quantum differentiability of essentially bounded functions ... · pdf filebounded functions on...

Quantum Dif-ferentiabilityof EssentiallyBounded

Functions onEuclideanSpace

F. Sukochev,E. McDonald

Quantum Differentiability of EssentiallyBounded Functions on Euclidean Space

F. Sukochev, E. McDonald

UNSW Australia

August 25, 2016

Introducing Non-commutative Geometry

What is noncommutative geometry?

The study of noncommutative algebras which resemble algebrasof functions on geometric spaces, using the methods and thelanguage of geometry.

Non-commutative geometry

Non-commutative geometry in analysis is usually the study ofalgebras of operators on Hilbert space. Algebras of operatorsare considered the generalisation of algebras of functions.

Example

The algebra L∞(R) is an algebra of bounded operators onL2(R). The full algebra B(L2(R)) is the “noncommutativeextension” of the study of functions on R.

(Briefly) Quantised Calculus

Broadly speaking, calculus is the study of functions on Rd .

Philosophy

Every locally integrable function f on Rd determines a(potentially unbounded) operator Mf on L2(Rd).In fact virtually every object of classical calculus arises in thisway.For example, differentiations ∂j are operators on L2(Rd) and

[∂j ,Mf ] =M∂j f .

Quantised Calculus

Suppose we take this point of view seriously, that calculus isthe study of operators on L2(Rd).This means that the class of “functions” is now expanded toL(L2(Rd)).L(L2(Rd)) contains many new objects which have no analoguein usual real analysis.(In fact we will typically expand our point of view to operatorson L2(Rd ,CN))

Infinitesimals

Infinitesimals have been a part of mathematics since nearly thevery beginning (since at least the 5th Century BC, from theEleatic school in Greek Italy)

Definition (Infinitesimal, informal)

An infinitesimal ε is a quantity such that for all positiveintegers n,

0 ≤ ∣ε∣ < 1

Expressed geometrically, this means that a line segment oflength ε, when appended end to end any finite number oftimes, will never exceed 1 in length.

Obviously the only infinitesimal in R is zero.

Uses of nonzero infinitesimals

Mathematicians historically liked to pretend that R has nonzeroinfinitesimals. They found the concept extremely useful, suchas in the following definitions:

Definition

A function f ∶ R→ R is said to be continuous at x if for allinfinitesimals ε, the difference f (x + ε) − f (x) is infinitesimal.

Definition

A function f ∶ R→ R is said to be differentiable at a point x iffor all infinitesimals ε, there is a constant c such that

f (x + ε) − f (x) − cε

is infinitely smaller than ε. The difference f (x + ε) − f (x) isusually called df (x).

Infinitesimals as operators

Alain Connes noticed that B(H) contains elements not unlikeinfinitesimals:

Definition (Infinitesimal Operators)

An operator T ∈ B(H) is called infinitesimal if for any ε > 0there is a finite dimensional subspace E ⊂ H such that

∥T ∣E⊥∥ ≤ ε.

It is not hard to see that the set of infinitesimal operators isjust the set of compact operators.

What properties should infinitesimals have?

Following the traditions of pre-rigorous analysis, we expectinfinitesimals to have the following properties:

1 If f ∶ R→ R is a function, then f is continuous if df isinfinitesimal

2 If f is smoother than g , then df is somehow “smaller”than dg

3 If x is a positive infinitesimal, then x2 is “smaller” than x

4 If f is a differentiable function, then we can writedf = f ′dx (provided that “sufficiently small” infinitesimalsare ignored.

Provided that these vague properties are intepreted correctly,quantised calculus can give rigorous meaning to all of them.This talk focuses especially on 1 and 2.

Quantised Calculus and Infinitesimals

To make sense of differentiability, we need to figure out what itmeans for one infinitesimal to be infinitely small compared toanother.

Definition

The sequence of singular values {µ(n,T )}∞n=0 associated to acompact operator T is defined as

µ(n,T ) = inf{∥T − R∥ ∶ rank(R) ≤ n}.

They are also the eigenvalues of ∣T ∣ in non-increasing order.

The rate of decay of {µ(n,T )}∞n=0 is a reasonable measure ofthe size of T .

Sizes of Infinitesimals

We can quantify the sizes an infinitesimal T by placingconditions on the rate of decay of {µk(T )}∞k=0.

The smallest infinitesimals have {µk(T )}∞k=0 of finitesupport. Then T is of finite rank.

We say that T ∈ Lp if {µk(T )}∞k=0 ∈ `p.

We say that T ∈ Lp,∞ if µk(T ) = O(k−1/p).

We say that T ∈ Lp,q if {k1/p−1/qµk(T )}∞k=0 ∈ `q.

We say that T ∈M1,∞ if { 1log(k+1) ∑

kn=0 µk(T )}∞k=0 ∈ `∞.

These last four conditions in fact correspond to ideals of B(H).

Sizes of Infinitesimals

Theorem

Let T be a compact operator in B(H). Then for every k ≥ 0,

µk(T 2) ≤ ∥T ∥µk(T ).

Hence, if T is an infinitesimal, then T 2 is a smallerinfinitesimal.

Notations

H denotes any complex separable Hilbert space.

B(H) is the algebra of bounded operators on H.

Lp,q(H) is the Schatten-Lorentz space of operators Twith {µ(n,T )}∞n=0 ∈ `p,q.

We work over Rd , and always d > 1.

L∞(Rd) acts on L2(Rd) by pointwise multiplication asMf .

S(Rd) is the Schwartz space of functions on Rd .

We denote −i∂j as Dj .

Riesz Transforms

For j = 1, . . . ,d , define

Rj =∂j√

∂21 + ∂22 +⋯ + ∂2d.

Rj is called the jth Riesz transform. By functional calculus Rj

is bounded on L2.

Quantised Differentials

In noncommutative geometry, we have a new object that is notpresent in classical analysis called a quantised derivative orquantised differential.

Definition

Let f ∈ L∞(R). Mf is the operator on L2(R) of pointwisemultiplication: Mf g(x) = f (x)g(x) for almost all x ∈ R. Fdenotes the Hilbert transform:

Fg(x) ∶= 1√2π∫R

sgn(ξ)e ixξg(ξ) dξ.

(Defined at least initially for g a smooth function of compactsupport, then extended by continuity to g ∈ L2(R)). Then wedefine:

d f ∶= [F ,Mf ].

What is a quantised differential?

d f is supposed to represent the “infinitesimal variation”, likedf in classical analysis.

Warning:

d f is not a one-form, or a derivative. It is an infinitesimaldeviation. It should be thought of as f (x + h) − f (x) forinfinitesimal epsilon.

It is very hard to motivate the definition of d f . Instead, we willshow that is satisfies a number of “headline properties” of aclassical differential.

A dictionary

Classical Analysis Non-commutative Analysis

Function OperatorRange SpectrumInfinitesimal Compact OperatorDifferential Quantised differential

Rediscovering Classical Definitions

Let f ∶ R→ R be Borel. f is continuous at x ∈ R if and only iff (x +T ) − f (x) is compact for all compact self-adjointoperators T .

Lipschitz continuity and Lp spaces

If f ∶ R→ R is Lipschitz and ε is an informal infinitesimal, oneshould think that f (x + ε)− f (x) is an infinitesimal of the samesize. The following can be viewed as a rigorous justification:

Theorem (Potapov and Sukochev, 2001)

Let p ∈ (1,∞). A function f ∶ R→ R is Lipschitz if and only iffor all self adjoint T ∈ Lp and all x we have

∥f (x +T ) − f (x)∥Lp ≤ cabs∥f ′∥∞∥T ∥Lp .

Another approach to differentiability

Recall the classical differential df of a function f , defined as aninfinitesimal variation in f .Connes defines the following “quantised differential”

Definition (Quantised Differential on R)

Let f ∈ L∞(R).

d f = i[sgn(−i ddx

) ,Mf ].

The origin of the definition is in noncommutative geometry andis beyond the scope of this talk.

Quantised differentials

Let us just take for granted the definition of d f , and try towork with it.

Conjecture

d f should be infinitesimal (i.e., compact) if f is continuous.

Conjecture

The size of d f (i.e., the rate of decay of the singular values)should somehow decrease with the smoothness of f .

Revision of Classical Fourier Analysis and Notation

We define T = {ζ ∈ C ∶ ∣ζ ∣ = 1}. Let z ∶ T→ T be the identityfunction. Denote the normalised Haar (or arc length) measureon T by m.For f ∈ L1(T,m), define for n ∈ Z,

f (n) ∶= ∫Tz−nf dm.

Recall that any f ∈ L2(T,m) can be written as

f = ∑n∈Z

f (n)zn.

The sum converges in the L2 sense. This effects an isometricisomorphism between L2(T) and `2(Z).

Revision of Classical Fourier Analysis and Notation

The closed linear span of {zn}∞n=0 in L2 is denoted H2(T), andthe orthogonal complement is denoted H2

−(T).We define the space of polynomials P(T) to be the finite linearspan of {zn}n∈Z. PA(T) = span{zn}n≥0.

Differentials on T

Defining quantised differentials for functions on T is exactly likefor functions on R.

Definition

The Hilbert transform, for g ∈ L2(T), is defined to be

Fg ∶= ∑n∈Z

sgn(n)g(n)zn.

We define the quantised differential:

d f ∶= [F ,Mf ].

Differentials on T

Hence for f ∈ L2(T),Ff = ϕ ∗ f

ϕ = ∑n∈Z

sgn(n)zn = 1

1 − z− z−1

1 − z−1= 2

1 − z.

(d f )g = ([F , f ]g)(t)

= 2 limε→0∫∣τ−t ∣>ε

f (t) − f (τ)t − τ g(τ) dm(τ).

Finite rank differentials

Let f ∶ T→ C. The strictest condition we can put on thesmoothness of f is that f is a rational function. The strictestcondition we can put on the size of d f is that d f is finite rank.These two conditions are equivalent.

Theorem (Kronecker)

If f ∶ T→ C, then d f is finite rank if and only if f is a rationalfunction.

Bounded differentials

Let f ∶ T→ C. The weakest condition that we can place on d fis that d f is bounded.

Definition

Let f ∶ T→ C be measureable. We say that f is of boundedmean oscillation if for an arc I ⊆ T, define

m(I ) ∫I f dm

m(I ) ∫I ∣f − fI ∣ dm <∞

where the supremum runs over all arcs I . The set of functionswith bounded mean oscillation is denoted BMO(T).

Bounded differentials

Theorem (Nehari)

Let f ∶ T→ C. Then d f is bounded if and only if f ∈ BMO(T).

Compact differentials

We define the space VMO(T):

Definition

We say that f ∈ VMO(T) if f ∈ BMO(T) and

limm(I)→0

m(I ) ∫I ∣f − fI ∣ dm = 0.

Theorem

If f ∶ T→ C, then d f is compact if and only if f ∈ VMO(T).

Can we do better?

We seek a more precise characterisation of the relationshipbetween the smoothness of f and the size of d f . To this end,we define the Besov classes Bs

(Aside) Besov Spaces

Let f ∈ L∞(Rd).

Central Dogma of Harmonic Analysis

Smoothness of f corresponds to the rate of decay of f (andvice versa).

The Besov space Bsp,q(Rd) consists of functions f such that f

can be written asf = ∑

n∈Zfn

converging in Lp such that supp(fn) ⊆ B(0,2n+1) ∖B(0,2n−1)and

∑n∈Z

2∣n∣sq∥fn∥qp <∞.

Results of Peller (and others)

These conjectures have amazingly nice answers:

Theorem

If f ∈ C0(R) +C then d f is compact.

Theorem

For p ∈ (0,∞), d f ∈ Lp if and only if f ∈ B1/pp,p (R).

There are similar (but less easily stated) theorems givingnecessary and sufficient theorems for d f ∈ Lp,q for p ∈ (0,∞)and q ∈ (0,∞].

Classical Limits

How can we justify the name “quantised differential”?Let f ∈ C0(R) +C. Then ∣d f ∣ has a sequence of eigenvaluesλ0 ≥ λ1 ≥ ⋯ ≥ 0, with limn→∞ λn = 0.Quantum mechanically, the numbers {λn}∞n=0 correspond toobservable values of ∣d f ∣. So their asymptotics mustcorrespond to a classical limit (somehow)... Let ω be a dilationinvariant extended limit on `∞(N). Then,

ωn→∞ (λ0 + λ1 +⋯ + λnlog(2 + n) ) = cabs∫

R∣f ′(x)∣dx .

So d f corresponds to f ′ in the classical limit (in some sense).

Historical introduction:the normal trace and singular traces

Normal trace

Let B(H) be an algebra of all bounded linear operators on aseparable Hilbert space H.

Canonical trace:

Tr(A) =∞

∑k=0

λ(k,A),

the sequence {λ(k ,A)}∞k=0 is an eigenvalue sequence of acompact operator A.

The normal trace is essentially unique.

Question: Are there any other traces besides the canonical?

The Dixmier’s construction (1966)

Let A ∈ B(H) be a positive compact operator s.t.

supn>0

log(1 + n)n−1

∑k=0

λ(k ,A) <∞.

t(A) ∶= lim1

log(1 + n)n−1

∑k=0

λ(k ,A), 0 ≤ A.

Two problems: (i) convergence, (ii) additivity.

Setting lim to be a linear form invariant under the group ofaffine transformations t ↦ at + b on R solves both of theseproblems.

Non-normal traces

If ω is a dilation invariant singular state on the space l∞ ofbounded sequences, (that is,ω(x0, x1, . . . ) = ω(x0, x0, x1, x1, . . . )) the functional

Trω(A) ∶= ω ( 1

log(1 + n)n−1

∑k=0

λ(k ,A),) , 0 ≤ A,

is positively homogeneous and additive.

1. Trω is a trace;2. Trω is non-trivial: Trω(diag { 1

n+1}) = 1;

3. Trω(A) = 0 if A is finite rank operator.

Hence, Trω is a non-trivial, non-normal trace.

Dixmier traces and applications

Extended limits

To introduce the revised construction of Dixmier traces werequire some preparations.

Definition

A linear functional ω on L∞ = L∞(0,∞) is called an extendedlimit if

(i) ω(x) ≥ 0, whenever 0 ≤ x ∈ L∞;

(ii) ω(χ(0,∞)) = 1;

(iii) ω(x) = limt→∞ x(t) (if the limit exists).

Conditions (i)-(iii) are equivalent to the fact that ω is aHahn-Banach extension of the usual limit functional.

Such functionals are singular, that is they vanish on compactlysupported functions.

Invariant extended limits

Definition

An extended limit ω on L∞ is called

● translation invariant if ω(Thx) = ω(x) for every h > 0;

● dilation invariant if ω(σβx) = ω(x) for every β > 0,

● exponentiation invariant if ω(Pax) = ω(x) for every a > 0,where

(Thx)(t) ∶= x(t+h), (σβx)(t) ∶= x(t/β), (Pax)(t) = x(ta).

Later on another type of invariant extended limits will beintroduced.

Singular value function

For every A ∈ B(H) a generalized singular value functionis defined by the formula

µ(t,A) = inf{∥Ap∥∞ ∶ p is a projection in B(H) with τ(1−p) ≤ t}.

If A is compact, then µ(k − 1,A) is the k-th largest eigenvalueof an operator ∣A∣ = (A∗A)1/2, k ∈ N and

µ(t,A) =∞

∑n=0

µ(n,A)χ[n,n+1)(t), t > 0.

Dixmier traces

Definition

The classical Dixmier-Macaev ideal:

M1,∞ ∶= {A is compact ∶ supt>0

log(1 + t) ∫t

0µ(s,A) ds <∞} .

For an arbitrary dilation invariant extended limit ω thefunctional

Trω(A) ∶= ω ( 1

log(1 + t) ∫t

0µ(s,A) ds) , 0 ≤ A ∈M1,∞,

extends to a non-normal trace (a Dixmier trace) on M1,∞1.

1A.Carey, F.Sukochev, Dixmier traces and some applications tononcommutative geometry, Uspekhi Mat. Nauk, 2006.

Dixmier trace is singular

If 0 ≤ A ∈ L1, then it follows from the definition of Dixmiertrace that

Trω(A) = ω( O(1)log(2 + n)) = 0.

In particular, Dixmier trace vanishes on every finite rankoperator.One can now see that Trω is no normal. Indeed,

diag({ 1

k + 1}0≤k≤n) ↑ diag({

k + 1}0≤k)

as n →∞. However,

0 = Trω(A)(diag({ 1

k + 1}0≤k≤n)) /→ Trω(A)(diag({ 1

k + 1}0≤k)) = 1.

Non-commutative integral

Let (A,H,D) be a spectral triple. Here, A is a ∗-algebra ofbounded operators on the Hilbert space H andD ∶ Dom(D)→ H is an unbounded self-adjoint operator, suchthat D2 is a positive operator with compact resolvent.Assume that (1 +D2)−d/2 ∈M1,∞ for some d ∈ R.Hence, a ⋅ (1 +D2)−d/2 ∈M1,∞ for every a ∈ A.

If all Dixmier traces coincide on a ⋅ (1 +D2)−d/2, then onedefines

⨏ a ∶= Trω(a ⋅ (1 +D2)−d/2).

Application: Connes Trace Theorem

Let ∆ = ∑dn=1

∂x2ndenotes the Laplace operator on Rd .

If f ∈ C∞c (Rd), then Mf (1 −∆)−d/2 is a classical

pseudo-differential operator of order −d .

Hence2, the operator Mf (1 −∆)−d/2 belongs to M1,∞ (thedomain of Dixmier traces) and

⨏ f = Trω(Mf (1 −∆)−d/2) = VolSd−1

d(2π)d ⋅ ∫Rd

f (u)du,

for every Dixmier trace Trω.

2A. Connes, The action functional in noncommutative geometry,Comm. Math. Phys., 1988.

Generalisation of Connes Trace Theorem

The extension of CTT can be done in few directions:- expand the class of functions f ;- extend the class of traces beyond Dixmier traces.

Theorem (3,4)

If f ∈ L2(Rd) is compactly supported, thenMf (1 −∆)−d/2 ∈M1,∞ and

⨏ f = τ(Mf (1−∆)−d/2) = τ(diag{1/n})VolSd−1

d(2π)d ⋅∫Rd

f (u)du,

for every trace (that is, unitarily invariant linear functional) τon M1,∞.

3N.Kalton, S.Lord, D.Potapov, F.Sukochev, Traces of compactoperators and the noncommutative residue, Adv. Math., 2013

4S.Lord, F.Sukochev, D.Zanin, Singular Traces: Theory andApplications, 2012.

Generalisation of Connes Trace Theorem 2

Let X be a d-dimensional closed Riemannian manifold withmetric g . Let ∆g denotes the Laplace-Beltrami operator onC∞(X ). (Laplace-Beltrami operator is an extension of thenotion of Laplace operator to manifolds. It is defined as thedivergence of the gradient, as in the classical case.)

Theorem (5,6)

If f ∈ L2(X ), then Mf (1 −∆g)−d/2 ∈M1,∞ and

⨏ f = τ(Mf (1−∆g)−d/2) = τ(diag{1/n})VolSd−1

d(2π)d ⋅∫Xf (u)du,

for every trace τ on M1,∞.5S.Lord, D.Potapov, F.Sukochev, Measures from Dixmier traces and

zeta functions, JFA, 20106S.Lord, F.Sukochev, D.Zanin, Singular Traces: Theory and

Applications, 2012.

Generalisation of Connes Trace Theorem 3

Let T2 = R2/Z2 be a torus. Let ∆ = 14π (

∂x21+ ∂2

∂x22) denotes the

Laplace operator on T2.

Theorem (7,8)

If f ∈ L∞(T2), then Mf (1 −∆)−1 ∈M1,∞ and

⨏ f = τ(Mf (1 −∆)−1) = c ∫T2

f (u)du,

for every trace τ on M1,∞, where c > 0 does not depend on τ .

7S.Lord, D.Potapov, F.Sukochev, Measures from Dixmier traces andzeta functions, JFA, 2010

8S.Lord, F.Sukochev, D.Zanin, Singular Traces: Theory andApplications, 2012.

Application: Banach space geometry

G. Pisier and Q. Xu9 raised the question on the differentiabilityof the norms of non-commutative Lp-spaces associated with anarbitrary von Neumann algebra M. They asked whether theLp-norm of a noncommutative Lp-space has the samedifferentiability properties as the norm of a classical(commutative) Lp-space.

When the algebra M is of type I this question has been fullyresolved10, but the general case required new ideas.

9G. Pisier and Q. Xu, Non-commutative Lp-spaces, Handbook of thegeometry of Banach spaces, 2003.

10D. Potapov and F. Sukochev, Frechet differentiability of Sp norms,Adv. Math., 2014.

Application: Banach space geometry 2

Theorem (11)

Let H ∈ Lp(M). The function H ↦ ∥H∥pp is

(i) infinitely many times Frechet differentiable, if p ∈ 2Z;

(ii) (p − 1)-times Frechet differentiable, if p ∈ 2Z + 1;

(iii) [p]-times Frechet differentiable, p ∉ Z.

Here, Frechet differentials are defined via singular traces onweak non-commutative L1-spaces associated with semifinite(non-finite) von Neumann algebras.

11D. Potapov, F. Sukochev, A. Tomskova, D. Zanin, Frechetdifferentiability of the norm of Lp-spaces, submitted manuscript.

Higher Dimensions 1

What is the appropriate analogue of d f when f is a functionon Rd?Let N = 2⌊d/2⌋. {γj}dj=1 are N ×N self-adjoint matricessatisfying γjγk + γkγj = 2δj ,k , where δ is the Kronecker delta.The precise choice of matrices satisfying this relation isunimportant so we assume that a choice is fixed. Using thischoice of gamma matrices, we can define the d-dimensionalDirac operator.

Higher Dimensions 2

Let D be the operator on L2(Rd ,CN) defined as

∑j=1

γj ⊗ (−i∂j).

This is a linear operator on the Hilbert space CN ⊗ L2(Rd)initially defined with dense domain CN ⊗ S(Rd), where S(Rd)is the Schwartz space of functions on Rd . It is easily seen thatD is symmetric on this domain. Taking the closure we obtain aself-adjoint operator which we also denote D.

Higher Dimensions 3

We then define the sign of D as sgn(D) = D ∣D ∣−1,D = ∑k γk ⊗Dk , ∣D ∣ = 1⊗ (∑k D

2k)1/2. Thus,

sgn(D) ∶=d

∑j=1

γj ⊗Dj√

D21 +D2

2 +⋯ +D2d

This is defined through the Borel functional calculus.

Specifically, the operator Dj/√

D21 +D2

2 +⋯ +D2d is the result

of applying the function x ↦ xj/∥x∥ to D. Consequentlysgn(D) extends to a bounded operator on CN ⊗ L2(Rd).Suppose f ∈ L∞(Rd). The operator 1⊗Mf is a bounded linearoperator on CN ⊗ L2(Rd), where 1 here is considered as theidentity operator on CN . Thus we can define,

d f ∶= i[sgn(D),1⊗Mf ].

Smoothness in higher dimensions 1

What is the relationship between smoothness of f ∈ L∞(Rd)and the size of d f ∶= i[sgn(D),1⊗Mf ]?In one dimension, necessary and sufficient conditions onf ∈ L∞(R) such that [sgn(−id/dx),Mf ] ∈ Lp,q wherep,q ∈ [0,∞] are provided by Peller.Janson and Wolfe (1982), Connes, Sullivan and Teleman (1994)have studied necessary and sufficient conditions for d f ∈ Lp,qwith p,q ∈ [0,∞] in the higher dimensional case, d > 1.

Theorem (Janson and Wolff)

if 0 < p ≤ d , we have that d f ∈ Lp if and only if f is constant.

For p > d , we have d f ∈ Lp if and only if f ∈ Bd/pp,p (Rd).

The case of p ≠ q with p,q ∈ [1,∞) was answered by Rochbergand Semmes in 1989. Necessary and sufficient conditions onf ∈ L∞(Rd) are given so that d f ∈ Lp,q with p ∈ [1,∞) andq ∈ [1,∞). These conditions are given in terms of the meanoscillation of f , and it is not obvious whether an equivalentcondition could be given in terms of more familiar functionspaces.

It is of interest in Connes’ quantised calculus to determineconditions on f such that d f ∈ Ld ,∞(CN ⊗ L2(Rd)).The asymptotic behaviour of the singular values of thequantised derivative denote the dimension of the infinitesimalin the quantised calculus. That the sequence of singular valuesbelongs to the weak space Ld ,∞ when the dimension of theEuclidean space is d indicates analogous behaviour betweenquantum derivatives and differential forms. Specifically, aproduct of d derivatives lies in the space L1,∞ which is the onlyweak space admitting a non-trivial trace that acts as theintegral.

1 What conditions on f ensure that d f ∈ Ld ,∞?

2 Is there some “classical limit” relating d f to ∇f ?

We can now answer these questions!

Commutators with order zero pseudodifferentialoperators

There is an interesting theorem in the Appendix of a 1994paper of Connes, Sullivan and Teleman:

Theorem (Connes, Sullivan and Teleman)

Let T be an order zero pseudodifferential operator that istranslation and dilation invariant, and let f be a function onRd , d > 1. Then the commutator [T ,Mf ] is in Ld ,∞ if andonly if f ∈ L1loc(Rd) and ∇f ∈ Ld(Rd ,Cd).

They provided a sketch proof. We reprove this result using newmethods.

Our results 1

When we write ∇f ∈ Ld(Rd ,Cd) we implicitly assume that theessentially bounded function f has odd weak partial derivativesand that the Bochner norm of ∇f in Ld(Rd ,Cd) given by

∥∇f ∥Ld(Rd ,Cd) = (∫Rd

∥∇f (x)∥dddx)1/d

=⎛⎝∫Rd

∑j=1

∣Dj f (x)∣ddx⎞⎠

is finite.

Our results 2

Let f ∈ L∞(Rd). Then we have:

Theorem

d f ∈ Ld ,∞ if and only if ∇f ∈ Ld(Rd ,Cd), and

∥∇f ∥Ld(Rd ,Cd) ≲ ∥d f ∥d ,∞ ≲ ∥∇f ∥Ld(Rd ,Cd).

Theorem

Let ∣d f ∣d have eigenvalues λ0 ≥ λ1 ≥ ⋯ ≥ 0. Then for a dilationinvariant extended limit ω,

ωn→∞ (λ0 +⋯ + λdlog(2 + n) ) = cd∥∇f ∥dLd(Rd ,Cd)

where cd > 0 depends on d .

How are these results obtained?

Our proofs use the techniques:

1 Double Operator Integrals

2 Pseudodifferential Operators

3 Singular traces

The plan of proof

The proof is done in three steps:

1 First we prove that ∥d f ∥d ,∞ ≤ Kd∥∇f ∥Ld(Rd ,Cd)

2 Then we obtain the formula for ϕ(∣d f ∣d)3 From the trace formula we finish the proof by obtaining

kd∥∇f ∥Ld(Rd ,Cd) ≤ ∥d f ∥d ,∞

The first step

Now we show how to prove that:

∥d f ∥d ,∞ ≤ Kd∥∇f ∥Ld(Rd ,Cd).

First under the assumption that f ∈ S(Rd), then extended tof ∈ L∞(Rd) and the right hand side being finite.

Method of proof

Let g(t) = t(1 + t2)−1/2. Then g(D) is a genuinepseudodifferential operator. First we show that,

∥[sgn(D) − g(D),1⊗Mf ]∥d ≤ K ′d∥f ∥d

using Cwikel estimates Then we obtain the bound

∥[g(D),1⊗Mf ]∥d ,∞ ≤ Kd∥∇f ∥Ld(Rd ,Cd)

using double operator integrals.

Using Cwikel estimates

For j = 1, . . . ,d and x = (x1, . . . , xd) ∈ Rd let

hj(x) =xj

∥x∥ −xj√

1 + ∥x∥2.

Then all we need to do is show that ∥hj(−i∇)Mf ∥d ≤ K ′d∥f ∥d .

From Cwikel estimates, this follows from hj ∈ Ld , which is easilychecked.

Double Operator Integrals

We also get the formula

[f (A),B] = T A,A

f [1]([A,B])

when [A,B] ∈ L2.Hence if we know that [A,B] ∈ Lp, then to conclude that

[f (A),B] ∈ Lp it is enough to show that T A,A

f [1]is bounded from

Lp to Lp.

Back to the proof

Our goal is to show that

∥[g(D),1⊗Mf ]∥d ,∞ ≤ Kd∥∇f ∥Ld(Rd ,Cd).

We start by considering the transfomer T D,Dg [1]

The function g

Recall that g(t) = t(1 + t2)−1/2. Then a whole lot of algebrashows that

g [1](λ,µ) = ψ1(λ,µ)ψ2(λ,µ)ψ3(λ,µ).

Where,

ψ1(λ,µ) ∶= 1 + 1 − λµ(1 + λ2)1/2(1 + µ2)1/2

ψ2(λ,µ) ∶=(1 + λ2)1/4(1 + µ2)1/4

(1 + λ2)1/2 + (1 + µ2)1/2

ψ3(λ,µ) ∶=1

(1 + λ2)1/4(1 + µ2)1/4.

Note that ψ1 and ψ3 are in L∞(R)⊗ L∞(R).

On ψ2

The following lemma is crucial:

T D,Dψ2is bounded from Ld ,∞ to Ld ,∞.

Hence we have,

[g(D),1⊗Mf ] = T D,Dg [1]([D,1⊗Mf ])

= T D,Dψ1T D,Dψ2

((1 +D2)−1/4[D,1⊗Mf ](1 +D2)−1/4).

The Bikchantaev Conjecture

We have the following result.

Theorem (Sukochev, 2013)

Let A > 0 be an operator on H, and B ∈ B(H). Then for everyθ ∈ (0,1), we have the submajorisation,

BθAB1−θ ≺≺ AB.

Hence,∥B1/2AB1/2∥d ,∞ ≲ ∥AB∥d ,∞.

Finishing the proof

[g(D),1⊗Mf ] ≺≺ [D,1⊗Mf ](1 +D2)−1/2.

Combining all our results so far, we get

∥d f ∥d ,∞ ≤ ∥d f − i[g(D),1⊗Mf ]∥d ,∞ + ∥i[g(D),1⊗Mf ]∥d ,∞≲ ∥f ∥d + ∥∇f ∥Ld(Rd ,Cd).

Then we can use a “dilation trick” to eliminate the dependenceon ∥f ∥d .

Expanding the class of f

Everything that we have done so far relies on the assumptionthat f ∈ S(Rd).To extend to f ∈ L∞(Rd), we use the fact that for allf ∈ L∞(Rd) such that ∇f ∈ Ld(Rd ,Cd) there exists a sequence{fn}∞n=0 ⊂ S(Rd) such that fn → f uniformly on bounded setsand ∥∇fn −∇f ∥Ld(Rd ,Cd) goes to 0.

Next Steps

1 Next we need to prove our “classical limit” trace formula,

ϕ(∣d f ∣d) = cd ∫Rd

∥∇f ∥d2 dx .

2 To then wrap everything up by proving that

∥∇f ∥Ld(Rd ,Cd) ≤ Cd∥d f ∥d ,∞.

The constant cd can be computed by following the argumentcarefully.

General Strategy of proof

Firstly assume that f ∈ C∞c (Rd) for now (we will remove this

assumption later). Furthermore assume that f is real valued.It is hard to compute ϕ(∣d f ∣d) directly: ∣d f ∣d is certainly not apseudodifferential operator in general.Instead we find a pseudodifferential operator A such that

d f ∈ A(1 +D2)−1/2 +L2d/3,∞.

Quite general computations then lead us to:

ϕ(∣d f ∣d) = ϕ(∣A∣d(1 +D2)−d/2)

which is a bit better.

Connes’ trace formula

Recall that if σ is the symbol of a pseudodifferential operatorTσ and σ has an asymptotic expansion,

σ(x , ξ) ∼−∞

∑j=−m

pj ,Tσ(x , ξ)

as ∥ξ∥→∞ and pj is homogeneous of order j in ξ, the functionp−m is called the principal symbol of Tσ, and Tσ is of order −m.If such an asymptotic expansion exists, then Tσ is calledclassical.Note that σ is matrix valued!

Properties of ΨDO’s

A pseudodifferential operator T on S(Rd ,Cd) is said to becompactly based if there is φ ∈ C∞

c (Rd) such that

(1⊗Mφ)T = T

and compactly supported if there is also ψ ∈ C∞c such that

(1⊗Mφ)T = T (1⊗Mψ) = T .

More properties of ΨDO’s

The principal symbol is multiplicative,

pn+m,AB = pn,Apm,B .

if A and B are of order n and m respectively. This has somesurprisingly strong consequences, such as:

Let T be a compactly supported pseudodifferential operator oforder −β. Then T ∈ Ld/β,∞.

Let φ ∈ C∞c (Rd) be such that T (1⊗Mφ) = T .

Then T (1 +D2)β/2 is bounded, so there is a bounded operatorA such that

T = A(1 +D2)−β/2.

Hence T = A(1 +D2)−β/2(1⊗Mφ) and so the result followsfrom Cwikel estimates.

A commutator lemma

To warm up, we prove the following:

Let A be a compactly based classical pseudodifferentialoperator with self-adjoint extension to L2(Rd ,Cd). Then for allβ ≥ 0 and 0 < α < d − 1,

[(1 +D2)−α/2, ∣A∣](1 +D2)−β/2 ∈ Ld/(α+β+1),∞.

First we let φ ∈ C∞c (Rd) and we prove that,

[(1 +D2)−α/2,1⊗Mφ](1 +D2)−β/2 ∈ Ld/(α+β+1),∞.

Pick ψ ∈ C∞c (Rd) such that φψ = φ. Then,

[(1 +D2)−α/2, (1⊗Mφ)(1⊗Mψ)](1 +D2)−β/2 =[(1 +D2)−α/2,1⊗Mφ](1⊗Mφ)(1 +D2)−β/2

+ (1⊗Mφ)[(1 +D2)−α/2,1⊗Mψ)](1 +D2)−β/2.

So the result follows from our previous lemma and Cwikelestimates.

Now let A be compactly based, and (1⊗Mψ)A = A. Then,

[(1 +D2)−α/2, (1⊗Mψ)A](1 +D2)−β/2 =[(1 +D2)−α/2,1⊗Mφ]A(1 +D2)−β/2

+ (1⊗Mφ)[(1 +D2)−α/2,A](1 +D2)−β/2.

So we get [(1 +D2)−α/2,A](1 +D2)−β/2 ∈ Ld/(α+β+1),∞.Then we get the final result from a DOI computation.

Connes’ trace formula

The following is a consequence of Connes’ trace theorem.

Theorem

Let T be a compactly based pseudodifferential operator oforder 0 on S(Rd ,Cd) with self-adjoint extension to L2(Rd ,Cd)and principal symbol p0,T . Then there is a constant kd > 0such that for every continuous normalised trace ϕ on L1,∞,

ϕ(T (1 +D2)−d/2) = kd ∫Rd∫Sd−1

tr(p0,T )(x , ξ)dξdx .

F. Sukochev,E. McDonald It turns out that A(1 +D2)−d/2 is a pseudodifferential operator

with principal symbol p0,A(x , ξ) 1∥ξ∥d2

Then we simply use the fact that any trace ϕ onL1,∞ ⊗MN(C) splits as a product ϕ = ϕ′ ⊗ tr.

Connes’ trace formula, cont.

Recall that we wanted to compute a trace likeϕ(∣A∣d(1 +D2)−d/2).Since principal symbols are multiplicative, we easily get that forevery polynomial h with h(0) = 0,

ϕ(h(A)(1 +D2)−d/2) = kd ∫Rd∫Sd−1

tr(h(p0,A(x , ξ)))dξdx .

The above is also true for continuous functions h, and thefollowing can be shown by approximating h uniformly withpolynomials:

If h is continuous on the spectrum of A, and h(0) = 0, then

ϕ(h(A)(1+D2)−d/2) = kd ∫Rd∫Sd−1

tr(h(p0,A(x , ξ)))(x , ξ)dξdx .

Finding a good candidate for A

Now what we need to do is find some good compactly basedself-adjoint pseudodifferential operator A of order 0 such that

d f ∈ A(1 +D2)−1/2 +L2d/3,∞.

Replacing sgn with g

g(D) is a pseudodifferential operator. It is much nicer to workwith i[g(D),1⊗Mf ] than d f .Fortunately, we have the following:

(g(D) − sgn(D))(1⊗Mf ) ∈ Ld/2

from a previous talk.So it’s enough to find A such that

i[g(D),1⊗Mf ] ∈ A(1 +D2)−1/2 +L2d/3,∞.

Defining A

We can in fact do better.Let φ ∈ C∞

c (Rd) be arbitrary. For k = 1,2, . . . ,d , define

Ak ∶=M∂k f −1

∑j=1

MφDjDk

1 −∆M∂j f +Mφ

1 −∆M∂j f

and then,

A ∶=d

∑k=1

γk ⊗Ak .

With the above definition of A

Theorem

i[g(D),1⊗Mf ] ∈ A(1 +D2)−1/2 +Ld/2,∞.

(Proof omitted, for length and for being uninteresting)

Taking the trace

Now we prove the key result of this section, which is actually alittle more general than needed.

If T is an operator on L2(Rd ,Cd) and A is a compactlysupported self-adjoint pseudodifferential operator of order 0with self-adjoint bounded extension and

T ∈ A(1 +D2)−1/2 +L2d/3,∞

thenϕ(∣T ∣d) = ϕ(∣A∣d(1 +D2)−d/2).

Let K = (1 +D2)−1/2.So we have T −AK ∈ L2d/3,∞. Hence, T ∗ −KA ∈ L2d/3,∞. Sowe get,

∣T ∣2 +K ∣A∣2K −KAT −T ∗AK ∈ Ld/3,∞.

∣T ∣2 −K ∣A∣2K ∈ KA(AK −T ) + (KA −T ∗)AK +Ld/3,∞.

So, since AK ∈ Ld ,∞,

∣T ∣2 −K ∣A∣2K ∈ L2d/5,∞.

Now we split the proof into d = 2 and d > 2.For d = 2, we are done since L4/5,∞ ⊂ L1.Now suppose d > 2.Now,

[Kα, ∣A∣]Kβ ∈ Ld/(α+β+1),∞.

So setting α = β = 1/2,

K ∣A∣ ∈ K 1/2∣A∣K 1/2 ∈ Ld/2,∞.

So,K ∣A∣2K ∈ (K 1/2∣A∣K 1/2)2 +Ld/3,∞.

So putting this all together,

∣T ∣2 ∈ (K 1/2∣A∣K 1/2)2 +L2d/5,∞.

So using the Birman-Koplienko-Solomyak formula,

∣T ∣ ∈ K 1/2∣A∣K 1/2 +L4d/5,∞.

So,∣T ∣ ∈ ∣A∣K +L4d/5,∞.

It then follows that,

∣T ∣d ∈ (∣A∣K)d +L1.

So far we have established that

ϕ(∣T ∣d) = ϕ((∣A∣K)d).

So we will be done if we can show that

(∣A∣K)d ∈ ∣A∣d−2Kd ∣A∣2 +L1.

This is shown by a fairly convoluted induction on j , that for all0 ≤ j ≤ d − 2,

(∣A∣K)d ∈ ∣A∣jK j(∣A∣K)d−j +L1.

Nearing the end

Now we can apply Connes’ trace theorem to get:

ϕ(∣d f ∣d) = ∫Rd∫Sd−1

tr(∣p0,A(x , ξ)∣d)dxdξ.

We can then compute p0,Ak:

p0,Ak(x , ξ) = ∂k f (x) −

∑j=1

φ(x)∂j f (x)ξkξj

∥ξ∥2

And so in fact:

tr(∣p0,A(x , ξ)∣d) = ∥∇f (x) − ξ(ξ,∇f )∥ξ∥2 ∥d

So then we get:

ϕ(∣d f ∣d) = kd ∫Rd∫Sd−1

∥∇f (x)∥d ∥ ∇f∥∇f ∥2

− ξ (ξ, ∇f∥∇f ∥2

dxdξ.

Since the inner integral is rotationally invariant,

ϕ(∣d f ∣d) = kd ∫Rd

∥∇f (x)∥d2 dx ∫Sd−1

∥e1 − ξξ1∥d2 dξ.

So set cd = kd ∫Sd−1 ∥e1 − ξξ1∥d2 dξ and we are done.

We’re not really done

So far we have proved that

ϕ(∣d f ∣d) = cd ∫Rd

∥∇f (x)∥d2 dx

under the assumption that f ∈ C∞c (Rd). We can do better.

Broadening the class of possible f

Now we can prove the trace formula for f ∈ L∞(Rd) such that∇f ∈ Ld(Rd ,Cd).To do this: note that we can choose {fn}∞n=0 ⊂ C∞

c (Rd) suchthat fn → f uniformly on compact sets.Hence, d fn converges to d f in the strong operator topology,and {d fn}∞n=0 is Cauchy in the Ld ,∞ quasinorm.

Finishing things off

So we have proved that ϕ(∣d f ∣d) = cd ∫Rd ∥∇f ∥d2 dx whenf ∈ L∞(Rd) and ∇f ∈ Ld(Rd ,Cd).To be really thoroughly finished, we should show that theequality holds when either side of the equation is defined. Thuswe wish to prove that if f ∈ L∞(Rd) is such that d f ∈ Ld ,∞,then ∇f ∈ Ld(Rd ,Cd).Specifically, we will prove that

∥∇f ∥Ld(Rd ,Cd) ≤ Cd∥d f ∥d ,∞

For some constant Cd > 0.

Using the trace formula

Note that we already have:

∥∇f ∥Ld(Rd ,Cd) ≤ Cd∥d f ∥d ,∞

for f a priori such that ∇f ∈ Ld(Rd ,Cd).

Using Banach-Alaoglu

We use the following common sort of argument in the theory ofdistributions:

Let f ∈ S ′(Rd) be a tempered distribution, and let {φj}dj=1 bean approximate identity of C∞

c function, and ψn(t) = ψ(t/n) isa sequence of cut-off functions. Let 1 < p <∞. If,

supn≥0

∥∇((φn ∗ f )ψn))∥p <∞

then ∇f ∈ Lp(Rd ,Cd).

F. Sukochev,E. McDonald Basically:

The sequence ∇(φn ∗ f )ψn is bounded in Lp, which is reflexivefor 1 < p <∞. So by the Banach-Alaoglu theorem there is aweakly converging subsequence to some element of Lp. Generaldistributional theory implies then that the limit point is ∇f .

With this in mind, let {φn}∞n=0 and {ψn}∞n=0 be as above.If f ∈ L∞(Rd), then (f ∗ φn)ψn ∈ C∞

c (Rd).So we have:

∥∇((f ∗ φn)ψn)∥Ld(Rd ,Cd) ≤ Cd∥d((f ∗ φn)ψn)∥d ,∞.

Straightforward computations show that the right hand side isbounded, and so we are (essentially done).

Future Prospects

Conjecture

If (A,H,D) is a p-summable spectral triple and a ∈ A, and ϕ isa continuous normalised trace,

ϕ(∣[sgn(D), a]∣p) = cϕ(∣[D, a]∣p(1 +D2)−p/2). (1)

The End.

Thank you for listening!

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