Proofs of Theorem 4.8 and 4.10

aa

aa

aa aa aa

aa

aa

aa

aa

aa

aa aa

F₉

F₁₀

F₁₁ F₁₂ F5

F₆

F₇

F8

F1

F₂

F4

F3

Figure 4.4.: Depiction of a recursive n-polyconvex combination

bounded below by a n-polyconvex function and the expression

f Σ^I_i=1λ_iF_i
≤ Σ^I_i=1λ_if (F_i) (4.9) holds whenever (λ_i, F_i)_1≤i≤I satisfy (H_Iⁿ).

It is trivial to prove that f is n-polyconvex if (4.9) is true for all (λ_i, F_i)_1≤i≤I satisfying (H_Iⁿ). The reverse implication follows the same reasoning as for the case of rank-one convexity (n = 1) and we will omit it as it does not provide any new insights. The proposition will be useful later since it provides an intuition as to how to compute the n-polyconvex envelope of a given function (see Section 4.4.1).

4.2.1. Proofs of Theorem 4.8 and 4.10

In both proofs we will largely follow the proof given by Dacorogna [20, Th. 5.6] which we generalise to the case of n-polyconvexity. A part of the proof is to show that co(T (R^d×D)) = R^{τ (d,D)}. In our case this will take a more general form and we will present it as a separate result.

Lemma 4.14. Let V be a simple rank-n subspace of R^d×D and F ∈ R^d×D. Then there exists a subspace T_V ⊆ R^d×D s.t.

co(T (F + V )) = T (F ) + TV,

i.e. the convex hull of the minors of the coset F + V ⊆ R^d×D is itself a coset in R^{τ (d,D)}. Furthermore, dim(T_V) ≤ τ (n) :=

n

P

i=1 d i

_D

i
(with n ≤ d ∧ D).

Proof. The proof consists of two parts. The first part is to show that co(T (F + V )) is an affine space and the second is that the subspace TV defining this affine space has dimension less than τ (n) =Pn

i=1 d i

_D

i
. Define

TV := span{T (FV) − T (F ) : FV ∈ F + V }.

We then show that C := co(T (F + V )) − T (F ) = T_V. It is easy to see that C ⊆ T_V since C = co(T (F + V )) − T (F ) = co(T (F + V ) − T (F )) and the latter representation naturally lies in the span of T (F + V ) − T (F ). Therefore, assume for a contradiction that C 6= T_V. Then by the separation theorem A.2 there exists α ∈ T_V, α 6= 0 and β ∈ R s.t.

hα, Xi ≤ β for all X ∈ C. (4.10)

Due to the way TV was defined there also exists FV ∈ F + V s.t.

hα, T (F_V) − T (F )i 6= 0.

(Otherwise hα, Xi = 0 for all X ∈ TV, implying that α = 0 since α is an element of T_V itself.) Since F_V ∈ F + V and V is simple subspace there exist u_i, v_i ∈ R^d, λ_i ∈ R, i = 1, . . . , m s.t.

F_V = F +

m

X

i=1

λiui⊗ v_i.

Now define the matrices

F_l:= F +

l

X

i=1

λiui⊗ v_i

for l = 0, . . . , m. Then F_l and F_l−1 are rank-one connected for l = 1, . . . , m. Furthermore Fm= FV and F0 = F and thus

hα, T (F_m) − T (F )i 6= 0

and the map to minors is affine on rank-one lines we then obtain

T ( bF (λ)) = T (F

for all λ ∈ R. This is clearly a contradiction since this cannot be true for all λ ∈ R.

It remains to show that dim(TV) ≤ τ (n). Again, we will defer the proof of this to a later stage, namely Section 4.2.2 when we learn more about the structure of the space T_V.

Note that it is an integral part of the proof that V is a simple subspace. The following example shows that the result is false if that condition is violated.

Example 4.15. Consider R^2×2and let V = span{e1⊗e₁+e2⊗e₂} =

and so it cannot be a subspace.

Proof of Theorem 4.8. The implication that (4.7) holds if f is n-polyconvex is trivial, since it would then hold for the particular choice I = τ (n) + 1. We now want to show that I can always be taken to be equal to τ (n) + 1. Thus, let X ∈ co(T (F + V )), Fi ∈ F + V and λ_I∈ Λ_I s.t.

X =

I

X

i=1

λiT (Fi).

Similarly to Dacorogna’s proof we first show that I = τ (n) + 2 is sufficient. We define T (epi f |_{F +V}) := {(T (F ), µ) ∈ T (F + V ) × R : f (F ) ≤ µ} ⊆ co(T (F + V )) × R, where now co(T (F + V )) × R = (T (F ) + TV) × R = (T (F ), 0) + TV × R with TV defined as in Lemma 4.14. Recall that T_V is a subspace of R^d×D with dimension no more than τ (n).

Thus, co(T (F + V )) × R is an affine space with dimension dim(TV) + 1 ≤ τ (n) + 1. Denote τ = dim(Tb V). Then by applying the usual Carath´eodory Theorem, see Theorem A.3, for the coset we obtain I =τ + 2. A further step in the proof is to show that this numberb can be further reduced to I =bτ + 1. The reasoning is completely analogous to the proof given in Theorem 5.6 in [20] and hence we will omit it here. In Lemma 4.14 we claim that bτ ≤ τ (n) =Pn

i=1 d i

_D

i
(proof to follow) so for the purposes of the theorem we may choose I = τ (n).

Proof of Theorem 4.10. The implication ‘⇐’ is relatively straightforward. Let F ∈ R^d×D and V be a simple rank-n subspace of R^d×D. Then there exists a function gF +V, s.t. gF +V

is convex on co(T (F + V )) ⊆ R^{τ (d,D)} with f = g_{F +V} ◦ T on F + V . Thus, for c = g_{F +V} the inequality (4.6) is satisfied. Then for F₁, . . . , F_{τ (n)+1} ∈ F + V satisfying (4.8) we use the convexity of gF +V on its coset F + V and obtain (4.7).

The implication ‘⇒’ requires more work and we will use the results of Theorem 4.8 to prove the assertion. Assume f is n-polyconvex. Then (4.7) holds for all simple rank-n subspaces V ⊆ R^d×D and F + V ∈ R^d×D/V and F1, . . . , F_{τ (n)+1}∈ F + V satisfying (4.8) and let V and F + V be fixed. We then need to show that there exists a convex function g_{F +V} : co(T (F + V )) → R ∪ {+∞} with f |F +V = g_{F +V} ◦ T . Let I ≥ τ (n) + 1 be an integer and define the function gI: co(T (F + V )) → R ∪ {+∞} such that

g_I(X) = inf (

λ_if (F_i) :

I

X

i=1

λ_i = 1, λ_i ≥ 0,

I

X

i=1

λ_iT (F_i) = X and F₁, . . . , F_I ∈ F + V )

.

Along the lines of the proof in [20] we will show that, without loss of generality, I can be taken to be equal to τ (n) + 1. We then take g_{F +V} = g_{τ (n)+1} and show that g_{F +V} is convex and satisfies f (F ) = gF +V(T (F )). Note that in the original version of this proof for the polyconvex case g_I was defined on R^{τ (d,D)}instead of its corresponding version for n = d ∧ D here, where it is defined on co(T (R^d×D)). Using g_I defined on R^{τ (d,D)} requires to check whether gI is actually well defined, i.e. whether for each X ∈ R^{τ (d,D)+1} there exist I ∈ N and F1, . . . , F_I ∈ R^d×D, λ ∈ Λ_I s.t. X =PI

i=1λ_iT (F_i), or in other words, whether R^{τ (d,D)}= co(T (R^d×D)). We avoid this step since we define g_I on co(T (F + V )) straight away.

We now show that g_{F +V} is convex. Let X, Y ∈ co T (F + V ) and µ ∈ [0, 1]. We want to prove that

µg_{F +V}(X) + (1 − µ)g_{F +V}(Y ) ≥ g_{F +V}(µX + (1 − µ)Y ). (4.11) Fix ε > 0. Then from the considerations above there exist λ, bλ ∈ Λ_{τ (n)+1} and F_i, bF_i ∈ F + V s.t.

µg_{F +V}(X) + (1 − µ)g_{F +V}(Y ) + ε ≥ µ

τ (n)+1

X

i=1

λ_if (F_i) + (1 − µ)

τ (n)+1

X

i=1

bλ_if ( bF_i)(4.12) with

τ (n)+1

X

i=1

λ_iT (F_i) = X,

τ (n)+1

X

i=1

bλ_iT ( bF_i) = Y. (4.13) Upon redefining for 1 ≤ i ≤ τ (n) + 1

eλi = µλi Fei = Fi

eλ_{i+τ (n)+1}= (1 − µ)bλi Fe_{i+τ (n)+1} = bFi

both (4.12) and (4.13) can be written as

µgF +V(X) + (1 − µ)gF +V(Y ) + ε ≥

2τ (n)+2

X

i=1

eλif ( eFi) (4.14)

with eλ ∈ Λ_{2τ (n)+2} and

2τ (n)+2

X

i=1

eλiT ( eFi) = µX + (1 − µ)Y.

Then, taking the infimum over (eλi, eFi) in (4.14), and noticing that ε was arbitrary we do indeed have (4.11), i.e. g_{F +V} is convex.

The final step is to prove that f = gF +V ◦ T on F + V . Take eF ∈ F + V . Because we assume (4.7) holds for all F1, . . . , F_{τ (n)+1}∈ F + V such that (4.8) holds for eF , taking the infimum on both sides of the inequality we immediately obtain that f ( eF ) ≤ g(T ( eF )).

Since also for X = T ( eF ) in the evaluation of gF +V(X = T ( eF )) a trivial candidate convex combination is eF itself, we also obtain g_{F +V}(T ( eF )) ≤ f ( eF ), and hence, f = g_{F +V}◦T .

The proof of Theorem 4.8 includes a reference to a particular choice of the convex representative g_{F +V}. This is the purpose of the following theorem.

Theorem 4.16. Let f : R^d×D → R be n-polyconvex. Then for any F ∈ R^d×D and V ⊆ R^d×D simple rank-n we define g_{F +V} : co T (F + V ) → R ∪ {+∞} by

g_{F +V}(X) := inf







τ (n)+1

X

i=1

λ_if (F_i) : λ ∈ Λ_{τ (n)+1},

τ (n)+1

X

i=1

λ_iT (F_i) = X, F_i∈ F + V





 .

(4.15) Then g_{F +V} is convex on T (F + V ) and

f ( eF ) = gF +V(T ( eF )) for all eF ∈ F + V . Moreover, for every X ∈ co T (F + V )

g_{F +V}(X) = sup{G(X) : G : co T (F + V ) → R ∪ {+∞} convex and f = G ◦ T on F + V }.

Note that gF +V defined by (4.15) is also called the Busemann representative of the function eg_{F +V} : T (F + V ) → R with eg_{F +V}(T ( eF )) = f ( eF ) for all eF ∈ F + V . The function eg_{F +V} is defined on the nonconvex set T (F + V ) (if n > 1) and according to Busemann et al. [16] the convex representative may not be unique. However, this particular

choice of representative is the largest of all possible choices, which simply follows from the definition of g_{F +V} directly. To see this let G : co T (F + V ) → R ∪ {+∞} be another representative, i.e. f = G ◦ T on T (F + V ) and G convex. Then for X ∈ co T (F + V ) we have in particular that G(X) ≤Pτ (n)+1

i=1 λ_if (F_i) for all F_i ∈ F + V and λ ∈ Λ_{τ (n)+1} such X =Pτ (n)+1

i=1 λiFi. Hence, by taking the infimum we obtain G(X) ≤ g_{F +V}(X).

Since the first part of this proof is basically included in the proof of Theorem 4.8 there is nothing more to show.

In document On established and new semiconvexities in the calculus of variations. (Page 78-84)