An inequality in noncommutative $L_p$-spaces

0.11 MB·

by  Éric Ricard

AN INEQUALITY IN NONCOMMUTATIVE L -SPACES
p
E´RICRICARD
Abstract. Weprovethatforany(trace-preserving)conditionalexpectationE onanoncommu-
tativeLp withp>2,Id−E isacontractiononthepositiveconeL+p.
6
1
0 1. Introduction
2
It is plain that for positive real numbers a,b>0 and p>2, one has
n
a (a−b)(ap−1−bp−1)>|a−b|p.
J
7 Integrating the above inequality on some measure space (Ω,µ) implies that for f,g ∈Lp(Ω,µ)+,
2 f(x)−g(x) fp−1(x)−gp−1(x) dµ(x)> f −g p.
Z p
Ω(cid:0) (cid:1)(cid:0) (cid:1) (cid:13) (cid:13)
] (cid:13) (cid:13)
A In [3], Mustapha Mokhtar-Kharroubinotices that this inequality may be used to get contractivity
O results on the positive cone of Lp(Ω,µ). This note originates from the question whether its non-
commutative analogue remains true. We provide a proof in the next section. We hope that the
.
h techniques involved there may be useful for further studies. We end up by making explicit some
t
a results from [3] for noncommutative Lp-spaces.
m We refer the reader to [4] for the definitions of L -spaces associatedto semifinite vonNeumann
p
[ algebras or more general ones. We also freely use basic results from [1].

1 2. Results
v
6 Let(M,τ)be asemifinite vonNeumannalgebra. We denotebyf :M+ →M+,thepth-power
p
5 map and by M++ the set of positive invertible elements. We will often refer to positivity of the
4 trace for the fact that if a,b∈M+∩L (M), then τ(ab)>0.
7 1
0 Theorem 2.1. Let p>2 and a, b∈Lp(M)+, then
1. τ |a−b|p 6τ (a−b)(ap−1−bp−1) .
0 (cid:0) (cid:1) (cid:0) (cid:1)
Proof. First, as for any sequence of (finite) projections p going to 1 strongly in M and any
6 i
1 x∈Lq(M) (16q <∞) kpixpi−xkq →0, we may assume that M is finite. Next by replacing a
: and b by a+ε1 and a+ε1 for some ε > 0, we may also assume that [a,b] ⊂ M++ to avoid any
v
unnecessary technical complication.
i
X We write a=b+δ. To prove the result, we distinguish according to the values of p.
r
a Case 1: p∈[2,3]
Case 1.a: a>b, i.e. δ >0.
As p−1=1+θ with θ ∈[0,1], we use the well known integral formula
tθs2 dt s2 t2
(1) s1+θ =c , =s−t+ .
θZR s+t t s+t s+t
+
Hence
dt
τ δ(a1+θ−b1+θ) =c tθτ δ δ+t2(b+δ+t)−1−t2(b+t)−1 .
(cid:0) (cid:1) θZR+ (cid:16) (cid:0) (cid:1)(cid:17) t
Recall the identity (b+δ+t)−1 −(b+t)−1 = −(b+δ+t)−1δ(b+t)−1. Using positivity of the
trace with δ(b+δ+t)−1δ 6δ(δ+t)−1δ and (b+t)−1 6t−1:
τ δ δ+t2(b+δ+t)−1−t2(b+t)−1 >τ δ2−tδ(δ+t)−1δ =τ δ3(δ+t)−1 .
(cid:16) (cid:0) (cid:1)(cid:17) (cid:0) (cid:1) (cid:0) (cid:1)
2010 Mathematics Subject Classification: 46L51;47A30.
Key words: Noncommutative Lp-spaces.
1
2 E´.RICARD
Integrating, we get the desired inequality τ δ(a1+θ −b1+θ) >τ δ2+θ .
Case 1.b: δ arbitrary with decomposition(cid:0) δ+−δ− into p(cid:1)ositiv(cid:0)e and(cid:1)negative parts. We reduce
it to the previous case by introducing α=a+δ− =b+δ+, so that α>a,b. We have
τ (a−b)(ap−1−bp−1) = τ (a−α)(ap−1−αp−1) +τ (a−α)(αp−1−bp−1)
(cid:0) (cid:1) (cid:0)+τ (α−b)(αp−1−bp−(cid:1)1) +(cid:0)τ (α−b)(ap−1−αp−(cid:1)1) .
The first and the third terms are bigge(cid:0)r than τ δp and τ δ(cid:1)p b(cid:0)y Case 1.a. Hence it(cid:1)suffices
− +
to check that the two remaining terms are posit(cid:0)ive.(cid:1)We use(cid:0)ag(cid:1)ain the integral formula (1) and
δ+δ− =0 and positivity of the trace
dt
τ(cid:0)−δ−(αp−1−bp−1)(cid:1) = −cθZR+tθτ(cid:16)δ−(cid:0)δ++t2(b+δ++t)−1−t2(b+t)−1(cid:1)(cid:17) t
dt
= cθZR+tθt2τ(cid:16)δ−(cid:0)(b+t)−1−(b+δ++t)−1(cid:1)(cid:17) t >0.
The last term is handled similarly.
Case 2: p > 3. First, for any n ∈ N, n > 1, one easily checks by induction that we have the
following identity
n
τ (a−b)(ap−1−bp−1) =τ δ (b+δ)p−1−n−bp−1−n bn + τ δ(b+δ)p−1−kδbk−1 .
(cid:0) (cid:1) (cid:0) (cid:0) (cid:1) (cid:1) Xk=1 (cid:0) (cid:1)
Let n>1 be so that p−1−n=1+θ with θ ∈[0,1[. By positivity of the trace, we get
τ (a−b)(ap−1−bp−1) >τ δ (b+δ)1+θ −b1+θ bn +τ δ2(b+δ)p−2).
(cid:0) (cid:1) (cid:0) (cid:0) (cid:1) (cid:1) (cid:0)
By the same computations as above thanks to (1)
dt
τ δ (b+δ)1+θ −b1+θ bn = c tθτ δ δ+t2(b+δ+t)−1−t2(b+t)−1 bn
(cid:0) (cid:0) (cid:1) (cid:1) θZR+ (cid:16) (cid:0) (cid:1) (cid:17) t
dt
= c tθτ δ δ−t2(b+δ+t)−1δ(b+t)−1 bn
θZR+ (cid:16) (cid:0) (cid:1) (cid:17) t
dt
> c tθτ δ δ−tδ(b+t)−1 bn
θZR+ (cid:16) (cid:0) (cid:1) (cid:17) t
dt
= c tθτ δ2b(b+t)−1bn
θZR (cid:16) (cid:17) t
+
= τ δ2bn+θ =τ δ2bp−2 ,
where we used again positivity of the tra(cid:0)ce with(cid:1)(b+(cid:0)δ+t)−1(cid:1)6t−1 and 06(b+t)−1bn.
To conclude let E to be the conditional expectation onto the subalgebra N = {δ}′′. As N is
commutative, the Jensen inequality is valid; for any α > 1 and x ∈ M+: E(xα) > (Ex)α. With
α=p−2>1,
τ δ2bp−2 =τ δ2E(bp−2) >τ δ2(Eb)p−2 , τ δ2(b+δ)p−2 >τ δ2(E(b+δ))p−2 .
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
But with the usual decomposition δ = δ+ −δ−, as a, b > 0, Eb > δ− and E(b+δ) > δ+. By
commutativity of N, we can conclude
τ (a−b)(ap−1−bp−1) >τ δ2(δp−2+δp−2) =τ |δ|p .
− +
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
(cid:3)
We provide an alternative proof when p∈[3,4].
DenotebyR andL therightandleftmultiplicationoperatorsbyx∈MdefinedonallL (M)
x x p
(1 6 p 6 ∞). When p = 2, for any x ∈ Msa, the C∗-algebra generated in B(L (M)) by L and
2 x
R is commutative and isomorphic to C(σ(x)×σ(x)) where σ(x) is the spectrum of x.
x
Lemma2.2. Forp>1,themapf isFr´echetdifferentiableonM++. ForMfinite,thederivative
p
is given by the formula in L (M):
2
1 p−1
∀x∈M++,∀h∈Msa, D f (h)=p tL +(1−t)R (h)dt.
x p Z0 (cid:16) x x(cid:17)
AN INEQUALITY IN NONCOMMUTATIVE Lp-SPACES 3
Proof. Assume x>3δ for some δ >0. Taking h∈Msa with khk<δ, we may compute f (x+h)
p
using the holomorphic functional calculus by choosing a curve γ with index 1 surrounding the
spectrum of x with γ ⊂{z |Rez >0} and dist(γ,σ(x))>2δ:
1 zp
(x+h)p = dz
2iπ Z z−(x+h)
γ
Hence
(x+h)p−xp = 1 zp z−(x+h) −1h z−x −1dz
2iπ Z
γ (cid:0) (cid:1) (cid:0) (cid:1)
It follows directly that f is Fr´echet differentiable with derivative
p
Dxfp(h)= 21iπ Z zp z−x −1h z−x −1dz = 21iπ Z zpL(z−x)−1R(z−x)−1(h)dz.
γ (cid:0) (cid:1) (cid:0) (cid:1) γ
It then suffices to check that the two formulas coincide when M is finite; as M ⊂ L (M), we do
2
it for h ∈L (M). But in B(L (M)), this boils down to an equality in C(σ(x)×σ(x)) so that we
2 2
need only to justify that
∀a,b∈R+∗, 1 zp dz =p 1 ta+(1−t)b p−1 dt.
2iπ Z (z−a)(z−b) Z
γ 0 (cid:0) (cid:1)
The above computations yield that the left-hand side is ap−bp if a 6= b and pap−1 if a = b which
a−b
clearly coincide with the right-hand side. (cid:3)
Assuming M finite anda=b+δ,b∈M++ asabove,the alternativeproofwhenp∈[3,4]relies
on Lemma 2.2:
1 1 p−2
τ (a−b)(ap−1−bp−1) = p τ δ tL +(1−t)R (δ) dtdu
(cid:0) (cid:1) Z0 Z0 (cid:16) (cid:16) b+uδ b+uδ(cid:17) (cid:17)
1 1 p−2
= pZ0 Z0 hδ,(cid:16)tLb+uδ+(1−t)Rb+uδ(cid:17) (δ)iL2(M) dtdu
As p−2 ∈ [1,2], fp−2 is operator convex, so that for any m ∈ B(L2(M))+ and any projection
E ∈B(L (M)), we have Emp−2E > EmE)p−2. We choose E to be the L -conditional expectation
2 2
onto the subalgebra generated by δ.(cid:0)
p−2 p−2
hδ,(cid:16)tLb+uδ+(1−t)Rb+uδ(cid:17) (δ)iL2(M) = hδ,E(cid:16)tLb+uδ+(1−t)Rb+uδ(cid:17) E(δ)iL2(M)
p−2
> hδ,(cid:16)tELb+uδE +(1−t)ERb+uδE(cid:17) (δ)iL2(M)
p−2
= hδ,(cid:16)tLE(b)+uδE +(1−t)RE(b)+uδE(cid:17) (δ)iL2(M)
p−2
= hδ,(cid:16)tLE(b)+uδ+(1−t)RE(b)+uδ(cid:17) (δ)iL2(M),
where in the last equality we have used that R and E commute if x ∈ δ′′. Tracking back the
x
equalities, we obtain
τ δ((b+δ)p−1−bp−1) >τ δ((E(b)+δ)p−1−E(b)p−1) >τ |δ|p ,
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
where the last inequality comes from the result in the commutative case.
Remark 2.3. We point out that, for p ∈]2,3[, the result cannot be reduced to the commutative
case as in the alternative proof. Indeed, t7→tp−2 is operatorconcaveandthe first inequality right
above reverses.
Remark 2.4. Letϕ:Lp(M)→Lp′(M)bethedualitymapsothathx,ϕ(x)iLp(M),Lp′(M) =kxkpp
and kϕ(x)kp′ = kxkpp−1. When restricted to Lp(M)+, it is exactly fp−1, so the result can be
written as: for a, b∈L (M)+
p
ha−b,ϕ(a)−ϕ(b)iLp(M),Lp′(M) > a−b pp.
(cid:13) (cid:13)
In this form, the inequality extends to general L -spaces in t(cid:13)he sen(cid:13)se of Haagerup, see [5, 2] for
p
the arguments.
4 E´.RICARD
Corollary2.5. Let(M,τ)beasemifinitevonNeumannalgebraandE :M→Mbeaτ-preserving
conditional expectation, then for all p>2 and x∈L (M)+,
p
(2) x−Ex 6 x .
p p
(cid:13) (cid:13) (cid:13) (cid:13)
Proof. Apply the above theorem with a(cid:13)= x an(cid:13)d b =(cid:13)E(cid:13)x, as τ (x−Ex)(Ex)p−1 = 0, the Ho¨lder
inequality gives: (cid:0) (cid:1)
x−Ex p 6τ (x−Ex)xp−1 6 x−Ex x p−1.
p p p
(cid:13) (cid:13) (cid:0) (cid:1) (cid:13) (cid:13) (cid:13) (cid:13) (cid:3)
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
Remark 2.6. The inequality (2) does not hold for p<2; a counterexample with M=ℓ2 can be
∞
foundin[3]. There,aslightextensionof (2)is given;onecanreplaceE byanypositivecontractive
projection C on L (M).
p
As explained in [3], the main inequality applies more generally to semigroups.
Corollary 2.7. Let (M,τ) be a semifinite von Neumann algebra and (T ) be a trace preserving
t t>0
unital positive strongly continuous semigroup on M. For λ > 0, let R = ∞e−λtT dt be its
λ 0 t
resolvent. Then for all p>2, λ>0 and x∈L (M)+, R
p
x−λR x 6 x .
λ p p
(cid:13) (cid:13) (cid:13) (cid:13)
Proof. We proceed as in Corollary 2.5(cid:13). We apply(cid:13)Theo(cid:13)re(cid:13)m 2.1 with a=x and b=λR x to get
λ
x−λR x p 6 τ (x−λR x)xp−1 −τ (x−λR x)(λR x)p−1
λ p λ λ λ
(cid:13) (cid:13) (cid:0) (cid:1) (cid:0) (cid:1)
(cid:13) (cid:13) 6 x−λR x x p−1−τ (x−λR x)(λR x)p−1 .
λ p p λ λ
(cid:13) (cid:13) (cid:13) (cid:13) (cid:0) (cid:1)
To conclude, it suffices to note tha(cid:13)t τ (x−λ(cid:13)R(cid:13)x)(cid:13)(R x)p−1 >0.
λ λ
Itcanbe checkedbyapproximation(cid:0)sthanks tothe resolve(cid:1)ntformula: x−λRλx=limt→∞t(1−
tR )R x. Indeed, recall that tR is positive unital and trace preserving and hence a contraction
t λ t
on L so that
p
τ tR (R x).(R x)p−1 6 R x p and τ t(1−tR )R x (R x)p−1 >0.
t λ λ λ p t λ λ
(cid:0) (cid:1) (cid:13) (cid:13) (cid:0)(cid:0) (cid:1) (cid:1)
(cid:13) (cid:13) (cid:3)
References
[1] Rajendra Bhatia. Matrix analysis, volume 169 of Graduate Texts in Mathematics. Springer-Verlag, New York,
1997.
[2] UffeHaagerup,MariusJunge,andQuanhuaXu.AreductionmethodfornoncommutativeLp-spacesandappli-
cations.Trans. Amer. Math. Soc.,362(4):2125–2165, 2010.
[3] MustaphaMokhtar-Kharroubi.ContractivitytheoremsinrealorderedBanachspaceswithapplicationstorelative
operatorbounds,ergodicprojectionsandconditionalexpectations.Preprint,https://hal.archives-ouvertes.fr/hal-
01148968.
[4] Gilles Pisier and Quanhua Xu. Non-commutative Lp-spaces. In Handbook of the Geometry of Banach Spaces,
Vol. 2,pages1459–1517. North-Holland,Amsterdam,2003.
[5] E´ric Ricard. H¨older estimates for the noncommutative Mazur maps. Arch. Math. (Basel) 104 (2015), no. 1,
37–45.
Laboratoire de Math´ematiques Nicolas Oresme, Universit´e de Caen Normandie, 14032 Caen Cedex,
France
E-mail address: [email protected]