J. Spectr. Theory 1 (2011), 87–109
DOI 10.4171/JST/4
Journal of Spectral Theory
© European Mathematical Society
Geometrical versions of improved Berezin–Li–Yau inequalities
Leander Geisinger, Ari Laptev and Timo Weidl1
Abstract. We study the eigenvalues of the Dirichlet Laplace operator on an arbitrary bounded,
open set in Rd , d 2. In particular, we derive upper bounds on Riesz means of order
3=2, that improve the sharp Berezin inequality by a negative second term. This remainder
term depends on geometric properties of the boundary of the set and reflects the correct order
of growth in the semi-classical limit.
Under certain geometric conditions these results imply new lower bounds on individual
eigenvalues, which improve the Li–Yau inequality.
Mathematics Subject Classification (2010). Primary 35P15; Secondary 47A75.
Keywords. Dirichlet-Laplace operator, semi-classical estimates, Berezin-Li-Yau inequality.
1. Introduction
Let Rd be an open set and let denote the Dirichlet Laplace operator on
L2 ./, defined as a self-adjoint operator with form domain H01 ./. We assume that
the volume of , denoted by jj, is finite. Then the embedding H01 ./ ,! L2 ./
is compact and the spectrum of is discrete: it consists of positive eigenvalues
0 < 1 2 3 accumulating at infinity only.
Here we are interested in upper bounds on the Riesz means
X
.ƒ k /C D Tr. ƒ/ ; 0;
k
where we use the notation x˙ D .jxj ˙ x/=2. In 1972 Berezin proved that convex
eigenvalue means are bounded uniformly by the corresponding phase-space volume,
see [2]: for any open set Rd , 1, and all ƒ > 0
“
1
Cd=2
Tr. ƒ/ .jpj2 ƒ/ dp dx D Lcl
;
(1)
;d jj ƒ
.2/d Rd
1
Financial support from the German Academic Exchange Service (DAAD) and the British Council
(ARC Project 50022370) is gratefully acknowledged. The first author wants to thank for the hospitality at
Imperial College London.
88
L. Geisinger, A. Laptev and T. Weidl
see also [27], where the problem is treated from a different point of view. Here Lcl
;d
denotes the so-called Lieb–Thirring constant
Lcl
;d D
. C 1/
:
. C 1 C d=2/
.4/d=2
The Berezin inequality (1) captures, in particular, the well-known asymptotic limit
that goes back to Hermann Weyl [34]: for Rd and 0 the asymptotic identity
Cd=2
C o.ƒ Cd=2 /
Tr. ƒ/ D Lcl
;d jj ƒ
(2)
holds true as ƒ ! 1. From this follows that the Berezin inequality is sharp, in the
sense that the constant in (1) cannot be improved. However, Hermann Weyl’s work
stimulated further analysis of the asymptotic formula and (2) was gradually improved
by studying the second term, see [4], [19], [21], [28], [31], [22], and references within.
The precise second term was found by Ivrii [21]: under appropriate conditions on the
set and its boundary @ the relation
Cd=2
Tr. ƒ/ D Lcl
;d jj ƒ
C o.ƒ C.d 1/=2 /
1 cl
L
j@j ƒ C.d 1/=2
4 ;d 1
(3)
holds as ƒ ! 1. To simplify notation we write jj for the volume (the d -dimensional Lebesgue measure) of , as well as j@j for the .d 1/-dimensional surface
area of its boundary. Since the second term of this semi-classical limit is negative,
the question arises, whether the Berezin inequality (1) can be improved by a negative
remainder term.
Recently, several such improvements have been found, initially for the discrete
Laplace operator, see [11]. The first result for the continuous Laplace operator is due
to Melas [29]. From his work follows that
Tr. ƒ/ Lcl
;d jj ƒ Md
jj Cd=2
; ƒ > 0 ; 1;
I./ C
(4)
where Md is a constant depending only on the dimension and I./ denotes the second
moment of , see also [20], [36], and [37] for further generalisations. One should
mention, however, that these corrections do not capture the correct order in ƒ from
the second term of the asymptotics (3). This was improved in the two-dimensional
case in [23], where it is shown that one can choose the order of the correction term
arbitrarily close to the correct one.
In this paper we are interested in the case 3=2. For these values of it is
known, [33], that one can strengthen the Berezin inequality for any open set Rd
with a negative remainder term reflecting the correct order in ƒ in comparison to
the second term of (3). However, since one can increase j@j without changing the
individual eigenvalues k significantly, a direct analog of the first two terms of the
Geometrical versions of improved Berezin–Li–Yau inequalities
89
asymptotics (3) cannot yield a uniform bound on the eigenvalue means. Therefore
– without further conditions on – any uniform improvement of (1) must invoke
other geometric quantities.
In the result from [33] the remainder term involves certain projections on (d-1)-dimensional hyperplanes. In [16] a universal improvement of (1) was found, that holds
for 3=2 with a correction term of correct order, depending only on the volume
of .
The proof of the aforementioned results relies on operator-valued Lieb–Thirring
inequalities [26] and an inductive argument, that allows to reduce the problem to
estimating traces of the one-dimensional Dirichlet Laplace operator on open intervals.
In this paper we use the same approach, but with new estimates in the onedimensional case, in order to make the dependence on the geometry more transparent.
The new one-dimensional bounds involve the distance to the boundary of the interval
in question and are related to Hardy–Lieb–Thirring inequalities for Schrödinger operators, see [7]. There it is shown that, for 1=2 and potentials V 2 L C1=2 .RC /,
given on the half-line RC D .0; 1/, the inequality
Z 1
d2
1 C1=2
L
V .t/ 2
dt
Tr 2 V
dt
4t C
0
holds true, with a constant L independent of V. For further developments see [10]
and [13].
We start this paper by analysing the special case of the Dirichlet Laplace operator
given on a finite interval I R, with the constant potential V ƒ. For 1 we
establish that the estimate
Z d2
1 C1=2
Tr 2 ƒ
Lcl
ƒ
dt
;1
dt
4ı.t/2 C
I
is valid with the sharp constant Lcl
;1 , where ı.t/ denotes the distance to the boundary
of I. This is done in Section 2.
Then we can use results from [26] and [33] to deduce bounds in higher dimensions:
in Section 3 we first derive improvements of .1/, which are valid for any open set
Rd , d 2. These improvements depend on the geometry of . In view of
the asymptotic result (3) one might expect, that this geometric dependence can be
expressed in terms of the boundary of . To see this, we adapt methods which were
used in [5], [6], and [18] to derive geometric versions of Hardy’s inequality. Here the
result gives an improved Berezin inequality with a correction term of correct order
depending on geometric properties of the boundary.
If is convex and smooth this dependence can be expressed only in terms of jj,
j@j, and the curvature of the boundary. In particular the first remainder term of the
estimate is very similar to the second term of the semiclassical asymptotics (3): it
shows the same order in ƒ and it depends only on the surface area of the boundary.
In Section 4 we return to the general case, where Rd is not necessarily
convex or smooth, and obtain lower bounds on individual eigenvalues k . Under
90
L. Geisinger, A. Laptev and T. Weidl
certain conditions on the geometry of , these results improve the estimate
k Cd
d k 2=d
;
d C 2 jj
(5)
from [27], where Cd denotes the semi-classical constant 4 .d=2 C 1/2=d .
Finally, in Section 5, we specialise to the two-dimensional case, where we can
use the foregoing results and refined geometric considerations to further improve and
generalise the inequalities. In particular, we avoid dependence on curvature, thus we
do not require smoothness of the boundary.
The question whether such improved estimates can be generalised to 1 < 3=2
remains open.
2. One-dimensional considerations
Let us consider an open interval I R of length l > 0. For t 2 I let
ı.t/ D inffjt sj W s … I g
be the distance to the boundary. The eigenvalues of d 2 =dt 2 subject to Dirichlet
boundary conditions at the endpoints of I are given by k D k 2 2 = l 2 . Therefore
the Riesz means equal
Tr X
d2
k 2 2 ƒ
D
ƒ
:
dt 2
l2 C
k
To find precise bounds on the Riesz means in the one-dimensional case, it suffices to
analyse this sum explicitly. Our main observation is
Lemma 2.1. Let I R be an open interval and 1. Then the estimate
Z d2
1 C1=2
Tr 2 ƒ
Lcl
ƒ
dt;
;1
dt
4ı.t/2 C
I
holds true for all ƒ > 0. The constant 1=4 cannot be improved.
The remainder of this section deals with the proof of this estimate. First we need
two rather technical results. The proof is elementary but not trivial and therefore will
be given in the appendix.
Lemma 2.2. For all A 1=
X
k
k2 2
1 2
C
A
3
Z
A
1
1
1 3=2
ds:
s2
(6)
Geometrical versions of improved Berezin–Li–Yau inequalities
91
Lemma 2.3. Let I R be an open interval of length l > 0. Then for 1 and
c>0
Z 1 p X
c C1=2
2 k 2 ƒ
dt ƒ 2
D
c ƒ C o.ƒ /
Lcl
;1
2
C
C
ı.t/
l
2
I
k
holds as ƒ ! 1.
Proof of Lemma 2.1. Note that one can always assume I D .0; l/, where l > 0
denotes the length of the interval. First we deduce the estimatepfor D 1 from
Lemma 2.2. Assume ƒ l 2 and apply Lemma 2.2 with A D l ƒ= to get
X
d2
2k2 2
1 2
ƒ
Tr 2 ƒ D ƒ
C
dt
l ƒ
3
k
Z
l
1
p
ƒ
1
1 3=2
ds:
s2
p
Substituting s D 2t ƒ, we find that
Z l=2
Z l=2
d2
4
1 3=2
1 3=2
cl
Tr 2 ƒ
ƒ
dt
D
2L
ƒ
dt
1;1
dt
3 1=.2pƒ/
4t 2
4t 2 C
0
holds for all ƒ l 2 . Note that this inequality is trivially true for 0 < ƒ < l 2 ,
since the left hand side is zero. Finally, we use the identities
Z l=2
Z l Z
3=2
1 3=2
1
1 l
1 3=2
ƒ 2
dt D
ƒ
dt
D
ƒ
dt
4t C
4.l t/2 C
2 0
4ı.t/2 C
0
l=2
to finish the proof for D 1.
To deduce the claim for > 1 we can apply a method from [1]. Writing
Z 1
X
X
1
ƒ k
D
2
ƒ k d ;
C
C
B.2; 1/ 0
k
k
we estimate
d2
1
Lcl
Tr 2 ƒ
dt
B.2; 1/ 1;1
Z Z
I
1
2 ƒ 0
3=2
1
d dt
C
4ı.t/2
Z
1 C1=2
B.5=2; 1/ ƒ
dt
D Lcl
1;1
B.2; 1/ I
4ı.t/2 C
cl
and the result follows from the identity Lcl
1;1 B.5=2; 1/ D L;1 B.2; 1/.
The claim, that the constant 1=4 cannot be improved, follows from Lemma 2.3:
for c D 1=4 the leading term of the asymptotics in Lemma 2.3 vanishes. For any
constant c > 1=4 the leading term is negative, thus the estimate from Lemma 2.1
must fail in this case, for large values of ƒ.
92
L. Geisinger, A. Laptev and T. Weidl
Figure 1 illustrates the results of Lemma 2.1 and Lemma 2.3 for l D 2 and
D 1 with the sharp constant c D 1=4: the function
Z X
1 3=2
cl
2
ƒ
dt
ƒ
k
f .ƒ/ D L1;1
C
4 ı.t/2 C
0
k
is plotted for 1 < ƒ < 112, so that the first ten minima are shown.
2.5
2.0
1.5
1.0
0.5
20
40
60
80
100
Figure 1. The function f .ƒ/, illustrating the results from Section 2.
3. Results in higher dimensions
In this section we use the one-dimensional result to prove uniform eigenvalue estimates for the Dirichlet Laplace operator in bounded open sets in higher dimensions.
These estimates – refinements of the Berezin inequality (1) – depend on the geometry
of the set, in particular on properties of the boundary.
3.1. Arbitrary open sets. First we provide general estimates, valid for any open
subset Rd , d 2. Let
Sd 1 D fx 2 Rd W jxj D 1g
denote the unit-sphere in Rd. For an arbitrary direction u 2 Sd 1 and x 2 set
.x; u/ D infft > 0 W x C tu … g;
d.x; u/ D inff.x; u/; .x; u/g;
Geometrical versions of improved Berezin–Li–Yau inequalities
93
and
l.x; u/ D .x; u/ C .x; u/:
Theorem 3.1. Let Rd be an open set and let u 2 Sd 1 and 3=2. Then for
all ƒ > 0 the estimate
Z Cd=2
1
cl
ƒ
dx
(7)
Tr. ƒ/ L;d
4 d.x; u/2 C
holds true.
Remark 3.2. Let us define
l0 D inf
sup l.x; u/ :
(8)
u2Sd 1 x2
Then Theorem 3.1 implies the following improvement of Melas-type: for 3=2
and all ƒ > 0 the estimate
Tr. ƒ/ Lcl
;d jj ƒ 1 Cd=2
l02 C
(9)
holds. In convex domains l0 is the minimal width of the domain, see [3]. In this case
1= l02 is bounded from below by a multiple of jj2=d, [35], while no such bound
holds for the improving term jj=I./ in Melas’ inequality (4).
The proof of Theorem 3.1 relies on a lifting technique, which was introduced in
[25], see also [26], [8], [33], and [9] for further developments and applications.
Proof of Theorem 3.1. We apply the argument used in [33] to reduce the problem to
one-dimensional estimates. Fix a Cartesian coordinate system in Rd, such that the
given direction u corresponds to the vector .0; : : : ; 0; 1/.
For x 2 Rd write x D .x 0 ; t/ 2 Rd 1 R and let r 0 and 0 denote the gradient
and the Laplace operator in the first d 1 dimensions. Each section .x 0 / D ft 2
R W .x 0 ; t/ 2 g consists of at most countably many open intervals Jk .x 0 / R,
k D 1; : : : ; N.x 0 / 1.
We consider the quadratic form of ƒ on functions ' from the form core
C01 ./ and write
Z
Z
2
2
0
2
0
dx
.j@ t 'j2 ƒj'j2 /dt:
kr'kL2 ./ ƒk'kL2 ./ D kr 'kL2 ./ C
Rd 1
.x 0 /
The functions '.x 0 ; / satisfy Dirichlet boundary conditions at the endpoints of each
interval Jk .x 0 / forming .x 0 /, hence
Z
2
2
.j@ t 'j ƒj'j /dt D
.x 0 /
0
N.x
X/ Z
kD1
.j@ t 'j2 ƒj'j2 /dt
Jk .x 0 /
94
L. Geisinger, A. Laptev and T. Weidl
0
N.x
X/
hVk .x 0 /'.x 0 ; /; '.x 0 ; /iL2 .J
kD1
k .x
0 //
;
where the bounded, non-negative operators Vk .x 0; ƒ/ D .@2t ƒ/ are the negative
parts of the Sturm–Liouville operators @2t ƒ with Dirichlet boundary conditions
on Jk .x 0 /. Let
0
N.x
M/
0
Vk .x 0; ƒ/
V .x ; ƒ/ D
kD1
0
ƒ on .x / subject to Dirichlet boundary conditions
be the negative part of
on the endpoints of each interval Jk .x 0 /, k D 1; : : : ; N.x 0 /, that is on @.x 0 /. Then
Z
.j@ t 'j2 ƒj'j2 /dt hV .x 0 ; ƒ/'.x 0 ; /; '.x 0 ; /iL2 ..x 0 //
@2t
.x 0 /
and consequently
Z
2
2
0
2
kr'kL
ƒk'k
kr
'k
2 ./
L2 ./
L2 ./
dx 0 hV '.x 0 ; /; '.x 0 ; /iL2 ..x 0 // :
Rd 1
Now we can extend this quadratic form by zero to C01 .Rd n @/, which is a form
core for .Rd n / ˚ . ƒ/. This operator corresponds to the left hand side
of the equality above, while the semi-bounded form on the right hand side is closed
on the larger domain H 1 .Rd 1; L2 .R//, where it corresponds to the operator
0 ˝ I V .x 0 ; ƒ/ on L2 .Rd 1; L2 .R//:
(10)
Due to the positivity of Rd n we can use the variational principle to deduce that
for any 0
Tr. ƒ/ D Tr..Rd n / ˚ . ƒ//
Tr.0 ˝ I V .x 0; ƒ// :
Now we apply sharp Lieb–Thirring inequalities [26] to the Schrödinger operator (10)
with the operator-valued potential V .x 0 ; ƒ/ and obtain that for 3=2
Z
cl
Tr V .x 0; ƒ/ C.d 1/=2 dx :0
(11)
Tr. ƒ/ L;d 1
Rd 1
To estimate the trace of the one-dimensional differential operator V .x 0; ƒ/ we
apply Lemma 2.1. Our choice of coordinate system implies that for x D .x 0 ; t/ 2
Jk .x 0 / the distance of t to the boundary of the interval Jk .x 0 / is given by d.x; u/.
Hence, Lemma 2.1 implies
0
C.d 1/=2
Tr V .x ; ƒ/
0
N.x
X/
D
Tr kD1
C.d 1/=2
d 2 ˇˇ
ˇ
ƒ
dt 2 Jk .x 0 /
Geometrical versions of improved Berezin–Li–Yau inequalities
Z
Lcl
C.d 1/=2;1
.x 0 /
ƒ
95
Cd=2
1
dt
4 d..x 0 ; t/; u/ C
Lcl
D Lcl
.
and the result follows from (11) and the identity Lcl
;d 1 C.d 1/=2;1
;d
We proceed to analysing the geometric properties of (7). Note that the left hand
side of (7) is independent of the choice of direction u 2 Sd 1, while the right hand
side depends on u and therefore on the geometry of . For a given set one can
minimise the right hand side in u 2 Sd 1. The result, however, depends on the
geometry of in a rather tricky way. In order to make this geometric dependence
more transparent, we average the right hand side of (7) over u 2 Sd 1. Even though
the resulting bound is – in general – not as precise as (7), it allows a more appropriate
geometric interpretation.
To analyse the effect of the boundary, one would like to estimate d.x; u/ in terms
of the distance to the boundary. See [5], [6], and [18], where this approach is used
to derive geometrical versions of Hardy’s inequality. To avoid complications that
arise, for example, if the complement of contains isolated points, we use slightly
different notions: for x 2 let
.x/ D fy 2 W x C t.y x/ 2 ; 8t 2 Œ0; 1
g
be the part of that “can be seen” from x and let
ı.x/ D inffjy xj W y … .x/g
denote the distance to the exterior of .x/.
For fixed " > 0 put
A" .x/ D fa 2 Rd n .x/ W jx aj < ı.x/ C "g
and for a 2 A" .x/ set Bx .a/ D fy 2 Rd W jy aj < jx ajg and
a .x/ D
jBx .a/ n .x/j
;
!d jx ajd
where !d denotes the volume of the unit ball in Rd. To get a result, independent of
a and ", set
.x/ D inf sup a .x/:
">0 a2A" .x/
Note that R n .x/ is open, hence a .x/ > 0 and .x/ > 0 hold for x 2 . Finally,
we define
Z
M .ƒ/ D
.x/ dx;
d
R .ƒ/
p
where R .ƒ/ denotes the set fx 2 W ı.x/ < 1=.4 ƒ/g. The main result of
this section allows a geometric interpretation of the remainder term.
96
L. Geisinger, A. Laptev and T. Weidl
Theorem 3.3. Let Rd be an open set with finite volume and 3=2. Then for
all ƒ > 0 we have
Cd=2
d C1 Cd=2
Lcl
ƒ
M .ƒ/:
Tr. ƒ/ Lcl
;d jjƒ
;d 2
(12)
The function .x/ depends on the behaviour of the boundary close to x 2 .
For example, .x/ is small close to a cusp. On the other hand .x/ is larger than
1=2 in a convex domain. By definition, the function M .ƒ/
p gives an average of this
behaviour over R .ƒ/, which is like a tube of width 1=.4 ƒ/ around the boundary.
Note that M .ƒ/ tends to zero as ƒ tends to infinity. This decay in ƒ is of
the order .ıM d /=2, where ıM denotes the interior Minkowski dimension of the
boundary, see e.g. [24], [14], and [12] for definition and examples. If d 1 ıM < d
and if the upper Minkowski content of the boundary is finite, then the second term
of the asymptotic limit of the Riesz means equals O.ƒ CıM =2 / as ƒ ! 1, see
[24]. Therefore the remainder term in (12) reflects the correct order of growth in the
asymptotic limit.
In particular, if the dimension of the boundary equals d 1, we find
M .ƒ/ D j@j ƒ1=2 C o.ƒ1=2 /
as ƒ ! 1 and the second term in (12) is in close correspondence with the asymptotic
formula (3).
Proof of Theorem 3.3. We start from the result of Theorem 3.1 and average over all
directions to get
Z Z
Cd=2
1
Cd=2
Tr. ƒ/ Lcl
ƒ
1
d.u/ dx; (13)
;d
2
4 ƒ d.x; u/ C
Sd 1
where d.u/ denotes the normed measure on Sd 1 .
For x 2 and a … .x/ let ‚.x; a/ Sd 1 be the subset of all directions
u 2 Sd 1, satisfying x C su 2 Bx .a/ n .x/ for some s > 0. For such s we have
s 2 jx aj :
By definition of a .x/ and ‚.x; a/ we find
d
a .x/ !d jx aj D jBx .a/ n .x/j hence
Z
(14)
Z
d.u/ !d .2jx aj/d;
‚.x;a/
d.u/ 2d a .x/ :
(15)
‚.x;a/
Using (14) we also see that for u 2 ‚.x; a/ the estimate d.x; u/ s 2 jx aj
holds.
97
p
Now fix ƒ > 0 and choose 0 < " < 1=.4 ƒ/ and a 2 A" .x/. By definition of
A" .x/ it follows that for all u 2 ‚.x; a/
Geometrical versions of improved Berezin–Li–Yau inequalities
d.x; u/ 2jx aj < 2.ı.x/ C "/:
(16)
The set ‚.x; a/ must be contained in one hemisphere of Sd 1 which we denote by
d 1
SC
. Using that d.x; u/ D d.x; u/ we estimate
Z
Z
Cd=2
Cd=2
1
1
1
d.u/
D
2
1
d.u/
2
2
1
4ƒd.x; u/ C
4ƒd.x; u/ C
Sd 1
Sd
C
Z
12 ˚
p d.u/:
1
u2Sd
W d.x;u/1=.2 ƒ/
C
p
Assume that ı.x/ 1=.4 ƒ/ ". From (16) it follows that
p
‚.x; a/ fu 2 SdC1 W d.x; u/ 1=.2 ƒ/g;
hence, using (15), we conclude
Z
Z
Cd=2
1
1
d.u/
1
2
d.u/ 1 21d a .x/ :
2 C
d
1
4ƒd.x;
u/
S
‚.x;a/
Since a 2 A" .x/ was arbitrary we arrive at
Z
Cd=2
1
1
d.u/ 1 21d .x/;
4ƒd.x; u/2 C
Sd 1
p
for all x 2 with ı.x/ 1=.4 ƒ/ " and we can take the limit " ! 0.
It follows that
Z Z
Cd=2
1
1
d.u/ dx
4ƒd.x; u/2 C
Sd 1
Z
jj 21d ˚
p .x/ dx
x2 W ı.x/<1=.4 ƒ/
and inserting this into (13) yields the claimed result.
3.2. Convex domains. If Rd is convex, we have .x/ D and
.x/ > 1=2
(17)
for all x 2 . Thus we can simplify the remainder term, by estimating M .ƒ/.
Corollary 3.4. Let Rd be a bounded, convex domain and let t D fx 2
W ı.x/ > tg be the inner parallel set of . Then for 3=2 and all ƒ > 0 we
have
Cd=2
d 2
Lcl
j@1=.4pƒ/ j ƒ C.d 1/=2 :
Tr. ƒ/ Lcl
;d jjƒ
;d 2
98
L. Geisinger, A. Laptev and T. Weidl
Proof. Inserting (17) into the definition of M .ƒ/ yields
Z
Z
1
1
dx D
M .ƒ/ >
dx
˚
2 R .ƒ/
2 x2 W ı.x/<1=.4pƒ/
D
1
2
(18)
p
1=.4 ƒ/
Z
0
j@ t jdt:
In view of (12) estimating
p
1=.4 ƒ/
Z
0
j@ t jdt 1
p j@1=.4pƒ/ j
4 ƒ
proves the claim.
If more is known about the geometry of we can refine the estimate.
Corollary 3.5. Let Rd be a bounded, convex domain with smooth boundary and
assume that the curvature of @ is bounded from above by 1=R. Then for 3=2
and all ƒ > 0 we have
Cd=2
Tr. ƒ/ Lcl
;d jjƒ
Z 1
d 1 d 2
C.d 1/=2
Lcl
2
j@j
ƒ
1
p s ds:
;d
4R ƒ C
0
Note that one can estimate the integral in the remainder term to get the simplified
bound
Cd=2
d 2
Lcl
j@j ƒ C.d 1/=2 1 Tr. ƒ/ Lcl
;d jjƒ
;d 2
d 1 :
p
8R ƒ C
Proof of Corollary 3.5. To estimate M .ƒ/ we can use Steiner’s Theorem, see [15]
and [32], namely
d 1 t j@j :
j@ t j 1 (19)
C
R
From (18) it follows that
1
M .ƒ/ > j@j
2
p
1=.4 ƒ/
Z
0
1
d 1 j@j
t dt D p
C
R
8 ƒ
Inserting this into (12) completes the proof.
Z 1
d 1 1
p s ds:
4R ƒ C
0
Geometrical versions of improved Berezin–Li–Yau inequalities
99
4. Lower bounds on individual eigenvalues
In order to further estimate the remainder terms, in particular to show that the remainder is negative for all ƒ 1 ./ one needs suitable bounds on the ground state
1 ./. We point out the following consequence of the proof of Theorem 3.1 which
might be of independent interest.
Corollary 4.1. For any open set Rd the estimate
2
l02
1 ./ holds, where l0 is given in (8).
Remark 4.2. One can compare this bound with
1 ./ inf
x2
d ;
4m.x/2
see [5]. Here m.x/2 denotes the average of d.x; u/2 over all directions u 2 Sd 1.
The relation between these bounds depends on the geometry of . In particular, the
bound from Corollary 4.1 is stronger, if the width of is small along an arbitrary
direction u 2 Sd 1 such that l0 is small enough.
Proof of Corollary 4.1. Fix " > 0 and choose a direction u0 2 Sd 1, such that
supx2 l.x; u0 / < l0 C ". We write x D .x 0 ; t/ 2 Rd 1 R, where the t-axes
is chosen in the direction of u0 . Let us recall inequality (11) from the proof of
Theorem 3.1: for any 3=2
Z
Tr V .x 0 ; ƒ/ C.d 1/=2 dx 0 ;
Tr. ƒ/ Lcl
;d 1
Rd 1
where V .x 0 ; ƒ/ denotes the negative part of the operator @2t ƒ on .x 0 / D
0/
[N.x
J .x 0 / with Dirichlet boundary conditions at the endpoints of each interval
kD1 k
0
Jk .x /. This inequality can be rewritten as
Z
Tr. ƒ/
Lcl
;d 1
0
N.x
X/X Rd 1
ƒ
kD1 j 2N
2 j 2 C.d 1/=2 0
dx :
jJk .x 0 /j2 C
Our choice of coordinate system implies jJk .x 0 /j supx2 l.x; u0 / < l0 C " for
all k D 1; : : : ; N.x 0 / and all x 0 2 Rd 1. It follows that the right hand side of the
inequality above is zero for all ƒ 2 =.l0 C "/2. Thus by taking the limit " ! 0
we find
X
.ƒ n /C D Tr. ƒ/ D 0;
n2N
for all ƒ 2 = l02 and 1 2 = l02 follows.
100
L. Geisinger, A. Laptev and T. Weidl
From (9) we obtain similar bounds on higher eigenvalues using a method introduced in [25].
Corollary 4.3. For any open set Rd with finite volume and any k 2 N the
estimate
k ./ Cd
holds, with
12 1=d
..d C 3/=2/ 2=d k 2=d
d
1
C 2
1C1=d
2=d
.d=2 C 1/
.d C 3/
jj
l0
Cd D 4 .d=2 C 1/2=d :
Proof. Let N.ƒ/ D Tr. ƒ/0 denote the counting function of the eigenvalues
below ƒ > 0. In [25] it is shown that for > 0, and all ƒ > 0, > 0
N.ƒ/ .ƒ/ Tr. .1 C /ƒ/ :
(20)
If we apply this inequality with D 3=2, we can use (9) to estimate
d=2
N.ƒ/ Lcl
3=2;d jj ƒ
.d C3/=2
1
.1 C /.d C3/=2 1
:
3=2
ƒ.1 C /l02 C
Minimising the right hand side in > 0 yields min D 3.ƒl02 1/=.dƒl02 / and
inserting this we find
N.ƒ/ Lcl
3=2;d jj
.d C 3/.d C3/=2 1 d=2
ƒ
:
33=2 d d=2
l02 C
This is equivalent to the claimed result.
Remark 4.4. Applying the same method to (1) with D 1 we recover the Li–
Yau inequality (5). In the proof of Corollary 4.3 we have to start from D 3=2,
therefore the result is not strong enough to improve (5) in general, i.e. the coefficient
of k 2=d =jj2=d in Corollary 4.3 is less than Cd d=.d C 2/ for all d 2. However,
one gets an improvement of (5) for low eigenvalues whenever l0 is small.
5. Further improvements in dimension 2
In this section we further improve Corollary 3.5 and generalise it to a large class of
bounded convex domains R2. Here we do not require smoothness, therefore
we cannot use (19) to estimate inner parallels of the boundary. To find a suitable
substitute let w denote the minimal width of and note that for l0 given in (8) the
identity
w D l0
Geometrical versions of improved Berezin–Li–Yau inequalities
101
holds true, see e.g. [3]. In the remainder of this section we assume that for all t > 0
3t j@j :
(21)
w C
This is true for a large class of convex domains, including the circle, regular polygons
and arbitrary triangles and equality holds for the equilateral triangle. Actually we
conjecture that (21) holds true for all bounded convex domains in R2.
Furthermore we need a lower bound on the ground state. From Corollary 4.1 we
obtain that for all convex domains R2
j@ t j 1 2
(22)
w2
holds. One should mention, that this follows immediately from the variational principle and the fact that 1 ./ D 2 =w 2 if is the infinite strip with width w. Moreover,
the same estimate can be obtained from the inequality 1 ./ 2 =.4ri2n /, see [30],
where ri n is the inradius of .
Using (21) and (22) we can derive a more precise version of Theorem 3.3, where
the correction term depends only on j@j and is apart from that independent of the
geometry of .
1 ./ Theorem 5.1. Let R2 be a bounded, convex domain, satisfying (21). Then for
3=2 we have
Tr. ƒ/ D 0
and
C1
C1=2
Cco Lcl
Tr. ƒ/ Lcl
;2 jj ƒ
;1 j@j ƒ
if ƒ 2 =w 2
if ƒ > 2 =w 2 ;
with a constant
Cco 4 11
3
2
ln
> 0:0642:
9 2 20 4 5 2
3
Proof. The first claim follows directly from (22), thus we can assume ƒ > 2 =w 2.
First we prove the result for D 3=2. Again we can start from (13) and we need to
estimate d.x; u/ in terms of ı.x/, which is just the distance to the boundary, since is convex.
Fix x 2 . Since is convex and smooth we can choose u0 2 Sd 1, such
d 1
that d.x; u0 / D ı.x/. We can assume u0 D .1; 0; : : : ; 0/ and put SC
D fu 2
d
R W u1 > 0g.
Let a be the intersection point of the semi-axes fx C tu0 ; t > 0g with @ and for
d 1
let bu be the intersection point of fx C tu; t > 0g with the plane
arbitrary u 2 SC
through a, orthogonal to u0 . We find d.x; u/ jx bu j and if u denotes the angle
between u0 and u, we find
d.x; u/ jx bu j D
jx aj
ı.x/
D
:
cos u
cos u
102
L. Geisinger, A. Laptev and T. Weidl
Using (13) and taking into account that d.x; u/ D d.x; u/ we get
Z
Z
cos2 ./ 5=2
2 =2
3=2
cl
5=2
Tr. ƒ/ L3=2;2 ƒ
1
d dx;
4ƒı.x/2
0
p
p
p
where 0 D 0 if ı.x/ 1=.2 ƒ/ and 0 D arccos.2ı.x/ ƒ/ if ı.x/ < 1=.2 ƒ/.
To obtain a simple bound one could estimate
1
cos2 ./ 5=2
cos2 ./
1
:
4ƒı.x/2
4ƒı.x/2
However, to get a result as precise as possible we use
1
cos2 ./ 5=2
cos2 ./
cos4 ./
1
C
4ƒı.x/2
2ƒı.x/2
16ƒ2 ı.x/4
and get
Z Z =2
ƒ5=2
cos2 ./
cos4 ./ Tr. 1
C
d dx: (23)
5 2 0
2ƒı.x/2
16ƒ2 ı.x/4
p
Set u0 D min.1; 2ı.x/ ƒ/ and calculate
q
Z =2
arcsin.u0 / u0 1 u20
cos2 ./
cos4 ./ 1
C
d D 0 2ƒı.x/2
16ƒ2 ı.x/4
2
4ƒı.x/2
0
q
3 arcsin.u0 / u0 2u20 C 3 1 u20
C
:
128ƒ2 ı.x/4
ƒ/3=2
Inserting this back into (23) yields
Tr. ƒ/3=2
Z
where
I1 D ˚
p
ı.x/1=.2 ƒ/
and
1
jj ƒ5=2 ƒ5=2 .I1 C I2 /;
10
3
1
1 dx
10 4ƒı.x/2 128ƒ2 ı.x/4
p
p
arcsin.2 ƒı.x//
1 arccos.2 ƒı.x// C
I2 D ˚
p 2
4ƒı.x/2
ı.x/<1=.2 ƒ/ 5
p
p
1 4ƒı.x/2 3 arcsin.2 ƒı.x//
p
128ƒ2 ı.x/4
2 ƒı.x/
p
.8ƒı.x/2 C 3/ 1 4ƒı.x/2 C
dx:
64ƒ3=2 ı.x/3
Z
(24)
Geometrical versions of improved Berezin–Li–Yau inequalities
First we turn to
I1 D
1
10
Z
1
p
1=.2 ƒ/
ˇ
ˇ
ˇ
ˇ
ˇ@ t ˇ
103
3
1
dt:
2
2
4
4ƒt
128ƒ t
p
Note that the term in brackets is positive, thus after substituting s D 2 ƒt we can
use (21) and ƒ > 2 =w 2 to obtain
Z
1 j@j 1
3s 1
3 I1 p
1
ds
20 ƒ 1
2 C s 2 8s 4
D
2 1 j@j 7
3
39
27
:
ln
p
20 ƒ 8 32 128 3 2
3
Similarly we can treat I2 and get
1 j@j 557 7
3
p
I2 2
10
ƒ 192 16 2
Z
1
0
arcsin.s/ ds :
s
In view of Lcl
D 3=16 we can write
3=2;1
2 3
2
2
j@j 11
I1 C I2 Lcl
p
ln
ln.2/
3=2;1
3
5 2
ƒ 9 2 20 4 5 2
and inserting this into (24) yields the claim in the case D 3=2.
To prove the estimate for > 3=2 we again refer to [1] and use the identity
Z 1
3=2
1
5=2 Tr .ƒ /
d ;
Tr. ƒ/ D
B. 3=2; 5=2/ 0
from which the general result follows.
Now we can apply the same arguments that lead to Corollary 4.3 to derive lower
bounds on individual eigenvalues.
Corollary 5.2. Let R2 be a bounded, convex domain, satisfying (21). Then for
k 2 N and any ˛ 2 .0; 1/ the estimate
s
2 j@j2
15 Cco j@j
225 2 Cco
k ./
k
3=2 k
10 ˛ 3=2
10 ˛
C
C
1˛
jj
8
jj
jj
256
jj2
C
2
225 2 Cco
j@j2
128
jj2
holds, with the constant Cco given in Theorem 5.1.
104
L. Geisinger, A. Laptev and T. Weidl
Proof. Applying (20) and Theorem 5.1 with D 3=2 yields
N.ƒ/ Lcl
3=2;2 jj ƒ
p .1 C /2
.1 C /5=2
cl
C
L
j@j
ƒ
co
3=2;1
3=2
3=2
for any > 0 and ƒ > 2 =w 2. With D ˛=.1 ˛/, ˛ 2 .0; 1/, this is equivalent to
the claimed estimate.
Remark 5.3. Given a fixed ratio j@j=jj one can optimise the foregoing estimate
in ˛ 2 .0; 1/, depending on k 2 N. As mentioned in the remark after Corollary 4.3,
the result cannot improve the Li–Yau inequality (5) in general, since we have to apply
(20) with D 3=2 instead of D 1. However, the estimates obtained from Corollary
5.2 are stronger than (5) for low eigenvalues and the improvements depend on the
ratio j@j=jj.
In particular, one can use the isoperimetric inequality, namely j@j 2.jj/1=2
for all R2, to derive general improvements of the Li–Yau inequality (5) for low
eigenvalues. From (5) we get
2k
;
k ./ jj
and one can supplement these estimates with the Krahn–Szegö inequality [17]
2 ./ 2
2j0;1
jj
:
(25)
Optimising the estimate of Corollary 5.2 with j@j D 2.jj/1=2 , we find that for
any convex domain, satisfying (21),
6 ./ >
2
2j0;1
40:50
12
>
>
;
jj
jj
jj
7 ./ >
46:74
14
>
;
jj
jj
::: ;
23 ./ >
144:58
46
>
:
jj
jj
In this way we can improve (5) and (25) in convex domains for all eigenvalues k ./
with 6 k 23.
Finally let us make a remark about the square Ql D .0; l/ .0; l/ R2 , l > 0.
Using the methods introduced in Section 2 one can establish the following twodimensional version of Lemma 2.1: choose a coordinate system .x1 ; x2 / 2 R2 with
axes parallel to the sides of the square and for x 2 Ql put
ı.xi / D min.xi ; l xi /; i D 1; 2:
Geometrical versions of improved Berezin–Li–Yau inequalities
105
Then for 1 and all ƒ > 0 the estimate
Tr. ƒ/ D
X
m;n2N
Lcl
;2
2 2
2
n
C
m
C
l2
ƒ
Z l Z l
0
0
ƒ Csq
1 2 C1
1
dx1 dx2
C
C
ı.x1 /
ı.x2 /
holds with a constant Csq > 1=10.
A. Proof of Lemma 2.2 and Lemma 2.3
A.1. Proof of Lemma 2.2. For A 2 R let Ax and Az denote the integer and fractional
part of A respectively. Then we can calculate
X
k
Ax X
k2 k2 1 Ax 3
Ax 2
Ax
1 2
D
1 2 D Ax 2
C
C
A C
A
A
3
2
6
kD1
2A 1
1
z A/
z 1 C A.1
z 3Az C 2Az 2 / 1 :
C A.1
3
2 6A
A
6A2
p
z A/
z 1=4 and A.1
z 3Az C 2Az 2 / 3=18
From 0 Az < 1 we conclude A.1
and we get
p
X
3
k2 2A 1
1
1 2
:
(26)
C
C
C
A
3
2
12A
108A2
D
k
To estimate the right hand side of (6) note that
Z 1 3=2
1 p 2
3
1
1 2
ds D 1 C 2
s 1 C arctan p
;
s
2s
2
s2 1
thus
p
Z A
1
1
2
1 3=2
2 2 A2 C 1 2 A2 1 1
1 2
ds D
:
C
arctan
p
3 1
s
3 2 A
A
2
2 A2 1
Now we can insert the elementary estimates
1
1
arctan p
A
2 A2 1
and
p
a
1
2 A2 1
1
A
2 2 A2 A4
106
L. Geisinger, A. Laptev and T. Weidl
p
both valid for A 2, where we write a D 16 2= 2 8 4 2 1= for simplicity.
We get
Z
2 A
1 3=2
2A 1
a 2
1 1
1
1 2
ds C
(27)
C 2 3 1
s
3
2
A 6 4 A3 3 A3
2 A5
for all A 2. From (26) and (27) we deduce that (6) holds true for all A 2, since
p
1
1 4
2a 2
a
3 3 1
A
C
A 0
A
2 12
108
6 4
3
3 2
for all A 2.
Note that (6) is trivial for 1= A 1, since the left hand side equals zero. The
remaining case 1 A 2 can be checked by hand.
p
A.2. Proof of Lemma 2.3. We assume I D .0; l/, substitute t D s c=ƒ and write
Z l
ƒ
0
p
Z p
p l ƒ=.2 c/
c C1=2
1 C1=2
dt
D
2
c
ƒ
1
ds:
ı.t/2 C
s2
1
The claim of the lemma follows, if we show that
Z l pƒ=.2pc/
X
p cl
1 C1=2
2 k 2 1 p
2 cL;1
1 2
ds 1
D c C o.1/
2
s
ƒl C 2
1
k
p
as ƒ ! 1. With A D l ƒ= this is equivalent to
Z A=.2pc/
X
p cl
1 C1=2
k 2 1 p
1 2
ds 1 2 D c C o.1/ (28)
2 c L;1
s
A C
2
1
k
as A ! 1.
It is easy to see that
X
k
1
k 2 A 1 1
D
C
1;
C o.1/
B
A2 C
2
2
2
as A ! 1. Moreover, we claim
Z A=.2pc/
3
1 C1=2
A
1 1
1 2
ds D p C B ; C
C o.1/
s
2
2
2
2 c
1
(29)
as A ! 1 and (28) follows from Lcl
;1 D B. C 1; 1=2/=.2/, if we can establish
(29). Let us write
1
X k m 1
1 m X
k m
2k
D
.1/
s
D
s 2k
s2
k
k
k0
k0
Geometrical versions of improved Berezin–Li–Yau inequalities
107
for m 1 and note that the sum is finite if m 2 N, while the sum converges uniformly
on s 2 Œ1; 1/ if m … N. Hence we have
Z y
X k m 1 1
1 m
1 2 ds D y C
C o.1/
s
2k 1
k
1
k0
as y ! 1. Using that
X k m 1
k0
we obtain
k
1
1 1
D B ; mC1
2k 1
2
2
Z y
1 m
1 1
1 2 ds D y C B ; m C 1 C o.1/
s
2
2
1
as y ! 1, which is equivalent to (29). This finishes the proof of Lemma 2.3.
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Received October 23, 2010, revised November 19, 2010
Leander Geisinger, Universität Stuttgart, Pfaffenwaldring 57, 70569 Stuttgart, Germany
E-mail: [email protected]
Ari Laptev, Imperial College London, 180 Queen’s Gate, London SW7 2AZ, UK
E-mail: [email protected]
Timo Weidl, Universität Stuttgart, Pfaffenwaldring 57, 70569 Stuttgart, Germany
E-mail: [email protected]