Table Of ContentRefinements of Hermite-Hadamard inequality on
simplices
Flavia-Corina Mitroi and Cătălin Irinel Spiridon
2
1
0
Some refinements of the Hermite-Hadamard inequality are obtained in the case of continuous
2
convexfunctionsdefinedonsimplices.
n
a AMS2010 Subject Classification: 26A51.
J
Key words: Hermite-Hadamardinequality,convex function,simplex.
1
]
A
1 Introduction
C
.
h The aim of this paper is to provide a refinement of the Hermite-Hadamard
t
a inequality on simplices.
m
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
7 the existence of a unique point b ∈ K (called the barycenter of µ) such that
µ
v
3
4 x′(b ) = x′(x)dµ(x)
µ
0 ZK
5
. for all continuous linear functionals x′ on E. The main feature of barycenter
5
0 is the inequality
1
f(b )≤ f(x)dµ(x),
1 µ
: ZK
v
valid for every continuous convex function f : K → R. It was noted by several
i
X
authors that this inequality is actually equivalent to the Jensen inequality.
r
The following theorem due to G. Choquet complements this inequality and
a
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 ExtK, and verifies
the double inequality
(1) f(b ) ≤ f(x)dµ(x) ≤ f(x)dν(x)
µ
ZK ZExtK
1
for all continuous convex functions f :K → R.
Here ExtK denotes the set of all extreme points of K.
The details can be found in [7], pp. 192-194. See also [6].
Intheparticularcaseofsimplicesonecantakeadvantageofthebarycentric
coordinates.
Suppose that ∆ ⊂ Rn is an n-dimensional simplex of vertices P ,...,P .
1 n+1
InthiscaseExt∆ = {P ,...,P }andeachx ∈ ∆canberepresenteduniquely
1 n+1
as a convex combination of vertices,
n+1
(2) λ (x)P = x,
k k
k=1
X
where the coefficients λ (x) are nonnegative numbers (depending on x) and
k
n+1
(3) λ (x) = 1.
k
k=1
X
Each function λ : x → λ (x) is an affine function on ∆. This can be easily
k k
seen by considering the linear system consisting of the equations (2) and (3).
The coefficients λ (x) can be computed in terms of Lebesgue volumes (see
k
[1], [5]). We denote by ∆ (x) the subsimplex obtained when the vertex P is
j j
replaced by x ∈ ∆. Then one can prove that
Vol(∆ (x))
k
(4) λ (x) =
k
Vol(∆)
for all k = 1,...,n + 1 (the geometric interpretation is very intuitive). Here
Vol(∆)= dx.
∆
Intheparticularcasewheredµ(x)= dx/Vol(∆)wehaveλ b =
R k dx/Vol(∆)
1 for every k = 1,...,n+1.
n+1 (cid:0) (cid: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
P ,...,P and µ be a Borel probability measure on ∆ with barycenter b .
1 n+1 µ
Then for every continuous convex function f : ∆ → R,
n+1
f(b ) ≤ f(x)dµ(x)≤ λ (b )f (P ).
µ k µ k
Z∆ k=1
X
Proof. In fact,
n+1 n+1
f(x)dµ(x)= f λ (x)P dµ(x) ≤ λ (x) f(P )dµ(x)
k k k k
Z∆ Z∆ k=1 ! Z∆k=1
X X
n+1 n+1
= f(P ) λ (x) dµ(x) = λ (b )f (P ).
k k k µ k
k=1 Z∆ k=1
X X
2
On the other hand
n+1 n+1
λ (b )f(P )= fd λ (b )δ
k µ k k µ Pk
k=1 ZExt∆ k=1 !
X X
and n+1λ (b )δ is the only Borel probability measure ν concentrated at
k=1 k µ Pk
the vertices of ∆ which verifies the inequality
P
f(x)dµ(x) ≤ f(x)dν(x)
Z∆ ZExt∆
for every continuous convex functions f : ∆ → R. Indeed ν must be of the
form ν = n+1α δ , with b = n+1α P = b and the uniqueness of
k=1 k Pk ν k=1 k k µ
barycentric coordinates yields the equalities
P P
α = λ (b ) for k = 1,...,n+1.
k k µ
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 con-
tinuous convex functions defined on intervals:
1 b 1 f (a)+f(b) a+b
f(x)dx ≤ +f .
b−a 2 2 2
Za (cid:18) (cid:18) (cid:19)(cid:19)
See [7], p. 52.
THEOREM 2. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices
P ,...,P ,endowedwiththenormalizedLebesguemeasuredx/Vol(∆).Then
1 n+1
for every continuous convex function f : ∆ → R and every point P ∈ ∆ we
have
n+1
1 1
(5) f(x)dx ≤ (1−λ (P))f (P )+f(P)
k k
Vol(∆) n+1
Z∆ k=1 !
X
In particular,
n+1
1 1 n
(6) f(x)dx ≤ f (P )+f b .
k dx/Vol(∆)
Vol(∆) n+1 n+1
Z∆ k=1 !
X (cid:0) (cid:1)
3
Proof. Consider the barycentric representation P = λ (P)P ∈ ∆ .
k k k
According to Corollary 1,
P
1 1
f(x)dx ≤ f (P )+f(P) ,
k
λ (P)Vol(∆) n+1
i Z∆i(P) k6=i
X
where ∆ (P) denotes the simplex obtained from ∆ by replacing the vertex P
i i
by P. Notice that (4) yields λ (P)Vol(∆) = Vol(∆ (P)). Multiplying both
i i
sides by λ (P) and summing up over i we obtain the inequality (5).
i
In the particular case when P = b all coefficients λ (P) equal
dx/Vol(∆) k
1/(n+1).
REMARK 1. Using [1, Theorem 1] instead of Corollary 1 of the present
paper, one may improve the statement of Theorem 2 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
P ,...,P endowed with the Lebesgue measure and let ∆′ ⊆ ∆ a subsimplex
1 n+1
of vertices P′,...,P′ which has the same barycenter as ∆. Then for every
1 n+1
continuous convex function f : ∆ → R, and every index j ∈ {1,...,n+1}, we
have the estimates
1 1
(7) f(x)dx ≤ λ P′ f (P )+f P′
Vol(∆) n+1 i k i j
Z∆ k6=j i
XX (cid:0) (cid:1) (cid:0) (cid:1)
1
≤ f (P )
i
n+1
i
X
and
1 1
(8) f(x)dx ≥ λ P′ f P +P′
Vol(∆) i j n+1 k j
Z∆ i k6=i
X (cid:0) (cid:1) X
1
≥ f P
i
n+1
!
i
X
Proof. We will prove here only the inequalities (7), the proof of (8) being
similar.
By Corollary 1, for each index i,
1 1
f(x)dx ≤ f(P )+f P′ ,
λi Pj′ Vol(∆)Z∆i(Pj′) n+1 Xk6=i k (cid:0) j(cid:1)
(cid:16) (cid:17)
4
where∆ P′ denotes thesimplexobtainedfrom ∆by replacingthevertexP
i j i
by P′. No(cid:16)tice(cid:17)that (4) yields λ P′ Vol(∆)= Vol(∆ P′ ). By multiplying
j i j i j
both sides by λ P′ and summ(cid:16)ing(cid:17)over i we obtain (cid:16) (cid:17)
i j
(cid:16) (cid:17)
1 1
f(x)dx ≤ λ P′ f (P )+f P′
Vol(∆) n+1 i j k j
Z∆ i k6=i
X (cid:0) (cid:1)X (cid:0) (cid:1)
1
(9) = λ P′ f(P )−f(P ) +f P′
n+1 i j k i j
! !
i k
X (cid:0) (cid:1) X (cid:0) (cid:1)
1
= 1−λ P′ f (P )+f P′ .
n+1 i j i j
!
i
X(cid:0) (cid:0) (cid:1)(cid:1) (cid:0) (cid:1)
Furthermore, since ∆′ and ∆ have the same barycenter, we have
1−λ P′ = λ P′
i j i k
k6=j
(cid:0) (cid:1) X (cid:0) (cid:1)
for all i= 1,...,n+1. Then
1 1
f (x)dx ≤ λ P′ f(P )+f P′ .
Vol(∆) n+1 i k i j
Z∆ k6=j i
XX (cid:0) (cid:1) (cid:0) (cid:1)
The fact that ∆′ and ∆ have the same barycenter and the convexity of the
function f yield
f(P ) = λ P′ f(P )
j i k i
j k i
X XX (cid:0) (cid:1)
= λ P′ f(P )+ λ P′ f(P )
i k i i j i
k6=j i i
XX (cid:0) (cid:1) X (cid:0) (cid:1)
≥ λ P′ f(P )+f λ P′ P
i k i i j i
!
k6=j i i
XX (cid:0) (cid:1) X (cid:0) (cid:1)
= λ P′ f(P )+f P′ .
i k i j
k6=j i
XX (cid:0) (cid:1) (cid:0) (cid:1)
This concludes the proof (7).
When ∆′ 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).
5
An immediate consequence of Theorem 2 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
1 b (1−λ)f(a)+λf(b)+f (λa+(1−λ)b)
f(x)dx ≤
b−a 2
Za
f (a)+f (b)
≤
2
and
1 b a+(1−λ)a+λb b+(1−λ)a+λb
f(x)dx ≥λf +(1−λ)f
b−a 2 2
Za (cid:18) (cid:19) (cid:18) (cid:19)
a+b
≥f .
2
(cid:18) (cid:19)
Proof. Apply Theorem 2 for n = 1 and ∆′ the subinterval of endpoints
(1−λ)a+λb and λa+(1−λ)b.
Anotherrefinement oftheHermite-Hadamard inequality inthecaseofsim-
plices is as follows.
THEOREM 4. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices
P ,...,P endowed with the Lebesgue measure and let ∆′ ⊆ ∆ be a subsim-
1 n+1
plexwhosebarycenterwithrespecttothenormalizedLebesguemeasure dx
Vol(∆′)
is P = λ (P)P . Then for every continuous convex function f : ∆ → R,
k k k
1
P
(10) f(P)≤ f(x)dx≤ λ (P)f (P ).
Vol(∆′)Z∆′ j j j
X
Proof. Let P′, k = 1,...,n+1 be the vertices of ∆′. By Corollary 1,
k
1 1
f (P) ≤ f(x)dx ≤ f P′ ,
Vol(∆′) Z∆′ n+1 k k
X (cid:0) (cid:1)
The barycentric representation of each of the points P′ ∈ ∆ gives us
k
λ (P′)P =P′
j j k j k
.
λ P′ = 1
( Pk k j
(cid:16) (cid:17)
Since f is a convex function, P
1 1
f P′ = f λ P′ P
n+1 k n+1 j k j
k k j
X (cid:0) (cid:1) X X (cid:0) (cid:1)
1
≤ λ P′ f(P ) = λ (P)f(P )
n+1 j k j j j
!
j k j
X X (cid:0) (cid:1) X
6
and the assertion of Theorem 2 is now clear.
Asacorollary ofTheorem 2wegetthefollowing resultdueVasićandLack-
ović [10], and Lupaş [2] (cf. J. E. Pečarić et al. [8]).
COROLLARY3. Letpandq betwopositivenumbersanda ≤ a ≤ b ≤ b .
1 1
Then the inequalities
pa+qb 1 A+y pf(a)+qf(b)
f ≤ f(x)dx ≤
p+q 2y p+q
(cid:18) (cid:19) ZA−y
hold for A = pa+qb, y > 0 and all continuous convex functions f : [a ,b ] → R
p+q 1 1
if and only if
b−a
y ≤ min{p,q}.
p+q
The right-hand side of the inequality stated in Theorem 2 can be improved
as follows.
THEOREM 5. Suppose that ∆ is an n-dimensional simplex of vertices
P ,...,P and let P ∈ ∆. Then for every subsimplex ∆′ ⊂ ∆ such that
1 n+1
P = jλj(P)Pj = bdx/Vol(∆′) we have
P
1 1
f(x)dx ≤ n λ (P)f (P )+f (P)
Vol(∆′) Z∆′ n+1 j j j
X
for every continuous convex function f : ∆ → R.
Proof. Let P′, k = 1,...,n + 1 be the vertices of ∆′. We denote by ∆′
k i
the subsimplex obtained by replacing the vertex P′ by the barycenter P of the
i
normalized measure dx/Vol(∆′) on ∆′.
According to Corollary 1,
1 1 1
f(x)dx≤ f P′ + f (P)
Vol(∆′i) Z∆′i n+1 k6=i k n+1
X (cid:0) (cid:1)
1 1
= f λ P′ P + f(P)
n+1 j k j n+1
k6=i j
X X (cid:0) (cid:1)
1 1
≤ λ P′ f(P )+ f (P)
n+1 j k j n+1
j k6=i
X X (cid:0) (cid:1)
1 1
= λ (P)− λ P′ f(P )+ f (P)
j n+1 j i j n+1
j (cid:18) (cid:19)
X (cid:0) (cid:1)
7
for each index i = 1,...,n+1. Summing up over i we obtain
n+1
f(x)dx
Vol(∆′)Z∆′
1
≤ (n+1) λ (P)f (P )− λ P′ f (P )+f (P)
j j n+1 j i j
j i j
X X X (cid:0) (cid:1)
= n λ (P)f(P )+f (P).
j j
j
X
and the proof of the theorem is completed.
Of course, the last theorem yields an improvement of Corollary 2. This
was first noticed in [8, pp. 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.
Wąsowicz, wheretheresultwearegoingtopresentwas obtainedbysomemore
general approach (see [11, Theorems 2 and Corollary 4]).
THEOREM 6. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices
P ,...,P endowed with the Lebesgue measure and let M ,...,M be points
1 n+1 1 m
in ∆ such that b is a convex combination β M of them. Then for
dx/Vol(∆) j j j
every continuous convex function f :∆ → R, the following inequalities hold:
P
m n+1
1
f b ≤ β f (M )≤ f(P ).
dx/Vol(∆) j j i
n+1
j=1 i=1
(cid:0) (cid:1) X X
Proof. The first inequality follows from Jensen’s inequality. In order to
establish the second inequality, since we have
1
b = β λ (M )P = β λ (M ) P = P ,
dx/Vol(∆) j i j i j i j i i
n+1
!
j i i j i
X X X X X
we infer that β λ (M ) = 1 for every i and
j j i j n+1
P
β f (M )= β f λ (M )P ≤ β λ (M ) f (P )
j j j i j i j i j i
!
j j i i j
X X X X X
1
= f(P ).
i
n+1
i
X
This completes the proof.
Acknowledgement. The first author has been supported by CNCSIS
Grant 420/2008. The authors are indebted to Szymon Wąsowicz for pointing
out to them the references [11] and [4].
8
References
[1] M. Bessenyei, The Hermite–Hadamard Inequality on Simplices, American
Mathematical Monthly, volume 115, April 2008, pages 339–345.
[2] A.Lupaş, A generalization of Hadamard inequalities for convex functions,
Univ.Beograd.Publ.Elektrotehn.Fak.Ser.Mat.Fiz.(1976),no.544-576,
115-121.
[3] A. El Farissi, Simple proof and refinement of Hermite-Hadamard inequal-
ity, J. Math. Inequal., 4 (2010), No. 3, 365–369.
[4] A. Guessab and G. Schmeisser, Convexity results and sharp error esti-
mates in approximate multivariate integration, Math. Comp., 73 (2004),
pp. 1365–1384.
[5] E. Neuman, J. E. Pečarić, Inequalities involving multivariate convex func-
tions, J. Math. Anal. Appl. 137 (1989), no. 2, 541-549.
[6] C. P. Niculescu, The Hermite-Hadamard inequality for convex functions
of a vector variable, Math. Inequal. Appl., 4 (2002), 619–623.
[7] C. P. Niculescu, L.-E. Persson, Convex Functions and their Applications.
A Contemporary Approach, CMSBooksinMathematicsvol.23,Springer-
Verlag, New York, 2006.
[8] J. E. Pečarić, F. Proschan, Y. L. Tong, Convex Functions, Partial Order-
ingsandStatisticalApplications,MathematicsinScienceandEngineering,
vol. 187, 1992.
[9] T. Trif, Characterizations of convex functions of a vector variable via
Hermite-Hadamard’s inequality, J. Math. Inequal., 2 (2008), 37–44.
[10] P.M.VasićandI.B.Lacković, Some complements to the paper: On an in-
equality for convex functions, Univ. Beograd. Publ. Elektrotech. Fak. Ser.
Mat. Fiz. No. 461-497 (1974), 63-66; Univ. Beograd. Publ. Elektrotehn.
Fak. Ser. Mat. Fiz. (1976), no. 544-576, 59-62.
[11] S. Wąsowicz, Hermite-Hadamard-type inequalities in the approximate in-
tegration, Math. Inequal. Appl., 11 (2008), 693–700.
[12] S. Wąsowicz, A. Witkowski, On some inequality of Hermite-Hadamard
type, Submitted.
Department of Mathematics
University of Craiova
Street A. I. Cuza 13, Craiova, RO-200585
Romania
[email protected], [email protected]
9
Refinements of Hermite-Hadamard inequality on
simplices
Flavia-Corina Mitroi and Cătălin Irinel Spiridon
2
1
0
Some refinements of the Hermite-Hadamard inequality are obtained in the case of continuous
2
convexfunctionsdefinedonsimplices.
n
a
J AMS2010 Subject Classification: 26A51.
1
] Key words: Hermite-Hadamardinequality,convex function,simplex.
A
C
. 1.INTRODUCTION
h
t
a The aim of this paper is to provide a refinement of the Hermite-Hadamard
m
inequality on simplices.
[
Suppose that K is a metrizable compact convex subset of a locally convex
7 Hausdorff space E. Given a Borel probability measure µ on K, one can prove
v
the existence of a unique point b ∈ K (called the barycenter of µ) such that
3 µ
4
0 x′(b ) = x′(x)dµ(x)
µ
5
ZK
.
5 for all continuous linear functionals x′ on E. The main feature of barycenter
0
1 is the inequality
1
f(b )≤ f(x)dµ(x),
: µ
v ZK
Xi valid for every continuous convex function f : K → R. It was noted by several
r authors that this inequality is actually equivalent to the Jensen inequality.
a
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 ExtK, and verifies
the double inequality
(1) f(b ) ≤ f(x)dµ(x) ≤ f(x)dν(x)
µ
ZK ZExtK
1
