Download SPECTRAL AND FUNCTIONAL INEQUALITIES Let us consider two

SPECTRAL AND FUNCTIONAL INEQUALITIES
A BSTRACT. We shall derive a number of spectral inequalities including: Spectral estimate for the negative eigenvalues of Schrödinger operators Dirichlet and
Neumann Laplacians in domains of finite measure, spectrum of non-self- adjoint
operators. Many of these inequalities are related to special types of Sobolev and
Hardy inequalities.
1. S PECTRUM OF D IRICHLET AND N EUMANN L APLACIANS
Let us consider two spectral problems for the Laplace operator in
Q = (0, a) × (0, a),
a > 0.
The first one is the Dirichlet problem:
Z
2
−∆u(x, y) = λu(x, y),
u ∈ L (Q) = {u :
|u(x, y)|2 dxdy < ∞},
Q
u
= 0,
(D).
∂Q
The second one is the Neumann problem:
−∆v(x, y) = µv(x, y),
v ∈ L2 (Q),
∂v (N).
= 0,
∂n ∂Q
It is easy to find solutions of these two problems:
(D)
unm (x, y) = sin(πxn/a) sin(πym/a),
λnm = π 2 (n2 +m2 )/a2 ,
n, m = 1, 2, 3, . . . ,
(N)
vnm (x, y) = cos(πxn/a) cos(πym/a),
µnm = π 2 (n2 +m2 )/a2
n, m = 0, 1, 2, . . . .
Exercise 1.1. Show that
Z
Z
1 2
1
N (λ, D) = #{(n, m) : λnm < λ} ≤
dx
dξ,
a λ=
(2π)2
(2π)2 Q
|ξ|2 <λ
1 2
lim λ−1 N (λ, D) =
a,
Weyl0 s asymptotics.
λ→∞
(2π)2
Exercise 1.2. Show that
Z
Z
1 2
1
dx
dξ,
N (µ, N) = #{(n, m) : µnm < µ} ≥
a µ=
(2π)2
(2π)2 Q
|ξ|2 <µ
1 2
lim µ−1 N (µ, N) =
a,
Weyl0 s asymptotics.
2
µ→∞
(2π)
1
2
SPECTRAL AND FUNCTIONAL INEQUALITIES
Let (x)+ = (|x| + x)/2 be the positive part of x.
Exercise 1.3. Show that
∞
∞
X
X
(λ − λnm )+ =
n,m=1
n,m: λnm
Exercise 1.4. Show that
∞
∞
X
X
(µ − µnm )+ =
n,m=0
n,m: µnm
1
(λ − λnm ) ≤
(2π)2
<λ
Z
Z
1
(µ − µnm ) ≥
(2π)2
<µ
Z
(λ − |ξ|2 )+ dξ.
dx
R2
Q
Z
dx
(µ − |ξ|2 )+ dξ.
R2
Q
Pólya’s Conjecture (George Pólya, 1887-1985)
For any domain Ω ∈ R2 such that the Lebesgue measure |Ω| < ∞, the counting
function N (λ, D) of the spectrum of the Dirichet Laplacian satisfies the inequality
Z
1
N (λ, D) ≤
dξ.
|Ω|
(2π)2
|ξ|2 <λ
Pólya has proved this conjecture for tiling domains.
We can prove the following result
Theorem 1.5. Let Ω ⊂ R2 , such that |Ω| < ∞ and let {λ} are eigenvalues of the
Dirichlet Laplacian in L2 (Ω). Then
Z
X
−2 2
(1 − |ξ|2 )+ dξ.
(λ − λk )+ ≤ (2π) λ |Ω|
R2
Proof. Let ϕk be the orthonormal basis in L2 (Ω) consisting of eigenfunctions of the
Dirichlet Laplacian which is denoted by A. Let ϕ̂ be the Fourier transform of ϕ.
Then by using Parceval formula we find
Z
X
X
X
−2
(2π)
(λ − |ξ|2 ) |ϕ̂k |2 dξ
(λ − λk )+ =
λ − (Aϕk , ϕk ) + =
k
k
≤ (2π)−2
k
= (2π)
XZ
−2
k
−2
Z
= (2π)
X
2
(λ−|ξ| )+
R2
R2
+
R2
k
XZ
(λ − |ξ|2 )+ |ϕ̂k |2 dξ
R2
Z
2
(λ − |ξ| )+ ei(x,ξ) ϕk (x) dx dξ
2
Ω
i(·,ξ)
k(e
2
, ϕk )k dξ = (2π)
−2
Z
R2
k
(λ−|ξ|2 )+ dξ kei(·,ξ) k2 .
| {z }
=|Ω|
Exercise 1.6. Show that
2
N (λ, D) ≤
|Ω|
(2π)2
Z
dξ.
|ξ|2 <λ
SPECTRAL AND FUNCTIONAL INEQUALITIES
Hint: use the fact that for any η > λ
X
1≤
k: λk <λ
3
1 X
(η − λk )+
η−λ k
and then minimise the right hand side w.r.t. η.
2. N EGATIVE SPECTRUM OF S CHR ÖDINGER OPERATORS
Consider the spectral problem for a Schrödinger operator H in Rd , d = 1, 2, 3, . . .
u ∈ L2 (Rd ).
Hu = −∆u(x) − V (x)u(x) = λu(x),
If V → 0 as |x| → ∞ rapidly enough, the operators H typically has the so-called
continuous spectrum that equals [0, ∞) and also might have a finite or an infinite
number of the negative eigenvalues.
Note that if V ≡ 0, then
Z
Z
|∇u(x) dx
(−∆u(x))u(x) dx =
Rd
Z
−d
= (2π)
|ξ|2 |û(ξ)|2 dξ ≥ 0.
(Hu, u)L2 (Rd ) =
Rd
Rd
Here û is the Fourier transform of u
Z
u(x)e−ixξ dx.
u(ξ) =
Rd
This means that if V ≡ 0, then H ≥ 0.
Lieb-Thirring inequalities.
Let V > 0 and let {−λk } be the negative eigenvalues of the operator H. Then
Lieb-Thirring inequalities state
Z
ZZ
X γ
γ+d/2
−d
λk ≤ Lγ,d
V
(x) dx = Rγ,d (2π)
(V (x) − |ξ|2 )γ+ dξdx.
k
Rd
R2d
Here
Lγ,d = Rγ,d (2π)
−d
Z
(1 − |ξ|2 )γ+ dξ.
Rd
The case γ = 1, d = 3 is the most important case as it is related to the energy of
quantum mechanical systems. The sharp values of the constants Lγ,d play important
role and the sharp value L1,3 is unknown.
For more information see [LT] and [LW].
4
SPECTRAL AND FUNCTIONAL INEQUALITIES
3. H ARDY INEQUALITIES
Let
Z
n
o
2
2
H (R ) = u :
(|∇u(x)| + |u(x)| ) dx < ∞ .
1
d
Rd
Theorem 3.1. Let d ≥ 3. Then
Z
Z
(d − 2)2
|u(x)|2
dx ≤
|∇u(x)|2 dx.
4
|x|2
Rd
Rd
holds true for all u ∈ H 1 (Rd ).
Remark 3.2. In particular, if d = 3 this inequality becomes
Z
Z
Z
1
|u(x)|2
2
dx ≤
|∇u(x)| dx =
(−∆u(x)) u(x) dx
4 R3 |x|2
R3
R3
This implies that in R3 we have that
−∆ −
1 1
≥0
4 |x|2
is a non-negative operator.
Proof. Consider the quadratic form
Z 2
x
I=
∇u(x) + α 2 u(x) dx ≥ 0.
|x|
Rd
Elementary computations give us
Z x 2 1
2
2
|u(x)| dx
I=
|∇u(x)| + α 2 ∇u(x)u(x) + u(x)∇u(x) + α
|x|
|x|2
Rd
Z x
2
2
2 1
2
|u(x)| dx
=
|∇u(x)| + α 2 ∇ |u(x)| + α
|x|
|x|2
Rd
Z x
2
2 1
2
2
|u(x)| + α
|u(x)| dx
=
|∇u(x)| + α ∇
|x|2
|x|2
Rd
Z d−2
2
2 1
2
2
=
|∇u(x)| + α
|u(x)| + α
|u(x)| dx.
|x|2
|x|2
Rd
Minimising w.r.t. α we choose α = −(d − 2)/2. Substituting this value into the
latter integral we find
Z d − 2)2 1
2
2
I=
|∇u(x)| −
|u(x)| dx ≥ 0.
4
|x|2
Rd
The proof is complete.
SPECTRAL AND FUNCTIONAL INEQUALITIES
5
4. S OBOLEV INEQUALITY
Theorem 4.1. Let d ≥ 3. Then there is a constant Cd > 0, such that
1/2
Z
Z
d−2
2d
2d
2
≥ Cd
|u(x)| d−2 dx
|∇u(x)|
.
Rd
Rd
Remark 4.2. In the case d = 3 we have
Z
Z
2
2
|∇u(x)| ≥ Cd
R3
6
1/3
|u(x)| dx
.
Rd
Consider in R3 the quadratic form of the Schrödinger operator H = −∆ − V ,
V ≥ 0. Using the above inequality and also the Hölder inequality we find
Z
Z
2
(Hu, u)L2 (R3 ) =
|∇u(x)| dx −
V (x)|u(x)|2 dx
R3
≥
Cd2
Z
R3
6
1/3
|u(x)| dx
Z
−
V
3/2
2/3 Z
|u(x)| dx
(x) dx
1/3
.
R3
R3
Rd
6
If now
Cd2
Z
≥
V
3/2
2/3
(x) dx
,
R3
then the operator H ≥ 0.
The proof of Theorem 4.1 is reduced to the following statement:
Theorem 4.3. Hardy’s inequality implies Sobolev’s inequality.
(See for the detailed proof [S], page 8 and also [LS].)
Proof. We prove the Sobolev inequality for the case when u ≥ 0, depends only on
|x| and is non-increasing. Then for any y ∈ Rd , 0 < a < 1 and q > 1 we have
Z
Z
q
u (x) dx ≥
uq (x) dx ≥ C|y|d uq (y)
Rd
where
(1 − ad )
C=
ωd ,
d
a|y|≤|x|≤|y|
d−1
andwhere ωg = |S
Z
|=
ds.
|x|=1
Therefore
Z
2/d
q
u (x) dx
u2 (y) |y|−2 ≥ C 2/d |y|2 (u(y))2q/d u2 (y) |y|−2 = C 2/d (u(y))2q/d+2 .
Rd
Let us now choose q = 2d/(d − 2). Then 2q/d + 2 = 4d/d(d − 2) + 2 = 2d/(d − 2)
and
Z
2/d Z
Z
u2 (y)
2/d
2d/(d−2)
2d/(d−2)
C
(u(y))
dy ≤
(u(x))
dx
dy.
2
Rd
Rd
Rd |y|
6
SPECTRAL AND FUNCTIONAL INEQUALITIES
By using now the Hardy inequality we find
Z
1−2/d
2d/(d−2)
2/d
(u(y))
dy
≤
C
Rd
4
(d − 2)2
Z
|∇u(x)|2 dx.
Rd
The proof is complete.
The proof of Sobolev inequality for arbitrary functions can be reduced to the case
considered in the latter proof by using rearrangements, see [LL].
5. L IEB -T HIRRING INEQUALITIES FOR 1D -S CHR ÖDINGER OPERATORS
Let {ψj }nj=1 be in orthonormal system of function in L2 (R) and let
ρ(x) =
n
X
ψj2 (x).
j=1
We first prove a so-called generalised 1D Sobolev inequality:
Theorem 5.1.
Z
3
ρ (x) dx =
Z X
n
R
2
|ψj (x)|
3
dx ≤
j=1
n Z
X
j=1
|ψ 0 j (x)|2 dx.
R
Proof. We first derive a so-called Agmon inequality
1/2
1/2
kψkL∞ ≤ kψkL2 kψ 0 kL2 .
Indeed
1 |ψ(x)| = 2
2
Z
x
Z
2 0
∞
|ψ | dt −
−∞
x
Z
|ψ | dt ≤ |ψ||ψ 0 | dt ≤ kψkL2 kψ 0 kL2 .
2 0
Let now ξ = (ξ1 , ξ2 , . . . , ξn ) ∈ Cn . Then by Agmon inequality
n
n
n
X
X
1/4 X
1/4
¯
ξj ψj (x) ≤
ξj ξk (ψj , ψk )
ξj ξ¯k (ψj0 , ψk0 )
j=1
j,k=1
≤
n
X
j,k=1
|ξj |2
1/4
j=1
n
X
1/4
.
ξj ξ¯k (ψj0 , ψk0 )
j,k=1
If we set ξj = ψj (x) then the latter inequality becomes
n
n
X
1/4
X
2
1/4
0
0
ρ(x) =
|ψj (x)| ≤ ρ (x)
ψj (x) ψk (x)(ψj , ψk )
.
j=1
j,k=1
Thus
3
ρ (x) ≤
n
X
j,k=1
ψj (x)ψk (x)(ψj0 , ψk0 ).
SPECTRAL AND FUNCTIONAL INEQUALITIES
7
Integrating both sides we arrive at
Z X
n
n Z
3
X
2
|ψj (x)|
dx ≤
|ψj0 |2 dx.
j=1
j=1
Spectrum of 1D-Schrödinger operators
Let {ψj }∞
j=1 be the o.n.s.
Schrödinger operator
−
corresponding to the negative eigenvalues of the
d2
ψj (x) − V (x)ψj (x) = −λj ψj (x),
dx2
V ≥ 0,
Then
Z X
n
2
|ψj (x)|
3
dx −
Z
V
3/2
dx
n
2/3 Z X
j=1
2
|ψj (x)|
3
dx
1/3
j=1
≤
XZ X
|ψj0 |2 − V (x)|ψj |2 dx = −
λj .
j
j
Denote
X=
n
Z X
2
|ψj (x)|
3
dx
1/3
,
j=1
then the latter inequality can be written as
Z
2/3
X
3
X −
V 3/2 dx
X≤−
λj .
j
1/3
V 3/2 dx
. This implies
Maximizing the left hand side we find X = √13
Z
Z
Z
X
1
2
1
3/2
3/2
√
V dx − √
V dx = − √
V 3/2 dx ≤ −
λj
3 3
3
3 3
j
R
P
and we finally obtain j λj ≤ 3√2 3 V 3/2 dx.
R
This is the best known constant in Lieb-Thirring’s inequality.
R EFERENCES
[LW]
[LT]
[LL]
[LS]
[S]
A. Laptev and T. Weidl, Sharp Lieb-Thirring inequalities in high dimensions, Acta Mathematica, 184 (2000), 87–111.
E.H. Lieb and W. Thirring, Inequalities for the moments of the eigenvalues of the
Schrödinger Hamiltonian and their relation to Sobolev inequalities, Studies in Math.
Phys., Essays in Honor of Valentine Bargmann., Princeton, (1976), 269–303.
E.H. Lieb and M. Loss, Analysis, AMS, Graduate Studies in Mathematics, 14, 2000.
E.H. Lieb and R. Seiringer, Stability of matter in Quantum Mechanics, Cambridge University Press 2010.
R. Seiringer, Inequalities for Schrödinger operators and applications to the stability of
matter problem Lectures given in Tucson, Arizona, March 16-20, 2009.