Exponential convexity of Petrović and related functional

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Abstract

We consider functionals due to the difference in Petrović and related inequalities and prove the log-convexity and exponential convexity of these functionals by using different families of functions. We construct positive semi-definite matrices generated by these functionals and give some related results. At the end, we give some examples.

1 Introduction

First time exponentially convex functions are introduced by Bernstein . Independently of Bernstein, but some what later Widder  introduced these functions, as a sub-class of convex functions in a given interval (a, b), and denoted this class by Wa,b. After the initial development, there is a big gap in time before applications and examples of interest were constructed. One of the reasons is that, aside from absolutely monotone functions and completely monotone functions, as special classes of exponentially convex functions, there is no operative criteria to recognize exponential convexity of functions.

Definition 1. [, p. 373] A function f : (a, b) → is exponentially convex if it is continuous and

$∑ i , j = 1 n ξ i ξ j f ( x i + x j ) ≥ 0$
(1)

for all n and all choices ξ i and x i + x j (a, b), 1 ≤ i, jn.

Proposition 1.1. Let f : (a, b) → . The following propositions are equivalent.

1. (i)

f is exponentially convex.

2. (ii)

f is continuous and

$∑ i , j = 1 n ξ i ξ j f x i + x j 2 ≥ 0$

for every ξ i and every x i (a, b), 1 ≤ in.

Proposition 1.2. If f is exponentially convex, then the matrix

$f x i + x j 2 i , j = 1 n$

is positive semi-definite. In particular,

$det f x i + x j 2 i , j = 1 n ≥ 0$

for every n , x i (a, b), i = 1, ..., n.

Proposition 1.3. If f : (a, b) → (0, ∞) is an exponentially convex function, then f is log-convex which means that for every x, y (a, b) and all λ (0, 1)

$f ( λ x + ( 1 - λ ) y ) ≤ f ( x ) λ f ( y ) 1 - λ .$

We consider functionals due to the differences in the Petrović and related inequalities. These inequalities are given in the following theorems [, pp. 152-159].

Theorem 1.4. Let I = (0, a] be an interval, (x1, ..., x n ) In, and (p1, ..., p n ) be a non-negative n-tuple such that

$∑ i = 1 n p i x i ∈ I a n d ∑ i = 1 n p i x i ≥ x j f o r j = 1 , … , n .$
(2)

If f : I be a function such that f (x)/x is an increasing for x I, then

$f ∑ i = 1 n p i x i ≥ ∑ i = 1 n p i f ( x i ) .$
(3)

Remark 1.5. Let us note that if f(x)/x is a strictly increasing function for x I, then equality in (3) is valid if we have equalities in (2) instead of the inequalities, that is, x1= = x n and $∑ i = 1 n p i =1$ .

Theorem 1.6. Let I = (0, a] be an interval, (x1, ..., x n ) In, such that 0 < x1x n , (p1, ..., p n ) be a non-negative n-tuple and f : I be a function such that f(x)/x is an increasing for x I.

1. (i)

If there exists an m(≤ n) such that

$0 ≤ P ̄ 1 ≤ P ̄ 2 ≤ ⋯ ≤ P ̄ m ≤ 1 , P ̄ m + 1 = ⋯ = P ̄ n = 0 ,$
(4)

where$P k = ∑ i = 1 k p i , P ̄ k = P n - P k - 1 ( k = 2 , … , n )$and$P ̄ 1 = P n$, then (3) holds.

1. (ii)

If there exists an m(≤ n) such that

$P ̄ 1 ≥ P ̄ 2 ≥ ⋯ ≥ P ̄ m ≥ 1 , P ̄ m + 1 = ⋯ = P ̄ n = 0 ,$
(5)

then the reverse of inequality in (3) holds.

Theorem 1.7. Let I = (0, a] be an interval, (x1, ..., x n ) In, and x1 - x2 - ... - x n I. Also let f : I be a function such that f(x)/x is an increasing for x I. Then

$f x 1 - ∑ i = 2 n x i ≤ f ( x 1 ) - ∑ i = 2 n f ( x i ) .$
(6)

Remark 1.8. If f(x)/x is a strictly increasing function for x I, then strict inequality holds in (6).

Theorem 1.9. Let I = (0, a] be an interval, (x1, ..., x n ) In , (p1, ..., p n ) and (q1, ..., q n ) be non-negative n-tuples such that (2) holds. If f : I be an increasing function, then

$∑ i = 1 n q i f ∑ i = 1 n p i x i ≥ ∑ i = 1 n q i f ( x i ) .$
(7)

Remark 1.10. If f is a strictly increasing function on I and all x i 's are not equal, then we obtain strict inequality in (7).

Theorem 1.11. Let I = [0, a] be an interval, (x1, ..., x n ) In, and (p1, ..., p n ) be a non-negative n-tuple such that (2) holds.

If f is a convex function on I, then

$f ∑ i = 1 n p i x i ≥ ∑ i = 1 n p i f ( x i ) + 1 - ∑ i = 1 n p i f ( 0 ) .$
(8)

Remark 1.12. In the above theorem, if f is a strictly convex, then inequality in (8) is strict, if all x i 's are not equal or $∑ i = 1 n p i ≠ 1$ .

Theorem 1.13. Let I be an interval, 0 I, f be a convex function on I, h : [a.b] → I be continuous and monotonic with h(t0) = 0, t0 [a, b] be fixed, g be a function of bounded variation and

$G ( t ) : = ∫ a t d g ( x ) , G ¯ ( t ) : = ∫ t b d g ( x ) .$

(a) If$∫ a b h ( t ) d g ( t ) ∈I$and

$0 ≤ G ( t ) ≤ 1 f o r a ≤ t ≤ t 0 , 0 ≤ G ¯ ( t ) ≤ 1 f o r t 0 < t ≤ b ,$
(9)

then we have

$∫ a b f ( h ( t ) ) d g ( t ) ≥ f ∫ a b h ( t ) d g ( t ) + ∫ a b d g ( t ) - 1 f ( 0 ) .$
(10)

(b) If$∫ a b h ( t ) d g ( t ) ∈ I$and either

there exists an st0such that G(t) ≤ 0 for t < s,

$G ( t ) ≥ 1 f o r s ≤ t ≤ t 0 a n d G ¯ ( t ) ≤ 0 f o r t > t 0$
(11)

or

there exists an st0such that G(t) ≤ 0 for t < t0,

$G ¯ ( t ) ≥ 1 f o r t 0 < t < s , a n d G ¯ ( t ) ≤ 0 f o r t ≥ s ,$
(12)

then the reverse of the inequality in (10) holds.

In this paper, we consider certain families of functions to prove log-convexity and exponential convexity of functionals due to the differences in inequalities given in Theorems 1.4-1.13. We construct positive semi-definite matrices generated by these functionals. Also by using log-convexity of these functionals, we prove monotonicity of the expressions introduced by these functionals. At the end, we give some examples.

2 Main results

Let I be an interval and f : I be a function. Then for distinct points u i I, i = 0, 1, 2, the divided differences in first and second order are defined as follows:

$u i , u i + 1 , f = f ( u i + 1 ) - f ( u i ) u i + 1 - u i i = 0 , 1 , u 0 , u 1 , u 2 , f = u 1 , u 2 , f - u 0 , u 1 , f u 2 - u 0 .$
(13)

The values of the divided differences are independent of the order of the points u0, u1, u2 and may be extended to include the cases when some or all points are equal, that is

$[ u 0 , u 0 , f ] = lim u 1 → u 0 [ u 0 , u 1 , f ] = f ′ ( u 0 ) ,$

provided that f' exists.

Now passing through the limit u1u0 and replacing u2 by u in (13), we have [, p. 16]

$[ u 0 , u 0 , u , f ] = lim u 1 → u 0 [ u 0 , u 1 , u , f ] = f ( u ) - f ( u 0 ) - ( u - u 0 ) f ′ ( u 0 ) ( u - u 0 ) 2 , u ≠ u 0 ,$

provided that f' exists. Also passing to the limit u i u (i = 0, 1, 2) in (13), we have

$[ u , u , u , f ] = lim u i → u [ u 0 , u 1 , u 2 , f ] = f ″ ( u ) 2 ,$

provided that f″ exists.

One can note that if for all u0, u1 I, [u0, u1, f] ≥ 0, then f is increasing on I and if for all u0, u1, u2 I, [u0, u1, u2, f] ≥ 0, then f is convex on I.

(M1) Under the assumptions of Theorem 1.4, with all x i 's not equal, we define a linear functional as

$P 1 ( f ) = f ∑ i = 1 n p i x i - ∑ i = 1 n p i f ( x i ) .$

(M2) Under the assumptions of Theorem 1.6, with all x i 's not equal and (4) is valid, we define a linear functional as

$P 2 ( f ) = P 1 ( f ) .$

(M3) Under the assumptions of Theorem 1.6, with all x i 's not equal and (5) is valid, we define a linear functional as

$P 3 ( f ) = - P 1 ( f ) .$

(M4) Under the assumptions of Theorem 1.7, with all x i 's not equal, we define a linear functional as

$P 4 ( f ) = f ( x 1 ) - ∑ i = 2 n f ( x i ) - f x 1 - ∑ i = 2 n x i .$

(M5) Under the assumptions of Theorem 1.9, with all x i 's not equal, we define a linear functional as

$P 5 ( f ) = ∑ i = 1 n q i f ∑ i = 1 n p i x i - ∑ i = 1 n q i f ( x i ) .$

(M6) Under the assumptions of Theorem 1.11, with all x i 's not equal, we define a linear functional as

$P 6 ( f ) = f ∑ i = 1 n p i x i - ∑ i = 1 n p i f ( x i ) - 1 - ∑ i = 1 n p i f ( 0 ) .$

(M7) Under the assumptions of Theorem 1.13, such that (9) is valid, we define a linear functional as

$P 7 ( f ) = ∫ a b f ( h ( t ) ) d g ( t ) - f ∫ a b h ( t ) d g ( t ) - ∫ a b d g ( t ) - 1 f ( 0 ) .$

(M8) Under the assumptions of Theorem 1.13, such that (11) or (12) is valid, we define a linear functional as

$P 8 ( f ) = - P 7 ( f ) .$

Remark 2.1. Under the assumptions of (M k ) for k = 1, 2, 3, 4, if f(u)/u is an increasing function for u I, then

$P k ( f ) ≥ 0 , f o r k = 1 , 2 , 3 , 4 .$

If f(u)/u is strictly increasing for u I and all x i 's are not equal or$∑ 1 n p i ≠1$then strict inequality holds in the above expression.

Remark 2.2. Under the assumptions of (M5), if f is an increasing function on I, then

$P 5 ( f ) ≥ 0 .$

If f is strictly increasing function on I and all x i 's are not equal, then we obtain strict inequality in the above expression.

Remark 2.3. Under the assumptions of (M k ) for k = 6, 7, 8, if f is a convex function on I, then

$P k ( f ) ≥ 0 f o r k = 6 , 7 , 8 .$

If f is strictly increasing function on I and all x i 's are not equal, then we obtain strict inequality in the above expression for$P 6 ( f )$.

The following lemma is nothing more than the discriminant test for the non-negativity of second-order polynomials.

Lemma 2.4. Let I be an interval. A function f : I → (0, ∞) is log-convex in J-sense on I, that is, for each r, t I

$f ( r ) f ( t ) ≥ f 2 t + r 2$

if and only if, the relation

$m 2 f ( t ) + 2 m n f t + r 2 + n 2 f ( r ) ≥ 0$
(14)

holds for each m, n and r, t I.

To define different families of functions, let I and (c, d) be intervals. For distinct points u0, u1, u2 I we suppose

D1 = {f t : I | t (c, d) and t [u0, u1, F t ] is log-convex in J-sense, where F t (u) = f t (u)/u}.

D2 = {f t : I | t (c, d) and t [u0, u0, F t ] is log-convex in J-sense, where F t (u) = f t (u)/u and $F t ′$ exists}.

D3 = {f t : I | t (c, d) and t [u0, u1, f t ] is log-convex in J-sense}.

D4 = {f t : I | t (c, d) and t [u0, u0, f t ] is log-convex in J-sense, where $f t ′$ exists}.

D5 = {f t : I | t (c, d) and t [u0, u1, u2, f t ] is log-convex in J-sense}.

D6 = {f t : I | t (c, d) and t [u0, u0, u2, f t ] is log-convex in J-sense, where $f t ′$ exists}.

D7 = {f t : I | t (c, d) and t [u0, u0, u0, f t ] is log-convex in J-sense, where $f t ″$ exists}.

In this theorem, we prove log-convexity in J-sense, log-convexity and related results of the functionals associated with their respective families of functions.

Theorem 2.5. Let $P k$ be the linear functionals defined in(M k ), associate the functionals with D i in such a way that, for k = 1, 2, 3, 4, f t D i , i = 1, 2, for k = 5, f t D i , i = 3, 4 and for k = 6, 7, 8, f t D i , i = 5, 6, 7. Also for k = 7, 8, assume that the linear functionals are positive. Then, the following statements are valid:

(a) The functions$t↦ P k ( f t )$are log-convex in J-sense on (c, d).

(b) If the functions$t↦ P k ( f t )$are continuous on (c, d), then the functions$t↦ P k ( f t )$are log-convex on (c, d).

(c) If the functions$t↦ P k ( f t )$are derivable on (c, d), then for t, r, u, v (c, d) such that tu, rv, we have

$B k , i ( t , r ; f t ) ≤ B k , i ( u , v ; f t ) ,$

where

$B k , i ( t , r ; f t ) = P k ( f t ) P k ( f r ) 1 t - r , t ≠ r , exp d d t ( P k ( f t ) ) P k ( f t ) , t = r .$
(15)

Proof. (a) First, we prove log-convexity in J-sense of $t↦ P k ( f t )$ for k = 1, 2, 3, 4. For this, we consider the families of functions defined in D1 and D2.

Choose any m, n , and t, r (c, d), we define the function

$h ( u ) = m 2 f t ( u ) + 2 m n f t + r 2 ( u ) + n 2 f r ( u ) .$

This gives

$u 0 , u 1 , H = m 2 u 0 , u 1 , F t + 2 m n u 0 , u 1 , F t + r 2 + n 2 u 0 , u 1 , F r ,$

where H (u) = h(u)/u and F t (u) = f t (u)/u.

Since t [u0, u1, F t ] is log-convex in J-sense, by Lemma 2.4 the right-hand side of above expression is non-negative. This implies h(u)/u is an increasing function for u I.

Thus by Remark 2.1

$P k ( h ) ≥ 0 for k = 1 , 2 , 3 , 4 ,$

this implies

$m 2 P k ( f t ) + 2 m n P k ( f t + r 2 ) + n 2 P k ( f r ) ≥ 0 .$
(16)

Now [u0, u1, F t ] > 0 as it is log-convex, this implies f t (u)/u is strictly increasing for all u I and t (c, d). Also all x i 's are not equal and therefore by Remark 2.1, $P k ( f t )$ are positive valued, and hence, by Lemma 2.4, the inequality (16) implies log-convexity in J-sense of the functions $t↦ P k ( f t )$for k = 1, 2, 3, 4.

Now we prove log-convexity in J-sense of $t↦ P 5 ( f t )$. For this, we consider the families of functions defined in D3 and D4. Following the same steps as above and having H(u) = h(u), we have the log-convexity in J-sense of $P 5 ( f t )$ by using Remark 2.2 and Lemma 2.4.

At last, we prove log-convexity in J-sense of $t↦ P k ( f t )$ for k = 6, 7, 8. For this, we consider the families of Functions defined in D i for i = 5, 6, 7.

Choose any m, n , and t, r (c, d), we define the function

$h ( u ) = m 2 f t ( u ) + 2 m n f t + r 2 ( u ) + n 2 f r ( u ) .$

This gives

$u 0 , u 1 , u 2 , h = m 2 u 0 , u 1 , u 2 , f t + 2 m n u 0 , u 1 , u 2 , f t + r 2 + n 2 u 0 , u 1 , u 2 , f r .$

Since t [u0, u1, u2, f t ] is log-convex in J-sense, by Lemma 2.4 the right-hand side of above expression is non-negative. This implies h is a strictly convex function on I.

Thus by Remark 2.3

$P k ( h ) ≥ 0 for k = 6 , 7 , 8 ,$

this implies

$m 2 P k ( f t ) + 2 m n P k ( f t + r 2 ) + n 2 P k ( f r ) ≥ 0 .$
(17)

Since $P k ( f t )$ are positive valued, we have by Lemma 2.4 and inequality (17) the log-convexity in J-sense of the functions $t↦ P k ( f t )$ for k = 6, 7, 8.

(b) If $t↦ P k ( f t )$ are additionally continuous for k = 1, ..., 8 and D i 's associated with them, then these are log-convex, since J-convex continuous functions are convex functions.

(c) Since the functions log $P k ( f t )$ are convex for k = 1, ..., 8, and D i 's associated with them, therefore for tu, rv, tr, uv, we have [, p.2],

$log P k ( f t ) - log P k ( f r ) t - r ≤ log P k ( f u ) - log P k ( f v ) u - v ,$

concluding

$B k , i ( t , r ; f t ) ≤ B k , i ( u , v ; f t ) .$

Now if t = ru, we apply limrt, concluding,

$B k , i ( t , t ; f t ) ≤ B k , i ( u , v ; f t ) .$

Other possible cases are treated similarly.

In order to define different families of functions related to exponential convexity, let I and (c, d) be any intervals. For distinct points u0, u1, u2 I we suppose

E1 = {f t : I | t (c, d) and t [u0, u1, F t ] is exponentially convex, where F t (u) = f t (u)/u}.

E2 = {f t : I | t (c, d) and t [u0, u0, F t ] is exponentially convex, where F t (u) = f t (u)/u and $F t ′$ exists}.

E3 = {f t : I | t (c, d) and t [u0, u1, f t ] is exponentially convex}.

E4 = {f t : I | t (c, d) and t [u0, u0, f t ] is exponentially convex, where $f t ′$ exists}.

E5 = {f t : I | t (c, d) and t [u0, u1, u2, f t ] is exponentially convex}.

E6 = {f t : I | t (c, d) and t [u0, u0, u2, f t ] is exponentially convex, where $f t ′$ exists}.

E7 = {f t : I | t (c, d) and t [u0, u0, u0, f t ] is exponentially convex, where $f t ″$ exists}.

In this theorem, we prove the exponential convexity of the functionals associated with their respective families of functions. Also we define positive semi-definite matrices for these functionals and give some related results.

Theorem 2.6. Let$P k$be the linear functionals defined in (M k ), associate the functionals with E i in such a way that, for k = 1, 2, 3, 4, f t E i , i = 1, 2, for k = 5, f t E i , i = 3, 4 and for k = 6, 7, 8, f t E i , i = 5, 6, 7. Then, the following statements are valid:

(a) If$t↦ P k ( f t )$are continuous on (c, d), then the functions$t↦ P k ( f t )$, are exponentially convex on (c, d).

(b) For every q and t1, ..., t q (c, d), the matrices

$P k ( f t l + t m 2 ) l , m = 1 q$

are positive semi-definite. In particular

$det P k ( f t l + t m 2 ) l , m = 1 s ≥ 0 f o r s = 1 , 2 , … , q .$

(c) If$t↦ P k ( f t )$are positive derivable on (c, d), then for t, r, u, v (c, d) such that tu, rv, we have

$ℭ k , i ( t , r ; f t ) ≤ ℭ k , i ( u , v ; f t )$

where$ℭ k , i ( t , r ; f t )$is defined similarly as in (15).

Proof. (a) First, we prove exponential convexity of $t↦ P k ( f t )$ for k = 1, 2, 3, 4. For this, we consider the families of functions defined in E1 and E2.

For any n , ξ i and t i (c, d), i = 1, ..., n, we define

$h ( u ) = ∑ i , j = 1 n ξ i ξ j f t i + t j 2 ( u ) .$

This gives

$u 0 , u 1 , H = ∑ i , j = 1 n ξ i ξ j u 0 , u 1 , F t i + t j 2 ,$

where H (u) = h(u)/u and F t (u) = f t (u)/u.

Since t [u0, u1, F t ] is exponentially convex, right-hand side of the above expression is non-negative, which implies h(u)/u is an increasing function on I.

Thus by Remark 2.1, we have

$P k ( h ) ≥ 0 , for k = 1 , 2 , 3 , 4 ,$

thus

$∑ i , j = 1 n ξ i ξ j P k f t i + t j 2 ≥ 0 .$

Hence $t↦ P k ( f t )$ is exponentially convex for k = 1, 2, 3, 4.

Now we prove exponential convexity of $t↦ P 5 ( f t )$. For this, we consider the families of functions defined in E3 and E4. Following the same steps as above and having H (u) = h(u), we have the exponential convexity of the $P 5 ( f t )$ by using Remark 2.2.

At last, we prove exponential convexity of $t↦ P k ( f t )$ for k = 6, 7, 8. For this, we consider the families of functions defined in E i for i = 5, 6, 7.

For any n , ξ i and t i (c, d), i = 1, ..., n, we define

$h ( u ) = ∑ i , j = 1 n ξ i ξ j f t i + t j 2 ( u ) .$

This gives

$u 0 , u 1 , u 2 , h = ∑ i , j = 1 n ξ i ξ j u 0 , u 1 , u 2 , f t i + t j 2 .$

Since t [u0, u1, u2, f t ] is exponentially convex therefore right-hand side of the above expression is non-negative, which implies h(u) is a strictly convex function on I.

Thus by Remark 2.3, we have

$P k ( h ) ≥ 0 for k = 6 , 7 , 8 ,$

thus

$∑ i , j = 1 n ξ i ξ j P k f t i + t j 2 ≥ 0 .$

Hence $t↦ P k ( f t )$ are exponentially convex for k = 6, 7, 8.

(b) It follows by Proposition 1.2.

(c) Since $t↦ P k ( f t )$ are positive derivable for k = 1, ..., 8 with E i 's associated with them, we have our conclusion using part (c) of the Theorem 2.5.

3 Examples

In this section, we will vary on choices of families of functions in order to construct different examples of log and exponentially convex functions and related results.

Example 1. Let t and φ t : (0, ∞) → be the function defined as

$φ t ( u ) = u t t - 1 , t ≠ 1 , u log u , t = 1 .$
(18)

Then φ t (u)/u is strictly increasing on (0, ∞) for each t . One can note that t [u0, u0, φ t (u)/u] is log-convex for all t . If we choose f t = φ t in Theorem 2.5, we get log-convexity of the functionals $P k ( φ t )$ for k = 1, 2, 3, 4, which have been proved in [5, 6].

Since φ t (u)/u)' = ut-2= e(t - 2) log u, the mapping t (φ t (u)/u)' is exponentially convex . If we choose f t = φ t in Theorem 2.6, we get results that have been proved in [6, 8]. Also we get $ℭ 1 , 2 ( t , r ; φ t ) = A t , r 1 ( x ; p )$ for t, r ≠ 1. By making substitution $x i ↦ x i s$, t t/s, r r/s and s ≠ 0, t, rs, we get $ℭ 1 , 2 ( t , r ; φ t ) = A t , r s ( x ; p )$ for t, rs, where $A t , r s ( x ; p )$ is defined in .

Similarly, $ℭ 4 , 2 ( t , r ; φ t ) = C t , r 1 ( x )$for t, r ≠ 1, and by substitution used above $ℭ 4 , 2 ( t , r ; φ t ) = C t , r s ( x )$ for t, rs, where $C t , r s ( x )$ is defined in .

Example 2. Let t and β t : (0, ∞) be the function defined as

$β t ( u ) = u t t , t ≠ 0 , log u , t = 0 .$
(19)

Then, β t is strictly increasing on (0, ∞) for each t . One can note that t [u0, u0, β t ] is log-convex for all t . If we choose f t = β t in Theorem 2.5, we get log-convexity of the functional $P 5 ( β t )$, which have been proved in .

Since $β t ′ ( u ) = u t - 1 = e ( t - 1 ) log u$, the mapping $t↦ β t ′$ is exponentially convex . If we choose f t = β t in Theorem 2.6, we get results that have been proved in . Also we get 5,4 (t, r; β t ) = H t,r (x; p; q) for t, r ≠ 0, where H t,r (x; p; q) is defined in .

Example 3. Let t (0, ∞) and δ t : [0, ∞) → be the function defined as

$δ t ( u ) = u t t ( t - 1 ) , t ≠ 1 , u log u , t = 1 ,$
(20)

with a convention that 0 log 0 = 0. Then δ t is convex on [0, ∞) for each t (0, ∞). One can note that t [u0, u0, u0, δ t ] is log-convex for all t (0, ∞). If we choose f t = δ t in Theorem 2.5, we get log-convexity of the functionals $P k ( δ t )$ for k = 6, 7, 8, which have been proved in .

Since $δ t ″ ( u ) = u t - 2 = e ( t - 2 ) log u$, the mapping $t↦ δ t ″$ is exponentially convex . If we choose f t = δ t in Theorem 2.6, we get results that have been proved in [8, 10]. Also we get $ℭ 6 , 7 ( t , r ; δ t ) = B t , r 1 ( x ; p )$ for t, r ≠ 1. By making substitution $x i ↦ x i s$, t t/s, r r/s and s ≠ 0, t, rs, we get $ℭ 6 , 7 ( t , r ; δ t ) = B t , r s ( x ; p )$ for t, rs, where $B t , r s ( x ; p )$ is defined in .

Similarly, $ℭ 7 , 7 ( t , r ; δ t ) = F t , r 1 ( a , b , h , g )$ for t, r ≠ 1 and by substitution used above $ℭ 7 , 7 ( t , r ; δ t ) = F t , r s ( a , b , h , g )$ for t, rs, where $F t , r s ( a , b , h , g )$ is defined in .

Example 4. Let t (0, ∞) and ζ t : (0, ∞) → be the function defined as

$ζ t ( u ) = u t - u - log t , t ≠ 1 , u 2 , t = 1 .$
(21)

One can note that t [u0, u0, ζ t (u)/u] is log-convex for all t (0, ∞). If we choose f t = ζ t in Theorem 2.5, we get log-convexity of the functionals $P k ( ζ t )$ for k = 1, 2, 3, 4.

Since ζ t (u)/u)' = t-u, the mapping t (ζ t (u)/u)' is exponentially convex . If we choose f t = ζ t in Theorem 2.6, we get exponential convexity of the functionals $P k ( ζ t )$ for k = 1, 2, 3, 4.

For $P 1 ( f t )$ using the function ζ t in Theorem 2.6, 1,2 (t, r; ζ t ) in this particular case looks like

$ℭ 1 , 2 ( t , r ; ζ t ) = log r x ̃ n t - x ̃ n - ∑ i = 1 n p i x i t - x i log t x ̃ n r - x ̃ n - ∑ i = 1 n p i x i r - x i 1 t - r , t ≠ r , t , r ≠ 1 , x ̃ n t - x ̃ n - ∑ i = 1 n p i x i t - x i - log t x ̃ n 2 - ∑ i = 1 n p i x i 2 1 t - 1 , t ≠ r = 1 , exp - 1 t log t - x ̃ n 2 t - x ̃ n - ∑ i = 1 n p i x i 2 t - x i t x ̃ n t - x ̃ n - ∑ i = 1 n p i x i t - x i , t = r , t , r ≠ 1 , exp x ̃ n 3 - ∑ i = 1 n p i x i 3 - 2 x ̃ n 2 - ∑ i = 1 n p i x i 2 , t = r = 1 ,$

where $x ̃ n = ∑ i = 1 n p i x i$.

For $P 4 ( f t )$ using the function ζ t in Theorem 2.6, 4,2 (t, r; ζ t ) in this particular case looks like

$ℭ 4 , 2 ( t , r ; ζ t ) = log r x 1 t - x 1 - ∑ i = 2 n x i t - x i - x ^ n t - x ^ n log t x 1 r - x 1 - ∑ i = 2 n x i r - x i - x ^ n r - x ^ n 1 t - r , t ≠ r , t , r ≠ 1 , x 1 t - x 1 - ∑ i = 2 n x i t - x i - x ^ n t - x ^ n - log t x 1 2 - ∑ i = 2 n x i 2 - x ^ n 2 1 t - 1 , t ≠ r = 1 , exp - 1 t log t - x 1 2 t - x 1 - ∑ i = 2 n x i 2 t - x i - x ^ n 2 t - x ^ n t x 1 t - x 1 - ∑ i = 2 n x i t - x i - x ^ n t - x ^ n , t = r , t , r ≠ 1 , exp x 1 3 - ∑ i = 2 n x i 3 - x ^ n 3 - 2 x 1 2 - ∑ i = 2 n x i 2 - x ^ n 2 , t = r = 1 ,$

where $x ^ n = ( x 1 - ∑ i = 2 n x i )$.

Example 5. Let t (0, ∞) and θ t : (0, ∞) → be the function defined as

$θ t ( u ) = t - u - log t , t ≠ 1 , u , t = 1 .$
(22)

One can note that t [u0, u0, θ t ] is log-convex for all t (0, ∞), and if we choose f t = θ t in Theorem 2.5, we get log-convexity of the functional $P 5 ( θ t )$.

Since $θ t ′ ( u ) = t - u$, the mapping $t↦ θ t ′ ( u )$ is exponentially convex function . If we choose f t = θ t in Theorem 2.6, we get exponential convexity of the functional $P 5 ( θ t )$.

For $P 5 ( f t )$ using the function θ t in Theorem 2.6, 5,4 (t, r; θ t ) in this particular case looks like

$ℭ 5 , 4 ( t , r ; θ t ) = log r ∑ i = 1 n q i t - x ̃ n - ∑ i = 1 n q i t - x i log t ∑ i = 1 n q i r - x ̃ n - ∑ i = 1 n q i r - x i 1 t - r , t ≠ r , t , r ≠ 1 , ∑ i = 1 n q i t - x ̃ n - ∑ i = 1 n q i t - x i - log t ∑ i = 1 n q i x ̃ n - ∑ i = 1 n q i x i 1 t - 1 , t ≠ r = 1 , exp - 1 t log t - ∑ i = 1 n q i x ̃ n t - x ̃ n - ∑ i = 1 n q i x i t - x i - t ∑ i = 1 n q i t - x ̃ n - ∑ i = 1 n q i t - x i , t = r , t , r ≠ 1 , exp ∑ i = 1 n q i x ̃ n 2 - ∑ i = 1 n q i x i 2 - 2 ∑ i = 1 n q i x ̃ n - ∑ i = 1 n q i x i , t = r = 1 ,$

Where $x ̃ n = ∑ i = 1 n p i x i$.

Example 6. Let t (0, ∞) and λ t : (0, ∞) → be the function defined as

$λ t ( u ) = u e - u t - t .$
(23)

One can note that t [u0, u0, λ t (u)/u] is log-convex for all t (0, ∞). If we choose f t = λ t in Theorem 2.5, we get log-convexity of the functionals $P k ( λ t )$ for k = 1, 2, 3, 4.

Since $λ t ( u ) ∕ u ′ = e - u t$, the mapping t (λ t (u)/u)' is exponentially convex function . If we choose f t = λ t in Theorem 2.6, we get exponential convexity of the functionals $P k ( λ t )$ for k = 1, 2, 3, 4.

For $P 1 ( f t )$ using the function λ t in Theorem 2.6, 1,2 (t, r; λ t ) in this particular case looks like

$ℭ 1 , 2 ( t , r ; λ t ) = r x ̃ n e - x ̃ n t - ∑ i = 1 n p i x i e - x i t t x ̃ n e - x ̃ n r - ∑ i = 1 n p i x i e - x i r 1 t - r , t ≠ r , exp - 1 2 t - x ̃ n 2 e - x ̃ n t - ∑ i = 1 n p i x i 2 e - x i t 2 t x ̃ n e - x ̃ n t - ∑ i = 1 n p i x i e - x i t , t = r ,$

Where $x ̃ n = ∑ i = 1 n p i x i$.

Now for $P 4 ( f t )$ using the function λ t in Theorem 2.6, 4,2 (t, r; λ t ) in this particular case looks like

$ℭ 4 , 2 ( t , r ; λ t ) = r x 1 e - x 1 t - ∑ i = 2 n x i e - x i t - x ^ n e - x ^ n t t x 1 e - x 1 t - ∑ i = 2 n x i e - x i r - x ^ n e - x ^ n r 1 t - r , t ≠ r , exp - 1 2 t - x 1 2 e - x 1 t - ∑ i = 2 n x i 2 e - x i t - x ^ n 2 e - x ^ n t 2 t x 1 e - x 1 t - ∑ i = 2 n x i e - x i t - x ^ n e - x ^ n t , t = r ,$

Where $x ^ n = ( x 1 - ∑ i = 2 n x i )$.

Example 7. Let t (0, ∞) and ξ t : (0, ∞)→ , be the function defined as

$ξ t ( u ) = e - u t - t .$
(24)

One can note that t [u0, u0, ξ t ] is log-convex for all t (0, ∞). If we choose f t = ξ t in Theorem 2.5, we get log-convexity of the functional $P 5 ( ξ t )$.

Since $ξ t ′ ( u ) = e - u t$, the mapping $t↦ ξ t ′ ( u )$ is exponentially convex function . If we choose f t = ξ t in Theorem 2.6 we get exponential convexity of the functional $P 5 ( ξ t )$.

For $P 5 ( f t )$ using the function ξ t in Theorem 2.6, 5,4 (t, r; ξ t ) in this particular case looks like

$ℭ 5 , 4 ( t , r ; ξ t ) = r ∑ i = 1 n q i e - x ̃ n t - ∑ i = 1 n q i e - x i t t ∑ i = 1 n q i e - x ̃ n r - ∑ i = 1 n q i e - x i r 1 t - r , t ≠ r , exp - 1 2 t - ∑ i = 1 n q i x ̃ n e - x ̃ n t - ∑ i = 1 n q i x i e - x i t 2 t ∑ i = 1 n q i e - x ̃ n t - ∑ i = 1 n q i e - x i t , t = r ,$

Where $x ̃ n = ∑ i = 1 n p i x i$.

Example 8. Let t and ψ t : (0, ∞) → be the function defined as

$ψ t ( u ) = u e u t t , t ≠ 0 , u 2 , t = 0 .$
(25)

One can note that t [u0, u0, ψ t (u)/u] is log-convex for all t . If we choose f t = ψ t in Theorem2.5, we get log-convexity of the functionals $P k ( ψ t )$ for k = 1, 2, 3, 4.

Since (ψ t (u)/u)' = e ut , the mapping t (ψ t (u)/u)' is exponentially convex function . If we choose f t = ψ t in Theorem 2.6 we get exponential convexity of the functionals $P k ( ψ t )$ for k = 1, 2, 3, 4.

For $P 1 ( f t )$ using the function ψ t in Theorem 2.6, 1,2 (t, r; ψ t ) in this particular case looks like

$ℭ 1 , 2 ( t , r ; ψ t ) = r x ̃ n e x ̃ n t - ∑ i = 1 n p i x i e x i t t x ̃ n e x ̃ n r - ∑ i = 1 n p i x i e x i r 1 t - r , t ≠ r , t , r ≠ 0 , x ̃ n e x ̃ n t - ∑ i = 1 n p i x i e x i t t x ̃ n 2 - ∑ i = 1 n p i x i 2 1 t - 1 , t ≠ r = 0 , exp - 1 t + x ̃ n 2 e x ̃ n t - ∑ i = 1 n p i x i 2 e x i t x ̃ n e x ̃ n t - ∑ i = 1 n p i x i e x i t , t = r , t , r ≠ 0 , exp x ̃ n 3 - ∑ i = 1 n p i x i 3 2 x ̃ n 2 - ∑ i = 1 n p i x i 2 , t = r = 0 ,$

Where $x ̃ n = ∑ i = 1 n p i x i$.

Now for $P 4 ( f t )$ using the function ψ t in Theorem 2.6, 4,2 (t, r; ψ t ) in this particular case looks like

$ℭ 4 , 2 ( t , r ; ψ t ) = r x 1 e x 1 t - ∑ i = 2 n x i e x i t - x ^ n e x ^ n t x 1 e x 1 r - ∑ i = 2 n x i e x i r - x ^ n e x ^ n r 1 t - r , t ≠ r , t , r ≠ 0 , x 1 e x 1 t - ∑ i = 2 n x i e x i t - x ^ n e x ^ n t x 1 2 - ∑ i = 2 n x i 2 - x ^ n 2 1 t - 1 , t ≠ r = 0 , exp - 1 t + x 1 2 e x 1 t - ∑ i = 2 n x i 2 e x i t - x ^ n 2 e x ^ n t x 1 e x 1 t - ∑ i = 2 n x i e x i t - x ^ n e x ^ n t , t = r , t , r ≠ 0 , exp x 1 3 - ∑ i = 2 n x i 3 - x ^ n 3 2 x 1 2 - ∑ i = 2 n x i 2 - x ^ n 2 , t = r = 0 ,$

where $x ^ n = ( x 1 - ∑ i = 2 n x i )$.

Example 9. Let t and ω t : (0, ∞) → be the function defined as

$ω t ( u ) = e u t t , t ≠ 0 , u , t = 0 .$
(26)

One can note that t [u0, u0, ω t ] is log-convex for all t . If we choose f t = ω t in Theorem 2.5, we get log-convexity of the functional $P 5 ( ω t )$.

Since $ω t ′ ( u ) = e u t$, the mapping $t↦ ω t ′ ( u )$ is exponentially convex function . If we choose f t = ω t in Theorem 2.6 we get exponential convexity of the functional $P 5 ( ω t )$.

For $P 5 ( f t )$ using the function ω t in Theorem 2.6, 5,4 (t, r; ω t ) in this particular case looks like

$ℭ 5 , 4 ( t , r ; ω t ) = r ∑ i = 1 n q i e x ̃ n t - ∑ i = 1 n q i e x i t t ∑ i = 1 n q i e x ̃ n r - ∑ i = 1 n q i e x i r 1 t - r , t ≠ r , t , r ≠ 0 , ∑ i = 1 n q i e x ̃ n t - ∑ i = 1 n q i e x i t t ∑ i = 1 n q i x ̃ n - ∑ i = 1 n q i x i 1 t - 1 , t ≠ r = 0 , exp - 1 t + ∑ i = 1 n q i x ̃ n e x ̃ n t - ∑ i = 1 n q i x i e x i t ∑ i = 1 n q i e x ̃ n t - ∑ i = 1 n q i e x i t , t = r , t , r ≠ 0 , exp ∑ i = 1 n q i x ̃ n 2 - ∑ i = 1 n q i x i 2 2 ∑ i = 1 n q i x ̃ n - ∑ i = 1 n q i x i , t = r = 0 ,$

Where $x ̃ n = ∑ i = 1 n p i x i$.

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Acknowledgements

The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions. This research was partially funded by Higher Education Commission, Pakistan. The research of the first author was supported by the Croatian Ministry of Science, Education and Sports under the Research Grant 117-1170889-0888.

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Correspondence to Saad I Butt.

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The authors declare that they have no competing interests.

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All authors jointly worked on the results and they read and approved the final manuscript.

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