Rev. Anal. Num´er. Th´eor. Approx., vol. 37 (2008) no. 2, pp. 127–132 ictp.acad.ro/jnaat
A SIMPLE PROOF OF POPOVICIU’S INEQUALITY
MIHALY BENCZE∗and FLORIN POPOVICI†
Abstract. T. Popoviciu [5] has proved in 1965 the following inequality relating the values of a convex functionf :I →Rat the weighted arithmetic means of the subfamilies of a given family of pointsx1, ..., xn∈I:
X
1≤i1<···<ip≤n
(λi1+· · ·+λip)fλi
1xi1+···+λipxip λi1+···+λip
≤ n−2p−2
"
n−p p−1
n
X
i=1
λif(xi) +
n
X
i=1
λi
!
f λ1xλ1+···+λnxn
1+···+λn
# .
Heren≥3, p∈ {2, ..., n−1}andλ1, ..., λn are positive numbers (representing weights). The aim of this paper is to give a simple argument based on mathe- matical induction and a majorization lemma.
MSC 2000. Primary 26A51, 26D15; Secondary 26B25.
Keywords. Popoviciu’s inequality, convex function, convex combination.
T. Popoviciu [5] has proved in 1965 the following inequality relating the values of a convex functionf :I →Rat the weighted arithmetic means of the different subfamilies of a given family of points x1, ..., xn∈I:
X
1≤i1<···<ip≤n
(λi1 +· · ·+λip)fλi1xλi1+···+λipxip
i1+···+λip
≤ n−2p−2
"
n−p p−1
n
X
i=1
λif(xi) +
n
X
i=1
λi
!
fλ1xλ1+···+λnxn
1+···+λn
# .
Heren≥3, p∈ {2, ..., n−1}andλ1, ..., λnare positive numbers (representing weights);I is a nonempty interval.
The inequality above (denoted (Pn,p) in what follows) is nontrivial even in the case of triplets (that is, when n = 3 and p = 2). Several alternative approaches of (P3,2) are discussed in the recent book of C. P. Niculescu and L.-E. Persson [2]. See [4] and [3] for additional information.
∗Aprily Lajos College, Bra¸sov, Romania, e-mail: [email protected]
†Grigore Moisil College, Bra¸sov, Romania, e-mail: [email protected]
Theoretically, Popoviciu’s inequality is a refinement of Jensen’s inequality since it yields
f
n
P
i=1
λixi n
P
i=1
λi
≤ 1
n−1 p−1
n
P
i=1
λi
X
1≤i1<···<ip≤n
(λi1+· · ·+λip)fλi1λxi1+···+λipxip
i1+···+λip
≤ n−pn−1
n
P
i=1
λif(xi)
n
P
i=1
λi
+n−1p−1 f
n
P
i=1
λixi n
P
i=1
λi
≤
n
P
i=1
λif(xi)
n
P
i=1
λi
.
The aim of the present paper is to offer a simple argument of (Pn,p) based on mathematical induction and the following variant of the majorization in- equality:
Lemma 1. Let f : [a, b]→R be a convex function. If x1, ..., xn∈[a, b] and a convex combination Pn
k=1
µkxk of these points equals a convex combination λ1a+λ2b of the endpoints, then
n
X
k=1
µkf(xk)≤λ1f(a) +λ2f(b).
Proof. This can be established easily by using the barycentric coordinates (in our case the fact that every pointxk∈[a, b] can be expressed uniquely as a convex combination ofaand b).
A second argument is based on the geometric meaning of convexity. Denot- ing byA(x) the affine function joining (a, f(a)) with (b, f(b)),we have
n
X
k=1
µkf(xk) ≤
n
X
k=1
µkA(xk) =A
n
X
k=1
µkxk
!
= A(λ1a+λ2b) =λ1A(a) +λ2A(b)
= λ1f(a) +λ2f(b).
It is worth to mention that Lemma 1 still works (with obvious changes) within the framework of convex functions on simplices.
We pass now to the proof of Popoviciu’s inequality, by considering first the case where n∈N, n≥3 and p=n−1 :
(Pn,n−1) X
1≤i≤n
λif(xi) + (n−2)
X
1≤i≤n
λi
f
P
1≤i≤n
λixi
P
1≤i≤n
λi
≥ X
1≤j≤n
X
1≤i≤n, i6=j
λi
f
P
1≤i≤n, i6=j
λixi
P
1≤i≤n, i6=j
λi
.
Clearly, we may assume
x1 ≤x2 ≤ · · · ≤xn. Choose k∈ {1, ..., n−1}such that
xk≤
n
P
i=1
λixi n
P
i=1
λi
≤xk+1
and put
yj = P
1≤i≤n, i6=j
λixi
P
1≤i≤n, i6=j
λi
forj= 1, ..., n.Then it is clear that
(1)
n
P
i=1
λixi n
P
i=1
λi
≤yj ≤
n
P
i=k+1
λixi n
P
i=k+1
λi
for all j∈ {1, ..., k}.
We have
k
P
j=1
P
1≤i≤n, i6=j
λi
yj
k
P
j=1
P
1≤i≤n, i6=j
λi
=
k
P
j=1
P
1≤i≤n, i6=j
λi
P
1≤i≤n, i6=j
λixi
P
1≤i≤n, i6=j
λi
k
P
j=1
P
1≤i≤n, i6=j
λi
=
(k−1)
n
P
i=1
λixi+
n
P
i=k+1
λixi
(k−1)
n
P
i=1
λi+
n
P
i=k+1
λi
=
(k−1)
n P
i=1
λi
n
P
i=1
λixi n
P
i=1
λi
+
n P
i=k+1
λi
n
P
i=k+1
λixi n
P
i=k+1
λi
(k−1)
n
P
i=1
λi+
n
P
i=k+1
λi
so that by (1) and Lemma 1 we infer the inequality
k
X
j=1
X
1≤i≤n, i6=j
λi
f
P
1≤i≤n, i6=j
λixi
P
1≤i≤n, i6=j
λi
≤
≤(k−1)
n
X
i=1
λi
! f
n
P
i=1
λixi n
P
i=1
λi
+
n
X
i=k+1
λi
f
n
P
i=k+1
λixi n
P
i=k+1
λi
.
Or, by Jensen’s inequality,
k
X
j=1
X
1≤i≤n, i6=j
λi
f
P
1≤i≤n, i6=j
λixi
P
1≤i≤n, i6=j
λi
≤
≤(k−1)
n
X
i=1
λi
! f
n
P
i=1
λixi n
P
i=1
λi
+
n
X
i=k+1
λif(xi)
and
n
X
j=k+1
X
1≤i≤n, i6=j
λi
f
P
1≤i≤n, i6=j
λixi
P
1≤i≤n, i6=j
λi
≤
≤(n−k−1)
n
X
i=1
λi
! f
n
P
i=1
λixi n
P
i=1
λi
+
k
X
i=1
λif(xi)
whence we may conclude (Pn,n−1).
Consider now the case where n∈N, p≥3.We will prove that (Pn,p)⇒(Pn,p−1)
that is, if Popoviciu’s inequality works for families of n weighted points by grouping them into subfamilies of size p∈ {3, ..., n−1}then it also works by grouping them into subfamilies of sizep−1.
By Lemma 1,
λi1f(xi1) +· · ·+λipf(xip) + (k−2)(λi1+· · ·+λip)fλi1λxi1+···+λipxip
i1+···+λip
≥
≥
p
X
j=1
X
1≤k≤p, k6=j
λik
f
P
1≤k≤p, k6=j
λikxik
P
1≤k≤p, k6=j
λik
whence X
1≤i1<···<ip≤n
(λi1+· · ·+λip)fλi1λxi1+···+λipxip
i1+···+λip
≥ p−21 − n−1p−1
n
X
i=1
λif(xi)
+(n−p+ 1) X
1≤i1<···<ip−1≤n
(λi1+· · ·+λip−1)f
λi1xi1+···+λip−1xip−1 λi1+···+λip−1
.
By our hypothesis we get
n−2 p−1
n−p p−1
n
X
i=1
λif(xi) +
n
X
i=1
λi
! f
n
P
i=1
λixi n
P
i=1
λi
≥
≥ p−21 − n−1p−1
n
X
i=1
λif(xi) + (n−p+ 1)
X
1≤i1<···<ip−1≤n
(λi1 +· · ·+λip−1)f
λi1xi1+···+λip−1xip−1 λi1+···+λip−1
,
that is, n−2
p−2
n−p
p−1 + n−1p−1p−21
n
X
i=1
λif(xi) + n−2p−2
n
X
i=1
λi
! f
n
P
i=1
λixi n
P
i=1
λi
≥ n−p+1p−2 X
1≤i1<···<ip−1≤n
(λi1+· · ·+λip−1)f
λi1xi1+···+λip−1xip−1 λi1+···+λip−1
. Since
n−2 p−2
n−p
p−1 + n−1p−1p−21 = n−2p−2n−p+1p−2 and
n−2 p−2
= n−2p−3n−p+1p−2 we can restate the last inequality as follows:
n−2 (p−1)−2
n−(p−1) (p−1)−1
n
X
i=1
λif(xi) +
n
X
i=1
λi
! f
n
P
i=1
λixi n
P
i=1
λi
≥ X
1≤i1<···<ip−1≤n
(λi1 +· · ·+λip−1)f λ
i1xi1+···+λip−1xip−1 λi1+···+λip−1
, which proves to be precisely (Pn,p−1).
The proof of Popoviciu’s inequality is now complete.
Remark2. The induction step is not necessary in deriving the unweighted case of the inequalities (Pn,2) :
(nPn,2) (n − 2)f(x1)+···+f(xn n) + f x1+···+xn n ≥ 2n X
1≤j<k≤n
fxj+x2 k
for all x1, ..., xn in the domain off.
In fact, assuming that
x1 ≤ · · · ≤xn
we will consider first the case where
x1+xn
2 ≤ x1+···+xn n. Then, by Lemma 1 we get
(M) n1 f(x1) +f x1+x2 2+· · ·+f x1+x2 n ≤ 12 f(x1) +f x1+···+xn n while from Jensen’s inequality we infer that
(J) 2n X
2≤j<k≤n
fxj+x2 k≤ 2nn−22
n
X
i=2
f(xi).
Summing up (M) and (J) we get (nPn,2).The case where
x1+xn
2 ≥ x1+···+xn n
can be treated in a similar way (changing the role of the indices 1 and n in
(M)).
At first glance Popoviciu’s inequality is a one real variable result. This impression is strongly supported by the existence of counterexamples even in the two real variables context. For example, think at an upsidedown regular triangular pyramid (viewed as the graph of a convex function). Besides, all known arguments of (Pn,p) make use of the ordering of R.
However, as Professor Constantin P. Niculescu called to our attention, it is possible to develop a higher dimensional theory of convexity based on (P3,2).
This makes the objective of our joint paper [1].
REFERENCES
[1] Bencze, M., Niculescu, C. P.andPopovici, F.,Convexity according to Popoviciu’s inequality, submitted.
[2] Niculescu, C. P. and Persson, L.-E., Convex Functions and their applications. A Contemporary Approach, CMS Books in Mathematics, vol. 23, Springer-Verlag, New York, 2006.
[3] Niculescu, C. P.andPopovici, F.,A Refinement of Popoviciu’s Inequality, Bull. Soc.
Sci. Math. Roum.,49(97), no. 3, pp. 285–290, 2006.
[4] Peˇcari´c, J. E., Proschan, F.andTong, Y. C.,Convex functions, Partial Orderings and Statistical Applications, Academic Press, New York, 1992.
[5] Popoviciu, T.,Sur certaines in´egalit´es qui caract´erisent les fonctions convexes, Analele S
¸tiint¸ifice Univ. “Al. I. Cuza”, Ia¸si, Sect¸ia Mat.,11, pp. 155–164, 1965.
Received by the editors: May 8, 2008.
