Symmetric norms and the Leibniz property

0.2 MB·

by  Zoltan Leka

SYMMETRIC SEMINORMS AND THE LEIBNIZ PROPERTY
ZOLTA´NLE´KA
Abstract. We show that certain symmetric seminorms on Rn satisfy the
Leibniz inequality. As an application, we obtain that Lp norms of centered
boundedrealfunctions,definedonprobabilityspaces,havethesameproperty.
6 Even though this is well-known for the standard deviation it seems that the
1 complete resulthas never been established. Inaddition, weshallconnect the
0 resultswiththedifferentialcalculusintroducedbyCiprianiandSauvageotand
2 Rieffel’snon-commutativeRiemannmetric.

n
u
J

8 1. Introduction

Let(S, ,µ)denoteaprobabilityspaceandlet1 p< . Theseminormgiven
]
A by the pthFabsolute central moment of a random var≤iable f∞: S R is
→
F p 1/p
. σ (f;µ)= f Ef = f f dµ dµ .
h p p
k − k −
t (cid:18)ZS(cid:12) ZS (cid:12) (cid:19)
a (cid:12) (cid:12)
One of the most used quantity in probability(cid:12) theory and(cid:12)statistics is the standard
m (cid:12) (cid:12)
deviation(whenp=2). RecentlyM.A.Rieffelobservedthatthestandarddeviation
[ in ordinary and non-commutative probability spaces satisfies the strong Leibniz

3 inequalityandevenmatricialseminormshavethesameproperty[21]. Tobeprecise,
v we say that a seminorm L on a unital normed algebra ( , ) is strongly Leibniz
0 if (i) L(1 )=0, (ii) the Leibniz property A k·k
A
4
4 L(ab) a L(b)+ b L(a)
≤k k k k
0
holds for every a,b and, furthermore, (iii) for every invertible a, L(a−1)
0 ∈ A ≤
. a−1 2L(a) follows. For an ordinary probability space (S, ,µ), this means that
1 fkor evkery f and g L∞(S,µ), we have the inequalities F
0 ∈
6 fg E(fg) g f Ef + f g Eg
2 ∞ 2 ∞ 2
1 k − k ≤k k k − k k k k − k
: and
v f−1 E(f−1) f−1 2 f Ef if f−1 L∞(S,µ).
i k − k2 ≤k k∞k − k2 ∈
X
The study of strongly Leibniz seminorms regardedas non-commutative metrics on
r quantum metric spaces was initiated by M. Rieffel in his seminal papers [18], [20],
a
[19]. They played a crucial role in the development of a quantum theory for the
Gromov–Hausdorffdistance. A quantized version of this theory was established in
the recent papers by Li and Kerr [11], W. Wu [26], and a thorough survey is [12].
The most natural sources of strongly Leibniz seminorms are normed first-order
differential calculi. We recall now that a normed first order differential calculus is
a couple (Ω,∂), where Ω is a normed bimodule over such that
A
aωb a ω b for all a,b and ω Ω,
Ω
k k≤k kk k k k ∈A ∈
2010 Mathematics Subject Classification. Primary15A60, 46N30, 60E15 ; Secondary 26A51,
60A99.
Key words and phrases. standard deviation, central moments, Leibniz seminorm, symmetric
norm,derivation.
This study was supported by the Marie Curie IF Fellowship, Project ’Moments’, Grant no.
653943andbytheHungarianScientificResearchFund(OTKA)grantno. K104206.
1
2 ZOLTA´NLE´KA
and∂: ΩisaderivationwhichsatisfiestheLeibnizrule∂(ab)=∂(a)b+a∂(b).
A→
Readily,
L(a)= ∂a
Ω
k k
is a (strongly) Leibniz seminorm on (see [21, Proposition 1.1]).
A
A prototype of Leibniz seminorms is the Lipschitz number
L (f)=sup f(x) f(y)/ρ(x,y): x=y
ρ
{| − | 6 }
ofcomplex-valuedcontinuousfunctionsdefinedonanycompactmetricspace(X,ρ)
([25, Proposition 1.5.3]). Interestingly, one can obtain L by means of a normed
ρ
first order differential calculus ([24, Proposition 8], [20, Example 11.5]). We direct
the interested reader to [6], [17], [24] and [25] for a comprehensive study of general
Lipschitz seminorms, Lip-norms, and the associated Lipschitz algebras. Although,
we are unaware of any characterization of the Leibniz property [17, Question 6.3],
the lattice inequality L(f g) L(f) L(g), for all real f and g, is sufficient to
∨ ≤ ∨
conclude that a Lip-norm L is Leibniz ([17, Theorem 8.1]).
It is important to notice that any symmetric Dirichlet form (D(E),E) defined
on a dense domain D(E) of the real Hilbert space L2(S,µ) satisfies the Leibniz
inequality, see e.g. [8, Theorem 1.4.2], [3, Corollary 3.3.2]. See [14] and [13] for
Dirichlet forms on finite sets, graphs and fractals. Furthermore, F. Cipriani and
J.–L.Sauvageot[7]showedthateveryregularC∗-Dirichletformcanberepresented
asaquadraticformassociatedtoaderivationtakingitsvaluesinaHilbertmodule,
which is a direct link to the Leibniz rule.
Back to the standard deviation, one can present a direct simple proof of its
strong Leibniz property (see [20], [1]). More interestingly, the (quantum) standard
deviationcompletelyfits intothe aforementionedmachineryofdifferential1-forms.
This idea was exploited independently from [7] in some depth in Rieffel’s papers
[20], [21] and based on the observation that the variance of any random variable
f L2(S,µ) can be written as
∈
1
f Ef 2 = f(x) f(y)2dµ(x)dµ(y)
k − k2 2 | − |
ZZS×S
(that is, it is a Dirichlet form on L2(S,µ)). His differential calculus is defined
through the concept of spectral triples, introduced by Alain Connes, and what he
callsnon-commutativeRiemannmetric,see[6]and[20]. Henceonecansaythatthe
standarddeviation,commutativeornot,sharesaflavorofConnes’noncommutative
geometry.
The main goal of this paper is to show that the Leibniz inequality
fg Efg g f Ef + f g Eg
p ∞ p ∞ p
k − k ≤k k k − k k k k − k
is satisfied for all 1 p = 2 < and real f,g L∞(S,µ), which does not
≤ 6 ∞ ∈
seem to have been noticed previously. We remark that the end-point case p =
has already been settled in the recent paper [1]. First, we prove the result
∞
for the finite state space S = 1,...,n endowed with the uniform probability
n
{ }
measure. To get a friendly approach to the subject, we shall replace the ℓp norms
with symmetric norms on Rn. It should be stressed here that the essential part
of the paper works with symmetric norms and the uniform case. In Section 3
we shall investigate the results in terms of different differential calculi, including
the Cipriani–Sauvageot algebraic construction of differential 1-forms, and a very
briefconnectionwith Rieffel’s non-commutativeRiemannmetric. In Section4,the
failureofthenormedbimodulepropertywillleadustoafinitedimensionalexample
of a Leibniz seminorm that is not strongly Leibniz. Lastly, in Section 5, we shall
derive the Leibniz inequality for arbitrary probability measures by applying our
earlier results on Leibniz seminorms in probability spaces [1]. Our paper is in part
SYMMETRIC SEMINORMS AND THE LEIBNIZ PROPERTY 3
anattempttorevealapossiblelinkbetweennormeddifferentialcalculiandabsolute
central moments of bounded functions (random variables). Our future plan is to
studythecorrespondingresultsinnon-commutativematrixandC∗-algebrasaswell
as the case of complex-valued functions.
2. Leibniz inequality for symmetric seminorms
Atfirst,wecollectafew notations werequireinorderto provethe mainresults.
2.1. Symmetric norms. We say that a norm on Rn is symmetric if it is in-
k·k
variantundersign-changesandpermutationsofthecomponents. Symmetricnorms
are monotone which means that
x y if x↓ y ↓,
k k≤k k | | ≤| |
where x↓ denotes the usual non-increasing rearrangement of the vector x. Fur-
| | | |
thermore, the norm is absolute so
k·k
x = x
k k k| |k
for every x Rn (see [2, Section 2]).
∈
The vector k-norms (or Ky Fan k-norms) are special examples of symmetric
norms. Indeed, the vector k-norm of x is defined by
k
x = x ↓.
(k) i
k k | |
i=1
X
In the case when k = n and k = 1, we obtain the usual ℓ1 and ℓ∞ norms on Rn,
denoted by and , respectively. We recallnow that the dual normof any
1 ∞
k·k k·k
symmetric norm is symmetric as well. This follows easily from the duality relation
x =max (x,y): y 1 .
∗
k kA celebra{ted theokremk≤of K}y Fan says that, for any x,y Rn, the inequalities
∈ +
x y
(k) (k)
k k ≤k k
hold for every 1 k n, that is, x is weakly majorized by y, if and only if
≤ ≤
x y
k k≤k k
for every symmetric norm on Rn (see [2] or [23, Chapter 15]). Hence one can
k·k
look upon the vector k-norms as the cornerstones of symmetric norms.
Following Barry Simon’s terminology in [23, p. 248], let us introduce a class of
real matrices.
Definition. We say that a matrix A M (R) is real substochastic if
n
∈
n
a 1, j =1,...,n,
ij
| |≤
i=1
X
n
a 1, i=1,...,n.
ij
| |≤
j=1
X
ItissimpletoseethatAisrealsubstochasticifandonlyifAisacontractionon
Rn endowed with the ℓ1 norm and the ℓ∞ norm; i.e. Ay y and Ay
1 1 ∞
y for all y Rn. Additionally, if one can guaranteke akpro≤pekr lkinear coknnekctio≤n
∞
bketkween the ve∈ctors x and y, i.e. Ay = x for some A M (R), we can use
n
∈
interpolation methods. Actually, the Caldero´n–Mityagin theorem (see [5], [15] or
[23, Theorem 15.17]) says that if A is a real substochastic matrix then
Ay y
k k≤k k
follows for all y Rn and symmetric norms .
∈ k·k
4 ZOLTA´NLE´KA
Let x=(x ,...,x ) Rn. Let us introduce the symmetric matrix Θ with zero
1 n x
∈
row and column sum defined by
1 (x +x ) if i=j
(Θ ) = 2n i j 6
x ij
(− k:k6=i(Θx)ik if i=j.
Let us define the matrix P
I =I +Θ ,
x n x
where I denote the n n identity matrix. Throughout the section we shall use
n
the notation Ef = 1 ×n f 1, where f = (f ,...,f ) Rn and 1 stands for the
constant 1 vector. Mn orei=ov1eri, we shall consist1ently unse∈f and g for vectors of Rn
P
and fg for their pointwise product.
Ourfirstpropositionlinkstheproductoftwovectorsf,g Rn withthematrices
∈
If+1, Ig+1.
Proposition 2.1. For any f, g Rn,
∈
If+1(g Eg)+Ig+1(f Ef)=E(fg) fg.
− − −
Proof. Clearly, it is enough to show that
I (g Eg)+I (f Ef)=E((f 1)(g 1)) (f 1)(g 1)
f g
− − − − − − −
holds. A straightforwardcalculation gives for every index 1 m n that
≤ ≤
n(I (g Eg)+I (f Ef))
f g m
− −
1
= (f +f )(g g )
i m i j
2n −
1≤i6=m≤n1≤j≤n
X X
1
+ 1 (f +f ) (g g )
m i m j
− 2n −
(cid:16) 1≤iX6=m≤n (cid:17)1≤Xj≤n
1
+ (g +g )(f f )
i m i j
2n −
1≤i6=m≤n1≤j≤n
X X
1
+ 1 (g +g ) (f f )
m i m j
− 2n −
(cid:16) 1≤iX6=m≤n (cid:17)1≤Xj≤n
1
= (f +f ) (g g ) (g g )
i m i j m j
2n − − − !
1≤i6=m≤n 1≤j≤n 1≤j≤n
X X X
1
+ (g +g ) (f f ) (f f )
i m i j m j
2n − − − !
1≤i6=m≤n 1≤j≤n 1≤j≤n
X X X
+ (g g )+ (f f )
m i m i
− −
1≤i≤n 1≤i≤n
X X
1
= ((f +f )(g g )+(g +g )(f f ))
i m i m i m i m
2 − −
1≤i≤n
X
+ (g g +f f )
m i m i
− −
1≤i≤n
X
= (f g f g +g g +f f )
i i m m m i m i
− − −
1≤i≤n
X
= (f 1)(g 1) n(f 1)(g 1)
i i m m
 − − − − −
1≤i≤n
X
=n(E((f 1)(g 1)) (f 1)(g 1)) ,
m
− − − − −
SYMMETRIC SEMINORMS AND THE LEIBNIZ PROPERTY 5
which is what we intended to have. (cid:3)
Let us remember that the dual norm of the vector k-norm is
x
kxk(k)∗ =max kxk∞,kkk1 x∈Rn
(cid:18) (cid:19)
(e.g. [2, Ex. IV.2.12]).
Let B(k)∗ = x Rn: x (k)∗ 1 denote the closed unit ball of the dual space
(Rn, (k))∗. T{he∈n the ksetkof ex≤tre}me points of B(k)∗ can be readily described.
k·k
The result is well-known,but we sketcha short proof for the sake of completeness.
Lemma 2.2.
extB(k)∗ = ei: S 1,...,n and S =k ,
( ± ⊆{ } | | )
i∈S
X
where e -s denote the standard basis elements of Rn.
i
Proof. Denote K the points of the n-cube [ 1,1]n which has at most k non-zero
0
−
coordinates. It is not difficult to see that
conv K0 =B(k)∗.
Infact,pickapointvinB(k)∗ whichhasatmostk+1non-zerocoordinates. Denote
v a coordinate of v which has the smallest non-zero modulus. Obviously, v 1.
i i
| |≤
Now choose a vector c 1,0,1 n such that the support of c has cardinality k,
∈ {− }
i supp c andsign c =sign v for every j supp c. Then it is simple to see that
j j
∈ ∈
v v c
1−|vi| ∈B(k)∗.
i
−| |
Iterating the previous process, we arrive a point which has at most k non-zero
coordinates. This point is the convexcombinationofvertices ofa proper k-cube in
[ 1,1]n. (cid:3)
−
Now we are ready to prove the following proposition.
Proposition 2.3. For every f [ 1,1]n and 1 k n, the operator
∈ − ≤ ≤
If∗+1: (Rn,k·k(k)∗)→(Rn,k·k(k)∗)/R, x 7→If+1x+R
is a contraction.
Proof. First, to get an upper bound on the norm of I∗ , it is enough to calculate
f+1
thenormoftheclassIf+1v foreveryextremepointv oftheunitball(Rn, (k)∗).
k·k
From Lemma 2.2, we can assume that
v = e e
i i
−
iX∈S+ iX∈S−
forsomedisjointsetsS ,S Z suchthat S + S =k.Foranyx,y Rn and
+ − n − +
⊆ | | | | ∈
0 s 1, we have I∗ = sI∗+(1 s)I∗. Furthermore, since the quotient
≤ ≤ sx+(1−s)y x − y
norm is convex, one has
kIf+1vk(k)∗ =mλ∈iRnkIf+1v−λ1k(k)∗
≤x∈m[0a,2x]nmλ∈iRnkIxv−λ1k(k)∗
=x∈m{0a,2x}nmλ∈iRnkIxv−λ1k(k)∗.
Next, pick an x 0,2 n. Set
∈{ }
1
r = x,v .
v
nh i
6 ZOLTA´NLE´KA
In order to prove that I v is in the unit ball of the quotient space, it is enough to
x
show that
I v r 1 1.
k x − v k(k)∗ ≤
In fact,
I v r 1 = max I e n−1x,v
x v ∞ x i
k − k 1≤i≤n −
≤1m≤ia≤xn(cid:12)(cid:12)(cid:10)(Ix−n−1x⊗1(cid:11))(cid:12)(cid:12)ei (k)kvk(k)∗
max (cid:13)(I n−1x 1)e (cid:13) .
≤1≤i≤n(cid:13) x− ⊗ i(cid:13)1
(cid:13) (cid:13)
Let s=card i:xi =2 . For any 1(cid:13) i n, note that(cid:13)
{ } ≤ ≤
n n
1 1
(I n−1x 1)e = 1 (x +x ) + x x
x− ⊗ i 1 (cid:12) − 2n i j (cid:12) 2n | i− j|
(cid:13) (cid:13) (cid:12) Xj=1 (cid:12) Xj=1
(cid:13) (cid:13) (cid:12)(cid:12) s + n−s if x(cid:12)(cid:12) =2,
=(cid:12) n n (cid:12)i
( 1− ns + ns if xi =0
=1.
(cid:0) (cid:1)
Thus
I v r 1 1.
x v ∞
k − k ≤
Now, let P denote the projection n x e x e on Rn, where S =
S i=1 i i 7→ i∈S i i
S S is the support of v. Then
− +
∪ P P
n
1
I v r 1 = P I e x ,v
x v 1 S x i
k − k − n
i=1(cid:12)(cid:28) (cid:18) (cid:19) (cid:29)(cid:12)
X(cid:12) (cid:12)
(cid:12)n 1 (cid:12)
≤ (cid:12) PS Ixei− nx (cid:12)kvk(k)∗
i=1(cid:13) (cid:18) (cid:19)(cid:13)(k)
X(cid:13) (cid:13)
n (cid:13) 1 (cid:13)
(cid:13)P I e x (cid:13)
S x i
≤ − n
i=1(cid:13) (cid:18) (cid:19)(cid:13)1
X(cid:13) (cid:13)
(cid:13) n (cid:13)
(cid:13) 1 (cid:13) 1
= 1 (x +x ) + x x
(cid:12) − 2n i j (cid:12) 2n | i− j|
i∈S (cid:12) j=1 (cid:12) j∈S
X (cid:12) X (cid:12) X
(cid:12)1 (cid:12) 
+ (cid:12) x x (cid:12)
(cid:12)2n | i− j| (cid:12)
i6∈S j∈S
X X
n n
1 1
= 1 (x +x ) + x x ,
(cid:12) − 2n i j (cid:12) 2n | i− j|
i∈S (cid:12) j=1 (cid:12) j=1
X (cid:12) X (cid:12) X
that is, (cid:12)(cid:12) (cid:12)(cid:12) 
(cid:12) (cid:12)
I v r 1 (I n−1x 1)e
k x − v k1 ≤ x− ⊗ i 1
i∈S
X(cid:13) (cid:13)
= S .(cid:13) (cid:13)
| |
Hence
Ixv rv1 (k)∗ 1,
k − k ≤
and the proof is complete. (cid:3)
Let us define the hyperplane
n
X := x Rn: Ex= x =0 Rn.
0 i
{ ∈ }⊆
i=1
X
SYMMETRIC SEMINORMS AND THE LEIBNIZ PROPERTY 7
Obviously, the dual of the Banach space (X , ) is the quotient space (Rn,
0 (k)
(k)∗)/R. In fact, X0 is a one co-dimensional sku·bkspace of R, whilst y,x Ex =k0·
hkolds for every y R1. Clearly, If+11 = 1. Hence the adjoint ohf If+−1: (Xi0,
) (Rn, ∈) is the operator k·
(k) (k)
k → k·k
If∗+1 : (Rn,k·k(k)∗)→(Rn,k·k(k)∗)/R, x7→If+1x+R,
of Proposition 2.3. Since If+1 X0 = (If+1 X0)∗ (see e.g. [16, Proposition
k | k k | k
2.3.10]), a straightforwardcorollary is the following statement.
Proposition 2.4. For every f [ 1,1]n, the operator If+1 is a contraction on
the normed space (X , ). ∈ −
0 (k)
k·k
Furthermore, this leads us to the next proposition.
Proposition 2.5. For every symmetric on Rn and f [ 1,1]n, If+1 is a
contraction on (X , ). k·k ∈ −
0
k·k
Proof. For every x X and 1 k n, Proposition 2.4 says that
0
∈ ≤ ≤
k k
|If+1x|↓i ≤ |x|↓i.
i=1 i=1
X X
Thus the vector If+1x is weakly majorized by x. Now the absolute property of
| | | |
and Ky Fan’s theorem for symmetric norms give that
k·k
If+1x = If+1x x = x ,
k k k| |k≤k| |k k k
which is what we intended to have. (cid:3)
NowonecanreadilyprovethefollowingLeibnizinequalityforsymmetricnorms.
Theorem 2.6. Let be a symmetric norm on Rn. For every f,g Rn, we have
k·k ∈
fg E(fg) g f Ef + f g Eg .
∞ ∞
k − k≤k k k − k k k k − k
Proof. Withoutlossofgenerality,wecanassumethat f = g =1.Applying
∞ ∞
k k k k
Proposition 2.1 and Proposition 2.5, it follows that
fg E(fg) = If+1(g Eg)+Ig+1(f Ef)
k − k k − − k
If+1 X0 g Eg + Ig+1 X0 f Ef
≤k | kk − k k | kk − k
= g Eg + f Ef ,
k − k k − k
and the proof is complete. (cid:3)
Remark. OnecangiveadirectproofofProposition2.5viatheCaldero´n–Mityagin
interpolation result as we briefly indicate. For an x [0,2]n, let us consider the
∈
matrix
1
L =I x 1.
x x
− n ⊗
We note that the off-diagonal part of L is skew-symmetric: (L ) = (L )
x x i,j x j,i
−
for every i = j, hence LT = LT . From the proof of Proposition 2.3,
6 k xk1→1 k xk∞→∞
it follows that
LT 1 and LT 1.
k xk1→1 ≤ k xk∞→∞ ≤
Moreover,foranysymmetricnorm ,theadjointofI : (X , ) (Rn, ),v
x 0
k·k k·k → k·k 7→
I v, is the operator
x
I∗: (Rn, ) (Rn, )/R,
x k·k∗ → k·k∗
where
I∗v =I v+λ1
x x
8 ZOLTA´NLE´KA
1
and denotes the dual norm. Again, for any v Rn, let r = x,v . Then
∗ v
k·k ∈ nh i
1
I v r 1 = I v x,v
x v ∗ x ∗
k − k k − nh ik
1
= (I x 1)e ,v
x i i ∗
kh − n ⊗ i k
= LTv .
k x k∗
Since the dual norm is symmetric, the Caldero´n–Mityagintheorem says that
∗
k·k
min I v λ1 LTv v .
λ∈Rk x − k∗ ≤k x k∗ ≤k k∗
That is,
I∗ 1,
k xk≤
and the operator I is a contraction on (X , ).
x 0
k·k
Remark. Perhapsitisappropriatetonotethatifx [0,1]n thenI isdoublysto-
x
∈
chastic. Hence, the Birkhoff–von Neumann theorem gives that I 1 for
x k·k→k·k
any permutationinvariantnorm onRn. Now assume that fk,g akre nonn≤egative
k·k
and f = g =1 Then
∞ ∞
k k k k
I−f+1(Eg g)+I−g+1(Ef f)=E(fg) fg,
− − −
and the matrices I−f+1, I−g+1 are doubly stochastic as well. A simple corollaryis
the following statement.
Theorem 2.7. Let be a permutation invariant norm on Rn. For any nonneg-
ative vectors f and gki·nkRn, we have
+
fg E(fg) g f Ef + f g Eg .
∞ ∞
k − k≤k k k − k k k k − k
3. Derivations and the Leibniz inequality
To have a description of the Leibniz inequality in terms of derivations, we shall
need to introduce the fundamental concepts of Laplacians and related Dirichlet
forms on finite sets [14].
3.1. Laplacians and Dirichlet forms. We recallthat a Laplacianmatrix ∆ is a
non-positive definite matrix such that its kernel is the subspace R1 and all of its
off-diagonalsarenon-negative. LetusrememberthateveryLaplacian∆determines
a Dirichlet form E (u,v) = u,∆v on Rn Rn. To be precise, for any f Rn
∆
−h i × ∈
let us define the vector
0 if f 0,
i
≤
f = f if 0<f <1,
i  i i
1 if 1 fi.
≤
AsymmetricbilinearformE isaDirichletformifitsatisfiesthefollowingproperties:

(i) E(f,f) 0,
(ii) E(f,f)=≥0 if and only if f R1,
(iii) E(f,f) E(f,f) (Markovia∈nproperty).
≤
Actually, there is a one-to-one correspondence between the Laplacians and the
Dirichlet forms on finite sets, see [14, Proposition2.1.3].
On the other hand, it is simple to see that
f g f g = f g f g +f g f g
i i j j i i j i j i j j
| − | | − − |
g f f + f g g ,
∞ i j ∞ i j
≤k k | − | k k | − |
hence the Leibniz inequality
E1/2(fg,fg) g E1/2(f,f)+ f E1/2(g,g)
∞ ∞
≤k k k k
SYMMETRIC SEMINORMS AND THE LEIBNIZ PROPERTY 9
follows immediately (see [13, p. 281]).
Now let f Rn and P denote the orthogonal projection f 1 n f 1 with
respect to the∈usual inner product on Rn. Then the operator 7→ n i=1 i
P
1 n 1 ... 1
−
1 1 1 n 1
∆u =P −In = n ... − ... ... 
 
 1 ... 1 1 n
 − 
 
is a Laplacian and
E (f,f)= f Ef 2,
∆u k − k2
where denotes the usual ℓ2-norm.
k·k2 n
3.2. Derivations and the Leibniz inequality. Letµbeingeneralaprobability
measure on the set S . Then the variance of any random vector f can be written
n
as the Dirichlet form
1
σ2(f;µ)= f,∆ f = (f(x) f(y))2µ(x)µ(y),
2 −h µ i 2 −
x,Xy∈Sn
where the off-diagonal part of the Laplacian ∆ is (∆ ) =µ(i)µ(j), 1 i=j
µ µ i,j
≤ 6 ≤
n. Then the deviation σ (f;µ) can be represented as the L2-norm of the gradient
2
vector∂ f, where ∂ is the universalderivation∂ f =f 1 1 f, in the Hilbert
u u u
⊗ − ⊗
space L2(S S ,µ µ).
n n
× ⊗
To obtain σ as a norm of a derivation we need a refined approach. Let us
p
consider the matrix algebra M (R)=ℓ∞ ℓ∞ endowed with the Hilbert–Schmidt
n n ⊗ n
inner product as a bimodule over the finite dimensional algebra ℓ∞, where the left
n
and right actions are defined by linearity from
a(b c)d=ab cd.
⊗ ⊗
Define the derivation ∂: ℓ∞ M (R) by
n → n
1
(3.1) ∂f = (f 1 1 f),
√2n ⊗ − ⊗
whichsatisfiestheLeibnizequality,i.e. ∂(fg)=∂f g+f ∂g.Theadjointoperator
∂∗, defined by Tr(AT∂f)= ∂∗A,f for any A M· (R),·is the operator
n
h i ∈
1
(3.2) (∂∗A) = (A(1 1) (1 1)A) .
i ii
−√2n ⊗ − ⊗
Indeed, let ι denote the canonical embedding of the algebra ℓ∞ into M (R) as the
n n
diagonal algebra. Then one has that
1
∂f = ((ιf)1 1 1 1(ιf)).
√2n ⊗ − ⊗
Since the extended derivation d: A A(1 1) (1 1)A is a skew adjoint map
on M (R) with respect to the Hilber7→t–Schm⊗idt i−nner⊗product, we get ∂∗ = ι∗d.
n
−
Anelementarycalculationimpliesthefollowinglemma,whenceweomititsproof.
Lemma 3.1. One has the decomposition
∆ =∂∗∂.
u
−
Then the following definition is quite natural.
Definition. Fix a symmetric norm on Rn and let denote its dual norm.
∗
We define a seminorm on the matrixka·lkgebra M (R) byk·k
n
A =max Tr(AT∂f): f 1 .
∂ ∗
k k { k k ≤ }
10 ZOLTA´NLE´KA
The next propositionlinks the differential calculus (M (R),∂) over ℓ∞ with the
n n
norms of centered vectors.
Proposition 3.2. Let f =(f ,...,f ) ℓ∞ and be a symmetric norm on Rn.
1 n ∈ n k·k
Then the equality
n
1
f f 1 = ∂f
i ∂
− n k k
(cid:13) Xi=1 (cid:13)
holds. (cid:13) (cid:13)
(cid:13) (cid:13)
Proof. From Lemma 3.1 and duality
∆ f = ∂∗∂f =max ∂∗∂f,g : g 1
u ∗
k k k k {h i k k ≤ }
=max Tr(∂fT∂g) : g 1
∗
{ k k ≤ }
= ∂f .
∂
k k
(cid:3)
The next theorem shows a certain module property of the seminorm .
∂
k·k
Theorem 3.3. For any f and g Rn,
∈
∂f g g ∂f ,
∂ ∞ ∂
k · k ≤k k k k
g ∂f g ∂f .
∂ ∞ ∂
k · k ≤k k k k
Proof. First, we have
(∂f g) =g (∂f) and (g ∂f) =g (∂f) .
ij j ij ij i ij
· ·
For any a Rn, note that I +a 1=I holds on the subspaceX =(I P)Rn.
g g 0 n
∈ ⊗ −
Thus
1
Ig+1 =Ig+1 1=Θg on X0.
− n ⊗
From Proposition 2.5
Θ h g h for any h X .
g ∞ 0
k k≤k k k k ∈
On the other hand, a direct calculation shows that
n n
1 1
h,Θ f = (f f )(g +g )h = (f f )g (h h )
g i j i j i i j i i j
h i 2n − 2n − −
i,j=1 i,j=1
X X
=Tr((g ∂f)T∂h).
·
Thus
g ∂f =max h,Θ f : h 1
∂ g ∗
k · k {h i k k ≤ }
= Θ f
g
k k
= Θ ((I P)f Pf)
g
k − ⊕ k
= Θ (I P)f
g
k − k
g (I P)f
∞
≤k k k − k
= g ∂f .
∞ ∂
k k k k
The same argument gives that ∂f g g ∂f holds, hence the proof is
∂ ∞ ∂
complete. k · k ≤ k k k k (cid:3)
WesawinTheorem2.6thattheLeibnizinequalityholdswithsymmetricnorms.
Now one can provide a transparent reformulation of the proof relying upon the
previous results.