# H

Authors: Yansheng Zhong
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We show that a local minimizer ofΦ in the C1 topology must be a local minimizer in the H1∩Lp topology, under suitable assumptions for the functional Φ=(1/2)∫Ω|∇u|2+(1/p)∫Ω|u|p−∫ΩF(x,u) with supercritical exponent p>2∗=2n/(n−2). 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.


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## Body

## 1. Main Results for Supercritical Exponent

We consider the following functional:(1)Φ=12∫Ω|∇u|2+1p∫Ω|u|p-∫ΩF(x,u),
where Ω⊂ℝn with smooth boundary, supercritical exponent p>2*=2n/(n-2), and  F(x,u)=∫0uf(x,s)ds satisfies the growth condition:
(2)|f(x,u)|⩽C(1+|u|ℓ)withℓ<p,
as well as the usual assumptions that f is measurable in x and continuous in u.Our main results are the following.Theorem 1.
Assuming thatu0∈H01(Ω)∩Lp(Ω) is a local minimizer of Φ in the C1 topology, there is some r>0, such that
(3)Φ(u0)⩽Φ(u0+v),∀v∈C01(Ω¯)with∥v∥C1⩽r.
Then u0 is also a local minimizer of Φ in the H01(Ω)∩Lp(Ω) topology; that is, there exists ϵ0>0, such that
(4)Φ(u0)⩽Φ(u0+v),∀v∈H01(Ω)∩Lp(Ω)nnnnnnnnnnnnnnnwith∥v∥H01(Ω)∩Lp(Ω)⩽r,
where the topology X≜H01(Ω)∩Lp(Ω) given by ∥·∥X=∥·∥H01(Ω)+∥·∥Lp(Ω).It will be noted that anX neighbourhood is much bigger than a C01 neighbourhood. The proof depends on the special structure of Φ, and the claim clearly would be false for a general function Φ.In order to prove this theorem, the following preparatory steps are critical. We begin with a theorem concerning the topology ofX.Theorem 2.
LetX be defined as in the above theorem; then X is a reflexive and strictly convex Banach space with the duality X*⊂H-1(Ω)⊕Lq(Ω)((1/p)+(1/q)=1).Proof of Theorem2.
Now we give a detailed proof for the reader’s convenience.
At first we show that the definition of∥·∥X is actually a norm. Obviously, separate points are as follows: if ∥x∥X=0, that is, ∥x∥H01(Ω)+∥x∥Lp(Ω)=0, then x=0. And positive homogeneity is ∥αx∥X=∥αx∥H01(Ω)+∥αx∥Lp(Ω)=|α|[∥x∥H01(Ω)+∥x∥Lp(Ω)]=|α|∥x∥X. The triangle inequality is, for any x,y∈X,
(5)∥x+y∥X=∥x+y∥H01(Ω)+∥x+y∥Lp(Ω)⩽[∥x∥H01(Ω)+∥x∥Lp(Ω)]+[∥y∥H01(Ω)+∥y∥Lp(Ω)]=∥x∥X+∥y∥X.
And then it shows that the space X is complete; that is to say, any cauchy sequence {un} in ∥·∥X will converge. From the definition of the norm ∥·∥X, we know that un is also the cauchy sequence in H01(Ω) and Lp(Ω). By the completion of the Banach space H01(Ω) and Lp(Ω), we know that un will converge to u1 in H01(Ω) and un will converge to u2 in Lp(Ω). And, since H01(Ω)↪L2(Ω), we know that un→u1 in L2(Ω) and also, due to Lp(Ω)⊆L2(Ω), we also know that un→u2 in L2(Ω), and, based on the uniqueness of the limit in L2(Ω), we have u1=u2 (denoted by u). With this result, we have un converge to u in X, which implies that X is complete. Thus, X 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 ifx≠y and ∥x∥H01(Ω)+∥x∥Lp(Ω)=∥y∥H01(Ω)+∥y∥Lp(Ω)=1, then ∥x+y∥H01(Ω)+∥x+y∥Lp(Ω)<2, which can be done by the following inequality:
(6)∥x+y∥H01(Ω)+∥x+y∥Lp(Ω)⩽(∥x∥H01(Ω)+∥y∥H01(Ω))+(∥x∥Lp(Ω)+∥y∥Lp(Ω))=(∥x∥H01(Ω)+∥x∥Lp(Ω))+(∥y∥H01(Ω)+∥y∥Lp(Ω))=2.
And the fact “=” in (6) is true if and only if x=cy with the constant c>0 in consequence of the uniformly convex space H01(Ω) and Lp(Ω)(1<p<∞) (P97 [1]). And, combining with ∥x∥X=∥y∥X=1, we can get the constant c=1, which contradicts the assumption x≠y. Therefore, the Banach space X is strictly convex.
For the reflexive, we need the following lemmas (see P63, P105 [2]).Lemma 3.
LetX1,…, Xn be normed space. Then X1⊕⋯⊕Xn is a Banach spaces if and only if each Xj is a Banach space; furthermore, X1⊕⋯⊕Xn is reflexive if and only if each Xj is reflexive.Lemma 4.
Every closed subspace of a reflexive space is reflexive.Therefore, setting a spaceE=H01(Ω)⊕Lp(Ω) with the norm ∥·∥X=∥·∥H01(Ω)+∥·∥Lp(Ω). It follows from Lemma 3 that E is a reflexive Banach space. Obviously, our space X=H01(Ω)∩Lp(Ω)(1<p<∞) with the norm ∥·∥X=∥·∥H01(Ω)+∥·∥Lp(Ω) can be seen as a closed subspace of E by the embedded mapping u→(u,u) (denoting (u,u) by i(u) in the following). Thus, based on Lemma 4, X is a reflexive Banach space.For the dual, we need the following lemma (see P91 [2]).Lemma 5.
LetX1,…,Xn be normed spaces. Then there is an isometric isomorphism that identifies (X1⊕⋯⊕Xn)* with X1*⊕⋯⊕Xn*, such that, if the element y* of (X1⊕⋯⊕Xn)* is identified with the element x1*,…,xn* of X1*⊕⋯Xn*, then
(7)y*(x1,…,xn)=∑j=1nxj*xj
whenever (x1,…,xn)∈X1⊕⋯⊕Xn.From the Lemma5, we know that the dual space E* of E=H01(Ω)⊕Lp(Ω) will be E*=H-1(Ω)⊕Lq(Ω) with (1/p)+(1/q)=1. And, if our space X=H01(Ω)∩Lp(Ω) can be seen as a closed subspace of E, then, X*⊆H-1(Ω)⊕Lq(Ω) (in the sense of restriction). At the same time, from the Hahn-Banach theorem, we know that, for any f∈X*, we can extend f to be a bounded linear functional f~ on E, such that
(8)〈f~,i(u)〉E*,E=〈f,u〉X*,X∀u∈X.
And, from (7), we have
(9)〈f~,i(u)〉E*,E=〈f1,u〉H-1(Ω),H01(Ω)+〈f2,u〉Lq(Ω),Lp(Ω).
Hence,
(10)〈f,u〉X*,X=〈f~,i(u)〉E*,E=〈f1,u〉H-1(Ω),H01(Ω)+〈f2,u〉(LqΩ),Lp(Ω)
which implies that f∈E*. Therefore, X*⊂H-1(Ω)⊕Lq(Ω) and the proof of Theorem 2 is completely finished.Also, for the property of weak converge inX, we have the following.Lemma 6.
Ifun⇀v in X as n→∞, then
(11)un⇀vinH01(Ω),un⇀vinLp(Ω)asn⟶∞(12)un⟶vinLt(Ω)∀2⩽t<pasn⟶∞Proof.
In fact, for (11), from Theorem 2, we know that, for any f∈X*, there exists f1∈H-1(Ω) and f2∈Lq(Ω), such that
(13)〈f,u〉X*,X=〈f1,u〉H-1(Ω),H01(Ω)+〈f2,u〉Lq(Ω),Lp(Ω).
Now, choosing f2=0 in (13) (noting the fact that H-1(Ω)×{0}⊂X*) and combining with un⇀v in X, we know that, for any f1∈H-1(Ω),
(14)〈f,un〉X*,X=〈f1,un〉H-1(Ω),H01(Ω)+〈0,un〉Lq(Ω),Lp(Ω)=〈f1,un〉H-1(Ω),H01(Ω)⟶〈f1,v〉H-1(Ω),H01(Ω)
which implies that un⇀v in H01(Ω). Similarly, choosing f1=0 in (13) (also noting the fact that {0}×Lq(Ω)⊂X*), we can get un⇀v in Lp(Ω) and finish the proof of (11).
And, for (12), from the interpolation inequality,
(15)∥u-v∥Lt(Ω)⩽∥u-v∥L2(Ω)θ∥u-v∥Lp(Ω)1-θ
and H01(Ω)↪L2(Ω) 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 p>1Lemma 7.
Both of operators-Δ:u→-Δu from H01(Ω) to H-1(Ω) and F:u→|u|p-2u from Lp(Ω) to Lp′(Ω)=Lp/(p-1)(Ω) are bijective, where ∥u∥H012=∫Ω∇u∇u.Proof.
First, for anyu∈H01, it follows that -Δu∈H-1 from
(16)〈-Δu,v〉H-1,H01=∫Ω∇u∇v,∀v∈H01.
If u≠v∈H01, it follows from the maximum principle (see P179 Theorem 8.1 [3]) that -Δu≠-Δv∈H-1, which implies that it is an injection. Whereas, by Riesz’s Lemma, we know that, for any f∈H-1, there exists a u∈H01(Ω), such that ∥f∥H-1=∥u∥H01 and
(17)〈f,v〉H-1,H01=(u,v)H01=∫Ω∇u∇v=∫Ω-Δuv∀v∈H01(Ω)
which implies that f=-Δu and ∥-Δu∥H-1=∥f∥H-1=∥u∥H01. Hence -Δ:H01(Ω)→H-1(Ω) is bijective (indeed, isometric).
Secondly, the mapF:u→|u|p-2u is clearly bounded, continuous, and also injective; namely, if u≠v∈Lp(Ω), then |u|p-2u≠|v|p-2v∈Lp′, which can be obtained by the following inequality 〈|u|p-2u-|v|p-2v,u-v〉⩾(1/p)|u-v|p. For surjective, by applying the James Theorem in Banach space (see [4]) to the strictly convex space Lp and Lp′, for any ∥w∥Lp′=1, there is only one unique supporting functional ∥u∥Lp=1, such that 〈w,u〉=1, which implies that w=|u|p-2u. So F is bijective.
For the regularity of solution of (1), we have the following.Lemma 8.
Assumingu0∈X satisfies in the weak sense
(18)-Δu+|u|r-2u=f(x,u)inΩu=0on∂Ω
then one has u0∈C1,α(Ω¯), forallα<1.Proof.
Indeed, we set the corresponding evolution equation(19)ut-Δu+|u|r-2u=f(x,u)inΩu=0on⁡∂Ω
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 Theorem1.Proof.
Suppose the conclusion (3) does not hold. Then
(20)∀ϵ>0,∃vϵ∈BϵsuchthatΦ(vϵ)<Φ(0)
where Bϵ={u∈X;∥u∥X⩽ϵ}.Claim 1.
min
⁡
B
ϵ
Φ is achieved at some point (still denoted by vϵ).
Indeed, it is clear that there exists a constantC, such that ∥Φ(u)∥⩽C  forallu∈Bϵ. Hence, there exists a minimizing sequence un∈Bϵ, and there is by Lemma 6 a subsequence (also denoted by un) such that un⇀u in H01(Ω), Lp(Ω). Combining with the lower semicontinuity of norm, we have liminf⁡n→∞∥un∥H01(Ω)⩾∥u∥H01(Ω), liminf⁡n→∞∥un∥⩾∥u∥Lp(Ω), and lim⁡n→∞F(x,un)→F(x,u). Hence, liminf⁡n→∞Φ(un)⩾Φ(u) and Claim 1 is completely proved.
Now we will prove thatvϵ→0 in C1, but (3) and (20) are contradictory (also see [6]). The corresponding Euler equation for vϵ involves a Lagrange multiplier μϵ⩽0; namely, vϵ satisfies
(21)〈Φ′(vϵ),ζ〉X*,X=μϵ〈i(vϵ),ζ〉X*,X,∀ζ∈X,
where i(vϵ)=-2Δvϵ+p|vϵ|p-2vϵ due to Lemma 7.
That is,(22)∫Ω∇vϵ∇ζ+∫Ω|vϵ|p-2vϵζ-∫Ωf(x,vϵ)ζ=2μϵ∫Ω∇vϵ∇ζ+μϵp∫Ω|vϵ|p-2vϵζ.
This means that
(23)-(1-2μϵ)Δvϵ+(1-pμϵ)|vϵ|p-2vϵ=f(x,vϵ).
Recalling thatμϵ⩽0 and combining with Lemma 8, one may bootstrap the bound ∥vϵ∥X⩽C to ∥vϵ∥C1⩽C (independent of ϵ), since vϵ→0 in X, vϵ→0 in C1 (by Ascoli). This concludes the proof.


## 2. Application of Theorem1

Next, we present a simple and useful application of Theorem1.ConsideringΦ in Theorem 1 with f, such that, for some constant k,
(24)f(x,u)+kuisnondecreasingufora.e.x.
Assume that there are C(Ω¯) sub- and supersolutions u_ and u¯; that is, in the distribution sense,
(25)-Δu_+|u_|p-2u_-f(x,u_)⩽0⩽-Δu¯+|u¯|p-2u¯-f(x,u¯)inΩu_⩽0⩽u¯on∂Ω.
Moreover, neither u_ nor u¯ is a solution to (18).Theorem 9.
Under the assumption (2) there is a solution u0 to (18), u_<u0<u¯, such that, in addition, u0 is a local minimum of Φ in X.The proof relies on Theorem1 as well as on the following.Theorem 10.
LetΩ be a bounded domain in ℝn with smooth boundary ∂Ω. Let u∈Lloc⁡p-1(Ω) and assume that, for some k⩾0, u satisfies
(26)-Δu+|u|p-2u+ku⩾0inΩ.u⩾0onΩ
Then either u≡0 or there exists ϵ>0, such that
(27)u(x)⩾ϵdist⁡(x,∂Ω)inΩ.
Moreover, if k is replaced by the nonnegative continuous function c(x)∈C(Ω¯), then the conclusion is also valid.Proof.
The measureμ=-Δu+|u|p-2u+ku is nonnegative in Ω. We may assume u≢0.Case 1.
Considerμ≡0. In this case, u∈C∞(Ω) by induction applies to Lemma 8:
(28)-Δu+|u|p-2u+ku=0,u⩾0inΩ.
Since u≢0, we have u⩾δ>0 in some closed ball B in Ω. Let Ωj be an expanding sequence of subdomains of Ω with smooth boundaries and ⋃jΩj=Ω; suppose B⊂Ωj, forallj. Let hj be the solution in Ωj∖B of
(29)-Δhj+|hj|p-2hj+khj=0,hj⩾0inΩj∖Bhj=δon⁡∂Bhj=0on⁡∂Ω.
In order to compare u with hj, we need the following comparison principle for the operator L=-Δ+|·|p-2· defined in Lemma 7Lemma 11.
Letu,v∈C1,α(Ω¯) satisfy, for some k⩾0,
(30)Lu+ku⩾Lv+kvinΩu⩾voin∂Ω;
then, u⩾v in Ω.Proof.
Indeed, setting(31)Lu-Lv=-Δu+|u|p-2-(-Δv+|v|p-2v)
and defining w=u-v, it is noted that the derivative expression (|s|p-2s)′=(p-1)|s|p-2⩾0. Then, by the mean value theorem, there is ξ=θu+(1-θ)v(0<θ<1) satisfying
(32)-Δw+(p-2)|ξ|p-2w+kw⩾0inΩw⩾0on⁡∂Ω.
Applying the weak maximum principle, Theorem 8.1 P179 [3] by choosing c(x)=(p-2)|ξ|p-2, we know that w⩾0 and complete the proof.Sinceu(x),h(x)∈C1,α(Ωj∖B¯) in Lemma 8, then, by the virtue of Lemma 11, u⩾hj in Ωj∖B. As j→∞, we find
(33)u⩾hinΩ∖B,
when h solves
(34)-Δh+|h|p-2h+kh=0,h⩾0inΩ∖Bh=δon⁡∂Bh=0on⁡∂Ω.
By the Hopf lemma 3.4 P34 [3] with c(x)=|h|p-2+k, one obtains for some ϵ>0(35)h(x)⩾ϵdist⁡(x,∂Ω)inΩ∖B.
The conclusion of Theorem 10 then follows directly.Case 2.
Considerμ≢0. Let ζ∈C0∞(Ω) be a cutoff function, 0⩽ζ⩽1, such that ζμ≢0. Let v be the solution of
(36)(L+k)v=ζμinΩv=0on⁡∂Ω.
Since v is smooth outside a compact set K⊂Ω, it follows and applies to the Hopf lemma as above for some ϵ>0,
(37)v(x)⩾ϵdist⁡(x,∂Ω)inΩ∖B.
The conclusion of Theorem 10 is a direct consequence of the following.Claim 2.
One hasu⩾v in Ω.Proof of Claim2.
Given anyα>0, we will prove that
(38)u¯=u+α⩾vinΩ.
The claim then follows.
Note thatw=u¯-v satisfies
(39)(-Δ+k)w+|u¯|p-2u¯-|v|p-2v=(1-ζ)μ+|u¯|p-2u¯-|u|p-2u+kα⩾0inΩ(40)w⩾0inNη={x∈Ω;dist⁡(x,∂)<η}
provided η is sufficiently small (depending on α). The property (39) follows from the inequality |u¯|p-2u¯-|u|p-2u⩾0, a.e.x∈Ω, when u¯=u+α>u. The last property (40) follows from the fact that v is smooth near ∂Ω and v=0 on ∂Ω.
Letρj be a sequence of mollifiers with supp⁡ρj⊂B(0,1/j) and set wj(x)=∫Ωρj(x-y)w(y).
Clearlywj is smooth, and, by (39) and the mean value theorem with ξ=θu¯+(1-θ)v, we have
(41)(-Δ+k+(p-1)|ξ|p-2)wj⩾0inΩ∖N¯1/j.
On the other hand, we deduce from (40) that
(42)wj⩾0on⁡∂(Ω∖N¯1/j)
provided η>2/j. The maximum principle (Corollary 3.2 P33 [3]) of choosing c(x)=(p-1)|ξ|p-2⩾0 implies that
(43)wj⩾0inΩ∖N¯1/j
when η>2/j. Passing to the limit as j→∞, we see that
(44)w⩾0inΩ
which is the desired conclusion. The similar argument is also true as k is replaced by the nonnegative continuous function c(x)∈C(Ω¯) and the proof of Theorem 10 is completely finished.Now we are in a position to prove Theorem9.Proof of Theorem9.
On the basis of our above results, we can prove Theorem9 by the similar argument as [7] and rewrite it here for the reader’s convenience. We introduce an auxiliary function. Set
(45)f~(x,s)={f(x,u_(x))ifs<u_(x)f(x,s)ifu_(x)⩽s⩽u¯(x)f(x,u¯(x))ifs>u¯(x);
it is continuous in s. Then set
(46)F~(x,u)=∫0uf~(x,s)ds,Φ~(u)=12∫Ω|∇u|2+1p|u|p-∫ΩF~(x,u).
By the similar argument as Claim 1, there is a minimum u0∈X satisfying
(47)-Δu0+|u0|p-2u0=f~(x,u0)inΩ.
And, with Lemma 8, we can get u0∈W2,p, forallp<∞. We claim that u_⩽u0⩽u¯; we will just prove the first inequality. Indeed, we have
(48)L(u_)-L(u0)⩽f(x,u_)-f~(x,u0)
and in particular
(49)L(u_)-L(u0)⩽0inA={x∈Ω;u0(x)<u_(x)}.
Since u_-u0⩽0 on ∂A, it follows from the comparison principle (i.e., Lemma 11) that u_-u0⩽0inA. Therefore, A=∅ and the claim is proved.
Returning to (48), we have
(50)L(u_)-L(u0)+k(u_-u0)⩽(f(x,u_)+u_)-(f(x,u0)+ku0)⩽0.
Since u_ is not a solution, it follows from Theorem 10 that there is some ϵ>0, such that
(51)u_(x)-u0(x)⩽-ϵdist⁡(x,∂Ω),∀x∈Ω.
Similarly, for u¯, we have
(52)u_(x)+ϵdist⁡(x,∂Ω)⩽u0(x)⩽u¯(x)-ϵdist⁡(x,∂Ω),000000000000000000000000000000000000000∀x∈Ω.
It follows that, if u∈C01(Ω¯) and ∥u-u0∥C1⩽ϵ, then
(53)u_⩽u⩽u¯inΩ.
Finally, we apply the fact that F~(x,u)-F(x,u) is a function of x alone for u∈[u_(x),u¯(x)]. In particular, Φ(u)-Φ~(u) is constant for ∥u-u0∥C1⩽ϵ. Hence, u0 is a local minimum of Φ in the C1 topology (since it is a global minimum for Φ~). So, from Theorem 1, we know that u0 is also a local minimum of Φ in the X topology and the proof of Theorem 9 is finished.



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*Source: 646145-2014-04-28.xml*