REFINEMENTS OF HERMITE-HADAMARD
INEQUALITY ON SIMPLICES
FLAVIA-CORINA MITROI and CĂTĂLIN IRINEL SPIRIDON
Communicated by the former editorial board
Some refinements of the Hermite-Hadamard inequality are obtained in the case of
continuous convex functions defined on simplices.
AMS 2010 Subject Classification: 26A51.
Key words: Hermite-Hadamard inequality, convex function, simplex.
1. INTRODUCTION
The aim of this paper is to provide a refinement of the Hermite-Hadamard
inequality on simplices.
Suppose that K is a metrizable compact convex subset of a locally convex
Hausdorff space E. Given a Borel probability measure µ on K, one can prove
the existence of a unique point bµ ∈ K (called the barycenter of µ) such that
Z
x0 (bµ ) =
x0 (x) dµ(x),
K
for all continuous linear functionals x0 on E. The main feature of barycenter
is the inequality
Z
f (bµ ) ≤
f (x) dµ(x),
K
valid for every continuous convex function f : K → R. It was noted by several
authors that this inequality is actually equivalent to the Jensen inequality.
The following theorem, due to G. Choquet, complements this inequality
and relates the geometry of K to a given mass distribution.
Theorem 1 (The general form of Hermite-Hadamard inequality). Let µ
be a Borel probability measure on a metrizable compact convex subset K of a
locally convex Hausdorff space. Then, there exists a Borel probability measure
ν on K which has the same barycenter as µ, is zero outside Ext K, and verifies
MATH. REPORTS 15(65), 1 (2013), 69–78
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Flavia-Corina Mitroi and Cătălin Irinel Spiridon
2
the double inequality
Z
Z
(1)
f (x) dµ(x) ≤
f (bµ ) ≤
f (x) dν(x),
Ext K
K
for all continuous convex functions f : K → R.
Here Ext K denotes the set of all extreme points of K.
The details can be found in [7], pp. 192–194. See also [6].
In the particular case of simplices one can take advantage of the barycentric coordinates.
Suppose that ∆ ⊂ Rn is an n-dimensional simplex of vertices P1 , . . . ,
Pn+1 . In this case Ext ∆ = {P1 , . . . , Pn+1 } and each x ∈ ∆ can represented
uniquely as a convex combination of vertices,
n+1
X
(2)
λk (x)Pk = x,
k=1
where the coefficients λk (x) are nonnegative numbers (depending on x) and
n+1
X
(3)
λk (x) = 1.
k=1
Each function λk : x → λk (x) is an affine function on ∆. This can be easily
seen by considering the linear system consisting of the equations (2) and (3).
The coefficients λk (x) can be computed in terms of Lebesgue volumes
(see [1], [5]). We denote by ∆j (x) the subsimplex obtained when the vertex
Pj is replaced by x ∈ ∆. Then, one can prove that
(4)
λk (x) =
Vol (∆k (x))
,
Vol (∆)
for all k =R 1, . . . , n + 1 (the geometric interpretation is very intuitive). Here
Vol (∆) = ∆ dx.
In the particular case where dµ(x) = dx/ Vol (∆) we have λk bdx/ Vol(∆)
1
= n+1
, for every k = 1, . . . , n + 1.
The above discussion leads to the following form of Theorem 1 in the
case of simplices:
Corollary 1. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices
P1 , . . . , Pn+1 and µ be a Borel probability measure on ∆ with barycenter bµ .
Then, for every continuous convex function f : ∆ → R,
Z
n+1
X
f (bµ ) ≤
f (x) dµ(x) ≤
λk (bµ )f (Pk ) .
∆
k=1
3
Refinements of Hermite-Hadamard inequality
71
Proof. In fact,
Z
Z
f (x) dµ(x) =
∆
f
∆
=
n+1
X
k=1
n+1
X
!
dµ(x) ≤
λk (x)Pk
k=1
Z
λk (x) dµ(x) =
f (Pk )
∆
Z n+1
X
λk (x) f (Pk ) dµ(x)
∆ k=1
n+1
X
λk (bµ )f (Pk ) .
k=1
On the other hand,
n+1
X
Z
Ext ∆
k=1
and
n+1
P
fd
λk (bµ )f (Pk ) =
n+1
X
!
λk (bµ )δPk
k=1
λk (bµ )δPk is the only Borel probability measure ν concentrated at the
k=1
vertices of ∆ which verifies the inequality
Z
Z
f (x) dµ(x) ≤
∆
f (x) dν(x)
Ext ∆
for every continuous convex functions f : ∆ → R. Indeed, ν must be of
n+1
n+1
P
P
the form ν =
αk δPk , with bν =
αk Pk = bµ and the uniqueness of
k=1
k=1
barycentric coordinates yields the equalities
αk = λk (bµ ),
for k = 1, . . . , n + 1. It is worth noticing that the Hermite–Hadamard inequality is not just a
consequence of convexity, it actually characterizes it. See [9].
The aim of the present paper is to improve the result of Corollary 1, by
providing better bounds for the arithmetic mean of convex functions defined
on simplices.
2. MAIN RESULTS
We start by extending the following well known inequality concerning
the continuous convex functions defined on intervals
Z b
1
1 f (a) + f (b)
a+b
f (x) dx ≤
+f
.
b−a a
2
2
2
See [7], p. 52.
Theorem 2. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P1 , . . . ,
Pn+1 , endowed with the normalized Lebesgue measure dx/ Vol (∆) . Then, for
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4
every continuous convex function f : ∆ → R and every point P ∈ ∆ we have
(5)
1
Vol (∆)
Z
∆
1
f (x) dx ≤
n+1
n+1
X
!
(1 − λk (P )) f (Pk ) + f (P ) .
k=1
In particular,
(6)
1
Vol (∆)
Z
∆
1
f (x) dx ≤
n+1
!
n+1
n X
f (Pk ) + f bdx/ Vol(∆) .
n+1
k=1
Proof. Consider the barycentric representation P =
According to Corollary 1,
1
λi (P ) Vol (∆)
f (x) dx ≤
∆i (P )
1 
n+1
k
λk (P ) Pk ∈ ∆.


Z
P
X
f (Pk ) + f (P ) ,
k6=i
where ∆i (P ) denotes the simplex obtained from ∆ by replacing the vertex Pi
by P . Notice that (4) yields λi (P ) Vol (∆) = Vol (∆i (P )). Multiplying both
sides by λi (P ) and summing up over i, we obtain the inequality (5).
In the particular case when P = bdx/ Vol(∆) , all coefficients λk (P ) equal
1/(n + 1). Remark 1. Using [1, Theorem 1] instead of Corollary 1 of the present
paper, one may improve the statement of Theorem 3 by the cancellation of the
continuity condition. Such statement is proved independently, by a different
approach, in [12].
An extension of Theorem 2 is as follows:
Theorem 3. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P1 , . . . ,
Pn+1 endowed with the Lebesgue measure and let ∆0 ⊆ ∆ a subsimplex of
0
vertices P10 , . . . , Pn+1
which has the same barycenter as ∆. Then, for every
continuous convex function f : ∆ → R, and every index j ∈ {1, . . . , n + 1}, we
have the estimates


Z
X
X
1
1 
(7)
f (x) dx ≤
λi Pk0 f (Pi ) + f Pj0 
Vol (∆) ∆
n+1
k6=j
≤
1
n+1
X
i
i
f (Pi )
5
Refinements of Hermite-Hadamard inequality
73
and

(8)
1
Vol (∆)
Z
f (x) dx ≥
X
∆
0
λ i Pj f 
i

1 
Pk + Pj0 
n+1
k6=i
!
1 X
Pi
n+1
≥f

X
.
i
Proof. We will prove here only the inequalities (7), the proof of (8) being
similar.
By Corollary 1, for each index i,

λi
1
Pj0 Vol (∆)
Z
1 
n+1
f (x) dx ≤
∆i (Pj0 )

X
f (Pk ) + f Pj0  ,
k6=i
where ∆i (Pj0 ) denotes the simplex obtained from ∆ by replacing the vertex Pi
by Pj0 . Notice that (4) yields λi (Pj0 ) Vol (∆) = Vol(∆i (Pj0 )). By multiplying
both sides by λi (Pj0 ) and summing over i we obtain

1
Vol (∆)
Z
f (x) dx ≤
∆
(9)
1 
n+1

X
λi Pj0
X
f (Pk ) + f Pj0 
i
k6=i
1
=
n+1
X
X
0
1
=
n+1
X
!
!
λ i Pj
i
0
f (Pk ) − f (Pi )
+ f Pj
k
!
1−
λi Pj0
f (Pi ) + f
Pj0
.
i
Furthermore, since ∆0 and ∆ have the same barycenter, we have
X
1 − λi Pj0 =
λi Pk0 ,
k6=j
for all i = 1, . . . , n + 1. Then

1
Vol (∆)
Z
f (x) dx ≤
∆
1 
n+1

XX
k6=j
i
λi Pk0 f (Pi ) + f Pj0  .
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6
The fact that ∆0 and ∆ have the same barycenter and the convexity of the
function f yield
X
XX
f (Pj ) =
λi Pk0 f (Pi )
j
k
=
i
XX
k6=j
X
λi Pk0 f (Pi ) +
λi Pj0 f (Pi )
i
i
!
≥
XX
k6=j
=
This concludes the proof (7).
λi Pk f (Pi ) + f
i
XX
k6=j
0
X
0
λi Pj Pi
i
λi Pk0 f (Pi ) + f Pj0 .
i
When ∆0 as a singleton, the inequality (7) coincides with the inequality
(6). Notice that if we omit the barycenter condition then, the inequality (9)
translates into (5).
An immediate consequence of Theorem 3 is the following result due to
Farissi [3]:
Corollary 2. Assume that f : [a, b] → R is a convex function and
λ ∈ [0, 1]. Then
Z b
f (a)+f (b)
1
(1 − λ) f (a) + λf (b) + f (λa + (1 − λ) b)
f (x) dx ≤
≤
b−a a
2
2
and
Z b
b+(1−λ) a+λb
1
a+(1−λ) a+λb
f (x) dx ≥ λf
+(1−λ) f
b−a a
2
2
a+b
≥f
.
2
Proof. Apply Theorem 3 for n = 1 and ∆0 the subinterval of endpoints
(1 − λ) a + λb and λa + (1 − λ) b. Another refinement of the Hermite-Hadamard inequality in the case of
simplices is as follows.
Theorem 4. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P1 , . . . ,
Pn+1 endowed with the Lebesgue measure and let ∆0 ⊆ ∆ be a subsimplex
dx
whose barycenter with respect to the normalized Lebesgue measure Vol(∆
0 ) is
P
P = k λk (P )Pk . Then, for every continuous convex function f : ∆ → R,
Z
X
1
f (x) dx ≤
λj (P )f (Pj ) .
(10)
f (P ) ≤
0
Vol (∆ ) ∆0
j
7
Refinements of Hermite-Hadamard inequality
75
Proof. Let Pk0 , k = 1, . . . , n + 1 be the vertices of ∆0 . By Corollary 1,
Z
1
1 X
0
f (P ) ≤
f
(x)
dx
≤
.
f
P
k
Vol (∆0 ) ∆0
n+1
k
The barycentric representation of each of the points Pk0 ∈ ∆ gives us
( P
0
0
j λj (Pk ) Pj = Pk
P
.
0
k λk (Pj ) = 1
Since f is a convex function,


X
X
1 X
1
f Pk0 =
f
λj Pk0 Pj 
n+1
n+1
j
k
k
!
X
X
1 X
0
λj (P ) f (Pj )
≤
λj Pk f (Pj ) =
n+1
j
j
k
and the assertion of Theorem 4 is now clear.
As a corollary of Theorem 4, we get the following result due to Vasić and
Lacković [10], and Lupaş [2] (cf. J.E. Pečarić et. al. [8]).
Corollary 3. Let p and q be two positive numbers and a1 ≤ a ≤ b ≤ b1 .
Then, the inequalities
Z A+y
pa + qb
1
pf (a) + qf (b)
f
≤
f (x) dx ≤
p+q
2y A−y
p+q
hold for A = pa+qb
p+q , y > 0 and all continuous convex functions f : [a1 , b1 ] → R
if and only if
b−a
y≤
min {p, q} .
p+q
The right-hand side of the inequality stated in Theorem 4 can be improved as follows.
Theorem 5. Suppose that ∆ is an n-dimensional simplex of vertices
P1 , . .P
. , Pn+1 and let P ∈ ∆. Then, for every subsimplex ∆0 ⊂ ∆ such that
P = j λj (P ) Pj = bdx/ Vol(∆0 ) . Then


Z
1
1  X
f (x) dx ≤
n
λj (P )f (Pj ) + f (P ) ,
Vol (∆0 ) ∆0
n+1
j
for every continuous convex function f : ∆ → R.
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Flavia-Corina Mitroi and Cătălin Irinel Spiridon
8
Proof. Let Pk0 , k = 1, . . . , n + 1 be the vertices of ∆0 . We denote by ∆0i
the subsimplex obtained by replacing the vertex Pi0 by the barycenter P of the
normalized measure dx/ Vol (∆0 ) on ∆0 .
According to Corollary 1,
Z
1 X
1
1
0
f
(x)
dx
≤
f
P
+
f (P )
k
Vol (∆0i ) ∆0i
n+1
n+1
k6=i


1 X X
1
=
f
f (P )
λj Pk0 Pj  +
n+1
n+1
j
k6=i


X
X
1
1

≤
λj Pk0  f (Pj ) +
f (P )
n+1
n+1
j
k6=i
X
1
1
λj (P ) −
λj Pi0 f (Pj ) +
f (P ),
=
n+1
n+1
j
for each index i = 1, . . . , n + 1. Summing up over i we obtain
Z
n+1
f (x) dx ≤
Vol (∆0 ) ∆0
X
X 1 X
≤ (n + 1)
λj (P )f (Pj ) −
λj Pi0 f (Pj ) + f (P )
n+1
j
i
j
X
=n
λj (P )f (Pj ) + f (P ) .
j
and the proof of the theorem is completed.
Of course, the last theorem yields an improvement of Corollary 3. This
was first noticed in [8, p. 146].
We end this paper with an alternative proof of some particular case of a
result established by A. Guessab and G. Schmeisser [4, Theorem 2.4]. Before
we start, we would like to turn the reader’s attention to the paper [12] by
S. Wasowicz, where the result we are going to present was obtained by some
more general approach (see [11, Theorem 2 and Corollary 4]).
Theorem 6. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P1 , . . . ,
Pn+1 endowed with the Lebesgue measure and
P let M1 , . . . , Mm be points in ∆
such that bdx/ Vol(∆) is a convex combination j βj Mj of them. Then, for every
continuous convex function f : ∆ → R, the following inequalities hold:
m
X
f bdx/ Vol(∆) ≤
βj f (Mj ) ≤
j=1
n+1
1 X
f (Pi ) .
n+1
i=1
9
Refinements of Hermite-Hadamard inequality
77
Proof. The first inequality follows from Jensen’s inequality. In order to
establish the second inequality, since we have


!
X X
X
X
1 X

bdx/ Vol(∆) =
βj
λi (Mj ) Pi =
βj λi (Mj ) Pi =
Pi ,
n+1
j
we infer that
X
i
i
P
j
j
X
for every i and


!
X X
X

βj λi (Mj ) f (Pi )
λi (Mj ) Pi ≤
βj f
j
=
1
n+1
i
i
X
This completes the proof.
i
1
n+1 ,
βj λi (Mj ) =
βj f (Mj ) =
j
j
f (Pi ) .
i
Acknowledgements. The first author has been supported by CNCSIS Grant
420/2008. The authors are indebted to Szymon Wasowicz for pointing out to them
references [4] and [11].
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[11] S. Wasowicz, On some extremalities in the approximate integration. Math. Inequal.
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Received 23 May 2011
University of Craiova
Department of Mathematics
13 A.I. Cuza Street
200585 Craiova, Romania
[email protected]
catalin [email protected]