Reﬁnements of some classical inequalities via superquadraticity

Some new reﬁned versions of the Jensen, Minkowski, and Hardy inequalities are stated and proved. In particular, these results both generalize and unify several results of this type. Some results are also new for the classical situation.


Introduction
Classical inequalities are of great importance for the development of several areas both within the mathematical sciences and beyond. Hence, it is not surprising that the area "Inequalities" has been developed to an independent area of increasing interest. Several wonderful generalizations, sharpening, and applications have been presented. In particular, fairly lately even refinements of these inequalities have been derived. See e.g. [1-4, 12, 14, 20] and the references given there.
In this paper we derive new such refinements of some classical inequalities. More exactly, the main content of this paper is as follows: In Sect. 2 some new refinements of Jensen's inequality can be found (see Theorems 2.3 and 2.4). In Sect. 3 we state, prove, and apply some new refinements of the Minkowski inequality and new Beckenbach-Dresher type inequality (see Theorems 3.3 and 3.6). In Sect. 4 we derive some corresponding refinements of Hardy's inequality even in a Banach function space setting. Finally, in Sect. 5 we give some concluding remarks and results, which in particular put our results to a more general context.
In all our results we use the concept of superquadratic function, so we finish this section with the following crucial information.
We say that f is subquadratic if -f is superquadratic.
We cite the following result, which is very useful in the proofs of our main results (see [2,3], and [21] for further details). If ϕ is a nonnegative superquadratic function, then ϕ is convex (see [2, Lemma 2.2]) and inequality (1) is a refinement of the Jensen inequality for a convex function which states Convention. Throughout this paper we assume that f is a measurable function on the considered measure space.

Refinements of Jensen's inequality
We need the following useful special case of Theorem 1.2. Lemma 2.1 Let ϕ be a superquadratic function, and let t be a nonnegative measurable function such that T = t(s) ds. The inequality Proof Set dμ(s) = t(s) T ds. Then (2) follows from (1). The proof is complete.
Example 2.2 Let ϕ be a superquadratic function, let x 1 , x 2 be two nonnegative real numbers and λ ∈ [0, 1]. Then Denote Our first main result in this section reads as follows.
holds, where Moreover, (4) holds in the reversed direction if ϕ is subquadratic.
Proof Set x 1 = f Q , x 2 = f R , and λ = Q T . It is clear that Then from Example 2.2 it follows that Moreover, using the fact that we obtain inequality (4). The proof is complete since the proof of the reversed inequality is similar to the proof above, and we can omit the details.
By making a further restriction of ϕ, we can also state the following version of Theorem 2.3.

Theorem 2.4 Let ϕ : [0, ∞) → R be a nondecreasing and superquadratic function such that
Then the following refined variant of Jensen-type inequality Proof We proved the first inequality ϕ(f ) ≤ I in Theorem 2.3, so we only need to prove the second inequality. By applying Lemma 2.1 in the first two terms of I, we get that where To finish the proof, it is enough to prove that Now, by using the triangle inequality, the nondecreasing property of ϕ, and (5), we obtain that Hence (6) follows as a sum of the above two inequalities. The proof is complete.
Theorem 3.2 Let f be a nonnegative measurable function on X × Y with respect to the measure μ × ν, and let p ≥ 1. Then Our first main result in this section is the following refinement of the Minkowski inequality.

Theorem 3.3 Let f be a nonnegative measurable function on X
where Let p ≥ 2 and 1 p + 1 q = 1. Using Lemma 3.1, by replacing f (x) and g(x) with f (x, y) and H p-1 (x), respectively, we get that where We integrate inequality (9) over Y , apply Fubini's theorem on the left side of the inequality to find that Since The proof of the case 1 < p ≤ 2 is similar so we omit the details and the proof is complete.
Next, we point out the following discrete version of the above theorem. where If 1 < p ≤ 2, then (11) holds in the reversed direction.
, for all i = 1, . . . , n, and let dν = dy be the Lebesgue measure. Define Then and h(x, y) = f -H X fH p-1 dμ Therefore, by applying Theorem 3.3, one can complete the proof.
The continuous form of the Beckenbach-Dresher inequality was first derived in [9, Theorem 3.1] (see also [22]). It has the following form.

Theorem 3.5 Let f and u be nonnegative measurable functions on X × Y with respect to the measures μ × ν and λ × ν, respectively, and let
provided all occurring integrals exist.
Our new result related to the continuous Beckenbach-Dresher inequality reads as follows. Theorem 3.6 Let f and u be nonnegative measurable functions on X × Y with respect to the measures μ × ν and λ × ν, respectively, and let 1 < q ≤ 2 ≤ p, s ≥ 1. Then Proof Let 1 < q ≤ 2 ≤ p. Then, in view of Theorem 3.3, for p ≥ 2 and 1 < q ≤ 2, we have that In the last inequality we used the reverse Hölder inequality for two functions a and b when one exponent (1s) is negative and the other exponent s is positive. The proof is complete. By using Theorem 3.6 and similar arguments as those in the proof of Corollary 3.4, we can also derive the following discrete version.
As an application of Corollary 3.7, by making the substitution s = p p-q , p = q, we obtain the following Beckenbach-Dresher type inequality.
where h i and r i are as in Corollary 3.7.

Refinements of Hardy's inequality
The results in this section may be seen as complements and further generalizations of some results in [14] and [19]. In [6, Theorem 2.1] the following Hardy-type inequality was given. let ϕ be a positive convex function on (a, c) and E be a Banach function space on [0, b). If E has the Fatou property and a < f (x) < c, then provided that both sides have sense.
To prove our main results, we need the following lemma (see [13,18]).

Lemma 4.2 (See [18]) Assume that the Banach function space E has the Fatou property.
Let f (x, t) ≥ 0 on × T and let for almost every t ∈ T, f (x, t) ∈ E. If the function f r (x, t) 1 r E is integrable on T, then, for r ≥ 1, Our first main result in this section reads as follows.  [0, b). If E has the Fatou property and a < f (x) < c, then provided that both sides have sense.
By using Theorem 1.2, the lattice property of E, Lemma 4.2 with r = 1, and (13), we find that The proof is complete.
Here we just give one example of application of Proof It is known that E = L 1 ((0, b), u(x) x dx) satisfy the Fatou property (see e.g. [7]). Moreover, Therefore, (14) follows from (15) and Theorem 4.3. The proof is complete.
Next we state a "dual" version of Theorem 4.3. Note that the natural dual operator of the Hardy operator H : t dt, but here we use its alternative H * : Then By using (16) and the same arguments as in the proof of Theorem 4.3 we obtain that The proof is complete.
We give the following example of application of  u(x) dx, t ∈ (b, ∞).
If the real-valued function ϕ is positive and superquadratic on (a, c), 0 ≤ a < c ≤ ∞, then the inequality holds for all f with a < f (x) < c, x ≥ b.