Modulus of Convexity for Operator Convex Functions, Which Can Be Expressed in Terms of the Matrix Bregman Divergence

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Petz noted that a similar treatment can be carried out even when p and q are promoted to operators, provided that the function f is operator convex. The resulting matrix-valued divergence is the matrix Bregman divergence. Interestingly, the strengthening is only applicable to operator convex functions; the inequality is false for a convex function that is not operator convex. As an application of this result, we prove an inequality that extends Pinsker’s inequality. Recall that Pinsker’s inequality asserts that the relative entropy D(ρkσ) := Tr(ρ(log ρ−log σ)) between two normalized positive semideﬁnite operators, ρ and σ, is lower bounded by their trace distance: 1 D(ρkσ) ≥ kρ − σk2. 2 1 A simple corollary of our main result is the following inequality: 1 S(cρ + (1 − c)σ) − cS(ρ) − (1 − c)S(σ) ≥ c(1 − c)kρ − σk2, c ∈ [0, 1], 2 1 where the underlying Hilbert space is ﬁnite-dimensional. The rest of the paper starts by describing the main result in Section II. We shall also show that the main result cannot be generalized to convex functions by providing a simple argument. Section III describes the key technical result of this paper, which is a strengthening of the well-known Arithmetic-Harmonic inequality. Using this strengthening, we prove the main result in Section IV.

II. AN INEQUALITY BETWEEN THE MODULUS OF CONVEXITY AND BREGMAN DIVERGENCE FOR OPERATOR CONVEX FUNCTIONS

In order to describe the main result, we set the notations first. Definition 1. Modulus of convexity of a function f(x) is c Cf (A, B) := cf(A)+(1 − c)f(B) − f(cA + (1 − c)B), c ∈ [0, 1]. Note that the modulus of convexity of an operator convex function is always nonnegative, as long as A and B are self-adjoint operators whose spectrum lie on the domain of f. The matrix Bregman divergence is nonnegative due to the same reason. Following Petz[10], we define the matrix Bregman divergence as follows.

Deﬁnition 2. Bregman divergence Df (A, B) is

−1 Df (A, B)= f(A) − f(B) − lim t (f(B + t(A − B)) − f(B)). t→0+ The main result asserts that the Bregman divergence provides an operator-valued lower bound for the modulus of convexity. Theorem 1. For A,B > 0, if f(x) on [0, ∞) is operator convex and 0

A. Convex vs. operator convex functions

A natural question is whether Theorem 1 can be extended to operator convex functions of order n. We will give a simple argument that such an extension cannot exist for n = 1. Recall that operator convex functions of order 1 refer 1 2 to all the convex functions. One can easily check that the function g(x)= 2 x − (1 + x) log(1 + x) is convex for x> 0. Our claim is that Eq.2 cannot hold for such g(x). If Eq.2 holds for f(x) = g(x), it implies that Eq.2 holds for f(x) = −(1 + x) log(1 + x) as well; this follows from a simple observation that Eq.2 holds with an equality if f(x) = x2. Since f(x) = −(1 + x) log(1 + x) as well as f(x) = (1+ x) log(1 + x) satisﬁes Eq.2, one can conclude that the inequalities in Eq.2 must be satisﬁed with an equality for such function. Clearly, this is not the case, and we arrive at a contradiction. Therefore, Eq.2 cannot be extended to operator convex functions of order 1. 3

III. STRENGTHENING OF THE ARITHMETIC-HARMONIC INEQUALITY

In this section, we prove a strengthening of the well-known Arithmetic-Harmonic(AH) inequality. AH inequality states that 1 1 4 + ≥ (3) A B A + B for positive deﬁnite matrices A and B. It is well known that any operator convex function has a unique integral representation that can utilize Eq.3. For example, the following theorem was recently proved by Hiai et al.[12] Theorem 2. [12] A continuous real function f on [0, ∞) is operator convex iﬀ there exists a real number a, a nonnegative number b, and a nonnegative measure µ on [0, ∞), satisfying ∞ 1 dµ λ < , 2 ( ) ∞ Z0 (1 + λ) such that ∞ x λ f(x)= f(0) + ax + bx2 + ( − 1+ )dµ(λ), x ∈ [0, ∞). (4) Z0 1+ λ x + λ Moreover, the numbers a,b, and the measure µ is uniquely determined by f. The existence of the canonical form for operator convex functions is the main motivation behind the strengthening of AH inequality. Our key lemma is the following: Lemma 1. For A,B > 0, 1 1 1 2 1 1 1 ( + ) − ≥ 2 (A − B) (A − B) (5) 2 A B A + B A + B A + B A + B

1 1 1 Proof. Deﬁne C = A− 2 BA− 2 . Applying a left and right multiplication of A 2 on both sides of Eq.5, the left hand side 2 2 (1−C) (1−C) 2 can be expressed as 2C(1+C) , while the right hand side can be expressed as 2 (1+C)3 . Using the fact that (1+C) ≥ 4C, one can establish the inequality.

IV. PROOF OF THE MAIN RESULT

A well-known approach for proving operator Jensen inequality involves (i) proving the inequality at the midpoint and (ii) making a judicious choice of matrices in an enlarged Hilbert space.[1] We shall follow a similar approach. Under elementary manipulations, one can show that Theorem 1 for a general operator convex function follows by 1 proving it for a special family of functions, namely fλ(x)= λ+x . Without loss of generality, we shall prove Theorem 1 1 for f(x)= λ+x . The case for the linear and quadratic terms are trivial, so we omit the proof for them. 1 First, we consider the c = 2 case of Theorem 1. f(A)+ f(B) A + B ∞ 1 1 1 2 − f( )= ( ( + ) − )dµ(λ) 2 2 Z0 2 A + λ B + λ A + B +2λ ∞ 1 1 1 ≥ 2 (A − B) (A − B) dµ(λ) Z0 A + B +2λ A + B +2λ A + B +2λ 1 ∞ 1 1 1 = A+B (A − B) A+B (A − B) A+B dµ(λ) 4 Z0 2 + λ 2 + λ 2 + λ 1 ∞ d2 1 = | dµ(λ) dx2 A+B x=0 8 Z0 2 + x(A − B)+ λ 1 d2 A + B = f( + x(A − B))| (6) 8 dx2 2 x=0 Away from the midpoint, we use the following choice of operators:[13]

1 1 c 2 I −(1 − c) 2 W = 1 1 , (1 − c) 2 I c 2 I 4 and A 0 T = . 0 B

† Each of the entries in the matrices correspond to a block of square matrices of the dimension. Setting T1 = W T W , † T2 = W T W , and applying it to the midpoint convexity result, one can obtain the desired result. More precisely, the convexity at the midpoint is the following:

f(T1)+ f(T2) T1 + T2 cf(A)+(1 − c)f(B) 0 cA + (1 − c)B 0 − f( )= − f . 2 2 0 cf(B)+(1 − c)f(A) 0 cB + (1 − c)A One can also easily check the following facts:

T1 + T2 cA + (1 − c)B 0 = . 2 0 cB + (1 − c)A

T1 − T2 0 c(1 − c)(A − B) = . 2 c(1 − c)(A − B)p 0 p By taking one of the blocks,

∞ c 1 1 1 Cf (A, B) ≥ c(1 − c) (A − B) (A − B) dµ(λ) Z0 M(c)+ λ M(1 − c)+ λ M(c)+ λ 1 ∞ 1 1 1 = c(1 − c) 2 (M(c) − M(1 − c)) (M(c) − M(1 − c)) dµ(λ). (2c − 1) Z0 M(c)+ λ M(1 − c)+ λ M(c)+ λ (7)

One can check that ∞ 1 1 1 Df (A, B)= (A − B) (A − B) dµ(λ), (8) Z0 B + λ A + λ B + λ completing the proof.

V. APPLICATION TO THE VON NEUMANN ENTROPY

Now we discuss an application of Theorem 1. Corollary 1. For density matrices ρ, σ on a ﬁnite-dimensional Hilbert space, 1 S(cρ + (1 − c)σ) − cS(ρ) − (1 − c)S(σ) ≥ c(1 − c)kρ − σk2 (9) 2 1 Proof. For f(x)= x log x, Petz showed that

Tr(Df (A, B)) = D(AkB), (10) where D(AkB) = Tr(A(log A − log B)) is the relative entropy between A and B.[10] Hence, the following inequality immediately follows. 1 S(cρ + (1 − c)σ) − cS(ρ) − (1 − c)S(σ) ≥ c(1 − c) D(cσ + (1 − c)ρkcρ + (1 − c)σ) (11) (1 − 2c)2

1 for c 6= 2 . Applying Pinsker’s inequality, 1 S(cρ + (1 − c)σ) − cS(ρ) − (1 − c)S(σ) ≥ c(1 − c)kρ − σk2. (12) 2 1

1 Eq.12 should be true for c = 2 as well by some continuity argument, which is discussed below. 5

Recall that Fannes’ inequality asserts that

|S(ρ) − S(σ)|≤ ǫ log d − ǫ log ǫ, (13) where ǫ = kρ − σk1 and d is the dimension of the Hilbert space.[14] Deﬁne S(c) as S(c)= S(cρ + (1 − c)σ) − cS(ρ) − (1 − c)S(σ).

Using Fannes’ inequality, 1 1 |S( ) − S( + δ)|≤ ǫδ(log d − log ǫδ)+2δ log d. (14) 2 2 Denoting the right hand side of Eq.14 as ∆(δ, ǫ, d) ≥ 0, 1 1 S( ) ≥ S( + δ) − ∆(δ, ǫ, d) 2 2 1 1 ≥ kρ − σk2 − δ2kρ − σk2 − ∆(δ, ǫ, d). 8 1 2 1 Taking the δ → 0 limit, we obtain: 1 1 S( ) ≥ kρ − σk2 2 8 1

VI. DISCUSSION

We have obtained a lower bound for the modulus of convexity for operator convex functions, which can be expressed in terms of the matrix Bregman divergence. We also gave a simple argument that the inequality cannot be extended to general convex functions. For the operator convex function f(x) = x log x, the trace of the matrix Bregman divergence reduces to quantum relative entropy. In this case, the inequality reduces to the strict concavity of Von Neumann entropy. It will be interesting to ﬁnd an application of this inequality. Another important question is to ﬁnd a strengthening of the operator Jensen inequality. Since many of the nontrivial results in quantum information theory can be essentially derived from the operator Jensen inequality, its strengthening will be undoubtedly useful in many contexts.

Acknowledgments

I would like to thank Andreas Winter and Alexei Kitaev for many helpful discussions which motivated this work. I would also like to thank Jon Tyson, Mary Beth Ruskai, Fernando Brand˜ao for helpful discussions. I would also like to thank Lin Zhang for pointing out an error in the original manuscript. Lastly, I thank the anonymous referee who suggested to investigate whether the main result holds for operator convex function of ﬁnite order. This research was supported in part by NSF under Grant No. PHY-0803371, by ARO Grant No. W911NF-09-1-0442, and DOE Grant No. DE-FG03-92-ER40701. Research at Perimeter Institute is supported by the Government of Canada through Industry Canada and by the Province of Ontario through the Ministry of Economic Development and Innovation.

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