A new proof of compactness in G(S)BD

Theorem 1.1.

Let U ⊆ R n be an open bounded subset of R n and let u k ∈ GBD ⁡ ( U ) be such that

Then there exists a subsequence, still denoted by u k , such that the set

has finite perimeter, i.e. u k → u a.e. in U ∖ A for some function u ∈ GBD ⁡ ( U ) with u = 0 in A. Furthermore,

where J u k σ := { x ∈ J u k : | [ u k ⁢ ( x ) ] | ≥ σ } .

Definition 3.1.

Let Ξ = { ξ 1 , … , ξ n } denote an orthonormal basis of ℝ n . We define

Given δ > 0 , we define the δ-neighborhood of S Ξ , 0 by

Definition 3.2.

In order to simplify the notation, given a family 𝒦 and a positive natural number m, we denote by 𝒦 m the set consisting of all subsets of 𝒦 containing exactly m-elements of 𝒦 , i.e.

Lemma 3.3.

Let M ∈ N be such that M ≥ n and consider a family K := { Ξ 1 , … , Ξ M } of orthonormal bases of R n such that for every Z ∈ K n ,

Then there exists a further orthonormal basis Σ = { ξ 1 , … , ξ n } such that for every Z ∈ K n - 1 ,

Proof.

First of all, notice that whenever 𝒵 ∈ 𝒦 n is such that

then we have

Indeed, let us suppose by contradiction that (3.3) does not hold for some 𝒳 ∈ 𝒵 n - 1 . Since for Ξ ∈ 𝒳 we have that each S Ξ , 0 is a finite union of ( n - 1 ) -dimensional subspaces of ℝ n intersected with 𝕊 n - 1 , the equality

implies that

As a consequence, we get

Hence, if we denote by Ξ ¯ the basis contained in 𝒵 ∖ 𝒳 , then, by using Grassmann’s formula,

which is valid for each couple V , W of vector subspaces of ℝ n , we deduce

Hence,

which is a contradiction to the assumption (3.1).

Fix an orthonormal basis { e 1 , … , e n } of ℝ n and let SO ⁡ ( n ) be the group of special orthogonal matrices. It can be endowed with the structure of an ( n 2 - n 2 ) -dimensional submanifold of ℝ n 2 . We can identify an element O ∈ SO ⁡ ( n ) with an ( n × n ) -matrix whose columns are the vectors of an orthonormal basis Ξ written with respect to { e 1 , … , e n } and vice versa.

In order to show the existence of Σ satisfying (3.2), we prove the following stronger condition: given 𝒵 ∈ 𝒦 n - 1 , for ℋ ( n 2 - n ) / 2 -a.e. choice of Σ we have that

This easily implies the existence of an orthonormal basis Σ satisfying (3.2), as the choice of 𝒵 ∈ 𝒦 n - 1 is finite. To show (3.4), for every i ∈ { 1 , … , n } let us define the smooth map

by

where ξ j denotes the j-th column vector of the matrix representing Σ. In order to show (3.4), we claim that it is enough to prove that for every x ∈ 𝕊 n - 1 we have

where

is the canonical projection map. Indeed, if Σ does not belong to π SO ⁡ ( n ) ⁢ ( { Λ i - 1 ⁢ ( x ) } ) for every x ∈ ⋂ Ξ ∈ 𝒵 S Ξ , 0 and for every i ∈ { 1 , … , n } , then, by using the definition of the map Λ i , we deduce immediately that Σ satisfies

Therefore, if (3.5) holds, then the set (remember that ⋂ Ξ ∈ 𝒵 S Ξ , 0 is a discrete set)

is of ℋ ( n 2 - n ) / 2 -measure zero and (3.4) holds true. Thus, ℋ ( n 2 - n ) / 2 -a.e. Σ satisfies (3.2).

To prove (3.5) it is enough to show that the differential of Λ i has full rank at every point

Indeed, this implies that Λ i - 1 ⁢ ( x ) is an ( n 2 - n - 2 2 ) -dimensional submanifold for every x ∈ 𝕊 n - 1 , which ensures the validity of (3.5) since

Notice that Λ i is the restriction to SO ⁡ ( n ) × { y ∈ ℝ n - 1 : | y | = 1 } of the map Λ ~ i : 𝕄 n × ℝ n - 1 → ℝ n defined by

where θ j is the j-th column vector of the matrix Θ ∈ 𝕄 n . To show that the differential of Λ i has full rank everywhere, it is enough to check that for every z ∈ SO ⁡ ( n ) × { y ∈ ℝ n - 1 : | y | = 1 } the differential of Λ ~ i restricted to Tan ⁡ ( SO ⁡ ( n ) × { y ∈ ℝ n - 1 : | y | = 1 } , z ) has rank equal to n - 1 . By using the relation

valid for every M ∈ 𝕄 n , we can reduce ourselves to the case z = ( I , y ¯ ) , where I denotes the identity matrix and y ¯ ∈ ℝ n - 1 is such that | y ¯ | = 1 . It is well known that

where 𝕄 skw n denotes the space of skew symmetric matrices. Using that ℝ n 2 + n - 1 ≅ 𝕄 n × ℝ n - 1 , we identify a point Z ∈ ℝ n 2 + n - 1 as

A direct computation shows that the differential of Λ i at the point ( I , y ¯ ) acting on the vector Z is given by

It is better to introduce the matrix P i ∈ 𝕄 n × ( n - 1 ) defined by

Roughly speaking, given X ∈ 𝕄 l × n , the product X ⁢ P i is the matrix in 𝕄 l × ( n - 1 ) obtained by removing from X the i-th column, while given Y ∈ 𝕄 ( n - 1 ) × l , the product P i ⁢ Y is the matrix in 𝕄 n × l obtained by adding a new row made of zero entries at the i-th position. With this definition, the linear map d ⁢ Λ i ⁢ ( I , y ¯ ) ⁢ ( ⋅ ) can be rewritten more compactly as

Given O ∈ SO ⁡ ( n - 1 ) such that O ⁢ e ~ 1 = y ¯ (where { e ~ 1 , … , e ~ n - 1 } denotes the reference orthonormal basis of ℝ n - 1 ), we can rewrite the system as

Hence, by the well-known relation

valid for every linear map α : V → W and all finite-dimensional vector spaces V and W, if we want to prove that d ⁢ Λ i ⁢ ( I , y ¯ ) has full rank, i.e.

since

(where the first inequality comes from Im ⁡ [ d ⁢ Λ i ⁢ ( I , y ¯ ) ] ⊂ Tan ⁡ ( 𝕊 n - 1 , Λ i ⁢ ( I , y ¯ ) ) ), it is enough to show that

Again by relation (3.6), we can reduce ourselves to find the dimension of the kernel of the map

But this dimension can easily be computed to be

which immediately implies (3.7). ∎

Remark 3.4.

By a standard argument from linear algebra, it is possible to construct n orthonormal bases of ℝ n , say 𝒦 = { Ξ 1 , … , Ξ n } , satisfying

Moreover, given U ⊂ SO ⁡ ( n ) open, Ξ i can be chosen in such a way that

Therefore, Lemma 3.3, and in particular condition (3.4), tells us that for every M ∈ ℕ ( M ≥ n ) we can always find a family of orthonormal bases of ℝ n , say 𝒦 = { Ξ 1 , … , Ξ M } , satisfying (3.1) and

Lemma 3.5.

Let A ⊂ R n be a measurable set with L n ⁢ ( A ) < ∞ , let ( B k ) k = 1 ∞ be measurable subsets of A, and let ( v k ) k = 1 ∞ be measurable functions v k : B k → S n - 1 . Then, given a sequence ϵ h ↘ 0 , there exist a sequence δ h ↘ 0 with δ h > 0 , a map ϕ : N → N , and an orthonormal basis Ξ of R n such that, up to passing through a subsequence on k,

Proof.

We claim that for every natural number N ≥ n , for every j ∈ { 0 , 1 , … , n - 1 } , for every ε > 0 , and for every open set U ⊂ SO ⁡ ( n ) there exist δ > 0 and a family of orthonormal bases 𝒦 := { Ξ 1 , … , Ξ N } ⊆ U such that, up to subsequences on k, we have

Clearly, the pair ( δ , 𝒦 ) depends on ( N , j , ε ) , but we do not emphasize this fact. We proceed by induction on j. The case j = 0 : given N ∈ ℕ , ε > 0 , and any open set U ⊂ y ⁢ ( n ) , we can make use of Lemma 3.3 and Remark 3.4 to find N orthonormal bases 𝒦 = { Ξ 1 , … , Ξ N } ⊆ U such that

Since the S Ξ , 0 are closed sets, there exists δ > 0 such that

Hence, (3.8) is satisfied with j = 0 and (3.9) holds true.

We want to prove the same for 0 < j ≤ n - 1 . For this purpose, we fix a natural number M ≥ n , a parameter ε > 0 , and an open set U ⊂ SO ⁡ ( n ) . By using the induction hypothesis, we may suppose that (3.8) and (3.9) hold true for j - 1 . This means that, given N ≥ n and ε ~ > 0 (to be chosen later), we find δ > 0 and orthonormal bases 𝒦 = { Ξ 1 , … , Ξ N } such that (3.8) and (3.9) hold true for j - 1 . Choose 𝒵 ∈ 𝒦 M and consider the set

which is the union of all possible ( n - j ) -intersections of sets of the form S Ξ , δ for Ξ ∈ 𝒵 .

We recall the following identity valid for any finite family of subsets of A, say ( B ) l = 1 L :

Now we partition 𝒦 into N M disjoint subsets (without loss of generality, we may choose N to be an integer multiple of M) each of which belongs to 𝒦 M . We call this partition 𝒫 . By construction, any l-intersection of sets of the form S 𝒵 , δ n - j with 𝒵 ∈ 𝒫 can be written as the union of ( M n - j ) l sets each of which, thanks to the fact that (we use that 𝒫 is a partition)

is the intersection of at least n - ( j - 1 ) different sets of the form S Ξ , δ with Ξ ∈ 𝒦 . Taking this last fact into account, if we replace the sets B j by v k - 1 ⁢ ( S 𝒵 , δ n - j ) and L = N M in identity (3.11), we obtain

where we have used the inductive hypothesis (3.8) for j - 1 to estimate the remaining terms in the right-hand side of (3.11).

Now suppose that for every 𝒵 ∈ 𝒦 M it holds true for some k that

Then inequality (3.12) implies

Therefore, if we choose N sufficiently large in such a way that

and ε ~ > 0 such that

then (3.14) implies that for every k there exists 𝒵 k ∈ 𝒫 for which (3.13) does not hold, i.e.

where we have used that B k , the domain of v k , is contained in A. Since 𝒫 is a finite family, we may suppose that, up to subsequences on k, we find a common 𝒵 ∈ 𝒫 for which

Taking into account the definition of S 𝒵 , δ n - j (see (3.10)), formula (3.15) gives our claim for j. Finally, by induction, this implies the validity of our claim for every j ∈ { 0 , … , n } .

Now we prove the lemma. For j = n - 1 , the claim says in particular that we find an orthonormal basis Ξ 0 and δ 0 > 0 such that, up to passing to a subsequence on k, we have

Notice that, by using a continuity argument, we find a neighborhood U 0 of Ξ 0 in SO ⁡ ( n ) such that

By applying again the claim, we find an orthonormal basis Ξ 1 ∈ U 0 and δ ~ 1 > 0 such that, up to passing to a further subsequence on k, we have

Hence if we set δ 1 := min ⁡ { δ ~ 1 , δ 0 / 2 } , we obtain as well

Proceeding again by induction, we find for every h = 1 , 2 , … an orthonormal basis Ξ h , δ h > 0 , and a subsequence ( k ℓ h ) ℓ such that

If we denote with abuse of notation the diagonal sequence ( k h h ) h simply as k, then we can find a map ϕ : ℕ → ℕ such that

Since the family ( S Ξ h , 0 ) h is made of compact subsets of 𝕊 n - 1 , then it is relatively compact with respect to the Hausdorff distance. This means that, up to a subsequence on h, we find an orthonormal basis Ξ such that

By using (3.17) and the fact that S Ξ h , δ h are relatively open subsets of 𝕊 n - 1 , this last convergence tells us that for every h the compact inclusion S Ξ , 0 ⋐ S Ξ h , δ h holds true. But this implies that, up to defining suitable δ h ′ > 0 with δ h ′ ≤ δ h , we can write

Finally, with abuse of notation, we set δ h := δ h ′ for every h. Then (3.16) implies

This gives the desired result. ∎

Remark 3.6.

Given U ⊂ ℝ n , u ∈ GBD ⁡ ( U ) , and σ ≥ 1 , we have that

Indeed, given ϵ > 0 , one can consider a partition of 𝕊 n - 1 into a finite family of measurable sets { S 1 , … , S M } such that for every m = 1 , … , M there exists an orthonormal basis Ξ m = { ξ 1 m , … , ξ n m } with ξ ⋅ ξ i m ≥ 1 4 for every ξ ∈ S m and for every i , j ∈ { 1 , … , n } and m ∈ { 1 , … , M } . Consider then the partition of J u σ given by { B 1 , … , B M } , where

We then have

where we have used that | [ 4 ⁢ u ^ y ξ ] ⁢ ( t ) | ≥ 1 for every t ∈ J 4 ⁢ u ^ y ξ ∩ ( B m ) y ξ for ℋ n - 1 -a.e. y ∈ Π ξ with ξ ∈ Ξ m .

Remark 3.7.

Let U ⊂ ℝ n and u ∈ GBD ⁡ ( U ) . Given ξ ∈ 𝕊 n - 1 and σ > 1 , if we introduce the map μ ^ σ ξ : ℬ ⁢ ( U ) → ℝ ¯ as

then we have μ ^ σ ξ ∈ ℳ b + ⁢ ( U ) . More precisely, for ℋ n - 1 -a.e. y ∈ Π ξ we have

(notice that for ℋ n - 1 -a.e. y the right-hand side is a finite measure thanks to Remark 3.6). By using the inclusion J v ^ y ξ 1 ⊂ ( J v 1 ) y ξ , valid for every v ∈ GBD ⁡ ( U ) for every ξ ∈ 𝕊 n - 1 , and for ℋ n - 1 -a.e. y ∈ Π ξ , we deduce

Finally, Remark 3.6 and the definition of μ ^ ξ (see [6, Definition 4.10]) imply that the right-hand side of (3.18) is a finite measure, and so is μ ^ σ ξ .