J. SIMON
Representation of distributions
and
explicit antiderivatives up to the
boundary.
This paper is concerned with the existence, the explicit construction, and the regularity of an antiderivative of a given distribution q = (q1 , . . . , qd ) in an open set Ω of
IRd , that is f such that ∂i f = qi for all i.
An explicit antiderivative of a distribution which is orthogonal to divergence free
test functions is constructed up to the boundary of a lipschitz domain. It provides an
easy proof of de Rham’s existence theorem, and of regularity properties. It relies on
a representation formula of a distribution only in terms of it’s derivatives.
1. Introduction
Existence of an antiderivative. J.-L. Lions [6], p. 67, noticed that a powerful theorem
of G. de Rham [9], thm. 17’ p. 114, provides the existence of an antiderivative of any
q which is orthogonal to all test functions with null divergence, that is
hq, ψi = 0 ∀ψ ∈ D(Ω)d such that ∇ . ψ = 0 .
The method of J.-L. Lions to infer this result is explained by R. Temam [18], foot-note
(1) p. 74.
In actual fact, the theorem of G. de Rham is much more general – it deals with currents on manifolds – and it requires hard material. The existence of an antiderivative
playing an essential part in existence proofs for pressure in Navier-Stokes equations,
see [2], [4], [6], [17] or [18], many authors provided simpler proofs in simpler cases.
Proofs were given by O.A. Ladyzhenskaya [4], theorem 1 p. 28, for q ∈ L2 (Ω)3 ,
and by L. Tartar [17] for q ∈ H −1 (Ω)d . The method of L. Tartar is used in the book
of V. Girault & P. A Raviart [2] theorem 2.3 p. 25. Outlines of the methods of G.
de Rham and L. Tartar are given in [11]. All these works prove the existence of an
antiderivative, but they do not provide an antiderivative.
Construction of an antiderivative. Recently [12], the author of these lines gave a
simpler proof of the existence result in the general case q ∈ D0 (Ω)d , by recollecting a
sequence of explicit local antiderivatives. Using a parametrix γ, that is a distribution
with support in a ball B satisfying −∆γ = δ − η, η being a smooth function, a local
antiderivative in any star-shaped set ω such that ω + B ⊂ Ω was defined by
R1
P
? η) a + t(x − a) dt .
(1)
f (x) = i −(qi loc
? ∂i γ)(x) + 0 (x − a)i (qi loc
? w)(x) =
R It uses a local convolution of distributions, which, for functions, yields (v loc
v(x
−
y)
w(y)
dy.
Such
a
formula
cannot
provide
a
derivative
up
to
the
boundary
B
since it is defined only for x such that x + B ⊂ Ω.
In the present paper, we construct explicitly an antiderivative up to the boundary.
Using kernelsPζi with supports in a conical sector of ball C, instead of the ball B,
satisfying − i ∂i ζi = δ − %, we define a local antiderivative in any connected set ω
such that ω + C by
(2)
f (x) =
P
i (qi
R
ζi )(x) + a x q % . d`
where ax
 is any path included in ω. We use an operation
on distribution, so-called
R
ponderation, which for functions yields (v w)(x) = C v(x + y) w(y) dy. Contrarily
to (local) convolution, there is no symmetry on y since it will not be pleasant here
because it changes an inner cone in an outer one.
If Ω is Lipschitz and connected, we cover it by a finite number K of connected ωk
such that ωk + Ck ⊂ Ω for suitable Ck , and we obtain a global explicit antiderivative
by recollecting corresponding local antiderivatives fk . We obtain a finite number of
suitable Ck , even if Ω is unbounded, by using a uniform Lipschitz property, which
provides a uniform partition of unity related to the ωk .
Representation formulas. Many representation formulas have been obtained by various authors, and they have been used to study properties of functions and distributions, such as regularity or extension.
One of the earlier representation formula, due to Kryloff [3], was extended by L.
Schwartz to distributions on all of IRd with compact support. Using a parametrix γ,
f was represented up to a smooth rest f ? η, [10] (VI, 6, 8) p. 184, by
(3)
f =−
P
i qi
? ∂i γ + f ? η .
We obtain formula (1) by representing the rest f ? η itself in terms of qi .
The earlier formula with kernels in a cone
Pis due to Sobolev [15]. Using kernels λi
with support in a cone K which satisfy − i ∂i λi = δ, he represented a function f ,
[5] p. 62, in any ω such that ω − K ⊂ Ω, by
f =−
P
i qi
? λi .
Unfortunately in many cases, and in particular if Ω is bounded, ω = ∅ since K is
unbounded. Therefore, this formula is generally used after a localisation step, which
adds a rest to the right hand side. Similar formulas were extended to distributions,
in particular by K. T. Smith [14].
A representation formula in a bounded sector of cone C, instead of K, is obtained
by using the kernels ζi (x) = θ(x) λi (−x) , θ being a smooth function with compact
support. It yields
(4)
f=
P
i qi
ζi + f % .
Formula (3) is a particular case, for which the kernel is radial and C = B. We obtain
formula (2) by representing the smooth rest f % itself in terms of qi .
202
Regularity of antiderivatives. For the Navier-Stokes equation, the above existence
result provides a pressure which is a distribution satisfying ∇f ∈ H −1 (Ω)d . By a
result of E. Magenes and G. Stampacchia, [7] footnote (27) p. 320, it follows that
f ∈ L2 (Ω) if Ω is bounded with a C 1 boundary. This was extended by J. Nečas, [8],
to Lipschitz domains.
More generally, if ∇f ∈ W −1,r (Ω)d , 1 < r < ∞, then f ∈ Lr (Ω) if Ω is Lipschitz
and bounded. Here we infer this property from representation formula (4), by using
integrability properties of kernels due to Calderon and Zygmund [1]
In fact, all the results of this paper are proved for vector valued distributions, which
are useful for evolution problems. More precisely, the distributions are valued in a
Banach space E. Results in more general spaces will be given in [13].
2. Representation formulas for functions.
In all the sequel, E is a Banach space and Ω is an open set in IRd . We denote
∂i = ∂/∂xi and ∂ = ∂/∂t.
Let C be a sector of cone, of summit 0, axis b, |b| = 1, angle ϑ, 0 < ϑ ≤ , and
radius r > 0, that is
C = {x ∈ IRd : x . b ≥ |x| cos ϑ, |x| ≤ r} .
Let θ ∈ C ∞ (0, ∞) such that θ(t) = 1 if t ≤ r/2 and θ(t) =
R 0 if t ≥ r. Let
σ ∈ C ∞ (S d−1 ), where S d−1 is the unit sphere of IRd , such that S d−1 σ ds = |S d−1 |
and σ(s) = 0 if s . b ≤ cos ϑ.
We define functions ζi ∈ L1 (IRd ), i = 1, . . . , d, with support in C, by
ζi (x) = −
x xi
1
1
σ( )
θ(|x|) d−1
|S d−1 | |x| |x|
|x|
∀x 6= 0 ,
and we define a function % ∈ C ∞ (IRd ) with support in C by
%(x) = −
x
1
1
σ( ) ∂θ(|x|) d−1
|S d−1 | |x|
|x|
∀x 6= 0 .
Given f ∈ C(Ω; E) and a weight ∈ L1 (C), a continuous function on
ΩC = {x : x + C ⊂ Ω}
is defined by
Z
(f )(x) =
f (x + y) (y) dy .
y∈C
A representation of f by its derivative, up to a smooth function f %, is as follows.
Lemma 1. If f ∈ C 1 (Ω; E),
(5)
f =f %+
d
X
∂i f ζi
i=1
203
in ΩC . u
t
Proof. Given x ∈ ΩC and s ∈ S d−1 such that s . b ≥ cos ϑ one has x + ts ∈ Ω for
0 ≤ t ≤ r and therefore, since θ(0) = 1 et θ(r) = 0,
Z
r
d
f (x + ts) θ(t) dt
0 dt
Z r
d
X
=−
f (x + ts) ∂θ(t) +
∂i f (x + ts) si θ(t) dt .
f (x) =
0
i=1
An integration with respect to s yields, since
s . b ≤ cos ϑ ,
S d−1
σ ds = |S d−1 | and σ(s) = 0 if
Z
1
f (x) = −
R
f (x) σ(s) ds
|S d−1 | s.b≥cos ϑ
Z
Z r
1
= − d−1
f (x + ts) ∂θ(t) σ(s)
|S
| s.b≥cos ϑ 0
+
d
X
∂i f (x + ts) si θ(t) σ(s) ds dt
i=1
and the variable change y = ts, therefore t = |y|, s = y/|y|, ds dt = |y|1−d dy, yields
f (x) =
Z f (x + y) %(y) +
C
d
X
∂i f (x + y) ζi (y) dy . u
t
i=1
Remark. This is similar to Sobolev’s formula [15] which corresponds to an unbounded
cone K = {x ∈ IRd : x . b ≥ |x| cos ϑ} and to θ ≡ 1 and therefore % ≡ 0, that is, [5] p.
46,
(6)
f (x) = −
1
|S d−1 |
Z X
d
∂i f (x + y) σ(
K i=1
y yi
)
dy .
|y| |y|d
In the identity (5) we use a parametrix, that is a kernel ζi with bounded support
which is equal to Sobolev’s kernel used in (6) in a neighborhood of 0. u
t
A representation of f only by it’s derivatives, up to a constant, is as follows.
Lemma 2. 2 If ω ⊂ ΩC is connected, given a ∈ ω,
(7)
f = (f %)(a) + h(∇f ) +
d
X
i=1
204
∂ i f ζi
in ω
where h(∇f ) ∈ C(ω; E) is defined at x ∈ ω by
Z
(8)
h(∇f )(x) =
∇f % . d`
a
x
where ax
 is any C 1 by parts path included in ω. u
t
Proof. It follows from lemma 1, since f % ∈ C 1 (ΩC ; E),
Z
Z
.
(f %)(x) − (f %)(a) =
∇(f %) d` =
∇f % . d` . u
t
a
a
x
x
Remark. A simpler representation of a C 1 function f by it’s derivatives is
Z
f (x) = f (a) +
∇f . d` ,
a
x
but, contrary to (7), this formula will not extend to a distribution f . u
t
3. Extension to distributions.
First, the weighted mean value, so called ponderation, is extended to distributions as
follows.
0
(IRd ) (that is ∈ D0 (IRd ) and supp ⊂ D),
Definition 3. Let f ∈ D0 (Ω; E) and ∈ DD
d
where D is a compact subset of IR . Then, a distribution
f ∈ D0 (ΩD ; E) on ΩD = {x ∈ IRd : x + D ⊂ Ω}
is defined by : ∀ϕ ∈ D(ΩD ),
hf , ϕiΩD = f (z), h(y), ϕ(y − z)iy∈IRd z∈Ω . u
t
This definition is permissible since, as proved in [13], the map {ϕ 7→ φ}, where
φ(z) = h(y), ϕ(y − z)iy∈IRd , is continuous from D(ΩD ) into D(Ω). The next result
states that, for functions, we find again the operation defined in section 1.
Lemma 4. If f ∈ C(Ω; E) and ∈ L1 (D), then f ∈ C(ΩD ; E), and
Z
(f )(x) =
f (x + y) (y) dy . u
t
D
Proof. The change of variable x = z − y yields
Z
Z
hf , ϕi =
f (z)
(y) ϕ(z − y) dy dz
z∈Ω
y∈D
Z
Z
=
f (x + y) (y) dy ϕ(x) dx . u
t
x∈ΩD
y∈D
The following result provides some properties of ponderation which are proved in
[13]. We denote ˇ the symmetry, that is ϕ̌(x) = ϕ(−x) for a function.
205
Proposition 5. (i) The map (f, ) → f is sequentially continuous from D0 (Ω; E) ×
0
DD
(IRd ) into D0 (ΩD ; E),
(a) ∂i (f ) = ∂i f = −f ∂i ,
(b) f δ = f , δ f = fˇ ,
(c) supp f ⊂ supp f − supp ,
(d) if ∈ DD (IRd ), then f ∈ C ∞ (ΩD ; E) . u
t
Remark. Instead of ponderation, we could define as well a generalized convolution
such that, for continuous functions,
Z
Z
f (x − y) (y) dy =
(f ? )(x) =
f (x + y) (y) dy .
−C
C
For Ω = IRd it provides commutativity, f ? = ? f , but for Ω 6= IRd the symmetry
on C is not pleasant, since an inner cone is changed in an outer one. These two
operations are related by f ? = f ˇ. u
t
Now, using the density of D(Ω; E) into D0 (Ω; E) and the sequential continuity of
ponderation, we can extend the identities of section 1, as follows.
Lemma 6. Let f ∈ D0 (Ω; E). Then, identity (5) is satisfied in D0 (ΩC ; E), and
identity (7) – (8) is satisfied in D0 (ω; E) ; in this last identity, ∂i f % and h(∇f ) lie
in C ∞ (ω; E). u
t
From identity (5) for distributions, we obtain the following equality.
Lemma 7. In D0 (IRd ),
%−
d
X
∂i ζi = δ . u
t
i=1
Proof. Identity (5) for f = δ yields
δ =δ%+
thus % −
P
i
P
i
∂i δ ζi = δ (% −
P
i
∂i ζi ) = (% −
P
i
∂i ζi )ˇ
∂i ζi = δ̌ = δ. u
t
Remark. Another
P way to prove the identity (5) is to calculate first the ∂i ζi and to
check that − i ∂i ζi + % = δ, and then to use
f =f δ =f %−
P
i
f ∂i ζi = f % +
206
P
i
∂i f ζi . u
t
4. Distributions which are orthogonal to divergence-free test
functions.
Given a distribution q ∈ D0 (Ω; E)d , a necessary condition for it to have an antiderivative is
(9)
hq, ψiD0 (Ω;E)d ×D(Ω)d = 0 ∀ψ ∈ D(Ω)d such that ∇ . ψ = 0 .
Indeed, if q = ∇f , then hq, ψi = h∇f, ψi = −hf, ∇ . ψi = 0 when ∇ . ψ = 0.
Conversely, we will prove in the next section that this condition is sufficient, by
using the following results.
Lemma 8. Condition (9) implies
(10)
∂i qj = ∂j qi ,
∀i, ∀j . u
t
Proof. Given ϕ ∈ D(Ω) and i 6= j, we define ψ ∈ D(Ω)d by ψi = ∂j ϕ, ψj = −∂i ϕ,
ψk = 0 otherwise. One has ∇ . ψ = ∂i (∂j ϕ) + ∂j (−∂i ϕ) = 0.
Then, (9) yields hq, ψi = hqi , ∂j ϕi + hqj , −∂i ϕi = 0, that is h−∂j qi + ∂i qj , ϕi = 0.
This is satisfied for all ϕ, whence (10). u
t
Condition (10) is also necessary for the existence of an antiderivative. Indeed, if
q = ∇f , it yields ∂i ∂j f = ∂j ∂i f .
Lemma 9. Let L be a lace in Ω and ρ ∈ D(D) such that L + D ⊂ Ω. Then (9)
implies
Z
(11)
q ρ . d` = 0 . u
t
L
R
Proof. A distribution T ∈ D0 (IRd )d is defined by RhTi , ϕi = L ϕ d`i . It’s support is L,
and ∇ . T = 0, since h∇ . T, ϕi = −hT, ∇ϕi = − L ∇ϕ . d` = 0 .
Let α ∈ D(Ω) such that α ≡ 1 on L + D. Then, on L, α
fq ρ = q ρ, where the
extension by 0 is defined by hf
αq, ϕiIRd = hq, αϕiΩ . Therefore,
Z
Z
.
q ρ d` =
α
fq ρ . d` = hT, α
fq ρiIRd .
L
L
For all distributions T and on IRd , hT, ρi = h, T ρ̌i where ρ̌(x) = ρ(−x). This
is easy for continuous functions, and it follows by continuity for distributions. Thus,
Z
q ρ . d` = hf
αq, T ρ̌iIRd = hq, α(T ρ̌)iΩ .
L
The function ψ = T ρ̌ is smooth and it’s support is included in supp T − supp ρ̌ =
L+D, whence, ψ ∈ D(Ω)d , and αψ = ψ, Moreover, ∇ . ψ = (∇ . T ) ρ̌ = 0. Therefore
assumption (9) yields
Z
q ρ . d` = hq, ψiΩ = 0. u
t
L
207
5. Explicit local antiderivative.
In this section, given a distribution q satisfying (9), we construct explicitly an antiderivative in a connected open set ω ⊂ ΩC . If such an antiderivative exists, by
identity (7) – (8) it is necessarily given, up to a constant, by
(12)
f = h(q) +
d
X
qi ζi
in ω
i=1
Z
h(q) (x) =
(13)
q % . d`
a
x
where a is a fixed point in ω and ax
 is any C 1 path included in ω.
The right hand side is defined since, by proposition 5, qi % ∈ C ∞ (ω; E). It defines a
function of x since it does not depend on the choice of the path. Indeed, the difference
of the values related to two paths ax
 and ax
ˆ is the integral on the lace L = ax
 ∪ xˆa,
which is zero by lemma 9.
Theorem 10. Let q ∈ D0 (Ω; E)d satisfy (9). Then, (12) – (13) defines an antiderivative of q in any connected open subset ω of ΩC , that is
∇f = q
in ω . u
t
Proof. By lemma 8, ∂i qj = ∂j qi , therefore
∂j (qi ζi ) = ∂j qi ζi = ∂i qj ζi = − qj ∂i ζi .
On other hand, denoting ej the j th basis vector,
1
∂j h(q) (x) = lim
t→0 t
Z
q % . d` = (q %)j (x) .
[x,x+tej ]
Therefore, by lemma 7,
∂j f = qj % −
d
X
∂i ζi = qj δ = qj . u
t
i=1
Remark. If Ω = IRd we can choose ω = IRd , which yields a global antiderivative. u
t
Remark. Sobolev’s formula (6) yields another explicit antiderivative, which is
f=
P
i qi
λi
where Sobolev’s kernel λi (y) = − (1/|S d−1| ) σ(y/|y|) yi /|y|d has it’s support in the
unbounded cone K.
208
This formula holds in ΩK . Unfortunately, this is useless for bounded Ω since then
ΩK = {x : x + K ⊂ Ω} = ∅ . u
t
6. Global antiderivative in a Lipschitz domain.
Here, Ω will be covered by a finite number of domains ΩCk , in each of which a
local antiderivative will be defined. This is possible if Ω is Lipschitz, that is if it
is “uniformly“ locally the graph of a Lipschitz function. More precisely, we use the
following definition.
Definition 11. An open subset Ω of IRd is Lipschitz if : there exists real numbers
a > 0, κ < ∞ and, for all z ∈ ∂Ω, there exists
— a system of cartesian coordinates (x1 , . . . , xd ) with origin at z,
— a real function ψ defined on U? = {x? ∈ IRd−1 : |x? | < a} where x? = (x1 , . . . , xd−1 )
such that, for all x? and y? , |ψ(x? ) − ψ(y? )| ≤ κ |x? − y? | and, for all x in U = {x ∈
IRd : |x? | < a, |xd | < a},
x ∈ Ω ⇐⇒ xd < ψ(x? ) ,
x ∈ ∂Ω ⇐⇒ xd = ψ(x? ) . u
t
Remark. If Ω is bounded, it is not necessary to assume that a and κ are uniform.
Indeed, the existence for all z of a(z) > 0 and κ(z) < ∞ imply, by a compactness
argument, the existence of a > 0 and κ < ∞ which does not depend on z. u
t
The existence of a finite number of spheric sectors of cones Ck , all equal up to a
rotation, such that the ΩCk cover Ω is provided by the next result, which is proved
in appendix.
Lemma 12. Let Ω be a Lipschitz open subset of IRd .
(a) There exists 0 < ϑ ≤ , r > 0, an integer K and, for k = 1, 2, . . . K, vectors
bk , |bk | = 1, spheric sectors of cones Ck = {x ∈ IRd : x . bk ≥ |x| cos ϑ , |x| ≤ r}, and
functions αk such that
∪k ΩCk = Ω ,
∞
αk ∈ C (Ω),
supp αk ∩ Ω ⊂ ΩCk ,
P
0 ≤ αk ≤ 1,
k αk ≡ 1 in Ω ,
sup sup |∂ β αk (x)| = cβ < ∞ .
x∈Ω k
(b) If Ω is connected, r may be chosen such that in addition, for any k, there exists
a connected open set ωk such that
supp αk ∩ Ω ⊂ ωk ⊂ ΩCk .
(c) If, in addition, Ω is bounded, the family of ωk may be chosen such that, in
addition, any points a and x in ωk may be connected by a path ax
 ⊂ ωk of length
|ax|
 ≤ c(Ω) < ∞. u
t
209
Given q satisfying (9), an antiderivative Fk in each ωk is explicitly provided by (12)
–R (13), from theorem 10. Let Λ be a ball included in ∩k ωk and ξ ∈ D(Λ) such that
ξ 6= 0. Another antiderivative in ωk is given by
Λ
fk = Fk − hFk , ξiωk .
Theorem 13. Let q ∈ D0 (Ω; E)d satisfy (9) and Ω be Lipschitz. Then, the fk can be
recollected in a unique f ∈ D0 (Ω; E) which is an antiderivative of q, that is f |ωk = fk
and
∇f = q in Ω . u
t
Remark. Using the partition of unity given by lemma 12, we get f =
is
Z
K
X
hf, ϕiΩ =
hFk , αk ϕiωk − hFk , ξiωk
αk ϕ . u
t
P g
k αk fk , that
Ω
k=1
Proof. By the recollecting theorem, it suffices to prove that fk = fl in ωk ∩ ωl . Since
∇(fk − fl ) = 0, fk − fl is constant in any connected component of ωk ∩ ωl .
In particular, fk − fRl is constant in Λ, say cΛ . By construction, hfk , ξi = 0 therefore
0 = hfk − fl , ξi = cΛ Λ ξ, whence cΛ = 0.
Now, let x be any point in ωk ∩ ωl and cx = (fk − fl )(x). Let z ∈ Λ, z xωk a path
included in ωk and z xωl a path included in ωl . We choose a closed ball B small
R enough
for z xωk + B ⊂ ωk and z xωl + B ⊂ ωl , and we choose ρ ∈ D(B̊) such that B ρ = 1.
Then, x + B is included in the connected component of ωk ∪ ωl which contains x,
whence fk − fl = cx in it, and
Z
(fk ρ − fl ρ)(x) =
cx ρ(y) dy = cx .
B
In the same way, (fk ρ − fl ρ)(z) = cΛ = 0.
On other hand, ∇(fk ρ) = ∇fk ρ = q ρ on z xωk , therefore
Z
(fk ρ)(x) − (fk ρ)(z) =
q ρ . d`
z xωk
and in the same way,
Z
(fl ρ)(x) − (fl ρ)(z) =
z xωl
q ρ . d`
thus, by substraction,
Z
q ρ . d`
z xωk ∪x zωl
is a lace, this is 0 by lemma 9, which proves that fk − fl ≡ 0 in
cx =
Since z xωk ∪ x zωl
ωk − ωl . u
t
7. Regularity of antiderivatives
Now we are interested in the regularity of an existing antiderivative.
210
Theorem 14. Let Ω be Lipschitz and bounded, and let f ∈ D0 (Ω; E)d satisfy, for all
i,
∇f ∈ W −1,r (Ω; E)d ,
(14)
1 < r < ∞.
Then, f ∈ Lr (Ω; E), and there exists c(Ω, r) independent on f such that
1
|Ω|
kf −
(15)
R
Ω
f kLr (Ω;E) ≤ c(Ω, r) k∇f kW −1,r (Ω;E)d .
0
More generally, for all g ∈ Lr (Ω), where 1/r0 + 1/r = 1, such that
kf −
(16)
R
Ω
Ω
g = 1,
f gkLr (Ω;E) ≤ c(Ω, r, g) k∇f kW −1,r (Ω;E)d ,
kf kLr (Ω;E) ≤ c(Ω, r, g) k∇f kW −1,r (Ω;E)d + k
(17)
R
R
Ω
f gkE . u
t
Proof. Integrability
of f . By assumption (14), there exists fij ∈ Lr (Ω; E) such that
P
∂i f = f0i + j ∂j fij . Then, formula (5) yields, for any sector of cone C, in ΩC ,
∂j fij ζi
P
P
= f % + i fi0 ζi − ij fij ∂j ζi .
f =f %+
P
i
fi0 +
P
j
Now, we will use ponderation properties, which are similar to convolution ones,
since f ρ = f ? ρ̌. One has ζi ∈ L1 (C), thus fi0 ζi ∈ Lr (Ω; E). In addition,
|∂j ζi (x)| ≤ c(d, σ)|x|−d and the integral of ∂j ζi on any sphere {x : |x| = c} is null,
therefore, by Calderon-Zygmund theorem, [1] or [16], theorem 2 p. 35, fij ∂j ζi ∈
Lr (Ω; E). Finally, by proposition 5, f % ∈ C ∞ (ΩC ; E). Therefore, f ∈ Lr (ω; E) for
any subset such that ω ⊂ ΩC .
By lemma 12, there exists a finite family of such subsets ωk which cover Ω, therefore
f ∈ Lr (Ω; E).
Calculus of the norm. In addition, we have
k
P
i
fi0 ζi −
P
ij
fij ∂j ζi kLr (ΩC ;E) ≤ c(d, C, r)
P
ij
kfij kLr (Ω;E)
therefore, considering the infimum with respect to all possible fij , this yields
(18)
kf − f %kLr (ΩC ;E) ≤ c(d, C, r) k∇f kW −1,r (Ω;E)d .
Let ω ⊂ ΩC be connected and such that all points x and a in ω may be connected
by a path of length |ax|
 ≤ c(ω) < ∞ ; then,
R
k(f %)(x) − (f %)(a)kE = k a x ∇f % . d`kE ≤ |ax|
 sup k(∇f %)(y)kE d .
y∈ω
211
Moreover k(∇f %)(y)kE d ≤ k∇f kW −1,r (Ω;E)d k%kW 1,r0 (C) , thus, denoting e = (f %)(a),
kf % − ekLr (ω;E) ≤ c(ω) |ω|1/r k∇f kW −1,r (Ω;E)d k%kW 1,r0 (C) .
With (18) it yields
kf − ekLr (ω;E) ≤ c(d, ω, C, r) k∇f kW −1,r (Ω;E)d .
By lemma 12, there exists a finite family of such ωk which cover Ω, then
kf − ek kLr (ωk ;E) ≤ c1 (Ω, r) k∇f kW −1,r (Ω;E)d .
(19)
Let Λ = ∩k ωk . Then,
kek − e1 kLr (Λ;E) ≤ kf − e1 kLr (Λ;E) + kf − ek kLr (Λ;E)
≤ 2c1 (Ω, r) k∇f kW −1,r (Ω;E)d .
Using (19) and kek − e1 kE = |Λ|−1/r kek − e1 kLr (Λ;E) , we obtain
kf − e1 kLr (ωk ;E) ≤ c1 (Ω, r) 1 + 2
|Ω|1/r k∇f kW −1,r (Ω;E)d
|Λ|1/r
whence, the ωk covering Ω,
kf − e1 kLr (Ω;E) ≤ c2 (Ω, r) k∇f kW −1,r (Ω;E)d .
R
R
Since Ω g = 1, one has f − Ω f g = (f − e1 ) − Ω (f − e1 ) g and therefore
(20)
R
kf −
R
Ω
f gkLr (Ω;E) ≤ kf − e1 kLr (Ω;E) + k
R
(f − e1 ) gkLr (Ω;E)
R
= kf − e1 kLr (Ω;E) + |Ω| k Ω (f − e1 ) gkE
= kf − e1 kLr (Ω;E) 1 + |Ω|1/r kgkLr0 (Ω)
Ω
1/r
whence, using (20), we obtain (16). This ends the proof, since (15) is a particular
case of (16), and (17) is an easy consequence of (16). u
t
8. Appendix : proof of lemma 12.
Proof of part (i). We choose r = a/4 and 2 ϑ = arctg κ. That is, the angle of the
cones Ck equals half the angle of the cone {x : xd ≥ κ |x? | }.
There exists a finite number of unit vectors bk such that the corresponding Ck cover
the ball {x : |x| ≤ a/4} ; they can be chosen explicitly, or by a compactness argument.
Therefore, given z ∈ ∂Ω and a corresponding system of coordinates x = (xk1 , . . . , xkd )
defined at definition 9, the vector (0, . . . , 0, a/4) is included in one of the Ck , say
Ck(z) . Then, Ck(z) ⊂ {x : xkd ≥ κ |xk? | } whence definition 11 yields
ball (z, 3a/4) ∩ Ω + Ck(z) ⊂ Ω .
212
It follows easily that any x ∈ Ω is included in one of the ΩCk . In order to obtain
a uniform unit partition we will prove a little more : for any x ∈ IRd , there exists k
such that
ball (x, a/4) ∩ Ω ⊂ ΩCk .
(21)
If ball (x, a/2) ⊂ Ω, or if ball (x, a/2) ⊂ IRd \ Ω this is satisfied for all k since
ball (x, a/4) + Ck ⊂ ball(x, a/2). On other hand, if ball (x, a/2) ∩ ∂Ω 6= ∅, there exists
z ∈ ∂Ω such that x ∈ ball (z, a/2), therefore ball (x, a/4) ∩ Ω ⊂ ball (z, 3a/4) ∩ Ω ⊂
ΩCk (z) , which proves (21).
Now, we consider functions αk = 1Uk ? ρ, where
Vk = {x ∈ IRd : ball (x, a/4) ∩ Ω ⊂ ΩCk } , Uk = Vk \ V1 ∪ . . . ∪ Vk−1 ,
Z
ρ ∈ D(IRd ) ,
ρ = 1 , supp ρ ⊂ ball (0, a/8) .
IRd
supp αk P
∩ Ω ⊂ ΩCk .
They satisfy supp αk ⊂ Vk + ball (0, a/8) therefore
P
On other hand, the Uk cover IRRd , whence k 1Vk = 1 and k αk = 1 in all of IRd .
Finally, |∂ β αk | = |1Vk ? ∂ β ρ| ≤ IRd |∂ β ρ| ≤ cβ . u
t
Proof of part (ii). Each cone Ck defined in part (i) contains a ball centered at bk r/2
of radius λ = r sin ϑ/2. Let
ωk = Λk ∪ ΩB(λ) ,
B(λ) = ball (0, λ) ,
Λk = ΩCk (r) + {s bk : 0 ≤ s ≤ r/2} ,
Ck (r) = cone (0, bk , ϑ, r) .
One has supp αk ∩Ω ⊂ ΩCk (r) ⊂ ωk . On the other hand, Λk ⊂ ΩCk (r/2) ⊂ ΩCk (λ) thus
ωk ⊂ ΩCk (λ) . To prove part (ii) (for r = λ), it remains to prove that ωk is connected.
Given x ∈ ωk , there exists x̂ ∈ ΩB(λ) such that the segment [x, x̂] is included in ωk .
Indeed, if x ∈ ΩB(λ) it is satisfied for x̂ = x, else x = x0 + s bk with x0 ∈ ΩCk (r) and
0 ≤ s ≤ r/2 and it is satisfied for x̂ = x0 + bk r/2.
We assume for a moment that ΩB(λ) is connected. Given x and y in ω there exists
a path L, from x̂ to ŷ, which is included in ΩB(λ) . Then, the path [x, x̂] ∪ L ∪ [ŷ, y] is
included in ωk , which proves that ωk is connected.
It remains to prove that ΩB(λ) is connected. To this end we define, for any x ∈ Ω,
+
Tk (x) = x + αk (x) − dk (x) bk
where dk (x) is the smallest positive number such that x − dk (x) bk ∈ ∂Ω and 0 < ≤
a/8 will be chosen small enough. It satisfies
(22)
dist t Tk (x) + (1 − t) x, ∂Ω ≥ dist (x, ∂Ω)
∀t ∈ [0, 1] .
This is obvious if Tk (x) = x. Else, αk (x) 6= 0 whence ball(x, a/8) ∩ Ω ⊂ ΩCk (r) , and
dk (x) < a/8 whence δx = dist (x, ∂Ω) ≤ a/8 ; denoting xt = t Tk (x) + (1 − t) x and
+
s = t αk (x) − dk (x) , this yields ball(xt , δx ) = ball(x, δx ) + s bk ⊂ ball(x, a/8) ∩
Ω + Ck (r) ⊂ Ω which proves (22).
213
We define T = TK ◦ TK−1 ◦ . . . ◦ T1 . It follows from (22) that, if x ∈ ΩB(λ) , then the
linear by parts path connecting the points x, T1 (x), T2 ◦ T1 (x), . . . , TK−1 ◦ . . . ◦ T1 (x),
and T (x) is included in ΩB(λ) . This path is denoted P x, T (x) . If y is another point
in ΩB(λ) , since Ω is connected, there exists a path L connecting x to y in Ω. We
assume for a moment that
(23)
dist (T (x), ∂Ω) ≥ λ
∀x ∈ Ω .
Then T ◦ L is a path connecting T (x) to T (y) in ΩB(λ) , and P x, T (x) ∪ T ◦ L
∪P T (y), y is a path connecting x to y in ΩB(λ) . This proves that ΩB(λ) is connected,
provided that we check (23).
Since the sum of the αk (x) equals 1, one
√ of them satisfies αk (x) ≥ 1/K. Denoting
c = supx,j |∂j αk (x)|, we choose ≤ 1/(2 c d K 2 ), and we denote xj = Tj ◦ . . . ◦ T1 (x),
√
thus xj = Tj (xj−1 ) . Then, |xj −xj−1 | ≤ whence |αk (xj )−αk (xj−1 )| ≤ c d ≤ 1/2K 2
and by iteration |αk (xj ) − αk (x)| ≤ 1/2K which yields αk (xj ) ≥ 1/2K. Therefore,
+
xk = xk−1 + αk (xk−1 ) − dk (xk−1 ) bk = xk−1 + sk bk .
If dk (xk−1 ) ≤ /2, then /4K ≤ sk ≤ ; moreover xk−1 ∈ ΩCk (r) since αk (xk−1 ) 6=
0, thus cone(xk−1 , bk , ϑ, r) ⊂ Ω. This cone contains ball(xk , sin ϑ/4K), whence
dist (xk , ∂Ω) ≥ sin ϑ/4K. By (22) for t = 1 and k = j, dist (xj , ∂Ω) ≥ dist (xj−1 , ∂Ω)
hence, by iteration, T (x) = xK satisfies
(24)
dist (T (x), ∂Ω) ≥
4K
sin ϑ .
Else, that is if dk (xk−1 ) > /2, by definition of dk , xk−1 − bk /2 ∈ Ω. Then, x0k−1 =
xk−1 − bk /2 satisfies αk (x0k−1 ) > 0, therefore x0k−1 ∈ ΩCk (r) , and
+
xk = x0k−1 + αk (xk−1 ) − dk (xk−1 ) +
bk = x0k−1 + s0k bk
2
where /2 ≤ s0k ≤ . Now, the above calculus yields dist (xk , ∂Ω) ≥ sin ϑ/2, then (24)
is again satisfied.
This proves (23) for all λ ≤ sin ϑ/4K, that is for all r ≤ /2K, which ends the proof
of the part (ii) of lemma 12. u
t
Proof of part (iii). Now, Ω is assumed to be bounded. By definition of ωk , any of its
points x may be connected to a point x̂ ∈ ΩB(2λ) by a path of length d = sin ϑ/2λ
included in ωk , provided d ≤ r/2 that is for λ ≤ r/4 sin ϑ.
By a compactness argument, ΩB(2λ) may be covered by a finite number N of balls
of radius λ, which are included in ΩB(λ) . Given another point y ∈ ωk , then x̂ and ŷ
may be connected by a path of length 2N λ included in ΩB(λ) and therefore included
in ωk . Then, x and y may be connected by a path of length 2d + 2N λ included in
ωk . u
t
214
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Université Blaise Pascal (Clermont-Ferrand 2) et C. N. R. S.,
63177 Aubière cedex.
215