Proofs of Stronger Results

CONCENTRATION OF EIGENVALUE SUMS AND GENERALIZED LIEB’S CONCAVITY THEOREM

5.4 Proof of Concavity Theorems

5.4.2 Proofs of Stronger Results

To continue, we use definitions (5.46), identity (5.50) and the Alexandrov–Fenchel inequality (Theorem 5.6.1) to obtain

TrM₀^(k)(B)TrM₂^(k)(T B,T B,B)]

= n!

k!(n−k)!D(B,· · · ,B

| {z }

k

,In,· · ·,In

| {z }

n−k

)

× n!

(k−2)!(n−k)!D(T B,T B,B· · · ,B

| {z }

k−2

,In,· · · ,In

| {z }

n−k

)

≤ k−1 k

n!

(k−1)!(n−k)!D(T B,B· · ·,B

| {z }

k−1

,In,· · ·,In

| {z }

n−k

)2

= k−1

k TrM₁^(k)(T B,B)2

= k−1

k (g⁰(0))². We therefore have proved that

g(0)g⁰⁰(0) ≤ k−1

k (g⁰(0))². The concavity of fH,k(A)onH⁺⁺_n then follows.

Next we prove the equivalence of (i) the concavity of the functions f_H,k(A)onH⁺⁺_n and (ii) the concavity of the functions ˜fH,k =log Trkexp H+logA onH⁺⁺_n . (i)

⇒(ii) is trivial. To prove (ii)⇒(i), we need the following lemma.

Let x = (x₁,x₂) ∈ (0,+∞)². Define f(x) = Trkexp H + log(x₁A₁+ x₂A₂) . One can easily verify that fH,k(A) being concave on H⁺⁺_n is equivalent to f(x)^1k being concave on(0,+∞)²for arbitrary but fixed choice of A₁,A₂ ∈H⁺⁺_n ,H ∈Hn. Similarly, ˜fH,k(A) being concave onH⁺⁺_n is equivalent to log f(x) being concave on (0,+∞)² for arbitrary but fixed choice of A₁,A₂ ∈ H⁺⁺_n ,H ∈ Hn. Using the definition of the k-trace Trk, it is easy to check that f(x)is homogeneous of order k. By Lemma 5.6.9, we know f(x)^1k is concave if and only if log f(x)is concave.

Therefore we have (i)⇔(ii).

Our choice of the holomorphic functionsG(z)in the following proof is inspired by Lieb’s constructions in [67] for the use of maximum modulus principle. Recall that we will writeφ(·)= Trk[]^1k for notational simplicity.

Proof of Lemma 5.3.2. Note that for s ∈ [0,1], the concavity of (5.18) is a direct consequence of the facts that (i)φis monotone increasing and concave onH⁺_n, and (ii) X 7→ X^r and X 7→ X^s are operator monotone increasing and operator concave onH⁺_n. So in what follows we may assume that 1 ≤ s ≤ ¹_r. We need to show that, for any A,B ∈H⁺_n and anyτ ∈[0,1],

τφ (K^∗A^rK)^s+(1−τ)φ (K^∗B^rK)^s ≤ φ (K^∗C^rK)^s,

whereC = τA+ (1− τ)B. We may assume that A,B ∈ H⁺⁺_n and K is invertible.

Once this is done, the general result for A,B ∈H⁺_n andK ∈C^n×ncan be obtained by continuity. Letw = ¹_s ∈ [r,1] and ˆr = r s ∈ [0,1], sor = rˆw. Let M = C^r2K, and let M = Q|M| be the polar decomposition of M for some unitary matrixQ. Since C,Kare both invertible,|M| ∈H⁺⁺_n . We then define two functions fromStoC^n×n:

GA(z) = A^{r zˆ2} C^{−r zˆ2}Q|M|^wz, GB(z) = B^{r zˆ2} C^{−r zˆ2}Q|M|^wz, z ∈ S,

whereSis given by (5.61). In what follows we will useX for AorB. We then have φ (K^∗X^rK)^s =φ (M^∗C^−r2X^rC^−r2M)^s

=φ (|M|Q^∗C^{−r wˆ2} X^{r wˆ2} X^{r wˆ2} C^{−r wˆ2} Q|M|)^w1

=φ |GX(w)|^w2.

Since A,B,C,M are now fixed matrices in H⁺⁺_n , GA(z) and GB(z) are apparently holomorphic in the interior ofSand continuous on the boundary. Also, it is easy to check thatkGA(z)kandkGB(z)kare uniformly bounded onS, since Re(z) ∈[0,1].

Therefore we can use inequality (5.65) withθ= w,pθ = _w² to obtain φ(|GX(w)|^w2)

≤ Z ₊∞

−∞

dt2(1−w)

wp₀ β_1−w(t)φ |GX(it)|^p0+ 2

p₁β_w(t)φ |GX(1+it)|^p1 . We still need to choose some p₀,p₁ ≥ 1 satisfying ¹_p^−w₀ + _p^w₁ = _p¹_w = ^w₂ to proceed.

Note thatGX(it) = X^{ir t2ˆ} C^{−ir t2ˆ} Q|M|^{i tw} are now unitary matrices for allt ∈ Rsince X,C,|M| ∈ H⁺⁺_n , and thus |GX(it)|^p0 = In for all p₀. Therefore we can take p₀ →+∞,p₁=2 to obtain

φ(|GX(w)|^w2) ≤ Z ₊∞

−∞

dtβ_w(t)φ |GX(1+it)|².

Further, for eacht ∈R, we have φ |GX(1+it)|²

=φ GX(1+it)^∗GX(1+it)

=φ |M|^{(1−i tw )}Q^∗C^{−rˆ(1−i t)2} X^rˆC^{−rˆ(1+i t2 )}Q|M|^{(1+i t)w}

=φ |M|^w1Q^∗C^{−rˆ(12−i t)}X^rˆC^{−rˆ(1+i t)2} Q|M|^w1,

where we have used the cyclicity of φ (|M|^{i tw} is unitary) for the last equality.

Therefore we have

τφ |GA(1+it)|² +(1−τ)φ |GB(1+it)|²

=τφ |M|^w1Q^∗C^{−rˆ(1−i t)2} A^rˆC^{−rˆ(1+i t)2} Q|M|^w1

+(1−τ)φ |M|^w1Q^∗C^{−rˆ(1−i t)2} B^rˆC^{−rˆ(1+i t)2} Q|M|^w1

≤ φ |M|^w1Q^∗C^{−rˆ(12−i t)}(τA^rˆ+(1−τ)B^rˆ)C^{−rˆ(1+2i t)}Q|M|^w1

≤ φ |M|^w1Q^∗C^{−rˆ(12−i t)}C^rˆC^{−rˆ(1+2i t)}Q|M|^w1

=φ |M|^w2

=φ (M^∗M)^w1.

The first inequality above is due to the concavity of φ, the second inequality is due to (i) thatφis monotone increasing onH⁺_n and (ii) thatX 7→ X^rˆis operator concave onH⁺_n for ˆr ∈ (0,1]. Finally, sinceφ (M^∗M)^w1 is independent oft, and β_w(t) is a density onR, we obtain that

τφ (K^∗A^rK)^s+ (1−τ)φ (K^∗B^rK)^s

=τφ |GA(w)|^w2 +(1−τ)φ |GB(w)|^w2

≤ φ (M^∗M)^w1

=φ (K^∗C^rK)^s.

So we have proved the concavity of (5.18) onH⁺_n. Our next proof, using essentially Hölder’s inequalities for the k-trace, is adapted from Zhang’s proofs of Theorem 1.1 and Theorem 3.3 in [136].

Proof of Theorem 5.3.3. Without loss of generality, we may assume that m = n. Otherwise we can replace Aby *

, A 0 0 0

+ -

and K by * ,

K 0

+ -

if n < m; or replace B

by* ,

B 0 0 0

+ -

and K by

K 0

isn > m. By the consistency ofφ, these changes of variables will not affect whether the function (5.19) is jointly concave in (A,B) or not. We write X = A^p2 andY = K B^q2. Let s₁ = ^p+_p^qs,s₂= ^p+_q^qs, so ¹_s = _s¹₁ + _s¹₂. Then for any Z ∈ C^n×n that is invertible, we have by Hölder’s inequality ((v) in Proposition 5.2.1) that

φ (B^q2K^∗A^pK B^q2)^s =φ |X Z Z⁻¹Y|^2s

≤ φ |X Z|^2s1_s^s₁φ |Z⁻¹Y|^2s2_s^s₂

≤ s

s₁φ (Z^∗X^∗X Z)^s1+ s

s₂φ (Y^∗(Z⁻¹)^∗Z⁻¹Y)^s2

= s

s₁φ (Z^∗X^∗X Z)^s1+ s

s₂φ (Z⁻¹YY^∗(Z⁻¹)^∗)^s2. We have used the fact thatφ f(|M|) = φ f(|M^∗|) for any matrix M ∈ C^n×nand any function f, sinceφis only a function of eigenvalues and the spectrums of f(|M|) and f(|M^∗|) are the same. Let (XY)^∗ = Q|(XY)^∗| be the polar decomposition of (XY)^∗, where Q ∈ C^n×n is unitary. So we have XY Q = |(XY)^∗|. If X andY are invertible, we can particularly chooseZ =Y Q|(XY)^∗|⁻

s1

s1+s2 to have X Z = XY Q|(XY)^∗|⁻

s1

s1+s2 = |(XY)^∗|

s2 s1+s2, and Z⁻¹Y = |(XY)^∗|

s1 s1+s2Q^∗, which yields the equality

s

s₁φ (Z^∗X^∗X Z)^s1+ s

s₂φ (Z⁻¹YY^∗(Z⁻¹)^∗)^s2 =φ |(XY)^∗|

2s1s2

s1+s2 = φ |XY|^2s. Now for general X,Y that are not necessarily invertible, we can always find two sequences of invertible matrices {Xj}⁺_j=1^∞,{Yj}⁺_j=1^∞ such that (i) Xj → X, Yj → Y and (ii) X^∗_jXj X^∗X, YjY_j^∗ YY^∗. Such sequences can be easily obtained by perturbing the singular values of X andY. For each pair of (Xj,Yj), we can find some invertible Zj so that the above equality holds. Also, for any invertible Z, we haveZ^∗X^∗_jXjZ Z^∗X^∗X Z, Z⁻¹YjY_j^∗(Z⁻¹)^∗ Z⁻¹YY^∗(Z⁻¹)^∗, and thus

φ (Z^∗X^∗X Z)^s1 ≤ φ (Z^∗X^∗_jXjZ)^s1, φ (Z⁻¹YY^∗(Z⁻¹)^∗)^s2 ≤ φ (Z⁻¹YjY_j^∗(Z⁻¹)^∗)^s2

by Theorem 5.7.6, which we will prove in Section 5.7. Then we obtain a sequence of inequalities,

φ(|XY|^2s) ≤ inf{ s

s₁φ (Z^∗X^∗X Z)^s1+ s

s₂φ (Z⁻¹YY^∗(Z⁻¹)^∗)^s2 : Z invertible}

≤ s

s₁φ (Z^∗_jX^∗X Zj)^s1+ s

s₂φ (Z⁻_j¹YY^∗(Z⁻_j¹)^∗)^s2

≤ s

s₁φ (Z^∗_jX^∗_jXjZj)^s1 + s

s₂φ (Z⁻_j¹YjY_j^∗(Z⁻_j¹)^∗)^s2

=φ(|XjYj|^2s).

But sinceφ(|XY|^2s) =limj→+∞φ(|XjYj|^2s)by continuity, the first inequality above must be an equality. Therefore, by substituting X = A^p2,Y = K B^q2, we obtain that

φ (B^q2K^∗A^pK B^q2)^s

=inf{ s

s₁φ (Z^∗A^pZ)^s1 + s

s₂φ (Z⁻¹K B^qK^∗(Z⁻¹)^∗)^s2 : Z invertible}.

Note thats ∈[0, _p+¹_q] implies s₁ ∈[0,¹_p],s₂∈[0, ¹_q]. By Lemma 5.3.2, the map (A,B) 7−→ s

s₁φ (Z^∗A^pZ)^s1 + s

s₂φ (Z⁻¹K B^qK^∗(Z⁻¹)^∗)^s2

is jointly concave in (A,B) for every invertible Z, which then implies the joint

concavity of the infimum over all invertible Z.

Proof of Theorem 5.3.4 (Part I). We first prove the theorem for m = 1. Let r = p₁ ∈[0,1], andK^(N) = (K^(N))^∗ = exp ₂¹_NH,N ≥ 1. Then using the Lie product formula

N→lim+∞ exp 1

2NYexp 1

NXexp 1 2NY

!N

=exp(X +Y), X,Y ∈Hn, we have

N→lim+∞φ

(K^(N))^∗A^Nr K^(N)N

= lim

N→+∞φ

exp 1

2NHexp r

N logAexp 1

2NHN!

=φ exp(H+rlogA). By Theorem 5.3.3, for each N ≥ 1, φ

(K^(N))^∗A^Nr K^(N)N

is concave in A, thus the limit functionφ exp(H+rlogA) is also concave in A.

To go from m = 1 to m > 1 in Theorem 5.3.4, we need to use the convexity of the map A 7→ φ(exp(A)), which we will prove via the following lemmas. They are the k-trace extensions of the Araki–Lieb–Thirring inequality [4], the Golden–

Thompson inequality and a variant of the Peierls–Bogoliubov inequality (see, e.g., [24, Theorem 2.12]).

Lemma 5.4.3 (k-trace Araki–Lieb–Thirring Inequality). For any A,B ∈ H⁺_n, the function

t 7→ Trk

(B^t2A^tB^t2)^1t is monotone increasing on(0,+∞), that is

Trk

(B^2tA^tB^2t)^1t ≤ Trk

(B^s2A^sB^s2)^1s, 0 <t ≤ s. (5.24) Proof. Using the definition and properties of the operatorM₀^(k) in Section 5.6.2, we have that

Trk

(B^2tA^tB^t2)^1t =TrM₀^(k) (B^2tA^tB^2t)^1t

=Tr

(M₀^(k)(B))^t2(M₀^(k)(A))^t(M₀^(k)(B))^2t1_t.

Since A,B ∈H⁺_n, M₀^(k)(A)andM₀^(k)(B)are both Hermitian and positive semidef- inite. Then inequality (5.24) follows immediately from the original Araki–Lieb–

Thirring inequality [4] for normal trace.

Lemma 5.4.4(k-trace Golden–Thompson Inequality). For any A,B ∈Hn,

Trkexp(A+B) ≤ Trkexp(A)exp(B), (5.25) with equality holds if and only if AB = B A.

Proof. We here only prove the inequality. The condition for equality will be justified in an alternative proof of this lemma in Section 5.6.2. For any A,B ∈Hn, we have

Trkexp(A+B) = lim

m→+∞Trk

f

exp 1

2mBexp 1

mAexp 1

2mB^mg

≤ Trkexp 1

2Bexp Aexp 1 2B

=Trkexp Aexp B .

The first equality above is the Lie product formula, and the inequality is due to

Lemma 5.4.3.

Lemma 5.4.5(k-trace Peierls–Bogoliubov Inequality). The function

A 7−→ log Trkexp(A) (5.26)

is convex onHn.

Proof. For any A,B ∈Hn,τ∈ (0,1), by Lemma 5.4.4 we have Trkexp(τA+(1−τ)B) ≤Trkexp τAexp (1−τ)B

≤Trkexp(A)τTrkexp(B)1−τ. The second inequality above is Hölder’s. Therefore

log Trkexp(τA+(1−τ)B) ≤ τlog Trkexp(A) +(1−τ)log Trkexp(B). We remark that Lemma 5.4.5 can also be proved using the operator interpolation in Lemma 5.6.8. Lemma 5.4.5 immediately implies that A 7→ logφ exp(A) =

1

k log Trkexp(A) is convex, and thus A 7→ φ exp(A) is convex. This will help us prove improve fromm= 1 tom ≥ 1 in Theorem 5.3.4.

Proof of Theorem 5.3.4 (Part II). Given any {A^(j)}^m_j=1,{B^(j)}^m_j=1 ⊂ H⁺⁺_n , and any τ ∈[0,1], letC^(j) = τA^(j)+(1−τ)B^(j),1≤ j ≤ m. Since the mapX 7→ φ(exp(X)) is convex onHn, the map X 7→ φ(exp(L+ X)) is also convex onHnfor arbitrary L ∈Hn. Now define

L= H+

m

X

j=1

pjlogC^(j), r =

m

X

j=1

pj ≤ 1.

Ifr = 0, the result is trivial; so we may assume thatr > 0. We then have that φ exp(H+

m

X

j=1

pjlogX^(j))

=φ

exp H+r

m

X

j=1

pj

r (logX^(j)−logC^(j))+

m

X

j=1

pjlogC^(j)

=φ

exp L+r

m

X

j=1

pj

r (logX^(j) −logC^(j))

≤

m

X

j=1

pj

r φ exp(L+rlogX^(j)−rlogC^(j)), X^(j) = A^(j),B^(j).

For each j, by the concavity of (5.20) form =1, we have τφ exp(L+rlogA^(j)−rlogC^(j))

+(1−τ)φ exp(L+rlogB^(j)−rlogC^(j))

≤ φ exp(L+rlog(τA^(j)+(1−τ)B^(j))−rlogC^(j))

=φ exp(L). Therefore we obtain that

τφ exp(H+

m

X

j=1

pjlogA^(j))+ (1−τ)φ exp(H +

m

X

j=1

pjlogB^(j))

≤

m

X

j=1

pj

r φ exp(L)

=φ exp(H+

m

X

j=1

pjlogC^(j)),

that is, (5.20) is jointly concave on (H⁺⁺_n )^×m for allm ≥ 1.