Lower bounds for pseudodifferential operators with a radial symbol

J. Math. Pures Appl. 103 (2015) 1157–1162

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Lower bounds for pseudodifferential operators with a radial symbol Laurent Amour ∗ , Lisette Jager, Jean Nourrigat Laboratoire de Mathématiques, FR CNRS 3399, EA 4535, Université de Reims Champagne-Ardenne, Moulin de la Housse, B.P. 1039, F-51687 Reims, France

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Article history: Received 19 April 2014 Available online 14 October 2014

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. © 2014 Elsevier Masson SAS. All rights reserved.

MSC: primary 35S05 secondary 28C20, 35R15, 81S30

r é s u m é Keywords: Pseudodifferential operators Lower bounds Gårding’s inequality

Dans cet article, on établit une borne inférieure pour les opérateurs pseudodifférentiels avec un symbole radial. La démonstration utilise des techniques classiques du calcul de Weyl ainsi que des propriétés connues vérifiées par les polynômes de Laguerre. © 2014 Elsevier Masson SAS. All rights reserved.

1. Introduction If a function F defined on R2d is smooth and has bounded derivatives, the Weyl calculus associates with it a pseudodifferential operator OpWeyl (F ) which is bounded on L2 (Rd ) and satisfies, for all f and g in S(Rd ), h   Weyl  −d Oph (F )f, g = (2πh) F (Z)Hh (f, g, Z)dZ, (1.1) R2d

where Hh (f, g, ·) is the Wigner function  Hh (f, g, Z) =

e

− hi t·ζ

   t t dt g z− f z+ 2 2 

Z = (z, ζ) ∈ R2d .

Rd

* Corresponding author. E-mail addresses: [email protected] (L. Amour), [email protected] (L. Jager), [email protected] (J. Nourrigat). http://dx.doi.org/10.1016/j.matpur.2014.10.008 0021-7824/© 2014 Elsevier Masson SAS. All rights reserved.

(1.2)

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For this form of the definition, see [15], [10] or [3, Chapter II, Proposition 14]. The different variants of Gårding’s inequality prove that, if F ≥ 0, the operator OpWeyl (F ) is roughly h ≥ 0. More precisely, according to the classical Gårding’s inequality (see [7] or [10]), the non-negativity of F implies the existence of a positive constant C, independent of h, such that, for all sufficiently small h and for all f in S(Rd ):  (F )f, f ≥ −Chf 2L2 (Rd ) . OpWeyl h



(1.3)

See [11] for other similar results. This inequality holds for systems of operators, whereas the more precise Fefferman–Phong inequality [4] is valid only for scalar operators. Fefferman–Phong’s inequality states that, under the same hypotheses as Gårding’s inequality, one has, for all h in (0, 1) and all f in S(Rd ): 

 OpWeyl (F )f, f ≥ −Ch2 f 2L2 (Rd ) . h

(1.4)

See [13] for these semiclassical versions. Sometimes the non-negativity of F implies the exact non-negativity of the operator, for example in the simple case when F depends on x or on ξ only. It is possible, too, to apply Melin’s inequality [14]. To take only one example, let F ≥ 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 of OpWeyl (F ) for a sufficiently small h. See [2] or [9] for cases when the h difference between F (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 R such that   F (x, ξ) = Φ |x|2 + |ξ|2

(x, ξ) ∈ R2d .

(1.5)

Moreover, we suppose that Φ is nondecreasing on [0, ∞) and such that F is smooth, with bounded derivatives. (F ). The In this case, we aim at giving an explicit lower bound on the spectrum of the operator OpWeyl h main result of this paper is the following theorem. Theorem 1.1. Let F be a smooth function defined on R2d , bounded as well as all its derivatives. Assume that F is of the form (1.5), where Φ is a non-decreasing function defined on [0, ∞). Then for all f in S(Rd ), 

 OpWeyl (F )f, f h

1 ≥ h

∞

Φ(t)e− h dt t

f 2L2 (Rd ) .

(1.6)

0

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, assuming that Φ is infinitely differentiable at the origin (on the right side), let m ≥ 1 be the smallest integer for which Φ(m) (0) = 0. Then one can see that 1 h

∞

  t Φ(t)e− h dt = Φ(0) + Φ(m) (0)hm + O hm+1 .

0

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 ) and T f (u) = ei(ξ0 /h)(u−x0 ) f (u − x0 ), then 

   OpWeyl (τ F )f, g = OpWeyl (F )T f, T g . h h

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Using metaplectic transformations, a lower bound could also be deduced from Theorem 1.1 when the symbol is given by   F (x, ξ) = Φ Q(x − x0 , ξ − ξ0 ) , where Q is a quadratic form on R2d , symplectically equivalent to |x|2 + |ξ|2 . We are greatly indebted to N. Lerner for reference [1]. 2. Proof of Theorem 1.1 We denote by (Hn )(n≥0) the sequence of the Hermite functions. It is a Hermitian basis of L2 (R), satisfying  2  D + x2 Hn = (2n + 1)Hn .

(2.1)

For each multi-index α = (α1 , ..., αd ), we set: uα (x) =

d

Hαj (xj ).

(2.2)

j=1

These functions form a Hermitian basis of L2 (Rd ). We shall need the Laguerre polynomials as well, which are defined by ex dn  n −x  x e . n! dxn

Ln (x) =

(2.3)

One has L0 (x) = 1

L1 (x) = 1 − x

L2 (x) =

x2 − 2x + 1. 2

(2.4)

Theorem 1.1 is a consequence of the following proposition, in which the parameter h is equal to 1 and the Weyl operator OpWeyl (F ) is denoted by OpWeyl (F ). 1 Proposition 2.1. Under the hypotheses of Theorem 1.1 one has, for all multi-indices α and β such that α = β: 

 OpWeyl (F )uα , uβ = 0.

(2.5)

For each multi-index α: 

Op

Weyl





−d

(F )uα , uα = 2

1 Φ(0)Vα (0) + 2

∞

 

Φ (t/2)Vα (t)dt ,

(2.6)

0

with Vα (X) = 4e− 2

X

d−1

k Cd−1 T|α|+k (X),

(2.7)

k=0

where we set, for all integer n,

n−1  (−1)n k Ln (X). Tn (X) = (−1) Lk (X) + 2 k=0

(2.8)

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Proof of (2.5). Let α and β be two different multi-indices and let j ≤ d be such that αj = βj . Set Pj = Dj2 + x2j . According to (2.1) we have:       2(αj − βj ) OpWeyl (F )uα , uβ = OpWeyl (F )Pj uα , uβ − OpWeyl (F )uα , Pj uβ . ∂F ∂F The fact that F is radial implies that xj ∂ξ − ξj ∂x = 0 which, in turn, implies that OpWeyl (F ) and Pj j j commute, thanks to properties of the Weyl calculus. Consequently, the right term of the above inequality is equal to 0, which proves (2.5). 2

Proof of (2.6). For each multi-index α, the Wigner function H(uα , uα ), (where the parameter h, equal to 1, is omitted), satisfies: H(uα , uα )(x, ξ) = 2d (−1)|α| e−(|x|

2

+|ξ|2 )

d

   Lαj 2 x2j + ξj2 .

(2.9)

j=1

See, for example, [6] or [8]. Hence, if F is as in Theorem 1.1, 

 OpWeyl (F )uα , uα = (2π)−d 2d (−1)|α|



d      2 2 Φ |x|2 + |ξ|2 e−(|x| +|ξ| ) Lαj 2 x2j + ξj2 dxdξ. j=1

R2d

The change of variables tj = 2(x2j + ξj2 ) allows to write: 

 OpWeyl (F )uα , uα = (2π)−d 2d (−1)|α| (π/2)d



d   1 Φ (t1 + ... + td )/2 e− 2 (t1 +...+td ) Lαj (tj )dt1 ...dtd . j=1

[0,∞)d

This equality can be written as 

Op

Weyl



−d d

(F )uα , uα = (2π)

∞ d

2 (π/2)

Φ(X/2)Uα (X)dX, 0

with: 

Uα (X) = (−1)|α| e− 2

X

Lαd (X − t1 − ... − td−1 )

d−1

Lαj (tj )dt1 ...dtd−1 ,

j=1

Ωd (X)

where

 Ωd (X) = (t1 , ..., td−1 ), tj > 0, t1 + ... + td−1 < X .

2

The equality (2.6) will be a consequence of an integration by parts, which is justified since t → is bounded on (0, +∞), together with the following lemma:

√

tΦ (t)

Lemma 2.2. We have: Uα (X) = −Vα (X) where Vα is defined by (2.7) and (2.8).

(2.10)

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Proof of Lemma 2.2. One knows (cf. [12], Section 5.5.2) that X Lα1 (t)Lα2 (X − t)dt = Lα1 +α2 (X) − Lα1 +α2 +1 (X).

(2.11)

0

It follows, by induction on d, that  Lαd (X − t1 − ... − td−1 )

d−1

Lαj (tj )dt1 ...dtd−1 =

j=1

Ωd (X)

d−1

k Cd−1 (−1)k L|α|+k (X).

k=0

Hence Uα (X) = e− 2

X

d−1

k Cd−1 (−1)|α|+k L|α|+k (X).

k=0

Using the recurrence relation Lk+1 (t) = Lk (t) − Lk (t), we prove (for example by induction) that for all integer n: t (−1)n+1 d −t e 2 Tn (t) = Ln (t)e− 2 . dt 4

The equality (2.10) of the lemma follows from (2.7) and from the above identities. 2 End of the proof of Theorem 1.1. We shall begin by proving (1.6) for h = 1. Set Sn (X) =

n

(−1)k Lk (X).

(2.12)

k=0

Using the recurrence relation Lk+1 (t) = Lk (t) − Lk (t), one verifies, by induction, that for all n: Tn (X) =

1 Sn−1 (X). 2

Since Ln (0) = 1 for all n, we see that Tn (0) = 1/2 and that 1 1 Tn (X) = + 2 2

X Sn−1 (t) dt. 0

According to [1, Theorem 12] (see [5] as well), Sn (X) ≥ 0 for all n ≥ 0 and for all X ≥ 0. Therefore Tn (X) ≥ 1/2 for all n and X, and, using (2.7): Vα (X) ≥ 2d e− 2 . X

(2.13)

Since Tn (0) = 1/2, Vα (0) = 2d . Hence, if Φ ≥ 0, one gets: 1 Φ(0)Vα (0) + 2

∞



∞

Φ (t/2)Vα (t)dt ≥ 2 0

d

Φ(t)e−t dt.

(2.14)

0

The inequality (1.6), for h = 1, follows from (2.5), (2.6) and (2.14). For an arbitrary h > 0, it suffices to apply the above result to the function Fh (x, ξ) = F (h1/2 x, h1/2 ξ), that is to say, to the function Φh (t) = Φ(th). 2

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