Research Article | Open Access
An Integral Mean Value Theorem concerning Two Continuous Functions and Its Stability
The aim of this paper is to investigate an integral mean value theorem proposed by one of the references of this paper. Unfortunately, the proof contains a gap. First, we present a counterexample which shows that this theorem fails in this form. Then, we present two improved versions of this theorem. The stability of the mean point arising from the second result concludes this paper.
The mean value theorems represent some of the most useful mathematical analysis tools. The first known result is due to Lagrange (1736–1813). In the years that followed, more mathematicians investigated this subject. As consequences of this fact, now we can find similar results, more generalizations, or extensions. Sahoo and Riedel’s book  presents a large collection of old and new mean value theorems. The readers can consult , , , or  to find some recent results. Reference  leads this subject to a new direction.
In , the following theorem was presented.
Theorem 1. Let and be two nonnegative continuous functions on the interval and . There exists such that
This result is very generous. Moreover, the authors obtained some interesting consequences. Unfortunately, the proof contains a gap. The aim of this paper is to present a counterexample which shows that the result from the previous theorem is not necessarily valid under this hypothesis. Afterwards, we present some conditions for which equality (1) holds. Finally, we include a stability result.
2. A Counterexample for Theorem 1
Let us consider the functions defined, for any , by and . These functions are continuous and nonnegative. We have We choose . Then for any . Hence Then, relation (1) fails.
3. Two Integral Mean Value Theorems
In this section we present two valid versions of Theorem 1, which were suggested by their original proofs.
Theorem 2. Let and be two nonnegative continuous functions on the interval . One supposes that there exists such that , , , and . Then, for any , there exists such that equality (1) holds.
Proof. First, for every , we have Hence, is nonnegative; then and we obtainEquality in any side of (6) holds if is constant or Similarly, we obtainWe obtain equality in any side of (7) if or is constant.
Let Let us consider the continuous function defined, for any , by ThenFrom (6) and (7), we obtain In the same mode, we have Then, there exists between and such that We obtain (1).
It remains to show that If , then the conclusion is clear. We suppose that Using (6) and (7), we conclude that and are constant functions. We obtain , for any Then, we can replace with any point from A similar conclusion is obtained if Now, the proof is complete.
As consequences of the previous theorem, we obtain the following result.
Theorem 3. Let and be two nonnegative continuous functions on the interval and monotone of the same type. Let Then there exists such that equality (1) holds.
Proof. We suppose that and are increasing functions. We have , , , and The conclusion follows by applying the previous theorem.
4. A Stability Result
The parents of the stability concept are considered to be the mathematicians Ulam and Hyers (see [8–10]). This notion is associated with the functional equations, the differential equations, or the linear recurrences. Starting with , a new direction was created. It is about the stability of the point arising from the mean value theorems. Reference  or  is relevant.
Theorem 4. Let Let and be two nonnegative continuous functions on the interval and monotone of the same type. One assumes that there exists a unique which is satisfying equality (1). Then, for any , there exists with the following property: for any continuous functions such that and , for any , there exists such that and
Proof. We assume that the functions and are increasing. We define the function by for any We have Moreover, The uniqueness of goes to Similar arguments show that Then , for any , and , for any
Now, let For any continuous functions such that and , for any , we define the function by for any Further, and we obtainfor any In the same mode, we obtainfor any Moreover, we have soBy using (15), (16), and (18), we obtainfor any
Now, let and such that Similarly, let such that Then We consider such that We consider such thatFurther, , so Similarly, we have , so Then, there exists such that From , we obtain The equality goes to and the proof is complete.
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
This work has been funded by the Sectoral Operational Programme Human Resources Development 2007–2013 of the Ministry of European Funds through the Financial Agreement POSDRU 187/1.5/S/155420.
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