m. junge and q. xu arxiv:0709.0433v1 [math.fa] 4 sep 2007 · 2020-01-23 · in our next step we...

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COUNTEREXAMPLES FOR THE CONVEXITY OF CERTAIN

MATRICIAL INEQUALITIES

M. JUNGE∗ AND Q. XU

Abstract. In [CL99] Carlen and Lieb considered Minkowski type inequalities in the

context of operators on a Hilbert space. More precisely, they considered the homogenous

expression

fpq(x1, ..., xn) =(tr((

n∑

k=1

xqk)

p/q))1/p

defined for positive matrices. The concavity for q = 1 and p < 1 yields strong subaddi-

tivity for quantum entropy. We discuss the convexity of fpq and show that, contrary to

the commutative case, there exists a q0 > 1 such that f1q is not convex for all 1 < q < q0.

This is achieved by constructing a family of interesting channels on 2× 2 matrices.

1. Introduction

The theory of convexity and concavity of matrix valued functions is is full of surprises.

For examples, the notion of convexity with matrix valued coefficients instead of scalars

imposes very strong conditions on the underlying function. We refer to the nice book of

Bhatia [Bha97] for a precise statement and more information. In many applications it

suffices however to prove convexity after composition with the trace. Holder’s inequality

for Schatten p-classes and Kosaki’s Wigner-Yanase-Dyson-Lieb provide good examples. In

their proof of the strong subadditivity for quantum entropy (SSA), Carlen and Lieb intro-

duced a noncommutative analogue for mixed ℓp(ℓq) norms by introducing the expression

fpq(x1, ..., xn) = [tr((

n∑

k=1

xqk)p/q)]1/p

defined for positive matrices in Mm. In their proof of the SSA they used the concavity

of the function f1,q (or more precisely Fpq(x) = tr((id ⊗ tr(xq))1/q)). Despite some effort

however, the basic question whether fpq is convex on the cone of sequence of positive

matrices remained open. Here is the information which has been known so far:

∗ The author is partially supported by the National Science Foundation Foundation DMS 05-56120.1

http://arxiv.org/abs/0709.0433v1

2 M. JUNGE AND Q. XU

• fp2 is convex for all 1 ≤ p ≤ 2.

• Carlen/Lieb [CL99] proved that fpq is not convex for q > 2.

• Hiai [Hia01] provided concrete counterexamples for q > 2, showing that

fq,φ(x) = (φ(xq))1/q .

is no longer convex for arbitrary states as long as q > 2.

• The question whether fpq is convex for 1 ≤ p, q ≤ 2 remained open.

In the context of operator spaces Pisier introduced vector-valued Lp spaces. These spaces

provide an alternative noncommutative generalization for the commutative spaces ℓp(ℓq).

With the help of these norms one can also show the SSA (see [DJKR06]), now using the

convexity of L1(Lp) for p > 1:

Theorem 1.1 (Pisier). Let 1 ≤ p ≤ q, r ≤ ∞ such that 1p= 1

q+ 1

r. The expression

‖x‖Lp(Lq) = infx=(a⊗1)y(b⊗1)

‖a‖L2r(Mn)‖y‖Lq(Mnm)‖b‖L2r(Mn)

is a norm on Mnm.

We may restrict Pisier’s definition to block matrices x =

x1 0 · · · 0

0 x2 · · · 0

. . .

0 0 · · · xn

. For

commuting matrices xk we find exactly the expression fpq. Pisier also provides a formula

for the dual norm Lp(Lq)∗ = Lp′(Lq′) such that

‖x‖Lp(L1) = ‖(id⊗ tr)(x)‖pholds for all positive matrices x ∈ Mnm. In view of these competing expression for a non-

commutative version of Lp(Lq) it seems important to decide Carlen/Lieb’s problem and

determine the convexity of fpq. Our first result makes efficient use of Hiai’s counterexam-

ples.

Theorem 1.2. Let 1 ≤ p, q <∞ and ‖ ‖pq be a norm on the space Mnm,sa of sequences of

selfadjoint matrices such that

c−1pq ‖(x1, ..., xn)‖pq ≤ fpq(x1, ..., xn) ≤ cpq‖(x1, ..., xn)‖pq

for constants c(p, q) independent of m and n. Then q = 2.

A crucial tool in this theorem is the use of the central limit theorem. This allows us to

replace the summation sign∑

by more general general expressions. Indeed, we show that

COUNTEREXAMPLES FOR CONVEXITY 3

if fp,q is convex and p 6= q, then

Fpq(x) = [tr(Φ(xq)p/q)]1/p

is convex for every completely positive map Φ between matrix algebras. We have some

positive results in this direction.

Theorem 1.3. Let Φ : A → B be a completely positive map. The map Fp,q is convex

under the following assumptions.

i) 1 ≤ q ≤ p ≤ ∞;

ii) A or B is commutative.

Part ii) might be interesting for operator-valued measures. The conditions ii) seem to be

too restrictive. However, it may happen that commutative results have no noncommutative

analogue:

Theorem 1.4. There exists q0 > 1 such that for every 1 < q < q0 there exists a completely

positive map Φ : M2 → M2 such that F1,q is not convex.

In order to solve Carlen/Lieb’s question in the negative we use the following transference

principle.

Lemma 1.5. Assume that there exists a constant cpq such that fpq is cpq equivalent to

a convex function. Then Fpq is convex for every completely positive map between matrix

algebras.

Apart from Stinespring’s dilation theorem we use the central limit procedure and a

tensor trick in this argument. The drawback of our negative to solution of Carlen/Lieb’s

conjecture is the fact that we use matrices of humongous size.

Corollary 1.6. There exists an interval (1, q0) such that f1,q is not convex.

The paper is organized as follows. We first discuss the ingredients for the proof of

Lemma 1.5 in section 1). In section 2) we recall the basic facts about differentiation of

matrix valued functions based on beautiful formulae of Birman/Solomjak. In section 3)

we construct the family of counterexamples. It is possible that these techniques can be

implemented numerically and provide further counterexamples. We did not pursue this

direction of research. For basic notation on matrix-valued functions we refer to [Bha97].

Further information on vector-valued Lp spaces can be found in [Pis98, Pis03] but that is

not needed here. We refer to Paulsen’s book [Pau02] for standard notation and facts on

completely positive maps.


2. From sums to channels

In the following section we want to study the relation between the interpolation norms

Lp(Lq) and the mixed Lp norms defined by partial traces. We will discuss several variants

of these norms. The simplest version of mixed Lp norms is given by

‖(xi)‖Lp(lq) = ‖(∑

i

xqi )1

q ‖p .

This is of course a special case of an Lq-norm with respect to a partial trace for x ∈ N⊗Mn

‖x‖Lp(Lq) = ‖(id⊗ tr(xq))1

q ‖p .In the literature one also finds tr2 = id⊗ tr. Writing the trace as an average it is shown in

[CL99, Theorem 4] that the convexity of the first expression implies the second. We may

go further and replace the trace by a state and find

‖x‖Lp(Lq(id⊗φ)) = ‖(id⊗ φ(xq))1

q ‖p .Clearly id⊗ φ is a conditional expectation. For conditional expectation we define

‖x‖Lp(Lq(E)) = ‖E(xq) 1

q ‖p .The most general case however is given by a (completely) positive map Φ : N →M and

‖x‖Lp(Lq(Φ)) = ‖Φ(xq) 1

q ‖p .

Lemma 2.1. Let Φ : Mn → Mn be a completely positive map. Then Lp(Lq(Φ)) embeds

isometrically into Lp(Lq(id⊗ φ)) for a suitable φ.

Proof. We may assume that Φ(1) ≤ 1. By Stinespring’s dilation we find V : ℓn2 → ℓnm2such that

Φ(x) = V ∗(1m ⊗ x)V .

Note that

‖V ‖ = ‖V ∗V ‖1/2 = ‖Φ(1)‖implies that V is a contraction. There is a standard trick to dilate V to a unitary. Indeed,

let V = ur be the polar decomposition. Here we use the standard inclusion ℓn2 ⊂ ℓnm2 and

consider V as a map on ℓnm2 . Then

U =

(u 0

0 1

)(r −

√1− r2√

1− r2 r

)=

(ur −u

√1− r2√

1− r2 r

)

is a unitary such that

Φ(x) = (e11 ⊗ 1nm)U∗

(x⊗ 1m 0

0 0

)U(e11 ⊗ 1nm) .


In particular, we find

Φ(xq) = (e11 ⊗ 1nm)U∗(e11 ⊗ 1m ⊗ xq)U(e11 ⊗ 1nm)

= (e11 ⊗ 1nm)(U∗(e11 ⊗ 1m ⊗ x)U)q(e11 ⊗ 1nm)

This allows us to define the (non-faithful state) φ(y) = y11 and we obtain

Φ(xq) = (φ⊗ 1m)((U∗(e11 ⊗ 1m ⊗ x)U)q) .

Therefore the embedding is given by the (non-unital) ∗-homomorphism π : Mn → M2nm,

π(x) = (U∗(e11 ⊗ 1m ⊗ x)U).

Remark 2.2. 1) For a reader more familiar with Kraus operators

Φ(x) =

m∑

j=1

R∗jxRj

we have V = (R1 . . . Rm)t and the unitary U is constructed as above.

2) Let Φ : N →M be a normal completely positive contraction such thatN∗ is separable.

Then we may also assume that M∗ is separable. The Kasparov dilation theorem yields

Φ(x) = e11π(x)e11

where π : N → B(ℓ2)⊗M is a normal (in general not unital) representation (see [Jun02]

or [Rua04] for details, the argument are closely related to Paschke’s work [Pas73]). Thus

Lemma 2.1 works under the additional separability assumption, even in full generality

using the theory of W ∗-Hilbert modules (see [Rua04]).

3) The state φ from Lemma 2.1 is non-faithful, but in the separable case can be perturbed

to a faithful stated so that the following result apply by approximation.

In our next step we will reduce problems for conditional expectations of the form id⊗φ

to sums. This argument is a simple application of the central limit theorem. We will

assume for a moment that N is a semifinite von Neumann algebra and φ : N → C is a

normal faithful state. For a given natural number n ∈ N, we consider N⊗n and the faithful

state φn = φ⊗n. Then we find ∗-homomorphisms πi : N → N⊗n defined by

πi(x) = 1⊗ · · · ⊗ 1⊗ x︸︷︷︸i-th position

⊗1⊗ · · · ⊗ 1 .

We will use the same notation for the amplification id⊗ πi :M ⊗N →M ⊗N⊗n .


Lemma 2.3. Let M be a finite von Neumann algebra with trace τ . Let Dφ be the density

of τ ⊗ φ on M ⊗N and Dn be the density of τ ⊗ φ⊗n. Then

limn

‖ 1n

n∑

i=1

D1/2pn πi(x)D

1/2pn ‖p = ‖id⊗ φ(x)‖Lp(M)

holds for all 0 < p <∞ and all x ∈M ⊗N .

Proof. Let us write E(x) = id⊗φ(x). We start with p = 2 and note that by orthogonality

‖n∑

i=1

D1/4n πi(x−E(x))D

1

4

n ‖22 =n∑

i=1

‖D1

2

φ (x− E(x))D1

2

φ ‖22 ≤ 4n‖x‖2∞ .

This implies for 0 < p ≤ 2 that

‖D1/2pn E(x)D1/2p

n − 1

n

n∑

i=1

D1/2pn πi(x)D

1/2pn ‖p = ‖ 1

n

n∑

i=1

D1/2pn πi(x− E(x))D1/2p

n ‖p

≤ ‖ 1n

n∑

i=1

D1/4n πi(x−E(x))D1/4

n ‖2 ≤ 2n−1/2‖x‖∞ .

Here we used that τ ⊗φ(1) = 1 with the corresponding contractive inclusion of Lp spaces.

Note that

‖D1/2pn E(x)D1/2p

n ‖p = ‖D1/2pφ E(x)D

1/2pφ ‖Lp(M⊗N) = ‖id⊗ φ(x)‖Lp(M)

and hence the assertion follows. For p > 2 we deduce from interpolation (see [Kos84]) that

‖D1/2pn (

1

n

n∑

i=1

πi(x−E(x)))D1/2pn ‖p

≤ ‖ 1n

n∑

i=1

πi(x− E(x))‖1−2

p∞ ‖D1/4

n (1

n

n∑

i=1

πi(x− E(x)))D1/4n ‖

2

p

2

≤ [2‖x‖∞]1−2

p [2n−1/2‖x‖∞]2

p .

Thus again we find 0 in the limit and the assertion follows.

Remark 2.4. More precise estimates on the difference D1/2pn ( 1

n

∑ni=1 πi(x − E(x)))D

1/2pn

are the subject of the Burkholder/Rosenthal inequalities in [JX03] and more precisely [JX].

Proposition 2.5. Let 0 < q, p <∞ and φ be a normal faithful state with density Dφ. Let

ψp,q(x) = tr(D1−q/pφ x)

Let M be a finite von Neumann algebra. The Lp(Lq(id⊗ψp,q)) is isometrically isomorphic

to a subcone of an ultraproduct∏

n,U Lp(lnq ).


Proof. Let us start with x =∑

j aj ⊗ bj ∈ Lp(M ⊗N) and consider

xi =∑

j

aj ⊗D1

p

φ ⊗ · · · ⊗ bj︸︷︷︸i-th position

⊗D1

p

φ ⊗ · · · ⊗D1

p

φ = πpi (x) .

For i = 1 we have simply

x1 = x⊗ [D1

p

φ ]⊗n−1

and hence

xq1 = xq ⊗ ([D1

p

φ ]⊗n−1)q = xq ⊗ [D

qp

φ ]⊗n−1 = D

q/2p

φ⊗n

((1⊗D

−q/2pφ xqD

−q/2pφ )

)Dq/2p

φ⊗n

= Dq/2pn π1(D

−q/2pφ xqD

−q/2pφ )Dq/2p

n .

Since shuffling is a homomorphism we find

xqi = Dq/2pn πi(D

−q/2pφ xqD

−q/2pφ )Dq/2p

n .

Therefore Lemma 2.3 implies

limn

‖( 1n

n∑

i=1

xqi )1/q‖qp = lim

n‖ 1n

n∑

i=1

xqi‖p/q

= limn

‖ 1n

n∑

i=1

Dq/2pn πi(D

−q/2pφ xqD

−q/2pφ )Dq/2p

n ‖p/q = ‖id⊗ φ(D−q/2pφ xqD

−q/2pφ )‖Lp/q(M) .

However, we have

id ⊗ φ(D−q/2pφ (a⊗ b)D

−q/2pφ ) = a⊗ tr(DφD

−q/2pφ bD

−q/2pφ )

= a⊗ tr(D1−q/pφ b) = ψp,q(a⊗ b) .

This implies

‖((id⊗ ψp,q)(xq))1/q‖p = lim

n‖( 1n

n∑

i=1

xqi )1/q‖p .

For any free ultrafilter U we find an isometric embedding.

The next argument allows us to ignore equivalence constants


Lemma 2.6. Let 0 < q, p <∞.

(1) Assume that for every completely positive map Φ : Mn → Mn the expression

‖x‖Lp(Lq(Φ)) is c(p, q) equivalent to a convex function. Then ‖x‖Lp(Lq(Φ)) is con-

vex for every Φ.

(2) Assume that Φ : Mn → Mk the expression ‖x‖Lp(Lq(Φ)) is c(p, q)-equivalent to the

restriction of a norm on the vector space of selfadjoint operators. Let Φ : Mn → Mk

be a completely positive map and y be such that Φ(yq)1/q = 0. Then

‖Φ((x+ ty)q)1/q‖p ≥ ‖Φ(xq)1/q‖pfor all t ≥ 0.

Proof. For the proof of i) we consider a channel Φ : Mn → Mn and positive elements x, y.

We will use the channel Φm = Φ⊗n : Mmn → M

mn which is still completely positive. Then

we consider

ξ = (x

2+y

2)⊗m = 2−m

∑

A⊂{1,...,m}

xA ⊗ yAc .

Here xA ⊗ yB = z1 ⊗ · · · ⊗ zm is such that zi = x for i ∈ A and zi = y for i ∈ Ac. By

assumption we have

‖Φm(ξq)1/q‖p ≤ c(p, q)2−m∑

A

‖Φm(xqA ⊗ yqAc)1

q ‖p

= c(p, q)2−m∑

A

‖Φ(xq)A ⊗ Φ(yq)Ac‖1/qp/q

= c(p, q)2−m∑

A

‖Φ(xq)1/q‖|A|p ‖Φ(yq)1/q‖|Ac|p

= c(p, q)(‖Φ(xq)1/q‖p + ‖Φ(yq)1/q‖p

2)m .

Clearly, we also have

‖Φm(ξq)1/q‖p = ‖Φ((x+ y

2)q)1/q‖mp .

Thus taking the m-th root we find

‖Φ((x+ y

2)q)1/q‖p ≤ c(p, q)1/m

‖Φ(xq)1/q‖p + ‖Φ(yq)1/q‖p2

.

Sending m to infinity we obtain the assertion. The proof of ii) is similar. Assume Φ(yq) =

0. We consider again Φ⊗m and assume that there is a norm such that

c−11 ‖x‖ ≤ ‖Φm(ξq)1/q‖p ≤ c2‖x‖


with c1c2 ≤ c(p, q). Then we note that

c2‖Φm(((x+ ty)⊗m)q)1/q‖p ≥ ‖(x+ ty)⊗m‖

= ‖x⊗m +∑

|Ac| ≥ 1

t|A|c

xA ⊗ yAc‖

≥ ‖x⊗m‖ −∑

|Ac| ≥ 1

t|A|c‖xA ⊗ yAc‖

≥ ‖x⊗m‖ − c1∑

|Ac| ≥ 1

t|A|c‖Φ(xq)1/q‖|A|p ‖Φ(yq)1/q‖|Ac|

p

= ‖x⊗m‖≥ c−1

1 ‖Φm(ξq)1/q‖p = c−11 ‖Φ(ξq)1/q‖mp .

This implies

‖Φ(x+ ty)q)1/q‖p = ‖Φ⊗m(((x+ ty)⊗m)q)‖1/mp ≥ (c2c1)− 1

m‖Φ(xq)1/q‖p .Thus for m→ ∞ the assertion follows.

Remark 2.7. The reader might object that we are working with non-faithul channels.

However, we may consider the perturbed faithful channel Φm,ε = Φ⊗m + εψ, where ψ is a

fixed faithful state. Let us assume that ‖ξ‖m,ε = ‖Φm(ξq)1/q‖p satisfiesc(p, q)−1‖ξ‖ε ≤ ‖ξ‖m,ε ≤ ‖ξ‖ε

for some norm ‖ξ‖ε. Performing the calculations for ‖ ‖m,ε and the passing to the limit

for ε→ 0 still provides us with

‖Φ(((x+ ty)⊗m)q)1/q‖p ≥ (c1c2)−1‖Φ((x⊗m)q)1/q‖p .

Thus taking the m-th root yields the assertion.

Lemma 2.8. Let 1 < q <∞ and

x(t) =

(12

12

12

12+ t

).

Then

x(t)q =

(1+( q

2−1)t

2

1+ q2t

21+ q

2t

2

1+(1+ q2)t

2

)+O(tmin{q,2}) .

Proof. Using the determinant we find

λj =1 + t

2+ εj

√414+ (1

2− (1

2+ t))2

2=

1 + t

2± 1 + t2

2=

1

2+ εj

1

2+t

2+O(t2)


where λ1/2 are the eigenvalues and ε1 = 1, ε2 = −1. Let x(t) = u(t)Dλ(t)u(t)∗ be such

that u(t) is a unitary with u11(t) > 0 and u22(t) > 0, uj = (u1j(t)

u2j(t)) is an eigenvector for

λj. Then it is easy to deduce from u2j1 + u2j2 = 1 and

uj12

+uj22

= λjuj1

that

u11 =1− t

2

2+O(t2) , u21 =

1 + t2

2+O(t2)

and

u12 = −1 + t2

2+O(t2) , u22 =

1− t2

2+O(t2) .

Then we observe that for q > 1 we have λq2 = ( t2+O(t2))q = O(tmin{q,2}). This implies

x(t) = u(t)Dλ(t)u(t)∗ = (1 +

t

2)qu(t)e11u(t)

∗

= (1 +t

2)q(

1−t2

12

12

1+t2

)+O(tmin{q,2}) .

Using (1 + t2)q = 1 + qt

2+O(t2), we deduce the assertion.

Theorem 2.9. Let 1 < q <∞, 1 ≤ p ≤ ∞ such that q 6= 2 and q 6= p.

a) The expression ‖ ‖Lp(lq) is not equivalent to a restriction of a norm.

b) If moreover, q > 2. Then ‖ ‖Lp(lq) is not equivalent to a convex function.

Proof. We consider the scalar-valued channel Φ : M2 → C ⊂ M2 defined by φ(x) = x11.

Then the expression

‖x‖q = φ(xq)1

q .

This corresponds to Lp(Lq(φ)) for a scalar valued completely positive map. We consider

x = 12

(1 1

1 1

)and y = e22. Clearly, we have φ(yq) = 0. For q > 2 we deduce from

Lemma 2.8 that

(2.1) 2−1

q (1 + (1

2− 1

q)t) ≤ φ((x+ ty)q)

1

q +O(t2)

Therefore ‖ ‖q is not convex. If it were, we would have

‖x+ ty‖q ≤ ‖x‖q + t‖y‖q = φ(xq)1

q + tφ(yq)1

q ≤ 2−1

q .(2.2)

This contradicts (2.1). Therefore it suffices to show that if ‖ ‖Lp(lq) is equivalent to a convex

function, then (2.2) holds. Now, we assume that for arbitrary matrices the expression


Lp(lq) is d(p, q) equivalent to a convex function. We will show that then for every faithful

state Ψ : Mk → C the expression Ψ(ξq)1/q is convex. Indeed, let a be the density of Ψ

such that tr(aξ) = Ψ(ξ). We define a positive functional state Φ0 : Mk → C by

Φ0(ξ) = tr(aq−pp ξ)

and Φ(ξ) = Φ0(1)−1Φ0(ξ). This implies DΦ = tr(a

q−pp )−1a

q−pp and

D1− q

p

Φ = tr(aq−pp )

qp−1a .

We apply Proposition 2.5 forM = C and find an embedding of Lp(Lq(ψp,q)) in∏

n,U Lp(lnq ).

By homogeneity we deduce that f(ξ) = Ψ(ξq)1/q is d(p, q) equivalent to a convex function

for every normal faithful state Ψ. Using Lemma 2.6 (and Remark 2.7), we deduce that

the original expression φ(xq)1/q is convex. This contradicts 2.1 and concludes the proof of

b). The proof of a) follows the same pattern. Assuming that Lp(lq) is d(p, q) equivalent

to the restriction of the norm. Then Proposition 2.5 implies that f(ξ) = tr(D1−p/qξq)1/q

is d(p, q) equivalent to a restriction of a norm for every faithful positive density D. Since

p 6= q this implies that f(ξ) = tr(aξq)1/q is d(p, q)-equivalent to a semi-norm for every (not

necessarily) faithful density a. According to Lemma 2.6 (and Remark 2.7) we deduce that

φ(yq) = 0 implies

(2.3) φ((x+ ty)q)1

q ≥ φ(xq)1

q .

However, for 1 < q < 2 we know from Lemma 2.8 that

φ((x+ ty)q)1

q ≤ φ(xq)1

q (1 + (1

2− 1

q)t) +O(t2) .

This contradicts (2.3) for t small enough.

For our following application we state the result for channels:

Proposition 2.10. Let q 6= p. If the expression ‖ ‖Lp(Mk,lnq ) is c-equivalent to a convex

function uniformly in k and n, then ‖ ‖Lp(Lq(Φ)) is convex for every completely positive map

Φ : Mk → Mk.

Proof. This follows the same pattern as the proof Theorem 2.9 a). We assume that

‖ ‖Lp(Mk ,lnq ) is c-equivalent to a convex function uniformly in k and n. According to Propo-

sition 2.5 this implies that ‖ ‖Lp(Lq(idMm⊗ψ)) is c-equivalent to a convex function for every

faithful state ψ. Here we use again the change of density argument

ψ(x) = tr(ax) = tr(aq−pp )1−

qp tr(D

1− qp

Φ x) , DΦ = tr(aq−pp )−1a

q−pp


from above. By perturbation, we deduce c-equivalence for every even non-faithful state.

Then Lemma 2.1 implies that ‖ ‖Lp(Lq(Φ)) is c-equivalent to convex function. By Lemma

2.6 we then obtain convexity.

Remark 2.11. 1) For q ≤ p and 1 ≤ q ≤ 2 the expression

‖x‖Lp(Lq(Φ))

is convex.

2) Counterexamples for q > 2 can also be found in [CL99] and [Hia03].

Proof of 1). We first note that

‖x‖Lp(Lq(Φ)) = ‖Φ(xq)‖1/qp/q = sup{tr(DΦ(xq)1/q) : ‖D‖(p/q)′ ≤ 1} .

Then φ(x) = tr(DΦ(x)) is a positive functional. It therefore suffices to show that the

homogenous function f(x) = φ(xq)1/q is convex. By homogeneity it suffices to show that

Bφ = {x ≥ 0 : φ(xq) ≤ 1} is convex. Let x1, x2 ∈ B. The function x 7→ xq is operator

convex for 1 ≤ q ≤ 2. Therefore

(λx1 + (1− λ)x2)q ≤ λxq1 + (1− λ)xq2 .

By positivity of φ this implies

φ((λx1 + (1− λ)x2)q) ≤ λφ(xq1) + (1− λ)φ(xq2) ≤ 1 .

Hence λx1 + (1− λ)x2 ∈ Bφ. The assertion follows.

3. Differentiation of matrix valued functions

In this section we will recall some basic facts about differentiation of matrix valued

functions and obtain the relevant formulae for the second derivative of

fq(x) = tr(Φ(xq)1/q) .

A good reference for the matrix case is the book by Bahtia, sufficient for our purpose. Let

us mention, however, that the formulas we are using go back to the work of Birman and

Solomjak. We fix a selfadjoint operator x with spectral decomposition x =∫RtdEt. Let

f : R → R be a differentiable function. Using the terminology of [BS73] we have

Df(x; y) =d

dtf(x+ ty)|t=0 =

∫

R2

f [1](s, t)dEsydEt


where

f [1](s, t) =

{f(s)−f(t)

s−ts 6= t

f ′(t) s = t.

Note that the right hand side converges for y in the Hilbert Schmidt operators, for more

information see also [Pel06]. A similar formula holds for the second derivative we use

f [2](s, t, r) =f [1](s, r)− f [1](t, r)

s− t=

f [1](s, t)− f [1](s, r)

t− r,

again defined as a limit for s = t. Then by [BS73] we have

D2f(x; y, z) = limt→0

Df(x+ tz; y)−Df(x; y)

t

=

∫

R3

f [2](s, t, r)dEsydEtzdEt +

∫

R3

f [2](s, t, r)dEszdEtydEt .(3.1)

All these formulas are easily checked for monomials. We are interested in the particular

function hq(x) = xq defined for selfadjoint x ∈ Mm. In the special case of a diagonal

operator x = Dλ we find

(3.2) Dhq(x; y) =∑

i

qλq−1i yiieii +

∑

i 6=j

λqi − λqjλi − λj

yijeij .

Here we assumed λi 6= λj for i 6= j.

Lemma 3.1. Let A ⊂ Mm be a ∗-subalgebra and Φ : A → Mm be a completely positive

map. Let fq(x) = tr(Φ(xq)1/q). Then

Dfq(x; y) = (1/q − 1) tr(Φ(xq)1/q−1Φ(Dhq(x; y))) .

Proof. We first note that

(x+ ty)q = xq + tDhq(x; y) +O(t2) .

This implies

Φ((x+ ty)q)1/q = [Φ(xq) + tΦ(Dhq(x; y)) +O(t2)]1/q

= Φ(xq)1/q + tDh1/q(Φ(xq); Φ(Dhq(x; y))) +O(t2) .

Let us assume for simplicity that Dµ = Φ(xq) is a diagonal operator. Then it is obvious

that

tr(Dh1/q(Dµ; Φ(Dhq(x; y))) =∑

i

(1/q − 1)µ1/q−1ii Φ(Dhq(x; y))ii

= (1/q − 1)tr(D1/q−1µ Φ(Dhq(x; y)) .

Using the tracial property, we obtain assertion in full generality.


For the second derivative we deduce from (3.1) the following formula (see also [Bha97])

(3.3) D2hq(x; y, y) = 2∑

i,j,k

h[2]q (λi, λj, λk)yijyjk eik .

Proposition 3.2. Let A ⊂ Mn and Φ : A → Mn be a (completely) positive map with

associated function fq(x) = tr(Φ(xq)1/q). Let x ∈ A with Φ(xq) > 0. Then

q

q − 1D2fq(x; y, y)

= tr

(Dh1/q−1

(Φ(xq); Φ(Dhq(x; y))

)Φ(Dhq(x; y))

)+ tr

(Φ(xq)1/q−1Φ(D2hq(x; y, y)

).

In particular, x > 0 and fq convex implies

tr

(Dh1/q−1


)Φ(Dhq(x; y))

)+ tr

(Φ(xq)1/q−1Φ(D2hq(x, y, y))

)≥ 0

for all selfadjoint y.

Proof. We have to find the derivative of

G(x) = Dhq(x; y) = (1/q − 1) tr(Φ(xq)1/q−1Φ(Dhq(x; y))

).

For fixed z we define

g1(t) = (1/q − 1) tr(Φ((x+ tz)q)1/q−1Φ(Dhq(x; y))

)

and

g2(t) = (1/q − 1) tr(Φ(xq)1/q−1Φ(Dhq(x+ tz; y))

).

According to the chain rule we have

DG(x; z) = g′1(0) + g′2(0) .

By linearity we have

g′2(0) = (1/q − 1)tr(Φ(xq)1/q−1Φ(D2hq(x; y, z)

).

For the derivative of g1 we follow the same procedure as in Lemma 3.1. Indeed, we have

Φ((x+ tz)q)1/q−1 = (Φ(xq) + tΦ(Dhq(x; z)) +O(t2))1/q−1

= Φ(xq)1/q−1 + tDh1/q−1(Φ(xq); Φ(Dhq(x; z))) +O(t2) .

This yields the desired formula. Assuming now that fq is convex and x > 0. Then x

is an interior point in the cone of selfadjoint matrices. More precisely, for every y there

exists a t0 > 0 such that x + ty ≥ 0 for all |t| ≤ t0. By convexity of fq, we deduce that

h(t) = fq(x+ ty) is convex and hence 0 ≤ h′′(t) = D2fq(x; y, y).


Theorem 3.3. Let A ⊂ Mn be a unital operator system and Φ : A → Mn be a positive

map and fq : A+ → R defined by fq(x) = tr(Φ(xq)1/q).

i) D2fq(1; y, y) ≥ 0 for all selfadjoint y.

ii) Let y ∈ A such that x and y commute. Then D2fq(x; y, y) ≥ 0.

Proof. For x = 1 the derivatives simplify significantly. First we note that by functional

calculus (i.e. by pointwise equality on the spectrum) we have

(1 + tx)q = 1 + tqx+ q(q − 1)t2

2x2 +O(t3) .

This yields Dhq(1; y) = qy and D2hq(1; y, y) = q(q − 1)y2. According to Proposition 3.2

the assertion i) amounts to showing that

(3.4) tr(Dh1/q−1(Φ(1); Φ(y))Φ(y)) + q(q − 1)tr(Φ(1)1/q−1(Φ(y2))) ≥ 0 .

We have to reformulate this inequality taking the special role of Φ(1) into account. By

approximation we may assume that Φ(1) > 0 and define the unital channel Φ(x) =

Φ(1)−1/2Φ(x)Φ(1)−1/2. We think of Φ(1) = Dµ and the simplified formula 3.2 for 1/q − 1

and then (3.4) becomes

(3.5) tr(Φ(1)Dh1/q−1(Φ(1); Φ(y))Φ(1)Φ(y)) + q(q − 1)tr(Φ(1)1/q(Φ(y2))) ≥ 0 .

Note that the case Φ = id is not excluded. We prefer to prove the inequality

(3.6) − tr(ξDh1/q−1(ξ; z)ξz) ≤ q − 1

qtr(ξ1/qz2)

for all selfadjoint z and positive ξ. Of course there is no loss of generality to assume that

ξ = Dµ is a diagonal operator. Let us write z = Dγ + w where w is the off-diagonal part

of z. Note that Dh1/q−1(Dµ;Dγ) = (1/q − 1)D1/q−2µ Dγ is still a diagonal operator. Thus

we get

tr(ξDh1/q−1(ξ; z)ξz) = (1/q − 1)tr(ξ2ξ1/q−1D2γ) + tr(ξDh1/q−1(ξ;w)ξw)

= −q − 1

qtr(D1/q

µ D2γ) + tr(DµDh1/q−1(ξ;w)Dµw) .

On the other hand we have

D1/qµ z2 = D1/q

µ (D2γ +Dγw + wDγ + w2) .

Note that D1/qµ (Dγw + wDγ) has 0’s on the diagonal. Thus (3.6) will follow from

(3.7) − tr(DµDh1/q−1(ξ;w)Dµw) ≤ q − 1

qtr(D1/q

µ w2)


for selfadjoint w with 0 on the diagonal. By approximation we can and will assume that

µi 6= µj holds for i 6= j. So that we have to show

∑

i 6=j

µiµjµ1/q−1j − µ

1/q−1i

µi − µj|wij|2 ≤ q − 1

q

∑

i 6=j

µ1/q−1i |wij|2 .

This means that (3.6) will follow from the homogeneous equation

2µ1µ2µ1/q−11 − µ

1/q−12

µ2 − µ1

≤ q − 1

q[µ

1/q1 + µ

1/q2 ] .

By homogeneity we may assume µ1 = 1 and µ > 1. Then we have to prove that

f(µ) = (1− 1

q)µ1/q+1 + (1 +

1

q)µ1/q − (1 +

1

q)µ− (1− 1

q)

is positive for µ ≥ 1. Note that f ′(µ) = (1 − 1/q)(1 + 1/q)µ1/q + 1/q(1 + 1/q)µ1/q−1 −(1 + 1/q) has a unique minimum on (0,∞). However, f ′′(1) = 0 so that the minimum is

attained for µ = 1. Thus f ′(1) = 0 implies f ′(µ) ≥ 0 for µ ≥ 1. With f(1) = 0 we

deduce f(µ) ≥ 0 for µ ≥ 1. This concludes the proof of (3.7) and thus the proof of i).

For the proof of ii) we assume by approximation that x and Φ(xq) are invertible invertible.

Then we may define Φ(z) = Φ(xq)−1/2Φ(xq/2zxq/2)Φ(xq)−1/2. Note that Φ is positive and

unital. By Kadison’s inequality, we deduce

(3.8) Φ(x−1/2yx−1/2)2 ≤ Φ(x−1/2yx−1yx−1/2) = Φ(y2x−2) .

Here we use xy = yx. Moreover, the derivatives simplify again due to commutativity.

Indeed, we have

(x+ ty)q = xq + qxq−1y + q(q − 1)t2

2xq−2y2 +O(t3) .

This implies Dhq(x; y) = qxq−1y = qxq/2yx−1xq/2 and D2hq(x; y, y) = q(q − 1)xq−2y2. We

may now apply (3.6) for ξ = Φ(xq) and z = Φ(yx−1). This yields together with (3.8) that

− tr

(Dh1/q−1


)Φ(Dhq(x; y))

)

= −q2tr(ξDh1/q−1(ξ, z)ξz

)≤ q(q − 1) tr(ξ1/qz2)

= q(q − 1) tr(Φ(xq)1/qΦ(x−1/2yx−1/2)2) ≤ q(q − 1) tr(Φ(xq)1/qΦ(y2x−2))

= q(q − 1)tr(Φ(xq)1/q−1Φ(xq−2y2)) = tr(Φ(xq)1/q−1Φ(D2hq(x; y, y))) .

Adding these terms up we deduce that D2fq(x; y, y) ≥ 0 according to Proposition 3.2.


Remark 3.4. By approximation equation ii) generalizes to positive maps Φ : A → Mn

defined on a commutative C∗-algebra A. Thus for commutative A the function

fq(x) = tr(Φ(xq)1/q)

is convex for all 1 ≤ q < ∞. Thus for 1 ≤ q ≤ 2 having either a commutative domain or

range implies that fq is convex. For commutative range this follows from Remark 2.11

4. Counterexamples

The rest of this section will be concerned with showing that convexity fails by considering

partial derivatives in non-commuting directions. Let us recall that a completely positive

map Φ : M2 → M2 is given by

(4.1) Φ(a) = a11A+ a12C + a21C∗ + a22B

where

MΦ =

(A C

C∗ B

)≥ 0

is a positive matrix. We will consider channels of a particular form represented by

MΦ =

(ρ1 0

0 1

) (0 0

1 0

)

(0 1

0 0

) (1 0

0 σ2

)

.

This is clearly a positive matrix because

‖A−1/2CB−1/2‖ = ‖(

0 0

1 0

)‖ ≤ 1 .

Let us write this down more explicitly

Φ(e11) = ρ1e11 + e22 ,(4.2)

Φ(e12) = e21 ,(4.3)

Φ(e21) = e12 ,(4.4)

Φ(e22) = e11 + σ2e22 .(4.5)

We will differentiate at x = Dλ in direction y = e12 + e21 =

(0 1

1 0

). We have

(4.6) Dhq(x; y) =λq1 − λq2λ1 − λ2

y .


For the second derivative we compute h[2]q :

h[2]q (λ1, λ2, λ1) =h[1]q (λ1, λ2)− h

[1]q (λ1, λ1)

λ2 − λ1=

λq2−λq

1

λ2−λ1− qλq−1

1

λ2 − λ1

=λq2 − λq1 − qλq−1

1 (λ2 − λ1)

(λ2 − λ1)2=

(q − 1)λq1 + λq2 − qλq−11 λ2

(λ2 − λ1)2.

It is very easy to check that h[2]q (λ1, λ2, λ1) ≥ 0. Using the special structure of y we find

that

D2hq(x; y, y) =2

[λ1 − λ2]2α1,q(λ)e11 + α2,q(λ)e22 ,

where

(4.7) α1,q(λ) = (q − 1)λq1 + λq2 − qλq−11 λ2 , α2,q(λ) = (q − 1)λq2 + λq1 − qλq−1

2 λ1 .

For simplicity we will assume λ2 = 1 and λ1 = λ < 1. This gives

α1,q(λ) = (1− q)λq + 1− qλq−1 ,

α2,q(λ) = (1− q) + λq − qλ .

Note that Φ(y) = y and hence Φ(y) is an eigenvector for the linear map z 7→ Dh1/q−1(Φ(xq); z).

It is now straight forward to determine the relevant terms

I(q) = tr(Dh1/q−1(Φ(x

q); Φ(Dhq(x; y)))Φ(Dhq(x; y)))

=

(λq − 1

λ− 1

)2

tr(Dh1/q−1(Φ(x

q); Φ(y))Φ(y))

=

(λq − 1

λ− 1

)2(λqρ1 + 1)1/q−1 − (λq + σ2)

1/q−1

λqρ1 + 1− (λq + σ2)tr(y2)

= 2

(λq − 1

λ− 1

)2(λqρ1 + 1)1/q−1 − (λq + σ2)

1/q−1

λqρ1 + 1− (λq + σ2).

For the other term we define

Dµ = Φ(xq) = (λqρ1 + 1)e11 + (λq + σ2)e22 .

We obtain

II(q) =2

(1− λ)2(α1,q(λ)tr(D

1/q−1µ Φ(e11)) + α2,q(λ)tr(D

1/q−1µ Φ(e22))

)

=2

(1− λ)2(α1,q(λ)[(λ

qρ1 + 1)1/q−1ρ1 + (λq + σ2)1/q−1] + α2,q(λ)[1 + (λq + σ2)

1/q−1σ2]).


After cancelation of some obvious common factors the convexity of fq implies (with

limρ1→0) that

[1− λq]2(λq + σ2)

1/q−1 − 1

1− (λq + σ2)

≤(α1,q(λ)(λ

q + σ2)1/q−1 + α2,q(λ)[1 + (λq + σ2)

1/q−1σ2])

?(4.8)

holds for all 0 < λ < 1 and σ2 > 0. The following interesting choice of λ has been found

by comparing α1,q(λ) and α2,q(λ).

Lemma 4.1. Let 0 < λq < 1 be the solution of

λq +q

2− qλq−1 = 1 .(4.9)

Let p > 0. Then there exists a constant cp such that

i) α1,q(λq) ≤ (q − 1) + cp(q − 1)p.

ii) α2,q(λq) ≤ (q − 1) + cp(q − 1)p.

iii) λq ≤ cp(q − 1)p for all p > 0.

Proof. First of all we note that for fixed q the function f(λ) = λq+ q2−q

λq−1 is continuous,

monotone increasing and satisfies f(0) = 0, f(1) = 1 + q2−q

> 1. Thus a unique solution

λq of (4.9) exists. Now, we assume that for p > 0 we have

lim infq→1

λq(q − 1)p

> C .

For a suitable subsequence we find

1 = λqnqn +qn

2− qnλqn−1qn ≥ Cqn(qn − 1)pqn + Cqn−1 qn

2− qn(qn − 1)p(qn−1) .

However, limq→1 q ln(q− 1) = −∞ and limq→1(q− 1) ln(q− 1) = limt→0 t ln t = 0. Thus in

the limit we get

1 ≥ C + 1 .

This implies C = 0. Thus limq→1λq

(q−1)p= 0 for all p > 0. This clearly implies iii). In

particular, we deduce that

limq→1

α2,q(λq) = limq→1

(q − 1) + λqq − qλq = 0 .

Moreover, by iii), we have α2,q(λq) ≤ (q−1)+cp(q−1)p for every p > 0. For the assertion

i) we use

λq−1q =

2− q

q− 2− q

qλqq .


This implies

α1,q(λq) = (q − 1)λqq + 1− qλq−1q = (q − 1)λqq + 1− ((2− q)− (2− q)λqq)

= (q − 1) + [(q − 1) + (2− q)]λqq = (q − 1) + λqq .

Thus the assertion follows again from iii).

The next observation is geared towards an optimal choice for σ2.

Lemma 4.2. Let βq =(

2−qq

) qq−1

and σq be the solution to

2− q = qσ1− 1

q + (q − 1)σ .

i) limq→1 β(q) = e−2.

ii) βq ≥ σq ≥ βq2. In particular, lim infq σq > e−3.

iii) Let σ < σq. Then

qσ1− 1

q + (q − 1)σ ≤ (2− q)− 2(q − 1)(σq − σ) .

Proof. The proof of i) is easy. Let us introduce t = 1− 1q. Then

ln βq =1q−1q

ln(2

q− 1) =

ln(2− 2t− 1)

t=

ln(1− 2t)

t.

Thus in the limit we find limq→1 lnβ(q) = limt→0ln(1−2t)

t= −2. Since the function f(σ) =

qσ1−1/q + (q− 1)σ is increasing and f(0) = 0 there is a solution f(σq) = 2− q. Let us first

show that σq ≤ βq. Indeed, assume σq > βq. Then we have

2− q = qβ1−1/qq < qσ1−1/q

q + (q − 1)σq = 2− q .

Thus necessarily σq ≤ βq. Then we observe that

qβ1−1/qq = 2− q = qσ1−1/q

q + (q − 1)σq

implies

(q − 1)σq = qβ1−1/qq − qσ1−1/q

q = q

∫ βq

σq

(1− 1/q)t−1/qdt

≥ (q − 1)β−1/qq (βq − σq) .

This yields

βq ≤ σq + β1/qq σq ≤ (1 + β1/q

q )σq ≤ 2σq .


For the proof of iii) we apply a similar argument. Consider σ < σq. Note that σq ≤ βq ≤ 1.

Hence

(2− q)− (qσ1− 1

q + (q − 1)σ) = σ1− 1

qq + (q − 1)σq − (σ1− 1

q + (q − 1)σ)

= (q − 1)

∫ σq

σ

t−1

q dt+ (q − 1)(σq − σ) ≥ (q − 1)(σ− 1

qq + 1)(σq − σ)

≥ 2(q − 1)(σq − σ) .

Thus iii) follows.

We are now ready to discuss the validity of (4.8).

Proposition 4.3. There exists a q0 > 1 and for 1 < q < q0 there are λq and σ2,q such that

(α1,q(λq)(λ

qq + σ2,q)

1/q−1 + α2,q(λq)[1 + (λqq + σ2,q)1/q−1σ2,q]

)< [1− λq]2

(λqq + σ2)1/q−1 − 1

1− (λqq + σ2,q)

Proof. Let λq be chosen as in Lemma 4.1. Let us define γq = σq2∈ (0, 1). We define σ2,q

such that

λqq + σ2,q = γq .

From lim infq σq ≥ e−3 and limq λqq = 0 we find q0 such that 0 < σ2,q < 1 holds for

1 < q < q0. This implies (with 1− γq < 1) that

γ1−1/qq [1− λq]2

(λqq + σ2)1/q−1 − 1

1− (λqq + σ2,q)≥ [1− λqq]

2(1− γ1−1/qq )

≥ (1− γ1−1/qq )− 2λqq(1− γ1−1/q

q ) ≥ (1− γ1−1/qq )− 2λqq .

Consider p > 1 and εp(q) = Cp(q−1)p−1. We note that σ2,q ≤ γq. We deduce from Lemma

4.1 and Lemma 4.2iii) for σ = γq < σq that

γ1−1/qq

(α1,q(λq)(λ

qq + σ2,q)

1/q−1 + α2,q(λq)[1 + (λq + σ2,q)1/q−1σ2,q]

)

≤((q − 1)(1 + εp(q)) + (q − 1)(1 + εp(q))(γ

1−1/qq + γq)

)

= (1 + εp(q))((q − 1)− γ1−1/q

q + qγ1−1/qq + (q − 1)γq

)

≤ (1 + εp(q))((q − 1)− γ1−1/q

q + 2− q − 2(q − 1)(σq − γq))

= (1 + εp(q))((1− γ1−1/qq )− 2(q − 1)(σq − γq))

≤ (1− γ1−1/qq ) + εp(q)− 2(q − 1)(σq − γq)

≤ γ1−1/qq [1− λq]2

(λqq + σ2)1/q−1 − 1

1− (λqq + σ2,q)+ 2λqq + εp(q)− (q − 1)σq .


Here we used γq =σq2. We choose p = 3 and find εp(q) = Cp(q−1)2. Applying Lemma 4.1

again we know that λqq ≤ C2(q − 1)2. From Lemma 4.2 we recall σq ≥ e−3 for q < q2.

Thus we can find q0 < min(q1, q2) such that

2λqq + εp(q)− (q − 1)σq ≤ C3(q − 1)2 + C2(q − 1)2 − e−3(q − 1) < 0

for all 1 < q < q1.

We have proved the following result.

Theorem 4.4. There exists a non-empty interval (1, q1) such that for every 1 < q < q0there exists a completely positive map Φ : M2 → M2 such that

fq(x) = tr(Φ(xq)1/q)

is not convex.

Theorem 4.5. There exists a non-empty interval interval (1, q1) such that the functions

f(x1, ..., xn) = tr((∑

i

xqi )1/q)

are not convex (not even uniformly equivalent to convex functions) for all 1 < q < q0.

Proof. This follows immediately from Proposition 2.10 and Theorem 4.4.

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Department of Mathematics, University of Illinois, Urbana, IL 61801, USA

E-mail address, Marius Junge: [email protected]

Departement de Mathematiques, Universite de France-Comte, 16 Route de Gray, 25030

Besancon Cedex, France

E-mail address, Quanhua Xu: [email protected]

m. junge and q. xu arxiv:0709.0433v1 [math.fa] 4 sep 2007 · 2020-01-23 · in our next step we...

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