Abstract
In this study, we extend some “sneak-out” inequalities on time scales for a function depending on more than one parameter. The results are proved by using the induction principle and time scale version of Minkowski inequalities. In seeking applications, these inequalities are discussed in classical, discrete, and quantum calculus.
1. Introduction
Bennett and Grosse-Erdmann [1] introduce the “sneak-out” principle concerned with the equivalence of two series. Bohner and Saker [2] extended the sneak-out principle on time scales and proved some new dynamic sneak-out inequalities and their converses on time scales which, as special cases, with , contain the discrete inequalities obtained by Bennett and Grosse-Erdmann (Section 6 in [1]). However, the sneak-out principle on time scales can be applied to formulate the corresponding integral inequalities by choosing . The paper aims to extend the work given by Bohner and Saker in [2] for functions depending on more than one parameter. Some other inequalities, such as Hardy-type, Hardy-Copson, and Copson-Leindler-type inequalities, are also studied for functions of more than one parameter [3–5] via time scales’ calculus. Some literature concerning with time scale can be seen in [6–13].
The paper is organized as follows. Section 2 provides some basics from time scales’ calculus. Section 3 features two dynamic inequalities of the Copson type, which are needed to prove further results. In Section 4, we present sneak-out inequalities on time scales for functions depending on more than one parameter.
2. Preliminaries
A time scale as well as close set in are nonempty [14, 15]. Some examples of time scales are , , and Cantor set. Assume that , where is empty set and . A time-scale interval is denoted by , for
The operators defined by and defined by are forward as well as backward jump operators, respectively, for . The point is right-scattered if it satisfies , and left-scattered if . The points which are at the same time left-scattered as well as right-scattered are called isolated. Furthermore, the point is right-dense if it satisfies and , and left-dense if it satisfies and ; furthermore, the point is called dense if it is left-dense as well as right-dense at the same time. A function , defined by is called the graininess function.
If a function is continuous at all right-dense points, the left-hand limits exist and are finite at left-dense points in ; then, it is right-dense continuous (rd-continuous) on . The set denoted by contain all rd-continuous functions on .
Consider a function , and define the number if it exists with the property that, for given , there is a neighborhood of which satisfiesthen is delta derivative of function at .
: for any function .(1)Product and quotient rule for delta derivative (Theorem 1.20 in [14]): assume are differentiable; then,(2)Integration by parts formula (Theorem 1.77 in [14]): for two delta differentiable functions and , we have(3)Minkowski inequality (Theorem 6.16 in [14]): for three rd-continuous functions , , we have where and .(4)Fubini’s theorem [16]: let there exist two time scales’ measure spaces and which have finite dimensions. If is a which is an integrable function and the function exists for almost every and exists for almost every then is integrable on , is integrable on , and
:
Some preliminary inequalities [2]: suppose is differentiable. Let is monotone, ., either always negative or always positive, thenand if is positive, then
3. Dynamic Copson-Type Inequalities for Finite Numbers of Parameters
We assume throughout that all the functions are nonnegative and the integrals considered exist. For , , let be time scales. Presume 1:
Theorem 1. Assume . Suppose is such thatis well defined and . Then,
Theorem 2. Assume . Suppose is such thatis well defined. Let and Then,
Proofs of Theorem 1 and Theorem 2 are after sneak-out inequalities.
4. Dynamic Sneak-Out Inequalities for Finite Numbers of Parameters
Let and Presume 2:
Lemma 1. Let be the time scales for under and , and we have
Proof. For , (16) is true by Theorems 4.1 in [2], .,Suppose (16) is true for To prove for , by using , we have defined asDenoteUse integration by parts’ formula (3) in (19) to obtainUse the right-hand side part of inequality (8) with in (20)Substitute (21) in (18):Use (5) “p times” in second term of (22):Use induction hypothesis for with fix (instead for ) in (23) and (25) to obtainBy applying (5) “p times” on (25) and making simplification, we obtainHence, by using (17) for , we obtainThus, by mathematical induction, (16) holds for all , which completes the proof.
Remark 1. If we chose in Lemma 1, then (16) becomes the following inequality:
Theorem 3. Assume , , and . Then,
Proof. We prove the result by using mathematical induction. For , statement is true by Theorems 4.1 in [2]. Assume for (29) holds. To prove the result for , take L.H.S of (29) in the following form:Using (27) in (30) for ,whereApply Minkowski’s inequality (4) on (31) to obtainwhereDenoteand one has thatUse Theorem 1 in (36) by taking to obtainSubstitute (37) in (33) and take power on both sides to obtainThus, by mathematical induction, (29) holds for all .
Example 1. Let and and . In this case, (29) in Theorem 3 takes the formwhereNote that (39) is extension of Example 4.4 in [2].
Example 2. Let , in Theorem 3. In this case, (29) takes the formwhereNote that (41) is extension of Example 4.3 in [2].
Example 3. Let , , and , in Theorem 3. In this case, (29) takes the formwhere
Lemma 2. Let be the time scales for under and , we have
Proof. For , (45) is true by Theorems 4.6 in [2], ,Suppose (45) is true for To prove for , by using , we haveDenoteUse integration by parts formula (3) in (48) to obtainUse and the right-hand side part of inequality (8) with in (49)Substitute (50) in (47)Use (5) “p times” on (51):Use induction hypothesis for with fix (instead for ) in (52) to obtain