integral inequalities for concave functions

7/23/2019 Integral Inequalities for Concave Functions

1/15

INTEGRAL INEQUALITIES FOR CONCAVE FUNCTIONS

SORINA BARZA AND CONSTANTIN P. NICULESCU

Abstract. The aim of this paper is to extend to the context of several vari-ables a number of results related to the Favard-Berwald inequalities.

1. Introduction

The aim of this paper is to prove extensions and refinements of some integralinequalities for concave functions of several variables, that is, for those real-valuedfunctionsfdefined on convex subsets K ofRn such that

f((1 )x + y)(1 )f(x) + f(y)for all x, y K and all [0, 1]. The following examples show that the class ofconcave functions covers a large spectrum of important functions:

(1) (x1,...,xn)(x1x2 xn)1/n , on the positive orthantRn+={(x1,...,xn) | x1,...,xn0} ;

this example extends to e1/kk , where ek is the kth elementary symmetric

function ofn variables (1kn),e1= x1+ x2+

+ xn

e2=i< j

xixj

...

en = x1x2 xn.(2) (x1,...,xn) (xp1 xp2 ...xpn)1/p, on the subset of Rn+ where xp1

xp2+ + xpn. Here p >1.(3) A(det A)1/n ,on the coneSn++of allnndimensional positively defined

matrices. The same is true for the functions (det A/ det Ak)1/(nk) , where

Ak denotes the principal submatrix ofA formed by taking the first k rowsandk columns ofA. Because every concave function is also log-concave, we

infer that log (det A) is also concave on Sn++.

(4) Aminx=1Ax,x,on the the subset of all n ndimensional Hermitianmatrices in Mn(R). Letting

1(A)2(A)...n(A)

2000 Mathematics Subject Classification. Primary 26D15.Key words and phrases. Concave function, superharmonic function, Favard-Berwald type

inequality.The first author was partially supported by GrantKAW2000.0048.The second author was partially supported by Wenner-Gren Foundations (Grant 25 12 2002).

1


2/15

2 SORINA BARZA AND CONSTANTIN P. NICULESCU

be the sequence of eigenvalues ofA in decreasing order, this function asso-ciates to each matrix A its smallest eigenvalue n(A). More generally, all

functionsAk(A) + + n(A) are also concave.

For details, see the classical book of E. F. Beckenbach and R. Bellman [4]. Whilethe one variable case has received a great deal of attention, the literature concerningthe peculiar properties of concave functions of several variables is quite scarce. Infact, leaving out those results which can be obtained by a change of sign fromsimilar ones, for convex functions, what remains counts few significant facts. Themost prominent is the following theorem due to L. Berwald [5]:

Theorem 1. LetKbe a compact convex subset ofRn of positive volume, and letf, f1,...,fm: K R+ be continuous concave functions. Then:

i) The function

t

t + n

n 1|K| Kf

t(x) dV1/t

is decreasing on(0, );ii) For every positive constants1,...,m the following inequality holds

1

|K|K

f11 (x) fmm (x) dV1+nn

m+nn 1++m+n

n

mk=1

1

|K|K

fkk (x) dV

.

HeredVdenotes the volume measure in Rn (that is, the Lebesgue measure) and|K| denotes the volume ofK.

Theorem 1 extends an earlier result due to J. Favard [7], which asserts that

1b a

b

a

fp(x)dx1/p

2

(p + 1)1/p 1

b a b

a

f(x)dxfor all continuous concave functions f : [a, b] R+ and all parameters p >1. Thiscomplements a well known consequence of the Rogers-Holder inequality,

1

b a ba

f(x)dx

1

b a ba

fp(x)dx

1/p.

The limiting case (forp ) of Favards inequality gives us1

2 supx[a,b]

f(x) 1b a

ba

f(x)dx.

Theorem 1 extends this conclusion to all continuous concave functions f :K

R+

defined on an arbitrary compact convex subsetK Rn of positive volume:

(FB) 1

n + 1 supxK

f(x) 1|K|K

f(x) dV.

The inequality (FB) (called in what follows the Favard-Berwald inequality) has avery simple geometrical meaning: the volume of every conoid of base Kand heightf(x) (for every x K) does not exceed the volume of the cylindroid of base K,bounded above by the hypersurfacev = f(u).From this geometrical interpretationone can infer immediately the equality case in (FB).


3/15

INTEGRAL INEQUALITIES FOR CONCAVE FUNCTIONS 3

Advanced Calculus allows us to complement (FB) using the barycenter of K,that is,

xK= 1|K|K

xdV.

In fact, as an easy consequence of Jensens inequality we get

(J) 1

|K|K

f(x) dVf(xK).

The conjunction of (FB) and (J) is a powerful device even in the 1-dimensionalcase. For example, they yields Stirlings inequality,

1 + 1

1 + 2x

1 +

1

x

x< e 0.In this paper the inequality (FB) will be the object of several generalizations and

refinements. In Section 2 we shall describe the connection of (FB) and (J) withthe topics of Choquets theory. In Section 3 we shall prove an extension of (FB),while in Section 4 we shall show that a reverse counterpart of Berwalds inequality(mentioned in Theorem 1) yields a multiple (FB) inequality:

(MFB) 1

|K|K

mj=1

fj(x)

dV C(n, m) m

j=1

supxK

fj(x)

.

Here C(n, m) is a positive constant that depends only on m and n. In the case offunctions of one real variable, the inequality (MFB) was previously noticed by J.L. Brenner and H. Alzer [6], who in turn extended the limiting case (for p, q )of a result due to D. C. Barnes [2]: If p, q 1 and the functions f and g are

non-negative, concave and continuous on [a, b], then1

b a ba

f(x)g(x)dx (p + 1)1/p

(q+ 1)1/q

6 fp gq.

Of course, an inequality like (MFB) is not possible without certain restrictions.However, a remarkable result due to C. Visser [23] offers the alternative of passingto subsequences. More precisely, if (X, , ) is a probability space and (fn)n is asequence of random variables such that 0fn1 and

X

fnd >0, then forevery >0 there exists a subsequence, say (gn)n, such that

X

gn1 gnsd(1 )s

for every string of indices n1< ... < ns. See G. G. Lorentz [13] for a nice combina-

torial argument.In the last section we discuss the generalization of our results to the context of

superharmonic functions.

2. The Favard-Berwald Inequality Within Choquets Theory

In what follows we shall prove a number of estimates from above and from belowof the integral mean value

M(f) = 1

|K|K

f(x)dV,


4/15


of a concave function f defined on a compact convex subset K Rn of positivevolume. For each such function f,

(E) inf xK

f(x) = inf xExtK

f(x),

whereExt Kdenotes the set of all extreme points ofK. Recall that a point xKis said to be an extreme point ofKif it admits no representation of the form

x= (1 )u + v with u, vK, u=v and(0, 1).The equality (E) is a consequence of the celebrated Krein-Milman theorem, whichasserts thatKis the closed convex hull ofExtK.

By (FB), applied to the non-negative concave function f infxKf(x), we get1

|K|K

f(x)dV 1n + 1

supxK

f(x) + n

n + 1 infxK

f(x),

so that, taking into account the relations (E) and (J) we arrive at the following

result:Proposition 1. For every continuous concave function f defined on a compactconvex subsetK Rn of positive volume,

1

n + 1 supxK

f(x) + n

n + 1 infxExtK

f(x) 1|K|K

f(x)dVf(xK).

In the 1-dimensional case, when K = [a, b], the result above represents an im-provement of the classical Hermite-Hadamard inequality,

(HH) f(a) + f(b)

2 1

b a ba

f(x)dxf

a + b

2

.

See [17], [20], [21].It is worth to notice that the left hand inequality in (HH) can be strengthened

as

(LHH) f(a) + f(b)

2 1

2

f

a + b

2

+

f(a) + f(b)

2

1

b a ba

f(x)dx.

In fact, we may assume thatf0 (replacingfbyfinfx[a,b] f(x) if necessary),which allows us to interpret an equivalent form of (LHH),

(b a) fa+b2

2

+ba

2 f(a)

2 +

ba2 f(b)

2

ba

f(x)dx,

in terms of areas: the sum of the areas of the triangles P AB, P MA and P BN(with basis of lengths b a, f(a) and respectively f(b)) does not exceeds the areaof the subgraph off. See Fig. 1.

Using the same geometrical idea, one can prove the following refinement of

Proposition 1:

Theorem 2. Suppose that K Rn is a compact convex set of positive volume,with piecewise smooth boundary. Then for every continuous concave function f :K R+,

1

n + 1supyK

f(y) +

1

|K|K

d(y, TxK)f(x)dS

1|K|

K

f(x)dVf(xK).

Here TxK is the tangent hyperplane at x to the boundary of K and dS is the(n 1)-dimensional surface measure induced by the Lebesgue measure.


5/15

INTEGRAL INEQUALITIES FOR CONCAVE FUNCTIONS 5

A C B

M

N

P

Figure 1. A polygonal approximation of the subgraph of a con-cave function.

y

f(x)

dS

Figure 2. A hint for the surface integral appearing in Theorem 2.

Corollary 1. Under the assumptions of Theorem2,

1

n + 1

f(xK) +

1

|K|K

d(xK, TxK)f(x)dS

1|K|

K

f(x)dVf(xK).

The next example gives us an idea how good is the estimate offered by Proposi-tion 1.

Example 1. Let us consider the functionf(x1,...,xn) = (x1

xn)1/n , when re-

stricted to the domain

Dn ={x1,...,xn0 | x1+ + xn1} .By a well known formula due to Liouville,

...

Dn

(x1+ + xn)xp111 xpn1n dV(LF)

= (p1) (pn)(p1+ +pn)

10

(u) up1++pn1 du


6/15


(that works for all p1,...,pn > 0), we easily deduce that the volume ofDn is1/n!and

1|Dn|Dn

f(x)dV =

n

(1 + 1/n)n + 1 .

Notice that limnn(1 + 1/n) = e = 0.561 46...

Proposition1 yields

1

n(n + 1)

integral inequalities for concave functions

Documents