Abstract

We show that a local minimizer of in the topology must be a local minimizer in the topology, under suitable assumptions for the functional with supercritical exponent . This result can be used to establish a solution to the corresponding equation admitting sub- and supersolution. Hence, we extend the conclusion proved by Brezis and Nirenberg (1993), the subcritical and critical case.

1. Main Results for Supercritical Exponent

We consider the following functional: where with smooth boundary, supercritical exponent , and  satisfies the growth condition: as well as the usual assumptions that is measurable in and continuous in .

Our main results are the following.

Theorem 1. Assuming that is a local minimizer of in the topology, there is some , such that Then is also a local minimizer of in the topology; that is, there exists , such that where the topology given by .

It will be noted that an neighbourhood is much bigger than a neighbourhood. The proof depends on the special structure of , and the claim clearly would be false for a general function .

In order to prove this theorem, the following preparatory steps are critical. We begin with a theorem concerning the topology of .

Theorem 2. Let be defined as in the above theorem; then is a reflexive and strictly convex Banach space with the duality .

Proof of Theorem 2. Now we give a detailed proof for the reader’s convenience.
At first we show that the definition of is actually a norm. Obviously, separate points are as follows: if , that is, , then . And positive homogeneity is . The triangle inequality is, for any , And then it shows that the space is complete; that is to say, any cauchy sequence in will converge. From the definition of the norm , we know that is also the cauchy sequence in and . By the completion of the Banach space and , we know that will converge to in and will converge to in . And, since , we know that in and also, due to , we also know that in , and, based on the uniqueness of the limit in , we have (denoted by ). With this result, we have converge to in , which implies that is complete. Thus, is a Banach space.
For strictly convex, which is based on the definition of the strictly convex of Banach space, we need to show that if and , then , which can be done by the following inequality: And the fact “” in (6) is true if and only if with the constant in consequence of the uniformly convex space and (P97 [1]). And, combining with , we can get the constant , which contradicts the assumption . Therefore, the Banach space is strictly convex.
For the reflexive, we need the following lemmas (see P63, P105 [2]).

Lemma 3. Let ,…, be normed space. Then is a Banach spaces if and only if each is a Banach space; furthermore, is reflexive if and only if each is reflexive.

Lemma 4. Every closed subspace of a reflexive space is reflexive.

Therefore, setting a space with the norm . It follows from Lemma 3 that is a reflexive Banach space. Obviously, our space with the norm can be seen as a closed subspace of by the embedded mapping (denoting by in the following). Thus, based on Lemma 4, is a reflexive Banach space.

For the dual, we need the following lemma (see P91 [2]).

Lemma 5. Let be normed spaces. Then there is an isometric isomorphism that identifies with , such that, if the element of is identified with the element of , then whenever .

From the Lemma 5, we know that the dual space of will be with . And, if our space can be seen as a closed subspace of , then, (in the sense of restriction). At the same time, from the Hahn-Banach theorem, we know that, for any , we can extend to be a bounded linear functional on , such that And, from (7), we have Hence, which implies that . Therefore, and the proof of Theorem 2 is completely finished.

Also, for the property of weak converge in , we have the following.

Lemma 6. If in as , then

Proof. In fact, for (11), from Theorem 2, we know that, for any , there exists and , such that Now, choosing in (13) (noting the fact that ) and combining with in , we know that, for any , which implies that in . Similarly, choosing in (13) (also noting the fact that ), we can get in and finish the proof of (11).
And, for (12), from the interpolation inequality, and with (11), it is easy to prove (12) and complete the proof of Lemma 6.

For the operators generated by (1), we have, for any

Lemma 7. Both of operators from to and from to are bijective, where .

Proof. First, for any , it follows that from If , it follows from the maximum principle (see P179 Theorem 8.1 [3]) that , which implies that it is an injection. Whereas, by Riesz’s Lemma, we know that, for any , there exists a , such that and which implies that and . Hence is bijective (indeed, isometric).
Secondly, the map is clearly bounded, continuous, and also injective; namely, if , then , which can be obtained by the following inequality . For surjective, by applying the James Theorem in Banach space (see [4]) to the strictly convex space and , for any , there is only one unique supporting functional , such that , which implies that . So is bijective.
For the regularity of solution of (1), we have the following.

Lemma 8. Assuming satisfies in the weak sense then one has , .

Proof. Indeed, we set the corresponding evolution equation and apply the same argument by the Moser iteration as Lemma  5.20 in [5], and, with the fact that the solution of (18) is the equilibrium point of (19), it is easy to show Lemma 8.
Now, we are in a position to prove Theorem 1.

Proof. Suppose the conclusion (3) does not hold. Then where .

Claim 1. is achieved at some point (still denoted by ).
Indeed, it is clear that there exists a constant , such that   . Hence, there exists a minimizing sequence , and there is by Lemma 6 a subsequence (also denoted by ) such that in , . Combining with the lower semicontinuity of norm, we have , , and . Hence, and Claim 1 is completely proved.
Now we will prove that in , but (3) and (20) are contradictory (also see [6]). The corresponding Euler equation for involves a Lagrange multiplier ; namely, satisfies where due to Lemma 7.
That is, This means that
Recalling that and combining with Lemma 8, one may bootstrap the bound to (independent of ), since in , in (by Ascoli). This concludes the proof.

2. Application of Theorem 1

Next, we present a simple and useful application of Theorem 1.

Considering in Theorem 1 with , such that, for some constant , Assume that there are sub- and supersolutions and ; that is, in the distribution sense, Moreover, neither nor is a solution to (18).

Theorem 9. Under the assumption (2) there is a solution to (18), , such that, in addition, is a local minimum of in .

The proof relies on Theorem 1 as well as on the following.

Theorem 10. Let be a bounded domain in with smooth boundary . Let and assume that, for some , satisfies Then either or there exists , such that Moreover, if is replaced by the nonnegative continuous function , then the conclusion is also valid.

Proof. The measure is nonnegative in . We may assume .

Case 1. Consider . In this case, by induction applies to Lemma 8: Since , we have in some closed ball in . Let be an expanding sequence of subdomains of with smooth boundaries and ; suppose , . Let be the solution in of In order to compare with , we need the following comparison principle for the operator defined in Lemma 7

Lemma 11. Let satisfy, for some , then, in .

Proof. Indeed, setting and defining , it is noted that the derivative expression . Then, by the mean value theorem, there is satisfying
Applying the weak maximum principle, Theorem 8.1 P179 [3] by choosing , we know that and complete the proof.

Since in Lemma 8, then, by the virtue of Lemma 11, in . As , we find when solves By the Hopf lemma 3.4 P34 [3] with , one obtains for some The conclusion of Theorem 10 then follows directly.

Case 2. Consider . Let be a cutoff function, , such that . Let be the solution of Since is smooth outside a compact set , it follows and applies to the Hopf lemma as above for some , The conclusion of Theorem 10 is a direct consequence of the following.

Claim 2. One has in .

Proof of Claim 2. Given any , we will prove that The claim then follows.
Note that satisfies provided is sufficiently small (depending on ). The property (39) follows from the inequality , , when . The last property (40) follows from the fact that is smooth near and on .
Let be a sequence of mollifiers with and set .
Clearly is smooth, and, by (39) and the mean value theorem with , we have On the other hand, we deduce from (40) that provided . The maximum principle (Corollary 3.2 P33 [3]) of choosing implies that when . Passing to the limit as , we see that which is the desired conclusion. The similar argument is also true as is replaced by the nonnegative continuous function and the proof of Theorem 10 is completely finished.