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Lower bounds for pseudodifferential operators with a radial symbol
Laurent Amour, Lisette Jager, Jean Nourrigat
To cite this version:
Laurent Amour, Lisette Jager, Jean Nourrigat. Lower bounds for pseudodifferential operators with a radial symbol. Journal de Mathématiques Pures et Appliquées, Elsevier, 2015. �hal-01881688�
Lower bounds for pseudodifferential operators with a radial symbol.
Laurent Amour, Lisette Jager and Jean Nourrigat Universit´e de Reims
ABSTRACT. In this paper we establish explicit lower bounds for pseudodifferential operators with a radial symbol. The proofs use classical Weyl calculus techniques and some useful, if not celebrated, properties of the Laguerre polynomials.
2010 Mathematical Subject Classification Primary 35S05; Secondary 28C20,35R15,81S30.
Key words and phrases : Pseudodifferential operators, lower bounds, G˚arding’s inequality.
1. Introduction.
If a functionF defined on IR2d is smooth and has bounded derivatives, the Weyl calculus associates with it a pseudodifferential operatorOpW eylh (F) which is bounded onL2(IRd) and satisfies, for allf andg inS(IRd), (1.1) < OpW eylh (F)f, g >= (2πh)−d
Z
IR2d
F(Z)Hh(f, g, Z)dZ, whereHh(f, g,·) is the Wigner function
(1.2) Hh(f, g, Z) = Z
IRd
e−hit·ζf
z+ t 2
g
z−t
2
dt Z= (z, ζ)∈IR2d. For this form of the definition, see [U], [L] or [C-R], Chapter II, Proposition 14.
The different variants of G˚arding’s inequality prove that, if F ≥0, the operatorOpW eylh (F) is roughly≥0.
More precisely, according to the classical G˚arding’s inequality (see [HO] or [L]), the non negativity of F implies the existence of a positive constantC, independent of h, such that, for all sufficiently small hand for allf inS(IRd):
(1.3) < OpW eylh (F)f, f >≥ −Chkfk2L2(IRd).
See [L-N] for other similar results. This inequality holds for systems of operators, whereas the more precise Fefferman-Phong inequality [F-P] is valid only for scalar operators. Fefferman-Phong’s inequality states that, under the same hypotheses as G˚arding’s inequality, one has, for allhin (0,1) and allf in S(IRd):
(1.4) < OpW eylh (F)f, f >≥ −Ch2kfk2L2(IRd).
See [MAR] for these semiclassical versions. Sometimes the non negativity ofF implies the exact non neg- ativity of the operator, for example in the simple case when F depends onxor on ξ only. It is possible, too, to apply Melin’s inequality. To take only one example, letF ≥0 attain its minimum only once, for a nondegenerate critical point. In this case (and in other analogous situations), Melin’s inequality ensures the exact non negativity ofOpW eylh (F) for a sufficiently smallh. See [B-N] or [L-L] for cases when the difference betweenF(x, ξ) and its minimum is equivalent to a power, greater than 2, of the distance between (x, ξ) and the unique point where the minimum is attained.
In this article we are interested in the case when F is radial. We assume that there exists a function Φ defined on IR such that
(1.5) F(x, ξ) = Φ(|x|2+|ξ|2) (x, ξ)∈IR2d.
Moreover, we suppose that Φ is nondecreasing on [0,∞) and such thatF is smooth, with bounded derivatives.
In this case, we aim at giving an explicit lower bound on the spectrum of the operator OpW eylh (F). The main result of this paper is the following theorem.
Theorem 1.1Let F be a smooth function defined on IR2d, bounded as well as all its derivatives. Assume that F is of the form (1.5), whereΦis a non decreasing function defined on [0,∞).
Then for allf in S(IRd),
(1.6) < OpW eylh (F)f, f >≥ 1 h
Z ∞
0
Φ(t)e−htdt kfk2L2(IRd).
Remarks
1 - We do not need to assume that Φ≥0 to ensure the non negativity of the operator. The non negativity of the integral suffices.
2 - In the case when Φ is not flat at the origin, let m≥ 1 be the smallest integer for which Φ(m)(0)6= 0.
Then one can see that 1 h
Z ∞
0
Φ(t)e−htdt= Φ(0) + Φ(m)(0)hm+O(hm+1).
3 - The result can be applied to symbols F depending on the distance from another point (x0, ξ0) for, if τ F(x, ξ) =F(x+x0, ξ+ξ0) andT f(u) =ei(ξ0/h)(u−x0)f(u−x0), then
< OpW eylh (τ F)f, g > =< OpW eylh (F)T f, T g > .
We are greatly indebted to N. Lerner for the reference [A-G].
2. Proof of Theorem 1.1.
We denote by (Hn)(n≥0)the sequence of the Hermite functions. It is a Hermitian basis ofL2(IR), satisfying
(2.1) (D2+x2)Hn= (2n+ 1)Hn.
For each multi-indexα= (α1, ...αd), we set :
(2.2) uα(x) =
d
Y
j=1
Hαj(xj).
These functions form a Hermitian basis ofL2(IRd).
We shall need the Laguerre polynomials as well, which are defined by
(2.3) Ln(x) =ex
n!
dn
dxn xne−x .
One has :
(2.4) L0(x) = 1 L1(x) = 1−x L2(x) = x2
2 −2x+ 1.
Theorem 1.1 is a consequence of the following proposition, in which the parameter his equal to 1 and the Weyl operatorOpW eyl1 (F) is denoted byOpW eyl(F).
Proposition 2.1 Under the hypotheses of Theorem 1.1 one has, for all multi-indices α and β such that α6=β:
(2.5) < OpW eyl(F)uα, uβ>= 0.
For each multi-indexα:
(2.6) < OpW eyl(F)uα, uα>= 2−d
Φ(0)Vα(0) +1 2
Z ∞
0
Φ′(t/2)Vα(t)dt
,
with
(2.7) Vα(X) = 4e−X2
d−1
X
k=0
Cd−1k T|α|+k(X),
where we set, for all integern,
(2.8) Tn(X) =
"n−1 X
k=0
(−1)kLk(X)
#
+(−1)n 2 Ln(X).
Proof of (2.5). Let α and β be two different multi-indices and let j ≤ d be such that αj 6= βj. Set Pj=Dj2+x2j. According to (2.1) we have :
2(αj−βj)< OpW eyl(F)uα, uβ >=< OpW eyl(F)Pjuα, uβ>−< OpW eyl(F)uα, Pjuβ> . The fact that F is radial implies that xj∂F
∂ξj −ξj∂F
∂xj = 0 which, in turn, implies that OpW eyl(F) and Pj
commute, thanks to properties of the Weyl calculus. Consequently, the right term of the above inequality is equal to 0, which proves (2.5).
Proof of (2.6). For each multi-indexα, the Wigner functionH(uα, uα) (where the parameterh, equal to 1, is omitted), satisfies:
(2.9) H(uα, uα)(x, ξ) = 2d(−1)|α|e−(|x|2+|ξ|2)
d
Y
j=1
Lαj(2(x2j+ξj2)).
See, for example, [FO] or [J-L-V]. Hence, ifF is as in Theorem 1.1,
< OpW eyl(F)uα, uα>= (2π)−d2d(−1)|α|
Z
IR2d
Φ(|x|2+|ξ|2)e−(|x|2+|ξ|2)
d
Y
j=1
Lαj(2(x2j+ξj2))dxdξ.
The change of variablestj= 2(x2j+ξj2) allows to write :
< OpW eyl(F)uα, uα>= (2π)−d2d(−1)|α|(π/2)d Z
[0,∞)d
Φ((t1+...+td)/2)e−12(t1+...+td)
d
Y
j=1
Lαj(tj)dt1...dtd.
This equality can be written as
< OpW eyl(F)uα, uα>= (2π)−d2d(π/2)d Z ∞
0
Φ(X/2)Uα(X)dX,
with :
Uα(X) = (−1)|α|e−X2 Z
Ωd(X)
Lαd(X−t1−...−td−1)
d−1
Y
j=1
Lαj(tj)dt1...dtd−1, where
Ωd(X) ={(t1, ..., td−1), tj>0, t1+...+td−1< X}.
The equality (2.6) will be a consequence of an integration by parts using the following lemma.
Lemma 2.2We have:
(2.10) Uα(X) =−Vα′(X)
whereVα is defined by (2.7) and (2.8).
Proof of Lemma 2.2. One knows (cf [M-O-S], section 5.5.2) that (2.11)
Z X
0
Lα1(t)Lα2(X−t)dt=Lα1+α2(X)−Lα1+α2+1(X).
It follows, by induction ond, that Z
Ωd(X)
Lαd(X−t1−...−td−1)
d−1
Y
j=1
Lαj(tj)dt1...dtd−1=
d−1
X
k=0
Cd−1k (−1)kL|α|+k(X).
Hence
Uα(X) =e−X2
d−1
X
k=0
Cd−1k (−1)|α|+kL|α|+k(X).
Using the recurrence relation L′k+1(t) = L′k(t)−Lk(t), we prove (for example by induction) that for all integern:
d
dte−2tTn(t) =(−1)n+1
4 Ln(t)e−t2.
The equality (2.10) of the Lemma follows from (2.7) and from the above identities.
End of the proof of Theorem 1.1. We shall begin by proving (1.6) forh= 1. Set
(2.12) Sn(X) =
n
X
k=0
(−1)kLk(X).
Using the recurrence relationL′k+1(t) =L′k(t)−Lk(t), one verifies, by induction, that for alln:
Tn′(X) =1
2Sn−1(X).
SinceLn(0) = 1 for alln, we see thatTn(0) = 1/2 and that Tn(X) =1
2 +1 2
Z X
0
Sn−1(t)dt.
According to [A-G], Theorem 12 (see [F] as well ), Sn(X)≥0 for alln≥0 and for all X ≥0. Therefore Tn(X)≥1/2 for allnandX, and, using (2.7):
(2.13) Vα(X)≥2de−X2.
SinceTn(0) = 1/2,Vα(0) = 2d. Hence, if Φ′≥0, one gets :
(2.14) Φ(0)Vα(0) +1
2 Z ∞
0
Φ′(t/2)Vα(t)dt≥2d Z ∞
0
Φ(t)e−tdt.
The inequality (1.6), forh= 1, follows from (2.5), (2.6) and (2.14). For an arbitraryh >0, it suffices to apply the above result to the functionFh(x, ξ) =F(h1/2x, h1/2ξ), that is to say, to the function Φh(t) = Φ(th).
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Laboratoire de Math´ematiques, FR CNRS 3399, EA 4535, Universit´e de Reims Champagne-Ardenne, Moulin de la Housse, B. P. 1039, F-51687 Reims, France,
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