Extensions and improvements of Sherman’s and related inequalities for n-convex functions

Remark 1.1

For n = 1 we take the sum in (4) to be zero.

Lemma 2.1

Let G_k(·, s), s ∈ [α, β], k = 2, 3, 4, be defined as in (8)-(10). Then for every φ ∈ C² ([α, β]), it holds that

Proof

Utilizing integration by parts we have

from which (11) follows immediately. Similarly, we can prove other two identities. ☐

Lemma 2.2

Let 𝐱 ∈ [α, β]^l, 𝐲 ∈ [α, β]^m, 𝐚 ∈ ℝ^l and 𝐛 ∈ ℝ^m be such that (2) holds for some matrix 𝐀 ∈ ℳ_ml(ℝ) whose entries satisfy the condition ∑j=1laij=1 for i = 1, ..., m. Let G_k(·, s), s ∈ [α, β], k = 1, 2, 3, 4, be defined as in (7)-(10). Then for every φ ∈ C² ([α, β]), it holds that

Proof

Let us consider Green’s function G₂ defined by (8). Applying (11) in the difference ∑j=1lajϕ(xj)−∑i=1mbiϕ(yi) , we have

Since (2) holds, then we have

and

Moreover, after interchanging the order of summation in (15), we easily get

Analogously, we can prove the identities for other three Green’s functions. ☐

Lemma 2.3

Let 𝐱 ∈ [α, β]^l, 𝐲 ∈ [α, β]^m, 𝐚 ∈ ℝ^l and 𝐛 ∈ ℝ^m be such that (2) holds for some matrix 𝐀 ∈ ℳ_ml(ℝ) whose entries satisfy the condition ∑j=1laij=1 for i = 1, ...,m. Let P(t, s) and G_k (·, s), s, t ∈ [α, β], k = 1, 2 , 3, 4, be defined as in (5) and (7)-(10), respectively. If a function f : [α, β] → ℝ is such that φ⁽ⁿ⁻¹⁾ is absolutely continuous on [α, β] for some n ≥ 3, then

Proof

Applying (4) for φ″, we get

By an easy calculation, applying (17) in (14), we get

After interchanging the order of summation and integration and applying Fubini’s theorem we get (16). ☐

Theorem 3.1

Let 𝐱 ∈ [α, β]^l,𝐲 ∈ [α, β]^m, 𝐚 ∈ ℝ^l and 𝐛 ∈ ℝ^m be such that (2) holds for some matrix 𝐀 ∈ ℳ_ml(ℝ) whose entries satisfy the condition ∑j=1laij=1 for i = 1, ...., m. Let G_k (·, s), s ∈ [α, β], k = 1, 2, 3, 4, be defined as in (7)-(10). Then the following statements are equivalent:

1. For every continuous convex function φ : [α, β] → ℝ, it holds that

   ∑i=1mbiϕ(yi)≤∑j=1lajϕ(xj). (18)

2. For every k = 1, 2, 3, 4 and s ∈ [α, β], it holds that

   ∑i=1mbiGk(yi,s)≤∑j=1lajGk(xj,s). (19)

Furthermore, the statements (i) and (ii) are also equivalent if one changes the sing of inequality in both (18) and (19).

Proof

(i) ⇒ (ii) Let (i) hold. Since G_k(·, s), s ∈ [α, β], k = 1, 2, 3, 4, is continuous and convex on [α, β], then also

(ii) ⇒ (i) Let (ii) hold. Let us consider the function G₂(·, s), s ∈ [α, β], defined by (8). For every function φ ∈ C²([α, β]) from (14) we have

If φ is convex, then φ″ ≥ 0 on [α, β]. Furthermore, if

then also (18) holds. Note that it is not necessary to demand the existence of the second derivative of the function φ. The differentiability condition can be directly eliminated by using the fact that a continuous convex function is possible to approximate uniformly by convex polynomials (see [8, p. 172]). The same conclusion we have for other three Green’s functions.

The last part statement of theorem can be proved analogously. ☐

Next, we develop Sherman’s type inequalities for n-convex functions.

Theorem 3.2

Let 𝐱 ∈ [α, β]^l, 𝐲 ∈ [α, β]^m, 𝐚 ∈ ℝ^l and 𝐛 ∈ ℝ^m be such that (2) holds for some matrix 𝐀 ∈ ℳ_ml(ℝ) whose entries satisfy the condition ∑j=1laij=1 for i = 1, ..., m. Let P(t, s) and G_k(·, s), s, t ∈ [α, β], k = 1, 2, 3, 4, be defined as in (5) and (7)-(10), respectively. Let φ: [α, β] → ℝ be n-convex function such that φ⁽ⁿ⁻¹⁾ is absolutely continuous on [α, β] for some n ≥ 3.

If

then

If the reverse of (20) holds, then the reverse of (21) holds.

Proof

Under the assumptions of theorem, (16) holds.

Since φ⁽ⁿ⁻¹⁾ is absolutely continuous on [α, β], then φ⁽ⁿ⁾ exists almost everywhere (see [10]). By assumption, φ is n-convex on [α, β], therefore φ⁽ⁿ⁾ ≥ 0 on [α, β]. Using this fact and the assumption (20) in combination with (16), we obtain (21). ☐

Let us denote

When we take in account Sherman’s condition of nonnegativity of vectors 𝐚, 𝐛 and matrix 𝐀, we obtain the following extensions.

Theorem 3.3

Let 𝐱 ∈ [α, β]^l, 𝐲 ∈ [α, β]^m, 𝐚 ∈ [0, ∞)^l and 𝐛 ∈ [0, ∞)^m be such that (2) holds for some row stochastic matrix 𝐀 ∈ ℳ_ml(ℝ). Let P(t, s) and G_k(·, s), s, t ∈ [α, β], k = 1, 2, 3, 4, be defined as in (5) and (7)-(10), respectively. Let a function φ : [α, β] → ℝ be n-convex such that φ⁽ⁿ⁻¹⁾ is absolutely continuous on [α, β] for some n ≥ 3.

1. If n is even, then (21) holds.

2. If n is odd, then for t ≤ s, the inequalities (20) and (21) hold, while for s ≤ t, the reverse inequalities in (20) and (21) hold.

3. If (21) holds and F_k, defined by (22), is convex on [α, β], then

   ∑j=1lajϕ(xj)−∑i=1mbiϕ(yi)≥∑w=0n−3n−w−2(β−α)w!×∫αβ(∑j=1lajGk(xj,s)−∑i=1mbiGk(yi,s))(ϕ(w+1)(β)(s−β)w−ϕ(w+1)(α)(s−α)w)ds≥0. (23)

4. If (21) holds and φ^(w+1)(α) ≤ 0, φ^(w+1)(β) ≥ 0 for even w and φ^(w+1)(α) ≤ 0, φ^(w+1)(β) ≤ 0 for odd w, then (23) holds.

Proof

(i)-(ii) Since G_k(·, s), s ∈ [α, β], is convex on [α, β], by Sherman’s theorem

If n is even, then

while for odd n, the reversed inequality in (24) holds while the inequality (25) remains the same. Now, applying Theorem 3.2, we conclude (i) and (ii).

(iii) If (21) holds, the right hand side can be written in the form

where F_k is defined as in (22). If F_k is convex, then by Sherman’s theorem we have

i.e. the right hand side of (21) is nonnegative and (23) holds.

(iv) For even w we have

while for odd w the first inequality remains the same while the second is reversed. So if for even w we have φ^(w+1)(α) ≤ 0, φ^(w+1)(β) ≥ 0 and for odd w we have φ^(w+1)(α) ≤ 0, φ^(w+1)(β) ≤ 0, then

i.e. the right hand side of (21) is nonnegative what we need to prove. ☐

Remark 3.4

Note that from (23), Sherman’s inequality (1) immediately follows. Moreover, (23) improves Sherman’s inequality for a different choice of Green’s functions G_k, k = 1, 2, 3, 4.

Remark 4.1

Note that under assumptions of Theorem 3.2, for every n-convex function φ : [α, β] → ℝ, we have T_k(φ) ≥ 0.

In this section, we present upper bounds for T_k(φ). We start with upper bounds when φ is such that its nth derivatives φ⁽ⁿ⁾ belongs to L_p[α, β].

Theorem 4.2

Assume that (p, q) is a pair of conjugate exponents, i.e. 1 ≤ p, q ≤ ∞, 1/p + 1/q = 1. Let 𝐱, 𝐲, 𝐚, 𝐛, 𝐀, P(t, s) and G_k(·, s), s, t ∈ [α, β], k = 1, 2, 3, 4, be as in Theorem 3.2. Let a function φ : [α, β] → ℝ be such that φ⁽ⁿ⁻¹⁾ is absolutely continuous and φ⁽ⁿ⁻¹⁾ ∈ L_p[α, β] for some n ≥ 3. Then

The constant on the right hand side is sharp for 1 < p ≤ ∞ and the best possible for p = 1.

Proof

Applying the well-known Hölder inequality to (16), we have

The proof of the sharpness is analogous to the one in proof of [5, Theorem 13]. ☐

Let us consider the Čebyšev functional Δ(f, g), of two Lebesgue integrable functions f, g : [α, β] → ℝ, defined by

Let 𝐱, 𝐲, 𝐚, 𝐛, 𝐀, P(t, s) and G_k(·, s), s, t ∈ [α, β], k = 1, 2, 3, 4, be as in Theorem 3.2. We define

Considering the functions ℬ_k we obtain the following estimations for the given remainders R_k.

Theorem 4.3

Let 𝐱, 𝐲, 𝐚, 𝐛, 𝐀, P(t, s) and G_k(·, s), s, t ∈ [α, β], k = 1, 2, 3, 4, be as in Theorem 3.2. Let φ : [α, β] → ℝ be such that φ⁽ⁿ⁾ is absolutely continuous with (· − α)(β − ·)(φ⁽ⁿ⁺¹⁾)² ∈ L[α, β] for some n ≥ 3. Then

where the reminder R_k(φ) is given in formula

and T_k(φ) and ℬ_k are defined by (26) and (29), respectively.

Proof

Comparing identities (31) and (16) we have

Applying [11, Theorem 1] on the functions ℬ_k and ɸ ⁽ⁿ⁾ we obtain (30). ☐

Theorem 4.4

Let φ : [α, β] → ℝ be such that φ⁽ⁿ⁾ is absolutely continuous with φ⁽ⁿ⁺¹⁾ ≥ 0 on [α, β] for some n ≥ 3. Then for R_k given by (31)

Proof

We have Rk(ϕ)=1(n−3)!Δ(Bk,ϕ(n)).

Applying [11, Theorem 2] on the functions ℬ_k and φ⁽ⁿ⁾ and using the identity

we obtain (32). ☐

Corollary 5.1

Let x, y, a, b, A, P(t,s) and G_k(·,s), s,t ∈ [α, β], k = 1, 2, 3, 4, be as in Theorem 3.3. Let a function φ : [α, β] → ℝ be 3-convex such that φ″ is absolutely continuous on [α, β].

1. If t ≤ s, then

   ∫αβ∑j=1lajGk(xj,s)−∑i=1mbiGk(yi,s)P(t,s)ds≥0(33)

   and

   ∑j=1lajϕ(xj)−∑i=1mbiϕ(yi)≥ϕ′(β)−ϕ′(α)β−α∫αβ∑j=1lajGk(xj,s)−∑i=1mbiGk(yi,s)ds.(34)

   Moreover, if in addition φ′(β) ≥ φ′(α), then the left hand side of (34) is nonnegative.

2. If s ≤ t, then the reverse inequalities in (33) and (34) hold. Moreover, if in addition φ′(β) ≤ φ′(α), then the left hand side ofthe reversed (34) is nonpositive.

Remark 5.2

Note that (34) improves Sherman’s inequality (1) when φ′(β) ≥ φ′(α). Moreover, if l = m and all weights a_j and b_i are equal, we get the following extension of weighted majorization inequality

i.e. we get improvements when φ′(β)≥ φ′(α).

If we denote Am=∑i=1mai and put y₁ = y₂ = ... = y_m = 1Am∑i=1maixi, we get the following Jensen’s type inequality

i.e. if in addition φ′(β) ≥ φ′(α), then we get double inequality

which improves Jensen’s inequality.

Especially, choosing Green’s function G₂, defined by (8) and setting m = 2, A₂ = 2, x₁ = α, x₂ = β, y₁ = y₂ = α+β2, we get