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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 368610, 7 pages
A New Extension of Serrin's Lower Semicontinuity Theorem
1School of Mathematics and Physics, Chongqing University of Posts and Telecommunications, Chongqing 400065, China
2Mathematical College, Sichuan University, Chengdu 610064, China
Received 18 February 2013; Revised 3 June 2013; Accepted 3 June 2013
Academic Editor: Yisheng Song
Copyright © 2013 Xiaohong Hu and Shiqing Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We present a new extension of Serrin's lower semicontinuity theorem. We prove that the variational functional defined on is lower semicontinuous with respect to the strong convergence in , under the assumptions that the integrand has the locally absolute continuity about the variable .
1. Introduction and Main Results
The aim of this paper is to give some new sufficient conditions for lower semicontinuity with respect to the strong convergence in for integral functionals where is an open set of , , defined on , denotes the generalized gradient of , and the integrand satisfies the following condition:(H1) is continuous in , and is convex in for any fixed .
The integral functional is called lower semicontinuous in with respect to the strong convergence in , if, for every , such that in , then It is well known that condition (H1) is not sufficient for lower semicontinuity of the integral in (1) (see book ). In addition to (H1), Serrin  proposed some sufficient conditions for lower semicontinuity of the integral . One of the most known conclusions is the following one.
Theorem 1 (see ). In addition to (H1), if f satisfies one of the following conditions:(a) when , for all ,(b) is strictly convex in for all ,(c)the derivatives , , and exist and are continuous for all .
then is lower semicontinuous in with respect to the strong convergence in .
Conditions (a), (b), and (c) quoted above are clearly independent, in the sense that we can find a continuous function satisfying just one of them but none of the others. Many scholars have weakened the conditions of integrand and generalized Theorem 1, such as Ambrosio et al. , Cicco and Leoni , Fonseca and Leoni [6, 7]. In particular Gori et al. [8, 9] proved the following theorems.
Theorem 2 (see [8, 9]). Let one assume that satisfies (H1) and that, for every compact set , there exists a constant such that and, for every compact set , there exists a constant such that Then is lower semicontinuous in with respect to the strong convergence in .
Theorem 3 (see [8, 9]). Let f satisfy (H1) such that, for every open set , there exists a constant such that, for every , , and , Then the functional is lower semicontinuous in with respect to the convergence.
Condition (5) means that is locally Lipschitz continuous with respect to , that is, the Lipschitz constant is not uniform for . This is an improvement of (c) of Serrin’s Theorem 1. Then a question arises, that is whether there are weaker enough conditions more than locally Lipschitz continuity. In this paper, we consider absolute continuity. Obviously, absolute continuity is weaker than Lipschitz continuity. The following theorems show that, in addition to (H1), the locally absolute continuity on about is sufficient for the lower semicontinuity of the variational functional.
Theorem 4. Let be an open set; satisfies the following conditions:(H1) is continuous on , and, is convex in for all ;(H2) is continuous on , and for every compact set of , is absolutely continuous about ;(H3) for every compact set , there exists a constant , such that Then the functional is lower semicontinuous in with respect to the strong convergence in .
Theorem 5. Let be an open set; satisfies (H1) and the following condition:(H4) for every compact set of , is absolutely continuous about .Then the functional is lower semicontinuous in with respect to the strong convergence in .
It is well know that a real function is called an absolutely continuous function on , if, , , such that for any finite disjoint open interval on , when , we have Obviously, if is Lipschitz continuous on , is absolutely continuous on .
Lemma 6 (see ). Let be an open set and a continuous function with compact support on , such that, for every is convex on . Then there exists a sequence , and , such that, if we let one has satisfying the following results:(i)for every is a continuous function with compact support on such that, for all is convex on . Moreover, for all , and (ii)for every , there exists a constant , such that, for all , and, for all , and ;
3. Proof of Theorem 4
We will divide four steps to complete the proof of Theorem 4.
Step 1. Let be a sequence of smooth functions satisfying (1)there exists a compact set , such that , for all ;(2)for every , for all ;(3), for all .
It is clear that, for each satisfies all the hypotheses in Theorem 4 and also vanishes if is outside ; thus By Levi’s Lemma, we have Thus, without loss of generality, we can assume that there exists an open set , such that
Let such that in . We will prove that Without loss of generality, we can assume that By (17), we have and ; thus we will only prove the following inequality:
Step 2. Let be a mollifier, and, for , define where . We have In the following, we denote the derivative of by . When , we know . By the properties of convolution, we know and That is, , such that
Step 3. Now, we estimate the right side of inequality (27).
By (6) and (24), we have Thus Since and are continuous functions, they are bounded functions on compact subset. By Lebesgue Dominated Convergence Theorem, we obtain Now, we will prove By Lemma 6, there exists a sequence of nonnegative continuous functions , such that is convex on , and, for all , By Levi’s Lemma, we obtain In order to prove (31), we only need to prove By (33), we have Thus (31) holds.
Step 4. Now, we need to prove Let By triangle inequality and (7), we have By (39), condition (H2) and , we know that is a locally absolute continuous function about . So is almost everywhere differentiable; that is, exists almost everywhere. Taking derivatives in both sides of (38), we obtain Because vanishes outside , we obtain By (40), we have where We note By Fubini’s Theorem, we have Since is absolutely continuous about , is integrable and absolutely integrable with respect to ; that is, By (17) and (46), we obtain Because of the absolute continuity of integral, we have By (42), we obtain Thus we just proved (36). By (29)–(31) and (36), we have Thus we deduce that the functional is lower semicontinuous in with respect to the strong convergence in . We complete the proof.
4. Proof of Theorem 5
Step 1. Similar to the first step of the proof in Theorem 4, without loss of generality, we assume that the integrand vanishes outside a compact subset of . Thus we assume that there exists an open set , such that
Let , such that in ; we need to prove By Lemma 6, there exists a function sequence , such that, for all , is a continuous function on , for all , is convex on , and, , Let be a mollifier, and define the ; that is, By (55), we have Choose . By (57), we have So By (53), (54), and Levi’s Lemma, we have Let By (59)–(61), we have Obviously, Thus Therefore , being the supremum of the family of functionals , will be lower semicontinuous if every is lower semicontinuous.
Step 2. In order to prove that, for all is lower semicontinuous in with respect to the strong convergence in , we will prove that, for all , the integrand satisfies all conditions of Theorem 4. Obviously, , satisfies condition (H1).
For all and , by (55), we have Thus So satisfies (6) in condition (H3) of Theorem 4.
Now, we will prove that satisfies (7) in condition (H3) of Theorem 4. By , we have By (55) and (67), we have where is a constant depending on . Thus satisfies (7). So we proved that satisfies condition (H3).
Step 3. Next we will prove that satisfies condition (H2).
By condition (H4), for every compact subset , is absolutely continuous about , that is, for all such that for any finite disjoint open interval in , when , we have By Step 1, satisfies (53)-(55) and the following property: where And, for all , are mollifiers satisfying , and , for all . By (71), without of loss generality, we assume that there exists , such that where are given by (72). By (70), we obtain where is a constant. Similar to the above proof, we have Thus Since belongs to a compact set, then . Choose sufficient small, so that is small enough. Thus is absolutely continuous about for all , which is a compact subset of . By (56) and (77), we have By (67) and (78), we obtain where are constants depending on and given by (69) (). By (79), for every compact subset on is absolutely continuous about . Thus satisfies condition (H2).
Now, we have proved satisfies all conditions in Theorem 4, so is lower semicontinuous in with respect to the strong convergence in . Thus has the same lower semicontinuity. This completes the proof of Theorem 5.
The authors would like to thank editors for their hard work and the anonymous referees for their valuable comments and suggestions. This article is supported partially by NSF of China (11071175) and the Ph.D. Programs Foundation of Ministry of Education of China.
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