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A reconsideration of Jensen’s inequality and its applications

Abstract

A finite form of Jensen’s inequality for a continuous convex function from a topological abelian semigroup to another topological ordered abelian semigroup is obtained under some assumption. As an application, a refinement of a mean inequality is also obtained.

MSC:39B62, 26B25, 26A51.

1 Introduction

This is the inheritance of the idea of [[1], Theorem 1] which gives a new interpretation of Jensen’s inequality by φ-mean. The finite form of Jensen’s inequality proved by Jensen in 1906 asserts that if t 1 ,, t n are positive numbers with i = 1 n t i =1 and f is a continuous convex (resp. concave) function on a real interval I, then

f ( i = 1 n t i x i ) i = 1 n t i f( x i ) ( resp.  f ( i = 1 n t i x i ) i = 1 n t i f ( x i ) )

holds for all x 1 ,, x n I.

We first introduce a concept called ‘(,)-convex (or concave)’ for a continuous function from a topological abelian semigroup (I,) to another topological ordered abelian semigroup (J,) and give an interesting example of such a function (see Remark 1). Our purpose of this paper is to give a finite form of Jensen’s inequality for such a function under some assumption (see Theorem 1). Also, as an application, we give a refinement of a mean inequality (see Theorem 2).

2 Terminology and main theorem

Let I be a topological space, and let be a topological abelian semigroup operation on I. For any xI and nN with n1, define the n th power x ( n ) of x recursively by x ( 1 ) =x and x ( n + 1 ) = x ( n ) x for n1. We assume that

( 1 ) any n th-power function x x ( n ) is a bijection of I onto itself.

By the assumption ( 1 ), for each xI and nN, there exists a unique element a of I such that a ( n ) =x. Denote by x ( 1 / n ) such an element a. Moreover, we define

x ( m / n ) = ( x ( 1 / n ) ) ( m )

for each m,nN. Then we can easily see that this definition is well defined, that is,

m n = m n ( x ( 1 / n ) ) ( m ) = ( x ( 1 / n ) ) ( m ) (xI).

In this case, we can easily show that the following power laws:

x ( p + q ) = x ( p ) x ( q ) , x ( p q ) = ( x ( p ) ) ( q ) and ( x y ) ( p ) = x ( p ) y ( p )
(1)

for all p,q Q + and x,yI. Here Q + denotes the set of all positive rational numbers. Moreover, we assume that

( 2 ) for each xI, the function p x ( p ) is continuous on Q + and it has a continuous extension to R + , say t x ( t ) .

Here R + denotes the set of all positive real numbers. Therefore power laws (1) hold for all p,q R + . Denote by A + (I) the set of all topological abelian semigroup operations on I satisfying the assumptions ( 1 ) and ( 2 ). Our assumption ( 1 ) leads to the following important concept called ‘mean’. For each x,yI, put

M (x,y)= ( x y ) ( 1 / 2 ) .

We call M (x,y) the mean of x and y with respect to the operation .

Moreover, let J be a topological ordered space with relation ≤, and denote by A + 0 (J,) the set of all operations A + (J) such that

( 1 ) abacbc (a,b,cJ)

and

( 2 ) ab a ( t ) b ( t ) (a,bJ, t R + ).

Let C(I,J) be the set of all continuous functions from I to J. Take A + (I), A + 0 (J,) and fC(I,J) arbitrarily. If f satisfies

f ( M ( x , y ) ) M ( f ( x ) , f ( y ) ) ( resp.  f ( M ( x , y ) ) M ( f ( x ) , f ( y ) ) )

for all x,yI, then we say that f is said to be (,)-convex (resp. concave).

The following theorem states a finite form of Jensen’s inequality for a (,)-convex (or concave) function.

Theorem 1 Let A + (I) and A + 0 (J,). If fC(I,J) is (,)-convex, then

f ( x 1 ( t 1 ) x n ( t n ) ) f ( x 1 ) ( t 1 ) f ( x n ) ( t n )

holds for all nN, x 1 ,, x n I and t 1 ,, t n R + with t 1 ++ t n =1.

If f is (,)-concave, then the inequality above is reversed.

Remark 1 Let R + 2 be the product space of R + with ordinary topology. Let be the operation on R + 2 defined by

(a,b)(c,d)=(ac,ad+bc)

for each (a,b),(c,d) R + 2 . Then is a topological abelian semigroup operation on R + 2 (cf. [[2], p.157-160]). In fact, ( R + 2 ,) is topologically isomorphic to an abelian subsemigroup of the semigroup of all 2×2 matrices with usual product under the following mapping:

(a,b) ( a b 0 a ) .

Note that

( a , b ) ( n ) = ( a n , n a n 1 b ) and ( a , b ) ( 1 / n ) = ( a 1 / n , b n a 1 1 / n )

for all nN. Then a simple calculation implies that

( a , b ) ( m / n ) = ( a m / n , m n a m n 1 b )

for each m,nN. Therefore we see that A + ( R + 2 ). In this case, we obtain from a simple calculation that

M ( ( a , b ) , ( c , d ) ) = ( a c , a d + b c 2 a c )

for each (a,b),(c,d) R + 2 .

Now let be the ordinary multiplication on R + . Since R + becomes a topological ordered space with the ordinary topology and the ordinary order ≤, we have that A + 0 ( R + ,) and M (x,y)= x y for each x,y R + . Let α and β be real numbers and put

f α , β (a,b)= a α b β

for each (a,b) R + 2 . Then f α , β is a continuous function from R + 2 to R + such that

f α , β ( M ( ( a , b ) , ( c , d ) ) ) = ( a c ) α β ( a d + b c 2 ) β

and

M ( f α , β ( a , b ) , f α , β ( c , d ) ) = ( a c ) α ( b d ) β

for each (a,b),(c,d) R + 2 . Therefore we can easily see that if β0 (resp. β0), then f α , β is (,)-convex (resp. (,)-concave).

Remark 2 Let E be a nontrivial real interval with the ordinary topology and the ordinary order ≤. In this case, Craigen and Pales [3] showed that if is a continuous cancellative semigroup operation on E, then there exists a continuous order-preserving bijection φ of E onto another (necessarily unbounded) real interval such that

ab= φ 1 ( φ ( a ) + φ ( b ) )

for all a,bE (cf. Aczel [4, 5]). Therefore we can easily see that if φ(E)= R + , then all continuous cancellative semigroup operations on E are in A + 0 (E,).

Remark 3 It is clear that a direct product of R + admits a semigroup operation in A + 0 that is given as the product of semigroup operations on each R + . However, the semigroup operation on R + 2 described in Remark 1 does not satisfy properties ( 1 ) and ( 2 ). So, it would be of interest to give an example of an ordered abelian semigroup with a semigroup operation in A + 0 , which is not isomorphic, as a topological semigroup, to the direct product of the topological semigroup R + .

3 Lemmas and proof of Theorem 1

Throughout this section, let I and J be as in Section 2 and suppose that A + (I), A + 0 (J,) and that fC(I,J) is (,)-convex.

Remark 4 If fC(I,J) is (,)-concave, then all inequalities in this section are reversed.

Lemma 1 The inequality

f ( x 1 ( 1 / 2 ) x n ( 1 / 2 n ) ) f ( x 1 ) ( 1 / 2 ) f ( x n 1 ) ( 1 / 2 n 1 ) [ f ( x n ( 1 / 2 ) ) ] ( 1 / 2 n 1 )

holds for all nN and x 1 ,, x n I.

Proof Let nN and x 1 ,, x n I. Since f is (,)-convex, it follows that

f ( x 1 ( 1 / 2 ) x 2 ( 1 / 2 2 ) x n ( 1 / 2 n ) ) = f ( ( x 1 x 2 ( 1 / 2 ) x n ( 1 / 2 n 1 ) ) ( 1 / 2 ) ) ( f ( x 1 ) f ( x 2 ( 1 / 2 ) x n ( 1 / 2 n 1 ) ) ) ( 1 / 2 ) = f ( x 1 ) ( 1 / 2 ) [ f ( x 2 ( 1 / 2 ) x n ( 1 / 2 n 1 ) ) ] ( 1 / 2 ) = f ( x 1 ) ( 1 / 2 ) [ f ( ( x 2 x 3 ( 1 / 2 ) x n ( 1 / 2 n 2 ) ) ( 1 / 2 ) ) ] ( 1 / 2 ) f ( x 1 ) ( 1 / 2 ) [ ( f ( x 2 ) f ( x 3 ( 1 / 2 ) x n ( 1 / 2 n 2 ) ) ) ( 1 / 2 ) ] ( 1 / 2 ) = f ( x 1 ) ( 1 / 2 ) f ( x 2 ) ( 1 / 2 2 ) [ f ( x 3 ( 1 / 2 ) x n ( 1 / 2 n 2 ) ) ] ( 1 / 2 2 ) f ( x 1 ) ( 1 / 2 ) f ( x 2 ) ( 1 / 2 2 ) f ( x n 1 ) ( 1 / 2 n 1 ) [ f ( x n ( 1 / 2 ) ) ] ( 1 / 2 n 1 ) .

Therefore we obtain the desired inequality. □

Let p and q be two nonnegative rational numbers. We define x ( p ) y ( q ) = y ( q ) if p=0, q>0 and x ( p ) y ( q ) = x ( p ) if q=0, p>0. We also define in the same way as for . Applying these notations, we show the following lemma.

Lemma 2 The inequality f( x ( p ) y ( 1 p ) )f ( x ) ( p ) f ( y ) ( 1 p ) holds for all x,yI and p R + with 0<p<1.

Proof Let x,yI and p R + with 0<p<1. By the binary system, we have the expansion p= i = 1 p i / 2 i , where p i {0,1} (i=1,2,). Since p i {0,1} (i=1,2,), it follows that

lim n [ f ( x ( p 1 ) y ( 1 p 1 ) ) ] ( 1 / 2 ) [ f ( x ( p n 1 ) y ( 1 p n 1 ) ) ] ( 1 / 2 n 1 ) = lim n [ f ( x ) ( p 1 / 2 ) f ( y ) ( ( 1 p 1 ) / 2 ) ] [ f ( x ) ( p n / 2 n 1 ) f ( y ) ( ( 1 p n 1 ) / 2 n 1 ) ] = lim n f ( x ) ( i = 1 n 1 p i / 2 i ) f ( y ) ( i = 1 n 1 ( 1 p i ) / 2 i ) = f ( x ) ( p ) f ( y ) ( 1 p ) .

So, putting

a n = [ f ( x ( p 1 ) y ( 1 p 1 ) ) ] ( 1 / 2 ) [ f ( x ( p n 1 ) y ( 1 p n 1 ) ) ] ( 1 / 2 n 1 )

for each nN, we have

lim n a n =f ( x ) ( p ) f ( y ) ( 1 p ) .
(2)

Put

N 1 = { n N : f ( ( x ( p n ) y ( 1 p n ) ) ( 1 / 2 ) ) = f ( x ( 1 / 2 ) ) }

and

N 0 = { n N : f ( ( x ( p n ) y ( 1 p n ) ) ( 1 / 2 ) ) = f ( y ( 1 / 2 ) ) } .

Since

f ( ( x ( p n ) y ( 1 p n ) ) ( 1 / 2 ) ) ={ f ( x ( 1 / 2 ) ) ( p n = 1 ) , f ( y ( 1 / 2 ) ) ( p n = 0 )

for each nN, it follows that N= N 1 N 0 . Then either N 1 or N 0 are infinite. Put

e n = [ f ( ( x ( p n ) y ( 1 p n ) ) ( 1 / 2 ) ) ] ( 1 / 2 n 1 )

for each nN. If N 1 is infinite, then we have

lim n N 1 e n f ( x ( 1 / 2 ) ) = lim n N 1 ( f ( x ( 1 / 2 ) ) ) ( 1 + 1 / 2 n 1 ) =f ( x ( 1 / 2 ) ) .
(3)

Therefore it follows from (2) and (3) that

f ( x ) ( p ) f ( y ) ( 1 p ) f ( x ( 1 / 2 ) ) = lim n N 1 a n e n f ( x ( 1 / 2 ) ) .
(4)

Also we have from Lemma 1 that

f ( x ( i = 1 n p i / 2 i ) y ( i = 1 n ( 1 p i ) / 2 i ) ) = f ( ( x ( p 1 ) y ( 1 p 1 ) ) ( 1 / 2 ) ( x ( p n ) y ( 1 p n ) ) ( 1 / 2 n ) ) [ f ( x ( p 1 ) y ( 1 p 1 ) ) ] ( 1 / 2 ) [ f ( x ( p n 1 ) y ( 1 p n 1 ) ) ] ( 1 / 2 n 1 ) [ f ( ( x ( p n ) y ( 1 p n ) ) ( 1 / 2 ) ) ] ( 1 / 2 n 1 ) = a n e n

for all n N 1 . Then it follows from ( 1 ) that

f ( x ( i = 1 n p i / 2 i ) y ( i = 1 n ( 1 p i ) / 2 i ) ) f ( x ( 1 / 2 ) ) a n e n f ( x ( 1 / 2 ) )
(5)

for all n N 1 . Letting n N 1 in (5), we obtain from (4) that

f ( x ( p ) y ( 1 p ) ) f ( x ( 1 / 2 ) ) f ( x ) ( p ) f ( y ) ( 1 p ) f ( x ( 1 / 2 ) ) .
(6)

Canceling f( x ( 1 / 2 ) ) in (6) by ( 1 ), we obtain the desired inequality. Similarly, the desired inequality is obtained in case that N 0 is infinite. □

Lemma 3 The inequality

f ( ( x 1 x n ) ( 1 / n ) ) ( f ( x 1 ) f ( x n ) ) ( 1 / n )

holds for all nN and x 1 ,, x n I.

Proof It is clear that the lemma holds for n=1. Suppose the lemma holds for n=k. Let x 1 ,, x k , x k + 1 I and put

a= ( f ( x 1 ) f ( x k ) ) ( 1 / k ) andx= ( x 1 x k ) ( 1 / k ) .

Then af(x) by hypothesis, and hence a ( k / k + 1 ) f ( x ) ( k / k + 1 ) by ( 2 ). It follows from ( 1 ) that

a ( k / k + 1 ) f ( x k + 1 ) ( 1 / k + 1 ) f ( x ) ( k / k + 1 ) f ( x k + 1 ) ( 1 / k + 1 ) .

Therefore we have from Lemma 2 that

( f ( x 1 ) f ( x k + 1 ) ) ( 1 / k + 1 ) = ( a ( k ) f ( x k + 1 ) ) ( 1 / k + 1 ) = a ( k / k + 1 ) f ( x k + 1 ) ( 1 / k + 1 ) f ( x ) ( k / k + 1 ) f ( x k + 1 ) ( 1 / k + 1 ) f ( x ( k / k + 1 ) x k + 1 ( 1 / k + 1 ) ) = f ( ( x 1 x k x k + 1 ) ( 1 / k + 1 ) ) .

In other words, the lemma holds for n=k+1. Then, by mathematical induction, the lemma holds for all nN. □

Lemma 4 The inequality

f ( x 1 ( p 1 ) x n ( p n ) ) f ( x 1 ) ( p 1 ) f ( x n ) ( p n )

holds for all x 1 ,, x n I, nN and p 1 ,, p n Q + with p 1 ++ p n =1.

Proof This result follows directly from Lemma 3. □

We are now in a position to prove Theorem 1.

Proof Let nN, x 1 ,, x n I and t 1 ,, t n R + with t 1 ++ t n =1. For each 1in, choose a sequence { p i k } k = 1 in Q + which converges to t i . Put q k = p 1 k ++ p n k for each kN. Then we have from Lemma 4 that

f ( x 1 ( p 1 k / q k ) x n ( p n k / q k ) ) f ( x 1 ) ( p 1 k / q k ) f ( x n ) ( p n k / q k ) .
(7)

Hence, after taking the limit with respect to k in (7), we obtain the desired inequality:

f ( x 1 ( t 1 ) x n ( t n ) ) f ( x 1 ) ( t 1 ) f ( x n ) ( t n ) .

Of course, if f is (,)-concave, the above inequality is reversed, as stated in Remark 4. This completes the proof of Theorem 1. □

4 Applications

Let K be a topological ordered space with order ≤, and let , and be three operations in A + 0 (K,) which have the following properties:

(ab)c=(ac)(bc)(a,b,cK),
(8)
eK:ex=x(xK),
(9)
ab=ab(ab)(a,bK).
(10)

In this case, it is clear that an element e in (9) is unique.

Lemma 5 The equality x ( t ) e= ( x e ) ( t ) holds for each xK and t R + .

Proof Take xK arbitrarily. By (8) and (9), we have (ab)e=(ae)(be) for all a,bK. By mathematical induction, we have

( x 1 x k )e=( x 1 e)( x k e)
(11)

for all kN and x 1 ,, x k K. In particular, we have

x ( k ) e= ( x e ) ( k )
(12)

for all kN. Then we have

( x e ) ( n / m ) = ( ( x e ) ( n ) ) ( 1 / m ) = ( x ( n ) e ) ( 1 / m ) by (12) = ( ( x ( n / m ) ) ( m ) e ) ( 1 / m ) = ( ( x ( n / m ) e ) ( m ) ) ( 1 / m ) by (12) = x ( n / m ) e

for all n,mN. Therefore x ( p ) e= ( x e ) ( p ) holds for all p Q + . Take t R + arbitrarily, and choose a sequence { p n } in Q + which converges to t. Then

x ( t ) e= lim n x ( p n ) e= lim n ( x e ) ( p n ) = ( x e ) ( t )

holds and so the proof is complete. □

Lemma 6 Suppose that M (a,b) M (a,b) holds for all a,bK. Then M (a,b) M (a,b) M (a,b) holds for each a,bK.

Proof Let a,bK. We first show that M (a,b) M (a,b). Since

( ( a b ) ( 1 / 2 ) ) ( 2 ) ab,

it follows from Lemma 5 that

( ( a b ) ( 1 / 2 ) e ) ( 2 ) = ( a b ) ( ( a b ) ( 1 / 2 ) ) ( 2 ) e ( a b ) ( a b ) e = ( a b ) e = ( ( a b ) ( 1 / 2 ) ) ( 2 ) e = ( ( a b ) ( 1 / 2 ) e ) ( 2 ) .

Therefore we obtain from ( 2 ) for that

( a b ) ( 1 / 2 ) e ( a b ) ( 1 / 2 ) e.

Canceling e in the above inequality, we obtain the desired inequality.

We next show that M (a,b) M (a,b). Since

ab ( ( a b ) ( 1 / 2 ) ) ( 2 ) ,

it follows from Lemma 5 that

( ( a b ) ( 1 / 2 ) e ) ( 2 ) = ( ( a b ) ( 1 / 2 ) ) ( 2 ) e = ( a b ) e = ( a b ) a b e ( ( a b ) ( 1 / 2 ) ) ( 2 ) a b e = ( ( a b ) ( 1 / 2 ) e ) ( 2 ) .

Therefore we obtain from ( 2 ) for that

( a b ) ( 1 / 2 ) e ( a b ) ( 1 / 2 ) e.

Canceling e in the above inequality, we obtain the desired inequality and so the proof is complete. □

The following result is a refinement of the mean inequality.

Theorem 2 Let K be a topological ordered space with orderand ,, A + 0 (K,) satisfying (8), (9) and (10). If M (x,y) M (x,y) holds for all x,yK, then

x 1 ( t 1 ) x n ( t n ) x 1 ( t 1 ) x n ( t n ) x 1 ( t 1 ) x n ( t n )

holds for all nN, x 1 ,, x n K and t 1 ,, t n R + with t 1 ++ t n =1.

Proof This follows immediately from Theorem 1 and Lemma 6. □

5 Examples

Throughout this section, let R + be an ordinary topological ordered space.

Example 1 Put x t y= ( x t + y t ) 1 / t for each x,y R + and tR{0}. Then each t is a topological abelian semigroup operation on R + such that x ( n ) t = n 1 / t x and x ( 1 / n ) t = ( 1 / n ) 1 / t x for all nN and x R + . Also, since

x ( n / m ) t = ( x ( 1 / m ) t ) ( n ) t = n 1 / t ( 1 / m ) 1 / t x= ( n / m ) 1 / t x

for each m,nN, tR{0} and x R + , it follows that t A + 0 ( R + ,) for each tR{0}. Let x 1 ,, x n , α 1 ,, α n R + with α 1 ++ α n =1. Since x ( α ) t = α 1 / t x for each x,α R + and tR{0}, we have that

x 1 ( α 1 ) t t t x n ( α n ) t = ( α 1 x 1 t + + α n x n t ) 1 / t
(13)

for all tR{0}. Note that

M s (x,y)= ( x s + y s 2 ) 1 / s ( x t + y t 2 ) 1 / t = M t (x,y)

for all x,y R + and s,tR{0} with st. Therefore Theorem 1 implies the following well-known inequality:

( α 1 x 1 s + + α n x n s ) 1 / s ( α 1 x 1 t + + α n x n t ) 1 / t ,

where s,tR{0} with st.

Example 2 Let t be a positive real number, and let t be the operation on R + defined in Example 1. Then we have t A + 0 ( R + ,). Also, if xy=xy (x,y R + ), then A + 0 ( R + ,) and x ( α ) = x α for all x,α R + . Hence we have

M (x,y)= x y ( x t + y t 2 ) 1 / t = M t (x,y)

for each x,y R + .

Let nN and x 1 ,, x n , α 1 ,, α n R + with α 1 ++ α n =1. Then it is clear that

x 1 ( α 1 ) x n ( α n ) = x 1 α 1 x n α n .
(14)

If xy=x t y t (xy) (x,y R + ), then A + 0 ( R + ,) and three operations , and on R + satisfy (8), (9) and (10). If x,α R + , we have from Lemma 5 that

x ( α ) t 1= ( x t 1 ) ( α ) = ( x t 1 ) α = ( x t + 1 ) α / t .
(15)

Then we have from (11) and (15) that

x 1 ( α 1 ) x n ( α n ) t 1 = ( x 1 ( α 1 ) t 1 ) ( x n ( α n ) t 1 ) = ( x 1 ( α 1 ) t 1 ) ( x n ( α n ) t 1 ) = ( x 1 t + 1 ) α 1 / t ( x n t + 1 ) α n / t ,

and hence

x 1 ( α 1 ) x n ( α n ) = ( ( x 1 t + 1 ) α 1 ( x n t + 1 ) α n 1 ) 1 / t .
(16)

Then we obtain from (13), (14), (16) and Theorem 2 that

x 1 α 1 x n α n ( ( x 1 t + 1 ) α 1 ( x n t + 1 ) α n 1 ) 1 / t ( α 1 x 1 t + + α n x n t ) 1 / t .

This is a refinement of the geometric-arithmetic mean inequality.

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Acknowledgements

The authors are grateful to the referee for careful reading of the paper and for helpful suggestions and comments. The second author is partially supported by Grant-in-Aid for Scientific Research, Japan Society for the Promotion of Science.

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YN carried out the design of the study and performed the analysis. ST participated in the sequence alignment and drafted the manuscript. All authors read and approved the final manuscript.

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Nakasuji, Y., Takahasi, SE. A reconsideration of Jensen’s inequality and its applications. J Inequal Appl 2013, 408 (2013). https://doi.org/10.1186/1029-242X-2013-408

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Keywords

  • Jensen’s inequality
  • mean
  • refinement
  • continuous associative semigroup operation