# Distance Critical Graphs

Joshua Cooper and Gabrielle Tauscheck

###### Abstract

In 1971, Graham and Pollak provided a formula for the determinant of the
distance matrix of any tree on $n$ vertices. Yan and Yeh reproved this by
exploiting the fact that pendant vertices can be deleted from trees without
changing the remaining entries of the distance matrix. Considering failures of
their argument to generalize invites the question: which graphs have the
property that deleting any one vertex results in a change to some pairwise
distance? We refer to such worst-case graphs as “distance critical”. This work
explores the structural properties of distance critical graphs, preservation
of distance-criticality by products, and the nature of extremal distance
critical graphs. We end with a few open questions.

##  1 Introduction

Graham and Pollak ([2]) famously showed that the determinant of the distance
matrix of a tree $T$ on $n$ vertices – the $n\times n$ matrix whose each
$(v,w)\in V(T)\times V(T)$ entry is the ordinary graph distance between $v$
and $w$ – depends only on $n$. In fact, they gave a formula which, strikingly,
does not depend on the tree except via $n$: $-(n-1)(-2)^{n-2}$. These results
spawned several generalizations and new directions in subsequent years; see
[1] for an extensive survey on the topic.

Some of this research is concerned with finding new proofs of the Graham-
Pollak Theorem. One such example that largely inspired the present work is
[5], providing an elegant reproof that relies on the fact that deleting
pendant vertices from a tree causes all remaining pairwise distances to remain
unchanged. In order to consider the limits of their methods, it is natural to
ask which graphs have the property that no vertex can be deleted without
altering the distance metric on the remaining vertices. The question can also
be viewed as determining whether there exists a maximal proper induced
subgraph $H$ of a graph $G$ such that $H$ embeds isometrically into $G$. This
paper introduces “distance critical” graphs, which are characterized by a lack
of such subgraphs, and studies some of their properties. In Section 2, we give
the formal definition and give a few preliminary results. Section 3 contains
more in-depth analysis of the structure of distance critical graphs, and
Section 4 investigates how distance criticality interacts with standard graph
products. Then, Section 5 delves into properties of extremal distance critical
graphs. Numerous questions remain unanswered, of which we list a few
interesting ones in the final Section 6.

##  2 Preliminaries

We begin with several useful definitions.

######  Definition 2.1.

Given a graph $G=(V(G),E(G))$, the distance $d_{G}(x,y)$ between two vertices
$x,y\in V(G)$ is the length of the shortest path whose endvertices are $x$ and
$y$, or $\infty$ if there is no such path. A shortest path between two
vertices is called a geodesic path.

######  Definition 2.2.

A graph $G$ is distance critical if, for each $v\in V(G)$, there exist $x,y\in
G-v$ so that $d_{G}(x,y)\neq d_{G-v}(x,y)$.

For an example of a distance critical graph, see Figure 1. Below, we present a
useful reformulation of distance criticality that draws on the notion of
“determining pairs”.

######  Definition 2.3.

A pair of vertices $a,b$ is a determining pair for $v$ if $a$ and $b$ are
distinct and nonadjacent, and $v$ is their unique common neighbor.

######  Proposition 2.4.

A connected graph $G$ is distance critical if and only if for all $v\in V(G)$,
$v$ admits a determining pair $\\{a,b\\}$ where $a,b\in V(G)$.

###### Proof.

Let $v\in V(G)$ be arbitrary in a distance critical graph $G$. Note that $v$
must be an internal vertex of some geodesic path $P$ which is the unique
geodesic path connecting its endvertices $x$ and $y$, or else its deletion
would not alter any vertex pair’s distance. Suppose $a$ and $b$ are the
neighbors of $v$ in $P$. Notice that $a$ is not adjacent to $b$; otherwise,
$P$ could be shortened by deleting $v$. Assume there exists another vertex,
$w\neq v$, such that $a$ is adjacent to $w$ and $b$ is adjacent to $w$. Then
the path $xPawbPy$ has the same length as $P$, contradicting the fact that it
was the unique geodesic between $x$ and $y$. Therefore, $v$ must be the unique
common neighbor of some two nonadjacent vertices, as desired.

For the reverse direction, suppose $G$ has the property that every vertex $v$
admits a determining pair $\\{a,b\\}$. Then $d_{G}(a,b)=2$, but
$d_{G-v}(a,b)>2$, so $G$ is distance critical. ∎

---
Figure 1: Dodecahedron

Proposition 2.4 immediately implies the following corollaries.

######  Corollary 2.5.

Distance critical graphs have minimum degree at least $2$.

######  Corollary 2.6.

A graph is distance critical if and only if all its connected components are
distance critical.

Table 1 depicts the number of connected distance critical graphs up to 11
vertices obtained using SageMath ([4]). Clearly, there are no distance
critical graphs on one or two vertices since there are not enough vertices to
form a determining pair. Further, there are no distance critical graphs on
three or four vertices given the restrictions imposed by Proposition 2.4. The
smallest distance critical graph is the cycle on five vertices.

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
---|---|---|---|---|---|---|---|---|---|---|---
$\textrm{A}349402(n)$ | 0 | 0 | 0 | 0 | 1 | 1 | 4 | 15 | 168 | 2,252 | 94,504
Table 1: The number $\textrm{A}349402(n)$ of connected distance critical
graphs on $n$ vertices ([3, A349402]).

##  3 Structural Properties

This section presents results on the structural properties of distance
critical graphs. The girth of a graph $G$ is the length of the shortest cycle
in the graph.

######  Lemma 3.1.

Let $g$ represent the girth of a graph. If graph $G$ has minimum degree at
least $2$ and girth $g>4$, then it is distance critical.

###### Proof.

Let $G$ be a graph of minimum degree at least $2$ which is not distance
critical. Since $G$ is not distance critical, Proposition 2.4 implies that
there exists a vertex $v$ which admits no determining pair. However, $v$ must
have at least $2$ neighbors, call them $a$ and $b$. If $a$ and $b$ are
adjacent, then the graph contains a triangle, so that $g=3$. Therefore, assume
$a$ and $b$ are not adjacent. Since $v$ does not have a determining pair,
there must be another vertex $w\neq v$ adjacent to both $a$ and $b$. Then $G$
contains the $4$-cycle $vawbv$, so that $g=4$. ∎

We say that a connected graph $G$ is $\kappa$-connected if $|V(G)|>\kappa$ and
removing any $\kappa-1$ vertices from $G$ does not disconnect it.

######  Lemma 3.2.

In a $2$-connected distance critical graph, every vertex is contained in a
cycle of length at least $5$.

###### Proof.

Let $v\in V(G)$. Proposition 2.4 guarantees that $v$ has some determining pair
$\\{a,b\\}$. Corollary 2.5 ensures that $a$ has degree at least $2$. Since $a$
and $b$ are nonadjacent, there must exist some other vertex, $x\in
V(G)\setminus\\{a,b,v\\}$, that is adjacent to $a$. This vertex $x$ is not
adjacent to $b$; otherwise, $v$ would not be the unique common neighbor of $a$
and $b$. We can use the same argument to show that there exists a vertex,
$y\in V(G)\setminus\\{a,b,v,x\\}$, that is adjacent to $b$ but not adjacent to
$a$. Now, $G$ is $2$-connected; therefore, there is still a path, $P$, between
$x$ and $y$ in $G-v$. The proof is completed via several cases.

  * •

If $a$ and $b$ do not lie on the path $P$, then $xavbyPx$ is already a cycle
in $G$ with length at least $5$.

  * •

Assume without loss of generality that $b$ lies on $P$ but $a$ does not. Since
$x$ and $b$ are non-adjacent in $G$ and therefore also in $G-v$, some other
vertex, call it $c$, lies between $x$ and $b$ on $P$. The graph $G$ then
contains the cycle $xavbPx$ of length at least $5$.

  * •

Assume both $a$ and $b$ lie on $P$. Since $a$ and $b$ are nonadjacent, some
other vertex, call it $e$, lies between $a$ and $b$ on $P$. If $e$ is the only
other vertex of $aPb$, then $v$ is not the unique common neighbor of $a$ and
$b$. Therefore, the path $aPb$ has length at least $3$ and $G$ contains the
cycle $avbPa$ of length at least $5$.

∎

A dominating vertex is a vertex that is adjacent to all other vertices.

######  Lemma 3.3.

If $G$ is distance critical, then $G$ has no dominating vertex.

###### Proof.

Assume $G$ is a distance critical graph with $v$ a dominating vertex. Let $w$
be some other vertex of the graph. By Proposition 2.4, $w$ has some
determining pair $\\{a,b\\}$. Note that $v\not\in\\{a,b\\}$, since otherwise
$a$ and $b$ would be adjacent. Since $v$ is a dominating vertex, $a$ and $b$
are also adjacent to $v$, contradicting the fact that $w$ is their unique
common neighbor. ∎

######  Lemma 3.4.

If $G$ is a distance critical graph with $x,y\in V(G)$ such that
$d_{G}(x,y)>3$, then $G+xy$ is distance critical as well.

###### Proof.

Suppose $G$ is distance critical and $G+xy$ is not. Then there exists a $v\in
V(G)$ such that $v$ has a determining pair, call it $\\{w,z\\}$, in $G$ but
not in $G+xy$. Two cases must be considered: (1) $wz\in E(G+xy)$ or (2) there
exists another vertex $u\in V(G)\setminus\\{v,w,z\\}$ such that $uw\in
E(G+xy)$ and $uz\in E(G+xy)$.

In case (1), $wz\in E(G+xy)$. Since $\\{w,z\\}$ is a determining pair for $v$
in $G$, $wz\not\in E(G)$, so without loss of generality we may assume $w=x$
and $z=y$. In $G$, $\\{w,z\\}$ is a determining pair for $v$; therefore,
$wv\in E(G+xy)$ and $vz\in E(G+xy)$. Since $w=x$ and $z=y$, this means that
$xvy$ is a $P_{3}$ in $G$ which implies that $d_{G}(x,y)\leq 2$, a
contradiction.

In case (2), both $uw\in E(G+xy)$ and $uz\in E(G+xy)$ for some vertex $u\neq
v$. Without loss of generality, assume $uw\notin E(G)$ but $uw\in E(G+xy)$.
Therefore, $uw=xy$. Since $uw\notin E(G)$, $uz\in E(G)$. Without loss of
generality, assume $u=x$ and $w=y$. Notice that $xzvy$ is a path of length $3$
in $G$, a contradiction. ∎

######  Lemma 3.5.

Let $G$ be a distance critical graph and $v\in V(G)$ such that $F=G-v$ is also
distance critical. If $\deg(v)\leq 3$, then $v$ is involved in some
determining pair.

###### Proof.

Corollary 2.5 ensures $\text{deg}(v)\geq 2$. Assume that $\text{deg}(v)=2$,
and label the neighbors of $v$ as $x$ and $y$. Since $G$ is distance critical,
$\\{x,y\\}$ is the determining pair of $v$; therefore, $xy\notin E(G)$. Since
$F$ is distance critical, $x$ has a determining pair that does not include
$v$, call this determining pair $\\{u,w\\}$ where $u,w\in
V(G)\setminus\\{x,y,v\\}$. We claimed that $\text{deg}(v)=2$; therefore,
$uv\notin E(G)$. This implies that $\\{u,v\\}$ is a determining pair for $x$
unless $uy\in E(G)$, which cannot occur, since otherwise $u$ would be a second
common neighbor of $x$ and $y$.

Assume $\text{deg}(v)=3$, and label the neighbors of $v$ as $x,y$, and $z$.
Since $G$ is distance critical, $v$ has a determining pair. Without loss of
generality, let $\\{x,y\\}$ be the determining pair for $v$; therefore,
$xy\notin E(G)$. Suppose that $v$ is not involved in any determining pair.
Then $x$ has a determining pair in $V(G)\setminus\\{v\\}$. Since $xy\notin
E(G)$, at least one of these vertices, call it $u$, must be distinct from
$\\{x,y,z,v\\}$. Then $u$ is nonadjacent to $y$; otherwise, $\\{x,y\\}$ would
not be a determining pair for $v$. The same argument implies that $y$ has a
neighbor $w\notin\\{x,y,z,v,u\\}$ such that $xw\notin E(G)$. By assumption,
$\text{deg}(v)=3$; therefore, $uv\notin E(G)$ and $wv\notin E(G)$. Since $v$
is not involved in any determining pair, $u$ must be adjacent to a neighbor of
$v$ other than $x$, or else $\\{u,v\\}$ would be a determining pair for $x$.
The only possibility is if $uz\in E(G)$. The same argument implies that $wz\in
E(G)$.

From here, we prove that at least one of the pairs, $xz$ or $yz$ is not an
edge. Indeed, assume $yz\in E(G)$. Then $xz\notin E(G)$ or else $\\{x,y\\}$
would not be a determining pair for $v$. Since $x$ has a determining pair that
does not include vertex $v$, then it has at least one other neighbor $p$
distinct from $\\{x,y,z,u,v,w\\}$. The fact that $\\{x,y\\}$ is a determining
pair for $v$ implies that $py\notin E(G)$. Further, $pv\notin E(G)$ since
$\text{deg}(v)=3$, and $p$ is adjacent to $z$ since $\\{v,p\\}$ is not a
determining pair for $x$. This implies that every neighbor of $x$ is also a
neighbor of $z$. Therefore, $x$ does not have a determining pair, a
contradiction. ∎

Note a useful observation that is employed at the end of the preceding proof:
in any distance critical graph, the family of neighborhoods of vertices is an
antichain with respect to the subset relation.

######  Lemma 3.6.

Let $G$ be a distance critical graph on $n$ vertices and let $S$ be the set of
vertices which are involved in some determining pair. Then $|S|>\sqrt{2n}$.

###### Proof.

Since $G$ is distance critical, each of the $n$ vertices has a determining
pair. The vertex corresponding to a determining pair is unique, so,
$\binom{|S|}{2}\geq n$. Thus, $|S|>\sqrt{2n}$ as desired. ∎

######  Lemma 3.7.

Let $G$ be a distance critical graph and $z\in V(G)$. If $z$ has no
determining pair in $G+xy$ for some $xy\not\in E(G)$, then either $\\{x,y\\}$
is the only determining pair for $z$ or every $z$-determining pair intersects
$\\{x,y\\}$.

###### Proof.

There are two ways for the addition of the edge $xy$ to disrupt the existence
of a $z$-determining pair: (1) $\\{x,y\\}$ is the only determining pair for
$z$, or (2) $xy$ interferes with all determining pairs for $z$, i.e. for every
determining pair $\\{u,v\\}$, $\\{u,v\\}$ is no longer a determining pair for
$z$ in $G+xy$.

In case (2), the edge $xy$ interferes with all determining pairs of $z$. If
the addition of $xy$ destroys the $z$-determining pair $\\{u,v\\}$, then $u$
and $v$ must have a common neighbor $w$ other than $z$ in $G+xy$. Since this
$w$ was not a neighbor of $u$ and $v$ in $G$, either $xy=uw$ or $xy=vw$, and
the conclusion follows. ∎

##  4 Graph Products

######  Definition 4.1.

The Cartesian product of two graphs $G$ and $H$, denoted $G\Box H$, is a graph
such that $V(G\Box H)=V(G)\times V(H)$. Two vertices $(x,y)$ and
$(x^{\prime},y^{\prime})$ are adjacent in $G\Box H$ if and only if either

  * •

$x=x^{\prime}$ and $y\sim y^{\prime}$ or

  * •

$y=y^{\prime}$ and $x\sim x^{\prime}$.

######  Lemma 4.2.

The Cartesian product of a distance critical graph and any other graph is
distance critical.

###### Proof.

Consider $(v,w)\in V(G\Box H)$ with $G$ distance critical and $H$ any other
graph. Suppose $v\in V(G)$ has determining pair $\\{a,b\\}$ in $G$. Then,
$(v,w)$ is adjacent to $(a,w)$ as well as to $(b,w)$. Further, $(a,w)$ and
$(b,w)$ are nonadjacent because $a$ and $b$ are nonadjacent in $G$. Assume
$(x,y)\neq(v,w)$ is also adjacent to $(a,w)$ and $(b,w)$. Since $(x,y)$ is
adjacent to $(a,w)$, either $x=a$ or $y=w$. If $x=a$, then $w$ is adjacent to
$y$. Since $(x=a,y)$ is adjacent to $(b,w)$, we have $y=w$ and $a$ is adjacent
to $b$, a contradiction. Therefore $x\neq a$, and instead, $y=w$. This implies
$x$ is adjacent to $a$; similarly, $x$ is adjacent to $b$. Since $\\{a,b\\}$
is the determining pair of vertex $v$, then $x=v$ and thus $(x,y)=(v,w)$.
Therefore, every vertex of $G\Box H$ has a determining pair and the Cartesian
product is distance critical as claimed. ∎

######  Definition 4.3.

The tensor product of two graphs $G$ and $H$, denoted $G\times H$, is a graph
such that $V(G\times H)=V(G)\times V(H)$, with two vertices $(x,y)$ and
$(x^{\prime},y^{\prime})$ adjacent in $G\times H$ if and only if $x\sim
x^{\prime}$ and $y\sim y^{\prime}$.

######  Lemma 4.4.

If $G$ and $H$ are distance critical, then $G\times H$ is as well.

###### Proof.

Assume both $G$ and $H$ are distance critical and consider $(v,w)\in V(G\times
H)$. Suppose $v\in V(G)$ has determining pair $\\{a,b\\}$, and suppose $w\in
V(H)$ has determining pair $\\{c,d\\}$. Since $a$ and $b$ are adjacent to $v$
and $c$ and $d$ are adjacent to $w$, this implies that $(v,w)$ is adjacent to
both $(a,c)$ and $(b,d)$. Further, $(a,c)$ and $(b,d)$ are nonadjacent because
$a$ and $b$ are nonadjacent in $G$. Assume there is another neighbor $(x,y)$
that is adjacent to both $(a,c)$ and $(b,d)$. This implies $x$ is adjacent to
$a$ and $b$, and $y$ is adjacent to $c$ and $d$. Since $\\{a,b\\}$ and
$\\{c,d\\}$ are determining pairs, this implies that $x=v$ and $y=w$.
Therefore, $(v,w)$ is the unique common neighbor between vertices $(a,c)$ and
$(b,d)$, and the tensor product is distance critical as claimed. ∎

######  Definition 4.5.

The strong product of two graphs $G$ and $H$, denoted $G\boxtimes H$, is a
graph such that $V(G\boxtimes H)=V(G)\times V(H)$, with two distinct vertices
$(x,y)$ and $(x^{\prime},y^{\prime})$ adjacent in $G\boxtimes H$ if and only
if

  * $\circ$

$x=x^{\prime}$ or $x\sim x^{\prime}$ and

  * $\circ$

$y=y^{\prime}$ or $y\sim y^{\prime}$.

######  Lemma 4.6.

If $G$ and $H$ are distance critical, then $G\boxtimes H$ is as well.

###### Proof.

Assume $G$ and $H$ are distance critical graphs and consider $(v,w)\in
V(G\boxtimes H)$. Suppose $v\in V(G)$ has determining pair $\\{a,b\\}$, and
$w\in V(H)$ has determining pair $\\{c,d\\}$. Since $a$ and $b$ are adjacent
to $v$ and $c$ and $d$ are adjacent to $w$, this implies that $(v,w)$ is
adjacent to both $(a,c)$ and $(b,d)$. Further, $(a,c)$ and $(b,d)$ are
nonadjacent because $a$ and $b$ are nonadjacent in $G$. Assume there is
another neighbor $(x,y)$ that is adjacent to both $(a,c)$ and $(b,d)$. Since
$(x,y)$ is adjacent to $(a,c)$, either $x=a$ or $x$ is adjacent to $a$. If
$x=a$, then $y$ must be adjacent to $c$. Further, we are assuming that
$(x=a,y)$ is adjacent to $(b,d)$, so $a=b$ or $a$ is adjacent to $b$, both
which contradict $\\{a,b\\}$ being a determining pair. Therefore, $x\neq a$,
and we can assume that $x$ is adjacent to $a$. The same argument implies that
$x$ is adjacent to $b$. Since $\\{a,b\\}$ is a determining pair, this implies
$x=v$. The same argument applies to the second coordinate so that $y=w$ and
$(v,w)$ will be the unique common neighbor between $(a,c)$ and $(b,d)$.
Therefore, $G\boxtimes H$ is distance critical as claimed. ∎

Note that we must assume $G$ and $H$ are both distance critical in the two
preceding results; indeed, $C_{5}\times C_{4}$ and $C_{5}\boxtimes C_{4}$ are
not distance critical, though $C_{5}$ is.

##  5 Extremal Results

The definition of distance critical relies on a local property (requiring that
each vertex has a determining pair), so it is unclear what the maximum edge
density of a distance critical graph is. Notice that a graph being distance
critical does not imply that it is (edge) maximal with respect to this
property. Consider, for example, the cycle on $8$ vertices. $C_{8}$ is clearly
a distance critical graph; however, edges can be added without disrupting its
determining pairs. See, for example, Figure 2.

---
Figure 2: Maximal distance critical graph on $8$ vertices containing $C_{8}$
as a proper subgraph.

For the following three lemmas, let $n$ represent the number of vertices in
the graph. A regular graph is a graph in which all vertices have the same
degree. First, we note that the fewest number of edges a distance critical
graph can have is $n$.

######  Lemma 5.1.

Every distance critical graph has a minimum of $n$ edges, and $C_{n}$ achieves
this bound for $n\geq 5$.

###### Proof.

Every vertex of a distance critical graph has a determining pair. Therefore,
the degree of every vertex is at least $2$. Thus, $|E(G)|\geq 2|V(G)|/2=n$. ∎

######  Lemma 5.2.

The maximum degree of any distance critical graph $G$ on $n\geq 6$ vertices is
at most $n-4$.

###### Proof.

Proposition 3.3 guarantees that no vertex has degree $n-1$.

Suppose $v\in V(G)$ had $\deg(v)=n-2$. Then $v$ is adjacent to every vertex
except one; label this exception as $u$. Let $S$ be the set of vertices that
are adjacent to $v$. Vertex $u$ has a determining pair $\\{x,w\\}$ since $G$
is distance critical. Since $u$ is not adjacent to $v$, $x,w\in S$. Every
vertex in $S$, however, is adjacent to $v$, so $\\{x,w\\}$ is not the
determining pair of $u$, and $v$ cannot have degree $n-2$.

Assume $\deg(v)=n-3$. Let $S$ be the set of $n-3\geq 3$ vertices adjacent to
$v$, and label the remaining two vertices $u_{1}$ and $u_{2}$. Let $w\in S$.
The options for a determining pair for $w$ are (a) $\\{v,u_{1}\\}$, (b)
$\\{v,u_{2}\\}$, (c) $\\{u_{1},u_{2}\\}$, (d) $\\{x,u_{1}\\}$, and
(e)$\\{x,u_{2}\\}$ where $x$ is some other vertex in $S$. Suppose $w$ has
option (a) for its determining pair. Then $w$ is adjacent to $u_{1}$ and
$u_{1}$ is nonadjacent to every other vertex in $S$. Therefore, the remaining
vertices of $S$ have determining pairs given by either options (b) or (e). In
either case, this implies $u_{2}$ is adjacent to one other vertex in $S$ and
nonadjacent to all other vertices of $S$. Therefore, the remaining vertices of
$S$ do not have a determining pair, so option (a) and, similarly, option (b)
are impossible.

Consider $u_{1}$. Since $u_{1}$ is not adjacent to $v$, therefore, if $u_{1}$
and $u_{2}$ are nonadjacent, $u_{1}$ must have a determining pair with both
vertices in $S$. This contradicts that $u_{1}$ is the unique neighbor of the
vertices in the determining pair, since $v$ is adjacent to every element of
$S$. Therefore, $u_{1}$ is adjacent to $u_{2}$ and some other vertex in $S$.
But then $\\{u_{1},u_{2}\\}$ cannot be a determining pair for $w$, and we may
rule out option (c) as well.

The vertices of $S$, therefore, only have determining pairs with either option
(d) or (e). Assume $w\in S$ with determining pair (d). Then $w$ is adjacent to
$u_{1}$ and $x$ where $x$ is another vertex of $S$ with determining pair of
type (e). Therefore, $x$ is adjacent to $u_{2}$ and $u_{1}$ is adjacent to
$u_{2}$ so $w$ is not their unique common neighbor. The same argument applies
if we start with a vertex in $S$ with determining pair of type (e). Therefore,
there exist vertices without a determining pair, a contradiction. ∎

The above bound is in fact tight for $n\geq 6$, as shown by the following
construction. Suppose $n\geq 8$ is even. Let $U=\\{u_{1},\ldots,u_{n/2-2}\\}$
and $U^{\prime}=\\{u^{\prime}_{1},\ldots,u^{\prime}_{n/2-2}\\}$, let
$V(G)=U\cup U^{\prime}\cup\\{v,w_{1},w_{2},w_{3}\\}$, and define

$\displaystyle E(G)$ $\displaystyle=\\{u_{j}u^{\prime}_{j}:u_{j}\in
U,u^{\prime}_{j}\in U^{\prime}\\}\cup\\{w_{1}u:u\in
U\\}\cup\\{w_{2}u^{\prime}:u^{\prime}\in U^{\prime}\\}$
$\displaystyle\qquad\cup\\{vw:w\in U\cup
U^{\prime}\\}\cup\\{w_{1}w_{3},w_{2}w_{3}\\}$

It is straightforward to check that $G$ is distance critical and
$\deg(v)=n-4$. See Figure 3.

---
Figure 3: Maximum degree construction for $n=12$.

If $n$ is odd, then the same construction works with an additional vertex
$w_{4}$ and new edges $vw_{4}$ and $w_{3}w_{4}$. If $n=6$, we can take
$G=C_{6}$, and for $n=7$, consider adding to $C_{6}$ a single vertex adjacent
to an antipodal pair. (If $n=5$, the graph $C_{5}$ is the only distance
critical graph, and its maximum degree is $n-3$.)

The $k^{th}$ power $G^{k}$ of an undirected graph $G$ is another graph on the
same set of vertices, in which two vertices are adjacent when their distance
in $G$ is at most $k$.

######  Lemma 5.3.

If $n\geq 5$, the maximum $d$ for which there is a $d$-regular distance
critical graph is $\lfloor\frac{n-1}{4}\rfloor+\lfloor\frac{n}{4}\rfloor$.

###### Proof.

First, we prove the upper bound. Every vertex in a distance critical graph,
$G$, has a determining pair. Let $\\{x,y\\}$ be the determining pair for some
vertex $v$. The neighborhoods of $x$ and $y$ only intersect at $v$. Therefore,
the remaining $n-3$ vertices are nonadjacent to at least one of $x$ or $y$.
Further, since $x$ and $y$ are both adjacent to $v$, we have that
$\deg(x)+\deg(y)\leq n-3+2=n-1$. In a regular graph, all vertices have the
same degree; therefore, $\deg(x)=\deg(y)$ so that $\deg(x)\leq\frac{n-1}{2}$
as desired. Note that, if $n\not\equiv 3\pmod{4}$, then
$\lfloor(n-1)/2\rfloor=\lfloor\frac{n-1}{4}\rfloor+\lfloor\frac{n}{4}\rfloor$.
On the other hand, if $n\equiv 3\pmod{4}$, then both $n$ and $(n-1)/2$ are
odd, so there is no $(n-1)/2$-regular graph on $n$ vertices, and we may
conclude that
$\deg(x)\leq\frac{n-1}{2}-1=\lfloor\frac{n-1}{4}\rfloor+\lfloor\frac{n}{4}\rfloor$.

Now, we construct a regular distance critical graph for each $n\geq 5$ to show
that this bound is tight. Let $G$ be the cycle on $n$ vertices where vertex
$i$ is adjacent to $i\pm 1\pmod{n}$ for $0\leq i\leq n-1$.

  * •

Assume $n\equiv 1\pmod{4}$, and consider $G^{\prime}:=G^{(n-1)/4}$. For every
vertex $i$ in $G^{\prime}$, $i$ is adjacent to $i\pm j\pmod{n}$ for $1\leq
j\leq\frac{n-1}{4}$. Therefore, we have a $\frac{n-1}{2}$-regular graph. Now
we show that this graph is indeed distance critical. Choose some $i\in V(G)$.
This vertex has the determining pair $\\{i+\frac{n-1}{4},i-\frac{n-1}{4}\\}$.
Indeed, $i+\frac{n-1}{4}$ is not adjacent to $i-\frac{n-1}{4}$ and by
definition, $i+\frac{n-1}{4}$ is adjacent to $i+\frac{n-1}{4}\pm j$ while
$i-\frac{n-1}{4}$ is adjacent to $i-\frac{n-1}{4}\pm j$ for $1\leq
j\leq\frac{n-1}{4}$. Therefore, $i$ is the only common neighbor between
$i+\frac{n-1}{4}$ and $i-\frac{n-1}{4}$, and the graph is distance critical as
claimed.

  * •

Assume $n\equiv 2\pmod{4}$, and consider $G^{(n-2)/4}$. This is a
$\frac{n-2}{2}$-regular graph where vertex $i$ has a determining pair
$\\{i+\frac{n-2}{4},i-\frac{n-2}{4}\\}$. Further, this is the maximum possible
degree since the first paragraph of the proof restricted the degree to at most
$\frac{n-1}{2}$.

  * •

Assume $n\equiv 3\pmod{4}$, and consider $G^{(n-3)/4}$. This is a
$\frac{n-3}{2}$-regular graph where vertex $i$ has the determining pair
$\\{i+\frac{n-3}{4},i-\frac{n-3}{4}\\}$. Further, this is the maximum possible
degree. Notice that $\frac{n-1}{2}$ is odd; therefore, an even number of
vertices is required to form a $\frac{n-1}{2}$-regular graph. However,
$n\equiv 3\pmod{4}$ implies that $n$ is odd.

  * •

Assume $4$ divides $n$, and consider $G^{(n-4)/4}$. This is a
$\frac{n-4}{2}$-regular graph where vertex $i$ has the determining pair
$\\{i+\frac{n-4}{4},i-\frac{n-4}{4}\\}$. We can; however, make this a
$\frac{n-2}{2}$-regular graph by adding an edge between vertices $i$ and
$i+\frac{n}{2}\pmod{n}$. This does not change the fact that vertex $i$ has
determining pair $\\{i+\frac{n-4}{4},i-\frac{n-4}{4}\\}$ because those two
vertices are still nonadjacent and their only common neighbor is still only
vertex $i$. Therefore, we have constructed a $\frac{n-2}{2}$-regular distance
critical graph. This is the maximum possible degree since the degree is at
most $\frac{n-1}{2}$.

∎

Next, we examine the properties of edge-maximal distance critical graphs,
i.e., distance critical $G$ so that, for all $e\not\in E(G)$, $G+e$ is not
distance critical.

######  Corollary 5.4.

Every edge-maximal distance critical graph is connected.

###### Proof.

This follows immediately from Lemma 3.4. ∎

See Table 2 for the number of edge-maximal distance critical graphs with $n$
small.

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
---|---|---|---|---|---|---|---|---|---|---|---
$\textrm{A}371674(n)$ | 0 | 0 | 0 | 0 | 1 | 1 | 2 | 4 | 14 | 82 | 557
Table 2: The number $\textrm{A}371674(n)$ of edge-maximal distance critical
graphs on $n$ vertices ([3, A371674]).

######  Corollary 5.5.

Let $S$ be the set of vertices which are involved in some determining pair in
an edge-maximal distance critical graph, $G$. Then every non-edge of $G$
intersects the set $S$.

###### Proof.

Let $xy\not\in E(G)$. Since $G$ is edge-maximal distance critical, there
exists a $z\in V(G)$ such that $z$ does not have a determining pair in $G+xy$.
Lemma 3.7 ensures that every $z$-determining pair contains either $x$ or $y$;
therefore, $xy$ intersects $S$ as desired. ∎

######  Corollary 5.6.

If $G$ is an edge-maximal distance critical graph and $S$ is the set of
vertices involved in some determining pair, then the set of vertices
$T=V(G)-S$ induces a clique.

###### Proof.

Assume not, i.e., instead there are two vertices $x$ and $y$ of $T$ which are
non-adjacent in $G$. We assumed $G$ is edge-maximal, and $xy\notin E(G)$ and
does not intersect $S$, which contradicts Corollary 5.5. ∎

In order to understand better the density of distance critical graphs, we
investigate the maximum size of their cliques. Since a determining pair
consists of nonadjacent vertices, the determining pair for vertices within the
clique have two options: $(1)$ the determining pair consists of one vertex
within the clique and one vertex outside the clique, and $(2)$ the determining
pair consists of vertices outside the clique. This leads to the following
results.

######  Definition 5.7.

Let $\Gamma_{m}$ be a graph which has a vertex set divided into $3$ sets: set
$A$ has $m(m-1)/2$ vertices labeled as $a_{ij}$ for $0\leq i<j<m$, set $B$ has
$m$ vertices labeled as $b_{j}$ for $0\leq j<m$, and set $C$ has $2m$ vertices
labeled as $c_{j}$ for $0\leq j<2m$ (understood modulo $2m$). The edge set
$E(\Gamma_{m})=K\cup X\cup Y\cup Z$, where $K$ consists of all pairs of
vertices from $A$; $X$ consists of all pairs $a_{ij}b_{i}$ and $a_{ij}b_{j}$,
where $0\leq i\neq j<m$; $Y$ consists of all edges of the form $b_{j}c_{j}$
and $b_{j}c_{j+m}$, where $0\leq j<m$; and $Z$ contains every edge of the form
$c_{j}c_{j+1}$ for $0\leq j<2m$.

The graph $\Gamma_{5}$ is shown in Figure 4. The white vertices lie in set
$A$, the gray vertices are in set $B$, and the black vertices are in set $C$.

---
Figure 4: $\Gamma_{5}$, with $A$ vertices white, $B$ vertices gray, and $C$
vertices black.

######  Theorem 5.8.

Among distance-critical graphs $G$ on $n$ vertices, the maximum possible
clique number of $G$ is $n-\Theta\left(\sqrt{n}\right)$.

###### Proof.

Consider the graph $\Gamma_{m}$ for $m\geq 3$. First, we argue that
$\Gamma_{m}$ is indeed distance critical by noting the determining pairs for
each vertex. Consider the vertex $a_{ij}$. This vertex has the determining
pair $\\{b_{i},b_{j}\\}$ because $b_{i}$ is not adjacent to $b_{j}$ and
$b_{i}$ is adjacent to $a_{ij}$, $c_{i},$ and $c_{i+m}$ while $b_{j}$ is
adjacent to $a_{ij}$, $c_{j}$, and $c_{j+m}$. Therefore, $a_{ij}$ is the only
common neighbor between $b_{i}$ and $b_{j}$. Consider vertex $b_{j}$. This
vertex has determining pair $\\{c_{j},c_{j+m}\\}$ because $c_{j}$ is not
adjacent to $c_{j+m}$ and $c_{j}$ is adjacent to $b_{j}$, $c_{j-1}$, and
$c_{j+1}$ while $c_{j+m}$ is adjacent to $b_{j}$, $c_{j+m-1}$, and
$c_{j+m+1}$. Therefore, $b_{j}$ is the only common neighbor between $c_{j}$
and $c_{j+m}$. Lastly, consider vertex $c_{j}$. This vertex has determining
pair $\\{c_{j-1},c_{j+1}\\}$ because $c_{j-1}$ is not adjacent to $c_{j+1}$,
and $c_{j-1}$ is adjacent to $c_{j}$, $c_{j-2}$, and $b_{j-1}$, while
$c_{j+1}$ is adjacent to $c_{j}$, $c_{j+2}$, and $b_{j+1}$. Therefore, $c_{j}$
is the only common neighbor between $c_{j-1}$ and $c_{j+1}$.

Next, we establish the clique number. By construction of $\Gamma_{m}$, a
clique of size $m(m-1)/2$ is induced by the vertex set $\\{a_{ij}\\}_{0\leq
i<j<m}$ with a remaining $3m$ vertices of the form $b_{j}$ or $c_{j}$.
Therefore, $n=|V(\Gamma_{m})|=\binom{m}{2}+3m=m^{2}/2+O(m)$ so that
$m=\sqrt{2n}(1+o(1))$. From here, we see that $\max_{G}\omega(G)\geq
n-(3+o(1))\sqrt{2n}=n-O\left(\sqrt{n}\right)$.

Now we establish a matching upper bound. Consider a distance critical graph
$G$ on $n$ vertices. Every vertex $v$ in a max clique $K$ of size $m$ must
have a determining pair, say, $\\{x_{v},y_{v}\\}$. Let $S=\bigcup_{v\in
K}\\{x_{v},y_{v}\\}$. Note that, for each $v\in K$,
$|\\{x_{v},y_{v}\\}\setminus K|=1$ or $2$, because if it were zero, then
$\\{x_{v},y_{v}\\}\subset K$ so that $x_{v}y_{v}\in E(G)$, contradicting that
$\\{x_{v},y_{v}\\}$ is a determining pair. Let $D$ be the subset of $V(K)$
with $|\\{x_{v},y_{v}\\}\setminus K|=1$ and $E=K\setminus D$. If $v\in D$,
wlog we assume $x_{v}\in K$ and $y_{v}\not\in K$. Note that the $y_{v}$ are
distinct across all $v\in D$, since, if $y_{v}=y_{w}$ for some $w\in D$, then
$x_{v}$ and $y_{v}$ have common neighbors $v$ and $w$, contradicting that they
form a determining pair for $v$. Thus,

$|V(G-K)|\geq|\\{y_{v}:v\in D\\}|\geq|D|.$

On the other hand, if $v\in E$, then the pair $\\{x_{v},y_{v}\\}$ is entirely
contained in $V(G-K)$. Since none of these pairs are repeated (or else they
could not be determining pairs), the vertex set $\bigcup_{v\in
E}\\{x_{v},y_{v}\\}$ admits at least $|E|$ vertex pairs and therefore at least
$\sqrt{2|E|}$ vertices, all of which lie outside $K$. Therefore,

$|V(G-K)|\geq\max\\{|D|,\sqrt{2|E|}\\}=\max\\{|D|,\sqrt{2(m-|D|)}\\}\geq\sqrt{2m+1}-1,$

since $0\leq|D|\leq m$. Then $G$ contains at least $\sqrt{2m+1}-1$ vertices in
addition to the clique $K$, and so $n\geq m+\sqrt{2m}+o(\sqrt{m})$ which
implies $m\leq n-\sqrt{(2+o(1))n}$, and we may conclude that $\omega(G)\leq
n-\Omega(\sqrt{n})$. ∎

The next result builds off of the preceding construction to show that distance
criticality is a highly non-local property: every graph is the induced
subgraph of some slightly larger distance critical graph.

######  Theorem 5.9.

Every graph is an induced subgraph of some distance critical graph.

###### Proof.

Let $G$ be any graph on $n$ vertices. Clearly, we may assume $n\geq 3$.
Modifying the construction of $\Gamma_{m}$ to obtain a $\Gamma_{m}^{\prime}$,
let $A$ be $V(G)$, and include $E(G)$ in $E(\Gamma_{m}^{\prime})$. An
additional $m$ vertices are needed to make up the set $B$ where
$m^{2}-m-2n\geq 0$; therefore, add $m=\lceil\frac{1+\sqrt{1+8n}}{2}\rceil$
additional vertices to account for the $B$ set. Label the vertices of $G$ as
$a_{ij}$ for the $n$ lexicographically least pairs $\\{i,j\\}$ with $0\leq
i<j<m$. Label the added vertices $b_{j}$ for $0\leq j<m$. Add edges
$a_{ij}b_{i}$ and $a_{ij}b_{j}$ for every element $a_{ij}$ of set $A$.
Similarly, add an additional $2m=\lceil 1+\sqrt{1+8n}\rceil$ vertices to make
up set $C$ and label these vertices as $c_{j}$ for $0\leq j<2m$ (understood
modulo $2m$). Add edges $b_{j}c_{j}$ and $b_{j}c_{j+m}$ for $0\leq j<m$.
Lastly, add edges $c_{j}c_{j+1}$ for $0\leq j<2m$. The argument that every
vertex of $\Gamma_{m}^{\prime}$ admits a determining pair is nearly identical
to the argument for $\Gamma_{m}$ in the proof of Theorem 5.8. ∎

We conclude by bounding the edge density of distance critical graphs.

######  Theorem 5.10.

Among distance critical graphs $G$ on $n$ vertices, the maximum edge density
of $G$ is between $1-O(1/\sqrt{n})$ and $1-\Omega(1/n)$.

###### Proof.

$\Gamma_{m}$ has $N^{2}-\Theta(N^{3/2})$ edges on $N:=m(m+5)/2$ vertices.
Further, every vertex of a distance critical graph has a determining pair,
ensuring at least $N$ non-edges. Therefore, the maximal edge density is
between $1-O(1/\sqrt{n})$ and $1-\Omega(1/n)$. ∎

##  6 Open Questions

To the best of our knowledge, the present work introduces distance critical
graphs. Therefore, open questions abound. We end by presenting a few
particularly interesting ones.

First, since there is a gap between the upper and lower bounds in Theorem
5.10, we ask the following.

######  Question 6.1.

What is the maximal edge density of distance critical graphs?

Next, although Theorem 5.9 can be used to obtain a lower bound on the number
of distance critical graphs, we were unable to determine if this bound is
tight.

######  Question 6.2.

What fraction of graphs are distance critical?

Finally, the property of distance criticality is easily generalized to finite
metric spaces, but the present work did not investigate which of the above
results translate easily to that setting, or which ones fail to generalize.
So, we offer the following open-ended question as well.

######  Question 6.3.

Let $(X,\rho)$ be a finite metric space with the property that the induced
metric space on $X-v$ for any $v\in X$ disagrees with $\rho$ on at least one
pair of points. What can be said about such spaces?

## References

  * [1] Mustapha Aouchiche and Pierre Hansen.  Distance spectra of graphs: A survey.  Linear Algebra and its Applications, 458:301–386, 2014.
  * [2] R. L. Graham and H. O. Pollak.  On the addressing problem for loop switching.  Bell System Tech. J., 50:2495–2519, 1971.
  * [3] OEIS Foundation Inc.  The On-Line Encyclopedia of Integer Sequences, 2024.  Published electronically at http://oeis.org.
  * [4] The Sage Developers.  SageMath, the Sage Mathematics Software System (Version x.y.z), YYYY.  https://www.sagemath.org.
  * [5] Weigen Yan and Yeong-Nan Yeh.  A simple proof of Graham and Pollack’s theorem.  Journal of Combinatorial Theory, 113:892 – 893, 2006.


Asymptotic distribution of certain degenerate V- and U-statistics with estimated parameters

Marija<EMAIL_ADDRESS>,  Bojana<EMAIL_ADDRESS>, Marko Obradović<EMAIL_ADDRESS>
Faculty of Mathematics, University of Belgrade, Studenski trg 16, Belgrade, Serbia

The asymptotic distribution of a wide class of V- and U-statistics with estimated parameters is derived in the case when the kernel is not necessarily differentiable along the parameter. The results have their application in  goodness-of-fit problems.

keywords: degenerate V-statistics, characterization, goodness-of-fit

MSC(2010): 62E20,  62G20

§ INTRODUCTION

V- and U-statistics frequently appear in inferential procedures as estimators and test statistics. In many cases they depend on nuisance parameters, and, in order to keep the
inference as broad as possible, estimators of such parameters are included in the statistic.

Although some earlier papers (see e.g. <cit.>) consider some particular cases,
the first general study devoted to U-statistics with estimated parameters was done in <cit.>, where the effect
of estimating parameters in asymptotically normal (non-degenerate) U-statistics was examined. De Wet and Randles
<cit.> studied a special case of degenerate V-statistics of order 2 obtained as an integrated square of a
V-statistic of order 1. They proved that, under some regularity conditions, the limiting distribution
is certain infinite linear combination of $\chi^2_1$ random variables. Their method was subsequently used for examining particular goodness-of-fit tests in several papers (see e.g. <cit.>).

Here we extend the result from <cit.> by considering the case when the V-statistic of order 1
is replaced with a V-statistic of an arbitrary order. The practical value of this result is reflected in the fact that
a class of characterization based test statistics are of this form. This class includes $L^2$-type tests employing V-empirical Laplace transforms of equidistributed random variables (see  <cit.>), but it is not limited to them. In fact, instead of  V-empirical Laplace transforms one could employ  V-empirical distribution functions (as e.g. in <cit.>), characteristic functions, densities, etc.
Exploring the asymptotics of such tests is closely related to V- and U-empirical functions with estimated parameters. Additionally, it enables calculation of the approximate Bahadur efficiency, one of important tools for test comparison (see e.g. <cit.>). The exceptional performance of the tests from <cit.> set the motivation to generalize the result and hence facilitate further research on this topic.

In addition, we present an analogous result for U-statistics. This result, using a different method of proof and slightly more general form, is also obtained in <cit.>.

Let $X_1,\ldots, X_n$ be a sequence of i.i.d. random variables with distribution function F  that depends on an unknown parameter $\lambda$. A V-statistic of order $m$
with estimated parameter, where $\widehat{\lambda}_n$ is a consistent estimator of $\lambda$, is

\begin{align}\label{Vocenjeno}
\end{align}

If its kernel $\Phi(x_1,...,x_m;\lambda)$ satisfies that

$\varphi_1(x;\lambda)=0,$ for all $x$, and
where $\varphi_1(x)=E_{\lambda}(\Phi(x,X_2,...,X_m;\lambda))$ and $\varphi_2(x,y;\lambda)=E_{\lambda}(\Phi(x,y,X_3,...,X_m;\lambda))$ are the first and second projection of the kernel $\Phi$,

then (<ref>) is a V-statistic with a weakly degenerate kernel.

If the parameter $\lambda$ is known, then the asymptotic distribution is (see e.g. <cit.>)

\begin{align}\label{raspodelaPoznato}
\end{align}

where $\{\upsilon_k\}, k=1,2,...,$ is the sequence of eigenvalues of the operator $A$ defined on $L^2(\mathbb{R},F)$ as

\begin{equation}\label{operatorA}
\end{equation}

and $Z_{k},k=1,2,...,$ are i.i.d random variables with standard normal distribution.

In <cit.> the following special case was considered

\begin{align*}
V_n(\widehat{\lambda})=\int_{-\infty}^\infty \Big(\frac1n \sum_{i=1}^n g(X_i,t;\widehat{\lambda}_{  n})\Big)^2 dM(t),
\end{align*}

where $g$ is a function satisfying some regularity conditions and $M(t)$ is a finite measure. The kernel of
$V_n(\widehat{\lambda})$ is then

\begin{align*}
\Phi(x_1,x_2,\widehat{\lambda})=\int_{-\infty}^\infty g(x_1,t;\widehat{\lambda}_{  n})g(x_2,t;\widehat{\lambda}_{  n})dM(t).
\end{align*}

§ MAIN RESULTS

Generalizing the idea from <cit.>, we consider the  V-statistics of order $2m$ with the following
symmetrized kernel

\begin{align}\label{jezgro}
\Phi(x_{1},...,x_{{2m}};\widehat{\lambda}_n)&=\frac{1}{(2m)!}\sum\limits_{\pi\in\Pi(2m)}\int\limits_{-\infty}^{+\infty}
g(x_{\pi(1)},...,x_{\pi(m)},t;\widehat{\lambda}_n) g(x_{\pi({m+1})},...,x_{\pi({2m})},t;\widehat{\lambda}_n)dM(t),
\end{align}

where $\Pi(2m)$ is the set of all permutations of $\{1,2,...,2m\}$ and $M(t)$ is a finite measure.
Without loss of generality, we assume that the function $g$ is a symmetric function of its first $m$ arguments.

Next we state some regularity conditions imposed on the smoothness of the  function $g$ and the rate of convergence  of the estimator $\widehat{\lambda}_n$.

Suppose that $\mu(t;\gamma)=E_\lambda(g(X_1,...,X_m,t;\gamma))$ exists for all $\gamma$ in the  neighbourhood of $\lambda$
and it satisfies $\mu(t;\lambda)\equiv 0$ for each $t$.
Suppose additionally that for all $\varepsilon>0$ there exists a  ball $S$ with finite radius in $\mathbb{R}^p$ centered at $\lambda$, such that
for $\gamma\in S$ it holds

\begin{equation*}
\frac{1}{||\gamma-\lambda||^2}\int\limits_{-\infty}^{+\infty}(\mu(t;\gamma)-d_1\mu(t;\lambda)^T(\gamma-\lambda))^2dM(t)<\varepsilon,
\end{equation*}

where $d_1\mu(t;\!\lambda)$ is the vector of partial derivatives
of $\mu(t;\gamma)$ at $\gamma=\lambda$,

\begin{equation*}
\int\limits_{-\infty}^{+\infty}(d_1\mu(t;\lambda)_r)^2dM(t)<\infty,
\end{equation*}

for $r=1,...,p$ where $d_1\mu(\cdot)_r$ is the $r-$th component of the vector $d_1\mu(\cdot)$.

\begin{align}
\widehat{\lambda}_n=\lambda+\frac{1}{n}\displaystyle\sum_{i=1}^n\alpha(X_i)+o_p\Big(\frac{1}{\sqrt{n}}\Big),\; n\to\infty,
\end{align}

where $E(\alpha(X_i)_r)=0$ and $E(\alpha(X_i)_r\alpha(X_i)_{r'})<\infty$ for all $1\leq r\leq r'\leq p$.

For any variation $(i_1,...,i_m)$ of indices from $\{1,2,....,n\}$, there exists a neighbourhood $K(\lambda)$ of $\lambda$ and a constant $C>0$, such that
for $\gamma\in K(\lambda)$ and for any ball $D(\gamma,d)$ centered at $\gamma$ with radius $d$ such that $D(\gamma,d)\subset K(\lambda)$, it holds

\begin{align*}
\int\limits_{-\infty}^{+\infty}(E(\sup_{\gamma'\in D(\gamma, d)}|g(X_{i_1},...,X_{i_m},t;\gamma')-g(X_{i_1},...,X_{i_m},t;\gamma)|))^2dM(t)\leq Cd^2.
\end{align*}

In addition, for every $\varepsilon>0$ there exists $d^*>0$ such that for $0<d<d^*$ holds

\begin{align*}
% \int\limits_{-\infty}^{+\infty}E(\sup_{\gamma'\in D(\gamma, d)}|g(X_{i_1},...,X_{i_m},t;\gamma')-g(X_{i_1},...,X_{i_m},t;\gamma)|^4)dM(t)<\epsilon.
\int\limits_{-\infty}^{+\infty}E(\sup_{\gamma'\in D(\gamma, d)}|g(X_{i_1},...,X_{i_m},t;\gamma')-g(X_{i_1},...,X_{i_m},t;\gamma)|)^4dM(t)<\varepsilon.
\end{align*}

Following <cit.> we define a new statistic, that depends on the true value of the parameter

\begin{equation}\label{Vpom}
\begin{split}
\\&\times\bigg(g(X_{i_{m+1}},...,X_{i_{2m}},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\sum_{i_k\in\{i_{m+1},..,i_{2m}\}}\alpha(X_{i_k})\bigg)dM(t),
\end{split}
\end{equation}

where the function $\alpha(\cdot)$ is defined in Condition <ref>. Let $\varphi^*_1(x;\lambda)$ and $\varphi^*_2(x,y;\lambda)$ be the first and second projection of the symmetrized version of the kernel $\Phi_*$. Using Conditions 1 and 2 it can be easily shown that $\varphi^*_1(x;\lambda)=0.$

\begin{align}\label{fi2}
&\varphi^*_2(x_1,x_{m+1};\lambda)=\frac{2m^2(2m-2)!}{(2m)!} E(\Phi_*(x_1,X_2,...,X_{m},x_{m+1},X_{m+2}...,X_{2m};\lambda))\nonumber\\&=
% \frac{2m^2(2m-2)!}{(2m)!}\color{black}E\bigg(\int\limits_{-\infty}^{+\infty}\bigg(g(x_1,X_{2},...,X_{m},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\Big(\alpha(x_1)+\sum_{j\in\{2,...,m\}}\alpha(X_{j})\Big)\bigg)\nonumber
%\\&\color{black}\hspace{1cm}\times\bigg(g(x_{m+1},X_{{m+2}},...,X_{{2m}},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\Big(\alpha(x_{m+1})+\sum_{k\in\{{m+2},..,{2m}\}}\alpha(X_{k})\Big)\bigg)dM(t)\bigg)\nonumber\\&\color{black}= \frac{2m^2(2m-2)!}{(2m)!}\color{black}\int\limits_{-\infty}^{+\infty}E\bigg(g(x_1,X_{2},...,X_{m},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\Big(\alpha(x_1)+\sum_{j\in\{2,...,m\}}\alpha(X_{j})\Big)\bigg)\nonumber
%\\&\color{black}\hspace{1cm}\times E\bigg(g(x_{m+1},X_{{m+2}},...,X_{{2m}},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\Big(\alpha(x_{m+1})+\sum_{k\in\{{m+2},..,{2m}\}}\alpha(X_{k})\Big)\bigg)dM(t)\nonumber\\&=
\frac{2m^2(2m-2)!}{(2m)!}\int\limits_{-\infty}^{+\infty}\bigg(g_1(x_1,t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\alpha(x_{1})\bigg)
\bigg(g_1(x_{m+1},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\alpha(x_{m+1})\bigg)dM(t),
\end{align}

where $g_1(x,t;\lambda)=E_\lambda(g(x,X_{2},...,X_{m},t;{\lambda}))$.
Since $V^*_n(\lambda)$ is a weakly degenerate V-statistic, it holds

\begin{equation}\label{raspodelaV*}
\end{equation}

where $\{\upsilon^*_k\}$ is the sequence of eigenvalues of  integral operator $A^*$ defined with

\begin{equation}\label{operatorA*}
\end{equation}

and $\{Z_k\}$ is the sequence of i.i.d. random variables with standard normal distribution.
The following theorem gives the asymptotic distribution of a V-statistic with kernel (<ref>).

Suppose Conditions <ref>-<ref> are satisfied and, additionally, that it holds $E(\Phi^2_*(\!X_1,\!...,X_{2m}; \lambda))\!<\!+\!\infty$ and
$E(\Phi_*(X_{i_1},...,X_{i_{2m}}; \lambda))<+\infty,$ for all variations $i_1,\ldots,i_m$, $1\leq i_1,...,i_{2m}\leq n$.

\begin{align}
\end{align}

Moreover, if $E(\varphi_2^*(X_1,X_2;\lambda))^2>0$, then

\begin{equation*}
\end{equation*}

where $\{\upsilon_k^*\}$ is the sequence of eigenvalues of the operator $A^*$ defined in (<ref>)

is the sequence of i.i.d. random variables with standard normal distribution.

The idea is to show the equidistribution of $nV_n(\widehat{\lambda}_{n})$ and $nV^*_n(\lambda)$, from where, taking into account (<ref>), will follow the statement of the theorem.

Without loss of generality we assume  $\int_{-\infty}^\infty dM(t)=1$.
Define an auxiliary statistic $Y_n$ as

\begin{equation*}
\end{equation*}

It is sort of midway from $V^*_n(\lambda)$ to $V_n(\widehat{\lambda}_n)$, depending both on the true value $\lambda$ and its estimator $\widehat{\lambda}_n$.
We show that $nY_n-nV^*_n(\lambda)\stackrel{P}{\to}0$ and $nV_n(\widehat{\lambda}_n)-nY_n\stackrel{P}{\to}0$.

Consider first the difference between $Y_n$  and $V^*_n(\lambda)$. Then, using the identity $a^2-b^2=(a-b)^2+2b(a-b)$ we get

\begin{align*}
&Y_n-V^*_n(\lambda)%=\int\limits_{-\infty}^{+\infty}\bigg(\frac{1}{n^{m}}\sum_{i_1,...,i_m}g(X_{i_1},...,X_{i_m},t;\lambda)+\mu(t;\widehat{\lambda}_n)\bigg)^2dM(t)\\&-\!\int\limits_{-\infty}^{+\infty}\!\bigg(\!\frac{1}{n^{m}}\!\sum_{i_1,...,i_m}\!\Big(\!g(X_{i_1},...,X_{i_m},t;\!\lambda)\!+\!d_1\mu(t;\!\lambda)^T\frac{1}{m}\!\sum\limits_{i_k\in\{i_1,...,i_m\}}\!\!\alpha(X_{i_k})\!\Big)\!\bigg)^2\!dM(t)    \\&
\int\limits_{-\infty}^{+\infty}\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)^T\frac{1}{n}\sum_{k=1}^n\alpha(X_{k})\bigg)^2dM(t)\\&+\!2\int\limits_{-\infty}^{+\infty}\!\bigg(\frac{1}{n^{m}}\sum_{i_1,...,i_m}\bigg(g(X_{i_1},...,X_{i_m},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\sum_{i_k\in\{i_1,...,i_m\}}\alpha(X_{i_k})\bigg)\!\bigg)\!\bigg(\!\mu(t;\widehat{\lambda}_n)\!-\!d_1\mu(t;\lambda)^T\frac{1}{n}\sum_{k=1}^n\alpha(X_{k})\bigg)dM(t).
\end{align*}

Therefore, using the Cauchy-Schwarz inequality, we obtain

\begin{equation}\label{Y-V}
\begin{split}
n(Y_n-V^*_n(\lambda))&\leq n\int\limits_{-\infty}^{+\infty}\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)^T\frac{1}{n}\sum_{k=1}^n\alpha(X_{k})\bigg)^2dM(t)
\\&+2\bigg(\!\int\limits_{-\infty}^{+\infty}n\bigg(\!\frac{1}{n^{m}}\sum_{i_1,...,i_m}g(X_{i_1},...,X_{i_m},t;\lambda)\!+\!d_1\mu(t;\lambda)^T\frac{1}{n}\!\sum_{k=1}^n\!\alpha(X_{k})\!\bigg)^2\!dM(t)\!\bigg)^\frac{1}{2}
\\&\times\bigg(\int\limits_{-\infty}^{+\infty}n\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)^T\frac{1}{n}\sum_{k=1}^n\alpha(X_{k})\bigg)^2dM(t)\bigg)^\frac{1}{2}
% n(Y_n-V_n(\lambda))&\leq        n\int\limits_{-\infty}^{+\infty}\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)(\widehat{\lambda}_n-\lambda)\bigg)^2dM(t)\\&+2n\bigg(\int\limits_{-\infty}^{+\infty}\bigg(\frac{1}{n^{m}}\sum_{i_1,...,i_m}g(X_{i_1},...,X_{i_m},t;\lambda)+d_1\mu(t;\lambda)(\widehat{\lambda}_n-\lambda)\bigg)^2dM(t)\bigg)^\frac{1}{2}\\&\times\bigg(\int\limits_{-\infty}^{+\infty}\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)(\widehat{\lambda}_n-\lambda)\bigg)^2dM(t)\bigg)^\frac{1}{2}.
\\&= n\int\limits_{-\infty}^{+\infty}\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)^T\frac{1}{n}\sum_{k=1}^n\alpha(X_{k})\bigg)^2dM(t)
\\&+2\sqrt{nV^{*}_n(\lambda)}\bigg(\int\limits_{-\infty}^{+\infty}n\bigg(\mu(t;\widehat{\lambda}_n)-d_1\mu(t;\lambda)^T\frac{1}{n}\sum_{k=1}^n\alpha(X_{k})\bigg)^2dM(t)\bigg)^\frac{1}{2}.
\end{split}
\end{equation}

From the Condition <ref> it follows that

\begin{equation}\label{prviClan}
\begin{split}
\end{split}
\end{equation}

Since, due to Condition <ref>, $\sqrt{n}(\widehat{\lambda}_n-\lambda)$ is bounded in probability, the first summand of (<ref>) tends to zero in probability.
The second summand also tends to zero due to the Slutsky theorem and the fact that $nV^{*}_n(\lambda)$ for V-statistics with kernels satisfying the conditions
of the theorem are bounded in probability.

Consider now the difference between $V_n(\widehat{\lambda}_n)$ i $Y_n$. Analogously to the previous case we get

\begin{align*}
\int\limits_{-\infty}^{+\infty}n\bigg(\frac{1}{n^{m}}\!\sum_{i_1,...,i_m}\!\Big(g(X_{i_1},...,X_{i_m},t;\widehat{\lambda}_n)\!-\!g(X_{i_1},...,X_{i_m},t;\lambda)\!\Big)-\mu(t;\widehat{\lambda}_n)\bigg)^2dM(t)\\&
\end{align*}

Since $nY_n$ is bounded in probability, it suffices to prove that

\begin{align*}
\int\limits_{-\infty}^{+\infty}\!n\bigg(\!\frac{1}{n^{m}}\!\sum_{i_1,...,i_m}\!\Big(g(X_{i_1},...,X_{i_m},t;\widehat{\lambda}_n)\!-\!g(X_{i_1},...,X_{i_m},t;\lambda{  )}\Big)\!-\!\mu(t;\widehat{\lambda}_n)\!\bigg)^2\!dM(t)
\end{align*}

converges to zero in probability.

\begin{equation*}
\begin{split}
Q_n(s,{r})\!=\!\!\int\limits_{-\infty}^{+\infty}\!\bigg(\!\frac{1}{n^{m}}\!\sum_{i_1,...,i_m}\!\Big(\!g\Big(X_{i_1},...,X_{i_m},t;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)\!-\!g\!\Big(X_{i_1},...,X_{i_m},t;{\lambda\!+\!\frac{r}{\sqrt{n}}}\!\Big)\!-\!\mu\!\Big(\!t;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)\!+\! \mu\Big(t;\lambda\!+\!\frac{r}{\sqrt{n}}\!\Big)\!\Big)\!\bigg)^2\!dM(t).
\end{split}
\end{equation*}

We need to show that $nQ_n(\sqrt{n}(\widehat{\lambda}_n-\lambda),{  0})\stackrel{P}{\rightarrow}0.$

Condition <ref> ensures the existence of a  ball $S$ in $\mathbb{R}^p$ such that

\begin{equation*}
P\{\sqrt{n}(\widehat{\lambda}_n-\lambda)\notin S\}\to 0,\;\text{as } n\to \infty.
\end{equation*}

Then for $\widetilde{\eta}>0$ it holds

\begin{equation*}
\begin{split}
P\{nQ_n(\sqrt{n}(\widehat{\lambda}_n-\lambda),{  0})\geq\widetilde{\eta}\}%&=P\{nQ_n(\sqrt{n}(\widehat{\lambda}_n-\lambda))>\varepsilon',\sqrt{n}(\widehat{\lambda}_n-\lambda)\in S\}\\&+P\{nQ_n(\sqrt{n}(\widehat{\lambda}_n-\lambda))>\varepsilon',\sqrt{n}(\widehat{\lambda}_n-\lambda)\notin S\}\\&
\leq P\{\sup\limits_{s\in S}nQ_n(s,{  0})\geq \widetilde{\eta}\}+P\{\sqrt{n}(\widehat{\lambda}_n-\lambda)\notin S\}.
\end{split}
\end{equation*}

Hence it is enough to prove that $\sup\limits_{s\in S}nQ_n(s,{  0})\stackrel{P}{\rightarrow}0$.
Following <cit.> (see also <cit.>), let $\delta>0$ and let $\{S_i\}$, $i=1,\ldots,i_{\max}$, be a collection of balls, centered at $r_i$ with radius $\delta$,
such that for all $s\in S$  there exists at least one $S_i$ such that $s\in S_i$.  The existence of $\{S_i\}$ is ensured
by the finiteness of $S$.

Then we have

\begin{align*}
P\{\sup_{s\in S} nQ_n(s,0)\geq \widetilde{\eta}\}\!\leq\! P\Big\{\bigcup_{i=1}^{i_{\max}}\{\sup_{s\in S_i} nQ_n(s,0)\geq \widetilde{\eta}\}\Big\}\!\leq\!
\sum_{i=1}^{i_{\max}} P\{\sup_{s\in S_i} nQ_n(s,0)\geq \widetilde{\eta}\}.
\end{align*}

Since the sum is finite, it is enough to prove that $P\{\sup_{s\in S_i} nQ_n(s,0)\geq \widetilde{\eta}\}\to 0$  for all $i$
as $n\to\infty$.
Consider the following decomposition

\begin{align*}
\\&=\int\limits_{-\infty}^{+\infty}\bigg(\frac{1}{n^{m}}\sum_{i_1,...,i_m}\!\Big(g\Big(X_{i_1},...,X_{i_m},t;\lambda+\frac{s}{\sqrt{n}}\Big)\!-\!
\\&+g\Big(X_{i_1},...,X_{i_m},t;\lambda+\frac{r_i}{\sqrt{n}}\Big)
\\&= Q_n(s,r_i)+Q_n(r_i,0)+2\widetilde{Q_n}(s,r_i,0),
\end{align*}

\begin{align}\label{mesoviti}
\begin{aligned}
\widetilde{Q_n}(s,r_i,0)&\!=\!\int\limits_{-\infty}^{+\infty}\!\bigg(\!\frac{1}{n^{2m}}\!\sum_{i_1,...,i_m}\!\Big(g\Big(X_{i_1},...,X_{i_m},t;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)\!-\!
\Big)\\&\times\Big(g\Big(X_{i_{m+1}},...,X_{i_{2m}},t;\lambda+\frac{r_i}{\sqrt{n}}\Big)
\end{aligned}
\end{align}

Then the next inequality holds

\begin{align}\label{dekompozicija}
\sup_{s\in S_i} nQ_n(s,0)\leq \sup_{s\in S_i} nQ_n(s,r_i) + nQ_n(r_i,0)+ 2\sup_{s\in S_i} \Big|n\widetilde{Q_n}(s,r_i,0)\Big|,
\end{align}

and we shall prove that each summand is smaller than $\eta=\frac{\widetilde{\eta}}{3}$ with probability one when $n\to\infty$.

\begin{align}\label{Hfunkcija}
\begin{aligned}
H(x_1,...,&x_{2m};S_i)=\sup_{ s\in S_i}\int\limits_{-\infty}^{+\infty}\bigg|\Big(g\Big(x_1,...,x_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
-g\Big(x_1,...,x_m,t;\lambda +\frac{r_i}{\sqrt{n}}\Big)+
\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\\&\times
\Big(g\Big(x_{m+1},...,x_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-g\Big(x_{m+1},...,x_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)
\end{aligned}
\end{align}

Then the first summand on the right hand side of (<ref>) can be bounded by

\begin{align}\nonumber
\sup_{s\in S_i} nQ_n(s,r_i)&\leq \frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}}H(X_{i_1},...,X_{i_{2m}};S_i)\\&
\begin{aligned}\label{dekompSupremuma}& =\frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}}\big(H(X_{i_1},...,X_{i_{2m}};S_i)-EH(X_{i_1},...,X_{i_{2m}};S_i)\big)
+ \frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}}EH(X_{i_1},...,X_{i_{2m}};S_i)).
\end{aligned}
\end{align}

We intend to choose $\delta$ such that the second summand of  (<ref>) is less than $\frac\eta2$
and the first one is less than $\frac\eta2$ with probability one when $n\to\infty$.

Consider the second summand of (<ref>). Partition the set of $2m$-tuples into
$\mathcal{I}_{c;2m}$, $c=0,1,\ldots,2m-1$, such that in $\mathcal{I}_{c;2m}$ there are $2m-c$ distinct entries. When $i_1,\ldots,i_{2m}\in\mathcal{I}_{0;2m},$ repeated application of the Cauchy-Schwarz inequality  gives

\begin{align*}
\!\leq\! \int\limits_{-\infty}^{+\infty}\!E\bigg(\!\sup_{ s\in S_i}\!\bigg|\!\Big(g\Big(X_1,...,X_m,t;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!-\!\mu\Big(t;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!-\!g\Big(X_1,...,X_m,t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\Big)\!+\!
\mu\Big(t;\lambda\!+\!\frac{r_i}{\sqrt{n}}\Big)\!\Big)\\&\times
\Big(g\Big(X_{m+1},...,X_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-g\Big(X_{m+1},...,X_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)
\\&\leq \int\limits_{-\infty}^{+\infty}E\bigg(\sup_{ s\in S_i}\Big|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
-g\Big(X_1,...,X_m,t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\Big)\!+\!
\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big|
\\&\times \sup_{ s\in S_i}\Big|g\Big(X_{m+1},...,X_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-g\Big(X_{m+1},...,X_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)
\\&= 	\int\limits_{-\infty}^{+\infty}\bigg(E\sup_{ s\in S_i}\Big|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big) 	-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big) 	-g\Big(X_1,...,X_m,t;\lambda +\frac{r_i}{\sqrt{n}}\Big)+	\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big|	\Bigg)^2dM(t).
\end{align*}

Applying the inequality $(a+b)^2\leq 2a^2+2b^2$, from Condition <ref>  we get

\begin{align*}
EH(X_1,...,X_{2m};S_i) &\leq \!2\!\int\limits_{-\infty}^{+\infty}\!\bigg(E\bigg[\sup_{s\in S}\Big|g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
\!-\!g\Big(X_1,...,X_m,t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\!\Big|\!\bigg]\!\bigg)^2dM(t)\\&+2\int\limits_{-\infty}^{+\infty}\Big(\sup_{s\in S}\Big|\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big|\Big)^2 dM(t) \\&
\leq 2\bigg(C\frac{ \delta^2}{n}+
\int\limits_{-\infty}^{+\infty}{\Big(}\sup\limits_{s\in S}\Big|d_1\mu(t;\lambda)^T\frac{s-r_i}{\sqrt{n}}\Big|{\Big)^2}dM(t)\bigg)
\leq 2\frac{ \delta^2}{n}(C+K),
\end{align*}

\begin{align}\label{constK}
\end{align}

which is finite due to Condition <ref>.

Since the cardianality of $\mathcal{I}_{0;2m}$ is $\binom{n}{2m}(2m)!$, we get

\begin{align*}
\frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}\in\mathcal{I}_{0;2m}}EH(X_1,...,X_{2m};S_i))&\leq \frac{n}{n^{2m}}\binom{n}{2m}(2m)!\cdot 2\frac{ \delta^2}{n}(C+K)
\sim 2\delta^2(C+K), \text{ as } n\to\infty.
\end{align*}

When $i_1,\ldots,i_{2m}\in\mathcal{I}_{c;2m},$ where $c\geq 1$, applying the same technique as before, we have

\begin{align*}
&EH(X_1,...,X_{2m};S_i)\!\leq\! \int\limits_{-\infty}^{+\infty}\!E\bigg(\!\sup_{ s\in S_i}\!\bigg|\!\Big(g\Big(X_1,...,X_m,t;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)
\!-\!\mu\Big(t;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)
\!-\!g\Big(X_1,...,X_m,t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\Big)\!+\!
\mu\Big(t;\lambda\!+\!\frac{r_i}{\sqrt{n}}\!\Big)\!\Big)\!\bigg|\\&\times\sup_{ s\in S_i}
\bigg|\Big(g\Big(X_{m+1},...,X_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-g\Big(X_{m+1},...,X_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)
%	\\&\color{green}\leq \int\limits_{-\infty}^{+\infty} \bigg(E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
%	-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)
%	\\&-g\Big(X_1,...,X_m,t;\lambda +\frac{r_i}{\sqrt{n}}\Big)+
%	\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2\bigg)^\frac12\bigg(E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_{m+1},...,X_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
%	\\&-\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)
%	-g\Big(X_{m+1},...,X_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)+
%	\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2\bigg)^\frac12dM(t)
\\&\leq \!\bigg(\!\int\limits_{-\infty}^{+\infty}\! E\bigg(\!\sup_{ s\in S_i}\bigg|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
\!-\!\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)
\!-\!g\Big(X_1,...,X_m,t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\!\Big)+
\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2dM(t)\bigg)^\frac{1}{2}\\&\times\!\bigg(\!\int\limits_{-\infty}^{+\infty}\!E\bigg(\!\sup_{ s\in S_i}\!\bigg|\!\Big(g\Big(X_{m+1},...,X_{2m},t;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!-\!\mu\Big(t;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!-\!g\Big(X_{m+1},...,X_{2m},t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\!\Big)\!+\!
\mu\Big(t;\lambda\!+\!\frac{r_i}{\sqrt{n}}\!\Big)\!\Big)\!\bigg|\!\bigg)^2dM(t)\!\bigg)^\frac12
%   \\&\color{green}\leq \bigg(\int\limits_{-\infty}^{+\infty} E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
%\\&-g\Big(X_1,...,X_m,t;\lambda +\frac{r_i}{\sqrt{n}}\Big)+
%\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2dM(t)\bigg)^\frac{1}{2}\bigg(\int\limits_{-\infty}^{+\infty}E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_{m+1},...,X_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
%	-g\Big(X_{m+1},...,X_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)+
%	\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2dM(t)\bigg)^\frac12
%\\&\leq \int\limits_{-\infty}^{+\infty} E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)-g\Big(X_1,...,X_m,t;\lambda +\frac{r_i}{\sqrt{n}}\Big)\bigg|\\&+\sup_{ s\in S_i}\bigg|\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2dM(t)
%	\\&\leq \bigg(2\int\limits_{-\infty}^{+\infty} E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
%-g\Big(X_1,...,X_m,t;\lambda +\frac{r_i}{\sqrt{n}}\Big)\bigg|\bigg)^2dM(t)\\&+2\int\limits_{-\infty}^{+\infty}\bigg(\sup_{ s\in S_i}\bigg|\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-
%	\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2dM(t)\bigg)^\frac{1}{2}\\&\times\bigg(2\int\limits_{-\infty}^{+\infty} E\bigg(\sup_{ s\in S_i}\bigg|\Big(g\Big(X_{m+1},...,X_{2m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
%-g\Big(X_{m+1},...,X_{2m},t;\lambda +\frac{r_i}{\sqrt{n}}\Big)\bigg|\bigg)^2dM(t)\\&+2\int\limits_{-\infty}^{+\infty}\bigg(\sup_{ s\in S_i}\bigg|\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)-
%	\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big)\bigg|\bigg)^2dM(t)\bigg)^\frac{1}{2}
\\&\leq \!\bigg(2\bigg(\int\limits_{-\infty}^{+\infty}\! E\bigg(\!\sup_{ s\in S_i}\bigg|\Big(g\Big(X_1,...,X_m,t;\lambda+\frac{s}{\sqrt{n}}\Big)
\!-\!g\Big(X_1,...,X_m,t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\Big)\bigg|\!\bigg)^4dM(t)\bigg)^\frac{1}{2}\\&+2\int\limits_{-\infty}^{+\infty}\bigg(\sup_{ s\in S_i}\bigg|d_1\mu\Big(t;\lambda\Big)^T\frac{s-r_i}{\sqrt{n}}\bigg|\bigg)^2dM(t)\bigg)^\frac{1}{2}\!\bigg(\!2\bigg(\!\int\limits_{-\infty}^{+\infty}\! E\bigg(\!\sup_{ s\in S_i}\!\bigg|\!\Big(\!g\Big(\!X_{m+1},...,X_{2m},t;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!\\&-\!g\Big(\!X_{m+1},...,X_{2m},t;\lambda \!+\!\frac{r_i}{\sqrt{n}}\!\Big)\!\bigg|\!\bigg)^4\!dM(t)\!\bigg)^\frac{1}{2}+2\int\limits_{-\infty}^{+\infty}\bigg(\sup_{ s\in S_i}\bigg|d_1\mu\Big(t;\lambda\Big)^T\frac{s-r_i}{\sqrt{n}}\bigg|\bigg)^2dM(t)\bigg)^\frac12\\&
\leq 2\Big(\varepsilon^\frac12+\frac{\delta^2}{n}K\Big).
\end{align*}

The cardinality of $\mathcal{I}_{c;2m}$ is proportional to $n^{2m-c}$ as $n\to\infty$,
hence  we get

\begin{align*}
\frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}\in\mathcal{I}_{c;2m}}EH(X_{i_1},...,X_{i_{2m}};S_i))=O\Big(\frac1{n^{c-1}}\Big).
\end{align*}

\begin{align*}
\frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}}EH(X_{i_1},...,X_{i_{2m}};S_i))\leq 2\delta^2(C+K) + 2\sqrt{\varepsilon} + O\Big(\frac1{n}\Big).
\end{align*}

Choosing $\delta<\sqrt{\frac{\eta}{4(C+K)}}$ the second sum of (<ref>) becomes smaller than
as $n\to\infty$. We now prove that, for this choice of $\delta$, the first summand of (<ref>)
is also smaller than $\frac{\eta}2$ with probability 1. Using the Chebyshev inequality we get

\begin{align*}
\big(H(X_1,...,X_{2m};S_i)-EH(X_1,...,X_{2m};S_i)\big)>\frac{\eta}{2}\Big\}
\\&\leq
\frac{4}{\eta^2}E\Big(\frac{n}{n^{2m}}\sum_{i_1,\ldots,i_{2m}}
\big(H(X_1,...,X_{2m};S_i)-EH(X_1,...,X_{2m};S_i)\big)\Big)^2
\\&= \frac{4}{\eta^2}\frac{n^2}{n^{4m}}\sum_{i_1,\ldots,i_{4m}}{\rm Cov}\big(H(X_1,...,X_{2m};S_i),
\end{align*}

Partition the set of all $4m$-tuples into sets $\mathcal{I}_{c;4m}$ defined as before. Define
[Strictly speaking, $\zeta(c)$ depends not only on $c$,  however, the bounds in
Condition <ref> hold for all variations which makes the notation justifiable.]

\begin{align*}
\zeta(c)={\rm Cov}\big(H(X_1,...,X_{2m};S_i),
H(X_{2m+1},...,X_{4m};S_i)\big), \;\; \{i_{1},\ldots,i_{4m}\}\in\mathcal{I}_{c;4m}.
\end{align*}

When $c=0$ the covariance
is equal to zero. When $c=1$, the covariance is non-zero only when one of the indices among the first $2m$ is
equal to one among the last $2m$. In this case we have

\begin{align*}
\zeta(1)&\leq
\\&{\leq}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}
E\bigg[\sup_{s\in S_i}\Big|g\Big(X_1,...,X_m,t_1;\lambda+\frac{s}{\sqrt{n}}\Big)-g\Big(X_1,...,X_m,t_1;{\lambda}+\frac{r_i}{\sqrt{n}}\Big)
+\mu\Big(t_1;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big|\\&\times\sup_{s\in S_i}\Big|g\Big(X_{m+1},...,X_{2m},t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&\times\!\sup_{s\in S_i}\!\Big|g\Big(\!X_1,X_{2m+1},...,X_{3m-1},t_2;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)\!-\!g\Big(\!X_1,X_{2m+1},...,X_{3m-1},t_2;{\lambda\!+\!\frac{r_i}{\sqrt{n}}}\!\Big)\\&-\mu\Big(t_2;\lambda+\frac{s}{\sqrt{n}}\Big)
\cdot\sup_{s\in S_i}\Big|g\Big(X_{3m},...,X_{4m-1},t_2;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&-\!g\Big(X_{3m},...,X_{4m-1},t_2;{\lambda}\!+\!\frac{r_i}{\sqrt{n}}\Big)\!-\!\mu\Big(t_2;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)
\!+\!\mu\Big(t_2;\lambda\!+\!\frac{r_i}{\sqrt{n}}\Big)\!\Big|\bigg]dM(t_1)dM(t_2).
\end{align*}

Applying the Cauchy-Schwarz inequality and grouping back the integrals we get

\begin{align*}
\int\limits_{-\infty}^{+\infty}\!\bigg(E\bigg[\sup_{s\in S_i}
\Big|g\Big(X_1,...,X_m,t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
E\bigg[\sup_{s\in S}\Big|g\Big(X_{m+1},...,X_{2m},t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&\times \int\limits_{-\infty}^{+\infty}
\bigg(E\bigg[\sup_{s\in S}\Big|g\Big(X_1,X_{2m+1},...,X_{3m-1},t_2;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&+\mu\Big(t_2;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big|\bigg]^2\bigg)^\frac{1}{2}
E\bigg[\sup_{s\in S}\Big|g\Big(X_{3m},...,X_{4m-1},t_2;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&-\mu\Big(t_2;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&
=\bigg(\int\limits_{-\infty}^{+\infty}\bigg(E\bigg[\sup_{s\in S_i}
\Big|g\Big(X_1,...,X_m,t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&\times\! E\bigg[\!\sup_{s\in S}\!\Big|g\Big(X_{m+1},...,X_{2m},t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\! -\!g\Big(X_{m+1},...,X_{2m},t_1;{\lambda\!+\!\frac{r_i}{\sqrt{n}}}\!\Big)
\!-\!\mu\Big(t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!+\!\mu\Big(t_1;\lambda\!+\!\frac{r_i}{\sqrt{n}}\Big)\!\Big|\!\bigg]dM(t_1)\!\bigg)^2
\\&\leq \int\limits_{-\infty}^{+\infty}E\bigg[\sup_{s\in S_i}
\Big|g\Big(X_1,...,X_m,t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\!-\!\mu\Big(t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)
\!+\!\mu\Big(t_1;\lambda\!+\!\frac{r_i}{\sqrt{n}}\Big)\Big|\bigg]^2\!dM(t_1) \!
\\&\times\! \int\limits_{-\infty}^{+\infty}\! \bigg(\!E\bigg[\sup_{s\in S}\!\Big|g\Big(X_{m+1},...,X_{2m},t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!-\!g\Big(X_{m+1},...,X_{2m},t_1;{\lambda\!+\!\frac{r_i}{\sqrt{n}}}\!\Big)
\!-\!\mu\Big(t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!+\!\mu\Big(t_1;\lambda\!+\!\frac{r_i}{\sqrt{n}}\!\Big)\!\Big|\!\bigg]\!\bigg)^2dM(t_1\!).
\end{align*}

Applying now the inequality $(a+b)^2\leq 2a^2+2b^2$ to both factors of the product above
we obtain

\begin{align*}
\zeta(1)&
%\leq\int\limits_{-\infty}^{+\infty}E\bigg[\sup_{s\in S}\Big|g\Big(X_1,...,X_m,t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
%\\&+\sup_{s\in S}\Big|\mu\Big(t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
%\\&\times \int\limits_{-\infty}^{+\infty}
%\bigg(E\bigg[\sup_{s\in S}\Big|g\Big(X_{m+1},...,X_{2m},t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
%\\&+\sup_{s\in S}\Big|\mu\Big(t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\leq \Bigg(2\int\limits_{-\infty}^{+\infty}E\bigg[\sup_{s\in S}\Big|g\Big(X_1,...,X_m,t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&+2\int\limits_{-\infty}^{+\infty}
\bigg(\sup_{s\in S}\Big|\mu\Big(t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&\times\!\Bigg(\!2\!\int\limits_{-\infty}^{+\infty}\!\bigg(\!E\bigg[\!\sup_{s\in S}\!\Big|g\!\Big(\!X_{m+1},...,X_{2m},t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\!\Big)
\!-\!g\Big(\!X_{m+1},...,X_{2m},t_1;\!{\lambda+\frac{r_i}{\sqrt{n}}}\Big)\!\Big|\!\bigg]\!\bigg)^2\!dM\!(t_1)
\\&+2\int\limits_{-\infty}^{+\infty}\bigg(\sup_{s\in S}\Big|\mu\Big(t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&\leq  \Big(2\varepsilon^\frac{1}{2}+2\frac{\delta^2}{n}K\Big)\Big(2\frac{\delta^2}{n}(C+K)\Big)
\sim4(C+K)\varepsilon^\frac{1}{2}\frac{\delta^2}{n}+o\Big(\frac{1}{n}\Big),
\end{align*}

where $K$ is defined in (<ref>).

Let $c\geq 2$. For any such variation $X'_1,\ldots,X'_{4m}$ it holds

\begin{align*}
\zeta(c)&\leq \big({\rm Var} H(X'_1,...,X'_{2m};S_i)\big)^{\frac12}
\big({\rm Var} H(X'_{2m+1},...,X'_{4m};S_i)\big)^{\frac12}\\
&\leq {\rm Var} H(X_{1},...,X_{2m};S_i)\leq EH^2(X_{1},...,X_{2m};S_i),
\end{align*}

where $\{X_{1},...,X_{2m}\}$ is the variation for which the function $H$ from (<ref>) has the maximal variance.
Applying the same techniques used for the previous case we get

\begin{align*}
\zeta(c)&\leq \int\limits_{-\infty}^{+\infty}E\bigg(\sup_{s\in S_i}
\bigg|g\Big(X_{1},...,X_{m},t;\lambda+\frac{s}{\sqrt{n}}\Big)
\sup_{s\in S}\bigg|g\Big(X_{{m+1}},...,X_{{2m}},t;\lambda+\frac{s}{\sqrt{n}}\Big)
E\bigg(\sup_{s\in S_i}\bigg|g\Big(X_{1},...,X_{m},t_1;\lambda+\frac{s}{\sqrt{n}}\Big)
\\&+\int\limits_{-\infty}^{+\infty}
\bigg(\sup_{s\in S}\bigg|\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)+\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\bigg|\bigg)^4dM(t_1)\bigg)^\frac12\\&\times\!\bigg(\!\int\limits_{-\infty}^{+\infty}\!
E\bigg(\!\sup_{s\in S_i}\bigg|g\Big(X_{m+1},...,X_{2m},t_1;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)
\!-\!g\Big(X_{m+1},...,X_{2m},t_1;\lambda\!+\!\frac{r_i}{\sqrt{n}}\Big)\!\bigg|\!\bigg)^4\!dM(t_1)
\\&+\int\limits_{-\infty}^{+\infty}
\bigg(\sup_{s\in S}\bigg|\mu\Big(t;\lambda+\frac{s}{\sqrt{n}}\Big)+\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\bigg|\bigg)^4dM(t_1)\bigg)^\frac12
\\&<8\Big(\varepsilon+\frac{\delta^4}{n^2}K\Big).
\end{align*}

Since the cardinality of $\mathcal{I}_c$ is proportianal to $n^{4m-c}$ (say $\kappa_c\cdot n^{4m-c}$)  we get

\begin{align*}
E(\sup_{s\in S_i}nQ_n(s,r_i))^2&\!\leq\! \frac{n^2}{n^{4m}}\!\sum_{i_1,\ldots,i_{4m}}\!{\rm Cov}\!\big(H(X_1,...,X_{2m};S_i),
% \\& =\frac{n^2}{n^{4m}} \sum_{c=0}^{2m}\binom{n}{4m-c}\frac{4m!}{c!} \zeta(c)\\
\sim\! \sum_{c=1}^{4m-1}\!  \kappa_c \cdot \frac{\zeta(c)}{n^{c-2}}
\!<\! {\rm const}\! \cdot \!(\varepsilon^\frac{1}{2}\delta^2 \!+\!\varepsilon) \!+\! o\Big(\frac1n\Big),
\end{align*}

and hence $P\Big\{\sup_{s\in S_i}nQ_n(s,r_i)<\frac\eta2\Big\}{\to}1$.

We now pass to the second summand of (<ref>). Applying the Chebyshev inequality we get

\begin{align*}
P\Big\{nQ_n(r_i,0)\geq \eta\Big\}&\leq \frac1{\eta^2}E(nQ_n(r_i,0))^2\\&
\!-\!\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\!\bigg)
\\&\times\bigg(g\Big(X_{i_{m+1}},...,X_{i_{2m}},t;\lambda+\frac{r_i}{\sqrt{n}}\Big)
\\&\times\bigg(g\Big(X_{i_{2m+1}},...,X_{i_{3m}},t;\lambda+\frac{r_i}{\sqrt{n}}\Big)
\\&\times\bigg(g\Big(X_{i_{3m+1}},...,X_{i_{4m}},t;\lambda+\frac{r_i}{\sqrt{n}}\Big)
\end{align*}

Since expectation of each factor in the sum above is zero, the expectation of the product will be zero whenever
there is at least one factor independent from all the others. Hence it is easy to see that for
$\{i_{1},\ldots,i_{4m}\}\in \mathcal{I}_{0;4m}\cup\mathcal{I}_{1;4m}$ the summands are equal to zero.
For the rest of summands, using the previously obtained bound we get an upper bound for the general term $\mathcal{G}$

\begin{align*}
\mathcal{G}\leq EH(X_{i_1},...,X_{i_{2m}};S)H(X_{i_{2m+1}},...,X_{i_{4m}};S)
\end{align*}

where   $d_S$ is the radius od $S$. Therefore,

\begin{align*}
E(nQ_n(r_i,0))^2<{\rm const}\cdot \varepsilon + o\Big(\frac1n\Big),
\end{align*}

and hence $nQ_n(r_i,0)$ converges to zero in probability.

For the last summand of (<ref>) defined in (<ref>), using the Cauchy-Schwarz inequality
we get

\begin{align*}
\sup_{s\in S_i}{|}\widetilde{Q_n}(s,r_i,0){|}&
\leq
\bigg(\frac{1}{n^{2m}}\sum_{i_1,...,i_{2m}}\!\int\limits_{-\infty}^{+\infty}\!{\Big(}\sup_{s\in S_i}\Big|g\Big(X_{i_1},...,X_{i_m},t;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)\!-\!\mu\Big(t;\lambda\!+\!\frac{s}{\sqrt{n}}\Big)-
\\&+\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\Big|{\Big)^{2}}
\!\bigg(
\frac{1}{n^{2m}}\!\sum_{i_1,...,i_{2m}}\!\int\limits_{-\infty}^{+\infty}\!
\Big(g\Big(X_{i_{m+1}},...,X_{i_{2m}},t;\lambda+\frac{r_i}{\sqrt{n}}\Big)\!\\&-\!\mu\Big(t;\lambda+\frac{r_i}{\sqrt{n}}\Big)
\Big)^2dM(t)\bigg)^\frac{1}{2}.
\end{align*}

Combining the steps used for $\sup_{s\in S_i}nQ_n(s,r_i)$ and $nQ_n(r_i,0)$, it is straightforward to show that
$2\sup_{s\in S_i}n\widetilde{Q_n}(s,r_i,0)$ converges to zero in probability, which ends the proof.

§.§ U-statistics

We now present the analogous theorem for U-statistics with estimated parameters. Define

\begin{equation*}
\end{equation*}

where $\Phi(x_{1},...,x_{{2m}};\widehat{\lambda}_n)$ is defined in (<ref>) and the corresponding auxiliary statistic

\begin{equation*}
\end{equation*}

where $\Phi_*(x_{1},...,x_{2m};\lambda)$ is defined in (<ref>).

Let $X_1,...,X_n$ be a random sample with distribution function $F$. Let all conditions of Theorem <ref> hold, and, additionally,

\begin{equation}\label{uslovTeoreme}
\end{equation}

\begin{equation*}
\end{equation*}

where $\{\upsilon_k^*\}$ and $\{\upsilon_k\}$, $k=1,2,...,$ are sequences of eigenvalues of operators $A^*$ and $A$, defined in (<ref>) and (<ref>), respectively, while $Z_{k},k=1,2,...,$ are i.i.d. standard normal random variables.

Using the result on limiting distribution of degenerate U-statistics (see e.g. <cit.>), we get that

Thus it is enough to prove that

\begin{align}\label{razlikaUstat}
n(U_n(\widehat{\lambda}_n)-U_n^*(\lambda))\overset{P}\to \binom{2m}{2}\sum\limits_{k=1}^\infty(\upsilon^*_k-\upsilon_k).
\end{align}

Consider the difference

\begin{align*}
n(U_n(\widehat{\lambda}_n)-U_n^*(\lambda))%\\&=\!\frac{n}{n(n\!-\!1)\!\cdots\!(n\!-\!2m\!+\!1)}\!\sum\limits_{\{i_1,\ldots,i_{2m}\}\in\mathcal{I}'}\!\Big(\! \Phi(X_{i_1},...,X_{i_{2m}};\!\widehat{\lambda}_n)\!-\!\Phi_*(X_{i_1},...,X_{i_{2m}};\!{\lambda})\!\Big)
\end{align*}

where the set $\mathcal{I}$ is the set of $2m$-tuples where at least one entry repeats.
From the proof Theorem <ref> we know that
$n(V_n(\widehat{\lambda}_n)-V_n^*(\lambda))\overset{p}{\to}0$, so it remains to find the limit in probability of

\begin{align*}
\Xi_n\!&=\!\frac{1}{(n\!-\!1)\!\cdots\!(n\!-\!2m\!+\!1)}\!\sum_{{\{i_1,\ldots,i_{2m}\}\in\mathcal{I}}}\!\Big(\!\Phi(X_{i_1},...,X_{i_{2m}};\!\widehat{\lambda}_n\!)\!-\!\Phi^*(X_{i_1},...,X_{i_{2m}};\!{\lambda})\!\Big)\\
\end{align*}

where $\mathcal{I}_1$ is the set of all $2m$-tuples of indices such that only one entry repeats only once and all the others are different.  For the first summand we have

\begin{align*}
\Xi_n^{(1)}\!&=\!
\frac{n}{n-2m+1}\binom{2m}{2}\frac{1}{\binom{n}{2m\!-\!1}}\!\sum\limits_{i_1\neq i_2\neq\cdots\neq i_{2m-1}}\!\bigg(\Phi(X_{i_1},X_{i_1},X_{i_2},...,X_{i_{2m-1}};\!\widehat{\lambda}_n)-\Phi_*(X_{i_1},X_{i_1},X_{i_2},...,X_{i_{2m-1}};\lambda)\bigg)\\&=\frac{n}{n-2m+1}\binom{2m}{2}\frac{1}{2m-1}\frac{1}{\binom{n}{2m-1}}\sum\limits_{i_1\neq i_2\neq\cdots\neq i_{2m-1}}\bigg(\phi(X_{i_1},X_{i_2},...,X_{i_{2m-1}};\widehat{\lambda}_n)-\phi_*(X_{i_1},X_{i_2},...,X_{i_{2m-1}};\lambda)\bigg),
\end{align*}

where $\phi(X_{i_1},...,X_{i_{2m-1}};\widehat{\lambda}_n)$ and $\phi_*(X_{i_1},...,X_{i_{2m-1}};{\lambda})$ are symmetrized versions of  $\Phi(X_{i_1},X_{i_1},X_{i_2},...,X_{i_{2m-1}};\widehat{\lambda}_n)$ and $\Phi_*(X_{i_1},X_{i_1},X_{i_2},...,X_{i_{2m-1}};{\lambda})$, respectively.

Using the law of large numbers for U-statistics (see e.g. <cit.>), and the law of large numbers for U-statistics with estimated parameters (see <cit.>), we get that

\begin{align*}
\Xi_n^{(1)} \overset{P}\to \binom{2m}{2}\frac{1}{2m-1}\big(E\phi(X_{i_1},...,X_{i_{2m-1}};\widehat{\lambda}_n)-E\phi_*(X_{i_1},X_{i_2},...,X_{i_{2m-1}};\lambda)\big).
\end{align*}

We then have

\begin{align*}
\frac{m^2(2m-1)!}{(2m)!}E\bigg(\int\limits_{-\infty}^{+\infty}\bigg(g(X_{1},X_2,...,X_{i_m},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\sum_{j\in\{1,...,m\}}\alpha(X_{j})\bigg)\\&\times\bigg(g(X_{{1}},X_{m+1},...,X_{{2m-1}},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\sum_{k\in\{1,{m+1},..,{2m-1}\}}\alpha(X_{k})\bigg)dM(t)\bigg)
%    \binom{2m}{2}\frac{m^2}{2m(2m-1)}\int\limits_{-\infty}^{+\infty}E\bigg(\bigg(g(X_{1},X_2,...,X_{i_m},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\sum_{j\in\{1,...,m\}}\alpha(X_{j})\bigg)
%   \\&\color{black}\hspace{3cm}\times\bigg(g(X_{{1}},X_{m+1},...,X_{{2m-1}},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\sum_{k\in\{1,{m+1},..,{2m-1}\}}\alpha(X_{k})\bigg)\bigg)dM(t)
\\&
%  =\frac{m^2}{2m}\int\limits_{-\infty}^{+\infty}E\bigg(g_1(X_{1},t;\lambda)+d_1\mu(t;\lambda)^T\frac{1}{m}\alpha(X_{1})\bigg)^2dM(t)\\&
=(2m-1)E\varphi_2(X_1,X_1; \lambda)=(2m-1)\sum\limits_{k=1}^\infty\upsilon_k^*,
\end{align*}

provided that the operator $A^*$ is nuclear, i.e. $\sum_{k=1}^\infty|\upsilon^*_k|<\infty$. This will be the case if $E|\varphi_2^*(X_1,X_1; \lambda)|<\infty$. Indeed,

\begin{align*}
&\leq 2E\int\limits_{-\infty}^{+\infty}\bigg|g_1(X_{1},t;\lambda)\bigg|^2dM(t)+\frac{2}{m^2}E\bigg|\alpha(X_{1})\bigg|^2\cdot\int\limits_{-\infty}^{+\infty}\bigg|d_1\mu(t;\lambda)^T\bigg|^2dM(t),
\end{align*}

which is finite due to (<ref>) and Conditions <ref> and <ref>.

Analogously, $E\phi(X_{i_1},...,X_{i_{2m-1}};{\widehat{\lambda}_{n}})=(2m-1)\sum_{k=1}^\infty\upsilon_k$,
and hence

\begin{align*}
\Xi_n^{(1)} \overset{P}\to \binom{2m}{2}\sum\limits_{k=1}^\infty(\upsilon_k-\upsilon_k^*).
\end{align*}

Next we show that $\Xi_n^{(2)}\overset{P}\to0$. Let $1<r<2m$ and let $2m-r$  of $2m$ indices in the expression for $\Xi_n^{(2)}$ be different.  Then, the corresponding terms of the sum are proportional to

\begin{equation*}
\begin{split}
\frac{1}{\binom{n}{2m\!-\!r}}\sum\limits_{i_1\neq i_2\neq\cdots\neq i_{2m-r}}\bigg(\phi^{(r)}(X_{i_1},X_{i_2},...,X_{i_{2m-r}};\widehat{\lambda}_n)-\phi^{(r)}_*(X_{i_1},X_{i_2},...,X_{i_{2m-r}};\lambda)\bigg),
\end{split}
\end{equation*}

where $\phi^{(r)}(X_{i_1},X_{i_2},...,X_{i_{2m-r}};\widehat{\lambda}_n)$ and $\phi^{(r)}_*(X_{i_1},X_{i_2},...,X_{i_{2m-r}};\lambda)$ are obtained using the appropriate symmetrization of initial kernels. Using the law of large numbers for U-statistics and the Slutsky theorem, we get that the term above tends to zero in probability. This holds for all $r$, hence  $\Xi_n^{(2)}\overset{P}\to0$
and (<ref>) follows.

§ APPLICATION

The main application of Theorems <ref> and <ref> lies in goodness-of-fit tests based on equidistribution-type characterizations. Such characterizations have the following form. Let $X_1,...,X_m$ be independent copies of random variable $X$ and let $\omega_1(\cdot)$ and $\omega_2(\cdot)$ be two functions such that

\begin{align}\label{eqChar}
\omega_1(X_1,...,X_m)\overset{d}{=}\omega_2(X_1,...,X_m)
\end{align}

if and only if the distribution of $X$ belongs to some family $\mathcal{F}$.
The tests based on $L^2$ distance of estimators of V- or U-empirical functions of $\omega_1(\cdot)$ and $\omega_2(\cdot)$ often have the form of degenerate V- or U-statistics of order higher than 2.

One of the examples are test statistics based on V-empirical Laplace transforms from <cit.>,<cit.>. There asymptotic distributions can be obtained using Theorem <ref>. However, the kernel in this case is differentiable, and it is possible to obtain the asymptotics directly using the mean value theorem. Here we present two examples when this is not possible and the application of our result is called for.

§.§.§ Example 1

Let $X_1,\ldots,X_n$  be a random sample from distribution $F$. Consider testing the composite null  hypothesis
$F(x)=F_0(x;\lambda)$ where $\lambda$ is the scale parameter.
A bunch of scale families of distributions can be characterized with equidistribution-type characterizations  of the form (<ref>) where
$\omega_1(\cdot)$ and $\omega_2(\cdot)$ are two  homogeneous functions, i.e.

\begin{align*}
\omega_i(\lambda X_1,...,\lambda X_m)=\lambda\omega_i(X_1,...,X_m),\;\;\lambda>0.
\end{align*}

Some examples of the characterization of this type are given in <cit.>,<cit.>,<cit.>.

The most natural approach to construct a test is to estimate distribution functions of $\omega_1(\cdot)$ and $\omega_2(\cdot)$ respectively and to  base a test on their difference. A natural approach which yields a degenerate V-statistics is to integrate the squared difference over an empirical measure, i.e.

\begin{align*}
\end{align*}

where $F_n(t)$ is empirical distribution function. Examples of such tests can be found in <cit.>.

Another possibility, currently not explored yet, is to consider

\begin{align*}
\widetilde{W}_n(\widehat{\lambda}_{n})&=\int(\widetilde{G}_n^{(1)}(t)-\widetilde{G}_n^{(2)}(t))^2dM(t),
\end{align*}

where $\widetilde{G}^{(i)}$ is U-(V-)  empirical distribution function of the  scaled sample and $\{M(t)\}$ is a finite measure. Without loss of generality we may assume that ${dM(t)}$ is a density function of some random variable.

Applying the Theorem <ref> we get that the distribution of $n\widetilde{W}_n(\widehat{\lambda}_{n})$ coincides with  $n\widetilde{W}_n(\lambda)$ which doesn't depend on $\lambda$.  This follows from the fact that

\begin{align*}
\widetilde{W}_n(\widehat{\lambda}_{n})&=\frac{1}{n^{2m}}\sum\limits_{i_1,...,i_{2m}}\int (I\{\widehat{\lambda}_{n}\omega_1(X_{i_1},...,X_{i_m})<t\}-I\{\widehat{\lambda}_{n}\omega_2(X_{i_1},...,X_{i_m})\}<t)\\&\times(I\{\widehat{\lambda}_{n}\omega_1(X_{i_{m+1}},...,X_{i_{2m}})<t\}-I\{\widehat{\lambda}_{n}\omega_2(X_{i_{m+1}},...,X_{i_{2m}})<t\})dM(t),
\end{align*}

which can be represented as

\begin{align*}
\widetilde{W}_n(\widehat{\lambda}_{n})&=\frac{1}{n^{2m}}\sum\limits_{i_1,...,i_{2m}}\int g(X_{i_1},...,X_{i_m},t;\widehat{\lambda}_n)g(X_{i_{m+1}},...,X_{i_{2m}},t;\widehat{\lambda}_n)dM(t),
\end{align*}

\begin{align*}
g(x_1,...,x_m,t;\lambda)\!=\!\frac{1}{m!}\!\sum_{\pi(m)}\! (I\!\{\!\lambda\omega_1(x_{\pi(1)},...,x_{\pi(m)})\!<\!t\}\!-\!I\!\{\!\lambda\omega_2(x_{\pi(1)},...,x_{\pi(m)})\}\!<\!t).
\end{align*}

Under $H_0$ it holds

\begin{align*}
\mu(t;\gamma)%&=E_\lambda(g(X_1,..,X_m,t;\gamma))\\&%=E_\lambda(I\{\gamma\omega_1(X_1,...,X_m)<t\}-I\{\gamma\omega_2(X_1,...,X_m)<t\})\\&
\end{align*}

Last equality holds for each $\gamma$ due to the characterization. Therefore the first derivative  $d_1\mu(t;\lambda)$ will be also equal to zero.

Consider now the particular case of the Puri-Rubin characterization <cit.>, i.e. when $\omega_1(X_1,X_2)=|X_1-X_2|$ and $\omega_2(X_1)=X_1$, which implies that $F$ is exponential distribution with some scale parameter $\lambda$, and let $dM(t)=e^{-t}dt$.

Then the corresponding test statistic is

\begin{align*}
\widetilde{W}^{\textup{PR}}_n(\widehat{\lambda}_{n})&=\frac{1}{n^{4}}\sum\limits_{i_1,...,i_{4}}\int (I\{\widehat{\lambda}_{n}|X_{i_1}-X_{i_2}|<t\}-I\{\widehat{\lambda}_{n}X_{i_1}<t\})(I\{\widehat{\lambda}_{n}|X_{i_3}-X_{i_4}|<t\}-I\{\widehat{\lambda}_{n}X_{i_3}<t\})e^{-t}dt,
\end{align*}

where $\widehat{\lambda}_{n}=\bar{X}^{-1}$.

According to the argument above, it is enough to obtain the asymptotic distribution of $n\widetilde{W}^{\textup{PR}}_n(\lambda)$. The symmetrized kernel of $\widetilde{W}^{\textup{PR}}_n(\lambda)$ is

\begin{align*}
\Phi^{\textup{PR}}_n(x_1,x_2,x_3,x_4;\lambda)&=\frac{1}{4!}\sum\limits_{\pi\in\Pi(4)}  \bigg(e^{-\lambda\max (\left| x_{\pi(1)}-x_{\pi(2)}\right| ,\left| x_{\pi(3)}-x_{\pi(4)}\right| )}- e^{-\lambda \max (x_{\pi(3)},\left| x_{\pi(1)}-x_{\pi(2)}\right| )}- e^{-\lambda\max (x_{\pi(1)},\left| x_{\pi(3)}-x_{\pi(4)}\right| )}\\&+e^{-\lambda\max (x_{\pi(1)},x_{\pi(3)})}\bigg).
\end{align*}

Since $\widetilde{W}^{\textup{PR}}_n(\widehat{\lambda}_n)$ is scale-free, we may assume that $\lambda=1$. The second projection is then equal to

\begin{equation*}
\begin{split}
\varphi^\textup{PR}_2(s,t)&=\frac{1}{18}+\frac{1}{2}(e^{-2s-t}+e^{-s-2t})-\frac{1}{4}(e^{-2t}+e^{-2s})-\frac{16}{9}e^{-s-t}+\frac{1}{9}e^{-\min(s,t)}(2-3\min(s,t))\\&+\frac{1}{18}e^{-\max(s,t)}(19-6\min(s,t)).
\end{split}
\end{equation*}

Conditions <ref> and <ref> obviously hold and Condition <ref> follows from the fact that the kernel is a linear combination of indicators.

Hence, the asymptotic distribution follows from (<ref>) with the corresponding integral operator being

\begin{equation*}
\end{equation*}

The eigenvalues of $A^{\textup{PR}}$ cannot be obtained analytically, however they can be approximated numerically using the method from <cit.>.

§.§.§ Example 2

The second example comes from testing hypothesis within the location family. In testing goodness-of-fit  based on equidistribution characterizations, it is often the case that estimating a location parameter, unlike the scale one,  changes the asymptotic distribution (see e.g. <cit.>).

Let $X_1,\ldots,X_n$  be a random sample from distribution $F$. Consider testing the composite null normality hypothesis
$F(x)=\Phi(\frac{x-\theta}{\sigma})$, where both $\theta$ and $\sigma$ are unknown, based on the famous Polya's characterization
<cit.>, arguably the first ever published equidistribution-type characterization. It states that if $X_1$ and $X_2$ are i.i.d. random variables with distribution function $F$, then the equidistribution

\begin{align}\label{poljaKara}
\frac{X_1+X_2}{\sqrt{2}}\overset{d}{=}X_1
\end{align}

implies that $F$ is normal with zero mean and arbitrary variance.
Some normality tests based on this characterization can be found in <cit.> and <cit.>.

Let $F_n(t)$ be the usual empirical distribution function and let

\begin{align*}
G_n(t)=\frac1{n^2}\sum_{i,j=1}^n {\rm I}\Big\{\frac{X_i+X_j}{\sqrt{2}}\leq t\Big\}
\end{align*}

be the V-empirical distribution function associated with the random variable in (<ref>).

Here we consider an $\omega^2$-type test

\begin{align}
T_n(\widehat{\theta})=\int_{-\infty}^\infty (\widetilde{G}_n(t)-\widetilde{F}_n(t))^2 d\widetilde{F}_n(t),
\end{align}

$\widetilde{F}_n(t)$ and $\widetilde{G}_n(t)$ are the aforementioned empirical d.f.'s applied to the shifted sample $X^*_i=X_i-\widehat{\theta}$, and $\widehat{\theta}=\bar{X}$,
the sample mean. It is easy to see that the statistic is location and scale free. After transformation we obtain

\begin{align*}
\int_{-\infty}^\infty(G_n(s-\widehat{\theta}+\widehat{\theta}\sqrt{2})-F_n(s))^2dF_n(s).
\end{align*}

However, in order to apply Theorem <ref> we need to get rid of the empirical measure.

\begin{align*}
\overline{T}_n(\widehat{\theta})=
\int_{-\infty}^\infty(G_n(s-\widehat{\theta}+\widehat{\theta}\sqrt{2})-F_n(s))^2dF(s),
\end{align*}

where $F(s)=\Phi(\frac{s-\theta}{\sigma})$, where $\theta$ and $\sigma$ are true parameter values. Since the test statistic is location-scale invariant, we assume $\theta=0$ and $\sigma=1$.

Put $W_n(s)=(G_n(s-\widehat{\theta}+\widehat{\theta}\sqrt{2})-F_n(s))$. Then, using the law of large numbers for
V-statistics, the fact that $\sqrt{n}W_n(s)$ converges to a centered Gaussian
process, continuous mapping theorem, and Donsker theorem, we get

\begin{align*}
\int\frac{1}{\sqrt{n}} nW^2_n(s)d(\sqrt{n}(F_n(s)-F(s))\stackrel{d}{\to} \int 0 d\mathbb{G}_F=0,
\end{align*}

where $\mathbb{G}_F=B\circ F$ and  $\{B(s)\}$ the standard Brownian bridge. Hence the statistics are asymptotically equivalent.

The statistic  $\bar{T}_n(\widehat{\theta})$ is a V-statistic with kernel of the form (<ref>)

\begin{align*}
\Psi(x_{1},...,x_{{4}};\widehat{\lambda}_n)=\frac{1}{4!}\sum\limits_{\pi\in\Pi(4)}\int\limits_{-\infty}^{+\infty}
\end{align*}

\begin{align*}
g(x_1,x_2,s;\gamma)&=I\bigg\{\frac{x_1+x_2}{\sqrt{2}}\leq s+\gamma(\sqrt{2}-1)\bigg\}-\frac{1}{2}I\{x_1\leq s\}-\frac{1}{2}I\{x_2\leq s\}.
\end{align*}

We now show that the conditions for applying Theorem <ref> are fulfilled. Taking into account that $\theta=0$,

\begin{align*}
\mu(s;\theta)=E_{\theta}( g(x_1,x_2,s;\gamma))|_{\gamma=\theta}=(\Phi(s+\gamma(\sqrt{2}-1))-\Phi(s))|_{\gamma=\theta}=0.
\end{align*}

In addition $d_1\mu(s;\gamma)=\phi(s+\gamma(\sqrt{2}-1))\cdot(\sqrt{2}-1),$

where $\phi(x)$ is the standard normal density.

\begin{equation*}
\begin{split}
&\int\limits_{-\infty}^{+\infty}(\phi(s)\cdot(\sqrt{2}-1))^2\phi(s)ds<\infty, \\
\end{split}
\end{equation*}

where the last inequality follows from the boundness of the function $\phi'(s)$. Hence, Condition <ref> is satisfied.
Condition <ref> holds obviously for the sample mean.
Condition <ref> is straightforward to verify using the properties of the normal density and the finiteness of the first and second moment of the kernel $\Psi_*$.
The  second projection is equal to

\begin{align*}
\psi^*_2(x_1,x_{3};\theta)%=\frac{16}{24} E(\Psi_*(x_1,X_2,x_3,X_{4};\theta))\\&%=   \frac{2}{3}\int\limits_{-\infty}^{+\infty}\bigg(g_1(x_1,s;\theta)+d_1\mu(s;\theta)\frac{x_1}{m}\bigg)   \bigg(g_1(x_{3},s;\theta)+d_1\mu(s;\theta)\frac{x_3}{m}\bigg)\phi(s) ds\\&
\frac{2}{3}\int\limits_{-\infty}^{+\infty}\bigg(\Phi(\sqrt{2}s-x_1)-\frac{1}{2}\Phi(s)-\frac{1}{2}I\{x_1<s\}+\phi(s)\cdot(\sqrt{2}-1)\frac{x_{1}}{m}\bigg)\\&\times
\bigg(\Phi(\sqrt{2}s-x_3)-\frac{1}{2}\Phi(s)-\frac{1}{2}I\{x_3<s\}+\phi(s)\cdot(\sqrt{2}-1)\frac{x_{3}}{m}\bigg) \phi(s)ds.%\\&=   \frac{2}{3}\int\limits_{-\infty}^{+\infty}\Phi(\sqrt{2}s-x_1)\Phi(\sqrt{2}s-x_3)\phi(s)ds-\frac{1}{3}\int\limits_{-\infty}^{+\infty}\Phi(\sqrt{2}s-x_1)\Phi(s)\phi(s)ds-\frac{1}{3}\int\limits_{-\infty}^{+\infty}\Phi(\sqrt{2}s-x_3)\Phi(s)\phi(s)ds\\&-\frac{1}{3}\int\limits_{-\infty}^{+\infty}\Phi(\sqrt{2}s-x_1)I\{x_3<s\}\phi(s)ds+\frac{2}{3}(\sqrt{2}-1)\frac{x_{3}}{m}\int\limits_{-\infty}^{+\infty}\Phi(\sqrt{2}s-x_1)\phi^2(s)ds+...
\end{align*}

The expression above can be calculated and expressed as a function of bivariate normal distributions using the formulae from <cit.>.
The asymptotic distribution of $nT_n(\widehat{\theta})$ now follows from Theorem <ref> and the corresponding eigenvalues can be obtained numerically using the method from <cit.>.

§ ACKNOWLEDGEMENT

The authors express their deep gratitude to two anonymous referees whose suggestions improved the quality of the paper.