Potential Analysis 11: 39–76, 1999. c 1999 Kluwer Academic Publishers. Printed in the Netherlands.
39
Relaxation of a Quadratic Functional Defined by a Nonnegative Unbounded Matrix JUAN CASADO-D´IAZ Departamento de Ecuaciones Diferenciales y An´alisis Num´erico, Facultad de Matema´ ticas, C. Tarf´ıa s/n, 41012 Sevilla, Spain. (e-mail:
[email protected]) (Received: 16 April 1996; accepted: 22 July 1997) Abstract.R We study the lower semicontinuous envelope in Lp (Ω), F¯ , of a functional F of the form F (u) = Ω A∇u∇u dx where A = A(x) is not strictly elliptic and not bounded. We prove that F¯ R √ √ may also be written as F¯ (u) = Ω B∇u ∇u dx with B = AP A for a matrix P which is the matrix of an orthogonal projection. In the one-dimensional case, we characterize the domain of F¯ and we explicit the matrix P . Mathematics Subject Classification (1991): 31b10. Key words: Relaxation, lower semicontinuity.
Introduction The goal of the present paper is to study the relaxation of the functional Z
F (u) =
A∇u∇u dx Ω
where Ω is an open bounded set of Rd and A a measurable function with values in the nonnegative symmetric matrices which can take an infinite value in arbitrary points of Ω and arbitrary directions of Rd (see Definition 1.1). Thus the functional F is neither coercive nor takes finite values on the space H 1 (Ω). We study the relaxation of F (i.e. its lower-semicontinuous envelope) with respect to the various Lp topologies: this is more convenient than considering the relaxation with respect to the Sobolev spaces since there is no estimate on the gradients. The problem is indeed related with the existence of solution (using the direct method of the calculus of variations) of the minimization problem for a functional of the type Z Ω
A ∇u ∇u dx +
Z
g(u) dx, Ω
where g is a convex function satisfying g(s) > α|s|p with α > 0, p > 1.
PDF JEFF/J.V. PREEPROOF: Pota 364 MATHKAP 146314.tex; 27/05/1999; 13:44; v.7; p.1
40
JUAN CASADO-D´ıAZ
In Section 1 we pose the problem and we establish some abstract results. Precisely we define
F (u) =
Z A∇u∇u dx
Ω
if u ∈ W 1,∞ (Ω), if u ∈ X\W 1,∞ (Ω),
+∞
¯ or C 0 (Ω) where X stays either for Lp (Ω) or Lploc (Ω), 1 6 p < ∞, or for C 0 (Ω) loc (we consider C 0 and not L∞ because the closure of W 1,∞ (Ω) in L∞ (Ω) is equal ¯ ). to C 0 (Ω) We then define F¯ the lower semicontinuous envelope of the functional F in X , and H as the space defined by H = {u ∈ X : F¯ (u) < +∞}.
Our concern is to characterize H and to find a representation of F¯ in the whole space H , and not only in a subspace because H is the space in which the functional F¯ is naturally defined. √ Let WA (Ω) be the space defined by WA (Ω) = {u ∈ √W 1,∞ (Ω): A∇u ∈ L2 (Ω)d } and let V be the closure in X × L2 (Ω)d of {(u, A∇u) ∈ WA (Ω) × L2 (Ω)d }. Theorem 1.1 in Section 1 gives the following characterization of F¯ and H which is the starting point which will allows us to arrive to an integral representation of F¯ in Section 2 2 d H = {u ∈ X : ∃v ∈ L (Ω) with (u, v) ∈ V } Z 2 ¯ F (u) = min |v| dx: (u, v) ∈ V ∀u ∈ H. Ω
In the remainder of Section 1, we study some properties of the unique function wu which gives the above minimum, showing that wu satisfies properties analogous to a gradient, and that H is therefore analogous to a Sobolev space. In particular, we prove that if S : R 7→ R is a lipschitz continuous function and if u ∈ H , then S(u) ∈ H and wS(u) = S 0 (u)wu , which ensures that F¯ is a Dirichlet form if A ∈ L1 (Ω)d×d and X = L2 (Ω)d (for an extensive study of Dirichlet forms and its applications we refer to [13] or [17]). In Section 2, a more detailed study of the structure of V allows us to obtain an integral representation for F¯ . More precisely we will see in Theorem 2.1 that there exists a measurable function P (which does not depend on the choice of X ) with values in the d × d matrices, such that for almost every x ∈ Ω, P (x) is an orthogonal projection on a subspace T (x) of Rd for which we have F¯ (u) =
Z Ω
|P v|2 dx, ∀(u, v) ∈ V.
146314.tex; 27/05/1999; 13:44; v.7; p.2
RELAXATION OF A QUADRATIC FUNCTIONAL
41
This implies in particular that F¯ (u) =
Z
Ω
B∇u∇u dx, ∀u ∈ WA (Ω),
√ √ where B is given by B = AP A. e We prove this result under the following restrictive hypothesis (H) ( e) (H
For any u ∈ L∞ (Ω) there exists a sequence un ∈ WA (Ω) which converges almost everywhere to u.
e is a hypothesis about the size of A, (H) e is clearly satisfied if A Hypothesis (H) 1 d×d belongs to L (Ω) , but also holds true for more general cases. It is not satisfied in general if A takes infinite values as we have assumed in Section 1. Theorem 2.1 is related to the results obtained by L. Carbone and C. Sbordone [5], who studied the Γ-convergence in X of the sequence of functionals1 Z
F ε (u) = Ω
f ε (x, u, ∇u) dx,
where f ε (x, s, p) are Carath´eodory functions which are convex in p and satisfy 0 6 f ε (x, s, p) 6 aε (x)|p|2 , with aε compact in the weak topology of L1 (Ω). Using an abstract theorem of G. Dal Maso [6], they proved the existence of a subsequence of F ε which Γ-converges to a functional F¯ which can be written in the form F¯ (u) =
Z
Ω
f (x, u, ∇u) dx, ∀u ∈ W 1,∞ (Ω),
where f is analogous to f ε . Note that the representation of the Γ-limit is given only on W 1,∞ (Ω) and not on X . In fact, we can find in [DG] several examples which show that the representation of F¯ outside of W 1,∞ (Ω) may be different. In the special case where f ε (x, s, p) = A(x)pp with A ∈ L1 (Ω)d×d is easy to see that the function f (x, s, p) does not depend on s and is quadratic in p, i.e, f is of the form f (x, s, p) = B(x)pp for some matrix B . In Section 3 we consider the one-dimensional case (d = 1), where Ω = (0, 1) and where the matrix A reduces to a nonnegative function a with values in [0, +∞]. e Previous results in this direction were obtained by We do not make hypothesis (H). M. M. Hamza [14], who considered the functional Z 2 du dλ e F (u) = Ω dx
+∞
if u ∈ C 1 (Ω) if u ∈ L2 (Ω)\C 1 (Ω),
Let us recall that if F ε is a constant sequence, i.e. F ε = F for some F and for any ε > 0, then F Γ-converges to the lower semicontinuous envelope of F . 1
ε
146314.tex; 27/05/1999; 13:44; v.7; p.3
42
JUAN CASADO-D´ıAZ
where λ is a given Radon measure on Ω. He proved that the quadratic form Fe is closable in L2 (Ω) if and only if λ is absolutely continuous with respect to the Lebesgue measure dx and if the density function a of λ with respect to Lebesgue measure (a ∈ L1loc (Ω)) vanishes outside of an open set in which the function a1 is locally integrable. This result has then been extended to a diagonal matrix in infinite dimensions by S. Albeverio and M. R¨ockner [1]. The representation of F¯ in the one-dimensional case was obtained by P. Marcellini [16] for u ∈ H 1 (Ω) in the case where a ∈ L∞ (Ω). Defining an open set G by G = {x ∈ (0, 1): ∃δ = δ(x) > 0 such that
1 ∈ L1 (x − δ, x + δ)}. a
P. Marcellini proved that F¯ (u) =
Z G
2 du a dx, ∀u ∈ H 1 (0, 1). dx
In the present paper we are interested to characterize H and to extend the representation of P. Marcellini to the whole space H . Actually we prove in Theorem 3.1 that for any measurable function a with values in [0, +∞] we have ( 1,1 H = u ∈ X ∩ Wloc (G)
Z
such that G
) 2 du a dx < +∞ , dx
Z 2 du F¯ (u) = a dx, ∀u ∈ H. dx G
Our proof is very different of P. Marcellini’s one (which is more constructive) and is based on the characterization of F¯ and H given by Theorem 1.1. By a duality √ 1,1 2 (Ω) with a du argument we show that for any u ∈ X ∩ Wloc dx ∈ L (G) we have √ (u, a(du/dx)χG ) ∈ V . e is satisfied in the one-dimensional case d = 1 In Section 4 we prove that (H) whenever a is almost everywhere finite. To prove this result we use the Stone– e is satisfied in other Weierstrass’ theorem which can also be used to show that H cases (remark that WA (Ω) is an algebra which contains the constant functions and ¯ if and only if it separate points). We also prove hence WA (Ω) is dense in C 0 (Ω) e is not necessarily in Section 4 that in the d-dimensional case with d > 1, (H) satisfied whenever the matrix A is almost everywhere finite. Actually we prove that for Ω = (0, 1)d , d > 1 there exists an almost everywhere finite nonnegative 1,1 (Ω) we have measurable function h defined on Ω such that for any u ∈ Wloc h∇u ∈ L1 (Ω)d ⇒ ∇u = 0.
Taking then A = h2 I , we get that H reduces to the constants functions and therefore e ) is not satisfied. (H
146314.tex; 27/05/1999; 13:44; v.7; p.4
RELAXATION OF A QUADRATIC FUNCTIONAL
43
The difference between the one-dimensional case and the d-dimensional case is essentially due to the fact that the image of a path has in general Lebesgue measure zero in Rd for d > 1. The idea of the construction of the function h is to consider a function which is equal to infinity at each point of Ω with a rational coordinate. Then if u is such that h∇u ∈ L1 (Ω)d , and if x1 , x2 ∈ Ω, we have that ∇u = 0 on a path joining two points as close as we want to x1 , x2 . Thus u(x1 ) = u(x2 ). Notation We denote by Ω a bounded open set of Rd . For 1 6 p < +∞, we denote by Lp (Ω), (resp. Lploc (Ω)) the space of functions whose power p is integrable (resp. integrable on the compact sets of Ω). The space of the functions essentially bounded is denoted by L∞ (Ω). For 1 6 p 6 +∞, we denote by W 1,p (Ω) the usual Sobolev spaces. The space 1,2 W (Ω) is also denoted by H 1 (Ω). ¯ , equipped with the ¯ the space of continuous functions in Ω We denote by C 0 (Ω) 0 ¯ topology of the uniform convergence on Ω. The space C (Ω) denotes the space of continuous functions in Ω with the topology of the uniform convergence on the compact sets of Ω. We denote by D(Ω) the space of the infinitely differentiable functions in Ω with compact support in Ω. For a set of vectors {λi } ⊂ Rd , we denote by span({λi }) the linear subspace spanned by {λi }. The Lebesgue measure of a set S ⊂ Rd is denoted by |S|. The characteristic function of a set S ⊂ Rd is denoted by χS , i.e. χS (x) = 1 if x ∈ S , χS (x) = 0 if x 6∈ S . For r > 0, the truncation Tr : R 7→ R is defined by r,
Tr (s) =
if r 6 s,
s,
if − r 6 s 6 r,
−r,
if s 6 −r.
The normalization function N : Rd 7→ Rd is defined by ξ
N (ξ) =
|ξ|
0
if ξ 6= 0, if ξ = 0.
The sign function sgn: R 7→ R is defined by sgn(s) = N (s) =
s
if s 6= 0,
0
if s = 0.
|s|
146314.tex; 27/05/1999; 13:44; v.7; p.5
44
JUAN CASADO-D´ıAZ
e u and Lu by Consider a function u ∈ L1loc (Ω). Then, we define the sets L ( eu = L
1 x ∈ Ω: ∃ lim r→0 |B(x, r)|
)
Z
u(y) dy . B(x,r)
Lu = x ∈ Ω: ∃h = h(x) ∈ R such that
1 r→0 |B(x, r)|
Z
|u(y) − h(x)| dy .
lim
B(x,r)
The set Lu is called the set of Lebesgue points of u. Note that Lu is contained in e u and L 1 r→0 |B(x, r)|
Z
h(x) = lim
B(x,r)
u(y) dy, ∀x ∈ Lu .
e u | = 0 and that It is well-known (see e.g. [F]) that |Ω\Lu | = |Ω\L
1 r→0 |B(x, r)|
Z
u(x) = lim
u(y) dy
a.e. in Lu .
B(x,r)
1. Position of the Problem and First Results Let Ω ⊂ Rd be a bounded open set. We consider a ‘measurable’ function A from Ω into the quadratic nonnegative forms on Rd , which can take infinite values in arbitrary directions of Rd . Exactly, A is defined as follows. DEFINITION 1.1. Let Ω be a bounded open set of Rd and let γ = (γ1 , . . . , γd ): Ω 7→ (Rd )d be a measurable function, such that for almost every x ∈ Ω, {γ1 (x), . . . , γd (x)} is an orthonormal basis of Rd . Let λ = (λ1 , . . . , λd ): Ω 7→ [0, +∞]d be measurable. For almost every x ∈ Ω, we define the quadratic form A(x): Rd 7→ [0, +∞] by A(x)ξξ =
d X
λi (x)(ξγi (x))2 , ∀ξ ∈ Rd ,
i=1
where ξγ denotes the scalar product in Rd . For almost every x ∈ Rd we define MA (x) = {ξ ∈ Rd : A(x)ξξ < +∞},
146314.tex; 27/05/1999; 13:44; v.7; p.6
RELAXATION OF A QUADRATIC FUNCTIONAL
45
which it is easily seen to be a linear subspace of Rd . From the quadratic form A(x) we define a linear map (which we also denote by A(x)) A(x): MA (X) 7→ Rd , by the formula A(x)ξ =
d X
λi (x)(ξγi (x))γi (x), ∀ξ ∈ MA (x).
i=1
√
We also define √
A(x)ξ =
A(x): MA (x) 7→ Rd by
d √ X
λi (x)(ξγi (x))γi (x), ∀ ξ ∈ MA (x).
i=1
The linear map √
√ A(x) satisfies
√ A(x)ξ A(x)ξ = A(x)ξξ, ∀ξ ∈ MA (x).
We will often not specify the point x ∈ Ω. The quadratic form A allows us to define the functional F as follows ¯ DEFINITION 1.2. Let X denote either Lp (Ω) or Lploc (Ω), 1 6 p < ∞, or C 0 (Ω) 0 or C (Ω) (X is a Fr´echet space, whose metric will be denoted by dx ). We define on X the functional F by
F (u) =
Z A(x)∇u∇u dx,
Ω
+∞,
if u ∈ W 1,∞ (Ω),
if u ∈ X\W 1,∞ (Ω).
The goal of this paper is to study the lower semicontinuous envelope F¯ of F in X , which, we recall, is defined by F¯ (u) = inf {lim inf F (un ): un ∈ X, un → u in X}, ∀u ∈ X,
or also, as the greater lower semicontinuous functional which is lower or equal than F . Let us remark that to consider the strong or the weak topology of X is equivalent. Indeed the function F is convex and so the epigraph epi(F¯ ) of F is the smallest convex closed set in X × R containing the epigraph epi(F ) of F , but the strong and weak convex closed sets are the same. To obtain a first characterization of F¯ , we need some definitions. DEFINITION 1.3. Let WA (Ω) be the space defined by WA (Ω) = {u ∈ W 1,∞ (Ω):
√
A∇u ∈ L2 (Ω)d } = {u ∈ X : F (u) < +∞},
146314.tex; 27/05/1999; 13:44; v.7; p.7
46
JUAN CASADO-D´ıAZ
and let H be the set defined by H = {u ∈ X : F¯ (u) < +∞}.
Let V be the linear space defined as the closure in X × L2 (Ω)d of the linear space √ {(u, A∇u): u ∈ WA (Ω)}. Let π1 , π2 be the projections from X × L2 (Ω)d into X and L2 (Ω)d respectively. For any u ∈ π1 (V ) we define the space Vu by Vu = {v ∈ L2 (Ω)d : (u, v) ∈ V }.
Remark that to consider the strong or the weak topology of X × L2 (Ω)d to define the space V is equivalent (use the same arguments as above). REMARK 1.1. Since Vu = π2 (({u}×L2 (Ω)d )∩V ) and since π2 is an isomorphism from {u} × L2 (Ω)d into L2 (Ω)d , Vu is a closed affine subspace of L2 (Ω)d for any u ∈ π1 (V ). In particular V0 is a closed linear subspace of L2 (Ω)d . For (u, v) ∈ V , we have Vu = v + V0 . The following theorem gives a characterization of H and F¯ . THEOREM 1.1. We have H = π1 (V ) = {u ∈ X : ∃v ∈ L2 (Ω)d with (u, v) ∈ V }.
For u ∈ H and w ¯ ∈ Vu we have Z
F¯ (u) = min Ω
|w|2 dx: w ∈ Vu
Z
= min Ω
|w ¯ + v|2 dx: v ∈ V0 . (1.1)
Proof. Let u ∈ π1 (V ). By the definition of V , there exists for any w ∈ Vu a sequence un ∈ WA (Ω) such that un → u in X, √ A ∇un → w in L2 (Ω)d .
Therefore F¯ (u) 6 lim
Z
n→∞ Ω
Z √ 2 | A ∇un | dx = |w|2 dx < +∞ Ω
which proves that π1 (V ) ⊂ H
(1.2)
146314.tex; 27/05/1999; 13:44; v.7; p.8
47
RELAXATION OF A QUADRATIC FUNCTIONAL
and that Z
F¯ (u) 6 inf
|w| dx: (u, w) ∈ V 2
Ω
, ∀u ∈ π1 (V ).
(1.3)
Conversely, if u ∈ H we have F¯ (u) < +∞ and thus
F¯ (u) = inf lim inf
Z
n→∞
Ω
√ | A ∇un |2 dx: {un } ⊂ WA (Ω), un → u
in X .
We can actually choose the sequence {un } ⊂ WA (Ω) such that F¯ (u) = lim
Z
n→∞ Ω
√ | A ∇un |2 dx.
(1.4)
√ e ∈ L2 (Ω)d Thus the sequence A ∇un is√bounded in L2(Ω)d and thus there exists w 2 d e . Therefore such that (for a subsequence) A ∇un converges weakly in L (Ω) to w √ e (un , A ∇un ) → (u, w) in X(strong) × L2 (Ω)d (weak), e ∈ V . Hence i.e. (u, w)
H ⊂ π1 (V ).
(1.5)
From (1.2) and (1.5) we deduce that H = π1 (V ). Using now the fact that the functional k w k2L2 (Ω)d is weakly lower semicontinuous in L2 (Ω)d , we have Z Ω
e 2 dx 6 lim |w|
Z
n→∞ Ω
√ | A ∇un |2 dx = F¯ (u).
With (1.3) this proves that Z
Z Ω
e 2 dx = min |w|
Ω
|w|2 dx: (u, w) ∈ V
This completes the proof of Theorem 1.1.
= F¯ (u), ∀u ∈ H. 2
DEFINITION 1.4. For u ∈ H we denote by wu ∈ L2 (Ω)d the unique function which gives the minimum in (1.1) (this function is well defined because the functional k w k2L2 (Ω)d is strictly convex in L2 (Ω)d ). REMARK√1.2. We have shown in Theorem 1.1 that for un satisfying (1.4), the sequence A ∇un converges weakly to wu in L2 (Ω)d and satisfies k
√ A ∇un k2L2 (Ω)d →k wu k2L2 (Ω)d .
146314.tex; 27/05/1999; 13:44; v.7; p.9
48
JUAN CASADO-D´ıAZ
Thus
√
A ∇un converges strongly to wu in L2 (Ω)d .
Although they are not needed to obtain the integral representation of the functional F¯ given in Theorem 2.1, we are going to prove some results which establish that the functional u 7→ wu satisfies properties analogous to a gradient. This means that the space H has a structure similar to a Sobolev space. PROPOSITION 1.1. For any u ∈ H , the function wu is characterized by (
wu ∈ Vu , R
Ω wu v dx
(1.6)
= 0, ∀ v ∈ V0 .
Proof. The result is clear since wu ∈ Vu gives the minimum in (1.1).
2
PROPOSITION 1.2. For any u ∈ H ∩ L∞ (Ω) and for any w ∈ Vu , there exists a sequence un such that (
un ∈ WA (Ω), un is bounded in L∞ (Ω), √ un → u in X, A ∇un → w in L2 (Ω)d .
(1.7)
If u further belongs to WA (Ω), then the sequence un can be chosen such that (
un ∈ WA (Ω), un → u in L∞ (Ω),
√
A ∇un → w in L2 (Ω)d .
(1.8)
In particular, for any v ∈ V0 there exists a sequence un such that (
un ∈ WA (Ω), un → 0 in L∞ (Ω),
√
A ∇un → v in L2 (Ω)d .
(1.9)
Proof. Assume u ∈ H ∩ L∞ (Ω). By the definition of V and Theorem 1.1, there exists a sequence zn ∈ WA (Ω) satisfying zn → u
in X,
√
A ∇zn → w
in L2 (Ω)d .
(1.10)
The sequence2 un = Tm (zn ) with m =k u kL∞ (Ω) satisfies then (1.7). If now u ∈ WA (Ω), consider k > 0. Since A ∇(zn − u) ∇(zn − u) is equiintegrable, there exists δk > 0 such that Z S 2
A ∇(zn − u) ∇(zn − u) dx <
1 , k
(1.11)
See in Notation the definition of Tm .
146314.tex; 27/05/1999; 13:44; v.7; p.10
49
RELAXATION OF A QUADRATIC FUNCTIONAL
for any n ∈ N and for any measurable subset S ⊂ Ω with |S| < δk . Since the strong convergence in X implies the convergence in measure, we can extract a subsequence nk such that Sk = {x ∈ Ω: |znk − u| > 1/k} satisfies |Sk | < δk as well as √ 1 A ∇znk − w kL2 (Ω)d < . k
k
(1.12)
Define now uk ∈ WA (Ω) by uk = u + T1/k (znk − u). Then |uk − u| < 1/k gives the strong convergence in L∞ (Ω) while √ √ √ A ∇uk − w = A ∇znk − w + χSk A ∇(u − znk ) gives the third assertion of (1.8) by (1.12), |Sk | < δk and (1.11).
2
REMARK 1.3. Proposition 1.2 shows that V0 is independent on X . PROPOSITION 1.3. The operator u 7→ wu satisfies the following properties: (i) (ii) (iii) (iv)
For any u ∈ H and for any λ ∈ R, we have λu ∈ H and wλu = λwu . For any u, v ∈ H , we have u + v ∈ H and wu+v = wu + wv . For any u, u ¯ ∈ H ∩L∞ (Ω), we have u¯ u ∈ H ∩L∞ (Ω) and wu¯u = u ¯wu +uwu¯ . ∞ For any u ∈ H ∩ L (Ω) and for any v ∈ V0 , we have uv ∈ V0 .
¯ in H ∩ L∞ (Ω) and w, Proof. We will only prove (iii) and (iv). Consider u, u w ¯ which respectively belong to Vu and Vu¯ . By Proposition 1.2, there exist two ¯n in WA (Ω) which are bounded in L∞ (Ω), which converge to sequences un and u u and u ¯ in X and which satisfy √ √ A ∇un → w in L2 (Ω)d , A ∇¯ un → w ¯ in L2 (Ω)d .
Then, zn = un u ¯n belongs to WA (Ω) and converges to u¯ u in X . On the other hand √ √ √ A ∇zn = u ¯n A ∇un + un A ∇¯ un converges to u ¯w + uw ¯ in L2 (Ω). Therefore u¯ u ∈ H and u ¯w + uw ¯ ∈ Vu¯u . This shows uVu¯ + u ¯Vu ⊂ Vu¯u .
(1.13)
¯ = 0 in (1.13), we deduce (iv). Taking u ¯wu + uwu¯ ∈ Vu¯u and then, by Proposition 1.1 In order to prove (iii), by (1.13) u it is enough to show Z Ω
(¯ uwu + uwu¯ )v dx = 0, ∀ v ∈ V0 .
(1.14)
146314.tex; 27/05/1999; 13:44; v.7; p.11
50
JUAN CASADO-D´ıAZ
Proposition 1.1 gives that Z
Z
Ω
wu v = 0,
Ω
wu¯ v = 0, ∀ v ∈ V0 .
(1.15)
¯v belong to V0 and then, (1.15) implies (1.14). 2 But by (iv), for any v ∈ V0 , uv and u
REMARK 1.4. In Lemma 2.1 we will improve (iv) of Proposition 1.3, with the help e (see Section 2). The integral representation of F¯ of Theorem 2.1 of hypothesis (H) will follow from that result. PROPOSITION 1.4. The operator H ⊂ X 7→ L2 (Ω)d defined by u 7→ wu is closed. Proof. Let (un , wun ) ∈ H × L2 (Ω)d be which converges to (u, w) in X × 2 L (Ω)d . By definition of V , there exists a sequence zn ∈ WA (Ω) such that dX (zn , un ) <
1 n
and
k
√
A ∇zn − wun kL2 (Ω)d <
1 . n
Therefore, u ∈ H and w ∈ Vu . In order to prove that w = wu we use Proposition 1.1 and prove that Z Ω
wv dx = 0, ∀ v ∈ V0 ,
which is clear passing to the limit in Z Ω
wun v dx = 0, ∀ v ∈ V0 .
2
COROLLARY 1.1. The space H is a complete metric space when equipped with the distance dH (u, u¯) = dX (u, u ¯)+ k wu − wu¯ kL2 (Ω)d .
So, H is a Fr´echet space. COROLLARY 1.2. If (un , wun ) ∈ H × L2 (Ω)d converges weakly in X × L2 (Ω)d to (u, w), then u ∈ H and wu = w. Proof. This is clear from Proposition 1.4 and Mazur’s Theorem. PROPOSITION 1.5. For any measurable set N ⊂ R of zero measure, the function wu vanishes almost everywhere on the set EN = {x ∈ Ω: u(x) ∈ N }. For any Lipschitz-continuous function S and for any u ∈ H , S(u) belongs to H and we have wS(u) = S 0 (u)wu .
(1.16)
146314.tex; 27/05/1999; 13:44; v.7; p.12
51
RELAXATION OF A QUADRATIC FUNCTIONAL
REMARK 1.5. A particular case of Proposition 1.5 is the case where N = {c} for c ∈ R and where S is a Lipschitz-continuous, piecewise C 1 function. The result of Proposition 1.5 is of course much more general. Proposition 1.5 is similar to the superposition result of functions of H 1 (Ω) by Lipschitz-continuous functions. Similarly to this result, observe that since S 0 is only a measurable function, which is defined only outside of a set NS 0 (the set of the points of R which are not Lebesgue points for S 0 ) of zero measure of R, the function S 0 (u) is not correctly defined on {x ∈ Ω: u(x) ∈ NS 0 }. But wu is zero almost everywhere on this set and thus the product S 0 (u)wu makes sense almost everywhere on Ω. In other terms, (1.16) should be interpreted as (
wu =
S 0 (u)wu on {x ∈ Ω: u(x) ∈ LS 0 },
0
on {x ∈ Ω: u(x) ∈ NS 0 },
where LS 0 (respectively NS 0 ) is the set of points of R which are (respectively are not) Lebesgue points of S 0 . Proof. The proof is based on the idea of A. Ancona to prove the analogous result for H 1 (Ω) (see [B M]). First step. We first prove the result assuming that the derivative of S is continuous. Let u ∈ H be, it is easy to see from the definition of V that S(u) belongs to H and that S 0 (u)w belongs to VS(u) , for any w ∈ Vu and thus S 0 (u)wu belongs to VS(u) . To conclude that wS(u) = S 0 (u)wu we now use Proposition 1.1 and show that Z
Ω
S 0 (u)wu v dx = 0, ∀ v ∈ V0 .
(1.17)
For any v ∈ V0 , wu + v belongs to Vu and so, S 0 (u)(wu + v) belongs to VS(u) . Since the difference of two elements of VS(u) belongs to V0 we find that S 0 (u)v = S 0 (u)(wu + v) − S 0 (u)wu belongs to V0 for any v ∈ V0 . Then (1.17) is deduced from Z
Ω
wu v dx = 0, ∀ v ∈ V0 .
Second step. If now, S is only a Lipschitz-continuous function and then the e 0 by derivative of S is only in L∞ (R), we define3 L S (
e S0 = L
1 s ∈ R: ∃ lim n→∞ n
Z
s+(1/2n)
)
S 0 (r) dr
s−(1/2n)
and we define Sn by 1 Sn (s) = S(0) + n 3
Z 0
s Z t+(1/2n)
S 0 (r) dr dt.
t−(1/2n)
See Notation.
146314.tex; 27/05/1999; 13:44; v.7; p.13
52
JUAN CASADO-D´ıAZ
Thus, we obtain a sequence Sn of Lipschitz-continuous functions with continuous derivative such that Sn0 is bounded in L∞ (R) and Sn0 (s) pointwise converges to e 0 , where S 0 is defined by S 0 (s) at every point s ∈ L S 1 S (s) = lim n→∞ n 0
Z
s+(1/2n) s−(1/2n)
e 0. S 0 (r) dr, ∀ s ∈ L S
Thus, for any k > 0 and for any s ∈ R with |s| < k, the Lebesgue dominated convergence theorem implies Z s Z 0 0 |S(s) − Sn (s)| = (S (t) − Sn (t)) dt 6 0
k
0
|S 0 (t) − Sn0 (t)| dt → 0,
i.e. Sn converges uniformly to S on the compact sets of R. By the first step, for any u ∈ H , the function Sn (u) belongs to H and wSn (u) = Sn0 (u)wu . The sequence Sn0 (u) is bounded in L∞ (R) and so (for a subsequence) e 0 ) = {x ∈ Sn0 (u) converges weakly-∗ in L∞ (Ω) to a function r . On the set u−1 (L S −1 e e Ω: u(x) ∈ LS 0 } we have (it is not difficult to see that u (LS 0 ) is measurable) that Sn0 (u(x)) converges almost everywhere to S 0 (u(x)) and thus e 0 }. S 0 (u)(x) = r(x) a.e. in {x ∈ Ω: u(x) ∈ L S
(1.18)
The uniform convergence of Sn to S on the compact sets on R implies that Sn (u) ¯ or C 0 (Ω) while the pointwise converconverges in X to S(u) when X is C 0 (Ω) gence of Sn , the inequality |Sn (s)| 6 |S(0)|+ k S 0 kL∞ (Ω) |s|, ∀ s ∈ R,
(which is easily deduced from the definition of Sn ) and the Lebesgue dominated convergence theorem imply that Sn (u) converges in X to S(u) when X is Lp (Ω) or Lploc (Ω) (1 6 p < ∞). Since we also have that wSn (u) = Sn0 (u)wu converges weakly in L2 (Ω) to rwu , Corollary 1.2 implies that S(u) belongs to H and the equality wS(u) = rwu .
(1.19)
Once we prove that for any set N ⊂ R of zero measure wu vanishes almost everywhere on u−1 (N ), then by (1.19) and (1.18) we will obtain that wS(u) = e 0 | = 0. It thus remains to prove the first S 0 (u)wu almost everywhere from |R − L S statement of Proposition 1.5. This will be carry out in the third step. Third step. Let N ⊂ R be a set of zero measure. For any n ∈ N there exists an open set On ⊂ R which contains N and which satisfies |On | < 1/n. We
146314.tex; 27/05/1999; 13:44; v.7; p.14
53
RELAXATION OF A QUADRATIC FUNCTIONAL
consider the decomposition of On as union of its connected components, i.e. S On = ∞ i=1 (ai , bi ), and define Z
Z
s
Rk (s) = 0
χ∪k
i=1
(ai ,bi ) dt,
R(s) = 0
s
χ∪∞ dt. i=1 (ai ,bi )
e 0 we deduce from (1.18) and (1.19) that Because every point of R is in L Rk wRk (u) = Rk0 (u)wu with Rk0 defined by S 1 if s ∈ ki=1 (ai , bi ), 1 if ∃ i, j with 1 6 i, j 6 k such that s = ai = bj , Rk0 (s) = 12 if s = ai for some i with 1 6 i 6 k and s 6∈ ∪kj=1 {bj }, 1 k 2 if s = bi for some i with 1 6 i 6 k and s 6∈ ∪j=1 {aj }, Sk
0 if s 6∈
i=1 [ai , bi ].
∞ Now |Rk (s) − R(s)| 6 Σ∞ i=k+1 (bi − ai ) and thus, since Σi=1 (bi − ai ) < +∞, we get
Rk (u) − R(u) → 0
in L∞ (Ω)
and thus Rk (u) converges to R(u) in X . Because we have pointwise convergence of Rk0 and |Rk0 (s)| 6 1, it is also clear that Rk0 (u) converges weakly-∗ in L∞ (Ω) (and strongly in Lp (Ω) for 1 6 p < ∞) to R0 (u) where R0 is defined by S 1 if s ∈ ∞ i=1 (ai , bi ), 1 if ∃ i, j such that s = ai = bj , S R0 (s) = 12 if s = ai for some i and s 6∈ ∞ j=1 {bj }, 1 ∞ 2 if s = bi for some i and s 6∈ ∪j=1 {aj }, S∞
0 if s 6∈
i=1 [ai , bi ].
Therefore, Rk0 (u)wu converges weakly in L2 (Ω) to R0 (u)wu and then, from Corollary 1.2 wR(u) = R0 (u)wu . S
This result was obtained for every open set On = ∞ i=1 (ai , bi ). Denote by Rn the function R corresponding to On . It is clear that Rn (u) converges in L∞ (Ω) (and hence in X ) to zero when n tends to infinite. Because Rn0 (u) is bounded in L∞ (Ω), we deduce from Corollary 1.2 that Rn0 (u)wu = wRn (u) converges weakly in L2 (Ω) to zero.
146314.tex; 27/05/1999; 13:44; v.7; p.15
54
JUAN CASADO-D´ıAZ
T
e be defined by N e = ∞ On . N e is a Borel set (in fact it is a G ) which Let N δ n=1 e = u−1 (N e ) is measurable and satisfies contains N and has measure zero. The set E Z
|wu | dx 6
Z
2
e E
|wu | dx 6
Z
2
u−1 (On )
Ω
Rn0 (u)wu wu dx → 0.
e. Therefore, wu is zero almost everywhere in E ⊂ E
2
REMARK 1.6. For X = L2 (Ω) and A ∈ L1 (Ω)d×d , Propositions 1.4 and 1.5 imply that the form F¯ is a Dirichlet form (see [13], [17]). ¯ 2. Integral Representation of F
In this Section, we use Theorem 1.1 to obtain an integral representation of the functional F¯ . For the whole of this section, we will make the following hypothesis ( e) (H
For any u ∈ L∞ (Ω) there exists a sequence un ∈ WA (Ω) which converges almost everywhere to u.
e is a hypothesis on the ‘size’ of A or in REMARK 2.1. In some sense, hypothesis (H) different terms on the size of the set where A is unbounded or infinite. If A belongs e clearly holds true (take any sequence un ∈ W 1,∞ (Ω) to L1 (Ω)d×d , hypothesis (H) which converges almost everywhere to u). We will see in Section 4 that in the onee holds true whenever A is finite almost everywhere and that dimensional case, (H) this does not hold true in the d-dimensional case for any d with d > 1. e holds true. Let v ∈ V0 and let r : Ω 7→ R be a LEMMA 2.1. Assume that (H) measurable function such that rv ∈ L2 (Ω)d . Then rv ∈ V0 . Proof. By Proposition 1.3(iv), Lemma 2.1 holds true, when r further belongs to H ∩ L∞ (Ω). e , a sequence r¯n ∈ If r only belongs to L∞ (Ω), there exists, because of (H) WA (Ω) which converges to r almost everywhere. Define m =k r kL∞ (Ω) and rn = Tm (¯ rn ). Then rn belongs to WA (Ω) ⊂ H ∩ L∞ (Ω), and thus rn v ∈ V0 . By the dominated convergence theorem, rn v converges to rv in L2 (Ω)d and then since V0 is closed in L2 (Ω)d , rv belongs to V0 . Finally, if r is only measurable with rv ∈ L2 (Ω)d , consider the sequence rn = Tn (r) and conclude using the dominated convergence theorem as above. 2 e holds true. If v ∈ V0 , then ve(x) = N (v(x)) is LEMMA 2.2. Assume that (H) also in V0 . 2 Proof. Apply Lemma 2.1 to r = (1/|v|)χ{v6=0} .
DEFINITION 2.1. Let {wn } be a countable set which is dense in V0 . For almost
146314.tex; 27/05/1999; 13:44; v.7; p.16
RELAXATION OF A QUADRATIC FUNCTIONAL
55
every x ∈ Ω, let T (x) = span({wn (x)}) and let Q(x) and P (x) = I − Q(x) be respectively the orthogonal projection from Rd on T (x) and T (x)⊥ . REMARK 2.2. The set T (x) is defined for any point x ∈ Ω which is a Lebesgue point for every wn , n ∈ N, and thus is defined almost everywhere in Ω. So are also P (x) and Q(x). The idea which leads to the definition of T (x) is to consider the linear space spanned by w(x) when w ∈ V0 . Unfortunately this definition does not allow one to define T (x) in a correct way, since one has to exclude all the points x which are not Lebesgue points for every w, i.e. a non countable set of sets of zero measure, which is not necessarily of zero measure. This is the reason of the introduction of the countable set {wn }. We will prove that T (x) does not depend on the choice of the countable set {wn } (see Lemma 2.3) and that P (x) and Q(x) belong to L∞ (Ω)d×d (see the comment after Lemma 2.5). Finally note that T (x), P (x) and Q(x) do not depend on X since V0 does not depend on X . The set T (x) and the matrix P will allow us to characterize the space V0 and F¯ (see Theorem 2.1). LEMMA 2.3. Except on a set of zero measure, T does not depend on the choice of the countable set {wn }. Proof. Let {wn0 } be another countable set which is dense in V0 and let T 0 (x) be defined by T 0 (x) = span({wn0 (x)}). For any wk0 there exists a subsequence {wkn } of the countable set {wn } which converges pointwise to wk0 except on a set Zk0 of zero measure. If x 6∈ Zk0 , wk0 (x) is dimensional and hence closed space. Therefore, thus in T (x), since T (x) is a finite S 0 0 the set Z 0 defined by Z 0 = ∞ k=1 Zk is of zero measure and T (x) ⊂ T (x) for 0 x ∈ Ω\Z . In the same way there exists a set Z of zero measure such that T (x) ⊂ T 0 (x) for x ∈ Ω\Z. Thus, T (x) = T 0 (x) if x 6∈ Z ∪ Z 0 . 2 e holds true. There exist (s1 , . . . , sd ) in V d and LEMMA 2.4. Assume that (H) 0 a measurable function q defined on Ω with values in {0, . . . , d} such that for almost every x in Ω
(i) sk (x) = 0 for every k > q(x), (ii) {s1 (x), . . . , sq(x) (x)} is an orthonormal basis of T (x). Proof. (We follow the idea of [9] pages 140 and 141 where a similar result is proved). Let {wn } be the set used in the definition of T . Let us define
y1 = N (w1 )
146314.tex; 27/05/1999; 13:44; v.7; p.17
56
JUAN CASADO-D´ıAZ
and for n > 1 wn −
yn = N
!
n−1 X
(wn yi )yi .
i=1
Then span({yn (x)}) = span({wn (x)})
a.e. in Ω
and {yn (x): yn (x) 6= 0} is an orthonormal basis of T (x) for a.e. x ∈ Ω. In the definition of yn , each term of the form (wn yi )yi belongs to L2 (Ω)d since |(wn , yi )yi | 6 |wn | almost everywhere. Using Lemma 2.1 and Lemma 2.2, we prove by induction that yn ∈ V0 . We will now reorder the yi (x) for almost every x ∈ Ω in order for the first ones to be nonzero. For that we define Kki , with i ∈ {1, . . . , d}, k ∈ N as the set of those points x ∈ Ω where yk (x) is exactly the ith nonzero element in the sequence {yn (x)}. Let us first prove that Kki is measurable. Let Sj be the measurable set Sj = {x ∈ Ω: yj (x) 6= 0}.
The function qk defined by qk (x) =
k X
χSj (x)
j=1
is measurable and qk is the number of those integers j ∈ {1, . . . , k} such that yj (x) is not zero. Note that qk (x) ∈ {0, . . . , d} for any k ∈ N and almost every x ∈ Ω. For i ∈ {1, . . . , d} let Cki be the measurable set defined by Cki = {x ∈ Ω: qk (x) = i}.
The set Cki ∩ Sk is measurable and coincides with Kki . Note that Kki ∩ Kki 0 = ∅ for k 6= k0 . Let us now define the functions q and s = (s1 , . . . , sd ) by q(x) =
∞ X
χSj (x),
j=1
si =
∞ X
yn (x)χKni ∀i
with 1 6 i 6 d.
n=1
The functions s and q clearly satisfy requirements (i) and (ii) of Lemma 2.4. Let us prove that s belongs to V0d . Define for m ∈ N sm i (x) =
m X
yn (x)χKni (x).
n=1
146314.tex; 27/05/1999; 13:44; v.7; p.18
57
RELAXATION OF A QUADRATIC FUNCTIONAL
Since yn belongs to V0 , Lemma 2.1 implies that sm i ∈ V0 and we have ∞ [ k2L2 (Ω) 6 χKni (x) = Kni = Ω n=m+1 n=m+1 Z
k si − sm i
∞ X
∞ X
|Kni | → 0,
n=m+1
2
as m tends to infinity, since Ω is bounded. Therefore s belongs to V0d .
e holds true. For any v ∈ L2 (Ω)d , Qv ∈ V0 . LEMMA 2.5. Assume that (H) This shows in particular that Q and thus P are measurable because Qe ∈ V0 ⊂ L2 (Ω)d for any e ∈ Rd . In fact P and Q belong to L∞ (Ω)d×d because they are projections, which implies |Q(x)| 6 1, |P (x)| 6 1. Proof. We have
Qv =
d X
(vsi )si .
i=1
Since |(vsi )si | 6 |v| implies (vsi )si ∈ L2 (Ω)d , Lemma 2.1 implies that (vsi )si ∈ V0 and so, Qv ∈ V0 . 2 We are now in a position to give a characterization of V0 and an integral representation for F¯ . e holds true. The space V0 is characterized by THEOREM 2.1. Assume that (H)
V0 = {v ∈ L2 (Ω)d : v(x) ∈ T (x)
a.e.}.
(2.1)
If u ∈ H , then for any w ∈ Vu we have wu = P w, Z |P w|2 . F¯ (u) =
(2.2)
Ω
√ u ∈ W (Ω) we can choose w = A ∇u in (2.2). REMARK 2.3. In the case where A √ Using the symmetry of P and A, and the equality P 2 = P , we have √ √ √ |P A ∇u|2 = AP A ∇u ∇u, ∀u ∈ WA (Ω).
We have then the following integral representation for F¯ (u) F¯ (u) =
Z Ω
B ∇u ∇u
with B =
√
√ AP A, ∀ u ∈ WA (Ω).
(2.3)
Proof of Theorem 2.1. Let S be defined by S = {v ∈ L2 (Ω)d : v(x) ∈ T (x) a.e.}.
146314.tex; 27/05/1999; 13:44; v.7; p.19
58
JUAN CASADO-D´ıAZ
To prove (2.1) we have to prove that V0 and S coincide. From Lemma 2.5 we have S = QS ⊂ V0 .
Conversely, if v ∈ V0 there exists some subsequence wkn of the countable set {wn } considered in Definition 2.1 which converges to v in L2 (Ω)d and almost everywhere in Ω. Thus v(x) ∈ T (x) almost everywhere. This shows first statement of Theorem 2.1. Let now u ∈ H and let w be such that (u, w) ∈ V . From (1.1) we obtain Z
F¯ (u) = min
|w + v| : v ∈ V0 . 2
Ω
For any v ∈ V0 , using the definition of Q we have |w − Qw| 6 |w + v|
a.e. in Ω.
Since by (2.1) −Qw belongs to V0 we obtain (2.2).
2
3. The One-Dimensional Problem In the one-dimensional case we will characterize completely H and P . We assume in the present section that Ω is the interval (0, 1) and that A is a measurable function a defined on Ω with values in [0, +∞]. Note that this hypothesis is more general than the hypothesis made in Section 2, which would imply that a is nonnegative and finite almost everywhere. In contrast, we can here have a(x) = +∞ for a subset of (0, 1) with Lebesgue measure strictly positive. DEFINITION 3.1. Let G be defined by G = {x ∈ (0, 1): ∃ δ(x) > 0
such that 1/a ∈ L1 (x − δ(x), x + δ(x))}.
REMARK 3.1. Note that G is an open set and 1/a belongs to L1loc (G); actually G is the biggest open set such that 1/a is locally integrable on it. THEOREM 3.1. We have (
H=
u∈X∩
1,1 Wloc (G)
Z
such that G
) 2 du a dx < +∞ dx
(3.1)
and for all u ∈ H we have the integral representation of F¯ (u) F¯ (u) =
Z G
2 du dx, ∀u ∈ H. dx
a
(3.2)
146314.tex; 27/05/1999; 13:44; v.7; p.20
59
RELAXATION OF A QUADRATIC FUNCTIONAL
REMARK 3.2. In the one-dimensional case the only orthogonal projections which exist are the zero function and the identity. So the matrix P in Section 2 must be now reduced to a characteristic function. Theorem 3.1 gives P = χG (more exactly P = χG∩S c , see the proof of Theorem 3.1). Proof. Let S be the measurable set defined by S = {x ∈ (0, 1): a(x) < +∞}
and let S c be the complementary of G defined by S c = {x ∈ (0, 1): a(x) = +∞}.
Let us assume |S| > 0. If not, a(x) = +∞ almost everywhere and Theorem 3.1 is trivial: F¯ (u) = +∞ for any u ∈ X , u nonconstant and F¯ (u) = 0 for u constant. Let us consider the space H defined by (
H=
u∈X∩
1,1 Wloc (G)
Z
such that G
) 2 du a dx < +∞ dx
and the functional F defined on X by Z 2 du a dx if u ∈ H, dx F(u) = G
+∞
if u ∈ X\H.
We shall show that F is lower semicontinuous. Once this will be proved, since F(u) 6 F (u), ∀ u ∈ X,
we will obtain F(u) 6 F¯ (u), ∀ u ∈ X.
Hence H⊂H Z G
and
(3.3)
2 du a dx 6 F¯ (u), ∀ u ∈ H. dx
(3.4)
In order to see that F is lower semicontinuous, consider a sequence un which converges to a function u in X . If F(un ) converges to infinity, clearly F(u) 6 lim inf F(un ). n
146314.tex; 27/05/1999; 13:44; v.7; p.21
60
JUAN CASADO-D´ıAZ
If not, there exists a subsequence unk of un such that lim F(unk ) = lim inf F(un ) < +∞. n
k
For k large enough unk belongs to H and from Z
lim k
G
d un k 2 dx < +∞, dx
a
we can suppose (extracting another subsequence if necessary) that √ d un k a *v dx
in L2 (G).
√ On the other hand, from the definition of G, 1/ a belongs to L2loc (G) and thus for any ϕ ∈ D(G), we have ϕ √ ∈ L2 (G). a
Hence Z G
d un k ϕ dx = dx
Z
Z √ d un k ϕ ϕ √ dx → a v √ dx, ∀ ϕ ∈ D(Ω). d x a a G G
But unk converges to u in X and hence in the sense of distributions in Ω and thus in G, we have then Z G
d un k ϕ dx = dx
d un k ,ϕ dx
D 0 (G),D(G)
→
du ,ϕ dx
D 0 (G),D(G)
and therefore v du =√ . dx a √ √ Since v belongs to L2 (G) and 1/ a belongs to L2loc (G) we have that v/ a belongs √ 1,1 (G) and a(du/dx) = v belongs to L2 (G) to L1loc (G) and thus u belongs to Wloc which means that u belongs to H. Now, since the norm in L2 (G) is a lower semicontinuous function for the weak topology of L2 (G), we have F(u) =
Z G
2 Z du dx = |v|2 dx dx
a Z
6 lim k
G
G
dunk 2 dx = lim inf F(un ), n dx
a
146314.tex; 27/05/1999; 13:44; v.7; p.22
61
RELAXATION OF A QUADRATIC FUNCTIONAL
which proves that F is lower semicontinuous. To complete the proof of the theorem, √ it is sufficient to show that for every u which belongs to H we have that (u, a du dx χG ) belongs to V . Indeed by Theorem 1.1, this implies that u belongs to H and that F¯ (u) = min
(Z
1
0
Z 6
G
)
|w|2 dx: w ∈ Vu
2 du dx = F(u) < +∞, ∀u ∈ H. dx
a
(3.5)
Hence H ⊂ H which combined with (3.3) gives H = H. Finally, Inequalities (3.4) and (3.5) give (3.2). Let us thus prove that
√ du u ∈ H ⇒ u, a χG dx
∈ V.
(3.6)
By Definition 1.3, V is the closure in E = X × L2 (0, 1) of
Γ=
√ du u, a : u ∈ Wa (0, 1) dx
and thus ¯ E =⊥ (Γ⊥ ), V =Γ
where Γ⊥ = {κ ∈ E 0 : hκ, γiE 0 ,E = 0, ∀γ ∈ Γ}
and ⊥
(Γ⊥ ) = {δ ∈ E : hδ, κiE,E 0 = 0, ∀κ ∈ Γ⊥ }.
Now, we note that (µ, z) ∈ Γ⊥
0 2 µ ∈ X , z ∈ L (0, 1) with Z Z 1 ⇐⇒ √ du u dµ(x) + z a dx = 0, ∀u ∈ Wa (0, 1), [0,1]
0
(3.7)
dx
R
where we have written hµ, uiX 0 ,X = [0,1] u dµ(x) since X 0 is always contained in the space of the Radon measures in [0, 1]. We will first prove (3.6) in the simpler case in which X = L2 (0, 1) (X = p L (0, 1) or X = Lploc (0, 1) are similar) and a ∈ L1 (0, 1). The general case will follow from the same argument but the proof is more technical.
146314.tex; 27/05/1999; 13:44; v.7; p.23
62
JUAN CASADO-D´ıAZ
For X = L2 (0, 1) and a ∈ L1 (0, 1) we have that Wa (0, 1) = W 1,∞ (0, 1) and that µ = g dx with g ∈ L2 (0, 1). Thus (3.7) may be written as Z
Z
1
1
gu dx + 0
0
√ du z a dx = 0, ∀u ∈ W 1,∞ (0, 1), dx
(3.8)
which implies that √ √ √ √ d(z a) z a ∈ H 1 (0, 1), = g, z a(0) = z a(1) = 0. dx √ In particular, z a is a continuous function and then for every x0 ∈√(0, 1) such √ that (z a)(x0 ) 6= 0, there exists λ > 0 and δ > 0 which satisfy |z a| > λ in (x − δ, x + δ). Since z belongs to L2 (0, 1) we have then Z
λ
x+δ
2
x−δ
1 dy 6 a
Z
1
z 2 dy < +∞
0
and thus x belongs to G. We have thus proved that √ {x ∈ (0, 1): (z a)(x) 6= 0} ⊂ G √ √ and then that z a = 0 in [0, 1]\G. Since g is the derivative of z a which belongs to H 1 (0, 1), this implies that g = 0 almost everywhere in [0, 1]\G and therefore (3.8) reads as Z
Z
gu dx + G
√ du z a dx = 0, ∀u ∈ W 1,∞ (0, 1), dx G
which by an approximation argument (see Step 9 below) implies that Z
Z
gu dx + G
√ du z a dx = 0, ∀u ∈ H, dx G
or since g = 0 almost everywhere in [0, 1]\G, that Z
Z
1
gu dx + 0
0
1
√ du z a χG dx = 0, ∀u ∈ H dx
and then that (u, (du/dx)χG ) ∈⊥ (Γ⊥ ) = V . Let us now prove the general case. The proof is rather long and will be divided in several steps. √ Step 1. The function azχS ∈ L∞ (0, 1) and for any r ∈ L1 (0, 1) we have Z
Z
x
Z
rχS dy dµ(x) + [0,1]
0
√
azr dx = 0.
(3.9)
S
146314.tex; 27/05/1999; 13:44; v.7; p.24
63
RELAXATION OF A QUADRATIC FUNCTIONAL
Proof. Consider the sets S n = {x ∈ (0, 1): a(x) < n}.
For rˆ ∈ L∞ (0, 1), let un be defined by Z
x
un (x) = 0
sgn(z)|ˆ r |χS n dy.
Then un belongs to Wa (0, 1) and (3.7) gives Z
Z [0,1]
x
sgn(z)|ˆ r |χS n dy dµ(x) +
0
Z Sn
√ a|zˆ r | dx = 0.
Thus Z
√
Sn
a|zˆ r | dx 6k µ k
Z 0
1
|ˆ r | dy,
R
where k µ k= [0,1] d|µ| denotes the total variation of µ. √ r is integrable on S and satisfies From the monotone convergence theorem, azˆ Z Z √ 6k µ k azˆ r d x S
0
1
|ˆ r| dy.
The functional rˆ →
Z
√
azˆ r dy
S
can to a continuous linear functional defined in L1 (0, 1) and thus √ thus be extended ∞ azχS belongs to L (0, 1). We now consider rˆ in L∞ (0, 1) and define Z
x
un (x) =
rˆχS n dy. 0
From (3.7) we have Z
Z [0,1]
x
Z
rˆχS n dy dµ(x) + 0
Sn
√
azˆ r dx = 0.
(3.10)
The dominated convergence theorem gives Z Sn
√
azˆ r dx →
Z
√
azˆ r dx,
S
146314.tex; 27/05/1999; 13:44; v.7; p.25
64
JUAN CASADO-D´ıAZ
while for the first term we have Z
max
x∈[0,1]
x
0
Z
rˆχS n dy −
x
rˆχS dy 6k rˆ kL∞ (0,1) |S\S n |,
0
R
which implies that un converges to u = 0x rˆχS dy in C 0 ([0, 1]). This allows us to pass to the limit in (3.10) obtaining that (3.9) holds true for rˆ ∈ L∞ (0, 1). If r ∈ L1 (0, 1), taking rˆ = Tn (r) in (3.9) and using Z max x∈[0,1]
x
0
rχS dy −
Z
x
0
Z Tn (r)χS dy 6
0
1
|r − Tn (r)| dy → 0,
we have that (3.9) holds true for r in L1 (0, 1). Step 2. Let ω be the distribution in (0, 1) defined by hω, ϕiD0 (Ω),D(Ω) Z
√
= S
azϕ dx −
Z
Z [0,1]
x 0
ϕχS c dy dµ(x), ∀ϕ ∈ D(0, 1).
(3.11)
Then ω belongs to BV (0, 1), the space of functions of bounded variation and satisfies √ dω = µ and ωχS = azχS . dx
(3.12)
Proof. The distribution ω is well defined by (3.11). In fact since the operator ϕ 7→ hω, ϕiD0 (0, 1),D(0, 1) is continuous for the topology of L1 (0, 1) we have that ω belongs to L∞(0, 1) and using in (3.11) as test function ϕ = φψn were φ ∈ D(0, 1) and were ψn ∈ D(0, 1) approximates χS in L1 (0, 1), we obtain √ ωχS = azχS . (3.13) If we now, calculate the derivative of ω in the sense of distributions, using (3.9) implies that for ϕ ∈ D(0, 1)
dω ,ϕ dx =−
D 0 (0, 1),D(0, 1)
Z S
Z
√
dϕ az dx + dx
Z
x
= [0,1]
Z
0
Z
= [0,1]
dϕ = − ω, dx
0
x
Z
Z [0,1]
dϕ χS c dy dµ(x) dy
dϕ χS dy dµ(x) + dy
D 0 (0, 1),D(0, 1)
x
0
dϕ dy dµ(x) = dy
Z
Z [0,1]
0
x
dϕ χS c dy dµ(x) dy
Z
ϕ dµ(x). [0,1]
146314.tex; 27/05/1999; 13:44; v.7; p.26
RELAXATION OF A QUADRATIC FUNCTIONAL
65
Thus ω ∈ BV (0, 1), and dω/dx = µ. To avoid the problems due to the fact that ω is only defined almost everywhere, we precise the value of ω(x) by taking for any x ∈ [0, 1] ω(x) = ω0 + µ((0, x]).
This defines ω as a right continuous function. Step 3. We have Z G
|ω|2 dx < +∞. a
(3.14)
Proof. We observe that a is strictly positive and finite almost everywhere in G ∩ S and thus, using (3.12), we have ω √ χS = zχS a
in G.
Since z ∈ L2 (0, 1) so is the left-hand side and since 1/a is zero outside of S , (3.14) is proved. Let G=
∞ [
(αi , βi )
(3.15)
i=1
be the decomposition of G in its connected components. We will now describe the behaviour of ω outside of G. Step 4. Let x0 ∈ [0, 1)\G. Then either ω(x0 ) = 0 or x0 = αi for some i; in the latest case, there exists δ > 0, (which depends on αi ) such that (1/a) ∈ L1 (αi , αi + δ). If x0 ∈ (0, 1]\G either ω(x0 −) = 0 (the left limit of ω in x0 ) or x0 = βi for some i; in the latest case there exists δ > 0, (which depends on βi such that (1/a) ∈ L1 (βi − δ, βi ). Proof. Let x0 ∈ [0, 1) be and suppose that ω(x0 ) 6= 0. Then, since ω is right continuous, there exists γ > 0 and δ > 0 (depending on x0 ) such that |ω(x)| > γ, ∀ x ∈ (x0 , x0 + δ).
(3.16)
√ and thus From √ (3.12) and (3.16) we have that γχS√ 6 a|z|χSc in (x0 , x0 + δ) √ γ/ a 6 |z| in S ∩ (x0 , x0 + δ). Since 1/ a = 0 in S we also have γ/ a 6 |z| in S c ∩ (x0 , x0 + δ). Hence Z
γ2
x0 +δ
x0
1 dy < +∞. a
146314.tex; 27/05/1999; 13:44; v.7; p.27
66
JUAN CASADO-D´ıAZ
Thus (x0 , x0 + δ) ⊂ G if ω(x0 ) 6= 0. If x0 6∈ G, this implies x0 = αi for some i. The proof for the left limit is analogous. This has the following corollary. Step 5. For any αi 6= 0 we have ω(αi ) = 0
or
ω(αi −) = 0.
For any βi 6= 1 we have ω(βi ) = 0
or
ω(βi −) = 0.
REMARK 3.3. If X is neither C 0 ([0, 1]) nor C 0 (0, 1) then ω is a continuous function and we can thus write ω(αi −) = ω(αi ) = ω(βi ) = ω(βi −) = 0.
Proof. We shall proof the first assertion, the second is analogous. If ω(αi ) 6= 0 and ω(αi −) 6= 0 then, from Step 4, there exists βj and δ > 0 with βj = αi and 1 ∈ L1 ((βj − δ, βj ) ∪ (αi , αi + δ)). a But then (αj , βi ) ⊂ G which is a contradiction with the fact that (αi , βi ) and (αj , βj ) are different connected components of G. Step 6. We have the following rule of integration by parts (recall that µ = dω/dx). For [α, β] ⊂ [0, 1] and u ∈ W 1,1 (α, β) we have Z
Z
β
u dµ(x) +
ω α
[α,β]
du dx = ω(β)u(β) − ω(α−)u(α), dx
(3.17)
where for α = 0 we define ω(0−) = ω0 − µ({0}). For the proof, see [F] Theorem 3.30. The behaviour of ω in x = 0 and x = 1 is given by Step 7. We have ω(0−) = ω0 − µ({0}) = 0,
ω(1) = 0.
(3.18)
Proof. Taking u = 1 in (3.7), we have 0 = µ([0, 1]) = ω(1) − ω0 + µ({0}) = ω(1) − ω(0−).
(3.19)
Taking r = 1 in (3.9) and taking into account (3.13), we have Z
Z
x
Z
χS dy dµ(x) + [0,1]
0
ω dx = 0.
(3.20)
S
146314.tex; 27/05/1999; 13:44; v.7; p.28
67
RELAXATION OF A QUADRATIC FUNCTIONAL
Using (3.17) in [0, 1] for u given by Z
x
u=
χS dy, 0
we have Z
Z [0,1]
x
Z
χS dy dµ(x) + 0
Z
1
ω dx = ω(1) S
χS dy = ω(1)|S|. 0
Since |S| > 0 and (3.20), we have ω(1) = 0. Now (3.19) gives ω(0−) = 0. Step 8. For any i > 1 we have |ω(αi )|, |ω(αi −)| 6 |µ|({αi }) |ω(βi )|, |ω(βi −)| 6 |µ|({βi }).
Proof. The result follows easily from Step 5, Step 7 and |µ|({αi }) = |ω(αi ) − ω(αi −)|,
|µ|({βi }) = |ω(βi ) − ω(βi −)|.
Step 9. Let (αi , βi ) be a connected component of G and let u be with 1,1 u ∈ Wloc (αi , βi ) ∩ X,
and ω
du ∈ L1 (αi , βi ). dx
Then u belongs to L1 ([αi , βi ], dµ) and Z
Z
βi
u dµ(x) +
ω
[αi ,βi ]
αi
du dx = u(βi )ω(βi ) − u(αi )ω(αi −), dx
(3.21)
where if X is neither C 0 ([0, 1]) nor C 0 (0, 1) (and then u is not well defined in {αi } neither {βi }) the right-hand side is defined by zero, since in this case, from Remark 3.3 and Step 7 we have that ω(βi ) = ω(α− i ) = 0. Proof. Let ci = (αi + βi )/2. Applying (3.17) in [αi , βi ] for un given by Z
x
un (x) = u(ci ) + ci
du dy, χ dy (αi +(1/n),βi −(1/n))
with
2 < βi − αi , n
we obtain Z
Z
[αi ,βi ]
un d µ +
βi −(1/n)
ω αi +(1/n)
du dx = un (βi )ω(βi ) − un (αi )ω(α− i ). (3.22) dx
Since |ω(du/dx)χ(αi +(1/n),βi −(1/n)) | 6 |ω(du/dx)| ∈ L1 (αi , βi ) almost everywhere in (αi , βi ), the dominated convergence theorem gives Z
βi −(1/n)
αi +(1/n)
du ω dx → dx
Z
βi
ω αi
du dx. dx
146314.tex; 27/05/1999; 13:44; v.7; p.29
68
JUAN CASADO-D´ıAZ
If X is C 0 ([R0, 1]) then un converges to u in C 0 [αi , βi ] and this permit to pass to the limit in [αi ,βi ] un dµ(x) and the right-hand side of (3.22). If X = Lp (0, 1) the right-hand side is zero while for the first integral in the lefthand side we have that un converges to u almost everywhere and |un | 6 |u| almost everywhere. The dominated convergence theorem thus implies that the sequence 0 un converges to u in Lp (αi , βi ) and then, since µ belongs to Lp (0, 1) we get Z [αi ,βi ]
un dµ(x) →
Z
u dµ(x). [αi ,βi ]
The cases X = C 0 (0, 1) and X = Lploc (0, 1) are analogous. Step 10. We have |µ| [0, 1]\
∞ [
!
[αi , βi ] = 0.
(3.23)
i=1
Proof. The proof is easy if X = Lp (Ω) or Lploc (Ω) with 1 6 p < ∞, in this 0 case ω ∈ W 1,p (Ω) and ω = 0 in [0, 1]\G. Then it is know that µ, which is the 0 derivative of ω , is of the form g dx where g belongs to Lp (Ω) and is zero almost everywhere in [0, 1]\G. In the case C 0 the proof is more involved. For any interval [α, β] contained in [0, 1], the total variation of µ in (α, β] can be calculated by |µ|((α, β]) = sup
q X
|ω(tj ) − ω(tj−1 )|: α = t0 < t1 < · · · < tq−1 < tq = β
j=1
.
Let T be a given partition of [0, 1]. T = {0 = t0 6 t1 6 · · · 6 tq−1 6 tq = 1}.
Since ∞ X
|µ|([αi , βi ]) < 2|µ|([0, 1]) < ∞,
i=1
for ε > 0, we can choose n large enough such that ∞ X
|µ|([αi , βi ]) < ε
(3.24)
i=n+1
and such that for any tj ∈ T with tj ∈ [αi , βi ] for some i, we have i 6 n.
146314.tex; 27/05/1999; 13:44; v.7; p.30
69
RELAXATION OF A QUADRATIC FUNCTIONAL
We consider a new partition T 0 defined by T 0 = T ∪ {αi , βi : i 6 n}.
Changing the notation we have
T 0 = 0 = t00 6 t01 < · · · < t0m−1 < t0m = 1 ,
m>q
and since T ⊂ T 0 q X
|ω(tj ) − ω(tj−1 )| 6
j=1
m X
|ω(t0j ) − ω(t0j−1 )|.
j=1
We establish the following properties of T 0 that we will use later S S (P1) If t0j ∈ T 0 and t0j 6∈ ni=1 (αi , βi ) then t0j 6∈ ∞ i=1 (αi , βi ). S To see this, let us remark that by the construction of T 0 , if t0j ∈ ∞ i=1 (αi , βi ) 0 0 then tj ∈ T and there exists i with tj ∈ (αi , βi ) but then i 6 n. S S (P2) If t0j 6∈ ni=1 (αi , βi ] then t0j−1 6∈ ni=1 [αi , βi ). This follows, since t0j−1 ∈ [αi , βi ) and βi ∈ T 0 implies αi 6 t0j−1 < t0j 6 βi and then t0j ∈ (αi , βi ]. S (P3) Let t0l , t0s be with t0l 6= t0s and t0l , t0s 6∈ ∞ i=1 (αi , βi ), then there does not exist any i such that t0l−1 , t0s−1 ∈ [αi , βi ). To show (P3), let us assume t0l < t0s , then we have t0l−1 < t0l 6 t0s−1 < t0s ; and then if t0l−1 and t0s−1 belong to [αi , βi ) we obtain that t0l belongs to (αi , βi ). Using (P1), we have m X
|ω(t0j ) − ω(t0j−1 )| = s1 + s2 + s3 + s4 ,
j=1
with s1 =
X Sn
t0j ∈
s2 =
|ω(t0j ) − ω(t0j−1 )|,
S∞ i=n+1
{βi }
X S∞
t0j ∈
s4 =
(αi ,βi ]
X t0j ∈
s3 =
i=1
|ω(t0j ) − ω(t0j−1 )|,
i=1
S∞
{αi }\
X S∞
t0j 6∈
i=1
i=1
|ω(t0j ) − ω(t0j−1 )|, {βi }
|ω(t0j ) − ω(t0j−1 )|.
[αi , βi ]
146314.tex; 27/05/1999; 13:44; v.7; p.31
70
JUAN CASADO-D´ıAZ
We shall estimate each term. s1 =
n X
X
|ω(t0j ) − ω(t0j−1 )|
i=1 t0j ∈(αi ,βi ]
6
n X
|µ|((αi , βi ]) 6 |µ|
i=1
∞ [
!
(αi , βi ] ,
i=1
where we have used that for every i with 1 6 i 6 n the set {t0j : t0j ∈ (αi , βi ]} ∪ {αi } ⊂ T 0 is a partition of [αi , βi ]. In order to estimate s2 , let t0j be such that there exists βl > n + 1, with βl = t0j , we have the following cases S 0 [α , β If t0j−1 6∈ ∞ i=1 i i ) then from Steps 4 and 7 we have that ω(tj−1 ) = 0 and thus by Step 8 we get |ω(t0j ) − ω(t0j−1 )| = |ω(βl )| 6 |µ|({βl }) 6 |µ|((αl , βl ]).
If there exists i = il such that t0j−1 ∈ [αil , βil ) then by Step 8 we have |ω(t0j ) − ω(t0j−1 )| = |ω(βl ) − ω(αl ) + ω(αl ) − ω(βil ) + ω(βil ) − ω(t0j−1 )| 6 |ω(βl ) − ω(αl )| + |ω(αl )| + |ω(βil )| + |ω(βil ) − ω(t0j−1 )|
+ |ω(t0j−1 ) − ω(αil )| 6 |µ|((αl , βl ]) + |µ|({αl }) + |µ|({βil }) + |µ|((αil , βil ]) 6 |µ|([αl , βl ]) + 2|µ|([αil , βil ]).
From (P2) il > n + 1 and by (P3) il 6= is if βl 6= βs . Then by (3.24) s2 6 3ε. S
To estimate s3 , if t0j = αl with αl 6∈ ∞ i=1 {βi } then as above we distinguish two cases S 0 If t0j−1 6∈ ∞ i=1 [αi , βi ), then ω(tj−1 ) = 0 and thus by Step 8 |ω(t0j ) − ω(t0j−1 )| = |ω(αl )| 6 |µ|({αl }).
If t0j−1 ∈ [αil , βil ) we have |ω(t0j ) − ω(t0j−1 )| 6 |ω(αl ) − ω(βil ) + ω(βil ) − ω(t0j−1 )|
146314.tex; 27/05/1999; 13:44; v.7; p.32
71
RELAXATION OF A QUADRATIC FUNCTIONAL
6 |ω(αl )| + |ω(βil )| + |ω(βil ) − ω(t0j−1 )| + |ω(t0j−1 ) − ω(αil )| 6 |µ|({αl }) + |µ|({βil }) + |µ|([αil , βil ]) 6 |µ|({αl }) + 2|µ|([αil , βil ]).
In this last case, (P2) and (P3) imply as for s2 that il > n + 1 and that il 6= is if αl 6= αs (αl 6= 0 since we are supposing that there exists t0j−1 ). Thus, we have the estimate X
s3 6
|µ|({αi }) + 2ε.
Si>1 αi 6∈ {βj }∪{0}
To estimate s4 , Steps 4 and 7 imply that ω(t0j ) = 0 when t0j 6∈ reasoning with t0j−1 as for s2 or s3 we have
S∞
i=1 [αi ,
βi ] and
s4 6 2ε.
Thus q X
|ω(tj ) − ω(tj−1 )| 6 |µ|
j=1
∞ [
!
[αi , βi ] \ {0} + 7ε.
i=1
Since this is true for any ε > 0 and for any partition T , we have |µ|([0, 1]) 6 |µ|
∞ [
!
[αi , βi ] .
(3.25)
i=1
Because the contrary inequality always holds true we deduce that (3.25) is in fact an equality from which it follows (3.23). Step 11.
√ du u ∈ H ⇒ u, a χG dx
∈ V.
√ √ Proof. Using (3.15) and that u ∈ H, we have that ω/ a and a(du/dx) 2 belong Cauchy–Schwartz inequality, ω(du/dx) = √ to √ L (G) and thus, by the (ω/ a) a(du/dx) belongs to L1 (G). If X is neither C 0 (0, 1) nor C 0 ([0, 1]) then, since µ belongs to X 0 we deduce that the points of [0, 1] have µ-measure zero and that u belongs to L1 (Ω, dµ). By Step 9 we have
146314.tex; 27/05/1999; 13:44; v.7; p.33
72
JUAN CASADO-D´ıAZ
Z
Z
u dµ(x) +
S∞ i=1
=
[αi ,βi ]
ω G
du dx dx Z
Z ∞ X
u dµ(x) +
[αi ,βi ]
i=1
βi
αi
!
du ω dx = 0. dx
(3.26)
¯ or C 0 (Ω), as above, u belongs to L1 ([0, 1], dµ) and by If X equals C0 (Ω) (3.21), we have Z Z du u dµ(x) + ω dx S∞ [α ,β ] G dx i=1 i i =
∞ X
βi
u dµ(x) +
[αi ,βi ]
i=1
=
Z
Z ∞ X
αi
!
du ω dx − dx
u(βi )ω(βi ) − u(αi )ω(α− i ) −
X
=
S
βi ∈
−
{αj }
X S
αi ∈
X
u(βi )ω(βi ) +
S
u(αi )ω(αi ) −
{βj }
u(αi )µ({αi })
{βj }
u(αi )(ω(αi ) − ω(αi −))
{βj }
u(βi )ω(βi )
S
βi 6∈
S
αi ∈
X αi ∈
i=1
X
{αj }
X S
αi 6∈
u(αi )ω(αi −).
{βj }
In the right-hand side the first and the third terms are the same with different sign and the second and the fourth terms are zero by Steps 4 and 7. Hence as for Lp we have (3.26). By (3.23), (3.13) and du/dx = 0 almost everywhere in S c ∩ G, equality (3.26) means Z Z 1 √ du u dµ(x) + aw χG dx = 0 dx 0 [0,1] √ 2 and hence that (u, a(du/dx)χG ) belongs to V . e 4. A Remark about (H) e holds always true in the one-dimensional case Let us see in this section that (H) if we assume that a is finite and that this result does not hold true for higher dimensions.
PROPOSITION 4.1. The space Wa (0, 1) is dense in C 0 ([0, 1]), when a is finite almost everywhere.
146314.tex; 27/05/1999; 13:44; v.7; p.34
73
RELAXATION OF A QUADRATIC FUNCTIONAL
Proof. The space Wa (0, 1) is an algebra which contains the constant functions. Hence from the Stone–Weierstrass Theorem, the proposition holds true if and only if Wa (0, 1) separate points, i.e. if for any points x1 , x2 ∈ [0, 1], with x1 6= x2 , there exists a function u ∈ Wa (0, 1) such that u(x1 ) 6= u(x2 ). This follows by taking Z
x
1 min √ , 1 dx. a
u(x) = 0
2
One could conjecture that the same result holds true for higher dimensions. Actually this conjecture is false since one deduces from following Theorem that in some cases WA (Ω) reduces to the constants THEOREM 4.1. Let Ω be the cube (0, 1)d , with d > 1. Then there exists a 1,1 (Ω) measurable function h: Ω 7→ R finite almost everywhere, such that if u ∈ Wloc with h ∇u ∈ L1 (Ω) then u is constant. h: (0, 1) 7→ R Proof. Let {rn } be the set of the rational numbers of (0, 1). Define e by ( e h(t) = max
∞ X
)
1
1 ,1 . n(d+2) |t − rn |d+1 2 n=1
Then, the function e h is almost everywhere finite. Indeed consider e Z = {t ∈ (0, 1): h(t) < +∞}
and for ε > 0, define Uε =
∞ [
rn −
n=1
ε ε , rn + n . n 2 2
If t 6∈ Uε , we have (
∞ X
1
1 e h(t) 6 max 1, n(d+2) |t − r |d+1 2 n n=1
) 61+
1
∞ X 1
εd+1
2n n=1
=1+
1 . εd+1
Then (0, 1)\Uε ⊂ Z and hence |Z| > 1 − |Uε | > 1 −
∞ X 2ε n=1
2n
= 1 − 2ε.
e is finite almost everywhere. Since ε is arbitrary, we have |Z| = 1, i.e., h
146314.tex; 27/05/1999; 13:44; v.7; p.35
74
JUAN CASADO-D´ıAZ
We now define h: Ω 7→ R by h(x) =
d Y
e i ). h(x
i=1
Then, since e h is finite almost everywhere, so is h. 1,1 We claim that h is the desired function. Let u ∈ Wloc (Ω) such that h|∇u| = f ∈ L1 (Ω).
(4.1)
We consider for u its Lebesgue representative. Remember (see e.g. [11]) that the set of points which are not lebesgue points of u is a set of Hausdorff dimension lower than d − 1. Set ut = ρt ∗ u, (u is assumed to be prolongated by zero outside of Ω) where ρt is a mollifier defined by
ρt (x) =
1 x ρ , td t
with ρ ∈ D(R ), d
supp(ρ) ⊂ B(0, 1),
Z
ρ > 0,
Rd
ρ(x) dx = 1.
It is well-known (see e.g. [12] pages 230–237) that ut ∈ C ∞ (Rd ) and that for t converging to zero, ut (x) converges to u(x) on the Lebesgue set of u. Let x be a point of Rd which has a rational component. This means that for some i with 1 6 i 6 n there exists a rational rj with xi = rj . Then, we have Z k ρ kL∞ (Rd ) Z |∇ut (x)| = ρ (x − y)∇u(y) dy 6 |∇u(y)| dy. B(x,t) t td B(x,t) e > 1, we have From (4.1) and h
|∇u(y)| =
f (y) f (y) 6 6 2j(d+2) |yi − rj |d+1 f (y). e h(y) h(yi )
Then since |yi − rj | = |yi − xi | < t, we have |∇ut (x)| 6k ρ kL∞ (Rd ) 2
Z j(d+2)
where Cj = 2j(d+2) k ρ kL∞ (Rd )
t B(x,t)
f (y)dy 6 Cj t,
(4.2)
Z
f (y) dy. Ω
146314.tex; 27/05/1999; 13:44; v.7; p.36
75
RELAXATION OF A QUADRATIC FUNCTIONAL
Writing x ∈ Ω ⊂ Rd with d > 2 as x = (x1 , x0 ) = (x1 , x2 , x00 ), we have |ut (rj , x0 ) − ut (r1 , x0 )| 6 |ut (rj , x0 ) − ut (rj , r1 , x00 )| + |ut (rj , r1 , x00 ) − ut (r1 , r1 , x00 )|
+|ut (r1 , r1 , x00 ) − ut (r1 , x0 )|.
Hence |ut (rj , x0 ) − ut (r1 , x0 )| Z 6
x2
00
|∇ut (rj , s, x )|ds +
r1
Z
r1
+ x2
Z
rj
r1
|∇ut (s, r1 , x00 )|ds
|∇ut (r1 , s, x00 )| ds
and therefore, by (4.2) we have |ut (rj , x0 ) − ut (r1 , x0 )| 6 (Cj + 2C1 )t.
Since ut (x) converges to u(x) on the Lebesgue’s set of u, we get |u(rj , x0 ) − u(r1 , x0 )| = 0
for a.e. x0 ∈ Ω0 = (0, 1)d−1 .
(4.3)
For ϕ in D(Ω0 ) define γ : (0, 1) 7→ R by Z
γ(x1 ) =
Ω0
u(x1 , x0 )ϕ(x0 ) dx0 .
The derivative in distributional sense of γ is the function x1 7→
Z Ω0
∂u (x1 , x0 )ϕ(x0 ) dx0 ∂x1
1,1 and hence by the Fubini’s Theorem γ belongs to Wloc (0, 1). In particular, we deduce that γ is a continuous function which takes the same value for any rational number ¯1 ∈ (0, 1) and thus, γ is constant. Since ϕ is arbitrary, we deduce that for any x1 , x we have
u(x1 , x0 ) = u(¯ x 1 , x0 )
a.e. x0 ∈ Ω0 .
The same argument used for the first component of x is valid for any component 2 of x and thus u is constant. REMARK 4.1. Taking Ω = (0, 1)d with d > 1 and A = h2 I with h given by
146314.tex; 27/05/1999; 13:44; v.7; p.37
76
JUAN CASADO-D´ıAZ
Theorem 4.1, we have that WA (Ω) is reduced to the set of the constant functions e does not holds true although h is finite almost everywhere. in Ω and thus H Acknowledgments I thank Franc¸ois Murat for interesting discussions and advices concerning this paper. This work has been partially supported by the Project EURHomogenization, Contrat SC1-CT91-0732 of the Program SCIENCE of the Commission of the European Communities and by the Project PB92-0696 of the DGICYT of Spain. References 1. Albeverio, S. and R¨ockner, M.: ‘Classical Dirichlet forms on topological vector spaces’. Closability and a Cameron–Martin formula, J. Funct. Anal. 88 (1990), 395–436. 2. Buttazzo, G.: ‘Semicontinuity, relaxation and integral representation in the calculus of variations’, Pitman Res. Notes Math. Ser. 207, Longman, Harlow, 1989. 3. Boccardo, L. and Murat, F.: ‘Remarques sur l’homog´en´eisation de certains probl`emes quasilin´eaires’, Portugaliae Mathematica 41(1–4) (1982), 535–562. 4. Casado-D´ıaz, J.: Sobre la homogeneizaci´on de problemas no coercivos y problemas en dominios con agujeros, Ph. D. Thesis, University of Seville, 1993. 5. Carbone, L. and Sbordone, C.: ‘Some properties of Γ-limits of integral functionals’, Ann. Mat. Pura Appl. IV 122 (1979), 1–60. 6. Dal Maso, G.: ‘Alcuni teoremi sui Γ-limiti di misure’, Boll. Un. Mat. Ital. 5 15B (1978), 182–192. 7. Dal Maso, G.: An Introduction to Γ-Convergence, Birkh¨auser, Boston, 1993. 8. De Giorgi, E.: ‘Convergence problems for functionals and operators’, in: Proceedings of the International Meeting on Recent Methods in Non Linear Analysis (Rome, May 8–12, 1978), ed. by E. De Giorgi, E. Magenes and U. Mosco, Pitagora, Bologna, 1979, pp. 131–188. 9. Dixmier, J.: Les alg`ebres d’op´erateurs dans l’espace hilbertien (Alg`ebres de Von Neumann), Gauthier-Villars, Paris, 1969. 10. Evans, L. C. and Gariepy, R. F.: Measure Theory and Fine Properties of Functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, 1992. 11. Federer, H. and Ziemer, W.: ‘The Lebesgue set of a function whose distribution derivatives are pth. power summable’, Indiana U. Math. J. 22 (1972), 139–158. 12. Folland, G. B.: Real Analysis. Modern Techniques and their Applications, John Wiley and Sons, New York, 1984. 13. Fukushima, M.: Dirichlet Forms and Markov Processes, North-Holland, Amsterdam, 1980. 14. Hamza, M. M.: Determination des formes de Dirichlet sur Rn , Th`ese 3´eme cycle, Universit´e d’ Orsay, 1975. 15. Kato, T.: Perturbation Theory for Linear Operators, Springer-Verlag, New York, 1966. 16. Marcellini, P.: ‘Some problems of semicontinuity and of Γ-Convergence for integrals of the calculus of variations’. In: Proceedings of the International Meeting on Recent Methods in Nonlinear Analysis (Rome, May 8–12, 1978), ed. by E. De Giorgi, E. Magenes and U. Mosco, Pitagora, Bologna, 1979, pp. 205–221. 17. Mosco, U.: Formes de Dirichlet et homog´en´eisation, Cours de 3´eme cycle a` l’Universit´e Paris VI, 1993. 18. Yoshida, K.: Functional Analysis, 6th edn, Springer-Verlag, Berlin, 1980. 19. Ziemer, W. P.: Weakly Differentiable Functions, Springer-Verlag, New York, 1989.
146314.tex; 27/05/1999; 13:44; v.7; p.38