Relaxation of functionals with linear growth: Interactions of emerging measures and free discontinuities

Proof of Proposition 2.6.

Clearly, it suffices to prove

the opposite inequality being trivial.

In order to achieve the desired conclusion, we argue as follows. For every A ∈ 𝒜 r ⁢ ( Ω ¯ ) , denote by ℱ ¯ w the localized sequentially weak W 1 , 1 ⁢ ( Ω ∩ A ; ℝ m ) × L 1 ⁢ ( Ω ∩ A ; ℝ d ) lower semicontinuous envelope of F in (2.11). It was proved in [10] that for every A ∈ 𝒜 r ⁢ ( Ω ¯ ) (so that A ∩ Ω can be any open subset of Ω),

Since for every ( u , v ) ∈ W 1 , 1 ⁢ ( Ω ∩ A ; ℝ m ) × L 1 ⁢ ( Ω ∩ A ; ℝ d ) ,

we infer that

for every ( u , v ) ∈ BV ⁡ ( Ω ∩ A ; ℝ m ) × ℳ ⁢ ( A ; ℝ d ) , which concludes the proof. ∎

Proof of Proposition 2.7.

It suffices to show “ ≤ ”, as “ ≥ ” is trivial.

For every fixed k ∈ ℕ and u k ∈ BV ⁡ ( Ω ∩ A ; ℝ m ) , we can choose a sequence ( w k , l ) l ∈ ℕ ⊂ W 1 , 1 ⁢ ( Ω ∩ A ; ℝ m ) such that, as l → ∞ , we have w k , l ⇀ * u k in BV ⁡ ( Ω ∩ A ; ℝ m ) and

(any W 1 , 1 recovery sequence). For instance, since W 1 , 1 is dense in BV with respect to area strict convergence, there exists ( w k , l ) ⊂ W 1 , 1 such that, as l → ∞ , we have w k , l → u k area-strictly, which yields (A.1) by Proposition 2.1. By (H3) (coercivity and growth of W), equation (A.1) implies that

with a constant C > 0 .

In addition, given any v k ∈ L 1 ⁢ ( Ω ∩ A ; ℝ m ) , we also have that

by dominated convergence, also using that w k , l → u k in L 1 , f 1 is bounded and f 2 ∗ ∗ ⁢ ( v k ) ∈ L 1 ⁢ ( Ω ∩ A ) .

If u k ⇀ * u in BV and v k ⇀ * v in ℳ , it is therefore possible to choose a diagonal sequence u ~ k := w k , l ⁢ ( k ) , with l ⁢ ( k ) → ∞ fast enough as k → ∞ , such that for

we have

and u ~ k → u in L 1 ⁢ ( Ω ∩ A ; ℝ m ) . In particular, u ~ k ⇀ * u in BV ⁡ ( Ω ∩ A ; ℝ m ) . This means that if

is an arbitrary admissible sequence for the infimum defining ℱ ¯ ∗ ∗ in (2.13), then

is admissible, too. Hence, (A.2) yields the assertion. ∎

Lemma A.1.

Let Ω ¯ be as above. Let λ : A r ⁢ ( Ω ¯ ) → [ 0 , + ∞ ) and μ be such that the following conditions hold:

1. μ is a finite Radon measure on Ω ¯ .

2. λ ⁢ ( Ω ¯ ) ≥ μ ⁢ ( Ω ¯ ) .

3. λ ⁢ ( A ) ≤ μ ⁢ ( A ) for all A ∈ 𝒜 r ⁢ ( Ω ¯ ) .

4. (Subadditivity) λ ⁢ ( A ) ≤ λ ⁢ ( A ∖ U ¯ ) + λ ⁢ ( B ) for all A , B , U ∈ 𝒜 r ⁢ ( Ω ¯ ) such that U ⋐ B ⋐ A .

5. For all A ∈ 𝒜 r ⁢ ( Ω ¯ ) and ε > 0 , there exists U ∈ 𝒜 r ⁢ ( Ω ¯ ) such that U ⋐ A and λ ⁢ ( A ∖ U ¯ ) < ε .

Then λ = μ on A r ⁢ ( Ω ¯ ) .

Proof.

The inequality λ ⁢ ( A ) ≤ μ ⁢ ( A ) for every A ∈ 𝒜 r ⁢ ( Ω ¯ ) follows from (iii).

For what concerns the other inequality, we can observe that, by the inner regularity of μ, we can find a relatively open subset of Ω ¯ , say A ′ ⋐ A , such that

Thus, letting ε → 0 , we obtain the desired conclusion. ∎

Proof of Lemma 2.8.

In order to prove that ℱ ¯ ⁢ ( u , v , A ) is the trace of a Radon measure, we refer to Lemma A.1, and define the increasing set function λ : 𝒜 r ⁢ ( Ω ¯ ) → [ 0 , + ∞ ] by

By the definition of ℱ ¯ , we know that there exists a sequence ( u h , v h ) ∈ W 1 , 1 ⁢ ( Ω ; ℝ m ) × L 1 ⁢ ( Ω ; ℝ d ) such that u h ⇀ ∗ u in BV ⁡ ( Ω ; ℝ m ) and v h ⇀ ∗ v in ℳ ⁢ ( Ω ¯ ; ℝ d ) such that

Next, denoting by λ h the measures ( f 1 ( u h ) f 2 ( v h ) + W ( ∇ u h ) ℒ n , we obtain that it converges weakly * in the sense of measures (duality with elements in C ⁢ ( Ω ¯ ) ), up to a subsequence (due to the bounds) to a measure μ. Now, due to the lower semicontinuity with respect to the weak* convergence, we have

Then, by the definition of λ, we have, for every A ∈ 𝒜 r ⁢ ( Ω ¯ ) ,

Now by the previous lemma, we would have that λ = μ if we prove inner regularity and subadditivity for ℱ ¯ ⁢ ( u , v , ⋅ ) .

For what concerns inner regularity (i.e., Lemma A.1 (v)), we employ the growth conditions in (H1)–(H3). It is enough to show that

with some constant β > 0 , because if we substitute A ∖ U ¯ instead of A in (A.3), the inner regularity of the upper bound measures provides the asserted inner regularity for ℱ ¯ . By the definition of ℱ ¯ and the growth conditions from above, we have that

for any sequence ( v k , u k ) k ⊂ L 1 ⁢ ( A ; ℝ d ) × W 1 , 1 ⁢ ( A ; ℝ m ) with v k ⇀ * v in ℳ and u k ⇀ * u in BV . It now suffices to choose ( v k , u k ) so that v k → v strictly in ℳ ⁢ ( A ; ℝ d ) , and u k → u strictly in BV ⁡ ( A ∖ ∂ ⁡ Ω ; ℝ m ) , because then

By the latter, (A.4) implies (A.3). Here, recall that we use the convention | D ⁢ u | ⁢ ( ∂ ⁡ Ω ) := 0 , so that we have | D ⁢ u | ⁢ ( A ) = | D ⁢ u | ⁢ ( A ∖ ∂ ⁡ Ω ) . By mollification similar to the classical approximation result of Meyers and Serrin, this choice of sequences is indeed possible even for arbitrary open A without boundary regularity; see, for instance, [3, Theorem 3.9] (to choose ( u k ) , while the choice of ( v k ) is easier and can be done similarly, or similarly to Lemma 2.5).

It remains to prove that ℱ ¯ ⁢ ( u , v , ⋅ ) is subadditive in the sense of Lemma A.1 (iv), i.e., it suffices to prove that

for all A , U , B ∈ 𝒜 r ⁢ ( Ω ¯ ) with U ⋐ B ⋐ A , u ∈ BV ⁡ ( Ω ; ℝ m ) and v ∈ ℳ ⁢ ( Ω ¯ ; ℝ d ) (see, e.g., [7, Lemma 4.3.4]). Without loss of generality, in view of Proposition 2.6, we can assume f 2 convex and W quasiconvex. Fix η > 0 and find ( w h ) ⊂ W 1 , 1 ⁢ ( ( A ∖ U ¯ ) , ℝ m ) and ( v h ) ⊂ L 1 ⁢ ( A ∖ U ¯ , ℝ d ) such that w h ⇀ ∗ u in BV ⁡ ( ( A ∖ U ¯ ) , ℝ m ) , v h ⇀ ∗ v in ℳ ⁢ ( A ∖ U ¯ , ℝ d ) and

Extract a subsequence still denoted by n such that the above upper limit is a limit. Let B 0 be a relatively open subset of Ω ¯ with Lipschitz boundary such that U ⋐ B 0 ⋐ B . Then there exist ( u h ) ⊂ W 1 , 1 ⁢ ( B 0 , ℝ m ) and ( v ¯ h ) ⊂ L 1 ⁢ ( B 0 ; ℝ d ) such that u h ⇀ ∗ u in BV ⁡ ( B 0 , ℝ m ) , v ¯ h ⇀ ∗ v in ℳ ⁢ ( B ¯ 0 , ℝ d ) and

Consider, for every D ∈ 𝒜 r ⁢ ( Ω ¯ ) , the set function 𝒢 ⁢ ( u , v , D ) := ∫ D ( 1 + | ∇ ⁡ u | ) ⁢ 𝑑 x + | v | ⁢ ( D ) . Due to (H1)–(H3), we may extract a bounded subsequence, that we will not relabel, from the sequences of measures

restricted to B 0 ∖ U ¯ , converging in the sense of distributions to some Radon measure ν defined on B 0 ∖ U ¯ . For every t > 0 , let

Define, for 0 < δ < η , the subsets

Consider a smooth cut-off function φ δ ∈ C ∞ ⁢ ( B η - δ , [ 0 , 1 ] ) such that φ δ = 1 on B η . As the thickness of the strip L δ is of order δ, we have an upper bound of the form ∥ ∇ ⁡ φ δ ∥ L ∞ ⁢ ( B η - δ ) ≤ C / δ . Define

Clearly, the sequences w h ′ and v h ′ weakly* converge to u in BV ⁡ ( A , ℝ d ) and to v in ℳ ⁢ ( A ; ℝ m ) as h → + ∞ , respectively, and

By the growth conditions (H1)–(H3), we have the estimate

Thus, passing to the limit as n → + ∞ and making use of the lower semicontinuity of ℱ ¯ ⁢ ( ⋅ , ⋅ , A ) (which is a consequence of its definition (2.12)), (A.6) and (A.7), we obtain

Now passing to the limit as δ → 0 + , we get

It suffices to choose a subsequence { η h } such that η h → 0 + and ν ⁢ ( ∂ ⁡ B η h ) = 0 , to conclude the proof of (A.5). ∎