Spectral properties of the operator of Bessel potential type

JID:YJMAA AID:18448 /FLA Doctopic: Functional Analysis [m3L; v 1.133; Prn:2/05/2014; 11:21] P.1 (1-18)J. Math. Anal. Appl. ••• (••••) •••–•••

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Spectral properties of the operator of Bessel potential type ✩

Milutin R. Dostanić †

Faculty of Mathematics, University of Belgrade, Studentski trg 16, P.O. Box 550, 11000 Belgrade, Serbia

a r t i c l e i n f o a b s t r a c t

Article history:Received 29 December 2013Available online xxxxSubmitted by D. Khavinson

Keywords:Bessel potentialEigenvaluesSingular valuesEigenvalues distribution functionRegularized trace

Let Ω be a convex bounded domain in Rm having regular boundary. In this paper,we study integral operators Bα

m on L2(Ω) of Bessel potential type. If N(λ) denotesthe number of eigenvalues of Bα

m that are ≥ λ, for λ > 0, we find the asymptoticsof the regularized eigenvalue distribution function λ �→

∫∞λ

N(μ)dμ when λ → 0+.As a consequence, we find the regularized traces of these operators.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

We study the operators

Bαm : L2(Ω) −→ L2(Ω), Ω ⊂ R

m

defined by

Bαmf(x) =

∫Ω

Gα(x− y)f(y)dy

where

Gα(x) = 2 2−m−α2

πm2 Γ (α2 )

·Km−α

2(|x|)

|x|m−α2

, α > 0.

✩ Partially supported by MNZZS Grant No 174017.E-mail addresses: [email protected], [email protected] (D.R. Jović).

† Deceased.

http://dx.doi.org/10.1016/j.jmaa.2014.04.0230022-247X/© 2014 Elsevier Inc. All rights reserved.

http://dx.doi.org/10.1016/j.jmaa.2014.04.023

http://www.ScienceDirect.com/

http://www.elsevier.com/locate/jmaa

mailto:[email protected]

mailto:[email protected]

http://dx.doi.org/10.1016/j.jmaa.2014.04.023

JID:YJMAA AID:18448 /FLA Doctopic: Functional Analysis [m3L; v 1.133; Prn:2/05/2014; 11:21] P.2 (1-18)2 M.R. Dostanić / J. Math. Anal. Appl. ••• (••••) •••–•••

Here x = (x1, x2, . . . , xm), |x| =√∑m

i=1 x2i , dx = dx1dx2 . . . dxm and Kν is the McDonald function:

Kν(z) = π

2 sin νπ

(I−ν(z) − Iν(z)

), ν /∈ Z,

Kn(z) = limν→n

Kν(z), n = 0,±1,±2 . . .

and

Iν(z) =∞∑k=0

( z2 )ν+2k

k!Γ (ν + k + 1) .

The domain Ω is convex, bounded with sufficiently regular boundary. By |Ω| we will denote the Lebesguemeasure of Ω. We call the convolution operators with kernel Gα Bessel potentials. They occur in severalplaces, including the theory of fractional integration, Operator Theory, Harmonic Analysis and MathematicalPhysics.

If α > 0, the operators Bαm are compact on L2(Ω). They are close to negative fractional power of the

operator I − Δ.The operators (−Δ)s appear in numerous fields (such as mathematical analysis, mathematical physics,

mathematical biology, ...). Two term asymptotics expansion of the sum of eigenvalues of fractional Laplacian(−Δ)s, when 0 < s < 1, has been found in the paper [9].

In [12], the author finds the two term Weyl type asymptotics for the eigenvalues of the one-dimensionalfractional Laplace operator (−Δ)s (0 < s < 1) on the interval (−1, 1).

In [1] it was proved that the second term in the asymptotic expansion of the trace of the semigroup ofa symmetric stable process (fractional powers of the Laplacian) of order α, for 0 < α < 2, in a Lipschitzdomain is given by the surface area of the boundary domain, as t → 0+.

In [16] the two term asymptotic formula for the eigenvalue distribution function for the n-th power ofthe Laplacian (and also for more general operators) has been given.

In this paper we denote by f and f the direct and inverse Fourier transform:

f(x) =∫Rm

e−ix·yf(y)dy

f(x) = 1(2π)m

∫Rm

eix·yf(y)dy

(with x · y, x, y ∈ Rm, denoting the inner product in R

m).It is well known (see, e.g., [17]) that Gα is a positive function and

Gα(x) = 1(1 + |x|2)α

2,∫

Rm

Gα(x)dx = 1.

It follows that the operator

f �→∫

Gα(x− y)f(y)dy
Rm

JID:YJMAA AID:18448 /FLA Doctopic: Functional Analysis [m3L; v 1.133; Prn:2/05/2014; 11:21] P.3 (1-18)M.R. Dostanić / J. Math. Anal. Appl. ••• (••••) •••–••• 3

is bounded on L2(Rm) and its norm is ≤ 1. Also, the operators Bαm are positive and Ker Bα

m = {0}. Letλn(Bα

m) denote the set of eigenvalues of the operator Bαm arranged in decreasing order, according to their

multiplicity and

N(λ) =∑

λn(Bαm)≥λ

1

(eigenvalues distribution function).The function λ �→

∫∞λ

N(μ)dμ will be called the regularized eigenvalues distribution function of theoperator Bα

m.In this paper we give precise asymptotics for the function

∫∞λ

N(μ)dμ when λ → 0+.In [13], the author proved a result concerning the asymptotics of the regularized eigenvalues distribution

function, for a class of integral operators with homogeneous kernel, under assumptions on the domain Ω

which included some conditions on normal curvatures of ∂Ω.The corresponding asymptotic formula contains only the first term and estimation of the second term

(when λ → 0+).In our case, the kernel Gα is not homogeneous and an assumption on curvatures of ∂Ω is not necessary.For a compact operator T (on a separable Hilbert space H with inner product 〈·,·〉), we denote by sn(T )

(singular values of the operator T ) the n-th eigenvalue of the operator |T | = (T ∗T ) 12 , i.e. sn(T ) = λn(|T |).

By cp we denote the set of compact operators T such that

|T |p =(∑

n≥1spn(T )

) 1p

< ∞.

For more information about cp and about properties of singular values of compact operators, we refer thereader to [10] or [18].

If p = 1, c1 is the set of nuclear operators. It is well known that if A ∈ c1 and (ϕn)∞n=1 is an orthonormalbasis in H, then the series

∑∞n=1〈Aϕn, ϕn〉 converges, its sum does not depend on the choice of (ϕn)∞n=1,

and

∞∑n=1

〈Aϕn, ϕn〉 =∑n≥1

λn(A)

holds.(If A does not have eigenvalues, the right-hand side of the above equality is considered to be 0.)If A ∈ c1 then the expression

∑n λn(A) is denoted by tr A and called the trace of the operator A. The

function tr(·) is a bounded linear functional on c1.If a compact operator A is not nuclear, then

∑n λn(A) does not necessarily converge. In that case,

for some suitable chosen sequence (dn)∞n=1, the series∑∞

n=1(λn(A) − dn) converges. Its sum is called theregularized trace of the operator A.

An excellent review of results concerning trace theory (the regularized trace) of differential and abstractdiscrete operators is given in [15], but in the literature there is almost nothing about regularized traces ofnon-nuclear integral operators.

Let λ > 0 and

ϕλ(x) ={

0; 0 ≤ x ≤ λ

x− λ; x ≥ λ.


We observe that ϕλ(Bαm) ∈ c1 and

tr ϕλ

(Bα

m

)=

∞∫λ

N(μ)dμ (∗)

(This follows from tr ϕλ(Bαm) = −

∫∞0 ϕλ(μ)dN(μ) by integration by parts.)

In the case m = α = 2, the exact value of regularized trace of the inverse of Dirichlet Laplacian forconvex planar domains was found in [7], using the known first two terms in the asymptotics of tr ϕλ(B2

2)when λ → 0+.

2. Main result

Theorem 1. If Ω ⊂ Rm (m ≥ 2) is a bounded convex domain with analytic boundary, then:

a) If m− 1 < α < m then

+∞∫λ

N(μ)dμ = cmλ1−mα + dm + o(1), λ → 0+

holds, where

cm = |Ω|2−m

πm2 Γ (1 + m

2 )· α

m− α

dm = − |Ω|π

m2

21−m Γ (α+2−m2 )

(m− α)Γ (α2 ) .

b) If α = m then

+∞∫λ

N(μ)dμ = c′m ln 1λ

+ d′m + o(1), λ → 0+

holds, where

c′m = |Ω|2−m

πm2 Γ (1 + m

2 )

d′m = |Ω|21−m

πm2 Γ (m2 )

(ln 2 + Am − 1

m

)

and

Am =∞∫0

xm−1 − (1 + x2)m−12

(1 + x2)m2

dx.

Remark 1. If m = 1 the statement of Theorem 1 is also true; in that case Ω is an interval.


Remark 2. From Widom’s result [19] and Ky–Fan theorem [11], it follows that

λn

(Bα

m

)∼ const. · 1

nαm, n → ∞

and hence, for α > m, the operators Bαm are nuclear.

Remark 3. The term tr ϕλ(Bαm) is expressed (with accuracy up to o(1) when λ → 0+) as the sum of

two terms which depend on |Ω|, because this representation is suitable for finding the regularized trace ofoperator Bα

m.

In order to prove Theorem 1 (and its consequences), we shall need the following lemmas.

Lemma 1. (See [14].) Let Ω be a convex, bounded domain in Rm (m ≥ 2) with analytic boundary ∂Ω and let

F (r · ξ) (r > 0, ξ ∈ Sm−1) denote Fourier transform of the characteristic function XΩ in polar coordinates.Then

supr>0

rm+1

2∣∣F (r · ξ)

∣∣ = g(ξ),

and for some p > 2 we have ∫Sm−1

∣∣g(ξ)∣∣pdσ(ξ) < ∞.

(Here Sm−1 is the boundary of unit (Euclidean) ball in Rm and dσ is the surface measure on Sm−1 and

r · ξ denotes the product of scalar r and vector ξ ∈ Sm−1.)

Lemma 2. (See [11].) If C and D are compact operators on a complex Hilbert space such that

sn(C) = an−α + O(n−β

), a > 0, 0 < α < β < α + 1,

sn(D) = O(n−β1

), β1 ≥ β

α + 1 − β

then

sn(C + D) = an−α + O(n−β

).

Lemma 3. If

R(x) =∫Rm

∣∣∣∣ 1(1 + |x− t|2)α

2− 1

(1 + |x + t|2)α2

∣∣∣∣dtthen

R(x) ={

O(|x|m−α); m− 1 < α < m

O(ln |x|); α = m

when |x| → +∞.


Proof. We consider first the case when m = 1 and 0 < α ≤ 1. Then, we have

R(x) =∫R

∣∣∣∣ 1(1 + (x− t)2)α

2− 1

(1 + (x + t)2)α2

∣∣∣∣dt.The function R is even and we have

R(x) = 2∞∫0

∣∣∣∣ 1(1 + (x− t)2)α

2− 1

(1 + (x + t)2)α2

∣∣∣∣dtand it is sufficient to find the growth of R when x → +∞.

If x > 0, then

R(x)2 =

∞∫0

(1

(1 + (x− t)2)α2− 1

(1 + (x + t)2)α2

)dt.

Now, we define the function (for fixed x > 0)

G(z) =∞∫0

(1

(1 + (x− t)2) z2− 1

(1 + (x + t)2) z2

)dt

The function G is analytic in domain {z : Re z > 0}.If Re z > 1 then the functions G1(z) =

∫∞0

1(1+(x−t)2)

z2dt and G2(z) =

∫∞0

1(1+(x+t)2)

z2dt are analytic

and

G(z) = G1(z) −G2(z) (Re z > 1).

Since

G1(z) −G2(z) =∞∫

−x

du

(1 + u2) z2−

∞∫x

du

(1 + u2) z2

= 2x∫

0

du

(1 + u2) z2

and the function z �→∫ x

0du

(1+u2)z2

is entire, we have that

G(z) ≡ 2x∫

0

du

(1 + u2) z2

Re z > 0

by the uniqueness theorem.So, if 0 < α ≤ 1 and x > 0 we have

R(x) = 4x∫

du

(1 + u2)α2. (1)

0


If α = 1 it follows from (1) that

R(x) = O(ln x), x → ∞.

If 0 < α < 1 then

R(x) ∼ 4x∫

0

du

uα= O

(x1−α

), x → ∞.

Now, consider the case when m ≥ 2 and m− 1 < α ≤ m.If we introduce the change of variable

t = r · ξ, r > 0, ξ ∈ Sm−1

in the integral

R(x) =∫Rm

∣∣∣∣ 1(1 + |x− t|2)α

2− 1

(1 + |x + t|2)α2

∣∣∣∣dt=

∫Rm

∣∣∣∣ 1(1 + |x|2 + |t|2 − 2x · t)α

2− 1

(1 + |x|2 + |t|2 + 2x · t)α2

∣∣∣∣dtwe obtain

R(x) = cm

∞∫0

rm−1dr

∫Sm−1

∣∣∣∣ 1(1 + |x|2 + r2 − 2rx · ξ)α

2− 1

(1 + |x|2 + r2 + 2rx · ξ)α2

∣∣∣∣dσ(ξ)

where cm is some constant depending only on m. Here rx denotes the product of scalar r and vector x ∈ Rm

and rx · ξ denotes the inner product of vectors rx and ξ ∈ Sm−1.Calculating the inner integral (using the formula

∫Sm−1

Ω(u · ξ)dσ(ξ) = 2πm−12

Γ (m−12 )

1∫−1

Ω(ρ|u|

)(1 − ρ2)m−3

2 dρ, u ∈ Rm

)

we get

R(x) = c′mR1(x)

where

R1(x) =∞∫0

rm−1dr

1∫0

(1 − u2)m−3

2

[1

(1 + |x|2 + |r|2 − 2r|x|u)α2− 1

(1 + |x|2 + |t|2 + 2r|x|u)α2

]du

(c′m-constant depending only on m).We will prove that

R1(x) ={

O(|x|m−α); m− 1 < α < m

O(ln |x|); α = m(2)

when |x| → +∞.


Now, we prove that the function R1 can be represented in the following way:

R1(x) = 2[m−1

2 ]∑k=0

(m− 1

2k

)Ak(x) + 2

[m2 ]∑k=1

(m− 12k − 1

)Bk(x) (3)

where

Ak(x) = |x|m−1−2k1∫

0

um−1−2k(1 − u2)m−32 du

u|x|∫0

v2kdv

(1 + v2 + |x|2(1 − u2))α2

and

Bk(x) = |x|2k−11∫

0

u2k−1(1 − u2)m−32 du

∞∫u|x|

vm−2kdv

(1 + v2 + |x|2(1 − u2))α2.

Let x ∈ Rm be fixed and

H(z) =1∫

0

(1 − u2)m−3

2 du

∞∫0

rm−1[

1(1 + |x|2 + r2 − 2r|x|u) z

2− 1

(1 + |x|2 + r2 + 2r|x|u) z2

]dr.

The function H is analytic in the domain {z : Re z > m− 1}. If we put

H1(z) =1∫

0

(1 − u2)m−3

2 du

∞∫0

rm−1dr

(1 + |x|2 + |r|2 − 2r|x|u) z2

and

H1(z) =1∫

0

(1 − u2)m−3

2 du

∞∫0

rm−1dr

(1 + |x|2 + |r|2 + 2r|x|u) z2

then we have that H1 and H2 are analytic functions on the domain {z : Re z > m} and

H(z) = H1(z) −H2(z)

for z : Re z > m. Having in mind that

1 + |x|2 + |r|2 ± 2r|x|u =(r ± |x|u

)2 + 1 + |x|2(1 − u2)

after the corresponding change variable we obtain

H1(z) =1∫

0

(1 − u2)m−3

2 du

+∞∫−u|x|

(v + u|x|)m−1dv

(1 + v2 + |x|2(1 − u2)) z2

and


H2(z) =1∫

0

(1 − u2)m−3

2 du

+∞∫u|x|

(v − u|x|)m−1dv

(1 + v2 + |x|2(1 − u2)) z2.

If Re z > m, we have

H1(z) −H2(z) =1∫

0

(1 − u2)m−3

2 du

u|x|∫−u|x|

(v + u|x|)m−1dv

(1 + v2 + |x|2(1 − u2)) z2

+1∫

0

(1 − u2)m−3

2 du

+∞∫u|x|

(v + u|x|)m−1 − (v − u|x|)m−1

(1 + v2 + |x|2(1 − u2)) z2

dv.

Applying the binomial formula to the function in inner integrals, after simplification, we have (for Re z > m):

H(z) = 2[m−1

2 ]∑k=0

(m− 1

2k

)|x|m−1−2k

1∫0

um−1−2k(1 − u2)m−32 du

u|x|∫0

v2k

(1 + v2 + |x|2(1 − u2)) z2dv

+ 2[m2 ]∑k=0

(m− 12k − 1

)|x|2k−1

1∫0

u2k−1(1 − u2)m−32 du

+∞∫u|x|

vm−2k

(1 + v2 + |x|2(1 − u2)) z2dv.

The first sum in the previous equality is an entire function; the second sum is analytic function in domain{z : Re z > m− 1}. So, the function

H(z) = 2[m−1

2 ]∑k=0

(m− 1

2k

)|x|m−1−2k

1∫0

um−1−2k(1 − u2)m−32 du

u|x|∫0

v2k

(1 + v2 + |x|2(1 − u2)) z2dv

+ 2[m2 ]∑k=0

(m− 12k − 1

)|x|2k−1

1∫0

u2k−1(1 − u2)m−32 du

+∞∫u|x|

vm−2k

(1 + v2 + |x|2(1 − u2)) z2dv

is analytic in domain {z : Re z > m− 1}. So, the function

H0(z) = 2[m−1

2 ]∑k=0

(m− 1

2k

)|x|m−1−2k

1∫0

um−1−2k(1 − u2)m−32 du

u|x|∫0

v2k

(1 + v2 + |x|2(1 − u2)) z2dv

+ 2[m2 ]∑k=0

(m− 12k − 1

)|x|2k−1

1∫0

u2k−1(1 − u2)m−32 du

+∞∫u|x|

vm−2k

(1 + v2 + |x|2(1 − u2)) z2dv

is analytic in domain {z : Re z > m− 1}.Since both the functions H and H0 are analytic in domain {z : Re z > m − 1} and H ≡ H0 in domain

{z : Re z > m}, by the uniqueness theorem, H(z) = H0(z) for every z : Re z > m− 1. Especially, if we putz = α where m− 1 < α ≤ m we obtain (3).

Now, we estimate Ak and Bk.


Let m be odd, m− 1 < α < m and k = m−12 . Then we have α− 2k = α− (m− 1) < 1 and so

Ak ≤1∫

0

(1 − u2)m−3

2 du

u|x|∫0

v2k−αdv

= |x|m−α 1m− α

1∫0

(1 − u2)m−3

2 um−αdu

= O(|x|m−α

), |x| → ∞.

Let m be odd, m − 1 < α < m and k < m−12 . Then we have α − 2k > m − 1 − 2k ≥ 2 and the integral∫ +∞

0v2k

(1+v2)α2dv converges.

Since,

Ak(x) = |x|m−1−2k1∫

0

um−1−2k(1 − u2)m−32 du

u|x|∫0

v2k

(1 + v2 + |x|2(1 − u2))α2dv,

by the change variable v = s ·√

1 + |x|2(1 − u2) (in inner integral) we get

Ak(x) <∞∫0

s2k

(1 + s2)α2ds · |x|m−1−2k

1∫0

um−1−2k(1 − u2)m−32

(1 + |x|2

(1 − u2))k+ 1−α

2 du. (4)

Having in mind that m − 1 < α < m, 2k ≤ m − 3 and 1(1+|x|2(1−u2))

α−2k−12

≤ 1|x|

α−2k−12

· 1(1−u2)

α−2k−12

,

from (4) it follows that

Ak(x) ≤ |x|m−α

∞∫0

s2k

(1 + s2)α2ds

1∫0

um−1−2k(1 − u2)m−α+2k−22 du

= O(|x|m−α

), |x| → ∞

(because∫ 10 um−1−2k(1 − u2)m−α+2k−2

2 du < ∞).So,

Ak(x) = O(|x|m−α

), |x| → ∞, (5)

for all k = 0, 1, 2, . . . , [m−12 ] and m is odd.

Let now m be even and m− 1 < α < m.Since k ≤ [m−1

2 ] we have k < m−12 , i.e. m > 2k+1 and so m−1 ≥ 2k+1. Then we have α > m−1 ≥ 2k+1

i.e. α− 2k > 1 and so∫∞0

s2k

(1+s2)α2ds < ∞.

Repeating the reasoning from the previous case, we obtain,

Ak(x) = O(|x|m−α

), |x| → ∞, (6)

for all k = 0, 1, 2, . . . , [m−1 ] if m is even.
2


From (5) and (6) we get

Ak(x) = O(|x|m−α

)(7)

for all k = 1, 2, . . . , [m2 ] (m− 1 < α < m).Now we estimate Bk, k = 0, 1, 2, . . . , [m−1

2 ] if m− 1 < α < m.Since 1 ≤ k ≤ [m2 ] we have α−m + 2k ≥ α−m + 2 > 1. From

Bk(x) = |x|2k−11∫

0

u2k−1(1 − u2)m−32 du

+∞∫u|x|

vm−2k

(1 + v2 + |x|2(1 − u2))α2dv,

having in mind that

+∞∫u|x|

vm−2k

(1 + v2 + |x|2(1 − u2))α2dv <

+∞∫u|x|

vm−2k−αdv = um−2k−α+1|x|m−2k−α+1

α + 2k −m− 1 ,

we obtain

Bk(x) ≤ |x|m−α

α + 2k −m− 1

1∫0

(1 − u2)m−3

2 um−αdu

So, we have

Ak(x) = O(|x|m−α

), |x| → ∞ (8)

for all k = 1, 2, . . . , [m2 ] (m − 1 < α < m). From (3), (7) and (8) it follows that (2) holds and also thestatement of Lemma 3 in the case when m− 1 < α < m.

In a similar way it can be proved that if α = m we have (when |x| → +∞)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

Bk(x) = O(1), k = 1, 2, . . . ,[m

2

]Ak(x) = O(1), k = 1, 2, . . . ,

[m− 1

2

]− 1

A0(x) = O(ln |x|

)A[m−1

2 ](x) = O(ln |x|

).

(9)

From (3) and (9), (2) follows. Lemma 3 is proved. �3. Proof of Theorem 1

Let Sαm, PΩ : L2(Rm) → L2(Rm) be the operators defined by

Sαmf(x) =

∫Rm

Gα(x− y)f(y)dy

PΩf(x) = XΩ(x) · f(x).

The operator ϕλ(Sαm) (which acts on L2(Rm)) is an integral operator with kernel


(2π)−m

∫Rm

ϕλ

(Gα(t)

)e−it·(x−y)dt.

Since the function ϕλ is convex, applying Berezin–Lieb inequality (see [2,3,13,18]), we obtain

tr ϕλ

(Bα

m

)= tr ϕλ

(PΩS

αmPΩ

)≤ tr PΩϕλ

(Sαm

)PΩ

= |Ω|(2π)m

∫Rm

ϕλ

(1

(1 + |t|2)α2

)dt.

Hence,

tr ϕλ

(Bα

m

)≤ |Ω|

(2π)m

∫Rm

ϕλ

(1

(1 + |t|2)α2

)dt. (10)

Now, we estimate tr ϕλ(Bαm) from below.

With 〈·,·〉 we denote the inner product in L2(Ω) i.e.

〈f, g〉L2(Ω) =∫Ω

f · gdx.

Let (ωn)∞n=1 be the orthonormal system of eigenfunctions of the operator Bαm corresponding to eigenvalues

λn(Bαm) and

ωn(t) =∫Ω

e−it·xωn(x)dx.

Then, we obtain

tr ϕλ

(Bα

m

)=

∑n

ϕλ

(λn

(Bα

m

))=

∑n

ϕλ

(λn

(Bα

m

)) ∫Ω

|ωn|2dx

= (by Parseval equality)

= 1(2π)m

∑n

ϕλ

(λn

(Bα

m

)) ∫Rm

|ωn|2dt

= 1(2π)m

∑n

ϕλ

(λn

(Bα

m

)) ∫Rm

ωn(t) ωn(t)dt

= 1(2π)m

∫Rm

(∑n

ϕλ

(λn

(Bα

m

)) ∫Ω

∫Ω

e−it·(x−y)ωn(x)ωn(y)dxdy)dt.

Let et(x) = e−it·x (t, x ∈ Rm) and Es (s ∈ R) be the spectral projection of the selfadjoint operator Bα

m.Then we have: ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

tr ϕλ

(Bα

m

)= 1

(2π)m

∫Rm

dt

∫R

ϕλ(s)d〈Eset, et〉L2(Ω)

= |Ω|(2π)m

∫dt

∫ϕλ(s) 1

|Ω|d〈Eset, et〉L2(Ω).(11)

Rm R


Having in mind that

1|Ω|

∫R

d〈Eset, et〉L2(Ω) = 1|Ω| ‖et‖

2L2(Ω) = 1,

by applying the Jensen inequality to the right-hand side of (11) we obtain∫R

ϕλ(s) 1|Ω|d〈Eset, et〉L2(Ω) ≥ ϕλ

(1|Ω|

∫R

sd〈Eset, et〉L2(Ω)

)

= ϕλ

(1|Ω|

⟨Bα

met, et⟩L2(Ω)

). (12)

From (11) and (12) we get

tr ϕλ

(Bα

m

)≥ |Ω|

(2π)m

∫Rm

ϕλ

(1|Ω|

⟨Bα

met, et⟩L2(Ω)

)dt. (13)

Let ht(x) = e−it·xXΩ(x), x, y ∈ R. Then we have⟨Bα

met, et⟩L2(Ω) = 〈Gα ∗ ht, ht〉L2(Rm)

= 1(2π)m 〈Gα ∗ ht, ht〉L2(Rm) (Parseval equality)

= 1(2π)m

∫Rm

1(1 + |u|2)α

2

∣∣∣∣ ∫Ω

e−ix·(u+t)dx

∣∣∣∣2du= 1

(2π)m

∫Rm

1(1 + |x− t|2)α

2

∣∣h(x)∣∣2dx

where h(x) =∫Ωe−it·xdt. (Here f ∗ g denotes convolution of the functions f and g.)

So,

1|Ω|

⟨Bα

met, et⟩L2(Ω) = 1

(2π)m|Ω|

∫Rm

|h(x)|2(1 + |x− t|2)α

2dx.

From the previous equality we conclude that

1|Ω|

⟨Bα

met, et⟩L2(Ω) −

1(1 + |t|2)α

2

= 1|Ω|(2π)m

∫Rm

(1

(1 + |x− t|2)α2− 1

(1 + |t|2)α2

)∣∣h(x)∣∣2dx (14)

Since |ϕλ(x) − ϕλ(x)| ≤ |x− y| (x, y, λ > 0) we have∣∣∣∣ϕλ

(1|Ω|

⟨Bα

met, et⟩L2(Ω)

)− ϕλ

(1

(1 + |t|2)α2

)∣∣∣∣≤

∣∣∣∣ 1 ⟨Bα


12 α

∣∣∣∣
|Ω| (1 + |t| ) 2


≤(according to (14)

)≤ 1

|Ω|(2π)m

∫Rm

∣∣∣∣ 1(1 + |x− t|2)α

2− 1

(1 + |t|2)α2

∣∣∣∣∣∣h(x)∣∣2dx

From Lemma 1 and Lemma 3 it follows that for m− 1 < α ≤ m∫Rm

dt

∫Rm

∣∣∣∣ 1(1 + |x− t|2)α

2− 1

(1 + |t|2)α2

∣∣∣∣∣∣h(x)∣∣2dx < ∞

and the function

t �→ ϕλ

( 〈Bαmet, et〉L2(Ω)

|Ω|

)− ϕλ

(1

(1 + |t|2)α2

)has an integral dominant. Then, according to the Lebesgue dominated convergence theorem, we obtain

limλ→0+

∫Rm

(ϕλ

(1|Ω|

⟨Bα

met, et⟩L2(Ω)

)− ϕλ

(1

(1 + |t|2)α2

))dt

=∫Rm

(1|Ω|

⟨Bα


1(1 + |t|2)α

2

)dt

=(according to (14)

)= 1

(2π)m|Ω|

∫Rm

dt

∫Rm

(1

(1 + |x− t|2)α2− 1

(1 + |t|2)α2

)∣∣h(x)∣∣2dx

= 1(2π)m|Ω|

∫Rm

∣∣h(x)∣∣2dx ∫

Rm

(1

(1 + |x− t|2)α2− 1

(1 + |t|2)α2

)dt

= 0

because ∫Rm

(1

(1 + |x− t|2)α2− 1

(1 + |t|2)α2

)dt = 0, α > m− 1.

So,

limλ→0+

∫Rm

(ϕλ

(1|Ω|

⟨Bα

met, et⟩L2(Ω)

)− ϕλ

(1

(1 + |t|2)α2

))dt = 0. (15)

From (10), (13) and (15) it follows that

tr ϕλ

(Bα

m

)= |Ω|

(2π)m

∫Rm

ϕλ

(1

(1 + |t|2)α2

)dt + o(1), λ → 0+.

Having in mind (∗) we obtain

+∞∫N(μ)dμ = |Ω|

(2π)m

∫ϕλ

(1

(1 + |t|2)α2

)dt + o(1), λ → 0+. (16)

λ Rm


Let m− 1 < α < m. Then we have∫Rm

ϕλ

(1

(1 + |t|2)α2

)dt =

∫|t|≤rα(λ)

(1

(1 + |t|2)α2− λ

)dt,

rα(λ) = (1 − λ2α ) 1

2

λ1α

=∫

|t|≤rα(λ)

(1

(1 + |t|2)α2− 1

|t|α)dt +

∫|t|≤rα(λ)

1|t|α dt− λ

∫|t|≤rα(λ)

dt

= 2πm2

Γ (m2 )

rα(λ)∫0

ρm−1(

1(1 + ρ2)α

2− 1

ρα

)dρ + 2πm

2

Γ (m2 )

rα(λ)∫0

ρm−1−αdρ− 2πm2

Γ (m2 )λrα(λ)∫0

ρm−1dρ.

Having in mind that rα(λ) → +∞ when λ → 0+ and

∞∫0

ρm−1(

1(1 + ρ2)α

2− 1

ρα

)dρ =

−Γ (m2 )Γ (α+2−m2 )

(m− α)Γ (α2 ) ,

simplifying the previous equality, from (16), we conclude the statement a) of Theorem 1.If α = m then∫

Rm

ϕλ

(1

(1 + |t|2)m2

)dt =

∫|t|≤rm(λ)

1(1 + |t|2)m

2dt− λ

∫|t|≤rm(λ)

dt

= 2πm2

Γ (m2 )

rm(λ)∫0

ρm−1 − (1 + ρ2)m−12

(1 + ρ2)m2

dρ + 2πm2

Γ (m2 )

rm(λ)∫0

dρ√1 + ρ2

− 2πm2

Γ (m2 ) · λ ·rm(λ)∫0

ρm−1dρ.

Since rm(λ) → ∞ when λ → 0+ and

Am =∞∫0

ρm−1 − (1 + ρ2)m−12

(1 + ρ2)m2

dρ

after simplification, we conclude from (16) the statement b) of Theorem 1. �Remark 4. Inequalities (10) and (13) hold for any bounded domain Ω ⊂ R

m and all α > 0.In order to prove Theorem 1 by the method recommended in this paper, it is necessary to estimate the

Fourier transform of the function XΩ . Theorem 1 can be proved under less restrictive conditions on domain Ω

(see [8], Th. 4.3, pp. 216–218) but with additional condition on Gaus curvature of the boundary ∂Ω. We havechosen more restrictive conditions in order to obtain simpler calculations.

4. Regularized traces of operators Bαm

From Theorem 1 we can obtain the regularized traces of the operators Bαm when m = 1,

√2− 1 < α < 1

and the operator Bmm .


Theorem 2. If m = 1, Ω = (−1, 1)√

2 − 1 < α < 1 then the series∑∞

n−1(λn(Bα1 ) − ( 2

nπ )α) converges and

∞∑n=1

(λn

(Bα

1)−(

2nπ

)α)= −

(2π

)α

ζ(α) − 2√π

11 − α

Γ (α+12 )

Γ (α2 )

(ζ(·) is the Riemann zeta function).

Proof. The kernel of the operator Bα1 is

k(x) = 2 1−α2

√πΓ (α2 )

K 1−α2

(|x|)|x| 1−α

2

and can be represented in the following way [5]

k(x) = |x|α−12 1−α2

2Γ (α) cos απ2

+ R1(x)

The function R1 generates the operator

R1 : L2(−1, 1) −→ L2(−1, 1)

defined by

R1f(x) =1∫

−1

R1(x− y)f(y)dy

which, according to Widom’s result [19], have the property

s1(R1) = O

(1

nα+2

). (17)

Let R : L2(−1, 1) → L2(−1, 1) be the operator defined by

Rf(x) =1∫

−1

|x− y|α−1

2Γ (α) cos απ2f(y)dy.

According to [6] it follows that

λn(R) = sn(R) =(

2nπ

)α(1 + O

(1nr

))(18)

for every r ∈ (0, 1).If we put r = 2

α+3 (√

2 − 1 < α < 1) then β = α + r = α2+3α+2α+3 > 1 and from Bα

1 = R + R1 and (18),according to Lemma 2 (case β = α + r, β1 = α + 2) we obtain

λn

(Bα

1)

=(

2nπ

)α

+ O

(1

nα+r

). (19)

So, the series∑∞ (λn(Bα

1 ) − ( 2 )α) converges.
n−1 nπ


From Theorem 1a), for m = 1 we get

∑λk(Bα

1 )≥λ

(λk

(Bα

1)− λ

)= 2

π

α

1 − αλ1− 1

α − 2√π

11 − α

Γ (α+12 )

Γ (α2 ) + o(1), λ → 0+. (20)

Putting in (20) λ = λn(Bα1 ) we obtain

n∑k=1

λk

(Bα

1)− nλn

(Bα

1)

= 2π

α

1 − αλ

1− 1α

n

(Bα

1)− 2√

π

11 − α

Γ (α+12 )

Γ (α2 ) + o(1), λ → 0+. (21)

Since

n∑k=1

1kα

= n1−α

1 − α+ ζ(α) + o(1), n → ∞

having in mind (19) it follows from (21) that

n∑k=1

[λk

(Bα

1)−(

2kπ

)α]= −

(2π

)α

ζ(α) − 2√π

11 − α

Γ (α+12 )

Γ (α2 ) + O

(1

nα+r−1

). (22)

Since α + r > 1 when n → ∞ the statement of Theorem 2 follows from (22). �Remark 5. In the case m = α = 1, a result similar in spirit to Theorem 2 was obtained in [6].

Theorem 3. For the operators Bmm the following asymptotic formula

λn

(Bm

m

)= c′m

n+ o

(1n

), n → ∞ (23)

holds. Also, the series∑∞

n=1(λn(Bmm) − c′m

n ) converges and

∞∑n=1

(λn

(Bm

m

)− c′m

n

)= (1 − γ)c′m + d′m − c′m ln c′m (24)

(γ is the Euler constant).

Proof. Formula (23) follows from a general result of Birman and Solomjak [4], pp. 75–76. From Theorem 1b)we get

∑λk(Bm

m)≥λ

(λk

(Bm

m

)− λ

)= c′m ln 1

λ+ d′m + o(1), λ → 0+.

Putting in previous equality λ = λn(Bmm) we obtain

n∑k=1

λk

(Bm

m

)= nλn

(Bm

m

)+ c′m ln 1

λn(Bmm) + d′m + o(1), n → ∞. (25)


Since∑n

k=11k = lnn + γ + o(1), then from (23) and (25) we get

n∑k=1

(λk

(Bm

m

)− c′m

k

)= (1 − γ)c′m + d′m + c′m ln 1

nλn(Bmm) + o(1), n → ∞. (26)

Since nλn(Bmm) → c′m, n → ∞ from (26) it follows that the series

∑nk=1(λk(Bm

m) − c′mk ) converges and we

obtain (24). �Remark 6. In the case α = m = 2 the relation (24) was obtained in [7].

References

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Spectral properties of the operator of Bessel potential type