Popoviciu�s Inequality, Fifty Years

Later

Constantin P. Niculescu

University of Craiova, Department of MathematicsE-mail: cpniculescu@gmail.com

The Eighth Congress of Romanian Mathematicians.Iasi, June 26-July 1, 2015.

1

Tiberiu Popoviciu (1906-1975)

Ph. D.: Sur quelques propriétés des fonctions d�une oude deux variables réelles, Université Paris IV-Sorbonne1933 (under the supervision of Paul Montel). FollowingE. Hopf, he developed the theory of n-convex functions.

Montel himself tackled with the algebraic variants of con-vexity in Sur les functions convexes et les fonctions soushar-moniques, Journal de Math. (9), 7 (1928), 29-60.

A Branch of the Mathematics Genealogy Tree ofPaul Montel

Borel and Lebesgue Supervisors

Paul Montel

Three other students of Montel: Henri Cartan, Jean Dieudonné and

Miron Nicolescu

1. Popoviciu�s Inequality [29]

If f : I ! R is a continuous convex function, then

X1�i1<���<ip�n

(�i1+� � �+�ip) f

0@�i1xi1 + � � �+ �ipxip

�i1 + � � �+ �ip

1A��n� 2p� 2

� 24n� p

p� 1

nXi=1

�i f(xi)

+

0@ nXi=1

�i

1A f

�1x1 + � � �+ �nxn

�1 + � � �+ �n

!35 :Here x1; :::; xn 2 I; n � 3; p 2 f2; :::; n � 1g; and�1; :::; �n are positive numbers (representing weights):

Basic case: Let f : I ! R be a continuous function.Then f is convex if, and only if,

f(x1) + f(x2) + f(x3)

3+ f

�x1 + x2 + x3

3

�� 2

3

�f

�x1 + x22

�+ f

�x2 + x32

�+ f

�x3 + x12

��for all x1; x2; x3 2 I:

Proof (of the basic case). We may assume the orderingx � y � z. If y � (x+y+z)=3, then (x+y+z)=3 �(x + z)=2 � z and (x + y + z)=3 � (y + z)=2 � z:This yields s; t 2 [0; 1] such that

x+ z

2= s � x+ y + z

3+ (1� s) � z

y + z

2= t � x+ y + z

3+ (1� t) � z;

whence (x+ y � 2z)(s+ t� 3=2) = 0.

If s+ t = 3=2, sum up the inequalities

f

�x+ z

2

�� s � f

�x+ y + z

3

�+ (1� s) � f(z)

f

�y + z

2

�� t � f

�x+ y + z

3

�+ (1� t) � f(z)

f

�x+ y

2

�� 1

2� f(x) + 1

2� f(y):

to conclude that

f

�x+ y

2

�+ f

�y + z

2

�+ f

�z + x

2

�� 1

2� (f(x) + f(y) + f(z)) +

3

2� f�x+ y + z

3

�: �

General case by mathematical induction.

Other form of the basic case:

f(x1) + f(x2) + f(x3)

3� f

�x1 + x2 + x3

3

�

� 2

24f�x1+x22

�+ f

�x2+x32

�+ f

�x3+x12

�3

�f�x1 + x2 + x3

3

��:

Second proof (the argument of Popoviciu)

By an approximation argument we may reduce to the caseof piecewise linear convex functions. Their general formis

f(x) = �x+ � +m�1Xk=1

ckjx� xkj;

so the proof ends by taking into account Hlawka�s in-equality:

jxj+ jyj+ jzj+ jx+y+ zj � jx+yj+ jy+ zj+ jz+xj

Third proof (using majorization)

Theorem 1. (The Hardy�Littlewood�Pólya inequality)Suppose that f is a convex function on an interval Iand consider two families x1; : : : ; xn and y1; : : : ; yn ofpoints in I such that

mXk=1

xk �mXk=1

yk for m 2 f1; : : : ; ng

andnXk=1

xk =nXk=1

yk:

If x1 � � � � � xn, then

nXk=1

f(xk) �nXk=1

f(yk);

while if y1 � � � � � yn this inequality works in the reversedirection.

Proof (of the basic case of Popoviciu�s inequality) Con-

sider the ordering x � y � z. Then

x+ y

2�z + x

2�y + z

2

and x � (x+ y + z)=3 � z:

Case 1: x � (x + y + z)=3 � y � z.

Consider the families

x1=x; x2= x3= x4=x+ y + z

3; x5= y; x6= z;

y1=y2=x+ y

2; y3= y4=

x+ z

2; y5= y6=

y + z

2

Case 2: x � y � (x + y + z)=3 � z.

Consider the families

x1= x; x2= y; x3= x4= x5=x+ y + z

3; x

6= z;

y1= y2=x+ y

2; y3 = y4 =

x+ z

2; y5 = y6 =

y + z

2: �

2. Simple Remarks

Cf. Mihai and Mitroi-Symeonidis [15].

Proposition 1. Suppose that f 2 C2 ([a; b]) and put

m = infnf 00(x) : x 2 [a; b]

oM = sup

nf 00(x) : x 2 [a; b]

o:

Then

M

36

�(x� y)2 + (y � z)2 + (z � x)2

��

f(x) + f(y) + f(z)

3+ f

�x+ y + z

3

�� 23

�f

�x+ y

2

�+ f

�y + z

2

�+ f

�z + x

2

��� m

36

�(x� y)2 + (y � z)2 + (z � x)2

�for all x; y; z 2 [a; b].

Indeed, both functions M2 x2�f(x) and f(x)�m

2 x2 are

convex.

The variant for uniformly convex functions (the functionsf such that f � C

2 x2 is convex for suitable C > 0) :

f(x) + f(y) + f(z)

3+ f

�x+ y + z

3

�

� C

36

�(x� y)2 + (y � z)2 + (z � x)2

�

� 2

3

�f

�x+ y

2

�+ f

�y + z

2

�+ f

�z + x

2

��:

Since ex � 12x2 for x � 0; this fact yields the inequality

a+ b+ c

3+

3pabc� 2

3

�pab+

pbc+

pca�

� 1

36

�log2

a

b+ log2

a

b+ log2

a

b

�;

for all a; b; c � 1:

Popoviciu�s inequality also works in the case of convexfunctions with values in a regularly ordered Banach space(in the sense of Davis [7]).

3. Convexity according to means [15]

Quasi-arithmetic mean (associated to a strictly monotonicfunction ') :

M'(x1; :::; xn;�1; :::; �n) = '�10@ nXk=1

�k'(xk)

1A :Other means:

L (a; b) =a� b

ln a� ln b(logarithmic mean)

I (a; b) =1

e

bb

aa

! 1b�a

(identric mean).

De�nition 1. (G. Aumann [3]). Given two means M andN de�ned respectively on the intervals I and J , a func-tion f : I ! J is (M;N)-convex if it is continuous onI and

f(M(x1; : : : ; xn)) � N(f(x1); : : : ; f(xn))

See Montel [16].

Theorem 2. Suppose that f : I ! J is�M';M

�-

convex. If strictly increasing,

M

�M (f(x); f(y); f(z)) ; f (M' (x; y; z))

��M (f (M' (x; y)) ; f (M' (y; z)) ; f (M' (z; x)))

for all x; y; z 2 I. The inequality works in the reverseway when is strictly decreasing.

A counterexample. The log-convex functions are also(A;L)-convex, because

f

�a+ b

2

�� exp

1

b� a

Z balog f(x) dx

!� 1

b� a

Z baf(x) dx

� L(f(a); f(b)) � f(a) + f(b)

2:

Logarithmic mean for triplets

L(a; b; c) =2a

log ab logac

+2b

log ba logbc

+2c

log ca logcb

:

Thus the analogue of Popoviciu�s inequality in the caseof (A;L)-convex functions should be

L (f(x); f(y); f(z))� f�x+y+z3

�logL (f(x); f(y); f(z))� log f

�x+y+z3

�� L

�f

�x+ y

2

�; f

�y + z

2

�; f

�z + x

2

��;

This does not work in the case of Gamma function.

Consider the points 1:40; 1:46; 1:47

4. The case of h-convex functions

I is an interval and h : (0; 1) ! (0;1) is a functionsuch that

(h1) h(1� �) + h(�) � 1 for all � 2 (0; 1):

De�nition 2. (Varo�anec [30]) A function f : I ! R iscalled h-convex if

f ((1� �)x+ �y) � h(1� �)f(x) + h(�)f(y)

for all x; y 2 I and � 2 (0; 1):

The role of the condition (h1) is to assure that the func-tion identically 1 is h-convex.

Special cases:

i) usual convex functions (when h is the identity func-tion).

ii) s-convex functions in the sense of Breckner [6] (h(�) =�s for some s 2 (0; 1]); see Hudzik and Maligranda [12]

and Pinheiro [24]. An example of an s-convex function(for 0 < s < 1):

f(t) =

(a if t = 0

bts + c if t > 0

where b � 0 and 0 � c � a: In particular, the functionts is s-convex on [0;1) if 0 < s < 1:

iii) the convex functions in the sense of Godunova-Levin[9] (when h(�) = 1

�): They verify the inequality

f ((1� �)x+ �y) � f(x)

1� �+f(y)

�

for all x; y 2 I and � 2 (0; 1): Every nonnegativemonotonic function (as well as every nonnegative con-vex function) is convex in the sense of Godunova-Levin.

iv) The h-convex functions corresponding to the caseh(�) � 1 are the P -convex functions in the sense ofDragomir, Peµcaric and Persson [8]. They verify inequali-ties of the form

f ((1� �)x+ �y) � f(x) + f(y)

for all x; y 2 I and � 2 (0; 1):

Theorem 3. (Mihai and Mitroi-Symeonidis [15]) If h isconcave, then every positive h-convex function f veri�esthe inequality

max�h

�1

2

�; 2h(1=4)

�(f(x) + f(y) + f(z))

+ 2h(3=4)f�x+ y + z

3

�� f

�x+ y

2

�+ f

�y + z

2

�+ f

�z + x

2

�for all x; y; z 2 I.

Application to the case f(t) = t1=2 (which is s-convexfor s = 1=2): In this case h(t) = t1=2; 2h(3=4) =

p3

and maxnh�12

�; 2h(1=4)

o= 1: Therefore

x1=2 + y1=2 + z1=2 +p3�x+ y + z

3

�1=2��x+ y

2

�1=2+�y + z

2

�1=2+�z + x

2

�1=2for all x; y; z � 0:

5. An Elementary Problem

Let a; b; c; d � 0 and a+ b+ c+ d = 4: Show thatX a

a3 + 8� 4

9:

Partial solution: f(x) = xx3+8

is concave for x 2 [0; 2]:By Jensen�s inequality,

1

4

�X a

a3 + 8

��

a+b+c+d4�

a+b+c+d4

�3+ 8

=1

9:

for all a; b; c; d 2 [0; 2] with a + b + c + d = 4: Thegeneral case is covered by the theory of relative convexity[20].

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­0.2

0.2

0.4

x

y

6. Convexity Relative to a Subset

f is a real-valued continuous function de�ned on the com-pact convex set K:

De�nition 3. (Niculescu and Roventa [20]) A point a inK is a point of convexity of the function f if there existsan in�nite convex subset V of K (called neighborhoodof convexity) such that a 2 V and

(J) f(a) �nXk=1

�kf(xk);

for every family of points x1; :::; xn in V and every familyof positive weights �1; :::; �n with

Pnk=1 �k = 1 andPn

k=1 �kxk = a:

If a is a point of convexity of f relative to the compactconvex neighborhood V; then

f(a) �ZVf(x)d�(x)

for every Borel probability measure � on V whose barycen-ter is a:

Theorem 4. (Niculescu and Roventa [20]) Suppose thatf is a real-valued function de�ned on an interval I. Ifa; b; c belong to I and a+b

2 ; a+c2 and b+c2 are points of

convexity of f relative to the entire interval I; then

f (a) + f (b) + f (c)

3+ f

�a+ b+ c

3

�� 2

3

�f

�a+ b

2

�+ f

�a+ c

2

�+ f

�b+ c

2

��:

The proof is done via an extension of Hardy-Littlewood-Pólya inequality of majorization.

7. The integral version

The only known result:

Theorem 5. (Niculescu [19]) For every convex function f

on [a; b];

1

b� a

Z baf(x)dx+

3

16f

�5a+ 3b

8

�+3

16f

�3a+ 5b

8

�� 1

b� a

Z (3a+5b)=8(5a+3b)=8

f(x)dx

� 4

(b� a)2

Z (5a+3b)=8a

Z xaf

�x+ t

2

�dtdx

+4

(b� a)2

Z (a+3b)=4(5a+3b)=8

Z (5a+3b)=4�xa

f

�x+ t

2

�dtdx

+4

(b� a)2

Z b(3a+5b)=8

Z x(3a+5b)=4�x

f

�x+ t

2

�dtdx:

The proof is based on Popoviciu�s idea to consider �rst

the case of piecewise linear convex functions.

8. Some open problems

� The real signi�cance of Popoviciu�s inequality (re-lated to the concentration of points with the samebarycenter) is still unclear, though some facts sug-gests to look at generalized entropies.

� Popoviciu�s inequality in the case of an arbitraryBorel positive measure is still open.

� Popoviciu�s inequality does not work in the gen-eral setting of convex functions of several variables(because it implies a stronger concept of convex-ity, called 2D-convexity). See the paper of Bencze,Niculescu and Popovici [4] from 2010.

However, the theory of 2D-convex functions is stillin infancy and need considerable more attention.

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