popoviciu™s inequality, fifty years...
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Popoviciu�s Inequality, Fifty Years
Later
Constantin P. Niculescu
University of Craiova, Department of MathematicsE-mail: [email protected]
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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