Jensen-type inequalities for m-convex functions

Definition 1

Let a < b and f ∈ L 1 ( ( a , b ) ; R ) . The right and left side Riemann-Liouville fractional integrals of order α , with Re ( α ) > 0 , are defined, respectively, by

and

with t ∈ ( a , b ) .

Definition 2

Let a < b and f ∈ L 1 ( ( a , b ) ; R ) . The right and left side Hadamard fractional integrals of order α , with Re ( α ) > 0 , are defined, respectively, by

and

with t ∈ ( a , b ) .

Definition 3

Let 0 < a < b , f : [ a , b ] → R an integrable function, and α ∈ ( 0 , 1 ) , ρ > 0 two fixed real numbers. The right and left side Katugampola fractional integrals of order α are defined, respectively, by

and

with t ∈ ( a , b ) .

Definition 4

Let 0 < a < b , g : [ a , b ] → R an increasing positive function on ( a , b ] with continuous derivative on ( a , b ) , f : [ a , b ] → R an integrable function, and α ∈ ( 0 , 1 ) a fixed real number. The right and left side fractional integrals in [11] of order α of f with respect to g are defined, respectively, by

and

with t ∈ ( a , b ) .

Definition 5

Let a < b and α ∈ R + . Let g : [ a , b ] → R be a positive function on ( a , b ] with continuous positive derivative on ( a , b ) and G : [ 0 , g ( b ) − g ( a ) ] × ( 0 , ∞ ) → R a continuous function which is positive on ( 0 , g ( b ) − g ( a ) ] × ( 0 , ∞ ) . Let us define the function T : [ a , b ] × [ a , b ] × ( 0 , ∞ ) → R by

The right and left integral operators, denoted, respectively, by J T , a + α and J T , b − α , are defined for each measurable function f on [ a , b ] as

with t ∈ [ a , b ] .

We say that f ∈ L T 1 [ a , b ] if J T , a + α ∣ f ∣ ( t ) , J T , b − α ∣ f ∣ ( t ) < ∞ for every t ∈ [ a , b ] .

Definition 6

Let a < b and α ∈ R + . Let g : [ a , b ] → R be a positive function on ( a , b ] with continuous positive derivative on ( a , b ) and G : [ 0 , g ( b ) − g ( a ) ] × ( 0 , ∞ ) → R a continuous function, which is positive on ( 0 , g ( b ) − g ( a ) ] × ( 0 , ∞ ) . For each function f ∈ L T 1 [ a , b ] , its right and left generalized derivatives of order α are defined, respectively, by

for each t ∈ ( a , b ) .

Definition 7

Let I ⊆ R be an interval containing zero, and let m ∈ ( 0 , 1 ] . A function φ : I → R is said to be m -convex if the inequality

holds for every pair of points x , y ∈ I and every coefficient t ∈ [ 0 , 1 ] .

Theorem 1

Let I ⊆ R be an interval containing the zero, and let ∑ k = 1 n w k x k be a convex combination of points x k ∈ I with coefficients w k ∈ [ 0 , 1 ] . If φ is an m-convex function on I, with m ∈ ( 0 , 1 ] , then

Theorem 2

Let μ be a probability measure on the space X . If I ⊆ R is an interval containing zero, f : X → I is μ -integrable, and φ is a continuous m-convex function on I , with m ∈ ( 0 , 1 ] , then

Theorem 3

Let x 1 ≤ x 2 ≤ ⋯ ≤ x n and let { w k } k = 1 n be positive weights whose sum is 1. If φ is a convex function on [ x 1 , x n ] , then

Lemma 4

Let I ⊆ R be an interval containing the zero, and let φ be an m-convex function on I with m ∈ ( 0 , 1 ] . For { x k } k = 1 n ⊂ I such that x 1 ≤ x 2 ≤ ⋯ ≤ x n , the following inequalities hold:

Proof

Let us consider y k = x 1 + m x n − m x k . Then x 1 + m x n = y k + m x k and so the pairs x 1 , m x n and y k , m x k have the same midpoint. Since x 1 ≤ y k and m x k ≤ m x n , we have x 1 ≤ y k , m x k ≤ m x n and there exists λ ∈ [ 0 , 1 ] such that

for 1 ≤ k ≤ n . From Definition 7 and its equivalent form (13) we obtain

and (16) follows.

Note that since m ∈ ( 0 , 1 ] , the hypothesis 0 ∈ I guarantees that m x k ∈ I .□

Theorem 5

Let I ⊆ R be an interval containing the zero, let { x k } k = 1 n ⊂ I with x 1 ≤ x 2 ≤ ⋯ ≤ x n , and let { w k } k = 1 n be positive weights whose sum is 1. If φ is an m-convex function on I, with m ∈ ( 0 , 1 ] , then

Remark 1

Theorem 3 gives that if m = 1 , the inequality in Theorem 5 also holds if we remove the hypothesis 0 ∈ I .

Proof

First, note that

and thus

Then it follows from (14) and (16) that

and this concludes the proof of the inequality.

Let us check that the hypothesis 0 ∈ I guarantees that m x 1 + m 2 x n − m 2 ∑ k = 1 n w k x k ∈ I :

Assume that I = [ a , b ] . Then

Also,

If x 1 ≤ 0 , then m x 1 − m 2 x 1 ≤ 0 and so

Assume now that x 1 > 0 . Let us consider the function v ( t ) = t x 1 − t 2 x 1 + t 2 x n . Since v ′ ( t ) = x 1 + 2 t ( x n − x 1 ) > 0 and m ∈ ( 0 , 1 ] , we have

If I is not a closed interval [ a , b ] , a similar argument gives the result.□

Theorem 6

Let a ≤ 0 ≤ b , let { y k } k = 1 n ⊂ [ a , b ] , and let { w k } k = 1 n be positive weights whose sum is 1. If φ is a continuous m-convex function on [ a , b ] , with m ∈ ( 0 , 1 ] , then

Proof

If we consider 0 < ε < 1 , y 0 = a , y n + 1 = b , w k ′ = ( 1 − ε ) w k ( 1 ≤ k ≤ n ) , w 0 ′ = ε / 2 , and w n + 1 ′ = ε / 2 , then ∑ k = 0 n + 1 w k ′ = 1 and Theorem 5 gives

Since φ is a continuous function on [ a , b ] , if we take ε → 0 + , we obtain (18).□

Theorem 7

Let μ be a probability measure on the space X and a ≤ 0 ≤ b real constants. If f : X → [ a , b ] is a measurable function and φ is a continuous m-convex function on [ a , b ] , with m ∈ ( 0 , 1 ] , then f and φ ( m f ) are μ -integrable functions and

If m = 1 , this inequality also holds if we remove the hypothesis 0 ∈ [ a , b ] .

Proof

Since a ≤ f ≤ b and φ is a continuous function on [ a , b ] , we have that f and φ ( m f ) are bounded measurable functions on X . And using that μ is a probability measure on X , we conclude that f and φ ( m f ) are μ -integrable functions.

For each n ≥ 1 and 0 ≤ k ≤ 2 n , let us consider the sets

Since f is a measurable function satisfying a ≤ f ≤ b , we have that { E n , k } k = 0 2 n are pairwise disjoint measurable sets and X = ∪ k = 0 2 n E n , k for each n . Thus,

for each n .

Since f is a measurable function satisfying a ≤ f ≤ b and { E n , k } k = 0 2 n is a partition of X , the sequence of simple functions

satisfies a ≤ f n ≤ b and f − 2 − n ( b − a ) < f n ≤ f for every n and so

Note that

Since { E n , k } k = 0 2 n is a partition of X , we have

Hence, Theorem 6 gives