Since the subpace of Hermitian matrices is provided with the order structure induced by the cone of positive semidefinite matrices, one can consider convexity of this map.. Introduction
Trang 1Volume 2010, Article ID 241908, 12 pages
doi:10.1155/2010/241908
Research Article
Trace-Inequalities and Matrix-Convex Functions
Tsuyoshi Ando
Hokkaido University (Emeritus), Shiroishi-ku, Hongo-dori 9, Minami 4-10-805, Sapporo 003-0024, Japan
Correspondence should be addressed to Tsuyoshi Ando,ando@es.hokudai.ac.jp
Received 8 October 2009; Accepted 30 November 2009
Academic Editor: Anthony To Ming Lau
Copyrightq 2010 Tsuyoshi Ando This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
A real-valued continuous function ft on an interval α, β gives rise to a map X → fX via functional calculus from the convex set of n × n Hermitian matrices all of whose eigenvalues
belong to the interval Since the subpace of Hermitian matrices is provided with the order structure induced by the cone of positive semidefinite matrices, one can consider convexity of this map
We will characterize its convexity by the following trace-inequalities: TrfB − fAC − B ≤ TrfC − fBB − A for A ≤ B ≤ C A related topic will be also discussed
1 Introduction and Theorems
Let ft be a real-valued continuous function defined on an open interval α, β of the real line The function ft is said to be convex if
f λa 1 − λb ≤ λfa 1 − λfb 0≤ λ ≤ 1; α < a, b < β. 1.1
We referee to 1 for convex functions Under continuity the requirement 1.1 can be
restricted only to the case λ 1/2, that is,
f
a b
2
≤ f a fb
2
α < a, b < β
It is well known that when ft is a C1-function, its convexity is characterized by the condition on the derivative
f b − fb − t
t ≤ fb ≤ f b t − fb
t
α < b − t < b t < β
Trang 2and, further when ft is a C2-function, by the condition on the second derivative
fb ≥ 0 α < b < β
On the other hand, it is easy to see that1.1 is equivalent to the following requirement
on the divided difference:
f b − fa
b − a ≤ f c − fb
c − b
α < a < b < c < β
or even to the inequality
f b − fac − b ≤f c − fbb − a α < a ≤ b ≤ c < β
LetMn be the linear space of n × n complex matrices, and H nitsreal subspace of n×n Hermitian matrices The identity matrix I will be denoted simply by 1, and correspondingly,
a scalar λ will represent λI For Hermitian A, B the order relation A ≤ B means that B − A is
positive-semidefinite, or equivalently
will mean that B − A is positive definite; that is, B ≥ A and B − A is invertible see 2 for basic facts about matrices.
Notice that for scalars α, β and Hermitian X the order relation α < X < β is equivalent
to the condition that every eigenvalue of X is in the interval α, β Denote by H n α, β the convex set of Hermitian matrices X such that α < X < β A continuous function ft, defined
onα, β, induces a nonlinear map X → fX from H n α, β to H n through the familiar
functional calculus, that is,
f X : U diagf λ1, , fλ nU∗ 1.8
with a unitary matrix U which diagonalizes X as
U∗XU diag λ1, , λ n . 1.9
The function ft is said to be matrix-convex of order n, or simply n-convex on the interval α, β
if the map X → fX is convex on H n α, β or more exactly
f λA 1 − λB ≤ λfA 1 − λfB 0≤ λ ≤ 1; A, B ∈ H n
α, β
1.10
see 3,4 This is a formal matrix-version of 1.1 In view of 1.7 this convexity means that
n
Trang 3Just as in the scalar case for the matrix-convexity the following matrix-version of1.2
is sufficient:
2fB ≤ fB X fB − X B ± X ∈ H n
α, β
This means that ft is n-convex if and only if the map t → fB tX is convex on −1, 1 when B ± X ∈ H n α, β.
The 1-convexity is nothing but the usual convexity of the function ft It is easy to see that n-convexity implies m-convexity for all 1 ≤ m ≤ n.
It is knownsee 3 that if ft is 2-convex then it is already a C2-function, andsee
5,6 that for each n there is an n-convex function which is not n 1-convex.
It should be mentioned here that in his original definition of n-convexity Kraus 3 restricted the requirement1.11 only for X ≥ 0 with rankX 1 We will return to this point
later
The corresponding matrix-versions of 1.5 and 1.6 have no definite meaning becausefB − fAC − B or fB − fAB − A−1is no longer Hermitian
On the spaceMn the most useful linear functional is the Trace, in symbol, TrX, which
is defined as the sum of diagonal entries of X with respect to any orhonormal basis The useful properties of the trace are commutativity, TrXY TrYX, and positivity, that is, X ≤
Y ⇒ TrX ≤ TrY .
We will use a characterization of positive semidefiniteness X ≥ 0 in terms of trace:
X ≥ 0 ⇐⇒ Tr XY ≥ 0 0 ≤ Y of rank-one. 1.12
Notice in this connection that if both X, Y are Hermitian, then TrXY is a real number.
Our main aim is to establish trace-versions of 1.5 and 1.6 The trace-version for
1.6 is quite natural
Theorem 1.1 A continuous function ft on an interval α, β is n-convex if and only if
Tr
f B − fAC − B ≤ Trf C − fBB − A A ≤ B ≤ C in H n
α, β
. †n
On the other hand, the trace-version for1.5 turns out quite restrictive.
Theorem 1.2 Let n ≥ 2 A continuous function ft on an interval α, β satisfies the condition
Tr
f B − fAB − A−1≤ Trf C − fBC − B−1
A < B < C in H n
α, β
‡n
if and only if it is of the form ft at2 bt c with a ≥ 0, and b, c ∈ R.
Trang 42 Preliminary
In order to prove theorems, we use a well-established regularization techniquesee 7 I-4
Take a nonnegative symmetric C∞-function ϕt on −∞, ∞ such that
ϕ t 0 |t| ≥ 1,
∞
and for > 0 let ϕ : ϕt−1−1 Then ϕ t is a nonnegative, symmetric C∞-function such that
ϕ t 0 |t| ≥ ,
∞
Given a continuous function ft on an interval α, β, setting ft 0 outside of the
intervalα, β, define f t as the convolution of this extended function f with ϕ , that is,
f t :f ϕ
t
∞
The following is well known
Lemma 2.1 The function f is a C∞-function, in fact,
d k
dt k f f d k
and f t converges to ft uniformly on each compact subset of the interval α, β as → 0.
Lemma 2.2 Let ft be a continuous function on an interval α, β.
i ft satisfies †n on α, β if and only if for small > 0 the function f t satisfies †n on
α , β − .
ii ft is n-convex on α, β if and only if for small > the function f t is n-convex on
α , β − .
Proof i Let ft satisfy †n on α, β Suppose that α < A ≤ B ≤ C < β − , then
Tr
f C − f BB − A − Trf B − f AC − B
−
Tr
f C − s − fB − s {B − s − A − s}
−Trf B − s − fA − s {C − s − B − s} ϕ sds ≥ 0,
2.5
Trang 5α < A − s ≤ B − s ≤ C − s < β |s| ≤ . 2.6
The converse statement is clear by the second half ofLemma 2.1
The proof ofii is also easy and omitted
In a similar way we have the following
Lemma 2.3 A continuous function ft satisfies ‡ n on α, β if and only if for small > 0 the
functin f t satisfies ‡n on α , β − .
When ft is a C1-function onα, β, B ∈ H n α, β and X ∈ H n , the map t → fB tX
is defined for small|t| and differentiable at t 0 The derivative of this map at t 0 will be
denoted byDfB; X, that is,
DfB; X : d
dt
t0
When B is daigonal as B diagλ1, , λ n, it is known see 2 V-3 and 8 6-6 that
DfB; X f1
λ i , λ j n i,j1◦x ij n i,j1 for X
x ij n i,j1 , 2.8
where ◦ denotes the Schur product entrywise product and f1s, t is the first divided
difference of f, defined as
f1s, t
⎧
⎪
⎪
f s − ft
s − t , if s / t,
fs, if s t.
2.9
Notice thatf1λ i , λ jn i,j1is a real symmetric matrix
In a similar way when ft is a C2-function, the second derivative of the map t → fB
tX at t 0 is written as see 2 V-3 and 8 6-6
d2
dt2
t0
f B tX 2
n
k1
f2
λ i , λ k , λ j
x ik x kj
n
i,j1 for X
x ij n i,j1 , 2.10
where f2s, t, u is the second divided difference of f, defined as
f2s, t, u f1s, t − f1t, u
s − u
⎧
⎪
⎪
⎪
⎪
⎪
⎪
f1s, t − f1t, u
s − u , if s / u,
fs − f1t, s
s − t , if s u / t,
fs
2.11
Trang 6Since the functional calculus is invariant for unitary similarity, that is, fV∗XV
V∗fXV , the formulas 2.8 and 2.10 well determine the forms of derivatives
Lemma 2.4 If ft is a C1-function on an interval α, β, then
Tr
DfB; X · Y TrDfB; Y · X B ∈ H n
α, β
; X, Y ∈ H n
Proof We may assume that B diagλ1, , λ n, then by 2.8 for X x ijn
i,j1 and Y
y ijn
i,j1
Tr
DfB; X · Y n
i,j1
f1
λ i , λ j
x ij y ji
n
i,j1
f1
λ j , λ i
y ji x ij TrDfB; Y · X.
2.13
3 Proofs of Theorems
By Lemmas2.3and2.4we may assume that ft in theorems is a C∞-function
Proof of Theorem 1.1 Suppose that the function ft satisfies †n on α, β Take B ∈ H n α, β
and 0 ≤ X, Y of rank-one such that C : B X and A : B − tY for small t > 0 belong to
Hn α, β Since A ≤ B ≤ C, by assumption †n we have
Tr
f B − fB − tYX
hence by2.7
Tr
Then it follows fromLemma 2.4that
Tr
Since 0≤ X, Y of rank-one are arbitrary, it follows from 3.3 and 1.12 that for any 0 ≤ X of rank one such that α < B ± X < β
and similarly
Trang 72fB ≤ fB X fB − X B ± X ∈ H n
α, β and 0≤ X of rank-one. 3.6
This means that the matrix-valued function t → fB tX is convex under the condition that
0≤ X is of rank-one.
At this point we proved the n-convexity in the sense of Kraus 3 as mentioned in
Section 1 The remaining part is essentially the same as Kraus’ approach3
Since for 0≤ X of rank-one and small t > 0 by 3.6
0≤ f B tX fB − tX − 2fB
dt2
t0
we can conclude from2.10 that for 0 ≤ X x ijn
i,j1of rank-one
n
k1
f2
λ i , λ k , λ j
x ik x kj
n i,j1
For each t > 0, consider a positive semidefinite matrix of rank-one
0≤ X x ij n i,j1: t ij−2 n
Then by3.8
0≤
n
k1
f2
λ i , λ k , λ j
x ik x kj
n
i,j1
diag1, t, , t n−1n
k1
f2
λ i , λ k , λ j
t2k−1
n
i,j1
diag
1, t, , t n−1
,
3.10
which implies
n
k1
f2
λ i , λ k , λ j
t2k−1
n i,j1
Letting t → 0, we have
C1 : f2
λ i , λ1, λ j
n
Trang 8In a similar way we can see that
C k: f2
λ i , λ k , λ j
n i,j1 ≥ 0 k 1, 2, , n. 3.13
Now since for any X x ijn
i,j1 ∈ Hn each matrixx ik x kjn
i,j1 k 1, 2, , n is
positive semidefinite and of rank-one, it follows from3.8 that
d2
dt2
t0
f B tX 2
n
k1
f2
λ i , λ k , λ j
x ik x kj
n i,j1
2n
k1
C k◦x ik x jk n i,j1 ≥ 0.
3.14
Here we used the well-known fact that the Schur product of two positive semidefinite matrices is again positive semidefinitesee 2 I-6 Therefore
d2
dt2
t0
f B tX ≥ 0 B ∈ H n
α, β
; X ∈ H n
which implies the convexity of the map t → fB tX whenever B ± X ∈ H n α, β This completes the proof of the n-convexity of the function ft.
Suppose conversely that ft is n-convex on the interval α, β, then by 1.3
f C − fB fB C − B − fB ≥ d
dt
t0
f B tC − B
DfB; C − B,
3.16
so that by1.12
Tr
f C − fBB − A ≥ TrDfB; C − B · B − A, 3.17 and similarly
Tr
f B − fAC − B ≤ TrDfB; B − A · C − B. 3.18 Now byLemma 2.4we can conclude
Tr
f B − fAC − B ≤ Trf C − fBB − A, 3.19
which shows that the function ft satisfies †n This completes the whole proof of
Theorem 1.1
In the above proof we really showed the following
Trang 9Theorem 3.1 A continuous function ft on an interval α, β is n-convex if and only if TrfB −
fAC − B ≤ TrfC − fBB − A, whenever A ≤ B ≤ C in H n α, β and rankB − A
rankC − B 1
Notice that Kraus3 cf 8, Theorem 6.6.52 really showed, for n ≥ 2, that ft is
n-convex on α, β if and only if it is a C2-function and
f2
λ i , λ1, λ j
n i,j1 ≥ 0 ∀λ1, , λ n∈α, β
For the proof ofTheorem 1.2, let us start with an easy lemma
Lemma 3.2 If condition ‡ n for ft is valid on α, β, so is condition (‡ m ) for 1 ≤ m < n.
Proof Given A < B < C in H m α, β, take λ ∈ α, β and small > 0 and consider the n × n
matrices
A :
0 λ − I n−m
B 0
0 λI n−m
, C :
0 λ I n−m
where I n−mis then − m × n − m identity matrix Then since A < B < C in H n α, β and
Tr
f
B− fA
B − A−1
Trf B − fAB − A−1 n − m f λ − fλ −
Tr
f
C
− fBC − B−1
Trf C − fBB − A−1 n − m f λ − fλ
3.22
by letting → 0 it follows from ‡n that
Tr
f B − fAB − A−1≤ Trf C − fBC − B−1. 3.23 This completes the proof
In view of Lemma 3.2 the essential part of the proof of Theorem 1.2 is in the next lemma
Lemma 3.3 If a C1- function ft satisfies (‡2) on α, β, then
fs ft 2f1s, t α < s, t < β
Trang 10Proof Take B diagt1, t2 with t1, t2 ∈ α, β Then for any 2 × 2 positive definite X, Y > 0 and small > 0 we have by assumption
Trf B − fB − X
−1≤ Trf B Y − fB
Letting → 0 by 2.8 this leads to the inequality
Tr
f1
t i , t j
2
i,j1 ◦ X
· X−1≤ Tr
f1
t i , t j
2
i,j1 ◦ Y
Replacing X and Y we have also
Tr
f1
t i , t j
2
i,j1 ◦ Y
· Y−1≤ Tr
f1
t i , t j
2
i,j1 ◦ X
Those together show that
Tr
f1
t i , t j
2
i,j1 ◦ X
· X−1 constant 0 < X ∈ H2. 3.28
It is easy to see that a 2× 2 positive definite matrix X with TrX 1 is of the form
X
a u
a 1 − a
u
a 1 − a 1− a
0 < a < 1; |u| < 1. 3.29 Now it follows3.28 and 3.29 that
Tr
f1
t i , t j
2
i,j1 ◦ X
· X−1
ft1 ft2 − 2|u|2
f1t1, t2
1− |u|2 constant |u| < 1,
3.30
which is possible only when3.24 is valid
Proof of Theorem 1.2 Suppose that a C2-function ft satisfies ‡n on α, β By Lemmas3.2
and3.3ft satisfies the identity 3.24 Therefore we have
ft fs t − s 2f t − fs α < s, t < β
Twice differentiating both sides with respect to t we arrive at
ftt − s 0 α < s, t < β
3.32
Trang 11which is possible only when ft is a quadratic function
Finally a ≥ 0 follows from the usual convexity of ft.
Suppose conversely that ft is of the form 3.33 with a ≥ 0 Take A < B < C in H n Then
Tr
f B − fAB − A−1 aTrB2− A2
and correspondingly
Tr
f C − fBC − B−1 aTrC2− B2
Since
we have
Tr
B2− A2
and correspondingly
Tr
C2− B2
Therefore we arrive at the inequality
Tr
f C − fBC − B−1− Trf B − fAB − A−1 a{TrC − TrA} ≥ 0. 3.39
This shows that ft satisfies ‡n for any n and on any interval α, β.
Acknowledgment
The author would like to thank Professor Fumio Hiai for his valuable comments on the original version of this paper
References
1 A W Roberts and D E Varberg, Convex Functions, vol 57 of Pure and Applied Mathematics, Academic
Press, New York, NY, USA, 1973
2 R Bhatia, Matrix Analysis, vol 169 of Graduate Texts in Mathematics, Springer, New York, NY, USA, 1997.
... Trang 9Theorem 3.1 A continuous function ft on an interval α, β is n-convex if and only if TrfB... 3.3 and 1.12 that for any ≤ X of rank one such that α < B ± X < β
and similarly
Trang 72fB... j
n
Trang 8In a similar way we can see that
C k: f2