Volume 9 (2008), Issue 3, Article 74, 8 pp.
INEQUALITIES OF JENSEN-PEČARIĆ-SVRTAN-FAN TYPE
CHAOBANG GAO AND JIAJIN WEN
C OLLEGE O F M ATHEMATICS A ND I NFORMATION S CIENCE
C HENGDU U NIVERSITY
C HENGDU , 610106, P.R. C HINA
kobren427@163.com
wenjiajin623@163.com
Received 22 January, 2007; accepted 06 July, 2008
Communicated by S.S. Dragomir
A BSTRACT. By using the theory of majorization, the following inequalities of Jensen-PečarićSvrtan-Fan type are established: Let I be an interval, f : I → R and t ∈ I, x, a, b ∈ I n . If a1 ≤
· · · ≤ an ≤ bn ≤ · · · ≤ b1 , a1 + b1 ≤ · · · ≤ an + bn ; f (t) > 0, f 0 (t) > 0, f 00 (t) > 0, f 000 (t) < 0
for any t ∈ I, then
fn,n (a)
fk+1,n (a)
fk,n (a)
f1,n (a)
A(f (a))
f (A(a))
=
≤ ··· ≤
≤
≤ ··· ≤
=
,
f (A(b))
fn,n (b)
fk+1,n (b)
fk,n (b)
f1,n (b)
A(f (b))
the inequalities are reversed for f 00 (t) < 0, f 000 (t) > 0, ∀t ∈ I, where A(·) is the arithmetic
mean and
X
1
xi1 + · · · + xik
fk,n (x) := n
f
, k = 1, . . . , n.
k
k
1≤i1 <···<ik ≤n
Key words and phrases: Jensen’s inequality, Pečarić-Svrtan’s inequality, Fan’s inequality, Theory of majorization, Hermite
matrix.
2000 Mathematics Subject Classification. 26D15, 26E60.
1. I NTRODUCTION
In what follows, we shall use the following symbols:
x := (x1 , . . . , xn );
A(x) :=
f (x) := (f (x1 ), . . . , f (xn ));
G(x) := (x1 x2 · · · xn )1/n ;
x1 + x2 + · · · + xn
; Rn+ := [0, +∞)n ; Rn++ := (0, +∞)n ;
n
I n := {x|xi ∈ I, i = 1, . . . , n, I is an interval};
This work is supported by the Natural Science Foundation of China (10671136) and the Natural Science Foundation of Sichuan Province
Education Department (07ZA207).
The authors are deeply indebted to anonymous referees for many useful comments and keen observations which led to the present improved
version of the paper as it stands, and also are grateful to their friend Professor Wan-lan Wang for numerous discussions and helpful suggestions
in preparation of this paper.
033-07
2
C HAOBANG G AO AND J IAJIN W EN
1
fk,n (x) := n
k
X
f
1≤i1 <···<ik ≤n
xi1 + · · · + xik
k
,
k = 1, . . . , n.
Jensen’s inequality states that: Let f : I → R be a convex function and x ∈ I n . Then
f (A(x)) ≤ A(f (x)).
(1.1)
This well-known inequality has a great number of generalizations in the literature (see [1] –
[6]). An interesting generalization of (1.1) due to Pečarić and Svrtan [5] is:
f (A(x)) = fn,n (x) ≤ · · · ≤ fk+1,n (x) ≤ fk,n (x) ≤ · · · ≤ f1,n (x) = A(f (x)).
(1.2)
In 2003, Tang and Wen [6] obtained the following generalization of (1.2):
fr,s,n ≥ · · · ≥ fr,s,i ≥ · · · ≥ fr,s,s ≥ · · · ≥ fr,j,j ≥ · · · ≥ fr,r,r = 0,
(1.3)
where
fr,s,n
n n
:=
(fr,n − fs,n ),
r
s
fk,n := fk,n (x),
1 ≤ r ≤ s ≤ n.
Ky Fan’s arithmetic-geometric mean inequality is (see [7]): Let x ∈ (0, 1/2]n . Then
A(x)
G(x)
≥
.
A(1 − x)
G(1 − x)
(1.4)
In this paper, we shall establish further extensions of (1.2) and (1.4) as follows:
Theorem 1.1. Let I be an interval. If f : I → R, a, b ∈ I n (n ≥ 2) and
(i) a1 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b1 , a1 + b1 ≤ · · · ≤ an + bn ;
(ii) f (t) > 0, f 0 (t) > 0, f 00 (t) > 0, f 000 (t) < 0 for any t ∈ I,
then
f (A(a))
fn,n (a)
fk+1,n (a)
fk,n (a)
f1,n (a)
A(f (a))
=
≤ ··· ≤
≤
≤ ··· ≤
=
.
f (A(b))
fn,n (b)
fk+1,n (b)
fk,n (b)
f1,n (b)
A(f (b))
(1.5)
The inequalities are reversed for f 00 (t) < 0, f 000 (t) > 0, ∀t ∈ I. The equality signs hold
if and only if a1 = · · · = an and b1 = · · · = bn .
In Section 3, several interesting results of Ky Fan shall be deduced. In Section 4, the matrix
variant of (1.5) will be established.
2. P ROOF OF T HEOREM 1.1
Lemma 2.1. Let f : I → R be a function whose second derivative exists and x ∈ I n ,
α ∈ Ωn = {α ∈ Rn+ : α1 + · · · + αn = 1}.
Writing
S(α, x) :=
1 X
f (α1 xi1 + · · · + αn xin ),
n! i ···i
1
where
P
i1 ···in
n
denotes summation over all permutations of {1, 2, . . . , n},
S(α, a)
F (α) := log
, a, b ∈ I n ,
S(α, b)
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I NEQUALITIES OF J ENSEN -P E ČARI Ć -S VRTAN -FAN T YPE
ui (x) := α1 xi1 + α2 xi2 +
vi (x) := α1 xi2 + α2 xi1 +
n
X
j=3
n
X
3
αj xij ,
αj xij ,
i = (i1 , i2 , . . . , in ).
j=3
Then there exist ξi (a) between ui (a) and vi (a), and ξi (b) between ui (b) and vi (b) such that
∂F
∂F
(2.1) (α1 − α2 )
−
∂α1 ∂α2
X f 00 (ξi (a))(ui (a) − vi (a))2 f 00 (ξi (b))(ui (b) − vi (b))2 1 X
=
−
,
n! i ···i 1≤i <i ≤i
S(α, a)
S(α, b)
n
n
3
1
2
P
where i3 ···in denotes the summation over all permutations of {1, 2, . . . , n} \ {i1 , i2 }.
Proof. Note the following identities:
1 X X
S(α, x) =
f (α1 xi1 + · · · + αn xin )
n! i ···i 1≤i 6=i ≤n
n
3
1
2
1 X X
=
[f (ui (x)) + f (vi (x))];
n! i ···i 1≤i <i ≤n
3
n
1
2
∂
∂
[f (ui ) + f (vi )] −
[f (ui ) + f (vi )] = [f 0 (ui ) − f 0 (vi )](xi1 − xi2 );
∂α1
∂α2
∂S
∂S
1 X X
(α1 − α2 )
−
=
[f 0 (ui ) − f 0 (vi )](α1 − α2 )(xi1 − xi2 )
∂α1 ∂α2
n! i ···i 1≤i <i ≤n
n
3
1
2
1 X X
(2.2)
[f 0 (ui ) − f 0 (vi )](ui − vi ).
=
n! i ···i 1≤i <i ≤n
3
n
1
2
By F (α) = log S(α, a) − log S(α, b) and (2.2), we have
∂F
∂F
(α1 − α2 )
−
∂α1 ∂α2
∂S(α, a)
−1 ∂S(α, a)
= (α1 − α2 ) [S(α, a)]
−
∂α1
∂α2
∂S(α, b)
−1 ∂S(α, b)
−[S(α, b)]
−
∂α1
∂α2
0
0
1 X X
[f (ui (a)) − f (vi (a))][ui (a) − vi (a)]
=
n! i ···i 1≤i <i ≤n
S(α, a)
n
3
1
2
[f 0 (ui (b)) − f 0 (vi (b))][ui (b) − vi (b)]
−
S(α, b)
00
1 X X
f (ξi (a))[ui (a) − vi (a)]2 f 00 (ξi (b))[ui (b) − vi (b)]2
=
−
.
n! i ···i 1≤i <i ≤n
S(α, a)
S(α, b)
3
n
1
2
Here we used the Mean Value Theorem for f 0 (t). This completes the proof.
Lemma 2.2. Under the hypotheses of Theorem 1.1, F is a Schur-convex function or a Schurconcave function on Ωn , where F is defined by Lemma 2.1.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 74, 8 pp.
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4
C HAOBANG G AO AND J IAJIN W EN
Proof. It is easy to see that Ωn is a symmetric convex set and F is a differentiable symmetric
function on Ωn . To prove that F is a Schur-convex function on Ωn , it is enough from [8, p .57]
to prove that
∂F
∂F
(2.3)
(α1 − α2 )
−
≥ 0, ∀α ∈ Ωn .
∂α1 ∂α2
To prove (2.3), it is enough from Lemma 2.1 to prove
(2.4)
f 00 (ξi (a))[ui (a) − vi (a)]2
f 00 (ξi (b))[ui (b) − vi (b)]2
≥
.
S(α, a)
S(α, b)
Using the given conditions a1 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b1 , f (t) > 0 and f 0 (t) > 0, we obtain
that aj ≤ bj (j = 1, 2, . . . , n) and the inequalities:
1
1
≥
> 0.
S(α, a)
S(α, b)
(2.5)
By the given condition (i) of Theorem 1.1 and 1 ≤ i1 < i2 ≤ n, we have
ai2 − ai1 ≥ bi1 − bi2 ≥ 0
and
(2.6)
[ui (a) − vi (a)]2 ≥ [ui (b) − vi (b)]2 ≥ 0.
From (2.5) and (2.6), we get
(2.7)
[ui (a) − vi (a)]2
[ui (b) − vi (b)]2
≥
≥ 0.
S(α, a)
S(α, b)
Note that a, b ∈ I n , ui (a), vi (a), ui (b), vi (b) ∈ I, and
min{ui (a), vi (a)} ≤ ξi (a)
≤ max{ui (a), vi (a)}
≤ min{ui (b), vi (b)}
≤ ξi (b)
≤ max{ui (b), vi (b)}.
It follows that
(2.8)
ξi (a) ≤ ξi (b)
(ξi (a), ξi (b) ∈ I).
If f 00 (t) > 0, f 000 (t) < 0 for any t ∈ I, from these and (2.8) we get
(2.9)
f 00 (ξi (a)) ≥ f 00 (ξi (b)) > 0.
Combining with (2.7) and (2.9), we have proven that (2.4) holds, hence, F is a Schur-convex
function on Ωn .
Similarly, if f 00 (t) < 0, f 000 (t) > 0 for any t ∈ I, we obtain
(2.10)
−f 00 (ξi (a)) ≥ −f 00 (ξi (b)) > 0.
Combining with (2.7) and (2.10), we know that the inequalities are reversed in (2.4) and (2.3).
Therefore, F is a Schur-concave function on Ωn . This ends the proof of Lemma 2.2.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 74, 8 pp.
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I NEQUALITIES OF J ENSEN -P E ČARI Ć -S VRTAN -FAN T YPE
5
Remark 1. When α1 6= α2 , there is equality in (2.3) if a1 = · · · = an and b1 = · · · = bn .
In fact, there is equality in (2.3) if and only if there is equality in (2.5), (2.8), (2.9) and the
first inequality in (2.6) or all the equality signs hold in (2.6). For the first case, by a1 ≤ · · · ≤
an ≤ bn ≤ · · · ≤ b1 , we get a1 = · · · = an , b1 = · · · = bn . For the second case, we have
ai1 − ai2 = 0 = bi1 − bi2 . Since 1 ≤ i1 < i2 ≤ n and i1 , i2 are arbitrary, we get a1 = · · · = an ,
b1 = · · · = bn . Clearly, if a1 = · · · = an , b1 = · · · = bn , then (2.3) reduces to an equality.
Proof of Theorem 1.1. First we note that if


α = αk := k −1 , k −1 , . . . , k −1 , 0, . . . , 0 ,
|
{z
}
k
we obtain that
S(αk , x) = fk,n (x)
and
F (αk ) = log
(2.11)
fk,n (a)
.
fk,n (b)
By Lemma 2.2, we observe that F (α) is a Schur-convex(concave) function on Ωn . Using
αk+1 ≺ αk for αk , αk+1 ∈ Ωn and the definition of Schur-convex(concave) functions, we have
[8]
(2.12)
F (αk+1 ) ≤ (≥)F (αk ),
k = 1, . . . , n − 1.
It follows from (2.11) and (2.12) that (1.5) holds. Since αk+1 6= αk , combining this fact with
Remark 1, we observe that the equality signs hold in (1.5) if and only if a1 = · · · = an ,
b1 = · · · = bn . This completes the proof of Theorem 1.1.
3. C OROLLARY OF T HEOREM 1.1
Corollary 3.1. Let 0 < r < 1, s ≥ 1, 0 < ai ≤ 2−1/s , bi = (1 − asi )1/s , i = 1, . . . , n, f (t) = tr ,
t ∈ (0, 1). Then the inequalities in (1.5) are reversed.
Proof. Without loss of generality, we can assume that 0 < a1 ≤ · · · ≤ an . By bi = (1 − asi )1/s
and 0 < ai ≤ 2−1/s (i = 1, . . . , n), we have
0 < a1 ≤ · · · ≤ an ≤ 2−1/s ≤ bn ≤ · · · ≤ b1 < 1.
Now we take g(t) := t + (1 − ts )1/s (0 < t ≤ 2−1/s ), so g 0 (t) = 1 − (1 − ts )(1/s)−1 ts−1 ≥ 0,
i.e., g is an increasing function. Thus
a1 + b 1 ≤ · · · ≤ an + b n .
It is easy to see that f (t) = tr > 0, f 0 (t) = rtr−1 > 0, f 00 (t) = r(r − 1)tr−2 < 0, f 000 (t) =
r(r − 1)(r − 2)tr−3 > 0 for any t ∈ (0, 1). By Theorem 1.1, Corollary 3.1 can be deduced. This
completes the proof.
Corollary 3.2. Let a ∈ (0, 1/2]n . Writing
Y
[AG; x]k,n :=
1≤i1 <···<ik
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 74, 8 pp.
xi1 + · · · + xik
k
≤n
!
1
(nk)
,
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6
C HAOBANG G AO AND J IAJIN W EN
we have
(3.1)
A(a)
[AG; a]n,n
=
A(1 − a)
[AG; 1 − a]n,n
[AG; a]k+1,n
[AG; a]k,n
≥ ··· ≥
≥
[AG; 1 − a]k+1,n
[AG; 1 − a]k,n
G(a)
[AG; a]1,n
≥ ··· ≥
=
.
[AG; 1 − a]1,n
G(1 − a)
Equalities hold throughout if and only if a1 = · · · = an . (Compare (3.1) with [7, 10, 11])
Proof. We choose s = 1 in Corollary 3.1. Raising each term to the power of 1/r and letting
r → 0 in (1.5), (3.1) can be deduced. This ends the proof.
Corollary 3.3. Let f : I → R be such that f (t) > 0, f 0 (t) > 0, f 00 (t) > 0, f 000 (t) < 0 for any
t ∈ I. Let Φ : I0 → I be increasing and Ψ : I0 → I be decreasing, and suppose that Φ + Ψ is
increasing and sup Φ ≤ inf Ψ. Then
R
R
−1
f |I0 |
Φdt
f (Φ)dt
I0
I0
R
(3.2)
≤
,
R
f (Ψ)dt
f |I0 |−1 I0 Ψdt
I0
where |I0 | is the length of the interval I0 . The inequality is reversed for f 00 (t) < 0, f 000 (t) > 0,
∀t ∈ I.
(a))
In fact, since (3.2) is an integral version of the inequality ff (A(a))
≤ A(f
, therefore (3.2)
(A(b))
A(f (b))
holds by Theorem 1.1.
According to Theorem 1.1, (1.5) implies inequalities (1.1), (1.2) and (3.1), and the implication (3.1) to (1.4) is obvious. Consequently, Theorem 1.1 is a generalization of Jensen’s
inequality (1.1), Pečarić-Svrtan’s inequalities (1.2) and Fan’s inequality (1.4). Note that Theorem 1.1 contains a great number of inequalities as special cases. To save space we omit the
details.
4. A M ATRIX VARIANT
P
Let A = (aij )n×n (n ≥ 2) be a Hermite matrix of order n. Then tr A = ni=1 aii is the trace
of A. As is well-known, there exists a unitary matrix U such that A = U diag(λ1 , . . . , λn )U ∗ ,
where U ∗ is the transpose conjugate matrix of U and the components of λ = (λ1 , . . . , λn ) are
the eigenvalues of A. Thus tr A = λ1 + · · · + λn . Let λ ∈ I n . Then, for f : I → R, we
define f (A) := U diag(f (λ1 ), . . . , f (λn ))U ∗ (see [9]). Note that diag(λ1 , . . . , λn ) = U ∗ AU .
Based on the above, we may use the following symbols: If, for A, we keep the elements on the
cross points of the i1 , . . . , ik th rows and the i1 , . . . , ik th columns; replacing the other elements
by nulls, then we denote this new matrix by Ai1 ···ik . Clearly, we have tr[U ∗ AU ]i1 ···ik = λi1 +
· · · + λik . Thus we also define that
X
1
λi1 + · · · + λik
fk,n (A) := n
f
k
k 1≤i1 <···<ik ≤n
X
1
1
∗
= n
f
tr[U AU ]i1 ···ik .
k
k 1≤i <···<i ≤n
1
k
In particular, we have
n
1X
1
f1,n (A) =
f (λi ) = tr(f (A));
n i=1
n
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I NEQUALITIES OF J ENSEN -P E ČARI Ć -S VRTAN -FAN T YPE
7
λ1 + · · · + λ n
1
fn,n (A) = f
=f
trA ;
n
n
n
trA − λi
E · trA − A
1
1X
f
fn−1,n (A) =
= trf
,
n i=1
n−1
n
n−1
where E is a unit matrix. In fact, from
trA − λ1
trA − λn
E · trA − A
∗
U
U = diag
,...,
,
n−1
n−1
n−1
we get
X
n
E · trA − A
trA − λi
trf
=
f
.
n−1
n
−
1
i=1
Based on the above facts and Theorem 1.1, we observe the following.
Theorem 4.1. Let I be an interval and let λ, µ ∈ I n . Suppose the components of λ,µ are the
eigenvalues of Hermitian matrices A and B. If
(i) λ1 ≤ · · · ≤ λn ≤ µn ≤ · · · ≤ µ1 , λ1 + µ1 ≤ · · · ≤ λn + µn ;
(ii) the function f : I → R satisfies f (t) > 0, f 0 (t) > 0, f 00 (t) > 0, f 000 (t) < 0 for any
t ∈ I, and we have
trf E·trA−A
f n1 trA
fk+1,n (A)
fk,n (A)
trf (A)
n−1
≤
≤
·
·
·
≤
≤
≤
·
·
·
≤
.
f
(B)
f
(B)
trf
(B)
f n1 trB
trf E·trB−B
k+1,n
k,n
n−1
The inequalities are reversed for f 00 (t) < 0, f 000 (t) > 0, ∀t ∈ I. Equalities hold
throughout if and only if λ1 = · · · = λn and µ1 = · · · = µn .
Remark 2. If I = (0, 1/2], 0 < λ1 ≤ · · · ≤ λn ≤ 1/2, B = E − A, then the precondition (i)
of Theorem 4.1 can be satisfied.
Remark 3. Lemma 2.2 possesses a general and meaningful result that should be an important
theorem. Theorem 1.1 is only an application of Lemma 2.2.
Remark 4. If f (t) < 0, f 0 (t) < 0 for any t ∈ I, then we can apply Theorem 1.1 to −f .
Remark 5. In [12, 13], several applications on Jensen’s inequalities are displayed.
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J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 74, 8 pp.
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