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Refinement of an integral inequality
Journal of Inequalities and Applications volume 2012, Article number: 103 (2012)
In this study, we generalize and sharpen an integral inequality raised in theory for convex and star-shaped sets and relax the conditions on the integrand.
Mathematics Subject Classification (2000): 26D15
In the study , which investigated convex and star-shaped sets, the following interesting result was obtained.
Theorem 1. ([, Lemma 2.1]) Let p : [0, T ] → ℝ be a nonnegative convex function such that p(0) = 0. Then for 0 < a ≤ b ≤ T and k ∈ ℕ+the inequality
In this note, we shall show that the convexity of the function p(t) may be replaced by the condition that is increasing, sharpen inequality (1), and obtain the following a general result using a monotone form of l'Hospital's rule, a elementary method, and Mitrinović-Pečarić inequality, respectively.
Theorem 2. Let p : [0, T ] → ℝ be a nonnegative continuous function such that p(0) = 0 andbe a monotone function on (0, T ]. LetThen for 0 < x ≤ b ≤ T and k ≥ 0 the double inequality
holds so that
(i) whenis increasing, we have, ;
(ii) whenis decreasing, we have, .
Furthermore, these paired numbers α and β defined in (i) and (ii) are the best constants in (2).
2 Two lemmas
Lemma 1. ([[2–5], A Monotone form of L'Hospital's rule]) Let f, g : [a, b] → ℝ be two continuous functions which are differentiable on (a, b). Further, let on(a, b). Ifis increasing (or decreasing) on (a, b), then the functions and are also increasing (or decreasing) on (a, b).
Lemma 2. ([, Mitrinović-Pečarić inequality]) If f is increasing function and p satisfies the conditionsfor x ∈[a, b], and for some, , , then we have
If f is decreasing the inequalities (3) are reversed.
3 A concise proof of Theorem 2
Let where , g1(t) = tk + 2, and 0 < t ≤ 1. Then .
When is increasing, we have is also increasing, and is increasing by Lemma 1. At the same time, , and . So we obtain(4)
are the best constants in (4).
Replacing t with x/b in (4), we have . Then let bs = u, we obtain , and
holds. Furthermore and are the best constants in (5).
When is decreasing, we obtain corresponding result by the same way.
4 New elementary proof of Theorem 2
Let for x ∈ (0, b]. Assume that is increasing. By a simple calculation and the inequality for 0 < t ≤ x ≤ b we have that
So F (x) is increasing and the chain inequality
holds. Then the double inequality (5) holds, α and β are the best constants in (6) or (2).
The decreasing case can be proved similarly.
5 Other proof of Theorem 2
In what follows, we also assume that is increasing.
Let p(t) = tk + 1,,c = x, and a = 0 in Lemma 2, we can obtain
The left-side inequality of (7) deduces
then the right-side inequality of (2) holds.
Let b → 0+ in the right-side inequality of (7), we can obtain
then the left-side inequality of (2) holds.
Let . Since and G(b) = β, we obtain that is α and β are the best constants in (2).
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The author declares that they have no competing interests.
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Zhu, L. Refinement of an integral inequality. J Inequal Appl 2012, 103 (2012). https://doi.org/10.1186/1029-242X-2012-103
- lower and upper bounds
- refinement of an integral inequality
- a monotone form of l'Hospital's rule
- Mitrinović-Pečarić inequality