# Universal Bounds for Fractional Laplacian
on the Bounded Open Domain in $\mathbb{R}^{n}$

Lingzhong Zeng  Lingzhong Zeng

School of Mathematics and Statistics

Jiangxi Normal University, Nanchang 330022, China<EMAIL_ADDRESS>
###### Abstract.

Let $\Omega$ be a bounded open domain on the Euclidean space $\mathbb{R}^{n}$
and $\mathbb{Q}_{+}$ be the set of all positive rational numbers. In 2017,
Chen and Zeng investigated the eigenvalues with higher order of the fractional
Laplacian $\left.(-\Delta)^{s}\right|_{\Omega}$ for $s>0$ and
$s\in\mathbb{Q}_{+}$, and they obtained a universal inequality of Yang
type(_Universal inequality and upper bounds of eigenvalues for non-integer
poly-Laplacian on a bounded domain, Calculus of Variations and Partial
Differential Equations, (2017) 56:131_). In the spirit of Chen and Zeng’s
work, we study the eigenvalues of fractional Laplacian, and establish an
inequality of eigenvalues with lower order under the same condition. Also, our
eigenvalue inequality is universal and generalizes the eigenvalue inequality
for the poly-harmonic operators given by Jost et al.(_Universal bounds for
eigenvalues of polyharmonic operator. Trans. Amer. Math. Soc. 363(4),
1821-1854 (2011)_).

††footnotetext: Key words and phrases: fractional Laplacian; eigenvalues with
lower order; universal, Euclidean space.††footnotetext: 2010 Mathematics
Subject Classification: 35P15, 53C23, 81T30, 83C57.††footnotetext: The author
partially supported by the National Natural Science Foundation of China (Grant
Nos. 11861036 and 11826213).

##  1\. Introduction

Let $\Omega$ be a bounded domain with piecewise smooth boundary
$\partial\Omega$ on an $n$-dimensional Euclidean space $\mathbb{R}^{n}$. The
Dirichlet eigenvalue problem of the poly-Laplacian with any order is described
by

(1.1) $\begin{cases}(-\Delta)^{l}u=\lambda u,\ \ &{\rm in}\ \
\Omega,\\\\[5.69054pt] u=\dfrac{\partial
u}{\partial\nu}=\cdots=\dfrac{\partial^{l-1}u}{\partial\nu^{l-1}}=0,\ \ &{\rm
on}\ \ \partial\Omega,\end{cases}$

where $\Delta$ is the Laplacian and $\nu$ denotes the outward unit normal
vector field of $\partial\Omega$. It is well known that the spectrum of the
eigenvalue problem (1.1) is discrete and satisfies

$0<\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{k}\leq\cdots\rightarrow+\infty,$

where each eigenvalue is repeated according to its multiplicity and
$\lambda_{k}$ denotes the $k^{th}$ eigenvalue. When $l=1$, eigenvalue problem
(1.1) is called the fixed membrane problem. For this case, in 1956, Payne et
al. [35] (Thompson-Colin [40] (for all $n\geq 3$)) proved a universal
inequality as follows:

(1.2) $\lambda_{k+1}-\lambda_{k}\leq\frac{4}{nk}\sum^{k}_{i=1}\lambda_{i}.$

Furthermore, in various settings, many mathematicians extended the universal
inequality given by Payne et al.. In particular, Hile and Protter [27] proved
the following universal inequality of eigenvalues:

(1.3)
$\sum^{k}_{i=1}\frac{\lambda_{i}}{\lambda_{k+1}-\lambda_{i}}\geq\frac{nk}{4},$

which is sharper than inequality (1.2) given by Payne et al.. Furthermore,
Yang [44] made an amazing contribution to eigenvalue inequality (cf. [20]),
and obtained a very sharp universal bound as follows:

(1.4)
$\sum^{k}_{i=1}(\lambda_{k+1}-\lambda_{i})^{2}\leq\frac{4}{n}\sum^{k}_{i=1}(\lambda_{k+1}-\lambda_{i})\lambda_{i}.$

From (1.4) and a celebrated recursion formula, Cheng and Yang gave the sharp
upper bounds for the eigenvalues with respect to $k^{\frac{2}{n}}$ ( cf.
[22]). In addition, for the case of degenerate elliptic operators, we refer
the readers to [13, 14] and references therein.

On the other hand, Payne et al. [35] investigated the lower order eigenvalues
and proved the following universal inequality in 1956:

(1.5) $\lambda_{2}+\lambda_{3}\leq 6\lambda_{1},$

for $\Omega\subset\mathbb{R}^{2}$, and they proposed a famous conjecture for
$\Omega\subset\mathbb{R}^{n}$ as follows:

(1.6) $\frac{\lambda_{2}+\lambda_{3}+\cdots+\lambda_{n+1}}{\lambda_{1}}\leq
n\frac{\lambda_{2}(\mathbb{B}^{n})}{\lambda_{1}(\mathbb{B}^{n})},$

where $\lambda_{i}(\mathbb{B}^{n})(i=1,2)$ denotes the $i^{th}$ eigenvalue of
Laplacian on the ball $\mathbb{B}^{n}$ with the same volume as the bounded
domain $\Omega$, i.e., ${\rm Vol}(\Omega)={\rm Vol}(\Omega^{\ast})$. Attacking
this conjecture, Brands [12] improved (1.5) to the following:
$\lambda_{2}+\lambda_{3}\leq\lambda_{1}(3+\sqrt{7}),$ when $n=2$. Furthermore,
Hile and Protter [27] proved $\lambda_{2}+\lambda_{3}\leq 5.622\lambda_{1}.$
In 1980, Marcellini [34] obtained
$\lambda_{2}+\lambda_{3}\leq(15+\sqrt{345})/6\lambda_{1}.$ In 2011, Chen and
Zheng [15] proved $\lambda_{2}+\lambda_{3}\leq 5.3507\lambda_{1}.$ For general
case, Ashbaugh and Benguria [5] established the following universal inequality

(1.7) $\frac{\lambda_{2}+\lambda_{3}+\cdots+\lambda_{n+1}}{\lambda_{1}}\leq
n+4,$

for $\Omega\subset\mathbb{R}^{n}$, in 1993. For more references on the
solution of this conjecture, we refer the reader to [3, 4, 27, 34, 37, 38].

In the case of $l=2$, problem (1.1) is also called the clamped plate problem
of bi-harmonic operator. There are similar inequalities as to (1.2), (1.3) and
(1.4), which were studied in [2, 17, 21, 29, 30, 28, 41], respectively. For
any positive integer $l$, more results can be found in [17, 29, 43]. In this
case, there is also an analogue of Yang type inequality for the poly-Laplacian
operator

(1.8)
$\sum_{i=1}^{k}\left(\lambda_{k+1}-\lambda_{i}\right)^{2}\leq\frac{4l(2l+n-2)}{n^{2}}\sum_{i=1}^{k}\lambda_{i}\left(\lambda_{k+1}-\lambda_{i}\right),$

which was obtained by Cheng et al. in [19] in 2009. In 1998, Ashbaugh [2]
announced two interesting inequalities without proofs as follows:

(1.9)
$\sum^{n}_{i=1}(\lambda^{\frac{1}{2}}_{i+1}-\lambda^{\frac{1}{2}}_{1})\leq
4\lambda^{\frac{1}{2}}_{1},$

and

(1.10) $\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})\leq 24\lambda_{1}.$

The proofs of (1.9)) and (1.10)) were given by Cheng et al. in [19] in 2009.
In fact, they considered more general case and proved

(1.11)
$\sum^{n}_{i=1}(\lambda^{\frac{1}{l}}_{i+1}-\lambda_{1}^{\frac{1}{l}})^{l-1}\leq(2l)^{l-1}\lambda^{\frac{l-1}{l}}_{1},$

for $l\geq 2$, and

(1.12) $\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})\leq 4l(2l-1)\lambda_{1},$

for any $l=1,2,\cdots$. It is easy to find that (1.11) and (1.12) respectively
become (1.9) and (1.10) when $l=2$. Moreover, (1.12) covers (1.7) when $l=1$.
In 2010, Cheng et al. [18] proved

(1.13)
$\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\leq[8(n+2)\lambda_{1}]^{\frac{1}{2}},$

which is generalized by Sun and the author in [39]. In 2011, Jost et al. [32]
derived the following inequality

(1.14)
$\sum^{n+1}_{i=2}\lambda_{i}+\sum^{n-1}_{i=1}\frac{2(l-1)i}{2l+i-1}(\lambda_{n+1-i}-\lambda_{1})\leq(n+4l(2l-1))\lambda_{1},$

which covers (1.7) when $l=1$ and improves (1.10) when $l=2$.

Furthermore, we consider the fractional Laplace operator restricted to
$\Omega$ and denote it by $\left.(-\Delta)^{s}\right|_{\Omega}$ with $s>0,$
which is defined as the pseudo-differential operator restricted to $\Omega$.
In other words, the fractional Laplacian can be defined by

$(-\Delta)^{s}u(x)=:\textbf{
P.V.}\int_{\mathbb{R}^{n}}\frac{u(x)-u(z)}{|x-z|^{n+2s}}dz,$

where P.V. denotes the principal value and
$u:\mathbb{R}^{n}\rightarrow\mathbb{R}$. We note that fractional Laplacian is
not a local operator when $s$ is not a integer. Define the characteristic
function $\chi_{\Omega}:t\mapsto\chi_{\Omega}(t)$ by $\chi_{\Omega}(t)=1$ when
$x\in\Omega$, while $\chi_{\Omega}(t)=0$ when
$x\in\mathbb{R}^{n}\backslash\Omega$, then the special pseudo-differential
operator can be represented as the Fourier transform of the function $u$ [33],
namely

$\left.(-\Delta)^{s}\right|_{\Omega}u:=\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left[u\chi_{\Omega}\right]\right],$

where $\mathscr{F}[u]$ denotes the Fourier transform of a function
$u:\mathbb{R}^{n}\rightarrow\mathbb{R}$

$\mathscr{F}[u](y)=\widehat{u}(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}e^{-\sqrt{-1}x\cdot
y}u(x)dx,$

and the notation $\mathscr{F}^{-1}$ is the inverse of Fourier transform. Here,
$\mathscr{F}^{-1}$ is given by

$\mathscr{F}^{-1}[w](x)=\widehat{w}(x)=\int_{\mathbb{R}^{n}}e^{\sqrt{-1}x\cdot
y}w(y)dy.$

It is well known that the fractional Laplacian operator $(-\Delta)^{s}$ can be
considered as the infinitesimal generator of the symmetric $s$-stable process
[11, 7, 8, 9, 10]. Suppose that a stochastic process $X_{t}$ has stationary
independent increments and its transition density (i.e., convolution kernel)
$p^{s}(t,x,z)=p^{s}(t,x-z),t>0,x,z\in\mathbb{R}^{n}$ is determined by the
following Fourier transform

$\operatorname{Exp}\left(-t|y|^{2s}\right)=\int_{\mathbb{R}^{n}}e^{\sqrt{-1}y\cdot
z}p^{2s}(t,z)dz,\quad t>0,\quad y\in\mathbb{R}^{n},$

then we can say that the process $X_{t}$ is an $n$-dimensional symmetric
$s$-stable process with order $s\in(0,1]$ in $\mathbb{R}^{n}$( also see [7, 8,
10]) There are two important examples of this process: one is the Cauchy
process; and the other is the Brownian motion. The fractional Laplacian
operator $(-\Delta)^{s}$ is the Riesz-Feller derivative of fractional order
$2s>0$, i.e. $(-\Delta)^{s}=\frac{\partial^{2s}}{\partial|x|^{2s}}$, which
appears in a wide class of physical systems including Lévy flights and
stochastic interfaces. Some detail information on Lévy process will be found
in a textbook [36] and a good survey [1]. We also refer the readers to an
excellent article [16], where some important and interesting literatures on
fractional Laplacian have been given by Chen and Zeng. We assume that $\phi\in
C_{0}^{\infty}(\Omega)\subset C_{0}^{\infty}\left(\mathbb{R}^{n}\right),$ then
the quadratic form
$\left\langle\left.(-\Delta)^{s}\right|_{\Omega}\phi,\phi\right\rangle=\int_{\Omega}\bar{\phi}(-\Delta)^{s}\phi
dx$ is positive on a dense subset of $L^{2}(\Omega),$ and it can be extended
to a unique minimal positive operator on $L^{2}(\Omega).$ In addition, the
fractional Laplacian operator $\left.(-\Delta)^{s}\right|_{\Omega}$ is self-
adjoint with purely discrete spectrum. Thus there is a sequence of discrete
eigenvalues which can be ordered, after counting multiplicity, as

$0<\lambda_{1}<\lambda_{2}\leq\cdots\leq\lambda_{k}\leq\cdots\rightarrow+\infty.$

Comparing with the case of Laplacian, the eigenvalue of fractional Laplacian
$\left.(-\Delta)^{s}\right|_{\Omega}.$ is much less studied, but it has
recently received more attention. Here, we only refer the reader to papers
[10, 16, 26, 42, 46, 47] and the references therein. In particular, Chen and
Zeng investigated the eigenvalues with higher order of fractional Laplacian.
See [16]. Applying an important result due to Ilias and Makhoul [31] and the
technique of commutators for operators, they proved an interesting result in
[16] in 2017 as follows:

######  Theorem (Chen and Zeng’s Universal Inequality).

Let $\Omega$ be a bounded open domain in $\mathbb{R}^{n}$. Suppose that
$\left\\{\lambda_{i}\right\\}_{i\geq 1}$ are eigenvalues of operator
$\left.(-\Delta)^{s}\right|_{\Omega},$ where $s=\frac{1}{m}$ with the positive
integer $m\geq 2$ or $s>\frac{1}{2}$ and $s\in\mathbb{Q}_{+}.$ Then,

$\displaystyle\left[\sum_{i=1}^{k}\left(\lambda_{k+1}-\lambda_{i}\right)^{2}\right]^{2}\leq\frac{4s(n+2s-2)}{n^{2}}\left(\sum_{i=1}^{k}\left(\lambda_{k+1}-\lambda_{i}\right)^{2}\lambda_{i}^{\frac{s-1}{s}}\right)\left(\sum_{i=1}^{k}\left(\lambda_{k+1}-\lambda_{i}\right)\lambda_{i}^{\frac{1}{s}}\right).$

By the above theorem and a variant of Chebyshev sum inequality, Chen and Zeng
[16] deduced an eigenvalue inequality of Yang type as follows:

(1.15)
$\sum_{i=1}^{k}\left(\lambda_{k+1}-\lambda_{i}\right)^{2}\leq\frac{4s(n+2s-2)}{n^{2}}\sum_{i=1}^{k}\left(\lambda_{k+1}-\lambda_{i}\right)\lambda_{i},$

which implies an upper bounds for the eigenvalues:

$\lambda_{k+1}\leq\left(1+\frac{4s(n+2s-2)}{n^{2}}\right)k^{\frac{2s(n+2s-2)}{n^{2}}}\lambda_{1}.$

As the above argument, Chen and Zeng only discussed the eigenvalues with
higher order for the fractional Laplacian. However, they have not addressed
the case of lower order. In this regard, it is natural to propose the
following problem:

######  Problem.

Can one obtain a universal inequality for the eigenvalues with lower order of
the fractional Laplacian on $\mathbb{R}^{n}$?

Under the same assumption as Chen and Zeng’s Theorem, we answer the above
problem. In details, we establish a key lemma in section 2, and by utilizing
this lemma, obtain an eigenvalue inequality in section 3. This is what the
following theorem says.

######  Theorem 1.1.

_(Universal inequality)_ Let $\Omega$ be a bounded open domain in
$\mathbb{R}^{n}$. Suppose that $\mathbb{Q}_{+}$ is the set of all positive
rational numbers and $\left\\{\lambda_{i}\right\\}_{i\geq 1}$ are eigenvalues
of operator $\left.(-\Delta)^{s}\right|_{\Omega},$ where $s=\frac{1}{m}$ with
the positive integer $m\geq 2$ or $s>\frac{1}{2}$ and $s\in\mathbb{Q}_{+}.$
Then, we have

(1.16)
$\sum_{i=1}^{n}\left(\lambda_{i+1}-\lambda_{1}\right)^{\frac{1}{2}}\leq[4s(n+2s-2)]^{\frac{1}{2}}\lambda_{1}^{\frac{1}{2}}.$

######  Remark 1.1.

Since the constant $[4s(n+2s-2)]^{\frac{1}{2}}$ in (1.16) is not dependent on
$\Omega$, eigenvalue inequality (1.16) is universal.

##  2\. A key lemma and its proof

In this section, we shall prove a key lemma, which will play an important role
in the proof of Theorem 1.1. Firstly, let us explain the operation rules of
the fractional Laplacian in a bounded domain. Let $u$ and $w$ is two functions
defined on $\Omega\subset\mathbb{R}^{n}$, and we recall that the fractional
Laplacian is defined by

$\left.(-\Delta)^{s}\right|_{\Omega}u:=\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left[u\chi_{\Omega}\right]\right],$

where

$\mathscr{F}[u](y)=\widehat{u}(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}e^{-\sqrt{-1}x\cdot
y}u(x)dx,$

and $\mathscr{F}^{-1}$ is the inverse of Fourier transform $\mathscr{F}$,
which is defined by

$\mathscr{F}^{-1}[w](x)=\widehat{w}(x)=\int_{\mathbb{R}^{n}}e^{\sqrt{-1}x\cdot
y}w(y)dy.$

By the definitions of Fourier transform and its inverse, we can conclude the
following green formula.

######  Proposition 2.1.

_(Green Formula)_ Let
$w_{i}:\Omega(\subset\mathbb{R}^{n})\rightarrow\mathbb{R}$ be two functions
with $w_{i}\in C^{\infty}(\Omega)$, where $i=1,2$. Then,

(2.1)
$\displaystyle\int_{\Omega}w_{1}(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{2}(x)dx=\int_{\Omega}w_{2}(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{1}(x)dx.$

###### Proof.

A straightforward calculation shows that

$\displaystyle\quad\int_{\Omega}w_{1}(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{2}(x)dx$
$\displaystyle=\int_{\mathbb{R}^{n}}(w_{1}\chi_{\Omega})(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{2}(x)dx$
$\displaystyle=\int_{\mathbb{R}^{n}}(w_{1}\chi_{\Omega})(x)\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left[(w_{2}\chi_{\Omega})(x)\right]\right]dx$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}(w_{1}\chi_{\Omega})(x)\mathscr{F}^{-1}\left[\int_{\mathbb{R}^{n}}|y|^{2s}e^{-\sqrt{-1}x\cdot
y}(w_{2}\chi_{\Omega})(x)dx\right]dx.$

Furthermore, the definition of inverse of Fourier transform implies that

$\displaystyle\quad\int_{\Omega}w_{1}(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{2}(x)dx$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}(w_{1}\chi_{\Omega})(x)\left\\{\int_{\mathbb{R}^{n}}e^{\sqrt{-1}x\cdot
y}\left[\int_{\mathbb{R}^{n}}|y|^{2s}e^{-\sqrt{-1}x\cdot
y}(w_{2}\chi_{\Omega})(x)dx\right]dy\right\\}dx$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}(w_{1}\chi_{\Omega})(x)e^{\sqrt{-1}x\cdot
y}|y|^{2s}e^{-\sqrt{-1}x\cdot y}(w_{2}\chi_{\Omega})(x)dxdydx$
$\displaystyle=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}(w_{2}\chi_{\Omega})(x)\left\\{\int_{\mathbb{R}^{n}}e^{\sqrt{-1}x\cdot
y}\left[\int_{\mathbb{R}^{n}}|y|^{2s}e^{-\sqrt{-1}x\cdot
y}(w_{1}\chi_{\Omega})(x)dx\right]dy\right\\}dx$
$\displaystyle=\int_{\mathbb{R}^{n}}(w_{2}\chi_{\Omega})(x)\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left[(w_{1}\chi_{\Omega})(x)\right]\right]dx$
$\displaystyle=\int_{\mathbb{R}^{n}}(w_{2}\chi_{\Omega})(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{1}(x)dx$
$\displaystyle=\int_{\Omega}w_{2}(x)\left.(-\Delta)^{s}\right|_{\Omega}w_{1}(x)dx,$

which gives (2.1). Thus, we finish the proof of Proposition 2.1. ∎

######  Lemma 2.1.

Suppose that $\\{x_{1},x_{2},\cdots,x_{n}\\}$ is an arbitrary Euclidean
coordinate. Let $\lambda_{j}$ and $u_{j}$ denote the $j$ -th eigenvalue and
the corresponding normalized eigenfunction of
$\left.(-\Delta)^{\alpha/2}\right|_{\Omega},$ respectively. Assume that

(2.2) $\int_{\Omega}h_{i}u_{1}u_{j+1}dv=0,\ \ for\ \ 1\leq j<i\leq n,$

where $h_{i}\in C^{\infty}(\Omega)$. Then,

(2.3)
$\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\Theta_{1}(h_{i})u_{1}dv\leq
2\left\\{\sum^{n}_{i=1}\int_{\Omega}\Theta_{2}(h_{i})u_{1}dv\sum^{n}_{i=1}\int_{\Omega}\Psi_{i}dv\right\\}^{\frac{1}{2}},$

where

(2.4) $\Theta_{1}(h_{i})=\frac{\partial}{\partial
x_{i}}(h_{i}u_{1})-h_{i}\frac{\partial}{\partial x_{i}}u_{1},$

(2.5)
$\Theta_{2}(h_{i})=\frac{1}{2}\left[2h_{i}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h^{2}_{i}(-\Delta)^{s}|_{\Omega}u_{1}-(-\Delta)^{s}|_{\Omega}(h_{i}^{2}u_{1})\right],$

(2.6) $\Psi_{i}=u_{1,i}^{2},$

and

$u_{1,i}=\frac{\partial u_{1}}{\partial x_{i}}.$

###### Proof.

Define function $\psi_{i}$ as follows:

$\psi_{i}=h_{i}u_{1}-\tau_{i}u_{1},$

where

$\tau_{i}=\int_{\Omega}h_{i}u_{1}^{2}dv.$

According to (2.2), one can conclude that

(2.7) $\int_{\Omega}\psi_{i}u_{j+1}dv=0,\ \ for\ \ 0\leq j<i\leq n.$

Moreover, it follows from (2.7) that

(2.8) $\int_{\Omega}\psi_{i}h_{i}u_{1}dv=\int_{\Omega}\psi_{i}^{2}dv.$

Since $\psi_{i}$ satisfies (2.7), by Rayleigh-Ritz inequality, we know that

(2.9)
$\lambda_{i+1}\leq\frac{\int_{\Omega}\psi_{i}(-\Delta)^{s}|_{\Omega}\psi_{i}dv}{\int_{\Omega}\psi_{i}^{2}dv}.$

Utilizing (2.7) and (2.8), we obtain

(2.10) $\displaystyle\int_{\Omega}\psi_{i}(-\Delta)^{s}|_{\Omega}\psi_{i}dv$
$\displaystyle=\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}\psi_{i}dv+\int_{\Omega}h_{i}\psi_{i}(-\Delta)^{s}|_{\Omega}u_{1}dv$
$\displaystyle=\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}\psi_{i}dv+\lambda_{1}\int_{\Omega}\psi_{i}^{2}dv.$

Substituting (LABEL:2.7) into (2.9), we yield

(2.11)
$(\lambda_{i+1}-\lambda_{1})\int_{\Omega}\psi_{i}^{2}dv\leq\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}\psi_{i}dv.$

It follows from the Green formula (2.1) that,

$\displaystyle\quad\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}u_{1}dv$
$\displaystyle=\int_{\Omega}h_{i}u_{1}(-\Delta)^{s}|_{\Omega}u_{1}dv-\int_{\Omega}h_{i}u_{1}(-\Delta)^{s}|_{\Omega}u_{1}dv$
$\displaystyle=0,$

which implies that

(2.12)
$\displaystyle\quad\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}\psi_{i}dv$
$\displaystyle=\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}h_{i}u_{1}dv.$

Consequently, plugging (LABEL:2.9) into (2.11), one has

(2.13)
$(\lambda_{i+1}-\lambda_{1})\int_{\Omega}\psi_{i}^{2}dv\leq\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}h_{i}u_{1}dv.$

By the formula of partial integration, we infer that

(2.14) $\int_{\Omega}u_{1,i}u_{1}dv=0,$ (2.15)
$2\int_{\Omega}u_{1}\frac{\partial(h_{i}u_{1})}{\partial
x_{i}}dv=\int_{\Omega}\left[\frac{\partial(h_{i}u_{1})}{\partial
x_{i}}-h_{i}\frac{\partial}{\partial x_{i}}u_{1}\right]u_{1}dv,$

and

(2.16)
$-2\int_{\Omega}u_{1,i}\psi_{i}dv=2\int_{\Omega}u_{1}\frac{\partial(h_{i}u_{1})}{\partial
x_{i}}dv+2\tau_{i}\int_{\Omega}u_{1,i}u_{1}dv.$

Hence, utilizing (2.14), (2.15) and (2.16), we obtain

(2.17) $\int_{\Omega}\left[\frac{\partial}{\partial
x_{i}}(h_{i}u_{1})-h_{i}\frac{\partial}{\partial
x_{i}}u_{1}\right]u_{1}dv=-2\int_{\Omega}\psi_{i}u_{1,i}dv.$

Multiplying both sides of (2.17) by
$(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}$ and noticing (2.13), one can infer
that

(2.18)
$\displaystyle(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\left[\frac{\partial}{\partial
x_{i}}(h_{i}u_{1})-h_{i}\frac{\partial}{\partial x_{i}}u_{1}\right]u_{1}dv$
$\displaystyle=-2(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\psi_{i}u_{1,i}dv$
$\displaystyle\leq\delta(\lambda_{i+1}-\lambda_{1})\int_{\Omega}\psi_{i}^{2}dv+\frac{1}{\delta}\int_{\Omega}u_{1,i}^{2}dv$
$\displaystyle\leq\delta\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}h_{i}u_{1}dv+\frac{1}{\delta}\int_{\Omega}u_{1,i}^{2}dv,$

where $\delta$ is a positive constant. Taking sum on $i$ from $1$ to $n$ in
(LABEL:2.14), we get

(2.19)
$\displaystyle\quad\delta^{2}\sum^{n}_{i=1}\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}h_{i}u_{1}dv$
$\displaystyle\quad-\delta\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\left[\frac{\partial}{\partial
x_{i}}(h_{i}u_{1})-h_{i}\frac{\partial}{\partial
x_{i}}u_{1}\right]u_{1}dv+\sum^{n}_{i=1}\int_{\Omega}u_{1,i}^{2}dv$
$\displaystyle\geq 0.$

Here, the left-hand side of (LABEL:2.15) is a quadratic polynomial of
$\delta$. By direct calculation, we show that

(2.20) $\displaystyle
2\int_{\Omega}\Big{[}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\Big{]}h_{i}u_{1}dv$
$\displaystyle=\int_{\Omega}h_{i}\left[(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}u_{1}\right]u_{1}dv$
$\displaystyle\quad-\int_{\Omega}\left[(-\Delta)^{s}|_{\Omega}(h_{i}^{2}u_{1})-h_{i}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})\right]u_{1}dv$
$\displaystyle=\int_{\Omega}\left[2h_{i}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h^{2}_{i}(-\Delta)^{s}|_{\Omega}u_{1}-(-\Delta)^{s}|_{\Omega}(h_{i}^{2}u_{1})\right]u_{1}dv.$

Combining (LABEL:2.15) with (LABEL:ineq-D-hu), we get

(2.21)
$\displaystyle\frac{\delta^{2}}{2}\sum^{n}_{i=1}\int_{\Omega}\left[2h_{i}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h^{2}_{i}(-\Delta)^{s}|_{\Omega}u_{1}-(-\Delta)^{s}|_{\Omega}(h_{i}^{2}u_{1})\right]u_{1}dv$
$\displaystyle-\delta\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\left[\frac{\partial}{\partial
x_{i}}(h_{i}u_{1})-h_{i}\frac{\partial}{\partial
x_{i}}u_{1}\right]u_{1}dv+\sum^{n}_{i=1}\int_{\Omega}u_{1,i}^{2}dv$
$\displaystyle\geq 0.$

From (2.11),(LABEL:2.9) and (LABEL:ineq-D-hu), we can conclude that

$\int_{\Omega}\left[2h_{i}(-\Delta)^{s}|_{\Omega}(h_{i}u_{1})-h^{2}_{i}(-\Delta)^{s}|_{\Omega}u_{1}-(-\Delta)^{s}|_{\Omega}(h_{i}^{2}u_{1})\right]u_{1}dv\geq
0,$

for $i=1,\cdots,n$, which means that its discriminant must be non-positive.
Hence, from (2.4), (2.5), (2.6), and (LABEL:2.15-1), we yield (2.3).
Therefore, we finish the proof of Lemma 2.1.∎

##  3\. Proof of Theorem 1.1

In this section, we shall give the proof of Theorem 1.1 by applying Lemma 2.1
proved in the previous section. Firstly, we need a result proved by Hook [30]
in 1990 as follows.

######  Proposition 3.1.

(See [30]) Let $\mathcal{H}$ be a real or complex inner product space. Let
$\mathcal{M}$ be a linear submanifold of $\mathcal{H}$ and
$\mathcal{L}:\mathcal{M}\rightarrow\mathcal{H}$ be a linear operator in
$\mathcal{H}$. Suppose $l$ is a positive integer and $u$ is a fixed vector
such that for all integers $0\leq r\leq q\leq l$

$\left|\left\langle\mathcal{L}^{q}u,u\right\rangle\right|=|\left\langle\mathcal{L}^{q-r}u,\mathcal{L}^{r}u\right\rangle|.$

Then, for all integers $0\leq r\leq q\leq l$, when $q$ is even, we have

(3.1)
$\left|\left\langle\mathcal{L}^{r}u,u\right\rangle\right|\leq\left|\left\langle\mathcal{L}^{q}u,u\right\rangle\right|^{r/q}\langle
u,u\rangle^{1-r/q}.$

This inequality is also true for $q$ odd and $0\leq r\leq q\leq l$. Moreover,
there is a finite collection of operators
$\left\\{\mathcal{D}_{j}\right\\}_{j=1}^{N}$ in $\mathcal{H}$ such that

(3.2)
$\left|\left\langle\mathcal{L}^{q}u,u\right\rangle\right|=\left|\sum_{j=1}^{N}\left\langle\mathcal{D}_{j}\mathcal{L}^{q-r}u,\mathcal{D}_{j}\mathcal{L}^{r-1}u\right\rangle\right|.$

holds for each pair of integers $r$ and $q$ with $1\leq r\leq q\leq l$.

######  Remark 3.1.

(See Remark (ii) in [30] or Remark 2.2 in [16]) If $\mathcal{L}$ is any (not
necessarily bounded) nonnegative self-adjoint operator, then (3.2) is
satisfied by taking $N=1$ and letting $\mathcal{D}_{1}$ be any square root of
$\mathcal{L}$. Hence, for nonnegative self-adjoint operators, the inequality
(3.1) holds for all integers $q$.

By Proposition 3.1, we shall prove the following lemma.

######  Lemma 3.1.

Assume that the function $\Theta_{2}$ is given by (2.5).Then, under the same
condition as Theorem 1.1, we have

(3.3)
$\displaystyle\sum_{i=1}^{n}\int_{\Omega}\Theta_{2}(x_{i})u_{1}dv=s(n+2s-2)\lambda_{1}^{\frac{s-1}{s}},$

where $x_{1},x_{2},\cdots,x_{n}$ are $n$ arbitrary coordinate functions on
$\mathbb{R}^{n}$.

###### Proof.

We consider the operator $\left.(-\Delta)^{s}\right|_{\Omega}$ and restrict it
to functions supported within bounded open domain
$\Omega\subset\mathbb{R}^{n}$. In what follows, some calculations are direct,
and we refer the reader to [16]. In addition, some more detailed calculations
also could be found in [42]. For example, by the definition of fractional
Laplacian and utilizing some properties of Fourier transformation, we can
prove

(3.4)
$\displaystyle\quad\Big{(}(-\Delta)^{s}|_{\Omega}x_{i}-x_{i}(-\Delta)^{s}|_{\Omega}\Big{)}u_{1}$
$\displaystyle=\chi_{\Omega}\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left(x_{i}u_{1}\chi_{\Omega}\right)\right]-\chi_{\Omega}x_{i}\mathscr{F}^{-1}\left[|y|^{2s}\hat{u}_{1}\chi_{\Omega}\right]$
$\displaystyle=\sqrt{-1}\chi_{\Omega}\mathscr{F}^{-1}\left[|y|^{2s}\frac{\partial\hat{u}_{1}}{\partial
y_{i}}-\frac{\partial}{\partial y_{i}}\left(|y|^{2s}\hat{u}_{1}\right)\right]$
$\displaystyle=-\sqrt{-1}\chi_{\Omega}\mathscr{F}^{-1}\left[2s|y|^{2s-2}y_{i}\hat{u}_{1}\right],$

where $u_{1}$ denotes the first eigenfunction of
$\left.(-\Delta)^{s}\right|_{\Omega}$. According to (2.5) and the similar
calculation as (LABEL:ineq-3-det), one can infer by putting $h_{i}=x_{i}$ that

(3.5) $\displaystyle\quad 2\Theta_{2}(x_{i})$
$\displaystyle=2x_{i}(-\Delta)^{s}|_{\Omega}(x_{i}u_{1})-x_{i}^{2}(-\Delta)^{s}|_{\Omega}u_{1}-(-\Delta)^{s}|_{\Omega}(x_{i}^{2}u_{1})$
$\displaystyle=2\chi_{\Omega}x_{i}\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left(x_{i}u_{1}\right)\right]-\chi_{\Omega}x_{i}^{2}\mathscr{F}^{-1}\left(|y|^{2s}\hat{u}_{1}\right)-\chi_{\Omega}\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left(x_{i}^{2}u_{1}\right)\right]$
$\displaystyle=\chi_{\Omega}\left[2\sqrt{-1}x_{i}\mathscr{F}^{-1}\left(|y|^{2s}\frac{\partial\hat{u}_{1}}{\partial
y_{i}}\right)-x_{i}^{2}\mathscr{F}^{-1}\left(|y|^{2s}\hat{u}_{1}\right)+\mathscr{F}^{-1}\left(|y|^{2s}\frac{\partial^{2}\hat{u}_{1}}{\partial
y_{i}^{2}}\right)\right]$
$\displaystyle=\chi_{\Omega}\mathscr{F}^{-1}\left[-2\frac{\partial}{\partial
y_{i}}\left(|y|^{2s}\frac{\partial\hat{u}_{1}}{\partial
y_{i}}\right)+\frac{\partial^{2}}{\partial
y_{i}^{2}}\left(|y|^{2s}\hat{u}_{1}\right)+|y|^{2s}\frac{\partial^{2}\hat{u}_{1}}{\partial
y_{i}^{2}}\right]$
$\displaystyle=\chi_{\Omega}\mathscr{F}^{-1}\left[\left(2s|y|^{2s-2}+2s(2s-2)|y|^{2s-4}y_{i}^{2}\right)\hat{u}_{1}\right].$

Since the support of $u_{1}$ is in $\Omega,$ from (LABEL:xax), we have

$\displaystyle\sum_{i=1}^{n}\int_{\Omega}\Theta_{2}(x_{i})u_{1}dv$
$\displaystyle=\frac{1}{2}\sum_{i=1}^{n}\int_{\Omega}\left[2x_{i}(-\Delta)^{s}|_{\Omega}(x_{i}u_{1})-x_{i}^{2}(-\Delta)^{s}|_{\Omega}u_{1}-(-\Delta)^{s}|_{\Omega}(x_{i}^{2}u_{1})\right]u_{1}dv$
$\displaystyle=\frac{1}{2}\sum_{i=1}^{n}\int_{\Omega}\mathscr{F}^{-1}\left[\left(2s|y|^{2s-2}+2s(2s-2)|y|^{2s-4}y_{i}^{2}\right)\hat{u}_{1}\right]u_{1}dv$
$\displaystyle=\sum_{i=1}^{n}\int_{\Omega}\left(s|y|^{2s-2}+s(2s-2)|y|^{2s-4}y_{i}^{2}\right)\hat{u}_{1}^{2}dv$
$\displaystyle=s(n+2s-2)\int_{\Omega}|y|^{2s-2}\hat{u}_{1}^{2}dv$
$\displaystyle=s(n+2s-2)\int_{\Omega}u_{1}\mathscr{F}^{-1}\left(|y|^{2s-2}\hat{u}_{1}\right)dv$
$\displaystyle=s(n+2s-2)\int_{\Omega}u_{1}(-\Delta)^{s-1}u_{1}dv.$

Next, recall that $\mathbb{Q}_{+}$ is the set of all of the positive rational
numbers and consider the following three cases.

Case I. We assume that $s=\frac{1}{m},~{}m\geq 2$, where $m$ is a positive
integer. Then, we have

$\displaystyle\int_{\Omega}u_{1}(-\Delta)^{s-1}u_{1}dv$
$\displaystyle=\int_{\Omega}u_{1}\left[(-\Delta)^{-\frac{1}{m}}\right]^{m-1}u_{1}dv$
$\displaystyle=\int_{\Omega}u_{1}\left(((-\Delta)^{s})^{-1}\right)^{m-1}u_{1}dv$
$\displaystyle=\left(\lambda_{1}^{-1}\right)^{m-1}$
$\displaystyle=\lambda_{1}^{\frac{s-1}{s}}.$

Case II. We assume that $\frac{1}{2}<s<1,$ where $s\in\mathbb{Q}_{+}$, and
denote $s=\frac{a_{2}}{a_{1}},$ where $a_{1},a_{2}$ are positive integers and
$a_{2}<a_{1}<2a_{2}.$ Meanwhile, putting

(3.6) $\mathcal{L}=(-\Delta)^{-\frac{1}{a_{1}}},$

it is well known that it is a positive and self-adjoint operator. Thus, the
operator $\mathcal{L}$ given by (3.6) satisfies the condition in Proposition
3.1. Furthermore, by Remark 3.1, then one can show that inequality (3.1) is
valid for all $0\leq r\leq q\leq l$ without parity condition on $q.$ Applying
$0<r=a_{1}-a_{2}<a_{2}$ and $q=a_{2}$ to (3.1), we have

$\displaystyle\int_{\Omega}u_{1}(-\Delta)^{s-1}u_{1}dv$
$\displaystyle=\int_{\Omega}u_{1}\left[(-\Delta)^{-\frac{1}{a_{1}}}\right]^{a_{1}-a_{2}}u_{1}dv$
$\displaystyle\leq\left[\int_{\Omega}u_{1}\left[(-\Delta)^{-\frac{1}{a_{1}}}\right]^{a_{2}}u_{1}dv\right]^{\frac{a_{1}-a_{2}}{a_{2}}}\left[\int_{\Omega}u_{1}^{2}dv\right]^{1-\frac{a_{1}-a_{2}}{a_{2}}}$
$\displaystyle=\left(\lambda_{1}^{-1}\right)^{\frac{a_{1}-a_{2}}{a_{2}}}$
$\displaystyle=\lambda_{1}^{\frac{s-1}{s}}.$

Case III. We assume that $s\geq 1$, where $s\in\mathbb{Q}_{+}$, and denote
$s=\frac{a_{2}}{a_{1}},$ where $a_{1},a_{2}$ are positive integers and
$a_{2}\geq a_{1}$. In Proposition 3.1, we shall choose an operator
$\mathcal{L}$ satisfying

$\mathcal{L}=(-\Delta)^{\frac{1}{a_{1}}}.$

Then, $\mathcal{L}$ is a positive and self-adjoint operator. According to
Remark 3.1, we know that inequality (3.1) is true for all $0\leq r\leq q\leq
l$ without parity condition on $q$. Applying (3.1) with $0\leq
r=a_{2}-a_{1}<a_{2}$ and $q=a_{2},$ we can deduce that

$\displaystyle\int_{\Omega}u_{1}(-\Delta)^{s-1}u_{1}dv$
$\displaystyle=\int_{\Omega}u_{1}\left[(-\Delta)^{\frac{1}{a_{1}}}\right]^{a_{2}-a_{1}}u_{1}dv$
$\displaystyle\leq\left[\int_{\Omega}u_{1}\left[(-\Delta)^{\frac{1}{a_{1}}}\right]^{a_{2}}u_{1}dv\right]^{\frac{a_{2}-a_{1}}{a_{2}}}\left[\int_{\Omega}u_{1}^{2}dv\right]^{1-\frac{a_{2}-a_{1}}{a_{2}}}$
$\displaystyle=\lambda_{1}^{\frac{a_{2}-a_{1}}{a_{2}}}$
$\displaystyle=\lambda_{1}^{\frac{s-1}{s}}.$

Hence, for all $s=\frac{1}{m}$ with the positive integer $m\geq 2$ or
$s>\frac{1}{2}$ with $s\in\mathbb{Q}_{+},$ we have

(3.7) $\displaystyle\sum_{i=1}^{n}\int_{\Omega}\Theta_{2}(x_{i})u_{1}dv$
$\displaystyle=s(n+2s-2)\int_{\Omega}u_{1}(-\Delta)^{s-1}u_{1}dv$
$\displaystyle\leq s(n+2s-2)\lambda_{1}^{\frac{s-1}{s}}.$

Therefore, we finish the proof of Lemma 3.1. ∎

In what follows, we assume that $x_{1},x_{2},\cdots,x_{n}$ are $n$ coordinate
functions on $\mathbb{R}^{n}$. In order to make use of Lemma 2.1, we construct
some functions satisfying (2.2) by using $\left\\{x_{i}\right\\}_{i=1}^{n}.$
We consider an $n\times n$ matrix

$Q=\left(\int_{\Omega}x_{i}u_{1}u_{j+1}dv\right)_{n\times n}.$

According to the QR-factorization theorem, we know that there exists an
orthogonal $n\times n$ matrix $P=\left(p_{ij}\right)_{n\times n},$ such that
$U=PQ$ is an upper triangle matrix. Namely, we have

$\sum_{k=1}^{n}p_{ik}\int_{\Omega}x_{k}u_{1}u_{j+1}dv=0,\quad\text{ for }1\leq
j<k\leq n.$

Define functions $\overline{x}_{i}$ by

(3.8) $\overline{x}_{i}=\sum_{k=1}^{n}p_{ik}x_{k}.$

Thus we infer

(3.9) $\int_{\Omega}\overline{x}_{i}u_{1}u_{j+1}dv=0,\quad\text{ for }1\leq
j<i\leq n.$

Moreover, because $P$ is an orthogonal matrix, by the orthogonality, we infer
that

(3.10)
$\displaystyle\sum^{n}_{i=1}\overline{x}_{i}(-\Delta)^{s}|_{\Omega}(\overline{x}_{i}u_{1})$
$\displaystyle=\sum^{n}_{i=1}\left(\sum_{k=1}^{n}p_{ik}x_{k}\right)(-\Delta)^{s}|_{\Omega}\left(\sum_{l=1}^{n}p_{il}x_{l}u_{1}\right)$
$\displaystyle=\sum^{n}_{i=1}\left(\sum_{k=1}^{n}p_{ik}x_{k}\right)\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left[\left(\sum_{l=1}^{n}p_{il}x_{l}u_{1}\right)\chi_{\Omega}\right]\right]$
$\displaystyle=\sum^{n}_{i=1}\left(\sum_{k=1}^{n}p_{ik}x_{k}\right)\sum_{l=1}^{n}p_{il}\left\\{\mathscr{F}^{-1}\left[|y|^{2s}\mathscr{F}\left[\left(x_{l}u_{1}\right)\chi_{\Omega}\right]\right]\right\\}$
$\displaystyle=\sum^{n}_{i=1}\left(\sum_{k=1}^{n}p_{ik}x_{k}\right)\sum_{l=1}^{n}p_{il}(-\Delta)^{s}|_{\Omega}(x_{l}u_{1})$
$\displaystyle=\sum_{k,l=1}^{n}\left(\sum^{n}_{i=1}p_{ik}p_{il}\right)x_{k}(-\Delta)^{s}|_{\Omega}(x_{l}u_{1})$
$\displaystyle=\sum^{n}_{i=1}x_{i}(-\Delta)^{s}|_{\Omega}(x_{i}u_{1}).$

Similarly, we can prove that

(3.11)
$\displaystyle\sum^{n}_{i=1}\overline{x}^{2}_{i}(-\Delta)^{s}|_{\Omega}u_{1}=\sum^{n}_{i=1}x^{2}_{i}(-\Delta)^{s}|_{\Omega}u_{1},$

and

(3.12)
$\displaystyle\sum^{n}_{i=1}(-\Delta)^{s}|_{\Omega}(\overline{x}_{i}^{2}u_{1})=\sum^{n}_{i=1}(-\Delta)^{s}|_{\Omega}(x_{i}^{2}u_{1}).$

From (3.10), (3.11), (3.12) and (3.3), we have

$\displaystyle\sum_{i=1}^{n}\int_{\Omega}\Theta_{2}(\overline{x}_{i})u_{1}dv$
$\displaystyle=s(n+2s-2)\lambda_{1}^{\frac{s-1}{s}}.$

Therefore, based on the above argument, we have the following lemma.

######  Lemma 3.2.

Assume that the function $\Theta_{2}$ is given by (2.5).Then, under the same
condition as Theorem 1.1, we have

(3.13)
$\displaystyle\sum_{i=1}^{n}\int_{\Omega}\Theta_{2}(\overline{x}_{i})u_{1}dv=s(n+2s-2)\lambda_{1}^{\frac{s-1}{s}},$

where $\overline{x}_{1},\overline{x}_{2},\cdots,\overline{x}_{n}$ are $n$
Euclidean coordinate functions satisfying (3.8).

$None$

Next, applying Lemma 2.1 and Lemma 3.2, we shall give the proof of Theorem
1.1. _Proof of Theorem_ 1.1. Assume that $\\{x_{1},x_{2},\cdots,x_{n}\\}$ is a
coordinate system given in Lemma 3.1 on $\mathbb{R}^{n}$. After a rotation
transformation for the coordinate system $\\{x_{1},x_{2},\cdots,x_{n}\\}$, we
can construct a new coordinate system
$\\{\overline{x}_{1},\overline{x}_{2},\cdots,\overline{x}_{n}\\}$ satisfying
equation (3.8). Furthermore, in (2.3), we take $h_{i}=\overline{x}_{i}$, where
$i=1,2,\cdots,n$. Then, under the new coordinate system
$\\{\overline{x}_{1},\overline{x}_{2},\cdots,\overline{x}_{n}\\}$, we have

(3.14)
$\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\Theta_{1}(\overline{x}_{i})u_{1}dv\leq
2\left\\{\sum^{n}_{i=1}\int_{\Omega}\Theta_{2}(\overline{x}_{i})u_{1}dv\sum^{n}_{i=1}\int_{\Omega}\Psi_{i}dv\right\\}^{\frac{1}{2}},$

and $\int_{\Omega}\overline{x}_{i}u_{1}u_{j+1}dv=0,\ \ {\rm for}~{}{\rm all}\
\ 0\leq j<i\leq n,$ which happens to satisfy condition (2.2) in Lemma 2.1.
From (2.4), we have

$\displaystyle\int_{\Omega}\Theta_{1}(\overline{x}_{i})u_{1}dv$
$\displaystyle=\int_{\Omega}\left[\frac{\partial}{\partial\overline{x}_{i}}(\overline{x}_{i}u_{1})-\overline{x}_{i}\frac{\partial}{\partial\overline{x}_{i}}u_{1}\right]u_{1}dv$
$\displaystyle=\int_{\Omega}\left[u_{1}\frac{\partial}{\partial\overline{x}_{i}}(\overline{x}_{i})+\overline{x}_{i}\frac{\partial}{\partial\overline{x}_{i}}u_{1}-\overline{x}_{i}\frac{\partial}{\partial\overline{x}_{i}}u_{1}\right]u_{1}dv$
$\displaystyle=\int_{\Omega}u_{1}^{2}dv=1,$

which implies that

(3.15)
$\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}\Theta_{1}(\overline{x}_{i})u_{1}dv=\sum^{n}_{i=1}(\lambda_{i+1}-\lambda_{1})^{\frac{1}{2}}.$

Moreover, we suppose $s>0$ with $s\in\mathbb{Q}_{+},$ and denote
$s=\frac{a_{2}}{a_{1}}$ as before, where $a_{1},a_{2}$ are two positive
integers. Furthermore, we assume that the operator $\mathcal{L}$ satisfies
that $\mathcal{L}=-\Delta.$ Then, it is clear that the conditions of
Proposition 3.1 are obviously satisfied by the above choice of the operator
$\mathcal{L}$. Therefore, for all $0\leq r\leq q\leq l$, inequality (3.1) is
true without parity condition on $q.$ Applying inequality (3.1) with $r=1$ and
$q=a_{2},$ it is not difficult to give the upper bound for the function
$\Psi_{i}$ as follows:

(3.16) $\displaystyle\sum^{n}_{i=1}\int_{\Omega}\Psi_{i}dv$
$\displaystyle=\int_{\Omega}u_{1}(-\Delta)u_{1}dv$
$\displaystyle\leq\left\\{\int_{\Omega}u_{1}(-\Delta)^{a_{2}}u_{1}dv\right\\}^{\frac{1}{a_{2}}}\left\\{\int_{\Omega}u_{1}^{2}dv\right\\}^{1-\frac{1}{a_{2}}}$
$\displaystyle=\left\\{\int_{\Omega}u_{1}\left((-\Delta)^{\frac{a_{2}}{a_{1}}}\right)^{a_{1}}u_{1}dv\right\\}^{\frac{1}{a_{2}}}$
$\displaystyle=\left\\{\int_{\Omega}u_{1}\Big{(}\left(-\Delta\right)^{s}\Big{)}^{a_{1}}u_{1}dv\right\\}^{\frac{1}{a_{2}}}$
$\displaystyle=\lambda_{1}^{\frac{a_{1}}{a_{2}}}=\lambda_{1}^{\frac{1}{s}}.$

Substituting (3.13), (3.15) and (3.16) into (3.14), we get the inequality
(1.16). Therefore, we finish the proof of Theorem 1.1.

$None$

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# On a metric on the space of monetary risk measures

Sh. A. Ayupov, A. A. Zaitov

###### Abstract

We introduce a metric on the space of monetary risk measure, which generates
the point-wise convergence topology and extends the metric on the initial
compactum.

2010 Mathematics Subject Classification. 28C20; 52A30; 54C10.

Key words and phrases: compactum, metric, monetary risk measure

###### Contents

  1. 1 Introduction
  2. 2 Basic constructions
  3. 3 On a metric on the space of monetary risk measures

##  1 Introduction

The financial market faces risks arising from many types of uncertain losses,
including market risk, credit risk, liquidity risk, operational risk, etc. In
1988, the Basel Committee on Banking Supervision proposed measures to control
credit risk in banking. A risk measure called the value-at-risk, acronym VaR,
became, in the 1990s, an important tool of risk assessment and management for
banks, securities companies, investment funds, and other financial
institutions in asset allocation and performance evaluation. The VaR
associated with a given confidence level for a venture capital is the upper
limit of possible losses in the next certain period of time. In 1996 the Basel
Committee on Banking Supervision endorsed the VaR as one of the acceptable
methods for the banks internal risk measure. However, due to the defects of
VaR, a variety of new risk measures came into being. This paper focuses on the
definition the distance between normed monetary risk measures. For an overview
of the subject we refer to [1].

Assume that all possible states and events that may occur at the terminal time
are known, namely, a measurable space $(X,\,\mathcal{F})$ is given. The
financial position (here refers to the wealth deducted investment cost) is
usually described by a measurable function $\varphi$ on $(X,\,\mathcal{F})$.
If we assume that a probability measure $\mathbb{P}$ is given on measurable
space $(X,\,\mathcal{F})$, the financial positions is usually described by a
random variables $\widetilde{\varphi}$ on $(X,\,\mathcal{F},\,\mathbb{P})$. In
order to facilitate the notation and description, we use
$\varphi=-\widetilde{\varphi}$ to denote the potential loss at the terminal
time of trading. Here the potential loss is relative to a reference point in
terms. If $\varphi$ is a negative value, it indicates a surplus.

A risk measure is a numerical value $\mu(\varphi)$ to quantify the risk of a
financial position (it may be a potential loss or a surplus as well)
$\varphi$. If we denote the set of financial positions to be considered by
$\mathcal{G}$, a risk measure $\mu$ is a map from $\mathcal{G}$ to
$\mathbb{R}$. Usually, they take $L^{\infty}(X,\,\mathcal{F},\,\mathbb{P})$ or
$L^{\infty}(X,\,\mathcal{F})$ as the set of all financial positions
$\mathcal{G}$ or $\mathcal{G}(\mathbb{R})$, where the former is the set of all
bounded $\mathcal{F}$ measurable functions on $(X,\,\mathcal{F})$, endowed
with the uniform norm $\|\cdot\|_{\infty}$, and the latter is the set of
equivalence classes of the former under probability $\mathbb{P}$. In the
former case, the states and the probabilities of the possible events are
unknown or are not consensus in the market, and then the risk measure is
called model-free. In the latter case, the risk measure is called model-
dependent. In the model-dependent case, naturally, we always assume that the
risk measure $\mu$ satisfies the following property: If $\varphi=\psi$,
$\mathbb{P}$-a.s., then $\mu(\varphi)=\mu(\psi)$.

###### Definition 1.1

A map $\mu$ from $\mathcal{G}$ to $\mathbb{R}$ is called a monetary risk
measure, abbreviated as risk measure, if it satisfies two conditions:

  * $(1)$

monotonicity: For all $\varphi,\,\psi\in\mathcal{G}$ satisfying
$\varphi\leq\psi$, it holds that $\mu(\varphi)\leq\mu(\psi)$.

  * $(2)$

translation invariance: For all $\varphi\in\mathcal{G}$ and any real number
$\alpha$, it holds that

$\displaystyle\mu(\varphi+\alpha)=\mu(\varphi)+\alpha.$

It is known that the algebra $L^{\infty}(X,\,\mathcal{B}(X),\,\mathbb{P})$ is
isomorphic to the algebra $C(X)$ of all continuous functions on a compact $X$
(to be more precise, $X$ is so called hyperstonean compact). Thus the above
notion of financial position can be interpreted as an element of the algebra
$C(X)$, while the monetary risk measure can be considered as a map from $C(X)$
to $\mathbb{R}$.

In the present paper we consider the problem in a more general setting, where
financial positions are interpreted as elements of the algebra $C(X)$, where
$X$ is a compact Hausdorff space. Then the algebra $C(X)$ plays a role of the
set $\mathcal{G}$ of all financial positions. A risk measure is a numerical
value $\mu(\varphi)$ to quantify the risk of a financial position $\varphi\in
C(X)$. A map $\mu$ from $C(X)$ to $\mathbb{R}$ is called a monetary risk
measure on $X$, if it satisfies the conditions $(1)$ and $(2)$ of Definition
1.1.

A monetary risk measure $\mu\colon C(X)\to\mathbb{R}$ is called normed if

  * $(3)$

$\mu(1_{X})=1$.

In works [2] – [20] normed monetary risk measures are called as order-
preserving functionals and the set of such functionals is denoted by $O(X)$.
The mentioned papers were devoted to study of $O(X)$. We consider $O(X)$ as a
subspace of the Tychonoff product $\mathbb{R}^{C(X)}$. The base of the induced
topology consists of the sets of the form

$\displaystyle\langle\mu;\,\varphi_{1},\,\dots,\,\varphi_{n};\,\varepsilon\rangle=\\{\nu\in
O(X):\,|\nu(\varphi_{i})-\mu(\varphi_{i})|<\varepsilon,\,i=1,\,\dots,\,n\\},$

where $\mu\in O(X)$, $\varphi_{i}\in C(X)$, $i=1,\,\dots,\,n$, and
$\varepsilon>0$. Note that the induced topology and point-wise convergence
topology coincide. For every compact Hausdorff space $X$ the space $O(X)$ is
also a compact Hausdorff space. $O(X)$ is a compact sublattice of
$\mathbb{R}^{C(X)}$.

A closed subset $A$ of $O(X)$ is called $O$-convex if for each $\mu\in O(X)$
with $\inf\,A\leq\mu\leq\sup\,A$ we have $\mu\in A$. It is known ([3], Lemma
3) that for each map $f\colon X\to Y$ and $\nu\in O(Y)$ the preimage
$O(f)^{-1}(\nu)$ is an $O$-convex subset of $O(X)$.

###### Proposition 1.1

Let $A_{1}$, $A_{2}$ be $O$-convex subsets of $O(X)$. Then $A_{1}\cap A_{2}$
is an $O$-convex subset of $O(X)$.

Proof. The proof consists of directly checking.

$\scriptstyle\blacksquare$

Let $X$, $Y$ be compact Hausdorff spaces, $f\colon X\to Y$ be a continuous
map. Then a map $O(f)\colon O(X)\to O(Y)$, defined as
$O(f)(\mu)(\varphi)=\mu(\varphi\circ f)$, $\varphi\in C(Y)$, is continuous.
So, the monetary risk measures $\mu$ and $O(f)(\mu)$ act the same rule. Let us
denote $\mu_{X}=\mu$, $\mu_{Y}=O(f)(\mu)$. Then the discussed situation is
bring to light: $\mu_{X}(\psi)=\mu(\psi)$, $\psi\in C(X)$, and
$\mu_{Y}(\varphi)=\mu(\varphi\circ f)$, $\varphi\in C(Y)$, i. e. $\mu_{X}$ and
$\mu_{Y}$ both act by the rule $\mu$.

Let $\mu_{i}\in O(X)$, $i=1,\,2$. We say that $\mu_{1}=\mu_{2}$ if
$\mu_{1}(\varphi)=\mu_{2}(\varphi)$ for all $\varphi\in C(X)$. The following
statement is rather obvious.

###### Proposition 1.2

Let $\mu_{i}\in O(X)$, $i=1,\,2$. Then $\mu_{1}=\mu_{2}$ if and only if
$\mbox{supp}\,\mu_{1}=\mbox{supp}\,\mu_{2}$ and $\mu_{2}=O(h)(\mu_{1})$, where
$h\colon\,\mbox{supp}\,\mu_{1}\rightarrow\mbox{supp}\,\mu_{2}$ is the identity
map.

Note that if $f$ is a surjective map then $O(f)$ is also a surjective map. If
$X$ is a closed subset of $Y$ and $f$ is an embedding then $O(f)$ is also an
embedding.

Let $X$ be a compact Hausdorff space and $\mu$ be a monetary risk measure.
$\mu$ is concentrated on a closed subset $A$ of $X$ if $\mu\in O(A)$. Note
that (Lemma 4, [2]) for a closed subset $A\subset X$ a monetary risk measure
$\mu\in O(X)$ is supported on $A$ if and only if for every pair
$\varphi,\,\psi\in C(X)$ such that $\varphi|_{A}=\psi|_{A}$ one has
$\mu(\varphi)=\mu(\psi)$. The smallest (with respect to inclusion) closed
subset $\mbox{supp}\mu$ of $X$ on which $\mu$ is concentrated is said to be a
support of monetary risk measure $\mu$. Evidently,

$\displaystyle\mbox{supp}\,\mu=\cap\left\\{A:\,A\,\mbox{ is a closed set in
}\,X\,\mbox{ and }\,\mu\in O(A)\right\\}.$

For a point $x\in X$ the Dirac measure $\delta_{x}$, defined by
$\delta_{x}(\varphi)=\varphi(x)$, $\varphi\in C(X)$, is a monetary risk
measure, concentrated at the singleton $\\{x\\}$, i. e.
$\mbox{supp}\,\delta_{x}=\\{x\\}$.

A subset $L\subset C(X)$ is called an $A$-subspace if $0_{X}\in L$ and for
every $\varphi\in L$ and every $c\in\mathbb{R}$ we have $\varphi+c_{X}\in L$.
According to the analog of the Hahn-Banach theorem (see, [2], [12]) for every
normed monetary risk measure $\mu\colon\,L\to\mathbb{R}$ there exists a normed
monetary risk measure $\widetilde{\mu}\colon\,C(X)\to\mathbb{R}$ such that
$\widetilde{\mu}|_{L}=\mu$.

The space $O(X)$ of monetary risk measure does not embed into any linear space
with finite algebraic dimension if $X$ consists more than one point.

###### Example 1.1

[20]. Let $X=\\{0,\,1\\}$ be a discrete two-point space. Then
$C(X)=\mathbb{R}^{2}$. Each functional $\mu\colon C(X)\to\mathbb{R}$ defined
by the equality

$\mu(\varphi)=\alpha_{1}\,\varphi(0)+\alpha_{2}\,\varphi(1)+\alpha_{3}\,\max\\{\varphi(0)+\lambda_{1},\,\varphi(1)+\lambda_{2}\\}+\\\
+\alpha_{4}\,\min\\{\varphi(0)+\lambda_{3},\,\varphi(1)+\lambda_{4}\\}+\alpha(\varphi)\,f(\varphi(1)-\varphi(0))$
(1)

is a normed monetary risk measure. Here
$\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=1$ with $\alpha_{i}\geq 0$,
$i=1,\,2,\,3,\,4$, $\lambda_{1},\,\lambda_{2}\in[-\infty,\,0]$ with
$\max\\{\lambda_{1},\,\lambda_{2}\\}=0$,
$\lambda_{3},\,\lambda_{4}\in[0,\,+\infty]$ with
$\min\\{\lambda_{3},\,\lambda_{4}\\}=0$,

$\alpha(\varphi)=\begin{cases}\min\\{\alpha_{1},\,\alpha_{2}\\},&\mbox{ if
}\alpha_{3}=\alpha_{4}=0,\\\
\min\\{\alpha_{1}+\alpha_{3}+\alpha_{4},\,\alpha_{2}\\},&\mbox{ if
}\varphi(0)+\lambda_{1}\geq\varphi(1)+\lambda_{2}\mbox{ and
}\varphi(0)+\lambda_{3}\leq\varphi(1)+\lambda_{4},\\\
\min\\{\alpha_{1}+\alpha_{3},\,\alpha_{2}+\alpha_{4}\\},&\mbox{ if
}\varphi(0)+\lambda_{1}\geq\varphi(1)+\lambda_{2}\mbox{ and
}\varphi(0)+\lambda_{3}>\varphi(1)+\lambda_{4},\\\
\min\\{\alpha_{1}+\alpha_{4},\,\alpha_{2}+\alpha_{3}\\},&\mbox{ if
}\varphi(0)+\lambda_{1}<\varphi(1)+\lambda_{2}\mbox{ and
}\varphi(0)+\lambda_{3}\geq\varphi(1)+\lambda_{4},\\\
\min\\{\alpha_{1},\,\alpha_{2}+\alpha_{3}+\alpha_{4}\\},&\mbox{ if
}\varphi(0)+\lambda_{1}<\varphi(1)+\lambda_{2}\mbox{ and
}\varphi(0)+\lambda_{3}>\varphi(1)+\lambda_{4},\end{cases}$ (2)

and, finally, $f\colon\mathbb{R}\to\mathbb{R}$ is a continuous non-decreasing
function such that

  * $(1^{*})$

$f(0)=0$,

  * $(2^{*})$

$t\leq f(t)\leq 0$ and concave at $t\leq 0$,

  * $(3^{*})$

$0\leq f(t)\leq t$ and convex at $t\geq 0$.

Since the set of functions $f$ considered in (1) and satisfying conditions
$(1^{*})-(3^{*})$, is uncountable, it follows that the space $O(X)$ of normed
monetary risk measure can not be embedded in any space with finite (even
countable) algebraic dimension if the compact $X$ contains more than one
point.

Example 1 shows that there exists extremely many monetary risk measures in
practice. Further, a question arises wether one can evaluate the difference
between distinct monetary risk measures. In other words, is it possible to
specify distance between monetary risk measures?

It is known [2] that for a compact Hausdorff space $X$ the equality
$w(X)=w(O(X))$ holds, where $w$ is the weight (i. e. the smallest cardinal
number which is the cardinality of an open base) of a topological space. From
here follows that if $X$ is a compactum then $O(X)$ is a compactum, i. e. the
space of normed monetary risk measures is metrisable compact space. Though for
a metrisable compact space $X$ the space $O(X)$ is metrisable, but the rule of
definition distance between monetary measures still was not known. In the
present paper for a given compact metric space $(X,\,\rho)$ we introduce a
metric $\rho_{O}$ on the space $O(X)$ of normed monetary risk measure, which
generates the point-wise convergence topology on $O(X)$. Besides, $\rho_{O}$
is an extension on $O(X)$ of the metric $\rho$.

##  2 Basic constructions

For a compact Hausdorff space $X$ we put

$\displaystyle X_{1}=X_{2}=X_{3}=X,\qquad X_{1\,2\,3}=X^{3}=X_{1}\times
X_{2}\times X_{3},\qquad X_{i\,j}=X^{2}=X_{i}\times X_{j},$

and let

$\displaystyle\pi^{1\,2\,3}_{i\,j}\colon X_{1\,2\,3}\to
X_{i\,j},\qquad\pi^{i\,j}_{k}\colon X_{i\,j}\to X_{k},\qquad 1\leq i<j\leq
3,\qquad k\in\\{i,\,j\\},$

be the corresponding projections.

Obviously, that

$\displaystyle O(\pi^{i\,j}_{i})^{-1}O(X_{i})\cap
O(\pi^{i\,j}_{j})^{-1}O(X_{j})=O(X_{i\,j}),\qquad 1\leq i<j\leq 3,$
$\displaystyle\bigcap\limits_{1\leq i<j\leq
3}O(\pi^{1\,2\,3}_{i\,j})^{-1}O(X_{i\,j})=O(X_{1\,2\,3}).$

###### Theorem 2.1

For every pair $(\mu_{1},\,\mu_{2})\in O(X_{1})\times O(X_{2})$ we have

$\displaystyle O(\pi^{1\,2}_{1})^{-1}(\mu_{1})\cap
O(\pi^{1\,2}_{2})^{-1}(\mu_{2})\neq\varnothing.$

Proof. At first we consider a particular case: let $\mu_{1}=\delta_{x}$ and
$\mu_{2}=\delta_{y}$, where $x\in X_{1}$, $y\in X_{2}$. The spaces
$O(X_{1}\times\\{y\\})$ and $O(X_{1})\times\\{\delta_{y}\\}$ are homeomorphic.
Indeed, one may determine the homeomorphism as the correspondence
$O(X_{1}\times\\{y\\})\ni\lambda_{y}\mapsto(\lambda,\,\delta_{y})\in
O(X_{1})\times\\{\delta_{y}\\}$, where
$\mbox{supp}\,\lambda_{y}=\\{(x,\,y):\,x\in\mbox{supp}\,\lambda\\}$, and
monetary risk measures $\lambda_{y}$ and $\lambda$ act the same rule, i. e.
$\lambda_{y}(\varphi)=\lambda(\varphi\circ\pi^{1\,2}_{1})$, $\varphi\in
C(X_{1}\times X_{2})$. Consequently,
$O(\pi^{1\,2}_{2})^{-1}(\delta_{y})=O(X_{1}\times\\{y\\})$. Similarly,
$O(\pi^{1\,2}_{1})^{-1}(\delta_{x})=O(\\{x\\}\times X_{2})$. It is easy to see
that $\delta_{(x,\,y)}$ is a unique monetary risk measure such that
$\delta_{(x,\,y)}\in O(\\{x\\}\times X_{2})\cap O(X_{1}\times\\{y\\})$. Thus,
$O(\pi^{1\,2}_{1})^{-1}(\delta_{x})\cap
O(\pi^{1\,2}_{2})^{-1}(\delta_{y})\neq\varnothing$.

Now we consider the general case. We construct a set

$\displaystyle\mathfrak{B}=\\{\varphi\circ\pi^{1\,2}_{i}+c_{X_{1}\times
X_{2}}:\,\varphi\in C(X),\,c\in\mathbb{R},\,i=1,\,2\\}.$

Then $\mathfrak{B}$ is an $A$-subspace in $C(X_{1}\times X_{2})$. Define a
functional $\widetilde{\mu}_{1\,2}\colon\mathfrak{B}\to\mathbb{R}$ as
following

$\displaystyle\widetilde{\mu}_{1\,2}(\varphi\circ\pi^{1\,2}_{i}+c_{X_{1}\times
X_{2}})=\mu_{i}(\varphi)+c,\qquad i=1,\,2.$

It is easy to see that $\widetilde{\mu}_{1\,2}$ is translation invariance and
normed.

We will show $\widetilde{\mu}_{1\,2}$ has monotonicity property. Taking
$\varphi\circ\pi^{1\,2}_{i}$, $\psi\circ\pi^{1\,2}_{i}$ $\in\mathfrak{B}$ with
$\varphi\circ\pi^{1\,2}_{i}\leq\psi\circ\pi^{1\,2}_{i}$, we obtain
$\varphi\leq\psi$, and thence

$\displaystyle\widetilde{\mu}_{1\,2}(\varphi\circ\pi^{1\,2}_{i})=\mu_{i}(\varphi)\leq\mu_{i}(\psi)=\widetilde{\mu}_{1\,2}(\psi\circ\pi^{1\,2}_{i}),\qquad
i=1,\,2.$

Take now $\varphi\circ\pi^{1\,2}_{1}$, $\psi\circ\pi^{1\,2}_{2}$
$\in\mathfrak{B}$ such, say, that
$\varphi\circ\pi^{1\,2}_{1}\leq\psi\circ\pi^{1\,2}_{2}$. Then

$\displaystyle\max\\{\varphi(x):\,x\in X\\}\leq\min\\{\psi(x):\,x\in X\\}.$

Choosing any $a\in[\max\\{\varphi(x):\,x\in X\\},\,\min\\{\psi(x):\,x\in
X\\}]$, we see

$\displaystyle\widetilde{\mu}_{1\,2}(\varphi\circ\pi^{1\,2}_{1})=\mu_{1}(\varphi)\leq
a\leq\mu_{2}(\psi)=\widetilde{\mu}_{1\,2}(\psi\circ\pi^{1\,2}_{2}).$

One similarly can establish the monotonicity of $\widetilde{\mu}_{1\,2}$ in
the case when $\varphi\circ\pi^{1\,2}_{1}\geq\psi\circ\pi^{1\,2}_{2}$.

Thus, $\widetilde{\mu}_{1\,2}$ is a normed monetary risk measure on the
$A$-subpace $\mathfrak{B}$. By the analog of the Hahn-Banach theorem (see Page
1.2 of the present paper) $\widetilde{\mu}_{1\,2}$ has an extension
$\mu_{1\,2}$ all over $C(X_{1}\times X_{2})$, which is a normed monetary risk
measure. We have

$\displaystyle
O(\pi^{1\,2}_{i})(\mu_{1\,2})=\mu_{1\,2}(\varphi\circ\pi^{1\,2}_{i})=\widetilde{\mu}_{1\,2}(\varphi\circ\pi^{1\,2}_{i})=\mu_{i}(\varphi),\qquad
i=1,\,2.$

Consequently, $\mu_{1\,2}\in O(\pi^{1\,2}_{1})^{-1}(\mu_{1})\cap
O(\pi^{1\,2}_{2})^{-1}(\mu_{2})$.

$\scriptstyle\blacksquare$

Denote $\Lambda(\mu_{1},\,\mu_{2})=O(\pi^{1\,2}_{1})^{-1}(\mu_{1})\cap
O(\pi^{1\,2}_{2})^{-1}(\mu_{2})$. An element
$\xi\in\Lambda(\mu_{1},\,\mu_{2})$ we call as a
$(\mu_{1},\,\mu_{2})$-admissible monetary risk measure.

###### Corollary 2.1

For normed monetary risk measures

$\displaystyle\mu_{2}\in O(X_{2}),\qquad\mu_{1\,2}\in
O(X_{1\,2}),\qquad\mu_{2\,3}\in O(X_{2\,3})$

such that

$\displaystyle
O(\pi^{1\,2}_{2})(\mu_{1\,2})=\mu_{2}=O(\pi^{2\,3}_{2})(\mu_{2\,3}),$

there exists a $\mu_{1\,2\,3}\in O(X_{1\,2\,3})$ satisfying the equalities

$\displaystyle O(\pi^{1\,2\,3}_{1\,2})(\mu_{1\,2\,3})=\mu_{1\,2}\qquad\mbox{
and }\qquad O(\pi^{1\,2\,3}_{2\,3})(\mu_{1\,2\,3})=\mu_{2\,3}.$

Really, to adopt this statement it is sufficient to note that

$\displaystyle O(\pi^{1\,2\,3}_{1\,2})^{-1}(\mu_{1\,2})\cap
O(\pi^{1\,2\,3}_{2\,3})^{-1}(\mu_{2\,3})\neq\varnothing.$

In this case we construct an $A$-subspace

$\displaystyle\mathfrak{B}=\\{\varphi\circ\pi^{1\,2\,3}_{i\,(i+1)}+c_{X^{3}}:\,\varphi\in
C(X^{2}),\,c\in\mathbb{R},\,i=1,\,2\\}$

in $C(X^{3})$ and repeat the analogous procedure as in the proof of Theorem
2.1.

###### Proposition 2.1

Let $X$ be a compactum and a sequence $\\{\mu_{n}\\}\subset O(X)$ converges to
$\mu_{0}\in O(X)$ with respect to point-wise convergence topology. Then for
every open neighbourhood $U$ of the diagonal $\Delta(X)=\\{(x,\,x):\,x\in
X\\}$ of $X^{2}$ there exists a positive integer $n$ and for each
$n^{\prime}\geq n$ there exists a $(\mu_{0},\,\mu_{n^{\prime}})$-admissible
monetary risk measure $\mu_{0\,n^{\prime}}\in O(X^{2})$ such that
$\mbox{supp}\,\mu_{0\,n^{\prime}}\subset U$.

Proof. The condition gives a sequence $\\{\mbox{supp}\,\mu_{n}\\}$ of closed
subsets of $X$. It is well known that for a compact Hausdorff space $X$ its
hyperspace $\exp\,X$ also is a compact Hausdorff space as well. Where
$\exp\,X$ is the set of all nonempty closed subsets of $X$, and $\exp\,X$ is
equipped with the Vietoris topology. So, the sequence
$\\{\mbox{supp}\,\mu_{n}\\}$ has a limit. Let
$A=\lim\limits_{n\to\infty}\mbox{supp}\,\mu_{n}$. Suppose
$A\neq\mbox{supp}\,\mu_{0}$. Then Proposition 1.2 implies that
$\mu_{0}\neq\lim\limits_{n\to\infty}\mu_{n}$. The got contradiction shows that
$\lim\limits_{n\to\infty}\mbox{supp}\,\mu_{n}=\mbox{supp}\,\mu_{0}$.

Consider open neighbourhoods $V_{x}$ of points $x\in\mbox{supp}\,\mu_{0}$ such
that $V_{x}\times V_{x}\subset U$. Since $\mbox{supp}\,\mu_{0}$ is a compact
set, its open cover $\\{V_{x}:\,x\in\mbox{supp}\,\mu_{0}\\}$ has a finite
subcover $\\{V_{k}:\,k=1,\,\dots,\,l\\}$, where $V_{k}=V_{x_{k}}$. Owing to
convergence $\\{\mbox{supp}\,\mu_{n}\\}$ to $\mbox{supp}\,\mu_{0}$ there
exists a positive integer $n$ such that
$\mbox{supp}\,\mu_{n^{\prime}}\in\langle V_{1},\,\dots,\,V_{l}\rangle$ for
every $n^{\prime}\geq n$. Here

$\displaystyle\langle
V_{1},\,\dots,\,V_{l}\rangle=\left\\{F\in\exp\,X:\,F\subset\bigcup\limits_{k=1}^{l}V_{k}\,\mbox{
and }\,F\cap V_{k}\neq\varnothing\,\mbox{ for each
}\,k=1,\,\dots,\,l\right\\}$

is a basic open neighbourhood of $\mbox{supp}\,\mu_{0}$ with respect to the
Vietoris topology in $\exp\,X$.

It is easy to see that $\mbox{supp}\,\mu_{n^{\prime}}\in\langle
V_{1},\,\dots,\,V_{l}\rangle$ if and only if

$\mbox{supp}\,\mu_{n^{\prime}}\subset\bigcup\limits_{k=1}^{l}V_{k}\quad\mbox{and
for every}\quad x\in\mbox{supp}\,\mu_{0}\quad\mbox{there exists}\quad
y\in\mbox{supp}\,\mu_{n^{\prime}}\\\ \quad\mbox{such that}\quad(x,\,y)\in
V_{k}\times V_{k}\quad\mbox{for some}\quad k\in\\{1,\,\dots,\,l\\}.$ (3)

It remains to show that for every $n^{\prime}\geq n$ there exists
$\mu_{0\,n^{\prime}}\in\Lambda(\mu_{0},\,\mu_{n^{\prime}})$ such that
$\mbox{supp}\,\mu_{0\,n^{\prime}}\subset\bigcup\limits_{k=1}^{l}V_{k}\times
V_{k}$. Suppose, that for some $n^{\prime}\geq n$ and for every
$\xi\in\Lambda(\mu_{0},\,\mu_{n^{\prime}})$ the intersection
$\mbox{supp}\,\xi\cap(X^{2}\setminus U)$ is nonempty. Let
$(x,\,y)\in\mbox{supp}\,\xi\cap(X^{2}\setminus U)$. Then $(x,\,y)\not\in
V_{k}\times V_{k}$ for all $k=1,\,\dots,\,l$, which contradicts (3). The
received contradiction finished the proof.

$\scriptstyle\blacksquare$

##  3 On a metric on the space of monetary risk measures

Let $(X,\,\rho)$ be a metric compact space. We suggest a distance function
$\rho_{O}\colon O(X)\times O(X)\to\mathbb{R}$ as follows

$\rho_{O}(\mu_{1},\,\mu_{2})=\inf\,\\{\max\\{\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\xi\\}:\,\xi\in\Lambda(\mu_{1},\,\mu_{2})\\}.$
(4)

###### Proposition 3.1

For every pair $\mu_{1}$, $\mu_{2}$ of monetary risk measures there exists a
$(\mu_{1},\,\mu_{2})$-admissible monetary risk measure $\mu_{1\,2}\in
O(X^{2})$ such that

$\displaystyle\rho_{O}(\mu_{1},\,\mu_{2})=\max\\{\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\mu_{1\,2}\\}.$

Proof. The Proof leans on Proposition 1.1 and Lemma 3 [3] (see Page 1.1 of the
present paper).

$\scriptstyle\blacksquare$

###### Theorem 3.1

The function $\rho_{O}$ is a metric on the space $O(X)$ of monetary risk
measures, which is an extension of the metric $\rho$ on $X$.

Proof. Since each $\xi\in O(X^{2})$ is monotone then the inequality $\rho\geq
0$ immediately implies $\rho_{O}\geq 0$. So, $\rho_{O}$ is nonnegative.
Obviously, $\rho_{O}$ is symmetric.

Let $\mu_{1}=\mu_{2}=\mu$. There exists a monetary risk measure $\mu_{1\,2}\in
O(\Delta(X))$ such that $\mu_{1\,2}\in O(\pi^{1\,2}_{1})^{-1}(\mu_{1})\cap
O(\pi^{1\,2}_{2})^{-1}(\mu_{2})$. Then

$\displaystyle\rho_{O}(\mu_{1},\,\mu_{2})\leq\max\\{\rho(x,\,x):\,(x,\,x)\in\mbox{supp}\,\mu_{1\,2}\\}=0.$

Inversely, let $\rho_{O}(\mu_{1},\,\mu_{2})=0$. Then there exists a
$\mu_{1\,2}\in O(\pi^{1\,2}_{1})^{-1}(\mu_{1})\cap
O(\pi^{1\,2}_{2})^{-1}(\mu_{2})$ such that $\rho(x,\,y)=0$ for all
$(x,\,y)\in\mbox{supp}\,\mu_{1\,2}$. Consequently, $\mbox{supp}\,\mu_{1\,2}$
must lie in the diagonal $\Delta(X)$. We have
$(\varphi\circ\pi^{1\,2}_{1})|_{\Delta(X)}=(\varphi\circ\pi^{1\,2}_{2})|_{\Delta(X)}$,
$\varphi\in C(X)$. From here with respect to Lemma 4 [2] (see Page 1.2 of the
present paper)

$\displaystyle\mu_{1}(\varphi)=\mu_{1\,2}(\varphi\circ\pi^{1\,2}_{1})=\mu_{1\,2}(\varphi\circ\pi^{1\,2}_{2})=\mu_{2}(\varphi),\qquad\varphi\in
C(X),$

i. e. $\mu_{1}=\mu_{2}$.

Let us show that the triangle inequality is true as well. Take an arbitrary
triple $\mu_{i}\in O(X)$, $i=1,\,2,\,3$. Let $\mu_{1\,2},\,\mu_{2\,3}\in
O(X^{2})$ be $(\mu_{1},\,\mu_{2})$\- and $(\mu_{2},\,\mu_{3})$-admissible
monetary risk measures such that
$\rho_{O}(\mu_{1},\,\mu_{2})=\max\\{\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\mu_{1\,2}\\}$
and
$\rho_{O}(\mu_{2},\,\mu_{3})=\max\\{\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\mu_{2\,3}\\}$,
respectively. Using Corollary 2.1 one can point out a $\mu_{1\,2\,3}\in
O(X_{1\,2\,3})$ which satisfies the equalities

$\displaystyle O(\pi^{1\,2\,3}_{1\,2})(\mu_{1\,2\,3})=\mu_{1\,2}\qquad\mbox{
and }\qquad O(\pi^{1\,2\,3}_{2\,3})(\mu_{1\,2\,3})=\mu_{2\,3}.$

We assume that $\mu_{1\,3}=O(\pi^{1\,2\,3}_{1\,3})(\mu_{1\,2\,3})$. Then
$\mu_{1\,3}$ is a $(\mu_{1},\,\mu_{3})$-admissible monetary risk measure. We
have

$\rho_{O}(\mu_{1},\,\mu_{2})+\rho_{O}(\mu_{2},\,\mu_{3})=\\\
=\max\limits_{(x_{1},\,x_{2})\in{\scriptsize\mbox{supp}}\,\mu_{1\,2}}\rho(x_{1},\,x_{2})+\max\limits_{(x_{2},\,x_{3})\in{\scriptsize\mbox{supp}}\,\mu_{2\,3}}\rho(x_{2},\,x_{3})=\\\
=\max\limits_{(x_{1},\,x_{2},\,x_{3})\in{\scriptsize\mbox{supp}}\,\mu_{1\,2\,3}}\rho(x_{1},\,x_{2})+\max\limits_{(x_{1},\,x_{2},\,x_{3})\in{\scriptsize\mbox{supp}}\,\mu_{1\,2\,3}}\rho(x_{2},\,x_{3})\geq\\\
\geq\max\limits_{(x_{1},\,x_{2},\,x_{3})\in{\scriptsize\mbox{supp}}\,\mu_{1\,2\,3}}\\{\rho(x_{1},\,x_{2})+\rho(x_{2},\,x_{3})\\}\geq\\\
\geq\max\limits_{(x_{1},\,x_{2},\,x_{3})\in{\scriptsize\mbox{supp}}\,\mu_{1\,2\,3}}\rho(x_{1},\,x_{3})=\\\
=\max\limits_{(x_{1},\,x_{3})\in{\scriptsize\mbox{supp}}\,\mu_{1\,3}}\rho(x_{1},\,x_{3})\geq\rho_{O}(\mu_{1},\,\mu_{3}),$

i. e.
$\rho_{O}(\mu_{1},\,\mu_{3})\leq\rho_{O}(\mu_{1},\,\mu_{2})+\rho_{O}(\mu_{2},\,\mu_{3})$.

For every pair of the Dirac measures $\delta_{x}$, $\delta_{y}$, $x,\,y\in X$,
the uniqueness of $(\delta_{x},\,\delta_{y})$-admissible monetary risk measure
$\delta_{(x,\,y)}\in O(X^{2})$ implies that

$\displaystyle\rho_{O}(\delta_{x},\,\delta_{y})=\max\\{\delta_{(x,\,y)}(\rho),\,(x,\,y)\in\mbox{supp}\,\delta_{(x,\,y)}\\}=\rho(x,\,y).$

From here we get that $\rho_{O}$ is an extension of $\rho$.

$\scriptstyle\blacksquare$

The following affirmation states one of remarkable properties of the metric
$\rho_{O}$.

###### Proposition 3.2

$\mbox{diam}(O(X),\,\rho_{O})=\mbox{diam}(X,\,\rho)$.

Proof. The proof is obvious.

$\scriptstyle\blacksquare$

###### Theorem 3.2

The metric $\rho_{O}$ generates point-wise convergence topology on $O(X)$.

Proof. Let $\\{\mu_{n}\\}\subset O(X)$ be a sequence and $\mu_{0}\in O(X)$.
Suppose the sequence converges to $\mu_{0}$ with respect to the point-wise
convergence topology but not by $\rho_{O}$. Passing in the case of need to a
subsequence, it is possible to regard that

$\displaystyle\rho_{O}(\mu_{n},\,\mu_{0})\geq a>0\qquad\mbox{ for all positive
integer }\quad n.$

Consider an open neighbourhood of the diagonal $\Delta(X)\subset X^{2}$:

$\displaystyle U=\left\\{(x,\,y)\in X^{2}:\,\rho(x,\,y)<\frac{a}{2}\right\\}.$

By virtue of Proposition 2.1 there exist a positive integer $n$ and a
$(\mu_{0},\,\mu_{n})$-admissible measure $\mu_{0\,n}\in O(X^{2})$ such that
$\mbox{supp}\,\mu_{0\,n}\subset U$, and consequently

$\displaystyle\rho_{O}(\mu_{n},\,\mu_{0})\leq\frac{a}{2}<a.$

The obtained contradiction finishes the proof. $\scriptstyle\blacksquare$

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