Absence of Lavrentiev gap for non-autonomous functionals with (p,q)-growth

Definition 2.1.

A function u ∈ W loc 1 , 1 ⁢ ( Ω ; ℝ N ) is a local minimizer of ℱ if and only if x ↦ f ⁢ ( x , D ⁢ u ⁢ ( x ) ) ∈ L loc 1 ⁢ ( Ω ) and

for any ϕ ∈ W 1 , 1 ⁢ ( Ω ; ℝ N ) with supp ϕ ⊂ ⊂ Ω .

Lemma 2.2.

Let u ∈ W 1 , p ⁢ ( B R ; R N ) be a function such that F ⁢ ( u , B R ) < + ∞ . Then L ⁢ ( u , B R ) = 0 if and only if there exists a sequence { u m } m ∈ N ⊂ W loc 1 , q ⁢ ( B R ; R N ) ∩ W 1 , p ⁢ ( B R ; R N ) such that

and

Theorem 3.1.

Let f : Ω × R n ⁢ N → R be a function satisfying the following conditions:

1. | z | p ≤ f ⁢ ( x , z ) ≤ L ⁢ ( 1 + | z | q ) , 1 < p < q < + ∞ ,

2. | f ⁢ ( x , z ) - f ⁢ ( x ~ , z ) | ≤ H ⁢ | x - x ~ | α ⁢ ( 1 + | z | q ) , 0 < α ≤ 1 ,

3. z ↦ f ⁢ ( x , z ) is convex for all x,

4. for B R ⊂ ⊂ Ω , ε 0 ∈ ( 0 , 1 ] such that B R + 2 ⁢ ε 0 ⊂ ⊂ Ω , x ∈ B R and ε ∈ ( 0 , ε 0 ) , there exists y ~ = y ~ ⁢ ( x , ε ) ∈ B ⁢ ( x , ε ) ¯ such that for z ∈ ℝ n ⁢ N and y ∈ B ⁢ ( x , ε ) ¯ , we have f ⁢ ( y ~ , z ) ≤ f ⁢ ( y , z ) .

Let u ∈ W loc 1 , p ⁢ ( Ω ; R N ) such that x ↦ f ⁢ ( x , D ⁢ u ⁢ ( x ) ) ∈ L loc 1 ⁢ ( Ω ) and assume that

Then L ⁢ ( u , B R ) = 0 for all B R ⊂ ⊂ Ω .

Remark 3.2.

Let us now explain condition (4). For every fixed z, y → f ⁢ ( y , z ) is continuous, so the minimization of f ⁢ ( y , z ) when y ∈ B ⁢ ( x , ε ) ¯ gives a minimizer y depending on x , ε and z. Condition (4) asks for independence on z, i.e., there exists a minimizer y ~ that works for every z. We will first give the proof of Theorem 3.1, and then we will show examples of densities f ⁢ ( x , z ) satisfying condition (4).

Remark 3.3.

Let us compare (1.5) with (3.1). When proving the absence of the Lavrentiev gap ℒ ⁢ ( u , B R ) = 0 , the borderline case q = p ⁢ ( n + α n ) is allowed but we need strict inequality (1.5) when proving higher integrability of minimizers, see [9, p. 32].

Proof.

Consider 0 < ε < ε 0 ≤ 1 as in hypothesis (4), then u ∈ W 1 , p ⁢ ( B R + 2 ⁢ ε 0 ; ℝ N ) and

Let us denote u ε ⁢ ( x ) := ( u * ϕ ε ) ⁢ ( x ) , the usual mollification, where x ∈ B R , and define

By definition, it follows that

where C = C ⁢ ( ∥ D ⁢ u ∥ L p ) > 1 . Moreover, by the Hölder continuity hypothesis (i.e., hypothesis 2), we have

Note that the left-hand side of hypothesis (1) gives

Now we observe that is possible to find K = K ⁢ ( p , q , ∥ D ⁢ u ∥ L p , H ) < + ∞ such that

Indeed, let us fix δ ∈ ( 0 , 1 ) and observe that, using (3.3), (3.4) and | z | ≤ C ⁢ ε - n p , we get

where the last estimate relies on the fact that q p ≤ n + α n , 0 < α ≤ 1 and 0 < ε < 1 . Then (3.5) follows choosing K = 1 δ = 1 + C q - p ⁢ H . Now, using hypothesis (4), Jensen’s inequality and (3.2), we obtain

Therefore, using (3.5), we have

Finally, since f ⁢ ( ⋅ , D ⁢ u ⁢ ( ⋅ ) ) ε ⁢ ( x ) → f ⁢ ( x , D ⁢ u ⁢ ( x ) ) strongly in L 1 ⁢ ( B R ) , by recalling that u ε → u in W 1 , p ⁢ ( B R ; ℝ N ) , and by using a well-known variant of Lebesgue’s dominated convergence theorem and Lemma 2.2, the proof is completed. ∎

Remark 3.4.

We note that our assumption (4) is very close to [22, assumption (2.3)]. Our proof is inspired by the one of [9, Lemma 13], which, in turn, is based on some arguments used in [22].

Remark 3.5.

Now we give some examples of functions for which Theorem 3.1 is valid.

1. f ⁢ ( x , z ) = b ⁢ ( z ) + a ⁢ ( x ) ⁢ c ⁢ ( z ) with the following conditions:

   1. a ∈ C 0 , α ⁢ ( Ω ¯ ) and a ⁢ ( x ) ≥ 0 for all x,

   2. b and c are convex functions such that

      | z | p ≤ b ⁢ ( z ) ≤ H ⁢ ( | z | q + 1 )   for  H ≥ 1 , 0 ≤ c ⁢ ( z ) ≤ L ⁢ ( | z | q + 1 )   for  L ≥ 1 .

   For instance, we can consider the following functions:

   1. f ⁢ ( x , z ) = b ⁢ ( z ) , independent of x.

   2. f ⁢ ( x , z ) = | z | p + a ⁢ ( x ) ⁢ | z | q . This example has been already dealt with in [9]; see also [22, 10, 8, 7, 1, 6].

   3. f ⁢ ( x , z ) = | z | p + a ⁢ ( x ) ⁢ | z | p ⁢ ln ⁡ ( e + | z | ) . This example is taken from [2, 1].

   4. f ⁢ ( x , z ) = | z | 2 + a ⁢ ( x ) ⁢ [ max ⁡ { z n , 0 } ] q , where q > 2 . This example is inspired by [21].

   5. f ⁢ ( x , z ) = | z | p + a ⁢ ( x ) ⁢ [ | z 1 - z 2 | q + | z 1 | q ] , where 1 < p < q < + ∞ . This example is inspired by [5].

2. f ⁢ ( x , z ) = ∑ i = 1 k [ b i ⁢ ( z ) + a i ⁢ ( x ) ⁢ c i ⁢ ( z ) ] , where k ∈ ℕ and a i , b i , c i verify the corresponding conditions of the previous example for all i ∈ { 1 , 2 , … , k } .

3. f ⁢ ( x , z ) = h ⁢ ( ∑ i = 1 k [ b i ⁢ ( z ) + a i ⁢ ( x ) ⁢ c i ⁢ ( z ) ] ) , where, in addition to the previous conditions, h is increasing, convex, Lipschitz and such that s ≤ h ⁢ ( s ) ≤ α ⁢ s + β .

4. f ⁢ ( x , z ) = h ⁢ ( a ⁢ ( x ) , z ) with the following conditions:

   1. t ↦ h ⁢ ( t , z ) is increasing,

   2. h is convex with respect to the second variable,

   3. a ∈ C ⁢ ( Ω ¯ ) ,

   4. f verifies assumptions (1) and (2) of Theorem 3.1.

   For example,

   f ⁢ ( x , z ) = | z | p + ( e + a ~ ⁢ ( x ) ⁢ | z | ) a + b ⁢ sin ⁡ ( ln ⁡ ( ln ⁡ ( e + a ~ ⁢ ( x ) ⁢ | z | ) ) ) ,

   where a ~ ∈ C 0 , α ⁢ ( Ω ¯ ) , a ~ ⁢ ( x ) ≥ 0 for all x, a ≥ 1 + 2 ⁢ b ⁢ 2 and b > 0 . In order to satisfy the non-standard ( p , q ) -growth condition, we can consider 1 < p < a + b ≤ q . This example is inspired by [11, 20], see also [16].

Remark 3.6.

Hypothesis (4) was used during the proof of Theorem 3.1 in order to obtain the second increase in (3.6). Now we want to show an example of a function for which hypothesis (4) fails. Let us consider Ω = B ⁢ ( 0 , 1 ) ⊂ ℝ 2 and the function f : B ⁢ ( 0 , 1 ) × ℝ 2 ⁢ N → ℝ such that

where, for x = ( x 1 , x 2 ) ,

In this case the minimum point of the function changes depending on the choice of z. Indeed, let us consider B R = B ⁢ ( 0 , 1 2 ) , ε 0 = 1 8 , x = 0 . Then we deal with the two cases: | z | = 0 and | z | = 2 .

When | z | = 0 , we have

and then the minimum value in B ⁢ ( 0 , ε ) ¯ is reached for y ~ = ( 0 , ε ) .

If | z | = 2 , then

and therefore, in this situation, y ~ is any point ( y 1 , y 2 ) such that y 2 ≤ 0 .

Corollary 3.7.

Let h : Ω × R n ⁢ N → R be a function verifying Theorem 3.1 and let f : Ω × R n ⁢ N → R be a function such that

where c 1 , c 2 ≥ 0 . We consider u ∈ W loc 1 , p ⁢ ( Ω ; R N ) such that x ↦ f ⁢ ( x , D ⁢ u ⁢ ( x ) ) ∈ L loc 1 ⁢ ( Ω ) and assume q ≤ p ⁢ ( n + α n ) . Then L ⁢ ( u , B R ) = 0 for all B R ⊂ ⊂ Ω .

Proof.

We follow the proof of [9, Theorem 6]. By the proof of Theorem 3.1, if u ∈ W loc 1 , p ⁢ ( Ω ; ℝ N ) is such that h ⁢ ( x , D ⁢ u ⁢ ( x ) ) ∈ L loc 1 ⁢ ( Ω ) , then there exists a sequence { u m } m ∈ ℕ ⊂ W 1 , q ⁢ ( B R ; ℝ N ) such that u m → u strongly in W 1 , p ⁢ ( B R ; ℝ N ) , D ⁢ u m ⁢ ( x ) → D ⁢ u ⁢ ( x ) a.e., h ⁢ ( x , D ⁢ u m ⁢ ( x ) ) → h ⁢ ( x , D ⁢ u ⁢ ( x ) ) a.e., and ∫ B R h ⁢ ( x , D ⁢ u m ⁢ ( x ) ) → ∫ B R h ⁢ ( x , D ⁢ u ⁢ ( x ) ) . Using a well-known variant of Lebesgue’s dominated convergence theorem, we have that

and then, by Lemma 2.2, ℒ ⁢ ( u , B R ) = 0 . ∎