Trace inequality – Wikipedia

before-content-x4

In mathematics, there are many kinds of inequalities involving matrices and linear operators on Hilbert spaces. This article covers some important operator inequalities connected with traces of matrices.[1][2][3][4]

after-content-x4

Basic definitions[edit]

Let

Hn{displaystyle mathbf {H} _{n}}

denote the space of Hermitian

n×n{displaystyle ntimes n}

matrices,

Hn+{displaystyle mathbf {H} _{n}^{+}}

denote the set consisting of positive semi-definite

n×n{displaystyle ntimes n}

Hermitian matrices and

after-content-x4
Hn++{displaystyle mathbf {H} _{n}^{++}}

denote the set of positive definite Hermitian matrices. For operators on an infinite dimensional Hilbert space we require that they be trace class and self-adjoint, in which case similar definitions apply, but we discuss only matrices, for simplicity.

For any real-valued function

f{displaystyle f}

on an interval

IR,{displaystyle Isubseteq mathbb {R} ,}

one may define a matrix function

f(A){displaystyle f(A)}

for any operator

AHn{displaystyle Ain mathbf {H} _{n}}

with eigenvalues

λ{displaystyle lambda }

in

I{displaystyle I}

by defining it on the eigenvalues and corresponding projectors

P{displaystyle P}

as

given the spectral decomposition

A=jλjPj.{displaystyle A=sum _{j}lambda _{j}P_{j}.}

Operator monotone[edit]

A function

f:IR{displaystyle f:Ito mathbb {R} }

defined on an interval

IR{displaystyle Isubseteq mathbb {R} }

is said to be operator monotone if for all

n,{displaystyle n,}

and all

A,BHn{displaystyle A,Bin mathbf {H} _{n}}

with eigenvalues in

I,{displaystyle I,}

the following holds,

where the inequality

AB{displaystyle Ageq B}

means that the operator

AB0{displaystyle A-Bgeq 0}

is positive semi-definite. One may check that

f(A)=A2{displaystyle f(A)=A^{2}}

is, in fact, not operator monotone!

Operator convex[edit]

A function

f:IR{displaystyle f:Ito mathbb {R} }

is said to be operator convex if for all

n{displaystyle n}

and all

A,BHn{displaystyle A,Bin mathbf {H} _{n}}

with eigenvalues in

I,{displaystyle I,}

and

0<λ<1{displaystyle 0

, the following holds

Note that the operator

λA+(1λ)B{displaystyle lambda A+(1-lambda )B}

has eigenvalues in

I,{displaystyle I,}

since

A{displaystyle A}

and

B{displaystyle B}

have eigenvalues in

I.{displaystyle I.}

A function

f{displaystyle f}

is operator concave if

f{displaystyle -f}

is operator convex;=, that is, the inequality above for

f{displaystyle f}

is reversed.

Joint convexity[edit]

A function

g:I×JR,{displaystyle g:Itimes Jto mathbb {R} ,}

defined on intervals

I,JR{displaystyle I,Jsubseteq mathbb {R} }

is said to be jointly convex if for all

n{displaystyle n}

and all

A1,A2Hn{displaystyle A_{1},A_{2}in mathbf {H} _{n}}

with eigenvalues in

I{displaystyle I}

and all

B1,B2Hn{displaystyle B_{1},B_{2}in mathbf {H} _{n}}

with eigenvalues in

J,{displaystyle J,}

and any

0λ1{displaystyle 0leq lambda leq 1}

the following holds

A function

g{displaystyle g}

is jointly concave if −

g{displaystyle g}

is jointly convex, i.e. the inequality above for

g{displaystyle g}

is reversed.

Trace function[edit]

Given a function

f:RR,{displaystyle f:mathbb {R} to mathbb {R} ,}

the associated trace function on

Hn{displaystyle mathbf {H} _{n}}

is given by

where

A{displaystyle A}

has eigenvalues

λ{displaystyle lambda }

and

Tr{displaystyle operatorname {Tr} }

stands for a trace of the operator.

Convexity and monotonicity of the trace function[edit]

Let f: ℝ → ℝ be continuous, and let n be any integer. Then, if

tf(t){displaystyle tmapsto f(t)}

is monotone increasing, so
is

ATrf(A){displaystyle Amapsto operatorname {Tr} f(A)}

on Hn.

Likewise, if

tf(t){displaystyle tmapsto f(t)}

is convex, so is

ATrf(A){displaystyle Amapsto operatorname {Tr} f(A)}

on Hn, and
it is strictly convex if f is strictly convex.

See proof and discussion in,[1] for example.

Löwner–Heinz theorem[edit]

For

1p0{displaystyle -1leq pleq 0}

, the function

f(t)=tp{displaystyle f(t)=-t^{p}}

is operator monotone and operator concave.

For

0p1{displaystyle 0leq pleq 1}

, the function

f(t)=tp{displaystyle f(t)=t^{p}}

is operator monotone and operator concave.

For

1p2{displaystyle 1leq pleq 2}

, the function

f(t)=tp{displaystyle f(t)=t^{p}}

is operator convex. Furthermore,

The original proof of this theorem is due to K. Löwner who gave a necessary and sufficient condition for f to be operator monotone.[5] An elementary proof of the theorem is discussed in [1] and a more general version of it in.[6]

Klein’s inequality[edit]

For all Hermitian n×n matrices A and B and all differentiable convex functions
f: ℝ → ℝ with derivative f ‘ , or for all positive-definite Hermitian n×n matrices A and B, and all differentiable
convex functions f:(0,∞) → ℝ, the following inequality holds,

Tr[f(A)f(B)(AB)f(B)]0 .{displaystyle operatorname {Tr} [f(A)-f(B)-(A-B)f'(B)]geq 0~.}

In either case, if f is strictly convex, equality holds if and only if A = B.
A popular choice in applications is f(t) = t log t, see below.

Proof[edit]

Let

C=AB{displaystyle C=A-B}

so that, for

t(0,1){displaystyle tin (0,1)}

,

varies from

B{displaystyle B}

to

A{displaystyle A}

.

Define

By convexity and monotonicity of trace functions,

F(t){displaystyle F(t)}

is convex, and so for all

t(0,1){displaystyle tin (0,1)}

,

which is,

and, in fact, the right hand side is monotone decreasing in

t{displaystyle t}

.

Taking the limit

t0{displaystyle tto 0}

yields,

which with rearrangement and substitution is Klein’s inequality:

Note that if

f(t){displaystyle f(t)}

is strictly convex and

C0{displaystyle Cneq 0}

, then

F(t){displaystyle F(t)}

is strictly convex. The final assertion follows from this and the fact that

F(t)F(0)t{displaystyle {tfrac {F(t)-F(0)}{t}}}

is monotone decreasing in

t{displaystyle t}

.

Golden–Thompson inequality[edit]

In 1965, S. Golden [7] and C.J. Thompson [8] independently discovered that

For any matrices

A,BHn{displaystyle A,Bin mathbf {H} _{n}}

,

This inequality can be generalized for three operators:[9] for non-negative operators

A,B,CHn+{displaystyle A,B,Cin mathbf {H} _{n}^{+}}

,

Peierls–Bogoliubov inequality[edit]

Let

R,FHn{displaystyle R,Fin mathbf {H} _{n}}

be such that Tr eR = 1.
Defining g = Tr FeR, we have

The proof of this inequality follows from the above combined with Klein’s inequality. Take f(x) = exp(x), A=R + F, and B = R + gI.[10]

Gibbs variational principle[edit]

Let

H{displaystyle H}

be a self-adjoint operator such that

eH{displaystyle e^{-H}}

is trace class. Then for any

γ0{displaystyle gamma geq 0}

with

Trγ=1,{displaystyle operatorname {Tr} gamma =1,}

with equality if and only if

γ=exp(H)/Trexp(H).{displaystyle gamma =exp(-H)/operatorname {Tr} exp(-H).}

Lieb’s concavity theorem[edit]

The following theorem was proved by E. H. Lieb in.[9] It proves and generalizes a conjecture of E. P. Wigner, M. M. Yanase and F. J. Dyson.[11] Six years later other proofs were given by T. Ando [12] and B. Simon,[3] and several more have been given since then.

For all

m×n{displaystyle mtimes n}

matrices

K{displaystyle K}

, and all

q{displaystyle q}

and

r{displaystyle r}

such that

0q1{displaystyle 0leq qleq 1}

and

0r1{displaystyle 0leq rleq 1}

, with

q+r1{displaystyle q+rleq 1}

the real valued map on

Hm+×Hn+{displaystyle mathbf {H} _{m}^{+}times mathbf {H} _{n}^{+}}

given by

  • is jointly concave in
  • is convex in

Here

K{displaystyle K^{*}}

stands for the adjoint operator of

K.{displaystyle K.}

Lieb’s theorem[edit]

For a fixed Hermitian matrix

LHn{displaystyle Lin mathbf {H} _{n}}

, the function

is concave on

Hn++{displaystyle mathbf {H} _{n}^{++}}

.

The theorem and proof are due to E. H. Lieb,[9] Thm 6, where he obtains this theorem as a corollary of Lieb’s concavity Theorem.
The most direct proof is due to H. Epstein;[13] see M.B. Ruskai papers,[14][15] for a review of this argument.

Ando’s convexity theorem[edit]

T. Ando’s proof [12] of Lieb’s concavity theorem led to the following significant complement to it:

For all

m×n{displaystyle mtimes n}

matrices

K{displaystyle K}

, and all

1q2{displaystyle 1leq qleq 2}

and

0r1{displaystyle 0leq rleq 1}

with

qr1{displaystyle q-rgeq 1}

, the real valued map on

Hm++×Hn++{displaystyle mathbf {H} _{m}^{++}times mathbf {H} _{n}^{++}}

given by

is convex.

Joint convexity of relative entropy[edit]

For two operators

A,BHn++{displaystyle A,Bin mathbf {H} _{n}^{++}}

define the following map

For density matrices

ρ{displaystyle rho }

and

σ{displaystyle sigma }

, the map

R(ρσ)=S(ρσ){displaystyle R(rho parallel sigma )=S(rho parallel sigma )}

is the Umegaki’s quantum relative entropy.

Note that the non-negativity of

R(AB){displaystyle R(Aparallel B)}

follows from Klein’s inequality with

f(t)=tlogt{displaystyle f(t)=tlog t}

.

Statement[edit]

The map

R(AB):Hn++×Hn++R{displaystyle R(Aparallel B):mathbf {H} _{n}^{++}times mathbf {H} _{n}^{++}rightarrow mathbf {R} }

is jointly convex.

Proof[edit]

For all

0<p<1{displaystyle 0

,

(A,B)Tr(B1pAp){displaystyle (A,B)mapsto operatorname {Tr} (B^{1-p}A^{p})}

is jointly concave, by Lieb’s concavity theorem, and thus

is convex. But

and convexity is preserved in the limit.

The proof is due to G. Lindblad.[16]

Jensen’s operator and trace inequalities[edit]

The operator version of Jensen’s inequality is due to C. Davis.[17]

A continuous, real function

f{displaystyle f}

on an interval

I{displaystyle I}

satisfies Jensen’s Operator Inequality if the following holds

for operators

{Ak}k{displaystyle {A_{k}}_{k}}

with

kAkAk=1{displaystyle sum _{k}A_{k}^{*}A_{k}=1}

and for self-adjoint operators

{Xk}k{displaystyle {X_{k}}_{k}}

with spectrum on

I{displaystyle I}

.

See,[17][18] for the proof of the following two theorems.

Jensen’s trace inequality[edit]

Let f be a continuous function defined on an interval I and let m and n be natural numbers. If f is convex, we then have the inequality

for all (X1, … , Xn) self-adjoint m × m matrices with spectra contained in I and
all (A1, … , An) of m × m matrices with

Conversely, if the above inequality is satisfied for some n and m, where n > 1, then f is convex.

Jensen’s operator inequality[edit]

For a continuous function

f{displaystyle f}

defined on an interval

I{displaystyle I}

the following conditions are equivalent:

  • For each natural number

for all

(X1,,Xn){displaystyle (X_{1},ldots ,X_{n})}

bounded, self-adjoint operators on an arbitrary Hilbert space

H{displaystyle {mathcal {H}}}

with
spectra contained in

I{displaystyle I}

and all

(A1,,An){displaystyle (A_{1},ldots ,A_{n})}

on

H{displaystyle {mathcal {H}}}

with

k=1nAkAk=1.{displaystyle sum _{k=1}^{n}A_{k}^{*}A_{k}=1.}

every self-adjoint operator

X{displaystyle X}

with spectrum in

I{displaystyle I}

.

Araki–Lieb–Thirring inequality[edit]

E. H. Lieb and W. E. Thirring proved the following inequality in [19] 1976: For any

A0,{displaystyle Ageq 0,}

B0{displaystyle Bgeq 0}

and

r1,{displaystyle rgeq 1,}

In 1990 [20] H. Araki generalized the above inequality to the following one: For any

A0,{displaystyle Ageq 0,}

B0{displaystyle Bgeq 0}

and

q0,{displaystyle qgeq 0,}

for

r1,{displaystyle rgeq 1,}

and

for

0r1.{displaystyle 0leq rleq 1.}

There are several other inequalities close to the Lieb–Thirring inequality, such as the following:[21] for any

A0,{displaystyle Ageq 0,}

B0{displaystyle Bgeq 0}

and

α[0,1],{displaystyle alpha in [0,1],}

and even more generally:[22] for any

A0,{displaystyle Ageq 0,}

B0,{displaystyle Bgeq 0,}

r1/2{displaystyle rgeq 1/2}

and

c0,{displaystyle cgeq 0,}

The above inequality generalizes the previous one, as can be seen by exchanging

A{displaystyle A}

by

B2{displaystyle B^{2}}

and

B{displaystyle B}

by

A(1α)/2{displaystyle A^{(1-alpha )/2}}

with

α=2c/(2c+2){displaystyle alpha =2c/(2c+2)}

and using the cyclicity of the trace, leading to

Effros’s theorem and its extension[edit]

E. Effros in [23] proved the following theorem.

If

f(x){displaystyle f(x)}

is an operator convex function, and

L{displaystyle L}

and

R{displaystyle R}

are commuting bounded linear operators, i.e. the commutator

[L,R]=LRRL=0{displaystyle [L,R]=LR-RL=0}

, the perspective

is jointly convex, i.e. if

L=λL1+(1λ)L2{displaystyle L=lambda L_{1}+(1-lambda )L_{2}}

and

R=λR1+(1λ)R2{displaystyle R=lambda R_{1}+(1-lambda )R_{2}}

with

[Li,Ri]=0{displaystyle [L_{i},R_{i}]=0}

(i=1,2),

0λ1{displaystyle 0leq lambda leq 1}

,

Ebadian et al. later extended the inequality to the case where

L{displaystyle L}

and

R{displaystyle R}

do not commute .[24]

Von Neumann’s trace inequality and related results[edit]

Von Neumann’s trace inequality, named after its originator John von Neumann, states that for any
after-content-x4