4}{2-\beta}}\Big)}.
\end{equation}

(ii) If $\dom$ is convex (or if (<ref>) holds), then there
exist positive finite constants $\overline{C}$, $\overline{c}$,
$\widetilde{c}$ such that for all $t > 0$, for all $x, x', y, y' \in
\dom$,

\begin{equation}\label{K-n-lb}
\mathcal{K}^\lambda_N(t, x, x', y, y')
\ge \frac{\overline{C} \lambda^2}{1\wedge t^d} e^{-16c_4 \frac{|x-x'|^2}{t}} e^{-12c_4\left(\frac{|x-y|^2}{t} + \frac{|x'-y'|^2}{t}\right)}
e^{t\Big(\overline{c}\lambda^2 + \widetilde{c} \lambda^{\frac{4}{2-\beta}}\Big)}.
\end{equation}

The proof is similar to that of Lemma <ref>.

§.§ Proof of Theorem <ref>

The correlation bounds (<ref>) and (<ref>) under Dirichlet
condition (or Neumann condition, respectively) follows immediately from
Propositions <ref> and <ref> (or <ref>
and <ref>, respectively).

§ THE CASE OF BOUNDED C ONE ALPHA-DOMAINS WITH DIRICHLET CONDITION AND SOME

In the case of $C^{1, \alpha}$-domains, one can expect better
moment estimates under Dirichlet boundary condition because the heat kernel
estimates in (<ref>) hold with $a_1 = a_2 = 1$. This implies that the
Dirichlet heat kernel estimates in (<ref>) are sharp, yielding matching
upper and lower bounds with the same factor $\left(1\wedge \frac{\Phi_1(x)}{1
\wedge t^{1/2}}\right)\left(1 \wedge \frac{\Phi_1(y)}{1 \wedge t^{1/2}}\right)$.

§.§ Better estimates for the resolvent kernel K

In this part, we study the case of bounded $C^{1, \alpha}$-domains with
Dirichlet boundary condition. For the points $x, x', y, y'$ that are close to
$\partial \dom$, the lemma below provides more precise estimate than what one
gets from the upper bound in Lemma <ref>.

If $\dom$ is a bounded $C^{1, \alpha}$-domain for some $\alpha > 0$, then
there exists a positive finite constant $C$ such that for all $t > 0$, for all
$x, x', y, y' \in \dom$,

\begin{align}\label{E:G^2}
&\int_0^t \ud s \iint_{\dom^2} \ud z \, \ud z' \, \Psi(t-s, x) \Psi(t-s, x') \Psi(t-s, z) \Psi(t-s, z')
\frac{e^{-2\mu_1 (t-s)}}{1 \wedge (t-s)^d} e^{-c_1 \frac{|x-z|^2 + |x'-z'|^2}{t-s}}\notag\\
&\qquad\times f(z-z') \Psi(s, z) \Psi(s, z') \Psi(s, y) \Psi(s, y') \frac{e^{-2\mu_1 s}}{1 \wedge s^d}
e^{-\frac{2c_1}{3} \frac{|z-y|^2 + |z'-y'|^2}{s}}\notag\\
&\le C \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y') \frac{e^{-2\mu_1 t}}{1 \wedge t^d}
e^{-\frac{2c_1}{3}\frac{|x-y|^2 + |x'-y'|^2}{t}} \int_0^t \left(1 \wedge \frac{(t-s)s}{t}\right)^{-\beta/2} \ud s,
\end{align}

where $\Psi(t, x)$ is as defined in (<ref>).

We first derive the following bound: for all $s, r > 0$ and $v, w \in \dom$,

\begin{align}\label{Psi-bd}
\Psi(s, v) \Psi(r, w) e^{-\frac{c_1}{6} \frac{|v-w|^2}{r}} \le C_0 \Psi(s, w).
\end{align}

Since $\dom$ is a $C^{1, \alpha}$-domain for some $\alpha > 0$, (<ref>)
holds with $a_1 = a_2 = 1$. By (<ref>) and the triangle inequality,

\begin{align*}
\Phi_1(v) \le c_0(\op{dist}(w, \partial \dom) + |v-w|) \le c_0^2\Phi_1(w) + c_0|v-w|.
\end{align*}

It follows that

\begin{align*}
\Psi(s, v) = 1 \wedge \frac{\Phi_1(v)}{1\wedge s^{1/2}}
& \le 1 \wedge \left(\frac{c_0^2\Phi_1(w)+c_0|v-w|}{\Phi_1(w)}\cdot \frac{\Phi_1(w)}{1 \wedge s^{1/2}}\right) \\
& \le \left(c_0^2 + \frac{c_0|v-w|}{\Phi_1(w)}\right) \Psi(s, w).
\end{align*}

\begin{align*}
\Psi(s, v) \Psi(r, w)
& \le \left(c_0^2 + \frac{c_0|v-w|}{\Phi_1(w)}\right) \left(1\wedge \frac{\Phi_1(w)}{1\wedge r^{1/2}}\right) \Psi(s, w) \\
& \le \left(c_0^2 + \frac{c_0|v-w|}{1 \wedge r^{1/2}}\right) \Psi(s, w).
\end{align*}

Moreover, since the function $xe^{-x^2}$ is bounded, we have

\begin{align*}
\Psi(s, v) \Psi(r, w) e^{-\frac{c_1}{6} \frac{|v-w|^2}{r}}
& \le \left(c_0^2 + c_0 |v-w| + \frac{c_0|v-w|}{r^{1/2}}e^{-\frac{c_1}{6} \frac{|v-w|^2}{r}}\right) \Psi(s, w) \\
& \le C_0 \Psi(s, w),
\end{align*}

where $C_0$ is a finite constant depending on $c_0$, $c_1$ and the diameter of
$\dom$. This proves (<ref>).

In order to prove the lemma, we split the integral over $[0, t]$ into the sum
of the integral over $[t/2, t]$ and the integral over $[0, t/2]$. Denote these
two integrals by $I_1$ and $I_2$, respectively. By (<ref>), we have the
following two bounds

\begin{align*}
\Psi(t-s, x)\Psi(s, z) e^{-\frac{c_1}{3}\frac{|x-z|^2}{t-s}}     & \le C_0 \Psi(s, x), \\
\Psi(t-s, x')\Psi(s, z') e^{-\frac{c_1}{3}\frac{|x'-z'|^2}{t-s}} & \le C_0 \Psi(s, x').
\end{align*}

These bounds and $\Psi(t-s, z) \Psi(t-s, z') \le 1$ imply that

\begin{align*}
I_1 &\le C_0^2 \int_{t/2}^t \ud s \, \Psi(s, x) \Psi(s, x') \Psi(s, y) \Psi(s, y') \\
& \quad \times \iint_{\dom^2} \ud z \, \ud z' \, \frac{e^{-2\mu_1(t-s)}}{1 \wedge (t-s)^d} e^{-\frac{2c_1}{3} \frac{|x-z|^2 + |x'-z'|^2}{t-s}} f(z-z') \frac{e^{-2\mu_1 s}}{1 \wedge s^d} e^{-\frac{2c_1}{3}\frac{|z-y|^2 + |z'-y'|^2}{s}}.
\end{align*}

Then, using the bound $\Psi(s, x) \le \Psi(t/2, x) \le \sqrt{2} \Psi(t, x)$
for $t/2 \le s \le t$ and Lemma <ref>, we obtain

\begin{align*}
I_1 \le C \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y) \frac{e^{-2\mu_1 t}}{1 \wedge t^d}
e^{-\frac{2c_1}{3} \frac{|x-y|^2 + |x'-y'|^2}{t}} \int_{t/2}^t \left(1 \wedge \frac{(t-s)s}{t}\right)^{-\beta/2} \ud s.
\end{align*}

For $I_2$, we use the bounds

\begin{align*}
\Psi(t-s, z)\Psi(s, y) e^{-\frac{c_1}{6}\frac{|z-y|^2}{s}}     & \le C_0 \Psi(t-s, y), \\
\Psi(t-s, z')\Psi(s, y') e^{-\frac{c_1}{6}\frac{|z'-y'|^2}{s}} & \le C_0 \Psi(t-s, y'),
\end{align*}

and $\Psi(s, z) \Psi(s, z') \le 1$ to get that

\begin{align*}
I_2 &\le C_0^2 \int_0^{t/2} \ud s\, \Psi(t-s, x)\Psi(t-s, x') \Psi(t-s, y) \Psi(t-s, y')\\
& \quad \times \iint_{\dom^2} \ud z \, \ud z' \, \frac{e^{-2\mu_1(t-s)}}{1 \wedge (t-s)^d} e^{-c_1 \frac{|x-z|^2+|x'-z'|^2}{t-s}} f(z-z') \frac{e^{-2\mu_1 s}}{1 \wedge s^d} e^{-c_1 \frac{|z-y|^2+|z'-y'|^2}{2s}}.
\end{align*}

For $0 \le s \le t/2$, we apply the identity (<ref>) with $t$ and $s$
replaced by $t' = t+s$ and $s' = 2s$ respectively, and with $v = z-x$ and $w =
y-x$. This gives

\begin{align*}
e^{-c_1 \frac{|x-z|^2+|x'-z'|^2}{t-s}}e^{-c_1 \frac{|z-y|^2+|z'-y'|^2}{2s}}
& = e^{-c_1 \frac{|x-y|^2 + |x'-y'|^2}{t+s}} e^{-c_1 \frac{|(z-x) - \frac{t'-s'}{t'}(y-x)|^2 + |(z'-x') - \frac{t'-s'}{t'}(y'-x')|^2}{\tau'}} \\
& \le e^{-\frac{2c_1}{3} \frac{|x-y|^2 + |x'-y'|^2}{t}} e^{-c_1 \frac{|(z-x) - \frac{t'-s'}{t'}(y-x)|^2 + |(z'-x') - \frac{t'-s'}{t'}(y'-x')|^2}{\tau'}},
\end{align*}

where $\tau' = (t'-s')s'/t'$.
It is easy to see that for $0 \le s \le t/2$,

\begin{align*}
(1 \wedge \tau')^{d-\beta/2}\le 2^d \left(1\wedge \frac{(t-s)s}{t}\right)^{d-\beta/2}.
\end{align*}

Hence, we can follow the proof of Lemma <ref>(i) and use the
bound $\Psi(t-s, x) \le \sqrt{2} \Psi(t, x)$ for $0 \le s \le t/2$ to deduce

\begin{align*}
I_2 \le C' \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y) \frac{e^{-2\mu_1 t}}{1 \wedge t^d}
e^{-\frac{2c_1}{3} \frac{|x-y|^2 + |x'-y'|^2}{t}} \int_0^{t/2} \left(1 \wedge \frac{(t-s)s}{t}\right)^{-\beta/2} \ud s.
\end{align*}

The proof of Lemma <ref> is complete.

We can now strengthen the bounds (<ref>) and (<ref>) for
$\mathcal{K}^\lambda_D(t, x, x', y, y')$ in the case of bounded $C^{1,
\alpha}$-domains. Note that $\Psi(t, x) = 0$ for all $x \in \partial \dom$,
hence (<ref>) and (<ref>) below provide more precise estimates
than (<ref>) and (<ref>) for $x, x', y, y'$ that are close to
$\partial \dom$.

If $\dom$ is a bounded $C^{1, \alpha}$-domain for some $\alpha > 0$. Then:

(i) There exist positive finite constants $C$, $c$, $c'$ such that for
all $t > 0$ and $x, x', y, y' \in \dom$,

\begin{align}\label{K-d-ub2}
\begin{split}
\mathcal{K}^\lambda_D(t, x, x', y, y')  \le
& \frac{C\lambda^2}{1 \wedge t^d} \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y') \\
& \times e^{-\frac{2c_1}{3}\left(\frac{|x-y|^2}{t} + \frac{|x'-y'|^2}{t}\right)}
e^{2t\left(c \lambda^2 + c' \lambda^{\frac{4}{2-\beta}}-\mu_1\right)}.
\end{split}
\end{align}

(ii) There exist positive finite constants $\bar C$, $\bar c$, $\tilde
c$ such that for all $t > 0$, for all $x, x', y, y' \in \dom$,

\begin{align}\label{K-d-lb2}
\begin{split}
\mathcal{K}^\lambda_D(t, x, x', y, y') \ge
& \frac{\bar C \lambda^2}{1 \wedge t^d} \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y') \\
& \times e^{-16c_2 \frac{|x-x'|^2}{t}} e^{-12c_2 \left(\frac{|x-y|^2}{t} + \frac{|x'-y'|^2}{t} \right)} e^{2t\left(\bar c\lambda^2 + \tilde c \lambda^{\frac{4}{2-\beta} - \mu_1}\right)}.
\end{split}
\end{align}

(i). Similarly to the proof of Proposition <ref>(i), the upper
bound (<ref>) can be proved by applying Lemmas <ref>,
<ref> and <ref>.

(ii). To prove the lower bound (<ref>), we claim that for each $n \ge

\begin{align}\label{E:Gn-d-lb}
\begin{split}
\widetilde{G}^{\triangleright (n+1)}_D(t, x, x', y, y') & \ge
C^{n+1} \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y') \\
& \quad \times \frac{e^{-2\mu_1 t}}{1 \wedge t^d} e^{-16c_2\frac{|x-x'|^2}{t}} e^{-12c_2 \frac{|x-y|^2+|x'-y'|^2}{t}}
\sum_{k=0}^n \frac{t^{n-k\beta/2}}{(n-k)!(k!)^{1-\beta/2}}.
\end{split}
\end{align}

Indeed, for $n = 0$, (<ref>) follows from Proposition <ref>(i).
For $n \ge 1$, we can prove (<ref>) by modifying the proof in
Proposition <ref>(ii) and we outline the major changes as follows.
Instead of $V(x)$ defined in (<ref>), we now consider

\begin{align*}
\widetilde{V}(x) =
\begin{cases}
\overline{\mathscr{C}(x+(\eps_0/4)\xi_{y_i}, \xi_{y_i}, \eps_0/4)} & \text{in case (1) of~\eqref{E:V(x)}}, \\
\overline{B(x, \eps_0/4)}                                          & \text{in case (2) of~\eqref{E:V(x)}},
\end{cases}
\end{align*}

so that $\widetilde{V}(x) \subset V(x) \subset \dom$ and there exists
$\delta_0 > 0$ such that for all $x \in \dom$, for all $z \in
\widetilde{V}(x)$, $\operatorname{dist}(z, \partial \dom) \ge \delta_0$.
Hence, (<ref>) implies that

\begin{align}\label{E:inf-Phi}
\inf_{x \in \dom} \inf_{z \in \widetilde{V}(x)}\Phi_1(z) \ge c_0^{-1}\delta_0 > 0.
\end{align}

For any $x \in \dom$, define $\tilde{x}$ by

\begin{align*}
\tilde{x} =
\begin{cases}
x+(\eps_0/4)\xi_{y_i} & \text{in case (1) of }\eqref{E:V(x)}, \\
x & \text{in case (2) of }\eqref{E:V(x)}.
\end{cases}
\end{align*}

Then, as in Lemma <ref>, we can show that there exists $C_0 > 0$ such
that for all $t > 0$, for all $x \in \dom$, for all $0 \le r \le s \le
\eps_0/4$, for all $z \in \widetilde{V}(x) \cap B(\tilde{x}, s)$,

\begin{align}\label{E:vol-Vtilde}
\operatorname{Vol}(\widetilde{V}(x) \cap B(\tilde{x}, s) \cap B(z, r)) \ge C_0 (1 \wedge r)^d.
\end{align}

For $n = 1$, similarly to the proof in Proposition <ref>(ii), we have

\begin{align*}
\MoveEqLeft \widetilde{G}_D^{\triangleright 2}(t, x, x', y, y')\\
& \ge \int_{t/4}^{t/2} \ud s \iint_{[\widetilde{V}(x) \cap B(\tilde{x}, \sqrt{t})]^2} \ud z\, \ud z'\,
\widetilde{G}_D(t-s, x, x', z, z')f(z-z') \widetilde{G}_D(s, z, z', y, y')\\
& \ge C \Psi(t, x) \Psi(t, x') \Psi(t, y) \Psi(t, y') \frac{e^{-2\mu_1 t}}{(1\wedge t^d)^2}
e^{-16 c_2 \frac{|x-x'|^2}{t}} e^{-12 c_2 \frac{|x-y|^2 + |x'-y'|^2}{t}}\\
& \quad \times \int_{t/4}^{t/2} \ud s \iint_{[\widetilde{V}(x) \cap B(\tilde{x}, \sqrt{t})]^2} \ud z\, \ud z' \, (1 \wedge \Phi_1(z))^2 (1 \wedge \Phi_1(z'))^2 |z-z'|^{-\beta}.
\end{align*}

Then, by using (<ref>) and (<ref>), we
obtain (<ref>) for $n=1$.

For $n \ge 2$, we first restrict the integrals of $s_1, s_2, s_3, \dots, s_n$
to the smaller intervals   $\left[(1-\frac{1}{4n})\frac t 2, \frac t
2\right]$, $\left[(1-\frac{1}{2n})\frac t 2, s_1-\frac{t}{8n}\right]$,
$\left[(1-\frac{2}{2n})\frac{t}{2}, s_2 - (s_1-s_2) \right]$, $\dots$,
$\left[(1-\frac{n-1}{2n})\frac t 2, s_{n-1}-(s_{n-1}-s_{n-1})\right]$,
respectively, so that $\frac{t-s_1}{5n} \le \frac{t}{8n} \le s_1-s_2 \le
s_2-s_3 \le \dots \le s_{n-1}-s_n$. We can then modify the proof of
Proposition <ref>(ii) by considering

z_1, z_1'     ∈A_1           V(x) ∩B(x̃, √((t-s_1)/(5n)) ∧(_0/4)),

z_2           ∈A_2           V(x) ∩B(x̃, √(s_1-s_2) ∧(_0/4)) ∩B(z_1 , √(s_1-s_2) ∧(_0/4)),

z_2'          ∈A_2'          V(x) ∩B(x̃, √(s_1-s_2) ∧(_0/4)) ∩B(z_1', √(s_1-s_2) ∧(_0/4)),

z_n           ∈A_n           V(x) ∩B(x̃, √(s_n-1-s_n) ∧(_0/4)) ∩B(z_n-1 , √(s_n-1-s_n) ∧(_0/4)),

z_n'          ∈A_n'          V(x) ∩B(x̃, √(s_n-1-s_n) ∧(_0/4)) ∩B(z_n-1', √(s_n-1-s_n) ∧(_0/4)).

Then, along the lines of the proof of Proposition <ref>(ii), we can
deduce (<ref>) using (<ref>) and (<ref>)

Finally, recall that

\begin{align*}
\mathcal{K}^\lambda_D(t, x, x', y, y')
= \lambda^2 \sum_{n=0}^\infty \lambda^{2n} \widetilde{G}_D^{\triangleright (n+1)}(t, x, x', y, y').
\end{align*}

By using (<ref>), interchanging the order of summation, and
using (<ref>), we can obtain the lower bound (<ref>) as in
the proof of Proposition <ref>(ii). This completes the proof of
Proposition <ref>.

§.§ Proof of Theorem <ref>

In Remark <ref>, we have seen that $\nu$ satisfies
condition (<ref>) if and only if $J_{2c_1/3}^* (t, x) < \infty$
for all $t > 0$ and $x \in \dom$. The proof of the existence, $p$-th moment
bounds and uniqueness of the solution is similar to the proof of
Theorem <ref> in Section <ref>, with the use of
Lemma <ref> instead of Lemma <ref> (i). The correlation
bounds follow immediately from Propositions <ref>
and <ref>.

§.§ Proof of Corollary <ref>

Note that the Dirichlet kernel kernel for $\mathscr{L}$ on $\dom = \prod_{i =
1}^m \dom_i$ is given by

\begin{align*}
G_D^\dom(t, x, y) = \prod_{i = 1}^m G_D^{\dom_i}(t, x_i, y_i)
\end{align*}

for $t > 0$ and $x = (x_1, \dots, x_m)$, $y = (y_1, \dots, y_m) \in \dom$,
where $G_D^{\dom_i}(t, x_i, y_i)$ is the Dirichlet heat kernel for
$\mathscr{L}_i$ on the $C^{1, \alpha_i}$-domain $\dom_i$. This and
Proposition <ref> applied to each $G_D^{\dom_i}(t, x_i, y_i)$ imply the
following heat kernel estimate

\begin{align*}
G_D^\dom(t, x, y) \le C_1 \Psi^* (t, x)\Psi^* (t, y) \frac{e^{-\mu_1 t}}{1 \wedge t^{d/2}}
e^{-c_1 \frac{|x-y|^2}{t}}
\end{align*}

with suitable constants $C_1$ and $c_1$, where $\Psi^*$ is as defined
in (<ref>) and $\mu_1$ is the sum of the leading eigenvalues of the
Dirichlet operators $\mathscr{L}_i$. By the proof of (<ref>) in
Lemma <ref>, for each $i$, we have

\begin{align*}
\left(1 \wedge \frac{\Phi_1^{\dom_i}(v_i)}{1 \wedge s^{1/2}}\right)
\left(1 \wedge \frac{\Phi_1^{\dom_i}(w_i)}{1 \wedge r^{1/2}}\right)
e^{-\frac{c_1}{6} \frac{|v_i-w_i|^2}{r}}
\le C_{0, i} \left(1 \wedge \frac{\Phi_1^{\dom_i}(w_i)}{1 \wedge s^{1/2}}\right)
\end{align*}

for all $s, r > 0$ and $v_i, w_i \in \dom_i$, where $C_{0, i}$ is a constant.
Taking products over $i \in \{1, \dots, m\}$ gives

\begin{align*}
\Psi^* (s, v) \Psi^* (r, w) e^{-\frac{c_1}{6} \frac{|v-w|^2}{r}} \le C_0 \Psi^* (s, w)
\end{align*}

for all $s, r > 0$ and $v, w \in \dom$, where $C_0 = \prod_{i = 1}^m C_{0,i}$.
Then, the same proof of Lemma <ref> shows that the
estimate (<ref>) holds with every $\Psi$ replaced by $\Psi^*$. This
implies that the estimate (<ref>) also holds with every $\Psi$
replaced by $\Psi^*$. Using these estimates, the statements (i) and (ii) in
Theorem <ref> with $\Psi$ replaced by $\Psi^*$ can be proved for the
product domain the same way they are proved in Theorem <ref>.
Condition (<ref>) ensures that $J^*_c(t, x) < \infty$ for all $t
> 0$ and $x \in \dom$.

§.§ Some auxiliary results related to examples in Section <ref>

For $\nu > -1$, let $J_\nu(\cdot)$ be the Bessel function of first kind and
of order $\nu$ and let $z_0$ be any positive zero of $J_\nu(\cdot)$. Then,

\begin{align*}
\sup_{r\in (0,1)} \frac{r^{-\nu} J_\nu(r z_0)}{1-r}<\infty.
\end{align*}

Set $f(r) = \frac{r^{-\nu} J_\nu(r z_0)}{1-r}$, which is a continuous
function on $(0,1)$. By Eq. 10.7.3 of <cit.>, we see
that $\lim_{r\to 0} f(r) = \Gamma(\nu)^{-1}
\left(\frac{z_0}{2}\right)^\nu<\infty$. Because all zeros of $J_\nu(\cdot)$
are simple (see Section 10.21 ibid), we see that $\lim_{r \to 1}f(r)
= -z_0 J_\nu'(z_0) <\infty$. Therefore, $f(r)$ is a continuous function on
$[0,1]$, from which the desired result follows.

In the setting of Example <ref>, the leading eigenfunction is given by

\begin{align*}
\Phi_1(x) = \frac{1}{C_d} |x|^{(2-d)/2} J_{(d-2)/2}\left(z_0|x|\right),
\end{align*}

where $z_0$ is the first positive zero of the Bessel function
$J_{(d-2)/2}(x)$, and $C_d$ is some normalization constant. One may refer to
34.2 in Chapter III of <cit.> for the case $d = 2$ and
Section H in Chapter 2 of <cit.> for the general case
$d \ge 3$. In particular, the leading eigenfunction $\Phi_1(x)$ corresponds to
$F_k^{lm}(x)$ in Theorem 2.66 (ibid.) with $k = 0$, $l = 1$ and $m = 1$.
Here are a few comments:

* The multiplicity for the leading eigenvalue $z_0$ is one since $d_k =
1$; see Corollary 2.55 (ibid.).

* $Y_0^1(x)$ in Theorem 2.66 (ibid.) is a constant function.

* In Figure <ref>, we have chosen the following normalization

\begin{align}\label{E:Cd}
C_d =
\lim_{r\to 0} r^{(2-d) /2} J_{(n-2) /2}(z_0 r) = 2^{-d/2} d\,
\Gamma\left(1+d/2\right)^{-1} z_0^{(d-2) /2},
\end{align}

where the limit is due to <cit.>, so that $\max_{|x|\le 1}\Psi(1,x) = 1$.

The following Lemma is used in Example <ref>.

$R_1$ and $R_2$ are two simple zeros for $Z(r)$ defined in Example <ref>.

It is straightforward to check that $R_1$ and $R_2$ are two zeros of $Z(r)$.
To show that they are simple, one needs to prove that $Z'(r)\ne 0$ for $r =
R_1$ and $R_2$. By Eq. 10.6.3 of <cit.>, we see that

\begin{align*}
z_0^{-1} Z'(r)
& =   J_0(R_1 z_0) Y_0'(r z_0)
- J_0'(r z_0)  Y_0(R_1 z_0)  \\
& = - J_0(R_1 z_0) Y_1(r z_0)
+ J_1(r z_0)  Y_0(R_1 z_0).
\end{align*}

By setting $r = R_1$ and applying Eq. 10.5.2 (ibid.), we have that

\begin{align*}
z_0^{-1} Z'(R_1)
& = - J_0(R_1 z_0) Y_1(R_1 z_0)
+ J_1(R_1 z_0) Y_0(R_1 z_0)
= \frac{2}{\pi R_1 z_0} \ne 0,
\end{align*}

which proves that $R_1$ is a simple zero of $Z(r)$.

Let $F(z)$ denote the function in (<ref>). By setting $r = R_2$, we

\begin{align}\label{E:Z'R2}
z_0^{-1} Z'(R_2)
& = - J_0(R_1 z_0) Y_1(R_2 z_0)
+ J_1(R_2 z_0) Y_0(R_1 z_0).
\end{align}

Because $J_0^2 (z) + Y_0^2 (z) > 0$ for all $z>0$ (see, e.g., Eq. 10.9.30
ibid), we see that if $Y_0(R_1z_0) = 0$, then $J_0(R_1z_0) \ne 0$.
Because all zeros of $Y_\nu(\cdot)$ are simple, we see that $Y_1(R_1z_0) =
-Y_0'(R_1z_0) \ne 0$. But since $z_0$ is a zero of $F(z)$, i.e.,

\begin{align*}
0 & = J_0(R_1z_0) Y_0(R_2z_0) -
J_0(R_2z_0) Y_0(R_1z_0)
= J_0(R_1z_0) Y_0(R_2z_0).
\end{align*}

Hence, $Y_0(R_2z_0) = 0$, which further implies that both $J_0(R_2z_0)\ne 0$
and $Y_1(R_2z_0) = - Y_0'(R_2z_0) \ne 0$. This proves Case 1 in
Table <ref>. Applying the same arguments with $Y_0(R_1z_0) = 0$
replaced by $Y_0(R_2z_0) = 0$, $J_0(R_1z_0) = 0$, and $J_0(R_2z_0) = 0$, we
see that only three cases can happen, which are listed in the following
Table <ref>:

[subtable]style = default, margin = 0pt, parskip = 0pt, hangindent = 0pt, labelformat = simple

Three cases in the proof of Lemma <ref>

[][Case 1.]

$J_0$       $J_1$     $Y_0$     $Y_1$
$R_1 z_0$     $\ne 0$     —       $= 0$     $\ne 0$

$R_2 z_0$     $\ne 0$     —       $= 0$     $\ne 0$

[][Case 2.]

$J_0$     $J_1$       $Y_0$       $Y_1$
$R_1 z_0$     $= 0$     $\ne 0$     $\ne 0$     —

$R_2 z_0$     $= 0$     $\ne 0$     $\ne 0$     —

[][Case 3.]

$J_0$       $J_1$     $Y_0$       $Y_1$
$R_1 z_0$     $\ne 0$     —       $\ne 0$     —

$R_2 z_0$     $\ne 0$     —       $\ne 0$     —

Cases 1 & 2: From the expression of $Z'(R_2)$
in (<ref>), we see that in these two cases, $Z'(R_2) \ne 0$.

Case 3: Since both $Y_0(R_1z_0)$ and $Y_0(R_2z_0)$ are
nonzero, and since $F(z_0) = 0$, we see that

\begin{align*}
J_0(R_1 z_0) = \frac{Y_0(R_1 z_0) J_0(R_2 z_0)}{Y_0(R_2 z_0)}.
\end{align*}

This and Eq. 10.5.2 (ibid.) imply that

\begin{align*}
z_0^{-1} Z'(R_2)
& = \frac{Y_0(R_1z_0)}{Y_0(R_2z_0)}
\left(-J_0(R_2z_0)Y_1(R_2z_0) + J_1(R_2z_0)Y_0(R_2z_0)\right)\\
& = \frac{Y_0(R_1z_0)}{Y_0(R_2z_0)} \frac{2}{\pi R_2z_0} \ne 0.
\end{align*}

Therefore, combining the above three cases, we have proved that $R_2$ is a
simple zero of $Z(r)$.

Acknowledgments C. Y. Lee is partially supported by Taiwan's National
Science and Technology Council grant NSTC111-2115-M-007-015-MY2.
L. Chen is partially supported by NSF DMS-2246850 and a
collaboration grant from Simons Foundation (959981). The authors would also
like to thank the anonymous referee for the careful reading and helpful
suggestions, which have greatly improved the quality of the paper.

Missing 'biblatex' package
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